Proof or disprove: There exists a real $n \times n$ matrix $A$ that satisfies the equation when $n$ is even. Prove or disprove: There exists a real $n \times n$ matrix $A$ with
$$
A^2+2\cdot A+5\cdot I_n = 0
$$
if and only if n is even.
I could not find a counterexample for an odd $n$. Therefore, I suspect that the statement is true, but I have not yet found a solution.

If $n$ is odd, then $A$ has a a real eigenvalue $\lambda$. But then $\lambda^2+2\lambda+5=0$ and no such real number exists.
If $n=2$, consider$$A=\begin{bmatrix}-1&-2\\2&-1\end{bmatrix}.$$It will work because its eigenvalues are $-1\pm2i$, which are the roots of $\lambda^2+2\lambda+5$.
Can you generalize this for $n$ even and greater than $2$?

A possible generalization of Santos's answer:
Note that $A^T$ also satisfies the equation. Hence $A^2+2A+5I_n = {A^T}^2+2A^T+5I_n$ => $A^2-{A^T}^2=-2(A-A^T)$ => $(A+A^T)(A-A^T)+AA^T-A^TA = -2(A-A^T)$ => $(A+A^T+2I_n)(A-A^T)=-AA^T+A^TA$. Now take transpose on both sides: we now have:
$-(A+A^T+2I_n)(A-A^T)=-AA^T+A^TA$. Hence $(A+A^T+2I_n)(A-A^T)=0$. Since $A\neq A^T$ as this would mean $A$ is symmetric and real eigen values. A guess (not the answer) is to assume $A$ is skew symmetric and hence we have: $A= -A^T-2I_n$ and hence we have $a_{ii}=-a_{ii}-2$. Hence $a_{ii}=-1$. This is a possibility.
Also we have from original equation: $A(A+2I_n)/5 = -I_n$. Hence we have $-5A^{-1}=A+2I_n$. If we substitute this in $A= -A^T-2I_n$ then we get $-5A^{-1}=-A^T$=> $5I_n = A^TA$. hence one possibility is to choose $A$ as a skew symmetric orthogonal matrix with diagonal entries $-1$ with norm if each column $5$. This matches the idea of solution given by Santos.


Is the following a correct logical proof? A → (F ∧ P)
~A → (S ∧ R)
~R

∴ P

    assume ~P
        assume A
        F ∧ P (1, 5, modus ponens)
        P (6, simplification)
    ~A (4, 7, reductio ad absurdum)
    S ∧ R (2, 8, modus ponens)
    R (9, simplification)
P (3, 10, reductio ad absurdum)

The focus for my doubt here is at step 8 where I use the initial assumption at line 4 and the derivation of its negation at line 7 to infer that ~A is in fact the case given the two assumptions. My question is really: am I allowed to use the outer assumption towards negating my inner assumption?

I realize this can also be done by assuming A in the proof and then going from there, I just wanted to make sure this technique is licensed in general.

Every step almost ends up as correct up to a point.

The last line needs to have the same scope as the set of premises {1, 2, 3}.  That is the only incorrect step.

Yes, you can use the outer assumption towards negating your inner assumption.  Natural deduction systems assume a rule of repetition which allow you to repeat any assumption from an outer domain in a domain within that outer domain (the term 'domain' here as I recall originally got used by Jaskowski).  Or if they don't assume a rule of repetition, they assume that any formula in an outer domain can get used to derive a formula in an inner domain.  This works more in natural deduction systems with axioms.

All natural deduction systems either have the formula (p $\rightarrow$ (q $\rightarrow$ p)) as an axiom or as a theorem.  This allows you to use an outer assumption $\alpha$ and re-derive $\alpha$ within the scope of the inner assumption using modus ponens twice.

Thus, you can derive the contradiction within the scope of your assumption 'A' and thus discharge it using reductio ad absurdum.  That is, every step of your proof comes as valid up to a point.

That doesn't quite answer if this consists of a correct proof though, because in order to determine whether or not we have a correct proof, we need to have the exact definition of what a proof consists of for this formal system.

Additionally, you have many steps which are NOT correct in another sense.  They qualify as correct up to a point, but I do say confidently that they don't fit the definition of a proof.  A proof consists of a sequence of well-formed formulas which follow the formation rules exactly.  But, many of the sequences of symbols that you have written are simply NOT well-formed formulas.  And thus, your proof only works as correct up to a point.

That said, I'm sure someone would try and argue that converting what got written above can get converted to a proof easily and thus the above comes as "convincing" enough to qualify as a proof.  But, a proof consists of something objective, and since you haven't written well-formed formulas and opted for abbreviations, you haven't written something which satisfies the definition of a proof.

The OP has the following question:

  The focus for my doubt here is at step 8 where I use the initial assumption at line 4 and the derivation of its negation at line 7 to infer that ~A is in fact the case given the two assumptions. My question is really: am I allowed to use the outer assumption towards negating my inner assumption?

Line 4 is outside the subordinate proof starting on line 5. Lines 4 and 7 are contradictory which is what we want. However, one needs to add a new line which shows that contradiction in the subordinate proof. Then one can discharge the assumption $A$ on line 4.

Line 8 is likely incorrect without that extra line. However, that brings up the next question the OP has:

  I realize this can also be done by assuming A in the proof and then going from there, I just wanted to make sure this technique is licensed in general.

One way to verify that a technique is correct is to use a proof checker. Such software will force the user to conform to its format requirements and rule justifications. However, that implementation of natural deduction inference rules need not be the only way those rules could be implemented.

Here is the proof in one proof checker:

Note that line 8 is replaced by a symbol for contradiction (⊥). Some proof checkers may ask that this be entered as P ∧ ¬P.


Optimization approaches to solving PDEs In modern numerical methods, a PDE is often recast into the form of a variational problem, which is sometimes equivalent to a minimization problem.
However in my courses on numerical analysis (say, finite element methods) the focus is not (apparently) on developing optimization techniques to minimize the arosen energy functional, but rather on approximating the variational problem on a smaller subspace.
Are there interesting approaches that focus on the minimization of the energy directly? Is research being done in this field, and could you maybe provide some reference?

I'll try to answer to your questions following the order of their appearance.

Are there interesting approaches that focus on the minimization of the energy directly?

Yes, there are several approaches to the solution of this problem: they are all known under the collective name of "Variational Method". This "method" has its root in the direct method in the calculus of variation which was pioneered by Leonida Tonelli, and aims to find the minimum of a functional by directly evaluating its value on a properly defined (sub)sequence of functions for which it is defined, without the need to calculate its functional derivative and solve the associated Euler-Lagrange DEs. Contrary to some widespread belief, the variational method is widely applicable to general (system of) PDEs, as we'll see below.
Let's start our description by saying that, when a functional $F$ which is equivalent to the given (system of) PDEs has been constructed, the general variational method can be described by mimicking the standard approach to direct method in the calculus of variation:$\DeclareMathOperator{\DM}{\mathrm{Dom}}$

show that the functional $F$ is (sequentially) lower semicontinuous (or upper semicontinuous, if we are searching for a maximum instead of a minimum) respect to a topology chosen on its domain of definition $\DM F$.

chose a minimizing sequence i.e. a sequence of functions $\{u_n\}$ contained in the domain $\DM F$ such that
$$
\lim_{n\to +\infty} F[u_n]=\inf_{u\in\DM F}F(u)
$$

Show that the chosen sequence $\{u_n\}$ admits a subsequence $\{u_{n_k}\}$ covering to $u\in \DM F$ respect to the given topology. Then $u$ is the sought for solution.

Differently from what happens in the direct method, where the functional is customarily known from the start, step 1. for the variational method is customarily done during the construction of $F$. For example, the very general method for giving a variational formulation to any operator equation worked out by Enzo Tonti (see for example [3] and [1], chapter IV, §17, pp. 168-182) produces the a functional which has a minimum $u$ if and only if $u$ solves the initial operator equation, implicitly proving its semicontinuity.
Finally, what are the differences between the various procedures that go under this collective name? They use different methods to construct the minimizing sequence and its subsequence: for example the method of orthonormal series, the Ritz (or Rayleigh-Ritz) Method, the Galerkin Method or its variants as the Faedo- Bunbnov- Galerking Methods, the Least Squares Method, The Courant Method and The Method of Steepest Descent are all variational methods.

Is research being done in this field, and could you maybe provide some reference?

Yes, there is currently a large deal of investigation on the variational method, due to its wide applicability to standard problems in engineering and physics: try googling one of the names given above, just to see the number of the results that you'll get. As a basic reference for the standard techniques, I'd suggest [2], part III, chapters 9-17, pp. 113-178, while the problem of constructing a functional "equivalent" to a given (system of) PDEs, technically called "the inverse problem of the calculus of variation", is comprehensively addressed in references [1] and [3]: the first one gives also a very general procedure for solving very general nonlinear equations (chapter IV, §18, pp. 182-198) due to A. D. Lyashko.
Final notes

The approach used in the solution of the inverse problem of the calculus of variation is briefly addressed also in this and this Q&A.
As I briefly alluded to in the answer to the first part of the question, it is a widespread belief that only some PDEs (or more generally some operator equations) admit a variational formulation, even if the works of Tonti, Filippov and others have shown that the inverse problem of the calculus of variation has a generalized solution for large classes of operator equations. However, the belief is not entirely unjustified since it is true that there are large classes of functional formulations of PDE problems for which the functional derivative does not lead to the Euler-Lagrange equations: for example Copson [A1] proves that the ordinary heat equation (and more generally equations of parabolic type) is not the Euler-Lagrange equation of any functional. But it also clear from the works [1] and [2] that such kind of equations do admit a variational formulation.

References
[1] Filippov, Vladimir Mikhailovich, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Transl. from the Russian by J. R. Schulenberger. Transl. ed. by Ben Silver, Translations of Mathematical Monographs, 77. Providence, RI: American Mathematical Society (AMS). pp. xiii+239 (1989), ISBN: 0-8218-4529-2. MR1013998, Zbl 0682.35006.
[2] Karel Rektorys, Variational methods in mathematics, science and engineering, Translated from the Czech by Michael Basch. 2nd ed. (English), Dordrecht - Boston - London: D. Reidel Publishing Company, pp. 571 (1980), ISBN: 90-277-1060-0, MR0596582,  Zbl 0481.49002.
[3] Tonti, Enzo, "Extended variational formulation", Vestnik Rossiĭskogo Universiteta Druzhby Narodov, Seriya Matematika 2, No. 2, 148-162 (1995). Zbl 0965.35036.
Addendum reference
[A1] Edward Thomas Copson, "Partial differential equations and the calculus of variations" Proceedings of the Royal Society of Edinburgh 46, 126-135 (1926), JFM 52.0509.01.

A good start to gather some insight into PDE is partial_differential_equation. This shows up that still the knowledge based start is most important. Identify what the PDE under consideration is. Retrieve the historical solution and improvements and then pose your own specifics in details.
There are existence and uniqueness consideration and a wellposed examination needed.
Following the successful and skillfull classification there is for a mathematician in need of reputation the calculation of analytically closed solution. Based on this for inhomogenities a variation problem is to be solved.
If this is not possible some numerical experiments can be conducted. Most of these method solve the pde on finite points and lead to an interpolation function as a result.
The energy method can be used to verify well-posedness of initial-boundary-value-problems. That is a subset of pde and one that does fit into the finite kompartment methodologies too. Energy principles in structural mechanics lists some of the successful applications. This again follows the knowlegde approach. The examples are of eternal importance but the references in the article are a little outdated.
Whether pde or energy principle is easier or less complex in the solution depends on the CAS or FEM system in use. Many of these have the knwolegde incorporated into their codes and the learning curves are steep and their are tutorials available that recompile the knowledge to the extent to provides solutions in these systems.
If there is a trend than in modularity. Most designers of course know that their may be two or more to be successfully sold to the interested ambitioned scientist. A comparison of both article shows that the pde methodologies are described to some depth while the energy principle is not. Why is this the case? The pde methodologies are used more often and considered more comfortable. They are indeed posed more to the community. The energy principle on the other is an overview page and have below each specification another more detailed page for the problem categories covered.
It is necessary to use such great tools a mind maps on the wikipedia graph of linkes articles to gain more overview. The energy principles does no exist independent of the pde page but the path is longer than reproduced in the given question. This path has to be studied and that is the designers of modern pde courses much to much content. They are in need to over the steep learning of the FEM systems or programmin languages too.
The interested and ambitioned student can use the literature to fill in the gaps. Put aside the question of moderness of the literature. The wikipedia links offer a sorrow and convinient introduction and give opportunity to speed plenty of time to start with correspondences up to cohesion.


Stochastic integral : $\int_0^T (W(s))^2dW(s)$ How to evaluate this integral $$\int_0^T(W(s))^2 \, dW(s)$$
where $W(s)$ is random variable associated with brownian motion.

I am new to this .Thanks in advance.

$\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$

In comments under the question, you say your main concern is how to evaluate $\int_0^T W\,dt$.

By symmetry, $\mathbb E\left(\int_0^T W\,dt\right)=0$, so the substantial part is finding $\var\left(\int_0^T W\,dt\right)$.  If linearity of covariance in each argument separately applies to integrals and not merely to the finite case (so you may have to either think about how this can be shown, or cite a theorem in some book), then we have
$$
\cov\left( \int_0^T W(t)\,dt , \int_0^T W(s)\,ds \right) = \int_0^T\int_0^T \cov(W(t),W(s))\,dt\,ds
$$
$$
= \int_0^T\int_0^T \min\{s,t\}\,dt\,ds.
$$

Now do two integrals: one for the part of the square below the line $t=s$, and the other for the part above it.  (But they're the same, by symmetry.)

Let $f(W_t,t)=\frac{W^3}{3}$.

Ito's Lemma for Brownian motion is $df = (f_t+\frac{f_{ww}}{2}) dt + f_w dW$.

So $df = W dt + W^2 dW$

So $\int df = \frac{W^3}{3} + C =  \int W^2 dW + \int W dt + C$

So $\int W^2 dW = \frac{W^3}{3} -\int W dt + C$.

(This begs the question of What does it mean to integrate a Brownian motion with respect to time?.  This is discussed thoroughly here by Jonathan Goodman.)

So

$\int_0^T W_s^2 dW_s = \frac{W_T^3}{3} -\int_0^T W_s ds$

So

$E[\int_0^T W_s^2 dW_s] = E[\frac{W_T^3}{3} -\int_0^T W_s ds]
= E[\frac{W_T^3}{3}] - E[\int_0^T W_s ds]
= \frac{T}{3} E[W_T] - \int_0^T E[W_s] ds=0$

and

$E[(\int_0^T W_s^2 dW_s)^2] = E[(\int_0^T W_s^4 ds)] = (\int_0^T E[W_s^4] ds)
= (\int_0^T s^2 ds) = \frac{T^3}{3}$

by Itô isometry and Fubini's Theorem.


Contradictions between the Alternating Series Test & Divergence Test? By the Alternating Test: The series $\sum_{n=1}^{\infty}(-1)^{n+1}\cdot b_n$ converges if all three of the following conditions are satisfied:

The $b_n$'s are all positive.
The positive $b_n$'s are (eventually) decreasing: $b_n\ge b_{n+1}$ $\forall n\ge N$.
$b_n \to 0$

But in the Divergence Test: Given $\sum_{n=1}^{\infty} a_n$, iff $\lim_{n \to \infty} a_n \neq 0 \implies$  $\sum_{n=1}^{\infty} a_n$ diverges

Now given an arbitrary alternating series:

$$S = \sum_{n=1}^{\infty}(-1)^{n+1}\cdot b_n$$

If we take the limit of $a_n$ in the series above

$$\lim_{n \to \infty}(-1)^{n+1}\cdot b_n = \underbrace{\left(\lim_{n \to \infty }(-1)^{n+1}\right)}_\text{This limit doesn't exist}\left(\lim_{n \to \infty} b_n\right)$$

Therefore by the Divergence test, $S$ should be a divergent series, regardless of the conditions needed for the alternating test. But by the Alternating Series Test, $S$ is a convergent series provided the three conditions stipulated initially are met.

So how is this seeming contradiction resolved, by the alternating series test?

if $a_n b_n$ converges, then this does not imply that $a_n$ converges. In case $(-1)^n b_n$ the first factor is bounded. In this case, if the second factor converges to $0$, the product also converges to $0$.

$$\lim_{n\rightarrow\infty} 1 = \lim_{n\rightarrow\infty} (-1)^n(-1)^n =\left(\lim_{n\rightarrow\infty}(-1)^n\right)\left(\lim_{n\rightarrow\infty} (-1)^n\right).$$


What is the limit of $\frac{\prod\mathrm{Odd}}{\prod\mathrm{Even}}?$ What is this limit
  $$
  \frac{1\times3\times5\times\cdots}{2\times4\times6\times8\times\cdots} = \lim_{n \rightarrow \infty}\prod_{i=1}^{n}\frac{(2i-1)}{2i}
$$

I remember that it was something involving $\pi$.

How can I compute it?

In addition; how can I compute it's sum series limit?

If you take the numerator product up to $2n-1$ and the denominator up to $2n,$ you have exactly
$$ \frac{(2n)!}{4^n (n!)^2}  $$

By Stirling's approximation this is asymptotic to
$$ \frac{1}{\sqrt {\pi n}} $$
and goes to $0$ slowly. If we were to add one more term to the numerator, that is $2n+1,$ the expression would go to $\infty$ slowly, as a constant times $\sqrt n.$

The only simple way to get a limit is to take the numerator up to $2n-1$ as before, but also multiply the numerator by a single factor of $\sqrt n.$ Then the limit would be $1/ \sqrt \pi$

Here's a formula which I found embedded in an old C program. I don't know where this comes from, but it converges to Pi very quickly, about 16 correct digits in just 22 iterations:

$\pi = \sum_{i=0}^{\infty}{
\frac{6(\prod{2j-1})}
{(\prod{2j})(2i+1)(2^{2i+1})}}$

(Each product is for j going from 1 to i. When i is 0, the products are empty, equivalent to 1/1. When i is 1, the products are 1/2. When i is 2, the products are 3/4. Etc.)

The limit of the quotient of the products is 0, but the limit of the sum as I give it above is exactly equal to $\pi$.

As for the limit of the sum of the quotients of the products, it diverges to +$\infty$.


positivity of a function Consider the function $G(\omega):=\omega^4+2r^2\omega^2+b$, where $r,b\in\mathbb{R}$. What constraints on the parameters are needed to guarantee the positivity of $G$ for all $\omega>0$?

First observe that $G(0)>0$ implies $b>0$. On the other hand, if $b>0$ then $G(\omega)\geq b>0$. Hence for $G$ to be positive, it is necessary and sufficient that $b>0$.


By writing 1.05 as (1 + 0.05) and expanding by using binomial Theorem, you can clearly see that in all terms there is positive power in 0.05 ( except the first one) which is making it smaller. So it can never be greater than 100 or even leave 100 it can never be greater than 50.


Pólya's urn scheme, proof using conditional probability and induction Problem

An urn contains $B$ blue balls and $R$ red balls. Suppose that one extracts successively $n$ balls at random such that when a ball is chosen, it is returned to the urn again along with $c$ extra balls of the same color. For each $n \in \mathbb N$, we define $R_n=\{\text{the n-th ball extracted is red}\}$, and $B_n=\{\text{the n-th ball extracted is blue}\}.$

Prove that $P(R_n)=\dfrac{R}{R+B}$.

I thought of trying to condition the event $R_n$ to another event in order to use induction. For example, if $n=2$, I can express $$P(R_2)=P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)$$$$=\dfrac{R+c}{R+B+c}\dfrac{R}{R+B}+\dfrac{R}{R+B+c}\dfrac{B}{R+B}$$$$=\dfrac{R}{R+B}.$$

Now, suppose the formula is true for $n$, I want to show it is true for $n+1$.

So, $P(R_{n+1}=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)P(B_n)$$$=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n)$$$$=P(R_{n+1}|R_n)\dfrac{R}{R+B}+P(R_{n+1}|B_n)(1-\dfrac{R}{R+B}).$$

I am having some difficulty trying to calculate $P(R_{n+1}|R_n)$ and $P(R_{n+1}|B_n)$. I would appreciate if someone could complete my answer or suggest me how can I finish the proof if what I've done up to now is correct.

The key is to condition by the composition of the urn at time $n$, say $X_n$ red balls and $Y_n$ blue balls, since $$P(R_{n+1}\mid X_n,Y_n)=Z_n,\qquad Z_n=X_n/(X_n+Y_n).$$ Obviously, $X_0=R$, $Y_0=B$, and, for every $n$, $$X_n+Y_n=R+B+nc,$$ which is deterministic. Conditionally on $(X_n,Y_n)$, one adds $c$ red balls with probability $Z_n$ and zero otherwise, hence $$E(X_{n+1}\mid X_n,Y_n)=X_n+cZ_n=(X_{n+1}+Y_{n+1})Z_n,$$ which implies $$E(Z_{n+1}\mid X_n,Y_n)=Z_n.$$ In particular, for every $n$, $$P(R_{n+1})=E(Z_n)=Z_0=R/(R+B).$$

$P(R_1)=\frac{r}{r+b}$ and $P(B_1)=\frac{b}{r+b}$

Applying theorem of total probability we have :

\begin{eqnarray*}
P(R_2)&=&P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)\\
&=& \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&=&\frac{r}{r+b}
\end{eqnarray*}

Now we prove for $P(R_3)$

Again apply theorem of total probability:

\begin{eqnarray*}
P(R_3)&=&P(R_3|R_1)P(R_1)+P(R_3|B_1)P(B_1)
\end{eqnarray*}

Now question is what is $P(R_3|R_1)$ and $P(R_3|B_1)$? We will show $P(R_3|R_1)=P(R_2|R_1)=\frac{r+1}{r+b+1}$. How? -- It is as follows:

Apply theorem of total probability on conditional probability.
\begin{eqnarray*}
P(R_3|R_1)&=&P(R_3\cap R_2|R_1)+P(R_3\cap B_2|R_1)\\
&=& P(R_3|R_2\cap R_1)P(R_2|R_1) + P(R_3|B_2 \cap R_1)P(B_2|R_1)\\
&=& \frac{r+2}{r+b+2}\frac{r+1}{r+b+1}+\frac{r+1}{r+b+2}\frac{b}{r+b+1}\\
&=&\frac{r+1}{r+b+1}
\end{eqnarray*}

Same way one can show $P(R_3|B_1)=\frac{r}{r+b+1}$.

Now under the induction hypothesis we have :

$P(R_{n-1}|R_1)=\frac{r+1}{r+b+1}$ and $P(R_{n-1}|B_1)=\frac{r}{r+b+1}$

Therefore,

\begin{eqnarray*}
P(R_{n})&=& P(R_{n-1}| R_1)P(R_1) + P(R_{n-1}| B_1)P(B_1)\\
&=& \frac{r+1}{r+b+1}\frac{r}{r+b}+\frac{r}{r+b+1}\frac{b}{r+b}\\
&=&\frac{r}{r+b}
\end{eqnarray*}

This is a classic example of Markov Chain


Existence of solution as a power series on non-empty interval Consider the following differential equation:

$$ x''(t) = a_1(t)\, x'(t)+a_0(t)\, x(t) $$

With $a_0,a_1$ functions defined for every real number and continuous on $\mathbb{R}$.

Question: Does there exist a solution which can be written as a converging power series in a non-empty open interval $I$?

To be honest I don't even know the answer yet to the question. It seems maybe unlikely in some sense that this is true because there are several functions that are smooth but not analytic.

This is directly from "Elementary Differential Equations, by Rainville and Bedient", who cites the rigorous proof from "Rainville's Intermediate Differential Equation":

If $ a_1(t),  a_2(t) $ are rational functions of x with denominators that do not vanish at 0

$$x''(t) = a_1(t)y'(t) + a_2(t)y(t)$$
$$x'''(t) = a_1'(t)x'(t) + a_1(t)x''(t) + a'_2(t)x'(t) + a_2(t)x(t) $$

So, $x'''(0)$ can be computed once $x''(0)$ is known.
Using $x'''(t)$, $x''''(t)$ can be computed.  With $x''''(t)$ known, $x''''(t)$ can be computed.

The above process can be continued so long as we wish (with conditions on the derivatives of $ a_1(t),  a_2(t) $, of course).  Therefore, a power series exists based upon Maclaurin's formula, and the initial values $x(0), x'(0)$, is given as:

$$ x(t) = x(0) + x'(0) + \sum_{n=2}^\infty x^{(n)}(0)\frac{x^n}{n!}$$

Where the x' values are computed as in above.


Prove the existence or the non-existence of a couple of numbers ($n$,$m$) such that $n^2=m!$ In recent days, while I was doing exercises on combinatorics, I thought if a number $m!$ could be a perfect square. I proved to demonstrate it through the prime factorization. My attempt: $$k=m!=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}....\cdot p_n^{a_n}$$
with $a_1,a_2,..,a_n$ that have to be even. Now if there exists a prime number $p_j$ between $\frac {k}{2}<p_j<k$ this prime number in the prime factorization the exponent ($a_j$) relative to the prime number $p_j$ is equal to $1$ ($a_j=1$) because the number $k$ isn't a multiple of $p_j$; for example $12!$ isn't a perfect square because the prime numbers $7$ and $11$ $6<7,11<12$ aren't divisors of $12$.

My question

There exist a prime number $p_j$ between $k$ and $\frac {k}{2}$ and how can I prove if there exist or not exist a number $m!=n^2$?

Let $m\ge 3$ (otherwise $(m,n)=(0,\pm 1),(1,\pm 1)$).

Assume for contr. $\ n^2=m!\ $ for some $n\in\Bbb Z$.

Then $\ \exists p\in\Bbb P\ $ with $\ m/2<p<m\ $ by Berdrand's postulate.

$\,\Rightarrow\, p\mid m!\,\Rightarrow\, p\mid n^2\,\Rightarrow\, p^2\mid n^2\,\Rightarrow p^2\mid m!$. But $m<2p$, contr.

gcd is an associative binary operation. gcd(m,m-1)=1. Thus gcd of terms involved in m! is 1. It follows that m! cannot be a perfect square.


what would a planetary orbit look like if gravity had constant magnitude? Consider a unit-mass particle that is always experiencing a single unit-magnitude
force towards the origin.  This is a central force, but it is not one of the
familiar ones, e.g. gravity whose magnitude is proportional to inverse distance squared,
or a spring force whose magnitude is proportional to distance.
So the particle is always accelerating towards the origin
with constant acceleration magnitude $1$.
Stated as a differential equation,
working in the $x$-$y$ plane, the particle's position as a function of time
$\mathbf{p}(t){=}(x(t),y(t))$ satisfies:
$$\ddot{\mathbf{p}}(t) = -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert.$$
If we are additionally given initial position $\mathbf{p}(0)$
and velocity $\mathbf{v}(0)=\dot{\mathbf{p}}(0)$,
then the function $\mathbf{p}$ is completely determined, and it can be easily computed
numerically to any desired accuracy by simply iterating the following with small enough timestep $dt$:
\begin{align}
    \mathbf{a} &\leftarrow -\mathbf{p}/\Vert\mathbf{p}\Vert \\
    \mathbf{v} &\leftarrow \mathbf{v} + \mathbf{a}\,\,dt \\
    \mathbf{p} &\leftarrow \mathbf{p} + \mathbf{v}\,\,dt
\end{align}
My question: is $\mathbf{p}(t)$ a well-known function,
and does it have a closed form?
Of course one case of this is a simple circular orbit of unit radius and speed:
$$\mathbf{p}(0){=}(1,0), \,\, \mathbf{v}(0){=}(0,1) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(\cos t,\sin t\right).$$
More generally, a uniform circular orbit of any radius $r$ and speed $\sqrt{r}$ can be obtained:
$$\mathbf{p}(0){=}(r,0), \,\, \mathbf{v}(0){=}(0,\sqrt{r}) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(r \cos\frac{t}{\sqrt{r}},r \sin\frac{t}{\sqrt{r}}\right).$$
and we check that the desired equation holds:
\begin{align}
     \dot{\mathbf{p}}(t) &= \left(-\sqrt{r} \sin \frac{t}{\sqrt{r}}, \sqrt{r} \cos \frac{t}{\sqrt{r}}\right) \\
    \ddot{\mathbf{p}}(t) &= \left(-\cos \frac{t}{\sqrt{r}}, -\sin\frac{t}{\sqrt{r}}\right) \\
                         &= -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert.
\end{align}
Another simple case is when the initial velocity is zero or collinear with the position and the origin;
in this case it's a 1-dimensional problem
and the position can easily be seen to be a simple piecewise quadratic function of time.
But what if the initial conditions are not so nicely aligned?
To get an idea of the shapes that are possible,
I've made some plots, using gnuplot, of simulations
using the simple evolution algorithm I described earlier, with
$dt = 1/10000$ degree $\approx .00000175$.
Figure 1 shows five different initial states,
each evolved from $t{=}0$ to $t{=}2 \pi$:
$\,\,\mathbf{p}(0){=}(1,0)$, $\mathbf{v}(0){=}(0,v_{0 y})$ for $v_{0 y}{=}0.5,1,1.5,2,2.5$.

Figure 2 shows the one with $\mathbf{v}(0){=}(0,2)$ evolved farther, to $t{=}20\pi$.

Figure 3 shows it evolved even farther, to $t{=}60\pi$.

CLARIFICATION: I am ultimately interested in finding the simplest way of expressing $\mathbf{p}$ as a function of $t$. I.e. I really want to know "what this function looks like" rather than "what the curve looks like".  Other parametrizations of the curve, and intuition about shape of the curve, are of interest only if they help lead to this answer.
UPDATE 2015/07/02:
It sure looks like a spirograph hypotrochoid, doesn't it?
http://mathworld.wolfram.com/Spirograph.html .
Exploring this possibility,
I found by binary search an initial velocity (0,1.662656) (probably accurate to only 4 decimal places or so)
yielding a closed orbit in the shape of a 7-petalled flower,
and then compared that sim result with the 7-petalled hypotrochoid
having the same min and max radii; see Figure 4.

Conclusion: It's really close, but it's not a hypotrochoid.
It moves too fast at the fast parts and too slow at the slow parts,
and stays a bit too close to the origin during the in-between parts.

This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit.  For conservative forces, it's generally convenient to define the scalar potential $V$.  The force will satisfy $\vec{F} = \vec{\nabla}V$, and your potential $V = r$ (yay, system is conservative).  Since the kinetic + potential energy is constant, and $V$ is unbounded above, this means that the particle will always be in a bounded orbit around the body.

The orbits you've found are bounded, but not closed (on themselves), resulting in the flower-pattern trajectory.  I do know that you only get closed orbits for potentials that are $\tfrac 1r$ (like ours), and $\tfrac 1{r^2}$ (where the force goes as $\tfrac{1}{r^3} \large)$, but I never really understood why (phase-space apparently).

A shallow cone opening upward, made of aluminum, makes a real potential well for a ball bearing to roll in, closely simulating a particle moving subject to a central force of constant magnitude, except for friction: the motion is not conservative. The orbit is a flower with 7 petals of diminishing size as the motion decays, with the seventh petal pointing the same way as the first. After about 10 petals the ball drops through a central hole.
A TI 83 calculator easily computes the trajectory (path) and the schedule (time table), requiring about a minute for the calculation of the improper integral of the unrefined quadrature. Integrals must be truncated at their upper and lower integrals to make them proper, with corrections added at the limits of integration, for efficient calculation.
The unrefined quadrature is an elliptic integral which can be reduced to a canonical form in which the integrals are proper. The trajectory theta(rho) is an elliptic integral of the third kind and the schedule t(rho) is a combination of elliptic integrals of the first and second kind. These formulas are not very complicated but took some doing to find the right transformations to accomplish the reduction. The reduced form runs on a TI 83 calculator in about 3 seconds, much faster than the unrefined quadrature.

Best,
Chuck


continuous function from $\mathbb{Q}$ to $\mathbb{R}$ Is there exist continuous function from $\mathbb{Q}$ to $\mathbb{R}$?

if such continuous function exist and contain more than one point then by intermediate value theorem contains uncountable points which is not possible f($\mathbb{Q}$) is atmost countable.

hence must be constant function

Let f($\mathbb{Q}$)=$\{a\}$ but singleton is closed in $\mathbb{R}$ and under continuous function inverse image of closed set is closed.Here inverse image of {a} is $\mathbb{Q}$. But $\mathbb{Q}$ is not closed hence there doesnot exist any continuous function from  $\mathbb{Q}$ to $\mathbb{R}$.

please correct me if i am wrong

Since $\Bbb Q$ is a subspace of $\Bbb R$, any continuous function $\Bbb R\to\Bbb R$, such as the identity $f(x)=x$, will be continuous when restricted to $\Bbb Q$.

Singletons are continuous as inverse of open sets in the image, under the subspace topology is open in the domain of $f$.


Topology - Compactness of $\mathbb{Z}\times\{0,1\}$ A question from my h.w.:

  Is the topological space $\mathbb{Z}\times\{0,1\}$
    (where $\mathbb{Z}$
    has the discrete topology and $\{0,1\}$
    the trivial one) compact? sequentially compact? or compact at accumulation points?

I'm not sure I understand the construction, but I don't see how this space has any notion of compactness - the open covering $\{z\}\times\{0,1\}_{z\in\mathbb{Z}}$
 doesn't have a finite sub covering, same goes for the sequence $\{z\times\{0\}\}_{z\in\mathbb{Z}}$
 that doesn't have a limit or an accumulation point. Did I misunderstand something?

Thanks for the help!

Well, you already got the answer but adding my point of view won't do any harm :)

Denote $X=\mathbb{Z}\times\{0,1\}$.

The space X is not compact: as you noticed $\{z\}\times \{0,1\}$ does not have a finite subcover.
The space X is not sequentially compact: consider the sequence $(a_n)_{n=1}^\infty$, $a_n=(n,1)$.

No subsequence of $a_n$ converges $x_0\in X$, for if such $(a_{n_j})_{j=1}^\infty$ exists with $lim_{j\to\infty}a_{n_j}=x_0$,

any open neighbourhood of $x_0$ (denoted $U$) has elements of the tail $(a_{n_j})_{j=l}^\infty$ in it, for any $l\in \mathbb{N}$.

In other words: $\forall l\in\mathbb{N},\forall U_{x_0}: U_{x_0}\cap\{a_{n_j}|j>l\}\neq\phi$.

But looking at $U_{x_0}=\{z\}\times\{0,1\}$ (where z is the first coordinate of $x_0$) and $l=z+1$ give the contradiction, hence X is not sequentially compact.
The space X is limit point compact: Let look at a more common definition of limit point compactness: X is said to be limit point compact if every infinite subset of X has a limit point in X, i.e a point x for which every neighbourhood of x contains at least one point of X different from x itself.

Let $S\subset X$ be an infinite set.
$i,j\in\{0,1\}$ and $i\neq j$.
Let $y=(k,i)\in S$ then any open neighbourhood of $x=(k,j)$ (even the smallest one $\{k\}\times \{0,1\}$) contains $y\in S$.


How to solve $\int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$ or $\int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$ I need to solve either of the following integrals, they are related to each other by integration by parts:

$$I_1 = \int_{0}^{\infty} e^{-ax} \frac{x}{\sqrt{1+x^2}}dx$$

or

$$I_2 = \int_{0}^{\infty} e^{-ax} \sqrt{1+x^2}dx$$

I have not been able to find them in integral tables and I wonder if it is possible to solve either of these integrals.

Any ideas on how to calculate $I_1$ or $I_2$?

For $\int_0^\infty e^{-ax}\dfrac{x}{\sqrt{1+x^2}}~dx$ ,

$\int_0^\infty e^{-ax}\dfrac{x}{\sqrt{1+x^2}}~dx$

$=\int_0^\infty e^{-a\sinh t}\dfrac{\sinh t}{\sqrt{1+\sinh^2t}}~d(\sinh t)$

$=\int_0^\infty e^{-a\sinh t}\sinh t~dt$

$=-\dfrac{d}{da}\int_0^\infty e^{-a\sinh t}~dt$

$=-\dfrac{\pi}{2}\dfrac{d}{da}\mathbf{K}_0(a)$ (according to https://dlmf.nist.gov/11.5)

For $\int_0^\infty e^{-ax}\sqrt{1+x^2}~dx$ ,

$\int_0^\infty e^{-ax}\sqrt{1+x^2}~dx=\dfrac{\pi\mathbf{K}_1(a)}{2a}$ (according to https://dlmf.nist.gov/11.5)

If you notice it, you will see that this is the form of Laplace transformation. Thus the first integration can be written as the Laplace transformation of (x/root(x^2 +1))
The second integral can be written by integration by parts
Which is e^-ax * root(x^2 +1) - integrate(-ae^-at * root(x^2 +1) dx)
Which is then equals to
e^-ax * root(x^2 +1) + a*laplace of (root(x^2 +1))

Hope this helped


Is zero in every ideal? Let $I \subseteq R$ be a subset of a ring $(R, +, \cdot)$ such that

$(I, + )$ is a group
$\forall x \in I, r \in R: xr \in I \text{ and } rx \in I$

Then $I$ is called an "Ideal"

Question: Is 0 in every ideal of any ring?

$(I,+)$ is a subgroup of $(R,+)$. What do you know about the identity of subgroups in relation to the containing group?

the interesting thing is that 0 is always an ideal in any ring.
the second interesting thing is a ring is always ideal for itself
meaning nothing is ideal for everybody
and each one is ideal for  itself


What is the difference between a function and a formula? I think that the difference is that the domain and codomain are part of a function definition, whereas a formula is just a relationship between variables, with no particular input set specified.

Hence, for two functions $f$ and $g$, $f(x)$ can be equal to $g(x)$ for all integers, say, but if the domain of $f$ is {2, 3, 4} and the domain of $g$ is {6, 7, 8, 9}, the two functions will be different.

And on the converse, if the functions 'do different things' - i.e. $f(x) = x$ and $g(x) = x^3$ - but the domains of $f$ and $g$ (these are the same) are set up such that the values of the functions are the same over the domain (this would work in this case for {-1, 0, 1}), then the functions are the same, even though the formulas are different.

Is this correct?

Thank you.

A formula is a string of symbols, arranged according to mathematical grammar.

A function is a mathematical object that plays a role in arithmetic operations like "evaluation" or "composition". A key point is that if $f$ and $g$ are expressions that denote two functions with the same domain and codomain, and we have $f(x) = g(x)$ for every $x$ in the domain, then $f$ and $g$ denote the same function.

So, in your example of functions $f$ and $g$ on the set $\{ -1, 0, 1 \}$, it is indeed true that $f=g$.

Some examples of mathematical grammar

Let $f$ be a variable that denotes a function on the reals. Let $x$ be a real-valued variable. Then:

$f$ is a function.
$f(x)$ is a real number.
$x^2 + 3$ is a real number. In particular, it is not a function. Unfortunately, people are frequently grammatically incorrect on this point. :(
$f(x) = x^2 + 3$ is an equation that relates two real numbers.

and all of the bullet points above are examples of formulas.

Recall that I mentioned functions were 'defined' pointwise: if two functions solve "$f(x) = x^2 + 3$ for all $x$" for $f$, then they must be the same function. Because of this, we can use equations like this as a way to specify functions.

But as you note, if two different formulas for the right hand side actually give the same values when you substitute values from the domain, then the functions so defined will be the same.

Incidentally, it is possible to define functions directly rather than pointwise, although it often isn't pleasant. e.g. the function $f$ defined above is given by

$$ f = p \circ ((\mu \circ \Delta), c_3) $$

where I'm using the notation

$\Delta$ is the diagonal function $\Delta(x) = (x, x)$
$\mu$ is the multiplication function $\mu(x,y) = xy$
$p$ is the addition function $p(x,y) = x+y$
$c_3$ is the constant function $c_3(x) = 3$
$(,)$ is a binary operation on functions; it's defining property is $(g,h)(x) = (g(x), h(x))$. I don't actually know of standard notation for this; sometimes I see $\times$ in place of $,$.
$\circ$ is composition of functions: $(g \circ h)(x) = g(h(x))$.

I was taking a refresher course on Calculus and as a pre-requisite, was reviewing algebra and some pre-algebra as well :). It was said, for a function to be a function, it must have a definite output corresponding to every input. Based on that, $f(x) = x$ is a function, but $f(x) = \sqrt{x}$ is not, as there are always two answers that could be the square root of a positive number (one positive and the other negative). However, both can be called formula. I believe this is at least one key differentiator of the two.


Discrete math: Big O notation proof with logarithms Determine whether or not the following is true: $\log((1/n) + n^2 )$ is $O(\log n)$.

I'm struggling to solve this. I'm especially confused because of the $O(\log n)$ and would appreciate some help or even a push in the right direction. I'm unconfident with logarithms in general which is why I'm struggling with this question in particular.

$\log((1/n) + n^2 )$

If we just focus on the term inside of the logarithm.

$ let x = 1/n + n^2$

As n grows, we notice that the 1/n term effectively becomes zero and the overpowering term is $n^2$

We can now state that $\log((1/n) + n^2 )$ has growth $log(n^2)$ when n gets large.

We know from logarithms that $log(a^b) = b* log(a)$

So $log(n^2)$ can be rewritten as $2*log(n)$ for large n we notice that the constant value of 2 is insignificant and can conclude that our original expression has growth $O(log(n))$


If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$ If $a,b,c$ are rational numbers and $a\sqrt2+b\sqrt3+c\sqrt5=0$, show that $a=b=c=0$

I can solve for two terms only but for additional $c\sqrt5$ term, I got stuck. I also want to know how to solve this for general $n$ terms i.e. $a\sqrt2+b\sqrt3+c\sqrt5+\ldots+n \text{ terms}=0$  will imply $a=b=c=\ldots=0$

Assume $a\sqrt 2+b\sqrt3+c\sqrt 5=0$. Then $5c^2=2a^2+2ab\sqrt 6+3b^2$. From the irrationality of $\sqrt 6$, we conclude that $ab=0$. If $b\ne 0$, this makes $a= 0$ and $5c^2=3b^2$, so $\sqrt{15}=\frac{5c}b$ rational, contradiction.
If $a\ne 0$, this makes $b=0$ and $5c^2=2a^2$, ao $\sqrt{10}=\frac{5c}{a}$ rational, contradiction. We conclude that $a=b=0$, but then also $c=0$.

The general statement is:

  Let $a_1,\ldots, a_n$ be different square-free positive integers. Then $\sqrt{a_1},\ldots, \sqrt{a_n}$ are linearly independent over the rationals.

Its proof involves but a bit deeper techniques. Edit: Found it, see my answer here

Multiply your equation by $\sqrt{2}$ to get,

$2a + \sqrt{6}b + \sqrt{10}c = 0 \implies a = 0,  \sqrt{6}b + \sqrt{10}c = 0$ since the rational and irrational parts both need to be zero.

Now multiply $\sqrt{6}b + \sqrt{10}c = 0$ by $\sqrt{6}$ and apply the same logic.


Find the minimum and maximum value of $\frac{1}{\sin{x} -3\cos{x} +5}$ Find the maximum value and minimum value of
$$\frac{1}{\sin{x} -3\cos{x} +5}.$$ Any tips?

First notice the following:

$\sin{x} -3\cos{x}$

$=\sqrt{10}(\frac{1}{\sqrt{10}}\sin{x} - \frac{3}{\sqrt{10}}\cos{x})$

$=\sqrt{10}(\cos{\alpha}\sin{x} +\sin{\alpha}\cos{x})$; (where $\tan{\alpha} = -3)$

$=\sqrt{10}\sin{(x+\alpha)}$

Now, since the 'sine' function lies between -1 and 1, the expression "$\sin{x} -3\cos{x}$" lies between $-\sqrt{10}$ and $\sqrt{10}$. Put these values for the epression in your overall expression to find it's range.


$p \to q$ vs. $p$ only if $q$ The question I wanted to ask is exactly as discussed on Math SE here. What's bugging me is why the first conversion from $p \to q$ to $p$ only if $q$ doesn't make sense in everyday speech.

Here's the accepted answer on that page:

  "I win the lottery only if I give you \$1 billion" is equivalent to "if
  I don't give you \$1 billion, then I won't win the lottery", and the
  latter statement is the contrapositive to "if I win the lottery, then
  I will give you \$1 billion."

  If you have doubts about why the contrapositive of a statement is an
  equivalent statement, have you tried making a truth table?

My problem is, sure the converted statement makes sense when I consider its equivalent and then contrapositive, but why doesn't it make sense in the original form?

Does this disconnect point to a fundamental difference between Boolean Logic and Human Language that human beings are not able to keep apart (like in my case) without adequate practice? Perhaps it's a question more related to philosophy or psychology, in which case I apologize for posting it here.

Another explanation I came across is this one. Here the statements "I'll stay home tomorrow only if I'm sick" and "If I stay home tomorrow then I'm sick" make perfect sense. I think that might be because in this example, the cause comes after the effect, whereas in the earlier example, the cause comes before the effect.

Honestly, I'm hopelessly confused at this juncture, and any guidance is much appreciated.

I think that the idea of causation  is the (or at least a) culprit here. In "most" implications in natural language, the implication is due to one thing causing another:

  If I eat thumbtacks, I will be unhappy.

Unfortunately, this doesn't always line up with how the formal implication is interpreted. For instance, let's look at "I will only give you a billion dollars if I have won the lottery." Then:

This sentence is equivalent to "If I (will) give you a billion dollars, then I (will have) won the lottery."
But there's no way that me giving you a billion dollars somehow causes me to win the lottery! In fact, the only reasonable causal mechanism here is the other way around: if I win the lottery, that causes me to - at the very least - be able to give you a billion dollars (hm! big lottery here :P).

The crucial test of how to interpret an implication is: Suppose I tell you $p\implies q$. Under what circumstances do you know I was lying to you? Clearly, the only way you can be certain that I lied to you is if $p$ is true and $q$ is not.

This raises the question of how to express causality in formal logic. One approach to this is via modal logic. In (propositional) modal logic, we have

A set of propositional constants, $p, q, r, . . .$ representing the basic things which can be true or false.
A set of worlds, representing possible states of affairs.
An accessibility relation $Sees(\cdot,\cdot)$ between worlds: $u$ is accessible from $v$ (written "$Sees(v, u)$," or "$v$ sees $u$") if $v$ is "possible", from the point of view of $u$.
Our language is then that of propositional logic, together with two new symbols (modalities) "$\Box$" and "$\Diamond$", read as "It is necessary that" and "It is possible that," respectively.
A satisfaction relation, telling us which propositions are true at which worlds. This basically amounts to an atomic satisfaction relation, telling us which propositional letters are true at which worlds, and is then extended to all propositions via induction: for instance, $w\models p\vee q$ if $w\models p$ or $w\models q$. The only new bit is how to define satisfaction for the operations $\Box$ and $\Diamond$; the rule is $$w\models \Box\varphi\iff \forall u(Sees(w, u)\implies u\models\varphi)$$ and dually $$w\models\Diamond\varphi\iff \exists u(Sees(w, u)\wedge u\models\varphi)\iff w\models\neg\Box\neg\varphi.$$

There are many different kinds of modal logics, and ways of interpreting them; for instance, we might consider worlds to be moments in time, and the relation $Sees(w, u)$ to be "$u$ is a later time than $w$." For instance, suppose $p$ means "I have won the lottery" and $q$ means "I have given you a billion dollars." We can express "the only way I'm going to give you a billion dollars is if I have won the lottery" as $$p\iff \Diamond q.$$ Note that this lines up with our usual understanding that me winning the lottery causes me to be able to give you a billion dollars. Meanwhile, at any given world, we have $q\implies p$, reflecting the truth-functional picture of things.

Personally I've been having the same issues between the language aspect and the mathematical writing.
from what I understand now:
When it is "if and only if" would be the implies in both directions, whilst "only if" would be q only if p which is equivalent to p implies q.
I agree that there is a problem between the boolean logic and the human language but when trying to work these out I would recommend doing a truth table.
I hope I have been of some help.


How to express the rest of division by three, with very elementary functions? Is it possible to express
$\; "\!n\pmod 3\!"\;$
with combinations of the functions plus, minus, multiplication, division and exponentiation in $\mathbb C$ or preferably in $\mathbb Z[i]$?

I'ts not allowed with modulo or integer parts, and the inverses are only valid if they can be expressed as above. Also all sums, products etc must be finite.

There is a simular question Closed-Form Modular Arithmetic where trigonometric functions and logarithms are allowed.

If it isn't possible, how to prove that?

If complex exponentiation is OK, then you may do it. The map $f: n \mapsto \exp(2\pi n/3)$ takes ${\mathbb Z}$ to the three cubic roots of $1$:  $1$, $\gamma=\exp(2\pi i /3)$ and $\overline{\gamma}$. Now pick a polynomial $P$ that maps $1$ to $0$, $\gamma$ to $1$ and $\overline{\gamma}$ to $2$. Then $P\circ f$ will do the job. Same principle for mod $N$.

Later edit: In the general case set $\gamma=\exp(2\pi i/N)$ and define $e_N(z)=\frac{1}{N} \left(z^{N-1}+z^{N-2} + \cdots +z +1\right) $. Then  $e_N$ maps $1$ to $1$ but every $\gamma^k$, $1\leq k<N$ to zero. The wanted function is then:
$$ h_N(n) = \sum_{k=1}^N k \ e_N \left( \ \gamma^{n-k}  \right), $$
which maps $n$ to $n\  (\mbox{mod}\; N)$. What is really happening here is that we are doing a forward + backward  Fourier transform of the modulus function.


trig substitution $\int_{2}^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$ $\int_{2}^{\infty}\frac{1}{x\sqrt{x^2-4}}dx$

I let $x = 2 \sec(\theta)$, then $dx = 2\sec(\theta) \tan(\theta) d \theta$

$$\int \frac{2\sec(\theta)\tan(\theta)}{2\sec(\theta)2\tan(\theta)}d\theta = \frac{1}{2}\theta + C = \frac{1}{2} \sec^{-1}\left(\frac{x}{2}\right) + C$$

Therefore the answer would be this after simplification?:

All I did is:

$$\int_{2}^{\infty} f(x)dx = \int_{2}^{3} f(x)dx + \int_{3}^{\infty} f(x)dx$$

$$\lim_{A\to2^+} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{A}^{3} + \lim_{B\to\infty} \frac{1}{2}\left(\sec^{-1}\left( \frac{x}{2}\right)\right)\bigg|_{3}^{B}$$

$$= \frac{1}{2} (sec^{-1} (\frac{3}{2}) - 0)+ \frac{1}{2}(\pi/2 - sec^{-1}(3/2)) = \pi/4$$ Converges.

Is this right?

POINT OUTS: You did not change limits. and it looks weird.
But to solve via your method you just need to compute:
$$\int^\frac{\pi}{2} _0 \frac{d \theta}{2}=\frac{\pi}{4}$$
My method: let $x=\frac{1}{t}$
$$\int^0_\frac{1}{2} \frac{dt}{\sqrt{1-4t^2}}$$
Which is equivalent to $$\int^0 _\frac{1}{2}
\frac{dt}{\sqrt{1-(2t)^2}}$$

Now i hope you know $$\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})+C$$

hint
the integral is convergent since

$$f (x)\sim \frac {1}{x^2}\;\;(x\to\infty) $$

$$f(x)\sim \frac{1}{4\sqrt{x-2}}\;\;(x\to2^+)$$

to find its value,
put $$x=2\cosh (t) =2\frac{e^t+e^{-t}}{2}.$$
 $x=2\implies t=0$

$x\to \infty\implies t \to \infty $.

on the other hand,

$$dx=2\sinh (t) $$
and
$$\cosh^2 (t)-\sinh^2 (t)=1.$$

thus

$$x\sqrt {x^2-4}=4\cosh(t)\sinh(t). $$

the integral becomes

$$\int_0^\infty \frac{dt}{e^t+e^{-t}} $$
$$\color {green}{\boxed {=[\arctan(e^t)]_0^\infty=\frac {\pi}{4}}}$$


Topology: Prove the intersection of these sets is non empty. The question is as follows:

  Let $C_n$ be a sequence of nonempty, closed subsets of a compact
  space $X$, such that, $C_n ⊃ C_{n+1}$ for each $n$. Show that the intersection
  of these sets is nonempty.

My question is where in my proof(below) am I wrong as I do not explicitly state that it is a compact space.
The unions below are infinite.

$Attempt:$

$C_0 \neq \emptyset $
$(X-C_0)\cup(\cup(X-C_i)) = X - \cap C_i$
$C_0$ is non empty and then $X-C_i$ is not X and is either a subset of X or the null set.
If it is not the null set then their difference is non empty and not X.
So:
$X-\cap C_i= A$, where A is a non empty subset of X.
Then the complement is non empty and thus our intersection is non empty.

If it's the null set then we reapply our original procedure for a subset in the sequence which is greater than the null set and arrive at the same result.

You're thinking along the right lines, but your proof needs to use the definition compactness somewhere, since it's not true without that assumption.

Here's a (big) hint. If it were the case that $\bigcap\limits_{n=0}^{\infty} C_n = \varnothing$, then we'd have
$$X = X - \varnothing = X - \bigcap_{n=0}^{\infty} C_n = \bigcup_{n=0}^{\infty} (X-C_n)$$
and so $\{ X-C_n \mid n \ge 0 \}$ is an open cover of $X$. By compactness, it has a finite subcover and so, in particular, there is some $k \in \mathbb{N}$ such that $\{ X-C_n \mid 0 \le n \le k \}$ is a cover of $X$.

Use this observation to derive a contradiction to one of the hypotheses about the sets $C_n$.

The intersection of $C_i$ with $C_{i+1}$ is $C_{i+1}$, therefore the intersection of all of the sets is $C_l$ where $l$ is either the maximum value of $n$ or the limit as $n$ goes to infinity. Since all $C_n$ are non-empty, $C_l$ is also non-empty and the intersection of all $C_n$ is also non-empty.


How to prove by contrapositive that for all integers $m$ and $n$, if $m - n$ is odd, then $m$ is odd or $n$ is odd? So far this is what I've come up with

$p:= m-n$ is odd
$q:= m$ is odd or $n$ is odd

since the contrapositive is defined as $\neg q \implies \neg p$, I have assumed that $m$ is even and $n$ is even (after applying De Morgan's law) and I am now proving that $m-n$ is even. However, this is where I am stuck. Since any even number can be represented as $2$ multiplied by an integer, I have $m = 2a$ and $n = 2b$ and plugging that in gives me $2a - 2b$ which, in the case of $a = b$ equals zero. Is there a way to rewrite this statement so that I can say it holds true in the case that $a$ does not equal $b$?

$2a-2b=2(a-b)$ is an even number since $a-b$ is an integer.

Note that $0$ is an even number as $0=2(0)$.

Well the contrapositive would be that if
 $m$ and $n$ are even, then $m-n$ is even.
Make $m=2a$ and $n=2b$
$$M-n= 2a-2b
=2(a-b)
=2k$$

make $k=a-b$. $2k$ is positive making the contrapositive true therefore making the original statement true.


Abelianization of GL_n Suppose $k$ is a field.

How to prove that the abelianization of $GL_2(k)$ is $GL_1(k)$ ?
Ditto for $GL_n(k)$.
Can we say the same thing, were we to replace $k$ with $\mathbb Z$ ?

For any field $k$, the abelianization of $GL_n(k)$ is $k^{\ast}$ except for $n=2$ and $k=\mathbb{F}_2$ or $\mathbb{F}_3$. I think this is in Lang's Algebra, Chapter XIII sections 8 and 9. (I say I think because I am relying on google books here, and some key pages are missing.)

EDIT: I just looked back at this old answer, and the abelianization of $GL_2(\mathbb{F}_3)$ is $\mathbb{F}_3^{\times}$ as well, so $GL_2(\mathbb{F}_2)$ is the only counterexample. I think that what I was thinking when I wrote this is that the abelianization of $SL_2(\mathbb{F}_3)$ in nontrivial.

As HenrikRueping points out, the determinant map $\det:\textrm{GL}_n(k) \longrightarrow k^\ast$ factors through the abelianization $\textrm{GL}_n(k)_{ab}$ of $\textrm{GL}_n(k)$. The induced morphism $\textrm{GL}_n(k)_{ab} \longrightarrow k^\ast$ can be shown to be an isomorphism as follows. It's clearly surjective. Thus, we have to show injectivity. To do so, you have to show that each element of $\textrm{SL}_n(k)$ is a commutator.

For $n=2$, I thought one could generate $\textrm{SL}_2(k)$ with the two matrices

$$\left( \begin{array}{cc} 1& 1 \\ 0 & 1 \end{array}\right)$$

$$\left( \begin{array}{cc} 0& 1 \\ -1 & 0 \end{array}\right)$$

So if you just show that these are commutators you're done, right?

Jim points out that these matrices do not generate $\textrm{SL}_2(k)$ but $\textrm{SL}_2(\mathbf{Z})$.


Expectation Maximization Algorithm for Gaussian Mixture Model Can we use the Expectation Maximization algorithm for estimation of Gaussian Mixture Model with full covariance matrices?

If yes then can you please give me a reference paper? So far all the machine learning books I have consulted describes the estimation of only the diagonal covariance matrices.

All you need is to know the formulas for the iterative estimates of your parameters $(\tau, \mu, \Sigma)$. Once you have it, coding the algorithm is just a simple loop.
Assuming your Gassian mixture is a mix of two gaussian. Let $\mathbf{x} = (\mathbf{x}_1,\mathbf{x}_2,\ldots,\mathbf{x}_n)$ be a sample of $n$ independent observations from a mixture of two multivariate normal distributions of dimension $d$, and let $\mathbf {z} =(z_{1},z_{2},\ldots ,z_{n})$ be the latent variables that determine the component from which the observation originates.
$X_{i}\mid (Z_{i}=1)\sim {\mathcal {N}}_{d}({\boldsymbol {\mu }}_{1},\Sigma _{1})$ and $X_{i}\mid (Z_{i}=2)\sim {\mathcal {N}}_{d}({\boldsymbol {\mu }}_{2},\Sigma _{2})$
where $\operatorname{P} (Z_i = 1 ) = \tau_1$ and $\operatorname {P} (Z_{i}=2)=\tau _{2}=1-\tau _{1}$
Initial Guess for unknown parameters:
Either give your own intuitive guess of $(\tau, \mu, \Sigma)$ or use K-mean algo to get a good estimate using sample mean, sample covariance, and $\tau = (1/2, 1/2)$
[E Step] Calculate expectation:
Calculate the expectation using previously calculated parameters $\theta^{(t)}$, it could be your initial guess $\theta^{(1)}$
$$
\begin{aligned}
Q(\theta \mid \theta ^{(t)})&=\operatorname {E} _{\mathbf {Z} \mid \mathbf {X} ,\mathbf {\theta } ^{(t)}}[\log L(\theta ;\mathbf {x} ,\mathbf {Z} )]\\
  &=\operatorname {E} _{\mathbf {Z} \mid \mathbf {X} ,\mathbf {\theta } ^{(t)}}[\log \prod _{i=1}^{n}L(\theta ;\mathbf {x} _{i},Z_{i})]\\
  &=\operatorname {E} _{\mathbf {Z} \mid \mathbf {X} ,\mathbf {\theta } ^{(t)}}[\sum _{i=1}^{n}\log L(\theta ;\mathbf {x} _{i},Z_{i})]\\
  &=\sum _{i=1}^{n}\operatorname {E} _{Z_{i}\mid \mathbf {X} ;\mathbf {\theta } ^{(t)}}[\log L(\theta ;\mathbf {x} _{i},Z_{i})]\\
  &=\sum _{i=1}^{n}\sum _{j=1}^{2}P(Z_{i}=j\mid X_{i}=\mathbf {x} _{i};\theta ^{(t)})\log L(\theta _{j};\mathbf {x} _{i},j)\\
  &=\sum _{i=1}^{n}\sum _{j=1}^{2}T_{j,i}^{(t)}{\big [}\log \tau _{j}-{\tfrac {1}{2}}\log |\Sigma _{j}|-{\tfrac {1}{2}}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{j})^{\top }\Sigma _{j}^{-1}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{j})-{\tfrac {d}{2}}\log(2\pi ){\big ]}.
\end{aligned}
$$
where:
$$
T_{j,i}^{(t)}:=\operatorname {P} (Z_{i}=j\mid X_{i}=\mathbf {x} _{i};\theta ^{(t)})={\frac {\tau _{j}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{j}^{(t)},\Sigma _{j}^{(t)})}{\tau _{1}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{1}^{(t)},\Sigma _{1}^{(t)})+\tau _{2}^{(t)}\ f(\mathbf {x} _{i};{\boldsymbol {\mu }}_{2}^{(t)},\Sigma _{2}^{(t)})}}
$$
[M Step] Maximize expectation and find the t+1 estimate
The linear form of $\tau_{i}$ and $\mu_{i}/\Sigma_i$ means we could maiximize $Q$ independently by maximizing $\tau_{i}$ and $\mu_{i}/\Sigma_{i}$ terms separately.
$\tau$ terms, with constrain $\tau_2 = 1 - \tau_1$:
$$
\begin{aligned}
{\boldsymbol {\tau }}^{(t+1)} &= {\underset {\boldsymbol {\tau }}{\operatorname {arg\,max} }}\ Q(\theta \mid \theta ^{(t)})\\
                              &={\underset {\boldsymbol {\tau }}{\operatorname {arg\,max} }}\ \left\{\left[\sum _{i=1}^{n}T_{1,i}^{(t)}\right]\log \tau _{1}+\left[\sum _{i=1}^{n}T_{2,i}^{(t)}\right]\log \tau _{2}\right\}.
\end{aligned}
$$
By taking derivative of each $\tau$ and equal to zero, we get for formula for t+1 estimate of $\tau _{j}^{(t+1)}$:
$$
\tau _{j}^{(t+1)}={\frac {\sum _{i=1}^{n}T_{j,i}^{(t)}}{\sum _{i=1}^{n}(T_{1,i}^{(t)}+T_{2,i}^{(t)})}}={\frac {1}{n}}\sum _{i=1}^{n}T_{j,i}^{(t)}
$$
For the next estimates of $(\mu_1, \Sigma_1)$:
$$
\begin{aligned}
({\boldsymbol {\mu }}_{1}^{(t+1)},\Sigma_{1}^{(t+1)}) &= {\underset { {\boldsymbol{\mu}}_{1},\Sigma_{1} }{\operatorname {arg\,max} }}  Q(\theta \mid \theta ^{(t)})\\
&= {\underset { {\boldsymbol{\mu}}_{1},\Sigma_{1} }{\operatorname {arg\,max} }} \sum _{i=1}^{n}T_{1,i}^{(t)}\left\{   -{\tfrac {1}{2}}\log |\Sigma _{1}|-{\tfrac {1}{2}}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{1})^{\top }\Sigma_{1}^{-1}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{1})  \right\}
\end{aligned}
$$
To derive the formula for $\mu_{i}^{(t+1)}$, we'll gonna make use of this formula: $\mathbf{ \frac{\partial w^T A w}{\partial w} = 2Aw}$, where A is symmetric.
$$
\begin{aligned}
  0 &= \frac{\partial{Q(\theta \mid \theta ^{(t)})}}{\partial{\mu_1}} \\
    &= \sum_{i=1}^{n} -T_{1,i}^{(t)} \Sigma_{1}^{-1} (\mathbf {x} _{i}-{\boldsymbol {\mu }}_{1})
\end{aligned}
$$
$$
\begin{aligned}
\boldsymbol{\mu}_1^{(t+1)} = \frac{\sum_{i=1}^n T_{1,i}^{(t)} \mathbf{x}_i}{\sum_{i=1}^n T_{1,i}^{(t)}} \\
\boldsymbol{\mu}_2^{(t+1)} = \frac{\sum_{i=1}^n T_{2,i}^{(t)} \mathbf{x}_i}{\sum_{i=1}^n T_{2,i}^{(t)}}
\end{aligned}
$$
To derive the formula for $\Sigma_{i}^{(t+1)}$, we'll gonna make use of the following equations:

Invariant under cyclic permutation $tr[ABC] = tr[CAB] = tr[BCA]$

Derivative of a trace: $\frac{\partial}{\partial{A}} tr[BA] = B^T$

$$
\begin{aligned}
  \frac{\partial}{\partial{a_{ij}}} tr[BA] &= \frac{\partial}{\partial{a_{ij}}} \sum_{k} \sum_{l} b_{kl}a_{lk} = b_{ji} \\
  \frac{\partial}{\partial{A}} tr[BA] &= B^T
\end{aligned}
$$

Derivative of a log determinant: $\frac{\partial}{\partial{A}}\log{|A|} = A^{-T}$.
To establish the result, note that: $\frac{\partial}{\partial{a_{ij}}}\log{|A|} = \frac{1}{|A|} \frac{\partial}{\partial{a_{ij}}} |A|$, and recall the formula for the matrix inverse: $A^{-1} = \frac{1}{|A|} C^T$, therefore $A^{-T} = \frac{1}{|A|} C$. All we need to prove is $\frac{\partial}{\partial{a_{ij}}} |A| = C$, However, $|A| = \sum_j{(-1)^{(i+j)}a_{ij}M_{ij}}$,
$$
  \frac{\partial}{\partial{a_{ij}}}|A| = \frac{\partial}{\partial{a_{ij}}}\left\{\sum_j{(-1)^{(i+j)}a_{ij}M_{ij}}\right\} = (-1)^{(i+j)} M_{ij}
  $$
Which is exactly the cofactor matrix $C$ at entry $(i,j)$. Therefore, $\frac{\partial}{\partial{a_{ij}}} |A| = C$.

Determinant of an inverse is the inverse of the determinant: $|A|^{-1} = |A^{-1}|$

Now we have all the formulas we need to derive the $\Sigma_1^{(t+1)}$.
Because $\frac{\partial}{\partial{A}} x^tAx = \frac{\partial}{\partial{A}} tr[x^tAx] = \frac{\partial}{\partial{A}} tr[xx^tA] = xx^T$, we get:
$$
\begin{aligned}
  0 &= \frac{\partial{Q(\theta \mid \theta ^{(t)})}}{\partial{\Sigma_1^{-1}}} \\
    &=\sum _{i=1}^{n}\sum _{j=1}^{2}T_{j,i}^{(t)}{\bigg [} \frac{\partial}{\partial{\Sigma_1^{-1}}} {\tfrac {1}{2}}\log |\Sigma _{j}^{-1}| -  \frac{\partial}{\partial{\Sigma_1^{-1}}} tr {\big [} {\tfrac {1}{2}}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{j})^{\top }\Sigma _{j}^{-1}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{j}) {\big ]} {\bigg ]} \\
    &=\sum _{i=1}^{n} T_{1,i}^{(t)} \tfrac{1}{2} \Sigma_1 - \sum _{i=1}^{n} T_{1,i}^{(t)} {\tfrac {1}{2}}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{1})(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{1})^{\top }
\end{aligned}
$$
Formula for estimating t+1 $(\Sigma_1^{(t+1)}, \Sigma_2^{(t+1)})$
$$
\begin{aligned}
  \Sigma_1^{(t+1)} &= \frac{\sum_{i=1}^n T_{1,i}^{(t)} (\mathbf{x}_i - \boldsymbol{\mu}_1^{(t+1)}) (\mathbf{x}_i - \boldsymbol{\mu}_1^{(t+1)})^\top }{\sum_{i=1}^n T_{1,i}^{(t)}} \\
\Sigma _{2}^{(t+1)} &={\frac {\sum _{i=1}^{n}T_{2,i}^{(t)}(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{2}^{(t+1)})(\mathbf {x} _{i}-{\boldsymbol {\mu }}_{2}^{(t+1)})^{\top }}{\sum _{i=1}^{n}T_{2,i}^{(t)}}}
\end{aligned}
$$
[Stop Step]
Stop the iteration when $\|Q(\theta_{m+1}|\hat{\theta}_m,\boldsymbol{x}) - Q(\theta_m|\hat{\theta}_m,\boldsymbol{x}) \| \le \epsilon$
Notes:
A more detail explanation can be found in here: https://youtu.be/GEdAP680w5Y
Pseudo code:
let tau = [1/2, 1/2]
let kmean = Kmean(x1, x2, ..., xn)
let mu = [kmean.sample[0].sampleMean, kmean.sample[1].sampleMean]
let sigma = [kmean.sample[0].sampleVar, kmean.sample[1].sampleVar]

for ( |Q[n+1] - Q[n]| < epsilon) {
  mu = Calculate mu[n+1] with the mu formula
  sigma = Calculate sigma[n+1] with the Sigma formula
}

If you are using python, all you need to do is add a parameter, covariance_type='full'.

For example,

m = mixture.GMM(n_components=3,covariance_type='diag')

Works for me.


Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$

I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?

Yes it is equivalent to

$$\frac{2x}{x^2-25}=\frac{2x+1}{x^2-25}\iff 2x=2x+1$$

which indeed has not solutions for $x\in \mathbb R$.

Rearrange the equation as

$$\frac{2x}{x^2-25}-\frac{1}{x-5}-\frac{1}{x+5}
=\frac{2x+1}{x^2-25}-\frac{2x}{x^2-25}= \frac{1}{x^2-25}=0$$

which leads to the solutions

$$x=\pm \infty$$


$a,b \in \mathbb N $ , $b$ odd $\implies$ $ \dfrac{2a^2-1}{b^2+2} \notin \mathbb Z $? If $a,b$ are positive integers and $b$ is odd , then is it ever possible that $ \dfrac{2a^2-1}{b^2+2} $ is an integer ?

Since $b$ is odd, $b^2 + 2\equiv 3\pmod{8}$. Thus $2$ is not a quadratic residue mod $(b^2 + 2)$. Hence $2a^2\not\equiv 1\pmod{b^2 + 2}$, from which the result follows.

Can this formula to use.
How to solve an equation of the form $ax^2 - by^2 + cx - dy + e =0$?
But we'll see this topic and select the desired formula.
Families of curves over number fields


Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0 The statement I'm trying to prove is:

$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0

I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.

I don't really understand how to deal with $k + 1$, so I'm a little lost.

I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.

The statement is not true. Take $n=1$ as a counter example.

Since it's not true, you won't manage to prove it (by induction or otherwise).

The given statement  $n^3 + 7n + 3$  is divisible by 3 for  $n ≥ 0$ is not true, as detailed below
P(n): n^3. + 7n + 3  is divisible by 3
P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
Hence given statement is not true.

I presume that statement may be n^3 - 7n + 3  is divisible by 3 for n ≥ 0
In that case Proof is detailed below.

P(n):   n^3 - 7n + 3   is divisible by 3
        P(1):   (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3

Assume P(k) is true for some positive integer k, i.e
P(k):  (k)^3 - 7k + 3   is divisible by 3
       We can write  (k)^3 - 7k + 3   = 3d, where d ∈ N ………….. (1)
       We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
       P(k+1):   (k+1)^3 - 7(k+1) + 3
        = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b)

    = k^3 + 1 +3k(k+1) - 7(k+1) + 3
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
    = 3d + 3(k^2 + k - 2 ).........{From (1)}
    = 3d + 3(k^2 + 2k - k - 2 )
    = 3d + 3 {k(k +2)-(k+2)}
    = 3d + 3 (k-1)(k+2)
    = 3 {d +(k-1)(k+2)}
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer

From the last line, we see  (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
Thus P(k+1) is true, whenever P(k) is true.
Hence, by the PMI,  the given statement is true for all natural numbers n ≥ 0


Simple Random Walk - Why are these two events the same? Let $S = (S_n)_{n \geq 1}$ be a simple random walk. We denote the hitting time of a point $b$ by $\tau_b = \min \{i \geq 1 : S_i \geq b\}$.

My text says that the events $\displaystyle\{\max_{k \leq n} S_k \geq b\}$ and $\{\tau_b\ \leq n\}$ are identical.

I'm having trouble reading the first set, which seems to be about the maximum value (the values on the $y$-axis of the simple random walk) of $S_k$, for $k \leq n$, but the second set seems to be about "time" (the values on the $x$-axis of the simple random walk). So how can the events be identical? Am I misunderstanding something here?

If $\tau_b\leqslant n$, then $S_j\geqslant b$ for some $1\leqslant j\leqslant n$, and hence $$\max_{1\leqslant k\leqslant n}S_k\geqslant S_j\geqslant b.$$
If $\max_{1\leqslant k\leqslant n}S_k\geqslant b$, then $\tau_b\leqslant \operatorname{argmax}_{1\leqslant k\leqslant n}S_k\leqslant n$.

It follows that the two sets are equal.

It turns out that what is trying to be said is that $\mathbb{P}\displaystyle\{\max_{k \leq n} S_k \geq b\}$ and $\mathbb{P}\{\tau_b\ \leq n\}$ are equal, which they indeed are.

The question otherwise makes no sense.


Angle in a triangle within a circle. A and B are two points on the circumference of a circle with centre O. C is a point on OB such that AC $\perp OB$. AC = 12 cm. BC = 5 cm. Calculate the size of $\angle AOB$, marked $\theta$ on the diagram.

The answer given in the textbook is $45.2 ^\circ$ (1dp)

Note: This is not a homework question. I'm just doing maths for my own interest.

let radius = $r$, then in triangle ACO, using Pythagoras:

$$\begin{align}
AO^2 &= AC^2+CO^2\\
\\
r^2 &= 12^2+(r-5)^2\\
\\
144+25-10r &= 0\\
\\
r &= 16.9\\
\end{align}$$

In triangle ACO,
$$\begin{align}
\sin\theta &= \dfrac{12}{r}\\
\\
\sin\theta &= \dfrac{12}{16.9}\\
\\
\theta &= 45.2^{\circ} (1dp)\\
\end{align}$$

Draw line AB, and observe that ABC is a (5,12,13) right-angled triangle. Since AB is a chord, the bisector of angle AOB also bisects AB, call this point T. Now triangle OTB is similar to triangle ACB. Therefore angle AOB is:

2 * arctan(5/12)

Which a handy calculator says is 45.2 degrees.

(I'm surprised someone marked this down: I did get the answer, which one other answer didn't, and I think it's neater than the other answer, since you need almost no calculation.)


Showing some transformation is linear Let $T: P_3(\mathbb{R}) \to P_3(\mathbb{R})$ be an operation defined by
  $$T(a+bx+cx^2+dx^3) = a + dx + (a+d)x^2  +(b-c)x^3$$
  Show that $T$ is linear

What I have done so far is look at it like this(since such transformations are homomorphic):

$$T(a+bx+cx^2+dx^3) = aT(1) + bT(x) + cT(x^2) + dT(x^3)$$

Now it would seem that the transformation does nothing, and rather the coefficients have been rearranged.

$\left( \begin{align} a\\b\\c\\d \end{align} \right)$ = $ a\left( \begin{align} 1\\-1\\0\\-1 \end{align} \right)$

Does this show that $T$ is linear? Or what am I meant to do?

It is not quite clear what you have done but then this is what you are supposed to do :

You have to show that $T(a+bx+cx^2+dx^3)=aT(1)+bT(x)+cT(x^2)+dT(x^3)$

It is given that $$T(a+bx+cx^2+dx^3) = a + dx + (a+d)x^2  +(b-c)x^3$$

So, we have :

$T(1)=T(1+0\cdot x+0 \cdot x^2+0\cdot x^3)=1+0\cdot x+(1+0)x^2+(0-0)x^3=1+x^2$

So, $T(1)=1+x^2$

$T(x)=T(0+1\cdot x+0 \cdot x^2+0\cdot x^3)=0+0\cdot x+(0+0)x^2+(1-0)x^3=x^3$

So,$T(x)=x^3$

$T(x^2)=T(0+0\cdot x+1 \cdot x^2+0\cdot x^3)=0+0\cdot x+(0+0)x^2+(0-1)x^3=-x^3$

So, $T(x^2)=-x^3$

$T(x^3)=T(0+0\cdot x+0 \cdot x^2+1\cdot x^3)=0+1\cdot x+(0+1)x^2+(0-0)x^3=x+x^2$

So, $T(x^3)=x+x^2$

Now, can you show that :

$$T(a+bx+cx^2+dx^3)=aT(1)+bT(x)+cT(x^2)+dT(x^3)$$

P.S : I have computed only $T(1)$ before and user142198 understood this and accepted this answer. But then another user Exam in 5 days was having issues in computing $T(x),T(x^2),T(x^3)$ and so i have computed that Only for him..

Given

$$
T\Big( a + bx + cx^2 + dx^3 \Big) = a + dx + ( a + d ) x^2 + ( b - c ) x^3
$$

You need to show that $T$ is linear, thus

$$
T\Big( a P_1 + b P_2 \Big) = a T\Big( P_1 \Big) + b T\Big( P_2 \Big),
$$

where

$$
P_k = a_k + b_k x + c_k x^2 + d_k x^3.
$$

and it is clear that $T$ is linear.

Note that

$$
T( 1 - x - x^3) = 1 - x - x^3
$$

and

$$
T( x + x^2 ) = 0
$$


Why can't $\sin(1/x)$ be differentiated? I've read in textbooks that $\sin(\frac 1x)$ can't be differentiated because it oscillates too rapidly, but so what? Why does oscillation stop it from being differentiated?

The function $f(x)=\sin(1/x)$ is trivially not differentiable at $0$ since it is not defined there. However we will prove that there is no possible value for $c$ so that
$$g(x) = \left\{\begin{array}{l l} \sin(1/x) & \qquad \text{if $x \neq 0$}\\
c & \qquad \text{if $x=0$}
\end{array}\right.$$
is continuous at $0$ (and hence, see (A) below, can't be differentiable at $0$). To do so we will show that the $\lim_{x\to 0} f(x)$ does not exist.

Proof: Let $\epsilon = \frac{1}{2}$. We will begin by showing that for each $\delta > 0$ there exists $x_1,x_2 \in (-\delta,\delta) \setminus \{0\}$ such that $|f(x_1)-f(x_2)| > 2\epsilon$. Let $N=\lceil \frac{1}{2\pi \delta} \rceil + 1$. It follows that $2\pi N + \frac{\pi}{2} > \frac{1}{\delta}$ and $2\pi N +  \frac{3 \pi}{2} > \frac{1}{\delta}$. This means that $x_1= \frac{1}{2\pi N + \frac{\pi}{2}}$, $x_2 = \frac{1}{2\pi N + \frac{3\pi}{2}}$ are in $(-\delta,\delta)\setminus \{0\}$. Since $f(x_1)=1$ and $f(x_2)=-1$ we see that $|f(x_1)-f(x_2)|=2>1=2\epsilon$ as desired. Now let $c$ be an element of $\mathbb{R}$ we have (by the triangle inequality) that $2\epsilon <  |f(x_1) - f(x_2)| \leq |f(x_1) -c| + |f(x_2) - c|$ so that at least one of $|f(x_1)-c|$ or $|f(x_2) -c|$ must be greater than $\epsilon$.

We have proved that for each $c$ in $\mathbb{R}$, there exists $\epsilon >0$ such that for all $\delta >0$ there exists $x$ in $(-\delta,\delta)\setminus \{0\}$ such that $|f(x)-c|>\epsilon$, and hence by definition that $\lim_{x\to 0} f(x)$ does not exist.

(A)
Let $f$ be a real valued function defined on an open interval containing $a$. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Proof: Suppose $f$ is differentiable at $a$. Then by definition there exists $L$ such that $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=L.$$
Using limit laws we have $$\lim_{x\to a} f(x) = \lim_{x\to a} f(x) -f(a) + f(a) = \lim_{x\to a} (f(x) -f(a)) + \lim_{x\to a} f(a) = \lim_{x\to a} \frac{f(x) -f(a)}{x-a}(x-a) + f(a)=\lim_{x\to a} \frac{f(x) -f(a)}{x-a}\lim_{x\to a}(x-a)+f(a)=L\times 0+f(a)=f(a)$$
and hence $f$ is continuous at $a$.

Differentiation is rigourously defined by taking limits; because of the increasing rapidity of oscillation of $sin(1/x)$ as one approaches the origin, the limit that defines the derivative there doesn't converge.


Conceptualization for proof of continuity of $sinx$ First noting that $|\sin(x)| < |x|$ and that $-1 \leq\sin(x) \leq 1$ and $-1 \leq \cos(x) \leq 1$
For all real $x$ and $a$:
$$\sin(x) - \sin(a) = 2 \sin(\frac{x-a}{2}) \cos(\frac{x+a}{2})$$
using the fact that $|\cos{\frac{x+a}{2}}| \leq 1$ we get:
$$|\sin(x)-\sin(a)| \leq 2 |\sin(\frac{x-a}{2})| \leq |x-a|$$
This apparently shows that sin is continuous everywhere. I don't see why. Is it trying to use squeeze theorem? If so, would this be the logical continuation (for any $a \in R$):
$$\lim\limits_{x \to a} |\sin(x)-\sin(a)| \leq  \lim\limits_{x \to a} 2 |\sin(\frac{x-a}{2})| \leq \lim\limits_{x \to a} |x-a|$$
$$ 0 \leq  \lim\limits_{x \to a} 2 |\sin(\frac{x-a}{2})| \leq 0$$
Hence we get $\lim\limits_{x \to a} |\sin(\frac{x-a}{2})| = 0$ showing that it is continous ? Could someone confirm this idea?

The squeeze theorem is really on
$$0 \le |\sin(x) - \sin(a)| \le |x-a|.$$
Since $\lim_{x \to a} |x-a| = 0$, we obtain
$$\lim_{x \to a} |\sin(x) - \sin(a)| = 0,$$
which is equivalent to
$$\lim_{x \to a} \sin(x) = \sin(a),$$
which is the definition of continuity for the function $f(x)=\sin(x)$.

The $\epsilon-\delta$ definition of continuity is:

A function $f$ is continuous at $x_0$ in its domain if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever x is in the domain of $f$ and $|x − x_0| < \delta$, we have $|f(x) − f(x_0)| < \epsilon$. Again, we say $f$ is continuous if it is continuous at every point in its domain.

Basically can we always find a range of $\delta>0$ around $x_0 = a$ where at every point the sin($x$) is within (as small as we choose) $\epsilon$-close to the sin($a$)? $\quad$ Yes.

If you pick the $\delta$ to be $\epsilon$ itself, then any $x$ such that $|x- a|<\epsilon$, will indeed satisfy
$|$sin($x$) - sin($a$)$| \leq |x - a| < \epsilon$
This follows from definition really.


Limit Evaluation - $\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$ $\lim_{x\to \infty} \frac{1-e^x}{e^{2x}}$

My guess is to evaluate by dividing all terms by $e^x$, which works and gives me Eulers identity.

But why should that be right? I thought we are only supposed to divide by the highest exponent term in the denominator? But when I do that, I cannot get a solution? How and when is it ok to divide by an exponent in the numerator?

You might see the limit more clearly by substituting

$y= e^x \Rightarrow \lim_{x\to \infty} \frac{1-e^x}{e^{2x}} = \lim_{y\to \infty} \frac{1-y}{y^2}$

So, you get
$$\frac{1-y}{y^2}= \frac{1}{y^2} - \frac{1}{y} \stackrel{y \to \infty}{\longrightarrow} 0$$

By using Stolz cesaro theorem $\lim_{x\to \infty}\frac{1-e^x}{e^{2x}}=\lim_{x\to \infty}\frac{1-e^{x+1}-1+e^x}{e^{2(x+1)}-e^{2x}}=\lim_{x\to \infty}\frac{e^x(1-e)}{e^{2x}(e^2-1)}=\lim_{x\to \infty}\frac{1-e}{e^x(e^2-1)}=0$


Why is this function defined on $\mathbb{R}$ ? I have to justify why from the equation $x^3-3x+2+ye^y=0$ a function $y=f(x)$ is implicitly defined on $\mathbb{R}$.

When we use the implicit function theorem, do we not get that the above holds in a subset of $\mathbb{R}$ ?

How can we justify that in this case?

It is not.

Note that $y \mapsto -y e^y$ has a $\max$ of $e$, hence if $x^3-3x+2 >e$, there is no value of $y$ that satisfies the equation.

Furthermore, when $x^3-3x+2$ lies in $(0,e)$ there are two values of $y$ that satisfy the equation.

Addendum:

Let $f(x) = x^3-3x+2$, $g(y) = -y e^y$. The idea is that given $x$, we want
to find a unique $y$ such that $f(x) = g(y)$.

A small amount of work shows (i) $\lim_{y \to -\infty} g(y) = 0$, (ii) $g$ is strictly increasing on $(-\infty,-1]$, (iii) $g$ is strictly decreasing on $[1,\infty)$ and (iv) $\lim_{x \to \infty} g(y) = -\infty$. Combining these, we see that the only values of $\alpha$ for which $g(y) = \alpha$ has a unique
solution are $\alpha \le 0$.

Consequently, we want to find values of $x$ such that $f(x) \le 0$.
A small amount of work shows that $f(x) \le 0$ iff $x \le -2$ or $x = 1$.

Hence we can uniquely define a function $h$ on $(-\infty, -2) \sup \{1\}$ that
satisfies $f(x) = g(h(x))$.

If we are willing to drop the uniqueness, we can extend the definition, then the
domain becomes $\{x | f(x) \le e \}$ which consists of two intervals.

Hint.

With the help of the so called Lambert function

$$
Ye^Y = X \Leftrightarrow  Y = W(X)
$$

we have

$$
y = W(3x-x^3-2)
$$

but this don't represents the full set associated to

$$
f(x,y)= x^3-3x+2+ye^y = 0
$$

Attached a plot showing in (blue $\cup$ red) the set associated to $f(x,y)=0$ and in red the set associated to $y = W(3x-x^3-2)$

Concluding, $f(x,y)=0$ does not represents an implicit function.


Prove that if $ \Gamma $ is inconsistent, then $\Gamma \vdash \beta $ for every formula $\beta$ Considering that $\Gamma$ is inconsistent if $ \Gamma \vdash ¬(\alpha \rightarrow \alpha) $ for some formula $\alpha$.

How to prove that if $ \Gamma $ is inconsistent, then $ \Gamma \vdash \beta $?

Do I need to assume that a proof of  $ ¬(\alpha \rightarrow \alpha) $ exists and try to find $\beta$ using axioms? Is it possible?

List of axioms:

Axioms

If you can prove $\Gamma \vdash ¬(\alpha \rightarrow \alpha)$, you can say $$\Gamma,\lnot \beta \vdash ¬(\alpha \rightarrow \alpha) \\ \Gamma \vdash (\lnot \beta \implies¬(\alpha \rightarrow \alpha)) \\ \Gamma \vdash \beta$$

Hint: Prove that, in Polish notation, $\vdash$CpCNpq.

More fully:

axiom 1 CpCqp
axiom 2 CCpCqrCCpqCpr
axiom 3 CCpqCCpNqNp
axiom 4 CNNpp
2 r/p 5 CCpCqpCCpqCpp
5, 1  6 CCpqCpp
6 q/Cqp 7 CCpCqpCpp
7, 1    8 Cpp
hypothesis  9 | p
hypothesis  10 || Np
1 q/Nq      11 || CpCNqp
1 q/Nq p/Np 12 || CNpCNqNp
11, 9       13 || CNqp
12, 10      14 || CNqNp
3 p/Nq, q/p 15 || CCNqpCCNqNpNNq
15, 14      16 || CCNqNpNNq
16, 14      17 || NNq
4 p/q       18 || CNNqq
18, 17      19 || q
10-19       20 | CNpq
9-20        21 CpCNpq
21 p/Cpp    22 CCppCNCppq
22, 8       23 CNCppq

Thus, given that $\Gamma$ $\vdash$ NCpp for some formula p, since 23 holds, it follows that $\Gamma$ $\vdash$ q.


Solve this: $\frac{dx}{dt} = [a-(b+c)]x-ax^2$ by separation of variables So I have this equation

$$\dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$

which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.

I have the initial condition of $x(0)=x_0$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:

NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.

$\frac{dX}{dT} = (a-(b+c))X-aX^2 \Rightarrow \frac{dX}{(a-(b+c))X-aX^2} = dT$
$\Rightarrow \int\limits_{x0}^x\frac{dX}{(a-(b+c))X-aX^2} =\int\limits_0^tdT$
$\Rightarrow t=\frac{\ln(x)-\ln((a-(b+c))-ax)}{a-(b+c)}-\frac{\ln(x_0)-\ln((a-(b+c))-ax_0)}{a-(b+c)}$

Finally isolate $x$ to have $x=f(t,a,b,c)$


How to calculate the angle between three points? If I have a circle with the center O(o1, o2), and three points A(x1, y1), B(x2, y2) and C(x3, y3) lying on its circumference, how can I find the angle AOC? The thing is, that there are two angles and I need the one which "contains" the point B.

I can use the law of cosines but I won't be able to tell if this is the angle that I need.

Edit:

I can check if OB and AC intersect. If yes, I need the angle that is less than pi, if no, I need the other one. Would that be correct?

We can use the two-dimensional analog of cross product here:
$$\sin\left(\angle AOC\right) = \frac{\vec{OA} \times \vec{OB}}{\lvert\vec{OA}\rvert \lvert\vec{OB}\rvert}$$
which is positive if $B$ is counterclockwise from $A$ in a right-handed coordinate system, negative if clockwise: $$-1 \le \sin\left(\angle AOB\right) \le +1$$or$$-180° \le \angle AOB \le +180°$$

Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be the vectors from the center of the circle to the respective points on the circumference:
$$\begin{array}
\; \vec{a} &= ( x_a, y_a ) = (x_1 - x_0, y_1 - y_0) \\
\vec{b} &= ( x_b, y_b ) = (x_2 - x_0, y_2 - y_0) \\
\vec{c} &= ( x_c, y_c ) = (x_3 - x_0, y_3 - y_0)
\end{array}$$
where $(x_0, y_0)$ is the center of the circle, and $\lvert\vec{a}\rvert = \lvert\vec{b}\rvert = \lvert\vec{c}\rvert = r$. Then,
$$\begin{array}
\; \sin\left(\angle AOB\right) &= \frac{x_a y_b - x_b y_a}{r^2} \\
\sin\left(\angle AOC\right) &= \frac{x_a y_c - x_c y_a}{r^2} \\
\sin\left(\angle BOC\right) &= \frac{x_b y_c - x_c y_b}{r^2}
\end{array}$$
So, if we calculate just
$$\begin{array}
\; \lambda_{AB} = r^2 \sin(\angle AOB) = x_a y_b - x_b y_a \\
\lambda_{AC} = r^2 \sin(\angle AOC) = x_a y_c - x_c y_a \\
\lambda_{BC} = r^2 \sin(\angle BOC) = x_b y_c - x_c y_b
\end{array}$$
we can use their signs to determine if $B$ is inside $\angle AOC$ or not. (Note that if all vectors have a nonzero length, the length multiplier only affects the scale; thus, this works, even if the points are not on the circumference of a circle.)

$B$ is inside $\angle AOC$ if and only if
$$\lambda_{AC} \ge 0 \text{ and } \lambda_{AB} \ge 0 \text{ and } \lambda_{BC} \le 0$$
or
$$\lambda_{AC} \le 0 \text{ and } \lambda_{AB} \le 0 \text{ and } \lambda_{BC} \ge 0$$
Otherwise $B$ is outside $\angle AOC$.

In pseudocode:

function inside(x_a, y_a, x_b, y_b, x_c, y_b):
    ab = x_a*y_b - x_b*y_a
    ac = x_a*y_c - x_c*y_a
    bc = x_b*y_c - x_c*y_b
    return ( (ac >= 0 && ab >= 0 && bc <= 0) ||
             (ac <= 0 && ab <= 0 && bc >= 0) )

or

function inside(x_0, y_0, x_1, y_1, x_2, y_2, x_3, y_3)
    ab = (x_1 - x_0)*(y_2 - y_0) - (x_2 - x_0)*(y_1 - y_0)
    ac = (x_1 - x_0)*(y_3 - y_0) - (x_3 - x_0)*(y_1 - y_0)
    bc = (x_2 - x_0)*(y_3 - y_0) - (x_3 - x_0)*(y_2 - y_0)
    return ( (ac >= 0 && ab >= 0 && bc <= 0) ||
             (ac <= 0 && ab <= 0 && bc >= 0) )

A formula you can use is as follows. Let $\alpha$ be the angle $\angle AOC$, then
$$cos(\alpha)=\frac{x_1x_3+y_1y_3}{\surd(x_1^2+y_{1}^2) \cdot \surd(x_3^2+y_{3}^2)}$$


How to compute determinant of $n$ dimensional matrix? I have this example:
$$\left|\begin{matrix}
-1 & 2 & 2 & \cdots & 2\\
2 & -1 & 2 & \cdots & 2\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
2 & 2 & 2 & \cdots & -1\end{matrix}\right|$$

When first row is multiplied by $2$ and added to second, to $nth$ row, determinant is:
$$\left|\begin{matrix}
-1 & 2 & 2 & \cdots & 2\\
0 & 3 & 6 & \cdots & 6\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
0 & 6 & 6 & \cdots & 3\end{matrix}\right|$$

Now using laplace expansion on first column:
$$-\left|\begin{matrix}
3 & 6 &  \cdots & 6\\
\vdots & \vdots & \ddots & \vdots \\
6 & 6 &  \cdots & 3\end{matrix}\right|$$

Is it possible to get recursive relation?
What to do next?

Thanks for replies.

Hint: the matrix $M = e e^T$ (where $e$ is a column vector consisting of $n$  $1$'s) satisfies $M^2 = n M$, so its eigenvalues are ...

You can row reduce the matrix to an upper triangular matrix then take the product of all the elements of the diagonal.


What is the Difference between Frequency and Density in a Histogram? I know count is simply the amount of times the observations occur in a single bin width, but what is density?

Illustrations:

Suppose $X_1, X_2, \dots, X_{100}$ is a random sample of size $n$ from a normal
distribution with mean $\mu=100$ and standard deviation $\sigma=12.$ Also, we have bins (intervals) of equal width, which we use to make a histogram.

The vertical scale of a 'frequency histogram' shows the number of observations
in each bin. Optionally, we can also put numerical labels atop each bar that show
how many individuals it represents.

The vertical scale of a 'density histogram' shows units that make the total
area of all the bars add to $1.$ This makes it possible to show the density
curve of the population using the same vertical scale.

From above, we know that the tallest bar has 30 observations, so this bar accounts for relative frequency
$\frac{30}{100} = 0.3$ of the observations. The width of this bar is $10.$ So
its density is $0.03$ and its area is $0.03(10) = 0.3.$ The
density curve of the distribution $\mathsf{Norm}(100, 15)$ is also shown superimposed on the histogram. The area beneath this density curve is also $1.$ (By definition, the are beneath a density function is always $1.)$ Optionally, I have added tick marks below the histogram to show the locations of the individual observations.

Definitions: If the frequency of the $i$th bar is $f_i,$ then its relative
frequency is $r_i = f_i/n,$ where $n$ is the sample size. Its density is $d_i = r_i/w_i,$ where $w_i$ is its width. Ordinarily, you should make a density
histogram only if each bar has the same width.

Notes: (1) Another type of histogram (that you did not ask about) would be a 'relative frequency' histogram with
relative frequencies (not densities) on the vertical scale. (2) The sample mean of the data shown is $\bar X =102.98$ and the sample standard deviation is $S = 15.37.$
(3) These histograms were made using R statistical software.

The area of the bars of frequency density histogram will NOT be 1. Even when the widths of the bars are the same, say, $w$, the area will be $1/w!$ Do any example.


When was $\pi$ first suggested to be irrational? When was $\pi$ first suggested to be irrational?

According to Wikipedia, this was proved in the 18th century.

Who first claimed / suggested (but not necessarily proved) that $\pi$ is irrational?

I found a passage in Maimonides's Mishna commentary (written circa 1168, Eiruvin 1:5) in which he seems to claim that $\pi$ is irrational. Is this the first mention?

Bhaskara I (the less famous of the two Bhaskara) wrote a comment on Aryabhata in 629, where he gives Aryabhata's approximation $\pi\approx {62832\over 20000}=3.1416$, and states that a nonapproximate value for this ratio is impossible.
See Kim Plofker: Mathematics in India, p. 140.

On the surface this would seem like a clear statement equivalent to the irrationality of $\pi$, but perhaps it is not quite so easy. The "reason" Bhaskara gives is that "surds (square roots of nonsquare numbers) do not have a statable size". It is believed that Aryabhata's method of approximating $\pi$ is essentially the same as Archimedes', computing the circumference of an inscribed regular polygon in a circle with 384 sides. This requires the computation of many surds, so perhaps Bhaskara only meant that he could not get an exact value of the circumference of the inscribed polygon.

From a non-wiki source:

  Archimedes [1], in the third century B.C. used regular polygons inscribed and circumscribed to a circle to approximate : the more sides a polygon has, the closer to the circle it becomes and therefore the ratio between the polygon's area between the square of the radius yields approximations to $\pi$. Using this method he showed that $223/71<\pi<22/7$.

Found on PlanetMath with a reference to Archimedes' work...


Best numerical method to integrate a differential equation like $\frac{\partial y}{\partial t}=f\left(t,\frac{\partial y}{\partial r}\right)$? What is the best numerical method to solve an equation like this:
$$\frac{\partial y}{\partial t}=f\left(t,\frac{\partial y}{\partial r}\right)\quad?$$

Can somebody give, at least, a reference?

This is quite difficult to answer without further details. Assuming that you have no clue about what to do, I would recommend you to look into finite difference methods. These are probably the easiest methods to understand. Other methods are finite elements (bit more complicated and in 1d often similar to finite differences, though boundary conditions are handled differently) and (pseudo)spectral methods (which may give you better accuracy if everything is smooth).

A good reference is Morton & Mayers, Numerical solutions of partial difference methods (though as Ross says, any numerical analysis book will have something about this, often in the last chapter). There is also a list of links at http://math.fullerton.edu/mathews/n2003/finitediffpde/FiniteDifferencePDEBib/Links/FiniteDifferencePDEBib_lnk_1.html .

I don't think we need numerical method to solve $\dfrac{\partial y}{\partial t}=f\left(t,\dfrac{\partial y}{\partial r}\right)$ .

Firstly, because EqWorld has already provided the complete integral of this type of PDE, see http://eqworld.ipmnet.ru/en/solutions/fpde/fpde3305.pdf for details:

$$y(r,t)=\int f(t,c_1)\,dt+c_1r+c_2$$

Secondly, for the general solution:

$\dfrac{\partial y}{\partial t}=f\left(t,\dfrac{\partial y}{\partial r}\right)$

$\dfrac{\partial^2y}{\partial r\partial t}=\dfrac{\partial f\left(t,\dfrac{\partial y}{\partial r}\right)}{\partial r}$

$\dfrac{\partial^2y}{\partial r\partial t}=\dfrac{\partial f\left(t,\dfrac{\partial y}{\partial r}\right)}{\partial\dfrac{\partial y}{\partial r}}\dfrac{\partial\dfrac{\partial y}{\partial r}}{\partial r}$

$\dfrac{\partial^2y}{\partial r\partial t}=\dfrac{\partial f\left(t,\dfrac{\partial y}{\partial r}\right)}{\partial\dfrac{\partial y}{\partial r}}\dfrac{\partial^2y}{\partial r^2}$

Let $u=\dfrac{\partial y}{\partial r}$ ,

Then $\dfrac{\partial u}{\partial t}=\dfrac{\partial f(t,u)}{\partial u}\dfrac{\partial u}{\partial r}$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dr}{ds}=-\dfrac{\partial f(t,u)}{\partial u}=-\dfrac{\partial f(s,u)}{\partial u}(u=u_0)$ , we have $r=F(u_0)-\int^s\dfrac{\partial f(s,u)}{\partial u}(u=u_0)~ds=F(u)-\int^t\dfrac{\partial f(s,u)}{\partial u}ds$


Roots of the characteristic polynomial of a symmetric matrix I'm looking for a proof (using basic tools : definition of the characteristic polynomial and its basic properties) of the following fact :
The roots of the characteristic polynomial of a symmetric matrix (with real coefficients) are reals.

Thank you.

$\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
Roots are the eigenvalues. So, we consider the eigenvalues equation of $M = M\+$. If an eigenvalue is zero, it is proved. So, we consider non null eigenvalues $\braces{\lambda}$:
$$
\left.%
\begin{array}{rcl}
M{\bf v} & = & \lambda{\bf v}\,,\
\mbox{with}\ {\bf v}\ \mbox{non null}
\\
{\bf v}\+\lambda^{*} & = & {\bf v}\+M\+
\end{array}\right\rbrace
\quad\imp\quad
0 = \pars{\lambda - \lambda^{*}}{\bf v}\+ M{\bf v}
=
\pars{\lambda - \lambda^{*}}\lambda\,{\bf v}\+{\bf v}
$$
Then, $\quad\color{#0000ff}{\large \lambda^{*} = \lambda}$.


What is the integral of powers of sines and cosines over one period? I am looking for an expression for the integral
\begin{equation*}
I_{n,m}=\int_0^{2\pi} \cos^n(x)\sin^m(x) \, dx
\end{equation*}
where $n,m$ are arbitrary positive integers. Thanks

HINT

The problem is how to reduce the cosine and sine with degrees to cosine and sine of multiple arguments, since $\cos(nx)$ and $\sin(mx)$ are orthogonal on the interval $(0,2\pi):$
$$\int_0^{2\pi}\cos(nx)\sin(mx)=0,$$
$$\int_0^{2\pi}\cos(nx)\cos(mx)=\int_0^{2\pi}\sin(nx)\sin(mx)=
\begin{cases}
2\pi\text{ for }n=m=1\\
\pi\text{ for }n=m\not=1\\
0\text{ for }n\not=m.
\end{cases}$$

To reduce, it's possible to use the formula
$$2\cos(nx)=z^n+\bar z^n,\quad 2i\sin(mx)=z^m-\bar z^m,\quad z=e^{ix},\quad \bar z=e^{-ix},$$
then
$$2^n\cos^n(x)=(z+\bar z)^n = \sum_{k=0}^h
\genfrac{(}{)}{0}{}{n}{k}\left(z^{n-2k}+\bar z^{n-2k}\right)+(n-2h-1)\genfrac{(}{)}{0}{}{n}{h}$$
$$=2\sum_{k=0}^h
\genfrac{(}{)}{0}{}{n}{k}\cos(n-2k)x + (n-2h-1)\genfrac{(}{)}{0}{}{n}{h},$$
where $h=\left[\dfrac{n}2\right]$;
$$(2i)^m\sin^m(x)=(z-\bar z)^n$$
$$
=\sum_{k=0}^{h'}(-1)^k\genfrac{(}{)}{0}{}{m}{k}\left(z^{m-2k}+(-1)^m\bar z^{m-2k}\right) + (-1)^{h'}(m-2h'-1)\genfrac{(}{)}{0}{}{n}{h'}$$
$$=
\begin{cases}
2\sum_{k=0}^l(-1)^k\genfrac{(}{)}{0}{}{m}{k}\cos(m-2k)x + (-1)^l\genfrac{(}{)}{0}{}{n}{l}\text{ for }m=2l\\
2i\sum_{k=0}^l(-1)^k\genfrac{(}{)}{0}{}{m}{k}\sin(m-2k)x\text{ for }m=2l+1
\end{cases}$$

These are dealt with in pretty much any Calculus text.
  If either sine or cosine is to an odd power, a simple substitution reduces it to an algebraic integral:

1) $\int \sin^n(x)\cos^{2k+1}(x)\, dx= \int \sin^n(x) \cos^{2k}(x) (\cos(x)\,dx)$
  Let $u= \sin(x)$ so that $du= \cos(x)\,dx$, $\cos^2(x)= 1- \sin^2(x)= 1- u^2$ and $\cos^{2k}(x)= (1- u^2)^k$.  The integral becomes $\int u^n(1- u^2)^k \,du$.

2) $\int \sin^{2k+ 1}(x)\cos^m(x)\,dx= \int \sin^{2k}(x)\cos^m(x)(\sin(x) \, dx)$
  Let $u= cos(x)$ so that $du= -\sin(x)\,dx$, $\sin^2(x)= 1- \cos^2(x)= 1- u^2$, and $\sin^{2k}(x)= (1- u^2)^k$.  The integral becomes $\int (1- u^2)^ku^m \,du$.

If both sine and cosine are to an even power use $\sin^2(x)= (1- \cos(2x))/2$ and $\cos^2(x)= (1+ \cos(2x))/2$, repeatedly if necessary, to reduce to a form in which either sine or cosine has an odd power.


In Geometric Algebra, is there a geometric product between matrices? Thanks for your help in advance.

I literally just started to self-study about geometric algebra.

I have some coursework background in linear algebra and was trying to make an educational bridge between what I know and what I'm trying to learn.

My question:  Is there a geometric product for matrices in geometric algebra, like there is a geometric product for vectors?  If so, how would one compute the geometric product between matrices?

Thanks

Let me address this more on the side of how linear algebra is presented in some GA material.

In traditional linear algebra, you use a lot of matrices and column/row vectors because this gives you an easy way to compute the action of a linear map or operator on a vector.  What I want to emphasize is that this is a representation.  It's a way of talking about linear maps, but it's not the only way.

In GA, there are reasons we don't often use matrices explicitly.  One reason is that we have a natural extension of a linear operator to all kinds of blades, not just vectors.  If you have a linear operator $\underline T$, and you want to compute its action on a bivector $a \wedge b$ with matrices, you would have to compute a totally different matrix from the one you would use just considering $\underline T$ acting on a vector (this matrix's components would describe its action on basis bivectors, not basis vectors).  This is one reason why matrices become rather useless.

Thus, since we tend to look at linear maps and operators merely as linear functions, we have to develop ways to talk about common linear algebra concepts without reference to matrices at all.  This is how we talk about a basis-independent of the determinant using the pseudoscalar $I$, saying $\underline T(I) = I \det \underline T$ for instance.  Texts on GA and GC also develop ways to talk about traces and other interesting linear algebra concepts without reference to matrices.

With all that in mind, since we don't talk about matrices when doing linear algebra in GA, we don't have to think about geometric products of matrices.  We just talk about compositions of maps (which would be represented through matrix multiplication) when applying several maps in succession.

Linear algebra with its vectors and matrices is made entirely obsolete by Clifford algebra which provides a better way. Good riddance!

Gone is the awkward distinction between "row vectors" and "column vectors". In Clifford there is no distinction. And many weird abstract concepts become concrete spatial concepts that are easy to visualize. The determinant becomes a three-dimensional volume between three vectors, and the volume goes to zero as the vectors become parallel.

In Clifford algebra a matrix is just an array of vectors that span a space. The geometric product has two effects: rotation and scaling. So a geometric product of a matrix will tend to rotate and scale the geometric shape.

I find that the most useful and interesting aspect of Clifford algebra is to try to picture all algebraic relationships as spatial structures, or operations by spatial structures on other structures.