Finding $\int\sqrt{\sec(x)}\ln(\sec(x))\tan(x)\,\mathrm dx$ by substitution $$\int\sqrt{\smash[b]{\sec(x)}}\ln(\sec(x))\tan(x)\,\mathrm{d}x$$

I started by making u-substitution $u = \sec(x)$:

$$\int\sqrt u\ln(u)\tan(x) \left(\frac{\mathrm{d}u}{\sec(x)\tan(x)}\right)$$

Now, does the $\tan(x)$ cancel? Then is integration by parts the appropriate method to use?

Yes! The $\tan(x)$ cancels.

I'm sure you know what to do with the remaining $\sec(x)$, given that $u=\sec(x)$.

Integration by parts is next.

Given $$\int\sqrt{\sec x}\ln(\sec x)\tan x\ dx$$
Apply integration by parts $u=\ln(\sec x),v^{\prime}=\sqrt{\sec x}\tan x$

$$=\ln(\sec x)\cdot2\sqrt{\sec x}-\int\tan x\cdot2\sqrt{\sec x} dx$$
Since $\int\tan x\cdot2\sqrt{\sec x}\ dx=4\sqrt{\sec x}$
$$=\ln(\sec x)\cdot2\sqrt{\sec x}-4\sqrt{\sec x}+C$$


How to determine if 3 points on a 3-D graph are collinear? Let the points $A, B$ and $C$ be $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$ respectively. How do I prove that the 3 points are collinear? What is the formula?

From $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$ we can get their position vectors.

$\vec{AB}=(x_2-x_1,y_2-y_1,z_2-z_1)$ and $\vec{AC}=(x_3-x_1,y_3-y_1,z_3-z_1)$.

Then $||\vec{AB}\times\vec{AC}||=0\implies A,B,C$ collinear.

If the distance between |AB|+|BC|=|AC| then A,B,C are collinear.


Random Walk on Clock Hands We do a random walk on a clock. Each step the hour hand moves clockwise or counterclockwise each with probability 1/2 independently of previous steps.

If you start at 1 what is the expected number before you hit 12?

Well, the answer is 11 according to the simulations I ran. I know to return to 1, or from any number back to itself, it would be 12 because of the properties of the invariant distribution; so if you start at a certain position in the long run you can expect to come back once every 12 times. The invariant distribution is uniform across the 12 states. I can't explain it beyond that case though. I got ~20 steps on average to get the hands to move from 1 to 11.

These are not problems on a circle but on the linear interval of integers $$L=\{0,1,2,\ldots,12\},$$ where the hand $12$ is represented by both states $0$ and $12$, and every other hand by the state with its number. Starting from $k$ in $L$, the mean number of steps $t_k$ that the symmetric simple random walk needs to hit $0$ or $12$ is $$t_k=k(12-k),$$ hence your simulations giving $t_1=11$ and $t_2=20$ (since hitting $11$ starting from $1$ on the clock is equivalent to hitting $12$ starting from $2$) are accurate.

To show this, the standard method is to consider the whole collection $(t_k)_{0\leqslant k\leqslant12}$ simultaneously. Then, $$t_0=t_{12}=0,$$ and the (simple) Markov property of the random walk after one step yields $$t_k=1+\tfrac12(t_{k-1}+t_{k+1}),$$ for every $1\leqslant k\leqslant11$. Solving this $13\times13$ affine system yields the desired formula.

from random import random
import matplotlib.pyplot as plt
import numpy as np
from __future__ import division
% matplotlib inline
r=12
lt=[0]*r
x,y=[],[]
count,pos=0,0
flag=False
n=10000
lc=[]
for i in range(n):
    while not flag:
        count+=1
        if random()<.5:
            pos+=1
        else:
            pos+=-1
        pos%=r
        lt[pos]=1
        if sum(lt)==12:
            flag=True
            lc.append(count)
            count=0
    flag=False
    lt=[0]*r
    pos=0
plt.plot(lc)
print np.average(lc)

~67.04525 as far as I found

How: I just built an array of 12 0's and a random variable uniformly distributed and I just defined when this variable is less than .5 it steps clockwise  one position or else the  other way around. (turning 0 in the n-position into 1)

Counting the number of steps it took for all the 0's to become 1 and repeating this 10000 times y arrived that result.

Feel free to run this code at jupyter notebook online


Showing that any of n balls drawn without replacement has the same probability of being a particular colour Suppose that $X\sim \text{HGeom}(w,b,n)$ represents the distribution of $w$ white and $b$ blue balls (where $w+b=n$) in an urn. Let $X_j$ represent the indicator random variable of the $j$-th ball being white if they are drawn without replacement. My question is how you can show that $E(X_j) = w/(w+b)$ for any $j$ by symmetry.
Clearly,
$$\begin{align}
E(X_1) &=\frac{w}{w+b}\\[3ex]
E(X_2) &= \frac{w}{w+b}\left(\frac{w-1}{w+b-1}\right)
+\frac{b}{w+b} \left(\frac{w}{w+b-1}\right)
=\frac{w}{w+b}
\end{align}
$$
Does this pattern continue for all $j$ up to $n$? Could it be extended if there were balls of $k$ different colours?

I believe there may be a number of elegant proofs, perhaps relaying on the linearity of expectation of random variables and indicator variables (expectation = probability). Joseph K. Blitzstein has a similar problem explain here, which would be paraphrased as follows with regard to the symmetry insight:
This is true by symmetry. The first ball is equally likely to be any of the $b + w$ balls, so the probability of it being white is $\frac{w}{w +b}.$ But the second ball is also equally likely to be any of the $b + w$ balls (there aren’t certain balls that enjoy being chosen second and others that have an aversion to being chosen second); once we know whether the first ball is $W$ we have information that affects our uncertainty about the second ball, but before we have this information, the second ball is equally likely to be any of
the balls. Alternatively, intuitively it shouldn’t matter if we pick one ball at a time, or take one ball with the left hand and one with the right hand at the same time. By symmetry, the probabilities for the ball drawn with the left hand should be the same as those for the ball drawn with the right hand.
If every possible result is a single-cycle permutation of $w$ and $b$ balls that can be considered otherwise distinguishable by their order of extraction, but that for every single result each ball could have equally have been extracted in one position posterior to the position it occupies, the actual result can be viewed as that sliding of one position forward of each one of the balls with a period of $w+b,$ so that every single different result is matched by $w+b$ results where the relative position doesn't change, and each ball occupies all possible extraction points.

In your calculation you get to $E(X_1)=\frac{w}{w+b}$ and $E(X_2)=\frac{w}{w+b}\left(\large \Box \right),$ where happily, $\large \Box =1,$ and hence, $E(X_1)=E(X_2).$ So what we want is that this patterns holds for all $X_i,$ such that $E(X_i)=\frac{w}{w+b}\left(\color{red}{\large \Box} \right)$ with $\color{red}{\large \Box}=1$ for all $i$'s.
And this pattern can possibly be teased out by just seeing what happens next -in the case of $X_3:$
$$ E(X_3) =\Tiny \left(\frac{w}{w+b}\right) \left(\frac{w-1}{w+b-1}\right)\left(\frac{w-2}{w+b-2}\right)
+2\left(\frac{b}{w+b}\right) \left(\frac{w}{w+b-1}\right)\left(\frac{w-1}{w+b-2}\right) + \left(\frac{b}{w+b}\right) \left(\frac{b-1}{w+b-1}\right)\left(\frac{w}{w+b-2}\right)$$
Clearly we'll always be able to extract the $E(X_1)=\frac{w}{w+b}$ as a factor in front of the sum since each $w$ and $w+b$ appear in each term in the numerator and denominator, respectively. What remains to be proven is that the sum multiplied by $E(X_1)$ is always equal to $1:$
$$\begin{align}
1=\Tiny{  \left(\frac{w-1}{w+b-1}\right)\left(\frac{w-2}{w+b-2}\right)
+2\left(\frac{b}{1}\right) \left(\frac{1}{w+b-1}\right)\left(\frac{w-1}{w+b-2}\right) + \left(\frac{b}{1}\right) \left(\frac{b-1}{w+b-1}\right)\left(\frac{1}{w+b-2}\right)}\implies \\[3ex]
\small{(w+b-1)(w+b-2)=(w-1)\;(w-2) + 2\;b\;(w-1) + b\;(b-1)\\
={2 \choose 0}(w-1)\;(w-2) +{2 \choose 1} b\; (w-1) + {2\choose 2} b\;(b-1)}
\end{align}$$
But the LHS is the 2-permutations of $(w - 1) + b$, while the RHS is the binomial expansion considering $w$ and $b$ to denote the number of elements in the set of class $\text W$hite and $\text{B}$lack, respectively.
This pattern will hold for any $X_i,$
$$\begin{align}
\left((w-1) + b\right)\left((w-2) + b\right)\cdots\left((w-i-1) +b\right)&\\[3ex]
=\small{ {i-1 \choose 0} (w-1)(w-2)\cdots (w-i-1)\\+\cdots + {i-1 \choose j} (w-1)\cdots (w-i-j)\,b\,(b-1)\cdots(b-j)\\+\cdots+{i-1 \choose i-1} b\,(b-1)\cdots (b-i-1)}
\end{align}$$

There are several perspective to recognize the fact without the tedious calculation. Here is some of my understanding can be shared with you.
Note that the usually "fallacy" that people have is that people think that when $j > 1$, they already obtained the information from $X_1, X_2, \ldots, X_{j-1}$ so the probability is changed from $w / (w + b)$. Although $X_j$ are dependent, now the question is asking about the marginal distributionof $X_j$, but not the conditional distribution of $X_j \mid X_1, X_2, \ldots, X_{j-1}$. The latter conditional probability is similar to  the experiment that we drawn the first $j-1$ balls and put on the table, observing their color, such that the probability of the next drawn will be dependent on the color of the balls we already drawn. However, in our case the marginal distribution is like we do not observe the color of the ball of the first $j-1$ balls, just put them into another bag immediately after drawn. So actually the marginal distribution of each ball is the same, assume they are all equally-likely to be drawn.
Actually we assign the order of the ball is like we assign a permutation. Imagine we assign the $n$ balls the number $1, 2, \ldots, n$. Assume tt is equally-likely for each permutation, and thus the probability of the $i$-th ball is put in the $j$-th position is equal to the number of favorable permutations divided by number of total permutations. Since for each favorable permutation, we can shift the balls such that the $i$-th ball is put in the other $k$-th position. (E.g. the original permutation is $(1, 3, 2, 4)$, we can shift it to $(4, 1, 3, 2), (2, 4, 1, 3), (3, 2, 4, 1)$. So by grouping the permutation in this way, we can claim that it is equally-likely for the $i$-th ball to be position in any one of the position. Now you paint the first $w$ ball to be white, and you can sum the probabilities to be $w/n$


How many three digit numbers exist such that the third digit is the geo mean How many three digit numbers exist such that one of the digits is the geometric mean of the other two?
  A 12, B 18, C 24, D other

So, $N = 100a + 10b + c$ let $c =\sqrt{ab}$.

$ab$ must be a perfect square, so $ab = 1, 4, 9, 16, 25, 36, 49, 64, 81$

Case: $ab = 1 \implies a=1, b=1$.

Case: $ab = 4 \implies a = 2, b=2$

Case: $ab = 9 \implies a=3, b=3$.

Case: $ab = 16 \implies (a=8, b=2), (a=4, b=4)$

Case: $ab = 25 \implies a=5, b=5$

Case: $ab = 36 \implies (a=9, b=4), (a=6, b=6)$

Case: $ab = 49 \implies (a=7, b=7)$

Case: $ab = 64 \implies (a=8, b=8)$

Case: $ab=81 \implies (a=9, b=9)$.

$(a=8, b=2, c = 4)$ also means that we could replace this with, $ (a, b, c) = (2, 4, 8) = (4, 2, 8) = (4, 8, 2)$

$(a=9, b=4, c=6)$ also means $(a, b, c) = (9, 4, 6) = (4, 6, 9) = (9, 6, 4) = (4, 9, 6) = (6, 9, 4) = (6, 4, 9)$.

So to me, the answer is B $18$

You asked two different questions.  The question in the title appears to be:

  "How many three digit numbers exist such that the third digit is the
  geometric mean of the other two?"

I will answer the question in the statement of the problem:

  "How many three digit numbers exist such that one of the digits is the
  geometric mean of the other two?"

Let $x, y \geq 0$.  The geometric mean of $x$ and $y$ is $\sqrt{xy}$.

A three digit positive integer is a number of the form $100a + 10b + c$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $b, c \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

We consider cases:

The geometric mean of the digits is $0 \implies 0 = \sqrt{a \cdot 0}$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  There are nine possibilities: $100, 200, 300, 400, 500, 600, 700, 800, 900$.
If $n \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, $n = \sqrt{n \cdot n}$.  There are nine such numbers.  They are $111, 222, 333, 444, 555, 666, 777, 888, 999$.
$2 = \sqrt{1 \cdot 4}$.  Then the number is one of the $3! = 6$ permutations of the digits $1$, $2$, and $4$.  They are $124, 142, 214, 241, 412, 421$.
$3 = \sqrt{1 \cdot 9}$.  Then the number is one of the $3! = 6$ permutations of the digits $1$, $3$, and $9$.  They are $139, 193, 319, 391, 913, 931$.
$4 = \sqrt{2 \cdot 8}$.  The number is one of the $3! = 6$ permutations of the digits $2$, $4$, and $8$.  They are $248, 284, 428, 482, 824, 842$.
$6 = \sqrt{4 \cdot 9}$.  The number is one of the $3! = 6$ permutations of the digits $4$, $6$ and $9$.  They are $469, 496, 649, 694, 946, 964$.

Hence, there are $2 \cdot 9 + 4 \cdot 6 = 42$ three digit numbers such that one of the digits is the geometric mean of the other two.

Third digit is $\sqrt{ab}$ implies square root should yield single digit so it can be $[0,9]$ thus now digits whose product is 0 so $100,..900$ , for 1 is only $1,1$ so number $111$ then for 2 so product is $4$ thus numbers are $222,142,412$ then $3$ so $333,193,913$ now $4$ so $284,824,444$ then $5$ so $555$ then $6$ so $496,946,666,$  for $7$ so $777$ then $8$ so $888$ and finally $9$ so $999$ thus total numbers are $9+1+3+3+3+1+3+1+1+1=26$ so its $D$


If $B$ if a finite Boolean algebra, then it contains an atom Let $B=\{a_1,\dots,a_n,0,1\}$ be a finite Boolean algebra. I want to show that there exists an atom $x\in B$. So I want to show that there exists $x\in B$, such that for each $a\in B$ for which $a<x$, we have $a=0$.

I was thinking of maybe looking at $a=\bigwedge\{a_1,\dots,a_n\}$. If $a\neq 0$, then it seems to me that $a$ is an atom. So assume $a=0$. Is it possible to shrink the size of $\{a_1,\dots,a_n\}$ so that we get $a\neq 0$? And for the biggest set for which $a> 0$ we could argue that $a$ is an atom?

(Short remark: I do know that every finite Boolean algebra equals $\mathcal P(\operatorname{At}(B))$, but to me it seems that we use the fact that $\operatorname{At}(B)$ is nonempty, or at least, that seems to be assumed. So I want to prove this from scratch.)

Suppose there is no atom. Then there exists $i_1$ such that $a_{i_1}<a_1$, and $a_{i_2}<a_{i_1}$, etc. Since there are only finitely many $a_i$'s, $i_k=i_{k'}$ for some $k\ne k'$. Thus, $a_{i_k}<a_{i_k}$ by transitivity, which is impossible.

Assume that $x\in B$ is not an atom. Then there are $x_a,x_b\in B$ such that $x_a\vee x_b=x$, $x_a,x_b\neq x$, and $x_a,x_b\neq 0$. If $x_a$ is not an atom, then we can find $x_{aa},x_{ab}$ such that $x_{aa}\vee x_{ab}=x_a$, $x_{aa},x_{ab}\neq x_a$, and $x_{aa},x_{ab}\neq 0$. Continuing in this way we find $$x=x_1\vee x_2\vee ...\vee x_n$$ such that $x_i\neq0$ and, because $B$ is finite, if $x_i=x_a\vee x_b$ with $x_a,x_b\neq 0,x_i$, then $x_a,x_b\in\{x_1,...,x_n\}$. When this happens $n$ can be decreased maintaining the same property.

Take $n$ minimum with the property above. Each of the $x_i$ must be atoms, otherwise one can split them and reduce $n$ further.


Is it possible in a group of seven people for each to be friends with exactly 3 others? Is it possible in a group of seven people for each to be friends with exactly 3 others?

I know that the sum of degrees of vertices in a graph must be even.

Let $G(V,E)$ represent our graph. Because we know every person must be friends with excatly three other people, we now: $\forall v \in V$. deg$(v) = 3$. By the hand-shaking lemma:
\begin{align}
2|E| = \sum_{v \in V} deg(v) \Rightarrow 2|E| = \sum_{n=1}^7 3 = 21.
\end{align}
However $|E|$ is an integer so we come to a contradiction, since we get $|E| = \frac{21}{2}.$

So you consider the people as vertices and friendship as edges between them. then sum of all degrees is 21 , which is a contradiction.


Find E(Y). Conditional Expectations Let $X$ be an exponential random variable with $\lambda =5$ and $Y$ a uniformly distributed random variable on $(-3,X)$. Find $\mathbb E(Y)$.

My attempt:

$$\mathbb E(Y)= \mathbb E(\mathbb E(Y|X))$$

$$\mathbb E(Y|X) = \int^{x}_{-3} y \frac{1}{x+3} dy = \frac{x^2+9}{2(x+3)}$$

$$ \mathbb E(\mathbb E(Y|X))= \int^{\infty}_{0} \frac{x^2+9}{2(x+3)} 5 e^{-5x} \, dx$$

Hint: $$\int^{x}_{-3} y \frac{1}{x+3} dy=\frac{\frac12\cdot y^2}{x+3}\bigg|_{-3}^x=\frac12\cdot \frac{x^2-(-3)^2}{x+3}=\frac12\cdot \frac{x^2-9}{x+3}$$

At the numerator you can use the second binomial formula.

You made a mistake in your calculation of $\mathbb E[Y|X]$. The correct calculation is
$$
E[Y|X=x] = \frac{1}{x+3}\int_{-3}^x{y dy} = \frac{1}{x+1} \left(\frac{1}{2}y^2\right)\bigg|_{-3}^x = \frac{x^2-9}{2(x+3)} = \frac{x-3}{2}.
$$

We then have that
$$
\mathbb E[Y] = \int_0^{\infty}\frac{x-3}{2}\cdot 5e^{-5x}dx = \left(\frac{1}{10}e^{-5x}(14-5x)\right)\bigg|_0^{\infty} = -\frac{7}{5}.
$$


Is the affine cone of a flat projective scheme again flat? I'm trying to solve Hartshorne Chap.III Ex.9.5.(c):

The biggest problem is to show the existence of a closed subscheme $\tilde{X}\subset \mathbb{P}_{T}^{n+1}$ such that $\tilde{X}_{t} = \operatorname{C}(X_{t})$.

I guess the projective cone $\operatorname{C}(X) \subset \mathbb{P}_{T}^{n+1}$ of $X$ works.

Note that an inclusion $\operatorname{C}(X_{t}) \subset \operatorname{C}(X)_{t}$ always holds but an equality doesn't necessarily hold in a general case.

However, one can easily see that $\operatorname{C}(X_{\eta}) = \operatorname{C}(X)_{\eta}$, where $\eta$ is the generic point of $T$.

Moreover, it is clear that the Hilbert polynomial of $\operatorname{C}(X_{t})$ is independent of $t \in T$.

Hence it suffices to show that the Hilbert polynomial of $\operatorname{C}(X)_{t}$ is independent of $t$, i.e., $\operatorname{C}(X)$ is flat over $T$.

(Then from the inclusion $\operatorname{C}(X_{t}) \subset \operatorname{C}(X)_{t}$ and the agreement of their Hilbert polynomials we conclude that $\operatorname{C}(X_{t}) = \operatorname{C}(X)_{t}$.)

Note that $\operatorname{C}(X)$ is almost everywhere an $\mathbb{A}^{1}-$bundle over $X$, so almost everywhere flat.

Therefore we only have to check the flatness of the affine cone $\operatorname{C}_{\operatorname{aff}}(X)$ of $X$.

So we have reduced the original problem to the following:

  Let $A$ be a noetherian domain and write $T:= \operatorname{Spec}A$.

  Let $X \subset \mathbb{P}_{T}^{n}$ be a closed subscheme which is flat
  over $T$.

  Then, is the affine cone $\operatorname{Spec} A[x_0,\cdots x_n] /
\operatorname{I}(X)$ flat over $T$?

Thank you!

WARNING: The below is not completely correct and I am in the process of attempting to fix it.

$\renewcommand{\Proj}{\operatorname{Proj}} \renewcommand{\Spec}{\operatorname{Spec}}$ Actually, $C(X_t)\cong C(X)_t$, always. It suffices to work affine-locally on $T$, so suppose $T= \Spec R$ and write $X=\Proj(S_X)$ where $S_X=R[x_0,\cdots,x_n]/\Gamma_*(\mathcal{I}_X)$.
By definition, $C(X) = \Proj(S_X[x_{n+1}])$, $X_t=\Proj(S_X)\times_T \Spec k(t)$, and $C(X)_t = \Proj(S_X[x_{n+1}])\times_T \Spec k(t)$. Since Proj is compatible with base extensions along the degree-zero part of our algebra, we have $X_t\cong \Proj(S_X\otimes_R k(t))$, which is Proj of $$k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n]).$$
Thus $C(X_t)$ is Proj of $$k(t)[x_0,\cdots,x_{n+1}]/(Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])),$$ which can be written as $$\left(k(t)[x_0,\cdots,x_n]/Im(I_X\otimes_R k(t)\to k(t)[x_0,\cdots,x_n])\right)[x_{n+1}],$$ or $(S_X\otimes_R k(t))[x_{n+1}]$, so $C(X_t)\cong \Proj((S_X\otimes_R k(t))[x_{n+1}])$.
On the other hand, $C(X)_t\cong \Proj((S_X[x_{n+1}])\otimes_R k(t))$. As $(S_X\otimes_R k(t))[x_{n+1}]\cong (S_X[x_{n+1}])\otimes_R k(t)$, we see that $C(X_t)\cong C(X)_t$, resolving your difficulty.


Number of ways to place $k$ non-attacking rooks on an $m\times n$ chessboard I need to calculate the number of ways to place $k$ non-attacking rooks on an $m \times n$ table where $k \leq n$ and $k \leq m$.  ("Non-attacking" means that no two rooks may share a row or column.)  My attempt:
Calculate the number of ways to place $k$ rooks on a $k \times k$ board ($k!$), then multiply by the number of ways to select a $k \times k$ board from an $m \times n$ board. (This is the part I can't calculate, if it is correct at all.)
My question:
Is my approach good and if so, how to calculate the second part?

It is a reasonable approach. The columns can be chosen in $\binom{m}{k}$ ways  and for each way of selecting columns the rows can be chosen in $\binom{n}{k}$ ways.

I think you can choose $k$ squares out of $nm$ in $\pmatrix{ nm\\k}$ different ways, and for each one of these choices there are $k!$ different ways to set the rooks, so the result is $k!\pmatrix{ nm\\k}$.


Is $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ an irreducible polynomial over $\mathbb{Q}[X]$? Let $a, b \in \mathbb{Q}$, with $a\neq b$ and $ab\neq 0$, and $n$ a positive integer.

  Is the polynomial $\bigl(X(X-a)(X-b)\bigr)^{2^n} +1$ irreducible over $\mathbb{Q}[X]$?

I know that $\bigl(X(X-a)\bigr)^{2^n} +1$ is irreducible over $\mathbb{Q}[X]$, but I have a hard time generalizing my proof with three factors.

PS: This is not homework (and may even be open).

Call $f$ your polynomial. Then $f$ is not irreducible if and only if there exists a root $x\in\mathbb{Q}$ of $f$. If a such root exists then, for $n\geq 1$ we have  $$0\leq(x(x-a)(x-b))^{2^n}=-1.$$

I leave the case $n=0$ to you.


Finding maximum value of $|f(z)|$ using Maximum modulus theorem? In the question asked here→ Maximum Modulus Exercise

I want to know, if we just want to find maximum value of $|f(z)|$, why 'Marlu' Sir in his answer (here https://math.stackexchange.com/a/325832/168676) has done calculations to show that there are no other Maxima?

My attempt: as $f(z)=z^2-3z+2$ is analytic inside and on $|z|=1$ hence by maximum modulus theorem, maximum value of $|f(z)|$ occurs on boundary!
and by traingle inequality,

$|f(z)|=|z^2-3z+2|≤|z^2|+3|z|+2$

$$≤6$$ (since on the boundary, $|z|=1$)

So from here, we know, maximum value of $|f(z)|$ cannot exceed $6$ and the fact that,  at point  $z=-1$ which is on boundary, $f(-1)=6$ confirms that, maximum value of $|f(z)|$ is $6$ is am I correct? Please help me...

No, 6 is just an upper bound but not the maximum modulus, check out the zill-complex analysis book.

Check out my annotated image if you still have issue.


MENSA IQ Test and rules of maths In a Mensa calendar, A daily challenge - your daily brain workout. I got this and put a challenge up at work.

The Challenge starts with..

Assume you are using a basic calculator and press the numbers in the order shown,
replacing each question mark ...continues... What is the highest number you can
possibly score?

Basically, only using $+,-, * ,\div$, once in place of a question mark.

  $5 ? 4 ? 7 ? 3 ? 2 =$

We all worked out the operators to be

  $5 + 4 $ x $ 7 - 3/2 =$

Except that I calculated the answer to be $31.5$ and the others argued $30$. THe answer sheet from MENSA says the calculated total is 30.

Nobody actually understood the first part about using a basic calculator. I initially thought the challenge was all about the rules of maths.
And when I asked why nobody else applied the rules of maths, they all forgot about it, not because the challenge said to use some "basic calculation"

I emailed MENSA and queried them about the challenge and they replied,

Thank you for your email.

On a basic calculator it will be:
5 + 4 = 9
9 x 7 = 63
63 – 3 = 60
60 ÷ 2 = 30

Kind regards,
Puzzle Team

My Reply,

 Thank you for your reply.

 Could you please define what a basic calculator is? I tried 4 pocket,
 £1 calculators, and all gave me 31.5.

And finally their definition.

I guess what the question really means, whether you do the sum manually or on a
calculator,  is don’t change the order of the numbers. The Casio calculators we have in
the office allow you to do the sum as it appears:

5 + 4 = 9
9 x 7 = 63
63 – 3 = 60
60 ÷ 2 = 30

Kind regards,
Puzzle Team

So they guess the challenge meant to do it that way. Why not just say ignore rules of maths. What is the point of this anyway?

My original question, on Maths Stack (this one) was why MENSA used 30 instead of 31.5. And initially I did not understand that using a basic calculator meant calculating left to right by pressing equals after each operation.

So what is going on here? If they wanted us the ignore rules of math they should of said taht. Because my basic calculator gives me 31.5 and not 30.0 (I dont have a special. Casio MENSA calculator though)

Windows standard calculator gives me 30. Why? None of my pocket, office, el cheapo calculators do this.

Google, or Windows Scientific give me 31.5 - As do ally my elelctornic calculators.

In response to edit of initial post, then answer is clearly $30$. Basic calculators are assumed to evaluate in order from left to right.

Original post I responded to

In a Mensa calander, IQ dialy challenge I got this and put a challenge up at work.

Using +,-,time and divide only once. Use the math operator only once to get the highest answer.

  5 ? 4 ? 7 ? 3 ? 2 =

We all worked out

  5 + 4 x 7 - 3 / 2 = 30

Except that my result answer was 31.5 and not 30, like in the answers of the MENSA calendar.

Why was I the only one that applied the rules of maths on this? ANd when I asked why nobody else applied the rule of maths, I got the weirdest looks. Nobody knew about multiplication before division, subtraction before adding?  I thought that was why the question was marked as the most difficutl to test if you knew this.

Response to original post

Sadly, many people forget the basic rules of arithmetic as they (a) don't view them as affecting their lives, (b) didn't like maths, and/or (c) know technology can handle the problem for them. The issue with the last point is that different technologies handle things differently. The Google calculator (much like most graphing calculators) will handle order of operations for you correctly. The standard Windows calculator appears to be operating like an old 4 function calculator which evaluates after every operation is completed as opposed to correct order of operations. Though this can also happen when users hit enter after every operation is finished as opposed to when the whole expression is finished. (Don't have access to a Windows calculator right now so can't tell which is the reason for the wrong answer.)

the answer is 31.5. you know it but you messed up the math.
with subtraction and addition it shouldnt matter which is carried out first; they can both be considered to be identical functions (addition of a positive and a negative no or vice versa)

the answer is therefore 28+5-1.5; whichever way you look at it.


Why do complex roots come in pairs? First time using this website so excuse me for using the body like this.

Complex roots of polynomials with real coefficients come in conjugate pairs because otherwise they wouldn't have real coefficients! Suppose a polynomial $p(z)=a_{0}+a_{1}z+ \ldots + a_{n}z^{n}$ has a complex root $\alpha$. Then $$\overline{p(\alpha)}=\overline{a_{0}+a_{1}\alpha+ \ldots + a_{n}\alpha^{n}}=\bar{a_{0}}+\bar{a_{1}}\bar{\alpha}+ \ldots + \bar{a_{n}}\bar{\alpha}^{n}\\=a_{0}+a_{1}\bar{\alpha}+ \ldots + a_{n}\bar{\alpha}^{n}=\bar{0}=0$$
So $\bar{\alpha}$ is also a root.

Because, in the complex plane, a root is also a rotation. In general, think of a complex number in polar representation:
$$z=r\cdot e^{i\phi}$$ Then taking the square root gives you:
$$\sqrt{z} = z^{-1/2} = \sqrt r\cdot e^{i\phi/2}$$
But - the angle $(\phi/2+\pi)$ also satisfies the demands for being half the angle $\phi$, because:
$$2\cdot(\phi/2+\pi) = \phi+2\pi = \phi$$

It all stems from the fact that rotations live in on a circle, which is a cyclic space (obviously). When you do arithmetic on angles you must remember that a $full\ rotation = no\ ratation$, and from that it follows that when solving simple equations like $2x=y$ you end up with more than a single solution.


Five tyres over $40000$ kilometres The five tyres of a car (four road tyres and one spare) were used equally in a journey of $40,000~\text{km}$. The number of kilometres of use of each tyre was

a. $40000$
b. $10000$
c. $32000$
d. $8000$

This is what I tried. I just want to know is my approach is correct or not. Or is there any better method to solve this type of question?

Total kilometers travelled by $4$ tyres $= 40000 \cdot 4 = 160,000$.
This has to be shared by $5$ tyres.
So each tyre capacity $= \frac{160,000}{5} = 32,000$.
After we travel $32,000~\text{km}$, we are left with $4$ worn tyres and one new tyre. But if the tyres are rotated properly after each $8000~\text{km}$,

You have to consider that every tire, in order to have equal use, had to at least once have been in the spare tire compartment. This means that there were 5 journeys in between the tire changes. Each journey was 8000km long (40,000/5). So, because each tire was on the road per journey segment except for 1, each tire was used a total of 32,000 times (8000x4)

You will switch tires $4$ times so every tire has a chance to run. So you will be making $5$ lags, and so each lag will be $8000$. So each of the $5$ tires rests in a lag of $8000$. Or, each tire runs a total of $32000$


Selecting 3 non-consecutive days of the week We need to find how many groups of $3$ days of the week there can be, given that no two days should be consecutive. The answer should be $7$, but I do not know how to get to the answer.
Thanks in advance.

Use 1010100 to represent that you 'pick' the first, third, and fifth day of the week. What ways are there to pick three non-consecutive days?

$1010100$

$1010010$

$1001010$

$0101001$

$0100101$

$0010101$

$0101010$

And that's it.

It turns out to be equal to the number of days in the week since as you can see there is one time where you have two days that you don't pick (the two 0's), and that 00 bit can start at any of the 7 days, fixing the rest of the sequence.

The general formula for the number of circular arrangements of $k$ elements over $n$ spots with no consecutive elements is
$$\binom{n-k}{k}+\binom{n-k-1}{k-1}$$
See Selection in Circular Table for a proof.


Why are sine and cosine vectors? I understand that sine and cosine functions are vectors, and I understand that what defines a vector is how it transforms under change of basis. Am I correct in understanding that means that for some angle $\alpha$ in the $(x,y)$ coordinate system and some $\alpha'$ in the rotated coordinate system $(x',y')$ $$\sin(\alpha')=\frac{\mathrm{d}\alpha}{\mathrm{d}\alpha'}\sin(\alpha)?$$

I've tried working with this definition for the transformation, which I've been led to believe is correct, but I can't seem to show that it holds. Is this even a correct starting point? If not, what is?

The word "vector" means slightly different things in various contexts. In linear algebra, a vector space is a structure satisfying a certain list of axioms, and a "vector" is just an element of whatever vector space you're talking about.

In that sense $\sin$  and $\cos$ are vectors; they're elements of for example $C(\Bbb R)$, the space of all continuous functions on the line. But this use of the word "vector" has more or less nothing to do with the thing you understand about how a "vector" is defined by how it changes under coordinate transformations.

That second sense of the word comes up in differential geometry, where for example it distinguishes vectors from "co-vectors".

I may be oversimplifying, but I believe that the word is being used in two different ways in "$\sin$ is a vector" and "what defines a vector is how it transforms under change of basis".


Prove/Disprove: if $x^2 = a^2$, then $x = a$ From Prof. Charles Pinter's "A Book of Abstract Algebra"'s Chapter 4 exercises:

  For each of the following rules, either prove that it is true in every group $G$, or give a counter-example.

$$ \text{if } x^{2}=a^{2}, \text{then } x=a$$

I believe that this is true by:

$$xx = aa$$

by cancellation,

$$x=a$$

Is that right? Also, when the word "prove" is used, does that mean to use theorems to prove?

What is it that you are canceling there? It looks to me like you used that $x=a$ to show that $x=a$, which is not a good plan.

Think: is this true even in the real numbers?

Cancellation is not possible, you must take the root of both sides. Square roots have the property $q = \pm\sqrt{q^2}$, therefore $a = \pm x$, which is not equivalent to $a = x$. The original statement has thus been disproved.


Compute $m_Z (t)$. Verify that $m'_Z (0)$ = $E(Z)$ and $m''_Z(0) = E(Z^2)$ Let $Z$ be a discrete random variable with $P(Z = z)$ = $1/2^z$ for $z = 1, 2, 3,...$

(b) Compute $m_Z (t)$. Verify that $m'_Z (0)$ = $E(Z)$ and $m''_Z(0) = E(Z^2)$

$E(Z) = \sum_{z=1}^{\infty} z P(Z = z) =\sum_{z=1}^{\infty} \frac{z}{2^z}$ Not sure how to proceed. $z^2$ for $E(Z^2)$

$m_z(t) = E[e^{tz}] = \sum_{z=1}^{\infty} \frac{e^{tz}}{2^z} = \sum_{z=1}^{\infty} (2e)^{tz - z} = \sum_{z=1}^{\infty} (2e)^{z(t-1)} = \frac{(2e)^{t-1}}{1-2e}$

Not sure if above is right.

I know afterward, I would just take the derivative of $m_z(t)$ once and twice and set $t = 0$ to verify but having trouble simplifying.

It is not true that
$$\frac{\exp(tz)}{2^z}=(2\text{e})^{tz-z}\,.$$
You can easily check that your version of $m_Z$ does not satisfy $m_Z(0)=1$.  The correct moment generating function should be $$m_Z(t)=\sum_{z=1}^\infty\,\frac{\exp({tz})}{2^z}=\sum_{z=1}^\infty\,\left(\frac{\exp(t)}{2}\right)^z=\frac{\left(\frac{\exp(t)}{2}\right)}{1-\left(\frac{\exp(t)}{2}\right)}=\frac{\exp(t)}{2-\exp(t)}\,.$$

It should be easy to find $m_Z'$ and $m_Z''$ now.  As for the calculations of $\mathbb{E}[Z]$ and $\mathbb{E}[Z^2]$, this link should be very useful: Proof of the equality $\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6$.

The mistake is $$\frac{e^{tz}}{2^z} = (2e)^{tz - z}$$

Actually

$$\frac{e^{tz}}{2^z} = (\frac {e^t} 2)^z \tag{*}$$

Anyhoo, right off the bat, you might recall the form $\frac 1 {2^z}$. It's a term of a geometric sequence. Do you know of the geometric distribution?

Anyhoo, let's compute moments with mgf: $\forall |k| < 1$, we have

$$\frac 1 {1-k} = \sum_{z=0}^{\infty} k^z$$

$$\implies \frac{d}{dk} \frac 1 {1-k} = \frac{d}{dk} \sum_{z=0}^{\infty} k^z$$

$$\implies \frac{d}{dk} \frac 1 {1-k} = \sum_{z=0}^{\infty} \frac{d}{dk} k^z$$

$$\implies \frac 1 {(1-k)^2} = \sum_{z=0}^{\infty} zk^{z-1}$$

$$\implies \frac 1 {(1-k)^2} = \sum_{z=1}^{\infty} zk^{z-1}$$

$$\implies \frac k {(1-k)^2} = \sum_{z=1}^{\infty} zk^{z}$$

$$\implies \frac 1 2 {(1-\frac 1 2)^2} = \sum_{z=1}^{\infty} z(\frac 1 2)^{z} = E[Z]$$

What's the moral lesson? We recognise $$\sum_{z=1}^{\infty} z(\frac 1 2)^{z} = E[Z]$$ as having to do with the derivative or integral of a geometric series. Similarly for the second moment we have, $$\sum_{z=1}^{\infty} z^2(\frac 1 2)^{z} = E[Z^2],$$ where we compute as follows:

$$\frac{d}{dk} \frac k {(1-k)^2} = \frac{d}{dk} \sum_{z=1}^{\infty} zk^{z}$$

$$\implies \frac{d}{dk} \frac k {(1-k)^2} = \sum_{z=1}^{\infty} \frac{d}{dk} zk^{z}$$

$$ = \sum_{z=1}^{\infty} \frac{d}{dk} zk^{z} = \sum_{z=1}^{\infty} z^2k^{z-1}$$

$$\therefore, ([k][\frac{d}{dk} \frac k {(1-k)^2}])|_{k=\frac 1 2} = \sum_{z=1}^{\infty} z^2k^{z}|_{k=\frac 1 2} = E[Z^2]$$

Now, how do we do this the mgf way? Let's go back to $(*)$

$$M_Z(t) = \sum_{z=1}^{\infty} \frac{e^{tz}}{2^z} = \sum_{z=1}^{\infty} (\frac {e^t} 2)^z = \frac{\frac {e^t} 2}{1 - \frac {e^t} 2}$$

This agrees with the mgf for geometric on Wiki with $p=\frac 1 2$.

So just differentiate and plug in:

$$M_Z'(t) = \frac{d}{dt} \frac{\frac {e^t} 2}{1 - \frac {e^t} 2}$$

$$\therefore, M_Z'(0) = [\frac{d}{dt} \frac{\frac {e^t} 2}{1 - \frac {e^t} 2}]|_{t=0}$$

$$M_Z''(t) = \frac{d}{dt} \frac{d}{dt} \frac{\frac {e^t} 2}{1 - \frac {e^t} 2}$$

$$\therefore, M_Z''(0) = [\frac{d}{dt} \frac{d}{dt} \frac{\frac {e^t} 2}{1 - \frac {e^t} 2}]|_{t=0}$$


Order of finite fields is $p^n$ Let $F$ be a finite field.
How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?

Let $p$ be the characteristic of a finite field $F$.${}^{\text{Note 1}}$ Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $q\neq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $x\in F$ whose order in $(F,+)$ is $q$.
Then $q\cdot x=0$. But we also have $p\cdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.
Thus $(ap+bq)\cdot x=x$. But $(ap+bq)\cdot x=a\cdot(p\cdot x)+b\cdot(q\cdot x)=0$, giving $x=0$, which is not possible since $x$ has order at least $2$ in $(F,+)$.
So there is no prime other than $p$ which divides $|F|$.

Note 1: Every finite field has a characteristic $p\in\mathbb N$ since, by the pigeonhole principle, there must exist distinct $n_1< n_2$ both in the set $\{1, 2, \dots, \lvert F\rvert +1\}$ such that $$\underbrace{1+1+\dots+1}_{n_1}=\underbrace{1+1+\dots+1}_{n_2},$$ so that $\underbrace{1+1+\dots+1}_{n_2-n_1}=0$. In fact, this argument also implies $p\le n$.

A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:

Let $F$ be a finite field (and thus has characteristic $p$, a prime).

Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.
A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.

The first claim is immediate, by the distributive property of the field. Let $x \in F, \ x \neq 0_F$. We have

\begin{align}
p \cdot x &= p \cdot (1_{F} x) = (p \cdot 1_{F}) \ x
\\
& = 0
\end{align}

This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.

The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).


Placing 5 people on chairs so that 2 people never sit together We have a family composed by 2 parents and 3 children that have to seat on 5 chairs (in line), two of the children don't want to stay together, how many are the ways that they can sit?
I was thinking about letting one of the 2 children sit first, now the second one can sit only on 3 chairs, the remaining parents and child now can sit on the other 3 chairs
So I came to that expression: 3*3! = 18 but I saw that this wasn't the correct answer...

If the chairs are in a line, then the calculation must take into account  where the first child sits.  There are two ways for him to take a seat on the end, then three ways for the other child to take a seat.  There are three ways for the first child to take a seat in the middle of the row, and then there are only two ways for the second child to take a seat.  So there are in all $$2\cdot3+3\cdot2=12$$ ways for the two of them to be seated.  There are then $3!$ ways to seat the rest of the family, so $72$ ways in all.

I think your idea is correct, you just miss some calculation.
I will assume here that the 5 chair are put in circle
So yes, let first choose the chair of one of the difficult child (5 choices, 4 chairs left)
Then the second difficult choose from the 2 chairs which are not together (1 chair as 2 neighbour, and one chair is already taken) (so 2 choices, and 3 chair left)
At least let the 3 other sit (3! choices)
So tu sums up, we have $5\times2\times 3! =60$ possibilities
Is this your expected answer, or did i miss something in the initial statement ?


Given a list of points on a rectilinear path, identify the corners You are provided list of co-ordinates in form of pair as shown below:
$$li=[(576, 64), (576, 192), (448, 192), (320, 192), (320, 320), (192, 320), (192, 448), (192, 576), (320, 576), (448, 576), (576, 576), (704, 576), (832, 576), (960, 576)]$$
After modifying it according to goal.
Goal is to first plot all points in co-ordinate plane then join them now carefully we need to find out all corner points.
$$li=[(576, 64), (576, 192),(320, 192), (320, 320), (192, 320), (192, 576), (960, 576)]$$

Please refer the image I need the points which are encircled by a blue rectangle, where as red circled points are points that are provided in the problem.

I tried brute force approach it worked but yes please I need something better than that.

Converting a comment (with some adjustments) to an answer, as OP is satisfied with its validity ...

For each index $i$ (starting with the second, and ending with the next-to-last), compare $p_{i-1}$ to $p_{i+1}$: if (and only if) their $x$-coordinates differ and their $y$-coordinates differ, then (and only then) $p_i$ is a corner.
This algorithm ignores the first and last points, but those are already "corners" by declaration.
As OP has noted, the algorithm has complexity $O(n)$, as it simply steps through the list once.

Something like
while i < N
    { Pi is a corner }
    while i < N and X[i] == X[i+1]
        i+= 1

    { Pi is a corner }
    while i< N and Y[i] == Y[i+1]
        i+= 1

Details need to be worked out.


What are natural boundary conditions in the calculus of variations? What do people mean, when they speak of natural boundary conditions in the calculus of variations?

How do natural boundary conditions relate to the Euler-Lagrange equations?

An example would be fantastic!

Basically two types of boundary conditions are used: Essential or geometric boundary conditions which are imposed on the primary variable like displacements, and Natural or force boundary conditions which are imposed on the secondary variable like forces and tractions. Essential boundary conditions are imposed explicitly on the solution but natural boundary conditions are automatically satisfied after solution of the problem.

Natural Boundary Conditions of the Simplest Kind:
Let $~J : C^2[x_0, x_1] → \mathbb R~$ be a functional of the form of $$J(y) =  \int^{x_1}_{x_0}f(x, y, y') dx$$ and assume that no boundary conditions have been imposed on $~y~$, then $~J~$ have an extremum $~y~$ if the following necessary conditions are satisfied:
$(i)~~$ The ordinate of the extremal satisfies the Euler-Lagrange Equation $$f_y - \frac{d}{dx} f_y' = 0$$
$(ii)~~$ At $~x = x_0~$ $$\left|\frac{\partial f}{\partial y'}\right|_{~x_0}=0$$
$(iii)~~$  At $~x = x_1~$ $$\left|\frac{\partial f}{\partial y'}\right|_{~x_1}=0$$

From the perspective of computational math: Solving a differential equation using e.g. finite elements you usually build a linear equation $Ax = b$ where $x$ is a discrete approximation of the solution. You set up $A$ and $b$ depending on different aspect of your problem, like how your domain looks like and also what kind of boundary conditions there are.

We speak of natural boundary conditions, if during assembly of the problem, this step of adding the information of boundary conditions is effectivly skipped. It's those boundary conditions that "naturally" arise when we don't pay attention to them.

See also here: Neumann = 0 boundary condition drops the term for boundary conditions completely, so we don't have to worry about them.


Find the value of $R$ such that $\sum_{n=0}^{\infty} \frac{x^n}{n! \cdot (\ln 3)^n} = 3^x, \forall x \in (-R,R)$ I am not quite sure how to finish this exercise:

  Find the value of $R$ such that $$\sum_{n=0}^{\infty} \frac{x^n}{n! \cdot (\ln
 3)^n} = 3^x, \forall x \in (-R,R)$$

I am honestly lost after I verify that the power series converges:

$$\begin{align*}
a_n &= \frac{x^n}{n! \cdot (\ln 3)^n}\\
a_{n+1} &= \frac{x^{n+1}}{(n+1)! \cdot (\ln 3)^{n+1}}
\end{align*}
$$
The ratio between the two and the limit are shown below:

$$\begin{align*}
\frac{a_{n+1}}{a_n} &= \frac{x}{(n+1) \cdot \ln 3}\\
\\
\lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert &= \frac{\vert x \vert}{\ln 3} \cdot \lim_{n \to \infty} \frac{1}{n+1} = 0
\end{align*}
$$

So as I mentioned, I am not sure what to do next. Any guidance is highly appreciated.

Thank you.

Hint: It's possible the problem is misstated as @MathLover suggested in a comment. If it's not misstated, recall that $\sum_{n=0}^{\infty}\frac{u^n}{n!}= e^u$ and let $u=x/\ln 3.$

The $\ln 3$ does not matter.
The series converges for
all $x$.

As you have shown,
$|a_{n+1}/a_n| \to 0$.
The $\ln 3$ does
not affect this,
so it still holds
with any constant.
This means that the series converges.


Is a strictly decreasing, invertible function necessarily continuous? We know that the inverse exists (therefore injective and surjective), and it is strictly decreasing. I'm not sure if continuity follows.

I don't think it necessarily follows. For example the split function:

$f: [-3;3] \rightarrow [-3;-1]\cup[0;4]$,

$f(x)= \mathrm{if}(x<1, -x+1, -x)$.

It is discontinuous at $x=1$, but it is still invertible and strictly decreasing.


$\infty\pm\infty$ on a Riemann sphere Here I use "$\infty$" to represent the infinite point on a Riemann sphere. I read that

(1) $z\cdot\infty = \infty$ for any complex non-zero $z$.
(2) $\infty+\infty=\infty$.
(3) $\infty-\infty$ is indeterminate.

(1) makes sense to me. I would expect that (2) and (3) are indeterminate. Although if  (2) is Inf, I expect that would mean that (3) is also Inf due to (1) and (2). If (1), (2), and (3) are correct as written above, why does it work that way?

----EDIT----

I take back (3) above, since I can't find a source that says that.

Howecer, (1) and (2) can be found in Mathematical Analysis, 2nd edition, by Tom M. Apostol. He includes it in his section on the extended complex plane. One response to my initial question and Wikipedia said $\infty+\infty$ is undefined. Does that mean not all mathematicians agree on this?

To say that $\infty-\infty$ is "indeterminate" means that if for every point $c$ in the Riemann sphere, there exist functions $f$ and $g$ and a point $a$ such that
\begin{align}
\lim_{z\to a} f(z) & =\infty, \\[6pt]
\lim_{z\to a} g(z) & = \infty, \\[6pt]
\lim_{z\to a} f(z)-g(z) & = c.
\end{align}
You can choose point $a$ can be chosen to be anything you like and make trivial modifications in $f$ and $g$ so that this all works for that value of $a$.

When working with real numbers rather than on the Riemann sphere, this does not work for $\infty+\infty$ nor for $c\cdot\infty$.  If $f(z)\to\infty$ and $g(z)\to\infty$ then $f(z)+g(z)\to\infty$.  You cannot choose $f$ and $g$ both approaching $\infty$ in such a way that $f(z)+g(z)$ approaches $19$, the way you can with $f(z)-g(z)$.  However, with the Riemann sphere, as some have noted in comments below, similar things happen.

This is not a mathematical issue, it's a memory trick or something when you want to find limits for instance.


what is the value of $\int \sin(x)\cos(x)dx$? $\frac{\sin^2(x)}{2}$ or $\frac{-\cos^2(x)}{2}$ or $\frac{-\cos(2x)}{4}$ $\int \sin(x)\cos(x)dx = \frac{\sin^2(x)}{2}$ because

$$\frac{d}{dx}\frac{\sin^2(x)}{2}=\sin(x)\frac{\sin(x)}{dx}=\sin(x)\cos(x)$$

but also

$$\frac{d}{dx}\frac{-\cos^2(x)}{2}=-\cos(x)\frac{\cos(x)}{dx}=\sin(x)\cos(x)$$

Moreover

$$\int \sin(x)\cos(x)dx = \int \frac{\sin(2x)}{2}dx = \frac{-\cos(2x)}{4} = \frac14 -  \frac{\cos^2(x)}{2} = \frac{\sin^2(x)}{2} - \frac14 $$

It's true that $\frac{\sin^2(x)}{2}$ and $\frac{-\cos^2(x)}{2}$ are not equal so what is that problem I made here.

You are forgetting the constant of integration.  The constant of integration will change depending on whether you chose to integrate the sine or the cosine.  The bottom line is that the integrals are the same.

$\displaystyle \int \sin x\cos xdx = \displaystyle \int \dfrac{\sin 2x}{2}dx = -\dfrac{\cos 2x}{4}+C$


Show that a polynomial p can't have both roots 1 and i I have to show that a polynomial $p\in P_3(\mathbb{R})$ where $p\neq0$, can't have both roots, $1$ and $i$, where $i^2=-1$. The polynomial has degree 2 or less.

$p(\alpha)=c_0+c_1\alpha+c_2\alpha^2$

I know that, when we want to find the roots of a function, we have to solve the function for $x$ so that $f(x)=0$

I don't know how to solve this kind of question, when $p$ cant be equal to $0$ and I have to show that the polynomial $p\in P_3(\mathbb{R})$ can't have both roots 1 and $i$.

Thanks in advance

If we have a polynomial with real coefficients, with $z$ as a root, one can check that $\overline{z}$ must also be a root of the polynomial. Thus, if $i$ is a root of a polynomial $p$, then so is $-i$. If we also require that $1$ be a root of $p$, then $p$ has at least $3$ roots. But how many roots can $p$ have, given that it's of degree at most $2$?

I don't understand the fuss... Just write $p(x)=(x-1)(x-i)$ and expand it to get $p(x)=x^2-(1+i)x+i$, which clearly does not belong to $P_{3}(\mathbb{R})$. Or is it?


How to prove that geometric distributions converge to an exponential distribution? How to prove that geometric distributions converge to an exponential distribution?

To solve this, I am trying to define an indexing $n$/$m$ and to send $m$ to infinity, but I get zero, not some relevant distribution. What is the technique or approach one must use here?

Recall pmf of the geometric distribution: $\mathbb{P}(X = k) = p (1-p)^k$ for $k \geq 0$.

The geometric distribution has the interpretation of the number of failures in a sequence of Bernoulli trials until the first success.

Consider a regime when the probability of success is very small, such that $n p = \lambda$, and consider $x = \frac{k}{n}$. Then, in the large $n$ limit:
$$
  1 = \sum_{k=0}^\infty \mathbb{P}(X = k) = \sum_{k=0}^\infty \lambda \left(\left(1 -  \frac{\lambda}{n} \right)^{n \cdot k/n} \frac{1}{n} \right) \stackrel{n \to \infty}\rightarrow \int_0^\infty \lambda \mathrm{e}^{-\lambda x} \mathrm{d} x
$$

Alternatively, you could look at the moment generating function for the geometric distribution:
$$
    \mathcal{M}(p, t) = \frac{p}{1-\mathrm{e}^t (1-p)}
$$
To recover the mgf of the exponential distribution consider the limit:
$$
   \lim_{n \to \infty} \mathcal{M}\left( \frac{\lambda}{n}, \frac{t}{n} \right) =
   \lim_{n \to \infty} \frac{\lambda}{n - \mathrm{e}^{t/n}\left(n - \lambda\right)} =
   \lim_{n \to \infty} \frac{\lambda}{n \left(1 - \mathrm{e}^{t/n}\right) + \lambda \mathrm{e}^{t/n} } = \frac{\lambda}{\lambda-t}
$$
which is the mgf of the exponential distribution with rate $\lambda$.

I think we can just expand 1-p in Taylor series for small p we get the exponential function.:)
1-p=exp(-p) for very small p (around p=0)such that np=constant. so we find:

p(x)=(1-p)^x.p is becoming p exp(-px) for a continuous random variable x, this in part has the continuous exponential distribution function:)


Does $\lim \frac{xy}{x+y}$ exist at $(0,0)$? Given the function $f(x,y) = \frac{xy}{x+y}$, after my analysis I concluded that the limit at $(0,0)$ does not exists.

In short, if we approach to $(0,0)$ through the parabola $y = -x^2 -x$ and $y = x^2 - x$ we find that $f(x,y)$ approaches to $1$ and $-1$ respectively. Therefore the limit does not exists.

I think my rationale is right. What do you think?

Alternatively, is there another approach for this problem?

I'll explain here how to approach limits of functions in two variables, with the example the OP proposed in mind. If the limit
$$\lim_{(x,y)\to (0,0)} \frac{xy}{x+y}$$
exists and equals $L$, then it also follows that if $\{(x_n,y_n)\}$ is a sequence of points with limit $(0,0)$, then
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=L.$$
Now we can choose a number of easy sequences $\{(x_n,y_n)\}$ with limit $(0,0)$, and calculate the limit. For instance, we can pick points in a line $y=\lambda x$, with slope $\lambda$, i.e., $(x_n,y_n) = (\frac{1}{n}, \frac{\lambda}{n})$. In this case:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{\lambda}{n^2}}{\frac{1}{n}+\frac{\lambda}{n}}=\lim_{n\to\infty} \frac{\lambda}{(1+\lambda)n}$$
and the limit is $0$ as long as $\lambda\neq -1$. Hence, if the limit exists, it must be $0$. But the problem with $\lambda=-1$ tells us that there may be a problem if we approach $(0,0)$ with a path that ends tangent to $y=-x$ (notice that the function is not defined at points with $y=-x$).
Thus, next we look at a sequence following a path on a curve with tangent line $y=-x$ at $(0,0)$. Examples of such curves include $y=x^2-x$, $y=-x^2-x$ or $y=e^{-x}-1$. Thus, we may consider sequences $(x_n,y_n)$ given by:
$$\left(\frac{1}{n},\frac{1}{n^2}-\frac{1}{n}\right),\quad \text{or} \quad \left(\frac{1}{n},-\frac{1}{n^2}-\frac{1}{n}\right), \quad \text{or} \quad \left(\frac{1}{n},e^{-1/n}-1\right).$$
For the first sequence we obtain:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1-n}{n}= -1.$$
But the limit was supposed to be $L=0$. Hence the limit cannot exist. Similarly, if we try the other two sequences listed above:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{\frac{1}{n}-\frac{1}{n^2}-\frac{1}{n}}=\lim_{n\to\infty} \frac{-\frac{1}{n^3}-\frac{1}{n^2}}{-\frac{1}{n^2}}=\lim_{n\to\infty} \frac{1+n}{n}= 1,$$
and
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n}(e^{-1/n}-1)}{\frac{1}{n}+e^{-1/n}-1}=\lim_{n\to\infty} \frac{e^{-1/n}-1}{1+ne^{-1/n}-n}=0.$$
These results are inconsistent, and therefore the limit cannot exist. Even more dramatic: let $\{x_n,y_n\}$ be a sequence following the curve $y=x^3-x$ towards the origin, for instance put $(x_n,y_n)=(\frac{1}{n},\frac{1}{n^3}-\frac{1}{n})$. Then:
$$\lim_{n\to\infty} \frac{x_ny_n}{x_n+y_n}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n}+\frac{1}{n^3}-\frac{1}{n}}=\lim_{n\to\infty} \frac{\frac{1}{n^4}-\frac{1}{n^2}}{\frac{1}{n^3}}=\lim_{n\to\infty} \frac{1-n^2}{n}= -\infty.$$

maybe I miss something but isn't it:

for $x=x,y=\dfrac{-x}{x+1}$

$\dfrac{xy}{x+y}=\dfrac{\dfrac{-x^2}{1+x}}{x-\dfrac{x}{1+x}}=\dfrac{\dfrac{-x^2}{x+1}}{\dfrac{x^2}{x+1}}=-1\to0$ when $(x,y)\to(0,0)$

and for $x\not=0,y=0$

$\dfrac{xy}{x+y}=0\to0$


Show that matrix $A$ is NOT diagonalizable. Let $A$ be a square matrix $A^2=0$ and $A\neq0$ and show that it is not diagonalizable.

I decided to use the sample matrix of $$A = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$$ which satisfies the conditions above.

So my question is: how would I prove this is not diagonalizable. The matrix leads to a eigenvalue of $\lambda=0$ with an algebraic multiplicity of $2$.

I know that if the algebraic multiplicity and geometric multiplicity are equal, then it is diagonalizable.

But I am kind of stuck from here since when I use $\det(A-0I)=0$, it just leads to get $x_2=0$ but then also that $x_2=t$ so I don't really know what to do. Any help would be appreciated.

It isn’t enough to prove that your particular sample matrix isn’t diagonalizable: you must show that every non-zero square matrix $A$ such that $A^2=0$ is non-diagonalizable.

HINT: Suppose that $A^2=0$ and $A$ is diagonalizable. Then there are an invertible matrix $P$ and a diagonal matrix $D$ such that $D=P^{-1}AP$.

What is $D^2$?
What does this tell you about $A$?
How does this prove the desired result?

You have $det(A^2)=0$ then $det(A)=0$. This implies A has a zero eigenvalues which gives you that A is not diagonalizable


Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $. Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2 $.

How do I prove it? I don't even intuitively get this... after so long time of trying to prove it.

For the intuition: take the primes $p_1=5$ and $p_2=11$. Then $p_1\cdot p_2=55$ and indeed the only divisors (among the natural numbers) of the product are $1,5,11$, and $55$. Now take not both numbers prime, say $p_1=24$ and $p_2=11$. Then $p_1\cdot p_2=264$ and the divisors of the product this time certainly include $1,24,11$, and $264$, but also $2,3,6,12$.

For the proof: suppose $p_1,p_2$ are prime and that $n\mid p_1p_2$, where $n\in \mathbb N$. If $n\ne 1$ then it has a prime divisor. Let $q$ be an arbitrary prime divisor of $q$. Then in particular, $q$ divides $p_1p_2$. Now, according to Euclid's Lemma, if a prime divides a product, then it divides one of the factors. Thus, either $q\mid p_1$ or $q\mid p_2$. Assume, without loss of generality, that $q\mid p_2$. But since $p_2$ is prime, it follows that $q$ is either $1$ or $q=p$. Since $q$ is prime, the possibility of $q=1$ is (explicitly in the definition of prime number) ruled out. We conclude that $q=p$.

Now consider $m=n/q=n/p_1$. It follows that $m\mid p_2$. Thus, either $m=1$ or $m=p$. In the case $m=1$, it follows that $n=p_1$. If $m=p_2$ it follows that $n=p_1p_2$.

To conclude, we just showed that the only possible prime divisor of the product $p_1p_2$ are $1,p_1,p_2$, and $p_1p_2$.

Let there be a divisor $q$ of $p_1p_2$ which is not in the list.
This means, that $q$ should be either divisor of $p_1$ or $p_2$ which meanse, that $p_1$ or $p_2$ is not a prime.


Find by integrating the area of the triangle vertices $(5,1), (1,3)\;\text{and}\;(-1,-2)$ Find by integrating the area of the triangle vertices $$(5,1), (1,3)\;\text{and}\;(-1,-2)$$

I tried to make straight and integrate, but it is very complicated, there is some better way?

You can try to simplify the calculations by using 1. Shift of origin ( say to (-1,-2) ), and 2. Rotation of axis. Then do the integrations.


Is there a general expression for the adjoint representation of $U(N)$ or $u(N)$? At least for low values of $N$ like $2$ or $3$ and such I would like to know if there are explicit matrices known giving the representation of $u(N)$ or $U(N)$ in the adjoint?

(..a related query: Is it for the Lie group or the Lie algebra of U(N) that it is true that the weight vectors in the fundamental/vector representation can be taken to be N N-vectors such that all have weight/eigenvalue 1 under its Cartan and the ith of them has 1 in the ith place and 0 elsewhere and for the conjugate of the above representation its the same but now with (-1)?..I guess its for the u(N) since they are skew-Hermitian but would still like to know of a precise answer/proof..)

I've only recently started learning about group theory so it would be worth double checking the following.

As I understand it, when we say something (for example a field, $X$) transforms under the adjoint rep of a group (say $U(N)$), the transformation can involve any matrix, $g$, belonging to the group $U(N)$. The adjoint rep doesn't put any constraints on the matrices we choose from $U(N)$, rather it tells us how to use those matrices in a transformation. When we say $X$ transforms in the adjoint rep of $U(N)$ it simply means it transforms as $X \rightarrow X' = gXg^{-1}$. The matrix $g$ can be any matrix belonging to $U(N)$. Note: $g^{-1}$ is the inverse of the matrix $g$. So when we say 'an object is in the adjoint representation of $U(N)$' this is a statement about how to act the matrices of $U(N)$ on an that object. It doesn't say anything about which matrices we choose from $U(N)$.

I don't know if it helps, but I think you can write
$$U(N)=SU(N)\times U(1)$$
usually (in physics) people consider semisimple groups, i.e. groups without U(1) factors (=tori). To give an explicit example, consider $U(2)=SU(2)\times U(1)$. The adjoint representation of the algebra can be defined by
$$(T_a)_{bc}=-if_{abc}$$
where $f_{abc}$ are the structure constants of the algebra. Perfoming an explicit calculation for $SU(2)\times U(1)$ and keeping in mind that the generator of $U(1)$ commutes with all other elements of the algebra we get the following generators for the adjoint representation:
$$
T_1=\begin{pmatrix}0&0&0&0\\0&0&-i&0\\0&i&0&0\\0&0&0&0\end{pmatrix}
T_2=\begin{pmatrix}0&0&i&0\\0&0&0&0\\-i&0&0&0\\0&0&0&0\end{pmatrix}
T_3=\begin{pmatrix}0&-i&0&0\\i&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}
T_4=0
$$
So you see, that with $T_1, T_2, T_3$ you can span a 3 dimensional real vector space, but with all four you can't since $T_4$ vanishes. This means, as I think, that while you can define an adjoint representation for $SU(2)$, you can't do so for $U(2)$ or $U(1)$.


Finding sum of factors of a number using prime factorization Given a number, there is an algorithm described here to find it's sum and number of factors.
For example, let us take the number $1225$ :

It's factors are $1, 5, 7, 25, 35, 49, 175, 245, 1225 $
and the sum of factors are $1767$.

A simple algorithm that is described to find the sum of the factors is using prime factorization.

$1225 = 5^2 \cdot 7^2$, therefore the sum of factors is $ (1+5+25)(1+7+49)  = 1767$

But this logic does not work for the number $2450$. Please check if it's working for $2450$

Edit :
Sorry it works for $2450$. I made some mistake in calculation.

Your approach works fine: $2450=2\cdot 5^2\cdot 7^2$, therefore the sum of divisor is $$(1+2)(1+5+25)(1+7+49)=5301=3\cdot 1767.$$

You are looking for the Formula For Sum Of Divisors, from there:

  Each of these sums is a geometric series; hence we may use the formula for sum of a geometric series to conclude
  $$
\sum_{d|n}d = \prod_{i=1}^k \frac{p_i^{m_i+1}-1}{p_i-1}
$$


Proof: $2\sqrt{m}-2 < \sum\limits_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1$ I know that problems similar to this one, involving either one of the two bounds, have been posted before, but I would like just a hint in the last part of the proof involving the upper bound, with which I have some troubles with being ''convincing'' enough.

Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ if $n\geq 1$. Then use this to prove that

$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if }\ m \geq 2$$

With some help I received here I got the following proof for the first set of inequalities:

Proof (direct):

Since

$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}}$$

then we have

$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}\qquad (1)$$

Likewise, since

$$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$$

then

$$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})\qquad (2)$$

as asserted.

For the second part I tried to prove the lower bound firt:

Proof (direct): From $(1)$ we have

$$\sqrt{n+1} - \sqrt{n} < \frac{1}{2\sqrt{n}}$$

By summing both sides we get

$$\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$

And after applying the telescoping property on the LHS, this becomes

$$\sqrt{m+1} - 1 < \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}$$

But since $\sqrt{m} - 1 < \sqrt{m+1} - 1$, by transitivity we have

$$\begin{align*}\sqrt{m} - 1 &< \sum_{n=1}^{m}\frac{1}{2\sqrt{n}}\\
2\sqrt{m} - 2 &< \sum_{n=1}^{m}\frac{1}{\sqrt{n}}\end{align*}$$

Then the upper bound:

Proof (direct): From $(2)$ we have

$$\frac{1}{2\sqrt{n}}< \sqrt{n} - \sqrt{n-1}$$

By taking the sum of both sides we get

$$\sum_{n=1}^{m}\frac{1}{2\sqrt{n}}< \sum_{n=1}^{m}(\sqrt{n} - \sqrt{n-1})$$

If we apply the telescoping property on the RHS, the inequality becomes

$$\begin{align*}\sum_{n=1}^{m}\frac{1}{2\sqrt{n}} &< \sqrt{n}\\
\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{n}\end{align*}$$

Here I'm not sure how to get the $-1$ on the RHS that I'm lacking. I could have started the summation at $n=2$ I guess, but don't think it's correct and it doesn't seem to be a convincing way to complete the proof.

Well you tagged induction so why not use it:

$$
\sum_{n=1}^{m+1} \frac{1}{\sqrt{n}} \lt 2\sqrt{m}-1 + \frac{1}{\sqrt{m+1}} \lt 2\sqrt{m}-1 + 2(\sqrt{m+1}-\sqrt{m}) = 2\sqrt{m+1}-1
$$

For the lower bound, proof (direct): from $(1)$:

$$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}$$

By summing both sides:

$$2\sum_{n=1}^{m}(\sqrt{n+1} - \sqrt{n}) < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$

After applying the telescoping property on the LHS, the inequality becomes

$$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}}$$

For the upper bound, proof (direct): from $(2)$:

$$\frac{1}{\sqrt{n}}< 2(\sqrt{n} - \sqrt{n-1})$$

By taking the sum of both sides from n=2:

$$\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< \sum_{n=2}^{m}2(\sqrt{n} - \sqrt{n-1})$$

Add 1 on both sides:

$$1+\sum_{n=2}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$

$$\sum_{n=1}^{m}\frac{1}{\sqrt{n}}< 1+2(\sum_{n=2}^{m}(\sqrt{n} - \sqrt{n-1}))$$

Apply the telescoping property on the RHS, the inequality becomes

$$\begin{align*}\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 1+2(\sqrt{m}-1)\\
\sum_{n=1}^{m}\frac{1}{\sqrt{n}} &< 2\sqrt{m}-1\end{align*}$$

So
$$2\sqrt{m+1} - 2 < \sum_{n=1}^{m}\frac{1}{\sqrt{n}} < 2\sqrt{m} - 1$$

This can be relaxed to:

$$2\sqrt{m}-2<2\sqrt{m+1}-2<\sum_{n=1}^{m}\frac{1}{\sqrt{n}}<2\sqrt{m}-1< 2\sqrt{m}$$


Can one describe conic sections using synthetic geometry? I was curious to know whether synthetic geometry is more powerful than analytical geometry. Take as an example, the conic sections. Can one describe conic sections without the co-ordinate system, without the equations of conic sections? How difficult would it be?

I havent read it but I think that this book might be what you are looking for :
https://archive.org/stream/conicsectionstre00besarich#page/n12/mode/1up

In the introduction we have

  The object of the following pages is to discuss the general forms and
  characetistics of these (conic sections) curves and to determine their
  most important properties by help of the methods and relations
  developed in the first six books , and in the eleventh book of
  Euclid's Elements and it will be found that for this purpose a
  knowledge of Euclid's Geometry is all that is necessary.

Also, I am also looking for books of this type so if you have found some yourself please post them as an answer or as a comment.

Yes, synthetic geometry works well for plan sections in conics in finding the focus points, the axes, the directors line positions. Danut Dragoi, PhD


Probability of ball hitting the fan Case 1: A person throws a ball upwards to hit a fan which is not rotating. Let the probability of the event of ball hitting the fan be $p_1$.

Case 2: Same person now does the same experiment, this time with a rotating fan. Let the probability of the event this time be $p_2$.

Then what is the relationship between $p_1$ and $p_2$?

$$P(2)>P(1)$$

In order to exist, the ball must have some radius $r$.

Now, as the ball passes through the plane on which the fan rotates, time passes (the amount depending on both the radius of the ball and its initial velocity).

Given that any amount of time passes while the ball is passing through the plane, even a terribly small amount of time, the fan rotates at least a very small distance, meaning it covers greater area than if angular velocity $\omega=0$.

Thus, the probability $P(2)$ must exceed that of a stationary fan $P(1)$ since having a nonzero radius is a necessary condition for the ball to exist.

$p(1)=p (2)$ same percent of ceiling is always covered by the fan at any time


Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$ Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$.
$$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
Could someone help here?

This limit can be evaluated by applying l'Hospital's rule twice. For the first time we differentiate

$$(1+x)^{\frac{1}{x}}=e^{\frac{\ln (1+x)}{x}}$$ to get

$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}
e^{\frac{\ln (1+x)}{x}}.$$

Now $$e^{\frac{\ln (1+x)}{x}}\rightarrow e$$ so we need the limit of
$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}$$ another application of l'Hosptial gives

$$\frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}=-\frac{1}{2}\frac{1}{(1+x)^2}\rightarrow -\frac{1}{2}$$

So the limit is $-\frac{e}{2}$.

You can think of the limit  $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
as the derivative of the function $f(x)=(1+x)^\frac1x$ at point $x=0$.

Did that idea gave you any help?

Edit: You should define the value of $f(x)$ at $x=0$ namely : $f(0)=e$


Find minimum value of the function If $g(x) = \max|y^2 - xy|$ for $0\leq y\leq 1$. Then the minimum value of $g(x)$ is?

I am not being able to proceed. Tried drawing the graph.

Computing the function $g(x) = \max{|y^2 - xy|}$ is equivalent to solving for the maximum value of the expression $|y^2 - xy| = f(y)$ with a fixed parameter $x$. Let's then imagine that $x$ is a constant. In general, the maximum value of a function (of one variable) can be found in either

Where the derivative is zero;
Where the derivative is not defined; or
At the boundaries of the domain.

Let's go through these cases one by one. Firstly, we want to find $\frac{d f(y)}{dy}=0$. For $y^2 - xy < 0$ the result is $y = \frac{x}{2}$, and for $y^2 -xy > 0$ there is no solution.

The derivative is not defined when $y^2 - xy = 0 \Rightarrow y= 0$ or $y = x$.

The boundaries of the domain were defined to be $y = 0$ and $y = 1$.

Inserting these results into $f(y)$ gives us $f(\frac{x}{2}) = \frac{x^2}{4}$; $f(0)=f(x)=0$; $f(1)=|1-x|$. We need the largest of these, which is

$g(x) = \begin{cases}
\left| 1 - x \right| , \text{if} -2(1+\sqrt{2}) \leq x \leq 2(\sqrt{2}-1)\\
x^2/4 , \text{otherwise}
\end{cases}$

This was found out by solving the equation $|1-x|>\frac{x^2}{4}$.
The minimum of $g(x)$ can be found at $x= 2(\sqrt{2}-1)$, which is $g(2(\sqrt{2}-1)) = 1- 2(\sqrt{2}-1) = 3-2\sqrt{2}$.

The maximum for the square of a non negative function is the square of the maximum , so find the maximum (x fixed) for  z= |...|^2=( y^2-xy)^2 = y^2(y-x)^2  this gets rid of the troublesome absolute value . Now study z as a function of y for fixed x . Determine where dz/dx is =,-,0 and sketch the graph .


Calculating $\int e^{-|x|} \, dx$. I have been trying to calculate
$\int e^{|x|} \, dx$, but merely splitting up in two cases for $x<0$ and $x>0$ does not give the desired result, which I got by calculating it on a CAS.
Suggestions would be very welcome.

Edit.
I made a mistake. It's $\int e^{-|x|} \, dx$.

Using $$\exp(|x|) =\begin{cases} \exp(x) & x \geqslant 0 \\ \exp(-x) & x < 0 \end{cases} $$
we can integrate branch-wise:
$$
   \int \exp(|x|) \mathrm{d}x = \begin{cases} \int \exp(x) \mathrm{d}x & x \geqslant 0 \\ \int \exp(-x) \mathrm{d}x & x < 0 \end{cases} = \begin{cases} \exp(x) + C_1 & x \geqslant 0 \\ -\exp(-x) + C_2 & x < 0 \end{cases}
$$

we can also write
$$\int e^{|x|} dx= sign(x) e^{|x|}+C$$


How to solve this pair of differential equations? The equations are

$$y' = y + 3z$$$$z' = -3y + z$$I vaguely remember I need to differentiate one to get the form of the other, but by differentiating the first and subbing in the second I get$$y'' = y' + 3z'=> y''- y' + 9y = 3z$$ I don't know what to do from here.

Edit:
From here I got to $y''-2y'+10y = 0$, so from $m^2-2m+10=0$ we get$$y=e^x(Acos(3x) + Bsin(3x))$$Where would I go from here?

From the first equation, $z = \frac{1}{3}(y'-y)$ and $y'' = y' + 3z'$. From the second eqution, $z' = -3y+z = -3y+\frac{1}{3}(y'-y)$. So, $$y'' = y'+ 3(-3y+\frac{1}{3}(y'-y)) \\ \Rightarrow y'' - 2y' + 10y = 0 \\ \Rightarrow y = (A\cos(3x)+B\sin(3x))\exp(x) $$

Recalling that $z = \frac{1}{3}(y'-y)$, we first calculate $$y' = y + (B\cos(3x)-A\sin(3x))\exp(x) $$

and so

$$ z = \left(\frac{B}{3}\cos(3x)-\frac{A}{3}\sin(3x)\right)\exp(x) $$

$(y,z)'=A\cdot(y,z)$ where $A=\begin{pmatrix}1&3\\-3&1\end{pmatrix}$. Hence
$$
(y,z)(t) = \exp(tA)\cdot(y,z)(0)
$$


Cards in a 5x5 grid- Probability of a diagonal of all hearts All the aces, 2’s, 3’s, 4’s, 5’s and 6’s, as well as the jack of diamonds are taken from a
regular deck of 52 playing cards, and then placed face up on a table in a 5 × 5 square
grid configuration randomly.

What is the probability that at least one of the diagonals in the array is all hearts?

I know that the probability of having all 5 hearts in a single row is 4!20!/24! but I am having trouble with this specific problem. Any help would be appreciated.

Since both diagonals can't be all hearts at the same time, the desired probability is simply twice the probability that one diagonal is all hearts, and thus

$$
2\cdot\frac{\binom65}{\binom{25}5}=2\cdot\frac{6!\cdot20!}{25!}=\frac2{8855}\;.
$$

There are two branches that must be considered. First, let's start with the center.  It has a 6/25 chance of being a heart.  Next, we'll traverse the diagonals.  The chance that the next spot we uncover is a heart is 5/24.  If it is a heart then we keep uncovering spots on that diagonal.  The chance that the next card is a heart is 4/23. If it is not a heart we still have a chance that the other diagonal is all hearts. So, there are two sub-branches here. If it is a heart then we have probabilities 3/22, and 2/21 until we uncover the entire diagonal.

There is a 19/23 chance that the third card uncovered is not a heart. If that occurs we switch to the other diagonal and require that all turns reveal a heart. We have probabilities 4/22, 3/21, 2/20, and 1/19 for this sub-branch.

After the center, there is a 19/24 chance that card we turned over is not a heart.  If that is the case then we switch to the other diagonal. The probabilities there are 5/23, 4/22, 3/21, and 2/20 until the entire diagonal is exposed.

So we have

P(total) = 6/25*(5/24*(4/23*3/22*2/21 + 19/23*4/22*3/21*2/20*1/19) + 19/24*5/23*4/22*3/21*2/20)

Recombining, we have

P(total) = 6/25 * (5!*20!/24! + 5!*19!/24! + 19*5!*19!/24!)
         = 6!/25! * (20! + 19! + 19 * 19!)
         = 6!/25! * (20! + 20!/20 + (19 * 20!)/20)
         = 6!*20!/25! * (1 * 1/20 + 19/20)
         = (2 * 6! * 20!)/25!

This is the same value as answer generated by my original solution. The original solution made some assumptions not made here.


Trouble understanding stated consequence of the division theorem Question: I'm having some trouble understanding the following Corollary (and construction of its proof) given in the book "An Introduction to Mathematical Reasoning".
The main reason being the condition $0 \le q \lt r$, which seems to me non-sense because it would imply that $0 \lt r$, so that $r$ couldn't be zero (contradicting the corollary itself!).
Also, regarding the second sentence of the construction of the proof, wouldn't $gcd(b, r)$ and $gcd(b, 0)$ be two other possible solutions?

Corollary: Let $a$ and $b$ be integers with $b \gt 0$ and suppose that $a = bq +r$ for integers $q$ and $r$ with $0 \le q \lt r$. Then $b$ divides $a$ if and only if $r = 0$.
Construction of the proof: The point here is that if $b$ divides $a$ then, for some integer $q_1$, $a=bq_1=bq_1+r_1$ with $r_1 = 0$. So if $a=bq+r$ with $0 \lt r \lt q$ then $gcd(q, r)$ and $gcd(q_1, 0)$ would be two distinct solutions to the division problem, contradicting the uniqueness part of the division theorem.

For completeness, I leave the statement of the division theorem as given in the book.

Division theorem: Let $a$ and $b$ be integers with $b \gt 0$. Then there are unique integers $q$ and $r$ such that
$$a = bq + r \quad and \quad 0 \le r \lt b$$

Thank you.

For your first question, yes, that's surely a typo and should be $0\le r<q$.

As for the second question: I'm pretty sure those gcds shouldn't be there—it seems that they want to say that the ordered pairs $(q,r)$ and $(q_1,0)$ are two distinct solutions to the division problem.


Max. distance of Normal to ellipse from origin How Can I calculate Maximum Distance of Center of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ from the Normal.

My Try ::  Let $P(a\cos \theta,b\sin \theta)$ be any point on the ellipse. Then equation of Normal at that point is

$ax\sec \theta-by\csc \theta = a^2-b^2$. Then How can I find Max. distance of Center of the ellipse from the Normal

So, the distance of the normal from the origin $(0,0)$ is $$\left| \frac{a^2-b^2}{\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}} \right|$$

So, we need to minimize $(a\sec\theta)^2+(-b\csc\theta)^2=a^2\sec^2\theta+b^2\csc^2\theta=f(\theta)$(say)

So, $\frac{df}{d\theta}=a^22\sec\theta\sec\theta\tan\theta+b^22\csc\theta(-\csc\theta\cot\theta)=2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}$

For the extreme value of $f(\theta),\frac{df}{d\theta}=0$

$\implies 2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}=0$ or $\tan^4\theta=\frac{b^2}{a^2}$

Assuming $a>0,b>0$, $\tan^2\theta=\frac ba$

Now, $\frac{d^2f}{d\theta^2}=2a^2\left(\frac1{\cos^2\theta}+\frac{3\sin^2\theta}{\cos^4\theta}\right)+2b^2\left(\frac1{\sin^2\theta}+\frac{3\cos^2\theta}{\sin^2\theta}\right)>0$ for real $\theta$

So, $f(\theta)$ will attain the minimum value at $\tan^2\theta=\frac ba$

So, $f(\theta)_\text{min}=a^2\sec^2\theta+b^2\csc^2\theta_{\text{at  }\tan^2\theta=\frac ba}=a^2\left(1+\frac ba\right)+b^2\left(1+\frac ab\right)=(a+b)^2$

So, the minimum value of $\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}$ is $a+b$

If $\tan\theta=\sqrt \frac ba, \frac{\sin\theta}{\sqrt b}=\frac{\cos\theta}{\sqrt a}=\pm\frac1{b+a}$

If $\sin\theta=\frac{\sqrt b}{a+b}\implies \csc\theta=\frac{a+b}{\sqrt b},\cos\theta=\frac{\sqrt a}{a+b}\implies \sec\theta=\frac{a+b}{\sqrt a}$

There will be another set $(\csc\theta=-\frac{a+b}{\sqrt b},\sec\theta=-\frac{a+b}{\sqrt a})$

There will be two more set of values of $(\csc\theta,\sec\theta)$ for $\tan\theta=-\sqrt\frac ba$

So, we shall have four normals having the maximum distance from the origin.

let a point p(acost,bsint) is on the ellipse.
x2/a2 y2/b2=1
dy/dx=-b2x/a2y

dy/dx of normal on(acost,bsint)
= a2y/b2x=a/btant
equestion of normal
y-bsint=b/atant(x-acost)
axsect-bycost-(a2-b2)=0

now
lenth from origin(0,0) of the normal
l=mode -(a2-b2)/ squat(a2sec2t b2cosec2t)
=a2-b2/squat(a2sec2t b2cosec2t)
then
Differential of this function respect t
dl/dt=a2-b2(b2cos4t-a2sin4t)/(a2sin2t b2cosec2t)3/2
for max. lenth
dl/dt=0
b2cos4t-a2sin4t =0
tant=squat(b/a)
now
d2l/dt2=(a2-b2)(-4sintcost)[(a2sin2t b2cos2t)3/2]
now
 d2l/dt2 on tant=squat(b/a)                                                  ' .'sint=squat(b/a b),cost=squat(a/a b)
    =  -4(a-b)a2b2   <0    ;.a>b
so the lenth of normal will max. on t=tan(inverse)squat(b/a)
now
 l=a2-b2/squat(a2sec2t b2cosec2t)
 l(max.)=  a2-b2/squat{a2(a b/a) b2(a b?b)           ;.sect=squat{(a b)/a,},cosect=squat{(a b)/b}   because tant=squat(b/a)
             =  a2-b2/squat{(a b)(a b)}
              = a2-b2/a b
               =a-b


How do I integrate the integral for the independent Poisson random variable? I don't know how to derive this (my red pen mark)

Possible explanation:
Since $\left| -\frac{1}{\sqrt n} \right | \approx 0,$ the interval of integration is very close to zero, and so $x^2 \approx 0 \implies e^{-x^2/2} \approx 1$.

When $x \approx 0$, we have $e^{-x^2/2} \approx 1$. Since the text states that we are doing a slightly less than rigorous proof, replace $e^{x^2/2}$ with $1$ and the answer follows immediately.


Finding a Lyapunov function for the differential system $x_1'=-8x_1^3-x_2$, $x_2'=-4x_2-4x_1^3$ I've got the following system of equations:
$$
x_1'=-8x_1^3-x_2 \qquad x_2'=-4x_2-4x_1^3
$$
I'm trying to check, if the equilibrium point in $(0,0)$ is stable or not. I am supposed to find so called Lyapunov function $L$, i.e. function which satisfies three following conditions:

1) $L(x_1,x_2)=0$ iff $(x_1,x_2)=(0,0)$

2) $\forall_{(x_1,x_2)\neq(0,0)}L(x_1,x_2)>0$

3) $\forall_{(x_1,x_2)}:\frac{dL}{dt}(x_1,x_2)\leqslant 0$ or $\frac{dL}{dt}(x_1,x_2)\geqslant 0$

The sign in 3) gives us the information about stability in the equilibrium point.

I was trying to find a polynomial in $\mathbb{R}[x,y]$ with only even exponents, but I was unable to find such function. Is there any way to do it without just "guessing"?

What is the Lyapunov function for this system of ODEs?

Hint: If $L(x_1,x_2)$ has a part $x_1^{2n}$, the stability condition involves some multiples of $x_1^{2n+2}$ and $x_1^{2n-1}x_2$. If $L(x_1,x_2)$ has a part $x_2^{2m}$, the stability condition involves some multiples of $x_2^{2m}$ and $x_2^{2m-1}x_1^3$. The mixed terms $x_1^{2n-1}x_2$ and $x_2^{2m-1}x_1^3$ are the same if $(n,m)=(2,1)$.

This suggest to check whether $L(x_1,x_2)=x_1^4+x_2^2$ is a Lyapunov function. If it was not (it is), one could have tried $L(x_1,x_2)=ax_1^4+x_2^2$ for some positive $a$, or even $L(x_1,x_2)=ax_1^4+bx_1^2x_2+x_2^2$ for some $(a,b)$ such that $b^2\lt4a$.

But differentiating your $L$, I get: $$\frac{dL}{dt}(x_1,x_2)=(x_1^4+x_2^2)'=4x_1^3 \cdot x_1'+2x_2\cdot x_2'=4x_1^3\cdot(-8x_1^3-x_2)+2x_2\cdot(-4x_2-4x_1^3)=-32x_1^6-4x_1^3x_2-8x_2^2-8x_1^3x_2=-4\cdot(8x_1^6+3x_1^3x_2+2x_2^2)$$ which doesn't have a constant sign. Differentiating $L=ax_1^4+x_2^2$ works only for $a=-2$, but it is not positive. I am going to try the remaining option.


Proof by contrapositive Prove that if the product $ab$ is irrational, then either $a$ or $b$ (or both) must be irrational.

How do I prove this by contrapositive? What is contrapositive?

The statement you want to prove is:

  If $ab$ is irrational, then $a$ is irrational or $b$ is irrational.

The contrapositive is:

  If not($a$ is irrational or $b$ is irrational), then not($ab$ is irrational).

A more natural way to state this (using DeMorgan's Law) is:

  If both $a$ and $b$ are rational, then $ab$ is rational.

This last statement is indeed true. Since the truth of a statement and the truth of its contrapositive always agree, one can conclude the original statement is true, as well.

If $a,b \in \Bbb{Z}^+$ and $a\neq b$, then $ax^2+bx+(b-a) = 0$ has no positive integer root (solution).


Determining stability of equilibrium point I want to determine the stability property of the equilibrium point (0,0) for the system $$x'=-xy^4-y\cos(x^2y) \\ y'=3x^5\cos(x^2y)-\sin(y) $$
I get the eigenvalues $-1$ and $0$ for the Jacobian evaluated at $(0,0)$. We have theorems regarding the case when either one eigenvalue has a real part, or when all the eigenvalues are negative. But since these eigenvalues ($0$ and $-1$) do not satisfy this, I feel lost. I know that one could show it by constructing
a Lyapanov function, but I wasnt able to do so. Could anyone help me out? How do I determine the stability property of origin?

The Lyapunov function is
$$
V(x,y)= \frac12 x^6+\frac12 y^2.
$$
Its derivative along the trajectories
$$
\dot V= 3x^5\dot x+y\dot y=3x^5(-xy^4-y\cos x^2y)+y(3x^5\cos x^2y-\sin y)
=-3x^6y^4-y\sin y
$$
is non-positive in some neighborhood of the origin, thus, the origin is stable.

In order to prove that the origin is asymptotically stable, we should show that the set
$$
S=\{ (x,y): \dot V(x,y)=0 \}= \{ (x,y): y=0 \}
$$
does not contain whole trajectories of the system except for the origin. This follows from the fact that
$$
\dot y|_{(x,y)\in S}= 3x^5
$$
is nonzero for any $x\ne 0$.

Assuming that near $(0,0)$ the dynamical system behaves as

$$
\cases{
\dot x = -x y^4-y\\
\dot y = 3 x^5-y
}
$$

and assuming for the center manifold $h(x) = \sum_{k=0}^n a_k x^{k+1}$ we have

$$
\dot h(x) = h'(x)(-x h^4(x)-h(x)) = \dot y =3x^5-h(x)
$$

and equating coefficients we have for $n=4$

$$
\left\{
\begin{array}{rcl}
 \left(a_0-1\right) a_0&=&0 \\
 \left(1-3 a_0\right) a_1&=&0 \\
 \left(1-4 a_0\right) a_2-2 a_1^2&=&0 \\
 \left(1-5 a_0\right) a_3-5 a_1 a_2&=&0 \\
 a_4-a_0^5-6 a_4 a_0-3 \left(a_2^2+2 a_1 a_3+1\right)&=&0 \\\end{array}
\right.
$$

with solution $h(x)=3 x^5$ for $n=4$. For $n = 2,3, h(x)=0$ and for $n = 8$ we have $h(x) = 45 x^9+3 x^5$

The flow along the central manifold in both cases is stable. Regarding the case $n = 4$ we have a flow given by

$$\dot x = -3 x^5 - 81 x^{21}
$$

which is stable.

Attached a stream flow showing in red part of the center manifold.

NOTE

The $h(x)$ coefficients determination involves two solutions. The plot shown is for $n = 10$


A question about polynomials as vectors The space $P_n(R)$ is very similar to the vector space $R^{n+1}$; indeed one can match
  one to the other by the pairing

  $a_nx^n + a_{n−1}x^{n−1} + \ldots + a_1x + a_0$ is equivalent to $(a_n, a_{n−1}, \ldots, a_1, a_0)$,

  thus for instance in $P_3(R)$, the polynomial $x^3 + 2x^2 + 4$ would be associated
  with the 4-tuple $(1, 2, 0, 4)$.

I am not sure I understand the equivalence. The left side of it looks like it could be summed up to a scalar - single number, while the right side is a vector in $R^n$.

The same goes for the cubic polynomial which could be summed up to be a scalar for some $x$ whereas the 4-tuple is a vector in $R^4$.

Please, explain all this.

Thanks.

In an arbitrary $n$-dimensional vector space with basis $\{\mathbf{f}_1,\ldots,\mathbf{f}_n\}$, it is usual to denote by the $n$-tuple $(x_1,\ldots,x_n)$ the vector

$$\sum_{i=1}^n x_i\mathbf{f}_i$$

If one fixes the following basis $\{1,x,\ldots,x^n\}$ of $P_n(\mathbb R)$, then the polynomial $x^3+2x^2+4$ can be written as

$$\mathbf{1}\cdot x^3 + \mathbf 2\cdot x^2 + \mathbf 0\cdot x +\mathbf 4 \cdot 1$$

Which (by the above) is represented by the tuple $(1,2,0,4)$, which we can identify with an element of $\mathbb R^{3+1} = \mathbb R^4$.

The explicit isomorphism is given by mapping the basis vector $x^i \in P_n(\mathbb R)$ to the $(n-i+1)$-th standard basis vector $e_{n-i+1}\in \mathbb R^{n+1}$.

The key thing here is that the powers of $x$ are not scalars, they are indeterminates in a polynomial ring. You should check that adding polynomials and multiplying them my scalars under this identification with an $(n+1)$-tuple works the same as in $\mathbb R^{n+1}$ if you're not yet convinced, but explicitly constructing (and checking) the map I defined above is an isomorphism of vector spaces should convince you.

As you already stated, this is an equivalence, no equation. So you actually can't set both sides as equal.

If you define a variable $x$ in $\mathbb{R}$, you can identify it for example as the tuple $(x,0)$ in $\mathbb{R}^2$. You have an equivalence between both elements, but they are not equal, as one lies in $\mathbb{R}$ and the other in $\mathbb{R}^2$. You can define a function

$$
f: \mathbb{R} -> \mathbb{R}^2  \\
f(x) = (x,0)
$$

that identifies one element with the other. Can you define a function that identifies the polynomial elements with the vectors in your equivalence? Or even an inverse function?


Representation of polynomials Let $A$ be a ring and $A[x_1, x_2, ... ,x_n] = A[x_1][x_2]...[x_n]$ be the ring of the polynomials of $n$ independent variables over $A$, i.e. every element $f \in A[x_1, x_2, ... x_n]$ is $$f = f_m x_n^m + f_{m-1}x_n^{m-1} + ... + f_0$$

for some $m \in \mathbb Z^{\geq 0}$ and $f_i \in A[x_1, x_2, ..., x_{n-1}]$ for $0 \leq i \leq m$. I'm trying to prove that every such polynomial $f$ is uniquely represented as a finite sum of monomials $x_1^{\alpha_1} x_2^{\alpha_2} ... x_n^{\alpha_n}$. What I have done so far is this:

First of all let $S = (\mathbb Z ^{\geq 0})^n$ and

$$G = \bigoplus_{(\alpha_1, \alpha_2, ..., \alpha_n) \in S} A$$

Then $\psi : G \longrightarrow A[x_1, x_2, ... , x_n]$ defined by

$$\psi (\{ a_{(\alpha_1, \alpha_2, ..., \alpha_n)}\} _ {(\alpha_1, \alpha_2, ..., \alpha_n) \in S}) = \sum _{(\alpha_1, \alpha_2, ..., \alpha_n) \in S} a_{(\alpha_1, \alpha_2, ..., \alpha_n)} x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$$

is a surjective group homomorphism and all I have to show is that it's kernel is $0$. My idea is to do this by induction over the number of variables $n$. The case when $n = 1$ is easy, but then I don't know how to use the inductive hypothesis. Please help me! Thanks in advance!

Suppose you have $f=f_m x_n^m + f_{m-1}x_n^{m-1} + ... + f_0 \in A[x_1, x_2, ... x_n]$.  By induction on $n$, each $f_i$ can be written uniquely as a sum of monomials in $x_1,\dots,x_{n-1}$.  Define an element $a\in G$ by saying that $a_{(\alpha_1,\dots,\alpha_n)}$ is defined to be $0$ if $\alpha_n>m$ and the coefficent of $x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}$ in the unique representation of $f_{\alpha_n}$ as a sum of monomials if $\alpha_n\leq m$.  Then $$\psi(a)=\sum_{(\alpha_1,\dots,\alpha_n)}a_{(\alpha_1,\dots, \alpha_n)} x_1^{\alpha_1}\dots x_n^{\alpha_n}=\sum_{\alpha_n}x^{\alpha_n}\sum_{(\alpha_1,\dots,\alpha_{n-1})}a_{(\alpha_1,\dots, \alpha_n)}x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}=\sum_{\alpha_n} x_n^{\alpha_n}f_{\alpha_n}=f.$$

This shows $\psi$ is surjective.  For injectivity, suppose that $a\in A$ is such that $\psi(a)=\sum_{(\alpha_1,\dots,\alpha_n)}a_{(\alpha_1,\dots, \alpha_n)} x_1^{\alpha_1}\dots x_n^{\alpha_n}=0$.  For each $i$, define $$f_i=\sum_{(\alpha_1,\dots,\alpha_{n-1})}a_{(\alpha_1,\dots, \alpha_{n-1},i)}x_1^{\alpha_1}\dots x_{n-1}^{\alpha_{n-1}}.$$
As above, we can write $\psi(a)=\sum_i x_n^i f_i$.  Since $f_i\in A[x_1,\dots,x_{n-1}]$, $\psi(a)=0$ implies each $f_i$ must be $0$.  By induction, this means that for each $i$, $a_{(\alpha_1,\dots,\alpha_{n-1},i)}=0$ for all $(\alpha_1,\dots,\alpha_{n-1})$.  But this just means that $a=0$.

Not much to prove, really, this is a non-problem.

Assume the result true for $n$ and study it for $n+1$. Note that $A[x_1, \dots, x_n, x_{n+1}] = A[x_1, \dots, x_n] [x_{n+1}]$, as you have noted yourself. To assume that $f$ has two distinct representations $S_1$ and $S_2$ as a sum of monomials is equivalent to showing that $S_1 - S_2$ is a representation not identically zero of the $0$ polynomial.

Assume that $0 \in A[x_1, \dots, x_n] [x_{n+1}]$ can be written as $f_m x_{n+1} ^m + \dots + f_0$ with the coefficients $f_0, \dots, f_m \in A[x_1, \dots, x_n]$. Then, equality of two polynomials in any ring means that they must be equal component-wisely, so $f_i = 0 \ \forall i = 0, \dots, m$, so every representation of $0$ as a sum of monomials must be identically zero. Therefore, your conclusion follows.


What common definition of norm on the space of analytic functions makes the basis $e_n=\frac{x^n}{n!}$ orthonormal? I mean, this basis, with factorials is very useful as it is the basis of Taylor expansion. But it is not orthonormal under usual definitions of norm ($\int_a^b \sqrt{f(x)^2}dx$). I wonder, how it should be modified so to make the basis normal.

You can extend it to complex domain and then u will have an upper traingular form for the matrix: $a_{ij}=\langle z^i,z^j\rangle $ since for $i>j$ $a_{ij}=\langle  z^i,z^j\rangle =0$ by cauchy integral theorem where the inner product is defined as $\langle z^i,z^j\rangle =\int_C z^{i-j}\,\mathrm{d}z$ for a closed curve $C$.
If u define the same inner product for a closed curve not including origin, it might work as diagonal matrix also.
You can convert ur intution into equation. Another definition of inner product for real analytic functions only would be: would be $<f,g>=\sum_i D^i(f)|_{x=a} D^i(g)|_{x=a}$ where $D$ is the derivative operator. In this inner product $(x-a)^i$ and $(x-a)^j$ are orthogonal.


Calculating permutations for a specific password policy The security researcher Troy Hunt posted an example of an obscure password policy and I've been trying to figure out how to calculate the possible permutations
(Source: https://twitter.com/troyhunt/status/885243624933400577)

The rules are:

The password must contain $9$ numbers (and only $9$ numbers)
It must include at least $4$ different numbers
It cannot include the same number more than three times

I understand the basic permutations will be $10^9$ ($0-9$ nine times) $= 1,000,000,000$

What I don't understand is how you factor in the reduction in permutations by enforcing $4$ different numbers and limiting repeats to $3$.

We calculate the number of valid passwords with the help of exponential generating functions.

  At first we are looking for strings of length $9$ built from the alphabet $V=\{0,1,\ldots,9\}$ which contain a digit from $V$ no more than three times.

  The number of occurrences of a digit can be encoded as
  \begin{align*}
1+x+\frac{x^2}{2!}+\frac{x^3}{3!}
\end{align*}

  In the following we denote with $[x^n]$ the coefficient of $x^n$ in a series. Since we have to consider ten digits building strings of length $9$ we calculate with some help of Wolfram Alpha
  \begin{align*}
9![x^{9}]\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^{10}=916\,927\,200\tag{1}
\end{align*}

The second condition requires at least four different digits for valid passwords.

  We respect the second condition by subtracting invalid words from (1).

  We observe there are no words of length $9$ which consist of one or two different digits whereby each digit does not occur more than three times.
  We conclude the only type of invalid strings of length $9$ counted in (1) is built of words with three different digits each digit occurring exactly three times.

There are $\binom{10}{3}$ possibilities to choose three digits out of $V$. The first digit can be placed in $\binom{9}{3}$ different ways, leaving $\binom{6}{3}$ different possibilities for the second digit and $\binom{3}{3}$ for the third digit.

  We finally conclude the number of valid passwords is
  \begin{align*}
9![x^{9}]&\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right)^{10}-\binom{10}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}\\
&=916\,927\,200-120\cdot 84\cdot 20\cdot 1\\
&=916\,927\,200-201\,600\\
&=\color{blue}{916\,725\,600}
\end{align*}

First, each number can appear at most 3 times. So we are choosing 9 elements from

1, 1, 1, 2, 2, 2, ..., 9, 9, 9.

Then, there must be at least 4 kinds of numbers. It's impossible to have 1, 2 kinds of numbers. The number of 3 kinds is 9 choose 3.

So 27 choose 9 - 9 choose 3 = 4686741 is the final answer.


When can I simplify an equation? Suppose I have a few ecuations:

$$\cos^2(x) = \sin(x)\cos(x)   \Rightarrow  cos(x) = sin(x) $$
$$ x^2  + 3x \ge 2x \Rightarrow  x(x+3) \ge 2x  \Rightarrow  x+3 \ge 2$$

Which of them are true and why?

Basically, when can one simplify an equation or an inequation with common terms that contain x (a variable) that we need to solve for? We may lose solutions.

For equations one can always apply a function on both sides, this includes dividing(unless of course you divide by $0$), squaring, subtracting etc. This is because of the obvious but nevertheless useful to note fact that $x = y$ implies $f(x) = f(y)$.

For inequalities this is not the case, because essentially $x \leq y$ does not imply $f(x) \leq f(y)$ for all functions, but there are a lot of functions for which it is known whether the function is decreasing or increasing. For example, multiplying by some $c>0$ is an increasing function, in other words: $x \leq y$ implies $cx \leq cy$.

Now we can apply the preceding discussion to your examples. If $\cos^2(x) = \sin(x) \cos(x)$, we can divide both sides by $\cos(x)$, obtaining $\cos(x) = \sin(x)$, if $\cos(x) \neq 0$. In the case $\cos(x) = 0$, the equation does not imply anything new. Now for your inequality, only the last step is dubious, because in the case $x<0$, $1/x< 0$, so multiplying both sides by $1/x$ turns around the sign and we get $x+3 \leq 2$. If $x>0$ the last step holds, but then this implies $x \geq -1$ which does not give us any new information, since $x> 0$. However, adding and subtracting is increasing, so $x^2 + 3x \geq 2x$ if and only if $x^2+x \geq 0$. We see this equation is satisfied if $x \geq 0$. if $x \leq 0$, we have $x \leq -1$. Conversely if $x \leq -1$ we have $x+3 \leq 2$, and multiplying both sides by $x$ and noticing $x<0$ we have $x(x+3) \geq 2x$. So the solutions of the inequality are all $x$ for which $x \geq 0$ and $x \leq -1$.

Finally a general tip for this kind of confusion: always remind yourself of what it is you want to do. Do you want to find $x$ such that $f(x) = g(x)$? Do you want to validate whether $f(x) = g(x)$ for all $x$? Do you want to see wheter $f(x) = g(x)$ implies some contradiction? Keep using logic, if you take a step, think about whether it is reversible (in other words, whether it is 'if and only if'), and always remember that an equation without an explanation means nothing.

If $f(x)$ is monotonic then you can simplify as if the equation is an equality.  Monotonic means the first derivative never changes sign.  Broadly speaking this means the function always goes up with $x$ or down with $x$ and never changes.

In your first example, $cos^2(x)$ isn't strictly $1:1$ because in some places it goes up with $x$ and in other paces it goes down.

In your 2nd, the quadratic form will again go both up and down due to positive and negative square roots.

As a rule, find the source of the duotonicity and this will lead you to find the multiple solutions.  In the case of the quadratic you will need to allow for the positive and negative square roots.  In the case of the trigonometric identities you need to account for the cyclicity of the functions with period $2\pi$.


Calculating the missing dimension? 1) a right cone has a surface area 12 m${^2}$ and radius 1.3 m

here is my answers:

1) $s$ = 1.28

(photo of how i got my answer)

textbook answer: 1.64 m

how did they get that??

Your error is in thinking that they want the vertical height of the cone. For a cone, $s$ usually represents the slant height. So the first thing you want to is to express the total surface are as

$$S=\pi r s+\pi r^2$$

where $\pi r s$ is the lateral area. Then

$$
s=\frac{S-\pi r^2}{\pi r}\approx1.638
$$

as in your text!

The formula to calculate the surface area of a cone is:

$$SA=\pi r(r+\sqrt{h^2+r^2})$$

We know the $SA$ and $r$. Now solve for $h$.

$$12=\pi(1.3)(1.3+\sqrt{h^2+(1.3)^2})$$

$$\frac{12}{1.3\pi}-1.3=\sqrt{h^2+(1.3)^2}\text{ moved stuff we already know to left side}$$

$$(\frac{12}{1.3\pi}-1.3)^2=h^2\text{ squared both sides}$$

$$\sqrt{\frac{12}{1.3\pi}-1.3}=h\text{ take square root to solve for $h$}$$

$h\approx 1.28$

I'm guessing they made a mistake.