Easy way of memorizing values of sine, cosine, and tangent My math professor recently told us that she wanted us to be able to answer $\sin\left(\frac{\pi }{2}\right)$ in our head on the snap. I know I can simply memorize the table for the test by this Friday, but I may likely forget them after the test. So is there a trick or pattern you guys usually use to remember it? For example, SOHCAHTOA tells us what sine, cosine, and tangent really mean.

If not, I will just memorize the table. But just wanted to know what memorization techniques you guys use. I feel this is the perfect place to ask, because I bet a bunch of people in the math stackexchange, also had to go through the same thing freshman year of college.

Oh here is a picture of the unit circle:

Note the pattern:

$$\sin 0^{\circ} = \frac{\sqrt{0}}{2}$$
$$\sin 30^{\circ} = \frac{\sqrt{1}}{2}$$
$$\sin 45^{\circ} = \frac{\sqrt{2}}{2}$$
$$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$$
$$\sin 90^{\circ} = \frac{\sqrt{4}}{2}$$

This is something of a mathematical coincidence as far as I know, so don't try to extend this to other angles; and it goes backwards for $\cos$.  Once you have these, you can find the other angles you want by drawing them on the unit circle and figuring out whether the values should be positive or negative, whether they should be $0$ or $1$, or otherwise whether they are "small" ($\frac{1}{2}$), "medium" ($\frac{\sqrt{2}}{2}$) or "large" ($\frac{\sqrt{3}}{2}$).

Consider the simple example first and then you can derive further values from it. Lets begin with the following table. Sin Cos Tan Table

All you need to remember about this table is the arrangement of rows and columns i.e. angles are placed in columns and trigonometric functions at the rows. So draw a table and start with the sine values.

Step 1 (red): In each of the empty cells for sine of 0,30,45,60 and 90 write (0,1,2,3,4) respectively.

Step 2 (green): Then divide each of them with 4 (i.e. 0/4, 1/4, 2/4, 3/4, 4/4).

Step 3(blue): The next step is to take square root of each (i.e. sqrt(0/4), sqrt(1/4), sqrt(2/4), sqrt(3/4), sqrt (4/4)).

Step 4(brown): Now cancel any terms from numerator and denominator if applicable and simplify the results. You end up with the values of sine 0,30,45,60 and 90.

Drawing Sin Cos Tan Table

Step 5(orange):Once you have values for sine function, invert them for cosine i.e( sin 90 = cos 0, sin 60 = cos 30, sin 45 = cos 45 and so on) and you get values for cosine function.

Step 6: For tangent, put sin/cos values and simplify.

Step 7: You can extend the table for further angles by using formulas such as

  sin(π−x)=sinx  sin(π−x)=sinx  and sin(π/2−x)=cosx sin(π/2−x)=cosx,
  and cos(π/2−x)=sinx cos(π/2−x)=sinx,


Is it true that if the set $A=\{x:f(x) =c\}$ is measurable for every $c$ in $\Bbb R$, then $f$ is measurable? Let $f: [0;1] \to \Bbb R$. Is it true that if the set $A=\{x:f(x) =c\}$ is measurable for every $c$ in $\Bbb R$, then $f$ is measurable?

I have this counter example
$f(x) =x$ if $x$ belong to $P$, $f(x) =-x$ otherwise, where $P$ is a non measurable set on $[0,1]$.

$A$ is measurable but $f$ is not measurable function.

I had proved that $f$ is not measurable, but how can I prove that $A$ is measurable and what is $A$ in this example ?

As you think, the answer to the question is no.  However, I am not sure how your example will work.  Maybe it does, but I am unable to show that there exists a measurable set $U$ such that $f^{-1}(U)$ is not measurable in your example.  I had a lapse of intelligence a little bit.  The comments both under the question and under my answer have explained how the OP's function works.  Nonetheless, I have a different counterexample.

Note that there exists (due to the Axiom of Choice) a non-measurable subset $X$ of $\left[0,1\right]$.  This set has the same cardinality as $\left[\dfrac12,1\right]$, so there exists a bijection from $g:X\to \left[\dfrac12,1\right]$.  Note also that $[0,1]\setminus X$ and $\left[0,\dfrac12\right)$ are also equinumerous, so there exists
a bijective function $h:\big([0,1]\setminus X\big) \to \left[0,\dfrac12\right)$.

Define $f:[0,1]\to\mathbb{R}$ to be
$$f(x):=\begin{cases}g(x)&\text{if }x\in X\,,\\h(x)&\text{if }x\in[0,1]\setminus X\,.\end{cases}$$
Note that $f^{-1}\big(\{c\}\big)$ is either empty or a singleton for each $c\in\mathbb{R}$ (it is empty if $c\notin[0,1]$, and is a singleton if $c\in[0,1]$).  However, for the measurable set $M:=\left[\dfrac12,1\right]$, $f^{-1}(M)=X$ is not measurable.  That is, $f$ is not a measurable function.


If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:

  If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is
  also an integer, does it implies that $k=1$?

This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$.

This is a fun problem! Where did you find it?

There are no solutions except for $k=1$. Assume from now on that $k \neq 1$. Since $k$ is clearly odd, it is also not $0$ and we deduce that $k(k-1)>0$. It is convenient to set $M = k(k-1)$. Also, we may assume WLOG that $x$ and $y\geq0$.

As you did, rewrite the equation to
$$ky^2 - 4 (k-1) x^2 = 1$$
or
$$(ky)^2 - k(k-1) (2x)^2 = k.$$
Set $Y=ky$ and $X=2x$ so the equation is
$$Y^2 - M X^2 = k. \quad (\ast)$$

We will study the equation $(\ast)$. In the end, we will see that there are no solutions with $X$ even and $Y$ divisible by $k$.

The rest of this proof works inside the ring $R:=\mathbb{Z}[\sqrt{M}]$. Note that $M=k(k-1)$ is not square, so this is an integral domain. For an element $\alpha = a+b \sqrt{M} \in R$, set $\bar{\alpha} = a - b \sqrt{M}$.

Set $\epsilon = (2k-1) + 2 \sqrt{M}$. Note that $\epsilon \bar{\epsilon} = (2k-1)^2 - 4 k(k-1) = 1$, so $\epsilon$ is a unit of $R$.

Set $\delta = Y+X \sqrt{M}$. Since $\delta$ is a positive real, and $\epsilon>1$, there is some integer $n$ such that $\epsilon^n \leq \delta < \epsilon^{n+1}$. Write $\delta = \gamma \epsilon^n$. Since $\epsilon$ is a unit, $\gamma$ is in the ring $R$ and, by construction,
$$1 \leq \gamma < \epsilon.$$
We have
$$\epsilon = 2k-1 + 2 \sqrt{k(k-1)} < 2k+ 2k = 4k.$$
So
$$1 \leq \gamma < 4k.$$

But, also $\gamma \bar{\gamma} = \delta \bar{\delta} =k$. So
$$\frac{1}{4} \leq \bar{\gamma} \leq k.$$
Write $\gamma = U + V \sqrt{M}$. So
$$\begin{matrix}
1 & \leq & U+V\sqrt{M} & < & 4k \\
\frac{1}{4} & \leq & U-V\sqrt{M} & < & k \\
\end{matrix}.$$
Solving for $V$, we have
$$\frac{-k}{2 \sqrt{M}} < V < \frac{2k}{\sqrt{M}}.$$
The LHS is $\approx -1/2$, and the right hand side is slightly larger than $2$. So $V$ is $0$, $1$ or $2$. We break into cases:

$\bullet$ If $V=0$, then the equation $\gamma \bar{\gamma} =k$ gives $U^2 =k$. So $k$ is a square, say $k=m^2$ and $M=m^2(m^2-1)$. We have
$$Y+X \sqrt{M} = m ((2k-1)+2 \sqrt{M})^n = m (2m^2-1 + 2 m \sqrt{m^2-1})^n.$$
An easy induction on $n$ shows that $Y \equiv (-1)^n m \mod m^2$ so, except when $m=1$, we do not have $Y$ divisible by $k=m^2$. Of course, the case $m=1$ corresponds to $k=1$.

$\bullet$ If $V=1$ then $U^2 - k(k-1) = k$ so $U=k$. We have
$$Y+ X \sqrt{M} = (k+\sqrt{M}) \cdot ((2k-1)+2 \sqrt{M})^n.$$
An easy induction on $n$ shows that $X$ is odd.

$\bullet$ If $V=2$ then $U^2 - 4k(k-1) = k$ so $4k^2-3k = U^2$. This gives $64k^2 - 48 k + 9= 64 U^2+9$ or $(8k-3)^2 - (8U)^2 = 9$. The only ways to write $9$ as a difference of squares are $3^2-0^2$ and $5^2-4^2$. The former gives $k=0$, but we saw that $k$ must be odd; the latter gives no solution since $8$ does not divide $4$.

Well notice that both the numerator and denominator are squares, so we may factor

\begin{equation}
\frac{4x^2 - 1}{4x^2 - y^2} = \frac{(2x+1)(2x-1)}{(2x-y)(2x+y)}.
\end{equation}

It is easy to see that $\gcd(2x + 1, 2x - 1) = 1$, so if $k$ is an integer, it must be the case that the denominator each divides one of the linear factors up top. Then there are only two cases for which this is possible: we must have either $y = 1$, or $2x-y = 1$ and $2x+y = (2x + 1)(2x - 1)$, since otherwise, the positive factor below will be larger than either of the numerator factors and thus cannot divide either of them.

Let's consider the case where $2x - y = 1$ and $2x + y = (2x + 1)(2x -1)$. From the first equation, we have $y = 2x + 1$, so

$$2x + y = 4x + 1 = 4x^2 - 1,$$

or

$$4x^2 - 4x - 2 = 0.$$

However, solving that for $x$ shows that $x$ is not an integer, so this case is impossible. Therefore, we are led to the conclusion that $k$ is also an integer when $y = 1$, and in that case, $k = 1$.

Edit: EuYu pointed out some details that I overlooked. Refer to the comments for the discussion.


Conditional Probability vs Conditional Probability wrt a RV I have some problems understanding how the following conditional probabilities relate.
Given X and Y two continuous rv, we can write the conditional probability of $X\leq x$ given Y as:
\begin{equation}
P[X\leq x|Y]= E[\mathbb{1}_{X\leq x}|Y]
\end{equation}

For the same two variables we can write the following conditional probability distribution:

\begin{equation}
P[X\leq x|Y\leq y]= \frac{P[X\leq x,Y\leq y]}{P[Y\leq y]}=\frac{E[\mathbb{1}_{X\leq x}\mathbb{1}_{Y\leq y}]}{E[\mathbb{1}_{Y\leq y}]}
\end{equation}
Question: What is the relation between the two definitions above and if possible how can we get from one to the other?
Thank you.

$P[X\leq x|Y]= E[\mathbb{1}_{X\leq x}|Y]$ is in fact for a constant $Y=y$, it would be more clear to write it as:

$P[X\leq x|Y]= E[\mathbb{1}_{X\leq x}|Y=y]$

Note that here you are taking the conditional expectaion when you fix $Y$.
But the second formula is just the application of Bayes rule to the first definition. In fact this gives you the intuition on how you can marginalize the joint distribution in a more flexible way.


Linear Programming Formulation of Traveling Salesman (TSP) in Wikipedia I am confused by Wikipedia's Linear Programming formulation of the Traveling Salesman Problem, in say the objective function.

Question: If there are n cities indexed 1,...,n, what is city with index 0?

This can be seen in the objective function which has the summation of $c_{ij}  x_{ij}$ from i=0...n. Also seen in the third and forth constraint.

It seems like Wikipedia's ILP formulation of the TSP is wrong/incomplete. What changes are needed to make it correct?

There is also an older/different version of the formulation, is this correct?

And another one found on Stackexchange

Example: Lp Formulation of TSP

Consider the following distance-matrix of 4-Cities TSP:

    To

From    1   2   3   4

1

13  21  26
2   10

29  20
3   30  20
    5
4   12  30  7

Exclude the the Diagonal Variables (X11, X22, X33, X44) from the Distance Table as shown above

The following table represents the Assignment Model + the Additional Constraints resulting from ui  – uj + n xij  <=  n-1; i,j = 2, . ., n & i ≠ j

n = 4
ui  – uj + n xij  <=  3; i,j = 2, . ., n & i ≠ j

n xij + ui  – uj  <=  3

4 X23 + U2 – U3 ≤ 3     (i = 2, j = 3)
4 X24 + U2 – U4 ≤ 3     (i = 2, j = 4)
4 X32 + U3 – U2 ≤ 3     (i = 3, j = 2)
4 X34 + U3 – U4 ≤ 3     (i = 3, j = 4)
4 X42 + U4 – U2 ≤ 3     (i = 4, j = 2)
4 X43 + U4 – U3 ≤       (i = 4, j = 3)

V’s X12 X13 X14 X21 X23 X24 X31 X32 X34 X41 X42 X43
    U2  U3  U4
OF  13  21  26  10  29  20  30  20  5   12  30  7

1
1   1   1                                                   =   1
2               1   1   1                                       =   1
3                           1   1   1                           =   1
4
                                    1   1   1               =   1
5               1           1           1                       =   1
6                               1           1                   =   1
7       1           1                           1               =   1
8
        1           1           1                           =   1
9                   4                               1   -1      ≤   3
10                      4                           1       -1  ≤   3
11                              1                   -1  1       ≤   3
12                                  1                   1   -1  ≤   3
13                                          4       -1      1   ≤   3
14                                              1       -1  1   ≤   3

The associated LP consists of the Assignment Model + the Additional Constraints.

Minimize    13X12 + 21X13 + 26X14 + 10X21 + 29X23 + 20X24 + 30X31 +
        20X32 + 5X34 + 12X41 + 30X42 + 7X43
s.to:
C1  1X12 + 1X13 + 1X14 = 1
C2  1X21 + 1X23 + 1X24 = 1
C3  1X31 + 1X32 + 1X34 = 1
C4  1X41 + 1X42 + 1X43 = 1

C5  1X21 + 1X31 + 1X41 = 1
C6  1X12 + 1X32 + 1X42 = 1
C7  1X13 + 1X23 + 1X43 = 1
C8  1X14 + 1X24 + 1X34 = 1

C9  4X23 + 1U2 - 1U3 <= 3
C10 4X24 + 1U2 - 1U4 <= 3
C11 4X32 - 1U2 + 1U3 <= 3
C12 4X34 + 1U3 - 1U4 <= 3
C13 4X42 - 1U2 + 1U4 <= 3
C14 4X43 - 1U3 + 1U4 <= 3
Integer: U2, U3, U4         Binary:   All Xs

Solution Summary for LP of TSP:

Decision    Solution    Unit Cost or
Variable    Value       Profit C(j)
1   X12     1.0000  13.0000
2   X13     0       21.0000
3   X14     0       26.0000
4   X21     0       10.0000
5   X23     1.0000  29.0000
6   X24     0       20.0000
7   X31     0       30.0000
8   X32     0       20.0000
9   X34     1.0000      5.0000
10  X41     1.0000      12.0000
11  X42     0       30.0000
12  X43     0       7.0000
13  U2      0       0
14  U3      1.0000      0
15  U4      2.0000      0

Objective Function (Min.) =  59.0000

The Route is:   X12 – X23 – X34 – X41 = 59


Why do we take $x=\cos t$, $y=\sin t$ for a parametric circle when we can take the opposite? Both $x = \cos t$, $y = \sin t$ and $x = \sin t$, $y = \cos t$ describe a circle

So why is the first parameterization so commonly used in mathematics, and not the second?

Because we usually
start the circle
at the right.
There,
the coordinates are
$(1, 0)$,
which is
$(\cos(0), \sin(0))$.

If we started it at the top,
we would probably use
$(\sin(t), \cos(t))$.

Using $t$ is the most ‘natural’  parametrisation  because $t$ is the polar angle  of the point $(x,y)$ in a standard polar coordinates system.


Proving $\log(b^a) = a \log(b)$ using calculus Sorry, this is a really simple question, but I'm trying to teach myself calculus and can't figure it out.
If we define $\log(b) = \frac{db^x}{dx}(0)$ how does one prove $\log(b^a) = a\log(b)$? I tried using the definition of derivative but got stuck at $\lim_{h \to 0}\frac{b^{ah} - 1}{h}$.

If $a\not=0$, let $k=ah$ and note that $h\to0$ iff $k\to0$, which gives

$$\lim_{h\to0}{(b^a)^h-1\over h}=a\lim_{h\to0}{b^{ah}-1\over ah}=a\lim_{k\to0}{b^k-1\over k}$$

If $a=0$, then $(b^a)^h-1=0$, so the limit is obviously $0$.

$$
\log(a^b) = \int_1^{a^b} \frac{dx}x.
$$
Suppose $w^{\,b}=x$.  Then $bw^{b-1}\,dw= dx$, so $b\dfrac{dw}{w} = \dfrac{dx}x$.  As $x$ goes from $1$ to $a^b$, then $w$ goes from $1$ to $a$.  Hence,
$$
\log(a^b) = \int_1^{a^b} \frac{dx}x = \int_1^a b \frac{dw}w = b\log a.
$$


Poincaré Inequality on Gaussian Measures So I have a working idea on Gaussian-Poincaré Inequality. Namely through the Ornstein-Ullenbeck Generator and Gaussian Integration by parts.
Recently I have stumbled across Sobolev Spaces and have seen there is a Poincaré Inequality defined there as well over an open set $\Omega$ and w.r.t the Lebesgue Measure. This seems more of a Distribution Theory approach which I admit I have just started learning.
My question is is there any connection between the classical poncare inequality defined for functions on a Sobolov Space and the Gaussian-Poincare Inequality.
Any reference or explanation is deeply appreciated.

A simple version of one of the several related results known as the Poincare inequality is the following:
Let $u\in H^1(\Omega)$, where $\Omega \subseteq B(0,R)$ for some $R>0$. Define $\langle u \rangle = \int_\Omega u(x) \, dx$. Then we have
$$ \int_\Omega \left(u(x) - \langle u \rangle \right)^2 \, dx\leq c R^2 \int_\Omega |\nabla u(x)|^2 \, dx$$
A Gaussian Poincare inequality is of a similar form. Let $X$ be a real valued r.v. with a unit normal distribution, and let $f\in C^1(\mathbb{R})$ such that $\mathbb{E}(f(X)^2)$ is finite. Then
$$ \text{Var}(f(X)) = \mathbb{E}\left((f(X) - \mathbb{E}(f))^2 \right) \leq \mathbb{E}(f'(X)^2)$$
You can see how the two results are really the same if you see the expectation operator as just being integration against a different measure.


Divisibility of $(4^{2^{2n+1}}-3)$ by 13. The question is to simply proof that $13\space|(\space4^{2^{2n+1}}-3)$ for all $n\in\mathbb{N}$. Since the chapter in my book is about the Euler-phi function and Euler's theorem, my guess would be to look at $4^{2^{2n+1}}\equiv3\space \pmod{13}$ and do something with Euler's theorem. Another guess is to solve this by induction, but in both approaches I am very stuck.

As you correctly note, to compute
$$
4^{2^{2n+1}} \pmod{13}
$$
you can compute
$$
2^{2n+1} \pmod{12}.
$$
It is advisable to use CRT, and compute separately modulo $3$ and modulo $4$.

Modulo $3$ you have
$$
2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pmod{3},
$$
whereas modulo $4$ you have
$$
2^{2n+1} \equiv 0 \pmod{4},
$$
for $n \ge 1$.

Therefore
$$
2^{2n+1} \equiv 8 \pmod{12}.
$$
(This usually implies solving the system of congruences
$$
\begin{cases}
x \equiv -1 \pmod{3}\\
x \equiv 0 \pmod{4}\\
\end{cases}
$$
but in this case it is just a matter of checking which of $x = 0, 4, 8$ is a solution.)

Thus
$$
4^{2^{2n+1}} \equiv 4^8 \equiv 3^4 \equiv 3  \pmod{13}
$$
as $4^2 = 16 \equiv 3 \pmod{13}$ and $3^3 = 27 \equiv 1 \pmod{13}$.


Boolean simplification gives two different results Here is the question I'm trying to solve:

  Use algebraic manipulation to find the minimum sum-of-products expression for the function $f=x_1x_3+x_1\overline{x}_2+\overline{x}_1x_2x_3+\overline{x}_1\overline{x}_2\overline{x}_3$

The way I attempted it was as follows:

f = ac + ab' + a'bc + a'b'c'
  = ac + b'(a + a'c') + a'bc   [distributive]
  = c(a + a'b) + b'(a + a'c')  [distributive]
  = c(a + b) + b'(a + c')      [using x + x'y = x + y]
  = ac + bc + ab' + b'c'

where a = x1, b = x2, c = x3

The correct answer, however is:

  $f=x_1x_3+x_2x_3+\overline{x}_2\overline{x}_3$

Can anyone tell me where I went wrong?

$\begin{align} &ac + bc + ab' + b'c'\\=~&bc +~ ac + ab' + b'c'\\ =~& bc+~(abc+ab'c)+(ab'c+ab'c')+(ab'c'+a'b'c')\\\vdots~~& \\ =~& bc+~ac+b'c'\\=~& ac+bc+b'c' \end{align} $

Complete the missing steps, and you are done.

Karnough maps are more standard and consistent than algebra, but algebra works too. You are almost there. Observe that bc + b'c' is true iff b$\equiv$c. If this is the case, then ac + ab'$\iff$a(c+b')$\iff$a. On the other hand, if bc + b'c' is false, then b$\equiv$c' and ac + ab'$\iff$a(c+b')$\iff$ac. So either the bc + b'c' term makes the expression true, or ac + ab' is equivalent to just ac.


Understanding dual spaces My professor asked me to give him an example about dual spaces from our real life so I was thinking if we take the set of all our actions as the space X can the space of all our thoughts be the dual space  ?!!

It is clear that the set of actions is a vector space :
     Adding two action give an action and multiplying an action by a scalar give an action ( the addition means doing more than one action and multiplication by a scalar means increase or decrease the power of the action).
        The norm of this space is the time we need to do an action. So if time separating a sequence of actions is very small we say that this sequence converges
        The reason I thing that the dual of this space should be the space of thought is that thoughts and actions are connected to each other and if we have a convergente sequence of actions we surely have a convergente sequence of thoughts but the inverse isn't true we can have a convergente sequence of thought without actions.

 What do you think about this example, should I explain it to my professor ??????

A possible example is the following. We have a shop with a (finite) set of articles $S$. Let $V$ be the vector space of prices for the objects in $S$, then a choice of things to buy would give you an element of the dual space $V^*$ (seen as a functional giving you back how much do you spend depending on the price tags).

A concrete example is the force applying on an object. The newton second law of motion said that Force = mass x acceleration. It is clear that here the mathematical representation of force is functional that map the set of masses to a real number. Then, if we take the set of masses as a space X. Then it dual would be the set of forces.


Second derivative of absolute value function proportional to Dirac delta function? I have recently discovered the relation

\begin{equation}
  \frac{\mathrm d^2}{\mathrm dx^2} \big| x \big| = 2\delta (x).
\end{equation}

I was very intrigued when I found this expression, and as it makes sense to me intuitively. However, I am aware that Dirac's delta "function" is defined as a distribution, not a function in the conventional sense, so there is probably more going on here than meets the untrained eye.

So my question is, what caveats are necessary for the above equation for it to be formally correct? Can it be taken literally?

I would have presented a definition of the delta function for completeness, but I am also not certain which definition is appropriate in this context. Am I correct in my understanding that the piecewise definition

\begin{equation}
  \delta (x) \equiv
  \begin{cases}
    \infty, & \operatorname{if} x = 0 \\
    0,      & \operatorname{if} x \neq 0
  \end{cases}
\end{equation}

Is only an informal, intuitive definition and actually an "abuse of notation" of proper functions?

I have knowledge up to multivariable/vector calculus and differential equations, but no formal or distribution theory, analysis, number theory, etc. training. Please accommodate if appropriate.

[I apologize for the formatting issues, I am genuinely trying to learn. Can someone please point me in the direction of a crash course for LaTeX (particularly math)? I've looked around a lot on SE and elsewhere on the web and I'm having trouble finding tutorials that explain the simplest of things, like when to use the dollar sign and when to use the double dollar, how to write dollar signs without them taking effect, when to use a package in the preamble, if the preamble is a specific location or just anything before the expression to be formatted, etc. Sorry for the meta question.]

The intuitive definition is just intuitive, mathematically it is wrong since it cannot differentiate between $\delta$ and $5 \delta$.

Formally, $\delta$ is understood as a function of functions:
$$\delta(f)=f(0)$$

Your relation
\begin{equation}
  \frac{\mathrm d^2}{\mathrm dx^2} \big| x \big| = 2 \delta .
\end{equation}
means that for all functions $f$ which are infinitely many times differentiable and have compact support we have
$$
\int_{-\infty}^\infty f''(x) |x| dx = 2 f(0) = 2 \delta(f)
$$

Added clarification If $g$ is a twice differentiable function, with $g''$ continuous, the for all functions $f$ which are infinitely many times differentiable and have compact support we have
$$
\int_{-\infty}^\infty g''(x) f(x) dx =\int_{-\infty}^\infty g(x) f''(x) dx
$$

Note that this identity doesn't necessary hold if $f$ is not compactly supported. This is the reason why we always test distributions on compactly supported functions.

The idea of distributions is the following:

Consider a function which is linear
$$u : C_c^\infty(\mathbb R ^d) \to \mathbb C$$
and continuous (we will ignore the topology we consider).

$\delta$ is one such distribution. If we have a continuous (integrable) and (locally) bounded function $g$ we can interpret it as a distribution the following way:
$$u_g(f) = \int_{-\infty}^\infty f(x) g(x) dx $$

This definition makes sense because $f$ has compact support.

Now, if $g$ is a function which is differentiable everywhere, we get by parts (and using support of $f$ compact):
$$\int_{-\infty}^\infty f(x)g'(x) dx =- \int_{-\infty}^\infty f'(x)g(x) dx$$
or
$$u_{g'}(f)=-u_g(f')$$

This suggest that we can differentiate any distribution:
$$u'(f):= -u(f')$$

With this definition $\delta$ is derivative of the $H(x)=1$ when $x \geq 0$ and $H(x)=0$ when $x <0$. Moreover the derivative of $\delta$ is
$$f \to -f'(0)$$

The concept of differentiability as distribution coincides with the classical concept for differentiable functions. But in this more general sense, many new functions become differentiable.

Note that much of what I wrote only makes sense if we use compactly supported functions....

Finally, there is a theorem which says that if $g$ has continuous derivative on $(0, \infty)$ and $(- \infty, 0)$ and has a jump discontinuity at $x=0$ then, in distribution sense the derivative is
$$g'+C \delta$$
where $g'$ is the derivative on $\mathbb R \backslash \{ 0\}$ and $C$ is the jump at $0$. Applying this result twice to $|x|$ you get the above formula.

Notice that the first derivative of $|x|$ is the sign function. This is why the second derivative is the Delta Dirac function.


How many parameters does the neural network have? We have a neural network with an input layer of ℎ0 nodes, hidden layers of ℎ1 , ℎ2 , ℎ3 , ..., ℎ−1 nodes respectively and an output layer of ℎ nodes.
        How many parameters does the network have?

Suppose the network has $784$ inputs, $16$ nodes in $2$ hidden layers and $10$ nodes in the output layer.
The amount of parameters (meaning weights and bias that make up the cost function) is then:
For the weights:
$$784\times 16+16\times16+16\times10=12960$$
For the bias components:
We have $32$ neurons in the hidden layers and $10$ in the output, so we have $$32+10 = 42$$ biases.
So in total, the amount of parameters in this neural network is $13002$.

for simplicity lets say the $l=5$
clearly we have,
           i=h0        //inputs
           h=[h1,h2,h3,h4]   //hidden layers
           o=h5             //output

therefore the no of params $= (h0*h1 + h1*h2 + h2*h3 + h4*h5) + [h1+h2+h3+h4+h5]$
=click here to view the expression


Provide a Venn diagram of a counter example of the converse of if A ⊆ B, then A ∪ C ⊆ B ∪ C? Provide a Venn diagram of a counter example of the converse of

$$A \subseteq B\rightarrow ((A \cup C) \subseteq (B \cup C))$$

First of all, we need to clearly state the converse of

$$ (A ⊆ B) \rightarrow [(A ∪ C) ⊆ (B ∪ C)].\tag 1$$

The converse of $(1)$ is as follows:

$$[(A ∪ C) ⊆ (B ∪ C)]\rightarrow (A\subseteq B)\tag{2}$$

One counterexample to the converse (2):  $$\text{Suppose } \;A = \{1\}, \;B=\{2\}, \;C=\{1,2\}$$

Then, it is true that $((A ∪ C) = C) ⊆ ((B ∪ C) = C)$ but $A\not \subseteq B$ because $\{1\} \not \subseteq \{2\}$

Consider  when B is a proper subset of A is a proper subset of C.


Solving $3|2-x| + |2x-4| = 8$ I'm not sure how to solve $3|2-x| + |2x-4| = 8$. I don't exactly need an answer, the method is what I'm interested in. I've done equations with one absolute value element. >_<

A generic approach:

A function that is a linear combination of absolute values of linear monomials, like yours, is piecewise linear and continuous. (General form of the equation: $f(x):=\sum_{k}a_k|b_k x+c_k|+d=0$.)

So it suffices to evaluate it at the "corners" (i.e. where the argument of some absolute value vanishes, $x=-c_k/b_k$) to detect changes of sign (you compute the $f(x_k)=\sum_{j}a_j|-b_j c_k/b_k+c_j|+d$).

We also need to look at the values at $\pm\infty$, the sign of which is given by the sign of $\sum_{k}a_k$.

Here we observe that there is a single corner, at $x=2$, and

$$3|2-2|+|2\cdot2-4|-8<0.$$

At infinity, the function is positive, so we have a root in $(-\infty,2)$ and another in $(2,\infty)$. To find the exact position of the roots, we remove the absolute values and supply the correct sign, to obtain ordinary linear equations. (I.e.$\sum_{k}\pm a_k(b_k x+c_k)-d=0$.)

Thus,

for the root in $(-\infty,2)$, $3(2-x)+(4-2x)-8=0$,
for the root in $(2, \infty)$, $3(x-2)+(2x-4)-8=0$.

Note that we didn't merge $|x-2|$ and $|4-2x|=2|x-2|$ for the sake of exposition of the method.


Can an arc from a circle be made a straight line? I want to write an algorithm that morphs or transforms an arc of a circle to a straight line joining both ends of the arc. I would like to understand the maths behind this such a transformation. How can this be done?

If $x(t)$ is a parametrization of your circle arc (see the other answers for how to find one), and $y(t)$ a parametrization of the straight line segment joining the endpoints of the arc, then

$$F_\lambda(t)=(1-\lambda)x(t)+\lambda y(t)$$

is the parametrizations for the circle arc when $\lambda=0$, the parametrization of the line segment when $\lambda=1$, and some intermediate form when $\lambda\in(0,1)$ is something in between. This technique is called linear interpolation and is one way (out of infinitely many) to achieve such a blend.

So draw $F_\lambda(t)$ for $\lambda=0$ and then slowly increase the value of $\lambda$ and repeatedly daw the resulting curve.

We can depict bending of  circular arcs of increasing radius. The math is elementary trig. Angle subtended at arc center goes on increasing when arc bends more. It is straight for infinite radius of curvature ($R$, $r$) at beginning..

When arc starts with horizontal tangent, the tangential rotation, arc length and radius respectively are $(\phi, s,R) .$

$$ \phi = s/R$$

$$  \frac{dx}{ds}= \cos \phi,\, \frac{dy}{ds}= \sin \phi $$

integrate,

$$ x= x_0 + R \sin (s/R),\, y= y_0 + R (1-\cos (s/R))\, ; $$

that gives co-ordinates as a function of distance $s$ from fixed point $ (x_0,y_0), $ the fixity point for any radius R. It can be animated. If only the endpoint is desired, we let $s=L_{max}$.

Lmax = 100.;
ParametricPlot[{r Sin[L/r], r (1 - Cos[L/r])}, {r, 20, 100}, {L,
  Lmax/10, Lmax}, Mesh -> {30, 20}, GridLines -> Automatic]


probability of illness There is a 1 in 350 chance of having an illness for a woman.

An ultrasound is 75% accurate in establishing whether the illness is present.

If the ultrasound is negative (i.e. suggests that the woman does not have the illness) what is the new probability of the woman actually having the illness?

I just got some results and am not sure how to connect the two probabilities.  I am as interested in filling my knowledge gap about how to calculate the probability as I am in the actual answer.

Thanks,

Ivan

Define events

$D$: the woman has this disease

$A$: the test result is positive

We have $\ \mathbb{P}(D)=1/350$, $\ \mathbb{P}(A|D)=\mathbb{P}(A^C|D^C)=0.75$

The problem is asking: what is the probability $\mathbb{P}(D|A^C)$ ?

\begin{align}
\mathbb{P}(D|A^C) =\dfrac{\mathbb{P}(D\cap A^C)}{\mathbb{P}(A^C)}
=\dfrac{\mathbb{P}(D\cap A^C)}{\mathbb{P}(D\cap A^C)+\mathbb{P}(D^C\cap A^C)},
\end{align}
where we have
\begin{align}
\mathbb{P}(D\cap A^C) = \mathbb{P}(D) - \mathbb{P}(D\cap A)
=\mathbb{P}(D)-\mathbb{P}(A|D)\,\mathbb{P}(D)
=\mathbb{P}(D)\big[1-\mathbb{P}(A|D)\big]
=\dfrac{1}{4}\times\dfrac{1}{350}
\end{align}
and
\begin{align}
\mathbb{P}(D^C\cap A^C)=\mathbb{P}(A^C|D^C)\,\mathbb{P}(D^C)
=\dfrac{3}{4}\times\dfrac{349}{350}
\end{align}
Thus
\begin{align}
\mathbb{P}(D|A^C)=\dfrac{1}{1+3\times349}
\end{align}

There's a $\frac{1}{4}$ chance the test is incorrect.

There's a $\frac{1}{350}$ chance the woman has the illness.

Assuming that these are independent events (there's no evidence in your question that they shouldn't be), so the probability both occurs is $\frac{1}{4}\times\frac{1}{350}=\frac{1}{1400}$.


Points Where Slope is Zero Problem:

  Find the coordinates of any point on the curve of $y^2-4xy=x^2+5$ for which the tangent is horizontal.

When I solve $\frac{dy}{dx}=0$, I end up getting $x=-2y$. So then I was thinking that I can put in any number for $x$ and get a corresponding $y$ where $(x,y)$ is a point where the slope is zero. So I plugged in $1$ for $x$ and got $-\frac{1}{2}$ for $y$, which is the point $(1,-\frac{1}{2})$. But this point can't be a point where the slope is zero because graphically the slope is zero only at $(2,-1)$ and $(-2,1)$. What am I missing?

After getting $x =-2y$ you also have to take into account the fact that $y^{2}-4xy=x^{2}+5$ because you have to find points ** on the curve** at which the derivative is $0$. If you solve these two equations simultaneously you will get the answer.

as you get x = -2y so it will satisfy the given function,
putting x = -2y in given equation

y^2 + 8y^2 = 4y^2 +5
5y^2 = 5
y^2 = 1
y = -1 and 1
and corresponding values of x will be 2 and  -2
so pints will be (-2,1) and (2,-1)


Is the p-adic sum of an infinite polynomial with aperiodic coefficients necessarily irrational? I'm studying sequences with the form $bp^{a_1} + b^2p^{a_2}+b^3p^{a_3}\dotsc$, where $p$ is a $p$-adic base and $b$ is some rational number (edit:) and the exponents $(a_1, a_2, a_3,\dotsc)$ are strictly increasing.
It's easy to see that the p-adic sum of an infinite sequence with this form must be rational when the sequence of exponents  is periodic: $(a_1,a_2,\dotsc ,a_1+j, a_2+j, \dotsc ,a_1+2j, a_2+ 2j ,\dotsc)$, or even eventually periodic: that is, periodic after some finite number of terms. By analogy with $p$-adic expansions generally I feel that the converse must also be true - that  the sum of the sequence must be irrational when the exponents are aperiodic. Am I correct, and if so how can I prove it?

Every $p$-adic number can be written as a quotient of two $p$-adic integers, so in that sense, every $p$-adic number is “rational”. In fact, for $p > 2$, $e^p \in \Bbb{Z}_p$, even though $e^p$ is not a rational number in $\Bbb{R}$.
Edit: @TorstenSchoeneberg in the comments tried to tell me that "$e^p$" (as defined by the usual convergent series $\sum_{n \geq 0} \frac{p^n}{n!}$) is not a number that exists in both $\Bbb{R}$ and $\Bbb{Q}_p$, because the metrics are different.
But the same logic he cites renders nonsensical his (and other people's) assertion that the OP is "obviously" looking for answers in "$\Bbb{Q} \cap \Bbb{Q}_p$", since the type of infinite series OP is discussing, $$b p^{a_1} + b^2 p^{a_2} + b^3 p^{a_3} + ...$$ with $0 < a_1 < a_2 < a_3 < ...,$ does not even converge in $\Bbb{Q}$. Commenters who think the OP wants an answer in "$\Bbb{Q} \cap \Bbb{Q}_p$" (whatever that is) are wrong, and have failed to appreciate this subtle but essential distinction.


Probability game with gates I am playing a game with gates and a walkway. The game is that I can only win if I get through the gate successfully. The gate works with two levers that need to be pulled. If both levers work correctly then the gate opens and you walk to the other side.

The first lever has a .01 probability of failure and the second lever has .02 probability of failure.

I need to find the probability that when pulling both levers I successfully get through the gate. If one lever fails the gate will not open and the levers not working are statically independent.

After looking over this problem I am having my doubts on how to approach this. I came up with four events that can happen:

A = both work

B = first fails

C = second fails

D = both fail

I need to solve for A with the other three probabilities but I am thinking that D does not matter because we only need one to fail.

Would I have to simply use the P(B) + P(C) - P($B \cap C$) to find this answer then take the compliment to find A?

The answer is $0.99\cdot 0.98=0.9702$ because you'll only able to go to the other side if both lever works.
Since both levers are independent, you can use the multiplication rule.


Intuition behind Conditional Expectation I'm struggling with the concept of conditional expectation. First of all, if you have a link to any explanation that goes beyond showing that it is a generalization of elementary intuitive concepts, please let me know.

Let me get more specific. Let $\left(\Omega,\mathcal{A},P\right)$ be a probability space and $X$ an integrable real random variable defined on $(\Omega,\mathcal{A},P)$. Let $\mathcal{F}$ be a sub-$\sigma$-algebra of $\mathcal{A}$. Then $E[X|\mathcal{F}]$ is the a.s. unique random variable $Y$ such that $Y$ is $\mathcal{F}$-measurable and for any $A\in\mathcal{F}$, $E\left[X1_A\right]=E\left[Y1_A\right]$.

The common interpretation seems to be: "$E[X|\mathcal{F}]$ is the expectation of $X$ given the information of $\mathcal{F}$." I'm finding it hard to get any meaning from this sentence.

In elementary probability theory, expectation is a real number. So the sentence above makes me think of a real number instead of a random variable. This is reinforced by $E[X|\mathcal{F}]$ sometimes being called "conditional expected value". Is there some canonical way of getting real numbers out of $E[X|\mathcal{F}]$ that can be interpreted as elementary expected values of something?
In what way does $\mathcal{F}$ provide information? To know that some event occurred, is something I would call information, and I have a clear picture of conditional expectation in this case. To me $\mathcal{F}$ is not a piece of information, but rather a "complete" set of pieces of information one could possibly acquire in some way.

Maybe you say there is no real intuition behind this, $E[X|\mathcal{F}]$ is just what the definition says it is. But then, how does one see that a martingale is a model of a fair game? Surely, there must be some intuition behind that!

I hope you have got some impression of my misconceptions and can rectify them.

Maybe this simple example will help. I use it when I teach
conditional expectation.

(1) The first step is to think of ${\mathbb E}(X)$ in a new way:
as the best estimate for the value of a random variable $X$ in the absence of any information.
To minimize the squared error
$${\mathbb E}[(X-e)^2]={\mathbb E}[X^2-2eX+e^2]={\mathbb E}(X^2)-2e{\mathbb E}(X)+e^2,$$
we differentiate to obtain $2e-2{\mathbb E}(X)$, which is zero at $e={\mathbb E}(X)$.

For example, if I throw a fair die and you have to
estimate its value $X$, according to the analysis above, your best bet is to guess ${\mathbb E}(X)=3.5$.
On specific rolls of the die, this will be an over-estimate or an under-estimate, but in the long run it minimizes the mean square error.

(2) What happens if you do have additional information?
Suppose that I tell you that $X$ is an even number.
How should you  modify your estimate to take this new information into account?

The mental process may go something like this: "Hmmm, the possible values were $\lbrace 1,2,3,4,5,6\rbrace$
but we have eliminated $1,3$ and $5$, so the remaining possibilities are $\lbrace 2,4,6\rbrace$.
Since I have no other information, they should be considered equally likely and hence  the revised expectation is $(2+4+6)/3=4$".

Similarly, if I were to tell you that $X$ is odd, your revised (conditional) expectation is 3.

(3) Now imagine that I will roll the die and I will tell you the parity of $X$; that is, I will
tell you whether the die comes up odd or even. You should now see that a single numerical response
cannot cover both cases. You would respond "3" if I tell you "$X$ is odd", while you would respond "4" if I tell you "$X$ is even".
A single numerical response is not enough because the particular piece of information that I will give you is itself random.
In fact, your response is necessarily a function of this particular piece of information.
Mathematically, this is reflected in the requirement that ${\mathbb E}(X\ |\ {\cal F})$ must be $\cal F$ measurable.

I think this covers point 1 in your question, and tells you why a single real number is not sufficient.
Also concerning point 2, you are correct in saying that the role of $\cal F$ in ${\mathbb E}(X\ |\ {\cal F})$
is not a single piece of information, but rather tells what possible specific pieces of (random) information may occur.

I was only able to understand the notion of the conditional expectation with respect to a sub-$σ$-algebra $\mathcal F$, when I realized that this game is only interesting when $\mathcal F$ is "not Hausdorff", meaning that there might be points $x$ and $y$ which cannot be separated by an $\mathcal F$-measurable set.  Any $\mathcal F$-measurable function must therefore coincide on $x$ and $y$, so $E(X|\mathcal F)$ tries to be the best photograph of the random variable $X$ which coincides on $x$ and $y$, as well as on any other similar pairs of points.
In the event that $\mathcal F$ is the smallest sub-$σ$-algebra possible, namely  $\mathcal F = \{\emptyset,  \Omega\}$,
only constant functions are measurable.   So $E(X|\mathcal F)$ must be a constant function, and that constant  turns out to be the
average of $X$, a.k.a. the expectation of $X$.

PS.  This is a comment I made in a recent question (Refference for conditional expectation) which in turn brought me here to this 10 year old question when I clicked on a "Related" post.  Reading the answers I did not find anyone referring to the above point of view, so I hope my little contribution will be useful to someone.


Dual Lefschetz Operator and Contraction with the Fundamental Form Let $M$ be a Kahler manifold, with metric $g$, Kahler form $\omega$, Lefschetz operator $L$, and dual Lefschetz operator $\Lambda$.
$\Lambda$ and contraction with $\omega$ both map $k$-forms to ($k-2$)-forms (for $k \geq 2$): are they equal?

Is it correct to argue this way: for any $k$-form $\alpha$,
$$\Lambda\alpha = g(\Lambda\alpha,1)=g(\alpha,L1)=g(\alpha,\omega)\ ?$$

No. What is $1$? The constant $0$-form $1$ is by definition orthogonal to any $k$-form with $k\neq 0$.

The correct answer is $L^* = \mathrm{sgn} *L*$, where $\mathrm{sgn}(\varphi)=(-1)^k$ on $k$-forms. To see that this is a contraction with $\omega^\sharp$, write $L$ and $\omega$ in holomorphic normal coordinates so that at a point $p$ we have $\omega(p) = \sum \mathrm{d}z^i \wedge \mathrm{d}\bar{z}^i$. Then use that $L$ is an algebraic operator.

Finally, to see that $L^*=\mathrm{sgn} *L*$, we compute for $\varphi\in\wedge^{k-2}$ and $\psi\in\wedge^{k}$
\begin{align*}
 \langle L \varphi , \psi\rangle \mathrm{vol}
&= L\varphi \wedge \bar * \psi \\
&= \omega \wedge \varphi \wedge *\bar \psi \\
&= \varphi \wedge \omega \wedge * \bar \psi \\
&= \varphi \wedge L*\bar\psi \\
&= \varphi \wedge (*\mathrm{sgn}*)(L*\bar\psi) \\
&= \varphi \wedge * (\mathrm{sgn}*L*\bar\psi) \\
&= \varphi \wedge * \overline{\mathrm{sgn} *L* \psi} \\
&= \langle \varphi , \mathrm{sgn} *L*\psi\rangle \mathrm{vol}.
\end{align*}
To see that $*\mathrm{sgn}* = \mathrm{id}$ on $k$-forms, just use that $*^2=(-1)^k$ on $k$-forms. A good reference is Wells Differential Analysis on Complex Manifolds.


Disjoint paths on grid graphs Let $f(G)$ be the smallest $m$, such that one can find $2m$ vertices in $G$ with the following property: pair up the vertices in any way, and find $m$ paths that join each pair. Then every set of path constructed will have a intersection. (Is there a name for this graph property?)

I have a feeling this property might be connected to the connectedness of a graph. but for large grid graphs, I can chose many vertices and still find disjoint paths connecting the pairs, when it is only 4 connected.

For a $n\times n$ grid graph $G$, what is $f(G)$?

Edit: I have found the name of the property. A graph is $k$-linked if there exist $2k$ elements, such that there are $k$ disjoint paths that pairs two of them.

Similar property name is bramble, from Diestel's textbook:

  A bramble is a set of mutually touching connected vertex sets in $G$;
  its order is the minimum number of vertices in a set meeting each
  member of the bramble.

If you looking for this, for $n\times n$ grid it's $n$.


How many resistors are needed? I was told about the following problem. Suppose you have infinite number of resistors with only value 1$\Omega$. Question is

  What minimal number of 1$\Omega$-resistors is needed to construct given fraction resistance $R$, i.e. $R\in \mathbb{Q}_+$.

Note that you can connect two resistors ($R_1$ and $R_2$) in two ways: parallel and series. The resulting resistance ($R_p$ and $R_s$ respectively) in those cases are:
$$R_s = R_1 + R_2$$
$$R_p = R_1 \oplus R_2 = \frac{1}{\frac{1}{R_1}+\frac{1}{R_2}},$$
where we introduced $\oplus$-operation in order to simplify notes. NOTE: only schemes which are able to be written in the form:
$$(... 1 + (1\oplus 1) ... )$$
are allowed. For example, this is not allowed:

If you know how to formulate this rule better, please, say.

For example, you can use the Euclidean algorithm.
$$\frac56 = \frac{1}{\frac65}=\frac{1}{1+\frac15} = 1 \oplus 5,$$
so you needed 6 resistors because
$$5 = \sum_{k=1}^{5}1$$
But it was not the minimum number of resistors because, for example,
$$\frac56 = \frac12 + \frac13 = (1\oplus 1)+(1\oplus 1\oplus 1),$$
where it is enough to use only 5 resistors.

I think that Euclidean algorithm often solves the problem. More over, it is needed to consider only one "half of $\mathbb{Q}_+$", for the other half it is enough to replace +'s by $\oplus$'s and vice-versa.

The one who told me about this problem adhered to the following notations.
$$[1,1]\quad \text{for}\quad 1\oplus 1.$$
$$(1,1)\quad \text{for}\quad 1 + 1,$$
so our previous example looks like
$$[1,(1,1,1,1,1)]\quad \text{and}\quad ([1,1,1],[1,1]).$$

I was able to find a bound for the number of resistors required.

Given a resistance $\frac{a}{b}$, at least $n$ resistors are required where $\phi_{n+2}$ is the first Fibonacci number greater than or equal to $a+b$.

Here's my thinking process and proof:

Naturally I started out by listing things to see if I could find a pattern.

\begin{align*}
n&=1 &&1/1\\
n&=2 &&1+1=2/1\\
& &&1\oplus1 = 1/2\\
n&=3 &&1+(1+1)=3/1\\
& &&1+(1\oplus1) = 3/2\\
& &&1\oplus(1+1) = 2/3\\
& &&1\oplus(1\oplus1) = 1/3\\
n&=4 &&1+(1+(1+1)) = 4/1\\
& &&1+(1+(1\oplus1)) = 5/2\\
& &&1+(1\oplus(1+1)) = 5/3\\
\end{align*}

At this point I got bored, but I noticed a curious pattern:
Let $R_n$ be the set of possible resistances for $n$ resistors. Let $M_n = \max\{a+b: a/b \in R_n\}$
It appears to be the case that the $n$th Fibonacci number, $\phi_n$, is equal to $M_{n-2}$. Define Max($R_n$) to be the subset of $R_n$ such that $a/b \in \mathrm{Max}(R_n)$ if $a+b = M_n$. Then the maximal resistances for $n$ resistors are any resistances in Max($R_n$). It can also be observed that Max($R_n$) always seems to have two elements, and these are of the form $\frac{\phi_{n+1}}{\phi_{n}}$ and $\frac{\phi_{n}}{\phi_{n+1}}$. (Note that the first "pattern" is a direct consequence of this.)

In other words, (if this is true), given a resistance $a/b$, at least $n$ resistors are required where $\phi_{n+2}$ is the first Fibonacci number greater than or equal to $a+b$.

Claim: Given $n$ resistors, the only two maximal resistances are of the form $\frac{\phi_{n+1}}{\phi_{n}}$ and $\frac{\phi_{n}}{\phi_{n+1}}$.

Proof

Note that from here on I will use $v(R) = a+b$, where $R = a/b$.

Base case: trivial.

Inductive step: Let $k\in \mathbb{N}$. Suppose the only maximal resistances for 1 through $k$ resistors have been of the form $\frac{\phi_{k+1}}{\phi_{k}}$ and $\frac{\phi_{k}}{\phi_{k+1}}$. It follows that for $k+1$ resistors, $R = \frac{\phi_{k+2}}{\phi_{k+1}}$ can be obtained, as $\frac{\phi_{k+2}}{\phi_{k+1}} = 1\oplus\frac{\phi_{k}}{\phi_{k+1}}$. Similarly the reciprocal can be obtained. Now it must be shown that $\frac{\phi_{k+2}}{\phi_{k+1}}$ and $\frac{\phi_{k+1}}{\phi_{k+2}}$ are maximal.

Suppose $R_k = a/b$ is a resistance such that $a+b+n = \phi_k+\phi_{k+1}, n > 0$. If $b\leq \phi_{k+1}$ and $a\leq \phi_{k+1}$ then $1+a/b = (b+a)/b = (\phi_{k+2}-n)/b$, so $v(1+a/b) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. Similarly, $1\oplus a/b = a/(b+a) = a/(\phi_{k+2}-n)$ and we get $v(1\oplus a/b) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. Now suppose $a > \phi_{k+1}$ or $b > \phi_{k+1}$. This is actually impossible, as $b = \alpha + \beta$ and $a = \eta + \mu$ for some $\alpha/\beta$ and $\eta/\mu$ resistances for $k-1$ resistors. However, by the inductive hypothesis, this sum is capped at $\phi_{k+1}$. There is a second way that a resistance $R_{k+1}$ can form: Rather than adding a resistance of 1 to $R_k$, this is adding two resistances $R_{m}$ and $R_{n}$ so that $m+n=k+1$. Using similar reasoning to the special case above and the fact that $\phi_s\phi_t < \phi_{s+t}$, one can prove that $v(R_{m}+R_{n}) < v(\frac{\phi_{k+2}}{\phi_{k+1}})$. It follows that for $k+1$ resistors, there are exactly two maximal resistances, given by $\frac{\phi_{k+2}}{\phi_{k+1}}$ and $\frac{\phi_{k+1}}{\phi_{k+2}}$.

I like the use of the "bracket - parenthesis" notation, but I will add following feature, namely to write $[3]$ for $[1,1,1]$ and $(5)$ for  $(1,1,1,1,1)$ likewise. I'd like to draw the attention of how continued fractions can be of use. Suppose we were to build a resistance of $\frac{5}{7} \Omega$. Brute force shows you that you can build this resistance as $([7][7][7][7][7])$, using $35$ resistors. If we write $\frac{5}{7}$ as a continued fraction we obtain:
$$
5/7 = 0+\frac{1}{1+\frac{1}{2+\frac{1}{2}}}
$$ Which shows that the resistance can be built as $[1,((2),[2])]$ using only $5$ resistances. It is not sure this method is really optimal, and probably poses the same logistic problem as the travelling salesman problem.


Find the value of $x$ in the displayed figure Find $x$ in the following figure.

$AB,AC,AD,BC,BE,CD$ are straight lines.

$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$

$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$

NOTE: figure not to scale.

By the Pythagorean theorem we have

$$\begin{equation*}
CE=\sqrt{10^{2}-\left( x-3\right) ^{2}}=\sqrt{91-x^{2}+6x}
\end{equation*}$$
and
$$\begin{equation*}
CD^{2}+AD^{2}=AC^{2}=\left( CE+AE\right) ^{2}
\end{equation*}.$$
So we have to solve the following irrational equation
$$\begin{equation*}
\left( x-3\right) ^{2}+\left( x+4\right) ^{2}=\left( \sqrt{91-x^{2}+6x}
+x\right) ^{2},\tag{1}
\end{equation*}$$
which can be simplified to
$$\begin{equation*}
x^{2}-2x-33=\sqrt{-x^{4}+6x^{3}+91x^{2}}.
\end{equation*}$$

After squaring both sides and grouping the terms of the same degree we get the quartic equation
$$\begin{equation*}
2x^{4}-10x^{3}-153x^{2}+132x+1089=0.\tag{2}
\end{equation*}$$

The coefficient of $x^{4}$ is $2=1\times 2$ and the constant term is $1089=1\times 3^{2}11^{2}$. To find possible rational roots of this equation, we apply the rational root theorem and test the numbers of the form
$$\begin{equation*}
x=\pm \frac{p}{q},
\end{equation*}$$
where $p\in \left\{ 1,3,9,11,33,99,121,363,1089\right\} $ is a divisor of $1089$ and $q\in \left\{ 1,2\right\} $ is a divisor of $2$. It turns out that $x=3$ and $x=11$ are roots. Now we divide the LHS by $x-3$
$$
\begin{equation*}
\frac{2x^{4}-10x^{3}-153x^{2}+132x+1089}{x-3}=2x^{3}-4x^{2}-165x-363
\end{equation*}$$
and this quotient by $x-11$
$$\begin{equation*}
\frac{2x^{3}-4x^{2}-165x-363}{x-11}=2x^{2}+18x+33.
\end{equation*}$$
So we have the equivalent equation
$$\begin{equation*}
\left( x-3\right) (x-11)\left( 2x^{2}+18x+33\right) =0\tag{3}
\end{equation*}$$
Since the solutions of $2x^{2}+18x+33$ are both negative and $x=3$ is not a solution of the original irrational equation, the solution is therefore $$x=11.
$$

"The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides".

By the Triangle Inequality Theorem,

$CE + (x-3) \gt 10$, $CE + x \lt (x+4) + (x-3)$

i.e. $ (13-x) \lt CE \lt (x+1)$

we now have, $(13-x) \lt (x+1)$

i.e. $x \gt 6$

from the, $\triangle EBC $ we have,

$x-3 \lt 10$

i.e. $x \lt 13$

we can conclude that, $6 \lt x \lt 13$


Why is $\infty^0$ indeterminate? In a recent test question I was required to us L'Hopital's rule to evaluate:

$$\lim_{x\to 0^+} x\ln{(e^{2x}-1)}$$

I assumed that anything multiplied by 0 would give an answer of 0. This turns out not to be the case. Is there a simple explanation as to why infinity multiplied by 0 is not 0?

Any number, when multiplied by 0, gives 0.  However, infinity is not a real number.  When we write something like $\infty \cdot 0$, this doesn't directly mean anything; rather, it's shorthand for a certain type of limit, where the first part approaches infinity.

Now, zero times anything approaching $\infty$ will still give a limit of zero.  However, that's not what the shorthand $\infty \cdot 0$ means.  It means something approaching infinity multiplied by something approaching zero.  And this doesn't have to be zero at all.

For a simple example, as $x \rightarrow \infty$, $x^2$ certainly approaches infinity.  And $\frac{1}{x^2}$ certainly approaches zero.  But $x^2 \cdot \frac{1}{x^2} = 1$, so when we multiply the two together we get something approaching 1 (because it is constantly 1).

In essence, solving these problems boils down to figuring out whether the part approaching infinity grows fast enough to "cancel out" the part approaching zero, or if it's the other way around, or if they grow/shrink at rates that perfectly match each other (as is the case with $x^2$ and $\frac{1}{x^2}$).

Here's very simple case: $\lim\limits_{x\to 0+} x\cdot\frac{6}{x}$.  Clearly $x$ goes to $0$.

But $x\cdot\frac{6}{x} = 6$ whenever $x\neq0$. So $\lim\limits_{x\to 0+} x\cdot\frac{6}{x} = \lim\limits_{x\to0+} 6 = 6$.  This limit is not $0$.


Prove that $ y''(0) = -1, x = \cos\left(\frac{t}{1+t}\right), y = \sin\left(\frac{t}{1+t}\right)$ My attempt:

$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

then I attempted to divide y' on x' which resulted in

$$-\cot\left(\frac{t}{1+t}\right)$$

When I try to take the double dash of this it gives me

$$\csc^2\left(\frac{t}{1+t}\right),$$

but this function is undefined at 0.

The second derivative of a parametric curves is
\begin{align*}
&\frac{d}{dt} \left( \frac{dy}{dx} \right) / \left( \frac{dx}{dt} \right)\\
=&\frac{d}{dt} \left( -\cot \left( \frac{t}{1+t} \right) \right) / \left( \frac{d}{dt} \left( \sin\left( \frac{t}{1+t} \right) \right) \right)\\
=&\frac{\csc^2\left( \frac{t}{1+t} \right)}{\left( 1+t \right)^2} \cdot \frac{\left( 1+t \right)^2}{-\sin\left( \frac{t}{1+t} \right)}\\
=&-\csc^3\left( \frac{t}{1+t} \right)
\end{align*}
This is not defined for $t=0$.

You can see this easily by $\frac{t}{1+t}\mapsto t$. Now you can more clearly see that it traces the unit circle with $t=0$ giving the point $(0,1)$. Intuitively, the slope is infinitely great at this point.

To prove y''(0)=-1

Take the derivative of the second equation using the product, quotient and chain rules. Then replace $t$ with $0$ and evaluate.

For the other two you can integrate:

$\int -\sin(\frac{t}{1+t})*\frac{1}{(1+t)^2}*dt$

$u=\frac{t}{1+t}$

$du=\frac{1}{(1+t)^2}*dt$

$\int -\sin(u)*du$

Which I think you can evaluate :)


How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)

Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?

I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$

So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$

However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$

Appreciate any and all help!

Thanks for reading,

A

  Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?

$$x\left(\frac{1}{1-(x+x^3)}\right)$$ will do it.


extremum under constraint question I was tasked with finding the extremum of $z=xy$ under the constraint $x+y=1$, here is what I did: $$z=xy$$ $$x+y=1$$ from the second line we get $y=x-1$ and we substitute that back in the first equation: $$z=x(1-x)= x-x^2$$
$$z'(x) = 1-2x$$ next $$z'(x) = 0$$ $$<=> $$ $$0 = 1-2x$$ $$x=0.5$$ $$=>y=0.5$$ $$=>$$ $$(0.5,0.5)$$ suspected as extrema.
$$z''(x)= -2$$ $$z''(0.5)= -2<0 $$ hence $$(0.5,0.5)$$ local maxima. $$=> z_{max} = z(0.5,0.5) = 0.5 * 0.5 = 0.25 $$

Is this ok?

hint: $(x-y)^2 \geq 0 \implies xy \leq \dfrac{(x+y)^2}{4}$. Can you take it from here?


Having trouble using law of sines I have a triangle like this and I must find the length of the unkown side with question mark:

And I am using law of sines like this:

$$ \dfrac{\sin 55^\circ}{2} = \dfrac{\sin 35^\circ}{x} $$
then we got this equation
$$ x = \dfrac{2\sin 35^\circ}{\sin 55^\circ} $$
and it gives me the answer $ x \approx 1.40 $

But when I use this equation:

$$ \tan 35^\circ = \dfrac{x}{2} $$
$$ x = \dfrac{2}{\tan 35^\circ} $$

I get this answer: $ x \approx 2.85 $

Did I miss something?

You're last step is incorrect.

$$\tan 35^\circ = \frac{x}{2}$$

is equivalent to

$$x = 2 \tan 35^\circ$$

which does give the same result.

You said:

  $$ \tan 35^\circ = \dfrac{x}{2} $$
  $$ x = \dfrac{2}{\tan 35^\circ} $$

  I get this answer: $ x \approx 2.85 $

  Did I miss something?

The answer is yes, you did.
$$ \tan 35^\circ = \dfrac{x}{2} $$
to remove the denominator, multiply both sides of the equation by the denominator thus:
$$ \frac 21 \times (\tan 35^\circ) = \frac 21 \times \biggr(\frac x2\biggr) $$
$$ 2 \tan 35^\circ =  x$$
$$ x \approx 2 \times 0.7002$$
$$x \approx 1.4004$$


How to check diagonalization of orthogonal matrix? Is an orthogonal matrix diagonalizable over  $\mathbb{R}$ or $\mathbb{C}$ ?

I know that if all eigenvalues of a matrix are distinct then it is diagonalizable.
G.m. of each eigenvalue = its A.m.
And so on but sometimes there are complex matrices and I'm unable to find eigenvalues.
Please help me.

The matrix $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ is orthogonal, but it is not diagonalizable over $\mathbb R$ (it has no real eigenvalues). However, it is diagonalizable over $\mathbb C$. In fact, every real orthogonal matrix is diagonalizable over $\mathbb C$; this follows from the spectral theorem.

Actually, if we consider rotation matrix of order $2\times 2$, then we know rotation matrix is orthogonal matrix.then rotation matrix have no real eigen value in $\mathbb{R}$, unless the rotation angle $A$ is either $0$ or  $\pi$.
That is  rotation matrix have no eigenvector.
Again $A$ linear operator on a real vector space need not have an eigenvector.
That is, every  orthogonal matrix is not always diagonalizable over $\mathbb{R}$, but it is always diagonalizable over $\mathbb{C}$,because every linear operator on a complex vector space has at least one eigenvector.


Why do we take a derivative? I don't have a fundamental  understanding of what taking a derivative means and I don't understand why would we do it. I just know how to mechanically do it.

If $f(x)$ is a (real) function (of a real variable), then its derivative $f'(a)$ at a point $a$ measures the sensitivity of $f$ to small changes in $x$ around $a$. For example, suppose $x$ is "the height of the sun in the sky" and $f(x)$ is "the length of the shadow cast by this tree." Then $f'(a)$ is a measure of how sensitive the length of the shadow is to a small change in the height of the sun if the height is around $a$. Experimentally, $f'(a)$ will be small if the sun is high in the sky ($a$ is large) while $f'(a)$ will get larger as the sun gets lower in the sky ($a$ is small). The derivative gives us a way to quantify this observation.

I decided not to use the term "rate of change" because I think some students have heard this term so many times that they have become immune to it, so perhaps the above will make more sense.

if you given a function and you asked to find the behave of the given function then you may take the derivative to the given function, but if you given a behave of any function you need to take the integral of the behave function, to find the function of the given behave.


Is the following logic of simplifying the complicated expression correct? Assume
I have three variables $s, \omega,\gamma$ and define a function $G(.)$.

Next consider the following function

\begin{align}
\prod_{{i>1}}\left(\int_{-\infty}^{+\infty}\Bigg(\frac{1}{1+s  G(\omega_i)\gamma_{i}^{-1} }\Bigg)\ d\omega_i\right)\\\nonumber
\end{align}

Notie that $\omega_i$ and $\gamma_i$ are both indexed by $i$.

Is it true that I can re-write the above as

\begin{align}
\prod_{{i>1}}h(\gamma_i,s) \\\nonumber
\end{align}

where

$$h(\gamma_i,s)= \int_{-\infty}^{+\infty}\Bigg(\frac{1}{1+s  G(\omega_i)\gamma_{i}^{-1} }\biggl) \ d\omega_i$$

It is the indexing under the $\prod$ that is confusing me, since both the integrand has index $i$ and the product outside too, is my logic correct?

Maybe it should be made clear what $\gamma_i$ and $\omega_i$ are, e.g. they are different possible values of $\gamma$ and $\omega$ (or components maybe)? In any case what you suggest here is simply writing the complicated-looking integral expression as a function with two parameters $s$ and $\gamma_i$. There are two possible explanations: either $\omega_i$'s are arbitrary dummy variables and the author wanted to emphasize that these are different integrals, or $\forall i$: $\gamma_i$ and $\omega_i$ are related, for instance componentwise. Depending on the context, it might disable you to first add the terms and then integrate.


Calculate the heigh of a square when drawing perspective to a vanishing point I know how to use a vanishing point to create the illusion of perspective, but how do I determine the height of a square if I draw two lines towards the vanishing point?

For this case, assume I have two parallel lines on a grid. One at y=0 and one at y=10. I select a point on the higher line to be my vanishing point (0, 10) and then draw two lines from the lower line to the higher line to give two "parallel" lines (not parallel, but look parallel). First line from (-3, 0) to (0, 10), and the second line from (3, 0) to (0, 10). How do I find out where on the y-axis to draw the horizontal line so that my illusionary shape appears to be a square?

Bonus: If I have a square paper with points drawn on it and I know the location of those points. How do map those points to the perspective square in the above scenario? For example I have a point at 20% width and 90% height (near the upper left corner), how do I draw that point at its exact location on the perspective square?

There is a way to calculate the location of the horizontal line that goes on top of the square.

There are actually two ways to do it, but I'm not an expert on matrices so I'm not going to go into detail on how 3D rendering/calculating.

The old way to do it is based off of the idea of having a pane of glass between your eye and the object and drawing straight lines from your eye to the object to find the intersect points on the pane of glass. This pane of glass (canvas plane) has the points correctly placed.

This post contains detailed information on this perspective drawing technique and proofs for it.

http://perspectiveresources.blogspot.com/p/principles.html

No numbers are actually required, just angles and straight lines. More euclidian geometry than anything.

Note: This technique works for all shapes and rotations of shapes from a plane into perspective. To make shapes have height, that takes more work.

The short answer is that the location of the image of the far edge of the square is indeterminate. There’s a range a possibilities for its distance from the near edge’s image and without some other point of reference (such as other objects in the scene), any of them will do.

The distance between the two edges in the image is a function of the size of the square and its position in the 3-D scene, but there’s an infinite combination of those factors that will result in the same image. Mathematically, this corresponds to the fact that every point on a ray through the origin is mapped to the same point on the image plane by a perspective projection. By the same token, if you fix the size of the square, its foreshortening in the image will increase as it moves away from the viewer.

Of course, if you have the perspective transformation in hand and can locate the square in 3-D space, it’s a simple matter to apply the transformation to the square. Unfortunately, establishing a single vanishing point isn’t enough to determine this transformation uniquely.

Once you’ve established an image for the square, however, there are simple techniques for locating points on it by, for instance, subdividing it. This tutorial on one-point perspective drawing illustrates the technique—see especially exercise 4. If you want to do this mathematically, there’s a fairly straightforward way to compute a planar perspective projection transformation which maps the unit square onto a given convex quadrilateral that I describe here. Doing this also requires that you first establish the image of the square in some other way, though.


Heavy-tailed distributions I have encountered the following two definitions of heavy-tailedness (right tail) for a $[0,\infty)$-valued random variable $X$ satisfying $\mathbb{E}[X]<\infty$:

(i) $\limsup_{x\to\infty}\frac{\mathbb{P}(X>x)}{e^{-\lambda x}}>0$ for all $\lambda>0$,

(ii) $\mathbb{E}[X-u|X>u]\to\infty$ as $u\to\infty$.

Are these two notions equivalent? If yes, how to prove equivalence? Thank you in advance!

We will need to show that $(i)\implies (ii)$ and $(ii) \implies (i)$

Part 1: $(i) \implies (ii)$

Lets assume that $P(X>x)$ is a smoothly decreasing function bounded below by $0$.

Assume (i): $\limsup_{x\to\infty}\frac{P(X>x)}{e^{-\lambda x}} >0 \;\forall \lambda>0.$ Since $e^{-\lambda x},P(X>x)$ are positive, monotonic decreasing functions, $(i) \implies \forall \lambda>0,\{\exists c>0:P(X>x)\geq e^{-\lambda x}\;\forall x>c\}$

Now, we also know that $P(X>y|X>u)=\mathbf{1}_{<u}(y)+\frac{\mathbf{1}_{\geq u}(y)P(X>y)}{P(X>u)}$

Let $g(u):=E[X-u|X>u]$ where $g(u)\geq 0$ by definition of the conditional probability, then the above result (combined with the fact that $X\in [0,\infty)$, means that $g(u)=\int_0^{u}\mathbf{1}_{<u}(x)dx+\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}-u=\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}$

Combining this with our implication from $(i)$ gives us:

$\forall (\lambda>0),\exists c: \left\{g(u):=\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}\geq\frac{\int_u^{\infty} e^{-\lambda x}dx}{e^{-\lambda u}}=\int_u^{\infty} e^{-\lambda (x-u)} dx=\frac{1}{\lambda}\;\forall u>c\right\}$

However, $\lim_{\lambda \to 0} \frac{1}{\lambda} = \infty$; therefore, $g(u)$exceeds any bound, hence $\lim_{u\to\infty} g(u) = \infty$

Thus $(i)\implies (ii)$

Part 2: $(ii) \implies (i)$

Now, let $X$ be an arbitrary, positive continuous, integrable random variable, as you've specified in your post. Define the shifted conditional tail expectation $E[X-u|X>u]=E[X|X>u]-u=\frac{\int_u^{\infty}x f_X(x)dx}{P(X>u)}-u:=g(u)$

Assuming $(ii)$, we know that $g(u)$ must grow without bound, hence $g'(u)=O(\frac{1}{u^p}), p\leq1$. The most stringent value for $p$ is $p=1$, as it is on the threshold of convergence. Then we get:

$g'(u)=O(u^{-1})\implies g(u)=O(\ln(u))$. For concreteness, lets take $g(u)=\ln(u)$. This implies:

$\frac{\int_u^{\infty}x f_X(x)dx}{P(X>u)}-u = \ln(u) \implies \int_u^{\infty}x f_X(x)dx=(1-F_X(u))(\ln(u)+u)=\ln(u)+u -F_X(u)\ln(u)-uF_X(u)$.

Taking the derivative wrt $u$, we get:

$-uF_X'(u)=u^{-1}+1-\frac{F_X(u)}{u}-\ln(u)F_X'(u)-uF_X'(u)-F_X(u)$

Simplifying, we get:

$F_X'(u)=(1-F_X(u))\left[\frac{1}{u\ln(u)}+\frac{1}{\ln(u)}\right]$

Suppressing the fact that the distribution is for $X$ with argument $u$ and separating differentials, we get:

$\frac{dF}{1-F}=\frac{1+u}{u\ln(u)}du$

Integrating both sides we get:

$-\ln(1-F)+A=\ln(\ln(u))-\Re(\Gamma(0,-\ln(u)))+B$

Exponentiating both sides and simplifying, we get:

$\frac{e^A}{1-F}=\frac{e^B\ln(u)}{e^{\Re(\Gamma(0,-\ln(u))})}$

A little more algebra to isolate $(1-F)$, and we get (after combining the two unknown constants of integration):

$\frac{Ce^{\Re(\Gamma(0,-\ln(u)))}}{\ln(u)}=1-F=P(X>u)$

Thus $F(x) = 1-\frac{Ce^{\Re(\Gamma(0,-\ln(u)))}}{\ln(u)}$. If we set $C=1$, this function is a valid distribution function for a random variable defined on  $[1.72719,\infty)$. Its plot is shown below:

Note that it crosses the $x-$axis at $x\approx 1.72719$, so the random variable's distribution function would be  $P(X<x)= \mathbf{1}_{>1.72719}(x)F(x)$

Unfortunately, $P(X>x)=1-\mathbf{1}_{>1.72719}(x)F(x)$ shrinks faster than an exponential, hence it will not satisfy $(i)$. Thus, $(ii) \nRightarrow (i)$

$\square$

Therefore, it appears that $(i)$ is a stronger definition than $(ii)$, in the sense that it implies $(ii)$, which is a fact that makes sense given a distribution is heavy tailed (i.e., $(ii)$ should apply to all heavy-tailed distribution). However, $(ii)$ also applies to distributions that do not satisfy $(i)$, which is also a reasonable criteria for a heavy-tailed distribution.

The definition I've seen most often is $(i)$, and I think the above demonstration shows why that's the case. It's a difference between what properties are necessary vs sufficient. $(i)$ is sufficient while $(ii)$ appears merely necessary.

Please let me know if any of the above steps do not make sense, or, as per @Did, there are gaps or oversights...I'm not a professional "proof-writer", but this problem seemed very interesting so I wanted to give it a go, especially since you haven't received other answers.

A comment about the given "definition" of heavy-tailedness. One can define anything however one wants for local use, but the general application of heavy-tailed distributions is to model outlier-prone processes. For that purpose, the stated "definition" of heavy tailedness is too limited. After all, one can have outlier-prone processes that are bounded, e.g., mix a U(-1,1) with a U(-100, 100), with mixing probabilities .99 and .01.


Calculating the Zeroes of the Riemann-Zeta function Wikipedia states that

  The Riemann zeta function $\zeta(s)$ is defined for all complex numbers $s \neq 1$. It has zeros at the negative even integers (i.e. at $s = −2, −4, −6, ...)$. These are called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, and states that: The real part of any non-trivial zero of the Riemann zeta function is $\frac{1}{2}$.

What does it mean to say that $\zeta(s)$ has a $\text{trivial}$ zero and a $\text{non-trivial}$ zero. I know that $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$ what wikipedia claims it that $\zeta(-2) = \sum_{n=1}^{\infty} n^{2} = 0$ which looks absurd.

  My question is can somebody show me how to calculate a zero for the $\zeta$ function.

You are going to need a bit of knowledge about complex analysis before you can really follow the answer, but if you start with a function defined as a series, it is frequently possible to extend that function to a much larger part of the complex plane.

For example, if you define $f(x)=1+x+x^2+x^3+...$ then $f$ can be extended to $\mathbb C\setminus \{1\}$ as $g(x)=\frac{1}{1-x}$.  Clearly, it is "absurd" to say that $f(2)=-1$, but $g(2)=-1$ makes sense.

The Riemann zeta function is initially defined as a series, but it can be "analytically extended" to $\mathbb C\setminus \{1\}$.  The details of this really require complex analysis.

Calculating the non-trivial zeroes of the Riemann zeta function is a whole entire field of mathematics.

Here's an extension method in c# to calculate the zeroes for Re(s) > 0. However, it is not very efficient for large values of t. Note, .5 in the calculation is the Zeta(1/2+it). Try any other number and you will not get a zero.

Also, one could easily modify the function to return an IEnumerable<Complex> and the user could create a query/filter against each term in the infinite sum. I found it interesting to plot each term on a graph and watch it converge in the plane. The zeroes are where the graph comes back to the origin. The Complex type is found in the System.Numerics namespace.

    /// <summary>
    /// Calculates the converged point for a Dirichlet series expansion.
    /// </summary>
    /// <param name="t">imaginary part of s. The first zero is at 14.134725</param>
    /// <param name="numberOfTerms">Use a higher number to find more accurate convergence.</param>
    /// <returns></returns>
    public static Complex CalcZetaZero(this double t, int numberOfTerms)
    {
        var range = Enumerable.Range(1, numberOfTerms);
        var zetaZero = Complex.Zero;

        foreach (int n in range)
        {
            var direction = n % 2 == 0 ? Math.PI : 0;
            var newTerm = Complex.Exp(new Complex(-Math.Log(n) * .5, -Math.Log(n) * t + direction));
            zetaZero += newTerm;
        }

        return zetaZero;
    }


Is this square root approximation correct? Playing around I found a series which looks to converge to the square root function.

$$\sqrt{p^2+q}\overset{?}{=}p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)$$

Is it correct?

No.

$$p\left(1-\sum_{n=1}^{+\infty}\left(-\frac q{2p^2}\right)^n\right)=p\left(1+\frac q{2p^2}\frac1{1+\dfrac q{2p^2}}\right)=p+\frac{pq}{2p^2+q}\ne\sqrt{p^2+q}.$$

No, it is not correct.  Your right-hand expression is a geometric series for $$p\left(1-\frac{-\frac q{2p^2}}{1+\frac q{2p^2}}\right),$$
which is a rational expression in $p$ and $q$.

The correct answer is given by Newton's "generalized binomial theorem", in the form $$\sqrt{p^2+q}=p\sqrt{1+q/p^2} = p\sum_{k\ge 0} \binom{1/2} k \left(\frac q {p^2}\right)^k,$$ which converges for $|q/p^2|<1$.
The first two values of $\binom{1/2} k$ are $1$ and  $1/2$, which match what you have.  But $\binom{1/2}{2}=-1/8$ which does not match your $-1/4$.


Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$ Let $$f(x) = \frac{\sin(x)}{x}$$ on $\mathbb{R}_{\ge 0}$.

$$\int |f| < + \infty\quad\text{iff}\quad \int f^+ < \infty \color{red}{\wedge \int f^- < \infty}$$

EDIT: So what I'm trying to do is to show that in fact $\int f^+ > \infty$ so that therefore $\int |f| = \infty$

Now consider the assertion that:
$$\int f+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx \le \sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$$

Two Questions:

(1) Is the first step of asserting that

$$
\int f^+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx
$$

correct, in the sense that you can partition a Lebesgue integral into an INFINITE series of integrals being added together (whose individual term domains cover all of the overall domain of the original integral s.t. they are also pairwise disjoint)?

(2) Is there a well known lower bound of $\sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$ that diverges whose existence establishes that $f^+$ is in fact not integrable?

You might want to start by showing that $f$ is conditionally convergent. This would show that $f$ is not Lebesgue integrable. Use the fact that
$$\int_{(n - 1)\pi}^{n\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{1}{n\pi}\int_{(n - 1)\pi}^{n\pi} \left| \sin x \right| dx = \frac{2}{n\pi}$$
to get
$$\int_0^{N\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{2}{\pi}\sum_{n = 1}^N \frac{1}{n}$$
which shows that
$$\int_0^\infty \frac{\left| \sin x \right|}{x} dx$$
is divergent.

It is well known that the integral $ \int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx $ does not converge. Now use the theorem "a function $f$ is Lebesgue integrable iff |f| is" to show, that $f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable..


How many tries on average before I see the same value $N$ times in a row? If I repeatedly roll a fair, $X$-sided die, on average how many rolls should I expect to make before I happen to roll the same value $N$ times in a row?

I've found questions on here with answers when $X=2$, or where $N=2$, or where you're looking for a specific result $N$ times in a row; I'm interested in the situation where I don't care what the specific value is, I just want it to be $N$ times in a row. The closest I can find is Suppose we roll a fair $6$ sided die repeatedly. Find the expected number of rolls required to see $3$ of the same number in succession., but I can't quite figure out how to generalize it beyond $N=3$.

Let $E_k$ denote the expected number of rolls that are still needed if $k$ rolls have passed with equal result. We write recurrence relations for $E_k,0\leq k\leq N-1$ and calculate the wanted expectation $E=E_0$ from them.

  We obtain
  \begin{align*}
E_0&=E_1+1\tag{1}\\
E_1&=\frac{1}{X}(E_2+1)+\frac{X-1}{X}(E_1+1)\\
E_2&=\frac{1}{X}(E_3+1)+\frac{X-1}{X}(E_1+1)\\
E_3&=\frac{1}{X}(E_4+1)+\frac{X-1}{X}(E_1+1)\tag{2}\\
&\ \ \ \vdots\\
E_{N-3}&=\frac{1}{X}(E_{N-2}+1)+\frac{X-1}{X}(E_1+1)\\
E_{N-2}&=\frac{1}{X}(E_{N-1}+1)+\frac{X-1}{X}(E_1+1)\\
E_{N-1}&=\frac{1}{X}+\frac{X-1}{X}(E_1+1)\tag{3}\\
\end{align*}

Comment:

In (1) we note that any roll of the $X$-sided die will do.
In (2) we are in the situation that a run of length $3$ is given. With probability $\frac{1}{X}$ we  get a run of length $4$ and with probability $\frac{X}{X-1}$ we get a different result and start with a run of length $1$.
In (3) we get a run of length $n$ with probability $\frac{1}{X}$ and can stop or we are back at the beginning and start again with a run of length $1$.

  This system can be easily solved. We iteratively obtain be respecting all equations from above besides the last one (tagged with (3)).
  \begin{align*}
E_1&=E_0-1\\
E_2&=E_0-(1+X)\\
E_3&=E_0-(1+X+X^2)\\
E_4&=E_0-(1+X+X^2+X^3)\\
&\ \ \ \vdots\\
E_{N-1}&=E_0-(1+X+X^2+\cdots+X^{N-2})\\
&=E_0-\frac{X^{N-1}-1}{X-1}\tag{4}\\
\end{align*}

  From equation (4) and (3) we finally conclude
  \begin{align*}
E_{N-1}&=E_0-\frac{X^{N-1}-1}{X-1}\\
(1-X)E_0+XE_{N-1}&=1
\end{align*}
  from which
  \begin{align*}
\color{blue}{E=E_0=\frac{X^N-1}{X-1}}
\end{align*}
  follows.

Note: This approach is a generalisation of @lulu's answer of this MSE question. In fact we get as small plausibility check with $N=3$ and $X=6$ the result
\begin{align*}
E=\frac{6^3-1}{6-1}=\frac{215}{5}=43.
\end{align*}
in accordance with @lulu's result.

  [Add-on] The strongly related problem: Rolling a fair $X$-sided die until we get a  run of length $N$ of  a  specific  value can be solved by the system above with the first equation (1) replaced with
  \begin{align*}
E_0&=\frac{1}{X}(E_1+1)+\frac{X-1}{X}(E_0+1)
\end{align*}
  which gives the result
  \begin{align*}
\color{blue}{E=E_0=\frac{X\left(X^N-1\right)}{X-1}}
\end{align*}
  This is also plausible compared with the result above, since the probability to roll a specific value is $\frac{1}{X}$.

$N_1$ - No of tosses for 1st Heads
$N_2$ - No of tosses for 2 consecutive Heads
$N_3$ - No of tosses for 3 consecutive Heads
$\mathbb E[N_1] = 2$
$\mathbb E[N_2] = (1+\mathbb E[N_1])2 = 6$
$\mathbb E[N_3] = (1+\mathbb E[N_2])2 = 14$
A general formula can be created
$\mathbb E(1) = X$
$\mathbb E(2) = (X+1)X$
$\mathbb E(3) = \big((X+1)X+1\big)X$
and so on.

$$\mathbb E(N) = \sum_{i=1}^N X^i = \frac{X(X^N-1)}{X-1} $$

I have made a video explanation and made a simple formula at the end
https://youtu.be/d72rcsBblRE


We roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by 6? So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?

Suppose that we roll a six-sided die ten times. What is the probability
that the total of all ten rolls is divisible by six?

Hint.

Roll $9$ times and let $x$ be the total.

For exactly one number $n\in\{1,2,3,4,5,6\}$  we will have $6 \mid (x+n)$ (i.e. $x+n$ is divisible by $6$).

Couldn't you also think of it as the maximum possible value you could get for rolling the die 10 times would be 60, how many numbers between 1 and 60 are divisible by 6? 10 numbers are divisible by 6(6*1, 2, 3, etc.)  So, 10 out of 60 possible values gives... 1/6.  I love math.


Arguing a heuristic solution to a linear PDE, in weak formulation I have a nasty question, so I'll do it in steps.

Imagine a simple PDE to solve, a mixed derivative equal to zero:
$$\dfrac{\partial^2 u(x,y)}{\partial x \partial y}=0$$
A candidate for the solution is an additively separable function: $$u(x,y)=f(x)+g(y)$$
I know how to prove it formally, without pitfalls, by integrating twice:

$$\dfrac{\partial u(x,y)}{\partial x}=h(x)$$

$$u(x,y)=\int_0^x h(z) dz + g(y)$$

Now the problem is that I need to do the same for $n>2$ and in "weak formulation".
That is, at the very least I need to solve this: $$\int \int \int u(x,y,z) \dfrac{\partial^3 \varphi(x,y,z)}{\partial x \partial y \partial z} dx dy dz=0$$ in some reasonable space, say $L^1$, for all test functions $\varphi$.
My guess is that the solution (loosely speaking) takes the form $$u(x,y,z)=f(x,y)+g(y,z)+h(z,x)$$
almost everywhere, and this is exactly what you get if you try to solve the thing heuristically, by first using the trick in part 3 repeatedly, and then forgetting that the function has to be smooth.
Now, I know how to deal with the problem in weak formulation for $n=2$, because then it is equivalent, with a change of variables, to the homogenous wave equation $$\dfrac{\partial^2 u}{\partial a^2}=\dfrac{\partial^2 u}{\partial b^2}$$ which was studied extensively in many textbooks. Basically, they show straightforwardly that the solution is indeed like that. However, for $n>2$ it is a high-order PDE and is not covered in even advanced textbooks.
Finally, The Question. I need to find a way to argue that the heuristic approach is valid. I heard some rumors that it is indeed true for linear PDE, but I fear that the proof of such a powerful result is complicated. On the other hand, for my particular PDE it might be possible to argue that with less blood, directly. If you guys have any wisdom about that, or you know a textbook/link, please tell me.

So I figured out the solution.

For my needs it is sufficient to focus on a smaller functional space, like left-continuous functions $u(.)$. This normalization allows to solve the weak-differential equation explicitly, rather than almost everywhere. The trick is to use continuity and a sequence of test functions to add a new "limit" test function to the arsenal.

For example, pick a weak-differential equation $$f'(x)=0 \text{ , or } \int f(x) \varphi'(x) dx=0$$ Typically we can not evaluate the functional at a test function $\varphi$ that takes value 1 on a segment $[a,b]$ and 0 elsewhere. However, with left-continuity you can approximate it with smooth test functions. It is then easy to see that the limit is $$f(b)-f(a)=0.$$ Similarly you can solve higher-order differential equations.

So the heuristic approach is valid, at least if you pick a proper normalization.


Find $\lim_{x\to 2}(x^2-3)$ using the definition of limit; also find $\delta$ for given $\epsilon$ Find the limit $L$  using the definition of limit.  Then find $\delta>0$ such that $|f(x)-L|<0.01$ whenever $0<|x-c|<\delta$.
  $$\lim_{x\to 2}(x^2-3)$$

I have gotten as far as $|x-2|=0.01/|x+2|$, and I need help with further steps, I do not know what more there is to do to find delta.

Source image

If you choose $$\delta:=\min\{1,\frac\varepsilon5\}$$
then for $|x-2|<\delta$ you have $|x-2|<1$, which implies
$1<x<3$ and $|x+2|<5$.

So you get
$$|x^2-2|=|x+2|\cdot|x-2|<\frac\varepsilon5\cdot 5 = \varepsilon.$$

$$\lim_{x \rightarrow 2} (x^2-3)=2^2-3=4-3=1$$

The limit $L$ is equal to $1$.

We have to find a $\delta>0$ such that $|f(x)-1|<0.01$ whenever $0<|x-2|<\delta$.

EDIT:
$$|x-2|<\delta \Rightarrow -\delta<x-2<\delta \Rightarrow 2-\delta<x<2+\delta$$

$$|f(x)-1|<0.01 \Rightarrow |x^2-3-1|<0.01 \Rightarrow |x^2-4|<0.01 \Rightarrow -0.01<x^2-4<0.01 \\ \Rightarrow 3.99<x^2<4.01 \Rightarrow \sqrt{3.99}<x<\sqrt{4.01}$$

Therefore, we could set $2+\delta=\sqrt{4.01} \Rightarrow \delta \approx  0.002$


Divergence formula on manifolds using a physical approach In rectangular coordinates, one can define the divergence of a vector field $\vec{V}$ as the following limit
$$ \text{Div } V = \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon^3} \oint_{\partial C_{\epsilon}(p)} \vec{V} \cdot d \vec{A} \: \: \: \: (1), $$
where $C_{\epsilon}(p)$ is the cube of length $\epsilon$ and with corners at $p$ and  $p+\epsilon(1,1,\ldots,1)$.
In the setting of Riemannian manifolds we have the formula
$$ \text{Div } V = \frac{1}{\sqrt{|g|}} \partial_i \left( \sqrt{|g|} V^i \right). \: \: \: \: (2)$$
By the divergence theorem (for manifolds) one also obtains a definition for $\text{Div } V$ as in $(1)$. My question is how to deduce $(2)$ from such a definition.

I'd like to have an answer which doesn't use differential forms or Lie derivatives.

The proof of (2) from (1) is entirely local, that is, it can be carried out in an arbitrarily small neighbourhood of $p$.

EDIT THE FOLLOWING IS WRONG (See comments)

Since all Riemannian manifolds are locally isometric to Euclidean space, if you can prove (2) from (1) in an arbitrary curvilinear coordinate system in $\mathbb R^n$ then you are done. And it seems to me that you have  already done this last step. (If not, then you can find it in books in vector calculus or fluid mechanics).


Why does substitution work in integrals Let's say I have this integral:

$$\int_0^\infty e^{-t} \, dt$$

And I make the substitution:

$$t = nu$$

Then why I can say that:
$$dt = n\,du$$
and then put this into my integral like this:

$$\int_0^\infty e^{-nu}n\,du$$

What's happening in the background that allow this to be done?
I'm asking this because I don't feel confortable threating diferencial operators as fractions and I don't know why this can be done.

It's just a way of looking at the chain rule.

The chain rule is differentiation by substitution.

One can write $$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x),$$ or one can look at
$$
\frac{d}{dx} f(g(x))
$$
and then do this substitution:
$$
u = g(x),\qquad \frac{du}{dx} = g'(x).
$$
Then one writes
$$
\frac{d}{dx} f(g(x)) = \frac{d}{dx} f(u) = \frac{df(u)}{dx} = \frac{df(u)}{du}\cdot\frac{du}{dx} = f'(u)\cdot g'(x) = f'(g(x))\, g'(x).
$$

In the same way, when one sees
$$
\int f'(g(x)) g'(x) \,dx,
$$
one does the substitution
$$
u=g(x),\qquad \frac{du}{dx} = g'(x),\qquad du = g'(x)\,dx.
$$
Then one has
$$
\int f'(g(x)) g'(x) \,dx = \int f'(u)\,du = f(u)+C = f(g(x))+C.
$$

So integration by substitution is the chain rule in reverse, just as integration by parts is the product rule in reverse.

The substitution in integrals are the reverse of the chain rule in derivatives.

The following thread gives a fairly good explanation on why does the substitution works

Why does substitution work in antiderivatives?


How many ways are there to put $N$ different balls in $M$ different boxes so that no box is empty. You are given $N$ different balls and $M$ different boxes. In how many ways can one distribute the balls so that no box is empty?

I've found it hard to answer, I've listed a few cases, but it seems to be much more complicated when $M$ is getting bigger. I don't know how to proceed. Any help appreciated.

From comment as requested:
If the boxes were indistinguishable, I would have thought the answer was Stirling numbers of the second kind $\left\{ {N \atop M} \right\}$.
Since they are distinguishable, you have to multiply this by $M!$ so $$M! \left\{ {N \atop M} \right\} = \sum\limits_{i=0}^{M} (-1)^{i} \binom{M}{i} (M-i)^N$$

The number of ways with no restrictions is $N^M$. We now use inclusion exclusion.
The number of ways to do so with at least $k$ boxes empty is ${M \choose k} \times N^{M-k}$ (the number of ways to choose the empty boxes, and the number of ways to distribute the remaining numbers).

So we obtain the answer $$\sum_{k=0}^{M-1} {M \choose k}N^{M-k} (-1)^k$$
which is simply $(N-1)^M$ by considering binomial expansion.


Vladimir Blinovsky's Union-Closed Sets Conjecture Proof Recently, Vladimir Blinovsky published an article (http://arxiv.org/pdf/1507.01270v6.pdf) claiming that he proved the union-closed sets conjecture (https://en.wikipedia.org/wiki/Union-closed_sets_conjecture).

At first, I was quite excited but then I realised that I understand almost nothing written in the paper. I was expecting a proof which is set-theoretic or algebraic but instead there were many computations (integration, summation etc.). Can someone explain the relation between Frankl's Conjecture and these computations? Are they used to prove the conjecture before or is it something new?

Long way to explain, it is original method which allows to solve many problems in extremal combinatorics on sequences, text contains some unaccurances, but the idea is (as usual) not difficult.

Author


Probability over time So given say, a deck of  cards, I draw cards until I pick out a specific one, in this case, the ace of spades. Once the ace of spades has been drawn, you restart. Now, this continues till infinity. On average, how many tries does it take to get the card? Is there a solution to a deck of size $n$?

Drawing cards serially

Assuming a 52-card deck, shuffled completely randomly, the Ace of Spades has a 1/52 chance of occupying any position in the deck.

Let $X$ denote the position of the ace of spades. Then, $P(X = x) = 1/52$ for all $x = 1, 2, \ldots, 52$.

The average value of $X$ is given by the formula for expectation of a discrete random variable:

$$E[X] = \sum_{i=1}^{52} i P(X = i) = \frac{1}{52} \sum_{i=1}^{52} i = \frac{52\cdot (52+1)}{52\cdot 2} = \frac{53}{2}$$

This generalizes to $n$ by replacing 52 with $n$.

Drawing cards randomly

The expected value formula is the same, but we modify $P(X=x)$ to reflect the random drawing of cards.

It is easy to see that the probability of drawing the Ace of Spades on the $i$th try is the same as the probability of $i-1$ failures followed by a single success.

The probability of $j-1$ failures, without replacement, can be written

$$ \frac{51}{52} \frac{50}{51} \cdots \frac{52-(j-1)-1}{52-(j-1)}.$$

It should be clear that diagonal terms cancel, leaving

$$\frac{52-j}{52}.$$

However, do note that because of the indexing, we have $j = i-1$. That is, if we pull the Ace of Spades on the first card ($i=1$), we have zero preceding failures.

After drawing $i-1$ cards, there are $52-(i-1)$ cards remaining in the deck, so the probability of success on the ith try is $1/(53-i)$.

Multiplying these together, one gets

$$P(X=i) = \frac{53-i}{52} \frac{1}{53-i} = \frac{1}{52}.$$

Hopefully it is more clear now that the result is the same as before.

This is well discussed in these forums itself, I've summarized it here. Hope that helps


Finding parameter values Problem:

What value of $a$ makes $$f(x)=x^2 + \frac {a}{x}$$ have (a) a minimum at $x=2$, (b) an inflection point at $x=1$?
(a) and (b) are separate and not dependent upon each other.

What I have done:
Not much. I wrote the derivative to try and find where $x=0$, but that is where I got stuck.
$$0=2x-ax^{-2} .$$
I'm not really sure what to do.

(a) Take first derivative of $f(x)$, set it equal to 0, replace $x$ by 2 , and solve for $a$.

(b) Take second derivative of $f(x)$, set it equal to 0, replace $x$ by 1, and solve for $a$.


Is it true that subset D is dense in space X iff every nonempty open set in X contains one point of D I have a proposition that I am not sure if it is right:

Let D be a subset of space X. D is dense in X if and only if every nonempty open set U of X contains at least one point of D.

When X is a metric space, then I could verify that this is true in both directions.

When X is a topological space, I could prove the ==> direction; however the <== direction is so difficult that I doubt the proposition is right. However, I cannot yet find a counter example.

This proposition is the generalization of a property of real numbers R:

Q is dense in R so every open interval of R contains a point of Q.

Thank you.

As to your definition: $D$ is dense when the closure of $D$ (a.k.a. $\overline{D}$) in $X$, equals $X$. (this does not mean in general that every point of $X$ is a limit of a convergent sequence from $D$; it does in metric spaces, but not in general topological spaces). So as the definition I will use $\overline{D} = X$.

Of course, now the solution will depend on the definition of the closure of a set $A$. Let's use the definition, or characterization, that $x \in \overline{A}$ iff for every open set $O$ of $X$ that contains $x$, $O$ intersects $A$.

Now the statement about the density of $D$ is pretty clear: if $D$ is dense and $O \neq \emptyset$ is open, then pick $p \in O$, then $p \in \overline{D}$, as $\overline{D} = X$, and so every open neighbourhood of $p$ intersects $D$; in particular $O$ intersects $D$, i.e. $O$ contains a point of $D$.

On the other hand, if every non-empty open set $O$ of $X$ contains a point of $D$, let $x \in X$. Let $O$ be any open set that contains $x$. Then $O$ is non-empty (because of $x$) and so contains a point of $D$, i.e. $O \cap D \neq \emptyset$. But this just says that $x \in \overline{D}$ by the definition above. As $x \in X$ was arbitrary, $X \subset \overline{D}$. As $\overline{D} \subset X$ by definition, we have equality.

Note that nothing at all of a metric is used. This is just a general fact for topological spaces.

I think your question is same with this question " A set D in a space X is dense in X iff every non-empty open set in X meets D", right?


Show the following equality Basically I want to show the following:
$$\sqrt{2}\ |z|\geq\ |\operatorname{Re}z| + |\operatorname{Im}z|$$
So what I did is the following:
Let $z = a + bi$
Consider the following:
$$2|z|^2 = 2a^2 + 2b^2 = a^2 + b^2 +a^2+b^2$$
Since $(a-b)^2\geq0$, hence $a^2+b^2\geq 2ab$
Thus $2|z|^2 \geq a^2+b^2+2ab = (a+b)^2$
Hence $\sqrt{2}|z| \geq a + b$ but $a = |\operatorname{Re}z|~,~ b = |\operatorname{Im}z|$
Did I make a mistake somewhere, if yes I would appreciate it if it could be pointed out and perhaps provide some guideline on how to prove this.

Let $z = r\cos\theta + ri\sin\theta$

Then $|Re(z)| + |Im (z)| = \pm\sqrt 2 r\sin(\theta \pm \frac{\pi}{4}) \leq \sqrt 2 r$

All four possible sign combinations will cover $\theta$ in all quadrants.

Observe
\begin{align}
|a|\cdot 1+|b|\cdot 1 \leq \sqrt{a^2+b^2}\sqrt{1^2+1^2}.
\end{align}


Probability of Snow in New York In New York, snow is reported 25% of days in February.  If this trend continues, what is the probability that it will snow exactly 9 days this coming February and is not a leap year?

Solve this problem by using
First)  The approximation of the binomial distribution
Second) The Binomial Distribution

How would I apply the Binomial Distribution and approximate distribution? I am having difficulty extracting the info from the problem.

So far I have the following
p = .25
q = .75
n = 28 days
k = 9

Would I apply that to C(n,k)(p^k)(q^n-k)?

Yes. Under the assumption that the binomial distribution is applicable (i.e. it snows on each day in February with probability $25\%$, independent of other days) the probability of exactly $9$ days with snow is
$$ \binom{28}{9} (.25)^9 (.75)^{19}. $$
There are $\binom{28}{9}$ ways to choose the $9$ days, the probability that is does snow on these days is $.25^9$ and the probability that it does not snow on all other days is $.75^{19}$.

But in reality this answer is not accurate because snow on each day is not independent of other days.  For example, it may start snowing at 11:30PM on Feb 1st so that would likely increase the chance that it would snow on Feb 2nd (right after midnight for example).  So the days are NOT independent of each other.

For things like randomly drawn cards and fair coin tosses, calculating probability is meaningful.  For this weather problem, it is only an estimate at best because of the problem stated above, and because past weather is not an assurance of future weather.