price and quantity after taxation Given that demand for a good X is equal to $q_D=393-2p$ and market supply is $q_S=p/4-12$. Find equilibrium price and quantity, consumer and producer surplus and draw a diagram illustrating the situation. Given that:

a) $T=20\% \pi$, total profit is taxed

b) $T=200$ tax does not depend on volume and value of goods sold

will it be simply $$393-2p=p/4-12-200$$
? Obviously i need to find $p$.

Obviously i have calculated the equilibrium price and quantity before taxation that is $p=180,q=33$. How to find situation after taxation in those two cases?


Find the height of a bar, given the lengths of shadows cast by it and another bar What is the height of the red bar?

My try: with respect to the picture, it seems for the green bar $\frac{h}{H}=\frac{2}{3}$. So, I think that ratio is the same for the red bar, and the height of the red bar is
$$\frac{h}{6+4}=\frac 23\qquad\to\qquad h_{red}=\frac{20}{3}$$

Is this correct?

Visualising the model is a bit tricky. The eye cannot see the tiny differences. One of the great examples that eyes can be trapped is the missing square puzzle. Therefore, we cannot rely on the visual presentation of the model to calculate the height.
Here, I will try to present the problem mathematically:

We know the length of $BC=6m$ which represents the horizontal shadow, $CD=4m$ which represents the vertical shadow, $BE=4m$, $AE=x$
height of the red bar is $AE + BE = 4m + x$
$\angle{AED}=90^\circ$ and $\angle{ADE}=60^\circ$ and $\angle{ADE}=30^\circ$
So, $AD=2x$ and $ED=x\sqrt{3}=6m \Rightarrow x = \frac{6}{\sqrt{3}}$
$AB = 4m + \frac{6}{\sqrt{3}}m \approx 7.464m$


Find two vectors orthogonal to $v = [1,2,0]$ Find two vectors orthogonal to $v = [1,2,0]$

Usually i see questions with asking you two find given two vectors find two orthogonal vectors for it. Then you would use cross product and then use the result to find the unit vector.

What i do not understand is how would i do this for a single vector if i'm trying to find two vectors orthogonal to it

Denote "$v$ is orthogonal to $w$" by $v\ \bot\ w$.  Then we define orthogonality by $$v\ \bot\ w \iff v\cdot w=0$$  where $v\cdot w$ is the dot product of $v,w \in \Bbb R^n$.

So a vector $(x,y,z)$ is orthogonal to $v$ if $$(x,y,z) \cdot (1,2,0)=x+2y=0$$

Clearly there are no restrictions on $z$ so you can pick any value of $z$.  But $x$ and $y$ are related by the equation $x=-2y$.

So just pick any values of $x$'s (or $y$'s) and $z$'s and use that equation to find the last coordinate of suitable vectors.

For instance $(2,-1,0)$, $(-10,5,3)$, $(2\pi,-\pi, e)$, $(0,0,\frac {11}{3})$, etc are all vectors orthogonal to $v$.  Confirm this for yourself.

"Orthogonal" also means perpendicular so all you would need is two vectors perpendicular to $v = [1,2,0]$, so the answers could be -v which is $v = [-1, -2, 0]$ and $[0,0,1]$


$\sup(a + B) = a + \sup B$ I believe my proof of this simple fact is fine, but after a few false starts, I was hoping that someone could look this over. In particular, I am interested in whether there is an alternate proof.

  For a real number $a$ and non-empty subset of reals $B$, define $a + B = \{a + b : b \in B\}$. Show that if $B$ is bounded above, then $\sup(a + B) = a + \sup B$.

My attempt:

  Fix $a \in \mathbb{R}$, take $B \subset \mathbb{R}$ to be nonempty and bounded above, and define
  $$a + B = \{a + b : b \in B\}.$$
  Since $B$ is nonempty and bounded above, the least-upper-bound axiom guarantees the existence of $\sup B$. For any $b \in B$, we have
  $$b \leq \sup B,$$
  which implies
  $$a + b \leq a + \sup B.$$
  As this is true for any $b \in B$, it follows that $a + \sup B$ is an upper bound of $a + B$, and hence $\sup(a + B)$ exists, by the completeness axiom, since $B \neq \emptyset$ implies immediately that $a + B \neq \emptyset$. I claim that $a + \sup B$ is in fact the least upper bound of $a + B$. As we have already shown it to be an upper bound, it suffices to demonstrate that $a + \sup B$ is the least of the upper bounds. Let $\gamma$ be an upper bound of $a + B$. Hence, for any $b \in B$,
  $$a + b \leq \gamma,$$
  which implies that
  $$b \leq \gamma - a.$$
  As this holds for all $b \in B$, $\gamma - a$ is an upper bound of $B$. Hence, by the definition of supremum,
  $$\gamma - a \geq \sup B,$$
  which implies that
  $$\gamma \geq a + \sup B,$$
  as desired.

I tried to write the proof initially be showing that $\sup(a + B) \leq a + \sup B$ and $\sup(a + B) \geq a + \sup B$, but didn't have any luck. If there is a trick to it, I would be interested in hearing it.

What you've done looks correct to me, but I think we can rework it more concisely using exactly the strategy that you suggest at the end of the question. Note that both suprema exist since the sets are non-empty.

First direction: Let $\lambda \in a + B$. Then $\lambda = a + b$ for some $b \in B$. Since a supremum is an upper bound, $b \leq \sup B$, so $\lambda \leq a + \sup B$. Since $\lambda \in a + B$ was arbitrary, $a + \sup B$ is an upper bound for $a + B$, hence $\sup(a + B) \leq a + \sup B$.

At this point it might be worth pausing to try the other direction yourself - the idea is similar, so it would be a good test of understanding.

Other direction: Let $b \in B$. Then $a + b \in a + B$, and since a supremum is an upper bound, $a + b \leq \sup(a + B)$. Rearranging, $b \leq \sup(a + B) - a$, so $\sup(a + B) - a$ is an upper bound on $B$, and hence $\sup B \leq \sup(a + B) - a$, and it follows that $\sup(a + B) \geq a + \sup B$.

Conclusion: It follows immediately that $\sup(a +B) = a + \sup B$.

$$\sup\left\lbrace x + y\right\rbrace = \sup \left\lbrace x \right\rbrace + \sup \left\lbrace x \right\rbrace $$
$$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$
For non negative numbers same properties we have for multiplication.

Proof for second:

$ \exists \inf \left\lbrace x \right\rbrace $ and $\exists \inf \left\lbrace y \right\rbrace \Rightarrow  \exists \inf\left\lbrace x + y\right\rbrace $ and $\inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace \leqslant x + y$.

For $ \forall \epsilon > 0$ $ \space \exists (x_1 + y_1) \in \left\lbrace x + y\right\rbrace  $ for which $$\inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace \leqslant (x_1 + y_1) < \inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace + \epsilon $$
This gives
$$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$


If Rolle's Theorem is assumed to be true, doesn't that prove the MVT? If we assume that Rolle's Theorem is true is it practical to say that it also proves the MVT?

My reasoning is that even though Rolle's Theorem is the special case for when $f(a)=f(b)$ and the secant line between $(a,f(a))$ and $(b,f(b))$ is horizontal, doesn't that mean you can 'rotate' and 'stretch' the function $f$ and thus 'stretch' and 'rotate' the straight line between $(a,f(a))$ and $(b,f(b))$ as well as the line tangent to $(c,f(c))$ to maintain its symmetry to the line secant between $(a,f(a))$ and $(b,f(b))$ to thus prove that for the straight line intersecting points $(a,f(a))$ and $(b,f(b))$, there must be a point $c$ on $[a,b]$ for which $f'(c)$ is parallel to the aforementioned line between $(a,f(a))$ and $(b, f(b))$?

I feel that this could be better explained with something like linear algebra.  I do not know it well enough to use it.

EDIT: My question was just as unclear as I thought it was.  I am going to try and explain it differently:

For a function $f$, if it is differentiable/continuous over $[a,b] $/$(a,b)$ etc. then there is a straight line between $(a, f(a))$ and $(b,f(b))$.  You can rotate $f$ around the origin until this straight line is parallel to the x-axis.  By Rolle's Theorem, there must exist a $c$ on the transformed $(a,b)$ such that $f'(c)=0$ and thus there is at least one line tangent to $f$ on $(a,b)$ that is parallel to the secant of $(a,f(a))$ and $(b,f(b))$.  Then, you can 'undo' the original rotation of $f$ and know that the lines proven to exist by Rolle's Theorem exist with the same symmetries with respect to $f$, meaning the MVT must hold even when the secant line is not horizontal.

A proof of MVT used Rolle's theorem explicitly.

Consider a continuous function $f:[a,b]\to\Bbb R$ that is differentiable on $(a,b)$. We are to prove MVT, i.e. to prove there exists a point $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, assuming that Rolle's theorem is true.

Define a new function $g:[a,b]\to\Bbb R$ by $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}x$. Verify yourself that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Verify also that $g(a)=g(b)$. So we can apply Rolle's theorem to this function $g$, to obtain a point $c\in(a,b)$ such that $g'(c)=0$. Note that the derivative of $g$ is $g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$. The point $c$ also satisfies $f'(c)-\frac{f(b)-f(a)}{b-a}=0$.

Not really.
Rolle's theorum is a special case of Legrange's MVT. Not the other way around. So just by assuming Rolle's theorum, MVT is not automatically proved. We do of course use RT in the MVT proof. But just because Rolle's theorum is true, MVT need'nt be.

If you are wondering why we use this special case to prove the MVT, it is because Rolle's theorum appeared first.


Uncorrelated and X given $Y = 0$ Is the following true or false?

Suppose that $X$ and $Y$ are two discrete random variables defined on the same probability space. If $E[X] = E[Y] = 0$ and $E[X | Y=y] = 0$ for all $y\in Y$, then $X$ and $Y$ are uncorrelated.

How does this tell me $E[XY] = 0$?

Note: it does not have to be zero.

Suppose for some constant $c$, $\forall y\in Y(\Omega): \mathsf E(X\mid Y=y)=c$.   That is $\mathsf E(X\mid Y)=c$

$\begin{align}\because
\mathsf {Var}(X,Y) & = \mathsf E(X\cdot Y)-\mathsf E(X)\cdot\mathsf E(Y)
\\ & = \mathsf E(\mathsf E(X\cdot Y\mid Y)) - \mathsf E(\mathsf E(X\mid Y))\cdot\mathsf E(Y)
\\ & = \mathsf E(\mathsf E(X\mid Y)\cdot Y)-\mathsf E(\mathsf E(X\mid Y))\cdot\mathsf E(Y)
\\ & = \mathsf E(c\cdot Y)-c\cdot \mathsf E(Y)
\\ \therefore \mathsf {Var}(X,Y) & = 0
\end{align}$


Patterns in Prime numbers, and the null hypothesis I've read about many attempts to find patterns in prime numbers.  First, is there a mathematical way to prove there is not a pattern to prime numbers?  Since there are ways to check if a number is prime or not, can these methods be combined into some large function that produces prime numbers?  And in case it isn't obvious, no, I never got passed pre-Cal.

I like Dirichlet's Theorem, which states that for relatively prime $a,d \in \mathbb{Z}^+$, there are infinitely many primes in the progression $\{a + nd \mid n \in \mathbb{Z^+}\}$.  Further, the proportion of primes in any relatively prime residue class of $d$ is about $1/\phi(d)$, where $\phi$ is the Euler phi function.

In essence, the primes can be seen as somewhat evenly distributed.  You can see more below.

http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions

(Amateur response. BS in MIS, MS in Curriculum & Instruction. Retired College Computer Instructor / IT Support. No higher mathematics classes. Non-standard expression of formulas and language. If this answer is of any value, I would consider giving a presentation about it.)

Prime Numbers are not random. Prime numbers are a completely patterned phenomenon. Prime numbers are a system of mechanistic chaos. Prime Numbers can be thought of as a collection of moiré-like interference patterns. Prime Numbers are the result of overlaying an infinite quantity of infinitely repeating finite patterns.

For this answer: Prime Numbers are Base$_{10}$ positive integers that have two and only two unequal positive integer factors.

$P_i$ is a specific Prime Number where $_i$ is a positive integer.
For example: $P_1$ = the first Prime Number, $2$. $P_2 = 3$, $P_3 = 5$, $P_4  = 7$, …

$P_i^2$ = the square of a specific Prime Number.
For example: $P_3^2$ = $5^2$ = $25$.

$P_i!$ or Prime Factorial is the multiplication of all Prime Numbers from $P_1$ through $P_i$.
For example: $P_4!$ = $P_1 * P_2 * P_3 * P_4$ = $2 * 3 * 5 * 7$ = $210$

$I$ is any positive integer

Because a Sieve is usually constructed as an array of rows and columns and because the only purpose of the sieve is to find prime numbers, patterns are very difficult, if not impossible, to discern. Instead of an array, construct the Sieve using a linear format.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 …

The integer one ($1$) will be eliminated because it does not have two unequal positive integer factors.

_ 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 …

The next available integer is two ($2$) and by direct examination is found to be Prime - it has two unequal positive integer factors: $1, 2$.

All multiples of $2$ or $P_1$ that are larger than $2$ will be eliminated.

_ 2 3 _ 5 _ 7 _ 9 _ 11 _ 13 _ 15 _ 17 _ 19 _ 21 _ 23 _ 25 _ 27 _ 29 _ 31 _ 33 …

To state the obvious, all Prime Numbers larger than $2$ will be odd. Also, it is important to observe that the first integer to be eliminated is $4$ or $2^2$ or $P_1^2$.

The elimination of multiples of $2$ creates the first pattern. Starting at $P_1$!$ + 1$ or $5$, is an infinitely repeating finite pattern with a length of $2$ or $P_1$!. Within each $P_1$! pattern, one element or $50$% of the pattern has been eliminated. Essentially, for the set of positive integers, $50$% of that set has been eliminated. The repeating pattern for $P_1$! is:

_ $I$

The next available integer is $3$ because it was skipped when the multiples of $2$ were eliminated. By direct examination - and because it was skipped - $3$ is a Prime Number. All multiples of $3$ or $P_2$ that are larger than $3$ and have not already been eliminated will be eliminated. The eliminated integers will be called Strikeouts.

_ 2 3 _ 5 _ 7 _ _ _ 11 _ 13 _ _ _ 17 _ 19 _ _ _ 23 _ 25 _ _ _ 29 _ 31 _ _ _ 35 _ 37 _ _ _ 41 _ 43 …

The first Strikeout is $9$ or $3^2$ or $P_2^2$. The elimination of multiples of $3$ creates the second pattern and it overlays the first pattern. Starting at $P_2$!$ + 1$ or $10$, is an infinitely repeating finite pattern with a length of $6$ or $2 * 3$ or $P_1 * P_2$ or $P_2$!. Within each $P_2$! pattern, one additional element or $16.6$% of the pattern has been eliminated. When combined with the 50% elimination of the previous pattern, then for the set of positive integers, $66.6$% of that set has been eliminated. The $P_2$! pattern is:

$I$ _ _ _ $I$ _

With this pattern, an infinite set of Prime Number groupings - twins, quads, etc. - has been created.

At this point, generalizations about Prime Numbers can be made.

For each Prime Number used to eliminate integers, the next Prime Number will always be known. In fact, the next several Prime Numbers will be known. As it will continue to be shown, Prime Numbers arise from a self-generating iterative process.

Every Prime Number has both a pattern and a Range of Influence (RoI). The RoI extends from the square of the Prime Number to one less than the square of the next Prime Number. Within the RoI, any remaining integers are Prime. For 2, its RoI extends from 4 to 8 - and the remaining integers are 3, 5, 7. For 3, its RoI extends from 9 to 24 - and the remaining integers are 11, 13, 17, 19, 23. Generalized, the RoI extends from $P_i^2$ to $P_{i+1}^2 - 1$. For every Prime Number and after its multiples have been eliminated, within its RoI, any remaining integers will be Prime.

The length of a Prime Number pattern is its Prime Number Factorial: $P_i$!. In progressing through the sequence of Prime Numbers, the Prime Number Factorial rapidly increases in size to extremely large values.

Subsequent Prime Number patterns will be less successful at eliminating integers at a rate that approaches zero. The Prime Number patterns, while repeating infinitely: have a finite length, always yield more Prime Numbers, as a percentage - each new pattern will continue to become less effective at eliminating integers, and no pattern through its entire length eliminates all the Prime Number groupings. Consequently, all Prime Number groupings are infinite.

The next available integer is $5$ because it is a remaining integer within the RoI of  $2$. All multiples of $5$ or $P_3$ that are larger than $5$ and have not already been eliminated will be eliminated.

_ 2 3 _ 5 _ 7 _ _ _ 11 _ 13 _ _ _ 17 _ 19 _ _ _ 23 _ _ _ _ _ 29 _ 31 _ _ _ _ _ 37 _ _ _ 41 _ 43 …

The first Strikeout is $25$ or $5^2$ or $P_3^2$. The elimination of multiples of $5$ creates the third pattern and it overlays the previous patterns. Starting at $P_3$!$ + 1$ or $26$, is an infinitely repeating finite pattern with a length of $30$ or $2 * 3 * 5$ or $P_1 * P_2 * P_3$ or $P_3$!. Within each $P_3$! pattern, two additional elements or $6.66$% of the pattern has been eliminated. When combined with the $66.6$% elimination of the previous pattern, then for the set of positive integers, $73.3$% of that set has been eliminated. The $P_3$! pattern is:

$I$ _ _ _ _ _ $I$ _ _ _ $I$ _ $I$ _ _ _ $I$ _ $I$ _ _ _ $I$ _ _ _ _ _ $I$ _

With the application of each new sieve, a new infinitely repeating finite pattern appears. Starting with the $P_3$! pattern, it is easy to discern that within the pattern – from the first integer through the next-to-last integer – is a symmetrical sequence of remaining integers and of new strikeouts. What can be observed in this pattern and all subsequent patterns is that the midpoint of the pattern is surrounded by a Prime Quadruple candidate.

These patterns and their effects can be found at every $IP_i$! – at every multiple of a Prime Factorial. The patterns become more predictive as $P_i$! becomes larger and the percentage of Strikeouts gets closer to zero.

Prime Number patterns can be used to locate rich sets of Prime Number candidates and the faster and more efficient discovery of yet-to-be-discovered Prime Numbers. This can be illustrated by looking at how the patterns appear adjacent to a very large Prime Factorial.

Create a linear sieve and eliminate all multiples of $2$ or $P_1$:

1 _ 3 _ 5 _ 7 _ 9 _ 11 _ 13 _ 15 _ 17 _ 19 _ 21 _ 23 _ 25 _ 27 _ 29 _ 31 _ 33 …

The sequence of integers is a counting line - enumerating the integers after a large Prime Factorial or after a multiple of a large Prime Factorial. The “$1$” means only that it is the first integer after the Prime Factorial. All the Prime patterns that comprise the Prime Factorial end simultaneously at the Prime Factorial and simultaneously begin again at the next integer after the Prime Factorial. Consequently, both the next integer after the Prime Factorial and the integer immediately preceding the Prime Factorial are Prime candidates. A Prime Factorial and any multiple of a Prime Factorial will always be surrounded by a Prime Twin candidate. The bigger the Prime Factorial, the more likely it is that the Prime Factorial is preceded by or followed by a Prime Number or even surrounded by a Prime Twin. The bigger the Prime Factorial, the less chance there is that either or both of the integers surrounding the Prime Factorial will be eliminated by subsequent patterns.

Eliminate all multiples of $P_2$ or $3$:

1 _ _ _ 5 _ 7 _ _ _ 11 _ 13 _ _ _ 17 _ 19 _ _ _ 23 _ 25 _ _ _ 29 _ 31 _ _ …

The first Strikeout will always be $P_i$ and the second Strikeout will always be $P_i^2$, in this case: 3 and 9. The next few Strikeouts will be the Prime Number multiplied by the following Prime Numbers up to the square of the next Prime Number. In this case, the first few Strikeouts are:

$3 * 1 = 3, 3 * 3 = 9, 3 * 5 = 15, 3 * 7 = 21$

This list stops here because the next multiplication would be $3 * 11 = 33$ which is larger than 25 or $5^2$. Remembering that these integers follow a very large Prime Factorial; the RoI says that any integers remaining within the RoI are Prime candidates.

After the P2 Strikeouts, it is easy to see that there will be infinite Prime Twin candidates and infinite Prime Quadruple candidates and an infinite quantity of candidates for all other Prime Number groupings. Remember, this follows a very large Prime Factorial. Cumulatively, the Strikeouts of all subsequent patterns will be too sparse to eliminate all of the Prime Number groupings.

Eliminate all multiples of $P_3$ or $5$:

1 _ _ _ _ _ 7 _ _ _ 11 _ 13 _ _ _ 17 _ 19 _ _ _ 23 _ _ _ _ _ 29 _ 31 _ _ …

Observe the increasing gap between the first integer on either side of the Prime Factorial and the next remaining integer on either side of the Prime Factorial. The minimum length of this Prime Gap increases in length according to the sequence of Prime Numbers. This Prime Gap can be longer if either or both of the Prime Candidates adjoining the Prime Factorial are eliminated by subsequent patterns. If both of the Prime Candidates adjoining the Prime Factorial are eliminated, then a very large Prime Gap has been discovered.

Observe the increasing length of the stable or consistent pattern between the first and second Strikeouts - between $P_i$ and $P_i^2$. While the pattern segments on either side and nearest the Prime Factorial lengthens according to the sequence of Prime Numbers, the next pattern segment lengthens according to the sequence of the squares of Prime Numbers.

By using a linear format for the Prime Number Sieve, the following observations can be made about prime numbers:

A unique finite pattern that repeats infinitely is associated with each prime number.
Within each pattern, the sequence from the first integer through the next-to-last integer is symmetrical.
Each successive Prime Number pattern has a length equal to its Prime Factorial.
Each successive Range of Influence (RoI) increases in length according to the sequence of Prime Number Squares. The length of an RoI is: $P_{i+1}^2 - P_i^2$.
The RoI of $2$ and $3$ contain complete patterns. All other RoI contain only a pattern segment. When examining only the sequence of Prime Numbers, discerning complete  patterns is impossible. The use of a horizontal sieve allows complete patterns to be discerned.
The pattern of $2$ displays the obvious – only odd integers larger than $2$ can be prime. The pattern of $3$ visually proves that the possible Prime Number Twins, Quads, and all other Prime Number groupings are infinite. Because the subsequent Prime Number patterns have a frequency of strikeouts that rapidly descends to a zero limit and a length that follows the sequence of Prime Factorials, not all Prime Number groupings can be eliminated.
Because the prevailing consensus has been that prime numbers have no foundational or overarching patterns, any patterns that were observed were assumed to be of a singular nature with limited range and usefulness. By using a horizontal sieve, it can be seen that Prime Numbers are derived from an infinite collection of infinitely repeating finite patterns. At least visually, this provides a better and more complete understanding of Prime Numbers. It is hoped that this visualization method will lead to answers for some of the outstanding questions about Prime Numbers.

What can be calculated:
* The length of each Prime Number pattern.
* The quantity and percentage of new or additional integers eliminated within each pattern (NEW STRIKEOUTS or NS).
* The total quantity and percentage of integers eliminated by all patterns (TOTAL STRIKEOUTS or TS) from $P_1$ through $P_i$.
* The total quantity and percentage of the remaining integers not eliminated by all patterns from $P_1$ through $P_i$.

By using Prime Number patterns, which provide natural breakpoints, there can be an exact computation and analysis of quantities, percentages, frequencies, and distributions of Prime Numbers within a pattern. Using breakpoints based on multiples or powers of 10 will yield results that are less than accurate or useful and might even be misleading.

When graphed, the percentage of new integers eliminated within each pattern (NEW STRIKEOUTS or NS) drops substantially with the first few prime numbers and continues to descend to a zero limit.

When graphed, the percentage of all non-prime integers increases substantially with the first few prime numbers then becomes increasingly more level. It has a limit of 100%.

The reason for this rapid leveling is more easily apparent visually than mathematically. The endpoint of the prime gap on either side of $IP_i$! is increasing incrementally through the sequence of Prime Numbers. The beginning of the RoI is increasing incrementally through the sequence of squares of Prime Numbers. Therefore, the distance between the prime gap endpoint and the RoI beginning point (where there are no strikeouts) is also increasing incrementally through the sequence of squares of Prime Numbers and becoming an increasingly longer area of stability. Because of the initial large descent of NS% toward zero percent and perpetual movement toward a near-zero limit, the internal portion of each successive pattern also becomes more stable though more slowly.

With a large value for $_i$, a $P_i$! pattern can be used to locate yet-to-be-discovered individual prime number candidates and prime group candidates that have a high probability of being Prime. A large $P_i$! eliminates in a single procedure almost all of the non-Prime integers and none of the Prime integers. The larger the value of $_i$, the better the probability of an associated integer candidate being a Prime Number. By providing rich sets of Prime candidates, this method will enable the faster discovery of new record-setting Prime Numbers, Prime Number groups, and making a more complete inventory of Prime Numbers.

$P_i$! $± 1$ are prime number candidates. This means that every multiple of a Prime Factorial or $IP_i$! is surrounded by a Prime Twin candidate.
$P_i$! $± P_j$ are prime number candidates, where $_j$ is any positive integer larger than $_i$.
The midpoint of every pattern is surrounded by a Prime Quad candidate:
$(P_i$! $± (P_i$!$/ 2)) ± 2$ and
$(P_i$! $± (P_i$!$/ 2)) ± 4$

The quantity and percentage of Strikeouts can be calculated without a sieve.
NS = NEW STRIKEOUTS PER PATTERN
TS = TOTAL STRIKEOUTS PER PATTERN

$NS_i = NS_{i-1}(P_{i-1} – 1)$
$TS_i = (P_i * TS_{i-1}) + NS_i$

In Excel$365$, when enhanced with xlPrecision (PrecisionCalc.com), calculations can have as many as $32,767$ significant digits. Those with access to applications and computers with larger capabilities and faster computations are invited and encouraged: to continue these iterations and determine when NS% is less than $10^{-5}, 10^{-6}, 10^{-7}$ …; when TS% is greater than $96$%, $97$%, $98$% …; and to compute and share very large Prime Factorials that will make the search for as-yet-undiscovered Prime Numbers be faster and more productive.

The NS% is less than $0.1$% at $P_{31}$ or $127$. The NS% is less than $.01$% at $P_{146}$ or $839$. The NS% is less than $.001$% at $P_{833}$ or $6,397$. The NS% is less than $.0001$% at $P_{5294}$ or $51,713$.

The TS% is greater than $90$% at $P_{55}$ or $257$. The TS% is greater than $91$% at $P_{93}$ or $487$. The TS% is greater than $92$% at $P_{181}$ or $1,087$. The TS% is greater than $93$% at $P_{429}$ or $2,971$. The TS% is greater than $94$% at $P_{1387}$ or $11,489$. The TS% is greater than $95$% at $P_{7388}$ or $74,891$.

For example: Where $_i = 4$
$P_4 = 7$

$NS = NS_{i-1} (P_{i-1} – 1) = NS_{4-1} (P_{4-1} – 1) = NS_3 (P_3 – 1)  = 2 * (5 – 1) = 2 * 4 = 8$

$TS = (P_i * TS_{i-1}) + NS_i = (P_4 * TS_{4-1}) + NS_4 = (P_4 * TS_3) + NS_4 = (7 * 22) + 8 = 154 + 8 = 162$

$P_4$! $= P_1 * P_2 * P_3 * P_4 = 2 * 3 * 5 * 7 = 210$

$NS$%$ = 8 / 210 = 3.8095238095238$%

$TS$%$ = 162 / 210 = 77.1428571428571$%

Finally, those with advanced math skills are invited to convert these observations into a correct mathematical form.


Intuition for spherical coordinates of a highly symmetric point Consider the sphere $\{(x,y,z)\in{\mathbb{R}^3}:x^2+y^2+z^2=1\}$ and the point on the sphere $(x,y,z)=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. Intuitively, I thought "oh, that point lies on the "dead center" of the sphere as far the first octant is concerned", so I thought in spherical coordinates the same point should be $\left(r,\theta,\varphi)=(1,\frac{\pi}{4},\frac{\pi}{4}\right)$, but later when I was getting some inconsistent calculations I came back to this statement and when I applied the formal conversions:
$$x(r,\theta,\varphi)=r\sin(\theta)\cos(\varphi)$$
$$y(r,\theta,\varphi)=r\sin(\theta)\sin(\varphi)$$
$$z(r,\theta)=r\cos(\theta)$$
I got that the point $(r,\theta,\varphi)=\left(1,\frac{\pi}{4},\frac{\pi}{4}\right)$ should actually be $(x,y,z)=\left(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\right)$, and this has been driving me insane! Either there's some really silly arithmetic error I'm making or my geometric intuition is failing me. Which one is it?

Your intuition is failing you.  If you do the conversion, the point $(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3})$ you first need $\cos (\theta) =\frac 1{\sqrt 3}$, which says $\theta\approx 0.955$.  You do have $\phi=\frac \pi 4$

The sphere intersects the $x$, $y$, and $z$ axis at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. To see if a point is in the "dead center" you can find the distance to each of these points and see if they are the same. For $(\frac{1}{2}, \frac{1}{2}, \frac{1}{\sqrt{2}})$, the distances are $1$, $1$, $\sqrt{2-\sqrt{2}}$, so it isn't in the "dead center". However, for $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$, the distances are all $\sqrt{2-\frac{2}{\sqrt{3}}}$.


Ordering real numbers between 0 and 1 by their decimal expansion? Writing only the decimal expansion of real numbers I could order them in this way:

1 < 2 < ... < 9 <
01 < 02 < ... < 09 < 10 < ... < 19 < 20 < ... < 98 < 99 <
001 < 002 < ... < 100 < ... < 200 < ... < 900 < ... < 999 <
...
0000000001 < ... < 9999999999
etc.

But where is my flaw of not counting all the reals ?
I mean the decimal expansion of square root of 2 is infinite, so can’t I just claim that I will count it after infinite number of steps ?

That is exactly Cantor's diagonal argument. Assume for the sake of eventual contradiction there is a list or all reals between 0 and 1 (exclude 1 itself), so you can talk about $r_n$, the $n$-th number of your list. They all start $0.\ldots$. For definiteness, forbid numbers ending in an infinite sequence of 9, as $0.59999\ldots = 0.6$ (this leaves out 1). Consider the number $d$, constructed from the list by making it's $k$-th digit different from the $k$-th digit of $r_k$, for example set it to 1 if the $k$-th digit of $r_k$ is not 1, and 2 if the $k$-th digit of $r_k$ is 1. Then $d$ is certainly in the range, it doesn't contain any 9 (let alone an infinite stretch of them at the end), and should be on our list. But it is different in at least one digit of each number on the list. Contradiction, such a list can't exist.

All numbers have infinite digits.

0.1 = 0.1000...
0.99 = 0.99000...

It's just we have a rule saying if a number ends with infinite zeros, we don't have to write it.

You can change it to 1, by doing so

0.1 = 0.1111...
0.99 = 0.99111...

And then you can not skip the zeros in
0.7000..., because 0.7 now means 0.7111...

Your list will only contain numbers that ends with infinite zeros with finite list size. Any number that doesn't end with infinite zeros will not in it.

Also, infinite digits means every number contains infinite size of data. Rational numbers that have a repeat pattern, therefore can be represented by finite data size are actually special case.


Recognizing when a tower of Galois extensions gives a Galois extension For the easiest case, assume that $L/E$ is Galois and $E/K$ is Galois. Under what conditions can we conclude that $L/K$ is Galois? I guess the general case can be a bit tricky, but are there some "sufficiently general" cases that are interesting and for which the question can be answered?

EDIT: Since Jyrki's reply seems to suggest that there is no general criterion on the groups. Can we say something if we put criterions on the fields? Assume say that $K=\mathbb{Q}$ or $K=\mathbb{Q}_p$?

This isn't an answer, nor is it very general, just an illustration with an example. Let $K$ be a number field, $E/K$ finite Galois, and $L/E$ the Hilbert class field of $E$ (the maximal unramified abelian extension of $E$, which I'll take to be inside some algebraic closure $\overline{K}$ of $K$). Then $L/E$ is Galois by definition, and using its characterizing property, it can be proved that $L/K$ is Galois. Namely, let $\sigma:L\rightarrow\overline{K}$ be a $K$-monomorphism. Then $\sigma(E)\subseteq E$ because $E/K$ is Galois, so $\sigma(E)=E$. Thus $E\subseteq \sigma(L)$, and $\sigma$ sets up an isomorphism between the extensions $L/E$ and $\sigma(L)/E$. I don't mean that $L$ and $\sigma(L)$ are $E$-isomorphic, because, while $\sigma(E)=E$, it needn't be the case that $\sigma$ fixes $E$ pointwise. But, for example, $\sigma(L)/E$ must be Galois, and we have an isomorphism $\mathrm{Gal}(L/E)\cong\mathrm{Gal}(\sigma(E)/E)$ induced by $\sigma$. One can also verify that $\sigma(L)/E$ is unramified because $L/E$ is. So $\sigma(L)/E$ is abelian and unramified. This means $L\sigma(L)$ (compositum inside $\overline{K}$) is abelian and unramified over $E$. By maximality, $L\sigma(L)=L$, which implies that $\sigma(L)=L$. So we can conclude that $L/K$ is Galois.

I don't really know if there is a way to make this formal. Similar arguments can be used to prove that various class fields of $E$ are Galois over $K$. I guess the informal idea is that $L/E$ should be maximal (inside $\overline{K}$) with respect to some properties which are preserved by $K$-embeddings of $L$ into $\overline{K}$ (which necessarily send $E$ onto itself) and which are preserved under compositum. I admit this is vague, but I'm not sure how to make it more precise.

Always.  Galois = normal + separable.  A tower of normal extensions is normal, and a tower of separable extensions is separable.

Edit: That's wrong.  Separability is transitive, but not normality.  See comments below.


From which set does the number $\sqrt[3]{-1}$ belong to? I was trying to draw the function $f(x)=\sqrt[3]{x^2(6-x)}$ by hand (I'm in my first year of engineering; having Calculus I; this drawing is actually an exercise given for my class) and used WolframAlpha to see if I've got it right. Well, I was expecting to miss a few things on my first try, but I didn't even consider the possibility that I would start getting its domain wrong (I thought it was all the real numbers, however, the program said it was all real numbers equal to or below $6$). So, trying to understand what I'd missed, I've concluded that Wolfram must see $\sqrt[3]{-1}$ as only as a complex number when I swore it was at least real (since the equation $x^3 = -1$ has at least a real number that solves it, i.e. $-1$). I mean, if Wolfram is right (and I'm supposing it is), then $f(x=7)$ does not return the "real number" $\sqrt[3]{-49}$ (I don't even know if this number is real anymore, and I'm getting more and more confused whilst writing this text), and my professor has put a wrong graph as an answer for the exercise above. Look, although I wanted the ability to understand what I'm not understanding so that I could be, at least, didactic, I'll try to resume my confusion with this question (and with it, I'll try to understand on my own why Wolfram didn't consider numbers like $x=7$ for $f(x)$ domain): what in heavens is this $\sqrt[3]{-1}$?
$\\$
PS: sorry if I've said anything wrong... I don't speak English fluently.

When working in the real numbers, $\sqrt[3]{x}$ denotes the unique number $a$ such that $a^3=x$. Since $(-1)^3=-1$, we have $\sqrt[3]{-1}=-1$.
When working in the complex numbers, the notation $\sqrt[3]{z}$ is imprecise. Every complex number has three cube roots, and there is no natural way of defining the 'principal' cube root of a complex number. It is sensible, however, to define $\sqrt[3]{z}$ in the complex numbers if $z$ also happens to be a real number. Then, $\sqrt[3]{z}$ can be defined in the way above.
That being said, sometimes we define the 'principal' $n$-th root of a complex number in the following way. If $z=re^{i\theta}$, where $\theta$ is the principal argument of $z$*, and $r$ is the magnitude of $z$, then
$$
\sqrt[n]{z} = \sqrt[n]{r} \cdot e^{i\theta/n} \, ,
$$
where $\sqrt[n]{r}$ denotes the principal real root of $r$. This means that $z \mapsto \sqrt[n]{z}$ is not a continuous function, but we still might choose to adopt the notation $\sqrt[n]{z}$ as a matter of convenience. This is probably the root that Wolfram Alpha was referring to:
$$
\sqrt[3]{-1} = \sqrt[3]{1e^{i\pi}} = \sqrt[3]{1} \cdot e^{(i\pi)/3} = e^{(i\pi)/3} \approx 0.5 + 0.866i \, .
$$

*Again, the 'principal' argument of a complex number involves a branch cut, where we require that $\theta \in (-\pi,\pi]$. Again, this does not define a continuous function.

You are completely right. $x^3=a$, with $a<0$ has a real root, so the problem must lay in the way you are writing it in Mathematica. Try using the function CubeRoot, and see if it works. If not,  I would advise you to share your code in the question.


How to get a basis of $U \cap V$ where U and V are the column space of $A$ and $B$.? Problem:

  Let $A = \begin{bmatrix}5& 2& -1\\3& 1& 0\\ -1& 0& -1 \end{bmatrix}$,
  $B = \begin{bmatrix} 4& -3\\ -2& 3\\ 1& -2\end{bmatrix}$, $U = C(A)$
  and $V = C(B)$ where U and V are the column space of $A$ and $B$. Find a basis of $\ U \cap V$.

What I have done:
Firstly, I calculate the bases of U and V, which are $\{ [ 5, 3, 1 ]^T, [2, 1, 0]^T  \}$ and $\{[4, -2, 1]^T, [-3, 3, -2]^T\}$.Then I want to get the intersection of it. But I find what I have done seems like out of logic, which makes me puzzle at how to solve it.
SO I check the answer:

  $C(A)\cap C(B) = span{(1,1,−1)}$
  Hint:  solve the system $\begin{bmatrix}A& B \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix}
= 0$ and use the fact that any nonzero solution will give an element in the intersection, namely Ax or By. Now just look for the right number of linearly independent elements in the intersection.

I think the hint want me to solve a linear system and get a nonzero solution that is the answer.
Do I think it right?
And I am wondering if there is a better way to get a basis of $C(A) \cap C(B)$?
If not mind, could anyone help me and give some inspirations?
Thanks in advance.

So $U$ is spanned by $\{(5,3,-1),(2,1,0),(-1,0,1)\}$ and $B$ is spanned by $\{(4,-2,1),(-3,3,-2)\}$.

Any vector in $U$ is of the form
$$a(5,3,-1)+ b(2,1,0)+ c(-1,0,1)= (5a+2b-c,3a+b,-a+c), \quad \textrm{for some } a,b,c \in \mathbb{R}$$

and any vector in $V$ is of the form

$$
x(4,-2,1)+ y(-3,3,-2)= (4x-3y,-2x+3y,x-2y), \textrm{for some } x, y \in \mathbb{R}.$$

Any vector in the intersection of $U$ and $V$ can be written in both forms so we must have $(5a+2b-c,3a+b,-a+c)= (4x-3y,-2x+3y,x-2y)$ or $5a+2b-c= 4x-3y$, $3a+b= -2x+3y$, and $-a+c= x- 2y$.  Now,

Add the first and last equations to get $4a+ 3b= 5x- 5y$.
Multiply the second equation by 3 to get $9a+ 3= -6x+9y$.
Subtract: $5a= (-6x+9y)-(5x-5y)= -11x+ 14y$ so $a= -(11/5)x+ (14/5)y$. Then,
$$b= -2x+ 3y- 3a= -2x+ 3y+ (33/5)x- (42/5)y= (23/5)x- (27/5)y$$
and
$$ c= x-2y+a= x-2y-(11/5)x+(14/5)y= -(1/5)x+ (4/5)y.$$
Put those for $a, b$, and $c$ in $a(5,3,-1)+ b(2,1,0)+ c(-1,0,1)$.
Multiply that last equation by 2 and add it to the second equation to get $a+ b+ 2c= -y$ or $y= -a-b-2c$.  Then $x= -a+c+2y= -a+c-2a-2b-4c= -3a-2b-3c$.

Any vector in the intersection of $U$ and $V$ is of the form
$$(-3a-2b-3c)(4,-2,1)+ (-a-b-2c)(-3,3,-2)= (-12a-8b-12c+3a+3b+6c,6a+4b+6c-3a-3b-6c,$$
(…)


Odds of the office raffle being rigged Hello mathemagicians,

We have this drawing for free tickets at my workplace and it just so happens that the same person has won twice in a row and this person has a close personal relationship with the raffle-drawer. I strongly suspect that the "random" raffle is not so random, but just to satisfy my own personal curiosity about how low the chances of this person winning twice in a row is I'd like to do the math but I ran into a snag, with which I'd like your help.

I'm trying to calculate the odds of this person winning twice in a row.

The first drawing had 31 entrants. All entrants have 1 ticket. 2 winners are drawn from this pool.

The second drawing had 25 entrants. All entrants, again, have 1 ticket. 1 winner is drawn from this pool.

What are the odds of this person winning both the first drawing and second drawing? I know it's not as straightforward as $\frac{1}{31}\ast\frac{1}{25}$ but I don't know how to transform the first probability to accurately capture the two drawings for two winners.

Thanks for your advice & expertise.

We might also consider what is the chance that someone (not a specific individual chosen in advance) wins twice in a row.  This is a bit tricky since there are different numbers of entrants.  But assuming that all $25$ people in the second contest were also in the first contest, the probability of someone winning both times is

$$1-\left(\frac{773}{775}\right)^{25}\approx .0626$$

So there is about a $6\%$ chance that someone would win twice in a row.

What my answer is about

I give the correct calculation to answer the question.
I point out that maybe the reason the event seems unlikely is because of the way we look at it.
I do some more calculation to show that similar events may be in fact common.

The correct calculation is the following :

For the first drawing, the probability of winning is the probability of winning the first draw, plus the probability of losing the first draw and winning the second:

$$p_1 = \frac{1}{31} + \frac{30}{31}\times\frac{1}{30} \approx 0.0645$$

For the second drawing, the probability is strightfoward to compute :

$$p_2 = \frac{1}{25} = 0.04$$

The total probability is the product of individual probabilities :

$$P \approx 0.06*0.04 \approx 0.026 = 2.6\%$$

This seems low.

However

When doing probabilities, you must be careful not to introduce bias! In your example, you include only the two drawings he won, but have not considered all the drawings that take place!

I will now show you that the event you consider suspicious might not be.

From your question, I will consider the following statement to be what you find suspicious :

"Mr. x won twice on 3 consecutive draws"

Now, say you have a drawing each week, with on average 30 people in each. Mr x participates every time.

The probability of him winning 2 times in 3 draws are :

$$p = 3\times (\frac{1}{30})^2 = \frac{1}{300}$$

The probability of this event happening in 52 weeks is the following :

$$P = 1- (1- \frac{1}{300})^{50} \approx 0.17 = 17\%$$

The term in parenthesis is the probability of not winning twice in any 3-week period, and there is 50 3-week periods in a year.

Those odds are a little bit less suspicious.

Furthermore

Now, if you are interested in the probability that anyone at some point in the year will win 2 times in 3 draws, the calculation is :

$$ P = 1- (1-0.17)^{30} \approx 0.99 = 99\%$$

(we consider 30 persons playing each week for a year)

It is very likely that someone, at some point, will win twice out of 3 consecutive draws.

Here, I made some assumptions, but the idea is right.


Does $\pi$ contain all possible number combinations? I came across the following image:

Which states:

  $\pi$ Pi

  Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that
  every possible number combination exists somewhere in pi.  Converted
  into ASCII text, somewhere in that infinite string of digits is the
  name of every person you will ever love, the date, time and manner
  of your death, and the answers to all the great questions of
  the universe.

Is this true? Does it make any sense ?

It is not true that an infinite, non-repeating decimal must contain ‘every possible number combination’. The decimal $0.011000111100000111111\dots$ is an easy counterexample. However, if the decimal expansion of $\pi$ contains every possible finite string of digits, which seems quite likely, then the rest of the statement is indeed correct. Of course, in that case it also contains numerical equivalents of every book that will never be written, among other things.

Yes and no.  Yes, any non-repeating infinite sequence can be translated into an ascii representation of random gibberish, which will of course randomly contain everything.  No, that isn't particularly amazing or useful, because whatever message you are looking for is also mistated and refuted an infinite number of times.

(For those that say that it isn't necessarily normal, that is unnecessary as the transformation into ascii can be as complex as you like in order to get the result you desire).


Maximize $x_1x_2+x_2x_3+\cdots+x_nx_1$ Let $x_1,x_2,\ldots,x_n$ be $n$ non-negative numbers ($n>2$) with a fixed sum $S$.
What is the maximum of $x_1x_2+x_2x_3+\cdots+x_nx_1$?

Fixed This is carlop's solution simplifyid, so please upvote that solution.

For $n=3$ we have by C-S or rearrangement

$$x_1x_2+x_2x_3+x_3x_1 \leq x_1^2+x_2^2+x_3^2$$

Adding $2(x_1x_2+x_2x_2+x_3x_1)$ we get

$$3(x_1x_2+x_2x_3+x_3x_1) \leq (x_1+x_2+x_3)^2 \,.$$

For $n \geq 4$ even we have

$$x_1x_2+x_2x_3+....+x_{2k}x_1 \leq (x_1+x_3+..+x_{2k} )( x_2+x_4+...+x_{2k-1})$$

since any term on the LHS appears on the RHS.

Thus

$$\sqrt{x_1x_2+x_2x_3+....+x_{2k}x_1} \leq \frac{x_1+x_3+..+x_{2k} + x_2+x_4+...+x_{2k-1}}{2} =\frac{S}{2}$$

For $n \geq 4$ odd

Since the LHS sum is cyclic, without loss of generality we can assume that $x_{2k+1}$ (otherwise reorder them) is the smallest term. Then $x_{2k+1} \leq x_4$ (here is where we need $n \geq 4$.)

$$x_1x_2+x_2x_3+....+x_{2k+1}x_1 \leq x_1x_2+x_2x_3+....+x_{2k}x_{2k+1}+x_{4}x_1  \leq (x_1+x_3+..+x_{2k} )( x_2+x_4+...+x_{2k-1})$$

Again BY AM_GM you get

$$\sqrt{x_1x_2+x_2x_3+....+x_{2k+1}x_1} \leq \frac{x_1+x_3+..+x_{2k+1} + x_2+x_4+...+x_{2k}}{2} = \frac{S}{2} $$

This is WRONG. It is kept here so that no one else will make the same blunder.

The answer is $\frac{S^2}{4}$. There are $nC_2$ answers that will give this value. All of them are the all-possible permutations of the vector $x=[\frac{S}{2},\frac{S}{2},0,0,\dots,0]^T$. I was able to prove this since I already knew the solution from the comments of user N.S. on user Ross Millikan's answer. So the credit goes to him. I will demonstrate the concepts using $5 \times 5$ matrix. The result is easily generalizable.

Define column vector $x=[x_1,x_2,x_3,x_4,x_5]^T$. Then the function in question can be written as
\begin{align}
f(x)=x_1x_2+x_2x_3+x_3x_4+x_5x_1=x^TQx
\end{align}
where
\begin{align}
Q=\begin{bmatrix}0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 & 0 \\\end{bmatrix}
\end{align}
Note that $Q$ is a permutation matrix and also non-symmetric. For every non-symmetric matrix $Q$, we have $x^TQx=x^TCx$ where $C=(1/2)*(Q+Q^T)$ and is a symmetric matrix. So we have
\begin{align}
f(x)=x^TCx
\end{align}
Also note that
\begin{align}
C=\begin{bmatrix}0 & .5 & 0 & 0 & .5 \\ .5 & 0 & .5 & 0 & 0 \\0 & .5 & 0 & .5 & 0 \\0 & 0 & .5 & 0 & .5 \\.5 & 0 & 0 & .5 & 0 \\\end{bmatrix}
\end{align}
Note that the diagonals on either side of the main diagonal are all .5 and values on top-right and left-bottom corner are also $.5$. Now $C$ is a symmetric matrix and hence
\begin{align}
\lambda_{min}(C)\leq x^TCx \leq \lambda_{max}(C)\text{ $\quad \forall x,$  $x^Tx=1$}
\end{align}
Now $C$ is a doubly stochastic matrix (column sum and row sum is one). So $1$ is always an eigenvalue. Most importantly, $C$ is an irreducible primitive matrix. This follows from perron-frobenius theory. Hence $1 $ is the largest eigenvalue. Now observe that $x=[\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,0]^T$ has $f(x)=x^TCx=1$. Also it has unit norm. In fact, all permutations of this vector (all possible rearrangements of its entries), has the same value for $f(x)$ and they all have unit norm. This $f(x)$ is always maximized in this unit-vectors direction. Now $x^Tx=1$ is not our requirement. But, sum of its elements should be $S$. From the above the discussion, one can conclude that all the elements except two of vector $x$ should be zero and the non-zero entries should be equal if $f(x)$ is to be maximized. By symmetry, it is see to that all permutations of the vector $x=[\frac{S}{2},\frac{S}{2},0,0,0]$ is the only set of vectors that satisfies the sum condition while giving the maximum of $f(x)$.


Find the following limit $\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$ and $\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$ Find the following limits
$$\lim_{x\to 0}\frac{\sqrt[3]{1+x}-1}{x}$$

Any hints/solutions how to approach this? I tried many ways, rationalization, taking out x, etc. But I still can't rid myself of the singularity. Thanks in advance.
Also another question.

Find the limit of
$$\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2}$$

I worked up till here, after which I got stuck. I think I need to apply the squeeze theore, but I am not sure how to.
$$\lim_{x\to 0}\frac{\cos 3x-\cos x}{x^2} = \lim_{x\to 0}\frac{-2\sin\frac{1}{2}(3x+x)\sin\frac{1}{2}(3x-x)}{x^2}=\lim_{x\to 0}\frac{-2\sin2x\sin x}{x^2}=\lim_{x\to 0}\frac{-2(2\sin x\cos x)\sin x}{x^2}=\lim_{x\to 0}\frac{-4\sin^2 x\cos x}{x^2}$$
Solutions or hints will be appreciated. Thanks in advance! L'hospital's rule not allowed.

Revised to avoid l’Hospital’s rule:

Your second one can be finished off like this:

$$\begin{align*}
\lim_{x\to 0}\frac{-2\sin 2x\sin x}{x^2}&=-2\left(\lim_{x\to 0}\frac{\sin 2x}x\right)\left(\lim_{x\to 0}\frac{\sin x}x\right)\\
&=-4\left(\lim_{x\to 0}\frac{\sin 2x}{2x}\right)\cdot1\\
&=-4\;.
\end{align*}$$

Try multiplying the fraction in your first limit by

$$\frac{(1+x)^{2/3}+(1+x)^{1/3}+1}{(1+x)^{2/3}+(1+x)^{1/3}+1}$$

and making use of the identity $(a^3-b^3)=(a-b)(a^2+ab+b^2)$.

When $x$ tends to $0$, $(1+x)^n = nx$
Therefore, $(1+x)^{1/3} = x/3$.
The limit becomes $\lim_{x\to 0} \frac {1+x/3-1}{x} = 1/3$


How to show the solution for $\dot{x}(t)=-cx^2(t)+\frac{x^2(t)}{1+x^2(t)}$ is bounded? How to show the following differential equation has a bounded solution.:

$$
\dot{x}(t)=-cx^2(t)+\frac{x^2(t)}{1+x^2(t)}
$$

where $c>0$.

My try:
Since $\frac{x^2(t)}{1+x^2(t)} \leq 1$ we can write

$$
\dot{x}(t)\leq-cx^2(t)+1
$$

I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.

If you assume at first that x(t) is unbounded as $t \to \infty$, the last term drops out. This leaves us with  $\dot{x}(t) = -cx(t)+1$. Using again that we assume x(t) to be unbounded, the term 1 can be disregarded, so we are in asymptotic approximation left with: $$\dot{x}(t) = -cx(t)^2$$

This has the following solution:

$$x(t) = \frac{1}{c}\left(\frac{1}{t+K_{1}}\right)$$

Clearly, this solution is bounded as $t \to \infty$, so therefore, we are in contradiction with our assumption.

The only problem we run into is that we require some initial condition on $x$. If we assume this initial condition to be bounded, then I think we are fine.

edit: Corrected $\dot{x}(t) = -cx(t)$ into $\dot{x}(t) = -cx(t)^2$


How many functions can be made with the following sets? Lets say we have a function $f:X\rightarrow Y$ where $X=\{a,b,c,d\}$ and $Y=\{x,y\}$. Can we assign each $x\in{X}$ to $y\in{Y}$ in $4^2$ different possible ways? that is, create $16$ possible functions? below is my answer. Is there an easier or quicker method to do this or is my answer sufficient for this type of question?

$$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$

You actually got the right answer for the wrong reason:   you want $\mid Y^X\mid=\mid Y\mid^{\mid X\mid}=2^4=16$.

Though it looks like you listed them correctly.

For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$.  In all we have: $2\cdot 2 \cdot 2\cdot 2 = 2^4 = 16$ ways to do that.

But you've explicitly written the list of all possible function assignments from $X\to Y$, which helps you understand this.

In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: X\to Y$ is equal to $|Y|^{|X|} = m^n$.


Prove $A^{2}=0$ iff $C\left(A\right)=R^{0}\left(A\right) $ I'm learning some Linear Algebra through a University Textbook and I've come across this question which I have a hard time solving:

Let There be a square Matrix A.
Prove that $A^{2}=0$ iff $C\left(A\right)=R^{0}\left(A\right) $

where $R^{0}\left(A\right)$ is the nullspace (kernel).

I just don't understand how to approach this question. I know how to find the four fundamental subspaces given a certain array, but not too much apart from that.

**EDIT: Apparently the term C(A) is not standard, in the textbook it refers to the subspace of the columns of A

Can anyone help? Thanks

Saying $A^2=0$ means that every column of $A$ is in the kernel $R^\circ(A)$, since the columns of $A^2$ are obtained by applying $A$ to the columns of$~A$. In other words it is equivalent to $C(A)\subseteq R^\circ(A)$.

But you cannot conclude from $A^2=0$ that $C(A)=R^\circ(A)$. The rank nullity theorem says that always $\dim(C(A))+\dim(R^\circ(A))=n$ (the size of the square matrix$~A$), and this can even prevent the two subspaces from ever being equal, if $n$ is odd. In any case, one can always (if $n>0$) choose $A$ such that $C(A)\subsetneq R^\circ(A)$; for instance $A=0$, or when $n>2$ any matrix with just one nonzero entry, not situated on the main diagonal.

It might not be as formal as a mathematician would put it, but this is how I would put it:

$Ker(A) =$ $\{x \in R^n: Ax=0\}$
$C(A) =$ $\langle a_i \rangle,$ where $A=[a_1 a_2 ... a_n]$  and thus, $C(A) =$ $\sum_i\alpha_ia_i$ for $\alpha_i \in \mathbb{R}$.

If:

$Ker(A) = C(A)$ $\leftrightarrow$ $\{x = \sum_i\alpha_ia_i \in R^n: Ax=0\}$

$\leftrightarrow$ $\{A (\sum_i\alpha_ia_i)=0\}$

$\leftrightarrow$ $\{A (A\alpha^T)=0\}$

$\leftrightarrow$ $\{A^2\alpha^T=0\}$

$\leftrightarrow$ $A^2=0$, assuming that $\alpha^T \in \mathbb{R}^n$ is a vector of arbitrary constants.

Hope it helps you!


Remainder when $-1$ is divided by $2$ The  two possible ways to find the reminder,

$-1 = -1 \times 2+1$

$-1 = - 0 \times 2-1$

From the above calculation, I have found  different quotients: $+1$ and $-1$. If I am asked to  tell  that if the reminder is greater than $0$ then how can I give the answer?

Remainders are usually defined to be positive. That is, the division algorithm says that for any integers $p$ and $q$, with $q\neq 0$, there exists integers $b$ and $r$ such that

$$p=bq+r$$

where $0\le r < q$.

The $0\le r<q$ is the key part here. As a matter of convenience, we often conform to this by writing, for example, $-1=(-1)(2)+1$, and the remainder is $1$.

You have found something along the lines of modular arithmetic.

Imagine a clock with twelve hours.  On this clock, 9 o'clock = 21 o'clock.

In math form this would be $9=21(\mod{12})$

Imagine a different clock with only 2 hours.

We would then have $$a=2n+a(\mod2)$$The $2n$ would disappear and make no difference because it simply represents an addition of $2$ $n$-times, which disappear in modular arithmetic.


Name of the Number series Is there are name for the number series which has the following pattern

Starts small, rises in the middle and goes down as in

1,2,3,4,3,2,1

To give you a background of why this question came up. I am relearning sorting algorithms and there was one case on quick sort (sorting is based on divide and conquer and the numbers lesser than the pivot goes to the left and the numbers greater than the pivot goes to the right) where if we chose the pivot as the median in every single slice, we get quadratic performance.

So, if the sequence is something like

1,4,2,5,1,4,2

and during each iteration, if the pivot is exactly the median, then in case of 1,4,2 in the second iteration, the left side is 1,2 and right side is empty. Finally the pivot is appended left side before the right side. If, throughout the entire bag of numbers, the left/right side is empty, then we suffer quadratic performance.

I am trying to blog about this and I am unable to give this series any name. Felt that I should leave it to the math experts.

Tried Googling and searching within the stackexchange but couldn't find (probably didnt know what to search for)


What is the simplest method to solve the following system? Find all the integers$( x,y,z,t)$ satisfying:
$$xz=a+uk$$
$$yt+xt-yz=-b+vk$$
$$xt+yz=c+wk$$
where $\gcd(a,b,c)=\gcd(u,v,w)=1$, and $ k\neq 0$

Above simultaneous equation shown below:

$xz=a+uk$

$yt+xt-yz=-b+vk$

$xt+yz=c+wk$

Above has numerical solution:

$(x,y,z,t)=(3,2,1,3)$

$(u,v,w,k)=(2,3,5,1)$

$(a,b,c)=(1,-10,6)$


Relationship between random and non random variables. Let $X$ and $Y$  be random variables and $k$ be a non-random constant.

Assume that $Y = kX$. It would be contradictory to write $k = Y / X$, since $k$ is non-random. In general, this must mean that (some) ordinary manipulations cannot be used when we have expressions mixed with random variables and non-random variables.

I know that $Y$ and $X$ are measurable functions but I am not knowledgable enough to say if it's because of this we cannot write $k = Y / X$.

For instance, consider two non-random functions $y(t)$ and $x(t)$ such that y(t) = k x(t) then we have that $k = y(t) / x(t)$...

Do someone have an explanation for this? Where can I learn more about this?

In this situation it is indeed risky to write $k=Y/X$ but this because random variable $X$ is actually a function that can take value $0$. In that case the RHS is not well-defined. If $X$ is a random variable that does now take value $0$ then there is no objection.

You could argue that on LHS we have a constant and on RHS we have a function, but that is not such a problem. We can just identify $k$ with a constant function on the same domain of $X$ and $Y$.

After all it does not hurt to write equalities like: $$1=\cos^2t+\sin^2t$$where the LHS shows a constant and the RHS a function on $t$.

Your objection "$k$ non-random" has the same character as the objection "$1$ is not a function".

Intuitively:

You can think of the randomness as how $\frac YX$ would be equal to $k$. For example, it could be $\frac {2k}2, \frac{5k}5, \frac{500k}{500}$, but the fraction would surely simplify to k.

Precisely (without measure theory):

All constant functions are random, by the definition of random variables. They just have a 'randomness' of zero, where 'randomness can be' made precise through a variance of zero.

For example, consider a fair coin flip where you get a payoff of 1 unit regardless of the outcome.

Our probability space is $(\Omega, \mathbb P)$

where $\Omega = \{H,T\}$ and $\mathbb P(H) = \mathbb P(T) \le 0.5$

The payoff can be modelled by a constant random variable

$$X=1 \ \text{if} \ \omega = H$$
$$X=1 \ \text{if} \ \omega = T$$

Observe that $E[X] = 1$ which coincides with $E[1] = 1$.

Observe that $Var[X] = 0$ which coincides with $Var[1] = 0$.

Precisely (with measure theory)

$X$ is a random variable in a probability space $(\Omega, \mathscr F, \mathbb P)$ if $\forall \ B \in \mathbb B$, $X^{-1}(B) \in \mathscr F$, that is for whatever value we want from X, the events that give rise to that value are going to be in the $\mathscr F$ of our probability space, where our choices for $\mathscr F$ are either $\mathscr F_0 := \{\Omega, \emptyset\}$ or $2^{\Omega} := \{\emptyset, \Omega, H, T\}$.

The possible $B$'s that we have are:

$\{B \in \mathscr B | 1 \in B \}$ eg $\{1\},(0,\infty),\{1,2,3,...\},\mathbb R$
$\{B \in \mathscr B | 1 \notin B \}$ eg $\{\pi\},(-\infty,0),\{-1,-2,-3,...\},\mathbb Q^C$

For the former kind of $B$'s, $X^{-1}(B) = \Omega \in \mathscr F$ for either of the two choices for $\mathscr F$, that is, the events that give rise to a value of $1$ are $H$, $T$ or $\Omega$ or simply $\Omega$ ('or' is $\cup$).

For the latter kind of $B$'s, $X^{-1}(B) = \emptyset \in \mathscr F$ for either of the two choices for $\mathscr F$, that is, the events that give rise to a value of anything but $1$ do not exist! They are known collectively as $\emptyset$.

Reference: Perhaps Introduction to Mathematical Statistics and Its Applications by Larsen & Marx or in Probability with Martingales by David Williams. If none, there may be some related concept. Also:

https://en.wikipedia.org/wiki/Degenerate_distribution
https://stats.stackexchange.com/questions/192179/


Every mere proposition is a set? In the HoTT Book, lemma 3.3.4 states that "every mere proposition is a set", where a mere proposition is a type $A$ such that $IsProp(A) = \prod_{x,y: A}(x=_Ay)$ holds, and a set is a type $A$ such that $IsSet(A) = \prod_{x,y:A} IsProp(x=_Ay)$ holds.

The proof of this lemma starts with "Fix $x:A$ [insert proof].", but never discharges the $x:A$ (as far as I can see).

So in fact, what the proof does is exhibit an element of type $\prod_{x:A}( isProp(A)\to IsSet(A))$, in other words of type $A\to isProp(A)\to IsSet(A)$ : as far as I can see, this not enough to get an element of type $isProp(A) \to IsSet(A)$ (without excluded middle) : am I missing something ?

If not, how can one fix the proof ?

It does discharge $x$. The statement being proved is $\prod_{x, y: A} \prod_{p, q: x = y} p = q$ (i.e. $\text{IsSet} A$) under the assumption of $f: \text{IsProp} A$. Along the way, the statement $\prod_{y, z: A} \prod_{p: y = z} p = g(y)^{-1} \cdot g(z)$ is proved, but then it's instantiated with $y := x$ and $z := y$ for the last line.

We can think of a proposition multiple ways:

We can think of a proposition of a statement that is either true or false

An example of a statement:

  Peter is eating

This whole statement is either true or false.

Next we can also see this as a proposition composed of one single statement without using any logical connectives.

For example

  Let $Q$ denote the proposition for the statement "Peter is drinking"
  This single proposition can be seen as the truth table of possible combinations of $Q$ being a single proposition that is not a composition of other propositions.

Truth table of Q, with on the left the given input values and on the right the result of the proposition.

Next we can use these non composed propositions in combination with our logical connectives to build more complex statements. This can either be by connecting the $Q$ in different ways or by using other defined statements.
For example

  We compose the following statement
  "Peter is drinking and Peter is not drinking."
  to be the proposition.

  $Q \land \neg(Q)$
  We can also define it with another Letter to denote this statement.

Notice we still are using one "building block" proposition. That is how we defined Q. Yet by using the and connective and the negation we now have a whole different outcome for each combination we originally assign Q to be. This new statement will always be false regardless of the value of the non composed statement of $Q$

Thus we can conclude we have two different composed propositions:

  Q as a proposition composed of one single proposition.
  $Q \land \neg(Q)$ as a proposition composed Q connected to negation of Q using the and connective to connect them $Q$ with $\neg(Q)$. And using $\neg$ to negate the $Q$

This will have the following truth table with possible inputs on the left column and the evaluation of all the possible outcomes of the proposition as whole:

To conclude:
We have a non composed proposition which is a single statement which can either be True or False.

We have a composed proposition without connectives which can have different possible inputs for True or False, yet the result of the statement depends on the input True or false.

We have composed propositions which are multiple propositions that are chained together using logical connectives. For each distinct non composed proposition we have can have input True or False. The chaining of these values will eventualy result in either True or False for the proposition as a whole depending on the given input values.


What is the intuition behind the product rule of probability? Let's say that I have two independent events $A$ & $B$ from two different experiments.

Assume that $P(A) = 3/13$ i.e. Out of the total 13 ways that an experiment can be performed, 3 ways make $A$ happen.

Assume that $P(B) = 7/13$ i.e. Out of the total 13 ways that another experiment can be performed, 7 ways yield $B$.

The product rule says $P(A \cap B) = P(A) * P(B)$. In this case the combined probability is $$P(A \cap B) = P(A) * P(B) = \frac {3}{13} * \frac {7}{13}$$ This makes sense to me that $13 * 13$ is actually the total number of ways that both experiments can be performed together & $3*7$ is the number of ways that $A$ and $B$ together can happen (by the fundamental rule of counting).

But this intuition quickly falls apart if $A$ & $B$ were dependent events of the same experiment. Let's say that experiment can be performed in 13 ways. $P(A) = \frac {3}{13}$ & $P(B) = \frac {7}{13}$ and lastly, $P(B|A) = \frac 13$. Now ,
$$P(A \cap B) = P(A) * P(B|A) =\frac {3}{13} * \frac {1}{3} = \frac {3}{39}$$
So $\frac {3}{39}$ shows 3 ways out of "39" ways the two events of the experiment happen. But the original experiment could only be done in "13" ways, what does 39 here represent?

I can understand if its unclear what I'm trying to convey. In essence, what is the intuition behind the product rule of probability?

Well, you've sort of found out that if your configuration space is $\Omega=\{1,2,...,n\},$ your probability distribution is
uniform (say you're rolling with a fair $n$-sided die as opposed to considering the sum of eyes on two fair six-sided dies, for instance), then any event $A$ such that $\mathbb{P}(A)=\frac{k}{n}$ with $k$ and $n$ co-prime can never be independent of any other non-trivial event - if $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=\frac{kj}{n^2},$  then there needs to be a set in $\Omega$ with probability $\frac{kj}{n^2}$, which can only happen in the above set-up if $n$ divides $kj$.

However, if $A$ and $B$ are events corresponding to different trials of experiments that are essentially the same, then you're back to the situation with a "natural" configuration space of cardinality $n^2$, and thus, you get independent events. This should be somewhat intuitively satisfying: It's not all that clear how many independent sub-events of a single experiment there really should be, unless the experiment involves independent components - say you're rolling two dice, or making measurements on two people and so on and so forth.

You can define indipendence in an other way
$$A \perp B \iff P(A|B)=P(A)$$
Which means that che event B doesn't affect A.
Now if you agree that the definition of
$$P(A|B)=P(A,B)/P(B)$$
Is intuitive, It becomes trivial that
$$A \perp B \iff P(A,B)=P(A)P(B)$$


Does an exponential series expansion exist? I know we can expand a function $f:\mathbb{R} \to \mathbb{R}$ with a power series
$$
f(x) = \sum_{n=0}^\infty a_n x^n,
$$
where we can find find $\{a_n\}$ given $f$ using its derivatives about $0$. What I would like to know is whether an "exponential series" exists for $f$. That is, is there a sequence $\{b_n\}$ such that
$$
f(x) = \sum_{n=0}^\infty b_n e^{x n},
$$
and if so, how would we find $b_n$ given a general $f$? Are there conditions on $f$ for which no sequence $\{b_n\}$ can exist?
Edit: To be clear, $x$ here is real, not imaginary, so this is not a Fourier series, and the $e^{xn}$ functions are not an orthonormal basis. Obviously some trivial series do exist, such as $f(x) = e^x$ for $b_n = \delta_{n1}$, but what about more interesting functions such as $f(x) = e^{x^2}$?

Such an expansion exists whenever $f$ is periodic. That is, say $f(x + 2\pi) = f(x)$ for every $x$. Obviously $2\pi$ is just a convention here. By rescaling, you can make it be whatever you want.
Then the Fourier Series associated to $f$ is the sum
$$f(x) = \sum_{-\infty}^\infty a_n e^{2 \pi i n x}$$
Under relatively general conditions (that are still technical enough for me to not want to mention) such a series exists, but periodicity is the important one. We can see periodicity is necessary since each of the $e^{2 \pi i n x}$ are periodic.
In much the same way that a taylor series lets you write $f$ in terms of its derivatives, a fourier series lets your write $f$ in terms of its "harmonics". Importantly, we still have a nice formula for the coefficients:
$$a_n = \int_{-\infty}^\infty e^{-2 \pi i n x} \cdot f ~ \mathrm{d}x$$
In many contexts, these $a_n$ are called $\hat{f}(n)$, and you'll see that notation scattered through the literature. This is a field that is very deep and very broad, so it should be exciting to get some exposure to it!
You can find a relatively polite treatment of the subject in these notes by Tom Leinster

I hope this helps ^_^


Determine the magnitude of the vector sum Determine the magnitude of the vector sum $V = V1 + V2$ and the angle $θx$ which $V$ makes with the positive x-axis.

I get $B$ to be 37.0414 but I am not sure what to add it do to get the correct angle.

Here's the easiest way: find the components of $V_1$ and $V_2$ explicitly to find $V$.  Once you have the components of $V$, it is easy to get its magnitude and the angle it makes with the $x$-axis.

Let $\theta_1$ be the angle $V_1$ makes with the x axis and $\theta_2$ be the same for $V_2$.

To start, we have
$$
\tan\theta_1=3/6=\frac12
$$
From there, do some quick manipulation to find $\sin\theta_1$ and $\cos\theta_1$.  We have
$$
V_1=(9\cos\theta_1,9\sin\theta_1)=(8.05,4.02)
$$
Now, $V_2$ makes an angle of $180˚-63˚=117˚=\theta_2$ with the x-axis.  So, we have
$$
V_2=(12\cos\theta_2,12\sin\theta_2)=(-5.45,10.69)
$$
From there, we have
$$
V=V_1+V_2=(2.60,14.71)
$$
Now, it's simply a matter of finding
$$
|V|=\sqrt{2.60^2+14.71^2}=14.9\\
\theta_x=\arctan(14.71/2.6)=80˚
$$

The angle is between the two vectors. So it is

$(90-{\arctan{{\frac{1}{2}}}})+(90-63)$,

such that the ${\arctan{{\frac{1}{2}}}}$ is due to the $v_{1}$


Two approaches to the method of integration by substitution I came across two approaches to the method of integration by substitution (in two different books).

Approach I

Let $I=\int f(\phi(x))\phi'(x) dx$

Let $z=\phi(x)$

$\therefore \phi'(x)dx=dz$

$\therefore I=\int f(z)dz$

Approach II

Let $I=\int f(x) dx$

Let $x=\phi(z)$

$dx=\phi'(z) dz$

$\therefore I=\int f(\phi(z))\phi'(z) dz$

My problem: While i can understand Approach I, I cannot understand Approach II. What is the difference between the two approaches. What is the difference in their applicability and usage? I am very confused. Please help.

A concrete example of approach 1 may be something like $\int\frac{1}{1+\sqrt x}\,\mathrm{d}x$ and you make the substitution $x=z^2$ in order to get rid of the square root. In this case our $\phi(z)=z^2$ and $\phi’(z)=2z\,\mathrm{d}z$, this makes our integral solvable by some trivial algebra and is already completely in terms of $z$ without any extra algebraic manipulation. Approach 2 on the other hand noticed that there is a derivative of a function on the outside such as $\int 2x\sin x^2\,\mathrm{d}x$ and one makes the substitution $z=x^2$. Both of these are ways to reverse the chain rule as you may recall $(f(g(x)))’=f’(g(x))g’(x)$, although the second approach is pretty much explicitly reversing the chain rule so is the first one in a different manner.

Neither approach as you have it is quite right, because each is missing the final step. In approach 1, suppose you can find an antiderivative $F(z)$ for $f(z).$  Are you done? No, remember you were looking for an antiderivative for $f(\phi(x))\phi'(x).$ The desired answer is $F(\phi(x)).$ Why? Because by the chain rule,

$$(F\circ \phi)'(x)F'(\phi(x))\phi'(x)=f(\phi(x))\phi'(x).$$

Approach 2 also follows from the chain rule, but it's more complicated. Here it's important that $\phi^{-1}$ exist and be differentiable. If we then find an antiderivative $g(z)$ for $f(\phi(z))\phi'(z),$ then $g\circ \phi^{-1}(x)$ is an antiderivative of $f(x),$ which is what we want. Let's see why: By the chain rule,

$$\tag 1(g\circ \phi^{-1})'(x)= g'(\phi^{-1}(x))\cdot(\phi^{-1})'(x)$$ $$ =f(\phi(\phi^{-1}(x)))\cdot\phi'(\phi^{-1}(x))\cdot(\phi^{-1})'(x).$$

Recalling $(\phi^{-1})'(x)= 1/[\phi'(\phi^{-1}(x))]$ (again by the chain rule), we see the last two factors on the right of $(1)$ cancel, leaving us with $f(\phi(\phi^{-1}(x))) = f(x)$ as desired.


How can you prove that the square root of two is irrational? I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. As squares of integers, both $q^2$ and $p^2$ have an even number of factors of two. $2q^2$ has an odd number of factors of 2, which means it can't be equal to $p^2$.

(Since p and q are coprime integers it is supposed to be in lowest form. But we demonstrated the existence of a fraction having lower positive integral denominator.)
It is proof by Contradiction


Double summation index problem I often meet the following situation:

$$\sum\limits_{n=0} ^\infty \sum\limits_{k=0} ^n f(k)g(n-k)=\sum\limits_{p=0} ^\infty \sum\limits_{q=0}^\infty f(p)g(q)$$

While intuitively this is very clear to me, I'm having problems to rigorously prove this. Could somebody please help me out?

So far my friends and I have come up with the map: $(n,k) \rightarrow (q,p) \text{ with } k\leq n$ where $q(n)=n-k$ and $p(n)=q+n=k$ Therefore we only need to prove  bijection.

Thanks

EDIT: The original problem was motivated by:

$$e^{(L_A+L_B)}=\sum\limits_{n=0} ^\infty \sum\limits_{k=0} ^n \frac{1}{k! (n-k)!} L_1^k  L_2^{n-k}=\sum\limits_{p=0} ^\infty \sum\limits_{q=0}^\infty \frac{1}{p! q!} L_1^p L_2^q=$$

where $L_A$ and $L_B \in \mathcal{B}$ and $\mathcal{B}$ is a Banach space. However, I encountered the same problem when dealing with double Fourier transforms

It seems that it is not the convergence that bothers you, but the description of the same set in two different ways. Draw a figure of ${\mathbb N}_{\geq0}\times {\mathbb N}_{\geq0}$, and everything becomes obvious. Anyway, here is a formal proof (all letters denote nonnegative integers):

For given $n\geq0$ one has
$$\{(k,n-k)\>|\>0\leq k\leq n\}=\{(p,q)\>|\>p\geq0, \ q\geq 0, \ p+q=n\}\ .$$
It follows that
$$\eqalign{\{(k,n-k)\>|\>n\geq0, \ 0\leq k\leq n\}
&=\bigcup_{n\geq0}\{(p,q)\>|\>p\geq0, \ q\geq 0, \ p+q=n\}\cr
&=\{(p,q)\>|\>p\geq0, \ q\geq 0\}\ .\cr}$$

Let us consider the term $f(p)g(q)$, where $p$, $q$ are fixed. The summation on the left-hand side contains the term $f(p)g(q)$ ($k=p$, and $n=p+q$). Moreover, the term appears only once in this summation. This implies the equality.


Proof that $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ is not countable by diagonalization. So, as the title says, I want to prove that $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ isn't countable - by diagonalization.
How can I create such $f$ that I cannot count?
edit:
Let's assume by contradiction that  $\left \{ f\in \mathbb{N}^\mathbb{N}\mid f\text{ is injective} \right \}$ is indeed countable.
Lets mark:
$f_0 = (a_{00},a_{01},...,a_{0n....}$
$f_1 = (a_{10},a_{11},...,a_{1n....}$
$f_2 = (a_{20},a_{21},...,a_{2n....}$
and so on...
Lets create $F' = (a_0,a_1,a_2,...,a_n...$ such that $a_{i}=\begin{cases}
a_{ii}+1 & if\;a_{ii}+1\notin \left \{ a_1,...,a_{i-1} \right \}\\
max\left \{ a_1,...,a_{i-1} \right \}+1 & otherwise
\end{cases}$
Is this build sufficient?

Note that $|S|\ge2^{\Bbb \aleph_0}\gt\aleph_0$  The second inequality by Cantor's diagonalization argument.

The first is not that hard to establish:  given a subset of $\Bbb N$, define an injective map by setting it equal to the identity on the subset, and extending to all of $\Bbb N$ by an element of the symmetric group on the complement with no fixed points.


Mary's current income y years after starting the job At her current job, Mary gets a $1.5$% raise twice per year. What is
  Mary's current income y years after starting the job at starting
  salary of s, in terms of s and y?

My answer is = s $(1.0075)$^2y. However, correct answer is = s$(1.015)$^2y, Why shouldn't I divide the rate by $2$, as it is being compounded twice per year?

Every time Mary gets a $1.5\%$ raise, her salaray is multiplied by $1.015$. After $y$ years, how many times has she got a raise? $2y$ times. Hence:
$$\text{new salary}=\text{old salaray}\cdot1.015^{2y}$$

The rate of interest is $~1.5\%~$ So according to compound interest formula, we can write it as $~(1 + 1.5/100)~$  which will be $~(1 + 0.015) = 1.015~$ and the time is $~6~$  months so yearly twice So we can write $~Y = 2Y~$.

So the final answer will be $~(1.015)^2~y~$.


Why is $-2-3(-1)^{2/3} = -5$? Have I gone mad? Consider this:

$$f(x) = -2-3x^{\frac {2}{3}}$$

Why would $f(-1) = -5$?

Doesn't make sense to me. $f(-1)$ equals $1$ when plugging it in on my calculator. But in my book and when plotting this using google's plot feature (just google this: "-2-3(x^(2/3))" and hover over $-1$), I get $f(-1) = -5$?

How do you define $(-1)^{2/3}$? That is the crux.

We can write  $-1^\frac23$ as $(-1^2)^\frac13$  which is equal to $\sqrt[3]1$ which is 1. So $-2 - 3(1) = -5$.

Note that it doesn't matter what you do first whether the cube root or the square. $\sqrt[3]{-1} = -1$ and ${(-1)}^2 = 1$. So you still get the same answer $-5$


Turning points of parametric curve Find the slope of the curve at $t=\frac{1}{4}\pi$.

$$\begin{cases}x=\sin t+\cos t \\ y=\frac{1}{2}\sin(2t)\end{cases}$$

$$\frac{dy}{dx}=\frac{\cos(2t)}{-\sin t+\cos t}$$

$$\left.\frac{dy}{dx}\right|_{\frac{1}{4}\pi}=\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}$$

Multiplying by the conjugate of the denominator won't help, so I applied L'Hôpitals Rule

$$\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}=\lim_{t\to \frac{1}{4}\pi}\frac{-2\sin(2t)}{-\cos t-\sin t}=\sqrt{2}$$

Is there another way to evaluate this limit?

Let  X=1+3sint   and  Y=2-5cost
determine the turning points of this curve in the interval (0;2)
2.classify the turning points in 4.1 as either maximum or minimum


What does the completed graph of a function mean zab said:

  the Levy metric between two distribution functions $F$ and $G$ is simply the Hausdorff distance $d_C$ between the closures of the completed graphs of $F$ and $G$.

I have difficulty in understanding the sentence. In particular, what does the completed graph of a function $F$ mean?  What are the two sets the Hausdorff distance is applied to?

Is the Levy-Prokhorov metric also similar to Hausdorff metric in some ways?
I haven't figured out a way to  understand the L-P metric either, so If you would reply to the linked question or here, I would appreciate it too!

Thanks and regards!

Sorry Tim, I just saw this post.

The completed graph of a function $F$ is simply a closed connected subset of $\mathbf{R}^2$ that consists of the set $\{ (x,F(x)), x \geq 0 \}$, together with the vertical line segments joining the points $(x,F(x-))$ and $(x,F(x+))$ whenever $F$ has a discontinuity at $x$. (Note that since $F$ is a probability distribution function, it is non-decreasing, so any discontinuities are jumps, and never type II discontinuities.)

Now going back to the original discussion here:

What is the motivation of Levy-Prokhorov metric?

Take two distribution functions $F$ and $G$, and let $\overline{F}$ and $\overline{G}$ represent the respective completed graphs of $F$ and $G$. Then the Levy-Prokhorov metric between $F$ and $G$ can indeed be interpreted as an appropriately defined Hausdorff distance between $\overline{F}$ amd $\overline{G}$.

The Levy distance is

$L(F,G) = \inf \{ \epsilon > 0 : F(x-\epsilon) - \epsilon \leq G(x) \leq F(x+\epsilon) + \epsilon, \forall x \in \mathbf{R} \}$.

In words, it means this: For $(x,y) \in \mathbf{R}^2$, define a square of sides $2 \epsilon$ centred at $(x,y)$ as $S_\epsilon(x,y)$. Then $\{ S_\epsilon(x,y), \forall (x,y) \in \overline{F} \}$ is the "tube" traced out by the square $S_\epsilon(x,y)$ as $(x,y)$ travels along the connected set $\overline{F}$. Now take the smallest $\epsilon$ for which the above "tube" contains $\overline{G}$. This gives the Levy-Prokhorov distance between $F$ and $G$.

Now let
$d_C((x_1,y_1),(x_2,y_2)) := |x_1-x_2| \vee |y_1-y_2|$. This is a metric on $\mathbf{R}^2$ called the Chebychev metric. Under this metric, an $\epsilon$-neigbourhood  $N_\epsilon(x,y)$ of the point $(x,y)$ is simply a square of sides $2 \epsilon$ centred at $(x,y)$. This is the same set as $S_\epsilon(x,y)$ defined above. Let $h_C(A,B)$ be the Hausdorff metric induced by $d_C$ on the space of closed subsets of $\mathbf{R}^2$. Then the $d_C(\overline{F},\overline{G})$ is the smallest $\epsilon$ for which the $\epsilon$-inflation of $ \overline{F}$ i.e., $ \cup \{ N_\epsilon(x,y), (x,y)  \in \overline{F}\}$ contains $\overline{G}$ and vice versa, i.e.,

$\inf \{ \epsilon > 0 :   \overline{G} \subseteq \cup\{ N_\epsilon(x,y), (x,y)  \in \overline{F}\} \text{ and } \overline{F} \subseteq \cup \{ N_\epsilon(x,y), (x,y)  \in \overline{G}\}  \}$.

Note further that strictly speaking, the Levy-Prokhorov metric is equal to

$\inf \{ \epsilon > 0 : F(x-\epsilon) - \epsilon \leq G(x) \leq F(x+\epsilon) + \epsilon \text{ and } G(x-\epsilon) - \epsilon \leq F(x) \leq G(x+\epsilon) + \epsilon, \forall x \in \mathbf{R} \}$,

but since $F$ and $G$ are increasing functions, $\overline{F}$ and $\overline{G}$ are rather special closed sets, and

$\inf \{ \epsilon > 0 : F(x-\epsilon) - \epsilon \leq G(x) \leq F(x+\epsilon) + \epsilon, \forall x \in \mathbf{R} \} = \inf \{ \epsilon > 0 : G(x-\epsilon) - \epsilon \leq F(x) \leq G(x+\epsilon) + \epsilon, \forall x \in \mathbf{R} \}$.

Note: $\cup\{ N_\epsilon(x,y), (x,y)  \in \overline{F}\}$  seemed cleaner than  $\cup_{(x,y)  \in \overline{F}} N_\epsilon(x,y)$.

Since you are talking about distance metrics, I assume the graphs under consideration are weighted. My guess is that you create a complete graph from an arbitrary graph by adding a 0-weight edges where is no edge in the original graph.


Check that that the sequence $(-1)^{n}\left ( 1-\frac{1}{n} \right )$ converge or not. Please check my proof and point the flaw in the proof :)
$(-1)^{n}\left ( 1-\frac{1}{n} \right )< \epsilon $

because n is even the sequence will be

$1\left ( 1-\frac{1}{n} \right )$

$\left ( 1-\frac{1}{n} \right )< \epsilon $

$\frac{n-1}{n}< \epsilon $

$n-1< \epsilon n$

since inequality hold Limit exist and convergence to some number.

we conclude for $|\frac{n-1}{n}|< \epsilon $

If n is odd

$(-1)^{2n+1} = -1$

$ -1\left ( 1-\frac{1}{n} \right )< \epsilon $

$ -1+\frac{1}{n}< \epsilon $

$ \frac{-n+1}{n}< \epsilon $

because the inequality hold limit exist at some number can converge to the same

then $\frac{-n+1}{n}< \epsilon $

But $|\frac{-n+1}{n}\neq |\frac{n-1}{n}|$

and limit is unique.

We conclude no limit exist for this sequence;divergent

How does the inequality $n-1<\epsilon n$ hold?

For example, what if $\epsilon = 0.001$ and $n=1000$? The inequality then becomes $999 < 1$, something you will probably agree is not true. In fact, if $\epsilon \leq \frac12$, then the inequality
$$n-1<\epsilon n$$

will only hold for $n=1$, since for $n\geq 2$, you have $$\epsilon n < \frac 12 n =n-\frac 12n  \leq n-\frac 12 \cdot 2 = n-1$$

i.e. the oposite of your inequality.

What you need to show is not that $1-\frac 1n$ is small (it is not), but that $1-\frac 1n$ is close to $1$. In other words, you have to show that for every $\epsilon>0$ there exists some $N$ such that if $n>N$, then the distance between $1$ and $1-\frac1n$ is smaller than $\epsilon$ (I'll leave it to you to fully formalize this statement).

That, along with showing that $-(1-\frac1n)$ is close to $-1$, is enough to see that the sequence has two convergence points and thus cannot converge.


Propositional formulas 3 What are the difference between:

$A \implies (B \implies C)$, $\ (A \implies B) \implies C$, $\ A\land B \implies C$, $\ A \implies B \land C$, $\ (A \implies B)\land (B \implies C)$?

Why we can not write $A \implies B \implies C$?

And the same questions about:

$A \iff(B \iff C)$, $\ (A \iff B) \iff C$, $\ A\land B \iff C$, $\ A \iff B \land C$, $\ (A  \iff B)\land(B \iff C)$?

Why we can not write $A \iff B \iff C$?

Can someone explain me this ?

You can work out the differences with truth tables, or you can note that $p\to q$ is equivalent to $q\hbox{ OR NOT }p,$ and for the second case, that $p\leftrightarrow q$ is equivalent to $(p\hbox{ AND }q)\hbox{ OR }(\hbox{ NOT }p\hbox{ AND NOT }q$).

Then, once you have established that brackets can change the meaning of an expression, writing something like $A\rightarrow B\rightarrow C$ becomes ambiguous.

For $$A\leftrightarrow B\leftrightarrow C$$ the story is slightly different. There we have that the truth value is independent of the order of evaluation, but both are distinct from the usual interpretation as "$A\leftrightarrow B$ AND $B \leftrightarrow C$".

The relation A -> (B -> C) says that C will come from B if A is present.  This makes no relation between A and B, it can be written as A=Policemen B=naughty boys C=good boys.  The policemen makes naughty boys become good boys.

(A->B) -> C would equate to Policemen make naughty boys, which give good boys as a result.


Finding minimum $\frac{c_1x + c_2y}{x+y}$ I am looking for a good strategy to tackle this non-linear, non-convex optimization problem:

Minimize $$\frac{c_1x + c_2 y}{x+y}$$

such that:

$x, y > 0$

$x + y \leq c_3$

$c_1, c_2, c_3$ are given

$c_3 > 0$

Does anyone have any suggestions?

I know this objective function is non-convex, but I was wondering if there were any smart ways to find (or at least approximate) the global optimum. If you suggest an approximation algorithm, please also share its approximation factor. Computational efficiency is not a major concern for me, but accuracy is.

Currently, my leading (though inelegant) idea is to fix the value of $x + y$ and iterate through possible combinations. In the context of my problem, it is an OK assumption to say there are a finite number of meaningful combinations of $x + y$, so I could conceivable iterate through them, but this would be very inefficient.

In case it helps, there is the option for the additional constraint that $c_1, c_2 > 0$, but it's preferable to retain the freedom for them to be either positive or negative.

Thanks in advance for your help.

Hint: for any $\forall x,y \gt 0$: $$\min(c_1,c_2) \le \frac{c_1x + c_2 y}{x+y} \le \max(c_1,c_2)$$
The bounds are asymptotically approached when $y \to 0$ and $x \to 0$ respectively, but never attained unless $c_1=c_2\,$.

\begin{eqnarray}
\inf_{x>0,y>0, x+y \le c_3} { c_1 x + c_2 y \over x+y } &=& \inf_{r \in (0,c_3]} \inf_{x>0,y>0, x+y = r} { c_1 x + c_2 y \over x+y } \\
&=& \inf_{r \in (0,c_3]} \inf_{x>0,y>0, x+y = r} { c_1 x + c_2 y \over r } \\
&=& \inf_{r \in (0,c_3]} \min_{x\ge 0,y \ge 0, x+y = r} { c_1 x + c_2 y \over r } \\
&=& \inf_{r \in (0,c_3]} \min(c_1,c_2) \\
&=& \min(c_1,c_2) \\
\end{eqnarray}
The fourth line follows because the minimisation of a linear functional
over the convex hull of a finite number of points (in this case, $(r,0), (0,r)$)
can be replaced by the minimisation over those points.


Are these simple field extensions normal? Learning for an upcoming exam on basic algebra I came across the following problem:

Test for $\alpha = \sqrt{2+\sqrt{2}}$ and $\alpha = \sqrt{1+\sqrt{3}}$ whether the field extension $\mathbb{Q}(\alpha)/\mathbb{Q} $ is normal.

I feel, that I should be able to solve it, but I am running against a wall here. Can you explain to me, how one would generally approach this kind of problem?

Best Regards

Take the first one, $\alpha =\sqrt {2+\sqrt 2}$.  Let's find it's minimal polynomial.  We remark that $$\alpha^2=2+\sqrt 2\implies \left(\alpha^2-2\right)^2=2$$  so we have $$\alpha^4-4\alpha^2+2$$  That is effectively quadratic and we can solve to get the roots $$\pm \sqrt {2\pm \sqrt 2}$$.

Letting $\beta=\sqrt {2-\sqrt 2}$ we easily see that $$\alpha \beta = \sqrt 2\implies \beta =\frac {\alpha^2-2}{\alpha}\implies \beta\in \mathbb Q(\alpha)$$

The other instance should follow along similar lines.

Since $\mathbb{Q}$ has characteristic zero, you only have to check that $\alpha$'s minimal polynomial has no multiple roots.


How to prove that all odd powers of two add one are multiples of three For example
  \begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}

I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?

Since $2 \equiv -1 \pmod{3}$, therefore $2^{k} \equiv (-1)^k \pmod{3}$. When $k$ is odd this becomes $2^k \equiv -1 \pmod{3}$. Thus $2^k+1 \equiv 0 \pmod{3}$.

There is exactly one integer that divides three among any three consecutive integers. Since $2^{61}-1$ is prime, $2^{61}$ trivially is not a multiple of $3$, $2^{61}+1$ is a multiple of $3$.


Two questions in spectral theory: the spectrum of the Fourier transform and the Hamiltonian of the hydrogen atom. I have the following two questions:

The Fourier transform defines a unitary (provided that it is normalized properly) map $\hat{\cdot}:L^2(\mathbf{R})\rightarrow L^2(\mathbf{R})$. I figured out its point spectrum, which is very easy; is it possible to determine the whole spectrum of this operator? I know $\sigma_p(\hat{\cdot})=\mu_4(\mathbf{C})$ ($4$-th roots of unity) already; is it the case $\sigma(\hat{\cdot})=\mu_4(\mathbf{C})$ also, maybe because $\sigma_p$ might be dense in $\sigma$?
Is there is a concise way (not taking more than two pages say) way to see that the closure $H$ of the Hamiltonian operator of the hydrogen atom (defined on $C^\infty_0$), viz.
$$-\frac{1}{2}\Delta-\frac{1}{\|x\|},$$
has domain $H^2(\mathbf{R}^3)$, is self-adjoint, and determine the spectrum.

Remarks on 2: I found a reasonably short proof of self-adjointness in Reed/Simon's Methods of modern mathematical physics, vol. II, Thm. X.15.

I look forward to your answers.

The spectrum is not as stated. The spectrum of the Hamiltonian for the non-relativistic Hydrogen atom has eigenvalues corresponding to the bound states of Hydrogen, and these are negative. The positive spectrum corresponds to unbound states, and is a continuous spectrum. In spherical coordinates, for a function which depends on the radius $r$ only, one has
$$
\begin{align}
   (-\Delta -\frac{1}{\|x\|})f & = -\Delta f -\frac{1}{r}f \\
             & = -\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{df}{dr}-\frac{1}{r}f.
\end{align}
$$
There is a purely radial eigenfunction of this operator which corresponds to the ground state of the Hydrogen atom. This eigenfunction has the form $f(r)=e^{-r/2}$ in this case where the physical constants are missing:
$$
\begin{align}
    (-\Delta-\frac{1}{r})e^{-r/2}
      & = -\frac{1}{r^{2}}\frac{d}{dr}(r^{2}(-\frac{1}{2})e^{-r/2})-\frac{1}{r}e^{-r/2} \\
      & = -\frac{1}{4}e^{-r/2}+\frac{1}{r}e^{-r/2}-\frac{1}{r}e^{-r/2}=-\frac{1}{4}e^{-r/2}.
\end{align}
$$
Therefore $-1/4 \in \sigma_{P}(-\Delta-\frac{1}{|x|})$. There are lots of other negative eigenvalues and eigenfunctions for this operator.

I already answered the first part of your question in the comment section. The spectrum of the Fourier transform is $\{ 1, i, -1, -i \}$.

Final Note: I see that you changed the question again. This was a complete and valid answer to the original question, and the answer is not a trivial one. Before that, the last paragraph above was a complete answer to the question. HAMMER, if you don't like the answer, it's best to ask a separate question, rather than invalidate a perfectly good answer that someone has put thought and effort into. You obviously learned something from my answers, and it's not polite to invalidate that in such a way that it makes my answer look completely irrelevant to the topic.

The answer to the first question is simply the following: take $f(z)=z^4$ in the spectral mapping theorem and use $\mathfrak{F}^4=1$.


Significance of negative eccentricity For any conic section, the eccentricity is the ratio of the distance to the focus and directrix. If the eccentricity were defined to be negative, would this have any significant meaning/application, or potentially be linked to complex numbers?

It actually doesn't change anything; if you look at the equations for finding the eccentricity, it is the square root of something, and since there are two square roots of a number, it can be both negative or positive.

If you make a plot of a conic section based on the eccentricity, it doesn't change at all. Plot it on desmos graphing calculator using the formula r=$\frac{d}{1-dcos(\theta)}$ where d is the eccentricity (polar coordinates). Nothing happens if you negate the eccentricity.

formula for eccentricity

We've been studying this exact question for all of 5 minutes. We have an answer. No.

This is why: The formula for eccentricity is $c/a$ which is the distance from the center to a foci divided by the distance from the center to a vertex. If eccentricity were negative, either the numerator or denominator would have to be negative. It is IMPOSSIBLE to have a negative distance for $a$ or $c$. Therefore, eccentricity cannot be negative.


Proving $a^2+b^2=c^2$ where $a,b,c$ are side lengths of a right triangle. Proving $a^2+b^2=c^2$ where $a,b,c$ are side lengths of a right triangle.

First, I have never done a proof before, sorry I am so poor here. I have spent many hours but my actions have mostly used wrong ideas. I will only say what I have done that I think is right. If you need my other writing I can write it out but it was mostly about fitting shapes inside the triangles and I think the big mistake is that if the equal statement $a^2+b^2=c^2$ were to be wrong then anything about the inside triangles cannot prove the statement to be wrong because  the algebra is connected to the geometry of the triangle and the shapes are inside the geometry of the triangle.

Start: I started by drawing a square with sides $d,e,f,g$ all the same as a.
The biggest possible right triangle has two sides the same length. This square can fit at least two right triangles inside it both no smaller than the other with one side equal to a. That is true because if you made one smaller than the other and pushed it into one corner of the square then the bigger triangle touching it would overlap the corners of the square which is wrong.

So then the sum $d+e+f+g=4a$ is more than $a+b+c$ where $a,b,c$ are the side lengths of one of the triangles and $4a>a+b+c$. Then I square

$$\begin{eqnarray*} 16a^2&>& (a+b+c)^2\\
&=&a^2+b^2+c^2+2(ab+bc+ac)\\
\end{eqnarray*}$$

The only thing I can do here is use the equation at the top so if $a^2+b^2=c^2$ isn’t true for the triangle then it’s $a^2+b^2=c^2+z$ for some $z$. But I subtracted that from the expression and it wasn’t wrong which I thought it would have to be since I added the $z$. I guess the extra $z$ can’t show the equation is wrong since any number could be equal to it including those like zero which it is not really equal to atleast if the equation were wrong. But if there is a rule that $z$ isn’t zero included somehow there should be a way to conclude having $z$ not be zero is wrong. But then I checked and found $a,b,c,z$ with $z$ not zero so that the inequality holds so I’m very confused. Maybe the squares are too simple and I should have chosen a rectangle or put squares around the outside of the triangle. I know a book that proves this but I want just some clues whether my actions are appropriate to solving this and otherwise a clue for a new way since I don’t care about it for a class I just like to think about the question.

Edit. I have corrected (a+b+c)^2 I think. I am also grateful for the answer below which gives a good clue and just as much as I wished for. It is certainly an interesting choice of shapes and different from what I was doing in that there are no gaps between the shapes. It helps clarify that as I am asking about ways that work I am also asking if an approach using gaps between shapes and estimations rather than exactness could work as an approach (perhaps just because I have spent some hours and wish to recover some value). Is that known? Or perhaps the notion of grouping proofs into approaches is just because the mind is simple and we can't talk about the correctness of approaches only whether one proof we are handed is right or wrong.

This is my favorite proof of the Pythagorean theorem. The algebra for this proof is really easy and straight forward, but it is harder to see the actual proportions that are stated in the Pythagorean theorem, if you want a proof that is clear from a visual stand point then I would look at the Lattice proof.

From a geometry standpoint it is important that you note WHY you are able to arrive to your conclusions in your proofs. In the first proof I mention, our proof is dependent on being able to show that the rhombus with sides c is in fact a square. We are able to do this because we are assuming that the sum of the angles in a triangle are 180 degrees, and that is a consequence of Euclid's 5th axiom. This is why the Pythagorean theorem fails for non-euclidean geometries.

This is a special case of the Law of Cosines, namely one where the cosine is zero because the angle between two of the sides is 90 degrees. Go from the Law of Cosines, and when you simplify the specific case of theta = 90 degrees, a^2+b^2=c^2 will line up.