Are there contradictions in math? Someone told me that math has a lot of contradictions.

He said that a lot of things are not well defined.

He told me two things that I do not know.

$1+2+3+4+...=-1/12$
what is infinity $\infty$?

Since I am not a math specialist and little. How to disprove the previous two?

How can I convince him?

Someone told me that math has a lot of contradictions.

Correct mathematics does not, as far as we know. However, mathematicians amuse themselves with little "proofs" whose conclusion is absurd. The game is to identify the error. It's important because the "proofs" usually rely on errors that people often make by accident. Finding the error helps mathematicians avoid making the same error themselves.

Probably the simplest such game is a "proof" that $1 = 0$:

Let $x = 1$ and $y = 0$. Then $x \cdot y = y \cdot y$. Dividing both sides by $y$, $x = y$.

The error of course is that you cannot divide both sides by zero and maintain an equality. The lesson is not to divide by something that is or might be $0$. In a more complex proof it might take some work to prove that the thing you want to divide by actually is never $0$, or if it could be $0$ to consider the case where it is separately from the case where it isn't.

Sometimes this game becomes more serious. There was a thing that is now called "naive set theory" that basically said, "any collection of sets that you can describe, is a set". This allows us to consider such things as "the set of all sets", and "the set of all sets that are elements of themselves". Of course, "the set of all sets" is an element of itself. The empty set is not an element of itself (no set is an element of the empty set, that's what empty means). So, what if I define S to be "the set of all sets that are not elements of themselves".

Oh dear. Now we have a contradiction. Is S an element of S or not? If it is, then it must satisfy the definition. But according to the definition anything in S is not an element of itself. And "itself" is S. That it to say, if it is an element then it isn't. And if it isn't an element of S, then by the definition it is an element of itself. So if it isn't an element then it is.

What this told mathematicians is that what is now called "naive set theory", just defining any old thing you like, isn't good enough. This is called Russell's Paradox, because Bertrand Russell published it first. Ernst Zermelo had previously discovered it but not published. Both Russell and Zermelo set about constructing systems that:

allowed everything that mathematicians needed to do with sets (actually Russell worked on "type theory" rather than "set theory", but I don't think the difference matters to this explanation).
prevented the paradox from occurring by limiting how you're allowed to define sets.

You could reasonably argue that prior to this activity, mathematics did contain contradictions. Fortunately not in a way that mattered, since as it happened there weren't any really important results that couldn't be brought onto more secure foundations. But for this and other reasons we cannot say for sure there are no contradictions, only that there are none we know about and haven't dealt with.

This might sound like a terrible crisis in mathematics, finding "math" to be contradictory. In some sense it was a crisis, in that it required a lot of re-checking of some basic assumptions and intuitions people had. It wasn't a disaster, since set theory as such had existed for less than 50 years at the time, and all the paradox said was that set theory wasn't quite right and needed improvement. Most of all mathematics ever done at the time had been without any particular appeal to this flawed set theory.

You can also think of it as an extended and surprising "proof by contradiction". A proof by contradiction says:

suppose X is true
deduce a contradiction
conclude that X is not true

Set theoreticians had:

constructed a theory of sets
deduced a contradiction
concluded that this theory of sets was no good

So mathematics does "involve" a lot of contradictions in the sense that a lot of proofs you see will end with one. Doesn't quite mean that it "contains" them :-)

  He said that a lot of things are not well defined.

This is arguable at a stretch. In mathematics pretty much everything is defined, but one interesting question is what it's defined in terms of. Foundations of mathematics is a large subject within mathematics.

  1+2+3+4+...=−1/12

Not true, but you can make it look true in a variety of ways. This is one of those amusements I mentioned above.

Given a divergent series, you can perform some incorrect but plausible-looking manipulations that result in it appearing to have any total you please. AFAIK the original historical reason for choosing -1/12 in particular, is that 1+2+3+4+... is the divergent Dirichlet series for -1, and -1/12 is the Riemann zeta function of -1. Now, when the Dirichlet series converges for a particular value, it is equal to the zeta function. In fact the zeta function can be defined as the "analytic continuation" of those convergent Dirichlet series. That means that it's the only function that's equal to the convergent Dirichlet series and also has another special property called "holomorphism".

But the Dirichlet series for -1 doesn't converge, as I said to begin with it's divergent. So -1/12 is the value of an important and interesting function that coincides with some Dirichlet series, but not this one, which doesn't have a value. So you can call it "the value of that Dirichlet series", but really it isn't. There is no contradiction.

  what is infinity ∞?

Since you're not mathematically trained you can't be expected to know what infinity "is" mathematically. Quite aside from the fact that the concept of "infinity" is somewhat mysterious in English, your friend has asked you an unfair question of mathematical general knowledge. He might as well have asked you what the Riemann zeta function is, or the definition of a metric space, and if you've never studied those things then you just don't know.

In mathematics you deal with the concept of infinity by defining very strict rules for how to handle it, and then following them. In different contexts mathematicians will define "infinity" differently. So you should think of ∞ as just being a symbol used to mean some specific thing that's defined somewhere else (hopefully the person who uses it can tell you where). It doesn't necessarily mean the same thing every time it's used in different places.

Back in that subject, "foundations of mathematics", there's plenty of interesting work on how to consider the infinity of the natural numbers 0, 1, 2, 3, ... Further, there's interesting work on whether mathematicians "should" be considering infinity as a quantity at all. Since it's foundational this work can in principle affect all of mathematics, in practice most areas of mathematics can just pick the version of "infinity" they need (if any) and stick with it.

  How to disprove the previous two?

I can think of several possibilities:

your friend has genuine questions about the foundation of mathematics. This is fine, but you, as a non-math-specialist, are not going to deal with his concerns. Refer him to a mathematician who will make him be more clear exactly what his concerns are: what does he feel is not well defined? How he has derived his contradictions?
your friend has encountered something he doesn't fully understand, and concluded that it is ill-defined and contradictory. That's a natural response, but in the case of mathematics there's an antidote, which is to ask genuine questions about the foundations of mathematics. See above.
your friend is messing with you. Since he has offered you no proof of what he says, there is no need to find a particular error. It's as if he said, "English contains a lot of contradictions. For example the word 'red' means 'orange'". Well firstly, no it doesn't, it means something different, you can just ignore him (and the sum of 1+2+3+4+... is not -1/12, that series doesn't have a sum). But secondly, OK, in certain contexts actually it does mean that for interesting historical reasons. In fact the word "orange" entered the English language long after certain orange things had already been named "red", such as redheads, red kites, and robins' red breasts. But this is not a contradiction (and in the context of Dirichlet series and the Riemann zeta function, -1/12 is in a strange sense what the sum of 1+2+3+4+... "would be if it converged").

Yes, it does. The mathematics with unrestricted infinite sets, before Zermelo rectified it after Russell's paradox, was contradictory. Why should we suppose it perfectly clean now then? Look to the zero and the infinity for contradictions since these two are the source of continual problems and directly related.

In my opinion the empty set (and zero) is contradictory. Here is a container {} representing non-existent walls even though you can see them as brackets. It contains no values. How can it contain no values if there are no values to contain and no walls to do the containing? So it is a wall free container of nothing, therefore it should not have brackets. {} represents nothingness.

There is no definition of nothingness possible except the circular nothingness = nothingness. If nothingness could be defined it would have location, and that is a contradiction. It is thought to be inside {}. But the undefinable cannot be contained.

However the real crux is this. Every single thing both imagined or real must have a location. Any concept or idea has a location where it is being thought about. Your dream is located inside you. Even as I write these words they are located on the screen, as is {}. But "no values" have not got a location if unwritten, and you can't locate them by thinking about them since they don't exist. {no values} is taken to be {} and this locates the unlocatable between the brackets as written, or inside the concept of {} in your mind. More of a contradiction I do not know. Nothingness is nowhere so it can't be represented at the location of {}. {} is just two located brackets representing two perfectly touching sizeless unlocatable brackets without space between. It's not a set and it is not empty because there is no represented space between the brackets to be empty.

Maybe this contradiction isn't provable but nor is the axiom of the empty set. You have to use common sense to see the axiom is flawed.


Are there spaces that are globally Euclidean, but locally not? I was reading this question, and I idly wondered if there could be spaces that were globally Euclidean, but locally not, and couldn't think of any circumstances in which that might be the case.

As the comments have suggested, the question is a bit under-determined, but let me supply some definitions in a plausible way and answer it based on those definitions.

Let $P$ be a property of topological spaces.  (Formally speaking this could be any "predicate", but it would only be asking for trouble if $P$ were not a topological property of topological spaces -- i.e., such that if $X$ and $Y$ are homeomorphic topological spaces, then $X$ has property $P$ iff $Y$ has property $P$.)

Then we say that $X$ is locally P if for every point of $x$ there is a neighborhood base $\{U_x\}$ each of whose elements $U_x$ has property $P$ in the subspace topology.

We say that $X$ is globally P if it has property $P$.  Why would we say this?  Well, it's a matter of emphasis, presumably by way of comparison to something which locally has property $P$.

Now define a space to be Euclidean if it is homeomorphic to $\mathbb{R}^n$ for some $n \in \mathbb{N}$.

Now I claim that any space which is globally Euclidean is also locally Euclidean.  In other words, I claim that for all $n \in \mathbb{N}$, every $x \in \mathbb{R}^n$ admits a neighborhood base, each of whose members is homeomorphic to Euclidean $m$-space for some $m$.  But indeed, for any $\epsilon > 0$, the open $\epsilon$-ball around $x$ is isomorphic to $\mathbb{R}^n$.

Note that in general "globally P" need not imply "locally P".  For instance a connected space need not be locally connected, and a quasi-compact (non-Hausdorff!) space need not be locally quasi-compact.

In fairness I should say that some mathematicians don't like the idea that globally $P$ need not imply locally $P$, so they weaken the definition of locally $P$ to "each point of $X$ admits some neighborhood which has property $P$".  I would call this weakly locally P and then it is trivial that globally $P$ implies weakly globally $P$: for
each $x$, take $X$ itself as a neighborhood which has property $P$.

Just another attempt could be:
Define sth is locally adjective if it admits a neighborhood base consisting of adjective neighborhoods at every point and sth is globally adjective if it admits a basis for the topology consisting of adjective opens...

However, that would be again just seeking for some consistency pressing in some technical dress, so hmm I think best is still to stick with the luckily almost widely accepted aggreement presented by Pete Clark - though I like Andrew's view on these things although it's kind of vague.


how to prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$ How can one prove that the circle $(x-a)^2+(y-b)^2=a^2+b^2$ is passing through point $(0,0)$?

I know that if i put: $x=y=0$,  i will get: $(0-a)^2+(0-b)=a^2+b^2=a^2+b^2$

But that's not a proof but checking.

Thanks.

The equation $$(x-x_M)^2+(y-y_M)^2=r^2$$ describes the circle (to be more specific the circle line) with center $(x_M,y_M)$ and radius $r$. This means, that every point $(a,b)$ that lies on the circle line fulfills this equation. Also the other way around, every point $(a,b)$ that fulfills this equation must lie on the circle line. In your situation you have $x_M=a,y_M=b$ and $r^2=a^2+b^2$.

So if you have to show that the circle passes through a specific point, it is indeed enough to just pluck the point in the given equation and see if everything turns out well.

I suppose this illustrates the difference between "showing" or "proving" as opposed to "verifying".

Just plugging in the values $x=0$ and $y=0$ and getting LHS=RHS is verifying the result.

To show or prove it you could show that the distance from the origin to the centre is the same as the radius, which is straightforward enough.


what is $\sqrt{i}+\sqrt{-i}$ what is $$\sqrt{i}+\sqrt{-i}$$ My book gave four possible answers:

$$\sqrt{i}+\sqrt{-i}=\pm\left(\frac{1+i}{\sqrt{2}}\right)\pm\left(\frac{1-i}{\sqrt{2}}\right)$$

So possible values of $$\sqrt{i}+\sqrt{-i}$$ are $\sqrt{2}$ ,$-\sqrt{2}$$\:$,$i\sqrt{2}$ $\:$,$\:$ $-i\sqrt{2}$

But i am getting an ambiguityif we take $i\sqrt{2}$:

$$\sqrt{i}+\sqrt{-i}=i\sqrt{2}$$ squaringboth sides we get

$$i+(-i)+2\sqrt{i} \times \sqrt{-i}=-2$$i.e.,

$$2=-2$$

can i know where i went wrong?

What is the value of $\sqrt{i}\times\sqrt{-i}$? For the value of $\sqrt{i}$ and $\sqrt{-i}$ you have chosen
$$\sqrt{i}=\frac{1+i}{\sqrt{2}}\qquad\text{ and }\qquad\sqrt{-i}=-\frac{1-i}{\sqrt{2}},$$
to get $\sqrt{i}+\sqrt{-i}=i\sqrt{2}$. That means
$$\sqrt{i}\times\sqrt{-i}=\frac{1+i}{\sqrt{2}}\times-\frac{1-i}{\sqrt{2}}=-\frac{1}{2}(1+i)(1-i)=-1,$$
so everything seems to work out fine. I'm guessing you went wrong by assuming
$$\sqrt{i}\times\sqrt{-i}=\sqrt{i\times-i}=\sqrt{1}=1,$$
but unfortunately square roots only obey such a rule when considering the square root as a single-valued function on the (nonnegative) real numbers. On the complex numbers the square root is not single valued, there is no canonical square root of a complex number.


Calculus: Cauchy's Mean Value Theorem I suppose I need to make use of Cauchy's mean value theorem to prove the statement but how am I able to do that?

Let $F(x) = \int_{a}^{x}f(t) dt$ and $H(x) = \int_a^{x}f(t)g(t) dt$. Both of these functions are continuous, and they are differentiable on $(a,b)$. Furthermore, $f(x) > 0$ on $[a,b]$ implies $F(a) \neq F(b)$.

Thus we can use Cauchy's mean value theorem: there exists some $c \in (a,b)$ such that $$ \frac{H'(c)}{F'(c)} = \frac{H(b) - H(a)}{F(b) - F(a)}. $$

Using the facts that $H'(c) = f(c)g(c), F'(c) = f(c), H(a)=F(a)=0$ gives you the desired result (after multiplying both sides by $F(b) = \int_a^bf(t)dt$).

Just a variation on simmilar ideas:

From Mean Value Theorem for Integrals:

$$\tag{1}\int^b_af(x)dx=f(c)(b-a) \rightarrow f(c)=\frac{\int^b_af(x)dx}{(b-a)}$$

$$\tag{2}\int^b_af(x)g(x)dx=f(c)g(c)(b-a)$$

From $(1)$ and $(2)$ we get:

$$\int^b_af(x)g(x)dx=\frac{\int^b_af(x)dx}{(b-a)}g(c)(b-a)=g(c)\int^b_af(x)dx$$


Finding the last digit of $3^{729}$ I'm practicing for my algebra exam but I stumbled on a question I don't know how to solve.

  Let $N = 3^{729}$. What is the last digit of $N$?

The example answer says

  Since $\gcd(3, 10) = 1$, check that
  $3^4 = 81 = 1 \pmod {10}$:
  Now, $729 = 182 \times 4 + 1,$ so we get we get (that might be a typo or missed a step)
  $3^{729} = 3 \pmod {10}$.

Can anybody help me with this question? Thanks so much!

Every digit 0 thru 9 has a pattern of 4 digits when raised to an exponential power. Simply divide the power by 4 and the remainder shows you where you are at in the pattern. Here are the patterns for all 10 digits. Note that remainder 1 corresponds to the first digit in the pattern and remainder 2 corresponds to the second digit in the pattern. Remainder 3 corresponds to the third digit and remainder 0 (the power is divisible by 4) corresponds to the fourth digit in the pattern. Try a few yourself using a calculator and you will get the hang of it.

0,0,0,0
1,1,1,1
2,4,8,6
3,9,7,1
4,6,4,6
5,5,5,5
6,6,6,6
7,9,3,1
8,4,2,6
9,1,9,1

Last Digit of $N = 3^{729}$ is equal to last digit of $3^9 = 3$


How can we know the answer to 1-1+1-1+1...? I was watching this video I noticed that the teacher said that 1-1+1-1+1... equals 1/2. How can we know that? The proof he uses doesn't make sense to me. We go from 1 to 0 to 1 and back again, etc. If it goes on like so forever, where does a fraction come into play?

The series does not converge

But if you treated it as a formal geometric series $1+r+r^2+\ldots= \dfrac{1}{1-r}$ and then let $r=-1$ you would get $\dfrac{1}{2}$

Similarly if $S=1-1+1-1+\cdots$ then you might set up

 S = 1 - 1 + 1 - 1 + ...
 S =     1 - 1 + 1 - ...

and adding the two vertically gives

2S = 1

For divergent series, you can use Cesàro summation to find your desired limit.

The idea is to take the mean of your partial sums, and we hope it converges:

1 1 1/1
2 0 1/2
3 1 2/3
4 0 2/4
5 1 3/5

etc. you can see the mean goes to $1/2$.


Exponential integral times exponential Let us define the exponential integral $E_{1}(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \mathrm{d}t$, with $x$ positive real.

Is there any way to simplify the following product?

$$e^{x} E_{1}(x)$$

I've tried to rewrite the exponential integral using the incomplete gamma function, i.e., $E_{1}(x) = \Gamma(0,x)$ but it does not simplify anything. I've also considered the "continued fraction expansion" in https://en.wikipedia.org/wiki/Exponential_integral#Approximations: if I multiply the exponential integral by the exponential, we get rid of the exponential in the numerator of the "continued fraction expansion", but the expression remains intractable.

If there was a known simplification, we would not be using the function $E_1$, but instead that simplified function times $e^{-x}$.

The only operation in the way to simplify this is integrating by parts the $E_1$:

$$E_1\left( x\right) = \int_x^\infty d\left(e^{-t}lnt\right) + \int_x^\infty e^{-t}lnt\ dt = -e^{-x}lnx + \int_x^\infty e^{-t}lnt\ dt$$

Then:

$$e^xE_1\left( x \right) = -lnx + e^x\int_x^\infty e^{-t}lnt\ dt$$


Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$ How to compute the Inverse Laplace of $\frac 1 {(s^2+a^2)^n}$?
I know that to compute Inverse Laplace $\frac 1 {(s^2+a^2)^2}$, the convolution Theorem is useful. but is there an interesting idea for general case?

We could, also, use the definition
\begin{align}
\mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s^2 + a^2)^n}ds\\
&= \sum\text{Res}
\end{align}
We have $n$ poles at $s = \pm ia$ so
$$
\mathcal{L}^{-1}\{F(s)\} = \lim_{s\to ia}\frac{1}{(n - 1)!}\frac{d^{n-1}}{ds^{n-1}}(s-ia)^n\frac{e^{st}}{(s^2+a^2)^n} + \lim_{s\to -ia}\frac{1}{(n - 1)!}\frac{d^{n-1}}{ds^{n-1}}(s+ia)^n\frac{e^{st}}{(s^2+a^2)^n}
$$


Find the probability distribution for the number of spades. Three cards are drawn in succession from a deck without replacement. Find the probability
distribution for the number of spades.

Let random variable $X$ be the number of spades. We want to calculate $\Pr(X=0)$, $\Pr(X=1)$, $\Pr(X=2)$, and $\Pr(X=3)$.

As a sample, we calculate $\Pr(X=2)$.

There are $\binom{52}{3}$ three-card hands, all equally likely. We count the number of two-spade hands.

The $2$ spades can be chosen in $\binom{13}{2}$ ways. For each of these ways, there are $\binom{39}{1}$ ways to choose the non-spade, for a total of $\binom{13}{2}\binom{39}{1}$.

Thus $\Pr(X=2)=\dfrac{\binom{13}{2}\binom{39}{1}}{\binom{52}{3}}$.

Another way: The event $X=2$ can happen in $3$ different ways: SSN (spade, then spade, then non-spade); SNS; NSS.

We calculate the probability of SSN. The probability the first card is a spade is $\frac{13}{52}$. Given that the first card was a spade, the probability the next is a spade is $\frac{12}{51}$. Given the first two cards are spades, the probability the third is a non-spade is $\frac{39}{50}$. Thus the probability of SSN is $\frac{13}{52}\cdot \frac{12}{51}\cdot \frac{39}{50}$.

Similarly, calculate the probability of SNS and NSS. (Or else note that it is clear that each is the same as the probability of SSN.) Add up.

Let the random variable X be number of spades then using bionomial distribution we can calculate

P(X=r) = probablilty that r spades comes out of n cards then we can write it as nCr * p^r * q^(n-r) where n is the total number of cards drawn and r is the number of spades and p is the probability of success and q is the probability of failure

Probability that the drawn card is spade = p = Probability of success i.e. 13/52

Probability that the drawn card is not spade = q = 1-p = Probability of failure i.e. 39/52

Therefore P(X=r) = 3Cr * (13/52)^r * (39/52)^(n-r)

Substituting values of r you can get probability of getting r spades which gives us the probability distribution for the number of spades.


Why does an infinite limit not exist? I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like  $$\lim_{x\to \infty}{\cos x}$$.

According to some presentations of limits, it is proper to write "$\lim_{x\to 0}\frac{1}{x^2}=\infty$."

This does not commit one to the existence of an object called $\infty$. The sentence is just an abbreviation for "given any real number $M$, there is a real number $\delta$ (which will depend on $M$) such that $\frac{1}{x^2}\gt M$  for all $x$ such that $0\lt |x| \lt \delta$." It turns out that we often wish to write sentences of this type, because they have important geometric  content. So having an abbreviation is undeniably useful.

On the other hand, some presentations of limits forbid writing "$\lim_{x\to 0}\frac{1}{x^2}=\infty$." Matter of taste, pedagogical choice. The main reason for choosing to forbid is that careless manipulation of the symbol $\infty$ all too often leads to wrong answers.

Ok ...
I think there is a difference between saying the limit doesn't exsist and equal to infinity
Limit f(x) as x approaches to x' (exsist ) iff lim f(x ) when x approaches to x' from the right = limit f(x)  as x reproaches to x' from the left
And   in the above limit in your question  is  exist because  limit from the right  equal the limit from the left equal to infinity +
  but if you take f(x)= 1\x  in this  case the limit doesn't exsist  the limit from the right =infinity+  but the limit from the left equal to infinity -
Please don't hesitate to discuss this with me.


What is the multiplication "$\times$" for the ordinary vector space $(\Bbb R^n, +, \times)$ If I recall correctly, a field $F$ is a set $V$ associated with two operators $+,\times$ st. $(V,+)$,$(V,\times)$ are both Abelian groups and in addition it satisfies distributivity of $\times$ over $+$, i.e. $\forall a,b,c\in V,a\times(b+c)=a\times b + a\times c$. A vector space over $F$ adds additional interaction with real number field $(\Bbb R, +,\times)$ where $+,×$ are overloaded to denote the ordinary multiplication of real numbers.

I am curious that such an abstract vector space must be a generalization of our ordinary vector space $(\Bbb R^n, +, \times)$ in which each vector consists of real-number components. I am wondering what is the appropriate multiplication here. It is obviously neither cross product nor inner product. Thank you!

You seem to be confusing two separate but related operations: (1) multiplication in the field, and (2) scalar multiplication in the vector space. The standard vector space $\mathbb R^n$ is defined over the field $\mathbb R$. Multiplication in $\mathbb R$ works exactly the way you expect. We can't properly speak of multiplication in $\mathbb R^n$, however. You don't multiply vectors with each other; rather, you multiply them by scalars. So if $\bf x \in \mathbb R^n$ and $c \in \mathbb R$, we can multiply them to get a new vector $c \bf x$. In this case, each co-ordinate of $\bf x$ is simply scaled by $c$.

the dot product will be used here and the symbol
$ \times $ does not mean for cross product as,

$\alpha ,\beta  \to {\mathbb{R}^n}$

$\begin{gathered}
  \alpha  = \left( {{a_1},{a_2}, \ldots ,{a_n}} \right), \\
  \beta  = \left( {{b_1},{b_2}, \ldots ,{b_n}} \right) \\
\end{gathered} $

then,

$\begin{gathered}
  \alpha \beta  = \left( {{a_1},{a_2}, \ldots ,{a_n}} \right)\left( {{b_1},{b_2}, \ldots ,{b_n}} \right) \\
   = \left( {{a_1}{b_1},{a_2}{b_2}, \ldots ,{a_n}{b_n}} \right) \\
\end{gathered} $

So, vector multiplication will be used.

Ii may be helpful for you.


Summand Noetherian is Noetherian? I want to prove if $S$ is a Noetherian ring, $R\subset S$ a subring, and there is an $R$-module homomorphism $\pi:S\to R$ such that $\pi$ is surjective and every element of $R$ is fixed by $\pi$ ($R$ is a summand of $S$), then $R$ is also Noetherian.

My attempt: It suffices to show that the kernel of $\pi$ is a submodule of $S$. Then since submodules and quotients of Noetherian modules are Noetherian, $\frac{S}{\ker\pi}\cong R$ is Noetherian.

$\ker\pi$ is an $R$-module, so we just need to show that it is also an $S$-module. This is where I'm not sure what to do since I can't just say $\pi(sr)=\pi(s)\pi(r)$ for some $r\in\ker\pi$ since $\pi$ doesn't necessarily preserve the multiplication by $s\in S$ action.

Am I missing something?

Assuming the rings are commutative, one can show with the assumption on $\pi$ that if $I \subseteq R$ is any ideal, and $SI$ is the ideal in $S$ generated by $I$, then $\pi(SI) = I$. From there, one can use the ascending chain condition for $S$-ideals in $S$ to get the a.c.c. for $R$-ideals in $R$.

you have a sequence $0\to ker\pi \to S \to R \to 0 $ then  $S$ is noetherian if and only if $ker\pi $ and $R$ noetherian.


Getting better at proofs So, I don't like proofs.

To me building a proof feels like constructing a steel trap out of arguments to make true what you're trying to assert.

Oftentimes the proof in the book is something that I get if I study, but hard to come up with on my own.  In other words I can't make steel traps, but I feel fine buying them from others.

How does one acquire the ability to create steel traps with fluency and ease?  Are there any particular reference books that you found helped you really get how to construct a proof fluently?  Or is it just practice?

I'd like to second one part of Qiaochu Yuan's answer: the recommendation to read Polya's book.  Unlike many other books I've seen (albeit none of the others recommended above), it actually does contain guidance on how to construct a proof "out of nothing".

And that's one problem with the "practise, practise, practise" mantra.  Practise what?  Where are the lists of similar-but-not-quite-identical things to prove to practise on?  I can find lists of integrals to do and lists of matrices to solve, but it's hard coming up with lists of things to prove.

Of course, practise is correct.  But just as with anything else in mathematics, there's guidelines to help get you started.

The first thing to realise is that reading others proofs is not guaranteed to give you any insight as to how the proof was developed.  A proof is meant to convince someone of a result, so a proof points to the theorem (or whatever) and knowing how the proof was constructed does not (or at least, should not) lend any extra weight to our confidence in the theorem.  Proofs can be written in this way, and when teaching we should make sure to present some proofs in this way, but to do it every time would be tedious.

So, what are the guidelines for constructing a proof?  You'll probably get different answers from different mathematicians so these should be construed as being my opinion and not a(n attempt at a) definitive answer.

My recommendation is that you take the statement that you want to prove and apply the following steps to it as often as you can:

Expand out unfamiliar terms.
Replacing generic statements by statements about generic objects.
Including implicit information.

Once you've done all that, the hope is that the proof will be much clearer.

Here's an example.

Original statement:

  The composition of linear transformations is again linear.

Replace generic statements:

  If $S$ and $T$ are two composable linear transformations then their composition, $S T$, is again linear.

It is important to be precise here.  The word "composable" could have been left out, as the statement only makes sense if $S$ and $T$ are composable, but until you are completely familiar with this kind of process, it is better to be overly precise than otherwise.  In this case, leaving in the word "composable" reminds us that there is a restriction on the domains and codomains which will be useful later.  (However, one has to draw the line somewhere: even the word "composable" is not quite enough since it leaves open the question as to whether it is $S T$ or $T S$!)
Include implicit information:

  If $S \colon V \to W$ and $T \colon U \to V$ are linear transformations then $S T \colon U \to W$ is again linear.

Here's where remembering that $S$ and $T$ are composable in the previous step helps keep things clear.  As $S$ and $T$ are composable, we only need $3$ vector spaces.  Then, since we explicitly have the vector spaces the fact that $S$ and $T$ are composable is plain, though some may prefer to keep that fact in the statement.  Also, some may like to have the fact that $U$, $V$, and $W$ are vector spaces explicitly stated.
Expand out definitions:

  If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$ for all $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.

Note that I have been careful not to repeat myself with the newly introduced symbols.  It would be technically alright to reuse $u_1$ and $u_2$ in place of $x_1$ and $x_2$ since these are local declarations (restricted by the phrases "for all ...").  However, humans are not good at differentiating between local and global declarations so it is best not to reuse symbols unless the scope is very clear.
Replace generic statements:

  If $S \colon V \to W$ and $T \colon U \to V$ are such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$, then whenever  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$,  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.

Up to now, the rephrasing has not taken into account the fact that there is a conclusion and a hypothesis.  This rephrasing modifies a part of the conclusion to turn it from a generic statement "$P(p)$ is true for all $p \in Q$" to a statement about a generic object "whenever $p \in Q$ then $P(p)$ is true".  We do not do this for the similar statements in the hypothesis.  This is because these two pieces are treated differently in the proof.
Replace generic statements, and reorganise to bring choices to the fore:

  Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  Then  $S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)$.

In this form, the distinction between hypothesis and conclusion is all the clearer.  Parts of the hypothesis use the word "Let", parts of the conclusion use the word "Then".

With this formulation, the proof essentially writes itself.  With all it's gory details:

Proof

Let $S \colon V \to W$ and $T \colon U \to V$ be such that $S(v_1 + \lambda v_2) = S(v_1) + \lambda S(v_2)$ and $T(u_1) + \mu T(u_2)$ for all $v_1, v_2 \in V$, $u_1, u_2 \in U$, and $\lambda, \mu \in \mathbb{R}$.  Let  $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.[^quick]  Then:

$$
S T(x_1 + \eta x_2) = S \big( T(x_1) + \eta T(x_2)\big)
$$

using the hypothesis on $T$ as $x_1, x_2 \in U$ and $\eta \in \mathbb{R}$.  So:

$$
S T(x_1 + \eta x_2) = S T(x_1) + \eta S T(x_2)
$$

using the hypothesis on $S$ as $T(x_1), T(x_2) \in V$ and $\eta \in \mathbb{R}$.  Hence the conclusion is true.

Notes:
1. This could be condensed, but the important thing here is how to find it, not what the final form should be.
2. Notice that I wrote "as $x_1, x_2 \in U$" rather than "with $u_1 = x_1$ and $u_2 = x_2$".  This is partly style, and partly because in the statement of linearity, $u_1$ and $u_2$ are placeholders into which we put $x_1$ and $x_2$.   So saying $u_1 = x_1$ is semantically incorrect as it equates a virtual vector with an actual vector.  This is a very minor point, though.

Finally, I would like to disagree with one part of Qiaochu's answer.  I actually like the imagery of a steel trap.  A proof is a bit like a trap: we want to capture the theorem in a trap so that it can't wriggle out.  We construct the proof so that there is no possibility of escape.  Eventually, yes, we want the proof to be beautiful but when it's first constructed we just want it to do the job.  Only once the theorem is caught can we spend a little time decorating the cage to make it look pretty and set it off to its best advantage.  So build the trap because theorems can be dangerous!  An escaped theorem can do untold damage, rampaging across the countryside, laying waste like an unchecked viking.

(Okay, not quite finally.  The step-by-step proof above was taking from a page I wrote for my students on the nature of proof.  The original can be found here.)

I recommend the book by Lay, "Analysis with introduction to proofs". The first chapter focuses on logic and proofs and personally I found it rather helpful. At least it helped me. Of course to master the art of proofs you have to keep practicing, but knowledge of some basics such that instead of proving the statement "A implies B you can prove contrapositive "not B implies not A", and how it differs from "Proof by contradiction", and the fact that negation of "A implies B" is "A and not B", and how to deal with existential and universal quantifiers, and so forth and so forth. Those are useful techniques one should master.

Incidentally, as an introduction to analysis, the aforementioned book is quite mediocre. I had a first hand experience with that book, since it was a textbook when I took the "low level introduction to analysis (wasn't sure if I was ready for Rudin's Principles of Analysis). So, this is only the chapter on logic and proof which is worth reading. Of course, purchasing the whole book just for a single chapter is funny, but may be you can find the used one or get one in the library.

But that chapter on proofs did help me. When subsequently I took a course on analysis and we used Baby Rudin as a textbook, I was in a very good shape and in fact did quite well.

And, of course, another suggestion: learn by example. See how other people write the proofs. Rudin's style, for instance, is very terse, you literally have to dig through his proofs, but after a while you start getting used to it, and still after a while you start writing similar proofs.

Good luck.


Is the union all numbers divisible by 5 and divisible 3 countable? Using the method where you determine if this set has the same cardinality as the positive integers where would you start? there appears to be no straightforward equation representing the set that contains all divisors of 3 and 5, can this be done with those diagonal charts?

insight into where to start or what to learn would be helpful

Use the Cantor Bernstien theorem:

https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem

The canonical injection $X \longrightarrow \mathbb{Z}$, together with $\mathbb{Z} \longrightarrow X$ defined by $n \mapsto 5n$, where $X$ is the set of all integers divisible by 5 or 3, fulfills the requirements for the CBT.

Hint: Can you construct a bijection from the natural numbers to your set?  If you can, what does that tell you?


Relationship between ordinary matrix product and hadamard product I would like to know is there any relationship between $(ABC)\circ D$ and $A(B\circ D)C$

Instead of general case, I would also like to know is there any special situational/ under any assumption that will make  $(ABC)\circ D$ and $A(B\circ D)C$ relate to each other.

I think it is not possible to perform any conversion. A counter-example:

Let's assume that A, B and C are matrices nxn of 1's. The rank of them is obviously one and also that of the product (ABC will be a nxn matrix of identical n^2 entries). After hadamrd-multiplying ABC by a random D, you can get anything, including rank $n$ matrices.

On the other hand, A (B * D) C is a rank one matrix. Therefore, little to do with it.


How many men required? 2000 men and 500 women can build a bridge in two weeks. If only 1500 men and 250 women are employed, it will take three weeks to build the same bridge.

How many men will be required if it has to be built in just one week?

[N.B. As Benji Altman points out in the comments to the OP, the question is unclear about the make-up of the work force.  The below assumes that the work force for the question consists solely of men.]

Basic approach. We assume, as is typical in these problems, that each person (possibly differentiated by gender) builds a certain fraction of the bridge per week.  Obviously, this is not truly realistic, but it's the characteristic fiction of this kind of problem.

If we let $m$ and $w$ stand for this fraction, then we have

$$
2000m+500w = \frac{1}{2}
$$

since the combined efforts of $2000$ men and $500$ women suffice to build half a bridge in a week.  Do the same with the other equation, and solve for $m$ and $w$.  Keep in mind that the desired answer—the number of men required to build the bridge in a week—is not $m$ itself, but its reciprocal.  That is to say, if $m = \frac{1}{5000}$ (it isn't; I just chose this number for example), then $5000$ men would be required to build the bridge in one week.

Incidentally, it looks to me as though someone was careful to avoid charges of discrimination.

Every $500$ women there are $2,000$ men for every two weeks. If we need the bridge built in one week then we should increase the amount of men by $150$ percent.


The difference of the focal semi axes of an ellipse and a hyperbola is equal to $4$.If the ratio of their eccentricities is $\frac{3}{7}$. An ellipse and a hyperbola have their principal axes along the coordinate axes and have a common foci separated by a distance $2\sqrt{13}$,the difference of their focal semi axes is equal to $4$.If the ratio of their eccentricities is $\frac{3}{7}$.Find the equation of these curves.

The distance between the foci of both ellipse and hyperbola is $2\sqrt{13}$.
So $2c_1=$distance between the foci of ellipse$=2\sqrt{13}$
$2c_2=$distance between the foci of hyperbola$=2\sqrt{13}$
$\frac{\text{eccentricity of ellipse}}{\text{eccentricity of hyperbola}}=\frac{3}{7}$
I do not understand what is the meaning of "the difference of their focal semi axes" and how to proceed further.

Focal semi axis is equal to the length of the semi axis passing through focus of ellipse or hyperbole.
Focal semi axis mean same as semi major axis.
I hope you can now get answers as:
For ellipse: $a^2=49$ and $b^2=36$.
For hyperbola: $a^2=9$ and $b^2=4$
Hope it helps.


Hexagon grid coordinate system If I had a grid of squares, they can be labeled with Cartesian coordinates such that given square $(x,y)$, you know it shares a boundary with squares $(x+1,y),(x-1,y),(x,y+1),(x,y-1)$.

Is there a way of labeling a tessellated hexagon grid, so that given any hexagon label you can work out it's neighbors?

In the interest of closing the question I've posted this answer, but full credit for the answer goes to Michael Lugo for his comment on my initial Post.

All information comes from this link. It goes into far more detail than this answer.

There are multiple approaches to labeling Hexagon grids, each with advantages and disadvantages.

Offset coordinates

  The most common approach is to offset every other column or row. Columns are named col or q. Rows are named row or r. You can either offset the odd or the even column/rows, so the horizontal and vertical hexagons each have two variants.

Cube coordinates

  Another way to look at hexagonal grids is to see that there are three primary axes, unlike the two we have for square grids. There’s an elegant symmetry with these.

Instead of having two values per coodinate, representing how far North and South the hexagon is, you can use three dimensions.

Axial Coordinates

  The axial coordinate system, sometimes called “trapezoidal”, is built by taking two of the three coordinates from a cube coordinate system. Since we have a constraint such as x + y + z = 0, the third coordinate is redundant. Axial coordinates are useful for map storage and for displaying coordinates to the user. Like cube coordinates, you can use the standard add, subtract, multiply, divide operations from cartesian coordinates.

So for my purposes, the most useful map system was cube coordinates, as they have the simplest and most consistent rules for changing coordinates based movement between cells.

Create the grid by drawing horizontal lines thru ALL hexagon centers and drawing vertical lines thru ALL hexagon centers.  The neighbors are:

(+1,-1), (+2,0), (+1,+1), (-1,+1), (-2,0), (-1,-1)

Note: the x-axis is scaled by a factor of sqrt(3).  This is ok.  Cartesian coordinates can have an axis that is scaled.

Note: the x,y does not match the hexagon label.  This is ok.

If we set the upper left hexagon with a coordinate of (0,0), the hexagon label is:  Math.floor(x/2)+1 plus y+1

(0,0) => 0101   (2,0) => 0201  (4,0) => 0301
     (1,1) => 0102     (3,1) => 0202    (5,1) => 0302


Given the characteristic equation, how to find the determinant of a matrix Take a look at this question:

Find $\det(A)$ given that A has $p(\lambda)$ as its characteristic polynomial.
$$
p(\lambda) = \lambda^3 - 2\lambda^2 + \lambda + 5
$$

My first step is to notice the following:
$$
\det(\lambda I -A) = \lambda^3 - 2\lambda^2 + \lambda + 5
$$
But $\det(\lambda I-A) \neq \det(\lambda I) - \det(A)$. Any idea or hint to what to do with the aforementioned equation to compute $\det(A)$? Do I have to compute the eigenvalues and substitute them?

By the fundamental theorem of algebra, we see that
\begin{align}
p(\lambda) = (\lambda-z_1)(\lambda-z_2)(\lambda-z_3)
\end{align}
where $z_1, z_2, z_3$ are the roots which means
\begin{align}
p(0) = (-1)^3z_1z_2z_3 = -\det(A).
\end{align}

By Caley-Hamilton theorem a matrix satisfies its own charateristic equation so Here we have
$$A^3-2A^2+A+5I=0~~~~(1)$$
A $3 \times 3$ matrix $A$ satisfies a unique monic cubic:
$$A^3-Tarce(A)~ A^2+Trace(adj~ A)~ A-\det|A| I ~~~~(2)$$ By comparision
of (1) with (2), we get $\det|A|=-5.$

Also, for example $I^3-I=0 \implies \det|I|=1$


Positive semidefiniteness of symmetric matrix with diagonal = 1 and non-diagonal elements less than 1 Does anyone know any useful results with respect to symmetric matrices with constant diagonals (specifically with respect to whether all eigenvalues are greater than $0$)? I am working on a set of general $n \times n$ matrices with diagonal entries all equal to $1$ and non-diagonal entries between $0$ and $1$  and I am attempting to figure out whether or not this is a subset of positive semidefinite matrices

In a symmetric matrix with positive entries, if the sum of off-diagonal entries in each row (column) is strictly less than the diagonal entry in this row (column), the matrix is called diagonally dominant and is strictly positive. See also Gershgorin theorem.

If the matrix is not diagonally dominant,  the eigenvalue with the greatest absolute value (this is spectral radius of the matrix) is strictly positive, and a corresponding eigenvector with strictly positive coordinates can be choosen - see Perron-Frobenius theorem.

The $3\times3$ matrix $\begin{pmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{pmatrix}$ is not p.s.d. but seems to satisfy your conditions.  So the answer to your question has to depend on what particular subsets of matrices you  actually have in mind.

Neither is the matrix  $\begin{pmatrix}1&1&1/2\\1&1&1\\1/2&1&1\end{pmatrix}$.

Most $n\times n$ symmetric matrices are not positive definite (after all, you are asking for all $n$ eigenvalues to be positive).  You should hook your set of matrices to one of the classical sources of psd-ness of symmetric matrices, for instance by exhibiting them as Gram matrices, or by connecting them via the Fourier Transform and Bochner's theorem to a positive measure.


Is there a function which has limit only given $a_1,a_2...a_n....$ infinite points. Find a function which has limit only in advanced marked points$(a_1,a_2,....a_n)$.
Here is my example.
$g(x) =
\begin{cases}
1,  & \text{if}~x\in\mathbb{Q} \\
0, & \text{if}~x\in\Bbb{R}\backslash \Bbb{Q}
\end{cases}$
$f(x) = (x-a_1)(x-a_2)...(x-a_n)g(x)$
But the problem I can't solve is that,is there a function which has limit only in advanced  given $a_1,a_2...a_n....$ infinite points.

An equivalent way of asking this question is whether there exists a function which is continuous only on a given sequence.
Such a function may fail to exists, for example when your sequence is dense in $\mathbb{R}$ (e.g. $\mathbb{Q}$). Because the set of continuity points must be a $G_\delta$ subset (and therefore cannot be dense and countable).
On the other hand, if your sequence has no limit points, then one can easily construct such a function by dividing $\mathbb{R}$ into intervals, each containing a single element of the sequence, and then define the function on the each interval to be continuous only on that element (for example as you suggested in your question).

Assume that there is an order preserving bijection between $a_n$ and N.
Define
$$ f(x)=\frac{1}{n}, \quad x \in[a_n, a_{n+1}).
$$
Then the function is discontinuous at every $a_n$ but has a right limit as $x \to a_n$.
If, however, you want double-sided limits then take a small $\delta$ and modify as
$f(x)=\frac{1}{n}$ for x $\in(a_n-\delta, a_{n}+\delta)$ and $f(x)=i$ inbeween (imaginary).
But in this case, you are going outside ot the real numbers.
From the perspective of the real numbers, the function is undefined between the intervals.


Does the operator $T(f)(t) := f(t) - f(0)$ preserve measurability? Denote by $\mathcal{B}$ the Borel field on $\mathbb{R}$, denote by $\mathbf{C}_{\left[0,\infty\right)}$ the set of continuous, real-valued functions over the domain $\left[0,\infty\right)$ and denote by $\mathcal{B}_{\left[0,\infty\right)}$ the minimal $\sigma$-algebra that renders each projection function $\pi_t:\mathbf{C}_{\left[0,\infty\right)}\rightarrow\mathbb{R}$
$$
\pi_t\left(f\right) := f\left(t\right)
$$
$\mathcal{B}_{\left[0,\infty\right)}/\mathcal{B}$-measurable.

Consider the following operator $T: \mathbf{C}_{\left[0,\infty\right)} \rightarrow \mathbf{C}_{\left[0,\infty\right)}$
$$
T\left(f\right)\left(t\right) := f\left(t\right) - f\left(0\right)
$$

Let $A \in \mathcal{B}_{\left[0,\infty\right)}$.

Is $T\left(A\right) \in \mathcal{B}_{\left[0,\infty\right)}$?
If the answer to the first question is: "No", is it still "No" if $A$ is a tail event? ($B \in \mathcal{B}_{\left[0,\infty\right)}$ is a tail event iff $B \in \bigcap_{t \in \left[0,\infty\right)}\sigma\left(\pi_s :\mid s \in \left[t, \infty\right)\right)$.)

My attempts at solving this problem

Attempt #1

If $T$ were defined instead like this:
$$
 T\left(f\right) := f + c
 $$
for some constant $c \in \mathbb{R}$, in other words, if $T$ were a rigid translation of $f$, then $T$ would be measurable and invertible, its inverse being itself a translation, and therefore $T\left(A\right)$ would indeed belong to $\mathcal{B}_{\left[0,\infty\right)}$. Unfortunately, $f\left(0\right)$ is not constant, so this approach fails.
Attempt #2

Suppose we know that for all $f \in A$, $f\left(0\right) \in \mathbb{Q}$. Then
$$
 A = \bigcup_{q \in \mathbb{Q}}\underbrace{A \cap \left\{\pi_0^{-1} \in \left\{q\right\}\right\}}_{=: A_q}
 $$
and therefore
$$
 T\left(A\right) = \bigcup_{q \in \mathbb{Q}}T\left(A_q\right)
 $$
For each $q \in \mathbb{Q}$, $A_q \in \mathcal{B}_{\left[0\infty\right)}$ and $T\left(A_q\right) = A_q - q$, so by attempt #1, $T\left(A_q\right) \in \mathcal{B}_{\left[0,\infty\right)}$ and hence $T\left(A\right) \in \mathcal{B}_{\left[0,\infty\right)}$. Unfortunately, it may not be the case that for every $f \in A$, $f\left(0\right) \in \mathbb{Q}$, in which case this attempt fails.

However, the rationals are dense in $\mathbb{R}$ and in addition, every $f \in \mathbf{C}_{\left[0,\infty\right)}$ is uniquely determined by the values $f$ assumes at the non-negative rationals. Can we take advantage of this additional structure, together with the result of attempt #2, to solve question 1 in the affirmative? (In which case question 2 is automatically resolved as well.)

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set.

Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$.  Then $A \in \mathcal{B}_{\left[0,\infty\right)}$.

How about $T(A)$?  In fact
$$
T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\}
$$
and is not Borel.

added Mar 10
Why is $T(A)$ not Borel?

First, note that $\mathcal{B}_{\left[0,\infty\right)}$ is the same as the Borel sets for the topology of uniform convergence on bounded sets for $\mathbf{C}_{[0,\infty)}$, a Polish space.

Suppose (for purposes of contracidtion) that $T(A)=\{g \in \mathbf{C}\colon g(0)=0 \text{ and } g(1) \in F\}$ is Borel.  Then so is its complement $T(A)^c := \{g \in \mathbf{C}\colon g(0) \ne 0 \text{ or } g(1) \in F^c\}$.  For Polish spaces, the continuous image of a Borel set is an analytic set.  Now $\pi_{01} \colon \mathbf{C}_{[0,\infty)} \to \mathbb R^2$ defined by
$$
\pi_{01}(f) = (f(0),f(1))
$$
is continuous.  So
$$
G_1:= \{(x,y) \in \mathbb R^2 \colon x=0 \text{ and } y \in F\},\qquad
G_2:= \{(x,y) \in \mathbb R^2 \colon x \ne 0 \text{ or } y \in F^c\},
$$
are both analytic sets in $\mathbb R^2$.  But then cross-sections
$$
\{y\colon (0,y) \in G_1\} = F \qquad\text{and}\qquad
\{y\colon (0,y) \in G_2\} = F^c
$$
are both analytic sets in $\mathbb R$, and therefore $F$ is Borel.  This contradiction shows that our assumption that $T(A)$ is Borel is wrong.

Hi I think it is true that the operator $T$ that you defined is such that the image of any measurable set of $\mathcal{B}([0,+\infty))$ is a measurable set or as claimed that $T$ preserves measurability.

The idea is to view things in a "topological" way first. If you "examine" carefully $T$ I think that you would shortly realize that it is a continuous linear application from one Banach spaces to another and that moreover it is surjective, from this, every ingredients are here for the Banach-Schauder theorem to apply unless mistaken.

This proves that the image of every open sets of the space of continuous functions equipped with the sup norm (on every compacts) are open sets of the space at arrival which is naturally embedded in the very same space (you could refine the argument by defining the space of continuous functions that start at 0 but this doesn't really matter).

Now let's reconcile topological and meausre theory, it is proven in Karatzas and Shreve "Brownian Motion and Stochastic Calculus" (may be as an exercise left to the reader I can't remember) that the $\sigma$-algebra defined the way you did by projection restricted to the space of continuous functions is the same $\sigma$-algebra that the Borelian $\sigma$-algebra of the space of continuous functions equipped with the sup norm (on every compacts).

Hence I think we are done because $T$ preserves open sets which is a collection of sets the generated our $\sigma$-algebra $\mathcal{B}([0,+\infty))$, an application of the Monotone Class Theorem should be enough to conclude I think and I leave this part to you.

Best regards


Intersection of n lines Let $n$ lines in the plane be given such that no two of them are parallel and no three of them have a common point. We want to choose the direction on every line so that the following holds: if we go along any line in its direction and put numbers from 1 to $n-1$ on the intersection points then no two equal numbers appear at the same point. For which numbers $n$ is it possible?

My guess is that, we cannot do it iff when $n$ is even. For the case $n$ is even, I think we should find a point $p$ of intersection of two lines which lies in the middle of two lines (for each of them half of the intersection points in one side of p and the others on the other side of p)

My guess is that, we cannot do it iff when n is even.

You could be right! But we'll need a proof ...

  For the case n is even, I think we should find a point p of intersection of two lines which lies in the middle of two lines

No, that won't work as a proof, since it is false. Here is a counterexample:

The midpoint for line 1 is where it crosses with line 6

The midpoint for line 2 is where it crosses with line 6

The midpoint for line 3 is where it crosses with line 6

The midpoint for line 4 is where it crosses with line 9

The midpoint for line 5 is where it crosses with line 2

The midpoint for line 6 is where it crosses with line 4

The midpoint for line 7 is where it crosses with line 3

The midpoint for line 8 is where it crosses with line 3

The midpoint for line 9 is where it crosses with line 3

The midpoint for line 10 is where it crosses with line 6

Or, as a graph, where an arrow from $i$ to $j$ means that the midpoint of line $i$ is where it crosses with line $j$:

So: you see that no two lines share the same midpoint!

In general, any $n$ random lines in the plane will be "almost surely" not crossing them in a common points with three lines.

This is because, $n$ lines have $2n$ variables (i.e. degrees of freedom, DOF), and having a common point shared with three lines imposes a constraint over 1 degree of freedom (for fixing the slope or crossing of just the third line), leaving you with $2n-1$ variables. Hence, the space of solutions of $2n-1$ DOFs  in a space of $2n$ DOFs solution. This shows that the required case is not "so common".

Hence, a lot of simple counterexample can be found trivially. For example if we take the tangents to the segment of the unit circle:

Take a unit circle,
Divide the hemi-circle $(0,\frac\pi2]$ in $\frac{\pi}{2n}$ angle segments, $n$ points,
Draw a tangent $\mathscr{L}_i$ at every division point $i=1...n$,
The lines $\mathscr{L}_i$ are the requested lines.

Proof:

The lines are:
$$
\mathscr{L}_i:\frac{y-\sin(\theta_i)}{x-\cos(\theta_i)}=\tan(\theta_i+\frac\pi2)\\
y=-\frac{x}{\tan(\theta_i)}+\frac{1}{\sin^2(\theta_i)}
$$
which are all with different slope for any $i$ in $(0,\frac\pi2]$.
The crossing are given by:
$$
y_{ij}=-\frac{x_{ij}}{\tan(\theta_i)}+\frac{1}{\sin^2(\theta_i)}=-\frac{x_{ij}}{\tan(\theta_j)}+\frac{1}{\sin^2(\theta_j)}
$$
which are all different for any given pair of $i,j$.
$$
\blacksquare
$$


Why is 11 times the 7th term of a fibonacci series equal to the sum of 10 terms? Why is 11 times the 7th term of a fibonacci series equal to the sum of 10 terms?

I was watching scam-school on youtube the other day and this number trick just astonished me. Can someone please explain why this works?

After a lot of searching, I've been stumbling onto slightly complicated mathematical explanations. An explanation of a simpler nature, one that a child can understand, would be much appreciated.

Also, Can you extend this to find the sum of n terms of a fibonacci type sequence?

As far as I know, it seems to be nothing more than coincidence. Say you have your starting numbers, $a$ and $b$. Your ten terms are
$a,b,a+b,a+2b,2a+3b,3a+5b,5a+8b,8a+13b,13a+21b,21a+34b$

the sum of which is $55a+88b$, which just happens to $11$ times the seventh term in your sequence.

If the first term is x, and the second term is y, then
(1) x
(2) y
(3) x+ y
(4) x + 2y
(5) 2x + 3y
(6) 3x +5y
(7) 5x +8y
(8) 8x + 13y
(9) 13x + 21y
(10) 21x + 34y

The sum of all of those numbers is 55x + 88y, which factors to 11(5x + 8y)
Notice that the 7th term is 5x + 8y, so multiply the 7th term times 11 and you will get the sum of all the numbers in the sequence.


Measure-theoretical isomorphism between interval and square What is an explicit isomorphism between the unit interval $I = [0,1]$ with Lebesgue measure, and its square $I \times I$ with the product measure? Here isomorphism means a measure-theoretic isomorphism, which is one-one outside some set of zero measure.

For $ x \in [0,1]$, let $x = .b_1 b_2 b_3 \ldots$ be its base-2 expansion (the choice in the ambiguous cases doesn't matter, because that's a set of measure 0).  Map this to
$(.b_1 b_3 b_5 \ldots,\ .b_2 b_4 b_6 \ldots) \in [0,1]^2$

See the unit square as made up by hyperbolas $yx=k$ with $k$ between 0 and 1. Consider the rectangle spanned by $(a,0)$ and $(0,b)$. Its NE vertex belongs to the hyperbola $xy=k_0$ and it can be located by its distance from the starting point $(k_0,1)$ obtained by descending along that hyperbola. Normalize such that the distance between $(k_0,1)$ and $(1,k_0)$ is $1-ab$ and call $d/(1-ab)$ the distance from $(k_0,1)$ to the NE vertex.
Send this rectangle to the interval $[d/(1-ab), d/(1-ab)+ab]$.
Does this determine a bijection from the unit square to the unit interval?


Proof of matrix exponential property $e^{\textbf A+\textbf B}=e^{\textbf A}e^{\textbf B}$ if $\textbf A \textbf B=\textbf B \textbf A$ I know how to prove this property with the matrix function $\pmb\Phi (t)=\exp((\textbf A+\textbf B)t)\exp(-\textbf Bt)\exp(-\textbf At)$ and its derivative $\pmb\Phi'(t)$, by proving that $\pmb\Phi'(t)=\textbf 0$ for all $t$.

It is also stated (Exercise 7.3.9, p. 539 of "Matrix Analysis and Applied Linear Algebra") that it could be proven with the function $\textbf F(t)=\exp((\textbf A+\textbf B)t)-\exp(\textbf At)\exp(\textbf Bt)$, proving that its derivative $\textbf F'(t)=\textbf 0$ so $\textbf F(t)$ is constant and since for $t=0$ it is zero so it is zero for all $t$.

I am trying to prove that $\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$ is zero for all $t$ but without success. Is my derivative function correct and if it is so how to prove it is identically zero?

Let's take as a starting point what you have calculated
$\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$

Then by substituting $\textbf F(t)$ to the previous expression we get
\begin{align*}
\textbf F'(t)&=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\
&=(\textbf A+\textbf B)(\textbf F(t)+\exp(\textbf At)\exp(\textbf Bt))-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\
&=(\textbf A+\textbf B)\textbf F(t)
\end{align*}
Now take any vector $v$ and multiply the previous equality
$$(\textbf F(t)v)'= (\textbf A+\textbf B)(\textbf F(t)v)$$
Which implies that
$$\textbf F(t)v =e^{(\textbf A+\textbf B)t}\textbf F(0)v=0,\, \forall t \in \mathbb{R}$$
Therefore
$$F(t)\equiv0$$


Compute Derivative in Polar Coordinates I have the function
$$\phi(r)=-192r^5+480r^4-440r^3$$ with $1/4\leq r\leq 3/4,$ in polar coordinates. So $r=\sqrt{x^2+y^2}.$ I want to compute $\phi'(r)$ and $\phi''(r).$ So, I use the chain rule,
$$\frac{d\phi}{dr}=\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}$$
for the first derivative. And the same technique for the second derivative. I am right?

There is no need to bring in x or y nor domain of $r$.

Just differentiate $\phi$ two times w.r.t. $r.$

I found the solution. Because $r=r(x,y)=\sqrt{x^2+y^2}$ then
$$\phi'(r)=\frac{d\phi}{dr}=\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}$$
and for the second derivative we must use chain rule again, so
$$\phi''(r)=\frac{d^2\phi}{dr^2}=\frac{d}{dr}\left(\frac{d\phi}{dr}\right)=\frac{d}{dr}\left(\frac{d\phi}{dx}\frac{dx}{dr}+\frac{d\phi}{dy}\frac{dy}{dr}\right)=\cdots=$$
$$=\cdots=\phi_x\frac{d^2x}{dr^2}+\phi_y\frac{d^2y}{dr^2}+\phi_{xx}\left(\frac{dx}{dr}\right)^2+2\phi_{xy}\left(\frac{dx}{dr}\right)\left(\frac{dy}{dr}\right)+\phi_{yy}\left(\frac{dy}{dr}\right)^2$$


on the boundary of analytic functions Suppose I have a function $f$ that is analytic on the unit disk $D = \{ z \in \mathbb{C} : |z| < 1 \}$ that is also continuous up to $\bar{D}$. If $f$ is identically zero on some segment of of the boundary (e.g. $\{ e^{it}, 0 \leq t \leq \pi/2 \}$ ), is it then true that $f$ is identically zero on the entire boundary?

I know that analytic functions that are zero at an accumulation point inside the domain of analyticity are identically zero throughout the entire domain, but I don't know what (if anything) can be said if something similar occurs on the boundary of the domain.

The only way the function can be zero on an arc of the circle is for the function to be identically zero. To see this, consider ${\displaystyle g(z) = \prod_{j = 0}^{n-1} f(e^{2\pi i j \over n}z)}$ for a large enough $n$ so that $g(z)$ is identically zero on the boundary of the disk. Then by the Cauchy integral formula, if $r < 1$ then for any $w$ in the interior one has
$$g(w) = {1 \over 2\pi i}\int_{|z| = r}{g(z) \over z - w}\,dz$$
Since $g(z)$ is continuous on the compact set $\bar{D}$, $g(z)$ is bounded, so one can apply the dominated convergence theorem and let $r$ go to $1$. One obtains
$$g(w) = {1 \over 2\pi i}\int_{|z| = 1}{g(z) \over z - w}\,dz$$
$$ = 0$$
Thus $g(z)$ is identically zero. Since the zeroes of a nonzero analytic function are isolated, we conclude that $f(z)$ is identically zero as well.

It is well known that if an analytic function assumes the value zero in some set which admit an accumulation point, is identically zero: as for your question there exist also  a more general result.
Theorem (Privalov). Let $u:\Bbb D \to \Bbb C$ be an analytic function on $\Bbb D$ ($\Bbb D\equiv$ unit disk in $\Bbb C$) which is continuous on $\operatorname{clo}(\Bbb D)$. If $u=0$ in some subset of  positive Lebesgue-measure of the boundary of $\Bbb D$, then $u$ is identically zero.


Probability of having at least one pair by drawing 4 shoes from 12 pairs. There are $12$ pairs of shoes in a cupboard. $4$ are drawn at random. What is the probability that there is at least one pair?

My first attempt: If we chose a pair at first and then draw any two at random from the rest, then there will be at least one pair.
We can choose one pair in ${12}\choose 4$ ways and chose any $2$ from the rest in ${22}\choose 2$ ways.
Therefore, the required probability= $\frac{{12\choose 4} \times {22\choose 2}}{{24\choose 4}}$ = $\frac{6}{23} =\frac{42}{161}$

But the given answer is $\frac{41}{161}$.

Another attempt: Each of the 4 shoes we choose, will come from one of the pairs. We can choose the four pairs in ${12\choose 4}$ ways and can select a shoe from each of the pairs in $2$ ways so that no pair is obtained. Therefore, required probability =$1-$ $\frac{{12\choose 4} \times 2^4}{{24\choose 4}}$ = $\frac{41}{161}$

What is wrong with the first attempt?

Calculate $1$ minus the probability of the complementary event:

The number of ways to choose $4$ out of $24$ shoes is:

Choose the $1$st shoe out of $24$ shoes
Choose the $2$nd shoe out of $23$ shoes
Choose the $3$rd shoe out of $22$ shoes
Choose the $4$th shoe out of $21$ shoes

The number of ways to choose $4$ out of $24$ shoes with no pairs is:

Choose the $1$st shoe out of $24$ shoes
Choose the $2$nd shoe out of $22$ shoes
Choose the $3$rd shoe out of $20$ shoes
Choose the $4$th shoe out of $18$ shoes

So the probability of choosing $4$ out of $24$ shoes with at least one pair is:

$$1-\frac{24\cdot22\cdot20\cdot18}{24\cdot23\cdot22\cdot21}$$

Please note that I've essentially taken into account the order of the shoes.

If I chose not to take it into account, then I would need to divide each result by $4!$.

But since this factor appears in both the numerator and the denominator, I can ignore it.

To select a pair, the order is irrelevant.
1. Select one shoe
2. Select one shoe - chance this is a pair of first = 1/23 = .04
3. Select one shoe - chance this is a pair of first = 1/22 or chance this is pair of second = 1/21 = 1/22+1/21 = .09
4. Select one shoe - chance this is a pair of first = 1/21 or of second = 1/20, or of third = 1/19 = 1/21+1/20+1/19 = .15
After four selections, probability is .04+.09+.15=.28


I am looking for solution of the following question? If

$x-\sqrt{\frac{3}{x}} =10$,

what is the result of the following expression?

$x-3\sqrt{x} =? $

thanks

The answer is 1.

Based on OP's comment copied above, the posted problem is most likely mistyped. The following answers the question assuming that the given equation was, instead:

  $$\;x-\frac{3}{\sqrt{x}} =10\,$$

Let $t=\sqrt{x} \gt 0\,$, then the equation becomes $t^3-10t-3=0\,$. Using the rational root theorem it is easy to find the root $t=-3\,$, then the LHS factors as:

$$(t+3)(t^2-3t-1)=0 \quad \iff \quad (\sqrt{x}+3)(x-3\sqrt{x}-1)=0 $$

The first factor is strictly positive $\sqrt{x}+3 \gt 0\,$, which leaves:

$$ x-3\sqrt{x}-1 = 0 \quad \iff \quad x-3\sqrt{x}=1$$

First, lets get everything in a nice order, so $\sqrt{3\over x} = x-10$

Now, let's square both sides to solve for x fully.

${3\over x} = (x-10)^2$ so this means that $3=x(x-10)^2$.

Solve for x here and then just plug it into the next equation.

This is of course based on the assumption these two are a system of equations, you have it tagged as saying system of equations so I went with this.


Fibonacci numbers of the form $5x^2+7$ Numerically I find the positive integer solution of the equation $F_n=5x^2+7$, where $F_n$ denotes the $n^\text{th}$ Fibonacci number, as $(n,x)=(16,14)$ and I guess that the only positive solution of it is $(n,x)=(16,14)$.

How can I solve this equation?

This isn't a full answer, but might give you enough information about methods of proving that this equation probably has only the one solution you have already.

In this online version of a paper published in the Fibonacci Quarterly in 1964, Cohn uses completely elementary methods such as congruences and properties of quadratic residues to show that there are no square Fibonacci numbers apart from 0, 1 and 144. He uses a long list of well-known identities connecting Fibonacci and Lucas numbers, which will almost certainly be  useful.

I suspect that these methods could probably be used to show that your equation has  only the one solution, as this sort of (basically) exponential diophantine equation usually has a finite number of solutions, which are almost always fairly small, unless you have been extremely unlucky.

the fibonacci sequence grows exponentially ($F_{n+1}/F_n\to\phi$) so you would only need to check finitely many instances.


Find all functions $f(x+y)=f(x^{2}+y^{2})$ for positive $x,y$ Find all functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that for any $x,y\in \mathbb{R}^{+}$ the following holds:
$$f(x+y)=f(x^{2}+y^{2}).$$

If $x+y=a$ then $x^2+y^2$ is onto the range $\left[\frac{a^2}{2},a^2\right)$. So $f$ must be constant on any interval $\left[b,2b\right)$ with $b\in\mathbb R^+$.

Now define $b_n=2^{n/2}$ for $n\in\mathbb Z$. Then show that $\bigcup_n [b_n,2b_n)=\mathbb R^+$ and that $b_{n+1}\in(b_n,2b_n)$. This means that $f$ must be constant on all of $\mathbb R^+$.

For any positives $a,b$ we have:
\begin{aligned} f(a) &=f\left(\frac{a+b}{2}+\frac{a-b}{2}\right) \\
 &=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{a-b}{2}\big)^2\right) \\
 &=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{b-a}{2}\big)^2\right) \\
 &=f\left(\frac{a+b}{2}+\frac{b-a}{2}\right) \\
 &=f(b)\end{aligned}


prove that sequence $\frac{n^2+1}{n+4}$ diverges Can somebody help me prove this sequence?

I've tried using Comparison Theorem and ended up with

$a_n = \frac{n^2+1}{n+4} < \frac{n^2}n = n > N = M$.

So I choose $N = M$.

But I don't think this is correct.

Hint:

$$\frac{n^2+1}{n+4} \geq \frac{n^2+1}{n+n} > \frac{n^2}{2n}=\frac{n}{2}.$$
The first inequality holds for $n \geq 4$. Also we know the behavior of $\left\{\frac{n}{2}\right\}$.

You could use L'Hôpital's rule:   $\lim_{n\to \infty}\dfrac {n^2+1}{n+4}\stackrel{L'h}=\lim_{n\to\infty}\dfrac {2n}1=\infty$.


Proving that $\lim_{x\rightarrow a}f(x) = \lim_{h\rightarrow 0} f(a+h) $ The Question:

Verify that if $\lim_{x\to a}f(x)$ exists, then $\lim_{x\to a}f(x)=\lim_{h\to 0}f(a+h)$.

My plan is to use the theorem showing that the limit of a sum is the sum of the limits.  By that theorem, $\lim_{h\to 0}f(a+h)=\lim_{h\to 0}f(a)+\lim_{h\to 0}f(h)=f(a)+0=f(a)=\lim_{x\to a}f(x)$.

Does this proof hold?  Please let me know if there is anything that I need to clarify.

The limit $\lim_{x\to a}f(x)$ exists and is equal to $b$: $$\lim_{x\to a}f(x)=b$$ if and only if for each $\varepsilon\in\mathbb R^+$ exists $\delta\in\mathbb R^+$ such that $$\forall x,0<|x-a|<\delta\implies|f(x)-b|<\varepsilon.$$

Hint:
Apply analogous definition for $\lim_{h\to0}f(a+h)$ and tell if you can derive one from the other.

Well first we have to prove that differentiability implies continuity. (We have to prove that $\lim_{x\to a}f(x)$ exists.)

Thus, suppose we have a differentiable function $f(x)$ such that $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$. This equals $f'(a)$.

Now, we multiply each side by $h$. Thus, we get $\lim_{h\to 0}hf'(a)$. (What is the answer here?)

Thus, we get $\lim_{h\to 0}f(a+h) \to f(a)$. (Why?)

Thus, we set $x = a + h$. Thus, $\lim_{x-a \to 0}f(x) \to f(a)$.

Thus, we have proven that such a $\lim_{x\to a}f(x) = f(a)$ exists.

The answer to your question:

   To answer your question, go backwards!


how to prove that: $|X| = 2^{\aleph_{0}}$, given $X = \{ A\in P(\mathbb{N}) \vert \ |A^{c}| = \aleph_{0} \}$ Let $X = \{ A\in P(\mathbb{N}) | \ |A^{c}| = \aleph_{0} \}$

Prove\ Disprove that: $|X| = \aleph_{0}$

My attempt:

I'd like to disprove that: $|X|=\aleph_{0}$, By proving that $|X|=2^{\aleph_{0}}$.

For every $A\in X\to A\in P(\mathbb{N})$,then $X\subseteq P(\mathbb{N})\to|X|\leq|P(\mathbb{N})|=2^{\aleph_{0}}$.

Now I'd like to prove that there exists a set, $K$,

such that $|K|=2^{\aleph_{0}}$, and $K \subseteq X$, which will finish the proof.

Suppose by contradiction that there exists no set $K$ as described,

such that: $K\subseteq X$.

Now, let $A=2\mathbb{N}$, then $A^{c}=\mathbb{N}\setminus2\mathbb{N}=\mathbb{N}_{odd}$,

and $|A^{c}|=|\mathbb{N}_{odd}|=\aleph_{0}\to A\in X$.

Notice that for every set $i$, $\alpha_{i}\subseteq2\mathbb{N}:$ $\mathbb{N}_{odd}\subseteq\alpha_{i}^{c}\subseteq\mathbb{N}\to|\alpha_{i}^{c}|=\aleph_{0}$,

therefore for every $i$ and for every $\alpha_{i}$ as descirbed, $\alpha_{i}\in X $.

the collection of all $\alpha_{i}$ exists: $\{\alpha_{i}\vert\ \forall i\in\mathbb{N}:\alpha_{i}\subseteq2\mathbb{N}\}$,hence

$P(2\mathbb{N})=\{\alpha_{i}\vert\ \forall i\in\mathbb{N}:\alpha_{i}\subseteq2\mathbb{N}\}$

hence $P(2\mathbb{N})\subseteq X\Longrightarrow \ 2^{\aleph_{0}}=|P(2\mathbb{N})|\leq|X|$

and this contradicts the assumption,

that there exists no $K\subseteq X:|K|=2^{\aleph_{0}}$.

by Cantor Bernstein we get that $|X|=2^{\aleph_{0}}$.

I'd like to disprove that: $|X|=\aleph_{0}$, By proving that $|X|=2^{\aleph_{0}}$.

For every $A\in X\to A\in P(\mathbb{N})$,then $X\subseteq P(\mathbb{N})\to|X|\leq|P(\mathbb{N})|=2^{\aleph_{0}}$.

Now I'd like to prove that there exists a set, $K$,

such that $|K|=2^{\aleph_{0}}$, and $K \subseteq X$, which will finish the proof.

Suppose by contradiction that there exists no set $K$ as described,

such that: $K\subseteq X$.

Now, let $A=2\mathbb{N}$, then $A^{c}=\mathbb{N}\setminus2\mathbb{N}=\mathbb{N}_{odd}$,

and $|A^{c}|=|\mathbb{N}_{odd}|=\aleph_{0}\to A\in X$.

Notice that for every set $i$, $\alpha_{i}\subseteq2\mathbb{N}:$ $\mathbb{N}_{odd}\subseteq\alpha_{i}^{c}\subseteq\mathbb{N}\to|\alpha_{i}^{c}|=\aleph_{0}$,

therefore for every $i$ and for every $\alpha_{i}$ as descirbed, $\alpha_{i}\in X $.

the collection of all $\alpha_{i}$ exists: $\{\alpha \vert\ \alpha \subseteq2\mathbb{N}\}$,hence

$P(2\mathbb{N})=\{\alpha\vert\ \alpha \subseteq2\mathbb{N}\}$

hence $P(2\mathbb{N})\subseteq X\to \ 2^{\aleph_{0}}=|P(2\mathbb{N})|\leq|X|$

and this contradicts the assumption,

that there exists no $K\subseteq X:|K|=2^{\aleph_{0}}$.

by Cantor Bernstein we get that $|X|=2^{\aleph_{0}}$.

therefore for every $\alpha \subseteq P(2\mathbb{N}),  \alpha \subseteq K$,

hence $P(2\mathbb{N}) \subseteq K \to  |P(2\mathbb{N})| \leq |K|$, by using Cantor Bernstein theorem we conclude that $|K| = \aleph$.


Volume of a wine barrel This is a famous calculus problem and is stated like this

  Given a barrel with height $h$, and a small radius of $a$ and
  large radius of $b$. Calculate the volume of the barrel
  given that the sides are parabolic.

Now I seem to have solved the problem incorrectly because here it seems 2
that the volume should be

$ \displaystyle \hspace{1cm}
   V(a,b,h) = \frac{h\pi}{3}\left(2b^2 + a^2\right)\,.
$

Below is my attempt. As in the picture I view the barrel from the side, and try to find a formula for the parabola. So i solve

$ \displaystyle \hspace{1cm}
    f(x) := A x^2 + B x + C
$

given $f(0) = f(h) = a/2$ and $f(h/2) = b/2$. This yields

$ \displaystyle \hspace{1cm}
f(x) =  \frac{2(a-b)}{h^2} \cdot  x^2 -
        \frac{2(a-b)}{h} \cdot x +
       \frac{a}{2}
$

Using the shell method integrating now gives the volume as

$ \displaystyle \hspace{1cm}
    V(a,b,h) := \pi \int_0^h \bigl[f(x)\bigr]^2\,\mathrm{d}x
              = \frac{\pi}{60} \cdot h (a+2b)^2 + \frac{\pi}{30} \cdot h(a^2+b^2)
$

Alas according to the formula above this seems incorrect! Where is my mistake?

Let $k=h/2$, and put the origin in the middle, where symmetry asks it to be.

Then the equation of the upper parabola is
$$y=b-\frac{b-a}{k^2}x^2.$$
The integral of $\pi y^2\,dx$ from $0$ to $k$ is
$$\pi k\left(b^2-\frac{2}{3}(b-a)b+\frac{1}{5}(b-a)^2\right).$$
This simplifies to
$$\frac{\pi k}{15}(3a^2+4ab+8b^2)$$
Replace $k$ by $h/2$ and multiply by $2$.

Volume of a barrel

V = 0.5 * πQuadrat * (π/18 + 1/6) * (r1Quadrat + r2Quadrat) * h parabolic bend
V = 0.5 * πQuadrat * (π/17 + 1/6) * (r1Quadrat + r2Quadrat) * h elliptic bend

Best regards
Hans-Jürgen Gläsel


Understanding some proofs-without-words for sums of consecutive numbers, consecutive squares, consecutive odd numbers, and consecutive cubes I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.

For the second and third images, I am completely lost.

For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?

How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)

The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+\dots +n^2)=\frac{n(n+1)}{2}\cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3\cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=\frac{6\cdot 5}{2}$ (see the first formula) and base $2\cdot 5+1$.

The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+\dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7\cdot 6$ is
$$4\cdot 1^2+8\cdot 2^2+12\cdot 3^2+16\cdot 4^2+20\cdot 5^2+24\cdot 6^2\\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$

For a diagram of the sum of consecutive odd nos, use the diagram in your sum of squares that shows a red L shape with an inner yellow L shape, with an inner blue L shape with an inner green L shape with a single green square.


Why is the damping force larger than the restoring force in this model? I'm reading the book "Differential equations with boundary-value problems" by D.Zill and I cannot understand the following:

In the Free Damped Motion Model,

we have a flexible spring suspended vertically from a rigid support.  We attach an object of certain mass $m$  to the spring so this is stretched $s$ units (in this situation, the object will be in an equilibrium position).   Assume that there is a resisting force due to the surrounding medium (as  a viscous medium, for example).  If $x(t)$ is the position of the object after $t$ time units, we have the ODE $$mx''=mg-\underset{Hooke\; Law}{\underbrace{k(x+s)}}-\underset{damping\; force}{\underbrace{\beta x'}},$$ where $g$ is the gravity acceleration , $k$ is the constant given by the Hooke Law (restoring force) and $\beta>0$ is the damping -constant given by the damping force.

The ODE has as solutions for its auxiliary equation:  $-\lambda\pm \sqrt{\lambda^2-\omega^2}$, where $\lambda=\beta/(2m), \omega^2=k/m$.

We have three cases to analyze, but it is one that I cannot get:

If $\lambda^2-\omega^2>0$, then * the system is said to be overdamped
because the damping coefficient b is large when compared to the spring constant k.*

Why does the author state that $b$ is large when compared to $k$?  I cannot get why he affirms this (which I guess he does in order to conclude that the damping force is large compared to the restoring force).  I know that the graph of the solutions helps to visualize the model but I want to understand this model just by looking at the equation and its variable involved (basically, a qualitative analysis of the problem).

Starting with the inequality $\lambda^2-\omega^2>0$, plug in the definitions of $\lambda$ and $\omega$ in terms of $\beta$, $k$, and $m$.  Then simplify the resulting inequality to get all the $\beta$'s on one side and all the $k$'s on the other.


Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$

My attempt is as follows:-

$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$

Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$

$$f(x)=\sqrt{13}z$$

Let's find the range of $z^2$

$$z^2=\dfrac{4}{13}\cdot\sin^2x+\dfrac{9}{13}\cdot\cos^2x-\dfrac{6}
{13}|\sin2x|$$
$$z^2=1-\dfrac{6}{13}\cdot|\sin2x|$$

$$z^2=1-\dfrac{6}{13}\cdot[0,1]$$
$$z^2=1-\left[0,\dfrac{6}{13}\right]$$
$$z^2=\left[\dfrac{7}{13},1\right]$$

As $0$ is not there in the range of $z^2$, so we can say $z\in \left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]$

Hence $y \in \sqrt{13}\left(\left[-1,-\dfrac{\sqrt{7}}{\sqrt{13}}\right]\cup \left[\dfrac{\sqrt{7}}{\sqrt{13}},1\right]\right)$

$$y\in[-\sqrt{13},-\sqrt{7}]\cup[\sqrt{7},\sqrt{13}]$$

But actual answer is $[-3,2]$

For $0\le x\le\dfrac\pi2,$

$$f(x)=\sqrt{2^2+3^2}\sin\left(x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\right)$$

Now $0\le x\le\dfrac\pi2\implies-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}$

Again,
$\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}=\arccos\dfrac3{\sqrt{2^2+3^2}}=\arcsin\dfrac2{\sqrt{13}}$

As $\sin$ is increasing in $\left[-\dfrac\pi2,\dfrac\pi2\right]$

$$\implies-\dfrac3{\sqrt{2^2+3^2}}\le\sin\left(x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\right)\le\dfrac2{\sqrt{13}}$$

Observe that $$f(\pi+x)=f(x)=f(\pi-x)$$

which cover $\left[\pi,\pi+\dfrac\pi2\right]$ and $\left[\dfrac\pi2,\pi\right]$

Your domain is 0 to $\frac{\pi}{2}$. That's where sine and cosine are positive. So at those points you are limited to the range of $[-3,2]$.

Unless you are given additional instructions, you are way over complicating it.


Intuitive explanation of $(a^b)^c = a^{bc}$ What is an intuitive explanation for the rule that $(a^b)^c = a^{bc}$. I'm trying to wrap my head around it, but I can't really do it.

When $b$ and $c$ are positive integers, we can make the following argument:
$$(a^b)^c=\underbrace{(a^b)\times(a^b)\times\cdots\times(a^b)}_{c\text{ times}}=$$
$$\underbrace{\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)\times\cdots\times \left(\underbrace{a\times a\times\cdots \times a}_{b\text{ times}}\right)}_{c\text{ times}}=$$
$$\underbrace{a\times a\times \cdots\times a}_{bc\text{ times}}=a^{bc}$$

It’s not a rule in the sense of a theorem, but of a definition. (In early mathematics training, even into calculus, there is a tendency, for administrative/teaching convenience, to conflate theorems and definitions, calling all such things “rules”.)
It is simply a matter of the notation being chosen such that it is what we might call “fortuitous”. Let’s consider three other such instances.

x – y = x + (-y), as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitous way in which x – y is DEFINED.
x/y = x(1/y), for y not equal to 0, as you can verify from numerical examples, but it is really a matter of definition: that is the fortuitious way in which x/y is DEFINED.
a raised to the power b = exp(b*log(a)), for positive a, as you can verify from numerical examples, but it is really a matter of definition: based on extensive numerical experience, that is the fortuitous way in which mathematicians decided to DEFINE a to the power b.

Now to your question. For positive a and arbitrary b and c (or arbitrary non-zero a and integer b and c), the expression a to the power of (bc) is DEFINED, for notational convenience, to be (a to the power of b) to the power of c. The fact that it is only a notational convenience can be seen if a is negative and not both b and c are integer . Presumably you would agree with the statement that in any (valid) mathematical expression, we should be able to reduce whatever fractions occur to lowest terms without affecting the value of the expression. But that does not happen if a is negative, say, -1, b is 2, and c = 1/4: ((-1 raised to the power 2))raised to the power (1/4) is 1, and this should equal (-1 raised to the (2/4) power), which equals (-1 raised to the 1/2 power), which is the imaginary unit, not unity.

So, the short answer to your question is: It is simply a fortuitous definition, based on extensive numerical experience. Your question arose simply from the fact that you were lacking a sufficient dose of that experience. It’s not enough to just think about it: you have to grok it.


The triangle must be obtuse if two of its medians are perpendicular? In problem #577679, the question says if two of the medians of triangle ABC are perpendicular, then .... ($5a^2 = b^2 + c^2$, the result).

In the course of solving it, I drew the following picture (slightly different to that question).

I found that angle ABC 'must be obtuse' before I can get a satisfactory (hand) drawing. (I did not use any tool from Wolfram.)

My statement is "Prove or disprove the triangle must be obtuse under the mentioned given".

I tried to use the cosine law to show that $cos ABC$ (or $a^2 + c^2 - b^2$) is negative, but was not successful.

Any idea?

(This old question came up again through an edit by @JonMarkPerry.)

Let's find all acute triangles with two perpendicular medians. We may assume the following situation (see the figure):
$$A=(0,0), \quad B=(c,0),\quad C=(u,v), \quad M_c=\left({c\over2},0\right), \quad M_b=\left({u\over2},{v\over2}\right)$$
with $v>0$. We want that $$BM_b\perp CM_c\ ,\tag{1}$$ which means that
$$\overrightarrow{BM_b}\cdot\overrightarrow{CM_c}=\left({u\over2}-c,{v\over2}\right)\cdot\left({c\over2}-u,-v\right)=0$$
or
$$\left(u-{5\over4}c\right)^2+ v^2={9\over16}c^2\ .$$
It follows that the points $C$ satisfying $(1)$ are lying on the circle $K$ with center $\bigl({5\over4}c,0\bigr)$ and radius ${3\over4}c$. The triangle  $ABC$ is acute iff $C$ lies outside of the Thales circle over $AB$ and to the left of the line $u=c$. This part of $K$ is drawn in bold red in the figure.

Going down the route of using cosine rule, if all the angles are less than $90$ then $c^2+a^2>b^2, c^2+b^2>a^2, b^2+a^2>c^2$. Additionally, $c^2+b^2=5a^2$. Hence one gets the inequalities $(6b^2)/5>(4c^2)/5/>(8b^2)/15$ and a similar one where $b$ and $c$ are interchanged. As an example, with $b=1, c=0.9, a=\frac{\sqrt{905}}{50}$, all the inequalities, including the triangle inequality are satisfied, providing an example of medians perpendicular to each other without being obtuse.


Given a (Deterministic Finite Automata) DFA that recognizes a language $L$, show how to construct another DFA that recognizes the language $\max(L)$. If $L$ is any language, then
$$\max(L)= \{w \mid \text{$w$ is in $L$ and there is no non-empty string $x$ such that $wx$ is in $L$} \}.$$

I am really confused about what this problem is asking and I would greatly appreciate some light.

It is easier to use the definition of a DFA in which the transition function is allowed to be partial. Let us say that a state $q$ is accessible if there is a path from the initial state to $q$ and coaccessible if there is a path from $q$ to some final state. A DFA is trim if every state is both accessible and coaccessible. It is easy to see that every DFA is equivalent to a trim DFA (just get rid of the non-accessible and non-coaccessible states).

Now, let start with a trim DFA ${\cal A} = (Q, A, i, ., F)$ recognizing $L$ and, as suggested by copper.hat, consider the automaton ${\cal A}' = (Q, A, i, ., F')$
where
$$
F' = \{q \in F \mid \text{there is at least one transition $q \xrightarrow{a} q'$ for some letter $a$\}}
$$
Then ${\cal A}'$ recognizes $\max(L)$.

This new language is composed of the "maximal elements" of $L$. For example, if $L = \{ab, ba, aba\}$, then max$(L) = \{ba, aba\}$. It doesn't include $ab$ since that can be extended to $aba$ and still be $L$.

So how can we construct a DFA recognizing this? Since $L$ is regular, it has a DFA $D = (Q, \Sigma, \delta, q_0, F)$ recognizing it. We should try to use this DFA to make a new one for max$(L)$, $D' = (Q', \Sigma, \delta', q_0', F')$. Given max$(L) \subseteq L$, it seems like a good idea to make $Q'$ the Cartesian product of $Q$ and another set. This will allow us to basically run two DFAs in parallel, and check in the first DFA if $w \in L$, and in the second that there is no way to extend $w$ to any $wx \in L$.

It may seem difficult to think of the second half of the tuple, but consider this: is there a way to cleverly define $F'$ using a simple second half, to make $F'$ do most of the work of "figuring out" if there is $wx \in L$? It may be useful to write out an actual DFA or two and see how you yourself would determine max($L$), and see if you can define an $F'$ that suffices. Remember that $Q$ is finite, so you (or a set-builder...) could do a graph search over it to find paths to accepting states. Once you have $F'$ and $Q'$, then $q_0'$ and $\delta'$ are straightforward. You could even drop the tuple entirely, but I find it easier to think about for this category of problem.


Divisibility of integer numbers If we have two integers $a$ and $b$ such that $a = \frac{5b}{6}$, is $a$ divisible by $5$?

If so, why is that?

I don't see it.

No prime factorisation is really required here: we have $5b=6a$,so $5$ divides $6a$. Now $5$ is prime, hence  by Euclid's lemma, it must divide one of the factors. As it does not divide $6$, it must divide $a$.

If $a$ is an integer then $b = 6k, k\in \mathbb{Z}$ so $$a = 5k \Rightarrow a \equiv 0 \pmod{5}$$


Word Problem - what percent of the women who apply are hired? At a certain factory, twice as many men as women apply for work. If 5% of the people who apply are hired and 3% of the men who apply are hired, what percent of the women who apply are hired?

Two thirds are men of which $3\%$ apply and hired, that makes $2\%$ of all.  As $5\%$ are hired this leaves $3\%$ of all which are women and hired.  Hence the percentage of women getting hired is $\frac{3\%}{1/3}=9\%$.

Work through these:

Total number applied:  $x$
Number of men applied: $m(x)$
Number of women applied:  $w(x)$
Total number hired:  $h(x)$
Total number of women hired:  $h_w(x)$
Total number of men hired:  $h_m(x)$


regularity of a pde after change of variables For pde $$u_t=x^2u_{xx},\; x\in [0,\infty) ,\; t\in (0,T], \; u(0,x)=u_0(x)$$ I can do change of variables as $y=ln(x)$ and arrive at another pde: $$v_t=v_{yy}-v_y,\; y \in (-\infty, +\infty),\; t\in (0,T],\; v_0(y)=u_0(e^y)$$ I can show the existence and uniqueness of the transformed pde and therefore imply the existence and uniqueness of the original pde. Does the results of the regularity transfer as well? My new equation might be infinitely smooth but I guess the statement of the smoothness would depend on change of variables via chain rule? Is that correct or there are some regularity that I can obtain for the original equation for free from the transformed equation?

Since $u(x,t) = v(\log x, t)$ for all $x, \ t$ and $v$ is $C^\infty$ smooth on $\mathbb{R} \times (0,\infty)$, so is $u$ on $(0,\infty) \times (0, \infty)$. However, there might be some problems for $x = 0, \, t > 0$ (where you can't prescribe data) and very likely at $x = t = 0$.