Permutations and school timetable If there are 6 periods in each working day of a school. In how many different ways can one arrange 5 subjects such that each subject is allowed at least one period? I tried this way- One of the six periods can be arranged in 5 ways and the remaining 5 periods in 5 factorial ways. Totally 600 ways

We assume that, sadly, there are no "spares." So some subject is taught twice, and the others once each.

Choose the subject that gets extra attention. This can be done in $5$ ways. For each such choice, the periods that this favoured subject gets can be chosen in $\binom{6}{2}$ ways, and for each choice the other subjects can be arranged in the remaining slots in $4!$ ways.

That gives  a total of $5\cdot\binom{6}{2}\cdot 4!$ ways.

Remark: For visualization, I like to think as follos. Imagine that the subjects are A (Art), B (Botany), C, D, and E. A timetable can be thought of as a word of length $6$ over this alphabet, in which one letter (subject)  occurs twice and the remaining letters occur once each. Our task is to count the number of such words. The letter that occurs twice can be chosen in $5$ ways. The $2$ positions it occupies in the word can be chosen in $\binom{6}{2}$ ways, and then the $4$ remaining letters can be put in the vacant positions in $4!$ ways.

There can be 5 cases as the mentioned case below
S1,S2, S3, S4, S5, S1 which can be arranged in 6!/2! ways.
Required no of ways will be
5× 6!/2! =1800


One question to know if the number is 1, 2 or 3 I've recently heard a riddle, which looks quite simple, but I can't solve it.

  A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "Yes", "No", or "I don't know," and after the girl answers it, he knows what the number is. What is the question?

Note that the girl is professional in maths and knows EVERYTHING about these three numbers.

EDIT: The person who told me this just said the correct answer is:

   "I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"

"I am thinking of a number which is either 0 or 1. Is the sum of our numbers greater than 2?"

Let the boy ask the girl "Subtract 2 from the number your are thinking and then take the square root of the result.Is the result positive?".
If the girl replies "yes" then the number is 3 because square root of 3-2=1 which is positive.
If the girl replies "no" then the number is 2 because square root of 2-2=0 which is not positive.If the girl replies "I don't know " then the number is 1 because square root of 1-2= is imaginary number which cannot be said whether it is positive or negative.
                                   OR

Let the boy ask the girl"Divide the number you have with the previous number.Ask if the result is a fraction?"If the girl replies "yes" the number is 3 because 3/2 is a fraction.If the girl replies "no" then the number is 2 because 2/2 is not a fraction.If the girl replies "I don't know" then the number is 1 because 1/0 is undefined.


For a semigroup $\{T(t)\}_{t \geq 0}$, $x=0$ is the only entire element if $T(s)=0$ for some $s>0$ This question is asked in a PDE course, but is primarily focused on semigroup theory.

On Banach space $X$, let $\{T(t)\}_{t \geq 0}$ be a strongly continuous semigroup with generator $A:D(A) \subset X \rightarrow X$.

We call an element $x \in \cap_{k=1}^{\infty} D(A^k)$ entire if $\sum_{n=0}^{\infty}\frac{|t|^n}{n!}\|A^n x\|_X < \infty$ for all $t \in \mathbb{R}$.

($D(A^k)=\{x \in D^{k-1} \ | \ A^{k-1}x\in D(A)\}$, inductively defined for integers $k\geq2$.)

Show that $x=0$ is the only entire element if $T(s)=0$ for some $s>0$.

Questions which I've already proven before this question:

I've already proven that $T(t)x=\sum_{k=0}^{\infty}\frac{t^k}{k!}A^kx$ for all $t \geq 0$ and $x\in X$ entire.
Also I've proven that if $x \in D(A^k)$, then $T(t)x \in D(A^k)$ for all $t\geq 0$.

I don't know whether these are relevant, but I posted it anyway. I keep trying to use continuity, or linearity or the semigroup properties to derive a contradiction, but it won't work. Can someone please help me? It bothers me that I can't seem to solve it when it looks kinda simple.

Let me just expand the answer by Tim.

Given a Banach space $E$.

Consider a semigroup:
$$T:\mathbb{R}_+\to\mathcal{B}(E):\quad T(t+t')=T(t)T(t')$$

By the assumption:
$$T(t_0)=0\quad(t_0>0)\implies T(t)=0\quad(t\geq t_0)$$

Regard the entire functions:
$$F_\varphi(\zeta):=\sum_{k=0}^\infty\frac{1}{n!}\varphi\left(A^nx\right)\zeta^n\quad(\varphi\in E')$$

By the identity theorem:
$$F_\varphi(t)=0\quad(t\geq t_0)\implies F_\varphi=0$$

By Cauchy's theorem:
$$F_\varphi=0\implies\varphi(A^kx)=0\quad(k\in\mathbb{N}_0)$$

By Hahn-Banach:
$$\varphi(1x)\quad(\varphi\in E')\implies 1x=0$$

Concluding the assertion.

Okay, so I found a answer to my own question. I however don't know whether this is the right way to post it. Correct me if I'm wrong.

Anyway, here is the answer:

$x=0$ is entire, which is easy to prove.

Assume $x \neq0$ is entire.

$T(s)=0$ for some $s>0$. Hence for all $t \geq s$: $T(t)=T(t-s+s)=T(t-s)T(s)=T(t-s)0=0$.

So for all $t \geq s: \, T(t)x=0=\sum_{n=0}^{\infty}\frac{t^n}{n!}A^n x $ (1).

Note $x \neq 0$ and $t>0$. Now we look at (1) as a 'polynomial of degree $\infty$' which has uncountable many zeros. This is either a contradiction with the number of zeros a polynomial can have (main theorem of algebra), or we have that for all $n \in \mathbb{N}$: $A^n x=0$. Which is to say that $y=0= \cap_{k=1}^{\infty} D(A^k)$. Hence 0 is the only entire element.


Tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$ I am trying to show that the tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$. This is from an exam, where there is a hint stating that this is more than showing that $TS^2$ is non-trivial.

I know how to show the hairy ball theorem, according to which $TS^n$ is non-trivial iff n is even.

I also know that a vector bundle $\pi:E\rightarrow M$ of rank $m$ on a smooth manifold $M$ is trivial (by definition) iff there exists a diffeomorphism $f:E\rightarrow M\times \mathbb{R}^m$ such that for every $p\in M$, $f$ induces a vector space isomorphism $f:\pi^{-1}(p)\rightarrow \{p\}\times \mathbb{R}^m$.

So I see that showing that $TS^2$ is non-trivial only guarantees that $TS^2$ is not diffeomorphic to $S^2\times \mathbb{R}^2$ via a diffeomorphism satisfying the property above, but it is not enough to conclude that there isn't any diffeomorphism.

How can I show this then?

Here's a proof that relies on a theorem of Kervaire with a somewhat-involved proof (I think the particular case of $n=k=2$ is probably much easier, but I couldn't find such a proof to confirm my suspicions.)

  Theorem: The normal bundle to an $n$-sphere embedded inside $\Bbb R^{n+k}$ is trivial if $k> \frac{n+1}2$

This implies any 2-sphere embedded in $S^2 \times \Bbb R^2$ has a tubular neighborhood isomorphic (as fiber bundles) to $  S^2 \times \Bbb R^2$, since $S^2 \times \Bbb R^2$ embeds as an open manifold of $\Bbb R^4$. If $f:TS^2\rightarrow S^2 \times \Bbb R^2$ is a diffeomorphism, then the image of the zero section has a trivial tubular neighborhood  $N$ by the above disscussion, but then $f^{-1}(N)$ is a trivial tubular neighborhood of the zero section in $TS^2$.

References:

Massey's  On the Normal Bundle of a Sphere Imbedded in Euclidean Space
Kervaire's An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant

I think that if suppose that exist a diffeomorphism $F$ then there is an correspondence between sections in $S^{2}\times\mathbb{R}^{2}$ and vector fields on $TS^{2}$. Take the constant section $s(x)=(x,(1,0))$ and by diffeomorphism there is $X\epsilon\mathfrak{X}(S^{2})$ such that $dF(s)=X$, but this section vanishing in south pole under stereographic proyection, that means $dF$ isnt isomorphism


Topological supremum of family of linear topologies In Hans Jarchow - Locally Convex Spaces 2.4.4 (c) it says:

The topological supremum of any family of linear topologies on a fixed vector space is linear.

I couldn't find a proof in the book and neither succeeded to proof it myself, any ideas?

OK, I didn't thought about it the way Tom Cruise mentioned, the supremum as the topology generated by the union.  I guess it's pretty much straight forward now, let's say we have the linear topologies $\tau_\alpha$ on the vector space $X$ and $\tau$ is the supremum of the $\tau_\alpha$, i.e. the topology generated by sets of the form $\cap_{i<n} A_{\alpha(i)}$ with $A_{\alpha(i)}\in\tau_{\alpha(i)}$.

To see that $\tau$ is linear we have to check that it is translation invariant and contains a neighbourhood base ${\cal U}$ around $0$ of balanced, absorbing sets such that $$\forall U\in{\cal U}.\exists V\in{\cal U}.V+V\subseteq U.$$

The basic open sets are translation invariant since $\cap_{i<n}A_{\alpha(i)}+x=\cap_{i<n}(A_{\alpha(i)}+x)$ and one can show similarly that the collection of arbitrary unions of basic open sets, i.e. $\tau$, is translation invariant as well.

Then choose neighbourhood bases ${\cal U}_\alpha$ of $0$ in $\tau_\alpha$ with the properties mentioned above and set ${\cal U}$ as the collection of sets $\cap_{i<n}U_{\alpha(i)}$ with $U_{\alpha(i)}\in{\cal U}_{\alpha(i)}$.  Obviously this is a neighbourhood base in $\tau$ around $0$ which is also balanced and absorbing and if we choose $V_{\alpha(i)}\in{\cal U}_{\alpha(i)}$ with $V_{\alpha(i)}+V_{\alpha(i)}\subseteq U_{\alpha(i)}$ then we get $$\cap_{i<n} V_{\alpha(i)}+\cap_{i<n} V_{\alpha(i)}\subseteq \cap_{i<n} U_{\alpha(i)}$$ which completes the proof.

Jarchow does not provide a proof of Proposition 2.4.4(c) because it follows directly from Proposition 2.4.1. Indeed, if $\{\mathscr T_j\}_{j \in J}$ is an arbitrary collection of linear topologies on $E$, then the topological supremum of this family is exactly the initial (i.e. "projective", "weak") topology induced by the family of linear maps $\{\text{id} : E \to (E,\mathscr T_j)\}_{j\in J}$.

Added later: my answer was downvoted without any kind of explanation, so I'm guessing that the real problem lies not with the deduction of Proposition 2.4.4(c) but with the proof of Proposition 2.4.1.
The proof is given directly before the statement of Proposition 2.4.1, and it is very terse. It is therefore understandable that one may have missed or misunderstood it. Here is an expanded version of Jarchow's argument:
Proposition 2.4.1. Let $E$ be a vector space, and let $\{T_j\}_{j\in J}$ be a family of linear maps $T_j : E \to E_j$, where each $E_j$ is a tvs. Then the initial topology on $E$ for the maps $\{T_j\}_{j\in J}$ is a linear topology.
Proof (following Jarchow's argument). We have to prove that addition $S_E : E \times E \to E$ and scalar multiplication $M_E : \mathbb{K} \times E \to E$ are continuous with respect to the corresponding product topologies. Jarchow proves this simply by pointing out the identities $S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$. The implication here is that these identities show that $T_j \circ S_E$ and $T_j \circ M_E$ are continuous for all $j\in J$, so that it follows from the characteristic property of the initial topology that $S_E$ and $M_E$ are continuous (which is what we had to show). Thus, to complete the proof, we must verify that:

$S_{E_j} \circ (T_j \times T_j)$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j)$ are continuous;
$S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$.

Claim 1. The map $T_j \times T_j : E \times E \to E_j \times E_j$, $(x,y) \mapsto (T_jx,T_jy)$ is continuous. Proof: by the universal property of the product topology, it suffices to show that $(x,y) \mapsto T_jx$ and $(x,y) \mapsto T_jy$ are continuous. But this is trivial, since these maps are given by the compositions $T_j \circ \pi_x$ and $T_j \circ \pi_y$ of continuous functions.
Claim 2. The map $\text{id}_{\mathbb{K}} \times T_j : \mathbb{K} \times E \to \mathbb{K} \times E_j$, $(\lambda,x) \mapsto (\lambda,T_jx)$ is continuous. Proof: analogous.
Since $S_{E_j}$ and $M_{E_j}$ are continuous (by assumption), it follows that $S_{E_j} \circ (T_j \times T_j)$ and $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j)$ are continuous. It remains to check the claimed identities.
Claim 3. One has $S_{E_j} \circ (T_j \times T_j) = T_j \circ S_E$. Proof: both compositions are maps $E \times E \to E_j$, so the domain and codomain match. Furthermore, for all $(x,y) \in E \times E$ we have
\begin{align*}
 (S_{E_j} \circ (T_j \times T_j))(x,y) &= S_{E_j}((T_j \times T_j)(x,y)) = S_{E_j}(T_jx,T_jy) = T_jx + T_jy\\
&= T_j(x+y) = T_j(S_E(x,y)) = (T_j \circ S_E)(x,y).
\end{align*}
Claim 4. One has $M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j) = T_j \circ M_E$. Proof: both compositions are maps $\mathbb{K} \times E \to E_j$, so the domain and codomain match. Furthermore, for all $(\lambda,x) \in \mathbb{K} \times E$ we have
\begin{align*}
 (M_{E_j} \circ (\text{id}_{\mathbb{K}} \times T_j))(\lambda,x) &= M_{E_j}((\text{id}_{\mathbb{K}} \times T_j)(\lambda,x)) = M_{E_j}(\lambda,T_jx) = \lambda T_j x\\
&= T_j(\lambda x) = T_j(M_E(\lambda,x)) = (T_j \circ M_E)(\lambda,x).\tag*{$\Box$}
\end{align*}


Is the kernel of the product of two commuting differential operators the sum of the kernels? Consider the wave equation in one dimension:

$$\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial x^2}=0.$$

The most general solution of this can be written as $F(x-t)+G(x+t)$ for arbitrary functions $F, G$. It is commonly said that this is a consequence of the factorization

$$\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}=\left( \frac{\partial }{\partial t}-\frac{\partial}{\partial x}\right)\left( \frac{\partial }{\partial t}+\frac{\partial}{\partial x}\right).$$

  Is this a general fact?

More precisely, assume that the differential operator $D$ factors into the product of two (commuting) operators $A, B$, that is
$$D=AB=BA.$$
Is it true that
$$
\{u\in C^{n}\ :\ Du=0\}=\{F+G\ :\ AF=0\ \text{and}\ BG=0\}?$$
Here $n$ is the order of $D$ (which is $2$ for $D=\partial^2_t-\partial^2_x$).

If I understand the question correctly, the answer is no, because it is
not even true for the square of an operator: solutions to
$\frac{d^2}{dx^2}f(x)=0$
are not just sums of two of the constant solutions of $\frac{d}{dx}f(x)=0$.

Clearly, $\mathrm{ker}(A)\cup\mathrm{ker}(B)\subset\mathrm{ker}(AB)$, and if the operators are linear, then $\mathrm{ker}(A)\oplus\mathrm{ker}(B)\subset \mathrm{ker}(AB)$. But take $f(x,t)=x$. Then, given $A=\partial_{t}-\partial_{x}$ and $B=\partial_{t}+\partial_{x}$, $f$ is in neither $\mathrm{ker}(A)$ nor in $\mathrm{ker}(B)$. But clearly, $f\in \mathrm{ker}(AB)$, so in general $\mathrm{ker}(A)\cup\mathrm{ker}(B)\subsetneq\mathrm{ker}(AB)$.


Relationship between elliptic integrals of first and second order. A colleague and I are making an issue where order to reach the result we are asked to prove identity between elliptic integrals first and second order.

This is:

$$\frac{V}{\pi}\left( \frac{\pi}{2}-\sqrt{1-k^2}K(k) \right)=\frac{V}{\pi}\left(E(k) -(1-k^2)K(k) \right)$$

Where $E (k)$ and $K (k)$ is the elliptic integral of second and first order, respectively.

In addition: $V$ is constant. Our question is: Is it true identity? What if so, as demonstrated? Any help is the grateful.

It's not an identity. Try a single value, $k\neq0$ and see.


dice probability - same 2 dice in 6 dice rolls I have this simple probability problem that I am not sure I solved correctly. I am not interested in formulas, but rather the thought process of how to solve it.

Suppose we roll six 6-sided dice that are equal. I want to find the probability that at least two dice have the same face.

Let's consider the following roll notation:
$$
[x] [y] [a][b][c][d]
$$
I first started to calculate the probability of the first two dice being the same in an individual 2 die roll, which is $$\frac{1}{6^2}$$
Now considering the next 4 dice as static, the number of permutations of x and y in the 6 die roll would be: $$\frac{6!}{(6-2)!} = 6*5 = 30$$
And with the total number of possible outputs for a->d being 6^4 we would have the probability of rolling the same two dice in a 6 dice roll: $$\frac{\frac{1}{6^2} * 30 * 6^4}{6^6} = \frac{6^2 * 30}{6^6} = \frac{30}{6^4} < \frac{1}{6^2}$$ which doesn't make sense to me, since rolling 6 dice gives you a greater probability of having doubles as opposed to when rolling only 2.

Can anyone pinpoint a correct method of finding this probability?

Hint: The probability that at least two dices have the same face is equal to
$1-P(\texttt{None of the dices have the same face})$

When you say "two with the same dice face" do you mean "at least two"?  If so then it is simply 1 minus the probability they are all different. The first one will come up with something.  The probability the second is different is $\frac{5}{6}$.  The probability the third is different from either of the first two is $\frac{4}{6}$, etc, until the probability the last is different from any of the first 5 is $\frac{1}{6}$.  The probability all 6 dice are different is the product of those, $\frac{5!}{6^6}$ and the probability of "at least two different" is $1-\frac{5!}{6^6}$.

If you mean "exactly two", again the first can be anything.  The probability the second is the same as the first is $\frac{1}{6}$.  Then the probability the third, fourth, fifth, and sixth are different from the all previous rolls is is $\frac{5}{6}$, $\frac{4}{6}$, $\frac{3}{6}$, and $\frac{2}{6}$ respectively.  The probability the first two are the same and the last four different is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6}\cdot\frac{2}{6} = \frac{5!}{6^6}$.

There are $\binom{6}{2} = \frac{6!}{2!\ \cdot\ 4!} = 15$ different ways to order 6 things, two of which are the same so the probability of two rolls the same, the other 4 different in any order is $15\cdot \frac{5!}{6^6}$.


How to prove B from a premise in which B does not occur using natural deduction? I am preparing for my first logic exam and in the test examples I've come across the following question:

Prove by natural deduction:

B from premise A ∧ ¬A

I am unsure how to proceed in formulating this proof as B does not bear any relation to the premise. Does someone know how to proceed with such a proof? What can I assume about B's relation to the premise?

Here is a proof in the logic software program Fitch:

Here are the questions:

  I am unsure how to proceed in formulating this proof as $B$ does not bear any relation to the premise. Does someone know how to proceed with such a proof? What can I assume about B's relation to the premise?

Since $B$ does not appear in the premises it may be that the premises are contradictory. From a contradiction anything follows (in particular, $B$). If one can show that the premises are contradictory, then one can use the rule of explosion (x) or ex falso quodlibet to derive $B$.

This is actually the case since $A$ and $¬A$ are contradictory.  Here is a proof of $A ∧ ¬A ∴ B$ which you can enter in this proof checker to verify for yourself:


Write the dual LP of the primal LP problem I have to find the dual of the lp problem given below
Minimise $$z=-x_1+\frac43 x_2$$
subject to∶
$$\begin{array}[t]{l}
2x_1+4x_2\le16\\
-\frac{1}2 x_1-x_2\le4\\
-3x_1+4x_2\ge-24\\
x_1≥0,x_2≤0
\end{array}
$$

Also find the solution to the dual problem without solving it.

By evaluating the objective function at each BFS of the primal, we see that $(x_1^\star,x_2^\star)=\left(\frac85,-\frac{24}5\right)$ is an optimal solution, with objective value $-8$. Consider the dual problem:
\begin{align}
\max & \quad 16y_1 +4y_2-24y_3\\
\mathrm{s.t.} &\quad 2y_1-\frac12 y_2-3y_3\geqslant -1\\
&\quad 4y_1 -y_2 +4y_3\leqslant \frac43\\
&\quad y_1,y_2,y_3\geqslant 0
\end{align}
At optimality, $x_1$ and $x_2$ are nonzero, so by complementary slackness, the corresponding constraints in the dual are binding. Also, the constraint in the primal corresponding to the variable $y_1$ in the dual is not binding, so $y_1=0$ at optimality. It follows  that $(y_2, y_3)$ is an optimal basis for the dual, with $y_2^\star = 0$, $y_3^\star = \frac13$, and objective value $8$. This is a feasible solution for the dual with the same objective value as an optimal solution for the primal, and therefore is an optimal solution for the dual.

The prime has infinite many optimal solution as long as x1 belongs to [8/5,8]. However, the dual solution is unique.
Note that the third constraint of the prime problem has to be tight to achieve the optimal value of -8 (dividing it by 3 and you obtain the objective function). Thus, y3 must not equal to 0.
The first and the second constraints of the prime problem are equivalent, and cannot be tight. Thus, y1=y2=0.


How do I show that $\sum_{i = 1}^n \frac 1{\sqrt{a_n}} \lt \frac {\sqrt 3}6$ for $a_n = 4n(4n + 1)(4n + 2)$? Let $a_n = 4n(4n + 1)(4n + 2)$, show that $$\sum_{i = 1}^n \frac 1{\sqrt{a_i}} \lt \frac {\sqrt 3}6 \quad \forall n \in \mathbb{N}^+.$$

I know I need to find an upper bound for $1/\sqrt{a_n}$ but I can't see how, especially with the square root. Any hints will be appreciated!

We may notice that:
$$ \forall n\geq 1,\qquad  \frac{1}{\sqrt{4n(4n+1)(4n+2)}}\leq \frac{1}{2}\left(\frac{1}{\sqrt{4n-1}}-\frac{1}{\sqrt{4n+3}}\right) $$
hence it follows that:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{4n(4n+1)(4n+2)}} < \left.\frac{1}{2\sqrt{4n-1}}\right|_{n=1}=\color{red}{\frac{1}{2\sqrt{3}}}$$
as wanted. Creative telescoping wins again.

you can write inequality as $2\sqrt{3}<\sqrt{4n(4n+1)(4n+2)}$ squaring we get it as $12<4n(4n+1)(4n+2)$ this the function is continuously increasing as $n$ is positive. Also $n\in N+$ this base case is $1$ so plugging in we get $120$ this $12<120$ so it's true for akin as function is monotonic and increasing


Understanding the rules of inference I'm struggling in my discrete mathematics class and need some clarification on a topic.
Recently we learned the truth table for P -> Q.
P   Q  (P → Q)
T   T   T
T   F   F
F   F   T
F   T   T

After that we learned about rules of inference - these are used to prove an argument to be true or false. However, I don't truly understand why these are necessary.
For example, one of the rules in Modus Ponens, which states this:
assume P → Q is true.
if p is true,
then Q is true as well.

When would that rule of inference ever be necessary in a real world application? Why would we need to assume a hypothesis such as P → Q to be true? What is the point of these "rules of inference" in a real world setting, or perhaps does anyone have a real world example where it's useful? (I'd like to know why these laws have a use so an example would be greatly appreciated. Something besides P= I ate my dog. and Q = there is a blizzard outside. I think my textbook is trying to confuse me with its examples...)
I apologize if this seems like an ignorant question - it's because I am ignorant and am seeking knowledge so I can dig myself out of this hole of confusion.
EDIT: I should probably add that my textbook uses weird example like p=the moon is made of cheese - assume it to be true. So I'm wondering how these rules are actually applicable when we are talking about things that are actually used in the real world - not moons made of cheese. I am a software engineering student, so I'm sure there's a use for this in coding and I just don't know what it is yet.

Here is a real-world example to consider: "If it is raining, then it is cloudy."
This does not mean that rain causes cloudiness. It means only that, at the moment, it is not the case that it is raining and not cloudy. It permits all other combinations of truth values:

It can be raining and cloudy

It can be cloudy and not raining

It can be not raining and not cloudy

Hence the entries in the truth table.

The use of bizarre falsehoods is often used in logic textbooks to demonstrate the principle of vacuous truth. In general, it states that for any propositions $A$ and $B$, we have the tautology:
$$A \implies (\neg A \implies B)$$
In words, all things follow from a falsehood. Sort of. If $A$ is true, then the implication $\neg A \implies B$ is also true (vacuously so). Fortunately, we cannot infer anything about the truth value of $B$ from this implication since the antecedent ($\neg A$) is assumed to be false.
To use a popular, bizarre example: "If pigs could fly then $X$."
Since pigs cannot fly, this implication will be true no matter what proposition we may substitute for $X$, be it true OR false. Of course, we cannot infer anything about the truth value of $X$ without more information.

Here is an example where you use Modus Ponens.
You know that every differentiable function at $ x=a$ is continuous at $ x =a$.
So, if $ g$ is special function differentiable at $ x=a$, you are Sure that $ g $ is continuous at $ x= a$.
Every man is mortal.
You are a man.
So, be sure, you are mortal.
My remark:
In mathematics, an implication never begins by a False statement.


Partial derivative in gradient descent for two variables I've started taking an online machine learning class, and the first learning algorithm that we are going to be using is a form of linear regression using gradient descent. I don't have much of a background in high level math, but here is what I understand so far.

Given $m$ number of items in our learning set, with $x$ and $y$ values, we must find the best fit line $h_\theta(x) = \theta_0+\theta_1x$ . The cost function for any guess of $\theta_0,\theta_1$ can be computed as:

$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)}) - y^{(i)})^2$$

where $x^{(i)}$ and $y^{(i)}$ are the $x$ and $y$ values for the $i^{th}$ component in the learning set. If we substitute for $h_\theta(x)$,

$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(\theta_0 + \theta_1x^{(i)} - y^{(i)})^2$$

Then, the goal of gradient descent can be expressed as

$$\min_{\theta_0, \theta_1}\;J(\theta_0, \theta_1)$$

Finally, each step in the gradient descent can be described as:

$$\theta_j := \theta_j - \alpha\frac{\partial}{\partial\theta_j} J(\theta_0,\theta_1)$$

for $j = 0$ and $j = 1$ with $\alpha$ being a constant representing the rate of step.

I have no idea how to do the partial derivative. I have never taken calculus, but conceptually I understand what a derivative represents. The instructor gives us the partial derivatives for both $\theta_0$ and $\theta_1$ and says not to worry if we don't know how it was derived. (I suppose, technically, it is a computer class, not a mathematics class) However, I would very much like to understand this if possible. Could someone show how the partial derivative could be taken, or link to some resource that I could use to learn more? I apologize if I haven't used the correct terminology in my question; I'm very new to this subject.

The answer above is a good one, but I thought I'd add in some more "layman's" terms that helped me better understand concepts of partial derivatives. The answers I've seen here and in the Coursera forums leave out talking about the chain rule, which is important to know if you're going to get what this is doing...

It's helpful for me to think of partial derivatives this way: the variable you're
focusing on is treated as a variable, the other terms just numbers. Other key
concepts that are helpful:

For "regular derivatives" of a simple form like $F(x) = cx^n$ , the derivative is simply $F'(x) = cn \times x^{n-1}$
The derivative of a constant (a number) is 0.
Summations are just passed on in derivatives; they don't affect the derivative. Just copy them down in place as you derive.

Also, it should be mentioned that the chain
rule is being used. The chain rule says
that (in clunky laymans terms), for $g(f(x))$, you take the derivative of $g(f(x))$,
treating $f(x)$ as the variable, and then multiply by the derivative of $f(x)$. For
our cost function, think of it this way:

$$ g(\theta_0, \theta_1) = \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0,
\theta_1)^{(i)}\right)^2 \tag{1}$$

$$ f(\theta_0, \theta_1)^{(i)} = \theta_0 + \theta_{1}x^{(i)} -
y^{(i)} \tag{2}$$

To show I'm not pulling funny business, sub in the definition of $f(\theta_0,
\theta_1)^{(i)}$ into the definition of $g(\theta_0, \theta_1)$ and you get:

$$ g(f(\theta_0, \theta_1)^{(i)}) = \frac{1}{2m} \sum_{i=1}^m \left(\theta_0 +
\theta_{1}x^{(i)} - y^{(i)}\right)^2 \tag{3}$$

This is, indeed, our entire cost function.

Thus, the partial derivatives work like this:

$$ \frac{\partial}{\partial \theta_0} g(\theta_0, \theta_1) = \frac{\partial}{\partial
\theta_0} \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0, \theta_1)^{(i)}\right)^2 = 2
\times \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0, \theta_1)^{(i)}\right)^{2-1} = \tag{4}$$

$$\frac{1}{m}
\sum_{i=1}^m f(\theta_0, \theta_1)^{(i)}$$

In other words, just treat $f(\theta_0, \theta_1)^{(i)}$ like a variable and you have a
simple derivative of $\frac{1}{2m} x^2 = \frac{1}{m}x$

$$ \frac{\partial}{\partial \theta_0} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_0} (\theta_0 + \theta_{1}x^{(i)} - y^{(i)}) \tag{5}$$

And $\theta_1, x$, and $y$ are just "a number" since we're taking the derivative with
respect to $\theta_0$, so the partial of $g(\theta_0, \theta_1)$ becomes:

$$ \frac{\partial}{\partial \theta_0} f(\theta_0, \theta_1) = \frac{\partial}{\partial \theta_0} (\theta_0 + [a \
number][a \ number]^{(i)} - [a \ number]^{(i)}) = \frac{\partial}{\partial \theta_0}
\theta_0 = 1 \tag{6}$$

So, using the chain rule, we have:

$$ \frac{\partial}{\partial \theta_0} g(f(\theta_0, \theta_1)^{(i)}) =
\frac{\partial}{\partial \theta_0} g(\theta_0, \theta_1) \frac{\partial}{\partial
\theta_0}f(\theta_0, \theta_1)^{(i)} \tag{7}$$

And subbing in the partials of $g(\theta_0, \theta_1)$ and $f(\theta_0, \theta_1)^{(i)}$
from above, we have:

$$ \frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)} \frac{\partial}{\partial
\theta_0}f(\theta_0, \theta_1)^{(i)} = \frac{1}{m} \sum_{i=1}^m \left(\theta_0 +
\theta_{1}x^{(i)} - y^{(i)}\right) \times 1 = \tag{8}$$

$$ \frac{1}{m} \sum_{i=1}^m \left(\theta_0 + \theta_{1}x^{(i)} - y^{(i)}\right)$$

What about the derivative with respect to $\theta_1$?

Our term $g(\theta_0, \theta_1)$ is identical, so we just need to take the derivative
of $f(\theta_0, \theta_1)^{(i)}$, this time treating $\theta_1$ as the variable and the
other terms as "just a number." That goes like this:

$$ \frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_1} (\theta_0 + \theta_{1}x^{(i)} - y^{(i)}) \tag{9}$$

$$ \frac{\partial}{\partial
\theta_1} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_1} ([a \ number] +
\theta_{1}[a \ number, x^{(i)}] - [a \ number]) \tag{10}$$

Note that the "just a number", $x^{(i)}$, is important in this case because the
derivative of $c \times x$ (where $c$ is some number) is $\frac{d}{dx}(c \times x^1) =
c \times 1 \times x^{(1-1=0)} = c \times 1 \times 1 = c$, so the number will carry
through. In this case that number is $x^{(i)}$ so we need to keep it. Thus, our
derivative is:

$$ \frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = 0 + (\theta_{1})^1
x^{(i)} - 0 = 1 \times \theta_1^{(1-1=0)} x^{(i)} = 1 \times 1 \times x^{(i)} =
x^{(i)} \tag{11}$$

Thus, the entire answer becomes:

$$ \frac{\partial}{\partial \theta_1} g(f(\theta_0, \theta_1)^{(i)}) =
\frac{\partial}{\partial \theta_1} g(\theta_0, \theta_1) \frac{\partial}{\partial
\theta_1} f(\theta_0, \theta_1)^{(i)} = \tag{12}$$

$$\frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)} \frac{\partial}{\partial
\theta_1}f(\theta_0, \theta_1)^{(i)} = \frac{1}{m} \sum_{i=1}^m \left(\theta_0 +
\theta_{1}x^{(i)} - y^{(i)}\right) x^{(i)}$$

A quick addition per @Hugo's comment below. Let's ignore the fact that we're dealing with vectors at all, which drops the summation and $fu^{(i)}$ bits. We can also more easily use real numbers this way.

$\require{cancel}$

Let's say $x = 2$ and $y = 4$.

So, for part 1 you have:

$$\frac{\partial}{\partial \theta_0} (\theta_0 + \theta_{1}x - y)$$

Filling in the values for $x$ and $y$, we have:

$$\frac{\partial}{\partial \theta_0} (\theta_0 + 2\theta_{1} - 4)$$

We only care about $\theta_0$, so $\theta_1$ is treated like a constant (any number, so let's just say it's 6).

$$\frac{\partial}{\partial \theta_0} (\theta_0 + (2 \times 6) - 4) = \frac{\partial}{\partial \theta_0} (\theta_0 + \cancel8) = 1$$

Using the same values, let's look at the $\theta_1$ case (same starting point with $x$ and $y$ values input):

$$\frac{\partial}{\partial \theta_1} (\theta_0 + 2\theta_{1} - 4)$$

In this case we do care about $\theta_1$, but $\theta_0$ is treated as a constant; we'll do the same as above and use 6 for it's value:

$$\frac{\partial}{\partial \theta_1} (6 + 2\theta_{1} - 4) = \frac{\partial}{\partial \theta_1} (2\theta_{1} + \cancel2) = 2 = x$$

The answer is 2 because we ended up with $2\theta_1$ and we had that because $x = 2$.

Hopefully the clarifies a bit on why in the first instance (wrt $\theta_0$) I wrote "just a number," and in the second case (wrt $\theta_1$) I wrote "just a number, $x^{(i)}$. While it's true that $x^{(i)}$ is still "just a number", since it's attached to the variable of interest in the second case it's value will carry through which is why we end up at $x^{(i)}$ for the result.

The idea behind partial derivatives is finding the slope of the function with regards to a variable while other variables value remains constant (does not change). Or what's the slope of the function in the coordinate of a variable of the function while other variable values remains constant.  As what I understood from MathIsFun, there are 2 rules for finding partial derivatives:

1.) Terms (number/s, variable/s, or both, that are multiplied or divided) that do not have the variable whose partial derivative we want to find becomes 0

example:
f(z,x,y) = z2 + x2y
f'z  = 2z + 0

2.) For terms which contains the variable whose partial derivative we want to find, other variable/s and number/s remains the same, and compute for the derivative of the variable whose derivative we want to find

example:
f(z,x,y,m) = z2 + (x2y3)/m
f'x  = 0 + 2xy3/m

For linear regression, for each cost value, you can have 1 or more input. For example for finding the "cost of a property" (this is the cost), the first input X1 could be size of the property, the second input X2 could be the age of the property.  We need to understand the guess function.  For linear regression, guess function forms a line(maybe straight or curved), whose points are the guess cost for any given value of each inputs (X1, X2, X3, ...).

For single input (graph is 2-coordinate where the y-axis is for the cost values while the x-axis is for the input X1 values), the guess function is:

Hθ = θ0 + θ1X1

For 2 input (graph is 3-d, 3-coordinate, where the vertical axis is for the cost values, while the 2 horizontal axis which are perpendicular to each other are for each input (X1 and X2). The 3 axis are joined together at each zero value:

Hθ = θ0 + θ1X1 + θ2X2

For 3 input (not easy to form a graph):

Hθ = θ0 + θ1X1 + θ2X2 + θ3X3

Note θ are variables and represents the weights. θ0 represents the weight when all input values are zero.  θ0 is base cost value, you can not form a good line guess if the cost always start at 0.  You can actually multiply θ0 to an imaginary input X0, and this X0 input has a constant value of 1.

To get the partial derivative the cost function for 2 inputs, with respect to θ0, θ1, and θ2, the cost function is:

$$ J = \frac{\sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)^2}{2M}$$

Where M is the number of sample cost data, X1i is the value of the first input for each sample cost data, X2i is the value of the second input for each sample cost data, and Yi is the cost value of each sample cost data

To compute for the partial derivative of the cost function with respect to θ0, the whole cost function is treated as a single term, so the denominator 2M remains the same.  We will find the partial derivative of the numerator with respect to θ0, θ1, θ2.  Using the combination of the rule in finding the derivative of a summation, chain rule, and power rule:

$$ f(x) = \sum_{i=1}^M (X)^n$$
$$ f'_x = n . \sum_{i=1}^M (X)^(n-1) . f'X $$

$$ So f'θ_0 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)^1 . f'θ_0 ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)}{2M}$$

$$  f'θ_0 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_0 ((\theta_0 + 0 + 0) - 0)}{2M}$$

$$  f'θ_0 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_0 (\theta_0)}{2M}$$

$$  f'θ_0 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . 1}{2M}$$

$$  temp_0 = \frac{\sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)}{M}$$

$$    f'θ_1 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)^1 . f'θ_1 ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)}{2M}$$

$$  f'θ_1 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_1 ((0 + X_1i\theta_1 + 0) - 0)}{2M}$$

$$  f'θ_1 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_1 (X_1i\theta_1)}{2M}$$

$$  f'θ_1 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . X_1i}{2M}$$

$$  temp_1 = \frac{\sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . X_1i}{M}$$

$$    f'θ_2 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)^1 . f'θ_1 ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i)}{2M}$$

$$  f'θ_2 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_1 ((0 + 0 + X_2i\theta_2) - 0)}{2M}$$

$$  f'θ_2 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . f'θ_1 (X_2i\theta_2)}{2M}$$

$$  f'θ_2 = \frac{2 . \sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . X_2i}{2M}$$

$$  temp_2 = \frac{\sum_{i=1}^M ((\theta_0 + \theta_1X_1i + \theta_2X_2i) - Y_i) . X_2i}{M}$$

Gradient Descent is

repeat until minimum result of the cost function {

// Calculation of temp0, temp1, temp2 placed here (partial derivatives for  θ0, θ1, θ1 found above)
$$   \theta_0 = \theta_0 - \alpha . temp0 $$
$$   \theta_1 = \theta_1 - \alpha . temp1 $$
$$   \theta_2 = \theta_2 - \alpha . temp2 $$
}

Where α is the learning rate


Evaluate the limit $\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\log }^2}(n + 1) - n{{\log }^2}n}}{{{{\log }^2}n}}$ Evaluate:
$$\mathop {\lim }\limits_{n \to \infty } \frac{{(n + 1){{\log }^2}(n + 1) - n{{\log }^2}n}}{{{{\log }^2}n}}$$

Intuitively, I feel that for large $n$, ${\log}(n+1) \approx \ {\log}(n)
$. So, the above limit should reduce to:

$$=\mathop {\lim }\limits_{n \to \infty } \frac{{\{ (n + 1) - n\} {{\log }^2}n}}{{{{\log }^2}n}} \ = 1$$

However, can someone please suggest how can one mathematically show this.

Thanks!

If you want to be strict, write $$\log(1+n)=\log(n)+\log(1+\frac 1n)$$ So, the numerator is $$A=(n + 1) \log^2(n + 1) - n\log^2(n)=(n+1)\left(\log(n)+\log(1+\frac 1n)\right)^2-n \log^2(n)$$ Expanding the square and grouping $$A=a^2 n+a^2+2 a n \log (n)+2 a \log (n)+\log ^2(n)$$ where $a=\log(1+\frac 1n)\approx \frac 1n $ is small when $n$ is large.

Is this what you are looking for ?

Edit

Alternatively, you could set $n=\frac 1x$ and use Taylor series around $x=0$. Doing so, the entire expression will be $$\left(1-\frac{2}{\log (x)}\right)+x \left(\frac{1}{\log ^2(x)}-\frac{1}{\log
   (x)}\right)+\frac{x^2}{3 \log (x)}+O\left(x^3\right)$$

It should be $$\lim_{n\to \infty} \left((n+1)(\frac{\ln(n+1)}{\ln n})^2 -n\right) = \lim_{n\to \infty} (n+1 - n) = 1.$$


Difficulty in solving challenging trig equation Find $\theta$ on $[0, 2\pi)$ such that

$$\cos{\theta}^{\sin{\theta}^{\cos{\theta}^{\dots}}} = 2 + 2\sec^2{\theta}\tan^2{\theta} - \sec^4{\theta} - \tan^4{\theta}$$

I'm not sure on how to tackle this problem. I've never really dealt with exponentiation of trig functions. Any help is helpful. Thank you very much

By inspection, $\Theta=0$ is a solution. There's little hope of solving this analytically unless you can find a closed form for the left-hand side for general $\Theta$.

Addendum: $\Theta=\pi$ is another solution because it produces the same RHS as $\Theta=0$, and the LHS $$(-1)^{0^{(-1)^0\cdots}}$$ converges to $1$ (the value of the RHS).

Answer is one $1$ but the answer given by the first vote is wrong because the expansion of (sin2o-1)2 is not equal to cos2o


What is the area of the shaded region in terms of $n$? We can see that shaded region is area of FEH minus the sector HGE.
To find the sector HGE i called the angle GHE as $\theta$ and used
$$
\pi\times\left(\frac{\theta}{360^{\circ}}\right)
$$
If FE is $x$, we see that $\cos(\theta)=\frac{1}{\sqrt{x^2+1}}$ therefore
$$
\theta=\arccos\left(\frac{1}{\sqrt{x^2+1}}\right)
$$
My attempt on this problem was by "brute-forcing" my way from the bottom of the figure to top, to find FE, using Law of Cosine on the way.

The shaded region will now be
$$
\frac{x}{2} - \pi\times\left(\frac{\arccos\left(\frac{1}{\sqrt{x^2+1}}\right)}{360^{\circ}}\right)
$$

To find $x$ i approached in two different ways:

Connect CD to AI by lengthening CD and connect HD. Do Law of Cosine on these triangles to get to FE.
Connect AD and HD, do law of cosines on these to get to FE.

Both of these resulted in immense work and, on the first case i was able to write $x$ in terms of $n$ but the equation barely fit a word page with font size 1. (and because of the immense work, it is likely to be wrong).

So, let's do it "classically".

With reference to the sketch

we have from the data given
$$
\left\{ \matrix{
  H = n\left( {\cos \alpha ,\;\sin \alpha } \right) \hfill \cr
  E = H + \left( {\cos \beta ,\;\sin \beta } \right) \hfill \cr
  D = \left( {n - \cos \left( {\alpha /2} \right),\,\sin \left( {\alpha /2} \right)} \right) \hfill \cr
  E - D = x\left( {\cos \left( {\beta  + \pi /2} \right),\;\sin \left( {\beta  + \pi /2} \right)} \right) \hfill \cr}  \right.
$$
in the unknown $x$ and $\beta$.
The points are written as position vectors wrt the origin.

Last line gives
$$
n\left( {\cos \alpha ,\;\sin \alpha } \right) + \left( {\cos \beta ,\;\sin \beta } \right)
 - \left( {n - \cos \left( {\alpha /2} \right),\,\sin \left( {\alpha /2} \right)} \right)
 = x\left( {\cos \left( {\beta  + \pi /2} \right),\;\sin \left( {\beta  + \pi /2} \right)} \right)
$$
which translates into
$$
\left\{ \matrix{
  n\cos \alpha  + \cos \beta  - n + \cos \left( {\alpha /2} \right) = x\cos \left( {\beta  + \pi /2} \right) =  - x\sin \beta  \hfill \cr
  n\sin \alpha  + \sin \beta  - \sin \left( {\alpha /2} \right) = x\sin \left( {\beta  + \pi /2} \right) = x\cos \beta  \hfill \cr}  \right.
$$
i.e.
$$
\left\{ \matrix{
  \cos \beta  + x\sin \beta  = n - n\cos \alpha  - \cos \left( {\alpha /2} \right) \hfill \cr
  x\cos \beta  - \sin \beta  = n\sin \alpha  - \sin \left( {\alpha /2} \right) \hfill \cr}  \right.
$$

To solve this, let's make a change of variable
$$
\left\{ \matrix{
  x = X\sin \eta  \hfill \cr
  1 = X\cos \eta  \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  X = \sqrt {1 + x^{\,2} }  \hfill \cr
  \eta  = \arctan \left( x \right) \hfill \cr}  \right.
$$
so that
$$
\left\{ \matrix{
  X\cos \left( {\eta  - \beta } \right) = n - n\cos \alpha  - \cos \left( {\alpha /2} \right) \hfill \cr
  X\sin \left( {\eta  - \beta } \right) = n\sin \alpha  - \sin \left( {\alpha /2} \right) \hfill \cr}  \right.
$$

Thereafter, I think you can conclude by yourself.

Maybe not much simpler than your answer, but here is a methodical way to approach it.

Extend CD to meet AF at X. Since you know AC, angle XAC and angle XCA, the triangle ACX can be solved completely in terms of n & $\alpha$. From law of sines:

AX / sin($\alpha$/2) = CX / sin($\alpha$) = (n-1) / sin($\pi$ - $3\alpha / 2$)

This should give you both AX and CX without much work which we will use later.

FX = n-1-AX, DX = CX-1 and you know the included angle FXD. From 1, you know AX and CX. So, you can now solve the triangle FXD completely. This is a bit more work, but is pretty straight forward as well.

FD = $\sqrt{FX^2 + XD^2 - 2.FX.XD.cos(3\alpha/2)}$

FD / sin($3\alpha / 2$) = DX / sin(XFD)

This should give you XFD (note that XFD is less than 90 degrees due to construction)

Now you know angle HFE = (angle XFD from 2). Thus, you know angle FHE = (90 - HFE). This helps you find area of sector FHE. Since FE = Tan(angle HFE), you can now find area of triangle FHE = 0.5 x 1 x FE.


When are we free in setting the free/basic variables in a linear system of equations Consider the following two linear system of equations of the form $b=Ax$
case 1:
$$
\begin{pmatrix}
4\\
5
\end{pmatrix}= \begin{pmatrix}
1& 2 & -1\\
2 & -4 & 0
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$
case 2:
$$
\begin{pmatrix}
0\\
5
\end{pmatrix}= \begin{pmatrix}
0 & 0\\
0 & 1
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
\end{pmatrix}
$$
The reduced row echelon form  of case 1 is
$$
\begin{pmatrix}
1 & 0 & -0.5 & 3.25\\
0 & 1 & -0.25 & 0.3750
\end{pmatrix}
$$
Hence, $x_1,x_2$ are basic variables, $x_3$ is free variable.
The reduced row echelon form  of case 2 is
$$
\begin{pmatrix}
0 & 1 & 5\\
0 & 0 & 0
\end{pmatrix}
$$
Hence, $x_2$ is basic variable, $x_1$ is free variable.

Question: There is a key difference between case 1 and case 2.
In case 1, I am free to decide which two variables among $\{x_1,x_2,x_3\}$ are set as basic and which one is set as free. The reduced row echelon suggests to set $x_1,x_2$ as basic variables and $x_3$ as free. However, I could equivalently set $x_1,x_3$ as basic variables and $x_2$ as free. Or, I could also set $x_3,x_2$ as basic variables and $x_1$ as free. This can be seen by doing naive calculations from the system of equations.
In case 2, I am not free to decide which variable between $\{x_1,x_2\}$ should be set as basic and which one should be set as free. Again, this can be seen by doing naive calculations from the system of equations.
More generally (and for more complicated systems) I am wondering whether there is an easy way to detect when I am (not) free in setting the free/basic variables from just looking at the reduced row echelon, without the need of doing other calculations. Is this a matter of zero rows or is there something more subtle?

In these sort of cases, the solution is a line or plane or something similar, embedded within the input space.  When describing this solution we can choose any variable that is not orthogonal to it as a free variable, and then once we have enough to fully describe a projection of the solution with the same number of dimensions, we can say any other variables we would need to get back to the solution set have you are "basic", forced by the selection of free variables to take on a particular value.
In case 1, the solution is $x = (\frac{109}{42}, \frac{1}{21}, -\frac{55}{42}) + (2,1,4)t$, and since none of $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are orthogonal to $2,1,4$, we can pick any one of $x_1$, $x_2$, and $x_3$ to be the free variable.
In case 2, the solution can be described as the line $x = (0,5) + (1,0)t$, where it is obvious that $(0,1)$ is orthogonal to the solution line, and so $x_2$ cannot be considered free.
I've done a bit of searching to figure out how to detect this in the RREF.  here's an example.
$$
\begin{pmatrix}
1\\
4
\end{pmatrix}= \begin{pmatrix}
4& -6 & 3\\
-2 & 3 & 4
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$
This has RREF
$$
\left(\begin{array}{ccc|c}
1 & -\frac{3}{2} & 0 & -\frac{4}{11}\\
0 & 0 & 1 & \frac{9}{11}
\end{array}\right)
$$
And it turns out that $x_3 = \frac{9}{11}$ all the time.  The solution set is $(-\frac{2}{11}, \frac{4}{33}, \frac{9}{11}) + (2,3,0)t$; this is clearly orthogonal to $(0,0,1)$ and so $x_3$ is fixed.
This provides guidance to the answer The second row of the RREF contains only columns from $x_3$ and the solution vector, so this one must be fixed; I can select from $x_1$ and $x_2$ to be free because they share a row.
Applying this wisdom to case 1: selecting one of $x_1$ or $x_2$ to be free fixes $x_3$ by the action of one of the rows, which then (since everything else in the other row is fixed) fixes the other one, and picking $x_3$ to be free fixes $x_1$ and $x_2$ by the respective actions of the two rows in themselves.
Applying similar wisdom to case 2: $x_1$ gets to be free because it affects absolutely nothing; $x_2$ does not because it is surrounded by already-fixed results.
So it looks like it goes something like this: if a variable is in a position in one of the rows where it can affect only variables that are either fixed or already chosen as free, then it must be fixed.  Otherwise, you may choose it to be free.

It depends on the rang of the row echelon matrix , say the rang of the row echelon matrix matches the dimension of the $x$ vecotr in your case, then you cannnot choose any of the $x_1,x_2\ldots x_n $ as free variables. In case 2 you can choose either $x_1$ or $x_2$ as a free variable , lets say $t\in \mathbb{R}$ and calculate the rest accordingly.
The amount of free varibales you can chosse is given by '$rows(x)-rang(A)$'. The varibale you choose to be free is up to you.


Length of a line segment A line segment of length 1 is cut into two pieces at a completely random point. What is the probability that the longer piece is at least three times the length of the shorter piece?

Attempt

Let X be the length of the shorter piece and Y be the length of the longer piece.

X: f(x) = 2, 0 < x < 0.5  or f(x) = 0 otherwise

Y: g(y) = 2, 0.5 < y < 1 or g(y) = 0 otherwise

I think the question is asking for P[Y>= 3 X] but I am getting a negative probability.

If the line was cut at 0.25, then the shorter piece would be 0.25 and the longer piece 0.75, 3 times longer. If the line is cut at a point less than 0.25, the longer piece will be more than 3 times the length of the shorter piece. So I think the answer is $\frac{1}{4}$.

Edit: As rightly pointed out by Badam Baplan, this answer doubles to $\frac{1}{2}$ with symmetry since we can cut the line at a point between 0 and 0.25 or between 0.75 and 1.

You just need to have only one probability function .

considering

  y = 3x
  y+x = 1
  3x+x = 1
   x  = 1/4

since the probability of one piece having length less than 1/4 answer is 1/4


A signed measure is bounded If $\nu$ is a signed measure defined over $(X,\mathcal{M})$ such that $\nu(E)\in\Bbb{R}$, for all $E\in\mathcal{M}$, then $\nu$ is bounded.

This looks weird to me. Of course, $\nu(E)\in\Bbb{R}$ implies that $\nu(E)<+\infty$, once we defines a signed measure to take values at $\Bbb{R}\cup\{-\infty,+\infty\}$. But only with this informations, how conclude that $\nu$ is bounded?

Assume that $E_1,E_2,...$ satisfy that $\nu(E_n)\to+\infty$. By excluding finitely many terms from the beginning, we can assume that $\nu(E_n)>0$ and by taking differences of sets of a subsequence for which $\nu(E_{n_k})>2^k$ that the $E_n$ are dijoint. Then $\mathbb{R}\ni\nu\left(\bigcup_nE_n\right)=\sum_{n=1}^{\infty}\nu(E_n)=+\infty$. Therefore, there is no such sequence.

Similarly, you can exclude the existence of a sequence with $\nu(E_n)\to-\infty$.


Correct way to address the famous "brothers" conditional probability paradox OK, this question will probably stir some controversy.

There is a well-known paradox (mentioned, among other sources, by Martin Gardner).

A) You know that I have two children, and the oldest is a daughter. What is the probability that both are daughters?
 (let's ignore the possibility of identical twins, which may muddle the question; and, for simplicity, assume that probability that a child is a daughter is exactly 1/2).
 This is very easy to answer.

B) Now, suppose you know that at least one child is a daughter (but you don't know if it's the oldest one or not).  What is the probability that both are daughters?

The common way to answer is: you got extra information that the number of daughters, let's call it D, is at least one.

Thus, $P(D=2 | D \ge 1) = P(D=2, D \ge 1)/P(D \ge 1) = (1/4)/(3/4) = 1/3$

The paradox is that the answer in (A) and (B) are different!

However, a student told me that this is a wrong way to look at it.
How do you know that at least one child is a daughter? Perhaps you overheard me saying "my daughter ... (just had a birthday or something)". But in this case you were more likely to overhear in the first place, if I had had two daughters, and not one. This messes up the probabilities calculated above.

To put it another way: let's generate the list of random 2-child families. Let's randomly "overhear" the gender of a child in each family on the list. Ignore suns and concentrate on daughters. Calculate which proportion of the families we picked had two daughters. Then your answer is 1/2 again.

Which answer is the right one for (B)? 1/3 or 1/2?

If you think it's 1/3, can you describe a realistic situation in which you can come up with the information "at least one girl"?

In calculating a conditional probability, the probability distribution of the given event is not under consideration.  It has already occurred.  To confuse this point by reintroducing some prior likelihood of having observed the given event is where your misunderstanding lies.

For instance, suppose I take a survey of people at random.  I ask them (1) how many children they have, and if they answer exactly two, then I ask them if both are boys.  If they say no, then I offer them $20 if they will come to a prearranged location at a specific time.

I then tell you to go to that prearranged location and time.  Then you will meet all these people, all of whom have exactly two children, of which at least one is female.  This situation meets your criteria:  you are given this sample.

Mathematically, you are conflating $\Pr[D = 2 \mid D \ge 1]$ with $$\Pr[D = 2 \cap D \ge 1] = \Pr[D = 2 \mid D \ge 1]\Pr[D \ge 1].$$  Once we are told $D \ge 1$, we are to treat the sample space as if all elementary outcomes for which the given condition is true have equal probability of occurrence.

Knowing that one child is a girl is different from hearing the gender of one child at random.

In the first case, say we ask a 2-child parent "Is one of your children a girl" and the answer is yes.  Then we are given nothing more than the fact that at least one child is a girl, and B applies.  Denote the event of a 'yes' answer as $X$.  Then $P(X) = 3/4$.

Now, if you "randomly overhear" that a child is a girl, call this event $Y$, you have $P(Y) = 1/2$, as it's the same as picking a random child from a random family, which is equally likely to be a boy or a girl.

If gg denotes the 2-girls event,
 you have $P(gg | Y) = \frac{P(gg, Y)}{Y} = \frac{1/4}{1/2} = 1/2$.


The Three Princesses (distinguishing truth-teller with 1 question) Is it possible to solve this problem:

A prince wish to marry a princess. There are 3 princesses, one is young, one is a little older and one is old. The prince is able to tell the princesses apart. One of the princesses always tells the truth, one never tells the truth and one sometimes tells the truth and sometimes not.

The prince only wish to marry a princess whom he can trust. Therefore it must be the princess that always tells the truth or the princess that never tells the truth (he can just negate her answers for the rest of their marriage).

Before he chooses the princess he wish to marry, he can ask one and only one princess a single question. She must only answer the question by yes or no.

Which question must he ask to be sure he marries one of the right princesses?

Edit: I was not expecting the question "who is more truthful", so consider this change of rules. Suppose we remove the "random princess", and instead insert an "evil princess". The evil princess can choose her strategy for answering, after she has seen which princess we are asking. So asking "Who is more truthful", does not make sense anymore, since the evil princess could choose to answer correct to every question.

Pick a random prince, then ask him:

  Is he [pointing to one of his brothers] older than him [pointing to the second one]?

If you receive "Yes" as an answer, choose the second brother, otherwise choose the first. This ensures you will not marry the middle one.

Explanation:

If the prince you asked is the eldest, he will answer truthfully, and you will pick the youngest.

If the prince you asked is the youngest, he will lie, and you will pick the eldest.

If the prince you asked is the middle one, he will answer randomly, and you will pick one of the two others, as desired.

Edit:

More information about the many variations of this puzzle can be found by searching for the term "Knights and Knaves", which is the name logician and author Raymond Smullyan used to describe these problems.

Of particular interest is a problem known as The Hardest Logic Puzzle Ever, which has the same basic setup as the "princes(ses)" puzzle – lier, truther and random answerer – but with more information needing to be extracted and the additional complication that instead of "Yes" and "No", equivalent words in some esoteric language are used and it is unknown which word means which.

You can't pick the one you ask the question to, as there is no yes/no question you can ask that will let you determine that the asker is not random.

From this, there are 3 cases to consider:

If the one you ask is random, it doesn't matter who you pick
If the one you ask is true, you want to pick the false one
If the one you ask is false, you want to pick the true one

We only need to worry about cases 2 and 3, since case 1 solves itself.  Thus, the question boils down to these 4 scenarios:

You get a true answer about false
You get a true answer about random
You get a false answer about true
You get a false answer about random

From these cases, let's constrain it so that you'll get a "yes" for 1 and 3, and a "no" for 2 and 4.  That way, you can pick the one they say "yes" about, or pick the other if they say "no".  There are many possibilities, but here's one such question:

If you are true is that brother false, otherwise is he random?


What is the kernel of the tensor product of two maps? Assume that $f_1\colon V_1\to W_1, f_2\colon V_2\to W_2$ are $k$-linear maps between $k$-vector spaces (over the same field $k$, but the dimension may be infinity). Then the tensor product $f_1\otimes f_2\colon V_1\otimes V_2\to W_1\otimes W_2$ is defined, and it's obvious that $\ker f_1\otimes V_2+ V_1\otimes \ker f_2 \subseteq \ker (f_1\otimes f_2)$. My question is whether the relation $\subseteq$ is in fact $=$.

If this does not hold, how about assuming all these vector spaces are commutative associated $k$-algebras with identity and that all the maps are $k$-algebra homomorphisms? Or can you give a "right" form of the kernel $\ker (f_1\otimes f_2)$?

Yes, that's true. Let $f_i : V_i \to W_i$ be two linear maps. Since $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $W_1 \otimes W_2$, we may assume that $f_1,f_2$ are surjective. But then they are split, so that we can assume that $V_i = W_i \oplus U_i$ and that $f_i$ equals the projection $V_i \to W_i$, with kernel $U_i$. Then $V_1 \otimes V_2 = W_1 \otimes W_2 \oplus W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2$ and $f_1 \otimes f_2$ equals the projection of $V_1 \otimes V_2$ onto $W_1 \otimes W_2$. Hence the kernel is $W_1 \otimes U_2 \oplus U_1 \otimes W_2 \oplus U_1 \otimes U_2 = U_1 \otimes V_2 + V_1 \otimes U_2$.

This shows even more: The kernel is the pushout $(\ker(f_1) \otimes V_2) \cup_{\ker(f_1) \otimes \ker(f_2)} (V_1 \otimes \ker(f_2))$.

By the way, this argument is purely formal and works in every semisimple abelian $\otimes$-category. What happens when we drop semisimplicity, for example when we consider modules over some commutative ring $R$? Then we only need some flatness assumptions:

Let $f_1 : V_1 \to W_1$ and $f_2 : V_2 \to W_2$ be two morphisms in an abelian $\otimes$-category (for example $R$-linear maps between $R$-modules). If $f_1,f_2$ are epimorphisms, then we have exact sequences $\ker(f_1) \to V_1 \to W_1 \to 0$ and $\ker(f_2) \to V_2 \to W_2 \to 0$. Applying the right exactness of the tensor product twice(!), we get that then also

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to W_1 \otimes W_2 \to 0$

is exact. If $f_1,f_2$ are not epi, we can still apply the above to their images and get the exactness of

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to \mathrm{im}(f_1) \otimes \mathrm{im}(f_2) \to 0.$

Now assume that $\mathrm{im}(f_1)$ and $W_2$ are flat. Then $\mathrm{im}(f_1) \otimes \mathrm{im}(f_2)$ embeds into $\mathrm{im}(f_1) \otimes W_2$ which embeds into $W_1 \otimes W_2$. Hence, we have still that the sequence

$\ker(f_1) \otimes V_2 \oplus V_1 \otimes \ker(f_2) \to V_1 \otimes V_2 \to  W_1 \otimes W_2$

is exact. In other words, we have a sum decomposition
$$\ker(f_1 \otimes f_2) = \alpha(\ker(f_1) \otimes V_2) + \beta(V_1 \otimes \ker(f_2)),$$ where $\alpha : \ker(f_1) \otimes V_2 \to V_1 \otimes V_2$ and $\beta : V_1 \otimes \ker(f_2) \to V_1 \otimes V_2$ are the canonical morphisms. In general, these are not monic! However, this is the case, by definition, when $V_1$ and $V_2$ are flat. So in this case we can safely treat $\alpha$ and $\beta$ as inclusions and write $$\ker(f_1 \otimes f_2) = V_1 \otimes \ker(f_2) + \ker(f_1) \otimes V_2.$$

The isomorphism
$$
\ker (f_1\otimes f_2)=V_1\otimes \ker f_2 + \ker f_1 \otimes V_2
$$
is equivalent to the isomorphism
$$
\operatorname{coker} (f_1\otimes f_2)=\operatorname{coker} f_1\otimes \operatorname{coker} f_2,
$$
since for any ideals $I,J\subset R$ we have
$$
R/(I+J)=R/I\otimes R/J
$$


Inverse of sum of scalar times matrix plus identity matrix I'm trying to figure out a way of seperating out my $n\times n$-dimensional matrix $A$ (to use in my Finite Difference scheme) from the following equation:

$(cA + I)^{-1}$,

where $c$ is a scalar and $I$ a $n\times n$ identity matrix.

Does anyone have a (reference to a) clue how to proceed and, if possible, under which assumptions a solution can be obtained?

I'm a bit confused; if you have a matrix $M$ and you want to write it in the form
$$M = cA + I$$
you simply directly solve for $A$:
$$A = (M-I)/c$$
where if you're not given the value of $c$, you can pick it arbitrarily ($c=1$, say).

If you're trying to compute the matrix inverse of $cA+I$, that is a different problem (and one without a closed-form solution, though you can use some common approximations, e.g. $$(I+cA)^{-1} \approx I - cA,$$ if $cA$ is small).


Why is the cross product of two vectors orthogonal? What is the intuition behind the fact that the cross product of two vectors is orthogonal? Every video I've seen just says it is orthogonal but they do not explain why. Since I have terrible memory, I need to understand things intuitively or I will just forget them.

By the way, I do not understand the intuition behind the calculation for the cross product either, so if that is necessary to understand why it is orthogonal, please explain.

Its not really an "intuition" thing, the cross product is defined that way. The following may help you understand why.

The short answer is that the cross product appears in many physical systems - most famously in calculating the force produced by an electrical current in a magnetic field. But this begs the question of why it appears so often in physics.

In physics, you often get linear relationships - in the example above doubling the current will double the force, as will doubling the magnetic field. So the current vector and magnetic vector are somehow multiplied together to find the force vector. You could potentially define vector multiplication in lots of ways. But there is an additional constraint in physics, that you must get the same answer for the force vector however you orient your co-ordinate system. Nature doesn't have a coordinate system, so the answer must be the same however you define what is the "x axis" and what is the "y axis". If you set up a coordinate system with the x axis along the direction of the magnetic field and the y axis along the current vector, then you should get the same answer for force as if you set up the co-ordinate system with x axis in the current direction and the y axis along the magnetic field direction.

All physical laws must follow this rule; nature doesn't have a preferred direction for x and y. The dot product (which presumably you have just learned about) follows this rule.

If you want to multiply two vectors in 3D space to form another vector, up to a constant term the only definition of vector  multiplication which has this property of being independent of the coordinate system is the cross product. It is the unique definition of vector multiplication which is independent of the choice of coordinate system.

So nature follows this rule whenever two (3D) vectors are multiplied together to form a 3rd vector. Any other rule would be depend on the choice of coordinate system. So natural processes in three dimensions are in a sense forced to use this rule, because its the only rule which is independent of the co-ordinate system. In 2D, there is no possible rule which is independent of the co-ordinate system. In higher dimensions there are more choices. But we live in a 3D Universe (spatial dimensions), so the cross product rule is forced onto nature and hence onto physics and maths.

The cross-product was defined by Gibbs to mimic the properties of quaternion multiplication without the negativity of the square.
Quaternions are very economic way to represent rotations in 3D and therefore many physical equations.

See for example here.

http://en.wikipedia.org/wiki/Geometric_algebra

http://en.wikipedia.org/wiki/History_of_quaternions

http://fexpr.blogspot.be/2014/03/the-great-vectors-versus-quaternions.html


Question clarification. This is an exercise question:

"Let $V$ be an open subset of $\mathbb{R}^3$. Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"

What exactly does it mean?

Is the question: "Given an open subset of $\mathbb{R}^3$, does there always exist a continuous injective function $f:V \rightarrow \mathbb{R}$?"?
$$
\forall V\subseteq \mathbb{R}^3:\ V \text{ is open } \Rightarrow \exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective}
$$

Or is it maybe:
"Does there exist a continuous injective function $f:V \rightarrow \mathbb{R}$? where $V$ is an open subset of $\mathbb{R}^3$.

$$
\exists f:\ V \rightarrow \mathbb{R} \text{ continuous and injective}, V\subseteq \mathbb{R}^3:\ V \text{ is open }
$$

I honestly do not have a clue.
How I would go about (dis)proving this would very much depend on my understanding.

The statement can be reformulated to

  Let $V\subset \mathbb R^3$ be open. Under what circumstances does there exist a continuous injective function $f:V\to\mathbb R$?

If you really want to make it formal, you can write

  Find a property $P$ such that
  $$\forall V \subset \mathbb R^3 \text{ open}: P(V) \implies \exists f:V\to \mathbb R\text{ continuous and injective}$$

The answer would then be $P(V) := V = \emptyset$.

If $V$ is the empty set, there is such an $f$ (called the empty function).
It doesn't map anything and since $\mathrm{dom}(f) = \emptyset$, all statements about preimages of $f$ are vacuously true, hence $f$ is injective.

If $V$ isn't empty, it contains some open ball $B_\epsilon(x_0) \subset \mathbb R^3$. Since there is no injective map from $B_1(0)$ to $\mathbb R$ (and any $B_\epsilon(x_0)$), there cannot be an injective map from $V$ to $\mathbb R$.

Thus, the answer to the question is:

  If $V\subset \mathbb R^3$ is open, there exists a continuous function $f:V\to\mathbb R$ if and only if $V=\emptyset$.

Use the following fact:

  There is no locally injective continuous function $f:[0,1] \to \Bbb{R}$ such that $f(0)=f(1)$.

Now, since $V$ is open, you can build a closed curve (for example, the equator of a sphere) inside some ball contained in $V$. This means that you are able to find some $\gamma: [0,1] \to V$ with $\gamma(0)=\gamma(1)$, which parametrizes your curve. You can suppose WLOG that $\gamma|_{[0,1)}$ is injective.

If $f: V \to \Bbb{R}$ is injective and continuous, then $f \circ \gamma$ is a locally injective continuous function and $f \circ \gamma(0) = f \circ \gamma(1)$: a contradiction.


If $\sum\limits_{n=1}^\infty a_n$ converges and $a_n\geqslant 0$ for every $n$ then $\sum\limits_{n=1}^\infty a_n^{1/4}n^{-4/5}$ converges If $a_n \in \mathbb{R}^+$ and $\sum\limits_{n=1}^\infty a_n$ is convergent, $\sum\limits_{n=1}^\infty a_n^{1/4}n^{-4/5}$ converges.

How to prove this?

Hint Use Hölder's inequality for series.

Edit: since this answer was accepted, I think @cmi found the solution, so here
is my version: Hölder's inequality gives

$$\sum _{n = 1}^{\infty } {a}_{n}^{\frac{1}{4}} \frac{1}{{n}^{\frac{4}{5}}}  \leqslant  {\left(\sum _{n = 1}^{\infty } {\left({a}_{n}^{\frac{1}{4}}\right)}^{p}\right)}^{\frac{1}{p}} {\left(\sum _{n = 1}^{\infty } {\left(\frac{1}{{n}^{\frac{4}{5}}}\right)}^{q}\right)}^{\frac{1}{q}}$$

when $\frac{1}{p}+\frac{1}{q} = 1$ and $p , q  \geqslant  1$. We choose $p = 4$ and
$q = \frac{4}{3}$. It gives

$$\sum _{n = 1}^{\infty } {a}_{n}^{\frac{1}{4}} \frac{1}{{n}^{\frac{4}{5}}}  \leqslant  {\left(\sum _{n = 1}^{\infty } {a}_{n}\right)}^{\frac{1}{4}} {\left(\sum _{n = 1}^{\infty } \frac{1}{{n}^{\frac{16}{15}}}\right)}^{\frac{3}{4}}$$

Both series on the RHS of this inequality converge, hence the series on the LHS also
converges.

This isn't a complete answer, since it requires monotonically of $a_n$
If $a_n$ is monotonic, then

Define $$b_n=\frac{a_n^{1/4}}{n^{\frac{4}{5}}}
, c_n=\frac{1}{n^{4/5+{1/4}}}$$

Then by the limit comparison test, define $ L :=\lim_{n\to \infty} \frac{b_n}{c_n}$, since $c_n$ converges, it is enough to show that $L \ne \infty $, (notice if $L=0$, $b_n$ still converges.)

$$\lim_{n\to \infty} \frac{\frac{a_n^{1/4}}{n^{\frac{4}{5}}}}{\frac{1}{n^{\frac{4}{5}+\frac{1}{4}}}}=\lim_{n\to \infty}a_n^{\frac{1}{4}}n^{\frac{1}{4}}=\lim_{n\to \infty} (a_nn)^{\frac{1}{4}}=0$$

Since $a_n$ converges, $(a_nn)^C \to 0$ for every $C$, (I will leave this part for you to prove, hint: Cauchy convergence test)


calculating $\int_{-\infty}^\infty e^{iat^2}\,{\rm d}t$ I want to show

$$\int_{-\infty}^\infty e^{ist^2}\,{\rm d}t=\pi^{1/2}e^{is\pi/4}$$

where $s$ is the sign of a function from a previous part of this question so just treat it as a constant (either $+1$ or $-1$).

My guess is to use contour integration and the residue theorem: I have taken a semicircular (radius $R$) clockwise contour $C$ tracing from $0$ to $Ri$ then taking an arc round to $-Ri$ then tracing back up to $0$. Then we consider the integral

$$\oint_Ce^{-iz^2}\,{\rm d}z$$

which is zero by Cauchy's theorem. Then separate $C$ into two parts namely

$$0 = \int_{-R}^Re^{-ist^2}dt + \int_{\pi/2}^{3\pi/2}\exp\left[-is(R^2e^{2i\theta})\right]iRe^{i\theta}\,{\rm d}\theta$$

but I'm stuck on what I should do next. Any help is great thanks. Also, I have seen this post but they don't seem to get the correct answer that agrees with what I am supposed to show so I have decided to re ask this question.

First integral, for $s=1$
\begin{align}
2\int \mathrm{e}^{it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (i-0) = \sqrt{i}\sqrt{\pi} = \mathrm{e}^{i\pi /4}\sqrt{\pi}
\end{align}

Second integral, for $s=-1$
\begin{align}
2\int \mathrm{e}^{-it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{-x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{-it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (1-0) = \frac{1}{\sqrt{i}} \sqrt{\pi} = \mathrm{e}^{-i\pi /4}\sqrt{\pi}
\end{align}

And we have
\begin{equation}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{ist^{2}} dt = \mathrm{e}^{is\pi /4}\sqrt{\pi}
\end{equation}
for $s=\pm 1$


Is there an object in reality that is proven to be uncountable? I've always wanted to come up with a fairly concrete example of an object or realistic set that could be uncountable. Most of the sets I can think about, even the hugest ones, are always countable. This could be useful to explain to a layperson the difference between countably and uncountably infinite sets.

Can anyone come up with a viable example?

That depends.

If by "realistic" you mean something that has to do with physical reality, then I defy you to come up with a set which has exactly $200^{200^{200}}$ elements.

If by "realistic" you mean something which comes up naturally in mathematics, then $\Bbb R$ is an uncountable set.

As for explaining the difference between them? That's not very easy, because first you need to be sure that the person understands the difference between sets of size $200^{200^{200}},200^{200^{200}}+1$ and $\aleph_0$. Which is most likely not going to be very easy. Sure, two of them have a finite number of elements, but it's so large it's infinite for all practical purposes. You couldn't even tell them apart if you put the two sets one right next to the other. If they can manage the difference there, then it's not difficult to explain what's "uncountable". Just infinite and not countable.

Unfortunately, mathematicians undergo a difficult training to work with definitions, rather than "common sense intuition" that we have before our studies. So explaining something that had to be earned by hard work is never easy. If it were, we wouldn't have to work so hard to get it.

Set of stars is infinite in reality. Infact it is countably infinite.

Another example of infinite set is "set of all points in a line segment is an infinite set"

See this it says the that set of integers is countably infinite whereas set of real numbers is uncountably infinite.

In countably infinite sets you cannot count the total number of elements but at least you can do the counting. Set of integers is an infinite set but you can count the elements in ascending or descending order(take any starting element).

In uncountably infinite sets you cannot even count because in between any two real numbers their are infinitely many reals. Here you cannot even do the counting! (suppose you are starting with 0 the you cannot find a number which is just greater then 0, if you think of 0.1 but 0.01 is closer to 0 to get more closer you have 0.001 and so on) this shows that you cannot do the counting.

(Remember you have to do the counting in either ascending or descending order without missing any element in between)

You can find a formal definition of uncountable sets here.

You can still compare the cardinality of two sets. If there exists a bijection between two sets then the cardinality of these two sets are equal. The cardinality of $\mathbb R$ and $\mathbb R^+$ is same! (See this)


Indefinite integral of absolute value When I looked up about absolute value on Wikipedia, I found that the antiderivative of $|x|$ is $\frac12 x|x|+C$. I am able to find the derivative of $|x|$ by treating the function as $\sqrt{x^2}$, but I am not able to integrate it.

When I put $\int_{-4}^{-1}|x|\,dx$ into Symbolab, the online calculator did not break the integral into piecewise function but calculate the indefinite integral first before using $F(b) - F(a)$. When I view the steps it used, it said:

If $\int f(x)\,dx = F(x)$
then $$\int \sqrt{(f(x))^2)}\,dx = \frac{\sqrt{f(x)^2}}{f(x)}$$ multiplied to $F(x)$
which becomes $\frac{\sqrt{x^2}}{x}$ multiplied to $\int x\,dx$

The function ${\rm abs}$ is continuous on all of ${\mathbb R}$, hence should have primitives $F$ defined on all of ${\mathbb R}$. Given that ${\rm abs}$ is "special" at $x=0$ we should look for the primitive obtained by integrating from $0$ to $x$. In this way we obtain
$$F(x)=\int_0^x |t|\>dt=\int_0^x t\>dt={x^2\over2}\qquad(x\geq0)$$
and
$$F(x)=\int_0^x |t|\>dt=\int_0^x (-t)\>dt=-{x^2\over2}\qquad(x\leq0)\ .$$
The two partial results can be condensed into the single formula
$$F(x)={x\>|x|\over 2}\qquad(-\infty<x<\infty)\ ,$$
and adding an arbitrary constant $C$ gives the general primitive of ${\rm abs}$.


$\# V(I) \le \text{dim}_K(K[X_1, \ldots, X_n]/I)$ Let $K$ be an algebraically closed field and $I \subseteq K[X_1, \ldots, X_n]$ be a zero-dimensional ideal, i.e. the set $V(I)$ is finite. This is equivalent to $\text{dim}_K (K[X_1, \ldots, X_n]/I)<\infty$. But the proof I know for this equivalence does not highlight any relation between $V(I)$ and $\text{dim}_K (K[X_1, \ldots, X_n]/I)$. I found recently that, in fact, one has the following inequality:
$$\# V(I) \le \text{dim}_K (K[X_1, \ldots, X_n]/I)$$
Here $\# V(I)$ denotes the number of points in $V(I)$. I am trying to prove this result. I know that the dimension of $K[X_1, \ldots, X_n]/I$ over $K$ is precisely the number of monomials from $K[X_1, \ldots, X_n]$ which do not belong to the initial ideal of $I$ (the classes of these monomials modulo $I$ form a $K$-basis of the vector space $K[X_1, \ldots, X_n]/I$). How can I relate these monomials with the number of solutions $(x_1, \ldots, x_n)$ of the set of equations $\{f(x_1, \ldots, x_n) = 0 \ | \ f\in I \}$?

At @KReiser's suggestion, I'll post the solution of this problem. Let me denote the ring $K[X_1, \ldots, X_n]$ with $R$.
The key observation is the following: Since $K$ is algebraically closed, all the maximal ideals of $R$ are exactly ones of the form $(X_1-c_1, \ldots. X_n-c_n)$, with $c_i \in K$, so there is a bijection between the elements of $V(I)$ and the maximal ideals of $R$ which contain $I$, namely $(a_1, \ldots, a_n) \rightarrow (X_1-a_1, \ldots. X_n-a_n)$.
Now $R/I$ is a finite dimensional $K$-algebra, and as is shown here using CRT, the number of maximal ideals of $R/I$ (which is the number of maximal ideals of $R$ which contain $I$) is less than or equal to $\text{dim}_K(R/I)$.

Here you get an approach to the problem over any field or Artinian ring:
Isomorphism of product of local ring of variety
I copy the post given in the above link here:
Let $B:=k[x_1,..,x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal.
Let $X:=Spec(A)$ with $A:=B/I$. By the Noether normalization lemma it follows there is a finite set of elements $y_1,..,y_d \in A$ with $d=dim(X)$ and the property that the sub-ring $k[y_1,..,y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1,..,y_d] \subseteq A$ is an integral extension of rings. If $d=0$ it follows from the same lemma the ring extension $k \subseteq A$ is integral and in particular it follows $A$ is an artinian ring. In particular it follows $A\cong R_1 \prod \cdots \prod R_l$ is a direct product of artin local rings $R_i$ with $\mathfrak{p_i} \subseteq R_i$ the maximal ideal (this is Atiyah-Macdonald's book Thm 8.7). Hence $Spec(A) \cong \prod Spec(R_i)$ is a disjoint union of the schemes $U_i:=Spec(R_i)$. We may write $X:=\{ \mathfrak{p_1}, \ldots , \mathfrak{p_l}\}$ as a disjoint union of $l$ distinct points, with $\mathfrak{p_i} \subseteq  R_i$ the unique maximal ideal in the local ring $R_i$. Since $U_i \subseteq X$ is an open subscheme it follows there are isomorphisms $\mathcal{O}_{X,\mathfrak{p_i}} \cong \mathcal{O}_{U_i,\mathfrak{p_i} } \cong (R_i)_{\mathfrak{p_i}} \cong R_i$ since $R_i$ is a local ring and every element in $R_i - \mathfrak{p_i}$ is a unit. Here $\mathcal{O}_{U_i}$ is the restriction of $\mathcal{O}_X$ to the open set $U_i$. Hence for any base field $k$ it follows from Atiyah-Macdonalds (AM) book "Commutative Algebra" that your ring $A:=B/I$ decompose as a finite direct sum of Artinian local rings $R_i$. At the level of sheaves you get the following formulas:
$\mathcal{O}_X \cong  \mathcal{O}_{U_1} \oplus \cdots \oplus \mathcal{O}_{U_l}$
and
$  \mathcal{O}_{X,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{X,\mathfrak{p_l}} \cong  \mathcal{O}_{U_1,\mathfrak{p_1}} \oplus \cdots \oplus \mathcal{O}_{U_l,\mathfrak{p_l}} \cong \mathcal{O}_{U_1}(U_1) \oplus \cdots \oplus \mathcal{O}_{U_l}(U_l) .$
These formulas answers the following questions:
Question 1: "But if k is not algebraically closed, is it still true that, if V(I) is finite, k[X1,…,Xn]/I is a direct product of local rings?"
Answer 1: Yes, this is true by the above argument.
Question 2: "And if so, then some factors are most likely given by localizations at the points of V(I), whereas other factors are not; how to describe these latter factors?"
Answer 2: All factors are given as localizations at points by the above argument.
Consider your example with $A:=\mathbb{Q}[x,y]/(x^2+y^2-3)$ and let $C:=Spec(A)$. By Zorn's Lemma (if you accept this) there is a maximal ideal $\mathfrak{m} \subseteq A$, and the residue field $\kappa(\mathfrak{m})$ is a finite extension of $\mathbb{Q}$. Hence $C$ - which as a set is the set of prime ideals in $A$ - is non-empty. If $k$ is the algebraic closure of $\mathbb{Q}$ and you pass to the fiber product $C\otimes k$ you get a curve. By Hartshorne Prop I.1.13 it follows $dim(C\otimes k)=1$ since it is defined by a hypersurface in $k[x,y]$.
Let $k$ be any field and let $A:=k[x]$ be the polynomial ring in the variable $x$. Let $f(x) \in A$ be any polynomial. We may write $f(x)=\prod_{i=1}^l p_i(x)^{d_i}$ where $p_i(x)$ is an irreducible polynomial and  $d_i \geq 1$ is an integer for every $i$. Here $p_i(x)$ and $p_j(x)$ are distinct for $i\neq j$. Let $B_i:=k[x]/(p_i(x)^{d_i})$ and let $B:=A/(f(x))$.
There is a canonical isomorphism
$\phi: B \rightarrow B_1\oplus \cdots \oplus  B_l$
defined by
$\phi(x + (f(x))) :=(x+(p_1(x)^{d_1}),..,x+(p_l(x)^{d_l}))$.
This is Prop 1.10 in the AM-book. The ideals $(p_i(x)^{d_i})$ and $(p_j(x)^{d_j})$ are coprime for $i\neq j$. Since the ideals $(p_i(x))$ are maximal, it follows the rings $B_i$ are Artinian local rings, hence the decomposition given by $\phi$ decompose the Artinian ring $B$ into a direct sum of Artinian local rings $B_i$. Hence in this case Prop 1.10 in AM implies Thm 8.7 in the AM book (the "stucture theorem on Artin rings").
Question 3: "And is there an higher-level explanation why it (apparently, not quite sure) works in one variable, that is, k[X]/f is a direct product of local rings for any f∈k[X], even if k is not algebraically closed, and even if V(f) is empty?"
Here you must understand the fact that any polynomial $f(x) \in k[x]$ factors as a product of powers of irreducible polynomials, and that an irreducible polynomial defines a maximal ideal in $k[x]$. Moreover if $p(x)$ is an irreducible polynomial, it follows the ring $k[x]/((p(x)^d)$ for $d\geq 1$ is an Artinian ring. Then you must read AM, Chapter1 where it is proved that the map $\phi$ is an isomorphism of rings.
Here the copied post ends.
This is an answer to your question: In your case you may write
$$D1.\text{   }A \cong \oplus_{i=1}^l R_i$$
where $R_i$ is a local ring with maximal ideal $\mathfrak{p}_i$. It follows
$$dim_k(A) :=d = \sum_{i=1}^l dim_k(R_i)$$
hence $l \leq d:=dim_k(A)$$.
Question: "I found recently that, in fact, one has the following inequality:
$$\# V(I)≤dim_k(k[x_1,…,x_n]/I)$$
I am trying to prove this result."
Answer: Here I interpret $\# V(I)$ as the "number of points" in the scheme $V(I)$. By the above argument it follows the number of maximal ideals in $A$ is less than or equal to $dim_k(A)$. Here $k$ is any field. Hence the general case (over any field) can be proved using methods from Atiyah-Macdonalds book "Commutative Algebra". In fact:
It seems to me the methods introduced above proves the result for any Artinian ring $A$ over any field $k$. Let $V(I) \subseteq Spec(A[x_1,..,x_n]):=\mathbb{A}^n_A$ and let $B:=A[x_i]/I$.  It follows
$$ \# V(I) \leq length_A(B)$$
where $length_A(B)$ is the "length" of $B$ as left $A$-module.
Formula $D1$ gives
$$D1.\text{   }B\cong R_1 \oplus \cdots \oplus R_l$$
where $R_i$ is a local ring, and each $R_i$ is an $A$-algebra. We get
$$length_A(B) = length_A(\oplus_{i=1}^l R_i)=\sum_{i=1}^l length_A(R_i).$$
Hence $\# V(I) \leq length_A(B)$.


How to prove $\sin 3A+\cos 3A=(\cos A-\sin A)(1+2\sin 2A)$ How to prove the following equation?
\begin{eqnarray}
\sin 3A+\cos 3A&=&\left(\cos A-\sin A\right)\left(1+2\sin 2A\right)\\
\end{eqnarray}

Let's start with the left hand side.
\begin{eqnarray}
LHS&=&sin 3A+\cos 3A\\
&=&\sin \left(2A+A\right)+\cos \left(2A+A\right)\\
&=&\sin 2A\cos A+\cos 2A\sin A+\cos 2A\cos A-\sin 2A\sin A\\
&=&\left(2\sin A\cos A\right)\cos A+\left(\cos ^2A-\sin ^2A\right)\sin A+\left(\cos ^2A-\sin ^2A\right)\cos A-\left(2\sin A\cos A\right)\sin A\\
&=&\cos ^3A-\sin ^3A-\sin ^2A\cos A+\sin A\cos ^2A-2\sin ^2A\cos A+2\sin A\cos ^2A\\
\end{eqnarray}

Than, I have no idea what I should do.

Just try to expand the RHS.

\begin{eqnarray}
RHS&=&\left(\cos A-\sin A\right)\left(1+2\sin 2A\right)\\
&=&\cos A-\sin A+2\sin 2A\cos A-2\sin 2A\sin A\\
&=&\cos A-\sin A+4\sin A\cos ^2A-4\sin ^2A\cos A\\
&=&?\\
\end{eqnarray}

Maybe there are something wrong.

Anyone can tell me what I should do?

Thank you for your attention.

Using$$\sin3A=3\sin A-4\sin^3A,\cos3A=4\cos^3A-3\cos A$$

$$\sin3A+\cos3A=3\sin A-4\sin^3A+4\cos^3A-3\cos A=3(\sin A-\cos A)-4(\sin^3A-\cos^3A)$$

Now, $\displaystyle\sin^3A-\cos^3A=(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)=(\sin A-\cos A)\left(1+\frac{\sin2A}2\right)$

LHS
sin(2A+A)+cos(2A+A)
= sin2A.cosA + cos2A.sinA + cos2A.cosA- sin2A.sinA
= 2sin A.cos^2A +(cos^2A-sin^2A).sinA + (cos^2A-sin^2A).cosA -2sin^2A.cosA
=cos^3A-sin^3A + sinA.cosA(2cosA+cosA- sinA-2sinA)
=(cosA-sinA)(cos^2A+cosA.sinA+sin^2A) +sinA.cosA(3cosA-3sinA)
=(cosA-sinA)(1+cosA.sinA)+3sinAcosA(cosA-sinA)
=(cosA-sinA)(1+4cosA.sinA)
=(cosA-sinA)(1+2sin2A)
=RHS


How many ways to add to $50$ with $1, 4$ and $6$ So I get how to find the number of ways of getting a number with $2$ sets of numbers (ex: using $1$s and $2$s to get $10$). But I have no clue where to start to figure out how many ways to get a number using $3$ sets of numbers:

  How many ways can you add to $50$ by using $1$s, $4$s, and $6$s?

Order matters and they count as a way ($1+4$ and $4+1$ is $2$ ways), but all $1$s or all $2$s in different orders count as one way ($1+1$ and $1+1$ is $1$ way).

Is there like a formula that I could use to figure this out? Thanks.

From the comments:

  I tried using brute force with 2 numbers and found a pattern where the next number of ways = number of ways now + previous number of ways. But using brute force with 3 numbers I'm cant seem to find a pattern.

You can write a recurrence.  You can get $50$ by getting $49$ and adding $1$, by getting $46$ and adding $4$, or by getting $44$ and adding $6$.  If $A(n)$ is the number of ways to get $n$ you have $$A(n)=A(n-1)+A(n-4)+A(n-6)\\A(0)=1\\A(n)=0 \quad n \lt 0$$
Unfortunately, the polynomial that comes out of this, $x^6-x^5-x^2-1,$ cannot be factored easily except for the factor $x+1$ so you are into using numeric approximations for the roots.

Added:  to do it with a calculator you make a column for $n$ with the numbers from $0$ to $50$ and a column for $A(n)$.  Start from the top, putting $A(0)=1$ because there is one way to add to $0$-don't add anything.  Then for each $n$ compute the sum of the lines $1, 4, \text {and} 6$ above.  The result up to $n=20$ is shown below.  For example, you get $A(9)=A(8)+A(5)+A(3)=10+3+1=14$ and keep going.  The image is from Excel, where I could just copy the formula down.

Hint:

Let $f(n)$ be the number of sums that evaluate to $n.$ So $f(5)=3$  because
$$5=1+1+1+1+1=1+4=4+1$$

The sums that evaluate to $50$ can be splittet into $3$ groups

the group of sums that starts with $1$
the group of sums that starts with $4$
the group of sums that starts with $6$

How can the size of these groups be expressed by $f$?

How can this be used to express $f(50)?$

Finally, calculate $f(50)$ but try to avoid to calculate values you have already calculated otherwise the number of calculations will explode.


Show that, in a chessboard it is possible to traverse to any given square from another given square using a knight. Show that, in a chessboard it is possible to traverse to any given square from another given square using a knight.

This was asked in a high school math competition.

There are no calculations necessary here, math is not all calculations. What you need to do is provide a proof (an irrefutable explanation) for why it's possible.

One possible proof (only works on an 8x8 board):

There is a knight's tour on the 8x8 board (a cycle of knight moves that reaches all squares):

[Insert knight's tour here]

To get from square A to square B, the knight can just start at A and follow the tour until he's at square B.

Another possible proof (using an intermediate step, but works on any board size 4x4 or greater and does not require knowing a knight's tour):

Claim: It is always possible for the knight to reach the top left square.

If the claim is true, then a square can go from square A to square B by navigating from square A to the top left square, then from that square to square B by moving backwards along the path from B to the top left square.

Proof that the claim is true:

The knight can always go up-up-left or up-up-right until he reaches the top two rows. Then, while staying within those top two rows, the knight can go left-left-up or left-left-down to reach the left two columns. Now the knight is in one of the four squares in the top left corner.

From those squares, it's easy to reach the top left square, using the paths shown in the following image:

I need to admit that I originally misread the question.
I think the easiest way to prove the statement is to show that given any square and any adjacent square there exists a sequence of moves such that the knight moves from the first to the second.

My feeling is that is enough (via some symmetry arguments) that starting on $a1$ the square $a2$ can be reached. I admit that I haven't thought this in detail, but it should be sufficient.
And this can be done via the route $a1-b3-c1-a2$.


What kind of curve is this? In fact, I am not good at maths but good at developing software. My current project involves implementing a method, for interpolating a curve.

But I can´t figure out what kind of curve this is. I have analyzed a curve, which I want to "reproduce". Maybe you guys can tell me, which type of curve this is (click to enlarge):

I've used following control points (don't get me wrong, this curve is generated by a different software, I wan't to reproduce this by implementing the method in my program):

cpx = [0, 9.48732, 37.7707, 53.0929, 83.0766, 124.333,128.004];
cpy = [0, -89.9931, 19.9962, -89.9986, 0, -89.9991,0];

There are several common ways to perform interpolation. The two most common methods use polynomials and splines (usually cubic splines). Both methods are described on this page.

Just from looking at it, I would guess that the example curve you showed is a cubic spline. But that's just a guess.

If you have a choice, I would suggest using cubic splines. If you have $n+1$ points, then, to use a polynomial interpolant, you'd need one of degree $n$. High degree polynomials are expensive to evaluate, and they tend to be "wiggly". Splines don't suffer from either of these problems.

There are plenty of software packages available for constructing cubic splines. A few of them are referenced on this page. The one you choose will depend partly on the programming language you're using.

Lagrange interpolation theorem states that (in programming parlance) given two arrays of  numbers of equal length $n$ there is a polynomial function of degree $n$ or less, such that applying that function to arguments from the first array outputs the second array.


For which values of $x$ is $ \sum_{n=1}^{\infty} \frac{(2n)!\,x^n}{2^{2n}(n!)^2}$ convergent or divergent? For which values of $x$ is $$ \sum_{n=1}^{\infty} \frac{(2n)!\,x^n}{2^{2n}(n!)^2}$$ convergent or divergent?

• I tried $x=-1$ and i tested it by using Comparison Test and Ratio Test but that was diverge. This series is not absolutely converge for $x=-1$ if I am right.

• Any suggestions?

If $x\ne0$, we have\begin{align}\lim_{n+\infty}\frac{\left|\frac{(2(n+1))!x^{n+1}}{2^{2(n+1)}((n+1)!)^2}\right|}{\left|\frac{(2n)!x^n}{2^{2n}(n!)^2}\right|}&=\lim_{n+\infty}\frac{(2n+2)(2n+1)}{4(n+1)^2}|x|\\&=|x|\end{align}and therefore the series converges absolutely if $|x|<1$ and diverges if $|x|>1$.

The sequence $\left(\frac{(2n)!}{2^{2n}(n!)^2}\right)_{n\in\Bbb N}$ decreases because, by the computation made above, if you divide the $(n+1)$th term by the $n$th, you get$$\frac{(2n+2)(2n+1)}{4(n+1)^2}=\frac{2n+1}{2n+2}<1.$$Also, it follows from the equality of the previous line that$$2^{2n}(n!)^2=\prod_{j=1}^n\frac{2j-1}{2j}.\tag1$$Now, note that$$\left(\frac{2j-1}{2j}\right)^2<\frac j{j+1},$$since this means that $4j^3-3j+1<4j^3$; in other words,$$\frac{2j-1}{2j}<\sqrt{\frac j{j+1}}.\tag2$$But then it follows from $(1)$ and $(2)$ that$$\frac{(2n)!}{2^{2n}(n!)^2}<\prod_{j=1}^n\sqrt{\frac j{j+1}}=\sqrt{\frac1{n+1}},$$and so, by the squeeze theorem, $\lim_{n\to\infty}\frac{(2n)!}{2^{2n}(n!)^2}=0$. Since it decreases and its limit is $0$, it follows from the Leibniz test that your series converges when $x=-1$.

But it diverges when $x=1$. You will find a proof here.

$${\sum_{n=1}^{\infty} \dfrac{(2n)!x^n}{2^{2n}(n!)^2}} = {\frac{1+{\sqrt{1-x}}}{\sqrt{1-x}}}$$

It's will be convergent if $\vert{x}\vert < 1$, by Ratio Test

Note that this can be checked using Wolfram Alpha, with this formula: series (((2n)!x^n)/(2^(2n)(n!)^2)) from 1 to inf


Show that if $ \ A \cup B = A$ and $ \ A\cap B = A$ then $ \ A = B$ Question:

Show that if $ \ A \cup B = A$ and $ \ A\cap  B = A$ then $ \ A = B$

My attempt:

Proof by contradiction:

Assume $ \ A \cup B = A$ and $ \ A\cap  B = A$ and $ \ A \neq B$

Case 1: $ \exists \ x \in A, x\notin B$

If $ x \in A \implies x\in A \cap B \implies x \in A \ and \  x\in B
\implies x\in B$, a contradiction.

Case 2: $ \exists \ x \in B, x\notin A$

If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.

Is this approach correct? Could someone please show me how to do a direct proof?

Your proof is correct.

Also,

$$A \cup B = A \implies B \subseteq A$$

$$A \cap B = A \implies A \subseteq B$$

Hence $A=B$.

Part 1: claim $A\subseteq B$.

Let $x\in A$. Claim $x\in B$.

Since $A\cap B=A$, $x\in A\cap B$. By the definition of intersection of two sets, we know that $x\in A$ and $x\in B$. Hence we have shown that $x\in B$. Since $x$ is arbitrary, for any $x$, if $x\in A$, then $x\in B$. Therefore, $A\subseteq B$.

  Part 2: claim $B\subseteq A$.

Let $x\in B$. Claim $x\in A$.

$x\in B$ implies that $x\in A$ or $x\in B$. By the definition of union of two sets, we know that $x\in A\cup B$. Since $A\cup B=A$, we know that $x\in A$. Since $x$ is arbitrary, for any $x$, if $x\in B$, then $x\in A$. Therefore, $B\subseteq A$.

Combine the two parts, we know that $A=B$.


Is $\{f \colon \mathbb{N}\mapsto\mathbb{N}: f \text{ is a bijection}\}$ numerable? Let $A :=\{f \colon \mathbb{N}\mapsto\mathbb{N}: f \text{ is a bijection}\}$. Is $A$ numerable?

I don't know how to attack this problem, but I think that this set is not numerable. However, I can't see how to start the demonstration.

Given $X\subset \Bbb N$ such that both $X$ and $\Bbb N\setminus X$ are infinite, we can define an $f\in A$ by letting $f$ map the $k$th smallest element of $X$ to the $k$th smallest element of $\Bbb N\setminus X$ and vice versa.
To see that $|A|\ge|\mathcal P(\Bbb N)|$, note that for every subset $Y$ of $\Bbb N$ we can pick $X=3Y\cup (3\Bbb N+1)$ to obtain a different $f$ for every choice of $Y$.

Hint: Find an injection of the powerset $\mathcal P(\mathbb N)$ into $A$. Now apply Cantor's theorem.


Converting to a complex integral and evaluating with Residue theorem So I'm running into this problem as I'm trying to evaluate this integral:
$$ \int_0^{\infty} \frac{dx}{x^4+4x^2+4} $$

My work:
$$ \int_0^{\infty} \frac{dx}{x^4+4x^2+4} \rightarrow \frac{1}{2}\int_{-\infty}^{\infty} \frac{dz}{z^4+4z^2+4} $$
The poles of $z^4+4z^2+4=0$ are: $z= -i \sqrt{2},-i \sqrt{2},i \sqrt{2},i \sqrt{2}$
$$ \rightarrow \frac{1}{2}\int_{-\infty}^{\infty} \frac{dz}{(z -i \sqrt{2})^2 (z +i \sqrt{2})^2} = \frac{1}{2}(2\pi i \sum \text{residues}) $$
$$ Res(i \sqrt{2}) = \lim_{z\rightarrow i \sqrt{2}}(z -i \sqrt{2}) f(z) = \lim_{z\rightarrow i \sqrt{2}} \frac{(z -i \sqrt{2})}{(z -i \sqrt{2})^2 (z +i \sqrt{2})^2} = \lim_{z\rightarrow i \sqrt{2}} \frac{1}{(z -i \sqrt{2}) (z +i \sqrt{2})^2} = -\infty $$

I don't think this is right? How should I be evaluating this integral so I don't get the nasty business of $\frac{1}{0}$?

Any advice on how to tackle this would be appreciated.

EDIT: I wasn't using the formula for higher order poles, thanks for pointing it out everyone

You have a pole of order $2$ in $i\sqrt{2}$ then the residue there is
$$\lim_{z\to i\sqrt{2}}\left((z-i\sqrt{2})^2\frac{1}{z^4+4z^2+4}\right)'=-i\dfrac{\sqrt{2}}{16}$$
do this with other poles!

writing your integral in the form:
$$\int_{0}^\infty \frac{1}{(x^2+2)^2}dx$$
Substitute $$x=\sqrt{2}\tan(u)$$ then we get
$$\sqrt{2}\int_{0}^{\pi/2}\frac{\cos^2(u)}{4}du$$ and this is equal to
$$\frac{1}{2\sqrt{2}}\frac{1}{2}\cos(2u)+\frac{1}{2}du$$
substutute $2u=s$ then we have
$$\frac{1}{8\sqrt{2}}\int_{0}^\pi\cos(s)ds+\frac{1}{4\sqrt{2}}\int_{0}^\frac{\pi}{2}1du$$ and this is equal to
$$\lim_{b\to \pi^-}\frac{\sin(b)}{8\sqrt{2}}+\frac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi}{2}}1du$$ since the first Limit is Zero we get
$$\lim_{b\to \frac{\pi^-}{2}}\frac{u}{4\sqrt{2}}|_{0}^b=\frac{\pi}{8\sqrt{2}}$$