Is there a power of 3 and power of 7 that are equal? I have been trying to find a set of values $(a,b)\in\mathbb Z^+$ which satisfy the following relationship:

$3^a = 7^b$

I haven not been able to find a value that can satisfy this and I also can't find a way disprove it.

Is there a theorem that I am missing?

The theorem you are missing is the fundamental theorem of arithmetic which states that every integer factors uniquely as a product of prime integers.

You can prove that differently as well.

Since $a,b$ are integers, $\frac{a}{b}$ is rational by definition.

But:

$$3^a=7^b \to a\log 3 = b\log 7 \to \frac{a}{b} = \frac{\log7}{\log3}$$ Which is an irrational number. Hence, there are not integers $a,b$ such that: $3^a=7^b$


$f(x)=x^m+1$ is irreducible in $\mathbb{Q}[x]$ if only if $m=2^n$. I have a question

Prove that : $f(x)=x^m+1$ is irreducible in $\mathbb{Q}[x]$ if only if $m=2^n$

Thanks for your helps!

It's not true. If you take $f(x)=x^{6}+1$, it's irreducible in $\mathbb{Q}[x]$ but $6$ is not a power of $2$. I think you want $f(x)$ irreducible iff $m=2n$.


Tetrahedron area not using triangle area Is there any formula for the area of a general tetrahedron that does not determine the area by summing up its 4 triangle areas but considers the tetrahedron as a unit? The tetrahedron is given by 4 points in $\mathbb{R}^3$.

Sometimes, you can embed the tetrahedron inside a rectangular prism, with the edges of the tetrahedron crossing the diagonals of the sides of the prisim.  The volume of the tetrahedron will be 1/3 the volume of the prism.

e.g. the tetrahedron with vertexes at:

$(1,1,1), (-1,-1,1), (-1,1,-1), (1,-1,-1)$

Can be embedded in a cube, with additional vertices at $(-1,1,1),(1,-1,1),(1,1,-1), (-1,-1,-1)$

The volume of this cube is 8.  So the volume of the tetrahedron is $\frac {8}{3}$

If you can place one vertex at the origin.  Arrange the other vertices as row vectors in a matrix.  The determinant of this matrix will be $6 \times$ the volume of the tetrahedron.

If we had a tetrahedron with vertices at:
$(0,0,0),(0,2,2),(2,0,2),(2,2,0)$

Set it up as a matrix like this:

$\begin{bmatrix} 0&2&2\\ 2&0&2\\2&2&0\end{bmatrix}$

Calculate the determinant (16)

$V = \frac {8}{3}$  Which shouldn't be a surprise as these two tetrahedra are congruent.


If something has a $12\%$ probability of occurring every decade, what is the probability that it will occur in $100$ years? If something has a $12\%$ probability of occurring every $10$ years, what is the probability that it will occur in $100$ years? And also in $150$ years?

Is the formula for this as simple as $1-{0.88}^{10}$ or is it more complicated?

One possible approach is to regard such rare events as Poisson.

If the Poisson rate is $\lambda_{10}$ for a ten year period and $X_{10}$ is the number
of events that occur in ten years, then we have
$$P(X_{10} \ge 1) = 1 - P(X_{10} = 0) = 1 - e^{-\lambda_{10}},$$
so that $e^{-\lambda_{10}} = 0.88$ and $\lambda_{10} = -\log_e (.88) =  0.1278.$

Then the rate for 100 years is $\lambda_{100} = 1.278.$ If $X_{100}$
is the number of events in 100 years, then
$P(X_{100} = 0) = e^{-1.278} = 0.2786$ and $P(X_{100} \ge 1) = 0.7214.$

Of course, this is about the same as your $1 - .88^{10} = 0.7215,$  but you might find it interesting to think about the Poisson approach.

1 - 0.12^10 * 100 = 100 - 6.1917364E-8 %

Try branching out the possibilities. in each branch multiply the percentage to get the odds of that branch happening.


Doubt on limits evaluation. Can anyone tell me what am I doing wrong here.

The limit provided is$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$

Method 1 (using standard limits)

$$= \ \ \ \  \dfrac{\displaystyle \lim_{x \to 0}\dfrac{xe^x}{x}-\lim_{x \to 0}\dfrac{\ln(1+x)}{x}}{\displaystyle \lim_{x \to 0}x}$$

$$= \ \ \ \  \dfrac{\displaystyle \lim_{x \to 0}e^x-\lim_{x \to 0} 1}{\displaystyle \lim_{x \to 0}x}$$

$$= \ \ \ \  \lim_{x \to 0} \dfrac{e^x-1}{x}$$

$$= \ \ \ \ 1$$

Method 2 (using  Maclaurin series )

$$\lim_{x \to 0}\dfrac{xe^x-\ln(1+x)}{x²}$$

$$= \ \ \ \ \dfrac{3}{2}$$

Even with L'Hopital rule I get $\dfrac{3}{2}$. Then what's wrong with method 1.

Some limit's properties

The error is quite simply described.

  Theorem. If $\lim_{x\to a}f(x)=l$ and $\lim_{x\to a}g(x)=m$ and $m\ne0$, then
  $$
\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{l}{m}
$$

This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.

In your case, $\lim_{x\to0}x=0$, so the theorem cannot be applied.

Why can't it? Because division by $0$ is not allowed.

$$\lim_{x \to 0}\frac{xe^x-\ln(1+x)}{x^2}=\lim_{x \to 0}\frac{xe^x}{x^2}-\lim_{x \to 0}\frac{\ln(1+x)}{x^2}=\lim_{x \to 0}\frac{e^x}{x}-\lim_{x \to 0}\frac{\frac{1}{1+x}}{2x}=\lim_{x \to 0}e^x-\lim_{x \to 0}\frac{1}{2x+2x^2}=1-\lim_{x \to 0}\frac{1}{2x+2x^2}$$

this final limit appears to approach infinity so the answer is $+\infty$ or $-\infty$ depending on whether we use $0^+$ or $0^-$


Finding a recursive formula/sum for $\pi$ I'm very much aware of the
$$\pi = 4 \left(1 - \frac13 + \frac15 - \frac17 + \frac19 -\cdots \right)$$ and
$$\pi = \sqrt{6\left(1 + \frac14 + \frac19 + \frac{1}{16}+ \cdots \right)}$$
and even less common $$\frac\pi2 = \frac21 \cdot \frac23 \cdot \frac43 \cdot \frac45 \cdot \frac65 \cdot \frac67 \cdot \frac87 \cdot \cdots$$

What I'm wondering is if there is a RECURSIVE (not summative) way to find $\pi$.

An example for phi ($\phi$) is: $$f(1) = 1; \quad f(n) = 1+\frac1{f(n-1)}$$

By recursive function, I mean a function that calculates its next value based on the current or previous (or even further back) value.

There is no rational recurrence of the form

$$\pi_{n+1}=\frac{P(\pi_n)}{Q(\pi_n)}$$ where $P,Q$ are polynomials with algebraic coefficients, because that would mean that $\pi$ is itself algebraic (it would be solution of $x Q(x)-P(x)=0$).

If transcendental functions are allowed, you can use the interesting

$$\pi_{n+1}=\pi_n+\sin\pi_n$$

but it is simpler to directly use $\pi_n=4\arctan 1$.

(This essentially means that there is no suitable recurrence.)

Your question is not very clear, in case of $\phi$ we calculate $\phi(n)$ but we do not calculate $\pi(n)$. $\pi$ is a constant, and you want recursive formula, so from the context i assume you want to calculate the value of $\pi$ using a recursion. To calculate value of $\pi$ using recursion you can use any of the formula listed above.
e.g.
$\pi = \displaystyle \prod_{n=1}^{\infty}  \cfrac{2n}{2n -1} \cfrac{2n}{2n - 1} $
You can do like this
$\pi =\lim_{z \to \infty} \displaystyle \prod_{n=1}^{z}  \cfrac{2n}{2n -1}
\cfrac{2n}{2n - 1} $
$\pi(z) = \displaystyle \prod_{n=1}^{z}  \cfrac{2n}{2n -1}
\cfrac{2n}{2n - 1} \implies \pi(z) = \pi(z-1)  \cfrac{2z}{2z -1}
\cfrac{2z}{2z - 1}$
More the value of  z more accurate value of $\pi$ you will get.


Deriving an equation How do we derive the following: $$ -\sum_{x} \sum_{y}  p(x,y) \left( \log_{2} p(x) + \log_{2} p(y) - \log_{2} p(x,y) \right)$$ $$= -\sum_{y} p(y) \log_{2} p(y)+\sum_{x} p(x) \sum_{y} p(y|x) \log_{2} p(y|x)$$

The first equation simplified is: $$-\sum_{x} \sum_{y} p(x,y) \log_{2} p(x)-\sum_{x}\sum_{y} p(x,y) \log_{2} p(x,y)+\sum_{x}\sum_{y} p(x,y) \log_{2} p(x,y)$$

Do we just use the fact that $p(x,y) = p(y)p(x|y) = p(x)p(y|x)$ and substitute in?


If all the numbers $(1^\alpha,\,2^\alpha,\,3^\alpha,\,\dotsc)$ are integer, then $\alpha$ is an integer. A theorem of Siegel asserts that

  If $\beta>0$ and $2^\beta,\,3^\beta,\,5^\beta$ are integers, then $\beta$ is an integer.

The following result is a beautiful consequence of this theorem

  If $\beta$ is a positive number such that $1^\beta,\,2^\beta,\,3^\beta,\,\dotsc$ are integers, then $\beta$ is itself is an integer.

I'm looking for a proof of this result.

Note. This result appeared as a problem in the 1972 Putnam Prize competition, and not one of more than 2000 university student competitors gave a solution; the solution, though not hard, could well elude even a professional mathematician for several hours (or days).

Well, k1.M, here an extension of important and beautiful Siegel Theorem (I did not know). It is clear $\beta$ must be integer or irrational and by another theorem of Siegel $\beta$ must be trascendental. I guess the statement goes for any three coprime integer x, y, z the triple 2, 3, 5 having been chosen for smallest.Thus the true spirit of the statement of Siegel would be "all trascendental as exponent of an integer can gives an integer at most twice".


Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists. Find an even number that can be represented as difference of two squares in exactly 3 ways or prove that none exists.

I tried many different numbers and nothing is working but I don't know how to prove that it isn't possible.


Infinite Dimensional Hilbert Space and Hamel Basis Currently I'm reading "A Course in Functional Analysis" by John B. Conway.In the section "Orthonormal Sets of Vectors and Bases" he writes "For an infinite-dimensional vector space , a basis is never a Hamel basis" without any further explanation. Can anyone throw some light on this matter?
Thanks in advance.

If $H$ is your Hilbert space and if $(e_j)_{j\in J}$ is an orthonormal basis of $H$, let $j(1),j(2),\ldots$ be a countable subset of $J$. Then$$\sum_{n=1}^\infty\frac1ne_{j(n)}\in H,$$but it cannot be expressed as a finite sum of elements of $(e_j)_{j\in J}$. Therefore, $(e_j)_{j\in J}$ is not a Hamel basis.

If H be finite dimensional space then each orthonormal bases will be Hamel bases, by Kronecker delta its easy to see. But for infinite dimensional spaces there is orthonormal bases that is not Hamel in general.


use Green theorem to evaluate line integral: In  the problem below,  parametrize the plane curves below in such a way that it  traversed only once , its unit normal vector towards the interior of the bounded region it enclose.

Then use Green theorem to evaluate line integral:

$$ \int_C y^2dx+xydy $$

where $ \ C \ $ is the boundary curve  of the region   lying between $ \ y=0 , \ y=\sqrt x \ $ and $ \ x=9 \ $

Answer:

Let $ \ y=t, \ x=t^2 \ , \ 0 \leq t \leq 3 \  $ be the parametrization.

Then,

$ F(x,y)=y^2 \hat i+xy \hat j, \\ r(t)=x \hat i+y \hat j=t \hat i+t^2 \hat j \\ \Rightarrow dr=(\hat i+2t \hat j)dt  \ \ and \ \ F(r(t))=t^2 \hat i+t^3 \hat j $

Thus,

$ \int_C y^2 dx+xy dy =\int_{0}^3 \ F(r) \cdot dr = \int_0^3 (t^2+2t^4) dt = 9+2 \cdot \frac{3^5}{5}$

Am I right ?

Help me

Greens Theorem states that
$$\int_C y^2 dx + xy dy = \iint_D \frac d {dx} xy - \frac d {dy} y^2$$
where $D$ is the region enclosed by $C$. This can be simplified to
$$= \iint_D y - 2y = \iint_D -y$$
It seems to me the problem wishes you to use Greens Theorem by evaluating this much simpler integral to get the integral over $C$

You seemed to try to evaluate the integral without using Greens Theorem which will get you the right answer too, however there are a couple errors:

You swapped the variables in $r(t)$. It should be that $r(t) = x \hat i + y \hat j = t^2 \hat i + t \hat j$. Thus $dr = 2t \hat i + \hat j$ and what you end up integrating is $\int_0^3 2t^3 + t^3\ dt$

You forgot to integrate over the $x=9$ and $y = 0$ portions of $C$.


$M$ is a maximal normal subgroup iff $G/M$ is simple $M$ is a maximal normal subgroup of G if and only if $G/M$ is simple.

I have a problem in "if" part. To prove ($\Leftarrow$) direction, assume that $N$ is a normal subgroup of $G$ properly containing $M$. Let $f:G\to G/M$ be the surjective canonical map. Then $f(N)=N/M$ is a nontrivial(since $M \subsetneq N$) normal subgroup of $G/M$. Since $G/M$ is simple, $N/M=G/M$. But how can I conclude that $N=G$? Isn't it be possible that a proper subgroup of $G$ may have the same image under factoring modulo $M$?

Reference: Fraleigh p. 150 Theorem 15.18 in A First
Course in Abstract Algebra 7th ed

Suppose that $N/M = G/M$ and there is some $g \in G \setminus N$; then there is an $n \in N$ such that $gM = nM$. In particular, $gm = n$ for some $m \in M$, whence $g = nm^{-1} \in N$, which is a contradiction.

$(G/M)/(N/M)$ is isomorphic to $G/N$. Since $G/M$ is equal to $N/M$. Then $G/N$ is isomorphic to the trivial subgroup. That means $G/N$ is isomorphic to the trivial subgroup too. Therefore $G=N$.


Harmonic series sum approximation Question: $ 1 + 1/2 + 1/3+\dots+ 1/n > 4.$
  Find the range of smallest value of n.

  Answer: $n$ lies in $(20,60)$.

  Source: KVPY 2017.

To the best of my knowledge I find this series to be divergent and thus its sum can not be calculated. I couldn't think of a way to approximate value of $n$. Help is appreciated!

Yes, the sequence $H_n=1 + 1/2 + 1/3+\dots+ 1/n$ is strictly increasing and divergent. The question is asking for the minimal positive integer $n$ such that $H_n>4$.

Hint. Note that for any $n\geq 2$,
$$\ln(n)=\int_{1}^n \frac{1}{x}dx  <  H_n<  1+\int_{1}^n \frac{1}{x}dx=1+\ln(n).$$
See for example Showing inequality for harmonic series.

You can approximate the harmonic series by using the fact that the nth partial sum of the harmonic series minus the natural logarithm of n approaches $\gamma$ as n approaches infinity


Mathematicians who started late, similar to my situation? I came here since I know this is the best place to ask a question.

I'm a first year student who changed his major to applied mathematics.  In middle school I was a garbage math student, but I realized the importance of math in high school when I was introduced to amazing teachers who truly loved what they did.  I put myself in tougher classes and eventually got to AP calculus.  There was an error in my idea, I never really got a deep understanding of the stuff I was doing and was struggling since I didn't understand the basics and never really did practice problems.

This year I began to start over from scratch from pre-algebra working to pre-calculus. Even though I have already took Calculus.

I'm in a introduction to research class this semester and we are preforming a meta-analysis of some random topic and then presenting at the end of the semester.  I'm really enjoying it, and I will definitely apply for more research as I progress through my undergraduate career.  (Urge to Compute)

I know I'll probably never win a field's medal, but I'm really intimidated and humbled by the near perfect SAT math scores and Math Olympiad participants.

It's too late for me to have that, but the best quality I have is sticking with the concepts and problems until I can explain them to my dog. (Basically until I understand it)

I'm really sorry for the long post / soft question, I've just been thinking about this since 11th grade but never asked anyone about it.

Basically I'm just wondering if I'm wasting my time, and if there have been mathematicians that were in a similar situation.  (Famous or not.)

Again, sorry for the soft question and thank you for taking the time to read this!

Let me tell you that it's not too late to enjoy the field of mathematics.  I too went through a circuitous path of getting into mathematics.  I was extremely lazy in high school, had mediocre SAT scores and had no idea what a math competition was.  I was a business major for undergrad and studied finance.  It wasn't until after I graduated that I made friends with a guy that had been a math major in his undergrad.  He eventually convinced me to take the honors calculus series at a local community college and promised he'd help tutor me.  I was fortunate, my teacher there was a PhD student at a local well known college and took a liking to me and became somewhat of a mentor.  I was a little neurotic and became obsessed with math.  I ended up quitting a full time job, moving back in with my parents and then taking classes at the local university (you could pay and take classes as a non-matriculated student).  I did this for two years, taking only math classes and then eventually applied to their MS program and was accepted.  I have since scored in the top 97% percentile of the GRE general math, and published one paper with another in progress.  I'm going to take the Math subject exam in the Fall and apply to Phd programs.

So, will I ever be a math wizard... no.  But do I love mathematics, yes.  I share this with you so that you know, there are people out here that started even later than you and are happy they did.

I'm not sure where you are in your education, but working problems is one of the (if not the) best thing you can do to solidify your foundation.  The book that made me want to become a mathematician:
http://www.amazon.com/The-USSR-Olympiad-Problem-Book/dp/0486277097

In middle school we used to complete math problems given by a computer system and then based upon our answers we would be promoted to a new level of problems or stay at our same area. I never made it to algebra until high school. However once I began studying algebra, I would do my homework in the minutes before class started and generally do well. I still felt incompetent in math. I did okay in some course and well in others. Algebra was my favorite as I would play on my calculator, graphing functions in order to see what they looked like such as high order polynomials, etc;. I never liked, and still don't particularly enjoy trigonometry. However I made it to calculus by my senior year. The summer before my senior year, I was at a smart kid camp and instead of hanging out and making friends like most of the others, I spent my time in the library and in my room learning about the concept of a derivative and an integral. Calculus changed my life in terms of my understanding of the power of mathematics to describe many aspects of our reality in such a beautiful and succinct manner. I attended college and immediately began as a finance major. However we were only required to take two high level math courses in calculus and applied mathematics such as system of equations (no matrix theory). I rejected doing the bare minimum and instead I opted for engineering calculus, which I found to be much easier than most people who attempted to talk me out of it did. I stopped listening to others after that and went after what interested me, mathematics were intimidating at that point but my curiosity to learn more about the subject would not go away. I proceeded to take calc 2 (integrals/series/sequences), calc 3 (multidimensional calc), and calc 4 (ode's) and I have fallen in love with the field of analysis. Since then I kept taking various math courses in probability theory, optimization, discrete mathematics (as I'm technically a computer scientist), etc;. I feel that I didn't truly begin to learn mathematics until the age of 17, I have never gone to math camp, competed in math competitions, but I have a love for math that is untouched by any other subject as it is present in nearly everything I have ever studied. Who cares when you start, if you have the passion for it nobody can take that from you and nobody can deny the beautiful truths that are consequences of  mathematical rigor and analysis.


Finishing the solution to this functional equation. I am trying to solve the below functional equation for $f$ (where $f: \mathbb{R} \to \mathbb{R}$):
$$ f(x^2+y)=f(f(x)-y) + 4f(x)y $$
(Source: https://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf: "Iran TST 1996")
Here's what I've got so far:
Substituting $y=0$ we have $f(x^2) = f(f(x))$
Substituting $y=f(x)$ we have
$$f(x^2+f(x)) = f(0) + 4f(x)^2$$
$$f(0) = 0 $$
Letting $x=0$ we can show the function is even.
Finally letting $y = -x^2$ we can simplify to yield
$$f(x) (x^2 - f(x)) = 0 $$
And this is where I'm stuck. Given that it's an Olympiad I'm fairly sure that actually the only solutions are $f(x) = x^2$ and $f$ is identically zero and that there will not be any pathological solutions that have discontinuities - though I have not been able to show this.
My ideas that I've tried:

Showing $f$ is injective when not identically zero (on $\mathbb{R}^+$ that is), this doesn't seem to go anywhere useful.
Showing that $f$ is surjective (again onto $\mathbb{R}^+$) when not identically zero, as if that is the case we always must choose $x^2$ instead of zero other we miss out a positive number. However I haven't been able to find any way to show this either.
Use inequalities to show that if $f$ is not identically zero then the inequality $f(x) \geq 0$ that we have from our current information (either $f$ is zero or the square of a real) is strict when $x \neq 0$, but again I couldn't do anything with this either.

We know that for each $x\in\mathbb R$, either $f(x)=x^2$ or $f(x)=0$.
Let $A=\{x\neq0\mid f(x)=0\}$ and $B=\{x\neq0\mid f(x)=x^2\}$. Then $A$, $B$ partition $\mathbb R\setminus\{0\}$.
Suppose both are non-empty. Take $a\in A$ and $b\in B$. Sub $x=a,y=-b$ to get $f(a^2-b)=b^2$. Since $b\neq0$, we must have $b^2=(a^2-b)^2\implies a^2=2b$ (as $a\neq0$).
This implies $\lvert A\rvert\leq2$ and $\lvert B\rvert=1$, a clear contradiction. So one of $A$, $B$ is empty, and we recover the two solutions.

Hint Take
\begin{equation}
y = \frac{f(x)-x^2}{2}
\end{equation}
This leads to $\forall x, f(x)\in \{0, x^2\}$, hence we can write $f(x) = x^21_x$ where $\forall x\in{\mathbb R}, 1_x\in\{0,1\}$. Let us rewrite the functional equation as
\begin{equation}
y^2(1_{x^2+y}-1_{x^2 1_x - y})+2x^2 y(1_{x^2+y}+1_{x^2 1_x-y}-2. 1_x)
+ x^4(1_{x^2+y}-1_{x^2 1_x-y}) = 0
\end{equation}
This equation must be satisfied for all $x, y$ but the coefficient of $y^2$ here can be only $-1,0 ,1$, the coefficient of $2x^2 y$  can be only $-2, -1, 0, 1, 2$ and the coefficient of $x^4$ can be only $-1,0,1$.
Let $x$ be given such that $x\not=0$ and $1_x=1$ (or equivalently $f(x) = x^2\not=0$). Clearly the coefficients of $y^2$ and $y$ can be non zero for at most a finite number of $y$'s because in this case, $y$ must be a solution of a polynomial equation of degree 1 or 2 chosen among 44 such non trivial equations. It follows that for all $y$ but a finite number of them one has
\begin{equation}
1_{x^2+y} = 1_{x^2 1_x-y} = 1
\end{equation}
Hence $1_z = 1$ for all $z\in{\mathbb R}$ except on a finite set and $f(z) = z^2$ except perhaps on this finite set.


Convert two polynomials into a target polynomials through operations You are provided with two polynomials $x^3-3x^2+2$ and $x^2-4x$ on a
piece of paper.
If there are polynomials $f(x)$ and $g(x)$ on the paper, you are allowed
to write $f(x) + g(x)$, $f(x) - g(x)$, $f(x)g(x)$, $f(g(x))$ and
$cf(x)$.
Can you obtain a linear polynomial (exactly one degree) given
two starting polynomials?

I feel like this could be a linear algebra problem, if the operations are just $f(x) + g(x)$, $f(x) - g(x)$ and $cf(x)$ Then we are looking for the linear dependence. But now sure how to handle $f(x)g(x)$, $f(g(x))$

As mentioned, subtraction decreases the degree if the leading coefficients are the same (which we can ensure with $cf(x)$), so all we need is polynomials of the same degree. This is the crux. See if you can work it out before reading on.

We currently do not have 2 polynomials of the same degree, so let's force them out and take their difference. I will work with the specific cases of $a(x) = x^3 - 3x^2 + 2$, $b(x) = x^2 - 4x$, but I suspect this will generalize (with certain conditions).
Step 1: Calculate $a(x)a(x) = x^6 - 6x^5 + 9x^4 + 4x^3 - 12x^2 + 4$.
Step 2: Calculate $b(x)b(x)b(x) = x^6 - 12x^4 + 48x^2 - 64  $. This gives us 2 polynomials of degree 6.
Step 3: Take their difference $a(x)^2 - b(x)^3 = 6x^5 -39 x^4 + 68x^3 - 12x^2 + 4 = c(x) $.
Step 4: Calculate $a(x) b(x) = x^5 - 7x^4 + 12x^3 + 2x^2 - 8x $. This gives us 2 polynomials of degree 5.
Step 5: Take their difference $ c(x) - 6a(x) b(x) = -15x^4 + 86x^3 - 96x^2 + 48x + 4 = d(x)$.
Step 6: Calculate $b(x) b(x)  = x^4 - 8x^3 + 16x^2$. This gives us 2 polynomials of degree 4.
Step 7: Take their difference $d(x) + 15 6 b(x)b(x) = -34x^3 + 144x^2 + 48x + 4 = e(x) $, which is a polynomial of degree 3.
Step 8: Take their difference $ e(x) + 34 a(x) = 42x^2 + 48x + 72 = f(x) $, which is a polynomial of degree (at most, and hopefully) 2.
Step 9: Take their difference $ f(x)  - 42 b(x) = 216x+72 $, which is a polynomial of degree 1.

Notes

As SenZen mentions, if our initial polynomials don't have a linear term, then none of the operations would have led to a linear term. So this isn't always doable.
As implied by SenZen, we might lower the degree by more than one if the ratio of the first few coefficients are equal. This has happened when I was using $ a(b(x)) , b(a(x))$ as my starting polynomials, and I didn't end up with a polynomial of degree 1.
This process is not unique. We could have made different choices for the polynomials of higher degrees.
I might have a calculation error. If so, let me know. This was way to much calculation for a contest-math problem, even after knowing the crux, so I'd like to know of an alternative solution.

No.
The degrees of both polynomials is greater that one and none of those operations decreases the degree.


Uniformly integrable for a random variable I am trying to show that for $k>0$ and $X$ a real valued r.v. with $E[| X|^k]<\infty$, we have:
$$
\lim_{M\to\infty} E[|X|^k 1_{\{|X|>M\}}]=0
$$

I was trying to apply Holder's inequality with $p=1,q=\infty$ to the expectation to separate it into two terms, then use Markov's inequality to get a finite expectation divided by M. However, I don't think I understand the case $q=\infty$ very well.

Question: is there another way to solve this without Holder?

If you have any tips on how to solve this, please comment below. Thanks for helping! :D

Consider the partition:
$$
\begin{align}
&\Omega=\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}=0\right\}\cup\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}\neq0\right\}\\
\end{align}
$$
Consider the set to right. A necessary condition for the limit inside it to be different than zero is that the indicator function is different than zero for infinitely many values of $M$. Since $M\to\infty$ and $X:\Omega\to\bar{\mathbb{R}}$, we must have $X(\omega)>M,\forall M$, which implies that $X(\omega)=\infty$. Thus:
$$
\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}\neq0\right\}\subset\left\{\omega\in\Omega:X(\omega)=\infty\right\}
$$
Now, if the size of that latter set is greater than zero:
$$
\begin{align}
&P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)>0\\
&\implies\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)\geq \left|X(\omega)\right|^\frac{k}{2}P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=\infty
\end{align}
$$
Which contradicts the hypothesis $\mathrm{E}\left(\left|X\right|^\frac{k}{2}\right)=\infty$. So:
$$
\begin{align}
&P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)=0\\
\end{align}
$$
Now using the first equation (partition):
$$
\begin{align}
1&=P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}\neq0\right\}\right)\\
&\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}=0\right\}\right)+P\left(\left\{\omega\in\Omega:X(\omega)=\infty\right\}\right)\\
&\leq P\left(\left\{\omega\in\Omega:\lim_{M\to\infty}\left| X(\omega)\right|^k 1_{\{\left |X(\omega)\right|>M\}}=0\right\}\right)
\end{align}
$$
So we have that:
$$
\begin{align}
\left| X\right|^k 1_{\{\left |X\right|>M\}}\overset{a.s.}{\to}0
\end{align}
$$
Since that sequence (in $M$) is bounded by $\left|X\right|^k$, which is bounded in expectation, the Dominated Convergence theorem gives the desired result.


Check if a point is within an ellipse I have an ellipse centered at $(h,k)$, with semi-major axis $r_x$, semi-minor axis $r_y$, both aligned with the Cartesian plane.

How do I determine if a point $(x,y)$ is within the area bounded by the ellipse?

The region (disk) bounded by the ellipse is given by the equation:
$$
\frac{(x-h)^2}{r_x^2} + \frac{(y-k)^2}{r_y^2} \leq 1. \tag{1}
$$
So given a test point $(x,y)$, plug it in $(1)$. If the inequality is satisfied, then it is inside the ellipse; otherwise it is outside the ellipse. Moreover, the point is on the boundary of the region (i.e., on the ellipse) if and only if the inequality is satisfied tightly (i.e., the left hand side evaluates to $1$).

I have another solution. In summary, transform everything so you can test whether a point is within a circle centered at (0,0). Since the ellipse is oriented orthogonally to the Cartesian plane, we can simply scale one of the dimensions by the quotient of the two axes.

First subtract (h,k) from both points.

(h,k) becomes (0,0).
(x,y) becomes (x-h,y-k).

Now we scale the second coordinate, normalizing it to the first.

Scaling (x-h,y-k) we get
$$
(x-h,(y-k) * \frac{r_x}{r_y})
$$

Now we simply need to test if
$$
|(x-h,(y-k) * \frac{r_x}{r_y})| <= r_x
$$


Difficult Expii Contest Question I came across this problem while writing a contest and I cant find the answer: The third dimension is another solution to traffic jams, if we first invent flying cars. Suppose that a metropolis is evenly distributed across a square of land which is 20 km on each side. Every morning, 1 million commuters each fly their car from one uniformly random point to another, by departing at a uniformly random time between 7 a.m. and 9 a.m., flying straight upward to a random altitude, then cruising in a straight line to the point above their destination, and flying straight downward, all at 100 km/hr. The flying cars are all spheres of diameter 5 meters.

To keep order, the cruising altitudes are always multiples of 10 meters. We say that a car encounters another if they are at the same altitude, and the centers of the cars pass the same point within 1 second of each other. How many different possible choices are needed for the random altitudes so that the expected number of other flying cars encountered by a given flying car is only 1? Any answer within 10% of the correct answer is accepted.

Some thoughts on how to tackle this. I'll read the problem definition quite literally, even if much of it doesn't make sense physically.

The size of the cars is irrelevant. The definition of “encounter” is for the centers of two cars to pass the same point within one second, which I'd read as “pass through the same point”. So you have some leeway in time but none in space.
This means that given two line segment trajectories you can decide whether two given cars will encounter one another by computing the point of intersection between the lines, checking whether it lies on the segments, then computing the time each car passes that point and checking whether the times are at most one second apart.
We can ignore encounters on the vertical legs. As there is no spatial leeway (as discussed above), any two cars will pass through a common point on their vertical journey with probability zero.
The difference between cruising altitudes is irrelevant. As we have no spatial leeway, cars on different altitudes can never encounter one another. Furthermore, although different altitudes mean different time till the cars reach these altitudes, that time is the same for all cars cruising at that altitude, so you can simply consider this a constant offset from the 2 hour departure time window, which doesn't matter since the arrival at cruising altitude still falls uniformly in a 2 hour window.
You essentially need to figure out the probability that two cars on the same altitude will encounter one another. Once you have that probability $p$, the expected number of encountered cars if there were $n$ cars at the same altitude would be $(n-1)p$, so you could solve $(n-1)p=1$ for $n$ and then divide the total poulation by this to obtain the number of required altitude levels.
Getting the pairwise encounter probability seems tricky to do symbolically, so you might want to try some Monte Carlo approach to this: Simulate pairs of cars at the same altitude until the probability value stabilizes, then perform your computation using that value.

If the question was meant differently, then it depends on what exactly the intended meaning was. Perhaps two cars enounter one another of their centers pass within 5 meters from a given point within 1 second from one another. How would that affect my approach?

Instead of intersecing line segments you'd essentially intersect long rectangles. Every point within such a rectangle will have a different time window when it is within 5m of one of the cars, so working out the exact geometry here will become a bit harder. You may want to simplify this somehow, or perform some simulations to e.g. turn this into an approximate time window within which the cars must pass through the same point.
Vertical legs become a problem as well. As do vertical legs encountering horizontal cruises. Hard work ahead! Perhaps you want to compute each of these probabilities separately (vertical vs. vertical, vertical vs. horizontal, horizontal vs. horizontal) for a given number of altitudes, and then tweak that number until you are satisfied.
If the cruise altitudes are exactly 10m apart, then points between them will be exactly 5m from the centers of cars cruising at adjacent altitudes. It is very unclear whether you'd consider this an encounter. If you do, cars on adjacent altitudes would suddenly start encountering one another.

On the whole, this makes the problem much harder, and less clearly specified as well.


Largest and smallest roots in a cubic equation How could I obtain the largest and smallest roots of this equation.

$$Y^3 - (1 - C_{1}) Y^2 + (A - C_{1} + C_{2}) Y - C_{2} = 0 $$

The three roots are real, positive, between $0$ and $1$.

I need to be sure they are the smallest and the largest in order to continue with really tedious work.

Any help wll be welcome. Thanks.

You can use the formula of Cardano for the roots of an equation.But before using you should make a substitution $y=x+\frac{1-C_{1}}{3}$.You will get an equation
$$x^3+px+q=0,$$ which can easy be dealed with in Maple,Mathematica,Matlab(choose one). There you can get the formula for the roots and minimize (maximize) them.


For $1<p<2, $ Fourier transform is not onto $L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$ For $1<p<2$ , Fourier transform $\mathscr{F}$ is not onto $L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$

For $1 \le p \le 2$, Hausdorff-Young inequality implies that $\mathscr{F}:L^p(\Bbb T) \to \ell^q(\Bbb Z)$ where $\frac{1}{p}+\frac{1}{q}=1$
Now for $p=1$ showing that $\mathscr{F}:L^1(\Bbb T) \to \ell^\infty(\Bbb Z)$ is not onto was easy due to Riemann-Lebesgue lemma (I know that it's image is in fact a dense sub-alegbra of $c_0(\Bbb Z)$).
But for $1<p<2$, I'm facing difficulty to show that it's not onto. I know that $\forall 1<p<2, \mathscr{F}(L^p(\Bbb T))\subset \ell^q(\Bbb Z)$ . Hence we have $$\mathscr{F}(L^p(\Bbb T))\subset \ell^q(\Bbb Z) \subsetneq c_0(\Bbb Z)$$ Also considering images of trigonometric polynomials under $\mathscr{F}$ we get that $c_{00}(\Bbb Z) \subsetneq \mathscr{F}(L^p(\Bbb T)$
So basically the scenario is, $$c_{00}(\Bbb Z) \subsetneq \mathscr{F}(L^p(\Bbb T)) \subset \ell^q(\Bbb Z) \subsetneq c_0(\Bbb Z)$$
How to produce an $\ell^q$ sequence which cannot be in the image of $\mathscr{F}$ ? Or if there are other functional analysis proofs which will assert the same?

Take $S=\sum_{n \ge 1} \frac{\cos (2^nx)}{\sqrt n}$. Note that the coefficients of $S$ are in $l^q(\mathbb Z)$ for all $q>2$ (but not in $l^2$!)
Then one can show that the partial series of $S$ and the Caesaro means of said partial series diverge ae on the unit circle so $S$ cannot be a Fourier series. Actually it is easier to show that the partial series/Caesaro means are unbounded ae using the fact that $\cos 2^nx$ and $\cos 2^mx$ are orthogonal for $n \ne m$ (and more generally any $k$ distinct such are orthogonal since for $n_1<n_2<..n_k, \pm 2^{n_1} \pm 2^{n_2}..\pm 2^{n_k} \ne 0$)

Clearly, $\mathscr F$ is one-to-one, and hence, if it were onto, then by virtue of the Open Mapping Theorem, it would be a homeomorphism.


Is the map $f^*: \mathrm{Spec}(S) \to \mathrm{Spec}(R)$ injective or surjective where $f: R \to S$ is a ring homomorphism. Is the map $f^*:\mathrm{Spec}(S) \to \mathrm{Spec}(R)$ injective or surjective where $f: R \to S$ is a ring homomorphism.

I know that if $P$ is a prime ideal in $S$, then $f^{-1}(P)$ is a prime ideal in $R$.

I think it is injective but cannot show why.

Take $$k\to k[x]$$ then $\rm{spec} \ k$ has one element, so it cannot be injective.

This from a book Algebraic geometry and arithmetic curves, Qing Liu, p.28.

  Lemma: Let $\varphi:A\to B$ be a ring homomorphism and $spec(\varphi):spec(B)\to spec(A)$ then if $\varphi$ is surjective then $spec(\varphi)$ induces a isomomorphism from $spec(B) $ onto the closed subset $V(ker \varphi)$ of $spec(A).$

EX: Let $A$ be a ring then the quotient homomorphism $\varphi:A\to A/I$ induces a homomorphism $spec(\varphi):spec(A/I)\to spec(A)$ and $ spec(A/I)\cong V(I)\subseteq spec(A)  $ where $V(I):=\{p\in spec(A); I\subseteq p  \}$


Average distance from a point in a sphere to the origin Consider a sphere $S$ of radius $a$ centered at the origin. Find the average distance between a point  in the sphere to the origin.

We know that the distance $d =  \sqrt{x^2+y^2+z^2}$.

If we consider the problem in spherical coordinates, we have a 'formula' which states that the average distance $d_{avg} = \frac{1}{V(S)}\iiint \rho dV$

I think that this is reminiscent of an average density function which I've seen in physics courses, and it is clear that the $\iiint \rho dV$ is equal to the volume of the sphere, but I'm not sure as to why we must integrate over the distance and then divide by the actual volume to calculate the average distance.

I am looking for a way to explain this to my students without presenting the solution as a formula, any insights would be appreciated.

If $\rho$ is the radius, the volume is in fact $\iiint dV$ without the  $\rho$.  The average distance is then as you have said.  This is an example of the general formula for the average value of a variable $X$ over a probability distribution $V$, which is $\bar X= \tfrac {\int X dV}{\int dV}$

All points are equidistant from the origin ... that is why it is a sphere.  Therefore, the average distance of a point on the sphere from its center is the radius.


is every function defined on natural numbers considered a sequence? I wonder if every function that operates over the set of natural numbers (or its finite subset ranging from $1$ to $k$) is considered a sequence?

It seems to me that the answer is yes, at least when one looks at the common definition of a sequence:

$$a: \mathbb{J} \to X$$
where: $\mathbb{J}$ is either $\mathbb{N}$ or $\{1, 2,..., k\}$

"We define and as "sequence" or "infinite sequence" any function :f : $ \mathbb N \rightarrow A, A \subseteq \mathbb R$. We also define as "finite sequence"  any function : $f : (1,2,..,n)\rightarrow A, A \subseteq \mathbb R $." Note that $(1,2,...,n)$ is a set $a \subseteq \mathbb N$.

From WolframMathWorld on Sequence:

  A sequence is an ordered set of mathematical objects.

From WolframMathWorld on ordered set:

  [An ordered set is a]n ambiguous term which is sometimes used to mean a partially ordered set and sometimes to mean a totally ordered set.

And I will not quote WolframMathWorld on totally ordered set.

However, there is a site of integer sequence, called the OEIS (On-Line Encyclopedia of Integer Sequences), which have sequences like this:

A000005 (the number of divisors of $n$):

  1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, ...

TL;DR:

From WolframMathWorld: the answer is no. A sequence must be increasing.
However, the OEIS (On-Line Encyclopedia of Integer Sequences) also record non-increasing sequences.


Different ways to prove there are infinitely many primes? This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.

Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.

Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that
$$
2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1),
$$
so that for $m < n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.

Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set
$$
\mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z
$$
should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.

Proof 4 : Let $a,b$ be coprime integers and $c > 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.

Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.

Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)

Do you have any other nice proofs?

The following proof is morally due to Euler. We have

$$\prod_{p \text{ prime}} \left( \frac{1}{1 - \frac{1}{p^2}} \right) = \zeta(2) = \frac{\pi^2}{6}.$$

The RHS is irrational, so the LHS must have infinitely many factors.

How about this?
Let x be a rational in (0,1). Then x is of the form x = m/M with M>m. If x has a non terminating decimal expansion then x must be of the form x = m/p where p is a prime number. Also, the period of x, say T(x), is less than p.
Let A = { x such that x is rational in (0,1) and has non terminating decimal expansion }
Let B = { y=T(x) such that x is an element of A }
We then have that for every y in B there is at least one prime p and natural m such that
y = T(x) = T(m/p) < p
Since B is unbounded so is the set of prime numbers


What is meant by $P(X = x)$? What is meant by the statement $P(X = x) = \theta$? As in, what is its English translation? I'm assuming that $X$ is a random variable and $x$ is a member of its sample space.

Is it just "the probability that event $x$ occurs for random variable $X$ is equal to $\theta$"?

If so, why use the "$=$" sign inside the parentheses? Isn't that technically misleading since the random variable $X$ is not equal to one of its sample space members $x$?

The statement is "The probability that the random variable $X$ takes the value $x$ is equal to $\theta$." The notation is a bit misleading. You are right that $X$ does not equal $x$, but rather, after a measurement of $X$ is made, the value $x$ is observed as a result.

The statement $\Bbb P(X=x)=\theta$ reads

  The probability that $X$ is equal to $x$ is $\theta$.

For example, consider the experiment of rolling one standard fair six-sided die once. Then our sample space is
$$
S=\{1,2,3,4,5,6\}
$$
Now, let $X:S\to\Bbb R$ be given by the formula
$$
X(x)=
\begin{cases}
1 & x\text{ not prime} \\
2 & x \text{ even prime} \\
3 & x\text{ odd prime}
\end{cases}
$$
Then
\begin{align*}
\Bbb P(X=1)&=\frac{1}{2}&\Bbb P(X=2)&=\frac{1}{6}&\Bbb P(X=3)&=\frac{1}{3}
\end{align*}

Note that the restrictions on the possible values of $\theta$ are quite severe if our random variable is continuous. Indeed, let $X$ be a random variable with probability density function $f$. Then
$$
\Bbb P(X=x)=\Bbb P(x\leq X\leq x)=\int_x^x f(t)\,dt=0
$$


Pythagoras theorem word problem A tree is 8 m north and 6 m east of another tree. One of the trees is 12 m tall, and the other tree is 17 m tall. Find the distance between the tops of the trees

You are asked to find the Euclidean distance between two points which are $8\,m$ apart on one axis, $6\,m$ apart on a second axis and $17-12\, m$ on a third. The axis are pairwise orthogonal.

Hint:
Your first task is modeling the given information into a geometric model.
The given numbers can be used to come up with Cartesian coordinates.

Here is a visualisation:

What could be $T_1$ and $T_2$?
What could be "TOP1" and "TOP1B", and "TOP2" and "TOP2B"?

Solution:

Let us model the information from the task:

  A tree is 8 m north and 6 m east of another tree.

So for the bases of the trees we can use relative coordinates
$T_1 = (0,0)$ and $T_2 = (6, 8)$, letting the $x$-axis point to the east and
and the $y$-axis point to the north.

  One of the trees is 12 m tall, and the other tree is 17 m tall.

This adds height to the picture. We abstract the trees as line segments from their base at $(x_i, y_i, 0)$ to their their top $(x_i, y_i, z_i)$.
So we have
$$
(0,0,z_1) \quad (6, 8, z_2)
$$
for the two top points.

The text gives the information about the heights, but leaves out which one has which height. So we have either
$$
\text{TOP}_1 = (0,0,12) \quad
\text{TOP}_2 = (6, 8, 17)
$$
or
$$
\text{TOP}_{1\text{B}} = (0,0,17) \quad
\text{TOP}_{2\text{B}} = (6, 8, 12)
$$

So lets us look at what to do:

  Find the distance between the tops of the trees

For the first scenario this would be
$$
d = \lVert(0,0,12) - (6, 8, 17)\rVert
$$
for the second scenario
$$
d = \lVert(0,0,17) - (6, 8, 12)\rVert
$$
where the Euclidean norm of a vector
$$
\lVert x \rVert
= \lVert (x_1, x_2, x_3) \rVert
= \sqrt{\sum_{i=1}^3 x_i^2}
$$
was used.
So or the first we get
$$
d^2 = (0-6)^2 + (0-8)^2 + (17-12)^2
$$
for the second
$$
d^2 = (0-6)^2 + (0-8)^2 + (12-17)^2
$$
so indeed we do not need the information which tree has what height height, both. In both scenarios we get
$$
d^2 = 36 + 64 + 25 = 125 \Rightarrow d = \sqrt{125} = 11.1803398\dotsb
$$


How to show that $n^{\ln(\ln(n))} = \ln(n)^{\ln(n)}$ I have verified that $n^{\ln(\ln(n))} = \ln(n)^{\ln(n)}$ by plugging in values for $n$, but do not understand why it is true. I am not aware of any $\log$ rules that can be used to simplify $n^{\ln(\ln(n))}$.

Note that $n=e^{\ln n}$ so

$$n^{\ln (\ln n)}=e^{\ln n\cdot \ln(\ln n)}=(\ln n)^{\ln n}$$

Following Cameron Williams' hint, we take the logarithm of both sides to get

$$\ln \left(n^{\ln\left(\ln (n)\right)}\right) = \ln \left( \ln (n)^{\ln (n)}\right)$$

Recall that $\log_b a^c = c\log_b a$. Use this property on both sides:

$$\left(\ln\left(\ln(n)\right)\right)\left(\ln(n)\right)=\left(\ln(n)\right)\left(\ln\left(\ln(n)\right)\right)$$

This is true because multiplication is commutative.


Discrete Mathematics Riddle, square grid problem I've been trying to solve this problem in my book to no avail:

I am even stuck on part a... to find a closed form of this I started with size 2x2 square, counted the number of paths, then used a 3x3 square, did the same but I feel that was a rather inefficient way to go about doing this problem. Can someone walk me through the solutions for parts (a), (b), (c)? Thanks.

You can write the number of ways to get to each intersection point while minimizing Taxicab distance by using the fact that the number of paths at any given intersection is equal to the sum of the number of paths that come directly from the left and the number of paths that come directly from below.

You might notice that this looks like Pascal's Triangle turned on its side. That's because it is exactly that. The reason these numbers are equal to binomial coefficients is because the number of paths is exactly the number of ways to arrange the rights and lefts in the path; to get to point $B$ you need $4$ lefts and $4$ rights. There are $$\binom{8}{4}=70$$ ways to choose the order of these rights and lefts.

You should be able to solve the second question using this method.

Good luck :)

EDIT: I've included how the numbers would be assigned for the third part. Notice the $10$ assigned to $(2,4)$.

For part (a), you have in total 8 moves to make, 4 of which has to be up and the other 4 right. You can rearrange the order of these moves in $\binom{8}{4} = 70$ ways. So that is the answer to part (a).

For part (b), you have in total $n+m$ moves to make, n of which has to be up and the other m right. You can rearrange the order of these moves in $\binom{n+m}{n}$ ways.


What is the set of positive integers $n$ such that $(6k+2)^n + 3$ and $(6k+2)^n + 5$ are both prime? That is, for some fixed $k\in N$, what is the set of positive integers $n$ such that $(6k+2)^n+3$ and $(6k+2)^n+5$ are both prime? At least, is this set finite?

See if you want the entire thing prime then the term should not have any prime factors. So $(6k+2)^n$ should not be divisible by 3 or 5. This term can never be divisible by 3, but can be divisible by 5. So k should not be 3, 8, 13, and so on.

I cannot think how can i find the set for n, but i think this might help.


Does $d(x,y)$ satisfy the triangle inequality? Let $d(x,y)=\sup_{j\in N}|\xi_j-\eta_j|$ where $x=(\xi_1,\xi_2,...)$ and $y=(\eta_1,\eta_2,...)$.

Does $d(x,y)$ satisfy the triangle inequality?

Yes, $d$ satisfies the triangle inequality. Reason: $|*|$ satisfies the triangle inequality. Now its your turn to give a proof.


Discrete Math (Proving by contradiction for sets) So, the question is:
Prove by contradiction that if A$\cap$B$\subseteq$C and x$\in$A-C, then x$\notin$B.

So My process is
Let

A:{X,Y,Z}

B:{Z,Y,V}

C:{Y,Z,Q}

A$\cap$B: {Z,Y}

A-C = {X,Y,Z} - {Y,Z,Q} = {X}

So it follows that x$\in$A-C but x$\notin$B.

So I'm not sure if this is the proper way to prove by contradiction as I have never done contradiction proofs with sets before

To prove $P \to Q$ by contradiction, you must assume both $P$ and $\lnot \, Q$ (not $Q$) are true and show that that leads to a contradiction.

In your particular case, you should start by assuming that $(A\cap B)\subset C$ and $x \in (A-C),$ as well as $x \in B.$ Then the last two conditions imply that $x \in (A \cap B).$ Can you see how to derive the contradiction from there?

Thanks everyone I got the answer

We assume that:

x∈A-C and x∈B

If, x∈A-C then x∈A

If, x∈A-C then x∉C

If, x∈B then X∈A∩B

But if X∈A∩B, then X∈C, since A∩B⊆C which is not possible, since x∉C

Therefore by contradiction x∉B.


Determine the greatest value of $n$ for which $b > a$ Let $a_n$ and $b_n$ be two recursive sequences so that:
$a_{n+1} = a_{n} + 2000$ where $a_{1} = 500$
and
$b_{n+1} = \frac{b_{n}}{1,001}$ where $b_{1} = 50000$

Determine the greatest value of $n$ for which $b > a$.

I have no idea of how to solve this. Perhaps the following is a start?

$a_{n+1} = a_{n} + 2000 \Rightarrow a_n = a_{n+1} - 2000$
$b_{n+1} = \frac{b_{n}}{1,001} \Rightarrow b_{n} = 1,001b_{n+1}$

And then you divide them and try to find a value for $n$ so that the fraction is greater than 1?

You should find a closed form for both sequences. I can tell you that
$$b_n = 5000 \cdot 1.001^{n-1} \\
a_n = 500 + 2000 \cdot (n-1) = 2000 n - 1500$$
Now you can solve $a_n = b_n$ for $n\in\mathbb R$ (an extension of the sequences). The resulting $n_0$ must then be rounded down for an answer to the question.

I believe you can simply work out a few cases by hand in a table for the answer.

Otherwise, find a simple formula for each term of the two sequences in this way:

Notice that $a_1 = 500, a_2 = 2000 + a_1, a_3 = 2000 + a_2 = 2 \times 2000 + a_1, \cdots$, which tells you that $a_n = 2000(n-1) + 500 = 2000n - 1500$.

Similarly, $b_n = \frac{b_1}{1001^{n-1}}$.

Now $$b_n > a_n \\ \implies \frac{50000}{1001^{n-1}} > 2000n - 1500 \\
\implies \frac{100}{1001^{n-1}} > 5n - 3$$ and so on.


Recursively defined sequences exercises. The sequence $ \{a_n \} $ defined recursively as,
$ a_1 = 1, a_2 = 2 $ and $ a_ {n + 2} = \dfrac{1}{2} (a_n + a_ {n + 1}) $ for $ n> 3 $.
I tried and found that the limit of that series is given by $ \dfrac {2} {3} \cdot 2+ \dfrac {1} {3} $ but I don't know if that result is good.

You can use characteristic equation to solve it. $$2x^2-x-1=0 \implies x = 1 \text{ or } x = -\frac{1}{2}$$ Therefore $a_n = (1)^n \cdot c_0+(-\frac{1}{2})^n\cdot c_1 \implies 1 = a_1 = c_0-\frac{1}{2}c_1$ and $2 = a_2=c_0+\frac{1}{4}c_1 \implies c_0=\frac{5}{3}$ and $c_1=\frac{4}{3}$, which means that $$a_n=\frac{5}{3}+(-\frac{1}{2})^n\cdot\frac{4}{3} \implies \lim_{n \to\infty}(a_n)=\frac{5}{3}$$

Using a standard method for
solving homogeneous linear recurrence relations with constant coefficients,
it can be seen that $a_n=\dfrac53+\dfrac43\left(-\dfrac12\right)^n$, so you are correct that $\lim\limits_{n\to\infty}a_n=\dfrac53$.


Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this

$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$

Someone can help me ?
Thanks.

Since

\begin{align}
\tan(2x)&=\frac{2\tan x}{1-\tan^2 x}&&&\text{and}&&\tan\left(\frac{\pi}{4}-x\right)&=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}
\end{align}
we have
\begin{align}
\lim_{x\to\frac{\pi}{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)&=\lim_{x\to\frac{\pi}{4}}\frac{(2\tan x)(1-\tan x)}{(1-\tan x)(1+\tan x)^2}\\
&=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}\\
&=\frac{2}{(1+1)^2}\\
&=\boxed{\color{blue}{\frac{1}{2}}}
\end{align}

hint: $\tan(2x) = \dfrac{\sin(2x)}{\cos(2x)}$ and write $P = \sin(2x)\cdot \dfrac{\tan\left(\dfrac{\pi}{4} - x\right)}{\cos(2x)}$ and use L'hospitale rule for the second factor.


Is $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$ continuous?. Where is bluff? I have a function defined by $f(x) = \sum^{\infty}_{n=1} \sqrt{x} e^{-n^2 x}$. The task is to check, whether $f(x)$ is continuous at $x = 0$. I have proposition of a solution and I would like someone to point out a bluff as there most likely is one - solution is suspisciously too short.
\begin{equation}
f(x) = \sqrt(x) \sum^{\infty}_{n=1} e^{-n^2x} = \sqrt(x) A_{n}
\end{equation}
I do it because x doesn't change when summing, I treat it as constant. The sum $A_{n}$ is obviously convergent, so there exist some finite M conforming $A_{n}$ < M. Thus, I can write $\lim_{x \rightarrow 0} f(x) < lim_{x \rightarrow 0} xM = 0$
well. that would be finished, but as I have mentioned before, I suspect bluff.

Here is an elementary approach to obtain the desired limit.

As $x$ tends to $0^+$, we have

  $$\sqrt{x}\sum_{n=1}^{+\infty}e^{\large -xn^2} \longrightarrow \frac{\sqrt{\pi}}{2}. \tag{*}$$

Proof.
Let $x>0$ and $n\geq 0$. Since $\displaystyle t \in [0,+\infty) \rightarrow e^{\large -xt^2}$ is a decreasing function, we have
$$
e^{-x(n+1)^2} \leq e^{-xt^2} \leq e^{-xn^2}, \quad t \in [n,n+1], \tag1
$$ integrating $(1)$ with respect to $t$, we get
$$
\int_n^{n+1}e^{-xt^2}dt \leq \int_n^{n+1}e^{\large -x n^2}dt = e^{\large -x n^2}\tag2
$$ and
$$
\int_{n-1}^{n}e^{\large -x n^2}dt = e^{\large -x n^2}\leq \int_{n-1}^{n}e^{-xt^2} dt. \tag3
$$
Then, summing $(2)$ for $n\geq0$, gives
$$
\int_0^{+\infty}e^{-xt^2}dt \leq 1+\sum_{n=1}^{+\infty}e^{\large -x n^2} \tag4
$$
and summing $(3)$ for $n\geq 1$, gives
$$
 \sum_{n=1}^{+\infty}e^{\large -x n^2} \leq \int_0^{+\infty}e^{-xt^2}dt. \tag5
$$Recalling the gaussian integral evaluation, we have
$$
\int_0^{+\infty}e^{-xt^2}dt=\frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}}, \quad x>0.\tag6
$$ Hence, combining $(4)$, $(5)$ and $(6)$, gives
$$
\frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}}-1 \leq \sum_{n=1}^{+\infty}e^{\large -x n^2} \leq \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}} \tag7
$$
leading, as $x$ tends to $0^+$, to the announced result $(*)$.

Note the summation when $x=0$ is infinite since it is an infinite sum of 1's, thus contradicting your argument of finite $M$.


What is $\Delta(1)$ for $1$ in $U(\mathfrak{g})$? Let $\mathfrak{g}$ be a semisimple Lie algebra and $U(\mathfrak{g})$ its universal enveloping algebra. Then $U(\mathfrak{g})$ is a hopf algebra. Is $\Delta(1) = 1 \otimes 1$ or $\Delta(1) = 1 \otimes 1 + 1 \otimes 1$? Here $1$ is the identity in $U(\mathfrak{g})$. Thank you very much.

One of the axioms of a bialgebra $B$ (in particular a Hopf algebra) is:

(taken from Wikipedia: in their notation $K$ is the base field and $B$ is the bialgebra).

Here $\eta$ is the unit and $\Delta$ is coproduct. In plain words, $\eta(\lambda \in K) = \lambda 1_B$. By the commutativity of the diagram, $$\Delta(1_B) = \Delta(\eta(1_K)) = (\eta \otimes \eta)(1_K \otimes 1_K) = 1_B \otimes 1_B.$$

As Hanno says you can interpret that by remembering that $\Delta$ is a algebra homomorphism in a bialgebra, so it has to preserve the unit.

The equation $\Delta(x) = 1 \otimes x + x \otimes 1$ is only valid for the elements of $\mathfrak{g}$ in $U(\mathfrak{g})$. Such an element is called a primitive element. You should be very careful when computing the coproduct in $U(\mathfrak{g})$, because it's tempting to believe that since $\Delta(x) = 1 \otimes x + x \otimes 1$ on generators, the same holds true for every element. That's not the case. The rule to use is again that $\Delta$ is a algebra homomorphism, so for example if $x, y \in \mathfrak{g}$,
$$\Delta(xy) = \Delta(x) \Delta(y) = (1 \otimes x + x \otimes 1) (1 \otimes y + y \otimes 1) = 1 \otimes xy + x \otimes y + y \otimes x + xy \otimes 1.$$
And a similar pattern holds true for products with more terms.

On my master thesis two year ago, i think i used  $\Delta(1) = 1 \otimes 1 + 1 \otimes 1$. By a fast googling, i found the same on wiki.

I hope that helps.


Integer $2 \times 2$ matrices such that $A^n = I$ An earlier question today motivates this slight variant:

For what natural numbers  $n$ does there exist a non-identity integer $2\times 2$ matrix $A$, such that $A^n = I$? (And let's say $A^k \ne I$ for $|k| < |n|$, too.)

Clearly there are solutions for $n=2,3$, because $(-I)^2 = I$, and for $n=3$ we have
$\left( \begin{smallmatrix} -2 & 1\\ -3 & 1 \\\end{smallmatrix} \right)$

The solution for $n = 3$ suggests to me that something involving the euclidean algorithm might come into play...and the fact the the determinant is $\pm 1$ suggests that this is really an $SL(2, \mathbb Z)$ (or $PSL(2, \mathbb Z)$)problem...but that's a group I'm woefully ignorant about.

For $n = 4$, there's $\left( \begin{smallmatrix} 0 & -1\\ 1 & 0 \\\end{smallmatrix} \right)$. And right about there I run out of ideas.

For $n=6$ there's $\left( \begin{smallmatrix} 2 & -1\\ 3 & -1 \\\end{smallmatrix} \right)$, not surprisingly. And these are the only such $n$, even if you allow rational entries instead of just integers.

Suppose that $A^n=I$ but $A^k\ne I$ for $k<n$. Then the minimal polynomial of $A$ divides $x^n-1$ but doesn't divide $x^k-1$ for $k<n$. It follows that the minimal polynomial of $A$ divides the $n$th cyclotomic polynomial $\Phi_n(x)$. But that polynomial is irreducible over the rationals, and so the minimal polynomial of $A$ equals $\Phi_n(x)$. Finally, the minimal polynomial of $A$ has degree at most $2$, since it divides the characteristic polynomial of $A$ by the Cayley-Hamilton theorem. Therefore the degre of $\Phi_n$ is at most $2$, which means that $\phi(n)$ is at most $2$, and so $n\in\{1,2,3,4,6\}$.

$$R_{\theta}=\begin{bmatrix}
cos \theta  & -sin \theta \\
 sin \theta& cos \theta
\end{bmatrix}\\$$if $$R_{\theta+2k\pi}=R_{\theta}\\(R_{\theta})^n=R_{n\theta}\\\theta=\frac{2k\pi}{n} \rightarrow (R_{\theta})^n=R_{n\theta}=R_{n\frac{2k\pi}{n}}=R_{2k\pi}=I $$


Solve the equation $\sqrt{x^2-(a+b)x+ab}+\sqrt{x^2-(b+c)x+bc}=0$ where $a$, $b$, and $c$ are real numbers So I realised that you can factor $x^2-(a+b)x+ab$ and $x^2-(b+c)x+bc$ to $(x-a)(x-b)$ and $(x-b)(x-c)$ and then you get  $$\sqrt{(x-a)(x-b)}+\sqrt{(x-b)(x-c)}=0$$ This is where I got stuck. I tried squaring both sides but I couldn't find the solution.

You almost done, $x=b$ is a solution, if $x\not = b$ then you can divide by $\sqrt{(x-b)}$ and you have $\sqrt{(x-a)}+\sqrt{(x-c)} = 0$ now a square is always positive so a sum of two squares is zero if both are zero, hence $x=a$ and $x=c$.

So conclusion: $x=b$ is always a zero, and if $a=c$ then $x=a$ is another solution.

Multiplying both sides with $\sqrt{(x-a)(x-b)}-\sqrt{(x-c)(x-b)}$, we see that necessarily
$$(x-a)(x-b)-(x-c)(x-b)=0,$$
i.e.,
$$ (x-b)(c-a)=0.$$
So $x=b$ (which we verify to actually be a solution), or we need $a=c$.
But with $a=c$, the original equation becomes
$$ 2\sqrt{(x-a)(x-b)}=0,$$
hence $x=a$ (or again $x=b$)


How can I get Maclaurin series for $\frac{x^2 + 3e^x}{e^{2x}}$? The answer for it is $$3 + \sum_{k=1}^n (3+k(k-1)2^{k-2})\frac{(-1)^k}{k!} x^k + o(x^n)$$
Well, I've tried to change every $e^x$ to $1 + x + \frac{x}{2!} + ... + o(x^n)$ and got nothing useful. I know how to get Maclaurin series for polynomials but I don't really know what should I do with fractions and how get that kind of result.

Hint: I would start by simplifying:
$$
\frac{x^2+3e^x}{e^{2x}}=x^2e^{-2x}+3e^{-x}.
$$

First Write the function as $3e^{-x}+x^2e^x$. Then add the Maclaurin expansions.


Given $1/x$ something? $$ \frac{1}{x} <E $$
$E, x$ always positive.

How can I get to expression like this:

$ x > \text{something} $

in one stage? If it's not possible in one stage (although this is what my teacher did) other answers are also good.

Multiplication by a (strictly) positive quantity conserves the (strict) inequality. So if $x>0$,

$$x\dfrac{1}{x}<Ex$$

Similarly for $E>0$, $1/E<0$ so $$\dfrac{1}{E}<x$$

Note that multiplication by a negative quantity reverses the sign of the inequality: $3>2$ but $-1\times 3< -1\times 2$.


Is this number divisible by $11$? Is  $73864589999999923243431$ divisible by $11$? Show and justify?

Usually I would take modulo 11, but it doesnt seem like a reasonable option here. I dont want the answer, just a hint?

Hint. Numbers are divisible by $11$ whenever the sum of the digits in odd positions differs from the sum of the digits in even positions by a multiple of $11$ (including zero and negative multiples of $11$).


To show equal cardinality To show two sets have equal cardinality is it always necessary to show bijection between the sets?

suppose I have two sets A and B isn't it enough to show that there exist one onto mapping from A to B and some another onto mapping form B to A. The reason why I am asking this question is that it sometimes becomes very difficult to construct such a function which gives bijection.

For example, when I wanted to show bijection between rationals and naturals.
It was difficult to construct such a function and later found out that
$f(m,n)=2^{m-1}(2n-1)$

will do the work.

Now,recently I came across another problem that show that [0,1] and (0,1) have equal cardinality.
I was able to construct onto mapping from (0,1) to [0,1] and also another onto mapping from [0,1] to (0,1) but still now I couldn't construct such a function which will biject the two sets.

It isn't necessary to construct an explicit bijection if you have a result at hand you can apply. For instance, the Schroeder-Bernstein Theorem states that if there are injections $f : A \to B$ and $g : B \to A$, then there exists a bijection $h : A \to B$.

In your last example, it isn't hard to find injections $f : [0,1] \to (0,1)$ and $g : (0,1) \to [0,1]$, so there must exist a bijection between the sets, although we didn't demonstrate a specific example of one.

Existence of onto map from $A$ to $B$ gives card$(A)\leq$ card$(B)$ and existence of onto map from $B$ to $A$ gives card$(B)\leq$ card$(A)$ . In either way we can say that card$(A)=$ card$(B)$  i.e. there must be a bijection between $A$ and $B.$
But generally we have to show that two sets are of equal cardinality or not, we are not intersting in bijective map and thats  not so simple to find such a bijection.


I'm confused on the limit of $\left(1+\frac{1}{n}\right)^n$ Okay so I read Richard Rusczyk's AoPS Volume 2 Book, and I stumbled upon the part where he informs very briefly that $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. But he doesn't really provide a rigorous proof as to why that's true (not criticizing him or anything).. It would really help if someone could provide me with the simplest proof possible as to why $\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n=e$. Thank you in advance!

Assume $e$ is defined by $e := \sum_\limits{k=0}^{\infty} \frac {1}{k!}$

Let

$$y=\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$$
$$\ln y=\lim_{n \rightarrow \infty} n\ln\left(1+\frac{1}{n}\right)$$
$$\ln y=\lim_{n \rightarrow \infty} \frac{\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$
using L'Hôpital's rule
$$\ln y = \lim_{n \rightarrow \infty} \frac{\frac{-1}{n^{2}}*\frac{1}{1+\frac{1}{n}}}{\frac{-1}{n^{2}}}$$
$$\ln y = \lim_{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}$$
$$\ln y = 1$$
$$y=e$$

This is of the form $1^\infty$ as $ n \to \infty$.

Now for any limit $\displaystyle\lim_{n\to \infty}f(x)^{g(x)}$ of the form $1^\infty$ we have direct formula to be : $e^{lim_{n\to\infty}(f(x)-1)(g(x))}$.

Now $f(x)=1+\frac{1}{n}$ and $g(x)=n$. Evaluating using the above formula the limit comes out to be : $$e^{lim_{n\to \infty}(1+\frac{1}{n}-1)n}=e^{lim_{n \to \infty}(\frac{1}{n})n}=e.$$


Probability that between two Bernoulli sequences, one will get 'ahead' and remain there to sequence end. As per title, given two Bernoulli sequences both of length $N$ with probability of success $P$ the same for both, what is the probability that one will 'get ahead' in its sum from 1st of the sequence to the current sequence position (or said another way, one of the two sequences has a first hit that gives them a lead before the other) and then stays ahead through sequence position $N$ (the end).

E.G., if the two sequences were $0,1,0,0,1$ and $1,1,0,1,0$ the latter 'got ahead' at the start, and stayed ahead.

Answers outright, hints, pointers to literature all appreciated, I have hit a wall beyond enumerating and counting simple examples.

I searched other queries here, found nothing similar.

Thanks.

This is my attempt to solve the problem and it could be a little different.

Let us assume that the sequence E gets ahead and stays ahead through the sequence. Another

way of interpretation is that its expected number of successes is always greater than that

of  the sequence G. Thus the required probability is Expected Success of E/(Expected

Successes of E+ Expected Successes of G).

Let us assume that E stays ahead after the first two positions of sequence. This can happen

in the following ways when compared to G. Thus E will have successes in the first two

positions. G which is lagging will have (Failure, success),(success, failure),

(failure,failure) and (success, success). The first three will characterize its lag and the

last will put the two positions unchanged with no clear winner of getting ahead.

Thus Expected Successes of E, $$\bar{E} =2*3p^2+(\bar{E}+1)*p$$.

Meaning three subsequent successes and hence $3p^2$  and one position after with a

probability of keeping the success.

Similarly Expected Successes of G,$$\bar{G}= [2.p.(1-p)+(1-p)^2].2+(\bar{G}+1).(1-p)$$ meaning

the probability of (Failure, success),(Success, failure) and (failure, failure) and one

position after with a probability of keeping the failure

$$\bar{E}=3p^2*2+p*\bar{E}+p$$ $$=> \bar{E}(1-p) = 6p^2+p$$  $$=> \bar{E}=\frac{(6p^2+p)}{(1-p)}$$

$$\bar{G}= [2p(1-p)+(1-p)^2]*2 + \bar{G}*(1-p)+ (1-p)$$  $$=>\bar{G}(1-1+p) = (1-p)*(4p+2-2p+1)$$

$$ =>\bar{G} = \frac{((1-p)*(2p+3))}{p}$$

Required Probability $$=\frac{[p^2(6p+1)]}{[(8p^3-4p+3)]}$$

Dynamics of the required probability given p to make sure $0\leq required Prob\leq 1$

0   0

0.1 0.006134969

0.2 0.038869258

0.3 0.125

0.4 0.284518828

0.5 0.5

0.6 0.711340206

0.7 0.86548913

0.8 0.952772074

0.9 0.990825688

1   1

Thanks

Satish


integrating a function within another function Is there a general solution for the following integration?

$$\int{f(g(x))}{dx}$$

Hypothesis (Probably wrong and you should ignore) from observations:

$$\int{f(g(x))}{dx}=\frac{F(g(x))}{g^{\prime}(x)}+c$$
Where $F(x)=\int{f(x)}{dx}$ and $c$ is a constant

No, there is no general solution for the primitive of $f\circ g$.

The derivative of your proposed solution is

$$\frac{g'(x) ( g'(x) f(g(x)) ) - F(g(x))g''(x) }{(g'(x))^2} = f(g(x)) - \frac{F(g(x))g''(x) }{(g'(x))^2}$$

If there was a formula with just the primitive of $f$, it would be easy to integrate any function $v$ : choose $f(x) = \ln(x)$ and define $g(x) = e^{v(x)}$. then $f(g(x)) = v(x)$

Another remark : $\exp$ and $-x^2$ are very simple functions to integrate, but the primitive of $e^{-x^2}$ cannot be expressed with a finite number of simple functions

This is incorrect as if you take derivative from both sides of your hypothesis, you get
$$f(g(x))=\frac{f(g(x))\cdot (g'(x))^2-F(g(x))\cdot g''(x)}{(g'(x))^2},$$
Which obviously does not hold.


How many 3 digit even numbers are there(No Repetition)? First find numbers ending with 0

So, 1's place-1 10's place-9 100's place-7 (2 digits are already consumed and 0 can't be used)

So 7*9*1.Im i doing the right thing?

First find the even numbers that are ending up in zero
so, for no. ending up with zero are
zero at one's place so 1 combination
now 9 at hundreds and 8 at tens
Now even numbers not ending with zero i.e. ending  in 2,4,6,8
so 4 at one's place only one is used so 8 at hundreds place since 0 can not be used and one number is 0 and now 8 are left for the tens place
so, total no. of digit sequences would be
=>9*8*1+8*8*4=72+256=328
When zero is used up there is no problem for using up tens or hundreds but when zero is not used up after putting one of 2,4,6,8 numbers at one's place we have to check if 0 is ending up in the most significant place or not. To avoid that we start with hundredth place. So we ensure this first.

The total 3 digit numbers are 999 including preceding zeros and there are 999/2 even numbers.
so total three digit even numbers are 499.


Are there any natural proofs of irrationality using the decimal characterization? Mathematicians typically define rational number to mean quotient of two integers.  It is not hard to show that a number is rational by that definition if and only if its decimal expansion terminates or repeats. Let us call that the “decimal characterization” of rationality. The proof of that characterization of rational numbers is obviously just as applicable to other bases as it is to base $10$.

Question: Are there any proofs of the irrationality of $\pi$ or $e$ or $\sqrt 2$ or $\log_2 3$ or any other naturally occurring number that use the decimal characterization, showing directly that the decimal expansion does not terminate or repeat, and that are at least as simple as any proof that uses the characterization that is conventionally taken to be the definition?

I just learnt this, it is not about the decimal expansion, but I think it is clearly related to your question :

A simple proof that $e$ is irrational is using its factorial expansion $2+\sum_{n=2}^\infty \frac{1}{n!}$.

The theorem is that
$$\sum_{n=1}^\infty \frac{a_n}{n!}, \qquad a_1 \in \mathbb{Z},\quad a_n \in \{0, \ldots, n-1\}$$
is irrational if and only if $a_n  >0$ infinitely often and $a_n < n-1$ infinitely often.

The proof is as follows :

any real number can be written as
$$\sum_{n=1}^\infty \frac{a_n}{n!}, \qquad a_1 \in \mathbb{Z},\quad  a_n \in \{0, \ldots, n-1\}$$
with the same algorithm as for the decimal expansion : $$a_n = \lfloor x_n n!\rfloor, \qquad x_{n+1} = x_n-  \frac{a_n}{n!}$$
with this algorithm, every rational number has a finite expansion, so if $x$  has a finite expansion then the algorithm finds it. More generally, if applied to $x = \sum_{n=1}^\infty \frac{c_n}{n!}, c_1\in \mathbb{Z},c_n \in \{0, \ldots, n-1\}$ where $c_n < n-1$ infinitely many times, then the algorithm finds back $a_n = c_n$.
when applied to $e-1$, the algorithm finds $a_n = 1$, thus $e-1$ doesn't have any such finite expansion, hence it is irrational.

The conclusion is that the factorial expansion (more generally the Cantor series expansion,  if someone knows a better reference ?) is much easier than the base $N$ expansion for proving the irrationality of a real number.

This is a suggestion rather than a proof, but it's too long for a comment.

What about $$\zeta(2)=\sum_1^\infty\frac{1}{k^2}=\frac{\pi^2}{6}\text{?}$$

If it is rational, it has period $n$. But there exists a number $N$ such that the term $\frac{1}{N^2}$ has a period $m$ which is not a divisor of $n$. If it were the only period-$m$ term in the series, we would have a contradiction, which by reductio ad absurdum means that $n$ cannot exist, and that $\zeta(2)$ can't have a period and so must be irrational.

The hand-waving part of this argument, which I'm sure can be made solid, is that the sum of all period-$m$ terms in the series is itself of period $m$. (It is only necessary, strictly speaking, that some values of $m$ should exist for which this is true: for instance, $m=2p$).

A similar argument with factorials instead of squares would show the irrationality of $e^1=e$. Less hand-waving would then be needed because the terms get smaller so fast that there is no possibility of the period-$m$ terms coalescing into something with a shorter period.


Find determinant for a matrix with $x$ on the diagonal Find determinant for a matrix with $x$ on the diagonal and $-1$ at the end of the first row and $-1$ at first of the last row other elements are $0$.
{ x  0 0 -1
  0  x 0  0
  .  0 x
  .  0 0  x
  -1 0 .  .  x
}

$$
\pmatrix{x & 0 & \cdots & 0 & -1\\
0 & x & \ddots & \vdots & 0\\
\vdots & \ddots & \ddots & 0 &\vdots\\
0& \cdots & 0 & x&0\\
-1&0&\cdots & 0 & x}
$$
I know if i add 1/xR1 to R2 the determinat will be $(x^{n-1})(x-1/x)$. I asked my professor she said what if $x$ is zero. Now i don't know what to do.

Your approach is perfectly fine. What you have shown is that in the case that $x \neq 0$, Your matrix $M$ satisfies
$$
\det M = x^{n-1}(x - 1/x) = x^{n-2}(x^2 - 1).
$$
All that remains is to show that this equation holds for $x = 0$ as well.
There are a few approaches to handle the case of $x = 0$. One is to note that $\det(M)$ must be a polynomial over $x$; since $\det(M) = x^{n-2}(x^2 - 1)$ holds for all $x \neq 0$ and polynomials are continuous, the equation must hold for $x = 0$ as well.
Alternatively, we can simply plug in $x = 0$. For the case of $n=2$, we have
$$
\det(M) = \det \pmatrix{0&-1\\-1&0} = 0^2 - (-1)^2 = -1,
$$
which agrees with the result of plugging $x = 0$ into $x^{2-2}(x^2 - 1) = x^2 - 1$, since $0^2 - 1 = -1$. For the case of $n > 2$, we see that $x = 0$ leads to a second row of zeros, which means that the determinant of $M$ is zero. This agrees with the result $x^{n-2}(x^2 - 1) = 0^{n-2}(0^2 - 1) = 0$.

One fast approach is to use the W-A (AKA Sylvester) identity. This matrix can be written as $M = xI + AB$, where
$$
A = \pmatrix{1&0\\0&0\\ \vdots & \vdots \\ 0&0\\0&1}, \quad B = -\pmatrix{0&0&\cdots & 0& 1\\1 & 0 & \cdots & 0 & 0}.
$$
For all $x \neq 0$, we have
$$
\begin{align}
\det(M) &= \det(xI_n + AB) = x^n \det(I_n + x^{-1}AB)
\\ &  = x^n \det(I_2 + x^{-1}BA)
\\ & = x^{n-2} \det(x I_2 + BA) = x^{n-2} \det \pmatrix{x & -1\\-1 & x}
= x^{n-2} (x^2 - 1).
\end{align}
$$

If $x=0$, then the determinant will be zero ( cause there is 1 row with all elements are zeros )


Difficulties with Modulus - Chinese Remainder Theorem, GCD and the Extended Euclidean Algorithm I am finding it quite difficult to deconstruct the steps for seemingly simple questions and decide which mathematical process to use. For example

If $x \equiv 2$ (mod $11$) and $ x \equiv 1$ (mod $17$), what is the value of $x$ (mod $187$)?

There are many online resources that patiently walk through questions such as this, but I find that some of the numbers suggested in the steps are either random guesses to get closer to an objective (without saying so) or there is no explanation as to how they were calculated or why they were chosen during a step. Then trying to follows the process onwards after that is unproductive because one is trying to puzzle out unknown step calculations instead of the objective. I'd appreciate any advice please. As in all of mathematics, it's more rewarding to solve it oneself, but sometimes productive learning hours turn to despair.

Saying that x is congruent to 2 modulo 11 means that x= 2+ 11m for some integer m.  Saying that x is congruent to 1 modulo 17 means that x= 1+ 17n for some integer n.
So 2+ 11m= 1+ 17n which is the same as 17n- 11m= 1.
Now 11 divides into 17 once with remaider 6: 17- 11= 6.
6 divides into 11  once with remainder 5:  11- 6= 5.
And, of course, 6- 5= 1.
6- 5= 6- (11- 6)= 2(6)- 11= 1.
2(17- 11)- 11= 2(17)- 3(11)= 1.
So one solution to this is n= 2, m= 3. x= 1+ 17(2)= 35 or x= 2+ 11(3)= 35.   Of course any number of the forM x= 35+ (11)(17)k=  35+ 187k is also a solution. So x= 35 (mod 187).


A simple Fourier Transformation I am a bit stuck with this small basic signal.

I have this

$$y(t)=\frac{\sin(200\pi\,t)}{\pi\,t}$$

and I want to take its Fourier Transformation. Obviously it looks like the sinc function. But that $200$ confuses me a lot.

In other words, the question is, how to rewrite that y(t), into a sinc(t) function , so i can use the FT table. –  RatzaJR

Elaborating on my comment, where I already answered this, we have that $y(t) = 200 \operatorname{sinc}(200t)$. Therefore, denoting the Fourier transform of a function $f$ by $\hat{f}$,
$$\hat{y}(\xi)=200 \frac{1}{200} \widehat{\operatorname{sinc}}\left( \frac{\xi}{200}\right) = \operatorname{rect} \left( \frac{\xi}{200}\right) = \begin{cases} 1 \qquad \text{if $|\xi|< 100$} \\ 0 \qquad \text{if $|\xi| \geq 100$}\end{cases},$$
where the first equality follows from the basic fact that for real numbers $a,b$ where $a \neq 0$ and a function $f$, the Fourier transform of $a f(bx)$ is given by $\frac{a}{|b|} \hat{f}\left(\frac{\xi}{b} \right)$. You should really try to understand this step, otherwise you're toasted in your exam...

$$Y(\omega)=\int_{-\infty}^{\infty}y(t) \cos(\omega t) d\omega=2\int_0^{\infty}y(t) \cos(\omega t) d\omega=(sign(200\pi-\omega)+sign(200\pi+\omega))/2$$


Differentiability of $f$. Given $0 < \beta < 1$ and $|f(x)|\geq |x|^ \beta $ and $f(0)=0$. Show that $f$ is not differentiable at $0$ This is what I have so far. I think it is the absolute value signs that are throwing me off. By the definition of derivatives,

$$
f'(0) =  \lim_{h \to 0} \frac {f(0 +h) - f(0)} {h}= \lim_{h \to 0} \frac {f(h)} {h}\\
$$

But now what? Since $f(0)=0$ and as $h$ tends to $0$, doesn't the the derivative read $0/0$?
I know that I should be getting the left and the right hand limit to be different at zero. I am just failing to carry the rest of the proof through. Will appreciate any help.

The result is wrong if $\beta=1$ (try $f(x)=x$). If $\beta\lt1$, this follows from
$$
\left|\frac{f(h)}h\right|\geqslant\frac1{|h|^{1-\beta}}\to\infty.
$$

Since as you stated:
$$f'(0) = \lim_{h\rightarrow 0} \frac{f(h)}{h}=\lim_{h\rightarrow 0} \frac{|f(h)|}{h}\ge\lim_{h\rightarrow 0} \frac{|h|^\beta}{h}$$
If we approach from the right, $h>0$. Assume $f(h)>0, |\beta|<1$:
$$f'(0)=\lim_{h\rightarrow 0} \frac{|f(h)|}{h}\ge\lim_{h\rightarrow 0} \frac{|h|^\beta}{h}=\lim_{h\rightarrow 0} h^{\beta-1}=\infty$$
If $f(h) <0$, we get the same divergence but to $-\infty$. The same proof follows for $h<0$.


Each user on a computer system has a password, which is six to eight characters long,$\dots$ Question:
Each user on a computer system has a password, which is six to eight characters long, where each character is an upper-case letter or a digit. Each password must contain at least one digit. How many possible passwords are there?

  I'm in the Basic of Counting section of my Discrete Mathematics book, and I have a problem with my reasoning with this. I will give you my reasoning and the books reasoning. Both give different answers, but I don't see a difference in the train of thought, so I need someone to point out the difference.

My Attempt:
Immediately I noticed 3 kinds of character length which allows me to break it down to three cases, respectively $P_6, P_7, P_8$, then add all of them. For $P_6$, $5$ will be made up of alpha numeric characters, and $1$ is made up of just digits due to the requirement of "at least one digit", thus

$$P_6 = (36)^5*10$$

Should be enough to show my train of thought, now the books solution.

Books Solution: The book did the same thing in dividing in 3 cases and adding them later so I'll go ahead and show you their train of thought for $P_6$.

$$P_6 = 36^6 - 26^6$$

Basically its the number of possible 6 alphanumeric minus just alpha numeric.

  I know that both give different answers, but I still can't tell why.

Your calculation for $P_6$ gives the number of words in which the first five characters can be letters or digits, and the last must be a digit.  But there is nothing in the rules to say that the last must be a digit.

Alternatively, you could interpret your answer to say that, for example, symbols $1,2,4,5,6$ can be anything, and the $3$rd must be a digit.  Same problem - there is no requirement that any specific symbol must be a digit.

The simplest solution to the problem is the one given in the book.

$36^6$ gives you the number of passwords  $6$ characters long, including passwords which are alphanumeric. But it also includes password which contains only alphabet, therefore subtract subtract $26^6$.

Similarly for passwords of length $7$ and $8$.


Average value of a function Maybe I am just bugging out, but I need a sanity check

The question:

  If $\int_0^4 f(x)dx = C$, what is the average value of $f(x)$?

The question does not include a body for $f(x)$ or a table of values/graph for the output of f(x), so can this question be answered?

I know that:
$$\int_a^b f(x)\,dx = F(b)-F(a),$$
will evaluate the integral across a given interval, but the question is asking what the average value of f(x) is, not its integral.

If I take the derivative of the above integral, I can get the function itself, but what do I need to do about the interval $[0, 4]$?

If I take the derivative of the integral, will I be left with $f(x) = 0$ (since the constant will disappear).

Am I thinking too hard about this?

Thanks for the help.

Remember that if $f(x)$ is continuous on $[a,b]$, then the average value of $f(x)$ on $[a,b]$ is defined to be
$$\text{average value of }f\text{ on }[a,b]  = \frac{1}{b-a}\int_a^b f(x)\,dx.$$

So if you need to know the average value of $f(x)$ on $[0,4]$ (as opposed to "the value of $f(x)$"), then just use that formula.

The $\int_0^4f(x).dx$ gives you the area, which is the same thing as the sum of all the values of f from 0 to 4,  multiplied by the dx (the horizontal distance between the values):

$\int_0^4f(x).dx$ = F(4)-F(0) = $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx$

divide this number by dx multiplied by the number of f's and you get the mean value.

$\frac{[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx}{(ndx)}$ where n is the number of f's (or points)

Remember, dx is just the (horizontal) length of the area you're integrating, divided by the number of points (as the number of points goes to infinity). So, ndx=n(horizontal length)/n is just the horizontal length of the area you're integrating over.

Simplifying the previous thing, you get $[f(0)+f(.00000...1)+f(.00000...2)+...f(3.999...9)+f(4)]dx/(ndx) = [F(4)-F(0)]/\text{(horizontal length)}= [F(4)-F(0)]/(4-0) = C/4$


How can one prove that the sequence of $n$ 1's followed by $n$ 0's is aperiodic Disclaimer: This is not a homework task. The question arises from genuine curiosity.
I'm trying to prove that the sequence of $n$ 1's followed by $n$ 0's is aperiodic.
Apparently, the corresponding OEIS entry is A118175.
The first few terms are as follows:
$1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, ...$
From the OEIS page, I've chosen what seems to be the simplest expression (not claiming that it's indeed the simplest one, with respect to this particuar proof) for the sequence, namely:
$a(n) = 1 - ceiling(sqrt(n+1)) + round(sqrt(n+1))$
My approach is proof by contradiction.
Assume that the sequence is periodic and, hence, a period $k$ for $k > 0$ exists.
This implies that:
$a(n) = a(n + k)$
The $1$s cancel out, and what remains is:
$round(sqrt(n+1)) - ceiling(sqrt(n+1)) = round(sqrt(n+k+1)) - ceiling(sqrt(n+k+1))$
For $k < 5$ this causes problems, however, because the results of each pair of LHS-RHS terms may be equal.
This is as far as I've been able to get.
Update: Perhaps the fact that all $0 < k < 5$ cases could be exhaustively proven by induction, as irrelevant, be utilized.
For example, as per the periodicity assumption, we have $a(0) = a(k) = 1$ and $a(1) = a(k + 1) = 0$. If $k = 1$, however, a contradiction emerges, because it implies that $a(0) = a(k) = a(1) = 1$.
Then, for $k \ge 5$, it'll follow as per the properties of the $ceiling$, $round$ and $sqrt$ functions.

It is a mistake to use the ceiling, round, and sqrt functions here -- they just complicate a simple situation. All you need is two facts:

for any $k$, the sequence contains $k$ consecutive $1$'s;
the sequence contains an infinite number of $0$'s.

These two facts taken together ensure that the sequence is not periodic with period $k$.

Notice that the sequence of $n$ ones starts at the position
$$2(1+2+\cdots + (n-1))+1 = (n-1)n+1$$
and the sequence of $n$ zeroes starts $n$ positions further, i.e. $(n-1)n+1+n$.
Let $n \in \Bbb{N}$ be arbitrary. If $a$ were periodic with period $n$ we would have $a(k+n) = a(k)$ for all $k \in \Bbb{N}$. However, the above remark implies that for $k = n(n-1)+1$ holds
$$a(n(n-1)+1) = 1, \quad a(n(n-1)+1+n) = 0$$
so $a$ is not periodic with period $n$.


How many ways are there to distribute pens between two girls and one guy? There are two girls and one guy and 121 pens. How many ways are there to distribute pens between two girls and one guy, so that the girls have the same number of pens. The pens all are identical.
Please can you explain a bit more deeply?

Choose an even number $n\in[0,\dots,121]$.

Give $n/2$ pens to each girl, and $121-n$ pens to the boy.

There are $61$ even numbers between $0$ and $121$, hence $61$ ways to distribute the pens.

mathematical modelling of this situation is $2x + 2y = 120$ where
$x$ is the number of pens for a girl, and $2y + 1$ is the number
of pen for the boy. therefore $x + y = 60$ with $y$ ranging from
$0$ to $59$, then  the number of way to do this task is $60$


Prove that that $S_n = 1+\lambda+\lambda^2+...+\lambda^n$ converges if $0 \le \lambda \lt 1$, $\lambda\in\mathbb{R}$ I need to prove this using the $\epsilon - N$ definition of convergence.

I know from the definition that $\forall \epsilon \gt 0, \exists N \in\mathbb{R}$ such that $\forall n \gt N, |S_n - L | \lt \epsilon$, where $L = \frac{1}{1-\lambda}$ in this case.

How do I find such an $N$?

We have $S_{n} = (1-\lambda^{n})/(1-\lambda)$ for all $n$; $|S_{n}-L| = |
\frac{{-\lambda}^{n}}{1-\lambda}| \leq \frac{\lambda^{n}}{1-\lambda}$ for all $n$. If $\delta > 0$, then $\lambda^{n}/(1-\lambda) < \delta$ iff $\lambda^{n} < \delta (1-\lambda)$, iff $n|\log \lambda| > |\log \delta(1-\lambda)|$. (Note that $\lambda$ and $1-\lambda$ are such that $\log$ is negative.) Can you do the rest from here on?

From the following Formula for finite power series, we can easily see that
$$S_n=\sum_{q=0}^n \frac{1-q^{n+1}}{1-q}$$

now if $0<q<1$ then we can easily verify that $S_n \to (1-q)^{-1}$ as $n\to \infty$.

Fix $\varepsilon > 0$, then

$$\left |S_n-L \right|= \left| \frac{1-q^{n+1}-1}{1-q} \right | = \left|\frac{1}{1-q}\right| \left |q^{n+1} \right|=c\left |q^{n+1} \right|$$

where $c$ is the constant term $(1-q)^{-1}$.

Do you think you can show that $|q^n|\to 0$?


Find parametric and symmetric equations of a line passing through two points? I am trying to find both the parametric and symmetric equations of a line passing through two points. This is for a study exam, so exact answers are not as helpful as detailed solutions.

$P(3,-1,1);$
$Q(-2,1,1)$

I found the points vectors then put them with respect to t:
$$r(t) = (3,-1,1)+t(-2,1,1)$$

Then set each x,y,z to t, which should be the parametric equation:
$$x(t)=3-2t$$

$$y(t)=-1+t$$

$$z(t)=1+t$$

Then to find the symmetric equation I set the points equal to giving me this:
$\frac{(3-x)}{2}=1+y=z-1$

I am having trouble finding if I went about this the wrong way, primarily when creating the vectors and putting them in the equation for a line with respect to $t.$

Your parameterization is not correct.

A parameterization of a line has the form $r(t)= P + t D$ where $P$ is a vector "touching" the line and $D$ is a direction vector for the line.

Given two points on the line, $P$ and $Q$, the equation $r(t) = P+ t Q$ is not the correct parameterization. This is where you are in error.  So, instead of using $Q$ as you did, use the displacement vector from $P$ to $Q$.  In your case, this would be
$$
Q-P = (-2,1,1) - (3,-1,1) =\bigl(-2-3 , 1-(-1), 1-1 \bigr)=(-5,2,0).
$$
The parameterization is then
$$
r(t)=(3,-1,1)+t(-5,2,0).
$$
(And don't forget to give a range of values for $t$.)

Your procedure for finding the symmetric equations looks ok.

Give a vector parametric equation for the line through the point (−4,1,1)(−4,1,1) that is parallel to the line ⟨−4−2t,2,−1−t⟩⟨−4−2t,2,−1−t⟩


show that $\lim \inf a_n = \inf a_n$? given that $m = \inf a_n$ and that $\forall n, a_n \not= m $ and $\exists M , |a_n| \leq M $

Prove that $\lim \inf a_n = \inf a_n$ ?

I proved by contradiction that $\inf a_n \not > \lim \inf a_n$

but i don't know how to prove $\inf a_n \not < \lim \inf a_n $

Also if you have a proof not by contradiction, please post it.

We have a bounded sequence which never equals its infimum. This implies that it is always greater than its infimum.

We must show that $$\lim \inf a_n = \lim_{k \to \infty} \left(\inf_{n > k} a_n\right) = \inf_{n > 0} a_n$$

It suffices to show that $\inf_{n > k} a_n = \inf_{n > 0} a_n$ for arbitrary $k$. Clearly $$\inf_{n > 0} a_n \leq \inf_{n > k} a_n$$ hence it remains to show $$\inf_{n > k} a_n \leq \inf_{n > 0} a_n$$

To prove this, note that if $\inf_{n > k} a_n > \inf_{n > 0} a_n$ we would have $\min(a_1, a_2 \cdots a_k) < \inf_{n > k} a_n $ which would in turn imply that $\min(a_1, a_2 \cdots a_k) = \inf_{n > 0} a_n$. This contradicts the second sentence of this answer.

$inf a_n \lt \lim\inf a_n$ is perfectly possible, unless $a_n$ is a monotone decreasing sequence.  Simple example $a_1=-1, a_n \gt 0,\ n\gt 1$.


Integral of $\cosh^3(x)$ What is the integral of $\cosh^3(x)$? And how exactly can I calculate it? I've tried setting $\cosh^3(x)=(\frac{e^x+e^{-x}}{2})^3$ but all I get in the end is one long fraction.

\begin{align}
\int\cosh^3x dx & = \int\left(\sinh^2x +1\right)\cosh x dx \\
& = \frac{1}{3}\int3\sinh^2x\cosh x dx +\int \cosh x dx \\
& = \frac{1}{3}\sinh^3 x + \sinh x + C\\
\end{align}


What could be the mathematical equation of the given signal? We know that Fourier series for periodic signal $y(t)$ is given by

$$ y(t) = \sum\limits_{m=0}^{+\infty} a_m \cos(w_m t) + \sum\limits_{m=0}^{+\infty}b_m \sin(w_m t). \quad (2)$$

Now,I want to find Fourier series of a periodic sinusoidal signal
$y(t)$ given below. But I don't understand how should I decide what are the harmonics present and how to calculate Fourier coefficients.

So what could be the mathematical equation of the given signal ?

Note: if there is any way than Fourier series to express the given signal in the form of mathematical equation  ,you can  explain with other way also.

Just looking at the signal, it seems to have components only up to the fourth harmonic.  You can read the $y(t)$ values at intervals of $\frac T8$ from the plot-I would pick the peaks of all the obvious waves, then use the orthogonality of the Discrete Fourier Transform to compute the coefficients.  You can do that in Excel easily.

Added:  As points out, the function is odd so only sines will be involved.  It appears four terms will be enough to get close, so the form would be $$y(t)=\sum_{i=1}^4 a_i \sin \frac {2 \pi i}T$$  You can just pick off the first four peaks from the graph, which seem to be at $\frac T{16}, \frac {3T}{16}, \frac {5T}{16}, \frac {7T}{16}$ and solve the four simultaneous equations for the $a_i$.  The FFT is easier if you learn how to do it, as it gives you each coefficient directly.

You can calculate it's Fast Fourier Transform, but choose to only calculate the frequencies you need.

To express Fourier series in terms of a continous Fourier transform, we need to introduce Theory of distributions as the fourier transform of sines and cosines are sums of Dirac distributions. If we are fine with considering the discrete fourier transform (DFT) then fourier series and the DFT coincide.

Combine this with the fact that the Fourier transform is linear and you will be done.


Is it true that $\sum_k a_k(n)\to0$ as $n\to\infty$ if $a_k(n)\to0$ for all $k$ and $\sum_k a_k(n)$ converges for all $n$? Let $a_k(n)$ be such that $\sum_k a_k(n)$ converges for all $n$ and $a_k(n)\to0$ as $n\to\infty$ for all $k$. Then I think it holds that $\sum_k a_k(n)\to0$ as $n\to\infty$.

For example, if $\displaystyle\sum_{k=1}^\infty a_k$ is any convergent series then $\displaystyle\sum_{k=1}^\infty a_k(n)=\sum_{k=1}^\infty \frac{a_k}{n}=\frac1n\sum_{k=1}^\infty a_k\to0.$

I guess one should write $a_k(n)=\delta(n)\varepsilon(k)$ where $\delta(n)$ and  $\varepsilon(k)$ respectively don't depend on $k$ and $n$... then one gets $\sum_k a_k(n)=\delta(n)\sum_k\varepsilon(k)\to0$ as $n\to\infty$, but how can we be sure that such $\delta$ and $\varepsilon$ exist?

If you consider $\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$, as $n$ gets large, the sequence $\langle\frac{1}{k^n}\rangle_{_n}\rightarrow0$ for all $k$ but the partial sums tend to $1$. While it may hold for certain sums it surely isn't applicable for all.

addendum: here I am taking $a_k(n)=\frac{1}{k^n}$


Show that $\inf \{ m+n\omega: m+n\omega>0,and,m,n\in{Z}\}= 0$, where $\omega>0 $ is irrational. Let $\omega\in\mathbb {R}$ be an irrational positive number. Set
$$A=\{m+n\omega: m+n\omega>0,and,m,n\in{Z}\}.$$

Show that $\inf{A}=0.$

How should I start this problem? I don't get this problem.

Hint

In this context ($\omega$ irrational) it can be shown that the set $\{n\omega-\lfloor n\omega\rfloor\mid n\in\mathbb Z\}$ is a dense subset of $(0,1)$. Also it can be shown to be a subset of $A$.

What $\inf A =0$ means is that, for any $\epsilon >0$ there is some multiple of $\omega$ that is within distance $\epsilon$ of some integer.

Hint 1: Pigeonhole principle.

For simplicity assume $\omega <1$. Then each interval $[0,1], [1,2], [2,3]$ contains a multiple of $\omega$, call them $n \omega, m \omega$ and $r\omega$. Each of those three intervals is divided into a "first half" and a "second half". For example $[0,1] = [0,1/2] \cup [1/2, 1]$. So two of $n \omega, m \omega$ and $r\omega$ are in the same half. For example $n\omega \in [1/2,1]$ and $r\omega \in [2+1/2,3]$. That implies $r\omega -2 \in [1/2,1]$ and so $r\omega -2$ and $n\omega$ are distance less than $1/2$ apart. Meaning $|(r\omega -2)-n\omega| < 1/2 \implies |(r-n)\omega -2| < 1/2$. So that $\inf A < 1/2$. How does this generalise?

Hint 2: (independent) Compactness of the circle


Queen of Hearts Odds for weeks 1 - 54 The queen of hearts is a weekly drawing where one card is selected each week until the queen of hearts is exposed.  The cards (52 plus the two jokers) are randomly ordered and placed face down and are numbered 1 - 54.

Each week a "chance" is drawn from those players participating  That ticket has one of the "available" numbers.  The card  matching that number is turned over and if the queen of hearts is shown the game is over and the person whose chance was selected wins.

It is possible for the game to continue for 54 weeks, my question is what are the odds that the queen will be found in week 1, week 2, .. week 53 & week 54?

I would suggest thinking about: "What are the odds that the Queen of Hearts hasn't been drawn in Week 34?" for example.

If you need more help, just let me know.

EDIT:

Let's just take week 35 as an example. It's the 35th drawing made, and the card hasn't been found in the 34 previous ones. Since it's week 35, there a still 20 cards left. The probability NOT to pick the Queen this week is $\frac{19}{20}$.

This then becomes a basic probability problem, where e.g. in week 35, you would have a

$\frac{53}{54}\frac{52}{53} ...\frac{20}{21}\frac{19}{20}$ % chance NOT to have drawn the Queen of Hearts at this time.

Can you figure out when the 50% chance mark is hit?

Obviously each week the queen of hearts is not picked your odds increase. I don't think that was the original question. Is it more likely to come up the 1st week or the 26th week or any other week. The answer is just as likely any week because your odds are 1/54. if it's not picked the 1st week, that's one of the 54 picks and each subsequent week is another of the 54 picks.