How much velocity can a canister of fuel give a spaceship? I've recently considered the issue of how much velocity a canister of fuel can provide a 'spaceship'. I assumed we could approximate a basic solution If we know the mass of the fuel $m$, the mass of the ship $M$ and the amount of energy in the fuel $E$. So the energy density of the fuel is, $\frac{E}{m} = p$, which is constant. So the amount of velocity gained by an increment of fuel is equal to $\sqrt{\frac{p\cdot{dm}}{M+m}} = dv$. Since $dE = p\cdot{dm} = \frac{1}{2}(m+M)dv^2$. In order to find $v$, I needed to integrate the left hand side from m equals the total mass of the fuel to $m=0$ (assuming the canister is massless).

Is this correct? How, (if at all) is it possible to evaluate this integral?

P.S. I think I'm way off from this derivation https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Can someone please lead me in the right direction?

The equation in the mentioned Wikipedia page is a special case of the general problem of motion equation for a system with variable mass that comes from the conservation of the momentum of the system. You can see the derivation of the pertinent equation at this page.

The  pure energetic approach in OP is not correct because the energy is conserved only in the full system, given by the rocket and the combusted fuels and also the heat produced in the combustion. The key parameter that determines the velocity of the roket is the expulsion velocity of the ejected propellant, that depends on the constructive characteristics of the roket.

${d\over dv}v^2=2v$
Hence $dv^2=2vdv$
Therefore your equation would be
Integral of $\int {p\cdot dm\over m+M}= \int vdv$


If $G = \langle a\rangle$ and $b$ $\in$ $G$, the order of $b$ is a factor of the order of $a$ Prof. Pinter's "A Book of Abstract Algebra" presents the following exercise from the "Cyclic Groups" chapter:

  If $G = \langle a\rangle$ is finite and $b$ $\in$ $G$, the order of $b$ is a factor
  of the order of $a$

I believe that the proof is given by Theorem $2$ of this chapter:

  Every subgroup of a cyclic group is a cyclic.

Does this theorem provide an adequate proof to this exercise?

The theorem you have indicated argues that every subgroup is cyclic. The question requires something slightly different.

Assuming $G$ is a finite group, what can you say about the order of the subgroup $H=\langle b \rangle$? Try and use Lagrange's theorem after that point.

Here's an answer more in line with the concepts learned in Pinter's book up to that point (ie. no Lagrange's theorem etc).

Let ord$(a) = n$. Therefore, there are $n$ elements in $G$.

Then since $b \in G$, we have $b = a^i$ for some $1 \leq i \leq n$. This is because since $G = \langle a \rangle$ is cyclic, $G$ contains all the powers of $a$ and nothing else.

So we have $b^n = (a^i)^n = (a^n)^i = e^i = e,$

showing that ord$(b) | n$.

That is, ord$(b) |$ ord$(a)$.


How treat a point where a function is not defined for indefinite integral I have a question maybe stupid but I don't know how solve it.
Solving the indefinite integral I have a doubt:
When I have a definite integral usually we check if  the interval of integration is finite or if the function to be integrated is well defined, in the interval of integration, if not I am talking of a improper integral.
For example $\int_0^x \frac{1}{x}$ is an improper integral since $\lim_{x\to 0^+}\frac{1}{x}=+\infty$.
$\textbf{My question:}$ when I have an indefinite integral, for instance $\int \frac{1}{x} dx$, and for $x=0$ the function $\frac{1}{x}$ is not defined, in particular $\lim_{x\to 0^+}\frac{1}{x}=+\infty$, I have to take into account this fact or not? How can I treat $x=0$ in this case?

The indefinite integral $\int f(x) \, dx$ is just a notation for the antiderivative, i.e. a function $F$ with $F' = f$. This implies that the domain of $F$ is a subset of the domain of $f$.
Your example function $f(x) = 1/x$ is defined on $\Bbb R\setminus \{ 0 \}$, therefore an antiderivative can only be defined on that set.
The notation $\int f(x) \, dx = F(x) + C$ is sometimes used to indicate that the “general antiderivative” is obtained by adding an arbitrary constant to $C$, e.g.
$$
\int \frac 1 x \, dx = \log |x| + C \, .
$$
As Hans Lundmark said, care must be taken if the domain of $f$ consists of multiple components. Strictly speaking, the general antiderivative of $1/x$ is
$$
 F(x) = \begin{cases}
  \log (-x) + C_1 & \text{ for } x < 0 \\
  \log (x) + C_2 & \text{ for } x > 0 \\
 \end{cases}
$$
with real constants $C_1, C_2$.
Back to your original question: If $f$ is not defined at a point $x_0$ then there is no antiderivative at $x_0$. But:

If the domain of $f$ consists of multiple intervals then the constant of integration can be independently chosen on each interval on which the antiderivative exists.

The fundamental theorem of calculus in the form $\int_a^b f(t) \, dt = F(b) - F(a)$ is only valid on each interval on which the antiderivative exists.
As an example, it would be wrong to say that $\int_{-1}^1 \frac 1x \, dx = \log|1| - \log |-1| = 0$.

An indefinite integral can be written as a definite one. E.g.
$$\int \frac{dx}x=\int_1^x\frac{dt}t+C.$$
Obviously, this results in a well-defined function for all $x>0$, and $x\le 0$ is still ruled out.


Justification of Proofs by Contradiction Is there a validation for the technique of proof by contradiction? Or do those who use it take its validity as an axiom?

The use of proof by contradiction is closely tied to the law of the excluded middle. The motivation for including this sort of proof in ordinary mathematics is that we think of each statement as being either true or false, and we think that if a statement is not true then its negation is true. If we are talking about abstract truth about a particular structure, then these are arguably self-evident principles.  For example, the structure might be the semiring of natural numbers, and the statement might say that there are infinitely many primes. In this structure, it seems clear that there are either infinitely many primes (meaning the set of primes is not bounded), or there are not, and that exactly one of these alternative holds. So when we take the structure as already given, and reason about which statements are true or false in the structure, classical logic is a good framework for this sort of reasoning.

However, there are mathematical frameworks that do not include the law of the excluded middle but still include forms of proof by contradiction. For example, the usual formalization of intuitionistic logic includes the ''ex falso quodlibet'' rule: if you can prove $P \land \lnot P$ then you can conclude $Q$, for any statements $P$ and $Q$. But this logic does not, in general, prove $P \lor \lnot P$.  Intuitionistic logic is a good framework if we think of a structure as only partially given, so that our knowledge of its overall properties may change over time as we discover more. To affirm a sentence in this logic, we need to know that the sentence is true based only on the part of the structure that we have seen so far. In this setting, we may not be able to affirm $P$ on the basis of the part of the structure we have seen, and we may not be able to affirm $\lnot P$ until the entire structure is known; in this case we can't affirm $P \lor \lnot P$ based only on our partial information. This logic still proves $\lnot (P \land \lnot P)$: it is impossible for the part of the structure we have seen to satisfy both $P$ and $\lnot P$.  One reason that ordinary mathematics is not done this way is that ordinary mathematicians don't think of structures such as the natural numbers as only partially specified; we think of the natural numbers as a completed object.

One logical framework that includes neither ''ex falso quodlibet'' nor the law of the excluded middle is known as "minimal logic". This framework is mostly of interest in proof theory, where it's used for the purpose of stating results in more generality.

Let "C" denote the material conditional, "N" negation, 0 falsity, and read via the Polish notational scheme.  Then proof-by-contradiction can get justified as follows:

We have the rule of inference of modus ponens.
We have CCNp0p as a true formula for our background logic.
We have a rule which goes "if we have derivation which starts with a proposition q and ends with a proposition r, we may infer Cqr".

Then, it follows that if we assume the negation of a proposition Np as true, and deduce a falsity, we may then infer that Np implies a falsity, or equivalently CNp0 as true.  Then since CCNp0p holds, and we have modus ponens as a rule of inference, we may infer the proposition p as true.

Of course, oftentimes, with K denoting conjunction, one takes the formula KpNp as a falsity. But, there do exist non-classical logics where KpNp isn't a falsity, and there still exist falsities in those logics.  So, proof-by-contradiction still can get used, say by deducing NCpp as the falsity, given that NCpp always has truth value of false, or if Cpp is a theorem, and the negation of a theorem qualifies as a contradiction.


In (relatively) simple words: What is an inverse limit? I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me.

The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction."

The Wikipedia article presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it.

Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then.

Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps)

(While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)

I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions).

We start with a motivating example.

Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment.

We use the notation $\mathbb{Z}_n$ for $\mathbb Z / n \mathbb Z$.
The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$.

An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get
$25y^2 + 20y = -5$.

That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$:
$$x = 5y+2 \equiv 5\cdot 1 + 2  = 7 \pmod{25}.$$

Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$.

Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues,
$$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$
each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms
$$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$

Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$.

Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution.

This is much along the lines of taking a Cauchy sequence of rationals, which "describes" a limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set.

This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$.

Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc.

The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from the $5$-adics.

Moreover, the ring of $5$-adic integers is universal for this property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps
$$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$

So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object."

More generally, assume that you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$), and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{rs}\circ f_{st} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner.

If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity".

Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this.

If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).

I'm currently dabbling around in a category where morphisms are refinements of computer programs, in the sense that they are functions $f : X \rightarrow Y$ from a program refinement (an implementation) $X$ to a program specification $Y$.

In this setting, a sequence (or net) of morphisms can be either describing an ever more elaborate specification $X_0 \rightarrow X_1 \rightarrow X_2 \rightarrow ...$ or it can be an ever more precise refinement $... \rightarrow X_2 \rightarrow X_1 \rightarrow X_0$. The first sequence has as a direct limit the 'ultimate' specification, while the second sequence has as an inverse (also called projective) limit the 'ultimate' refinement.

As an extra: maybe my own question about Directed and projective limit in Rel helps in looking at the notion in a different way. In Rel, where the two types of limit coincide, my intuition is that a net of relations simply results in 'lines' connecting points in a disjoint union, and the 'infinite' lines form a new object, the limit.


The myth of no prime formula? Terence Tao claims:

  For instance, we have an exact formula for the $n^\text{th}$ square
  number – it is $n^2$ – but we do not have a (useful) exact
  formula for the $n^\text{th}$ prime number $p_n$!
  “God may not play dice with the universe, but something
  strange is going on with the prime numbers.” (Paul Erdős,
  1913–1996)

However there exist an exact formula for the nth prime

  A double sum for the nth prime pn is $$p_n=1+\sum_{k=1}^{2(\lfloor n\ln n\rfloor+1)}\Biggl[1-\Biggl\lfloor\frac{\sum_{j=2}^k 1+\lfloor s(j)\rfloor}n\Biggr\rfloor\Biggr],\tag{13}$$ where $$s(j)\equiv-\frac{\sum_{s=1}^j \bigl(\bigl\lfloor\frac js\bigr\rfloor-\bigl\lfloor\frac{j-1}s\bigr\rfloor\bigr)-2}j\tag{14}$$

(See Prime formulas at MathWorld.)  There exist even an exact formula for prime counting function $\pi (x)$. So why are mathematicians trying to prove the Riemann hypothesis to find a better estimate for $\pi(x)$ and $p_n$ when they have those exact formulas?

If there is a mathematical body of knowledge that allows you to manipulate formulas (13) and (14) into a better form, then great!  The Riemann zeta function has already proved its worth in this field.  It was instrumental in the proofs by Hadamard and de la Vallée-Poussin of the Prime Number Theorem.  What allowed that to happen was that the zeta function is a meromorphic function, and so complex variable theory could be applied to its study.  One never knows with certainty whether further study of the zeta function will pay off, but people are betting that it will, based on past successes.

I don't know whether anyone is promoting (13) and (14) as useful tools in the study of prime numbers, but if so, the burden is on those people to exhibit mathematical tools that allow one to manipulate (13) and (14) effectively.

Because of all the floor functions, (13) and (14) don't have very nice analytical properties.  As other posters have pointed out, they encode an elementary algorithm for identifying primes.  In particular, the expression
$$
\left\lfloor\frac{j}{s}\right\rfloor-\left\lfloor\frac{j-1}{s}\right\rfloor
$$
that appears in the numerator of (14) equals $1$ if $s$ evenly divides $j$ and equals $0$ if it does not.  Summing this quantity for $s$ from $1$ to $j$ gives you the number of divisors of $j.$  Prime numbers have exactly two divisors; composite numbers all have at least three.  The numerator in (14) is therefore $0$ when $j$ is prime, and positive when it is composite.  Dividing by $j$ and multiplying by $-1$ results in the quantity $s(j),$ which is $0$ when $j$ is prime and lies in the interval $(-1,0)$ when $j$ is composite.  The floor of $s(j$) therefore equals $0$ when $j$ is prime and $-1$ when $j$ is composite.  Adding $1$ to $\lfloor s(j)\rfloor$ gives the characteristic function of the primes.

Notice that using the formula (14) requires one to do more divisions than even a naive brute-force algorithm to compute the number of divisors would require.  That price might be worth paying if the formula had nice mathematical properties that allowed one to extract additional information from it, but I don't see that this has been demonstrated.  To me, it looks like a cumbersome arithmetic encoding of an inefficient algorithm.  I would be happy to be proved wrong about this!

Formula (13) likewise looks like a cumbersome arithmetic encoding of a brute-force method for computing the $n^\text{th}$ prime, and seems likely to require more calculation than state-of-the-art algorithms.  Again, it would be nice if someone could show how to extract useful information from the formula, but it doesn't look very promising.

Added: I think it worth pointing out that some non-trivial analytic number theory enters into formula (13) in the choice of upper bound on the summation over $k.$  You need to ensure that the $p_n$ lies in the range of summation, but that $p_{2n}$ does not.  The reason is that the summand equals $1$ when $k<p_n,$ equals $0$ for $p_n\le k<p_{2n},$ and becomes negative for $k\ge p_{2n}.$  So you need the summation over $k$ to include all $p_n-1$ terms where the summand equals $1$, but none of the terms where it is negative.  This will ensure that the sum evaluates to $p_n-1.$

It appears to me that you need something like (3.12) and (3.13) in this paper of Rosser and Schoenfeld to justify the upper bound.  Formula (3.12) is Rosser's Theorem which Dietrich Burde's answer to this post suggests requires deep knowledge of zeta function zeros.  Interestingly, the author  of your formulas (13) and (14) (Sebastián Martín Ruiz) does not rigorously justify the upper bound in his article, giving only a heuristic argument and and stating that it is “very probable” that the necessary inequalities are satisfied.

Perhaps one could play with the constant in front of $\lfloor n\log n\rfloor$ and try to use Chebyshev's bounds (Theorem 3.1 here) instead of Rosser's Theorem.  (The derivation of Chebyshev's bounds doesn't require knowledge of zeta function zeros.)  But this seems possibly rather delicate, and may not work.

Second addition:   Some discussion of the formula, and Ruiz's view of it, can be found here.

Well in the context of computing the $n^{th}$ prime number, I can almost guarantee you from looking at that formula that if I now write the code to enumerate both it and another patch of code based entirely on checking the divisibility of the output of a random number generator, the RNG with beat the code written for that formula by a long shot.

Likewise for all of the studies I've been conducting on variations and alterations to Wilson's theorem, you see the computational expense of factorization is not going to disappear upon knowing explicit formula for the $n^{th}$ prime number, it is minimized by applying a number of considerations from a number of differing fields, but the one that is most familiar for me is Modular Arithmetic, particularly things like Euler's Theorem. In fact one thing I realised tonight, they will never ever run out of factorization exploits for which to ensure the security of encryption from those that all stem from that one theorem alone, and no doubt there are more out there that are too advanced for me at this point, so it really isn't as big of a deal as everyone makes it out to be in my opinion.


What are the differences between a fiber bundle and a sheaf? They are similar. Both contain a projection map and one can define sections, moreover the fiber of the fiber bundle is just like the stalk of the sheaf.

But what are the differences between them?

Maybe a sheaf is more abstract and can break down, while a fibre bundle is more geometric and must keep itself continuous. Any other differences?

If $(X,\mathcal{O}_X)$ is a ringed topological space, you can look at locally free sheaves of $\mathcal{O}_X$-modules on $X$.

If $\mathcal{O}_X$ is the sheaf of continuous functions on a topological manifold (=Hausdorff and locally homeomorphic to $\mathbb{R}^n$), or the sheaf of smooth functions on a smooth manifold, you get fiber bundles (the sheaf associated to a fiber bundle is the sheaf of "regular" (=continuous or smooth here) sections).

"In the geometric context the cast of characters changes: on the Galois side we now have vector bundles with flat connection on a complex Riemann surface X in the global case, and on the punctured disc D around a point of X in the local case. The definition of the objects on the other side of the geometric Langlands correspondence is more subtle. It is relatively well understood (after works of A. Beilinson, V. Drinfeld, G. Laumon and others) in the special case when the flat connection on our bundle has no singularities. Then the corresponding objects are the so-called “Hecke eigensheaves” on the moduli spaces of vector bundles on X. These are the geometric analogues of unramified automorphic functions. The unramified global geometric Langlands correspondence is then supposed to assign to a flat connection on our bundle (without singularities) a Hecke eigensheaf." https://math.berkeley.edu/~frenkel/loop.pdf


What does the integrand in a double integral represent? I see this problem and I'm not sure what the integrand here represents. Can someone tell me what it visually represents here?

Let $$S=\iint f(x,y)dxdy$$
When $ f(x,y)=1 $, $ S $ is a surface.
When $ f(x,y)\ne 1 $, $ S $ is a volume and $ f(x,y)dxdy $ is the elementary volume with $ dxdy$ as a base surface and $ f(x,y)$ as its height.


Counting candies in boxes There are $5$ boxes containing $80$ candies. After taking $\frac{1}{5}$ of the candies in the first box and putting them in the seconf one, we take  $\frac{1}{5}$ of the candies in the second box and putting them in the third one (after we added the $\frac{1}{5}$ candies from the first box) and so on untill we take $\frac{1}{5}$ of the candies in the fifth box and put it in the first one.
After this process, each box contained the smae number of candies.  How many candies were initially in each box?

So I can say $a+b+c+d+e=80$ and then slowly and brutally start calculating how many candies were in each box after the process- which will yield 6 equations (including the first one). But is there a better way?

Suggestion:

Can you work backwards?  At the end the distribution is $(16,16,16,16,16)$.  The step before, the distribution would be $(12,16,16,16,20)$.


Do formal polynomials make sense in arbitrary algebraic structures? Let $R$ denote a commutative ring with unity and $X$ a set of formal indeterminates. Then intuitively, the set of all formal polynomials in $X$ with coefficients in $R$ can be defined as the free commutative ring with unity generated by $R \cup X$ subject to the identities satisfied by $R$.

Question 1. Does this definition actually work?

If so, then I think that formal polynomials make sense in arbitrary algebraic structures. So my second question is:

Question 2. Irrespective of whether the above definition makes sense, do formal polynomials make sense in arbitrary algebraic structures?

Q1: Yes. But I would rather say that $R[X]$ is the free commutative $R$-algebra on $X$.

Q2: The notion of "free objects" can be defined in a very general category-theoretic setting, see Wikipedia or any book on category theory. In particular, we have the notion of a free algebraic structure of any given type. We have free monoids, free groups, free modules, free lie algebras, free boolean algebras, etc. But notice that the elements aren't called polynomials anymore, because they are just no polynomials.

Yes and you can call them polynomials. See Cox et al "Using Algebraic Geometry" for examples of polynomials over $\mathbb{Q}$, $\mathbb{Z}$ and other rings.


Explain why $E[XY\mid X]=XE[Y\mid X]$ I'm getting through the logic of $E[XY\mid X]=X(E[Y\mid X])$: since $E[X\mid X]=X$ is a constant, so we can pull the term $X$ out, and we get $X(E[Y\mid X])$. Is this the right way to think about this proposition?

This is a generalization of linearity. For conventional expectation, we have $E(XY)=XE(Y)$ when $X$ is a constant. Conventional expectation can be thought of as expectation with respect to the trivial sigma algebra, and a function is measurable w.r.t. the trivial sigma algebra iff it's constant. The general version is that $E(XY\mid F)=XE(Y\mid F)$ iff $X$ is $F$-measurable, where $F$ is some sigma algebra.

Intuitively, for $X$ to be $F$ measurable means that you "know" the value of $X$ given the information $F$, and $E(Z\mid F)$ is the "best guess" as to the value of $Z$ given the information $F$, so obviously if you know $X$ then your best guess for $XY$ is going to be $X$ times your best guess for $Y$.

When you write $\operatorname{E}(\cdot\mid X),$ then within that narrow context you're treating the value of $X$ as given, thus "constant" and not "random".


Probability $P(X>Y,X>Z)$ for independent normal random variables $X$, $Y$, $Z$ There are several answers already given for working out the probability of one random variable being greater than another, but I can't make the leap to working out the probability of one random variable being greater than several others. My random variables are independent and normally distributed.
For example:

  Let $X$, $Y$ and $Z$ be independent normal random variables. What is $P(X>Y,X>Z)$?

The obvious (to me) answer, being to just multiply the two probabilities $P(X>Y)$ and $P(X>Z)$ does not work because the difference random variables $(X-Y)$ and $(X-Z)$ are not independent.

Edit

For $P(X>Y)$ the answer is:
$$
{\rm P}(X  > Y )  = \Phi \left(\frac{\mu_X - \mu_Y }{\sqrt{\sigma_Y^2 + \sigma_Y^2}}\right).
$$
I'm hoping for a way of adding a third normal random variable to the equation. If this is possible I presume the answer can be easily expanded to add further random variables.

Can be written in the below form ,

$$P(X>Z,X>Y)= P(X> \min(Y,Z)) \ ( P(Y>Z) \text{ or } P(Y<Z))$$

$$ P(X> \min(Y,Z)) =P(X>Y) \ P(X>Z) $$

Hope it helps.


Is there a way to determine the prefactor of the trace of the product of two $SU(N)$ matrices? While renormalizing QCD I found the follow equation:
$$tr(T_aT_b) = n(r)\delta_{ab},$$
where $T_a$ is the generic SU(N) generator and $n(r)$ is a factor depending on the algebra representation.
In the defining representation of $SU(3)$ the generators are
$$T_a = \frac{\lambda_a}{2},$$
where $\lambda_a$ are the eight well know Gell-Mann matrices; in this case it's easy to verify that $n(r) = \frac{1}{2}$. I was thus wondering: what is the formal demonstration of this relation? And is there a way to determine $n(r)$ for generic matricial representation and generic N, or we are bound to know the explicit form of the generators in order to obtain it?

Yes, following the  discussion below, all such indices for all reps can be gotten from self-consistency in tensoring the defining/fundamental with itself, its conjugate, the adjoint, etc. You do not need to know the explicit form of the generator matrices in a given representation--only its provenance out of fundamentals and antifundamentals. Here is how it works for your case of the fundamental.

It is a one-line proof, but you are meant to be familiar with the language, or some equivalent thereof. Let's call F the fundamental representation N×N matrices, A the adjoint (N²—1)×(N²—1) ones, and 0 the 1-dim singlet rep ones.

The Kronecker product fundamental-antifundamental (N×N)⊗(N×N) representation matrices are given by the coproduct,
$$
F^a\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a = A^a  \oplus 0 = A^a,
$$
which you must check, indeed, satisfies the same Lie algebra,
$$
[(F^a\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a) ,(F^b\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^b)]=if^{abc}(F^c\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^c).
$$

Now multiply this with its a ⟶ b analog,
$$
(F^a\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a) (F^b\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^b) = A^a    A^b= \\
F^a F^b\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a \bar F^b +  F^a\otimes \bar F^b + F^b\otimes \bar F^a,
$$
and trace. The trace of the direct product amounts to a plain product of the separate traces of each tensor factor, of course, so the last two terms vanish, as the F s are traceless.

You thus establish in a blink that
$$
2N\operatorname{tr} F^a F^b = \operatorname{tr} A^a A^b .
$$
If the trace on the right is N, the trace on the left must be 1/2, your sought answer.

In physics, the r.h.s. is normalized to be N, as the structure constants providing the adjoint rep are chosen to be such. (Not in mathematics!)

Mutatis mutandis you know the trace of all irreps must be fixed w.r.t. the adjoint normalization:  scaling up the generators of the algebra results in an identical rescaling of the structure constants, which we already fixed.

The reason for these normalization choices, despised by mathematicians (who normalize the linchpin structure constant f  with 1), is because  physicists want all SU(N) s to include an isospin SU(2) with its conventional spin 1/2 normalizations, for ease in particle physics applications. (The mayhem started with flavor SU(3) and lives on.)

To  reassure yourself are in control of the algorithm, try composing ${\bf 3}\otimes {\bf 3}= {\bf 6}\oplus \bar{\bf 3}$ in SU(3) and compute this trace for the sextet: You should find 3 — 1/2=5/2. (It is half the representation index you find in math texts.)

Standard  physics (QFT) texts like Peskin & Schroeder, An Introduction To Quantum Field Theory (1995), §15.4, review the language.

You have to know the form of the generators, because it depends on them.

Simply put, if you assume $a=b$, then it's $tr(T_a T_b) = n(r)$ (which is obvious
but the closest you can get for a Formula for $n(r)$, where its implicitly assumed that $tr(T_a T_a) = tr(T_bT_b) \quad\forall a,b$, which is no problem after you scale the basis vectors (''Generators'') correctly).

I am not sure about the other Questions of yours (in the sense that I don't know
what answer you're seeking for).
What might help you is getting familiar with the so called ''Cartan-Weyl-Basis'',
and judging from the style in which your question was written, I assume you
are a physicist, so the following book might help you:
''Shattered Symmetry Group Theory from the Eightfold Way to the Perdiodic Table'',
by Pieter Thyssen and Arnout Ceulemans.

Finally, observe that in the equation you have written, there does not appear to be any given representation (besides, maybe, the fundamental one), so the asking for it in an arbitrary representation seems kind of pointless to me...


The extension of function $M$ is a smooth manifold and $A$ is a subset of $M$. If a function $f$ on $M$ is $C^\infty$ on $A$(that is, for every point $x\in A$, there is a open set $V_x$ and a $C^\infty$ function $f_x$ such that $f_x=f$ on $A \cap {V_x}$), then is there a $C^\infty$ function $f_1$ on $M$ such that $f_1=f$ on $A$?

Choose a $\mathbb{C}^{\infty}$ bump function $\rho$ which is supported in $A\cap V_x$ and identically $1$ in a nbd $V$ of $x$, Define $$f_1(x)=\{\rho(x)f(x) \text{for } x\in A\cap V_x$$ and $$0 \text{for x not in } A\cap V_x$$ As the product of two $\mathbb{C}^{\infty}$ function on $A\cap V_x$, $f_1$ is $\mathbb{C}^{\infty}$ on $A\cap V_x$. If $x$ is not in $A\cap V_x$, then $x$ is not in $supp\rho$, and so there is an open set containing $x$ on which $f_1$ is $0$ since $supp\rho$ is closed. So $f_1$ is also $\mathbb{C}^{\infty}$ at every point not in $A\cap V_x$ Finally since $\rho\equiv 1$ on $V$ the function $f_1$ agrees with $f$ on $V$


Proportional to 2 Separate Variables vs. Proportional to Product of 2 Variables I've always seen the following in physics and math textbooks but never understood the process by which it was mathematically deducted:

$A \propto B$ $\space$ and $\space$ $A \propto C \space\space\space \rightarrow \space\space\space A \propto BC$

Could someone walk me through how this is done? This has been bothering me for a while now :P

Thanks

Update: Here's something I found that explains how this works. (Page 387; "Proof" section). Still, this proof takes the two statements one after the other. The author uses $x \propto y$ when $z$ is constant, and then takes care of $x \propto z$ when $y$ is constant, where it left off from the first (going from $x$ to $x'$ and then $x_1$). Is this the only way it can be done?

Putting my comments into an answer: If we say $A \propto B$ when $A$ also depends on other things, what we mean is that holding everything else fixed, $A$ increases linearly with $B$. So $A = mB$ for some $m$, and $m$ is constant relative to $B$, but may vary depending on the other things.

This is getting a bit wooly, so let's be more explicit. Let's say $A$ is a function of $B$, $C$, and $D$. Then $A \propto B$ means $A(B,C,D) = f(C,D)\cdot B$ for some function $f$. On the other hand, if $A \propto C$, then $A(B,C,D) = g(B,D)\cdot C$ for some other $g$. When you can put those together and go through some algebra, you'll find that $A(B,C,D) = h(D)\cdot BC$, that is, $A \propto BC$.

Assuming $A\propto B$ means $A=kB$ for some constant $k$, then we can also say that $A=mC$ for some constant $m \Rightarrow A=kmBC \Rightarrow A\propto BC$.


Find maximal possible determinant value given constraint Task is to find maximal possible determinant value for 2x2 and 3x3 matrices given following constraint:
$$\sum_{i,j=1}^na_{ij}^2 \le 1$$
I was able to come up with solution, but I received the test result, where I scored zero for this task. I still hasn't got a chance to take a look at my examined paper, but for now I just want to find whether or not my solution was actually true.
Solution
First trivial idea is that optimum would be reached in the point of exact equality $\sum_{i,j=1}^na_{ij}^2 = 1$.
Then, terms which participate in det calculation with minus sign, must be negative or zero.
$$\left( \begin{matrix}  \frac 1 {\sqrt 2} & 0 \\
0 & \frac 1 {\sqrt 2} \end{matrix}\right)$$

$$\left( \begin{matrix} \frac 1 {\sqrt 3} & 0 & 0 \\
0 & \frac 1 {\sqrt 3} & 0 \\
0 & 0 & \frac 1 {\sqrt 3} \end{matrix}\right)$$
So, I came up with these matrices and respective det values are $\frac 1 2$ and $\frac 1 {3\sqrt 3}$.

Is that correct?
Thanks!

Here is a geometric argument that I would give:

It is always possible to decompose any matrix into a product of an orthogonal matrix and an upper triangular matrix. Orthogonal transformation is isometry, so $\sum\limits_{i,j=1}^{n}a_{ij}^{2}$ is preserved on the upper triangular matrix. Orthogonal matrix have determinant 1 or -1, so no effects on absolute value of determinant. Hence we need to look only at determinant of upper triangular matrix with the same constraint, but determinant there are only affected by the diagonal.

It's not obvious to me that it would be impossible to make the determinant larger in some way by having larger terms subtracted, if this caused larger terms to be added at the same time. Can you clarify why you think the terms participating with a minus sign would need to be zero or negative?

Edit:

Let the columns be denoted by $u, v, w$.

I agree about the maximum being attained on the surface where $|u|^2 + |v|^2 + |w|^2 = 1$.

The differential of $\det(u,v,w)$ is $(h,k,l) \mapsto \det(h,v,w) + \det(u,k,w) + \det(u,v,l) = (v \times w) \cdot h + (w \times u) \cdot k + (u \times v) \cdot l$.

The differential of $|u|^2 + |v|^2 + |w|^2$ is $(h,k,l) \mapsto 2u \cdot h + 2v \cdot k + 2w \cdot l$.

The Lagrange multiplier method implies that at a point that optimizes the determinant, these linear functionals need to be proportional. So the triple of vectors $(u,v,w)$ needs to be proportional to $(v \times w, w \times u, u\times v)$. This proves that $u$, $v$ and $w$ all need to be orthogonal, and after an (orthogonal) change of basis, you can assume your matrix is diagonal. From there I reach the same result as you.


Why do counits go that way? Imagine you want to motivate for an audience the definition of an adjunction in terms of unit and counit. So you can say: Often two functors
$\mathcal{C} \begin{array}{c} \stackrel{\large F}{\rightarrow} \\ \stackrel{\leftarrow}{G}\end{array} \mathcal{D}$ are not really inverse, up to isomorphism, but still, the compositions $F \circ G$ and $G \circ F$ are somehow "connected" to the identity functors. So we have natural transformations
$$\eta : \mathrm{id}_{\mathcal{C}} \to G \circ F,$$
$$\varepsilon : F \circ G \to \mathrm{id}_{\mathcal{D}},$$
which should be compatible in a suitable way. Then someone says:

"Wait, why not taking $\varepsilon : \mathrm{id}_{\mathcal{D}} \to F \circ G$ instead? What's wrong with that?"

What would you answer?

Just to be clear: I understand the concept of an adjunction and know that $\varepsilon$ has to go from $F \circ G$ to $\mathrm{id}_{\mathcal{D}}$, but I am not sure how to explain this  to a novice without going into the details of the notion of an adjunction, for example the triangle identities or the alternative hom-set-definition, which by the way I would like to exclude from the discussion here. A good answer should be intuitive.

Notice that it is not true that we cannot state any compatibility conditions between two natural transformations $\eta : \mathrm{id}_{\mathcal{C}} \to G \circ F$ and $\varepsilon : \mathrm{id}_{\mathcal{D}} \to F \circ G$. An obvious candidate would be
$$F \circ \eta = \varepsilon \circ  F : F \to F \circ G \circ F$$
and/or
$$\eta \circ  G = G \circ  \varepsilon : G \to G \circ F \circ G.$$
By the way, does this concept have a name?

The unit-counit definition of an adjunction makes sense in any 2-category, and so it makes sense in any monoidal category, where it recovers the definition of a dualizable object. So you can motivate this somewhat more general notion using the familiar special case of vector spaces: when a vector space $V$ has a dual $V^{\ast}$, there is a unit map $1 \to V^{\ast} \otimes V$, namely the coevaluation, and also a counit map $V \otimes V^{\ast} \to 1$, namely the evaluation.

Of course it's hard to distinguish between the two possible orders of the tensor product here since $\otimes$ is symmetric monoidal. At this point I would draw the triangle identities using string diagrams, which justifies using the orderings above, although if you're going to both avoid that and also the homset definition of an adjunction I don't know what to tell you.

Firstly, as tcamp also commented, there also exists a similar construction as adjunctions in any bicategory where the 'unit' and 'counit' go in the same direction, and these are the (dual) Morita contexts, and it is also worth to study, having already wide literature, at least in its origin world of rings.
(Note that adjoint equivalences are immediately Morita [and dual Morita] contexts by flipping the [co-]unit.)

Secondly, consider the two (pseudo) embeddings of bicategories ${\bf Cat}\to{\bf Prof}$ sending $F\mapsto F_*$ and $F\mapsto F^*$, where $F_*=(a,b)\mapsto\hom_B(Fa,b)$ and $F^*=(b,a)\mapsto\hom_B(b,Fa)$ for a functor $F:A\to B$.
Observe that both of these are contravariant in exactly one dimension ($F\mapsto F^*$ on arrows and $F\mapsto F_*$ on 2-cells).

Moreover, we can roll these two maps in one embedding of double categories ${\bf Q}({\bf Cat})\to{\bf Q}({\bf Prof})$ which is contravariant in exactly one direction, where ${\bf Q}(B)$ means the double category of Ehresmann quintets of the bicategory $B$.
Consequently, orthogonal adjoint pairs in ${\bf Q}({\bf Cat})\ $ (i.e. adjoint functors with counit-unit-definition) correspond to companion pairs in ${\bf Q}({\bf Prof})\ $ (which yields the "$L_*\simeq R^*$" definition).


What are these problems called, what progress has been made on them? $x^a + y^b = z^c$ and $ax^n + by^n = cz^n$ I was thinking about some generalizations of Fermat's Last Theorem, and I'm sure they have been studied before. The first one is looking for rational solutions of $$x^a + y^b = z^c$$ and trying to find a necessary and sufficient condition on the tuple $(a,b,c)$ such that solutions exist, whether finitely many or infinitely many, etc.

The other is the same question for the equation $$ax^n + by^n = cz^n$$
To what degree have these equations (and their natural synthesis) been studied? What progress has been made?

Edit: I have done some further research and learned that the $n=2$ case of the second problem I described is solved, it is Legendre's theorem on the ternary quadratic form. $ax^2 + by^2 =cz^2$ has nonntrivial solutions in the rationals iff $\left( \frac { - b c } { a } \right) = \left( \frac { - a c } { b } \right) = \left( \frac { a b } { c } \right) = 1$. More information here.

Look up the Beal Conjecture for the first equation. Solutions exist whereby x, y and z share a common factor. It is easy to generate one of an infinite number of solutions with this characteristic. Example:

From $2^3 + 3^3 = 35$

We can obtain $70^3 + 105^3 = 35^4$ by multiplying both sides by $35^3$

All terms share a common factor of $35$. The Beal Conjecture posits that no solutions exist without a common factor and to date no counter-example or proof of the conjecture exist.

Regarding the below mentioned equation:

$ax^n + by^n = cz^n$  ------(A)

The link below has an article about it:

http://celebrating-mathematics.com

Click on article titled "$ap^6+bq^6=cr^6$ ".

The article covers equation (A) for degree $n=6$ and $n=(2,3,4, 5)$

And regarding $(x^a + y^b = z^c)$, just like "Phil H" pointed out,

there is also an equation (on the internet) with common factor's

in (x,y,z) & is shown below:

(a^11*b^7*c^3)^2+(a^7*b^5*c^2)^3=(a^3*b^2*c)^7

Where,  $(a+b=c)$


Induction Proof of Algorithm [Greedy Graph Coloring] Having a $G = (V,E)$ with each vertex having a range $[a,b]$.
Every two vertices are connected only if $[a_1,b_1]$,$[a_2,b_2]$ have a common subrange.
Each range of vertex is:

$rV1 = [0,5]$
$rV2 = [1,3]$
$rV3 = [2,10]$
$rV4 = [4,9]$
$rV5 = [6,7]$
$rV6 = [8,12]$
$rV7 = [11,13]$

Graph based on above ranges.
The Algorithm:
ranges = rV1,rV2,rV3,rV4,rV5,rV6,rV7...

COLOR_INTERVAL_GRAPH(rV1,rV2,rV3,rV4,rV5,rV6,rV7...){
if(number of rVi > 0){
 C_m  = MAXIMAL_COLOR_CLASS(rV1,rV2,rV3,rV4,rV5,rV6,rV7...);
 //paint C_m vertices with color m.
 //new_ranges <- remove C_m from rV1,rV2,rV3,rV4,rV5,rV6,rV7...
 return {C_m} U COLOR_INTERVAL_GRAPH(new_ranges)
 else:
   return []

MAXIMAL_COLOR_CLASS(new_range){
C = []
i = 1
while(i <= new_range.size()){
  C = C U {Vi}
  j = i+1
  while(j <= new_range.size() AND rVi (not common subrange with rVj)){
  j = j+1
i=j
return C

How to prove using induction that the algorithm uses the fewest possible colors.
After searching a bit i found that the MAXIMAL_COLOR_CLASS function in line 4 extends the C set.
I have to prove that the optimum coloring of any graph (of this type) can be transformed in order the first chromatic class is the same as the output of MAXIMAL_COLOR_CLASS.
Then using the above (by induction) i can show that every optimum coloring is the same as the output of the MAXIMAL_COLOR_CLASS proving the algorithm correct.
Not sure how to use induction to prove that.

Induction proof proceeds as follows:
Is the graph simple? Yes, because of the way the problem was defined, a range will not have an edge to itself (this rules out one of the easiest ways to prove that a graph is not n-colorable). Does it hold for 0 or 1 vertex? Yes, trivially. How about two vertices (with one edge, because disconnected vertices do not affect chromatic number)? You should be able to work out the justification for this step, and it is arguably the first non-trivial case and should be used as your base case.
Suppose you know that it holds for n vertices. Those vertices correspond to particular ranges. You could choose to prove that it then holds for n+1 vertices by adding another range (and an unknown number of edges) but this seems hard and potentially complicated. Instead, suppose it holds true for n edges. There are two ways that we could add an edge: by extending the endpoint of a nearby range until they intersect or by adding a new range entirely. Consider the output of the algorithm in either scenario, and you will have your proof.
In general, the more of your work you post, the better help you will be able to receive. This site is not intended to solve homework problems without some demonstrated effort and understanding or lack thereof.


Bayes estimate of beta distribution I'm trying to derive Bayes estimate of a beta distribution.

We toss n coins i.i.d., and get X heads. Give the prior distribution for $\theta$ of the coin is beta(a,b), Bayes estimate maximize the poster distribution, which is also a beta distribution beta(a+X, b+n-X)

Basically we're maximizing:

$\int_0^1 \theta^{a+X-1}(1-\theta)^{b+n-X-1}\theta d\theta$

And the result is: $\theta=\frac{a+X}{b+n-X}$. How to derive this from the integral?

As you said, the posterior distribution is the beta distribution $(a+X, b+n-X)$. The mean of the beta distribution $(\alpha,\beta)$ is $\alpha/(\alpha+\beta)$ hence the posterior mean is $\hat\theta=(a+X)/(a+b+n)$.

Note that $0\leqslant X\leqslant n$ with full probability hence $0\lt\hat\theta\lt1$ with full probability. (By contrast the suggestion $\hat\theta=(a+X)/(b+n-X)$ in your post is absurd since if $X$ is large, $\hat\theta$ can be as large as $(a+n)/b$, which can be larger than $1$.)

The bayes estimate is simply the mean of beta distribution $E[\theta]$, which should be $\theta=\frac{a+X}{b+n-X}$.


In a sequence of positive terms, $\lim \frac{a_{n+1}}{a_n} = L \Rightarrow \lim (a_n)^{\frac{1}{n}} = L$. Here's the question:

  Let $(a_n)$ be a sequence of positive reals. Assume that $\lim
\frac{a_{n+1}}{a_n} = L$. Then show that $\lim (a_n)^{\frac{1}{n}} =
L$.

I tried everything including using the definition of convergence to prove it but nothing seems to work. There's no hint in the textbook as well. I'd appreciate if someone drops me some hint. I have been stuck in this problem for almost an hour!

try this: let $r_1=a_1$ and $r_n= \frac{a_{n}}{a_{n-1}}$ for $n>1$. Then the condition says $\lim_{n\to\infty} r_n = L$. Next express $a_n = r_1...r_n$. This should work well.


Method of characteristics for Burgers' equation with rectangular data Using the method of characteristics, find a solution to Burgers' equation
  \begin{cases}
u_t+\left(\frac{u^2}{2} \right)_x =0 & \text{in }\mathbb{R}\times(0,\infty) \\
\qquad \qquad \, \, u=g & \text{on } \mathbb{R} \times\{t=0\}
\end{cases}
  with the initial conditions
  $$g(x)=\begin{cases}
0 & \text{if }x < 0 \\
1 & \text{if }0 \le x \le 1 \\
0 & \text{if }x > 1
\end{cases}$$

First, I realized that the equation $u_t+\left(\frac{u^2}{2} \right)_x =0$ is equivalent to this form: $$u_t+uu_x =0$$

Then should I generally follow the method of solution as outlined in the answer of this page?

Note that this is not a duplicate question of that page. Rather I want to know if that page can be used for my problem, even though my ICs are different.

By the way, the solution printed in my book (PDE Evans, 2nd edition, page 142) is

  $$u(x,t) =
\begin{cases} 0 & \text{if } x < 0 \\
\frac xt & \text{if } 0 < x < t
\\ 1 & \text{if } t < x < 1 + \frac t2
\\ 0 & \text{if } x > 1 + \frac t2 \tag{$0 \le t \le 2$}
\end{cases}$$

The variable $u$ is constant along the characteristic curves, which satisfy
\begin{aligned}
x'(t) & = u(x(t),t) \, ,
\\
& = u(x(0),0) \, .
\end{aligned}
Thus, the latter are straight lines in the $x$-$t$ plane, determined by the initial data. Here, the initial data is piecewise constant, i.e. we solve Riemann problems. As displayed in the figure below,

characteristics separate in the vicinity of $x=0$, and a rarefaction wave occurs;
characteristics cross in the vicinity of $x=1$, and a shock-wave occurs.

The rarefaction wave is a continuous self-similar solution, deduced from the self-similarity Ansatz $u (x,t)=v (\xi)$ with $\xi = x/t$. Indeed,
$$
\partial_t
 u(x,t) + u(x,t)\, \partial_x u(x,t) = \left(v(\xi) - \xi\right) \frac{v'(\xi)}{t} \, ,
$$
and thus, $v(\xi) = \xi$ or $u(x,t) = x/t$.
The shock speed $s$ is given by the Rankine-Hugoniot jump condition:
$$
s = (1+0)/2\, .
$$
As long as the rarefaction and the shock don't interact, the solution is
therefore
$$
u(x,t) =
\left\lbrace
\begin{aligned}
& 0 &&\text{if }\; x\leq 0 \, , \\
& x/t &&\text{if }\; 0\leq x\leq t \, , \\
& 1 &&\text{if }\; t \leq x < 1+ t/2 \, ,\\
& 0 &&\text{if }\; 1+t/2 < x \, ,
\end{aligned}
\right.
$$
valid for times $t<t^*$ such that $t^* = 1 + t^*/2 = 2$. At the time $t^*$, both waves interact. The new shock speed is determined from the Rankine-Hugoniot condition
$$
x'(t) = (x(t)/t+0)/2 \, ,
$$
with initial shock speed $x'(t^*) = s$.
Hence, the solution for $t\geq t^*$ is
$$
u(x,t) =
\left\lbrace
\begin{aligned}
& 0 &&\text{if }\; x\leq 0 \, , \\
& x/t &&\text{if }\; 0\leq x< \sqrt{2t} \, , \\
& 0 &&\text{if }\; \sqrt{2t} < x \, .
\end{aligned}\right.
$$

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 $u=u_0$

$\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0s+f(u_0)=ut+f(u)$ , i.e. $u=F(x-ut)$

$u(x,0)=\begin{cases}0&\text{if}~x<0\\1&\text{if}~0\le x\le1\\0&\text{if}~x>1\end{cases}$ :

$\therefore u=\begin{cases}0&\text{if}~x-ut<0\\1&\text{if}~0\le x-ut\le1\\0&\text{if}~x-ut>1\end{cases}=\begin{cases}0&\text{if}~x<0\\1&\text{if}~0\le x-t\le1\\0&\text{if}~x>1\end{cases}$

Hence $u(x,t)=\begin{cases}0&\text{if}~x<0\\1&\text{if}~0\le x-t\le1\\0&\text{if}~x>1\\c&\text{otherwise}\end{cases}$


Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.

This is what I did

$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$

Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.

Is it correct? Is there another way to solve?

using the lemma $$t+\frac{1}{t}\geq 2$$ for $t>0$  we get the inequality $$\tan(x)+\frac{1}{\tan(x)}\geq 2$$
your way is also ok.

Even if your implications are correct, your logic is flawed.

Logically, what you have done is similar to

$$\begin{align}
+1&=-1
\\ \Rightarrow (+1)^2&=(-1)^2
\\ \Rightarrow 1&=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.

Deriving a true statement says nothing about the truth of the hypothesis.


Number of elements in a finite $\sigma$-algebra I have been asked to prove that the number of elements in a finite sigma algebra over a set $X$ is $2^n$ for some integer $n$. How do I go about this problem? I have no idea where to start. Thanks in advance for any ideas.

Do I need to prove that given a set $F$, $\sigma(F)$ is actually a power set of some set say $S$?

Let $\Sigma$ be the $\sigma$-algebra. Choose $x \in X$, and define $M_x = \cap_{M \in \Sigma, x \in M} M$. Clearly $M_x \neq \emptyset$, and $M_x \in \Sigma$.

Furthermore, the collection $F = \{M_x \} \subset \Sigma$ is a partition of $X$ (and finite, of course). To see this, suppose $M_x \cap M_y \neq \emptyset$. Then we must have $M_x = M_y$, or else either $M_x \setminus M_y $ or $M_y \setminus M_x $ would be strictly smaller sets contradicting the definition of either $M_x$ or $M_y$.

Furthermore, it is clear that if $M \in \Sigma$, then $M = \cup_{x \in M} M_x$, hence every element of $\Sigma$ is the (disjoint) union of members of $F$ (the empty set taken as the union of no members of $F$), hence $|\Sigma| = 2^{|F|}$.

I think all the above answers are wrong, since, in adittion to all the unions and intersections, you have to count all the possible differences and symmetric differences as well as their complements. For instance, for n=2, you will have 16 sets instead of 4. I think it should be 2^{2^n} instead of 2^n.


$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$ Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.

OK, since the hint seems not to be enough, here is the solution.

First,
Setting $g(x)=f(x)-f(\frac{x}{2})$ we have
$$g(x)-g(\frac{x}{2})=x^2 \\
f(x)-f(\frac{x}{2})=g(x) \tag{*}$$

We solve the first equation.

By induction
$$g(x)-g(\frac{x}{2^{n}})=x^2+\frac{x^2}{4}+\frac{x^2}{16}+...+\frac{x^2}{4^{n-1}}=x^2 \frac{1-\frac{1}{4^n}}{1-\frac{1}{4}}$$

Since $f$ is continuous at $x=0$, so is $g$. Taking the limit by $n$ you get
$$g(x)-g(0)=\frac{4}{3} x^2$$

Moreover, $f(x)-f(\frac{x}{2})=g(x)$ implies $g(0)=0$.

Now, we need to solve
$$f(x)-f(\frac{x}{2})=\frac{4}{3}x^2$$
which is exactly the equation above, multiplies by $\frac{4}{3}$. Therefore, solving exactly as above we get
$$f(x)-f(0)=\frac{4}{3}\frac{4}{3}x^2=\frac{16}{9}x^2$$

This shows that
$$f(x)=\frac{16}{9}x^2 +c$$
where $c=f(0)$
are all the solutions.

P.S. To make this more clear, the equation can be reduced via (*) to two equations of the type
$$h(x)-h(\frac{x}{2})=r(x)$$
with $h, r$ continuous. This equation can be solved as above: By induction
$$h(x)-h(\frac{x}{2^n})=\sum_{k=0}^{n-1}r(\frac{x}{2^n})$$
Using the continuity of $h$ at $x=0$ (we don't even need $h$ to be continuous at other points) we get
$$h(x)-h(0)=\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
(Note here that, if the series is divergent it implies that there is no solution which is defined at that $x$).

So, by calculating the series, you get the solution
$$h(x)=c+\sum_{k=0}^{\infty}r(\frac{x}{2^n})$$
where $c=h(0)$.

In this case, since the equation is linear and given the "clue" that the coefficient of $x^2$ is non-zero, then consider $f(x) = a + b x + c x^2$. This yields the possible polynomial solutions, but not necessarily all possible solutions.
\begin{align}
x^2 &= (a + b x + c x^2) - 2 \, \left( a + \frac{b x}{2} + \frac{c x^2}{4} \right) +  \left( a + \frac{b x}{4} + \frac{c x^2}{16} \right) \\
&= \frac{b x}{4} + \frac{ 9 \, c \, x^2}{16}.
\end{align}
This leads to $b=0$ and $c = 16/9$ and the general solution of the form
$$f(x) = a + \frac{16 \, x^2}{9}.$$


Laplace transform of $1/t$ Does the laplace transform of $1/t$ exist? If yes, how do we calculate it? Putting it in $$\int_0^\infty (e^{-st}/t) dt$$ won't solve.

Is there any other way? If not, why?

Thanks!

No, it doesn't exist. In general the Laplace transform of $t^n$ is $\frac{\Gamma(n+1)}{s^{n+1}}$, and $\Gamma(n)$ isn't defined on $0,-1,-2,-3...$ This integral is the definition of the Laplace transform, so the transform doesn't exist if the integral doesn't. While there are other integral transforms that could transform $\frac{1}{t}$ in a useful way, anything other than what you gave wouldn't be considered a Laplace transform anymore.

$L\left[\frac{f(t)}{t}\right](s)=\int_s^{+\infty}f(u)du$ so $L\left[\frac{1}{t}\right](s)=\int_s^{+\infty}\frac{du}{u}=\ln u|_s^{+\infty}=\ln(s)-\ln(+\infty)$ as $\ln(+\infty)$ is not defined, then $L\left[\frac{1}{t}\right]$ is also not defined but it can be solved when $L\left[\frac{1}{t}-\frac{\cos t}{t}\right]$ or more .


I need to show the validity of the below arguments by using a truth table I need to show the validity of

$P \rightarrow Q$

$P \rightarrow R$

$\therefore  P \rightarrow (R \wedge Q)$

Can i just show the truth table for $P \rightarrow Q$ and the truth table for $P\rightarrow R$ then the truth table for $P \rightarrow (R \wedge Q)$
I think the truth table for $P \rightarrow (R \wedge Q)$ is:
T
T
F
T
F
T

Use three columns, on the left most part of the truth-table, to keep track of the truth value assignments for $p, r, q$, one column per variable. So you need $2^3 = 8$ rows in your truth table, below the headings. Instead of creating three truth tables, create one truth-table, using a column for each premise, $p \rightarrow q$ and $p\rightarrow r$, and a final column for the conclusion: $p \rightarrow (q \land r)$.

For your argument to be valid, it must be the case that whenever all the premises are true, the conclusion must be true. So your truth table needs to show that whenever the premises $(p \rightarrow q)$ and $(p \rightarrow r)$ are both true, then so is be the conclusion $p \rightarrow (r \wedge q)$ true.

The following truth-table was generated by Wolfram Alpha. Be sure to check whether the condition of validity is met.

$p \rightarrow q \equiv \neg p \vee q$
$p \rightarrow r \equiv \neg p \vee r$
$(p \rightarrow q ) \wedge ( p \rightarrow r ) \equiv ( \neg p \vee q ) \wedge ( \neg p \vee r)$  [reverse distribution property]
$(\neg p \vee (q \wedge r))$
$p \rightarrow  (q \wedge r)$


Proof: The identity matrix is invertible and the inverse of the identity is the identity How can i show that: $II^{-1} = I = I^{-1}I$ (the identity matrix is invertible) for all cases. And then proof that: $I^{-1} = I$ (The inverse of the identity is the identity). I don't know how start both proof, any suggestion?

Suppose $A$ is the inverse of the identity matrix. Then $AI =IA =  I$. But $AI = IA = A$ as well so $A=I$.

AS INVESE OF ANY MATRIX IS GIVEN BY A FORMULA; A-1=[1/|A|]X[ADJ OF A]
BUT HERE WE PROOF BY GENERAL METHOD.
                                      LET  K IS INVERSE OF IDENTITY MATRIX I THEN
                         WE KHOW THAT AS, KI=IK=I
                                     ALSO,KI=IK=K
                                       SO,I=K  OR  [I=I-1]

SO INVERSE OF IDENTITY MATRIX IS IDENTITY MATRIX.

                                ** THANKS**


Can we say that an element is not a subset of a set? I am creating a multiple choices question for students that should be as follows:

Which of the following is true?

1) $8\in \mathbb{Z^{+}}$

2) $8\notin \mathbb{Z^{+}}$

3) $8\subset \mathbb{Z^{+}}$

4) $8 \not\subset \mathbb{Z^{+}}$

Should choices Number (1) and (4) be considered as valid answers?
My understanding is that we can't deal with an element with either $\subset$ or $\not\subset$.

Please give me references supporting your answer if there.

Yes, an element of a set can be a subset of that set.

Think of $\varnothing\in\{\varnothing\}$ and $\varnothing\subset\{\varnothing\}$, where $\varnothing$ denotes the empty set.

The answer to your multiple choice question depends on how $\mathbb Z_+$ and $8$ are defined.

A possible construction exists with $\mathbb Z_+=\{1,2,3,\dots,8,\dots\}$ and $8=\{1,2,3,4,5,6,7\}$.

In that case $8\in\mathbb Z_+$ and $8\subset\mathbb Z_+$.

For me the option 1 is valid, for options 3 and 4 proper notation of set isn't used. If we require to represent a set with elements then the elements should be surrounded by curly braces . Hence 8 doesn't represents a set therefore 3 and 4 options are invalid.

This wikipedia post suggests the notation used for set elements


Number theory question from today's Pre - RMO How many positive integers less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3?

By my comment above we are looking for three digits such that they sum to exactly $21$. This is possible as follows:

\begin{eqnarray}
9+9+3&=&21\\
9+8+4&=&21\\
9+7+5&=&21\\
\vdots&=& \vdots\\
9+3+9&=&21\\
8+9+4&=&21\\
8+8+5&=&21\\
\vdots&=&\vdots\\
8+4+9&=&21\\
7+9+5&=&21\\
\vdots&=&\vdots\\
7+5+9&=& 21\\
\vdots&=&\vdots\\
\end{eqnarray}

Can you spot the pattern in the list? It's not too difficult to count the number of such solutions without writing all of them down. There are $28$ solutions.

Answer is 21 because the first number which is multiple of both 7 and 3 is 21


Column Space vs Span and minimum size of a set of vectors to guarantee a vector is in span? Given a set A of vectors, is there a minimum number of vectors in A to guarantee that a particular vector b is in span(A)?

If we have a m x n matrix A - what is the difference between the column space and the span? Everywhere I read makes it seem like they both equal the same thing.

What I understand:

Col(A) is the span of the columns of A - thus Col(A) must have n vectors in it, since A is of size m x n. What differentiates this from the span(A)?

Also:

Given a set A of vectors, is there a minimum number of vectors in A to guarantee that a particular vector b is in span(A)?

Shouldn't A be of R$^n$ once again since there are n columns?

The notation $\operatorname{span}(X)$ should be used for a set $X\subseteq V$ that is contained in some vector space $V$. It is then defined as the intersection of all subspaces of $V$ containing $X$ or equivalently as the set of all linear combinations of elements from $X$.
Writing "$\operatorname{span}(A)$" when $A$ is a matrix, doesn't make any sense in terms of this definition. However, it might be an abuse of notation to denote the span of the set of columns of $A$, that is, the column space of $A$. This space is usually denoted $\operatorname{col}(A)$ though.
The question about a "minimum number of vectors in a set $X$ to guarantee $b\in\operatorname{span}(X)$" is unclear to me. For example in the plane $V=\mathbb R^2$ you can take the infinite set
$$
X = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 0 \end{bmatrix}, \dots \right\}
$$
with span
$$
\operatorname{span}(X) = \left\{ \begin{bmatrix} x \\ 0 \end{bmatrix} \,\middle|\, x\in\mathbb R\,\right\}
$$
the does not contain $b=\begin{bmatrix}0 \\ 1\end{bmatrix}$ despite $X$ having infinitely many elements.

For example

Let A=
\begin{bmatrix}
1 &0 &0&0\\
   0& 1 &0&1\\
   0 &0& 1&1\\
\end{bmatrix}.

Here $Col(A)= \mathbb{R}^3$ and observe that last column in A is sum of 2nd and 3rd columns.

Note that $Span(A)=Col(A)=Row(A)$


Show there exists a unique monic polynomial $f(x)$ in $F[x]$ of minimum degree such that $f(a) = 0$ Let $F$ be a field, $E$ an extension field of $F$, and a in $E$ an algebraic element. Show there exists a unique monic polynomial $f(x)$ in $F[x]$ of minimum degree such that $f(a) = 0$

All I have so far is that since a is algebraic over $E$, there exists some $p(x)$ in E such that $p(a) = 0$.

I'm really struggling with these ideas in general (fields, extension fields, and algebraic extensions).

Hint: Suppose that $p(x)$ and $q(x)$ are two monic polynomials of minimal degree such that $p(a) = q(a) = 0$.  What can we say about the polynomial $p(x) - q(x)$?

To summarize the comments below: $a$ is meant to be an element which is algebraic over $F$, rather than over $E$.

so $a \in E$ is algebraic over $F$ means there exists a monic minimal polynomial say $f(x) \in F[x]$ such that $f(a)=0$ but it is given $g(a)=0$ implies $f(x)$ divides $g(x)$ using division algorithm and minimality of $f$ having $a$ as a zero.Similarly g(x) divides f(x) , hence we get g(x)=h_1(x)*f(x)=h_1(x)*h_2(x)*g(x) implies h_1(x)*h_2(x)=1 but F[x] has only units same as F so h_1(x) and h_2(x) must be in F but f and g are minimal so h_1 and h_2 must be 1.


Find the real roots of $\sqrt{x-2}(x^2-4x+3)=0 $ I thought the roots of this equation are $1,2,3$.

But the teacher says 1 is not a real root because if we take $x=1$ we get
$$\sqrt{-1}(0)$$

and $\sqrt{-1}$ is not real.
Is this right?

While it is true $x=1$ is a root, the question is: what is the domain in which the equation "exists"?
The answer is that the square root only exists if you consider the inequality  $x-2\ge 0$.  Otherwise your solution does not compile, so to speak.
So the roots coexisting within your domain are $2$ and $3$.

We have indeed 3 roots: 1,2 and 3. But now we are looking for the real roots. And yes 1 is not a real root, because like you said $\sqrt -1$ is not real, it is imaginary.


Calculating the tangent in line given an implicitly given curve. Given the implicitly given curve $x^4y^2-2\frac{y}{x}=ln(y-1)$ and the point P = (-1, 0).

Calculate the tangent in P to the curve.

I've tried to calculate the partial derivatives for x and y, and then using the $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ to get the tangent line, but what's $z$?

Or am I trying to solve this in the wrong way?

The answer should be $x + y + 1 = 0$

Here's the original problem (in Dutch).


Are there random variables $X,Y$ such that $X,Y,X+Y \sim N(0,1)$? Let's suppose we have two, not necessarily independent, random variables $X,Y \sim N(0,1)$.
Then, is it possible that $X+Y \sim N(0,1)$?
What I have managed to show is that because $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ it would be necessary that $Cov(X,Y) = -1/2$.
But I'm still looking for an example or else a proof that it is not possible.

It is possible. Suppose
$$(X,Y)'\sim\mathcal{N}\bigg(\begin{pmatrix}0\\0\end{pmatrix},\begin{pmatrix}1&-\frac{1}{2}\\-\frac{1}{2}&1\end{pmatrix}\bigg)$$
And consider the vector $(1,1)$. Then $(1,1)\cdot (X,Y)'=X+Y$ is distributed Normal with mean
$$(1,1)\cdot\begin{pmatrix}0\\0\end{pmatrix}=0$$
and variance
$$(1,1)\cdot \begin{pmatrix}1&-\frac{1}{2}\\-\frac{1}{2}&1\end{pmatrix}\cdot\begin{pmatrix}1\\1\end{pmatrix}=1$$
Which is what you want.
Note: By a similar argument, using the vectors $(1,0)$ and $(0,1)$ you have that $X$ and $Y$ are standard normals on their own.
Note: One should be careful and make sure that the variance-covariance matrix is a valid one. One way of doing it is by directly checking it is possitive semi-definite, which is true in this case as the first principal minor is $1$ and the determinant is $5/4$, both strictly positive.
There is an easy second way to verify that a $2$ by $2$ matrix is a valid variance-covariance matrix. Call this matrix $\Sigma$. If you can verify that the diagonal entries are strictly positive, the matrix is symmetric and that there exists $\rho\in[-1,1]$, a valid correlation coefficient, such that $\Sigma_{21}=\Sigma_{12}=\rho\cdot(\Sigma_{11}\cdot\Sigma_{22})^{1/2}$, you are done. In this case $\rho=-\frac{1}{2}$.

The pdf of the sum of two absolutely continuous random variables is simply given by the convolution of the pdf's of the variables. In our case, $f_X(x) = f_Y(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, so $f_{X + Y}(x) = \int_{- \infty}^\infty f_X(t)f_Y(x - t) dt = \frac{1}{4} \int_{- \infty}^\infty \exp(- \frac{t^2 + (x - t)^2}{2}) dt = \frac{1}{2 \pi} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2})^2 - 2 t \sqrt{2} \frac{x}{\sqrt{2}} + \frac{x^2}{2} + \frac{ x^2}{2}}{2}) dt = \frac{1}{2 \pi} e^{-\frac{ x^2}{4}} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2} - \frac{x}{\sqrt{2}})^2}{2}) dt = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \int_{- \infty}^\infty e^{- \frac{u^2}{2}} du = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \sqrt{2 \pi} = \frac{1}{2 \sqrt{\pi}} e^{- \frac{x^2}{4}}.$ In particular, the sum of two normals of mean $0$ and variance $1$ is not a normal of mean $0$ and variance $1$. I hope this helps. :)


Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer? If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.

If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.

Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.

Hint $\ $ Since there is a unique denominator $\rm\:\color{#C00} {2^K}\:$ having maximal power of $2,\,$ upon multiplying all terms through by $\rm\:2^{K-1}$ one deduces the contradiction that $\rm\ 1/2\, =\, c/d \;$ with $\rm\: d \:$ odd, $ $ e.g.

$$\begin{eqnarray} & &\rm\ \ \ \  \color{green}{m}  &=&\ \  1 &+& \frac{1}{2} &+& \frac{1}{3} &+&\, \color{#C00}{\frac{1}{4}} &+& \frac{1}{5} &+& \frac{1}{6} &+& \frac{1}{7} \\
&\Rightarrow\ &\rm\ \ \color{green}{2m} &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &+&\, \color{#C00}{\frac{1}{2}} &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}\\
&\Rightarrow\ & -\color{#C00}{\frac{1}{2}}\ \ &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &-&\rm \color{green}{2m}  &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}
\end{eqnarray}$$

The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm\,d\, |\, 3\cdot 5\cdot 7$.

Note $\ $ I purposely avoided any use of valuation theory because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student.

if we consider highest prime upto $n$
then given sum can be written as $1/p + a/b$ where a is some integer
$b$ is also an integer not divisible by $p$.
so $b/p$ can not be an integer
and so $b/p + a$.
so the given sum can not an integer


Mathematicians ahead of their time? It is said that in every field there’s that person who was years ahead of their time. For instance, Paul Morphy (born 1837) is said to have retired from chess because he found no one to match his technique that very much resembled modern chess theory.
So, who was the Paul Morphy of mathematics?

My top vote would be for Ramanujan.  With severely limited resources, he was able to formulate deep number-theoretic identities that the top mathematicians in the field at the time hadn't the imagination to conceive, let alone the slightest clue to prove.  A close second would be Evariste Galois--dead at the age of 20, he had already established the foundations of what is now an entire algebraic theory named after him.  The world will never know what mathematics he could have discovered had he lived.

You can have people 'ahead of their time'.  What really happens is you have 'mainstream maths', and than a whole lot of eccentrics who are 'doing their thing', which may or may not be where maths will be in 20 or 30 years time.

There are several things one has to note here.  Some of us are just not really good at sitting down and writing theorms and proofs etc.  I have a degree in physics, which means you approach the world in a different way to what the mathematicians do.  None the same, you can do 'real maths', but do it without having to prove the tedia involved.

The sorts of things i needed for my studies in weights and measures, and number-bases (i use base 120 natively), etc, i had to go back something like 100 years and look at what the eccentrics were doing.

So, really, people really can be 20 or 30 years ahead of the mathematics, you just got to be in the right crowd of eccentrics.


show that the length of the curve $\alpha$ tends to infinity when $N \to \infty$ The following example shows that for any reasonable definition, the length of a curve $C ^0$ in a range defined closed may be unlimited. Let $ \alpha: [0,1] \to \mathbb {R}^2 $ given by
$$\alpha (t) =\left( t, \sin \frac {\pi} {t}\right) $$ if $ t\neq 0 $, and $\alpha (0) = (0.0) $. Shows geometrically that the arc length of the curve portion corresponding to
$ \frac {1} {n +1} \leq t \leq \frac {1} {n} $
is at least $\frac{2}{n + \frac{1}{2}}$. Use this to show that the curve length in the range $\frac {1} {N} \leq t \leq 1$ is greater than $ 2 \displaystyle \sum_{n = 1}^{N} \frac {1}{n +1} $, thus tending to infinity when $ N \to \infty $.

Do Carmo's exercise? But you miss a $t$, the parameter should be
$$
\alpha(t) = \left(t, t\sin\left(\frac{\pi}{t}\right)\right),
$$
when $t\neq 0$.


Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? I'm trying to evaluate $$
\lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2}
$$
over the domain $x>0$, $y>0$.

============
My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.

I tried to evaluate the limits "near" $(x,x)$ where, I thought, there's may be some problems:
$f(x, x-x^2)\to +\infty$.

Then I convinced myself the limit could be $+\infty$:
since $f(x,y)>0$ over the domain, I had to find such $g(x,y)$ that:
1. $f(x,y) \ge g(x,y)$
2. $ \lim_{(x,y)\to (0,0)} g(x,y)\to +\infty $

$$
f(x,y)=\frac{x-\sqrt{xy}}{x^2-y^2}=\frac{x-\sqrt{xy}+y-y}{x^2-y^2}=\frac{x-\sqrt{xy}+y}{x^2-y^2}-\frac{y}{x^2-y^2}=\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2}
$$
Where the last step follows by $(x-\sqrt{xy}+y) \ge 0$ with $x>0$, $y>0$. $$
\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2} = \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2}.
$$ From $\left[3\left(x-\sqrt{xy}+y\right)^2\right] \ge \left[x+xy+y^2\right]$, for every $(x,y)$ with $(x>y)$: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(x^2-y^2)}+\frac{y}{y^2-x^2}.
$$ From here I observated that $ \left[\lim_{(x,y)\to (0,0)} g(x,y)\to +\infty \right]$ and eventually $ \left[\lim_{(x,y)\to (0,0)} f(x,y)\to +\infty \right]$ for $(x>y)$.

I thought that for $(y>x)$, the inequality was formally equivalent when I replace $(x)$ with $(y)$ and viceversa: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(y^2-x^2)}+\frac{x}{x^2-y^2}.
$$
However I could see, through an online grapher, that it is false!!
So I remained without any chance to conclude the limit.

============
Is there anybody who knows why the last inequality isn't correct?
And also, has anybody some hints to evaluate the limit?

In order to disprove your conjecture that $\lim_{(x, y) \to (0, 0)} f(x, y) = +\infty$, take the limit along the curve $(x, y) = (t^5, t)$ as $t \to 0^+$.  Then we have:
$$f(t^5, t) = \frac{t^5 - t^3}{t^{10} - t^2} = \frac{t(1-t^2)}{1-t^8}$$
and from the last expression, we see that $f(t^5, t) \to 0$ as $t \to 0^+$.

If the limit exists, then it must be equal to the limit along any line, for example, $y=x/4$.

In that case,

$$ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} =\lim_{x\downarrow 0} \frac{8}{15}\frac{x}{x^2}=\frac{8}{15}\lim_{x\downarrow 0}\frac{1}{x}=\infty.$$


Dividing by 2 numbers at once, what is the answer? Let's say i have 4/1/5. or 4 divided by 1 divided by 5. Are there any rules that i am allowed to use to stop any mistakes?, for example this has 2 solutions, 4/5 , and 20.

Edit: Thanks for your responses. It seems the rule of thumb is to start from the beginning and follow through the numbers, while any brackets should be dealt with beforehand.

In which case something like 1/2/3/4 is in fact 1/24.

Edit2: There's a small bounty for whoever can prove that there solution is always the case, and any misuse of it always fails

In the USA and Canada, perhaps other places too, we use the BEDMAS system for interpreting expressions. First, we calculate the result of bracketed sub-expressions. Then exponentiation. Then from left to right, resolving divisions or multiplications as they show up. Then from left to right, resolving additions and subtractions as they show up.

So in the BEDMAS convention, we would deal with your 4/1/5 by doing the divisions left-to-right. 4/5.

Edit:

As the comments point out, some use PEDMAS instead. P for parenthesis instead of brackets ;)


What is the answer to the paradox of the infinitesimal? I just read this article on npr, which mentioned the following question:

  You can keep on dividing forever, so every line has an infinite amount
  of parts. But how long are those parts? If they're anything greater
  than zero, then the line would seem to be infinitely long. And if
  they're zero, well, then no matter how many parts there are, the
  length of the line would still be zero.

It further mentions that

  Today, mathematicians have found ways to answer that question so that
  modern calculus is rigorous and reliable.

Can anyone elaborate on the modern answers to this question?

The paradox has to do with the additivity with measure; in particular, naively switching from finite additivity to infinite additivity.

We're familiar with the fact that, if we split something up into two parts, the measure of the whole is the sum of the measure of the two parts.

If we repeat this with one of the individual parts, we've now split the original whole into three parts. The measure of the whole is the sum of the individual measures of the three parts.

And so forth; binary additivity of measure does extend to arbitrary, but finite additivity of measure.

Generally speaking, things have to change when you switch from finite to infinite. We can no longer justify additivity of measure when you have infinitely many parts, because you can never get to infinitely many parts by repeatedly splitting the whole into finitely many parts finitely many times.

A priori, there might not even be a reasonable notion of additivity of measure when you have infinitely many parts! However, experience has shown there is a useful extension of additivity to countably many parts, at least when studying a continuum.

i.e. if you split a whole into countably many parts (and in a measurable way), you can expect the measure of the whole to be the sum of the measures of the individual parts.

Note that "sum" must be meant in the sense of an infinite sum from calculus; e.g. as a limit of partial sums. Trying to literally interpret an infinite sum as repeated addition runs into all of the same problems we're trying to work around.

When you split the whole into more than countably many parts -- e.g. you split the number line into its individual points -- you now have too many parts for countable additivity of measure.

It states

  ...then the line would seem to be infinitely long.

It is infinitely long for a very very small person who walk exactly one real number with each step. For example if she/he is on the point 0.5, then she/he must find the next real point right after (or before) 0.5. Can she/he?

But for us, who use centimeter as a measuring unit, that line would not be infinitely long. So it basically depends on the unit of length that we consider.


Stuck on candy bowl fraction I am really stuck on this problem because I'm not even sure where to start.

Larissa has a bowl of candies. On the first day, she eats 1/2 of the candies plus one more. On the second day, she eats 1/3 of the remaining candies plus one more.  On the third day, she eats 1/5 of the remaining candies plus one more.  On the fourth day she eats the three remaining candies.  How many candies did she have at the start?

Hints

Let $N$ be the number of candies.

Day 1: she eats $\frac{1}{2}N + 1$: remainder $N_{1}$

Day 2: she eats $\frac{1}{3}N_{1} + 1$: remainder $N_{2}$

Day 3: she eats $\frac{1}{5}N_{2} + 1$: remainder $N_{3}$

Day 4: she eats $3$: $N_{3} = 3$

Can you go from here? (answer $=20$)

Step 1: English $\to$ algebra

Create a table which defined the key variables of $e_{k}$, the amount of candies eaten on day $k$, and $a_{k}$, the amount available at the end of day $k$.

Step 2: Solution criteria

We need to find $a_{0}$, the initial amount of candies, such that $a_{3}=3$.

Step 3: Solution strategy

We all agree that the solution involves expressing the intermediate variables $a_{k}$, and $e_{k}$, $k=1,2,3$ in terms of $a_{0}$.

The are a few ways to do this. We chose a top down approach.

Day 1

$$a_{1} = a_{0} - e_{1} = \tfrac{1}{2}a_{0}-1$$

Day 2

$$e_{2} = \tfrac{1}{3}a_{1} + 1 = \tfrac{1}{6}a_{0}+\tfrac{2}{3}$$
$$a_{2} = a_{1} - e_{2} = \tfrac{1}{3}a_{0}-\tfrac{5}{3}$$

Day 3

$$e_{3} = \tfrac{1}{5}a_{2} + 1 = \tfrac{1}{15}a_{0}+\tfrac{2}{3}$$
$$a_{3} = a_{2} - e_{3} = \tfrac{4}{15}a_{0}-\tfrac{7}{3}$$

Step 4: Solve

$$
a_{3} = 3
$$
implies
$$
\tfrac{4}{15}a_{0}-\tfrac{7}{3} = 3 \quad \implies \quad a_{0} = 20
$$

Hopefully, this is reflected in your class notes.

Check the solution

Let $x$ be the number of candies to start with ... then
\begin{eqnarray*}
\frac{1}{5} \left(\frac{1}{3} \left(\frac{x}{2} -1 \right) -1 \right) -1 =3.
\end{eqnarray*}
So ...


Proving the conjecture ∀x.x+0=x with some predefined axioms I saw this problem:

$$∀x.x+0=x $$

in a introduction to logic text book but I am unsure how to go about solving it.

The book provides the following axioms:

$A1: nat(0)$

$A2: nat(T) ⊢ nat(s(T))$

$A3: nat(T), nat(U) ⊢ nat(T+U)$

$A4: nat(T), nat(U) ⊢ nat(T · U)$

$A5: 1=s(0)$

$A6: ∀x. 0+x=x$

$A7: ∀x.∀y.s(x)+y=s(x+y)$

$A8: ∀x.0·x = 0$

$A9: ∀x.∀y.s(x)·y = y + x · y$

Here is my attempt so far, if it is totally on the wrong track feel free to start from scratch :)

    __________________
  1.| nat(a)          | assumption
  2.| nat(0)          | A1 nat(0)
  3.| ∀x1. 0+x1=x1    | A6 ∀x.0+x=x
  4.| 0+a=a           | ∀-elim 3,1
  5.| 0+a=0+a         | =-into
  6.| a=0+a           | =-elim-1 5,4
  7.| a=a             | =-intro
    | ...             |
  8.| a=a+0           |
  9.| a+0=a           | =-elim-1 7,8
     ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
 10.  ∀x.x+0=x          ∀-intro 1-7

Thank you guys especially @Brams28 I have managed to prove it using induction

Something is going wrong here. $\forall x.\ x + 0 = x$ is not provable in your system.

The given axioms are in effect a fragment of the standard weak "Robinson Arithmetic", so called Q (actually yours is the mirror image that you get by reversing the usual order of terms of an addition and of a multiplication, but let's not fuss about that).

Famously,not even full Q (in your mirror version), so with your axioms $\forall x.\ 0 + x = x$  etc., can prove $\forall x.\ x + 0 = x$.

There's a simple model-theoretic proof of this: see, for example, my Introduction to Godel's Theorems (CUP, 2nd edn), Theorem 10.8. (The idea is that you take the standard numbers, and add a couple of rogue elements to get a model that still makes axioms true, but the behaviour of the rogue elements makes addition non-commutative. But judging from the OP's comments on another answer, I don't think going into details will be helpful right here.)

To be complete there should be something to tell us the names of the naturals generated by $A2$, I'll call it $N2$ in the following (i.e. $2=s(1)$, $3=s(2)$, etc.)

But anyway, I think this can be done by repeated usage of $A6,A7$ (and of course $A1,A2,A3$ as they are constantly used by the process described below, I will not mention them explicitely).

$1+0\ \stackrel{N2}=\ s(0)+0\ \stackrel{A7}=\ s(0+0)\ \stackrel{A6}=\ s(0)\ \stackrel{N2}=\ 1\tag{B1}$

$2+0\ \stackrel{N2}=\ s(1)+0\ \stackrel{A7}=\ s(1+0)\ \stackrel{B1}=\ s(1)\ \stackrel{N2}=\ 2\tag{B2}$

$3+0\ \stackrel{N2}=\ s(2)+0\ \stackrel{A7}=\ s(2+0)\ \stackrel{B2}=\ s(2)\ \stackrel{N2}=\ 3\tag{B3}$

and so on either directly until we reach $x$...

or by induction: $s(x)+0\ \stackrel{A7}=\ s(x+0)\ \stackrel{induction}=\ s(x)$

@GrahamKemp

$1.x=x$ can be deducted from $1.y\ \stackrel{N2}=\ s(0).y\ \stackrel{A9}=\ y+0.y\ \stackrel{(*)}=\ y+0\ \stackrel{B\infty}=\ y$

And for $(*)\ :\ 0.y=0$ we do like we have done by induction for $+$ but this time for $\times$ starting from repeated usage of $A8,A9$.


Applied problem of multivariable calculus with integrals. How to solve this problem:

A medal has the shape of the portion of the plane $x + z = 1$ lying inside the cylinder $$x^2 + \frac{y^2}{2} = 1.$$ The edge of the medal is covered in gold and it costs $\$ 20$ per linear inch, while the medal's surface is covered in silver and it costs $\$10$ per square inch. How much does each medal cost?

The units for the variables are not stated explicitly.

I think this problem is solved with a couple of integrals, but really have no clue on how to set them up.

I think I found the cost of the edge

Find the angle the plane slices through the cylinder using the dot product of the normal vectors of the plane and the cylinder

n1(plane>=<1,0,1>

n2(cyl)=<0,1,0>

n1*n2=||n1||||n2||cos(theta)

Theta =45

Then use the formula for the circumferance of a cylinder

C=2*pi*sqrt((a^2+b^2)/2)

and make a right triangle with the angle theta to find the circumference of the slice

no idea about the area of the coin good luck hope this helps


Differential equation question Solve the following differential equation $(1+xy)ydx + (1-xy)xdy=0$.
I have been taught to solve differential equations of the following type:
1.Homogeneous
2.Variable separable
3.Linear differential equation
But, I'm unable to find to which category the given differential equation belongs.Please help. I would like some hints towards solving the question and please don't post the answer.

Set $\xi = xy$, $\eta = \frac{x}{y}$. Then
$$\begin{align}
\mathrm{d}\xi &= y\,\mathrm{d}x + x\,\mathrm{d}y
& \mathrm{d}\eta &= \frac{y\,\mathrm{d}x - x\,\mathrm{d}y}{y^2}
\end{align}$$
And therefore
$$\mathrm{d}\xi + \frac{\xi^2}{\eta}\mathrm{d}\eta = 0$$
I suppose, you can take it from there.

IF you have anything of form f(x.y)y dx + g(x.y)x dy = 0 ; then I.F. will be 1/(Mx-Ny) . Multiply this to make it exact differential equation.


What is the probability of two telephone numbers having 666 in them. My girlfriend and I both have phone numbers that are 10 digits long. They both happen to have three consecutive sixes in them (at different spots). We take it as a sign that we're meant for each other.

What is the probability that this actually occurs, if we can use digits 0-9. Is this reasoning correct?:

There are seven spots for the sixes to be placed:

The probability of placing the six in the first spot is $$\frac{1}{10}$$. Then the other two sixes need to be placed in the following spots, so the probability that this occuring, where the six is placed first is $$\frac{1}{10^3}$$

This can occur in seven more ways, so the total probability is $$\frac{7}{10^3}$$

Is that correct?

Denote by $D_n$ the set of decimal strings of length $n$. We are told to compute the probability $p$ that a random string from $D_{10}$ contains the substring $s_*:=666$. In order to avoid all sorts of inclusion-exclusion-traumata we argue as follows: Let

$a_n:\quad$the number of $s\in D_n$ not containing $s_*$, and not ending with $6$,

$b_n:\quad$the number of $s\in D_n$ not containing $s_*$,  and ending with a single $6$,

$c_n:\quad$the number of $s\in D_n$ not containing $s_*$, and  ending with $66$,

$d_n:\quad$the number of $s\in D_n$  containing $s_*$.

One easily sees that
$$a_3=900,\quad b_3=90,\quad c_3=9,\quad d_3=1\ .$$
The $a_n$, $b_n$, $c_n$, $d_n$ satisfy the following recursion:
$$\left[\matrix{a_{n+1}\cr b_{n+1}\cr c_{n+1}\cr d_{n+1}\cr}\right]=
\left[\matrix{9&9&9&0\cr 1&0&0&0\cr 0&1&0&0\cr 0&0&1&10\cr}\right]\ \left[\matrix{a_n\cr b_n\cr c_n\cr d_n\cr}\right]\ .$$
We therefore compute $$A^7\left[\matrix{900\cr 90\cr 9\cr 1\cr}\right]=(8942446170, 895052349, 89586081, 72915400)'$$and read off $d_{10}=72\,915\,400$. It follows that
$$p={d_{10}\over 10^{10}}\doteq0.00729\ .$$

Are you willing to accept, say, ABC-DEF-6666, that is are you counting FOUR sixes in a row as included in three sixes?  If so then, yes, this is correct.


Do these images both represent disk models of Euclidean geometry? Q: Do these images represent the same thing? The first one is a disk model of Euclidean geometry, and the second one is the warping of a regular grid through a glass ball. If they are the same, why? It seems like the second image is in the realm of spherical geometry. Why are the geodesics circular arcs if it's supposed to represent Euclidean geometry?

This is a disk model of Euclidean geometry (where geodesics are circular arcs).

It looks exactly like the grid on the glass ball in this image.

I'm not sure what ℝ²→ℝ² map your "disc model of Euclidian geometry" is using.
The "warped by glass ball" map is called refraction, and is explained by Ray Tracing math.
To be honest, I am not sure whether an ordinary refractive glass sphere would show the whole plane, but this can be determined using ray tracing.
How do you know the first one is all circular arcs?

They are not the same.  The lines all meet precisely on the perimeter of the disk model of Euclidean geometry, but not so in the glass orb, to take just one of several examples.


Multiplying a matrix with a scalar - dimensional consistency Let us represent an $n\times 1$ vector as $\vert .\rangle$ and a $1\times n$ vector as $\langle . \vert$. Consider the following expression where $X$ and $Y$ are $n\times n$ matrices.

$$E(X) = \langle i \vert X\vert i\rangle Y$$

Suppose $Y$ can be written as a product of two more $n\times n$ matrices $A$ and $B$. Then

$$E(X) = \langle i \vert X\vert i\rangle AB = A\langle i \vert X\vert i\rangle B,$$

where the change of order is allowed since $\langle i \vert X\vert i\rangle$ is just a number. I'm not sure how to think about the terms $A\langle i \vert$ and $\vert i\rangle B$. They seem dimensionally inconsistent, yet they should work out to be an $n\times n$ matrix. What is the issue here?

$A\langle i\vert$ is not an $N\times N$ matrix.  It is an $N\times N^2$ matrix.  More precisely, you should think of this (and correspondingly the other products in your question) as a tensor product of an $N\times N$ matrix with a $1\times N$ vector, that is,
$$ A\otimes \langle i\vert\ .$$

Notice that $\langle i |\cdot X\cdot |i\rangle$ is not actually “just a number,” but really a $1\times 1$ matrix (because the product of two matrices is always a matrix). Since $\langle i|X|i\rangle$ is not a scalar, the product $\langle i|X|i\rangle Y$ is not well-defined, since $i$ is $1\times 1$ and $Y$ is $n\times n$.


A question from G.Polya A question from G.Polya:

prove $$1 + \frac 1 4 + \frac 1 9 + \frac 1 {16} + \cdots
=  \log x \cdot \log (1 - x) + \frac {x+(1-x)} {1} + \frac {x^2 + (1-x)^2}{4}+\frac{x^3+(1-x)^3}{9}+\cdots$$
for $0<x<1$.

The answer says:

$\sum_{n=1}^\infty \frac{x^n}{n^2}=\int_0^x \sum_{n=1}^\infty \frac{t^{n-1}}{n}dt = - \int_0^{x} t^{-1}\log(1-t)dt;$
integrate by parts then introduce as new variable of integration $s=1-t$.

Following the answer, I got $$- \int_0^{x} t^{-1}\log(1-t)dt=-\log(1-x)\log x+\int _1^ {1-x}\frac{\log(1-t)}tdt.$$
But how to proceed?

Start from your last eqution:

$- \int_0^{x} t^{-1}\log(1-t)dt=-\log(1-x)\log x+\int _1^ {1-x}\frac{\log(1-t)}tdt$

Easy to realize that

$-Li_2(x)=-\ln x ln(1-x)+Li_2(1-x)$

where $Li_2(x)$ is polylogarithm function (Spence-function).

$Li_2(x)+Li_2(1-x)=\frac{\pi^2}{6}- \ln x ln(1-x)$

is one of the known $Li_2$ identities

As  $\sum\limits_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$  we are ready.


$E=\{e_k|k\in\mathbb{N}\}$ : characterization of $E$? Let $(X, \|.\|)$ be normed linear space consisting of sequences $a=\{a(n)\}_{n=1}^{\infty}$ such that the series $\sum_{n=1}^\infty a(n)$ is absolutely convergent,with $ \|a\|= \sum_{n=1}^\infty |a(n)|$. Let $e_k$ denote the sequence in $X$ whose $k$-th term is 1 and other terms are $0's$ and let $E=\{e_k | k \in \mathbb{N}\}.$ Then

$X$ is complete in the norm $\|.\|$
$E$ is bounded subset of $X$.
$E$ is closed subset of $X$.
$E$ is compact subset of $X$.

Let $\{ a_n \}_{n=1}^\infty$ be a Cauchy sequence in $X$, i.e. for each $\epsilon > 0$ there exists a $N \in \Bbb N$, such that
$$ \sum_{j=1}^\infty | a_n(j) - a_m(j) | = \| a_n - a_m \| < \epsilon \quad \text{for all } n,m \geq N \; . \tag{$\ast$}$$
Let $\epsilon > 0$ and $N \in \Bbb N$, such that $(\ast)$ holds. Note that for each $j \in \Bbb N^\times$ the sequence $\{ a_n(j) \}_{n=1}^\infty$ is a Cauchy sequence in $\Bbb R$, and since $\Bbb R$ is complete, the limit
$$ a(j) := \lim_{n \to \infty} a_n(j) $$
exists. Define $a := \{ a(j) \}_{j=1}^\infty$. We wnat to show that $a_n \to a$ in $X$, i.e. $\| a_n - a \| \to 0$, and $a \in X$. Note that for each $M \in \Bbb N^\times$
$$ \sum_{j=1}^M |a_m(j) - a_n(j) | \leq \sum_{j=1}^\infty | a_m(j) - a_n(j) | < \epsilon \quad \text{for all } m,n \geq N \; .$$
By letting $n \to \infty$, we conclude that
$$ \sum_{j=1}^M | a_m(j) - a(j) | \leq \epsilon \quad \text{for all $m \geq N$ and $M \in \Bbb N^\times$} \; ,$$
and now, by letting $M \to \infty$, we see that
$$ \| a_m - a \| = \sum_{j=1}^\infty | a_m(j) - a(j) | \leq \epsilon \quad \text{for all } m \geq N \; .$$
This shows that $a_n \to a$ in $X$. It remains to show that $a \in X$, i.e. $\Vert a \Vert < \infty$. Choose a $m \in \Bbb N^\times$, such that $\| a - a_m \| < 1$ (this exists, because $\| a_m - a \| \to 0$). Now we deduce by using the triangle inequality, that
$$ \|a \| = \| (a-a_m) + a_m \| \leq \underbrace{ \| a - a_m \| }_{< 1} + \underbrace{ \| a_m \| }_{< \infty} < \infty \; ,$$
so $a \in X$, and we have shown, that $X$ is complete.
This is clear, because $\| e_k \| = 1$ for each $k \in \Bbb N^\times$.
Observe, that for $e_k, e_\ell \in E$
$$ \| e_k - e_\ell \| = \sum_{j=1}^\infty | e_k(j) - e_\ell(j) | = \begin{cases} 2 \, , & k \neq \ell \\ 0 \, , & k = \ell \end{cases} \; .$$
With 1. we conclude, that a sequence $\{ a_n \}_{n=1}^\infty$ in $E$ converges, if and only if there exists a $N \in \Bbb N^\times$ and a $k \in \Bbb N^\times$, such that
$$ a_n = e_k \quad \text{for all } n \geq N \; ,$$
in which case
$$ \lim_{n \to \infty} a_n = e_k \in E \; .$$
This shows that $E$ is closed.
Note, that we can not use Heine-Borel here, because $X \neq \Bbb R^n$. But you can easily construct a sequence $\{ a_n \}_{n=1}^\infty$ in $E$, which has no convergent subsequence. This means, that $E$ is not compact.

Complete: for all $a=\{ a(n)\}_{n=1}^∞$ there is sequence $b(k) = \sum _ {i=1}^ k  a(i)*e_i.$ This sequence converges to $a$.
Bounded: for all $e_k: ∥e_k∥ = 1$
Closed: if $a !=e_k$ for any $k$ and we have at least two non-zero components $a(i)$ and $a(j)$. Let $a(i)$ be smaller than $a(j)$ by absolute value. Then $\mid (a(i))/2 \mid$ - surrounding of a doesn't belong to $E$ since it has 2 non -zero components. In case a has only one non-zero component $a(i)$ than $\mid(a(i)-1)/2\mid$ - surrounding of a doesn't belong to $E \Rightarrow$ addition to $E$ is open $\Rightarrow E$ is closed
$E$ is closed and bounded $\Rightarrow$ $E$ is compact.