Any linear transformation between two finite dimensional topological vector spaces is continuous. Theorem
Any linear transformation $T$ from one finite dimensional topological vector space $V$ into another finite dimensional topological vector $W$
space is necessarily continuous.
Unfortunately I don't be able to prove the statement so could someone prove it? Then if the statement is generally false is it false if $V=\Bbb R^m$ and $W=\Bbb R^m$ too? So could someone help me, please?

Let $T:V\longmapsto W$ a linear transformation, and $\mathcal{B}=\{e_1,\cdots,e_n\}$ a basis of V.
In finite dimension all norms are equivalents, let $||x||=\sum_{i=1}^n\vert x_i\vert$, then $$||f(x)||=||f(x_1e_1+\cdots+x_ne_n)|| \leqslant |x_1|||f(e_1)||+...+|x_n||f(e_n)||$$
Let $M=\underset{i\in\{1,...,n\}}{sup}||f(e_i)||$, then $$  |x_1|||f(e_1)||+...+|x_n||f(e_n)|| \leqslant M(|x_1|+...+|x_n|)=M||x|| $$
$\implies $ $$||f(x)|| \leqslant  M||x||$$


What are the morphisms and objects we must assume that are actually there? I'm reading Awodey's: Category Theory. I harbor a little confusion: When we speak about a category, say: $\mathbf{Set}$. In the book, usually he talks about this category but there is no notion of quantity of sets. How many sets are actually there?

Another thing that is getting me confused is this: Suppose that I have this category with $n$ sets, I must have the arrows, the composition, the identity, etc. But for these sets, should I assume that functions can be constructed from one set to every other set? That is: For a category with $3$ objects: $A,B,C$, should I assume that I have the following functions?

$$f_1:A\to B\\f_2:A\to C\\f_3:B\to C\\f_4:C\to B\\f_5:C\to A\\f_6:B\to A\\$$

I'm not sure if it's about what I have or what could be made.

First, "Set" is one specific category. Its objects are all the sets and its arrows are all the functions between sets.

In the second paragraph, you suppose you have a category with n sets. To define the category you have to say specifically which n sets the objects are.  You then have to say what the arrows are (they will be set functions in this case). Your specification of what the arrows are must satisfy the axioms for a category: (1) If a set S is an object, the identity function on S must be an arrow.  (2) If you have an arrow f:S\to T and an arrow g:T\to U then you must specify that the composite g\o f is also in the category.  If you do those things, then associativity (in this case) will be automatically satisfied, since composition of set functions is associative. Once you have done those things you have defined a category.

In your second paragraph you mention functions f_1:A to B and so on, but you didn't say what the functions are. Also, you mentioned f_1:A to B and f_3:B to C but you didn't mention any function from A to C, which is required by the axiom that says composites exist.

Also, you ask, "Can I assume that arrows can be constructed from one set to every other set?" That is wrong in two ways: (1) A category with two objects S and T is not required to have an arrow from S to T. (2) You can't "assume" there are arrows, you have to say specifically what they are.

How many sets are actually there?

The powerset of all sets P(SET) in the reals is infinite. Hence, from an infinite number of real numbers you can create an infinite number of sets.  This is not true of all powersets P(_), but true of SET.

  Suppose that I have this category with nn sets, I must have the arrows, the composition, the identity, etc. But for these sets, should I assume that functions can be constructed from one set to every other set?

P(n), the power set of all sets created up to "n" will give you all possible combinations.  Remember, SET doesn't have arrows (no structure beyond associativity and commutivity of elements) except from powerset to powerset.

  That is: For a category with 3 objects: A,B,CA,B,C, should I assume that I have the following functions?

Absolutely not; you should never "assume" any functions in CT. The only categorical assumption you are allowed of sorts is that if you assume you are dealing with a category, then you can assume that composition and associativity apply.

Instead, the powerset of 3 objects is:

$$P(3) = (a,b,c), (b,c,a), (c,a,b), (b,a,c), (a,c,b), (c,b,a), (a,c,b), (c,b,a), (b,a,c)$$

Thus the only arrows possible are those that take you from one set to the other:
eg. between (a,b,c) and (b,c,a) you have the following:

$$ f1:a \rightarrow b$$

$$f2: b\rightarrow c$$

$$f3: c\rightarrow a $$


Are any constants (from analysis) rational? There seems to me to be a meta-principle of analysis that all the "interesting" mathematical constants - $\pi$, $e$, $\gamma$, Bessel function zeros, etc. - are irrational (even transcendental).  I'm wondering if there are any counterexamples to this principle - are there any "interesting" constants arising in analysis that are rational?

What constitutes an interesting mathematical constant is of course murky.  I will propose the following guidelines to make the question somewhat more well-defined:

Small integers are excluded.  (These come up all the time for algebraic and combinatorial reasons.)  Common fractions like 3/2 should probably be excluded on the same grounds.
To be eligible, a constant should be "natural" in the sense that its definition does not depend on arbitrary choices for some other parameters.
I'm restricting this question to constants of analysis because some other branches of math, like group theory and combinatorics, have an abundance of large integer constants (e.g. order of the Monster Group).  To qualify as a constant of analysis, it should not be derivable from purely algebraic or combinatorial principles.

You might look here to see which of the listed examples relate to calculus.

Edit: In the link above, $\frac{32}{27}$ related to the closure of the set of all real zeros of all chromatic polynomials of graphs seems a good candidate to me.

There exists interesting natural (hence rational) constants in analysis that fulfill the conditions stated above.

Proof

Let $S$ be the natural $\lceil s\rceil$. Since $s$ is small, $S$ is also small. We can all agree that the number of interesting theorems in analysis is not small, therefore exists a $K>S$ with (analysis theorem $K$ is interesting) and $K$ not small. We all know that Gödel numbers are interesting. If mathematical theorems are ordered by their Gödel numbers and enumerated, we know that (Gödel number of theorem $n$) $\geq$ $n$. It follows that (Gödel number of analysis theorem $k$) $\geq$ $k$. Hence we know that (Gödel number of analysis theorem $K$) is interesting and not small, furthermore strongly analysis-related and rational. Q.E.D.


Substituting an even function for an odd function in an integral In a book I am reading it says while evaluating an integral it says you cannot substitute an odd function for an even function of another variable ( for example substituting sinx with cosu and integrating with respect to u) if the domain of integration of original integral contains 0.

Why is this so? Any help would be appreciated.

Hint:

For: $0<a<b$ we have:
$$
\int_{-a}^b f(x)dx=\int_{-a}^a f(x)dx+\int_{a}^b f(x)dx
$$
If $f(-x)=-f(x)$ is odd
$$
\int_{-a}^a f(x)dx=\int_{-a}^0 f(x)dx +\int_{0}^a f(x)dx=\int_{0}^a f(-x)dx+\int_{0}^a f(x)dx=0
$$

but this is no true (in general) if $f(x)$ is not an odd function.

I think the paper mentions the reason as well:

  ... because
  in that case a substitution by an even function would not work since
  an even function cannot assume both positive and negative values.


Why is "the set of all sets" a paradox, in Layman's terms? I've heard of some other paradoxes involving sets (i.e., "the set of all sets that do not contain themselves") and I understand how paradoxes arise from them.  But this one I do not understand.
Why is "the set of all sets" a paradox?  It seems like it would be fine, to me.  There is nothing paradoxical about a set containing itself.
Is it something that arises from the "rules of sets" that are involved in more rigorous set theory?

Let $|S|$ be the cardinality of $S$. We know that $|S| < |2^S|$, which can be proven with generalized Cantor's diagonal argument.

Theorem

  The set of all sets does not exist.

Proof

  Let $S$ be the set of all sets, then $|S| < |2^S|$, but $2^S$ is a subset of $S$, because every set in $2^S$ is in $S$. Therefore $|2^S| \leq |S|$. A contradiction. Therefore the set of all sets does not exist.

I think this paradoxon is not so complicated and can be explained simply.

If you put all existing sets S(1) to S(k) together in a set S(all), you create a new set S(k+1)={S(1),...,S(k)} and this set has to be in the set of all sets as well (because it should contain all sets). Thus, you put the new set S(k+1) in the set of all sets S(all). But this in turn creates a new set S(k+2)={S(1),...,S(k+1)} wich has to be inserted in the set of all sets as well. Putting S(k+2) in S(all), however, creates a new set and so on and so on.

In conclusion, everytime you put the new set into the set of all sets you create a new set. As a consequence, you can never build a set S(all) that contains all sets without creating a set which is not contained in S(all).


If each vector in a set $S_1$ is in the span of a set $S_2$, and each vector in $S_2$ is in the span of $S_1$, then they have the same span? Can someone expand on this? My book says just this one line but I'd like a bit of a longer explanation so I can understand why this is the case.

If $S_1\subseteq span(S_2)$ then it follows that $span(S_1)\subseteq span(S_2)$

To see why this is, recognize that any element in $span(S_1)$ can be written as a linear combination of elements in $S_1$, each of which since is an element of $span(S_2)$ can be written as a linear combination of things in $S_2$.

That is to say, any element in $span(S_1)$ can be written as a linear combination of linear combinations of things in $S_2$.  Thus, any element in $span(S_1)$ can be written as a linear combination of things in $S_2$ and is therefore an element of $span(S_2)$.

This proves that $span(S_1)\subseteq span(S_2)$.  The same argument can be applied in the other direction showing $span(S_2)\subseteq span(S_1)$ thus showing that the spans are equal.

Each vector in S1 is in the span of S2 so this is the span of S1 union S2. Similarly, the span of S1 is the span of hte union.


How to determine the dimension of a subspace I am working on a problem for my math class that I'm just not quite grasping:

Let S denote the set of all vectors (x, y, z) in V$_3$ whose components satisfy the equation x + y = 0. Determine whether S is a subspace of V$_3$. If S is a subspace, compute dim S.

I have determined S is a subspace of V$_3$ because the set is closed under addition and multiplication. Thus I must compute the dimension of S.

How do I go about finding the dimension of S? The only way I know of is to start with dimension n = 1 and increase n by one recursively if a set of n linearly independent elements of S spans V$_n$, but what does it mean for a three dimensional vector to span a two dimensional space? Does this mean the x and y components span two dimensions, the y and z components span two dimensions, x and z, any of the above, etc?

Since you know $x$ and $y$ must satisfy the relation $x + y = 0$, then you can rewrite your restriction as: $$x = -y$$
And your vector space will be defined as: $(-y, y, z)$. Now, there are several ways to justify that this vector space is bidimensional. I believe that the most simple one is to notice that it is defined by only two independent variables, so you need only two coordinates to uniquely define a vector in this vector space, thus making it bidimensional.
You can also show this by finding the canonical base to this vector space, and showing that it has only two vectors - $(-1, 1, 0)$ and $(0, 0, 1)$.
The geometrical meaning of a subspace of a three dimensional space being a two dimensional space is that all the vectors from that subspace are contained on a plane in the three dimensional space - besides the meaning of needing only 2 coordinates do be uniquely defined even on a three dimensional space, because the third coordinate is defined as a function of those two.

The set of $(x,-x,z)$ for all $x,z$, is a two dimensional subspace.  A basis is $B=\{(1,-1,0), (0,0,1)\}$.  The order of a basis is the dimension of the subspace. ..

You can also tell it's $2$-dimensional because there are $2$ free parameters $x$ and $z$.

Another way is that you start with a $3$-d space and impose $1$ restriction, $x+y=0$.  $3-1=2$...


Permutation group products (23)(12)(34)=(1243)? I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.

I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 \mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 \mapsto 3, 3 \mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 \mapsto 4$, giving us $(124$. Then since we know $4\mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 \mapsto 2, 2\mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 \mapsto 3, 3 \mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.

Thanks in advance for any of your clarifications.

I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.

If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:

  the product $\sigma\tau$ represents the permutation $\tau(\sigma(\cdot))$

meaning they apply permutations from left to right (first $\sigma$, then $\tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.

At least they are nice enough to tell you what convention they use.

Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.

Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]

Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]

Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]

Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).


What is the mean curvature of an infinitely long cylinder? Can someone help me understand how I can calculate the mean curvature of an infinitely long cylinder of radius $R$? I know the definition of mean curvature as
$H = \frac{1}{2}(\kappa_1 + \kappa_2)$
where $\kappa_i$ is the $i$th principal curvature. Since the cylinder is infinitely long, I think $\kappa_2 = 0$ (along the axis). Can someone confirm this?
Then the mean curvature of an infinitely long cylinder with radius $R$ would be simply
$H = \frac{1}{2R}$
Thank you for your help!

Your result is correct.
Pick a point on the cylindrical surface. You must convince yourself that one principal direction is perpendicular to the axis of the cylinder (but starting from your point of the surface). Along this direction, the surface looks like a circle with radius $R$, so the principal curvature for this direction is $\kappa_1=\frac1R$. The other principal direction is parallel to the cylinder axis, and along this direction, the surfaces looks (locally near your point) like a straight line, so $\kappa_2=0$. So from the formula $H=\frac12 (\kappa_1+\kappa_2)$ you get the mean curvature you mention.
As TonyK says, this is the same for any point you pick. So if you regard $H$ as a function, mapping each point on the surface to a real number, then $H$ is constant, for the cylindrical surface.
As we see, mean curvature is a local property, so it does not matter if the cylinder is infinitely long or not; as long as there is a neighborhood around the point you consider, where the surface is a cylinder, then the mean curvature at that point is $\frac1{2R}$.

Mean curvature is defined for each point on a surface. So in general, you can't calculate the mean curvature of a surface, infinite or not.
But in this case, the mean curvature of every point on the cylinder is the same, so I suppose the question makes some kind of sense. But it doesn't depend on whether the cylinder is infinite or not.


What is wrong with my proof: if $c^4$ is not divisible by 16, then c is odd? Prove that if $c^4$ is not divisible by $16$, then $c$ is odd.

Proof

By contraposition, let $c$ be an even integer, such that $c=2k$ for some integer $k$.

Then, by substitution,

$(2k)^4 = 16j$, for some integer j, by the definition of divisibility.

Then,

$ (2k)^4 = 16j$

$ = 16k^4 = 16j$

$ = 16b = 16j$, where $b$ is an integer and $b=k^4$

Hence, by divisibility, $b=j$, which are both integers.

Therefore, by contrapositivity, if $c^4$ is not divisible by $16$, then c is odd.

What is wrong with this proof? This was the answer I wrote down on an exam and only got partial credit. Where am I going wrong? Did I assign too many variables?

Thanks!

The essence of your argument is correct. You probably lost points for using too many variables. Ideally, a math proof should be as easy to understand as possible. That means using as little notation as possible.

Here's how I might write the proof:

  Let $c$ be an even integer. Then there exists an integer $k$ such that $c = 2k$. Thus, $c^4 = (2k)^4 = 16k^4$. Since $k$ is an integer, $k^4$ is an integer. Therefore, $c^4$ is divisible by $16$. This completes the proof.

You probably didn't get full credit on the exam because you didn't explicitly state that $c^4$ is divisible by $16$. That is the most important step of the proof. You proved this (you wrote $(2k)^4 = 16j$), but you should have explicitly written it out.

We can compute this using far fewer variables:

Set $c=2k$, and thus $c$ is even.

Now we can see that $(2k)^4=2^4k^4=16k^4$ which is clearly divisible by $16$

Now we set $c=2j+1$, and thus $c$ is odd.

Now we can show that $(2j+1)^4 = 16 j^4 + 32 j^3 + 24 j^2 + 8 j + 1$ which we can see is not divisible by $16$

Therefore, if $c$ is not divisible by $16$ then $c$ is odd.


how to calculate the remainder when the number is divided by 1000 $n=(2^{2014}-1)/3$ what is the remainder when it is divided by $1000$. i wrote $2$ as $3-1$ and proceeded but that only helped me prove $n$ is a integer but i dont know how to find the remainder on dividing it by 1000. well im guessing it involves some number theory theorom. please guide me.(i was thinking of using euler totient theorom but could not do so)

Without being clever, let's brute force our way through.

We want to understand $2^{2014} - 1 \pmod{1000}$. Notice that $\gcd(3,1000) = 1$, so $3$ has a modular inverse (a quick check shows that it's $667$).

Now, how do we find $2^{2014} - 1 \pmod {1000}$? As $\gcd(2,1000) = 2$, we should worry about the $2$ factor. So we think of $1000$ as $2^3 \cdot 5^3$.

Clearly $2^{2014} \equiv 0 \pmod {2^3}$. And as $\varphi(5^3) = 100$, we know that $2^{2014} \equiv 2^{14} \equiv 9 \pmod {125}$. Putting these together, perhaps with the Chinese Remainder Theorem or by quick brute force (there are only 8 things to check, afterall), we see that $2^{2014} \equiv 384 \pmod{1000}$.

Thus $2^{2014} - 1 \equiv 383 \pmod{1000} $. And thus $\frac{2^{2014}-1}{3} \equiv 667\cdot(2^{2014}-1) \equiv 667 \cdot 383 \equiv 461 \pmod{1000}$.

So the remainder is $461$.

Hint:

$(2^{2014}-1)/3=(1+2+2^2+ \cdots + 2^{2013})/3$ now calculate the congruence mod(1000).

Here $1+2 \equiv 0\pmod 3$, $2^2+2^3 \equiv 0\pmod 3$. Now proceed in a similar way & see what you will get. Then apply congruence mod(1000).


Fourier Series of $\ f(x)=1-x^2\ $ Verification Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by letting
  $$f(x)=1-x^2 \ \ \ \ \ \text{for} \ \ 0\leq x\leq 1$$
  and requiring that
  $$
f(x+2)=f(x) \ \ \ \ \ \text{and} \ \ \ \ f(-x)=f(x) \ \ \forall x\in\mathbb{R}
$$
  Find $Sf(x)$, the Fourier series of $f$.

My solution:

$$
    Sf(x) = \frac{a_0}{2}+\sum_{k = 1}^{\infty} a_k\ \text{cos}(k\pi x) \ \ \ \ \ \text{as} \ b_k=0\ \  (f\ \text{is an even function})
$$
Now for $a_0$,
$$
a_0=2\int_{0}^{1} 1-x^2 dx=\frac{4}{3}
$$
For $a_k$,
$$
a_k=2\int_{0}^{1} (1-x^2)\text{cos}(k\pi x) dx
$$
$$
a_k=2\int_{0}^{1} \text{cos}(k\pi x) dx-2\int_{0}^{1} x^2\text{cos}(k\pi x) dx
$$
$$
a_k=-\frac{4}{k\pi}\int_{0}^{1} x\ \text{sin}(k\pi x) dx dx
$$
$$
a_k=\frac{4}{k\pi}\Big(-\frac{\text{cos}(k\pi )}{k\pi}\Big)
$$
$$
a_k=\frac{4(-1)^{k+1}}{(k\pi)^2}
$$

Hence
$$
    Sf(x) = \frac{2}{3}+\frac{4}{\pi^2}\sum_{k = 1}^{\infty} \frac{(-1)^{k+1}}{k^2}\ \text{cos}(k\pi  x)
$$

Is my solution correct? I am unsure.

Based on feedback from user Dylan, I have to agree that the above from OP is correct, assuming that $f(-x)=f(x)$ and that $f(x+2)=f(x)$.


For what $\alpha$ does $\iint_D\frac{1}{(x+y)^{\alpha}}\ dxdy$ converge? For what values of $\alpha$ does $$\iint_D\frac{1}{(x+y)^{\alpha}}\
 dxdy$$ converge? $D=\{0\leq y \leq 1-x, \quad 0 \leq \ x \leq 1\}.$

The double integral can be written as

$$\int_{0}^{1}\left(\int_{0}^{1-x} \frac{1}{(x+y)^{\alpha}} \ dy\right)dx.$$

How does one find a primitive to the inner integral?

In general, the answer might depend on whether this is considered to be the integral over a region in $\mathbb{R}^2$ with respect to the product measure,

$$\tag{1}I = \int_D (x+y)^{-\alpha}$$

or as iterated integrals

$$\tag{2} J= \int_0^1 \left(\int_0^{1-x} (x+y)^{-\alpha} \, dy \right) \, dx $$

In this case it doesn't matter, but I will answer with reference to (1).

The diagonal boundary of the region of integration falls in  between circular arcs with radii $1/2$ and $1$.

Since the integrand is nonnegative,

$$\int_0^{\pi/2}\int_0^{1/2} \frac{r \, dr \, d\theta}{r^\alpha(\cos \theta + \sin \theta)^\alpha} \leqslant I \leqslant \int_0^{\pi/2}\int_0^{1} \frac{r \, dr \, d\theta}{r^\alpha(\cos \theta + \sin \theta)^\alpha}$$

Note that $\cos \theta + \sin \theta > 0$ for $\theta \in [0,\pi/2]$ which implies that

$$\int_0^{\pi/2} \frac{d \theta }{(\cos \theta + \sin \theta)^{\alpha}},$$

converges for all $\alpha$.

Thus, we have convergence of $I$ if $\alpha < 2$ and divergence if $\alpha \geqslant 2$, determined  by the convergence/divergence of

$$\int_0^c \frac{dr}{r^{\alpha-1}}$$.

hint for the inner $\int $

For $\alpha\ne 1$

$$\int_0^{1-x}\frac {dy}{(x+y)^\alpha}=$$
$$\Bigl [\frac {(x+y)^{1-\alpha}}{1-\alpha}\Bigr]_0^{1-x} =$$

$$\frac {1}{1-\alpha}(1-x^{1-\alpha}) =$$

The double integral is
$$\frac {1}{2-\alpha} $$

If $\alpha=1$, we find $1$.

The double integral exists if $\alpha\ne 2$.


If $\lim_{x \to \infty} f''(x) = k$, with $k > 0$, then $\lim_{x \to \infty} f(x) = \infty$ I'm having some trouble proving the following:

Let $f: \mathbb R \to \mathbb R$ be a function such that $\lim_{x \to \infty} f''(x) = k$, with $k > 0$. Prove that $\lim_{x \to \infty} f(x) = \infty$.

I did the following:
If $\lim_{x \to \infty} f''(x) = k > 0$, then this means that: $$\exists M\in \mathbb R: x > M \implies f''(x)>0$$
This means that the function $f'|_{(M,\infty)}: (M,\infty) \to \mathbb R$ with $f'|_{(M,\infty)}(x) = f'(x)$ is incresing, because it's derivative is allways $>0$.
Then my plan was to do the same and then conclude that $f$ was increasing that the limit is $+\infty$. But then I remembered the function can always be increasing but no go to $\infty$ as $x \to \infty$, like, for example:  $-1/x$ for $x > 0$.
How can I prove this?

$f''(x) > 0$ for large $x$ is not sufficient. But you have
$$
 f''(x) \ge \frac 12 k > 0
$$
for all sufficiently large $x \ge x_0$. It follows that
$$
 f'(x) \ge f'(x_0) + \frac 12 k (x-x_0)
$$
and then
$$
 f(x) \ge f(x_0) + f'(x_0) (x-x_0) + \frac 14 k (x-x_0)^2
$$
for all $x \ge x_0$. The right-hand side (the equation of an upward parabola) tends to $+\infty$ for $x \to +\infty$, which gives the desired conclusion.

We know that $f'(x)$ will be strictly increasing after sometime. If $f'(x)$ tends to a fixed value as $x$ tends to $\infty$ then clearly $\lim\limits_{x \to \infty} f''(x)=0$.  As $\lim\limits_{x \to \infty} f''(x) > 0$ therefore $f'(x)$ does not tends to a fixed value as
$x$ tends to $\infty$. Then $f'(x)$ will become  positive after sometime and so $f(x)$ will be strictly increasing after that ( for similar reason $f(x)$ will not tend to a fixed value as x goes $\infty$) and hence $\lim\limits_{x \to \infty} f(x) =\infty$ .


Flaw in my reasoning for the maximum of $ab$ if $a,b\ge0$ and $a+2b=3$? Problem statement: What is the maximum value of the product $ab$ if $a$,$b$ are non-negative numbers such that $a+2b=3$?
What is the flaw in my solution?
We know that $\sqrt{ab} ≤ \frac{ (a+b)}{2}$ and that $a=3-2b$. The product $ab$ will be maximum when it is equal to the square of the RHS of the inequality above. Plugging in for $a$, and squaring both sides we get the equation: $(3-2b)(b)=\frac{(((3-2b)+b)^2}{4}$. Which gives $b=1$. And plugging $b=1$ into the the equation in the problem statement gives $a=1$. So, the max product is $1*1=1$.
What am I doing wrong? What concepts could I be I misunderstanding? Can you please explain? The actual answer is $9/8$. Thank you.

Indeed: We have $$(3-2b)b \le\frac{(3-b)^2}{4}$$ is a true statement and in fact equality holds when $b=1$.
However, it is possibel for the LHS to get a larger value and the equality doesn't hold.
To maximize it notice that $(3-2b)b$ is a concave quadratic and the optimal value is attained when $$b = \frac{1.5}{2}=\frac34$$
That is the maximum value is $\left( \frac32\right) \cdot \left( \frac34\right)=\frac98.$

Here is the reason:
First, the maximum is achieved when $a = \frac{3}{2}, b = \frac{3}{4}$. you can check this would give the result you stated, and to get the answer you can use techniques like $Lagrange$ multiplier.
Then you can see that when $a = \frac{3}{2}, b = \frac{3}{4}$, we have that $\frac{a+b}{2} = \frac{9}{8}$, which is larger than $\frac{1}{2}$ (which is what you would get if $a=b=1$). So the inequality $\sqrt ab \leq \frac{a+b}{2}$ only tells you the LHS is less than or equal to RHS, but it does not imply that when LHS = RHS, you would get LHS as the maximum, because your RHS is changing.


A Banach algebra that cannot be represented on a Hilbert space Can someone give me an exemple of a Banach Algebra $A$ which does not have an isometric representation on a Hilbert Space? (With a proof or a reference to the proof.)

For me, the simplest example would be $A=M_2(\mathbb{C})$ equipped with the operator norm induced by any $\ell^p$ norm ($1\leq p\leq +\infty$, $p\neq 2$) on $\mathbb{C}^2$.

In short: $A$ is naturally a $*$-algebra and an isometric embedding into $B(H)$ for the latter norms is necessarily a $*$-homomorphism, turning $A$ into a $C^*$-algebra. And a $C^*$-norm is unique on a given $C^*$-algebra, as follows from the algebraic characterization of a $C^*$-norm: $\|x\|^2=\|x^*x\|=\rho(x^*x)$.

Here is a detailed elementary proof that $A$ can not be isometrically embedded in any $B(H)$. More precisely, we will see that if such an isometric embedding exists, the norm must be the one induced by the $\ell^2$ norm. Note that this argument can easily be generalized to $M_n(\mathbb{C})$. If you are familiar with matrix units and Peirce decomposition, the idea is simple: if a Banach algebra norm on $M_n(\mathbb{C})$  has $\|e_{jj}\|=1$ for every $j$ and if $\pi:M_n(\mathbb{C})\longrightarrow B(H)$ is an isometric embedding, then $\pi$ is unitarily equivalent to $x\longmapsto x\otimes 1_K$ into $M_n(\mathbb{C})\otimes B(K)$ for some subspace $K$ of $H$. Whence $\|x\|$ must be the unique $C^*$-norm of $M_n(\mathbb{C})$, namely $\|x\|=\sqrt{\rho(x^*x)}$.

  Note: with the same argument, we can show that if $\pi:M_n(\mathbb{C})\longmapsto B(H)$ is a nonzero contractive ($\|\pi(x)\|\leq \|x\|$) algebra homomorphism and if the matrix units (canonical basis of $M_n(\mathbb{C})$) and the identity all have norm not greater than $1$ (whence equal to $1$) as it is the case for any induced norm by $\ell^p$, then $\pi$ is actually a $*$-homomorphism (unitarily equivalent to a diagonal embedding) and $\|\pi(x)\|=\sqrt{\rho(x^*x)}$ is the spectral norm of $x$. So the initial norm on $M_n(\mathbb{C})$ can not be induced by an $\ell^p$ norm $(1\leq p\leq \infty)$ other than $p=2$ in this case as well. But note that any Schatten norm, in particular the Hilbert-Schmidt one, is dominated by the spectral norm, giving a contractive non isometric ($n\geq 2$) $*$-representation in $B(H)$.

I will use the following key properties of idempotents in $B(H)$.

Facts: if $p$ is an idempotent ($p^2=p$) is $B(H)$, then $p$ is self-adjoint ($p^*=p$) if and only if $\|p\|\leq 1$. If $1=p_1+p_2$ where $p_1,p_2$ are two projections (=self-adjoint idempotents) such that $p_1p_2=0$, then $K=\mbox{im} p_1=\ker p_2$ and $L=\mbox{im} p_2=\ker p_1$ are orthogonal and $H=K\oplus L$. Finally, $p_1$ and $p_2$ are Murray-von Neumann equivalent (i.e. there exists $u,v\in B(H)$ such that $p_1=uv$ and $p_2=vu$) if and only if their ranges are isometric.

Denote $1=I_2$ the unit and $e_j$ the idempotent matrix of $A$ whose $(j,j)$ coefficient is $1$ and other coefficients are $0$. So we have what is called an orthogonal decomposition $1=e_1\oplus e_2$ of the unit. Note that with the given norm of $A$, $\|e_1\|=\|e_2\|=\|1\|=1$.

Now assume there exists an isometric algebra homomorphism $\pi:A\longrightarrow B(H)$ for some Hilbert space $H$. Note $\pi$ is injective and that $\pi(1), \pi(e_1),\pi(e_2)$ are norm $1$ idempotents of $B(H)$ with norm $1$, whence projections. Since $e_1e_2=0$, we have $\pi(e_1)\pi(e_2)=0$. Since $\pi(1)$ commutes with $\pi(x)$ for every $x\in A$, $\pi(A)$ leaves the image of $\pi(1)$ invariant and is null on its nullspace. So without loss of generality, we can assume that $\pi(1)=1$ and $1=\pi(e_1)+\pi(e_2)$. Since there  exist $u_1,u_2\in A$ such that $e_1=u_1u_2$ and $e_2=u_2u_1$ (take $u_1=e_{12}$ and $u_2=e_{21}$ from the canonical basis), we see that $\pi(e_1)$ and $\pi(e_2)$ are Murray-von Neumann equivalent. So their ranges $K$ and $L$ give $H=K\oplus L$ an orthogonal decomposition of $H$ with $K$ and $L$ isometric. Thanks to this decomposition, $B(H)$ is $*$-isomorphic to $M_2(B(K))=M_2(\mathbb{C})\otimes B(K)$. With respect to this $2\times 2$ decomposition of $B(H)$, the embedding $\pi$ is simply the natural extension of the scalar embedding $\mathbb{C}\longmapsto B(K)$. Indeed, every $x\in A=M_2(\mathbb{C})$ can be uniquely written $x=e_1xe_1+e_1xe_2+e_2xe_1+e_2xe_2$ (the Peirce decomposition, which is just decomposition with respect to the canonical basis of $M_2(\mathbb{C})$. Applying $\pi$ yields the corresponding $2\times 2  $ decomposition in $B(H)\simeq M_2(B(K))$. So under this identification, $\pi$ is simply
$$
\pi:x=\pmatrix{a&b\\c&d}\longmapsto \pmatrix{a 1_K&b1_K\\c1_K&d1_K}=x\otimes 1_K.
$$
Therefore
$$
\|x\|_A=\|\pi(x)\|=\|x\otimes 1_K\|=\|x:(\mathbb{C}^2,\|\cdot\|_2)\longmapsto (\mathbb{C}^2,\|\cdot\|_2)\|=\sqrt{\rho(x^*x)}
$$
That is, the norm on $A$ must be the norm induced by the $\ell^2$ norm on $\mathbb{C}^2$. And this is obviously not equal to any norm induced by an $\ell^p$ norm for $p\neq 2$.

Precision: given $p_j=\pi(e_j)$ and $v_j=\pi(u_j)$, the identification between $B(H)=B(K\oplus L)$ and $M_2(B(K))$ is given by
$$
y=\pmatrix{a&b\\ c&d}\longmapsto\pmatrix{1_K&0\\0&v_1}\pmatrix{a&b\\ c&d}\pmatrix{1_K&0\\0&v_2}=V_1yV_2
$$
with $V_1:K\oplus L\longmapsto K\oplus K$ and $V_2:K\oplus K\longmapsto K\oplus L$ are such that $V_1V_2=1$ and $V_2V_1=1$. Now with the assumptions, $\|V_j\|\leq 1$. So both both are invertible contractions, with contractive inverse. This implies that the $V_j$ are surjective isometries whence unitaries, and $V_2=V_1^*$. So the identification is a unitary equivalence, in particular a $*$-isomorphism.

I should share an easy example of a Banach algebra that an entry level student can verify easily. On $\mathbb{C}^2$, with point wise linear operations, define the multiplication $$(x_1,x_2)(y_1,y_2) = (x_1y_1,0).$$ One verifies that with sup norm this makes $\mathbb{C}^2$ a Banach algebra. Naturally the sup norm does not satisfy the parallelogram law. Hence this cannot be isometrically isomorphic to any Hilbert space or it's subspace.


Evaluating the integral $\int_{-\infty}^{\infty} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}} dx$. How can one show that for $s_1,s_2 \in \mathbb{C}$ $$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}}  dx = &   2\sin(\pi(n+s_2))2^{1-s_1-s_2} \Gamma(1-n-s_2)\Gamma(s_1+s_2-1)/ \Gamma(s_1-n)  & \\ = & 2^{2-s_1-s_2}\pi \Gamma(s_1+s_2-1)/[\Gamma(n+s_2)\Gamma(s_1-n)].
\end{aligned}
$$

If I recall correctly this could be done with complex analysis noting that the integral is equal to
$$
\int_{C} \frac{(1-ix)^{n-s_1}}{(1+ix)^{n+s_2}}  dx,
$$

where $C$ starts from $+i \infty$ along the imaginary axis and goes around $i$ once in a small circle (positive direction) and then goes back to $+i\infty$.

However this doesn't help me since I don't know how we can deform the contour to $C$ and to evaluate this contour integral.

One considers the complex contour integral

$$\oint_C dz \frac{(1-i z)^{n-s_1}}{(1+i z)^{n+s_2}} $$

where $C$ is a semicircle of radius $R$ in the upper half plane, with a detour along either side of the imaginary axis and around the branch point $z=i$ with a circle of radius $\epsilon$.  We take as a branch cut the ray $\operatorname{Re}{z} =0$, $\operatorname{Im}{z} \gt 1$.  The contour integral is zero by Cauchy's theorem.  Writing the contour integral out, we get

$$\int_{-R}^R dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{(1-i R e^{i \theta})^{n-s_1}}{(1+i R e^{i \theta})^{n+s_2}} \\ + i \int_R^{1+\epsilon} dy \, \frac{(1+y)^{n-s_1}}{(1-y)_+^{n+s_2}} + i \epsilon \int_{\pi/2}^{-3 \pi/2} d\phi \, e^{i \phi} \frac{(2-i \epsilon e^{i \phi})^{n-s_1}}{(i \epsilon e^{i \phi})^{n+s_2}} \\+ i \int_{1+\epsilon}^R dy \, \frac{(1+y)^{n-s_1}}{(1-y)_-^{n+s_2}} + i R \int_{\pi/2}^{\pi} d\theta \, e^{i \theta} \frac{(1-i R e^{i \theta})^{n-s_1}}{(1+i R e^{i \theta})^{n+s_2}} = 0 $$

In the third and fifth intgerals, I use the subscripts $\pm$ to denote the effect of the different sides of the branch cut on the quantity that subscript modifies.  That is, $(1-y)_+^{n+s_2} = e^{i \pi (n+s_2)} (y-1)^{n+s_2}$ and $(1-y)_-^{n+s_2} = e^{-i \pi (n+s_2)} (y-1)^{n+s_2}$.

We take the limits as $R \to \infty$ and $\epsilon \to 0$.  Thus, we assume that $s_1$ and $s_2$ are defined such that the second and sixth integrals vanish in this limit, i.e., $\operatorname{Re}{(s_1+s_2)} \gt 1$.  Also, the fourth integral vanishes when $\operatorname{Re}{(n+s_2)} \lt 1$, which we will assume to be the case.  Thus, we have,

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} &= -i \int_{\infty}^1 dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2} e^{i \pi (n+s_2)}} -i \int_1^{\infty} dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2} e^{-i \pi (n+s_2)}}\\ &= i (-2 i \sin{[\pi (n+s_2)]})  \int_1^{\infty} dy \frac{(y+1)^{n-s_1}}{(y-1)^{n+s_2}}\\ &=  2 \sin{[\pi (n+s_2)]} \int_0^{\infty} dy \, y^{-(n+s_2)} (y+2)^{n-s_1}\\ &= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_0^{\infty} dy \, y^{-(n+s_2)} (y+1)^{n-s_1}\\ &= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_1^{\infty} dy \, (y-1)^{-(n+s_2)} y^{n-s_1}\\&= 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \int_0^1 du \, u^{s_1+s_2-2} (1-u)^{-(n+s_2)} \end{align}$$

We recognize that last integral as a beta integral.  Thus, we have

$$\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} = 2^{2-s_1-s_2} \sin{[\pi (n+s_2)]} \frac{\Gamma(s_1+s_2-1) \Gamma(1-n-s_2)}{\Gamma(s_1-n)} $$

Finally, using the reflection formula, i.e., $\pi/\sin{(\pi z)} = \Gamma(z) \Gamma(1-z)$, we get the result we seek in final form:

  $$\int_{-\infty}^{\infty} dx \frac{(1-i x)^{n-s_1}}{(1+i x)^{n+s_2}} = \frac{4 \pi \, \Gamma(s_1+s_2-1)}{2^{s_1+s_2} \Gamma(s_1-n) \Gamma(n+s_2)} $$

This is, as mentioned above, subject to the constraints $\operatorname{Re}{(s_1+s_2)} \gt 1$ and $\operatorname{Re}{(n+s_2)} \lt 1$.

ADDENUDUM

Here is a problem that is both a special case and a little bit of a generalization at the same time.  Note that there are two branch points rather than the single one here.

ADDENUDUM II
I made an error in the integral about the small circle about the branch point; there, $z=i+\epsilon \, e^{i \phi}$.  That error has been corrected and there is now an additional constraint that must be satisfied by the parameters $n$ and $s_2$ for the posted result to remain valid.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{-\infty}^{\infty}
{\pars{1 - \ic x}^{n - s_{1}} \over \pars{1 + \ic x}^{n + s_{2}}}\,\dd x =
-\ic\int_{-\infty\,\ic}^{\infty\,\ic}
{\pars{1 - x}^{n - s_{1}} \over \pars{1 + x}^{n + s_{2}}}\,\dd x =
-\ic\int_{1 - \infty\,\ic}^{1 + \infty\,\ic}
{\pars{2 - x}^{n - s_{1}} \over x^{n + s_{2}}}\,\dd x
\\[5mm] = &\
-2\ic\int_{1 - \infty\,\ic}^{1 + \infty\,\ic}
{2^{n - s_{1}}\pars{1 - x/2}^{n - s_{1}} \over
2^{n + s_{2}}\pars{x/2}^{n + s_{2}}}\,{\dd x \over 2}
\\[5mm] = &\
-2^{-s_{1} - s_{2} + 1}\ \ic
\color{#44f}{\int_{1/2 - \infty\,\ic}^{1/2 + \infty\,\ic}
x^{-n - s_{2}}\,\pars{1 - x}^{n - s_{1}}\,\dd x}\label{1}\tag{1}
\end{align}

  I'll take the principal branch of $\ds{z^{-n - s_{2}}}$ and the branch-cut of
  $\ds{\pars{1 - z}^{n - s_{1}}}$ 'along' $\ds{\left[1,\infty\right)}$ with
  $\ds{0 < \,\mrm{arg}\pars{1 - z} < 2\pi}$. The integration is performed by closing the contour in a 'big semicircle' $\ds{\,\mc{C}_{R}}$ of radius $\ds{R}$ with $\ds{\Re\pars{z} < {1 \over 2}}$. The integration along $\ds{\,\mc{C}_{R}}$ vanishes out, when $\ds{R \to \infty}$, whenever
  $\ds{\pars{~\Re\pars{-s_{1} - s_{2} + 1} < 0 \implies
\Re\pars{s_{1} + s_{2} > 1}~}}$

\begin{align}
&\color{#44f}{\int_{1/2 - \infty\,\ic}^{1/2 + \infty\,\ic}
x^{-n - s_{2}}\,\pars{1 - x}^{n - s_{1}}\,\dd x}
\\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} &\
-\int_{-\infty}^{-\epsilon}\pars{-x}^{-n - s_{2}}
\exp\pars{\ic\pi\pars{-n - s_{2}}}\pars{1 - x}^{n - s_{1}}\,\dd x
\\[5mm] & -
\int_{\pi}^{-\pi}\epsilon^{-n - s_{2}}\exp\pars{\ic\theta\pars{-n - s_{2}}}
\epsilon\expo{\ic\theta}\ic\,\dd\theta
\\[5mm] & -
\int_{-\epsilon}^{-\infty}\pars{-x}^{-n - s_{2}}
\exp\pars{-\ic\pi\pars{-n - s_{2}}}\pars{1 - x}^{n - s_{1}}\,\dd x
\\[1cm] = &\
-\pars{-1}^{n}\exp\pars{-\ic\pi s_{2}}\int_{\epsilon}^{\infty}x^{-n - s_{2}}
\pars{1 + x}^{n - s_{1}}\,\dd x
\\[5mm] & -
2\ic\,\pars{-1}^{n}\,{\sin\pars{\pi s_{2}} \over n + s_{2} - 1}\,
\epsilon^{-n - s_{2} + 1}
\\[5mm] & +
\pars{-1}^{n}\exp\pars{\ic\pi s_{2}}\int_{\epsilon}^{\infty}x^{-n - s_{2}}
\pars{1 + x}^{n - s_{1}}\,\dd x
\\[1cm] = &\
2\ic\pars{-1}^{n}\sin\pars{\pi s_{2}}\int_{\epsilon}^{\infty}x^{-n - s_{2}}
\pars{1 + x}^{n - s_{1}}\,\dd x -
2\ic\,\pars{-1}^{n}\,{\sin\pars{\pi s_{2}} \over n + s_{2} - 1}\,
\epsilon^{-n - s_{2} + 1}
\\[1cm] = &\
2\ic\pars{-1}^{n}\sin\pars{\pi s_{2}}\,\times
\\[5mm] &\ \bracks{%
{\epsilon^{-n - s_{2} + 1}\pars{1 + \epsilon}^{n - s_{1}} \over n + s_{2} - 1} -
{1 \over -n - s_{2} + 1}\int_{\epsilon}^{\infty}x^{-n - s_{2} + 1}
\pars{n - s_{1}}\pars{1 + x}^{n - s_{1} - 1}\,\dd x}
\\[5mm] & -
2\ic\,\pars{-1}^{n}\,{\sin\pars{\pi s_{2}} \over n + s_{2} - 1}\,
\epsilon^{-n - s_{2} + 1}
\\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to} &\
2\ic\pars{-1}^{n}\sin\pars{\pi s_{2}}\,
{n - s_{1} \over n + s_{2} - 1}\int_{0}^{\infty}
{x^{-n - s_{2} + 1} \over \pars{1 + x}^{s_{1} -n + 1}}\,\dd x\label{2}\tag{2}
\\[5mm] = &\
-2\ic\pars{-1}^{n}\,{\sin\pars{\pi s_{2}} \over n + s_{2} - 1}\,
\Gamma\pars{2 - n - s_{2}}\,{s_{1} - n \over \Gamma\pars{s_{1} - n + 1}}
\,\Gamma\pars{s_{1} + s_{2} - 1}
\\[1cm] = &\
-2\ic\pars{-1}^{n}\,{\sin\pars{\pi s_{2}} \over n + s_{2} - 1}\,
{\pi \over \Gamma\pars{-1 + n + s_{2}}\sin\pars{\pi\bracks{-1 + n + s_{2}}}}
\,{1 \over \Gamma\pars{s_{1} - n}}
\\[5mm] & \Gamma\pars{s_{1} + s_{2} - 1}
\\[1cm] = &\
-2\ic\pars{-1}^{n}\,\sin\pars{\pi s_{2}}\,
{\pi \over \Gamma\pars{n + s_{2}}\pars{-1}^{n - 1}\sin\pars{\pi s_{2}}}
\,{1 \over \Gamma\pars{s_{1} - n}}\,\Gamma\pars{s_{1} + s_{2} - 1}
\\[5mm] = &\
2\pi\ic\,
{\Gamma\pars{s_{1} + s_{2} - 1} \over
\Gamma\pars{n + s_{2}}\Gamma\pars{s_{1} - n}}\label{3}\tag{3}
\end{align}
In the last steps, I used the $\ds{\Gamma}$-Recurrence Property and the Euler Reflection Formula. The integral in \eqref{2} is a well know
Beta function integral representation which requires
\begin{align}
&\Re\pars{\bracks{-n - s_{2} + 1} + 1} > 0\,,\quad
\Re\pars{s_{1} + s_{2} - 1} > 0
\\[5mm] &
\implies
\Re\pars{s_{2} + n} < 2\ \mbox{and}\ \Re\pars{s_{1} + s_{2} > 1}
\end{align}

  With \eqref{1} and \eqref{3}:

\begin{align}
&\int_{-\infty}^{\infty}
{\pars{1 - \ic x}^{n - s_{1}} \over \pars{1 + \ic x}^{n + s_{2}}}\,\dd x =
\pars{-2^{-s_{1} - s_{2} + 1}\ \ic}\bracks{%
2\pi\ic\,
{\Gamma\pars{s_{1} + s_{2} - 1} \over
\Gamma\pars{n + s_{2}}\Gamma\pars{s_{1} - n}}}
\\[5mm] = &\
\bbx{\ds{{4\pi \over 2^{s_{1} + s_{2}}}\,
{\Gamma\pars{s_{1} + s_{2} - 1} \over
\Gamma\pars{n + s_{2}}\Gamma\pars{s_{1} - n}}}}\,,\qquad
\Re\pars{s_{1} + s_{2}} > 1\,,\quad\Re\pars{s_{2} + n < 2}
\end{align}


A little doubt about set notation. Is $ A \setminus B = A \setminus (A \cap {B})$.

I am learning sets and it is given that A minus B is {x $\in$ A such that x $\notin$ B}. Is it the same as A minus A intersection B. One word answer will do if "yes"; explain if "no".

Yes, it is: $A\setminus B=A\setminus (A\cap B)$.

Yes, that is correct. And here are some extra characters to make the answer long enough.


pointwise convergence of series? We have just been introduced to infinite series of functions, and immediately started working on uniform convergence and Weierstrass' M-test, and how the sum of the series behaves in terms of integration, differentiation, etc.

But what about 'pointwise' convergence of a series of functions, like we had for sequences? What's the definition for that? Do I simply pick and $x$, and show that the series I get (which is one of numbers $a_n$) is convergent, or is there more to it?

Yes, it is the product topology. The trouble with this thing is that it is rarely metrizable and it's pretty nasty.


Permutations of Intersections of Parabola and Circles What is the maximum number of points of intersection for four distinct parabolas and three distinct circles drawn on a sheet of paper?

Anyone good enough to solve this.?

Well considering the two different parabolas can intersect at 4 different points. WHen you bring a third one in, its even more.

Two circles intersecting bring 2 points. A circle and a parabola brings 4 I guess. So how many in total?

6 times 4 parabola-parabola intersections, 12 times 4 parabola-circle intersections, 3 times 2 circle-circle intersections.

Four parabolas can have a maximum of 24 points of intersection.  A circle can have four points of intersection with a parabola and hence 16 with four parabolas. Three circles can have have 48 points of intersection.

Each circle can intersect other circle at 2 points and three circles can intersect with each other at 6 points and hence I revise it to be 78.


How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$

Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we get$\implies x^2<\frac{3}{4} \implies \frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$Now, when we make set intersection with $x\geq-1 \wedge x\leq1$ we get the solution $\frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.
We also have to look at the possibility that $1-2x$ can be negative: $0>1-2x\implies x>\frac{1}{2}$We determine the solution with $x\geq-1 \wedge x\leq1$ and we get the complete solution $\frac{1}{2}<x<1$.

Now I want to get the solution for the original inequality and therefore I unite the final solutions from 1. and 2.case. I get the interval $$(-\frac{\sqrt{3}}{2}, 1]$$
This solution is wrong. The correct one is $[-1,1]$.   I would really appreciate if someone could check the complete process and point out where I made a mistake.  Also, is this solving process good and effective or I complicate things too much?

Note that if $x > 0$, then $\sqrt{1 + x} > 1$; similarly, when $x \leq 0$, we have that $\sqrt{1 - x} > 1$. As we have a sum of two positive numbers, of which at least one is larger or equal to one, on the LHS, it must always be larger than one.

Because of the existence of this particular argument, your solving process is too complicated indeed. However, removing square roots from an equation by squaring isn't a bad tactic in se. Just make sure the inequalities still hold after squaring.

Note that the domain of $\sqrt {1+x}+\sqrt {1-x}$ is $[-1,1]$

On $(0,1]$ we have $\sqrt{1+x}>1$

On $[-1,0)$ we have $\sqrt {1-x}>1$

At $x=0$ we have $\sqrt {1+x}+\sqrt{1-x}=2>1$

Thus $\sqrt {1+x}+\sqrt {1-x}>1$ on its domain.


Probability that $U_1 \geq U$, $U_2 \geq U$, $U_3 < U$, $U_4 \geq U$, $U_5 < U$, $U_6 \geq U$, $U_7 \geq U$, for i.i.d. uniform $U_k$s and $U$ Let $U,U_1,U_2,...$ be independant, on [0,1] uniform distributed random variables.

Let $E$ := {$U_1 \geq U,U_2 \geq U,U_3 < U,U_4 \geq U, U_5 < U,U_6 \geq U,U_7 \geq U$}.

Find the probabiliy $P(E)$ and then find the probability $P(U_8 \geq U |E)$.

I am struggling with this task and therefore need help, thanks!


Uniform convergence of $1/(1+n^2x)$ of course, we know that $\frac{1}{1+n^2x}$ converges uniformly on $[a,\infty] $ for a>0; for x>0.

I am having difficulty in proving for $x<0$. We have discontinuities at $x=1/k^2, \forall k\in Z^-.$ So for now I just wanted to prove it for $x<-1$

This is my attempt, if you could check it.

$$|1+n^2x|=|1-n^2|x||>n|1-n|x||$$so

$$\bigg|\frac{1}{1+n^2x}\bigg|<\bigg|\frac{1}{n(n|x|-1)}\bigg|<\bigg|\frac{1}{n(n-1)}\bigg| $$

and so the series converges uniformly for $x<-1$

Further, what do I do for $-1<x<0$ . Thank you

If a<0 ,then there exist a K such that for all n>K ,a>-1/n^2.At -1/n^2 ,function tends to infinity.This is true for all n>K.So sequence is not convergent in sup norm .So it isn't uniformily convergent


AES S-Box Cycles and Matrix Representation I would like to know how many cycles the Rijndael S-Box consists of. Ideally expressed as a product of k-cycles in standard notation.

Also, given that the AES cipher is totally linear without the permutation step, does it mean the decryption routine reduces to a simple matrix inversion?


Wrong initial intuition about the limit of $\frac{\sin(x)}{x}$ as $x$ goes to 0 I'm following the MIT OpenCourseware course on calculus, and when proving the derivative of $\sin(x)$ the following assumption is needed

$\lim\limits_{x\to0}  \frac{\sin(x)}{x} = 1$

The proof for this is given and I can follow it.

My question however relates to my initial intuition. Which is that the limit of $\sin(x)$ at 0 is defined: $\sin(0) = 0$. So Why is it wrong to just plug that in

$ \frac{\sin(0)}{0} = 0$

I would have never looked further because this seems like a simple problem. What is wrong with my intuition here?

You are making two fundamental mistakes:

First: after substituting $x=0$, you wind up dividing by $0$.
This is an undefined expression;
so, you should suspect that things have gone awry.
In your example, an even worse thing happens: you wind up with $0\over0$;
which (if it makes any sense) is even more "undefined" than, say, $1\over 0$.

For instance $\lim\limits_{x\rightarrow 0}{2x\over x}=2$
and $\lim\limits_{x\rightarrow 0}{ x\over 2x}={1\over2}$.

In fact a 0/0 form could possibly be anything.

Secondly, when taking limits, you should really avoid just plugging in the limit point into the expression.
This only works in certain cases (when the  limit exists and the function is continuous at the limit point),
and, more importantly,  hides the true nature of what a limit is:

Informally,
you write $$\lim_{x \rightarrow a} f(x)=L$$ if the values of $f(x)$ get closer and closer to $L$ as $x$ gets
closer and closer to, but not equal to, $a$.

Note the phrase "not equal to" in the above. This is an important restriction and is the reason why just
substituting the limit point into what you're taking the limit of can lead to trouble.

As an example of determining the limit of a 0/0 form,  let's try to find
$$
\tag{1} \lim_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)}
$$

Formally substitutiing $x=1$ in the above gives $0/0$.

If, however, $x$ is close to one, but different from 1, then  substitution into the
right hand side of (1) gives a well-defined number;   moreover, the $(x-1)$ terms will cancel and the final result will be just $x+1$.
Thus, if $x$ is close to 1, but different from 1,
the expression ${(x+1)(x-1)\over (x-1)}$ is close to 2.

As it turns out, the closer $x$ is to 1, but different from 1, the closer
${(x+1)(x-1)\over (x-1)}$ is to 2; and thus
$\lim\limits_{x\rightarrow 1}{(x+1)(x-1)\over (x-1)}=2$.

Finding your limit ($\thinspace\lim\limits_{x\rightarrow0}{\sin x\over x} \thinspace$)
is a bit more complicated than the above. You can find the standard derivation of this limit, which utilizes the
Squeeze Theorem
and simple geometry here (or at your course site, as you said).

You can just use L'Hôpital's rule.

$$\begin{align*}
\frac{d}{dx} \sin(x) &=\cos(x) \\
\frac{d}{dx} x &= 1 \\
\lim_{x\rightarrow0} \frac{\sin(x)}{x}&=\lim_{x\rightarrow0}\frac{\cos(x)}{1}=1
\end{align*}$$

Edit: I think this rule is very helpful to evaluate limits involving indeterminate forms. Not as proof but as a sanity check of an intuitive answer.


Find the field of intersection Let $\mathbb{F}_{p^{m}}$ and $\mathbb{F}_{p^{n}}$ be subfields of $\overline{Z}_p$ with $p^{m}$ and $p^{n}$ elements respectively. Find the field $\mathbb{F}_{p^{m}} \cap \mathbb{F}_{p^{n}}$.

Could you give me some hints how I could show that??

Often, when dealing with many field extensions, it is useful to draw a diagram depicting all of the fields that are/could be relevant and how they relate.

The case of finite extensions of finite fields is rather simple, however, since there is exactly one isomorphism class of field extensions of each degree (where the degree ranges over the positive integers).

In fact, the lattice consisting of all finite extensions of a finite field is isomorphic to the lattice of positive integers (ordered by divisibility), and translating questions from one lattice to another makes solving some problems very easy.

Hints:

For all prime $\;p\;$ and $\;n\in\Bbb N\;$ :

$$\begin{align}&\bullet\;\; \Bbb F_{p^n}=\left\{\;\alpha\in\overline{\Bbb F_p}\;:\;\;\alpha^{p^n}-\alpha=0\;\right\}\\{}\\
&\bullet\;\; \Bbb F_{p^n}\le\Bbb F_{p^m}\iff n\mid m\;\;\end{align}$$


Drawing marbles out of a bag... The infamous drawing marbles from a bag type question that I am used to seeing has became more complicated.  Given a bag with 2 green marbles, 5 red marbles, and 3 yellow marbles.  I then draw 2 marbles out without replacing them, what is the probability that there is at least one red marble?

I have worked out 7/9 as my answer on paper, but I would like to know the correct approach to solving this problem. Perhaps in a more abstract, but understandable form?

The easiest approach is to calculate the probability that you get no red marble and subtract that from $1$.

In order to get no red marble, you must (a) get a non-red marble with your first draw and then (b) get another non-red marble on your second draw. The probability of (a) is $\frac5{10}=\frac12$, since there are $2+3=5$ non-red marbles. Similarly, if (a) occurred, the probability of (b) is $\frac49$, since there are now only nine marbles in the bag, of which four are non-red. The combined probability of (a) and (b) occurring is therefore $$\frac12\cdot\frac49=\frac29\;,$$ and the probability of getting at least one red marble is $$1-\frac29=\frac79\;,$$ just as you calculated.

Of course you could work the problem directly, by calculating the probability of getting exactly one red marble and the probability of getting two red marbles and adding these probabilities. If you do it directly, you have $$\frac12\cdot\frac49=\frac29$$ as the probability of getting two red marbles; $$\frac12\cdot\frac59=\frac5{18}$$ as the probability of getting a red followed by a non-red marble; and $$\frac12\cdot\frac59=\frac5{18}$$ as the probability of getting a non-red followed by a red marble, for a total probability of $$\frac29+\frac5{18}+\frac5{18}=\frac79$$ of getting at least one red marble.

Clearly, though, it’s easier to consider the one ‘bad’ case of getting no red marbles than to have to work with the several ‘good’ cases. Had you been drawing three marbles instead of four, and had the problem asked for the probability of getting at least two red marbles, it would have been just about a toss-up, since there are two cases either way: $0$ or $1$ red marble (‘bad’) versus $2$ or $3$ red marbles (‘good’).

You can solve using combination as follows:

Probability at least one red means
$p(\text{at least 1 red}) = 1-p(\text{not red})$

$1-\frac{_5C_2}{_{10}C_2}$ because $_{10}C_2$ is the total out come=45
and $_5C_2 =10$ so

$$p(\text{at least one red})= 1-0.22222=0.777777$$


Are there any odd primes like this? Are there any odd primes $p, q, r$ such that

$$(p-1)(q-1)(r-1)\mid pqr-1$$

No, there are not.  Roughly speaking, because you can't have $(p-1)(q-1)(r-1)=pqr-1$, you need a factor of at least two between the sides, but the $pqr$ on each side is too big compared to the things you subtract.

To sketch a more rigorous argument, we can demand $p \le q \le r$  If $p=3$, we are asking that $2(q-1)(r-1)|3qr-1$ or $2qr-2q-2r+2|3qr-1$  It is clear the two sides cannot be equal, so we need the right at least double the left.  To be double, we would need $q+r \gt \frac {qr}2$, which cannot happen as $q,r$ are too large.  The same happens even more if $p \gt 3$

Pseudoprimes do just this.  They pass Fermat's little theorm for all non-divisors. but are composite.  They are rarer then primes, but there are some classical ones.  One happens before 1000, and three come to mind less than 2000.


Is $1$ the halfway point between $0$ and infinity? I am a math newbie, but I stumbled across this while I was working on a game in python. This is the proof I have, I believe it shows that the halfway point between $0$ and infinity is $1$:

If it is true, I'm probably not the first to figure this out, but even still, I just want some kind of confirmation.
This is very mind blowing for me right now :D

While I personally would not use the word "halfway", I don't find it totally unreasonable to say that 1 is in some sort of special position between $0$ and $\infty$. The sense I'm thinking of is that in the projective line (Wikipedia), over $\mathbb{R}$ or $\mathbb{C}$, the map $z\mapsto\frac{1}{z}$ flips all the numbers "below" $1$ to all the numbers "above" $1$ and vice versa (in particular, swapping $0$ and $\infty$), while fixing $1$.

Added: You can think of the projective line over $\mathbb{R}$ as a way of describing or capturing all the possible "slopes of lines" in the plane as a single mathematical object, and this object has a "symmetry" $z\mapsto \frac{1}{z}$ which makes $1$ stand out (though note that $-1$ stands out just as much). This directly relates to the idea behind your diagram, since you labeled your diagram with the slopes of the lines (the equations you wrote are incorrect though, as others have already pointed out.)

With regards to the projective line over $\mathbb{C}$, a.k.a. the Riemann sphere (Wikipedia), this is often manifested in illustrations as choosing the unit circle to be the "equator" between the "north pole" $\infty$ and "south pole" $0$. Here's the picture from Wikipedia:

(Of course, one can question to what extent the map $z\mapsto \frac{1}{z}$ is special.)

Short answer:

No, $1$ is the halfway point between $0$ and $2$.
The halfway point between $0$ and infinity is infinity.
(Unless you want to model a non-Euclidean world)

Long answer:

No, the image gives the values for $y'(x) = m$ of some lines $y(x) = m x$,
not the values for $y(x)$.

This value $m$ is not linear dependent on the angle $\alpha$ of the line $y(x)$ with the positive $x$-axis:

$$
m = \tan \alpha
$$

Your pick is the value at the mid angle $\pi/4$ (green line).

It is indeed in the middle of the $\alpha$ interval $[0,\pi/2]$ (the blue line is $x=\pi/2$), but not a good pick for a middle value of $y'$ between $\alpha=0, y'=m=0$ and $\alpha=\pi/2, y'=m=\infty$.
Almost all value of $y'$ is in the right half of the interval, in $[\pi/4,\pi/2]$.

Here the average is better suited.

In fact over the angles from $0$ to $\pi/2$ the average is
$$
\overline{y'} = \frac{\int\limits_0^{\pi/2}\tan \alpha\,d\alpha}{\int\limits_0^{\pi/2} d\alpha}
= \frac 2 \pi \left[ -\ln \lvert \cos \alpha \rvert \vphantom{\frac11} \right]_{\alpha=0}^{\alpha\to \pi/2} \to \infty
$$
as expected.


Prove that every matrix $\in\mathbb{R}^{3\times3}$ with determinant equal 6 can be written as $AB$, when $|B|=1$ and $A$ is the given matrix. Prove or disprove:

  Let $A$ be the matrix:

  $$
    \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
    \end{pmatrix}
$$

  Then every matrix $\in\mathbb{R}^{3\times3}$ with determinant equal to $6$ can be written as $AB$, when $|B|=1$ and $A$ is the given matrix.

My way so far:

Let $C\in\mathbb{R}^{3\times3}$ such as $|C|=6$.
Assuming $C=AB$ then $A^{-1}C=B$. And indeed $|B|=1$.

But does it prove that $C$ can be every real matrix with determinant equal to $6$? I mean how can I even make this $(C=AB)$ assumption, if that's what I need to prove?

Hint:

We have that $$A=\begin{pmatrix}
    1 & 0 & 0 \\
    0 & 2 & 0 \\
    0 & 0 & 3 \\
    \end{pmatrix}$$
Suppose $B$ is some matrix with $\det B=1$. Then what is $\det AB$?

Secondly, write $$C=\begin{pmatrix} c_{11}&c_{12}&c_{13}\\c_{21}&c_{22}&c_{23}\\c_{31}&c_{32}&c_{33}\end{pmatrix}$$ and try multiplying this by the matrix $A$. What does this give you? From here, how can you construct $B$?

Too easy. Let $C\in \mathbb R^{3\times 3}$ with $\det C=6$. Then $C=AB$, where  $B=A^{-1}C$. We have $\det C=\det A\det B$, hence $\det B=1$.


Proving that R is uncountable Is the following proof for the uncountability of $\Bbb{R}$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:\Bbb{N}\rightarrow(0,1)$

$$x_{1}=x_{11}x_{12}x_{13}\\x_{2}=x_{21}x_{22}x_{23}\\x_3=x_{31}x_{32}x_{33}\\.\\.\\.$$

Where $x_{ij}$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_{j}=x_{ii}+1$ when $x_{ii}\neq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_{1}$ in the first decimal, $x_2$ in the second, and so on. So $f:\Bbb{N}\rightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $\Bbb{R}$.

The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3\ldots$
But this is wrong. Let's say $x_1 = \frac{1}{2},$ but you choose the unusual representation $0.49999\ldots$ for it. Now it happens by accident that $x_{22}=x_{33}=x_{44}=\ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,\ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.

Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.

If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.

You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.

The binary version goes like this:

Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:

$$\{a_n\} \mapsto \sum_{j=1}^{\infty} \frac{a_j}{2^j}.$$

Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $\mathbb{R}$ must be uncountable.


How to show that a point is not an interior point? I understand that in order to show that a point, $x$, is an interior point of some set $A \subset B$, where $(B,d)$ is a metric space you just need to show that you can have an open ball around $x$ which is contained in $A$.
I was asked a question today, which just showed I don't really understand what an interior point is.

Let $X=(\mathbb{Q}\cap[0,3])$ be a metric space. Define $Y=\{y \in\mathbb{Q}:2\leq y \leq 3\}$
Is $2$ an interior point of $Y$?

I believe that it's not. The only way I thought of perphaps proving this is by taking an arbitrary open ball which is contained in $Y$ and show that 2 cannot be inside it. This however seems like too much work. Is there perhaps a shorter way of thinking and proving this?

Note that an open $\varepsilon$-radius ball $B(2,\varepsilon)$ around $2$ is of the form $(2-\varepsilon,2+\varepsilon)\cap\mathbb{Q}\cap [0,3]$. For any $\varepsilon>0$ take a rational point $q\in\mathbb{Q}\cap [0,3]$ with $2-\varepsilon<q<2$. Now since $q\in B(2,\varepsilon)$ and $q\notin Y$, then $B(2,\varepsilon)\not\subseteq Y$. Hence $2$ is not an interior point of $Y$.

Here $B=X$ and $A=Y$ in your definition of open set, and the topology is the induced topology on $Y$. Hence the $(\ 1,\ 3\ ) \cap Y$ is an open subset of $Y$ with the induced topology and it contains $2$, so $2$ is an interior point of $Y$.


Determine theta/radius line parameters from line segment endpoints I've been working on this for the past few hours and am quite stuck!

As part of a computer vision exercise I've build a Hough transform that maps between the (x,y) space of an image, and a parameter space (theta,radius). Each point in this (theta,radius) space maps to a single line in (x,y) according to the formula:

$$
r = x \sin{\theta} + y\cos\theta
$$

Geometrically, this is explained in this image where the line segment is in red.

The transform is working correctly in one directly. However I'm now looking at a type of probabilistic Hough transform in which I select two candidate points from the x/y space, draw a line through them and and determine the coordinate in (theta,radius) space that corresponds to this line.

Essentially this is solving the two equations with two unknowns (theta,radius):

$$
r = x_1 \sin{\theta} + y_1 \cos\theta
$$
$$
r = x_2 \sin{\theta} + y_2 \cos\theta
$$

I've been trying, unsuccessfully, to solve this geometrically, but I think an equation should be easier to reason about. However my mathematical ability isn't up to the task of solving these equations. I set them equal to each other, leaving one unknown (theta) but wasn't able to get the two instances out to one side of the equation.

The goal is to process video on a robot with limited computing resource, so efficiency is also important.

Can anyone please point me in the right direction?

$$x_1 \sin\theta + y_1 \cos\theta = x_2 \sin\theta + y_2 \cos\theta$$

$$(x_1 - x_2) \sin \theta = (y_2-y_1) \cos \theta$$

Divide over $$(x_1 - x_2)\cos \theta$$

You'll get

$$\tan \theta = (y_2 - y_1)/(x_1 - x_2)$$

Get the tan inverse to know $\theta$, then given $\theta$ solve for the following equation to know $r$ $$r = x_1 \sin \theta + y_1 \cos \theta$$

Abbreviations Explained:

Line is constructed through segment AB, given by coordinates A=[x1,y1] and B=[x2,y2]
Perpendicular to the line is constructed from the origin [x3,y3] = [0,0]
Point of contact [x4,y4] between the line and its perpendicular (source: https://www.cyberforum.ru/cpp-beginners/thread125838.html#post707538
)

x4=((x2-x1)(y2-y1)(y3-y1)+x1*(y2-y1)^2+x3*(x2-x1)^2)/((y2-y1)^2+(x2-x1)^2)

y4=(y2-y1)*(x4-x1)/(x2-x1)+y1;

theta = atan2(y4,x4);

rho = x1*cos(theta)+y1*sin(theta) = x2*cos(theta)+y2*sin(theta);


Show the usual metric on $C([0,1])$ does not give rise to a complete metric space. I know that I need to show there is a Cauchy sequence which converges to a point outwith $C[0,1]$ to show this. Is this the right way of going about it (I can only think of $f_n (x) = x_n = \frac{1}{n}$ as an example:

Let $x_n = \frac{1}{n}$. Then for $n \geq m > 0$ $|x_n - x_m| = \frac{1}{n} - \frac{1}{m} $

$= \frac{m - n}{nm} < \frac{m}{nm} = \frac{1}{n} <\epsilon$
Set $N = \frac{1}{\epsilon}$ then for $n \geq m > \epsilon$ $\frac{1}{n} \leq \frac{1}{m} < \epsilon$ Hence $\frac{1}{n}$ is Cauchy.

Taking the limit as $n \to \infty$ $f_n (x) = 0$ for $0<n<1$ and $1 for n = 1$. Thus there is a discontinuity at 1, hence it cannot converge.

I feel this is the right track but think I've gone wrong somewhere, can someone spot where? Thanks

Start with a nice discontinuous function $f$ - something like the step function at $1/2$.  Find the Fourier series for this function.  The series should converge to $f$ in your metric (and not to any other continuous function, since the discontinuity is essential).  But the sequence of partial sums is made up of continuous functions, and so is contained in $C(0,1)$.

Or, if you don't like Fourier series, try approximating the same step function with a sequence of piecewise continuous (or even piecewise linear) functions.


Confusion with properties of radians I have a problem that goes like this:

"In a circle with radius $6$, what is the measure (in degrees) of an arc whose length is $2\pi$?"

I thought that the answer would be $360$ degrees because $2\pi$ radians is one circle, but the answer in the book is $60$ degrees. The book solved it like this:

$$Circumference=2\pi r=12\pi$$
$2\pi$ is $1/6$ of the circumference:
$$1/6\times360=60$$
This route makes sense to me, but I am really confused on the part that it said the length of the arc was $2\pi$, and $2\pi$ radians is $360$ degrees. I drew the conclusion that it meant radians because I was told that you assume that a measure is in radians unless you are told that it is something else. I have just learned about radians, and I obviously made a mistake. Can someone explain to me where I went wrong?

By length they mean the arc length, not the angle measure.

First of all, a radius is a length (not a number).  Similarly, an arc (length) is a length (not a number).  We have to assume that, in the stated problem, the radius and the arc-length are both measured in the same length units, say LU.

Secondly, the (dimensionally correct) relationship between arc-length, s, radius, r, and central angle, theta, is:

s = r theta/rad

[not "s = r theta"--as is commonly (almost universally!) written.]

where rad is a reference angle of one radian.  [The quantity theta/rad is sometimes called the "radian measure of the central angle"--but unfortunately (and confusingly) written as "theta."]

In this case we have: s = 2 pi LU, r = 6 LU.  So:

2 pi LU = (6 LU) theta/rad

giving: theta = (2 pi)/6 rad = (2 pi rad)/6.  Since 2 pi rad = 360 degrees (= 1 rev), the answer is:

theta = 60 degrees.

--B P Leonard, Emeritus Professor of Mechanical Engineering, The University of Akron.


Is it analytic ? $f(z)=z^2\bar{z}$ Please find the question and its solution.

My answer gone wrong dont know why, I think the answer marked in the question is wrong.
My try:
I solved for CR equations and replaced x=y=0 and found CR equations satisfied. But not sure why in the solution it says CR equations doesn't satisfy.
Please let me know what should be the correct answer ?

You are right: since $u_x(0,0)=u_y(0,0)=v_x(0,0)=v_y(0,0)=0$, $(0,0)$ is a solution of the Cauchy-Riemann equations (and, since the partial derivatives are continuous at $(0,0)$,  $f$ differentiable at $0$).

Apart from the continuity of all partial derivatives $u_x, v_x, u_y, v_y$ along with $u$ and $v$ everywhere, note that $u_x=v_y \forall x,y$
But $u_y=-v_x$ holds for $xy=0\implies x=0$ or $y=0$ i.e. on coordinate axes. So

CR-equations hold at $z=0$ and the test says that $f(z)$ is differentiable at $z=0$ (not analytic though).

I would say that the question is poorly formatted as all the options stand incorrect.


Is there a largest number that is impossible to achieve by combining 6s, 9s and 20s? You might combine 6 and 9 to get 15, but it's impossible to get 16, or 17 and so on. Is there a largest number you can't achieve by adding some combination of only these 3 numbers? If there's not how do you prove it?

These are the "non-McNugget numbers", of which the largest is 43.

Consider the following cases:

$A_{44}=\{6,6,6,6,20\}$
$A_{45}=\{6,6,6,6,6,6,9\}$
$A_{46}=\{6,20,20\}$
$A_{47}=\{6,6,6,9,20\}$
$A_{48}=\{6,6,6,6,6,6,6,6\}$
$A_{49}=\{9,20,20\}$
$A_{n+6}=A_{n}\cup\{6\}$

So every $n\geq44$ can be represented as a sum of $6$s, $9$s and $20$s.

Now you just need to show that $43$ cannot, and you're done.

Here is a short Python script for asserting this:

def func(val,arr):
    if val > 0:
        for k in arr:
            if val >= k and func(val-k,arr):
                return True
    return val == 0

print func(43,[6,9,20])


Prove that $2^{2^{\sqrt3}}>10$ With a computer or calculator, it is easy to show that
$$
2^{2^\sqrt{3}} = 10.000478 \ldots > 10.
$$

How can we prove that $2^{2^{\sqrt3}}>10$ without a calculator?

I suspect any adequate answer to this question is going to be very computation-heavy.  Here's one.

First we claim that
$$
\sqrt{3} > \frac{3691}{2131} \tag{1}
$$
This fraction is not pulled out of a hat; rather, it is found by taking the continued fraction $\sqrt{3} = [1; 1, 2, 1, 2, 1, 2, \ldots]$ and carrying it out a while.  Anyway, the above is proved directly by
$$
3 \cdot 2131^2 = 3 \cdot 4541161 = 13623483 > 13623481 = 3691^2.
$$

We next claim that
$$
2^{{3691}/{2131}} > \frac{1465}{441} \tag{2}
$$

To prove this, one needs to show $2^{3691} 441^{2131} > 1465^{2131}$.
With repeated squaring, each term requires squaring about $12$ times, so this is doable$^{a}$ with a pencil and paper.

The last step is
$$
2^{1465/441} > 10 \tag{3}
$$
Again one uses repeated squaring, and if one is working in base 10 one only needs to count the number of digits in $2^{1465}$.  This should not take nearly as long as the previous computation.$^{b}$

From (1), (2), and (3),
$$
2^{2^{\sqrt{3}}} > 2^{2^{3691/2131}} > 2^{1465 / 441} > 10. \quad \square
$$

$^{a}$ Honestly, this is quite a stretch. I can't guarantee it won't take like a year of work.

$^{b}$ As a rough estimate, perhaps a day or so.

First take logs
$$2^{\sqrt3}>\log_2 10$$
Then again
$$\sqrt{3}>\log_2 \log_2 10$$
Now, $$\log_2(10)=\frac{\log 10}{\log 2}$$
With $\log 10\approx 2.302$ and $\log 2\approx 0.6931$, We find $$\log_2(10)\approx3.32131$$
And the $\log_2$ of that is $(\approx1.731)$ Now, $\sqrt 3\approx 1.732,$ and therefore we conclude that $$2^{2^{\sqrt3}}>10$$
Note that these values can be found with Newton's method and taylor series, though tedious to find and calculate.


How to solve this equation with error functions? I'd like to solve the following equation for $S$, in terms of $p, x,$ and $\sigma$:

$$
p = \int_0^x\sqrt{\frac{2}{\pi}} \frac{e^{\frac{-(S-\mu)^2}{2\sigma^2}}}{\mathrm{erf}\left(\frac{S}{\sqrt{2}\sigma}\right)- \mathrm{erf}\left(\frac{S-1}{\sqrt{2}\sigma}\right)} d\mu
$$

Solving the integral as Mhenni suggested yields:
$$
p = \frac{\frac{\sqrt{2}}{2}\left(\mathrm{erf}\left(\frac{S-x}{\sqrt{2}\sigma}\right) - \mathrm{erf}\left(\frac{S}{\sqrt{2}\sigma}\right) \right)}{{\mathrm{erf}\left(\frac{S}{\sqrt{2}\sigma}\right)- \mathrm{erf}\left(\frac{S-1}{\sqrt{2}\sigma}\right)}}
$$

The problem now is dealing with the $S$'s inside the error functions. I know there's an inverse error function ($\mathrm{erf}^{-1}$), but I don't see how this is ultimately solvable for $S$.

Is this possible?

Thanks for any help!

I suppose that there are some typo's.

The result I obtained is $$p=\sigma\frac{  \text{erf}\left(\frac{S}{\sqrt{2} \sigma
   }\right)-\text{erf}\left(\frac{S-x}{\sqrt{2} \sigma
   }\right)}{\text{erf}\left(\frac{S}{\sqrt{2} \sigma
   }\right)-\text{erf}\left(\frac{S-1}{\sqrt{2} \sigma }\right)}$$ I do not think that you could be able to solve analytically the equation for $S$. More than likely, only numerical methods could do it.

If you provide some typical values for $p,\sigma,x$, I could look at some numerical scheme.

You have not changed to limits of integration

  $$
p = \frac{ \sqrt{\frac{2}{\pi}} }{   {\mathrm{erf}\left(\frac{S}{\sqrt{2}\sigma}\right)- \mathrm{erf}\left(\frac{S-1}{\sqrt{2}\sigma}\right)}}
   \int_{S/(\sqrt{2}{\sigma}) }^{  (S-x) /(\sqrt{2}{\sigma}) } e^{-t^2} dt.
$$

Added: You need my answer to finish evaluating the integral.


Showing that $Aut~Z_p \simeq Z_{p-1}$ Okay. This is similar to a question I asked a few days ago. I am trying to show that $Aut~ Z_p$ is isomorphic to $Z_{p-1}$, where $Z_p$ denotes the congruence class of integers $\mod p$ and $p$ a prime. I have shown $Aut~ Z_p$ consists of $p-1$ elements, using the fact that a homomorphism is uniquely determined by how it maps $[1]_p$, the generator of $Z_p$; the only element it cannot be mapped to is $[0]_p$, otherwise $[1]_p \rightarrow [k]_p \neq [0]_p$ extends to an automorphism.

Now I am trying to show that $Aut~ Z_p$ is a cyclic, and show that mapping the generator of $Aut~ Z_p$ to $Z_{p-1}$ defines an isomorphism. However, I am having trouble showing this. I could use some hints.

Note: At this point, I don't know anything about fields (let alone rings), the order of an element, and related concepts.

Edit: I noticed that the question was asked here, but from my cursory reading I didn't see any ideas I could use yet.

Hint:

Let $\varphi$ the automorphism of $\mathbf Z_p$ mapping $[1]$ to $[r]$. What does $\varphi^k$ map $[1]$ to? How can you use Lil' Fermat?


Positive definiteness of $\sum_{i=1}^n \boldsymbol{x}_i \boldsymbol{x}_i^\top$ Let $\{\boldsymbol{x}_i\}_{i=1}^n$ be a set of $n$ linearly independent vectors in $\mathbb{R}^n$.

  My question: Is $\displaystyle\sum_{i=1}^n \boldsymbol{x}_i\boldsymbol{x}_i^\top$ always positive definite, i.e. $\displaystyle\sum_{i=1}^n \boldsymbol{x}_i\boldsymbol{x}_i^\top>0$?

Clearly, if the set $\{\boldsymbol{x}_i\}_{i=1}^n$ is composed by orthogonal vectors the answer is in the affirmative. What about the general case?

The answer is yes, and the $x_i$ need not be linearly independent so long as they span $\Bbb R^n$.  For any vector $y$, we have
$$
y^T\left(\sum_{i} x_ix_i^T \right) y = \sum_{i}(x_i^Ty)^2 \geq 0
$$
so clearly the matrix is positive semidefinite.  As for positive definiteness, note that the matrix can be written as $AA^T$, where
$$
A = \pmatrix{x_1 & x_2 & \cdots & x_N} \qquad (N \geq n)
$$
since $A$ has full row-rank, $AA^T$ must be invertible.  Also, for the first part, it suffices to note that
$$
y^T(AA^T)y = (A^Ty)^T(A^Ty) \geq 0
$$

You don't use your set anywhere in the question. The sum you write is just $\Vert x_i \Vert^2$ and yes, this is always greater than $0$ as long as $x_i \neq 0$.


Why is Euler's number $2.71828$ and not anything else? Why is Euler's number $\mathtt 2.71828$ and not for example $\mathtt 3.7589$?

I know that $e$ is the base of natural logarithms. I know about areas on hyperbola xy=1 and I know its formula: $$e =\sum_{n=0}^\infty \frac{1}{n!} \approx 2.71828$$
And I also know it has many other characterizations.
But, why is $e$ equal to that formula (which sum is approximately $\mathtt 2.71828$)?

I googled that many times and every time it ends in having "$e$ is the base of natural logarithms". I don't want to work out any equations using $e$ without understanding it perfectly.

Summary: I'm looking for the origin of $e$, if $\pi$ came from the radius of a circle with a unit diameter, then what is $e$ ???

$\sum\frac1{n!}$ is not that special.

$\lim_{n\to\infty}\left(1+\frac1n\right)^n$ is not really special.

$f'(x)=f(x)$ is a very simple differential equation, but unremarkable, really.

$\ln (x)$ is only marginally nicer than other logarithms, in that its derivative is $\frac1x$.

The fact that a single number connects all of these (and many, many others) as intimately as $e$ does is nothing short of a miracle. Oh, and also $e$ happens to have the decimal expansion $2.718\ldots$

One introduction of $e$ can be seen as a solution to $f'(x)=f(x)$. Indeed with $\lambda$ any constant, the function
$$ f \ : \ x \mapsto \lambda e^x$$
is so that at each point, its "increase" equals its value, hence $f'(x)=f(x)$


Visually stunning math concepts which are easy to explain Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

I think if you look at this animation and think about it long enough, you'll understand:

Why circles and right-angle triangles and angles are all related.
Why sine is "opposite over hypotenuse" and so on.
Why cosine is simply sine but offset by $\frac{\pi}{2}$ radians.

$\Bbb RP^2\sharp\Bbb RP^2\simeq \text{Klein Bottle}$


How to solve difficult positive integers and co-prime word problem? This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with derivative of algebra and prime numbers, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

The Roads Department in the state of Graphium uses graphs consisting of dots (called vertices) that represent towns, and lines (called edges) that represent roads between towns. To keep track of things they label the vertices with consecutive positive integers so that if two towns are joined by a road, then the numbers on the corresponding vertices are coprime, that is, their only common factor is 1. Such a labelled graph is called a COP-graph. Note that if two twons do not have a road joining them, then their labels may or may not be coprime. Here are two examples of COP-graphs, one with labels 1 to 5, the other with labels 1-6.

a) Show that the labels on COP-graphs A and B could be replaced with the integers 2 to 6 and 2 to 7 respectively so they remain COP-graphs.
b) Explain why COP-graph B could not remain a COP-graph if its labels were replaced with integers 5-10.
c) Explain why COP-graph A remains a COP-graph no matter what set of five positive consecutive positive integers label the vertices.
d) If a COP-graph has six vertices and labels 1 to 6, what is the maximum numbers of edges it can have? Explain.

a) Just write down the integers 2 to 6 (or 2 to 7) and join every pair of coprimes by edges. You should be able to see the graphs $A$ and $B$ in the results.

b) Each of vertices (originally) labelled $1,2,4,5$ has three neighbours. Among $5,6,7,8,9,10$, the number $6$ is coprime only to $5,7$ and $10$ is coprime only to $7,9$; hence $5,7,8,9$  must be used in some order for $1,2,4,5$ and $6,10$ replace $3,6$. and the two coprimes each has must correspond to their nieighbours in $B$. However, then $6$ and $10$ must have the common neighbour $7$ - but there is no common neighbour in $B$.

c) For odd $n$ $$\begin{matrix}&&n+4\\n&n+1&n+2&n+3\end{matrix} $$
works and for even $n$
$$\begin{matrix}&&n+2\\n&n+1&n+3&n+4\end{matrix} $$
works

d) Join coprime numbers by an edge and count edges. Any valid graph must be a subgraph of this and hence has at most just as many edges.

d is 15 edges, c is because COP grah a has only one town with two coprime neighbours


Find solution $\phi$ that satifies the differential equations and has a specific form I am looking at the following exercise:

Let $I=(0,1)$. Find the solution $\phi$ that has  a continuous derivative in $\mathbb{R}$ and satisfies :

$$y''=0 \text{ in } I \\y''+k^2y=0 \text{ apart from } I, \text{ where } k>0$$

and furthermore $\phi$ has the form:

$\phi(x)=\left\{\begin{matrix}
e^{ikx}+Ae^{-ikx} &, x \leq 0 \\
Be^{ikx} & , x \geq 1
\end{matrix}\right.$

That's what I have tried:

$$y''=0 \text{ in } I$$
$$\Rightarrow y'=c_1 \Rightarrow y(x)=c_1 x+c_2, x \in I, c_1, c_2 \in \mathbb{R}$$

Since $\phi$ satisfies $y''=0$ in $I$ it holds that $\phi(x)=c_1 x+ c_2, x \in I$.

$\phi$ has a continuous derivative in $\mathbb{R}$. So we have the following:

$$\lim_{x \to 1^{-}} \phi(x)=\lim_{x \to 1^{+}} \phi(x) \Rightarrow \lim_{x \to 1^{-}} (c_1 x+c_2)=\lim_{x \to 1^{+}} B e^{ikx} \Rightarrow c_1+c_2=Be^{ik} \\ \Rightarrow B=e^{-ik} (c_1+c_2) $$

$$\lim_{x \to 0^{-}} \phi(x)=\lim_{x \to 0^{+}} \phi(x) \Rightarrow \lim_{x \to 0^{-}} (e^{ikx}+Ae^{-ikx})=\lim_{x \to 0^{+}} (c_1x+c_2) \Rightarrow 1+A=c_2 $$

$$\lim_{x \to 0^{-}} \phi'(x)=\lim_{x \to 0^{+}} \phi'(x) \Rightarrow \lim_{x \to 0^{-}} (ik e^{ikx}-Aik e^{-ikx})= \lim_{x \to 0^{+}} c_1 \Rightarrow c_1=ik-Aik=(1-A) ik$$

$$\lim_{x \to 1^{-}} \phi'(x)=\lim_{x \to 1^+} \phi'(x) \Rightarrow \lim_{x \to 1^{-}} c_1=\lim_{x \to 1^+} Bik e^{ikx} \Rightarrow c_1=Bik $$

So we get:

$$Bik=(1-A) ik \Rightarrow B=1-A$$

$$B=e^{-ik} (Bik+1+A) \Rightarrow 1-A=e^{-ik} ((1-A)ik+1+A) \\ \Rightarrow 1-A=e^{-ik} ik-Aik e^{-ik}+1+A \Rightarrow A(2-ike^{-ik})=-e^{-ik} ik \Rightarrow A=\frac{-e^{-ik} ik}{2-ike^{-ik}} \\ \Rightarrow A=\frac{k^2 e^{-2ik}-2ik e^{-ik}}{4+k^2 e^{-2ik}}$$

$$B=1-A=1-\frac{k^2 e^{-2ik}-2ik e^{-ik}}{4+k^2 e^{-2ik}}=\frac{4+2ike^{-ik}}{4+k^2 e^{-2ik}}$$

Thus:

$\phi(x)=\left\{\begin{matrix}
e^{ikx}+ \frac{k^2 e^{-2ik}-2ik e^{-ik}}{4+k^2 e^{-2ik}}e^{-ikx} &, x \leq 0 \\
\frac{4+2ike^{-ik}}{4+k^2 e^{-2ik}}e^{ikx} & , x \geq 1 \\
\left(\frac{4+2ike^{-ik}}{4+k^2 e^{-2ik}} \right)ikx+ \frac{2-2ik e^{-ikx}}{2-ik e^{-ik}}
\end{matrix}\right.$

Could you tell me if it is right or if I have done something wrong?

can this work:
$y=\left\{\begin{matrix}
a\cos(kx - b) & x \leq 0 \\
(c\cos d - a\cos b)x + a\cos b&  0 \le x \le 1 \\
c\cos(kx - k + d) & 1 < x
\end{matrix}\right.$

where $a,b,c$ and $d$ are arbitrary constants.


How can I prove that F2 is a Field? I have to prove that the set {1,0} is a field. Do I have to go through every 8 possibilities when I want to show associativity or is there an easier way and how do you argue with an infinite set like $\mathbb{R}$ when you want to show that for ex. $(a + b) + c = a + (b + c)$, with $+$ being the normal addition.

I don’t know what you’re suppose to know or not.

It you write $\mathbb F_2= \mathbb Z /2 \mathbb Z$, then the result is clear. $\mathbb F_2$ is a field as it is the quotient of a ring over a maximal ideal and therefore is a field.

By the way, you’re almost forced to have this background. How do you define $\mathbb F_2$ without it?

Commutativity and associativity for both operations, as well as the distributive law all follow from the fact that these properties are satisfied in $\mathbb{R}$. After all, if you take $a,b,c\in\mathbb{F_2}$ then you can think of $a,b,c$ as of real numbers. Even in $\mathbb{R}$ we have $(a+b)+c=a+(b+c)$ so of course it is also true if you look at that number mod $2$. (if it is the same number then it also has the same remainder after division by $2$)


Infinity and the Maximum of Natural Number I am posting this question in the wake of my similar posting here. What I have concluded there is that infinity $\infty$ stands for something without limit, the $\max  (\mathbb N)$ and $\min (\mathbb N)$ do not exist, so do that of the $\mathbb R$. Having said that, let's take a look at this problem down here:

$$\begin{align}
A_t &= \{x \mid t^2 \leq x \leq (t+1)^2 \} \tag{1} \\
\bigcup_{t \in \mathbb N} A_t &= [1, 4] \cup [4, 9] \cup [9, 16]\cup [16, 25] \ldots \tag{2}\\
&= \tag{3}\ldots
\end{align}$$

My first reaction is to take $[1, \infty)$ as the answer since the right element moves to the right without limit. However, in the spirit of yesterday's posting, which do you think is the correct answer and what is your explanation:

(a) $[1, \infty)$, or
(b) $\emptyset$, since $\max (\mathbb N)$ does not exist?

And here is another beginner question:

$$\begin{align}
A_t &= \{x \mid -t \leq x \leq t \} \tag{4} \\
\bigcup_{t \in \mathbb N} A_t &= [-1, 1] \cup [-2, 2] \cup [-3, 3] \ldots \tag{5}\\
&= \tag{6}\ldots
\end{align}$$

Again, which do you think is the correct answer and what is your explanation:

(c) $(-\infty, \infty)$, or
(d) $\emptyset$, since $\max (\mathbb N)$ does not exist?

Thank you for your time and effort.

POST SCRIPT: ~~~~~~~~~~~~~~~~~~~~~
Questions edited per request from moderators. Note also that this question is on the first chapter of Measure & Integration, so the concept of Extended Real Number is not there yet.

First, $\Bbb{N}$ is the set of positive integers, so it is bounded below by zero. This means $\min{\Bbb{N}}=0$ or $1$, depending on whether or not you allow $0$ in the natural numbers. Your intuition on your first question is almost right, but you should find $$\bigcup_{t \in \Bbb{N}} \{x \mid t^2 \leq x \leq (t+1)^2 \} = [1,\infty)$$ because $\infty \notin \Bbb{R}$ as $\infty$ is not a number, and does not belong in the set $\Bbb{R}$ of all real numbers. The half open interval $[1,\infty)$ makes that clear. On your second problem, the answer should be $(-\infty, \infty)$ for similar reasons. For all $y\in \Bbb{R}$ you can find $t \in \Bbb{N}$ such that $|y| \leq t$, and hence $y \in [-t,t]$, which means $$\Bbb{R} \subseteq \bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \}$$ The reverse inclusion follows easily since a union of subsets of $\Bbb{R}$ is itself a subset of $\Bbb{R}$, so you have $$\bigcup_{t \in \Bbb{N}} \{x \mid -t \leq x \leq t \} = \Bbb{R} = (-\infty, \infty)$$

No.

$$
\bigcup_{t\in\mathbb N} A_t
$$
is just shorthand for
$$
\lim_{M\to\infty}\bigcup_{t=1}^{M} A_t
$$

So it doesn't matter that $\max\mathbb N$ doesn't exist.


Don't understand proof that interior of a set is open Say we want to show that the interior of a set $A$ is open.
If $x \in Int(A)$, then there exists an open ball $B_r(x) \subseteq A$.
Since $B_r(x)$ is open, $y \in B_r(x)$ also has an open ball $B_s(y) \subseteq B_r(x) \subseteq A$, so $y \in Int(A)$.
Now, somehow we have to show that the ball $B_r(x) \subseteq Int(A)$, and that would complete the proof.
All of the proofs I read say this is obvious, but I don't see how $B_r(x) \subseteq Int(A)$ immediately follows here.

$$\text{Int}(A)=\{x\in X\mid \exists r>0: B_r(x)\subset A\}.$$

Therefore
$$\forall x\in \text{Int}(A), \exists r_x>0: B_r(x)\subset A,$$
and thus
$$\text{Int}(A)=\bigcup_{x\in \text{Int}A}B_{r_x}(x).$$

We finally conclude that $\text{Int}(A)$ is open.

If $ x \in int(E) $ then $ \exists r >0 $ , $ B_r(x) \subset E $
Now $ \forall y \in B_r(x) \subset E $, $ \exists r_y = r - d(x,y) $, $ \ni B_{r_y} \subset B_r (x) \subset E $ Hence $ y \in int(E) $
Hence $ B_r(x) \subset int(E) $ or x is interior point of int(E)


Why do the coefficients of this series grow like a polynomial? Set-up

Let $f: \mathbb{N} \to \mathbb{C}$ be a multiplicative arithmetic function (meaning that $f(mn) = f(m)f(n)$ whenever $m$ and $n$ are relatively prime). Suppose that for every $\epsilon >0,$ that $$f(n) \ll_\epsilon n^\epsilon,$$ here I am using Vinogradov’s notation, this means that $|f(n)| \leq C_\epsilon n^\epsilon$ for some constant $C_\epsilon$ which depends on $\epsilon.$.

Suppose further that for each prime $p, f(p)=g $ for some $g \geq 1.$ This $g$ is thus independent of $p.$.

Consider now, for a prime p, the identity of formal power series $$1+gX+f(p^2)X^2 + \cdots = (1-X)^{-g} h_p(X).$$ It is then easy to see that $h_p(0)=1,h_p’(0)=0.$.

Question.
Let now $b_p(k)$ be the coefficient of $X^k$ in $h_p(X).$ I have seen it claimed that the coefficients $b_p(k)$ are bounded by a polynomial in $k$ that is independent of $p.$
Could someone explain why this holds?


If $X$ is a simple random variable, then $E(X)$ can be written in terms of the distribution function $F$ of $X$ $E(X) = \sum_{x\in\mathbb{R}}x[F(x)-F(x^-)]$;

I was thinking, as $X$ is a simple random variable, so it is defined as $X=\sum_{x=1}^{m}xI_{C_x}$. Then, the expected value is $\mathbb{E}(X)=\sum_{x=1}^{m}x\mathbb{P}({X=x})$. As a result from distribution function it has $\mathbb{P}(X=x)=F(x)-F(x^-)$. So I was wondering if as I just can use this result, and write $\mathbb{E}(X)=\sum_{x}x [F(x)-F(x^-)]$.

It seems right in the case of a discrete random variable.
For a continuous one, you would rather write $\int_{-\infty}^\infty x dF(x)$

This is terrible notation.  Why are you starting your sum from $i=1$?  What is $F(x^-)$? Do you mean the limit of $F(x)$ from the left?  If so, then $F(x)=F(x^-)$ so their difference is precisely $0$.  What are you even summing over? You are summing over values of $i$ apparently, but nowhere does $i$ appear in the summed expression.  Maybe you want to sum over a list of values of $x$, each of which differ by some small amount, and it seems like you want to let $x^-$ be that small amount less than $x$, then take the limit as that small amount goes to 0 after the sum is computed.  Still you run into trouble because as the difference $x-x^-$ gets smaller, the sum blows up, so you need to divide the entire sum by $(x-x^-)$ to get something that converges.  In that case, you've simply found the integral.

$$E(x)=\int_{-\infty}^{\infty}xF(x)dx$$

which is definition of expected value for continuous random variables.


Testing validity of Predicate Logic conditional statement without a Truth Tree. For the past half hour, I have been trying to prove the following statement in Predicate Logic without the use of a truth tree:

$$∃xPx∧∃xQx→∃x(Px∧Qx)$$

Which, of course, I know to be invalid. As such, I can easily show this using a truth tree.

Without a truth tree, here is my work thus far:
$$(∃xPx∧∃xQx)→∃x(Px∧Qx)$$
Then I assume the conditional to be false, and try my best to follow the proof methods I learned:
$$\neg[(∃xPx∧∃xQx)→∃x(Px∧Qx)]$$
$$(∃xPx∧∃xQx)\wedge\neg∃x(Px∧Qx)$$
$$(∃xPx∧∃xQx)\wedge\forall x\neg(Px∧Qx)$$
$$Pa$$
$$Qb$$
$$\neg(Pa∧Qa)$$
$$\neg(Pb∧Qb)$$

From here, I am lost. Since I don't study this formally, I may have made a mistake already.

I'm not sure if I should use De Morgan's Law and then Distributive laws.

You are attempting to prove a statement is not a tautology by proving the negation of the statement is a tautology.

This will not work.

Witness that $(a\to b)$ is not a tautology, and neither is its negation, $(a\wedge \neg b)$.

To prove that a statement is not a tautology, one provides a counter example: an interpretation of the predicates for which the statement is false.

For instance, let us take $P(x)$ to mean "$x$ is odd" and $Q(x)$ to mean "$x$ is even".   Under this interpretation, the statement then reads:

  "If there is an odd number, and there is an even number, then there is a number that is both odd and even."

$$(\exists x\;(2\not\mid x)\wedge \exists x\;(2\mid x)) \to \exists x\;((2\not\mid x) \wedge (2\mid x))$$

The basic problem is that your initial statement is confusing.

You have written:

$$\exists x P(x) \land \exists x Q(x) \implies \exists x (P(x)\land Q(x))$$

This is confusing because you should have made a difference between $P$'s $x$ and $Q$'s.

If we write:

$$\exists x P(x) \land \exists y Q(y) \implies \exists x (P(x)\land Q(x))$$

we can see the problem - there is no $y$ on the RHS!

The correct statement is:

$$\exists x P(x) \land \exists y Q(y) \land (x=y)\implies \exists x (P(x)\land Q(x))$$


Find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$ I want to find all $2\times 2$ matrices that commutes with the matrix $\pmatrix{a & b\\ c & d}$, with $bc\neq 0$.
For this, suppose $\pmatrix{e & f\\ g& h}$ such that
 $\pmatrix{a & b\\ c & d}\pmatrix{e & f\\ g& h}=\pmatrix{e & f\\ g& h}\pmatrix{a & b\\ c & d}$. Equating both sides, we can get the values of $e,f,g,h$. But this way is complicated. Is there any sophisticated way than the above.

The matrix equation $AX-XA=0$ is a special case of the Sylvester matrix equation $AX+XB=C$ for given $A,B,C$. There is indeed a nice way to solve this equation, see the references given above. For $2\times 2$ matrices, studying the system of polynomial equations is a good idea, too. It works quite well, even for matrices in $GL_2(\mathbb{Z})$, where you have Diophantine equations (and then, say $ab=1$ has only the solutions $a=b=1$ or $a=b=-1$).

First we have to find out which matrices commute with a Jordan block $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Since
$$ \begin{pmatrix}0&1\\0&0\end{pmatrix} \begin{pmatrix}x&y\\z&w\end{pmatrix} = \begin{pmatrix}z&w\\0&0\end{pmatrix}, \begin{pmatrix}x&y\\z&w\end{pmatrix} \begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&x\\0&z\end{pmatrix},$$
we find out that $z = 0$ and $x = w$, so the solution space is two-dimensional.

Second, we do the same for a diagonal matrix with two different eigenvalues: $\begin{pmatrix}0&0\\0&1\end{pmatrix}$:
$$ \begin{pmatrix}0&0\\0&1\end{pmatrix} \begin{pmatrix}x&y\\z&w\end{pmatrix} = \begin{pmatrix}0&0\\z&w\end{pmatrix}, \begin{pmatrix}x&y\\z&w\end{pmatrix} \begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}0&y\\0&w\end{pmatrix},$$
so the constraints are $y=z=0$, and again the solution space is two-dimensional.

Consider now the Jordan form of the given matrix. It is either a Jordan block of dimension 2, a diagonal matrix with different eigenvalues, or a scalar. In the first two cases, the commutator space has dimension 2, and in the last, it has dimension 4. This explains loup blanc's remark, which can be generalized for arbitrary square matrices.

As loup blanc remarks, since scalars commute with all matrices, the only matrices whose Jordan form is scalar are the scalars themselves, and our matrix can't be scalar since $bc \neq 0$. So the commutator must have dimension 2.

See the comments for why this argument unfortunately doesn't work. Though it almost does.

Let $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where $bc \neq 0$. In particular, $M$ is not a scalar, so its minimal polynomial has degree two. So $I,M,M^2$ are linearly independent, and span a linear space of dimension three of matrices commuting with $M$. Since the center of the matrix algebra consists only of the scalars, you obtain this way all matrices commuting with $M$.


Why was calculus invented? Can anyone show me any flaw for which there was a need of inventing calculus or give me an example where normal maths fail and we need calculus?

This is actually an excellent question and at the time calculus first emerged in the 17th century in the work of Leibniz and Newton (in alphabetical order, without implying any priority), many top-level scholars were sceptical and in fact issued a number of challenges.  For example, Huygens adopted the position that the problems Leibniz can solve with the new calculus are no different from the problems one can solve with the old methods, and the difference is only in the language.  Huygens may have been at least partly right if he was referring to earlier techniques such as Fermat's adequality which was indeed a powerful method that could solve many problems that today we typically think of as standard ware of the calculus: finding tangents to curves, maxima and minima, areas, centers of mass, etc. In fact, Lagrange considered Fermat to be the inventor of the calculus.

Half a century later, Bishop Berkeley was still vituperating against the new calculus, but the tide has clearly changed and the Fermat-Leibniz-Newton framework was an undeniable success in the hands of Euler.

Something similar is happening today, with Abraham Robinson's "legalisation" of infinitesimals in his 1966 book.  Half a century later today, opposing voices are still being heard, but the tide is clearly changing, with high-profile advocates like Fields medalist Terry Tao publishing applications and new insights based on Robinson's framework on an almost monthly basis.

Appreciating the advantage of the calculus over these earlier techniques involves detailed knowledge of the calculus and it is not that easy to give an elementary problem that can be solved by the calculus but not by Fermat's adequality, for example.  Using the calculus does simplify calculations significantly compared to the earlier methods.

One specific example that could be mentioned is finding the general tangent line of the cycloid curve.  Fermat was able to do it by means of an intricate geometric argument relying on his technique of adequality; see this publication for details.  Modern proofs using calculus are more straightforward.

Rectification of the ellipse (computation of the perimeter length) requires a special function known as the Elliptic Integral, which can only be defined in the frame of calculus.

https://en.wikipedia.org/wiki/Ellipse#Circumference