Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.

Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise say a 9 if the question required us to prove that the expression is divisible by 9. However in this case, I do not end up with such a thing.

My work so far below:

Inductive Hypothesis: $7^k(3k+1)-1=9N$ where $N\in\mathbb{N}$

Inductive Step:

$7^{k+1}(3k+4)-1 \\ =7\times 7^k(3k+1+3)-1 \\ =7\times \left [ 7^k(3k+1)+3\times 7^k \right ] -1 \\ = 7 \times \left [ 9N+1 + 3 \times 7^k \right ] -1 \\ = 63N+21\times 7^k+6 \\ = 3 \left [ 21N+7^{k+1}+2 \right ]$

So now I need to somehow prove that $21N+7^{k+1}+2$ is divisible by 3, but I'm not quite sure how to proceed from here...

${\rm mod}\ 9\!:\,\  \overbrace{7^n (1\!+\!3n) \equiv 7^n (1\!+\!3)^n}^{\rm\large   Binomial\ Theorem}\! \equiv 28^n\equiv 1^n\equiv 1 $

Remark $ $ We used only the first $2$ terms in the Binomial expansion, and this special case has an easy inductive proof whose inductive step amounts to multiplying by $\,1+a\pmod{\!a^2},\,$ namely
$\!\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\,\  (1+ a)^n\, \ \  \equiv&\,\ \ 1 + na\qquad\qquad\,\ \  {\rm i.e.}\ \ P(n)\\[1pt]
\Rightarrow\ \ (1+a)^{\color{}{n+1}}\! \equiv &\  (1+na)(1 + a)\quad\, {\rm by}\ \ 1+a \ \ \rm times\ prior\\
\equiv &\,\ \ 1+ na+a+n\color{#c00}{a^2}\\
 \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\ \ \ {\rm i.e.}\ \ P(\color{}{n\!+\!1})\\[2pt]
  \end{align}$
We could substitute this proof inline above (for $\,a=3)\,$ to get an explicit  proof by induction on $n\,$ (independent of the Binomial Theorem) but doing so would obfuscate the underlying arithmetic structure, i.e. we should call the Binomial Theorem by name (vs. call-by-value = inline) in order to highlight the key arithmetical structure. The proof is still inductive, but the induction has been encapsulated into a (ubiquitous) Theorem, with the benefit that we can easily reuse it later.

First, show that this is true for $n=1$:

$7^1\cdot(3\cdot1+1)-1=9\cdot3$

Second, assume that this is true for $n$:

$7^n\cdot(3n+1)-1=9k$

Third, prove that this is true for $n+1$:

$7^{n+1}\cdot(3(n+1)+1)-1=$

$7^{n+1}\cdot(3n+3+1)-1=$

$7^{n+1}\cdot(3n+1+3)-1=$

$7^{n+1}\cdot(3n+1)+7^{n+1}\cdot(3)-1=$

$7\cdot\color{red}{7^n\cdot(3n+1)}+3\cdot7^{n+1}-1=$

$7\cdot(\color{red}{9k+1})+3\cdot7^{n+1}-1=$

$63k+7+3\cdot7^{n+1}-1=$

$63k+3\cdot7^{n+1}+6=$

$\color{blue}{3(21k+7^{n+1}+2)}$

Now:

$7\equiv1\pmod3\implies$
$\forall{m}\in\mathbb{N}:7^m\equiv1\pmod3\implies$
$7^{n+1}\equiv1\pmod3\implies$
$7^{n+1}+2\equiv3\pmod3\implies$
$7^{n+1}+2\equiv0\pmod3\implies$
$3|7^{n+1}+2\implies$
$\exists{p}\in\mathbb{N}:7^{n+1}+2=3p\implies$
$3(21k+7^{n+1}+2)=3(21k+3p)=3(3(7k+p))\color{blue}{=9(7k+p)}$

Please note that the assumption is used only in the part marked red.


There are relatively prime positive integers $m,n$ such that $b-a = \frac{m}{n}$. Find $m+n$ $|x^2 - \frac{5}{2}| \le |x-8|$ if and only if $x$ is in the interval $[a,b]$. There are relatively prime positive integers $m,n$ such that $b-a = \frac{m}{n}$. Find $m+n$.
I have no clue how to approach it.Can anyone help? Thanks in advance.

Edit

After Googling on the problem I found that this problem is Purple Comet 2013 HS Problem 9 http://purplecomet.org/home/resource/543/2013HighSchoolProblems.pdf
Here is the solution of problem posted in AOPS
https://artofproblemsolving.com/community/c149h531513_purple_comet_2013_hs_problem_9


Do I need to understand Multi-Variable Calculus to study Linear Algebra? I'm currently studying Single-Variable Calculus independantly through MIT OCW. I can only focus on one course at a time independantly since it takes up so much time, and I really want to study Linear Algebra next instead of Multi-Variable Calculus.

My question is simply this: after understanding Single-Variable Calculus, would I be able to continue into and understand Linear Algebra, or should I do Multi-Variable Calculus first?

Multivariable calculus is helpful because it gives many applications of linear algebra, but it's certainly not necessary.  In fact, you probably need linear algebra to really start to understand multivariable calculus.

To wit, one of the central objects in multivariable calculus is the differential of a function.  In single-variable calculus, you are taught that the differential of a function $f:\mathbb{R}\to\mathbb{R}$ is a new map $f':\mathbb{R}\to\mathbb{R}$ which provides the slope of the tangent line to $f$ at each point in $\mathbb{R}$.  This is strictly correct, but it is not the best way to understand single-variable calculus if you want to easily generalize.

The better way to see single-variable calculus is to recall that the tangent line to $f$ at $x$ is the best affine-linear approximation to $f$ at $x$, i.e., $f$ is approximated by $f(y)\approx f'(x)(y - x) + f(x).$

This generalizes quite well!  If $f:\mathbb{R}^n\to\mathbb{R}^m$, the differential to $f$ at $x$, $df_x$, is the best linear approximation to $f$ at $x$: $f(y)\approx df_x(y-x) + f(x)$.  Now, we think of $x$ and $y$ as vectors in $\mathbb{R}^n$ and the differential $df_x$ is an $n\times m$ matrix.

Even more generally, we think of $df$ as a map from $\mathbb{R}^n$ into $Hom(\mathbb{R}^n,\mathbb{R}^m)$ which measures the best linear approximation of $f$ at each point $x\in\mathbb{R}^n$.

Generalizing further requires the notion, from differential geometry, of a smooth manifold.  Such manifolds carry objects called tangent bundles, which assign to each point of the manifold an abstract vector space.

You can see how linear algebra is a little more helpful for multivariable calculus than the other way around.

The two are almost completely independent.  In linear algebra terms, they are orthogonal to each other.  In linear algebra you're learning techniques and rules for matrix manipulation, whereas in multi-variable calculus you'll be learning about partial differentials and how to integrate over multiple variables.  Nothing in common except maybe coincidental overlaps of minor topics.


Is Turing completeness monotone with respect to Cook reductions? I think the post title is relatively clear assuming I worded it correctly, but since I was thinking of a specific example:

The language of Boolean expressions is Turing complete; Does this imply that every NP-Complete language is Turing complete?

Or alternatively:
The language of Horn clauses is Turing complete; Does this imply that every P-Hard language is Turing complete?

The question is a little confused, as the OP acknowledges, but it does have a straightforward answer in the context of complexity theory (although this answer might not be particularly illuminating for what the OP had in mind.)

The complexity class RE contains all languages (sets of strings) which can be enumerated by a Turing machine.  There are RE-complete languages, such as the language of the Halting Problem (set of strings encoding Turing machines which halt).  There is a fairly strong sense in which RE-completeness corresponds to Turing-completeness; a system is Turing-complete if it can enumerate the strings of an RE-complete language just as well as a Turing machine can.

On the other hand, the complexity class P (resp. NP) contains all languages which can be recognized by a Turing machine (resp. non-deterministic Turing machine) in polynomial time.

Both P and NP are known to be strictly contained within RE.  That means there are languages in RE which are not in either P nor NP; therefore there is no P-complete or NP-complete language which is also RE-complete.

(P-hard is a different matter, because that only says that we know a language is at least as complex as a P-complete language; it could presumably be much more complex and in fact RE-complete.)

There is some truth in what you are implying, but you need to sort things out.

Let a boolean expression be a combination of binary variables, the operators AND $\land$ , OR $\lor$ and NOT $\neg$ and the constants 0 and 1.

Then, every computable function can be expressed as a boolean expression.

$SAT$ is defined as the set of satisfiable boolean expressions. The $\mathbb{NP}$-complete problem is to decide if a boolean formula, that is given as input, is satisfiable or not.

Now, let's combine the two. Every problem can be decided by a boolean formula that determines membership in the problem's language. The SAT problem is $\mathbb{NP}$-complete, therefore, every problem is $\mathbb{NP}$-complete.

The above proof is of course false, since no single complexity class, that sets bounds on resources, can contain all problems. So, where's the catch?

Suppose that you have an instance $a$ of a problem $A$. In order to convert it to a boolean formula, i.e. reduce it to $SAT$, a process must be followed. This process requires additional time, therefore this process combined with an algorithm for SAT does not yield a polynomial time non-deterministic algorithm for every problem, but only for problems in $\mathbb{NP}$.

Suppose that we are not interested in time, but in computability. So, we don't care how long the algorithm takes to convert any instance of any problem to a boolean formula, but if it can do so in finite time. Since we know that this is possible, a computation model using only boolean formulas is Turing equivalent and can compute any function computable by a Turing machine.


The equation of the image of the circle $x^2+y^2+16x-24y+183=0$ by the line mirror $4x+7y+13=0$ is The equation of the image of the circle $x^2+y^2+16x-24y+183=0$ by the line mirror $4x+7y+13=0$ is:
$(A)x^2+y^2+32x-4y+235=0$
$(B)x^2+y^2+32x+4y-235=0$
$(C)x^2+y^2+32x-4y-235=0$
$(D)x^2+y^2+32x+4y+235=0$

I tried to solve it reflection formula.The image of any point on the circle $(x,y)$ in the line mirror $4x+7y+13=0$ is $(x',y')$.And it is given by
$\frac{x'-x}{4}=\frac{y'-y}{7}=\frac{-2(4x+7y+13)}{16+49}$
$x'=\frac{33x-56y-104}{65}$
$y'=\frac{-56x-33y-182}{65}$
Then i got stuck.It this not the proper way to find the image of a circle in the line mirror.What should be the correct method to solve it?Please help me.Thanks.

You need to follow the following procedure:

Find the center and radius of circle.
Find the image of the center.
Put the co-ordinates of image in place of center in the eqn. $(x-a)^2+(y-b)^2=r^2$

Do it yourself and if you still have problems I will do it then only.

First of all find the value of centre of circle. Then substitute the value of x and y in the equation. Then you will get the value of centre of required circle. Then substitute the values in the general equation of circle . That's it.


Prove that $\int _0^1x^a\left(1-x\right)^bdx$ = $\int _0^1x^b\left(1-x\right)^adx$, where $a,b\in \mathbb{R}$ Prove that $$\int _0^1x^a\left(1-x\right)^bdx = \int _0^1x^b\left(1-x\right)^adx$$

How can I even get started on this? I evaluate the integral with parts, but it just gets more and more tedious since I'm working with these constants here.

Substitute $u=(1-x)$. We then have $du=-dx$ and when $x=0$, we have $u=1$, $x=1$ gives $u=0$. Thus

$$\int_0^1x^a(1-x)^bdx=-\int_1^0(1-u)^au^bdu=\int_0^1(1-u)^au^bdu$$

No integration by parts or anything necessary, just a straight substitution.

One way to answer this is to note that this is the definition of the Beta function, which can be written as $$\operatorname{B}(a+1, b+1)\\= \frac{a!b!}{(a+b+1)!}\\= \frac{b!a!}{(b+a+1)!}\\= \operatorname{B}(b+1, a+1)$$


an affine transformation that maps an affine subspace on a parallel subspace is a dilation I thought that an affine transformation that maps an affine subspace on a parallel subspace is a dilation and in the other direction.
I tried to prove this.
So first I assumed that $F$ is an affine transformation that $S=p+V$ maps on $T=q+V$ (because $S$ and $T$ are parallel $(V=V)$). So $F(S)=T$.
Now I tried to prove that $F$ must be a dilation, so have an linear part $A=\lambda I$
I don't really know how to move on because I tried : $F(S)=A(p+V)+b=q+V$. So $Ap+AV+b=q+V$ and so $A=\frac{q+V-b}{p+V}$ and I'm not sure I can do this.
For the other way I assumed that $F(p)=\lambda Ip+b$ and I took an $S=p+V$. Now is $F(S)=\lambda (p+V)+b= \lambda p+b+(\lambda V)$. I'm not sure if this proves that $\lambda p+b+( \lambda V)$ is an affine subspace with the same direction as $S$.
Can someone correct and help me?

Notice that an affine transformation $F$ (such as a dilation) maps an affine supspace to an affine supspace. To be more concrete: if $S = p + V$ is an affine supspace of $\mathbb{A}^n$, then $F(S) = F(p) + F_*(V)$. Here $F_*$ denotes the linear part of the affine transformation.
First I will prove the other way around because this is rather trivial. Let $F$ be a dilation, i.e. $F_* = \lambda I$ for a $\lambda \in \mathbb{R}_0$. We have to prove that for every linear supspace $V$ of $\mathbb{R}^n$ we have $F_*(V) = V$. Let $x \in F_*(V)$. Then there exists a $v \in V$ such that $x = \lambda v$, and thus $x \in V$. For the other way around, let $x \in V$. Because $\frac{1}{\lambda}x \in V$, we have $ \lambda * \frac{1}{\lambda}x = x \in F_*(V)$. So we can conclude that $F_*(V) = V$ and thus every dilation maps an affine supspace to a parallel affine supspace.
Now for the "hard" part: Let $F$ be an affine transformation that maps every affine supspace to a parallel affine supspace. Let $v \in \mathbb{R}^n_0$. Define $V := \operatorname{span}(\{v\})$, which is a vector supspace of $\mathbb{R}^n$. Because $F$ maps every affine supspace to a parallel affine supspace, we have $F_*(V) = V$. Then $F_*(v) = \lambda v$ for some $\lambda \in \mathbb{R}$. Since $v$ was arbitrarily chosen, this means that every non-zero vector is an eigenvector of $F_*$. Now this precisely means that the linear part of the affine transformation is of the form $F_* = \lambda 'I$ with $\lambda' \in \mathbb{R}_0$. For a proof: Demonstration: If all vectors of $V$ are eigenvectors of $T$, then there is one $\lambda$ such that $T(v) = \lambda v$ for all $v \in V$.

Write $G(x)=F(x)-F(0)$ it is a linear map which maps a line to a line, it is of the form $G(x)=cx$.
Classical proof: Let $u,v$ independent, $G(u)=au, G(v)=bv, G(u+v)=c(u+v)=au+bv$ implies $(c-a)u+(c-b)v=0$, since $u$ and $v$ are linearly independent, $c=a=b$.


How can I calculate $\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}$ How should I get the result $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\sqrt2$$
without relying on Stirling's formula or the Central Limit Theorem?

I am totally clueless. Can somebody help me?

This is based on the following lemma.

  Lemma: If $f\in C^{1}[0,1] $ then $$\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx=\frac{f(1)-f(0)}{2}$$
  (proof available here)

In the above lemma it is sufficient to assume the Riemann integrability of $f'$.

If $a_n$ is the sequence in question and $L$ is the desired limit then we have
\begin{align}
\log L&=\log \lim_{n\to\infty}a_n\notag\\
&=\lim_{n\to \infty} \log a_n\notag \\
&=\lim_{n\to\infty} n\log\left(\frac{e}{4}\right)+\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\notag\\
&=\lim_{n\to\infty} \sum_{k=1}^{n}f\left(\frac{k}{n}\right)-n\int_{0}^{1}f(x)\,dx\notag
\end{align}
where $f(x) =\log(1+x)$. Using the lemma we see that $$\log L=\frac{\log 2}{2}$$ and hence $L=\sqrt{2}$.

Solution
According to Stirling's formula, $$n!=\sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta}{12n}}~~,(0<\theta<1)$$
Hence, $$\lim_{n \rightarrow \infty} \frac {e^n(2n)!}{(4n)^nn!}=\lim_{n \to \infty}\frac{e^n\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} \cdot e^{\frac{\theta_1}{24n}}}{(4n)^n \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \cdot e^{\frac{\theta_2}{12n}}}=\sqrt{2}.$$


How to draw a phase portrait of a two-dimensional ODE? If we're given:

$$\dot{x}=-x+y$$
$$\dot{y}=xy-1$$

How do I draw a phase portrait of this system? I don't understand which direction the arrows are supposed to point.

This is what I got so far:

I found the nullclines:

$$\dot{x}=0$$
$$x=y$$

and

$$\dot{y}=0$$
$$xy=1$$

Then I drew the lines  $x=y$ and $xy=1$.

Don't know what to do from here.

This three step process is a summary from the excellent book series "Differential Equations: A Dynamical Systems Approach, Higher-Dimensional Systems" by Hubbard and West.

$(1)$ Write the equations as:
$$\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{xy - 1}{y - x}$$

and sketch the isoclines of $(a)$ Horizontal Slope (where $y' = 0$) and $(b)$ Vertical Slope (where $x' = 0$).

$(2)$ In each region determined by these isoclines, put together the horizontal and vertical arrows and then sketch the resultant direction field.

Here you are using the above equation, choosing sample $(x, y)$ pairs and drawing the arrows that have direction and magnitude based on the slope.

$(3)$ Trace sample trajectories through the direction field.

Some other things that I find helpful are to determine the type of critical points which you see at the intersection of the nullclines, that is $(-1, -1)$ and $(1, 1)$ in this example. One is a stable spiral at $(-1, -1)$ and the other is an unstable saddle point $(1, 1)$. Additionally, you can look at the eigenvectors.

Putting all of these things together, we arrive at the phase portrait:

Lastly, it is worth noting, that these are not hard rules. Practice makes perfect and you'll develop your own approach in a way that is easy for you.

Note that your ODE is of the form $(\dot{x},\dot{y})=(F_1(x,y),F_2(x,y))$ and so you can plot the vector field $$(x,y)\mapsto(F_1(x,y),F_2(x,y)).$$


Fundamental group of a bouquet of circles Consider a bouquet of $n$ circles, centered at $(1,0), (2,0),...,(n,0)$ $$ W_n= \bigcup_1^nC_r$$ with $$C_r:(x-r)^2+y^2=r^2$$
I want to compute the fundamental group of this space.

$W_n$ is path connected. Moreover I know
$\pi_1(W_1)=\pi_1(C_1)=\mathbb{Z}.$ Seeing $W_n= W_{n-1}\cup C_n$ we
have both $W_{n-1}$ and $C_n$ are path connected and open. Moreover
$W_{n-1} \cap C_n $ is just the origin which has trivial fundamental
group. Hence
$$\pi(W_n)=\pi(W_{n-1})*\pi(C_1)=\pi(W_{n-1})*\mathbb{Z}=...=\pi_1(W_1)*\mathbb{Z}...*\mathbb{Z}=$$$$\pi(C_1)*\mathbb{Z}*...*\mathbb{Z}=\mathbb{Z}*...*\mathbb{Z}$$
so it is the $n$-times free product of $\mathbb{Z}$ with itself.

Does it work? First I want to know if my solution is correct and only secondly a different one thanks.

Your proof is correct but a bit sloppy. It is true that the sets $W_{n-1}$ and $C_1$ are path-connected, but they are not open. Clearly no real harm has occured, but you should be careful to meet the requirements of the exact statement of the Seifert-van Kampen Theorem which you clearly applying.
The space $W_n$ is a wedge of $n$ circle $\vee^nS^1$. There are slight more sophisticated van Kampen Theorems which can compute the fundamental group of such spaces directly (i.e. by working with a family of many path-connected open sets and their intersections rather than just two).
Another way to compute its fundamental group is to construct its universal cover. There is a standard way to describe the universal covers of wedge sums. The case for $n=1,2$ is clear, but I don't even want to think about what the space looks like for higher $n$.

Your answer is wrong, for example, it is not very difficult to see that this fundamental group is not abelian. A way to compute that is to show that the space $W_n$ is homotopy equivalent to the wedge sum of $n$ circles and then use the Seifert-Van Kampen theorem to show that its fundamental group is the free group with $n$ generators.


Why is there no explicit formula for the factorial? I am somewhat new to summations and products, but I know that the sum of the first positive n integers is given by:

$$\sum_{k=1}^n k = \frac{n(n+1)}{2} = \frac{n^2+n}{2}$$

However, I know that no such formula is known for the most simple product (in my opinion) - the factorial:

$$n!=\prod_{k=1}^n k = \prod_{k=2}^n k$$

I don't know if that is how products work, but I would really like to know!

So my question is why is there no explicit formula (in terms of n, other than n(n-1)...2*1) for the product of the first n integers? Is there a proof that says that one cannot exist or is it that one has not been discovered?

By explicit formula I mean a non-functional equation that does not require n amount of steps to calculate - just like the summation formula does not require n additions.

(The following is a joke.)

Put $a_n:={1\over 6}n(n+1)(n+2)$ and $b_n:={1\over2}n(n+1)$, and define the magic constant
$$\xi:=\sum_{n=1}^\infty {n!\over 2^{a_n}}\ =\ 0.630882266676063396815526621896\ldots\quad .$$
Then
$$n!=\bigl\lfloor\> 2^{a_n}\xi\>\bigr\rfloor\quad \bigl({\rm mod}\ \ 2^{b_n}\bigr)\qquad(n\geq1)\ .$$
Try it out!

If I am not mistaken, Manjul Bhargava generalized the factorial function to include numbers such as (pi)! and (2+√3)!


if $a \in \mathbb{Q}(2^{1/2},2^{1/4},2^{1/8},...)$ if $a \in \mathbb{Q}(2^{1/2},2^{1/4},2^{1/8},...)$ then why does it mean that $a$ can be expressed as a finite many different roots of $2$?

The key is to think about how you define something like $\mathbb{Q}(2^{1/2},2^{1/4},\ldots)$.

Defining that is sort of like defining $K[x_1,x_2,x_3,\ldots]$. Even though you've adjoined infinitely many variables, what does an element of $K[x_1,x_2,x_3,\ldots]$ look like? Well, it's a polynomial ring, so every element is a polynomial in arbitrarily many (but finitely so) variables (and with only finitely many terms!)

After all, $x_1 + x_2 + x_3 + x_4 + \cdots$ is not a polynomial.

The same is true for adjoining algebraic elements to a field, since after all the way you define a field extension of $K$ is by taking $K[x]/f(x)$, where $f(x)$ is an irreducible. Thus, your field is (by definition):

$$\mathbb{Q}[x_1,x_2,x_3,x_4,\ldots]/(x_1^2-2,x_2^4-2,x_3^8-2,x_4^{16}-2,\ldots)$$

Where modding out by the ideal $(x_1^2-2,x_2^4-2,x_3^8-2,x_4^{16}-2,\ldots)$ means that you should treat each $x_i$ as $\sqrt[2^i]{2}$.

Your field can be represented as a sequence of successive quadratic extensions $$\mathbb Q\subset\mathbb Q(\sqrt2)\subset\mathbb Q(\sqrt[4]2)\subset\mathbb Q(\sqrt[8]2)\subset\cdots\subset\mathbb Q(\sqrt[2^i]2)\cdots$$ from which you can see that $a$ given, necessarily there is a suitable $2^i$ so the element $a$ belongs to the subfield $\mathbb Q(\sqrt[2^i]2$) of finite degree.

Which is interesting to mention is that  this field is the only one where all construction with ruler and compass  is feasible. For instance the classic impossible problem of the trisection of the angle gives place to a cubic field (because of  the formula for third of the angle) which is not a subfield of your field.

Edit: I understand the downvote: I referred to the kind of field and number 2 obviously does not interfere in what I meant. It  was a distraction for me, it's true.


Number of possible eight digit number divisible by 9 An eight digit number divisible by 9 is o be formed by using 8 digits out of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. Find the number of ways in which it can be done.

I know divisible rule of 9 is sum of all the digits should be multiple of nine. But don't know how to use it in permutation.

As pointed out in the post, all that matters is the digit sum. The sum of all your digits is divisible by $9$. So our number is divisible by $9$ if and only if the two digits not chosen are $0,9$ or $1,8$, or $2,7$, or $3,6$, or $4,5$.

If the two digits not chosen are $0,9$, there are $8!$ possible numbers.

If the two digits not chosen  are any of the $4$ other pairs, then $0$ was chosen. Then there are in each case $7\cdot7!$ numbers. For $0$ cannot be the first digit of an $8$-digit number.

This gives  a total of $8!+(4)(7)(7!)$.


Can we always draw $n/3$ disjoint triangles from $n$ points in the plane in general position? Suppose we are given $n$ points in the plane, where $n$ is a multiple of $3$ and no three of these points lie on a line. Is it possible to group all of these points into sets of three, so that if we draw the triangles formed by these sets, then no two of the triangles intersect?

Hi guys, I can visualise this question, but I have no idea how to approach it mathematically, any help?

Given $n = 3k$ points, there are $\frac{n(n-1)}{2}$ ways to join them into lines and form unit vectors (2 for each line).
Pick a coordinate system such that none of these unit vectors are parallel to $y$-axis. The $x$-coordinates of these $n$ points will be distinct. You can sort the points according to their $x$-coordinates and group them to groups of 3.

There are finitely many slopes between pairs of points, so we may choose a line whose slope is not any of these. For ease of visualization we may choose it arbitrarily close to vertical.

Now sweep the line across the set of points from left to right. Since we will cross the points one at a time, we can cut the plane along this line as we pass each set of three points. Since each piece of the plane is convex, the triangle formed between the three points it contains lies within it. And since the pieces are disjoint, so are the triangles.


One-to-one mapping from $\mathbb{R}^4$ to $\mathbb{R}^3$ I'm trying to define a mapping from $\mathbb{R}^4$ into $\mathbb{R}^3$ that takes the flat torus to a torus of revolution.

Where the flat torus is defined by $x(u,v) = (\cos u, \sin u, \cos v, \sin v)$.
And the torus of revolution by $x(u,v) = ( (R + r \cos u)\cos v, (R + r \cos u)\sin v, r \sin u)$.

I think an appropriate map would be:
$f(x,y,z,w) = ((R + r x)z, (R + r x)w, r y)$ where $R$, $r$ are constants greater than $0$.

But now I'm having trouble showing this is one-to-one.

Compute the Jacobians of the two mappings.  I'll call your first mapping $X(u, v)$ so as not to re-use the variable $x$.  Let $R > r > 0$.

$$\frac{\partial X}{\partial(u, v)} = \left[ \begin{array}{cc}
-\sin u & 0 \\ \cos u & 0 \\ 0 & -\sin v \\ 0 & \cos v \end{array} \right] $$

$$\frac{\partial f}{\partial(x, y, z, w)} = \left[ \begin{array}{cccc}
rz & 0 & R + rx & 0 \\ rw & 0 & 0 & R + rx \\ 0 & r & 0 & 0 \end{array} \right] $$

The embedding in $\mathbb{R}^3$ is the composite $f \circ X$, whose Jacobian is the product of the Jacobians:

$$ \frac{\partial f}{\partial(x,y,z,w)}\frac{\partial X}{\partial(u,v)}
= \left[ \begin{array}{cc}
-rz\sin u & -(R + rx)\sin v \\ -rw\sin u & (R+rx)\cos v \\ r\cos u & 0  \end{array} \right]
= \left[ \begin{array}{cc}
-r\cos v\sin u & -(R + r\cos u)\sin v \\ -r\sin v\sin u & (R+r\cos u)\cos v \\ r\cos u & 0  \end{array} \right]

$$

(I used sage notebook www.sagenb.org for my computations.)

Now all of the questions concerning 1-to-1 can be phrased in terms of rank and nullity of these matrices.  For example, it easy to see the first and third matrices have rank 2 (so nullity=0) for all choices of $u, v$, (which shows both representations of the torus are immersive, meaning an appropriate restriction of domain will make those maps 1-to-1).

Now $f$ is 1-to-1 when restricted to $\mathrm{im} X$ (the flat torus) because $f \circ X$ is immersive and by restricting to $u, v \in [0, 2\pi)$, both $f$ and $f \circ X$ become 1-to-1.  In other words, if $f( X(u,v) ) = f(X(u', v'))$, then $X(u,v) = X(u', v')$.

Hope this helps!

Do you need to explicitly give the map or are you asked to prove both sets have the same cardinilty?

If not, it is a good exercise to prove $\mathbb{R}$ and $\mathbb{R}^2$ are both the same size (Schroeder-Bernstein theorem - consider decimal expansions to show a surjection) then via induction and the fact both sets are subsets of $\mathbb{R}^n$ we see there is a bijection between them. Explicitly finding a bijection is not clear to me, perhaps it is to someone else.


Determinant of a non-square matrix I wrote an answer to this question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function $\det$ defined on all real-valued matrices (not just the square ones) having the following properties.

$\det$ is real-valued
$\det$ has its usual value for square matrices
$\det(AB)$ always equals $\det(A)\det(B)$ whenever the product $AB$ is defined.
$\det(A) \neq 0$ iff $\det(A^\top) \neq 0$

Does such a function exist?

Such a function cannot exist. Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. Then, since both $AB$ and $BA$ are square, if there existed a function $D$ with the properties 1-3 stated there would hold
\begin{align}
\begin{split}
1 &= \det \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \det(BA) = D(BA) = D(B)D(A) \\
&= D(A)D(B) = D(AB) = \det(AB) = \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0.
\end{split}
\end{align}

This extension of determinants has all 4 properties if A is a square matrix, and retains some attributes of determinants otherwise.

$$|A|^2=|A^{T}A|$$

If you're willing to break the rules a little bit, this has a valid and useful geometric interpretation. If you have a space defined in a dimension higher than its own, this can still return the area it defines.

Since the square of the determinant of a matrix can be found with the above formula, and because this multiplication is defined for nonsquare matrices, we can extend determinants to nonsquare matrices. For example, take the 3 wide matrix A defined with column vectors, x y and z, where each have n components:

$$A=\begin{pmatrix}x|y|z\end{pmatrix}$$

You can dot each of the vectors with each other by right multiplying A by its transpose:

$$A^{T}A=\begin{pmatrix}x\\y\\z\end{pmatrix}\begin{pmatrix}x&y&z\end{pmatrix}=\begin{pmatrix}
x\cdot x & x\cdot y & x\cdot z\\
x\cdot y & y\cdot y & y\cdot z\\
x\cdot z & y\cdot z & z\cdot z
\end{pmatrix}$$

Taking the determinant of this, you get the square of A's determinant:
$$2 (x\cdot y) (x\cdot z) (y\cdot z)+(x\cdot x) (y\cdot y) (z\cdot z)-(x\cdot z)^2 (y\cdot y) - (x\cdot x )(y\cdot z)^2 - (x\cdot y)^2 (z\cdot z)$$

In this 3 vector example, the equation above returns the value of the volume defined by vectors x y and z.

You may take the positive square root of this to be the absolute value of the determinant. It's always positive because it doesn't make sense to define positive and negative areas for spaces defined in dimensions higher than the space itself. Depending on the perspective, a positive area can become a negative area if looked at from behind.


What is the rigorous justification for treating differentials as fractions while solving differential equations? For example, given the differential equation
$\dfrac{\mathrm{d}y}{\mathrm{d}x}=y\tag*{}$
We do,
$\dfrac{\mathrm{d}y}{y}=\mathrm{d}x\tag{*}$
At which point we go and integrate both sides.
Is there a rigorous justification for being able to go to $(*)?$

Basically you are using the fundamental theorem of calculus and the chain rule (combined together they constitute the rule of integration by substitution). When you write $f(y) dy = g(x) dx$ you are really writing $f(y(x)) \frac{dy}{dx} = g(x)$. You then integrate both sides over the same limits, obtaining

$$\int_a^b f(y(x)) \frac{dy}{dx} dx = \int_a^b g(x) dx$$

The left side is

$$\int_{y(a)}^{y(b)} f(y) dy$$

so that "dropping the integral signs" gives

$$f(y) dy = g(x) dx.$$

But in the standard framework this last equation does not really mean anything.

The definition of "differential" is $dy = f'(x) dx$.  Replace $f'(x)$ with its other notation $\frac{dy}{dx}$ to get $dy = \frac{dy}{dx} dx$.  So when you start with

$$\frac{dy}{dx} = y$$

you can multiply both sides by $dx$

$$\frac{dy}{dx} \; dx = y  \; dx$$

and then replace the left side by $dy$.


Show that $\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2 < 2$ I`m trying to show that this integral is converges and $<2$
$$\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2dx < 2$$
What I did is to show this expression:
$$\int^{1}_{0}\left(\frac{\sin(x)}{x}\right)^2dx + \int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx$$
Second expression :
$$\int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx < \int^{\infty}_{1}\left(\frac{1}{x^2}\right)^2dx = \lim\limits_{b\to 0}  {-\frac{1}{x}}|^b_0 = 1 $$
Now for the first expression I need to find any explanation why its $<1$ and I will prove it.

I would like to get some advice for the first expression. thanks!

Well, this likely isn't what you had in mind, but you could just evaluate the integral.  In this case, Parseval-Plancherel's theorem works:

$$\int_{-\infty}^{\infty} dx\, |f(x)|^2 = \frac{1}{2 \pi}\int_{-\infty}^{\infty} dk\, |\hat{f}(k)|^2$$

where $\hat{f}$ is the Fourier transform of $f$.  For $f(x)=\sin{x}/x$, we have

$$\int_{-\infty}^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 = \pi$$

so that

$$\int_0^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{\pi}{2} < 2$$

The answer is 0 < 2. Found using the congruity $\sin^2(x)=\frac {1-\cos(2x)}{2}$. And evaluating $\frac{-1-\cos(2x)}{2x}$ from 0 to infinity. This ends up being the limit of $\frac {\cos^2(x)}{x}$ as x goes to 0.


Why isn't the 3 body problem solvable? I'm new to this "integrable system" stuff, but from what I've read, if there are as many linearly independent constants of motion that are compatible with respect to the poisson brackets as degrees of freedom, then the system is solvable in terms of elementary functions. Is this correct? I get that for each linearly independent constant of motion you can reduce the degree of freedom by one, but I don't understand why the theorem

  Theorem (First integrals of the n-body problem) The only linearly independent integrals of the $n$-body problem, which are algebraic with respect to $q$, $p$ and $t$ are the $10$ described above. (http://en.wikipedia.org/wiki/N-body_problem#Three-body_problem)

implies that there is no analytic solution (I think this is synonymous with closed-form solution, and solution in terms of elementary functions). I've been trying to think about it, but I can't reason it, and apparently integrability implies no chaos, which I can't see either.

For the classical 3-body problem, the obstacle to a solution is, as you said, integrability.  This is also sometimes called separability, and when it fails, it means that there does not exist a manifold in phase space such that on that manifold, the equations for the independent degrees of freedom of the equation are separated into independent equations.  This is in turn related to being able to interchange mixed partial derivatives as you mention for the Poisson brackets, because if the equations separate, derivatives (and therefore integrals) can be performed in any order.

The relationship between this and chaos is that non-integrable systems are generically chaotic -- meaning "usually" or "observably" chaotic, the obstacle to separating the degrees of freedom being that there are intersecting stable and unstable manifolds of hyperbolic periodic points which cause the solutions to fold endlessly in phase space.  "Generic" has a definition here, it means true on a countable interesection of open dense sets -- in other words, for every solution, there is an open subset of solutions arbitrarily close which have this property.

Hope this helps.  There is a completely worked out solution for what is called the "restricted 3-body problem" (3 body problem in which one of the bodies has no mass) in Jurgen Moser's Stable and Random Motions in Dynamical Systems, which shows that even in this case, the motion of the massless body is chaotic for most initial conditions.

2-Body Problems also exist which have no specific solution such that there is a range of solutions for a given physical condition.  This means solvability is not based on the number of bodies but the state and representation of space.

Indian Journal of Science and Technology published a physical proof called, “Binary Precession Solutions based on Synchronized Field Couplings”

http://www.indjst.org/index.php/indjst/article/view/30008/25962

In this research, a generalized wave function with classical characteristics was isolated within the motion of binary stars.  The wave function provided the first tool for cracking the complex motion of DI Herculis and other binary stars that had several measured precession solutions.

http://xxx.lanl.gov/pdf/1111.3328v2.pdf

In this research, published about a year after the Indian Journal of Science and Technology publication, mathematicians from Imperial College London produced a proof for the physical existence of wave functions.  The research was published in Nature Magazine.


The difference between log and ln $$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$

Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?

The common logarithm is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $\log$.
The natural logarithm is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $\ln$.
In general, if $a\gt 0$, $a\neq 1$, then the inverse of the function $a^x$ is the "logarithm base $a$", $\log_a(x)$.
The "guiding formula" is
$$\log_a(b) = r\text{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:

$\log_a(xy) = \log_a(x)+\log_a(y)$: logarithm of a product is the sum of the logarithms.

Why? Say $\log_a(x) = r$ and $\log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $\log_a(xy) = r+s = \log_a(x) + \log_a(y)$.

$\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$.

Why? Again, say $\log_a(x) = r$ and $\log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$\frac{x}{y} = \frac{a^r}{a^s} = a^{r-s}$, which means $\log_a\frac{x}{y}=r-s = \log_a(x)-\log_a(y)$.

$\log_a(x^t) = t\log_a(x)$.

Why? If $\log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $\log_a(x^t) = rt = t\log_a(x)$.

$\log_a(a^r) = r$ and $a^{\log_a(x)} = x$. Because $\log_a(x)$ and $a^x$ are inverses of each other.

In particular, $\ln$, which is $\log_{e}$; and using $\log$ for $\log_{10}$, we have these properties:
$$\begin{align*}
\log(xy) &= \log(x)+\log(y) &\qquad \ln(xy) &=\ln(x) + \ln(y)\\
\log\left(\frac{x}{y}\right) &= \log(x) - \log(y) &\ln\left(\frac{x}{y}\right) &= \ln(x) - \ln(y)\\
\log(x^a) &= a\log(x) & \ln(x^a) &= a\ln(x)\\
\log(10^x) &= x & \ln(e^x) &= x\\
10^{\log(x)} &= x & e^{\ln(x)} &= x
\end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $\log_a(x)$ and $\log_b(x)$?
If $\log_b(x)=r$, then $b^r = x$. So
$$\log_a(x)= \log_a(b^r) = r\log_a(b) = \log_b(x)\log_a(b).$$
So we get that
$$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}.$$
As Henning points out below, while $\ln$ is not ambiguous (it always denotes logarithm base $e$), $\log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science, it is very often used to denote logarithm base $2$. For some applications, it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).

The symbol $\log$, by itself and without other conventions, is meaningless, as is the word "logarithm."
The phrase "logarithm of $x$ to the base $b$" has meaning if $b$ and $x$ are positive.  The meaning is defined as follows: the logarithm of $x$ to the base $b$ (denoted by  $\log_{b} (x)$) is the exponent to which $b$ needs to be raised to obtain $x$.
That is, if $y = \log_{b} (x)$, then $x = b^y$.
By convention, the "natural logarithm" is the logarithm to the base $e$, where $e$ is Euler's constant:
$$
\ln(x) = \log_{e}(x).
$$


Proving no real solutions Please read the question before marking as duplicate; this is proving there are NO SOLUTIONS OUTSIDE THE INTERVAL.

  Given that:

  $x^3−3x+3=\sin(x)$ show that there are no real solutions outside the
  interval $[−3,−2]$

am not trying to prove there is a solution; this has been done with intermediate value theorem; proving no real solutions outside interval

By derivation, the LHS has extrema at $x=\pm1$, namely $(-1,5)$ and $(1,1)$. We also have the points $(-3,-15)$ and $(-2,1)$.

We can write the following variation table:

$$\begin{array}&x&-\infty&-3&-2&-1&1&\infty\\LHS&-\infty&-15&1&5&1&\infty\end{array}$$

This proves that the LHS remains out of $[-1,1]$ outside $[-3,-2]$, except maybe at $x=1$. But then $\sin 1<1$ and there is no solution.

Nice! How's question 3c doing for you?

Essentially what you want to do is find the critical points and prove that the local minimum does not occur below the x-axis.
You would find the critical points by deriving f(x) to get f^(1)(x) = 3x^2-3-cos(x).
You know that -1<=cos(x)<=1, so in order for f^(-1)(x) to equal 0, the component |3x^2-3|<=1. Solving this, x<= sqrt(4/3), and x>=-sqrt(4/3)
Find the derivative of both, use either first or second derivative principals to determine nature, and check endpoints to see if they are strictly increasing, or strictly decreasing.

Simple...ish


Prove that for every real number $x$, if $x>2$ then there is a real number $y$ such that $y+1/y = x$. I'm trying to learn about proofs and I'm stuck in Velleman's book "How to Prove it".

This is the question (ex.6 p.118):

  Prove that for every real number $x$, if $x > 2$ then there is a real number $y$ such that $y+\frac{1}{y}= x$.

I couldn't do it so I went to see the answer, and it is written that we can assume that $y=(x+\sqrt{x^2-4})/2$.

How did the author get to this value for $y$?

Suppose that

$$y+\frac{1}{y}=x.$$

Then

$$y^2+1=xy,$$

and

$$y^2-xy+1=0.$$

It follows that

$$y=\frac{x\pm\sqrt{x^2-4}}{2}$$

either by completing the square, or by using the quadratic formula.

The author just pulled it out of nowhere - where it comes from isn't important to the proof, only that it works. For how they knew what to pull out of nowhere, try rearranging $y + \frac{1}{y} = x$ (which I suspect is what the formula was supposed to read?) to make $x$ the subject.


Solve for $x$ where $A = \sin (x) /B + \cos (x)/C$ where $A, B,$ and $C$ are known It's been a few years since I've done trig and I've reduced an equation down to the following:

$$A = \sin (x)/B + \cos (x)/C$$

Is it possible to solve for $x?$

Edit: Diagram and clarification added.

The problem I'm trying to solve is the angle of rotation for a scaled down rectangle inside another rectangle such that all 4 corners touch the original rectangle.

PS: Thank you all for the amazing responses!  This was clearly way above my head.  I'm super jealous and wish I had continued to pursue math after graduation.  I'll definitely have to start brushing up...

This is a linear equation in sine and cosine. Rewrite it as
$$
C\sin x+B\cos x=ABC
$$
and then divide both sides by $\sqrt{B^2+C^2}$:
$$
\frac{C}{\sqrt{B^2+C^2}}\sin x+\frac{B}{\sqrt{B^2+C^2}}\cos x=\frac{ABC}{\sqrt{B^2+C^2}}
$$
Set $s=ABC/\sqrt{B^2+C^2}$ and find an angle $\alpha$ such that
$$
\sin\alpha=\frac{C}{\sqrt{B^2+C^2}},
\qquad
\cos\alpha=\frac{B}{\sqrt{B^2+C^2}}
$$
which is surely possible. Then the equation becomes
$$
\cos x\cos\alpha+\sin x\sin\alpha=s
$$
or
$$
\cos(x-\alpha)=s
$$
that is elementary.

Of course, there's no solution if $|s|>1$.

Note about “surely possible”: the point with coordinates
$$
\left(\frac{B}{\sqrt{B^2+C^2}},\frac{C}{\sqrt{B^2+C^2}}\right)
$$
lies on the unit circle. Connecting it with the origin defines the required angle $\alpha$, considering as the other angle side the positive $x$-semiaxis.

$$
\cos(x - \phi) = \cos x\cos\phi + \sin x\sin \phi
$$
let
$$
\frac{1}{B} = \sin \phi\\
\frac{1}{C} = \cos \phi\\
$$
where $\vert\frac{1}{B}\vert \leq 1 $ and similarly for $\vert\frac{1}{C}\vert \leq 1 $.
thus
$$
\tan\phi = \frac{C}{B}
$$
provided that $C/B$ such that $|\tan \phi| <\infty$
then we have
$$
\cos(x-\phi) = A \text{for $\vert A\vert < 1$}
$$
solve for $x$ as usual.


What does 'real-valued' function mean in topology? If I have a topological space $X$ and a 'real-valued' function $f$ on $X$. Does this mean I have a map of the form: $f: X \rightarrow \mathbb R$ where $\mathbb R$ has the usual topology? Or something else?

This may seem obvious but somehow it's unclear to me.

I disagree with Mason. I have only ever seen the phrase "real-valued function on $X$" mean a function $f : X \to \mathbb{R}$. Wikipedia seems to support my claim.

'real valued' implies that the image is going to be a real number or vector in the n-dimensional reals. It doesn't necessarily imply that the image has the usual topology, or even that the image is all of the reals.


Finding kernel and range of a linear transformation We are given:

  Find $\ker(T)$, and $\textrm{rng}(T)$, where $T$ is the linear transformation given by

  $$T:\mathbb{R^3} \rightarrow \mathbb{R^3}$$

  with standard matrix

  $$ A = \left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]\textrm{.}
$$

The kernel can be found in a $2 \times 2$ matrix as follows:

$$ L = \left[\begin{array}{rrr}
    a & b\\
    c & d\\
    \end{array}\right] = (a+d) + (b+c)t
$$

Then to find the kernel of $L$ we set

$$(a+d) + (b+c)t = 0$$
$$d = -a$$
$$c = -b$$

so that the kernel of $L$ is the set of all matrices of the form
$$ A = \left[\begin{array}{rrr}
    a & b\\
    -b & -a\\
    \end{array}\right]
$$

but I do not know how to apply that to this problem.

$$
A = \left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]
$$
Consider a linear map represented as a $m × n$ matrix $A$ .
The kernel of this linear map is the set of solutions to the equation $Ax = 0$
$$
ker(A)=\{x \in R^n|Ax=0\}
$$
$$
det(A)=1(12+16)-(-1)(10+28)+3(20-42)=0
$$
Since $det(A)=0$ , $x\ne0$ and $0$ is a vector here.
$$
\left[\begin{array}{rrr}
    1 & -1 & 3\\
    5 & 6 & -4\\
    7 & 4 & 2\\
    \end{array}\right]
\left[\begin{array}{r}
   a\\b\\c
   \end{array}\right]
=\left[\begin{array}{r}
   0\\0\\0
   \end{array}\right]
$$
In row-reduced form,
$$
A = \left[\begin{array}{rrr}
    1 & 0 & \frac{14}{11}\\
    0 & 1 & \frac{-19}{11}\\
    0 & 0 & 0\\
    \end{array}\right]
$$
$$x=\frac{-14}{11}z$$
$$y=\frac{19}{11}z$$
$$
\left[\begin{array}{r}
   a\\b\\c
   \end{array}\right]
=\left[\begin{array}{r}
   -14\\19\\11
   \end{array}\right]z
$$
Similarly for $2×2$ matrix .

To find the range(image) of T, find the transpose of the matrix first and then reduce the transposed matrix to an echelon form, the remaining non zero matrix becomes the basis for the range and the dimension becomes the rank


Problem with square roots and squares Suppose we have
$$ x^{2} = (-x)^{2}.$$
I understand that this equation holds because
$$\begin{aligned}
(-x)^{2} & = (-1\cdot x)^{2} \\
& = (-1)^{2} \cdot x^{2} \\
& = x^{2},
\end{aligned}$$
and so $\sqrt{x^{2}} = \sqrt{x^{2}}$ becomes $x = x$ and everything is fine.
But what if we just take the square root first, like so:
$$\sqrt{x^{2}} = \sqrt{(-x)^{2}}.$$
Wouldn't this gives $x = -x$, which is a contradiction?

Since $\sqrt{x}$ denotes the nonnegative square root of $x$, it is not true in general that $\sqrt{x^2}=x$. For instance, if $x=-5$, then
$$
\sqrt{x^2}=\sqrt{25}=5=-x \, .
$$
So sometimes $\sqrt{x^2}$ equals $x$ and sometimes it equals $-x$, depending on the sign of $x$. In general,
$$
\sqrt{x^2} =
\begin{cases}
x \text{ if $x\geq0$} \\
-x \text{ if $x\leq 0$}
\end{cases}
\, .
$$
What makes things confusing is that if $x<0$, then $-x$ is positive.

i think here i'm not really sure if the outer square root can be applied first simply to be honest. I mean it looks like just in itself this example proves that the square root needs to follow certain rules to "make sense".


How do we know we get the right answer? The problem of ontology is one much discussed in mathematical philosophy with much categorization into different schools of thought, but the problem of epistemology seems to be less discussed; specifically the question not of how we 'know' mathematical objects in the first place, but of how we can be sure that a particular physical 'process' (whether a calculator or mental arithmetic) gives the 'right answer'.

My question arose after asking this  about Chaitin's constant $\Omega$, a number defined so that its digits represent the solution to an undecidable problem, and hence (assuming the Church-Turing principle or perhaps one slightly stronger, but not as strong as Church-Turing-Deutsch) we can never work out what its digits are (although we can find some). However, it is possible that a physical constant might have value $\Omega$, or $\Omega$ might even (as this answer put it) be 'engraved on a monolith buried on the moon alongside the axioms of ZFC'. It is standard to argue that if this were the case we could never know (e.g. see here). David Deutsch's response to this here is as follows:

  [It is not] obvious a priori that any of the familiar recursive functions is in physical
  reality computable. The reason why we find it possible to construct, say, electronic calculators, and
  indeed why we can perform mental arithmetic, cannot be found in mathematics or logic. The reason
  is that the laws of physics ‘happen to’ permit the existence of physical models for the operations of
  arithmetic such as addition, subtraction and multiplication. If they did not, these familiar operations
  would be non-computable functions. We might still know of them and invoke them in mathematical
  proofs ... but we could not perform them.... Chaitin (1977) has shown how the truth values of
  all ‘interesting’ non-Turing decidable propositions of a given formal system might be tabulated very
  efficiently in the first few significant digits of a single physical constant.
  But if they were, it might be argued, we could never know because we could not check the accuracy
  of the ‘table’ provided by Nature. This is a fallacy. The reason why we are confident that the
  machines we call calculators do indeed compute the arithmetic functions they claim to compute is not
  that we can ‘check’ their answers, for this is ultimately a futile process of comparing one machine
  with another: Quis custodiet ipsos custodes? The real reason is that we believe the detailed physical
  theory that was used in their design. That theory, including its assertion that the abstract functions of
  arithmetic are realized in Nature, is empirical. [bold emphasis mine]

Thus it appears that we have two different philosophical approaches to this problem, either claiming that we could never verify (or hence use) $\Omega$'s digits if we thought we had them, or else claiming that an empirical justification would be sufficient; I wonder whether there are other approaches to the problem? Some degree of empiricism seems to be required regardless. Note that this problem is not the same as the general epistemological problem of how we know anything. It is the problem of how, assuming we already know a mathematical object, we can be sure that a physical process corresponds to it. The ubiquity of computational devices (including our brains) in mathematics and the rise of computer-assisted proof seem to make this important.

My question is thus: What are the different mathematical schools of thought with regards to the problem of how we can be sure that physical processes give us the 'right' answer (i.e. correspond to the mathematical objects we think they do)?

I don't know that there's a clear universal definition of knowledge. I could define knowledge in such a way that somebody knows a mathematical statement is true after they mentally figure out that it's true. The problem with that definition is that people sometimes make careless mistakes in a mental computation. I think it's still possible to define knowledge in such a way that those with a different type of brain who never make careless mistakes really know that they figured out the right answer and other people just don't know that that person knows it. They could figure out a product mentally first then check whether the calculator gives the same result and then claim that they know the calculator gave the correct answer because they already figured it out, and then other people would agree that it's correct to say that person really knows it.


Equivalent conditions for almost sure convergence. Consider the definition of almost sure convergence:

  The sequence of random variables ${(X_n)}_{n \in \mathbb{N}}$ defined on the probability space $(\Omega, \mathcal{F}, P)$ converges almost surely to a random variable $X$ defined on the same probability space, if
  $$
P(\{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}) = 1.
$$

Set $A := \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$. Clearly, a.s. convergence implies that $P(A^{c}) = 1 - P(A) = 0$.

I am wondering if the definition of a.s. convergence is equivalent to any of the following statements:

$X_n \rightarrow X$ a.s. $\iff$ $\exists M \in \mathcal{F}$ with $P(M) = 1$, such that $\lim_{n \rightarrow \infty} X_n(\omega) = X(\omega) \quad \forall \omega \in M $.
$X_n \rightarrow X$ a.s. $\iff$ $\exists N \in \mathcal{F}$ with $P(N) = 0$, such that $\lim_{n \rightarrow \infty}X_n(\omega) = X(\omega) \quad \forall \omega \in N^c$

The first and the second statements follow from one another, provided that either of them is correct. To show "$\Rightarrow$" of the first statement, set $M := A$. To show "$\Leftarrow$", observe that $M \subset A$ and thus $P(A) > P(M) =1 \Rightarrow P(A)=1$.

I would like to know if the reasoning is correct or there is something that I neglect. Is the completeness of the probability space relevant?

With that definition of almost sure convergence then you will indeed need to assume completeness for your statements to be equivalent.
Indeed if the set $\Omega - \{ \omega \in \Omega: \lim_{n \rightarrow \infty} X_n(\omega) = X(\omega)\}$ is not measurable but negligible (i.e. can be included in a measurable set with measure 0), than its probability is not even defined, i.e. it does not converge a.s. as per the first definition. It does however converge a.s. as per your proposed definitions 1. and 2.

Moreover, on a probability space, almost sure convergence will be equivalent with almost uniform convergence (by Egoroff's theorem, see for instance: https://en.wikipedia.org/wiki/Egorov%27s_theorem#Statement_of_the_theorem).


Guessing a subset of $\{1,...,N\}$ I pick a random subset $S$ of $\{1,\ldots,N\}$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need?

An obvious upper bound is $N$ queries, since you can test each element individually.  On the other hand, it takes at least $\Omega(N/\log N)$ queries: $N$ bits of information are required to identify the target subset, and each query can yield at most $O(\log N)$ bits of information, since each query has only $O(N)$ possible answers.  To see that the upper bound is not sharp, consider the following strategy for $N=5$, which takes at most $4$ queries:

Guess $\{1,2,3,4,5\}$.  If the result is $0$ or $5$, we have the answer.  If the result is $1$ or $4$, bisection search (for the single member or the single missing element) gives the answer in three more queries.  Suppose the result is $2$ (the strategy for $3$ is the same by symmetry).
Guess $\{1,2\}$.  If the result is $2$, we have the answer.  If the result is $0$, then bisection search on $\{3,4,5\}$ (for the single missing element) gives the answer in two more queries.  Suppose the result is $1$.  Then we know the answer is $\{a,b\}$ for some $a \in \{1,2\}$ and $b \in \{3,4,5\}$.
Guess $\{1,3\}$.  If the result is $2$, we have the answer.  If the result is $0$, then the answer is $\{2,b\}$ for some $b\in\{4,5\}$, and one more query gives the answer.  Suppose the result is $1$.  Then we know the answer is $\{1,4\}$, $\{1,5\}$, or $\{2,3\}$.
Guess $\{1,4\}$.  The answer is $\{1,4\}$ if the result is $2$, or $\{1,5\}$ if the result is $1$, or $\{2,3\}$ if the result is $0$.

This example gives an improved upper bound asymptotic to $4N/5$.  It seems likely that the correct answer is strictly $o(N)$ (i.e., eventually less than $cN$ for any fixed $c$), but whether or not it's $\Theta(N/\log N)$, I can't say.

$M$ denotes the solution set, $|M|$ denotes its cardinality (the number of elements).

Note that if I guess $\{1\dots N\}$, then you'll have to answer with $|M|$.

Now I can just brute force your set by removing the elements of $\{1\dots N\}$ one-by-one. Every time your answer decreases I will have identified an element of the solution. By induction, your answer will have to reach zero eventually. By that time I will have identified all $x\in M$. It will take me one more guess to win the game.

Using the algorithm outlined above I would guess at most $N+1$ times.

I am pretty sure in that you cannot do better than $O(n)$ in the worst case (I offer no proof though).


Backgammon variant — two player double or nothing on random walk? Two players, A and B, play the following backgammon variant.

  A series of steps is labelled $-2$, $-1$, $0$, $1$ and $2$, and a
  chess piece is placed on step $0$. A fair coin is flipped, and it
  lands on heads, the chess piece is moved forward by one step, while if
  it lands on tails, the piece is moved back by one step. Player A wins
  if the piece lands on step $2$ first, while player B wins if the piece
  lands on step $-2$ first, and the loser must pay the winner some
  amount of money.

  The amount is originally set at \$1. Initially, either player can
  challenge the other player to double this stake at any time (before
  the game ends), and the other player must either accept this or
  forfeit and lose the game. Following that, two consecutive challenges
  cannot be raised by the same player (i.e. if player A challenges first
  and B accepts, A cannot challenge again until B challenges him), and
  only one challenge can be raised each turn.

  When should you challenge, and if challenged, should you accept? What
  is the expected value of your strategy?

My initial thought was that if challenged, I would only accept when at $0$ (indifferent) and $+1$ (if I was player A) or $-1$ (if I was player B). However, this would mean that the other player challenged one step before possibly losing. How do I reconcile this? Is there a reason why a player would challenge if one step away from potentially losing?

The analysis in quarague’s answer isn’t correct because it only takes the possibility of doubling once into account, whereas future doubling opportunities in fact increase the expected payoff.

Denote by $x_k$ the expected value of the game for $A$ when $A$ has score $k$ and has the right to challenge and $B$ doesn’t. We can guess that $A$ challenges when the score is $1$, and $B$ accepts, and then check whether this is self-consistent. Under these assumptions, we have

$$
2x_0=x_{-1}+x_1=x_{-1}-2x_{-1}=-x_{-1}
$$

and

$$
2x_{-1}=x_0+x_{-2}=x_0-1\;,
$$

and thus $x_{-1}=-\frac25$, $x_0=\frac15$ and $x_1=\frac45$. The assumptions turn out to be self-consistent, since the expected return for $A$ at score $1$ would only be $\frac12\left(\frac15+1\right)=\frac35\lt\frac45$ without challenging, and the expected return for $B$ upon refusing would be $-1\lt-\frac45$.

It remains to find the optimal strategy and expected return in the initial state of the game, where both players have the right to challenge. Denote these expected returns by $y_k$. Then $y_0=0$ by symmetry, and $y_1$ is $\frac12$ if $A$ doesn’t challenge, $1$ if $A$ challenges and $B$ refuses and $-2x_{-1}=\frac45$ if $A$ challenges and $B$ accepts, so $A$ challenges at score $1$ and $B$ accepts.

To summarize, the initial value of the game is $0$ by symmetry; a player challenges exactly if they have a score of $1$, the other player always accepts, and the value of the game for the player with the right to challenge is $-\frac25$, $\frac15$ and $\frac45$ at a score of $-1$, $0$ and $1$, respectively.

One can analyse the expected values at each stage. First the base case without doubling, at stage 0, player A has an expected return of $0\$$ due to symmetry. At stage $1$, he has a $50\%$ chance to win immediately and a $50\%$ chance to go back to stage $0$. This gives A an expected return of $0.5\$$. By symmetry at stage $-1$ A has an expected return of $-0.5\$$.

Now for doubling, by symmetry only consider whether A should offer and whether B should accept. At stage 0, A has zero expected return, so doubling doesn't change anything, neither does it for B. At stage 1, if A offers to double and B accepts, A's expected return doubles, if B refuses it also doubles to the current stake. So A should offer to double and B can take the double or not but it doesn't change his expected return. At stage -1, if A offers to double and B accepts, A's expected return goes from $-0.5$ to $-1$, if B refuses it goes to $1$. So B would accept it but A shouldn't offer it.


Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ?

Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$.  If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$.  So any odd number with more than one factorization of this type gives an example.

Thus from $m = 15 = 1 \cdot 15 = 3 \cdot 5$, we get $8^2 - 7^2 = 4^2 - 1^2$, or $1^2 + 8^2 = 4^2 + 7^2$.

From $m = 21 = 1 \cdot 21 = 3 \cdot 7$ we get $11^2 - 10^2 = 5^2 - 2^2$, or $2^2 + 11^2 = 5^2 + 10^2$.

Any Pythagorean triple $(A^2+B^2=C^2)$ provides a candidate where the $C^2$ can have $2$-or-more combinations of $A^2$ and $B^2$ that add up to it. These can be found by testing natural numbers of the form $(4n+1)$ with a range of $m$ values defined as shown below to see which, if any yield integers. We begin with Euclid's formula and solve the $C$-function for $n$ in terms of $C$ and $m$.
$$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$
$$C=m^2+n^2\implies n=\sqrt{C-m^2}\qquad\text{where}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$
The lower limit ensures $m>n$ and the upper limit ensures $n\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\text{ and we find} \quad m\in\{7,8\}\Rightarrow n\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\quad  F(8,1)=(63,16,65) $$
Here, we have $\qquad 33^2+56^2=65^2\qquad$ and $\qquad 63^2+16^2=65^2$
There infinitely many of these pairs of triples; the next that meet the criteria are
$\quad(13^2,84^2,85^2), (77^2,36^2,85^2)\qquad\qquad  (75^2,100^2,125^2), (117^2,44^2,125^2)\qquad \qquad (17^2,144^2,145^2), (143^2,24^2,145^2)\quad\quad ...$


Show there is a bijection between $R^n\ and\ R^N$ Let $N=\{1,2,3,...,n\}$, show there is a bijection between $R^n\ and\ R^N$

Define $f: R^n \rightarrow R^N$ by $f(x_1,..,x_n)=g_{(x_1,...x_n)}$ where the n-tuple is an element of $R^n$ and $g_{(x_1,...x_n)}:N \rightarrow \mathbb{R}$ is defined by:

$g_{(x_1,...x_n)}(i)=x_{i}$ for $i \in N$

What I want is g to map the elements in N to the corresponding real number in the n-tuple. But I'm not sure that I can define g in this way.

But if I can here is my proof this is injective.

Suppose $(x_1,...x_n) \neq (y_1,...y_n)$ then there is at least one $i \in N$ such that $x_i \neq y_i$.

Then the function $g_{(x_1,...x_n)}(i)=x_i \neq y_i=g_{(y_1,...y_n)}$ for some $i \in N$

Since $f(x_1,..,x_n)=g_{(x_1,...x_n)}$ then $f(x_1,...,x_n)\neq f(y_1,...,y_n)$

My issue and why I think this doesn't work is because when I want to prove this is surjective, I want to prove that for any g, there is an n-tuple s.t $g=g_{(x_1,...,x_n)}$ but the way I've defined g seems to make it depend upon this tuple already existing.

You don't have to worry too much about the tuple existing. You're looking at $\mathbb{R}^n$, which is the set of all $n$-tuples with real numbers in every entry. You are completely free to choose which real numbers to put in all $n$ entries.

If you look at the map $g \mapsto (g(1), g(2), \ldots, g(n))$, then this is a well-defined map from $\mathbb{R}^N$ to $\mathbb{R}^n$, and I think you'll find that it's the inverse of $f$.

I would just appeal to cardinal arithmetic.  We have $$\mathfrak c = |\Bbb R|=2^{\aleph_0}=2^{|\Bbb N|}\\
\mathfrak c = |\Bbb R|\le |\Bbb R^n|\le |\Bbb R^{|\Bbb N|}|=(2^{|\Bbb N|})^{|\Bbb N|}=2^{|\Bbb N|{|\Bbb N|}}=2^{|\Bbb N|}=\mathfrak c$$


Solve the following for $D$: $ABDB^{-1} = I$ So, here's what I know, 'I' is typically:

\begin{array}{ccc}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
  \end{array}

But, from that I really have no idea where to go with this question. I know basic matrix inversion rules, but I am unsure how to solve.

Hint: How would you solve this equation for $d$?

$$abdc=1$$ assuming $a,b,c \neq 0$. Let's also assume that $a,b,c,d$ do not commute. Secondly, the $I$ you mentioned above is often denoted $I_3$, as it is the $3\times 3$ identity matrix. You are working with $I_n$, which has a diagonal containing $n$ $1$'s.

$ ABDB^{-1} = I \quad\leadsto\quad D = B^{-1}A^{-1}IB $


Is it possible to have a symmetric and transitive relation on $\mathbb Z$ that isn't reflexive? I was wondering this the other day and I have been trying to come up with an example for a bit but can't produce one. If there is a proof that such a thing couldn't exist I would be interested in seeing it. Seeing an example of such a set would also be cool as well. Thanks in advance !

The empty relation - nothing relates to anything else - is symmetric and transitive, but not reflexive.

Well it is simple that is must be reflexive.Let R be relation in Z such that it is both symmetric and transitive.

Let xRy and yRx,but then by transitivity it must hold that xRx.

Thus such relation can not exist.


What if 'proof by contradiction' is not a valid method of proof? I've just been reading this question about the existence (or lack thereof) of contradictions in maths.

I've been wondering:

  What if 'proof by contradiction' is not a valid method to (dis)prove a statement? What if the 'absurdity' is actually a contradiction?

A proof by contradiction says that  to disprove a statement $P$, assume $P$ is true and show that it leads to some contradiction, therefore $P$ is false. But what if the contradiction in this proof was actually a valid contradiction and $P$ is true (and we are just dismissing the contradiction)?

I'll demonstrate my question with an example.

  $\underline{\text{Proof that the sum of a rational number and an irrational number is irrational}}$:

  Let $\frac{a}{b}$ be the rational number and let $x$ be the irrational number
  Assume for a contradiction that $\frac{a}{b}+x$ is rational. i.e. assume that $\frac{a}{b}+x=\frac{p}{q},$ for $p, q \in \mathbb{Z}$.

  Then $x=\frac{p}{q}-\frac{a}{b}=\frac{pb-aq}{qb}$ which is rational, a contradiction. Therefore, $\text{rational+irrational = irrational}. \square$

  But what if the sum of a rational and an irrational number is in fact rational (and this is a contradiction in mathematics)?

If anyone can trim this question to make it more concise and/or articulate, feel free!

The reality (existence) of the particular contradiction that you mention is excluded by the definition of rational or irrational numbers. The irrational numbers are defined to be precisely those real numbers that are not rational. You ask:

  But what if the sum of a rational and an irrational number is in fact rational (and this is a contradiction in mathematics)?

You have shown, in your proof, that the sum cannot be rational. The only way it could be is if $x$ (the irrational number) were a rational number. If $x$ were a rational number, it would, by definition, not be irrational and thus would not satisfy the hypothesis that $x$ is irrational.

Proof by contradiction is therefore a valid method in this instance precisely because we have not assumed that a contradiction cannot occur. On the contrary (ha ha), it is the strict definition of terms that prevents a contradiction. We are only authorised to name something a "contradiction" when we know that it is impossible. We are only authorised to claim to know that something is impossible because we have definitions.

Consider the claim "all red pencils are pencils". If you find a "red pencil" that is not a "pencil", I would claim that is not a "red pencil". This only works if we both agree at the outset that a "red pencil" is a "pencil" that happens to be "red". If we are in agreement with this definition, then the claim "all red pencils are pencils" is not an empirical claim. Similarly, the denial of contradictions in mathematics is not a question of evidence or lack thereof.

Well if the sum of a rational and an irrational number is rational, then you could have something like your example, x=p/q-a/b, and we already know that irrational numbers can't equal rational numbers.


Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$ Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$.

I am stuck with this proof. I thought at first that I need to use contradiction but that did not work. What do you suggest? I know I have to break them into cases basically when $n=1$ then $1=2^0 \cdot 1$ but what about the other case?

Induction argument hint. Do the basis step. Then for the inductive step, assume it holds for $1, \dots, n$. Now take two cases: If $m$ is odd, you are done (why?). If $n$ is even consider $(n+1)/2$.... you fill in the rest.

Hint:

Consider the factorisation of $n$ as a product of primes:
$$n=2^k\prod_{\substack{p_i\;\text{odd}\\\text{prime}}} p_i^{k_i}.$$


$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)!=0$. I am working on showing an integral is zero, and I have simplified to the following
for $m<n$.
 $\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)!$.

From the few test cases I have done, the sum above always seems to be zero. How can I show it is zero for $m<n$? What will it be when $m=n$?

Hint:

$$\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k(k+m)!}{k!}=m!\sum_{k=0}^n\binom{n}{k}\binom{k+m}{m}(-1)^k$$
$$=m![x^m]\sum_{k=0}^n\binom{n}{k}(1+x)^{k+m}(-1)^k$$

Where $[.]$ denotes the Coefficient operator. Can you proceed hereon?

\begin{align}
\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)! & = \sum_{k=0}^n \binom{n}{k}(-1)^km!\binom{k+m}{k} \\
&= m!\sum_{k=0}^n \binom{n}{k}\binom{-m-1}{k} \\
&= m!\binom{n-m-1}{n} \\
&= 0 \text{ when } n, m \in \textbf{N}.\
\end{align}
(If these steps seem a bit arcane, just know that they're all explained in Knuth's Art of Computer Programming, Vol. 1 where he gives a comprehensive list of ways to manipulate binomial coefficients.)


Suppose $f (x)$ is bounded and has a primitive on $[a,\,b]$, and $g\in C \big([a,\,b]\big) $. Must $f (x)\cdot g (x)$ have a primitive? Suppose $f (x)$ is bounded and has a primitive on $[a,\,b]$, and $g\in C \big([a,\,b]\big) $. Must $f (x)\cdot g (x)$ have a primitive?

Suppose it has a primitive, then it's necessary to be Riemann integrable in $[a,\,b]$.

With these hypotheses it is true that $fg$ has a primitive.

For $f$ to have a primitive on $[a,b]$ means that there exists a differentiable function $F$ such that $F'(x) = f(x)$ for all $x \in [a,b]$.

We first show that the function $x \mapsto xf(x)$ must also have a primitive. Taking $G(x) = xF(x)$ we have $G'(x) = xF'(x) + F(x) = xf(x) +F(x)$. Since $F$ is differentiable it is continuous and Riemann integrable on $[a,b]$, and, therefore, has a primitive $H' = F$. Thus $xf(x) = G'(x) - H'(x)$ and $x \mapsto xf(x)$ has a primitive.  It follows by induction that $x \mapsto p(x)f(x)$, where $p(x)$ is a polynomial, must have a primitive.

Since $g$ is continuous, it follows by the Weierstrass approximation theorem that there exists a sequence of polynomials $(p_n)$ that converges uniformly to $g$. Since $f$ is bounded, it follows that $p_nf$ converges uniformly to $fg$.

Finally, we apply the following lemma:

  If $h_n \to h$ uniformly and each $h_n$ has a primitive, then $h$ has
  a primitive.

Since each function $p_nf$ has a primitive and in view of the uniform convergence shown above, it follows that $fg$ has a primitive.

Alternatively consider

$$
H(x) = \int_a^x f(t)g(t)\, dt
$$

Then since $f$ has a primitive it is integrable, $g$ is bounded and continuous so the product $fg$ is integrable and $H$ is well-defined.


Non-uniformly continous function I have found a topic here (https://www.sciencedirect.com/topics/mathematics/uniformly-continuous-function) in Theorem 3.3.10 that: There is a continuous function $f$ on $[0, 1]$ which is unbounded, and
therefore not uniformly continuous.
Anyone, please help me to understand it.

The proof you link defines

Let $(I_n)_n$ be a covering of $[0, 1]$ by open intervals $(r_n, s_n)$ (...)

and then it says

But by the property of this cover, it follows that no finite part of it covers [0, 1]

Which is false: every open cover of $[0,1]$ has finite subcover, because $[0, 1]$ is compact. The proof is just wrong.


How do I solve $\ln(x^2 − 1) − \ln(x − 1) = 2$ I need to solve the following:

  $$\ln(x^2 − 1) − \ln(x − 1) = 2$$

But I am not entirely sure how to solve this.

I know that you can combine them into a fraction but then how would you simplify from there?

$\ln(x^2 - 1) = \ln((x + 1)(x - 1)) = \ln(x + 1) + \ln(x - 1), \tag 1$

assuming all the $\ln$ terms are well-defined; thus,

$\ln(x^2 - 1) - \ln(x - 1) = \ln(x + 1); \tag 2$

then

$\ln(x + 1) = 2, \tag 3$

or

$x + 1 = e^2, \tag 4$

yielding

$x = e^2 -1. \tag 5$

It is easy to see all the $\ln$ terms make sense here, since $e > 2.5 \Longrightarrow e^2 > 6.25 \Longrightarrow e^2 - 1 > 5.5$; also, $e > 2.5 \Longrightarrow e + 1 >3.5 \; \text{and} \; e - 1 > 1.5$ so all the logarithms are defined and positive.

A check:  with $x$ as in (5)

$x^2 = e^4 - 2e^2 + 1, \tag 6$

so

$x^2 - 1 = e^4 - 2e^2 =e^2(e^2 - 2) = (x + 1)(x - 1), \tag 6$

and $\ln$ takes over from here . . . we get

$\ln(x^2 - 1) = \ln(e^2(e^2 - 2)) = \ln e^2 + \ln(e^2 - 2) = 2 \ln e + \ln(x - 1), \tag 7$

etc . . .


Higher Order Differential Equation Solution The differential equation$$y′′′′+ay′′′+by′′+cy′+dy=0$$
has solution
$$ y=−3te^{2t}+2e^{−2t}\sin(5t)$$

Find $a, b, c$ and $d$.

I've tried looking online for problems similar to this but had no luck. From what I imagine since there is only 2 and -2 in exponents, we can write $$(r - 2)(r + 2) = r^2 - 4$$

Does that mean $a = 0, b = 1, c = 0, d = -4$?

Knowing the solution $$y=−3te^{2t}+2e^{−2t}\sin(5t)$$ as suggested by user61527,  you can just use brute force to get $y',y'',y''',y''''$ and plug their expressions into $$y′′′′+ay′′′+by′′+cy′+dy=0$$ After rearrangement, you should arrive to $$e^{-2 t} \left(-3 e^{4 t} (t (8 a+4 b+2 c+d+16)+12 a+4 b+c+32)+2 \sin (5 t) (142 a-21
   b-2 c+d+41)+10 \cos (5 t) (-13 a-4 b+c+168)\right)=0$$
Cancelling all coefficients leads to the following equations $$32 + 12 a + 4 b + c=0$$ $$16 + 8 a + 4 b + 2 c + d=0$$ $$168 - 13 a - 4 b + c=0$$ $$41 + 142 a - 21 b - 2 c + d=0$$ for which the solutions are $a=0,b=17,c=-100,d=116$.

Let D be the differential operator, that is, Dy = y'. The first term y1 in your equation for y satisfies (D-2)^2 y1 = 0. (The square takes care of the factor t.)

The second term y2 satisfies (D+2) y2 = 0.

The third term y3 satisfies (D^2+25) y3 = 0.

The sum y of all three terms satisfies

(D-2)^2 (D+2) (D^2+25) y = 0.

Multiply this out to get your answer.

Note: This is of order 5, not 4. I believe this is an error in the question. For example, if the second term were e^2t and not e^-2t then one would not need the D+2 factor.


What is a proof of Euler's formula $e^{ix}=\cos(x)+i \sin(x)$ that doesn't use differentiation or De Moivre's theorem, if it exists at all? The Taylor series expansions of $e^x$, $\sin(x)$ and $\cos(x)$ are the standard way to prove it, but does there exist an elementary way that doesn't use limits, calculus or De Moivre's theorem?

The number $e$ is transcendental, so it seems impossible to obtain any non-trivial result about it that does not involve infinite processes (limits, derivatives, integrals...). If you want to define the exponential function for irrational arguments using the basic operations, it indeed seems difficult to avoid polynomials or rational functions, and these must be of infinite degree. Finally, to establish the theorem, you need some definition of the trigonometric functions that is not based on the complex exponential. For this you need to resort to well-axiomatized geometry and to establish the equivalence with statements in calculus. In the end, it boils down to computing the arc length along a circle, hence establish $\pi$, another transcendental number.

The relationship between $e^x$ and $\sin(x)$ and $\cos(x)$
You know know the the $\text{exp()}$ function is a special one, and has many interesting properties among them is that expression $ e^{x} = \lim_{n \to \infty} (1+\frac{x}{n})^n $
If you expand this with binomial theorem, and put $n \rightarrow \infty$, you would get the series for $\text{exp()}$
$$ e^{x} = \lim_{n \to \infty} (1+\frac{x}{n})^n $$
$$ e^{x} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040} + \dots \dots $$
$$ e^{x}= \sum_{k=0}^{\infty}\frac{x^k}{k!}$$
From this series, replace $x$ with $xi$, where $i = \sqrt{-1}$ and split the result into real and imaginary section
$$ e^{x} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040} + \dots \dots $$
$$ e^{xi} = 1+xi+\frac{(xi)^2}{2}+\frac{(xi)^3}{6}+\frac{(xi)^4}{24}+\frac{(xi)^5}{120}+\frac{(xi)^6}{720}+\frac{(xi)^7}{5040} + \dots \dots $$
$$ e^{xi} = \{ 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+ \dots \} + \{ x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} + \dots \} \cdot i$$
$$\sin(x) = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} + \dots $$
$$\cos(x) = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+ \dots $$
Therefore $e^{xi} = \cos(x)+i\sin(x)$


Implicit Differentiation problem (Exponential Derivatives) Please help! Use the process of implicit differentiation to find $dy/dx$ given that:

$$x^2e^y − y^2e^x=0 $$

I am trying first to find $y$,

$$y^2e^x = x^2e^y$$

$$y^2 = (x^2e^y)/e^x$$

$$y = \sqrt{(x^2e^y)/e^x}$$

Is this correct? I have the feeling it is not.

The math is right, but if you are using implicit differentiation the point is not to solve for $y$. Instead you would differentiate each term with respect to $x$, assuming that $y$ is some function of $x$ whose derivative is $dy/dx$. For instance, the term $y^2e^x$ yields

$$2y\frac{dy}{dx}e^x+y^2e^x$$

Hopefully this helps as a starting point. (You will need to differentiate the equation term by term as shown, then solve for $dy/dx$.)

So, implicitly differentiating;
$$x^{2}\frac{dy}{dx}e^{y}+2x e^{y}-\left(y^{2}e^{x}-2y\frac{dy}{dx}e^{x}\right)=0$$
Collecting like terms
$$\frac{dy}{dx}\left(x^{2}e^{y}-2ye^{x} \right)=y^{2}e^{x}-2xe^{y}$$
From which it is readily seen that
$$\frac{dy}{dx}=\frac{y^{2}e^{x}-2xe^{y}}{\left(x^{2}e^{y}-2ye^{x} \right)}$$


Fourier transform problem using complex analysis? Find the Fourier transform of the following function:
  $$ \frac{x}{x^2+a^2} $$

Attempt: We want to solve
$$ \hat{F}(k) = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2}.$$
To solve for this real integral, I consider a semicircle contour. The origin is $z = 0$, and the half-circle will enclose the upper half plane. Call the total contour $C$ and the arc of the circle $C_R$. Then
$$\oint_C \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2} + \oint_{C_R} \frac{z e^{-ikz}}{z^2+a^2}. $$
I claim that the second integral on the righthand side goes to zero for large $R$. This is because

  If on a circular arc $C_R$ of radius R and center $z=0$ , $zf(z)\rightarrow 0$ uniformly as R → ∞, then
  $$ \oint_{C_R} f(z) = 0 $$

for us
$$ f(z) =\frac{z e^{-ikz}}{z^2+a^2} \Rightarrow zf(x) = \frac{z^2 e^{-ikz}}{z^2+a^2} $$
which should converge ($z^2/z^2$ approach 1, but the decaying exponential should drop the expression down to zero for large $R$). Thus
$$\oint_C \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2}$$
which implies
$$ \frac{z e^{-ikz}}{z^2+a^2} = \int_{-\infty}^{\infty} \frac{x e^{-ikx}}{x^2+a^2} = 2π i \,\,\, \text{Res}(f;ia) = π i e^{ak} = \text{The WRONG answer}$$

can someone help? I feel like my issue is with the $C_R$ integral…

I got $π^{-}$. We only integrate in the upper half plane, the only pole there is
$+ia$. Using the formula, gives you...


What are the surfaces of constant Gaussian curvature $K > 0$? Besides the sphere, is there any other surface with constant and positive Gaussian Curvature $K$?

The word "surface" is too vague. If it means a smooth subset of $\mathbb{R}^3$ that can be parametrized by two varibles it will "look" like the sphere of radius $1/\sqrt{K}$ near any point (will be locally isometric to a sphere in technical language). That could be any open piece of the sphere, or a disjoint collection of them.

If we additionally assume that this subset must be closed in $\mathbb{R}^3$ then the entire sphere is the only possibility, up to moving it around. If we still want it closed, but allow it to be a subset of $\mathbb{R}^n$ with large enough $n$, then there is a second possibility, the real projective plane. It still "looks" like the sphere near any point, but has a different overall shape.

The most abstract interpretation is to treat "surface" as not sitting inside any space (two-dimensional manifold), but still having some residual structure (called Riemannian metric), that makes the notion of curvature meaningful. However, even this does not lead to anything other than disjoint open pieces of spheres and real projective planes.

sieverts surface is a second example besides spherical surface. I guess that you want this answer.


about a difficult and weird Probability question Let W be the random variable that counts the number of tails before one gets r heads for a coin whose probability of heads is θ. Without using moment generating function, show that the mean and variance for W are [r(1-θ)]/θ and [r(1-θ)]/θ^2
Please help me with the complete proof. I have some ideas, but I can't write it in a formal way.
Thanks!

Let $X$ be the random number of tails that occur before $r$ heads are observed.  The probability that $X = x$ is given by the expression $$\Pr[X = x] = \binom{x + r - 1}{x} \theta^r (1-\theta)^x, \quad x = 0, 1, 2, \ldots,$$ because:

The total number of coin tosses must be $x + r$;
The final coin toss must be heads and must be the $r^{\rm th}$ occurrence of a head;
There are $\binom{x+r-1}{x}$ ways to arrange the $x$ tails among the $x+r-1$ tosses before the final toss;
The $r$ heads have a joint probability of $\theta^r$ of occurring and the $x$ tails have a joint probability of $(1-\theta)^x$ of occurring.

If we trust that $$\sum_{x=0}^\infty \Pr[X = x] = 1$$ for the PMF above, then it is a straightforward exercise to compute $$\sum_{x=0}^\infty x\Pr[X = x]$$ and $$\sum_{x=0}^\infty x(x-1) \Pr[X = x],$$ the first of which immediately gives the expected value, and the second indirectly leads to the variance with the help of the expectation.

As for evaluating the sum, note $$\begin{align*} \sum_{x=0}^\infty x \binom{x+r-1}{x} \theta^r (1-\theta)^x &= \sum_{x=1}^\infty \frac{(x+r-1)!}{(x-1)! \, (r-1)!} \theta^r (1-\theta)^x \\ &= \sum_{x=1}^\infty r \cdot \frac{(x+r-1)!}{(x-1)! \, r!} \theta^r (1-\theta)^x \\ &= r \sum_{x=0}^\infty \frac{(x+r)!}{x! \, r!} \theta^r (1-\theta)^{x+1} \\ &= r \sum_{x=0}^\infty \binom{x+(r+1)-1}{x} \theta^r (1-\theta)^{x+1} \\ &= r \frac{1-\theta}{\theta} \sum_{x=0}^\infty \binom{x+(r+1)-1}{x} \theta^{r+1} (1-\theta)^x, \end{align*}$$ and since this last sum is equal to $1$ (being the sum of the probabilities of all outcomes for a negative binomial distribution with parameter $r+1$ instead of $r$), the result follows.  A similar approach applies for the second sum, which I leave to you as an exercise.

$W$ is the sum of $r$ independent random variables $W_1,\ldots,W_r$ each of which counts the number of tails before one gets one head.  Therefore the mean and variance are respectively $r$ times the mean and variance of any of those independent random variables.

So $\Pr(W_1=k)=(1-\theta)^k\theta$ for $k=0,1,2,3,\ldots\,{}$.  One way to find $\operatorname{E}(W_1)$ is just applying the definition:
\begin{align}
\operatorname{E}(W_1) & = \sum_{k=0}^\infty k\Pr(W=k) = \sum_{k=0}^\infty k(1-\theta)^k\theta \\
& = \sum_{k=0}^\infty \sum_{j=1}^k (1-\theta)^k \theta \quad (\text{This last sum has $0$ terms when $k=0$.}) \\
& = \sum_{j=1}^\infty \sum_{k=j}^\infty  (1-\theta)^k \theta \\
& = \sum_{j=1}^\infty \frac{(1-\theta)^j\theta}{1-(1-\theta)} = \sum_{j=1}^\infty (1-\theta)^j = \frac {1-\theta} \theta.
\end{align}

To find $\operatorname{E}(W_1^2-W_1)$ is algebraically a bit simpler than to find $\operatorname{E}(W_1^2)$ directcly:
\begin{align}
\operatorname{E}(W_1^2-W_1) & = \sum_{k=0}^\infty k(k-1)(1-\theta)^k\theta \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty k(k-1)(1-\theta)^{k-2} \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty k(k-1)\eta^{k-2} \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty \frac{d^2}{d\eta^2} k\eta^{k-2} \\
& = \theta(1-\theta)^2 \frac{d^2}{d\eta^2} \sum_{k=2}^\infty \eta^k \\
& = \cdots
\end{align}
The last series is geoemtric.  Add it up; do the differentiation; then replace $\eta$ with $1-\theta$.  Then by using the known value of $\operatorname{E}(W_1)$, you can find $\operatorname{E}(W_1^2)$ and then $\operatorname{var}(W_1)=\operatorname{E}(W_1^2)-(\operatorname{E}(W_1))^2$.

Two issues:

That $\sum\dfrac{d}{d\eta}\cdots=\dfrac{d}{d\eta}\sum\cdots$ is easily shown when the sum has finitely many terms.  That it works with infinitely many terms is problematic.  A well-known theorem that's not that hard to prove says it does work when the sum is a power series and one is in the interior of the circle of convergence.
Perhaps it is inelegant for me to use two very different methods for the two expected values.  I have a preference (on this particular occasion of this particular posted question) for the first method because it avoids the issue in the first bullet point above.  (The second method might be somewhat more generally applicable and thus preferable in some other contexts.)  My excuse is that there are only 24 hours in a day and I must attend to some other things.


Examples of problems that are easier in the infinite case than in the finite case. I am looking for examples of problems that are easier in the infinite case than in the finite case. I really can't think of any good ones for now, but I'll be sure to add some when I do.

One can compute the value of
$$\int_{0}^{\infty}e^{-t^{2}}\,\mathrm{d}t$$
exactly. This is known as the Gaussian integral, and it has its own Wikipedia page. The answer turns out to be $\frac{1}{2}\sqrt{\pi}.$

But one cannot do the same with
$$\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t$$
because the antiderivative of the integrand is not an elementary function. This is why we gave a name to the error function $$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^{2}}\,\mathrm{d}t,$$ which also has its own Wikipedia page.

In that sense, the infinite case is easier than the finite case.

Addendum: The same phenomenon occurs for variants of this integral, in particular we can transform the integrand to evaluate $$\int_{-\infty}^{\infty}ae^{-(t-b)^{2}/(2c^{2})}\,\mathrm{d}t = \sqrt{2}a\lvert c\rvert \sqrt{\pi}$$ as detailed here on Wikipedia.

I don't believe nobody mentioned applications of the Central Limit Theorem (CLT).

For example calculating this:
\begin{align}
\int^{n+\sqrt[]{n}}_0 \frac{x^{n-1}}{(n-1)!}e^{-t}\,dt
\end{align}
is very easy when $n\to \infty$ using the CLT.

And yet another application of CLT itself is Wilk's Theorem very strong statement  about the infinity case.


The Fundamental group of Klein Bottle My question is if
$$\pi_{1}(KB)\cong\frac{\mathbb{Z}(a)}{\langle a^{2} \rangle}*\frac{\mathbb{Z}(b)}{\langle b^{2}\rangle}\cong\frac{\mathbb{Z}(a)*\mathbb{Z}(b)}{\langle a^{2}b^{2} \rangle}$$
and
$$\dfrac{\mathbb{Z}(c)}{\langle c^{2} \rangle} \cong \mathbb{Z}_{2},$$
where KB is the Klein Bottle, thus
$$\frac{\mathbb{Z}(a)}{\langle a^{2}\rangle}*\frac{\mathbb{Z}(b)}{\langle b^{2}\rangle}\cong \mathbb{Z}_{2}*\mathbb{Z}_{2}$$
and
$$\pi_{1}(KB)\cong \mathbb{Z}_{2}*\mathbb{Z}_{2} ?$$
It makes sense to me, but I'm not sure.

Claim: $\pi(K) = \mathbb{Z} \rtimes \mathbb{Z}$

By the application of Van Kampen's Theorem to two dimensional CW complexes we have:

$ \pi(K) = <a,b|abab^{-1}=1> $

Let $A$ be the subgroup generated by $a$ and $B$ be the subgroup generated by $b$.  Then since $bab^{-1}=a^{-1}$, we have that $B$ is a normal subgroup.

We can show that every element has a unique representation in the form $b^na^m$, so that $BA=\pi(K)$ and $B\cap A = 0$.

Now $\pi(K)$ is the internal semidirect product of $A$ and $B$, which are each isomorphic to $\mathbb{Z}$.


If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.

$\dfrac{1 + a + b + c + ab + bc + ac + abc}{7} > \dfrac{a + b + c + ab + bc + ac + abc}{7} \geq \sqrt[7]{a^4b^4c^4}$

where the second inequality follows from the AM-GM inequality.


Parallelogram identity in the wave equation Using the parallelogram identity, I need to solve the following initial boundary value problem
for a vibrating semi-infinite string with a nonhomogeneous boundary condition:

  $$ u_{tt} − u_{xx} = 0 , \ 0 < x < \infty, t > 0 $$
  $$u(0,t) = h(t)$$
  $$u(x,0) = f(x), \ u_{t}(x,0) = g(x)$$
  where $f, g, h ∈ C_2\{[0, ∞)\}$

I really have try to solve it, be I still dont know how to use the parallelogram identity. Thanks for your help.

Edit: The parallelogram identity is

$u(x_0 − a, t_0 − b) + u(x_0 + a, t_0 + b) = u(x_0 − b, t_0 − a) + u(x_0 + b, t_0 + a).
$

I have no idea about "parallelogram identity" in PDE, I only know that this PDE problem with those types of I.C.s and B.C.s can be found the solution exactly in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf.

Case $1$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds=h(t)$

$u(x,t)=\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds$

Case $2$: $\dfrac{f(t)+f(-t)}{2}+\dfrac{1}{2}\int_{-t}^t g(s)~ds\neq h(t)$

$u(x,t)=\begin{cases}\dfrac{f(x+t)+f(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}g(s)~ds&\text{when}~x>t\\\dfrac{f(x+t)-f(t-x)}{2}+\dfrac{1}{2}\int_{t-x}^{x+t}g(s)~ds+h(t-x)&\text{when}~x<t\end{cases}$


Is it always possible to split a complex function into the sum of its real + imaginary parts? I'm looking for any kind of proof or at least an intuitive explanation. I am curious because I want to know:

if I have an unknown complex function that I want to manipulate, can I assume it has the structure a(x) + ib(x).

It seems to me this should definitely be possible. I'm hoping someone can provide a lucid argument that makes the answer seem stupidly obvious

Of course you can.  The real and imaginary parts of a complex number uniquely determine it.  You can identify $\mathbb{C}$ with the plane $\mathbb{R}^2$ via $a + ib \mapsto (a,b)$.

Let $X$ be a set, and let $f$ be a function from $X$ to $\mathbb{C} = \mathbb{R}^2$.  Let $\pi_1: \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function $\pi(a,b) = a$, and let $\pi_2: \mathbb{R}^2 \rightarrow \mathbb{R}$ be the function $\pi_2(a,b) = b$.

Let $u: X \rightarrow \mathbb{R}$ be the composition $\pi_1 \circ f$, and let $v: X \rightarrow \mathbb{R}$ be the composition $\pi_2 \circ f$.  Then obviously

$$f(x) = (u(x),v(x))$$

for any $x \in X$.  Or in the usual notation of complex numbers,

$$f(x) = u(x) + i v(x)$$

for all $x \in X$.  This is written more succinctly as $f = u + iv$.

Hint: Any complex number $z$ can be written in the form $re^{i\theta}$...


Find the value of $\sqrt{10\sqrt{10\sqrt{10...}}}$ I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions $x=0,100$. I don't understand how $x=0$ is a possible solution, I know that squaring equations can introduce new, invalid solutions to equations and so you should check the solutions in the original (unsquared) equation, but doing that here doesn't lead to any non-real solutions or contradictions. I was wondering if anyone knows how $x=0$ turns out as a valid solution, is there an algebaric or geometric interpretation? Or is it just a "special case" equation?

A similar question says to find the limiting value of $\sqrt{6+5\sqrt{6+5\sqrt{6+5\sqrt{...}}}}$, and making a similar substituion for $x$ leads to
$$x=\sqrt{6+5x}$$
$$x^2=6+5x$$
which has solutions $x=-1,6$. In this case though, you could substitute $x=-1$ into the first equation, leading to the contradiction $-1=1$ so you could satisfactorily disclude it.

Is there any similar reasoning for the first question? I know this might be a stupid question but I'm genuinely curious :)

Denote the given problem as $x$, then
\begin{align}
x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\
&=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\
&=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\
&=10^{\large y}
\end{align}
where $y$ is an infinite geometric series in which its value is
\begin{align}
y
&=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\\
&=\frac{1}{1-\frac{1}{2}}\\
&=2
\end{align}
Therefore
\begin{equation}
x=10^{\large 2}=100
\end{equation}

have you noticed that $10$ is no special value. In this replace $10$ by $a$ i.e. any real number, equation will be still $x=a\sqrt{x}$, and $0$ will be a solution, so this can be regarded as a defect in such equations and thus always ignored, I don't think there is any geometrical or algebraic significance of this.


Intersection between three planes - geometric configurations for non-unique solutions I have three planes with equations

$$x+y+kz=-2$$ $$3x-y+14z=6$$ $$x+ky = -5$$There is no unique solution for $k=2$ and $k=\frac{7}{3}$, since that is when the determinant of the matrix of coefficients is $0$.

I want to look at the geometric configurations of the three planes for these values of $k$. For $k=2$, it is simple, as they form a sheaf and meet in line.

However, I am stuck with the case $k=\frac{7}{3}$. Do they form a prism or some other configuration? I am not sure on the conditions for them to form a prism. I have tried to plot these on a 3D plotter for this case, but I still can't seem to see if they form a prism or some other configuration.

Any help is appreciated.

Let’s look at it from a projective-geometric point of view. The intersection of the three planes is the null space of the matrix $$\begin{bmatrix}1&1&k&2\\3&-1&14&-6\\1&k&0&5\end{bmatrix}.$$ When $k=2$ this space is two dimensional, with basis $\{[1,-3,0,1]^T,[-4,2,1,0]^T\}$, i.e., the line $(1,-3,0)+\lambda(-4,2,1)$. Otherwise, the null space is one-dimensional and is spanned by $[6k+35,-21,-12,3k-7]^T$. When $k=\frac73$, this is the point at infinity $[49,-21,-12,0]^T$, which indicates that the three pairwise intersections are parallel lines in the direction of $(49,-21,-12)$. To find the specific lines, you’ll have to compute those pairwise intersections.

There is no solution when $k=7/3$. Even if there was a solution you can only get a line or a plane in the case when the determinant is zero.


A question on context-free languages from Sipser's computation book I'm trying to learn some computability theory, and I came across a question in Sipser's book that I can't figure out.  The exercise asks to show that there is an algorithm which will accept a context-free grammar $G$ and decide whether or not $\{1\}^*$ is a subset of the language generated by $G$.  I was trying do some kind of pumping while looking for patterns in the stack contents, but I couldn't get it to work.  After examining stack contents, is there some kind of maximal length after which you can stop checking any longer string of 1's?  Any hint would be appreciated.

This interesting problem seems to have no simple solution. It is tied to the fact that unary context-free languages (i.e., over a single letter alphabet) are actually regular. This is a quite advanced application of pumping, a special case of Parik's Theorem. For regular languages the property is decidable.
This should be contrasted to the two-letter case (can we generate every string in $\{0,1\}^*$) which is undecidable/

For a problem like this it is easier to work with the CFG than the PDA. You should start at the terminal $1$ and work your way back to the start variable $S$. By replacing variables with terminals, you eventually want to show that the rule

$S \rightarrow \epsilon \; |\; 1S\; |\;S1$

exists in the CFG.


Is there a trick to finding the number of odd numbers b/w two values? I know you could find the number of even numbers (since they are a multiple of two). For example the number of even numbers between $11$ and $30$ will be
$$n= \frac{28-12}{2} + 1 = 9  $$

I wanted to know is there a similar way to find the number of odd numbers b/w two extremes?

You can use the exact formula. What is number of odd numbers between $11$ and $30$?

$$n = (29-13)/2 + 1 = 16/2 + 1 = 9$$

Let's list them out to make sure $: 13, 15, 17, 19, 21, 23, 25, 27, 29.$ There are $9$ of them, so that is correct.
By the way, your formula was calculated wrong. $$n = (28-12)/2 + 1 = 16/2 + 1 = 8 + 1 = 9 (12, 14,16,18,20,22,24,26,28)$$

The Problem What is the number of EVEN or ODD integers between two numbers n and m?

(where m>n)

Solution:

Recall the formula (m-n+1)/2
Calculate it.
See below for how to interpret:

If m&n are mixed signs*, you're done. That's your answer.
If m&n are the same as the target**, round up.
If m&n are different than the target, round down.

Details:

*mixed signs = one is even and the other is odd.
**"same as the target = if you're looking for evens, then whether m&n are both even. If you're looking for odds, then whether m&n are both odd.