Definition of period of a decimal representation of a number I need to define the period of a decimal representation of a number!!

Thanks in advance!!

It's just how many digits in the repeated part i.e. 1.23123123... has period 3

PERIOD IN DECIMAL NUMBERS-
Period means that in recurring or in
terminating decimal the repeating decimal is known as period
Example-
     7.9999........ continuously goes
So,
Here 9 is period because only 9 repeat continuously.


How to find area of triangle from its medians The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is

a) $48$

b) $144$

c) $24$

d) $72$

I don't want whole solution just give me the hint how can I solve it.Thanks.

Area of a Triangle from the Medians

A triangle is divided in to $6$ equal areas by its medians:

$\hspace{2cm}$

In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.

In the case where the two red triangles share a common $\frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal vertically opposite angles, and they share a common base; therefore, equal areas.

Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:

$\hspace{3cm}$

The triangle in green has sides $\frac23a$, $\frac23b$, and $\frac23c$, and by Heron's formula has area
$$
\frac49\sqrt{s(s-a)(s-b)(s-c)}\tag{1}
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
\frac43\sqrt{s(s-a)(s-b)(s-c)}\tag{2}
$$

There is a direct formula:

Let
$$s = (m_1+m_2+m_3)/2,$$

Then
$$\text{area} = \frac{4}{3}\sqrt{s(s-m_1)(s-m_2)(s-m_3)}.$$

This gives answer of above question as $72$.


Differential $dX_t$ of the process $X_t=e^{tW_t}$ In a quant interview, he asked me to compute the differential $dX_t$ of the Ito process $X_t=e^{tW_t}$. Is it correct to do the following?

$f(t,W_t)=e^{tW_t}$

We know that:
$$d(f(t,W_t))=f_t(t,W_t)dt+f_w(t,W_t)dW_t+\dfrac{1}{2}f_{ww}(t,W_t)(dW_t)^2$$
we don't take $f_{tw}$ and $f_{tt}$, because:
$$(dW_t)^2=dt$$
$$dW_t\,dt=0$$
$$dt^2=0$$

So we have:

$$\begin{align}
d(e^{tW_t}) &= W_te^{tW_t}dt+te^{tW_t}dW_t+\dfrac{1}{2}t^2e^{tW_t}dt\\
&= \left(W_t+\dfrac{t^2}{2}\right)e^{tW_t}dt+te^{tW_t}dW_t
\end{align}$$

Naively $dX_t = X_{t+dt}-X_t$. Then we apply it to your formula

$$dX_t = X_{t+dt}-X_t = e ^{W_{t+dt}} -  e ^{W_{t}}$$

This is already looking good if I have an expression for the change of $W_t$.

$$W_{t+dt} = W_{t}+ dW_t$$

This gets places in your exponential and you expand:

$$ e^{x+\epsilon} =e^x \Big( 1+ \epsilon + \epsilon^2/2 + \dots\Big) $$

With $x= W_t$ and $ \epsilon= dW_t$.. This becomes something like Itô formula.


If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$ Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$

My attempt at the question:

Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore

$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}}  $$

Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.

The answer is indeed $\frac13$. To check this, let's go another way: set $g=x-\frac2x$, as you did. Then express $x$ through $g$:
$$x=\frac12\left(g\pm\sqrt{g^2+8}\right).$$

Then

$$f(g)=\sqrt{\frac12\left(g\pm\sqrt{g^2+8}\right)-1}.$$

Taking its derivative, we can find that it's real at $g=1$ for $\pm\to+$, and we get
$$f'(1)=\frac13.$$

(The other solution with $\pm\to-$ is $-\frac{i}{6\sqrt2}$).

Why does your source say the answer is $1$? Maybe it's just a mistake of evaluating $f(1)$ instead of $f'(1)$, since $f(1)=1$.

Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$

It is found that $x = 2$ or $x = -1$

Taking the positive value of $x$

Therefore $f(1) = \sqrt {x-1}$ and
Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$
$$\Rightarrow \frac{1}{2}\cdot(x-1)^{\frac{-1}{2}}$$
Putting the positive value $2$, we get
$f'(1) = \frac{1}{2}$


If $f(x)≤x$ , then $f′(x)≤1$? I'm studying Calculus and having a trouble solving this question.

1) If $f(x)\leq x$, then $f′(x)\leq 1$ for all $x$?

2) What if $f(0)=0$, $f′(x)$ exists for all $x$?

I could easily find the counter example for 1) (Therefore it is false)

But I'm not sure about 2)

If $f(0)=0$ and $f′(x)$ exists for all $x$  &
$f(x)\leq x$ , then $f′(x)\leq 1$ for all $x$?

Please leave a comment if you don't mind :)

Hint: Consider $$f(x)=x-A \sin^2 x $$
for large $A$.

Let for example $f(x)=x\sin x$, then $f(0)=0$ and for any real $x$ $f(x)\leq x$. Then the derivative will look like this: $f'(x)=\sin x+x\cos x$ and so will exist for all $x$. Taking $x=2\pi n$, where $n$ is positive integer we get $f'(x)=2\pi n$, which evidently has no upper limitation.


Finding the matrix of linear transformation $$T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}= \begin{pmatrix} 3x_1+5x_2\\ 7x_1-x_2\end{pmatrix}$$ $$T:{\mathbb R}^2 \rightarrow {\mathbb R}^2$$
I have no idea how to start this. Any help would be appreciated. I need to find the transformation matrix T.

One way to do this is to make an ansatz (educated guess) of the form of $T$.  You can assume that

$$
T \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right ]
$$

and now compute that

$$
T \left [ \begin{array}{c}
x_1 \\
x_2\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right ] \left [ \begin{array}{c}
x_1 \\
x_2\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{c}
ax_1 + bx_2 \\
cx_1 + dx_2 \\
\end{array} \right ]
$$

by ordinary matrix multiplication.  What now must be the values of $a,b,c,d$?

The matrix is: $A_{T} = \begin{pmatrix} 3 & 5 \\ 7 & -1 \end{pmatrix}$. The matrix has the columns $T(e_1),T(e_2)$. And what are $T(e_1) = T((1,0)), T(e_2) = T((0,1))$ ?


A man died. Let's divide the estate!!! How? This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with plain old algebra, which yields the shortest, simplest answers, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

A man dies and leave his estate to his sons.
The estate is divided as follows:
$1$st son gets 100 crowns $+$ $\frac{1}{10}$ of remainder of estate.
$2$nd son gets 200 crowns $+$ $\frac{1}{10}$ of remainder of estate ...
$(n)$$th$ son gets $100 × (n)$ crowns + $\frac{1}{10}$ of the remainder.
Each son receives same amount.
How many sons were there, what did each receive, and what was the estate?

Assuming all of his property was 'crowns', let the total amount he had be $\mathbf{T}$ crowns. Now the first son gets $\mathbf{100}$ crowns, so $\mathbf{T-100}$ crowns are left, of which he gets one-tenth, so his total inheritance is $\mathbf{\frac{T-100}{10} + 100}$ crowns. The second son first takes $\mathbf{200}$ crowns from the remaining estate, leaving us with $\mathbf{T-(200 + 100 + \frac{T-100}{10}) = \frac{9T - 2900}{10}}$ crowns, of which he takes one-tenth again. Gievn that each son receives an equal share, we equate these amounts with $\mathbf T$ being unknown $$\mathbf{\frac{T-100}{10} + 100 = \frac{9T - 2900}{100} + 200}$$ On solving, you are left with $\mathbf{T = 8100}$, so each son receives $\mathbf{900}$ crowns, i.e. there are $\mathbf9$ sons

Each son get 1100, and the total estate is 10100.
To find 10100 just resolve the equation $$100+0.1(x-100)=200+0.09(x-100)$$


Expectation of $Y := \min\{\frac{a}2, X\}$ if $X$ is uniform on $(0,a)$ The density of X is given by

$
f(x)=
\begin{cases}
 \frac{1}{a}&\, 0 \leq x < a\\
 0&\, otherwise\\
\end{cases}
$

Y := min{$\frac{a}{2}$, X} so Y belongs to [0, $\frac{a}{2}$]

Based on this solution, $E(Y) = \frac{3a}{8}$. Why is $E(Y) = \frac{3a}{8}$?

My Work

$
F_Y(y)=
\begin{cases}
 \frac{y}{a}&\, 0 \leq y < \frac{a}{2}\\
 1&\, y \geq \frac{a}{2}\\
\end{cases}
$

$
(F_Y(y))' = f_Y(y)=
\begin{cases}
 \frac{1}{a}&\, 0 \leq y < \frac{a}{2}\\
 0&\, y \geq \frac{a}{2}\\
\end{cases}
$

$
E(Y) = \int_{0}^{\frac{a}{2}} yf(y) dy = \int_{0}^{\frac{a}{2}} \frac{y}{a} dy = \frac{a}{8}
$

The cdf $F_Y(y)$ of variable $Y$ is not continuous, so $Y$ doesn't have a density. Thus differentiating $F_Y$ gives the wrong answer. The linked solution is using the formula
$$
E(h(X)) = \int_{-\infty}^\infty h(x) f(x)\,dx
$$
with $h(x):= \min\{\frac a2, x\}$. This gives
$$
E[Y]=E[h(X)]=\int_{-\infty}^{a/2}x \,f(x)\, dx + \int_{a/2}^\infty \frac a2 \,f(x)\,dx
$$
since $\min\{\frac a2, x\}$ equals $x$ when $x$ is less than $\frac a2$, and equals $\frac a2$ when $x$ exceeds $\frac a2$. Now plug in the density $f(x)$ for $X$.

A different approach would be:

$$EY=E\left[Y|X<\frac{1}{2}a\right]+E\left[Y|X\ge\frac{1}{2}a\right]= \int_0^{\frac{1}{2}a}x\frac{1}{a}dx= 0.125a+ 0.5a=\frac{3}{8}a$$

It relays on the properties of conditional expectations.


Satisfaction of Bessel equation by any other function. Is it possible that any function $y(x)$ other than Bessel group of functions, satisfy Bessel's equation?

$$x^2 \dfrac{d^2 y}{d x^2} + x \dfrac{d y}{d x} + (1-n^2/x^2) y = 0.$$

Neumann functions are what you'd be interested in. They form a set of solutions to the Bessel equation that have a singular point at the origin.


for any $k \in \mathbb{Z}^{+}$, how do we know $\omega^1$ is a primitive root of $x^k - 1$? I know that it is definitely a root.  The set $\{1, \omega, ..., \omega^{k-1}\}$ are the roots of unity.  But how are we guaranteed that it even exists and that furthermore it is a primitive root?

There is no need for any arguments specific to $ \mathbb{C} $ to see this. Let $ K $ be any field, and consider the polynomial $ X^n - 1 $. We first prove that if $ \alpha $ is a root, then its order must divide $ n $. Indeed, if $ k $ is the order of $ \alpha $, then we can divide $n $ by $ k $ to get $ n = kq + r $. Then, we have $ 1 = \alpha^n = \alpha^{kq + r} = \alpha^r $ where $ 0 \leq r < k $, so by definition of order we have $ r = 0 $.

We will need another result: let $ \alpha $ and $ \beta $ be primitive $ m $ and $ n $th roots of unity, respectively, where $ \gcd(m, n) = 1 $. Then $ \alpha \beta $ is a primitive $ mn $th root of unity. To see this, note that if $ \alpha^k \beta^k = 1 $, we may take $ m $th powers to obtain $ \beta^{mk} = 1 $, which means the order $ n $ divides $ mk $, and therefore $ k $. Repeating the same argument shows that $ k $ is divisible by $ mn $, showing the result. Note that this extends easily to the product of arbitrarily many numbers whose orders are pairwise coprime.

Now, let $ n = p^k $ be a prime power. Every root of $ X^n - 1 $ then has some order of the form $ p^i $ where $ 0 \leq i \leq k $, and there are less than or equal to $ p^i $ elements of order $ p^i $ by Lagrange's theorem. However, we have that

$$ \sum_{i=0}^{k-1} p^i = \frac{p^k - 1}{p-1} \leq p^k - 1 < p^k $$

so there must be some element $ \alpha $ with order $ p^k $, which is a primitive $n $th root of unity by definition.

To finish, let $ n = p_1^{q_1} p_2^{q_2} \ldots p_k^{q_k} $, and note that we may find $ \alpha_1, \alpha_2, \ldots, \alpha_k $ which are primitive $p_i^{q_i}$th roots of unity. By our second result,

$$ \alpha = \prod_{i=1}^{k} \alpha_i $$

then has order $ n $, finishing the proof.

It is because 1 is always coprime to k. If you looked at $\omega^i$ then it will be a primitive root iff $i$ is coprime to $k$. Think about generators of cyclic groups.


Find Domain of Log Function Need help determining the Domain of $$\log_2\left(2x+3\right)=\log_2\left(3x+4\right)-\log_2\left(3x+2\right)$$

Do not know if I should simplify first to get the domain, or to get the domain from the set logs that are in place?

This is an equation, and you can solve it. First of all we need to understand the single domains. That is easy. I will use the writing $\lg$ for the log base $2$.

$$\lg(2x+3)$$

This function exists iff $2x+3 >0$ hence $\boxed{x > -\frac{3}{2}}$.

$$\lg(3x+4)$$

This function exists if $3x+4 >0$ hence $\boxed{x > -\frac{4}{3}}$

$$\lg(3x+2)$$

This function exists if $3x+2 >0$ hence $\boxed{x >-\frac{2}{3}}$

Finally you can write th equation

$$\lg(2x+3) = \lg\left[(3x+4)/(3x+2)\right]$$

Where I used the well known property of the sum of log.

So now you can exponentiate in base $2$:

$$2x+3 = \frac{3x+4}{3x+2}$$

$$(2x+3)(3x+2) = 3x+4$$

$$6x^2 + 13x + 6 = 3x + 4$$

$$6x^2 + 10x + 2 = 0$$

Now you can solve it as a normal second degree equation

$$x = \frac{-10\pm\sqrt{100 - 48}}{12}$$

And check for the solutions, that don't have to be in contrast with the previous assumptions (necessarily).

I see an equation, not (just) a function?

In any case: $\log_a x$ (for a positive base $a$, different from $1$) is only defined for $x>0$. So for any logarithm appearing in your equation, the argument of the logarithm has to be strictly positive.

E.g. for $\log_2(3x+2)$ you need $3x+2>0 \Leftrightarrow x > -\tfrac{2}{3}$.

For all logarithms to be defined, you'll want the intersection of the 'domains' (sets where the different logarithms are defined), before simplifying (see below).

Notice that
$$\log x + \log x$$
is defined only for $x>0$ while (use $\log a + \log b = \log (ab)$)
$$\log \left( x^2 \right)$$
is defined for all $x\ne0$.


How to find the inverse/unit for mod? So for this problem, its asking to find the inverses of [4], [5], and [7]. I'm trying to skim through my textbook but I can't find an example.

Like for this problem, I don't understand why the Units are 1 and 2, the textbook I have is garbage.

Examples:

For a) you can find the inverses like this.

For example when we want to find the inverse  of $[4]$ in $\mathbb{Z}/13\mathbb{Z}$.
Since $13$ is a prime we know, that $\mathbb{Z}/13\mathbb{Z}$ is a field.

To find the inverse $[4^{-1}]$ we have to solve $[4][x]=[4x]=[1]$.
We can do this by brute force and just go all the way. There are just 13 possibilities for $x$.

One sees pretty fast, that $[4\cdot 10]=[40]=[1+13\cdot 3]=[1]$. Hence we have $[4^{-1}]=[10]$

This is a basic approach.
An other is to use the euclidean algorithm.

Now go find the other inverses!

For b) we can use the results in a).
The equations can be solved as always:

$[4]X=[7]$.

We want to isolate $X$. For that we multiply with $[4^{-1}]$, and get:

$[10][4]X=[10][7]$

$X=[70]=[5]$

Your question（２） is to solve equations
in mod13,
[1]4x≡７mod13
[2]5x≡１２mod13
[3]7x≡８mod13.

I will explain using [1]4x≡7mod13.

'To solve this equation' means what?

That is, 'To deformation the former equation[4x≡7mod13] to [x≡●mod13] and find  ●'.

What can we do to do such a thing?

Of couse, if only we could change 4 to 1 !

How can we do such a magical thing ?
No, No..., this is not a magical thing...

You can simply multiply on both side of equation from left side , 10 !
This number,10 is the number what we needed! And such a number　（satisfying ●ｘ≡１mod13
）is called inverse　of 13.

Generally, in mod p[p is prime], it is guaranteed that we can find the inverse in ｛0,1,2,...,p-1｝ uniquely .

Once you get　the inverse, you can solve equation in mod13 very easily like below.
 4x≡7mod13
4×10x≡7×10mod13
x≡70mod13＝5mod13 □

So,　if you say, problem（１）　is a very kind Hint for us to solve the equation problem（２）！

Then, I guess you understood you can solve the equation modp[p;prime] once you solve the inverse.

Then , how can we find The way to find the inverse　in concrete method?

The answer is quite simple ! The candidate of inverse only in 0,1,2,3,・・・,12 as I mentioned in the previous sentense !
So you just need to try and error and check these 13 number and find 10.

If you want to find the inverse more efficiently, you can omit the process and find the inverse a like below.

4a≡1mod13
12a≡3mod13
-a≡3mod13
a≡10mod13


Performing a differentiation on a Fourier series I'm tutoring a set of problem sheet to do with Fourier series and one problem is as follows:

  The Fourier series for a sawtooth wave is,

  $f(x)=x=-\sum^{\infty}_{n=1}\frac{2(-1)^n\sin(nx)}{n}$ for $-\pi < x<\pi$.

  If you differentiate this you get

  $1=-2\sum^{\infty}_{n=1}(-1)^n\cos(nx)$ again for $-\pi < x<\pi$

  What is wrong with this?

I have the solutions sheet and it says that it does not converge to 1 (fair enough, I plotted it to large $n$ and it sort of converges but oscillates between 0 and 2 in the interval) and then states ...

  An assumption has been made that you can interchange the order of summation
  and differentiation in the result stated.

It then goes to note that you can interchange the order of summation and integration

I don't understand the argument, can anyone shed some light on this?

As a complement: suitably interpreted, your computation is perfectly correct. That is, while the resulting series will not converge pointwise, there are many other possible (useful!) ways a Fourier series may converge. And, suitably interpreted, term-wise differentiation is always correct. [Edit: typo'd "pointwise" earlier, when I meant "termwise". Sorry!]

Recall that, for two Fourier expansions of "nice" functions $f(x)=\sum_n a_n e^{inx}$ and $g(x)=\sum_n b_ne^{inx}$, we have the Parseval-Plancherel theorem that ${1\over 2\pi}\int_0^{2\pi} f(x)\,\overline{g(x)}\,dx=\sum_n a_n\,\overline{b_n}$. And, yes, Fourier series do converge pointwise for infinitely-differentiable functions, and can be differentiated termwise...

Slightly changing your example, observe that the Fourier series $\delta(x)=\sum_n 1\cdot e^{inx}$ (the Dirac comb) certainly does not converge point-wise, because the terms do not go to $0$. Nevertheless, thinking in terms of Parseval-Plancherel, for nice function $f$
$$
f(0) \;=\; \sum_n a_n\,e^{in\cdot 0} \;=\; \sum_n a_n\cdot 1
$$
which has the same form as ${1\over 2\pi}\int_0^{2\pi} f(x)\cdot \overline{\delta(x)}\,dx$ if the latter were to make sense. That is, the evaluation-at-$0$ functional can be represented as a sort of inner product, under some constraints.

More precisely, for $s\in\mathbb R$, the $s$-th Sobolev space $H^s$ here consists of Fourier series $\sum_n b_ne^{inx}$ with $\sum_n |b_n|^2/(1+|n|)^s < \infty$. The Fourier series for the Dirac comb is in $H^{-{1\over 2}-\epsilon}$ for every $\epsilon>0$. For an "ordinary" function $f$ whose Fourier coefficients $a_n$ have sufficient decay to put $f$ in $H^{{1\over 2}+\epsilon}$, e.g., some smoothness of $f$ itself,
$$
|f(0)| \;=\; \Big|\sum_n a_n\cdot 1\Big| \;=\; \Big|\sum_n a_n\cdot (1+|n|)^{{1\over 2}+\epsilon}
\cdot {1\over (1+|n|)^{{1\over 2}+\epsilon}}\Big|
$$
and by Cauchy-Schwarz the square of this is dominated by
$$
\sum_n |a_n|^2(1+|n|)^{1+2\epsilon}
\cdot
\sum_n {1\over (1+|n|)^{1+2\epsilon}}
$$
That is, the evaluation functional $f\rightarrow f(0)$ is continuous in the $H^{{1\over 2}+\epsilon}$ metric topology, and the Dirac comb is a continuous linear functional on it. (This argument actually shows an instance of Sobolev imbedding, namely, that $H^{{1\over 2}+\epsilon}$ consists of continuous functions.)

Thus, a Fourier series in $H^{-s}$ (with $s> 0$), even though not pointwise convergent at all, directly gives a (continuous) linear functional on $H^s$ by extending Parseval-Plancherel:
$$
\Big(\sum_n b_n\,e^{inx}\Big)\bigg(\Big(\sum_n a_n e^{inx}\Big)\bigg)
\;=\; \sum_n a_n\cdot b_n
$$

That is, distributions have legitimate Fourier series expansions, though typically not converging pointwise at all. Termwise differentiation of not-very-convergent Fourier series is completely justifiable if construed as distributional derivatives, via integration by parts: after all, differentiation in the Fourier series is termwise multiplication by $in$.

Indeed, Sobolev and others looked at this sort of situation in the 1930s. A discussion in this direction, beginning more-or-less from scratch, is at http://www.math.umn.edu/~garrett/m/mfms/notes/09_sobolev.pdf

It bears repeating that there are many other types of convergence than pointwise.

"It then goes to note that you can interchange the order of summation and integration"

In your case an integration will get $n$ into the denominators, so the resulting series may converge.


Numerical solution to x = tan (x) I needed to find, using the bisection method, the first positive value that satisfy $x = \tan(x)$. So I went to Scilab, I wrote the bisection method and I got $1.5707903$. But after some reasoning I came to the conclusion that this value is wrong:

$\tan(1.5707903) \approx 1.6x10^5$. Not even close to $1.5707903$.
Forget for a moment the above. $x = \tan(x)$ is actually to find fixed points of $f(x) = \tan(x)$; $(x, f(x))$ must be in the line $y = x$. Here is the plot:

In $(0, \frac{3}{2}\pi)$ I can only see a fixed point to the right of $x = 4$, therefore $1.5707903$ is wrong.

Here comes the interesting part. If you go to Wolfram Alpha and type $x = \tan(x)$, you will see $1.5708$ in the Plot section:

However there is no $1.5708$ in the Numerical solutions section. Wolfram Alpha found $0, \pm 4.49340945790906, \ldots$.

But if you type $\tan(x) = x$, you will not see $1.5708$ in the Plot section!:

To summarize:

Is $4.49340945790906$ the first positive value that satisfy $x = \tan(x)$?
Do you know why Wolfram Alpha is showing $1.5708$ as a solution when you type $x = \tan(x)$ but not when you type $\tan(x) = x$?

Thanks.

As you see from the plot of $\tan x$, you're intercepting the asymptote, which is not really the desired behavior. Bisection is not the best method to use.

However, if you're required to use bisection, then instead note that $\tan x = \frac{\sin x}{\cos x}$, so, for relevant values of $x$,

$$x = \tan x \implies x\cos x - \sin x = 0$$

The latter function is continuous, and you should get the desired solution of $x \approx 4.49$.

I have a MathLab code, which is based on bisection method; Here it is:

clc
clear
% zeros of equation x = tan(x);
% The roots of equation x=tanx by means of bisection method
% z  represents the root
% epsilon is error
% NR is the number of roots that you want; number of roots = 2*NR+1
% Max is the number of bisections

epsilon = 0.0001;
NR = 2;
Max = 100;

g=@(x) x-tan(x);

for k = -NR:NR
    a = (2*k-1)*pi/2 + epsilon;
    b = (2*k+1)*pi/2 - epsilon;
    if feval(g,a)*feval(g,b) < 0
            for j=1:Max
             z = (a+b)/2;
                    if (feval(g,a)*feval(g,z) < 0)
                       b=z;
                    elseif (feval(g,a)*feval(g,z) > 0)
                       a=z;
                    end
            end
            if feval(g,z) < epsilon %error
            z
            end
    end
end

% outputs for NR=2 are :    -7.7253 ,    -4.4934 , 0 ,     4.4934 ,     7.7253


Prove that $p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\pmod{\! 4}$ Let a prime number $p$ divide $a^2+b^2$ with some  $a,b \in \left\{ 1,2, \ldots , p-1 \right\}$ Prove that  $p\equiv 1 \pmod{4}$. Is the converse true?

I know that $a^2+b^2\equiv 0 \pmod{p}$ and I don't know.

$1)$ $\ \,a^2\equiv -b^2\,\Leftrightarrow\, (a/b)^2\equiv -1\,\Rightarrow\, (a/b)^4\equiv 1\pmod{\!p}$, so $\text{ord}_p (a/b)=4$.

Fermat's little (FLT) $(a/b)^{p-1}\equiv 1\pmod{\!p}$ implies $4\mid p-1$ (proof below).

Theorem: $a^k\equiv 1\pmod{\!p}\,\Rightarrow\, \text{ord}_p a\mid k$.

Proof: If not, then $\,k=m\left(\text{ord}_pa\right)+r\,$ with $\,0<r<\text{ord}_p a$

But then $a^k\equiv (a^{\text{ord}_pa})^m(a^r)\equiv 1^ma^r\equiv a^r\equiv 1\pmod {\!p}$ - contradiction.

$2)\ $ By contradiction: if $\, p\equiv 3\pmod{\! 4}$,$\,$ then $\, a^2\equiv -b^2\,\Rightarrow\, (a/b)^2\equiv -1\,\stackrel{(p-1)/2}\Rightarrow$

$ (a/b)^{p-1}\equiv \color{#00F}{(-1)^{(p-1)/2}}\equiv \color{#00F}{-1}\, $ mod $p\,$ contradicts FLT.

Proof by exhaustion.
$$
p = a^2 + b^2 = 3 \pmod 4 \implies
$$
If
$$
p = 1^2 + 1^2 = 2, \\
p = 1^2 + 2^2 = 1, \\
p = 1^2 + 3^2 = 2, \\
p = 2^2 + 2^2 = 0, \\
p = 2^2 + 3^2 = 1, \\
p = 3^2 + 3^2 = 2, \\
\pmod 4
$$
Then we have a contradiciton.  By symmetry of variables $a,b$ (namely commutativity of $+$) we have covered all possibilities.
$\blacksquare$
The other proofs are impossible to understand easily.  This one is plain and simple.


Non-existence of bijective, continuous function from $(0,1)$ to $[0,1]$ The problem : Give example of a continuous, onto function from $(0,1)$ to $[0,1]$. Is it possible for such a function to be one-one?
My partial solution : For the $1^{st}$ part of the question, I came up with this example -
$f : (0,1) \to [0,1]$ given by $f(x)=\sin\big({2\pi x}\big)$
This is continuous and onto, but not one-one.
What I'm asking : For the $2^{nd}$ part of the question, I feel that it should be provable that there cannot exist a continuous, onto, one-one function from $(0,1)$ to $[0,1]$ (If not, we need a counter-example). Any help regarding this proof (or what would really surprise me, a counter example)?
Thanks in advance.

Bijective continuous map implies the function is monotone, now $f(x)=0$ for some $x\in (0,1)$.

We know there exists a real number $y\in (0,x)$ and a real number  $z\in (x,1)$.

Now since $f(x)=0$ and the function is bijective then both $f(y)$ and $f(z)$ are greater than $0$ so the function was decreasing somewhere in $(y,x)$ and increasing somewhere in $(x,z)$, contradicting the monotonicity of $f$.

If $f: A\rightarrow B$ is a continuous map, then $f$ pulls closed sets in $B$ to closed sets in $A$. Given, $f$ is onto and $[0, 1]$ is closed. So its pre-image should be closed. But its pre-image $(0,1)$ is not cloed in $\mathbb{R}$. A contradiction. So we can't have such $f$


How we can differentiate between the shapes of a parabola and a hyperbola? How we can differentiate between the shapes of the conics hyperbolas and parabolas?

Draw two parallel chords of your conic, and let $a$ be the line through the midpoints of the chords. Then draw another pair of parallel chord (not parallel to the previous pair) and let $b$ be the line through their midpoints. If lines $a$ and $b$ are parallel, then the conic is a parabola, otherwise it is a hyperbola (and the lines meet at the center of the hyperbola).

An example can be seen in figure below: on the left a parabola and on the right a branch of hyperbola.

A parabola is the intersection of a plane parallel to the surface of a cone (green) and a hyperbola the plane is parallel to the central axis of the cone (red).


Evaluating series with factorial denominator (sanity check). Is my approach to evaluating this series correct?

$$\sum_{n=1}^\infty \frac{n}{(n+1)!}$$

Has partial sum equivalent to:

$$S_m = \sum_{n=1}^m \frac{n}{(n+1)!} = \sum_{j=2}^{m+1} \frac{j-1}{j!} = \sum_{j=2}^{m+1} \frac{1}{(j-1)!} - \sum_{j=2}^{m+1} \frac{1}{j!} $$

For $j$ such that $m+1>j>2$ the terms of the left sum are cancelled by the terms of the right, leaving

$$ S_m =1-\frac{1}{(m+1)!}$$

Hence $ \lim_{m\rightarrow\infty} S_m  = 1$

Apologies for this one. The book I am using hasn't really offered anything on series with factorial denominators (yet). Thanks!

Your solution is not only correct but also a very nice one.

Once you know the closed form for the partial sum, you can also prove it by induction, starting with $S_1=1-1/2=1/2$ and taking the induction step

$$
\begin{align}
S_{m+1}&=S_m+\frac{m+1}{(m+2)!}
\\
&=1-\frac1{(m+1)!}+\frac{m+1}{(m+2)!}
\\
&=1+\frac{m+1-(m+2)}{(m+2)!}
\\
&=1-\frac1{(m+2)!}\;.
\end{align}
$$


definition of open sets in metric spaces For metric spaces, the definition of an open set $U\subset X$ is that it is a set which for any point $u\in U$ in the set there exists some $\epsilon>0$ such that the open ball $B_\epsilon(u)\subset U$.
My question is where is the effect of the mother metric space X?

The set of points in the open ball $B_\epsilon(u)$ depends on the ambient space $X$. That is, if we considered $Y \subset X$ with $U \subset Y$, then it is not necessary that $B_{\epsilon,X}(u) = B_{\epsilon,Y}(u)$. For example, if $Y = U$, then $B_{\epsilon,Y}(u)$ is by definition contained in $U$.

Consider the euclidean metric with

$U =[0,1)$ ; $X= \Bbb R_+$ including $0$
$U =[0,1)$ ; $X= \Bbb R$

In first case $U$ is open because of $X$. We cannot move outside $0$ to find the neighborhood around $0$ because of $X$. Thus there exist a positive $\varepsilon$ such that $( 0, +\varepsilon)$ lies entirely in $U$: this can be checked for other points in $U$ as well.

Second case, $U$ is not open as there is no $\varepsilon$ such that $(-\varepsilon,+\varepsilon)$ lies entirely in $U$.


Intuition on group homomorphisms So I'm studying for finals now, and came across the idea of homomorphisms again. This is not a new idea for me at all, having seen them in groups, rings, fields ect. However, on reevaluating them I realized suddenly that I really don't understand them on the same level as I thought. While isomorphisms seem to be very natural to think about, I can't visualize what's happening in homomorphisms. At this time, I only have a sketchy idea that the existence of a homomorphism between groups (or between groups and what they're acting on) means that they both somehow combine the same way..

Does someone mind sharing their intuition on the concept? As always, any help is appreciated, thanks

You know that if there is an isomorphism $h:A\to B$ from an algebraic structure (monoid, group, ring, etc.) $A$ onto another algebraic structure $B$ of the same kind, then $B$ is essentially just $A$ ‘in disguise’: the two structures are essentially the same structure. In other words, Isomorphisms are the maps that preserve the structure exactly.

Homomorphisms preserve some of the structure. (Here some may be all, since every isomorphism is a homomorphism. That is, it’s some in the sense of $\subseteq$, not $\subsetneqq$) They preserve the operations, but they may allow elements that ‘look enough alike’ to be collapsed to a single element. For instance, the usual group homomorphism from $\Bbb Z$ to $\Bbb Z/2\Bbb Z$ (for which you use the notation $\Bbb Z_2$) ‘says’ that all even integers are essentially the same and collapses them all to the $0$ of $\Bbb Z/2\Bbb Z$. Similarly, it ‘says’ that all odd integers are essentially the same and collapses them all to the $1$ of $\Bbb Z/2\Bbb Z$. It wipes out any finer detail than odd versus even. When you learn in grade school that even $+$ even $=$ even, odd $+$ even $=$ odd, and so on, you’re essentially doing the same thing.)

The kernel of the homomorphism is a measure of how much detail is wiped out: the bigger the kernel, the more detail is lost. In the example of the last paragraph, the kernel is the entire set of even integers: the fact that all even integers are in the kernel says that they’re all being seen as somehow ‘the same’, and even more specifically, ‘the same’ as $0$. An isomorphism has a trivial kernel: the only thing that it sees as looking like $0$ is $0$ itself, and no detail is lost.

Another way to put it is that a homomorphic image of an algebraic structure is a kind of approximation to that structure. If the homomorphism is an isomorphism, it’s a perfect approximation; otherwise, it’s more or less crude approximation. As the kernel of the homomorphism gets bigger, the crudeness of the approximation increases. In the case of groups, if the kernel is the whole group, then the homomorphic image is the trivial group, and all detail is lost: all that’s left is the fact that we started with a group.

This answer will be painfully similar to the one here , but in any case the details might be of help.
Say we have groups $ (G, *) $ and $ (G', \cdot) $ and a map $ \phi : G \rightarrow G' $. We can ask ourselves the following question : Under what kinds of $ \phi $ do we have that $ (\phi(G) , \cdot ) $ is a group, and that " relations in $ G $ give corresponding relations of images in $\phi(G)$ " i.e. that $ ( x*y = z \text{ in group } G ) \implies ( \phi(x) \cdot \phi(y) = \phi(z) \text{ in group } \phi(G) ) $ ?
Proceeding as in the answer linked above, we see maps satisfying $ \phi(x*y) = \phi(x)\cdot\phi(y) $ are precisely the ones we were looking for. Such maps are conventionally called "group homomorphisms".


Integrating $\int _0^\pi \frac{1}{1+\sin^2(\theta)}$ using Cauchy's formula I need to evaluate $\displaystyle \int _0^\pi \frac{d\theta}{1+\sin^2(\theta)}$ by using Cauchy's integral formula, and the substitution $z = e^{i\theta}.$ So far, I have that
$$d\theta = \frac{1}{iz}dz$$ and
$$1+\sin^2(\theta) = 1+ \frac{1}{4}\left(z^2 -2 +\frac{1}{z^2}\right) = \frac{1}{4}\left(4+ z^2 -2 +\frac{1}{z^2}\right) = \frac{1}{4}\left(2+ z^2 +\frac{1}{z^2}\right).$$

So the integral becomes

$$\int \frac{4dz}{iz \left(2+z^2+\frac{1}{z^2}\right)}= \int \frac{(4z)dz}{iz^4 +2iz^2+i}$$

Now, I'm not sure that last step was in the right direction, but it seems to me to be the only way I can factor the bottom so that I can apply Cauchy's integral formula. Now, the bottom factors as
$$(z-i)^2(z+i)(iz-i).$$
However, I'm a bit confused about this factorization because I did it by polynomial division. But when I used the quadratic formula, I got that $z = -i$ is also a double root, which would imply that it factorizes as $$(z-i)^2(z+i)^2.$$ So first of all, can someone help me clear up this confusion? And second, I'm not quite sure where to be evaluating the integral because both of the singularities lie on the unit circle, and I am only supposed to integrate from $0$ to $\pi,$ so should I just take half of the integral from $0$ to $2\pi$? If this is not always possible, I'd appreciate a brief explanation as to why.

One way to consider is to write:

$\displaystyle \int_{0}^{\pi}\frac{1}{1+\frac{1-\cos(2x)}{2}}dx$

$\displaystyle=\int_{0}^{\pi}\frac{2}{3-\cos(2x)}dx$

Let $t=2x, \;\ dx=1/2dt$

$\displaystyle\int_{0}^{2\pi}\frac{1}{3-\cos(t)}dt$

Now, use the complex cos rep.

$\displaystyle\int_{C} \frac{1}{3-\frac{z+z^{-1}}{2}}\cdot \frac{1}{zi}dz$

$\displaystyle\int_{C} \frac{2i}{z^{2}-6z+1}dz$

Only the pole $-2\sqrt{2}+3$ lies in the unit circle.

$\displaystyle 2\pi i Res(-2\sqrt{2}+3)=2\pi i \left(\frac{-i}{2\sqrt{2}}\right)=\frac{\pi}{\sqrt{2}}$

With $\sin \theta = \dfrac{z - 1/z}{2i}$ you should get $$1 + \sin^2 \theta = - \dfrac{z^2}{4} + \dfrac{3}{2} - \dfrac{1}{4z^2} $$
You certainly shouldn't have singularities on the unit circle since $1+\sin^2 \theta$ is never $0$ for real $\theta$.

You'll want to integrate over the whole circle to use the Residue Theorem, so
yes, take half of the integral from $0$ to $2 \pi$.


How does one formally define sampling from a probability distribution? Every introduction to probability that I have seen describes sampling from a distribution based on an intuitive notion of a fair coin flip.

I have never seen "fair coin flip" formally defined. Is the definition implied by the axioms of probability theory?

Is probability theory predicated on some notion of an atomic coin flip operation that is not further formalized?

Probability is formalized in terms of measure theory. One usually talks about a probability space $(\Omega, \mathcal{F}, P)$. Here $\Omega$ is the set of all "outcomes," $\mathcal{F}$ is the set of all "events," and $P:\mathcal{F} \to [0,1]$ is a probability measure, i.e., given an event $E \in \mathcal{F}$, $P(E)$ is the probability of $E$. These satisfy all sorts of axioms which I won't get into. It's good to have a running example, so let's say $\Omega$ is the set of possible outcomes when a red and a green die are rolled, so
$$\Omega = \{(1,1),(1,2),\dots,(6,6)\}.$$

Now a random variable is defined as a function $X:\Omega \to \mathbb{R}$. For instance, $X$ might be the sum of the red and the green dice. Thus $X(a,b) = a+b$. Usually we do not use function notation for $X$ because we are not concerned with the particular value $X(a,b)$, but rather the many values that $X$ could take on, and the distribution of those values. $X$ induces a probability measure on $\mathbb{R}$, namely if $A \subseteq \mathbb{R}$ then the measure of $A$ is $\mu(A) = P(X \in A)$.

This measure and the distribution of $X$ are completely captured by the distribution function $F(x)=P(X \leq x)$. In our example, $F(x)$ will be a step function. In general, distribution functions are nondecreasing, right-continuous, and have $\lim_{x \to -\infty}F(x)=0$ and $\lim_{x \to \infty}F(x)=1$. Moreover, every function with those properties is a distribution function.

I have just described how to go from a random variable to its distribution function, but we can go the other way. Namely, given a distribution, we can sample a random variable from it, by which we mean we choose a probability space $(\Omega,\mathcal{F},P)$ and a function $X:\Omega \to \mathbb{R}$ satisfying $P(X \leq x) = F(x)$ for all $x \in \mathbb{R}$. It is not obvious that such a random variable must exist! But in fact one does provided you have a valid distribution function.

In probability parlance, it means you are analyzing independent, identically distributed random variables with a given distribution.

However, when you get into the realm of statistics, the whole business of designing sampling schemes becomes a high and complex art.  In the real world, the assumption that observations are independent (or identically distributed) can be a muddy one.


Discontinuous solution to first order ODE with delta function coefficients Consider the following first-order ODE:

$$y'(x) = f'(x) y(x);$$

this has solution $y(x) = C e^{f(x)}$.  Now, consider taking

$$f_\lambda(x) = \frac{\alpha}{2}\left(1+\tanh(\lambda x)\right);$$

in the limit $\lambda \to \infty$, $f_\lambda(x) \to \alpha\Theta(x)$ (where $\Theta(x)$ is the Heaviside step function), and the solution becomes $y(x) = C e^{\alpha \Theta(x)}$, which is discontinuous at $x = 0$.  Presumably, there is a sense in which one can think of this as some kind of solution to the ODE with delta-function coefficient

$$y'(x) = \alpha \delta(x) y(x).$$

Now, the discontinuity in $y$ at $x=0$ is

$$\Delta y \equiv y(0^+) - y(0^-) = C(e^\alpha - 1).$$

My question is the following: is there a way to extract $\Delta y$ from the singular ODE directly without first finding a solution with smooth $f(x)$ and then taking an appropriate limit?  For instance, in analogy with what one does with second-order ODEs with delta function coefficients, one could try to integrate it:

$$\Delta y = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} y' dx = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} \alpha \delta(x) y(x) dx = \alpha y(0),$$

but this has the problem that $y(0)$ is not defined due to the discontinuity in $y(x)$.

A friend of mine figured out a solution, at least to this toy problem.  As commenters pointed out, the problem with naively integrating the ODE to get an expression for $\Delta y$ is that it is ill-defined to integrate a delta function against a discontinuous function.  The solution is to first divide through by $y$ to get

$$\frac{y'(x)}{y(x)} = \alpha\delta(x).$$

This equation can now be integrated across the delta function to yield

$$\Delta \ln y \equiv \ln y(0^+) - \ln y(0^-) = \alpha,$$

thus

$$\frac{y(0^+)}{y(0^-)} = e^\alpha,$$

which indeed agrees with the $f(x) \to \alpha \Theta(x)$ limit of the solution for general $f(x)$.

A cool way of solving this is using "symbolic derivatives".

Some mathematical background

Symbolic functions

Let $\phi(x) \in C_0^\infty (I)$ (a continuous function with continuous derivatives of all orders, all vanishing outside a finite interval $I$, also known as a test function). This set forms a vector space and, in this space, we say a sequence of testing functions $\phi_n(x)$ converges to zero if the functions $\phi_n(x)$ and all their derivatives converge uniformly to zero and if all the functions $\phi_n(x)$ vanish identically outside the same finite interval.

We sy $F(\phi)$ is a linear functional on the space of testing functions if to every testing function $\phi(x)$ a real or complex number is assigned such that
\begin{align}
F(\phi_1 + \phi_2) &= F(\phi_1) + F(\phi_2), \\
F(\lambda \phi) &= \lambda F(\phi),
\end{align}
where $\lambda$ is a scalar. A functional $F(\phi)$ is said to be continuous if the sequence of numbers $F(\phi_n)$ converges to zero whenever the sequence of testing functions $\phi_n(x)$ converges to zero in our sense. Here are some examples of continuous linear functionals:
\begin{align}
F_1(\phi) &= \phi'(0),\\
F_2(\phi) &= \int_0^1 \phi(x)\,dx.
\end{align}
Laurent Schwartz, in his famous work Théorie des distributions, calls any continuous linear functional on the space of testing functions a distribution.

We know that any continuous linear functional on a vector space having a scalar product can be expressed as a scalar product $[1]$, i.e.
$$
F(\phi) = \langle s(x), \phi(x) \rangle.
$$
With the natural choice of scalar product
$$
\langle \phi(x),\psi(x) \rangle = \int_{-\infty}^\infty \phi \psi\,dx,
$$
some linear functionals can be expressed as a scalar product and some not (this is because with this definition of scalar product, the space of testing functions is not complete). For example,
$$
F(\phi) = \int_0^1 \phi(x)\,dx = \int_{-\infty}^\infty s(x)\phi(x)\,dx,
$$
where $s(x) = 1$, $0 < x < 1$, and $s(x) = 0$ otherwise. There are other functionals that cannot be expressed this way; for example, $F(\phi) = \phi(0)$ or $F(\phi) = \phi'(0)$. However, we can introduce the $\delta(x)$ symbol such that
$$
F(\phi) = \phi(0) = \int_{-\infty}^\infty \delta(x) \phi(x)\,dx.
$$
In this context, the $\delta$-function is a symbolic function. In general, given a continuous linear functional $F(\phi)$ on the space of testing functions, we introduce a symbol, say $s(x)$, and write
$$
\int_{-\infty}^\infty s(x) \phi(x)\,dx = F(\phi).
$$
We say that $s(x)$ is a symbolic function. Note that a symbolic function need not have values. It produces values only when multiplied by a testing function and then integrated. A final observation is that every integrable function is a symbolic function (why?).

Derivatives of symbolic functions

To define the concept of the derivative of a symbolic function in such a way that it will be valid, we can use integration by parts. We say that $s'(x)$ is the derivative of $s(x)$ if
$$
\int_{-\infty}^\infty s'(x) \phi(x)\,dx = - \int_{-\infty}^\infty s(x)\phi'(x)\,dx
$$
for every testing function $\phi(x)$. Since the right-hand of this equation always exists, the left-hand side can be used to define the derivatives of $s(x)$. For example, $\delta'(0)$ is defined by
$$
\int_{-\infty}^\infty \delta'(x) \phi(x)\, dx = -\int_{-\infty}^\infty \delta(x) \phi(x)\,dx = -\phi'(0).
$$
Extending this argument, it is easy to show that $\delta^{(n)}(x)$ is such that
$$
\int_{-\infty}^\infty \delta^{(n)}\phi(x)\,dx = (-1)^n \frac{d^n \phi}{d x^n}\Bigg|_{x=0}.
$$

The problem at hand

Of course, the relation
$$
\frac{y'(x)}{y(x)} = a \delta(x)
$$
is formally nonsence, but if we write it in the context above, i.e.,
$$
\int_{-\infty}^\infty \frac{y'(x)}{y(x)} \phi(x)\,dx = a \int_{-\infty}^\infty \delta(x) \phi(x)\,dx = \phi(0),
$$
everything is well stated.

Before we proceed, let's derive a fantastic relation between the $\delta$-function and the Heaviside function. Given
$$
\int_{-\infty}^\infty H(x) \phi(x)\,dx = \int_0^\infty \phi(x)\,dx,
$$
we have
$$
\int_{-\infty}^\infty H'(x) \phi(x)\,dx = -\int_{-\infty}^\infty H(x) \phi'(x)\,dx = - \int_0^\infty \phi'(x)\,dx = \phi(0).
$$
Consequently $H'(x) = \delta(x)$ in the symbolic framework!

With this amazing result (for $y(x) > 0$),
\begin{align}
0 &= \int_{-\infty}^\infty \left[\frac{y'(x)}{y(x)} - a\delta(x)\right]\phi(x)\,dx \\ \\
  &= \int_{-\infty}^\infty \frac{d}{dx} \left[\log y(x) - a H(x)\right]\phi(x)\,dx \\ \\
  &= -\int_{-\infty}^\infty \left[\log y(x) - a H(x)\right] \phi'(x)\,dx
\end{align}
for every test function $\phi(x)$. Finally, using the du-Bois-Reymond theorem, the function
$$
y(x) = e^{a}H(x) + H(-x)
$$
satisfies the differential equation in the symbolic sense.

Nice, huh!


Unique steady state vector in relation to regular transition matrix Define a steady-state vector for a transition matrix $T$ as a probability vector $v$ such that $Tv = v$ ($1$ is the eigenvalue for $v$).

  Define a transition matrix $T$ as regular if there exist a $k \ge 1$ such that each entry of $T^k$ is non-zero.

Consider the following theorem:

Suppose $T$ is a regular transition matrix such that $x_k = T^kx_0$ (a discrete dynamic system), then there is a unique steady-state vector $v$ such that $Tv = v$. Furthermore, if $x_0$ is any probability vector: $$\lim_{k\to\infty} x_k = \lim_{k\to\infty} T^kx_0 = v$$

Here's the confusion:

Suppose $\lambda$ is an eigenvalue for the $x_0$ mentioned above, then $$T^kx_0 = \lambda^kx_0 \Rightarrow \lim_{k\to\infty} T^kx_0 = \lim_{k\to\infty} \lambda^kx_0$$
Suppose we have real entries and real eigenvalues, then we have few cases:

1) $\lim_{k\to\infty} \lambda^k = \pm\infty$ when $\lambda > 1$ or $\lambda < -1$

2) $\lim_{k\to\infty} \lambda^k = 0$ when $ -1 < \lambda < 1$

3) $\lim_{k\to\infty} \lambda^k = 1$ when $\lambda = 1$

4) $\lim_{k\to\infty}$ oscillate between $1$ and $-1$ when $\lambda = -1$

It seems to me that, besides the trivial case of 3), none of these cases can produce $\lim_{k\to\infty} \lambda^kx_0 = v$. Since $v$ is a unique steady-state vector, in the other $3$ cases, $v$ becomes $0$, $\pm\infty$, or oscillate between $x_0$ and $-x_0$.

I think I'm missing something but I don't know what it is. Can someone care to explain what I'm missing?

The Perron Frobenius theorem for non negative regular matrices states that there
is a real positive eigenvalue $\lambda$ of multiplicity one that has positive
left & right eigenvectors and for any other eigenvalue $\mu$ we have $|\mu| < \lambda$.

In this case, since $T$ is stochastic, we have $\|T\|_1 = 1$ and $e^T T = e^T$, hence we see that $\lambda = 1$.

By considering the above, if $V J V^{-1}$ is the Jordan form of $T$, we see that
$J^k \to \begin{bmatrix} 1 &   \\
& 0\end{bmatrix} $, where the $0$ is a block diagonal of sufficient size.
In particular, $T^k \to T_\infty$, where $T_\infty$ is a rank one matrix and
hence of the form $T_\infty = a b^T$ for some $a,b$. Since $e,v$ are left and
right eigenvectors and $v$ is positive, we see that $T_\infty = v e^T$.

In particular, $\lim_k T_k x_0 = T_\infty x_0 = v e^T x_0 = v$.

Suppose $\lambda$ is an eigenvalue for the $x_0$ mentioned above

If the matrix $T$ is a transition matrix and your assuming $x_0$ is a probability vector and an eingenvector associated to the eingenvalue $\lambda$, $x_0$ must be the steady-state vector $v$ associated with the aingenvalue $\lambda=1$.


Question regarding limits of multivariable functions - is it really zero? I am given the following problem in a textbook, and it's a solved problem.

  Find $\displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4}$ or justify its nonexistence.

The author justifies the nonexistence of the limit in the following way: he uses the paths $x = 0$ then $y = 0$ and then, to justify its nonexistence, $y = x$. These are the author's notes:

Since

\begin{align}
\lim_{(x,0) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(x,0) \to (0,0)} \frac{x^2 \cdot 0^2}{x^4 + 3 0^4} = \lim_{(x,0) \to (0,0)} \frac{0}{x^4} = 0
\end{align}

and

\begin{align}
\lim_{(0,y) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(0,y) \to (0,0)} \frac{0^2 \cdot y^2}{0^4 + 3 y^4} = \lim_{(0,y) \to (0,0)} \frac{0}{3y^4} = 0
\end{align}

but

\begin{align}
\lim_{(x,x) \to (0,0)} \frac{x^2 y^2}{x^4 + 3y^4} = \lim_{(x,x) \to (0,0)} \frac{x^2 \cdot x^2}{x^4 + 3 x^4} = \lim_{(x,x) \to (0,0)} \frac{x^4}{4x^4} = \frac{1}{4}
\end{align}

then the limit does not exist.

So my question is: are these limits

\begin{align}
\lim_{(0,y) \to (0,0)} \frac{0}{3y^4} \qquad \lim_{(x,0) \to (0,0)} \frac{0}{x^4}
\end{align}

really equal to zero? Shouldn't they be indeterminate, given that the variable on the denominator approaches zero?

No, both those limits are indeed defined and equal to zero.

The multivariable context is adding unnecessary "noise" here. Consider the simpler expression $$\lim_{x\rightarrow 0}{0\over x},$$ which already captures what's going on.

Remember the precise definition of a limit. Colloquially, we often say something like "$\lim_{x\rightarrow a}f(x)=L$ iff I can get $f$ to be as close as I want to $L$ just by requiring that $x$ be sufficiently close to $a$." However, this misses a very important point which is explicit in the precise formal definition: $$\lim_{x\rightarrow a}f(x)=L\iff \forall \epsilon>0\exists \delta>0\forall x((\color{red}{0<}\vert x-a\vert<\delta)\implies (\vert f(x)-L\vert<\epsilon)).$$ Note the "$0<$" clause in the left hand side: what this says is that we ignore the specific behavior of $f$ at exactly $a$.

This is why $\lim_{x\rightarrow 0}{0\over x}=0$, despite the bad behavior when $x$ is exactly $0$: for any $\epsilon>0$, let $\delta=17$ (say); then it's easy to check that for any $x$ within $17$ of $0$ but not actually equal to  $0$, we have $\vert{0\over x}-0\vert<\epsilon$.

A reasonable question at this point is why we define limits this way. Ultimately that's a question that deserves a serious answer, but very briefly the point is that we're interested in what $f$ "ought to be" at $a$, not what it literally is. Both the derivative and the integral give great examples of this: in each case, we really want to divide by zero but that gives nonsense, so instead we look at what happens when the change in $x$ or the width of our rectangles gets really close to zero without actually being zero. This also gives a nice spin on continuity: intuitively, a function is continuous if it always is what it ought to be.

Both limit

$$\begin{align}
\lim_{(0,y) \to (0,0)} \frac{0}{3y^4}=0 \qquad \lim_{(x,0) \to (0,0)} \frac{0}{x^4}=0
\end{align}$$

are correct since the function $f(x,y)$ is identically zero along the trajectories with either $x=0$ or $y=0$.

In other words, what we cannot consider is $x=y=0$ at the same time but all the trajectories with either $x=0$ or $y=0$ are admissible and consistent with the definition of limit.

To clarify better that point see also the related Why are we allowed to cancel fractions in limits?


Algorithm to find the maximum/minimum of a polynomial without graphing. For a quadratic equation of the form $y=ax^2+bx+c$ the max/min occurs at $x=-\frac{b}{2a}$. Is there any hard and fast equation like this for polynomials of degree $\geq 4$?. For such polynomials the solution which I got online suggested to plot the curve and find. How to find the absolute maxima without graphing?

If you have a third-degree one, you can derivate your expression (this is quite easy to do by a algorithm of your own, for polynomials). Then you get the max/min by x=-b'/2a'.

For a fourth-degree polynomial, just derivate again, and then use your previous programm for 3rd-deg to find the max/min !

Once you've got an algorithm for degree n, repeat the operation for degree n+1


Why are gauge integrals not more popular? A recent answer reminded me of the gauge integral, which you can read about here.

It seems like the gauge integral is more general than the Lebesgue integral, e.g. if a function is Lebesgue integrable, it is gauge integrable. (EDIT - as Qiaochu Yuan points out, I should clarify this to mean that the set of Lebesgue integrable functions is a proper subset of gauge integrable functions.)

  My question is this: What mathematical properties, if any, make the gauge integral (aka the Henstock–Kurzweil integral) less useful than the Lebesgue or Riemann integrals?

I have just a cursory overview of the properties that make Lebesgue integration more useful than Riemann in certain situations and vice versa. I was wondering if any corresponding overview could be given for the gauge integral, since I don't quite have the background to tackle textbooks or articles on the subject.

I would have written this as a comment, but by lack of reputation this has become an answer. Not long ago I've posed the same question to a group of analysts and they gave me more or less these answers:

1) The gauge integral is only defined for (subsets of) $\mathbb R^n$. It can easily be extended to manifolds but not to a more general class of spaces. It is therefore not of use in (general) harmonic analysis and other fields.

2) It lacks a lot of very nice properties the lebesgue integral has. For example $f \in \mathcal L^1 \Rightarrow |f| \in \mathcal L^1$ obviously has no generalization to gauge theory.

3) and probably most important. Afaik (also according to wikipedia) there is no known natural topology for the space of gauge integrable functions.

The gauge integral is as general as Lebesgue integral. It can be defined over arbitrary spaces. In fact it is absolutely necessary when we simplify calculus. There are simple proofs of Hakes theorem the study of improper integrals need not be carried out separately Feynman path integrals can be more naturally and easily defined using gauge integration rather than lebesgue and it is also easier to use gauge integral for Weiner integration.

The question regarding space of gauge integrable functions not being useful are irrelevant as one can always use the space $L^1$ for absolutely convergent gauge integrals. It is the same $L^1$ space, but gauge integrals give you added benefits of having better fundamental theorem of calculus , and Stokes theorem is given simpler proof. Fubini’s theorem proof is simpler and more natural. Further though monotone convergence theorem has simpler prof in lebesgue the simplicity is deceiving as first one has to extend measure to sigma algebra. In Henstock automatically measure is generated. It is simply the human inertia of a generation.


Am I solving dido's isoperimetric (variational) problem correctly? I am trying to solve dido's isoperimetric problem, more specifically, the version where we have to maximize the area under a curve, given that the two endpoints are on the x-axis, and given a fixed arclength:

That is, we have to maximize
$$J(y)=\int_a^by(x)dx$$
subject to constraint
$$C(y)=\int_a^b\sqrt{1+(y'(x))^2}dx$$
Where we assume $a$ is fixed, but $b$ is allowed to vary.

Am I solving this correctly?

Besides the boundary constraint (because of variable boundary $b$) $L_{y'}(b)=0$, we of course formulate the Euler-Lagrange equation $L_y=\frac d {dx}L_{y'}$, based on $J$ subject to the constraint $C$, so that the Lagrangian becomes:

$$L(y,y')=y(x)+\lambda\sqrt{1+(y'(x))^2}$$

Therefore:
$$L_y=1$$
$$L_{y'}=\lambda \frac {y'(x)} {\sqrt{1+(y'(x))^2}}$$
$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1} {\sqrt{1+(y'(x))^2}}-\frac {(y'(x))^2}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$

which is equal to

$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$

  This gives the Euler-Lagrange equation $$\lambda
y''(x)=\left (1+(y'(x))^2\right)^{\frac 3 2}$$

I have no idea how to solve this ODE, and moreover, it doesn't seem like this is what I should be getting.

Did I derive this result correctly? If so, How do I solve it?

note: I know that it is also possible to solve by parameterizing $x=x(t), y(x)=y(x(t))$. I want to do this as well, later, but I would first like to understand the approach I'm taking here.

First, integrate both sides with regards to $x$. Next, we can get an explicit function $x=f(y')+C_1$, $C_1$ is a constant. Here we have a fact :
                        $y'(x)=0$ if and only if $x=C_1$
Geometrically, curve $y$ reaches the maximum value when $x=C_1$. Hence, $x$ increases for $x<C_1$ and x decrease for $x>C_1$. So, $y'(x)$ will never be zero when $x$ in $(0,C_1) \cup (C_1,1)$. Finally, with some algebraic manipulations, we can get a circle equation (with radius $\lambda$).


How to calculate the spiral around a curve? I have a curve given by a set of points and want to build a spiral around it (like this).

I tried it by

STEP 1:
calculated the vector of each step by

$$v_n = \left<x_n,y_n,z_n\right> - \left<x_{n-1},y_{n-1},z_{n-1}\right>$$

STEP 2:
I assumed a vector/plane of the rotating trajectory as
$$ w_n = \left<\cos(i),1,\sin(i)\right>$$

where $i$ represents the intervals and increases by the point number.

And calculated the perpendicular vector to find the spiral points by

$$X = v_y \cdot w_z - v_z \cdot w_y $$
$$Y = v_z \cdot w_x - v_x \cdot w_z $$
$$Z = v_x \cdot w_y - v_y \cdot w_x $$

And, of course, I normalized the scale.

The spiral is formed, but I have discontinuity or deformation at sharp angles (probably because the vector direction is changed). It is not a matter of smoothness, as the spiral changes its direction to rotate in the opposite direction.

Where did I do wrong?

spiral will evolve according to helical angle $$\phi_n=\phi_{n-1}+2 \pi \frac{|\vec{x}_n-\vec{x}_{n-1}|}{\lambda}$$ where $\lambda$ is the pitch of the helix and is measured relative to an "up" orientation normal vector $\vec{u}$ which is the component of the acceleration vector $\vec{a}_n$ perpendicular to the velocity vector $\vec{v}_n$ $$\vec{u}_n=\hat{a}_n-(\hat{a}_n \cdot \hat{v}_n ) \hat{v}_n$$ where $$ \vec{v}_n=\vec{x}_n-\vec{x}_{n-1}$$ and $$\vec{a}_n=\vec{v}_n-\vec{v}_{n-1}$$ the location of the helix is then given by $$\vec{h}_n=\vec{x}_n+\hat{u}_nR\cos \phi_n+(\hat{u}_n\times\hat{v}_n)R\sin \phi_n$$ where $R$ is the radius of the helix

Probably your steps are too large.  You are rotating a full radian between steps.  If you use finer steps in both directions you will improve things.  If you just divide $i$ by $10$ in the $w$ equation the pitch will increase by a factor $10$.  Maybe you should interpolate $10$ steps between your $v$ points, then divide $i$ by $10$ in $w$.  The most challenging points are those where the axis of the helix bends most.  You need the steps in your helix rather small by comparison with that scale.


$2(n-1)! \equiv -1 \mod n+2 \iff n+2$ is a prime Problem: Show that $2(n-1)! \equiv -1 \mod n+2 \iff n+2$ is a prime.

I know that Wilson's theorem states that $(n-1)! \equiv -1 \mod p $ for $p$ a prime, so that is the important thing to know with these type of problems.

I know that $-1 \equiv n+1 \mod n+2$ ,
and that if $n+2$ is prime, then $(n+1)!  = -1 \mod n+2$. So this is all I have for now, any hints appreciated.

As per the comments since I know that $2 (n-1)! \equiv (-1)(-2)(n-1)! \equiv (n)(n+1)(n-1)! \equiv (n+1)! \equiv -1 \mod (n+2)$, then $n+2$ is prime.


Is $T$, the set of all polynomials with integer coefficients, countable? Is $T$, the set of all polynomials with integer coefficients, countable?

Here is my attempt at a proof:

Let $S$ = {$ (a_0, a_1, ..., a_k)$ | $a_i\in \Bbb Z$, $k \in \Bbb N$}

$T$ is the set in question. That is, $T$ = {$ (a_0, a_1x, ..., a_kx^{k})$ | $a_i \in \Bbb Z$, $k \in \Bbb N$}.

Then by the enumeration principle $S$ is countable, i.e., |$S$| = |$\Bbb N$|, because it can be labelled by

$L =$ {$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, , , (, )$} [Note: I've included "," as an element].

Now define an injection $f: T \to S$ by:

$f(a_0 + a_1x + ... +a_kx^{k}$) = ($a_0, a_1, ..., a_k$).

Then $|T| \le |S|$.

We can define a similar injection $g: S \to T$, establishing |$S| \le |T$|. So |$T$| = |$S$| = |$\Bbb N$| which is what we wanted to prove.

Is this good? I'm pretty new to set theory so I want to make sure these proofs are valid.

Any polynomial $p(x)$ over $\Bbb Z$ can be written as $p(x)=a_0+a_1x+\ldots +a_nx^n,n\in \Bbb N$.

Since $a_i\in \Bbb Z$ ,and $\Bbb Z$ is countable so for each $a_i$ we have countable choices .

EDIT:By "choices" I mean number of options available for us to select the $a_i's$ from.Since the polynomial is over $\Bbb Z$ so we have the option of selecting the $a_i's$ from the set of integers.

And since countable union of countable sets is countable so the set of all polynomials over $\Bbb Z$ is countable also

Your result only holds if you show that $|S| = |\Bbb N|$ which you have not done because you could not use the enumeration principle. $S$ has infinitely many elements of the form $(a, 0, \cdots, 0) $ and then infinitely many terms of the form $(0, b, 0, \cdots) $ and then infinitely many of the form $(a, b, 0, \cdots) $ and so on and so forth and therefore you cannot enumerate them with that ease. That is the problem with the demonstration. You would still have to build an explicit injection from $S $ on to $\Bbb N $ to prove $|S| = |\Bbb N|$.


Evaluate $x$ for $2x^2 - 3x - 2<0$ Solve $$2x^2-3x-2=0$$ we get $x=-0.5;2$

So, $$-0.5<x<2$$
The key answer is $$0.5<x<2$$
Which one is true?

Factorizing, we get $$2x^2-3x-2=(2x+1)(x-2).$$

Its roots are $-\frac12$ and $2$.  Using a tabular method, we have
\begin{array}{|c|c|c|c|}
\hline
\text{Interval} & 2x+1 & x-2 & \text{Sign of}\ f(x)\\
\hline
x<-\frac12 & - & - & +\\
\hline
x=-\frac12 & 0 & - & 0\\
\hline
\color{blue}{-\frac12 < x < 2} & + & - & \color{blue}{-}\\
\hline
x=2 & + & 0 & 0\\
\hline
x>2 & + & + & +\\
\hline
\end{array}

Therefore, the answer should be $\boxed{-\frac12<x<2}$.

The correct answer is $-0.5<x<2$.  There should be a negative sign in the key answer.

The inequality holds for all $x$ between the zeroes, including for example $x=0$.


Logarithm problem with two bases Given $$\log_x9+\log_9x=\dfrac{10}3.$$

How can I find the greatest value of $x$ that satisfies the equation above?

hint: $a = \log_{9}x$, then you have a quadratic in $a$: $a + \dfrac{1}{a} = \dfrac{10}{3}$.

HINT:

$$\log_x9+\log_9x=\dfrac{10}3$$

Now by Change of Base of Logarithm use $\log_x9=\dfrac{\log9}{\log x}=\dfrac1{\dfrac{\log x}{\log9}}=\dfrac1{\log_9x}$ to form a Quadratic equation in $\log_9x$

Now if $\log_9x=y,x=9^y>0$ for real finite $y$


What is the most efficient strategy to compute the area of the following region? $y=x^2-16$, $3y-16x=0$, $y=0$ for $y \geq 0$ Please keep in mind, I have included the original question as an image, for any references. So here is what I know: I know I am supposed to set the two equations equal to each other, and then integrate. But for the second equation, $3y-16x-16=0$ am I supposed to solve for $y$, and then set it equal to the first equation? If so, I get some weird numbers, and I am not sure if there is a more efficient way to do this. I know I integrate eventually. Also, what do I do with the $y=0$? Is that for the integration part?

Also, I'm new here, so I apologize in advance for any misconceptions I have, or for any formatting issues. Please notify me of any issues before choosing to downvote me.

You have been supplied 3 curves.

You should plot the 3 curves.  And you should algebraically find the points of intersection of the curves.

Then based on the shape of the curve, make a decision of whether it will be easier to integrate with respect to x or with respect to y.  Which curve will make your upper bound, your lower bound, and what will be your interval to integrate.

Finally you will integrate the result.

$y = x^2 - 16$ is a parabola.
$3y-16 x -16 = 0$ is a line
$y = 0$ is also a line.

$y = 0$ intercepts $y = x^2 - 16$ at $(-4,0),(4,0)$

$3y-16 x-16 = 0$ intercepts $y=0$ at (-1,0).

$y = x^2 - 16$ intercepts $3y-16 x -16= 0$

$y = \frac {16}{3} (x+1)\\
\frac {16}{3} (x+1) = x^2 - 16\\
x^2 - \frac {16}3 x - \frac {64}{3} = 0\\
x = \frac 83 + \frac {\sqrt {64+9\cdot 64}}{3} = \frac {8 + 8\sqrt {10}}{3}\\
y = \frac {176 + 128\sqrt {10}}{9}$

We only care about one of the points of intersection as $y<0$ at the other.

The region is vaguely triangular.  If you integrate with respect to x, you are going to need to break the region in two, if you integrate with respect to y, you do not.

$\int_0^{\frac {176 + 128\sqrt {10}}{9}} \sqrt {y+16} - \frac {3}{16} y + 1 \ dy$

or

$\int_1^4 \frac {16}{3} (x+1)\ dx + \int_4^{\frac {8 +8 \sqrt{10}}{3}} \frac {16}{3}(x+1) - (x^2 - 16)\ dx$

A segment of a parabola is known to occupy two thirds of its bounding rectangle.

By solving a quadratic equation, the intersection of the parabola and the straight line is at coordinates $\left(-\dfrac83,-\dfrac{80}9\right)$.

We can decompose the given domain in a large parabolic segment on the right (of area $\dfrac23\cdot4\cdot16$) and a smaller on the left (of area $\dfrac23\cdot\dfrac83\cdot\dfrac{64}9)$, plus a trapezoid (of area $\dfrac12\left(\dfrac83+1\right)\dfrac{80}9$).

Total: $\dfrac{5800}{81}$.


Intuitive reason for why $ \frac{1}{\ln(x)}$ has a change in concavity around $y = -\frac{1}{2}$ The graph for $ y = \frac{1}{\ln(x)}$ has a change in concavity right about where $y=-\frac{1}{2}$ Is there any intuitive reason for this, or way you could know with high school maths (other than calculus)?

Not very intuitive, but a nice alternative view based directly on the definition of $\log$ as an integral: Since the function $x \mapsto x^{-1}$ is convex for positive $x$ it follows that for any positive $a\leq x$ $$\log(x)-\log(a) = \int_a^x \frac{\mathrm{d}t}t \geq (x-a)\cdot \frac{1}{\frac{x+a}2}=2\frac{x-a}{x+a}.$$ For $0<x\leq a$ the reversed inequality holds  (with $\leq$ instead of $\geq$). In fact, the function $$\log(a) + 2\frac{x-a}{x+a}$$ is tangent to $\log(x)$ at $x=a$, less than $\log(x)$ for $x>a$, and greater than $\log(x)$ for $0<x<a$. It is a hyperbola with horizontal asymptote $\log(a) + 2$. In particular for $a = e^{-2}$, it “simplifies” to $$\frac{-4}{e^2 x + 1}.$$
Taking reciprocals it follows that the line $-\tfrac14(e^2x+1)$ is tangent to $\log(x)^{-1}$ at $x=e^{-2}$, greater for $x>e^{-2}$, and less for $x<e^{-2}$. In particular, $(e^{-2}, -\tfrac12)$ is an inflection point of $\log(x)^{-1}$.

Let's consider the behavior of $y$ for all positive real numbers $x$. For $0 < x < 1$, ln $x$ is negative and increasing from $-\infty$ to 0, so $y$ is negative and decreasing from 0 to $-\infty$. You have a vertical asymptote at $x = 1$. For $x > 1$, ln $x$ increases from 0 to $\infty$, so $y$ decreases from $\infty$ to 0. Knowing this and the shape of the graph of ln $x$, it should be easy to graph 1/ln $x$. I'm not exactly sure what you mean by change in concavity though.


Determinant of hermitian matrix Let $M=A+iB$ be a complex $n \times n$ Hermitian matrix. First of all we know that $$(\det M)^2=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}.$$ Also $\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$ is a polynomial in $n^2$ variables of degree $2n$. Is it true that $\det M$ is a polynomial, say D, of degree $n$ in this $n^2$ variables such that $D^2(M)=\det \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$? The explicit calculations for $n=1,2,3$ suggest so, yet I can't find the information if this is true neither proof this.

The matrix is similar to

$$
S = T \hat{M} T^{-1}=
\begin{pmatrix} I & iI \\ I & -iI \end{pmatrix}
\begin{pmatrix} A & -B \\ B & A \end{pmatrix}
\begin{pmatrix} \frac12I & \frac12 I\\-\frac i2I& \frac i2I\end{pmatrix} =
\begin{pmatrix} A+Bi &\\ &A-Bi \end{pmatrix}=
\begin{pmatrix} M &\\ &M^* \end{pmatrix}=\begin{pmatrix} M &\\ &M \end{pmatrix}
$$
hence $S$ and $\hat M$ have the same characteristic polynomial.


How many taxicabs are there in New York City? There is some unknown number $x$ which represents the amount of taxicabs in New York City. These taxicabs are randomly distributed throughout the city, so you are equally likely to find any of them.

Assume that the taxicabs were numbered sequentially as they were made, so the set of taxicab numbers is $\{1, 2, 3, \ldots, x\}$

While walking down the street, you see a taxicab with the number $n$. You deduce the following (call event $N$ seeing the taxicab with number $n$):

$$Pr(N \ |\ x < n) = 0$$
$$Pr(N \ |\ x = n) = \frac{1}{n}$$
$$Pr(N \ |\ x = n+1) = \frac{1}{n+1}$$
$$Pr(N \ |\ x = n+2) = \frac{1}{n+2}$$
$$\vdots$$

Since $n \geq 1$, you conclude that the most likely option is that $x = n$, and thus conclude that there are $n$ taxicabs in New York City.

I have two questions.

What is this logic called? It's not called probability matching, but there's some term for this idea of choosing the "highest probability" that's escaping me.

Is this logic fallacious? Is the above argument enough to believe that "most likely" there are $n$ taxicabs in New York City? If not, can someone explain why this is wrong?

This is an application of the so-called likelihood principle, called the maximum likelihood method. That is, if the distribution $P_x(Y=y)$ of an observable random variable $Y$ depends on an unknown parameter $x$, then $P_x(Y=y)$ is taken to be the "likelihood" of $x$ given an observation $Y=y$. As a function of $x$, $P_x(Y=y)$ is called the "likelihood function".  (Note the term "likelihood", not "probability", since this principle does not assign a probability distribution to a fixed parameter.)

This is what the Bayesian mathematical statistician Jimmie Savage called "attempting to prepare the Bayesian omelette without breaking the Bayesian eggs". That is, a Bayesian would assign a prior distribution to the unknown fixed parameter, and would make inferences about it based on the resulting posterior distribution:
$$P(X=x|Y=n) \propto  P(Y=n|X=x)\cdot P(X=x),
$$

which basically reduces to the likelihood function if the prior $P(X=x)$ is a uniform distribution.

This is equivalent to the German tank problem which faced mathematician in world war II. i.e. how many tanks does Germany have? which it's well documented.  The other thing I would suggest looking into are Fermi Aproximations. He used this to famously answer "how many piano tuners are there in Chicago", it uses a different probalistic approach.


Are the Elementary Binary Operations Analytic? Are the elementary binary operations of addition, multiplication, and exponentiation -- taken as multivariate functions over the real numbers -- analytic? That is, $f(a, b) = a + b$ and so forth. Does it change if $a, b$ are any complex numbers?

I don't know how to get started assessing multivariate analyticity. Any references on how to determine this would be appreciated.

Yes, all "elementary" operations
\begin{align*}
(z,w) \mapsto z+w \\
(z,w) \mapsto z-w \\
(z,w) \mapsto zw \\
(z,w) \mapsto \frac{z}{w}
\end{align*}
are analytic (holomorphic) as functions of two complex variables (and therefore also real-analytic as functions of two real variables), where defined. (Of course, for $z/w$ we have to exclude $w=0$).

For exponentiation, things are a little more tricky,
$$ (z,w) \mapsto z^w = \exp(w\log z) $$
is not a well-defined operation. If we fix a particular branch of $\log$, then $z^w$ is holomorphic as a function of two complex variables where defined, but this is more subtle. We can for example view $z^w$ as a holomorphic function on $\mathbb{C}^2$ minus the set where $z$ is ($0$ or) negative real.

Elementry binary operations are analytic. Binary operations, like. Addition, multiplication. For example :a+b=c,a/b{b should not equal to  zero to }


is $\ell_{2}$ space Banach algebra? Is $\ell_{2}$ space Banach algebra ? By $\ell_{2}$, we mean the space of all sequences of complex
numbers $( x_{n})$ for which $\sum|x_{k}|^{2}<\infty$.

The space l₂ is closed under multiplication. For let x, y be in this space. Then there is N such that ∣x_{i}∣,∣y_{i}∣<1,i≥N. Then ∑(x_{i}y_{i})²=∑_{i≥N}(x_{i}y_{i})²+∑_{i<N}(x_{i}y_{i})²
<∑_{i<N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣. For positive a,b we have a^{1/2}b^{1/2}≤(a+b)/2. Thus
((∣x_{i}y_{i}∣)/(∥x∥∥y∥))≤((∣x_{i}∣²/∥x∥²+∣y_{i}∣²/∥y∥²)/2). A little work shows that ∑_{i<N}(x_{i}y_{i})²+∑_{i≥N}∣x_{i}y_{i}∣<∞. Minkowwski's inequality holds. Thus the space is an algebra. But it does not contain 1 and so it is not a Banach algebra.


Propositional calculus logical proof help I am asked to prove:
(Y => Z) V X    ≡    (Y V X) =>(Z V X)
This is what I have so far,
(Y V X) =>(Z V X)
(Definition of => X,Y := Y V X, Z V X)
(Y V X) V (Z V X)≡ (Z V X)
(V SYMMETRY twice )
(X V Y) V (X V Z)≡ (Z V X)
(V/V X,Y,Z := X,Y,Z)
X V (Y V Z) = (Z V X)
How do I get rid of the (Z V X) so that I have the full proof Picture of the proofs I am using

$$\begin{align} (Y \to Z) \lor X &\equiv (Y\lor X) \to (Z\lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv \lnot (Y \lor X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \land \lnot X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X) \land \underbrace{(\lnot X \lor Z \lor X)}_{true}\\ \\\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X)\end{align}$$
I leave it to you to justify the steps.  If you are unsure of any of them, feel free to comment below the answer.  I've made use of the equivalence between $(P\to Q)\equiv (\lnot P \lor Q)$ (and more than once).  I've also used DeMorgan's law, and the fact that $\lnot X \lor X$ is always true ($\top$).  I used the fact that $\top \lor Z$ is always true, and the fact that $\top \land P \equiv P$.


Fourier transform of derivative for the bounded function I was going through the derivation of Wigner distribution properties, and encountered a certain step in the proof I could not justify.

Namely, the step requires the following equality to be true:

$\int d^2 \lambda \frac{\partial}{\partial \lambda} \left( \exp(-\lambda \alpha^* + \lambda^* \alpha) f(\lambda) \right) \equiv 0$,

where $\alpha$ and $\lambda = x + iy$ are complex numbers, $\int d^2 \lambda \equiv \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} dx dy$, and the differential is a Wirtinger one:

$\frac{\partial}{\partial \lambda} \equiv \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)$.

The function $f$ is bounded, continuous and infinitely differentiable (w.r.t. $x$ and $y$), but not necessarily going to zero on the infinity.

As far as I understand, this is a Fourier transform, and the equality would reduce to a known property (Fourier transform of derivative), if $\lim_{x \rightarrow \infty} f = 0$ and $\lim_{y \rightarrow \infty} f = 0$, which is, unfortunately, not the case. Could anyone point me to the idea behind the proof in such circumstances?

Returning to this question: it turned out that $f$ actually goes to zero on the infinity. In my case $f(\lambda) \equiv \mathrm{Tr} \{ \hat{A} \hat{D}(\lambda) \}$, where $\hat{A}$ is a Hilbert-Schmidt operator, and $\hat{D}$ is the displacement operator. This trace is square-integrable, as proved in K. Cahill and R. Glauber, Ordered Expansions in Boson Amplitude Operators, Phys. Rev. 177, 1857 (1969). With this amendment, the proof is trivial.

I have never seen the derivation of this but, if $f$ is entire and bounded (is not it?) then by Liouville's theorem it is constant. This seems to be enough to get the result you are looking for.

Coca


find all value of z belong to C Find all values of $z\in\mathbb{C}$ such that

  $$ e^z = -3i$$

My try:

$$e^z=e^{x+iy}$$
$$-3i = re^{i\cdot0}$$

take $\ln$ both sides

$$z =\ln(-3i)$$

stuck here, need help to solve it.

Note that
$$
-3i=3(-i)=3e^{i\frac32\pi+i2k\pi},\;\; k\in\Bbb Z
$$
and
$$
e^z=e^xe^{iy}.
$$
Thus
\begin{align*}
e^z=-3i\;\;
&\Longleftrightarrow\;\;e^x=3\;\;\vee\;\; e^{iy}=e^{i\frac32\pi+i2k\pi}\\
&\Longleftrightarrow\;\;x=\log3\;\;\vee\;\;y=\frac32\pi+2k\pi\\
&\Longleftrightarrow\;\;z=\log3+i\left(\frac32\pi+2k\pi\right),\;\;k\in\Bbb Z.
\end{align*}

Finally be careful using logarithims when you're dealing with complex numbers: here the log function is different than the one we use with real numbers. Take a look at the Complex Logarithm.

$z=a+ib$$e^z = e^a*e^{ib}$
$e^a*e^{ib}=-3i$
Magnitude of $e^{ib} = 1$
So, $e^a=3$, $a = ln3$
$e^{ib}=-i$
$b=$$(2n+1)\pi\over 2$ where $n$ is an integer.


Are there other kinds of bump functions than $e^\frac1{x^2-1}$? I've only seen the bump function $e^\frac1{x^2-1}$ so far.  Where could I find examples of functions $C^∞$ on $\mathbb{R}$ that are zero everywhere except on $(-1,1)$?

Are there others that do not involve the exponential function?  Are there any with a closed form integral?  Is there a preferred function?

There is even simpler example from this book of Loring Tu.

You simply start with
$$f(t)=\left\{\begin{array}{lr}
0&t\leqslant 0\\
e^{-1/t}&t>0
\end{array}\right..$$
Then you define
$$g(t)=\frac{f(t)}{f(t)+f(1-t)}$$
and shift it to the right by creating
$$h(t)=g(t-1).$$
To make it symmetric you put
$$k(t)=h(t^2)$$
and finally to make it look like bump function you put
$$\rho(t)=1-k(t).$$
It will look like this

$\hspace{4,5cm}$

As you can tell from the picture, Loring Tu covered more general case (he even generalised this construction to smooth manifolds of arbitrary dimension).

For more details and picutres go to Tus book.

BONUS I created plots of functions $f,g,h,k,\rho$ in this geogebra sheet. Enjoy.

Seems to me that if you take the second derivative of a bump function you get another bump function, which should be not so hard to show. In fact all even derivatives are bump functions. Then you linearly transform them (shift in the x or y axis and/or rescale the axes) to fit your criteria.


integral $\int \frac{1}{sin^{6}x+cos^{6}x} $ Hi I have checked my working out for numerous times but I could not find any error in it. Could you please have a look at my working out and point out my mistake. Underneath the working out is the correct answer.

Note

$$\tan^{-1} \frac{\tan x}{1-\tan^2x}
 =\tan^{-1} \frac12\tan 2x
=\tan^{-1} \frac1{2\cot 2x }\\
= \frac\pi2 - \cot^{-1}  \frac1{2\cot 2x }
= \frac\pi2 - \tan^{-1} (2\cot 2x ) $$

which is equivalent to the answer provided.


Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$

My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.As this sphere touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$.
So $1+4+4+2u+4v-4w+d=0\implies2u+4v-4w+d=-9.......(1)$
Also the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ passes through $(1,-1,0)$
So $1+1+2u-2v+d=0\implies2u-2v+d=-2......(2)$
I am stuck here.Please help me.

Hints:

(1) The given sphere is $\;(x+1)^2+(y-3)^2+z^2=9\;$

(2) The line through the above sphere's center and the given tangency point is $\;(-1,3,0)+t(1,2,-2)\;,\;\;t\in\Bbb R\;$

(3) In a similar way as in plane geometry, two tangent spheres' centers are joined by a line passing through the tangency point.

I have tried my best to answer the question if any doubt arises then post.
But i think its qite easy to understand provided you have the knowledge of directio ratios and cosine.
Thanks.
Happy learning
(https://i.stack.imgur.com/OXuSf.jpg)


Circle and Locus _ ONLY PEN AND PAPER ALLOWED. Q) Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $F_{1}$ be the set of all pairs of circles $(S_{1}$, $S_{2}$) such that T is tangent to $S_{1}$ at P and tangent to $S_{2}$ at Q, and also such that $S_{1}$ and $S_{2}$ touch each other at a unique point, say, M. Let $E_{1}$ be the set representing the locus of M as the pair ($S_{1}$, $S_{2}$) varies in $F_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $F_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$. Let $C$ be the circle $x^2+y^2+6(2y+7x)=53$. The number of times $C$ intersects $E_{1}$ and $E_{2}$ is (are):

As the diameter of the first is $PQ$ $$E_1=\{(x,y)| (x+2)(x-2)+(y-7)(y+5)=0\}-\{P,Q\}$$

and the diameter of the second is $DR$, with $D$ the midpoint of $PQ$
$$E_2=\{(x,y)| x(x-1)+(y-1)^2=0\}-\{(4/5,7/5),(36/37,43/37)\}$$

where the disallowed points stem from the intersections with $l_{PR}$ and $l_{QR}$.

Now

$$\#((V((x+2)(x-2)+(y-7)(y+5))\cup V(x(x-1)+(y-1)^2))\cap C)=4$$

but three points are disallowed, leaving the other intersection $(400/409,349/409)$ so

$$\#((E_1\cup E_2)\cap C)=1$$

Edit

$E_2$ is part of a circle: let $y-1=m(x-1)$ be lines through $R$, then together with the equation for the circle $E_1$ is a part of, we get the relation $(m^2+1)y^2+(-2m^2+2m-2)y-38m^2-2m+1=0$ that gives us solutions $y_{1,2}$ with corresponding $x_{1,2}$. What we're looking for are points $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{1}{m^2+1},\frac{m^2-m+1}{m^2+1})$ which implicitizes to $x^2-x+y^2-2y+1=0.$

Let the centers of $S_1$ and $S_2$ be $O_1$ and $O_2$ respectively. Note that $O_1P\perp PQ$ and also $O_2Q\perp PQ$.

Since $O_1M=O_1P$ and $O_2M=O_2Q$ then both $O_1PM$ and $O_2QM$ are isosceles triangles and
$$\begin{align}\angle MPO_1&=\angle PMO_1\\ \angle MQO_2&=\angle QMO_2\end{align}$$
This implies
$$\angle PMQ=\angle MPQ+\angle QPM$$
which means that $\angle PMQ$ is a right angle. So $M$ lies on a circle with diameter $PQ$ (that is called the set $E_1$).

The center of $E_1$ is located at $D(0,1)$ and its radius is $2\sqrt{10}$:
$$E_1=\{(x,y):x^2+(y-1)^2=40\}$$

Now imagine the set of all the line segments passing through $R(1,1)$ and crossing $E_1$ at two points $A$ and $B$, and note that the point $R$ is inside $E_1$ and $AB$ is in fact a chord of the $E_1$ circle.

Can you show that the loci of the midpoints of all these chords is a circle whose diameter is $RD$?


Global upper limit for function of two variables on unbounded domain I have the following function:

$f(a,b) = \frac{(a + a^2 + 6ab + 3b(1 + b))}{(a + a^2 + 2ab + 3b(1 + b))}, a \geq 0, b \geq 0$

I need to show, that the function has an upper limit / converges to a "global maximum" and also, what that upper limit is.

I am pretty certain from Mathematica use, that the upper limit is $\sqrt{3}$ (it could also be higher though, so I cannot presume, it is that value). Now I am looking for a way to prove it.

There is no point, for which the gradient of the function is zero (no point, for which all partial derivatives are 0 at the so time). So, formally, there is no local maximum.

Looking at the edges: The function for $a \rightarrow 0, a \rightarrow \infty$, a \rightarrow 0 and $b \rightarrow \infty$ yields function values of 1 respectively (doesn't matter, what value the other variables take). $a \rightarrow \infty ~ and ~ b \rightarrow \infty$ is indeterminate $(\frac{\infty}{\infty})$.

So, how I am getting to a solution though, is the following:

I set the partial derivative $\frac{\partial f}{\partial a} = \frac{4b(-a^2 + 3b(1 + b))}{a + a^2 + 2ab + 3b(1 + b))^2} = 0$, which gives me $a = \sqrt{3}\sqrt{b + b^2}$.

Now I take the partial derivative $\frac{\partial f}{\partial b} = \frac{(4a(a + a^2 - 3b^2))}{(a + a^2 + 2ab + 3b(1 + b))^2}$ and set $a = \sqrt{3}\sqrt{b + b^2}$. For $b \rightarrow \infty$ the value of the partial derivative converges to 0 (so the criterion of the gradient being zero is fulfilled for a local maximum, not checking second derivatives here).

If I set $a = \sqrt{3}\sqrt{b + b^2}$, the initial function becomes: $g(b) = \frac{6b^2 + \sqrt{3}\sqrt{b(1 + b)} + 6b(1 + \sqrt{3}\sqrt{b(1 + b)})}{(6b(1 + b) + \sqrt{3}\sqrt{b(1 + b)}*(1 + 2b)}$.

$g$ has no local maximum (first derivative test). $g(0)=0$, and for $b \rightarrow \infty$ it is $\sqrt 3$, which is my solution.

Now my question is: Does it work, the way I approached this? And if yes, why is that so? I actually cannot give the reasoning myself, why this actually gives us an upper limit. Basically what I'm doing is: I am looking at all points, for which partial derivates of a is zero (might be a local maximum) and look for the maximum of the function of those points. But why is this an upper limit?

Note: this also works the other way round (set partial derivative of b to 0, let a go to infinity).

Hint: It is $$f(a,b)\le \sqrt{3}$$
You will need to show, that $$\sqrt{3} a^2-a^2+2 \sqrt{3} a b-6 a b+\sqrt{3} a-a+3
   \sqrt{3} b^2-3 b^2+3 \sqrt{3} b-3 b\geq 0$$ for $a,b$ with $a\geq 0,b\geq 0$
Solving $$\frac{\partial f(a,b)}{\partial a}=0$$ and $$\frac{\partial f(a,b)}{\partial b}=0$$ we get
$$a=-\frac{3}{2},b=\frac{1}{2}$$
It is $$f(a,b)\geq 1$$ and the equal sign holds if $a=1,b=0$


Cardinality of rational number and irrational number Since there is a rational number between any two irrational numbers, so can we say that rational number and irrational number have the same cardinality?

No. $\mathbb Q$ is countable. Suppose that the irrational numbers have the same cardinality, then $ \mathbb R \setminus \mathbb Q$ is also countable. This gives

$ \mathbb R = \mathbb Q \cup (\mathbb R \setminus \mathbb Q)$ is countable, a contradiction.

Of course NOT.

The argument is not correct. For example, I can say there always exists a rational number between any two distinct real numbers, but apparently $\mathbb{Q}$ and $\mathbb{R}$ don't have the same cardinality.


Does the sequence $\frac{3^n}{n!}$ converge or diverge? Consider the sequence $\frac{3^n}{n!}$ where $n$ varies from $1$ to $\infty$. Does the corresponding  series converge or diverge?

  $$
\sum_{n=1}^\infty \frac{3^n}{n!}.
$$

I reduced this to $(e^3 - 1)$. But how should I decide if this converges/diverges?

Good news!

You have already solved the problem.

As you correctly note, this is $e^3 - 1$.

That's because the series for $e^x$ converges everywhere! So you are done, it converges (specifically to $e^3 - 1$).

If for all n, n not equals 0, then the following rules apply:
Let L = lim (n -- > inf) | an+1 / an |.
If L < 1, then the series sum (1..inf) an converges.
If L > 1, then the series sum (1..inf) an diverges.
If L = 1, then the test in inconclusive.
Here the limit lim (n -- > inf) | an+1 / an |. Is Zero. hence the series is converging.
This is the ratio test.


Formula for the summation of the series $a+2a^2+3a^3+...$ upto nth term? I need a formula for the summation of the series $a+2a^2+3a^3+...$ upto nth term. Is there any way to simplify it?

You request $\sum_{j=1}^n j a^j$.  Then we have \begin{align*}
    \sum_{j=1}^n j a^j &= a \sum_{j=1}^n j a^{j-1}  \\
        &= a \sum_{j=1}^n \frac{\mathrm{d}}{\mathrm{d}a} a^j  \\
        &= a \frac{\mathrm{d}}{\mathrm{d}a} \sum_{j=1}^n a^j  \\
        &= a \frac{\mathrm{d}}{\mathrm{d}a} \frac{a^{n+1} - a}{a-1}  \\
        &= a \frac{n a^{n+1} - n a^n - a^n + 1}{(a-1)^2}  \\
        &= \frac{a^{n+1}(an - n -1) + a}{(a-1)^2}  \text{.}
\end{align*}

$\frac{(a\,*\,(1+na^{(1+n)}\,-\,(1+n)*a^n))}{(-1+a)^2}$

Thanks for the guide on how to use latex.


Limit aproaching infinity $f(x) = -\infty$ ?!? On one of my homework problems it's asking for me to sketch a graph of a function which fulfills five requirements. All of the requirements make sense to me other than the following:

  $$\lim_{x\rightarrow \ \infty}  f(x) = {-\infty}$$

How could there be a limit approaching infinity at negative infinity? Doesn't one of the number need to be finite for me to know where to place the infinity limit? This makes no sense to me and is really throwing me off.

  $$f'(x) \ge 2 \ \text{when} -2 \lt x \lt 0 \ \text{and} \ \text{when} \ 2 \lt x \lt 4 $$

I don't even know how to approach this one. How do you apply information about the derivative/tangent line to the graph of the original equation?

Other requirements (purely for reference, I understand these):

  $$\lim_{x\rightarrow \ 0+}  f(x) = -2$$
  f(x) is an even function
  f(x) is differentiable everywhere but two x-values, and is continuous for all x

Thank you.

Saying that $$\lim\limits_{x\to\infty}f(x)=-\infty$$ just means that for each real number $M$ there is an $x_M$ such that $f(x)<M$ whenever $x\ge x_M$. Taking $M=-100$, for instance, there is an $x_{-100}$ such that $f(x)<-100$ whenever $x\ge x_{-100}$. A very simple example of such a function is the function $f(x)=-x$: for this function you can take $x_{-100}$ to be $101$ for instance, since $f(x)=-x<-100$ whenever $x\ge 101$.

For the other requirement, remember that $f\,'(a)$ is the slope of the tangent line to the curve $y=f(x)$ at $x=a$. The requirement that $f\,'(x)\ge 2$ when $2<x<4$ just means that the graph has a slope of at least $2$ over the entire interval $(2,4)$. Many functions have this property: $f(x)=ax$ with $a\ge 2$ (since $f\,'(x)=a$), $f(x)=x^2$ (since $f\,'(x)=2x$, which is certainly greater than $2$ when $2<x<4$, and many, many more.

Remember, you don’t have to write down a single expression for $f$: you can define it piecewise, with different definitions on different intervals.

How about $f(x)=-x^3$ or even easier, $f(x)=-x$


How to draw the image of a Möbius transformation? Let $f(z)=\frac{2z}{z+2}$. What is the image of the circle with radius 2 centered at 0?

I find that $f$ maps $2\to1$, $-2 \to \infty$. Hence, the image should be a line. Therefore, I pick another point on the circle, say $2i$, and find $f(2i)$. Then the line through $1$ and $f(2i)$ is the image?

Your function goes from the z in the w plane, so we need to find z to substitute in the circle equation.
$$w = \frac{2z}{z+2} \Rightarrow z = \frac{2w}{2-w}$$
The equation for the circle is $|z|=2 \Rightarrow |z|^2=4 \Rightarrow z\overline{z} = 4$. Substitute the inverse function:
$$\left(\frac{2w}{2-w}\right)\overline{\left(\frac{2w}{2-w}\right)} = 4$$
$$\left(\frac{2w}{2-w}\right)\left(\frac{2\overline{w}}{2-\overline{w}}\right) = 4$$
$$4w\overline{w} = 16 -8\overline{w} -8w + 4w\overline{w}$$
$$8w+8\overline{w} = 16$$
then let $w = u + vi$:
$$8(u+vi)+8(u-vi)=16$$
And the result is $u=1$

$f(2) = 1$ and $f(2i) = 1 + i$ so you get the same line.

First we know that this Möbius transformation maps the x-axis to x-axis. Then we know two points on the circle, 2 and -2, are mapped onto 1 and $\infty$. Since the Möbius transformation preserves angles, the image of the circle should intersect with the x-axis by 90 degrees. Also, the Möbius transformation maps circles to lines or circles. Since there is $\infty$ in the image, the image couldn't be a circle. Hence, the image should be a straight line passing through 1 and parallel to the y-axis. That is, $w=1+vi,v\in \mathbb{R}$ plus $\infty$.


Characterization of the sequence $x_1=\cos(x), x_{n+1}=\cos(x_n)$ where $x>0.$ Let $x>0$, $x_1=\cos(x),$ and $x_{n+1}=\cos(x_n), \forall n\geq1.$ Then the sequence  $(x_n)_{n\geq 1}$ is

  bounded but not monotone,
  bounded but not Cauchy,
  Cauchy,
  convergent.

It is a multiple select question , so it may have more than one correct option.

The above sequence is bounded, it's trivial. Also if the sequence converges then it will converge to a real root $\ \ l\ $ of $\ \ \ l-\cos(l)=0$.

I tried by taking a particular value of $x$. Suppose $x=\pi/2$, then the sequence will become $\{0,1,\cos(1),\cos(\cos(1)),\cos(\cos(\cos(1))),.....\}$. Without using calculator, I think we can't say whether this sequence is monotone or not.

Again while showing 'Cauchy' I considered $|x_m-x_n|\leq |x_{m-1}-x_{n-1}|\leq|x_{m-2}-x_{n-2}|\leq...$

I don't know how can we show whether the sequence Cauchy or not from the above relation.

  Any help is appreciated. Thank you.

Hint 1. Note that if $0\leq x_n\leq 1$, and $x_n\leq x_{n+1}=\cos(x_n)$ then $x_n\in[0,a]$ and $x_{n+1}\in [a,1]$ where $a\simeq 0.73908$ is the unique solution of the equation $\cos(x)=x$ . Therefore
$$x_{n+2}=\cos(x_{n+1})\leq x_{n+1}.$$
So the sequence has an alternating behaviour.

Hint2. If you show that the function $x\to \cos(\cos x)$ is strictly increasing in $[0,1]$ then it follows that the subsequences  $\{x_{2n}\}_n$ and $\{x_{2n+1}\}_n$ are strictly monotone and bounded. What may we conclude?

P.S. You may also note that $x\to \cos(\cos x)$ is a contraction in $[0,1]$ and Banach fixed-point theorem can be applied.

There's only one real root $y$ such that $\cos y=y$ (it's about $0.74$).  In the neighborhood of $y$,
$$
x_n=y+\varepsilon\implies x_{n+1}=\cos(y+\varepsilon)=\cos y - \varepsilon \sin y + O(\varepsilon^2)=y-\varepsilon\sqrt{1-y^2}+O(\varepsilon^2)...
$$
since $\vert{\sin y} \vert< 1$, this leads to exponential (and non-monotonic) convergence to $y$ if $x_n$ ever lands close enough to it.  So it seems like you'll sometimes have 1., 3., and 4. all true.  However, it's not obvious without further analysis that 3. and 4. have to hold... it might be that you can converge to some other stable limit cycle, or even if there were an unstable cycle, you might stay on it for some initial conditions.  (Update: per other answers, in this case there are no other limit cycles and the fixed point is universally attracting.)


How to prove that if each element of group is inverse to itself then group commutative? How can you prove that if each element of group is inverse to itself then the group is commutative?

Let $g \in G$; we know that $g = g^{-1}$, and that $g^{-1} \in G$ because groups contain the inverse of each of their elements. Now suppose $x,y \in G$. Then $xy \in G$ by closure under products, and so $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$, where we used the fact the $x=x^{-1}$ and $y = y^{-1}$ in the last step. This shows that $G$ is a commutative group.

say a and b belong to G
a=(a)^-1   and
b=(b)^-1

=> (axb) = ((a^-1) x (b^-1))

=>  axb = (bxa)^-1        -------------------------------(1)

let c=bxa

it is obvious from closure prop. that c belongs to G
               so c=(c)^-1
               or (bxa)=(bxa)^-1      ---------------------(2)

from (1) and (2)   axb=bxa
Hence it is a commutative group


basic Quantile proves Let this be my definition of a quantile funktion. X is a real-valued random variable. And let F be it's distribution function. then
\begin{align*}
F^{-1}(a):=\inf\{x\in \mathbb{R}: F(x) \ge a\}.
\end{align*}
Now I want to solve, how can I prove this simple equation:
\begin{align*}
F^{-1}(a)= x_0 \Leftrightarrow P(X \le x_o ) = a, \text{    almost surely}
\end{align*}

or what assumpations are needed in order to make it happen?

If $F^{-1}(a)=x_0$ and $F$ is continuous at $x_0$ then $F(x_0)=P(X \leq x_0)=a$. It can happen that $F$ is not continuous at $x_0$. In this case either $F(x_0)=a$ or else:

$\lim_{x \to x_0^-} F(x)<a$
$F(x_0)>a$
There is no $y$ such that $F(y)=a$.

The only way to have $F$ be discontinuous at $x_0$ is if $P(X=x_0)>0$.

There is a problem in the discrete case. If two consecutive points  $x_1<x_2$  are in the support of $F$ and $F(x_1)<a<F(x_2)$ then $F^{-1}(a)=x_2$  but $F(x_2)\neq a$.
What the problem asks is a special  case of the first line and holds only if $a$ is in the support of $F$.


A question about geodesics of hyperbolic space Let $ H^n $ be the the upper half space of $ R^n $ endowed with the conformal metric $ g=\frac{1}{x_{n}^{2}}|dz|^2 $ ($ |dz|^2 $ is the standard metric of $ R^n $). This space is the classical hyperbolic space and its Riemannian connection $ \overline{\nabla} $ satisfies

$$ \overline{\nabla}_{\partial_n}\partial_n =0 $$

where $ \partial_1, \ldots \partial_n $ is the standard frame of $ R^n $.  Note that $ \overline{\nabla} $ is not the standard connection of $ R^n $.

Now it is easy to see that the smooth curve $ \gamma(t)= (0, \ldots, 0,t) $ , $ t >0 $ is a geodesic in $ H^n $. This curve is clearly defined for $ t>0 $ and it is not defined for all t in $ R $. Moreover it is well known that $ H^n $ is complete. This fact apparently contradicts the Hopf Rinow theorem (every geodesic in a complete space is defined for every $ t \in R $). What is the mistake in this argument?

Thanks

Your $\gamma(t)$ is actually not a geodesic.  In fact, earlier you have a mistake in computing the Levi-Civitia connection.  I'll use $X$ to denote $\partial_n$, just to save on typing.  Via the formula on wikipedia, we have \begin{align*} \langle \overline{\nabla}_X X , X\rangle =& \frac{1}{2}\big(Xg(X,X) + Xg(X,X) - Xg(X,X)\\ &+ \quad g([X,X],X) - g([X,X],X) - g([X,X],X)\big)\\ \end{align*}

Since $X$ is a coordinate vector field, $[X,X] = 0$ so the last three terms vanish.  Hence, we see that \begin{align*}\langle \overline{\nabla}_X X, X\rangle &= \frac{1}{2} Xg(X,X)\\ &= \frac{1}{2}\partial_n \frac{1}{x_n^2}\\ &= -\frac{1}{x_n^3}\end{align*}

In particular, $ \overline{\nabla}_{\partial_n} \partial_n\neq 0$.

This in turn affects the computation of $\Gamma^k_{ij}$ which affects the geodesic equation.  The punchline is that your $\gamma$ has the same image as a geodesic, but is not constant speed.  This can be seen since, for example, $O(n-1)$ acts on $\mathbb{H}^n$ by rotating about the image of $\gamma$ which is clearly an isometry.  The fixed point set of this $O(n-1)$ action is thus a totally geodesic submanifold - that is, the image of $\gamma(t)$ is a geodesic.

On the other hand, the correct paramaterization is given by $\alpha(t) = (0,\ldots,0,e^t)$.  Then this has the same image as $\gamma$, hence is a geodesic iff it's constant speed.  But we have \begin{align*} \frac{d}{dt}\langle \alpha'(t),\alpha'(t)\rangle &= \frac{d}{dt} \langle e^t\partial_n, e^t\partial_n\rangle\\ &= \frac{d}{dt} e^{2t} \frac{1}{x_n^2}\\ &= \frac{d}{dt}e^{2t} \frac{1}{e^{2t}} \\ &= 0\end{align*} so the length squared, and hence the length, of $\alpha$ is constant.

The Hopf Rinow theorem is only valid in local (i.e. exponential) coordinates. The standard Riemannian metric you described is not the local metric, although it is extremely useful.


How to find the area between 2 curves given an interval Determine the area of the region within the interval $[0, 1]$ bounded by the curves $f(x)=2 \sin (x)$ and $g(x)=\displaystyle\frac{4-x}{2}$ correct to $2$ decimal places.

The first thing I did was:
$\int _0^{\pi }\:2\sin\left(x\right)-\left(\frac{4-x}{2}\right)$
and I got: $4-\frac{8\pi -\pi ^2}{4}$ or $0.18$
but my textbook answer gives: 4.29 units
I then graph this on Geogebra and get:

So is it the other way around? $g(x) - f(x)$? Also does this means that I should always visualise or graph the functions first?

Refer to the graph:
$\hspace{3cm}$
The area to be calculated is: $A1+A2+A3$. Sanity check: the area $A1$ is less than $1$ (compare with triangle side 2 and height 1); the area of $A2$ is slightly more than $A1$. So, the total area should be around $2$, not $4$.

To calculate exactly, first you need to find the intersection points:
$$2\sin x=2-\frac x2 \Rightarrow x_1\approx 0.89,x_2\approx 2.85$$
Next you need to break up the interval:
$$A=A1+A2+A3=\int_0^{0.89} \left(2-\frac x2-2\sin x\right)dx+\int_{0.89}^{2.85}\left(2\sin x-2+\frac x2\right)dx+\int_{2.85}^{\pi} \left(2-\frac x2-2\sin x\right)dx\approx 1.99.$$
WolframAlpha answers: Intersection points and the total area.

For the area between the curves you have the right idea, just you need to count all area as positive (unsigned area). Try using $$\int_0^\pi|f(x)-g(x)|dx$$