Is there an axiomatic approach to ordinal arithmetic? I've always wondered, is there an axiomatic approach to the arithmetic of ordinal numbers?

If so, I imagine it would be on par with set theory in terms of its proof-theoretic strength.

I haven't seen explicit axioms used before, but I belive the following should be trivially sufficient to start with:

Let s(x) be the successor function. s, <, +, * are taken as primitive, alongside = and the axioms of equality
Define Lim(x) as ~(∃z)(S(z)=x) [for simplicity, 0 is taken as a limit ordinal]
Define NextLim(x,y) as Lim(y)&(∀z)( z > x & Lim(z) => (z > y)v(z=y))
Define nextlim(x) as the y such that NextLim(x,y)
Define Anc(x,y) as nextlim(x) > y & Lim(x)
Define anc(x) for the y such that Anc(x,y), or 0 if there is no such y
Define ω as nextlim(0)
Axioms
1a. (∃y)( y < nextlim(x) & y = s(x) )
1b. (∃y)(y = nextlim(x))
2a. Lim(0)
2b. ~∃x(0 = nextlim(x))
3a. s(x)=s(y) => x=y
3b. anc(x)=anc(y)
3c. anc(x)=anc(s(x))
3d. nextlim(x)=nextlim(s(x))
4a. x < y & y < z => x < z
4b. x < y & y < z => x = z
4c. x < y v y < x v x=y
4d. x < s(x) & x < nextlim(x)
4e  x < y => s(x) < s(y)
5a. x+0=x
5b. x+(s(y))=s(x+y)
5c. x + nextlim(y) = nextlim(x+y)
6a. x*0=0
6a. s(x*y) = (x*y)+x
6b. x * nextlim(y) = nextlim(x*y)
7a. (b < a) & (∀x < a)((P(x) & P(anc(x))) => P(s(x))) => P(b)
7b. (∀y)((∀x)(x < y => P(x)) => P(y))) => P(a)

The basic intuition is to limit the induction scheme to induction under an ordinal with regard to successor, and then add transfinite recursion. When quantifiers are bounded under ω, Peano Arithmetic should follow trivially from the above axioms.

I may have forgotten one or two (possibly for the relation of nextlim and <), and I highly suspect that there are a few redundant ones, but the basic intuition is to limit the induction scheme to induction under an ordinal with regard to successor, and then add unbounded transfinite recursion. I believe exponentiation is definable from + and * in this context, but I might be wrong. In that case, just add explicit axioms for exponentiation. In either case, once exponentiation is defined, you should be able to use use a transfinite Godel numbering to give a model of the theory of types. Once that is done, you can manipulate that model to provide axioms for larger and larger ordinals.

I threw these together somewhat hastily, so if anyone spots any errors, I blame someone else.

Ordinal arithmetic falls out of manipulations in set theory. You can define order and the ordinal operations in basic set theory and then prove things about these operations.

How big the ordinals you can construct depends on which axioms of ZF you allow or add. The theory behind this is called Reverse Mathematics.

The intent of Reverse Mathematics is to show which theorems of (nominally non-logical) mathematics are of equivalent strength, that is, can allow you to prove each other, and what logical axioms fit.


Premise of Minkowski's theorem: "Irrational" Line If we take a straight line $L$ in $\Bbb{R}^2$ of the form $y=ax$ where $a\in\Bbb{R}\setminus{\Bbb{Q}}.$

Then we see that (except at the origin $(0,0)$) $$L\cap \Bbb{Z}^2=\emptyset.$$

But now if the line becomes a strip of width $r>0$ denoted $S$ how can we be sure that $$S\cap \Bbb{Z}^2\ne\emptyset?$$

I know about Minkowski's theorem but in this particular case is there a high school argument? Perhaps by writing the continued fraction of $a?$

Let $m\in\mathbb{N}$ be such that $1/m<r$.  Given $m+1$ real numbers, there must be two of them whose fractional parts are within an interval of length $1/m$, by pigeonhole.  In particular, there must be two different integers $n$ and $n'$ such that the fractional parts of $an$ and $an'$ differ by less than $1/m$.  Then $a(n-n')$ will be within $1/m$ of an integer $k$.  The point $(n-n',k)\in\mathbb{Z}^2$ is then in your strip $S$, since it is within $r$ of $(n-n',a(n-n'))$.

The strip is an open set so it contains a rational point $q = (a/b, c/d)$, because rationals points are dense in Euclidean space. Since it is a line, and in particular closed under scaling, it contains $(bd)q = (a,c)$ which is an integer point.

Edit: obviously this doesn't work, since it's not closed under scaling. I was being sloppy. Thanks to mercio for pointing that out. I'll leave this here to embarrass me and encourage me to check my work.


Average rate of change? How would I figure the following problem out?

  Find the average rate of change of $g(x)=x^2+3x+7$ from $x=5$ to $x=9$

My thought is that I would plug in 5 and 9 for the x values to get the y values. And the use the slope formula $\frac{y_2-y_1}{x_2-x_1}$.

Yes, your solution is correct.

Differentiating first is a detour.

The good thing here is that you have a quadratic equation, so differentiating w.r.t. $x$ , you get a linear equation...which makes it easy to find average rate of change.

Consider your equation $$g(x)=x^2+3x+7$$
differentiate it w.r.t. $x$ $$g'(x)=2x+3$$

The naive way of doing this question is finding $g'(5), g'(6),g'(7), g'(8),g'(9)$ and then finding its average. Alternatively, just find the average of $g'(5)$ and $g'(9)$. You will get 17.


For what $a$ the sequence $2x_{n+1}=x^2_n+a$ has finite limit? Let $(x_n)$ be the sequence defined recursively by the following formular $$x_0=0, \ 2x_{n+1}=x^2_n+a$$
Find $a$ such that the above sequence has finite limit?
With all of my effort, there is still one case left, say, $a<-3$, in which I don't even know how to prove or disprove the existence of the limit.

It's evident that if the sequence has a limit, this limit must satisfy the following equation:
$$2t=t^2 +a$$
and then if the sequence converge, it should converge to one of the 2 fixed point $p_1=1 +\sqrt{1-a}$,$p_2=1 -\sqrt{1-a}$.
Let's denote function $f(x)=\frac{1}{2}(x^2+a)$ with $a<-3$  (and so $x_{n-1}=f(x_n)$), we have
$$f'(p_1)=1 +\sqrt{1-a} >1$$
$$f'(p_1)=1 -\sqrt{1-a}<-1$$
then $|f'(p_{1,2})|>1$, according to this post, the both fixed points $p_1$ and $p_2$ are unstable. Because $x_0 \notin \{\pm p_{1,2}\}$ (4 values $p_1,p2,-p_1,-p_2$), then the sequence will never be able to hit the fixed points and so doesn't converge.
Conclusion: the sequence $x_n$ doesn't converge for $a<-3$.
PS: why these conditions $x_0 \notin \{\pm p_{1,2}\}$  are enough to conclude the divergence of the sequence? Let's prove by contractiction. Suppose $k$ is the smallest value that $x_k \in \{\pm p_{1,2}\}$. Then we have $f(x_{k-1})=f(p)$ with $p \in \{\pm p_{1,2}\}$. But this implies that $\frac{1}{2}(x_{k-1}^2 +a)= \frac{1}{2}(p^2+a)$ or $x_{k-1} \in \{\pm p_{1,2}\}$, hence $k-1$ which is smaller than $k$ is also satisfy the assumption(=> contradiction).

You could say $2x=x^2+a$ and then you get
the discriminante $D=4-4a$.
Now for $a < 1$ you get 2 real solutions.
For $a = 1$ you get 1 real solution and for $a > 1$ you get none.
Moreover you have to show that the sequence is monotonously falling and limited for certain intervalls.


Recurrence - finding asymptotic bounds for $T(n) = T(n-2) + n^2$ I've been working on a problem set for a bit now and I seem to have gotten the master method down for recurrence examples. However, I find myself having difficulties with other methods (recurrence trees, substitution). here is the question I am stuck on:
$$T(n) = T(n-2) + n^2$$
Is there a pattern as follows?
$$n^2 + T(n-2) + T(n-4) +...$$
where it goes until there is no more n left. so around n/2 times
and would that mean that
$$n^2 + (n-2)^2 + (n-i) ^2$$ so the asymptotic bound would be $\theta(n^2)$?

I am honestly taking a shot in the dark here, so I was hoping someone could help guide me in how to approach these questions.

Thank you,

Tyler

$$T(n) = T(n-2) + n^2 = T(n-4) + (n-2)^2 + n^2 = T(n-2k) + \sum\limits_{i = 0}^{k - 1}(n - 2i)^2$$

This goes down while $n - 2k \ge 0$. Assuming even $n$ (for asymptotic complexity, it does not really matter, and you can do similar calculations for odd $n$ also, with the same asymptotic results), we have $k = \frac{n}{2}$ at the end.

$$T(n) = T(0) + \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n - 2i)^2 = \sum\limits_{i = 0}^{\frac{n}{2} - 1}(n^2 - 4ni + 4i^2) + C$$
$$T(n) = n^2\cdot\left(\frac{n}{2}-1\right) - 4n\cdot\frac{1}{2}\cdot\frac{n}{2}\cdot\left(\frac{n}{2} - 1\right) + 4\cdot\frac{1}{6}\cdot\left(\frac{n}{2} - 1\right)\cdot\frac{n}{2}\cdot(n-1) + C$$
$$\therefore \ T(n) = \Theta(n^3)$$

let $n=log m$
$$T(\log m)=T(\log m-\log4)+(\log m)^2$$
$$T(\log m)=T(\log m/4)+(\log m)^2$$
$$s(m)=s(m/4)+m^2$$
here $a=1,b=4$ and $f(n)=m^2$
$$m^{\log a}=1$$
$$f(m)>m^{\log a}$$ (base 4)

case (iii) of masters theorem follows
hence $s(m)=\Theta(m^2)$
after comouting $c$ through $af(m/b)=cf(n)$
$$c=1/16<1$$
hence case (iii) totally followed
and $s(m)=\Theta(m^2)$
and $T(m)=\Theta(2^2n)$
m not confirm about the answer but the procedure will be the same


How to evaluate $\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$ I have to find the limit of following

$$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$

I have no idea how to start this one off.
How would I do it?

Do I just substitute the $0$? It doesn't look that easy and simple. The answer says it's negative infinity.

Please show me a solution without graphing(unless for better explanation).

An idea. We take $\;x\;$ very close to zero, say $\;|x|<10^{-4}\;$ :

$$x>0:\;\;\frac1x-\frac1{x^2}=\frac{x-1}{x^2}<\frac{-\frac12}{x^2}=-\frac1{2x^2}$$

and now you only have to show the rightmost expresion is unbounded below, which I think is pretty easy.

Here's another way to prove the following statement
$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$
Which is equivalent to the following
$$\forall N\gt0,\exists\delta\gt 0:0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -N$$
So whenever $0\lt\left|x\right|\lt\delta$, we have
$$\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}\lt\frac{\delta-1}{x^2}$$
Here we can make $\left|x\right|$ arbitrarily small by presupposing a bound of $\frac12$. We can then use this bound to potentially get a smaller $\delta$ later. Assuming that $|x|\lt\frac12$, we have
$$x^2\lt\frac14\Rightarrow-\frac{1}{x^2}\lt-4$$
Which implies that
$$\frac{\delta-1}{x^2}=-\frac{1-\delta}{x^2}\lt -4(1-\delta)=-N$$
Now we can set $\delta$ to the smallest of these two bounds
$$\delta =\min\left(\frac12, 1-\frac{N}{4}\right)$$
Putting it all together
$$0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -4(1-\delta)=-N$$
Therefore
$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$


How do we prove that if $M$ is complete, any sequence $x_n \in M$ satisfying $d(x_n,x_{n+1}) < 2^{-n}$ converges in $M$? I'd like to show that if $M$ is a metric space $(M,d)$, and is complete, any sequence $x_n \in M$ satisfying $d(x_n,x_{n+1}) < 2^{-n}$ converges in $M.

My proof is the following:

I think that this question is the same as showing that if all Cauchy sequences, say  $x_n \in M$ are convergent in $M$ (by definition of complete), then that sequence $x_n$ must also satisfy the condition $d(x_n,x_{n+1}) < 2^{-n}$. Is the proof as simple as choosing $\epsilon \leq 2^{-n}$ and then letting $m=n+1$? Since if we have that we can say:

Suppose $x_n \in M$ is a Cauchy sequence. Hence, it must be convergent. Then, by definition this means that given $\epsilon >0$, there exists an $N$ such that if $m,n \geq N$, $d(x_m,x_n)< \epsilon$.

Hence, just choose $\epsilon$ small enough so that $\epsilon \leq 2^{-n}$ and then let $m=n+1$. Then, pugging into the above, we will have the following statement:

given $0<\epsilon  \leq 2^{-n}$, there exists an $N$ such that if $m=n+1,n \geq N$, $d(x_m,x_n) = d(x_{n+1}, x_n) < \epsilon \leq 2^{-n}$. Hence, $x_n \in M$ satisfies

$d(x_{n+1}, x_n) < 2^{-n}$. And, since $x_n$ is a Cauchy sequence in a complete space, its also convergent.

I am wondering if the above makes sense. I am a bit wary of the part where I "choose" and "fix" the $\epsilon$. Thank you!

The proof as stated is wrong. A simple counterexample is $(1,0,\frac{1}{3},0,\frac{1}{5},0,\ldots)$. Moreover, it tries to prove the wrong implication. What you need to prove is the following: if $(x_n)$ is a sequence in $M$ such that $d(x_n,x_{n+1})<2^{-n}$ for all $n\in \mathbb{N}$, then $(x_n)$ is a Cauchy sequence. This suffices, since $M$ is complete.

A proof for this statement would look like this: for all $m,n\in \mathbb{N}$ with $n\ge m$ it is true that $d(x_m,x_n)\le \sum_{k=m}^n d(x_k,x_{k+1})$. You can use induction to prove this. Choose $N$ such that $\sum_{k=m}^n 2^{-k}<\epsilon$ for all $n,m\ge N$. You can do this, since the geometric series converges (absolutely). It follows that $d(x_m,x_n)\le \sum_{k=m}^n d(x_k,x_{k+1})\le \sum_{k=m}^n 2^{-k}<\epsilon$ for all $m,n\ge N$, hence $(x_n)$ is a Cauchy sequence.

Your proof is correct. For some large enough $N$, the distance between subsequent terms $x_n,x_{n+1}$ in any Cauchy sequence is less then $2^{-n}$, for any $n>N$.

However, this is not what you have to prove. You need the other implication.

Let $\{x_n\}^\infty$ be a sequence satisfying the property from your question. Let $\varepsilon>0$. Now choose a natural $N$ such that $\varepsilon>2^{-N}$. Such a number exists since the naturals are not bounded from above. Finally, consider any two natural numbers $n,m>N$. Do you see, by the triangle inequality, that $d(x_n,x_m)<2^{-N}<\varepsilon$? This makes the sequence a Cauchy sequence, and therefore it converges by completeness.

Q.E.D.


How to prove irreducibility of this polynomial? Let $p,q$ be primes.  Prove that $y^{n }-p$ is irreducible over $\mathbb{Q}(\sqrt[n ]q) $ .

I have tried for some time, but still feel confused about how to prove it. Can anyone help me?

Any root of a prime will be irrational. You can imagine trying to find roots of the polynomial in R and find there are no solutions in terms of radicals of other primes.


mathematical adjustment (IRR) Is here anybody who can help me with mathematical adjustment between equation 8 and 9 in the picture?

Ignore the text between equations

This is an instance of completing the square.

Note that $$(x-\frac{CF_1}{2CF_2})^2 = x^2 - \frac{CF_1}{CF_2}x + \frac{CF_1^2}{4CF_2^2}$$

Recognize that the right hand side of this equation is just the right hand side of $(8)$ + $\frac{CF_1^2}{4CF_2^2}$.

Thus if you add $\frac{CF_1^2}{4CF_2^2}$ to both sides of the equation in $(8)$, you get that $$\frac{y - CF_0}{CF_2} + \frac{CF_1^2}{4CF_2^2} = (x-\frac{CF_1}{2CF_2})^2$$ which gets you to $(9)$ after multiplying by $CF_2$.

It looks like they used the quadratic formula...


Find the total area between region and x-axis? I'm having some difficulty with this problem out of my Calculus Book:

  Find the total area between region and the $x$-axis: $y=x^3-x^2-6x$,  $-2 \le x \le 3.$

I know I start with setting the function equal to $0$:

$$0=x^3-x^2-6x$$
$$0=x(x-3)(x-2)$$
Hence $x=3$ or $x=2$.

I then need to take the integral of the Top Function minus the bottom function of each respective area:

$$\text{Area} = \int\limits_{a}^{b} (\text{Top}_f - \text{Btm}_f)\, dx$$

The problem I'm having is getting the initial function split in half? I need the function for the portion above the $x$-axis to the $x$-axis, PLUS the portion below the $x$-axis to the $x$-axis added together to get the entire area...?

Can someone walk me through what I am missing here?

You use thr Rienmann integral. Then $A=\int_{-2 }^3(x^3-x^2-6x) dx=[x^4/4-x^3/3-3x^2]_{x-2}^{x=3}$. This is the area with sign. If you want the area without sign: $A= \int_{-2} ^0(x^3-x^2-6x) dx- \int_{0} ^3(x^3-x^2-6x) dx $. The polynomial is negative in $[0,3]$.

You factored incorrectly. Note that $$f(x) = x^3 - x^2 - 6x = x(x-3)(x+2)$$ so the roots lie at $-2,0,3$. The plot looks like

If you want the real area between the curve and the $x$-axis, the region to the right of $0$ must be negated, since it is under the $x$-axis and you get $$\int_{-2}^0 f(x) dx - \int_0^3 f(x) dx$$


Determine if bounded solutions of ODE system are stable Please help me to handle the following problem:

Prove or disprove that bounded solutions of ODE system $$\dot x = 3x + y + 1; \dot y = 6x+2y+2$$ are stable using a theorem about stability of linear systems.

My progress:

What I was able to prove is that bounded solutions of this system have the form $x =c, y= -1-3c$, where $c \in \mathbb{R}$. I can post my solution about it upon request. But I do not see how to apply a theorem about the stability of linear systems here. For me it is clear that bounded solutions are stable: any bounded solution is fully determined by initial conditions and constant so you can just use the definition of stability. But I have to use a theorem about stability of linear system, and I do not see how to do it.

Thanks a lot for your hints and answers!

Step 1: Apply the Laplace transform to both equations and solve the system in the Laplace domain. The system becomes linear in the Laplace domain and it is, therefore, easy to solve.

Step 2: If the solution's poles are in the second or third quadrant the system is stable and therefore your solutions to the system are too.

I would apply the Routh–Hurwitz stability criterion-

https://en.wikipedia.org/wiki/Routh%E2%80%93Hurwitz_stability_criterion
http://pages.mtu.edu/~tbco/cm416/routh.html

The method above is normally used by Control Engineers to check the stability of LTI systems.


Number of primes less then $6000$ using $n/ \log n$ So I am trying to use this formula here and is giving me some trouble.

If I just substitute $6000$ into the formula, the answer is approximately $1500$.

But the number of primes under $6000$ is clearly half that. How do you use this formula properly?

$\pi(x)$ is the prime counting function, it gives the exact amount of primes below $x$

$$\pi(6000)=783$$

The formula provided is an approximation, so it gives:

$$\frac{6000}{\log_e 6000}=\frac{6000}{8.699514748}=689.69364083$$

While about $100$ from the actual answer, this is pretty accurate. It will increase in accuracy as $n$ goes to $\infty$

When you calculated the formula, you used $\log_{10}$ and not $\log_e$, that's why your value was so long from the actual value.

The prime number theorem doesn't say $\pi(n) = \frac{n}{\log n}$. It actually says $\lim_{n \to \infty} \pi(n) \large/ \frac{n}{\log n} = 1$, which means it gets more accurate as $n$ gets larger.


What will be $\lim_{x \to 0}\sin(1/x)$? What will be the following limits?

$\lim\limits_{x \to 0}\sin(\frac{1}{x})$

for $x \to 0 ,$ $\frac{1}{x} \to \infty$

Now what will be the $\sin$ value on $\infty$ ?

There is no definite value of $\sin$ at infinity. As you might know that infinity is not a number . It is a notion that shows that whatever the value we are considering it is big. So the limit does not exist.

One thing to note here is that the reason of the non-existing limit is not that the number in consideration is going to infinity. The reason is that we don't really know that value of $\sin$ at that angle. The value of sin can be anywhere between $-1$ and $1$.


Finding the X- Intercepts for this Equation How do you find the x intercepts for this problem? (Please show and explain how you solved it)

$$y= x^3-10x^2+31x-30$$

Another method is this:  if $$x^3+ax^2+bx+c=0$$ then you are looking to solve the following system.

$$r+s+t=a$$
$$rs+rt+st=b$$
$$rst=c$$

Should you find these $r,s,t$, these are your roots and your factored polynomial is just

$$(x-r)(x-s)(x-t)=0$$

However, not knowing these formulas is probably true.  In this case use the rational root theorem.  Test the possible rational roots (the factors of 30 in your case, $c$ in the general case) and then divide by the root or factor with long division or synthetic.

For your example, your possible rational roots are $1,2,3,5,6,10,15,30$.  You test both positive and negative factors.  You would find your first bit of luck with $2$ (the factor being $x-2$)...when you divide you will get a quotient of $x^2-8x+15$.  Then factor that.  See reckless reckoner's comment for further insight.

For X intercepts we have y=0 as shown in fig .You just have to findroots of polynomial .Thanks


A Charpit-type problem Solve in parametric form for $u(x,y)$: $$u + u_x^2 + u_y^2 - 2 = 0$$ with the data $u(0,y) = y$ for $0\leq y \leq 1$ and the restriction $u_x \geq 0$. Determine (and show on a sketch) the domain in which the solution is uniquely determined.

Progress: If $$F = u + u_x^2+ u_y^2 - 2,$$ $p = u_x$, $q = u_y$, then $$p_{\tau} = -p, ~ q_{\tau} = -q, ~ x_{\tau} = 2p, ~ y_{\tau} = 2q, ~ u_{\tau} = 2(p^2 + q^2) = 4-2u$$ so with the initial data (and $p = u_x \geq 0$) we get $x_0 = 0$, $y_0 = u_0 = s$ we have $$x = 2\sqrt{1-s}(1-e^{-\tau}), ~ y= s + 2(1-e^{-\tau}), \\ p=\sqrt{1-s}e^{-\tau}, ~ q=e^{-\tau}\\ u = 2 + (s-2)e^{-2\tau}.$$

Now to determine the domain note that $y = s + \frac{x}{\sqrt{1-s}}$ where $0 \leq s < 1$ (and $x = 0$ for $s=1$), that $x = 2\sqrt{1-s}(1-e^{-\tau}) \leq 2$
and that the envelope is determined by differentiating $y = s + \frac{x}{\sqrt{1-s}}$ with respect to $s$, i.e. $ 0 = 1 - \frac{x}{2}(1-s)^{-\frac{3}{2}}$, i.e. $s = 1 - (\frac{x}{2})^{\frac{2}{3}}$, so we obtain
$$y = 1 - (\frac{x}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}}x^{\frac{2}{3}}$$ as the envelope curve. Moreover $y=x$ at $s=0$.

But I don't think I can just say that the domain is between $y=x$, $x\leq 2$ and $y = 1 - (\frac{x}{2})^{\frac{2}{3}} + 2^{\frac{1}{3}}x^{\frac{2}{3}}$ as the lines $y = s + \frac{x}{\sqrt{1-s}}$, $0\leq s \leq 1$, do not cover all of it (this can be seen via a diagram).

So any advice?

First, for negative $x$ the characteristics intersect thus we can not have a unique solution

Thus, when searching for the envelope there is in fact no extremal point of $y(s)$. Therfore, the region is determined by the maximal $x$ values given by $\lim_{\tau\to\infty}$.
Taking the limit in $x$ and $y$ and substituting for $s$ yields $x\leq y\leq3-(x/2)^2$.

Second since the maps are continuous I see no reason why the domain should not be covered fully.

Third, you also need to think about the uniqueness which is given by the injectivity of the $y(x,s)$ map.


Introduction to logic starting with predicate logic The typical introduction to logic starts with propositional logic and later goes over to predicate logic, in the order syntax -- semantics -- possibly some calculus. My experience is that this is perceived as a huge skew in the learning curve; beginners tend to think that logic is just boring play with truth tables until suddenly there's predicate logic and everything stops to make sense entirely.
This got me curious as to how one might go about introducing logic by starting directly, but softly with predicate logic, then later treating propositional logic as a specialization. I imagine that, presupposing some elementary set theory (just the basics of what is a set, tuple etc.), one could come from some first idea of a structure and stepwise introduce formulas as a way to express what is going on in that structure, instead of starting with the inductive definition of a formal language and then building a semantics "on top" of it, in the hope that this may get the reader a better first intuition of what the semantics of predicate logic is about.
What I'm looking for is a textbook not too heavily aimed at proofs or mathematical logic/model theory, but a more generally/practically oriented introduction to the syntax and semantics of FOL that will simply put the reader in a position to make sense of and come up with simple predicate logic formulas, and get the idea of validity of an inference, but supported by rigorous definitions and not only informal descriptions.
(I figured there would be more folks here having an overview of the literature than over at Philosophy SE; feel free to migrate if you think it is a better fit there.)
So: Do you know of any good textbooks providing a general introduction to logic starting with predicate logic, perhaps (but not necessarily) introducing the concept of a structure very early on?

In my not-at-all-humble opinion beginning logic has to do with what follows from what.  phi (a formula) follows from Sigma (a set of formulae) iff Sigma u {~phi} is inconsistent.  If you can convince your students of that, then Smullyan truth trees (applied to Sigma u {~phi}) are obviously the way to go.
The truth tables follow from the trees, but that needs the extra assumption that there are only two truth values.  Because, if phi is true in the two-valued Boolean algebra then it will be true in all Boolean algebras (with 1 read as true).  That there are only two truth values is a silly prejudice that even the ancient Greek originators of logic did not believe.


The equation $3a+5b=n$. For a given number $n$, how can we find out whether we have any non-negative values for $a$ and $b$ for the equation
  $$3a+5b=n,$$
  where $1\le n\le 100,000$.

For example: If $n=5$, then $a=0, b=1$.

Your question is a special case of the Frobenius problem or coin problem. You want to know for which $n$ it is possible to find nonnegative integers $a,b$ such that $3a+5b=n$. Take a look at the two-variable case on the Wikipedia page: It is always possible if $n>3\cdot5-3-5$, that is, $n>7$.

For values smaller then or equal to $7$, it is known that exactly half of them can be written in the form $3a+5b$. All you have to do is check them manually:

$0=3\cdot0+5\cdot0$,

$3=3\cdot1+5\cdot0$,

$5=3\cdot0+5\cdot1$,

and $6=3\cdot2+5\cdot0$.

The other half ($1$, $2$, $4$ and $7$) indeed can't be written in the desired form.

These are explicit solutions:

If $n=3k$, then we have $n=3\cdot k+5\cdot0$.

If $n=3k+1$ with $k>2$, then we can write $n=3\cdot(k-3)+5\cdot2$.

If $n=3k+2$ with $k>1$, then $n=3\cdot(k-1)+5\cdot1$.

This covers all integers $n>7$.

All other solutions can now be derived from these using Bézout's identity.

You should caclulate the coeficient of $x^n$ for the following generating function:
$$
G(x)=\frac{1}{(1-x^3)(1-x^5)}.
$$
If $[x^n]G(x) \neq 0$  then the equation $3a+5b=n$ has  solutions.


Initial value problem $dy/dx=y^{1/3}$, $y(0)=0$ has one of the following solution I came across the following problem which says:

The initial value problem $y'=y^{1/3}$, $y(0)=0$ has:

(a) a unique solution,

(b) exactly two solutions,

(c)exactly three solutions,

(d)no solution.

The solution of the problem is given by: $2x=3y^{2/3}$. But I could not come to a conclusion. Clearly, (d) can not be true. But i am not sure about the other options. It will be helpful if someone throws light on it. Thanks in advance.

The answer is "none of the above". There are infinitely many solutions.

Pick any $\alpha > 0$ and define $f_\alpha (x) = 0$ for $x \le \alpha$ and $f(x) = (2/3)^{3/2} (x-\alpha)^{3/2}$ for $x > \alpha$. All these functions are solutions.

total $3$ solutions

1.$y=0$ is also a solution.

2.y=0 for $x \le 0$ and $y=(2/3x)^{(3/2)}$ (positive squ root)

3.similarly consider neg squ root.


Fibered category: The object is not in the pullback, what is it mapped to? Given two objects and their intersection which is the pullback, what is the pullback of any sub-object that is not in the intersection?
For instance given the category:
$
\begin{matrix}
    4 & \xrightarrow{p_{2}} & \{4,5,6,7 \}
    \\
     p_{1} \; \downarrow && \beta \downarrow
    \\
     \{1,2,3,4\} & \xrightarrow{\alpha} & c =\{1,2,3,4,5,6,7\}
\end{matrix}
$

The concept you need to consider is "diagram commutes". The pullback diagram creates a subset of $\alpha \times \beta$ which contains only the elements of $(a,b)$ which makes the diagram commute.


Stirling number of Second kind generating function I would like to prove that: $$f_m(x) = \dfrac{x^m}{(1-x)(1-2x)...(1-mx)}$$

Where $$f_m(x) = \sum_{n=0}^{\infty} S(n,m)x^n$$  and $S(n,m)$ is stirling number of 2nd kind

Multiplying the recurrence relation $S(n,k) = S(n-1,m-1) + m*S(n-1,m) $ by $x^n$ and simplifying (skipping some steps) I get:

$$f_m(x) = \dfrac{x}{1-mx}f_{m-1}(x)$$
The next step is what I am not sure of:
$$f_m(x)= (\dfrac{x}{1-mx})^m$$

$$ f_m(x) = \dfrac{x^m}{(1-x)(1-2x)...(1-mx)}$$

Is this correct? Or am I completely wrong?

You’re right to have doubts about that step: $f_{m-1}(x)$ isn’t $\left(\frac{x}{1-mx}\right)^{m-1}$, so from

$$f_m(x)=\frac{x}{1-mx}f_{m-1}(x)\tag{1}$$

you cannot conclude that

$$f_m(x)=\left(\frac{x}{1-mx}\right)^m\;.$$

(Note that this also doesn’t give you what you want. However, if you assume as an induction hypothesis that

$$f_{m-1}(x)=\prod_{k=1}^{m-1}\frac{x}{1-kx}\;,$$

then $(1)$ would let you conclude that

$$f_m(x)=\prod_{k=1}^m\frac{x}{1-kx}=\frac{x^m}{(1-x)(1-2x)\ldots(1-mx)}\;.$$

Alternately, you can use the identity that
$$S(n, k) = \frac{1}{k!}\sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^n.$$
Then is you apply the Norlund-Rice integral formula referenced here, you can obtain this generating function by residue calculus.


Showing $\dfrac{e^{-nx}}{n} < \epsilon$ I am trying to prove that $f_n(x)=\dfrac{e^{-nx}}{n}\rightarrow f(x)=0$ uniformly on $x\in[0, \infty)$. To do this I am trying to show that: $$\dfrac{e^{-nx}}{n} < \epsilon$$
I tried to do it this way, but I am not sure if it is correct:
\begin{align}
&\dfrac{e^{-nx}}{n} \\
&<\dfrac{1}{n} \quad \forall n>N_1 \ \exists N_1\in \mathbb{N} \\
&<\epsilon \quad \forall n>N_2 \ \exists N_2\in \mathbb{N} \\ \\
\therefore \dfrac{e^{-nx}}{n}&<\epsilon \quad \forall n>N=\max\{N_1, N_2\}
\end{align}

This in turn implies uniform convergence. Is this the correct way to do it?

Let $n \geq 1$ then $g(x)=e^{-nx}$ is a decreasing function defiend on $[0,+\infty)$

Thus $e^{-nx} \leq e^0=1$ thus $$\sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \frac{1}{n}$$

Thus  is $\epsilon >0$ then for $n \geq n_0=[\frac{1}{\epsilon}]+1$ we have that $$ \sup_{x \geq 0}|\frac{e^{-nx}}{n}| \leq \epsilon$$

You got the idea.

Need to show that for $\epsilon >0$ there is a

$N \in \mathbb{Z^+}$ such that for $x \in [0,\infty)$ and

$n \gt N$ we have $|\dfrac{e^{-nx}}{n} -0| \lt \epsilon.$

$\dfrac{e^{-nx}}{n} =\dfrac{1}{ne^{nx}} =$

$\dfrac{1}{n(1+nx+(nx)^2/2!+...)} \le \dfrac{1}{n}.$

Archimedes :

There is a $N \in \mathbb{Z^+}$ such that

$N >1/\epsilon.$

For $x  \in [0,\infty )$ and $n >N$ we have

$\dfrac{e^{-nx}}{n} \le 1/n \lt 1/N \lt \epsilon.$


$f(1+x) = f(x)$ for all real $x$, $f$ is a polynomial and $f(5) = 11$. What is $f\large(\frac{15}{2}\large)$? I was looking through a GRE math subject test practice test (here) and in particular I was confused regarding this question

I chose E because I thought that (since the problem didn't specify) it could be an infinite polynomial that becomes the Taylor expansion of $f(x) = A \sin (x\pi)+11$, where $A$ can be anything.

The correct answer is apparently C. Is there a reason (other than my probably unwarranted assumption) that the answer HAS to be C?

If $f$ isn't a constant polynomial then for every root $x_0$ of it $x_0+n$ is also a root for all $n$ and this contradicts the fundamental theorem of algebra. Conclude.

$$f(x) = a_0 + a_1x^1 + \cdots + a_nx^n$$

$$f(x+1) = a_0 + a_1(x+1)^1 + \cdots + a_n(x+1)^n$$

It can be shown by induction that

$$f(x) = f(x+1) \to a_1=a_2=\cdots=a_n=0$$

Note that

$$f(x) = a_0 + a_1x^1 + \cdots$$

is not a polynomial


let $f(a) = b.$  complete the rectangle with corners at $(0,0), (a,0), (a, b)$ and $(0,b).$ you see  that the rectangle is partitioned into two almost triangular looking regions:

(1) one bounded  by $y = 0, x = a$ and $y = f(x)$ which has area $\int_0^a f(x) \, dx$

(2) and the other one bounded  by $x = 0, x = f^{-1}(y)$ and $y = b$ which has area $\int_0^b f^{-1}(y) \, dy$

but these two areas make up the area of the rectangle, therefore we have
$$\int_0^a f(x) \, dx +  \int_0^b f^{-1}(y) \, dy = ab = af(a).$$


Heuristic in quotient rings with linear relations? Artin Algebra Chapter 11. Some solutions are here and here.

It looks like all I have to do is find an integer that equals $0$ if the answer turns out to be $\mathbb Z/(n)$, which is the case for the exercises above. I get

(a) Multiply 3 to get 15=18. $n=3$ (b) Multiply 2 to get 20=6. $n=14$ (3) Factor $\alpha^2+\alpha$ to get 1=0. $n=1$

What is the basis for this heuristic? Is this merely a way to guess? Is this a coincidence?
How can I do it for the exercises below (c) and (d)? (Adjoining elements is not introduced until the next section)

(c) $6=0$, but $2\alpha-1=0$.

It turns out we're not in $\mathbb Z/(6)$ but in $\mathbb Z/(n)$ for some $n$ where $6=0$, $n=1,2,3,6$.

Multiplying 3 I get $6\alpha-3=0=-3$. Then $n=3$.

It turns out $n=14$ in 5.4(b) was lucky and so actually requires more proof?

(d) How do I do similarly to show $7=0$ for $2\alpha^2-4=0=4\alpha-5$?

For 5.4.(a), (b), (c); it can be a way to guess the solution, but then you still have to prove that it works.

The fact that it works for these is not a coincidence, but the fact that each time, modulo the $n$ that you find, $\alpha$ is some integer, so you "haven't actually adjoined anything".

Indeed for (a), $3=0$, and $2\alpha =6$ so $2\alpha =0$ and so $\alpha = 3\alpha -2\alpha = 0$.
For (b), you have $\alpha =10$ that's given; and for (c) there's only one ring in which $1=0$.

For 4.3, you need to be careful : you won't always be in the situation where $\alpha$ (the class of $x$) is actually an integer; so you may have to deal with extensions of $\mathbb{Z}/(n)$


Numerical equality symbol for variables with different units Good day! I wonder, is there any equality symbol for variables with different units of measurements?

For instance, it is given $Ce = 3$ $[V*s/rad]$ which is back-EMF constant of a motor and $Cm$ - torque constant of a motor in $[N*m/A]$. For DC motor this constants are numericaly equal.

How symbolically indicate that they are equal? Because I think it is incorrect to say $Cm = Ce$, as they are variables with different units of measurements.

I don't think that exists a precise symbol for what you are looking for and sincerely I don't even see a need! You could come up with your notation if you want like stating that the magnitude of that two physical measurement are the same but it will be still imprecise if you want to be pedantic about it.

If you want to be creative you can use something like this $$C_m\cong C_e\;\;\;C_m\doteq C_e\;\;\;C_m\simeq C_e$$
You could even go to crazier places with something like that $$C_m\bowtie C_e\;\;\;C_m\triangleright C_e$$
sky is the limit: choose one that you like and define it as "[...] where this symbol is being use to indicate that  this constants are numericaly equal [...]"

It's rare, but some antique books used to show numerically equal as an equal with a "N" above.


show that if $\beta + n\alpha $ is a root for some integer $n$, then $\beta + n\alpha $ lies in the alpha string through beta. So I would like to show the following, which is,

If $\beta + n\alpha $ is a root for some integer $n$, then $\beta + n\alpha $ lies in the alpha string through $\beta$.

I'm guessing the fact that if $\beta -q\alpha, \ldots , \beta + p\alpha $ is an $\alpha$ string through $\beta$ then $\frac{q-p}{2}\alpha (x) = \beta (x)$ for any $x \in [L_{\alpha},L_{-\alpha}]$.

We have a hint for you:
consider $\alpha$ srting through $\beta$ and $\beta$ string through $\alpha$.


$\int_0^{\infty}\frac{dx}{(x^2+\sqrt{x})^p}$ convergence For which values of $p$ does integral $I$ converge? $$I=\int_0^{\infty}\frac{dx}{(x^2+\sqrt{x})^p}$$

My shot in the dark was $p\in(\frac{1}{2},2)$, since $\int_0^{1}\frac{dx}{x^{p/2}}<\infty \iff p/2<1$ and $\int_1^{\infty}\frac{dx}{x^{2p}}<\infty \iff 2p>1$. Sadly, I'm unable to prove that $I=\infty$ for $p<1/2$ or $p>2$ and no other way comes to my mind. But it doesn't look complicated at all, so there must be a simple soultion.

Sadly, I'm unable to prove that.

But it seems to me that you already did. You employed the comparison test, by forcefully factoring either $x^2$ or $\sqrt x=x^{^\frac12}$ outside the parentheses, then used the fact that the remaining term tends to $1$ as x tends to either $0$ or infinity, respectively. Either way, it's bound between $1+0=1$ and $1+1$ $=2$, assuming the split point is $1$.

As far as the actual evaluation is concerned, all integrals of the form $\displaystyle\int_0^\infty\frac{x^{^{k-1}}}{(a^{^n}+x^{^n})^{^m}}dx$ can be solved by letting $x=a~t$, and $u=\dfrac1{1+t^{^n}}$, then recognizing the expression of the beta function in the resulting integral, after which all that's left to do is employing Euler's reflection formula for the $\Gamma$ function in order to express it in terms of ordinary trigonometric functions. In this case, $k-1=$ $=-\dfrac p2,~n=\dfrac32,~m=p,~$ and $~a=1$.

$$\int_0^{\infty}\frac{dx}{(x^2+\sqrt{x})^p} = \frac{2 \Gamma(\frac{2-p}{3} )\Gamma(\frac{4p-2}{3})}{3\Gamma(p)}$$

This is the 'answer' supplied by Mathematica. It is clear that this only makes sense when $1/2 < p < 2$ as you've said. In fact Mathematica retuns this as conditional upon $p$ satisfying that inequality.

To actually prove this yourself you will need to use contour integration or some other advanced technique. For example you can see that the poles of the function lie at $x = 0$ and $x = -(1/2) + i \sqrt{3}/2$. So you can use the contour that (has no poles in it) and has the arcs: $x \in [-R, -\rho]$, $x \in [\rho, R]$, $x = \rho e^{i \phi}$ with $\phi \in [-\pi,0]$ and $x= R e^{i \theta}$ with $\theta \in [0,-\pi]$.

As $R\rightarrow \infty$ and $\rho \rightarrow 0$, It is not hard to show that the outer semicircle has no contribution, The "straight" one is the original integral. The small semicircle is possible to integrate, but it is no picnic. Some may consider this a long comment because I won't do it for you but I really don't see the point. The value comes out as above. The Gamma function diverges when its argument is a negative integer or zero. This bounds the value of $p$ as stated.


Why do we call $\sqrt{-1}$ imaginary and $-1$ real? Looking at the numbers I can see in nature only positive real numbers. Because many problems couldn't be solved using only positive real numbers a new number set, called "negative numbers" was introduced. Later on, for similar reasons, complex numbers were introduced and the term of "imaginary".

Can anyone please explain to me why $-1$ is "more real" than $i$? For me both seem to be a convention, a name given to the result of a function or a set of functions to the positive real numbers.

Disclaimer: I'm not a mathematician, I'm just trying to better understand some basic concepts and conventions in math

In mathematics, every modifier, such as "continuous, real, rational, ...", is reassigned meaning, lest there be unnecessary confusion such as "your 'continuous" is different than mine" that wastes time. So, if you come across a terminology in some math book, find its definition within the book; do not take a dictionary and check the meaning of the term...

I am not sure if you are asking about the origin of the reason why the word "imaginary" was chosen. If not, then leave the "secular" meaning of the word alone. If you do this, then it is crystal clear that "imaginary number" is no more than a word used to name the relatives of $\sqrt{-1}$. Note also that "positive" and "negative" in math have nothing to do with any sentiments.

Logically, you can call the real numbers "imaginary" and the imaginary numbers "real". Then in your book, the theorem "the square of every real number is $\geq 0$" becomes "the square of every imaginary number is $\geq 0$". Noticed? The name thing is rather superficial :), which in math serves as a mnemonic device and a shortcut of reference.

Whenever you have got a question about why we call something so.... refer to the definition.
According to Wikipedia:

A real number  is a value that represents a quantity along a line.
An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit $i$, which is defined by its property $i^2 = −1$.

There is no problem in understanding that why $-1$ is a real number as you can easily construct a line and represent $-1$ on that line. The problem lies in the definition of imaginary numbers. It is a real number times $i$.  Why $i$ ?? Because imaginary numbers are  defined that way and changing the definition is not so good I think.  :) :) :)


Fractional Sobolev spaces are Banach spaces Let $1\leq p\leq +\infty$, $0<s<1$ and $\Omega\subseteq \mathbb{R}^n$ an open set. The fractional Sobolev space $W^{s,p}(\Omega)$ is defined to be

$$ W^{s,p}(\Omega) = \left\{ u\in L^p(\Omega) : \frac{|u(x)-u(y)|}{|x-y|^{\frac{n}{p} + s}} \in L^p(\Omega\times\Omega) \right\} $$

equipped with the norm

$$ \|u\|_{W^{s,p}(\Omega)} = \left( \int_\Omega |u|^p \; dx + \int_\Omega\int_\Omega \frac{|u(x)-u(y)|^p}{|x-y|^{n+ sp}} \; dx dy \right)^{1/p}. $$

It is a well known result that this is a Banach space, but every reference I read says that this is true, without giving a proof.

Adam's Sobolev Spaces uses techniques from interpolation theory to prove this result, but I'm no familiar with the theory, so I'm asking for an "elementary proof", that is, as usual, proving that every Cauchy sequence has a limit in the space. Here is my "attempt":

Let $(u_n)$ be a Cauchy sequence in $W^{1,p}(\Omega)$, then $(u_n)$ is a Cauchy sequence in $L^p(\Omega)$, so there exists $u\in L^p(\Omega)$ such that $u_n\to u$ in $L^p(\Omega)$. I would like to prove that $u\in W^{s,p}(\Omega)$ and that $u_n\to u$ in $W^{s,p}(\Omega)$. But I don't know how to proceed here.

Thanks for the help!

Ok, I don't need the answer anymore. I found it in Demengel's Functional spaces for the theory of elliptic partial differential equations. Thank you anyway! :)


What is the proof that the total number of subsets of a set is $2^n$? What is the proof that given a set of $n$ elements there are $2^n$ possible subsets (including the empty-set and the original set).

Suppose you want to choose a subset. For each element, you have two choices: either you put it in your subset, or you don't; and these choices are all independent.

Remark: this works also for the empty set. An empty set has exactly one subset, namely the empty set. And the fact that $2^0=1$ reflects the fact that there is only one way to pick no elements at all!

Let us consider the following cases:

Let a set be $\{a\}$, then its subsets are: $\{a\}, {\emptyset}$ . Therefore the number of subsets is $2^1$
Let another set be $\{a,b\}$, then its subsets are: $\{a\},\{b\},\{a,b\}, {\emptyset}$ Therefore the number of subsets is $2^2$
Let another set be $\{a,b,c\}$, then its subsets are: $\{a\},\{b\},\{c\},\{a,b,c\},\{a,b\},\{b,c\},\{a,c\}, {\emptyset}$. Therefore, the number of subsets is $2^3$

$\therefore$  (1), (2) & (3) $\Rightarrow $ If there are $n$ sets, then the number of subsets is $2^n$


Characteristic polynomial and $p$-adic valuation Suppose I had a linear operator $L$ whose characteristic polynomial was $f(x) = x^{n} + a_{1}x^{n-1} + \cdots + a_{n-1}x + a_{n}$. Furthermore, I also know that the eigenvalues of $L$ have $p$-adic valuation $\geq 1$. Why does it follow that $a_{i} \in p^{i}\mathbb{Z}_{p}$ for all $i$? What if all the eigenvalues have $p$-adic valuation equal to 1?

The eigenvalues are the roots of the characteristic polynomial. Now you just need to recall the relation between the roots of a (monic) polynomial and its coefficients.

I can't comment at the moment - insufficient points. You need to go back to what the p-adic valuation means and basic facts about divisibility of integers by primes. I think it is a lot simpler than you realise.


Proof that if a matrix has real eigenvalues and orthogonal eigenvectors then it's Hermitian I saw many times that a Hermitian matrix has real eigenvalues and orthogonal eigenvectors and I pretty much know the proof of that.

What I was wondering is the other way around, if a matrix has real eigenvalues and orthogonal eigenvectors, does it make it Hermitian?

(I use Dirac's bra/ket notation)

Let $\left|v_1\right>$ and $\left|v_2\right>$ be two distinct orthogonal eigenvectors with distinct eigenvalues $\lambda$ and $\mu$ respectively for $A$, so we have $\left<v_1|v_2\right> = 0 = 0^\dagger = \left<v_2|v_1\right>$

$$\mu \left<v_1|v_2\right> = \left<v_1|A|v_2\right> = 0 $$
$$\lambda \left<v_2|v_1\right> = \left<v_2|A|v_1\right> = 0 = 0^\dagger = (\left<v_2|A|v_1\right>)^\dagger = \left<v_1\right|A^\dagger\left|v_2\right>$$

So we have $\left<v_1|A|v_2\right> = \left<v_1\right|A^\dagger\left|v_2\right>$. I don't think I can conclude that $A = A^\dagger$ that simply.

Am I missing something?

Thanks for your help.

EDIT

I was thinking of doing the following

$$ \left<v_1|A|v_2\right> \left|v_1\right> = \left< v_1 \right | A^\dagger \left | v_2 \right> \left| v_1 \right> $$
$$ \left<v_1|v_1\right> A\left|v_2\right> = \left< v_1 | v_1 \right> A^\dagger \left | v_2 \right> $$

Then the same for $\left|v_2\right>$ and get $||v_1||A||v_2|| = ||v_1||A^\dagger||v_2||$ and conclude that $A=A^\dagger$, but I'm not sure if I can reorder the bras and kets to draw this conclusion.

Let $A$ be an $N\times N$ complex matrix $[a_{n,m}]$ for which there is an orthonormal basis of column vectors $[(e_{j})_{n}]$ which are eigenvectors of $A$ with corresponding eigenvalues $\lambda_{1},\lambda_{n},\cdots,\lambda_{N}$ where the $\lambda_{j}$ are real and some of the $\lambda_{j}$ may be repeated. Then
$$
           A\left[\begin{array}{c}(e_j)_1 \\ (e_j)_2 \\ \vdots \\ (e_j)_{N}\end{array}\right] = \lambda_{j} \left[\begin{array}{c}(e_j)_1 \\ (e_j)_2 \\ \vdots \\ (e_j)_{N}\end{array}\right]
$$
That gives
$$
           A\left[\begin{array}{cccc}
                    (e_1)_1 & (e_2)_1 & \cdots & (e_N)_1 \\
                    (e_1)_2 & (e_2)_2 & \cdots & (e_N)_2 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    (e_1)_N & (e_2)_N & \cdots & (e_N)_N
                  \end{array}\right]\\
        = \left[\begin{array}{cccc}
                    (e_1)_1 & (e_2)_1 & \cdots & (e_N)_1 \\
                    (e_1)_2 & (e_2)_2 & \cdots & (e_N)_2 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    (e_1)_N & (e_2)_N & \cdots & (e_N)_N
                  \end{array}\right]
          \left[\begin{array}{cccc}
                    \lambda_1 & 0 & \cdots & 0\\
                    0         & \lambda_2 & \cdots & 0 \\
                    \vdots  & \vdots  & \ddots &  \vdots \\
                    0 & 0 & \cdots & \lambda_N
                  \end{array}\right]
$$
The matrix $U$ formed from the orthonormal columns vectors is an orthogonal matrix, which guarantees that $U^{\star}U=UU^{\star}=I$ (I'm using $\star$ to denote conjugate transpose.) Therefore,
$$
                        AU = UD \implies A = UDU^{\star}.
$$
The diagonal matrix $D$ has only real eigenvalues. So $D^{\star}=D$. Now you get what you're trying to prove:
$$
    A^{\star}= (UDU^{\star})^{\star}=(U^{\star\star}D^{\star}U^{\star})=UDU^{\star}=A.
$$

If you assume that the set of orthogonal eigenvectors is complete, such a matrix is Hermitian.
It is easy to show this. If not, such a matrix may be not Hermitian.


Is the local max point also a global max? Consider the following function $f$ of three variables, defined on $\mathbb{R^{3}}$: $$f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7$$

Is each of the critical points a local maximum? Is each of them a global maximum? Explain.

I've found the critical points after taking the first-order derivatives $$f_{x} = 15 + y - 8x = 0,$$ $$f_{y} = x - 4y + 2z = 0,$$ $$f{z}=-2z + 2y = 0.$$

So there's only one critical point $\left ( 2,1,1 \right )$.

To figure out whether the critical point is a local maximum I've computed the partial derivatives and have obtained:

$$H = \begin{bmatrix}
-f_{xx} &f_{xy}  &f_{xz} \\
 f_{yx}&f_{yy}  &f_{yz} \\
f_{zx} &f_{zy}  &zz
\end{bmatrix}$$

$$ \Rightarrow  H = \begin{bmatrix}
-8 &1  &0 \\
 1&-4  &2 \\
0 &2  &-2
\end{bmatrix}.$$

After using the Method of Leading principal minors, I've figured out that $H$ is negative definite so $\left ( 2,1,1 \right )$ is a local maximum.

However, I'm a bit confused about whether this local maximum point is also a global maximum point. I know that for a global maximum point, the first-order condition must be satisfied and $H$ (the matrix of partial derivatives) must be negative definite for all $x, y, z \in \mathbb{R^{3}}$.

So based on this, can I say that this local max point is also a global max because $H$ does not depend on the arguments?

The level surfaces of the function $ \ f(x,y,z) = 15x + xy - 4x^{2} - 2y^{2} - z^{2} + 2yz + 7 \ = \ c \ $ prove to be triaxial ellipsoids  (using the determinant method shown here ) with their axes not parallel to the coordinates axes and centered on $ \ (2  ,  1  ,  1 ) \ , $ the critical point found by OGC.  We can show that the maximum value of the function, $ \ f(2,1,1) = 15·2 + 2 - 4·2^2 - 2 - 1 + 2 + 7 \ = \ 22 \ $ is the global maximum of the function by an analysis of these surfaces.
The intersections of a level surface with "constant" planes through the center $ \ (2  ,  1  ,  1 ) \  $ are the ellipses
$$ \mathbf{x  =  2:  } \ \  30 \ + \ 2y \ - \ 16 \ - \ 2y^2 \ - \ z^2 \ + \ 2yz \ + \ 7 \ \ = \  -2y^2 \ + \ 2y \ - \ z^2 \ + \ 2yz \ + \ 21 \ \ = \  \ c \ \ ; $$
$$ \mathbf{y  =  1:  } \ \  15x \ + \ x \ - \ 4x^2 \ - \ 2 \ - \ z^2 \ + \ 2z \ + \ 7 \ \ = \  -4x^2 \ + \ 16x \ - \ z^2 \ + \ 2z \ + \ 5 \ \ = \  \ c \ \ ; $$
$$ \mathbf{z  =  1:  } \ \  15x \ + \ xy \ - \ 4x^2 \ - \ 2y^2 \ - \ 1 \ + \ 2y \ + \ 7 $$ $$ = \  -4x^2 \ + \ 15x \ - \ 2y^2 \ + \ 2y \  + \ xy \ + \ 6 \ \ = \  \ c \ \ . $$
The "cross-section" ellipse in the $ \ y \ = \ 1 \ $ plane does have its axes parallel to the $ \ x-$ and $ \ z-$ axes, so we can simply "complete the squares" to obtain
$$ 4·(x^2 - 4x + 4) \ + \ (z^2  +  2z  +  1) \ \ = \  \ 5 \ - \ c \ + \ 16 \ + \ 1  $$ $$ \Rightarrow \ \ 4·(x-2)^2 \ + \ (z-1)^2 \ \ = \  \ 22 \ - \ c \ . $$
This gives us an ellipse for $ \ c \ < \ 22 \ $ and "shrinks to a point" for $ \ c \ = \ 22 \ . $
The other two cross-sections must be rotated first, applying the standard method (coefficients shown are approximate):
$$ \mathbf{x  =  2:  } \ \ \left[ \tan 2\theta \ = \ \frac{2}{(-2)-(-1)}  =  -2 \right] \   -2.618y^2 \ + \ 1.701y \ - \ 0.382 z^2 \ + \ 1.051z \ + \ 21 \  \ = \ \ c \ \ , $$
$$ \mathbf{z  =  1:  } \ \ \left[ \tan 2\theta \ = \ \frac{1}{(-4)-(-2)}  =  -\frac12 \right]  \  -4.118x^2 \ + \ 14.139x \ - \ 1.882 y^2 \ + \ 5.393y \ + \ 6 \  \ = \ \ c \ \ ; $$
after which we "complete the squares" here (note: the centers of the ellipses are altered by these rotations):
$$ \mathbf{x  =  2:  } \quad 2.618·(y^2 \ - \ 0.650y \ + \ 0.1055) \ + \ 0.382·(z^2 \ - \ 2.751z \ + \ 1.892) $$ $$ = \ \ 21 \ - \ c \ + \ 0.276 \ + \ 0.723 \ \ = \ \ 22 \ - \ c \ \ , $$
$$ \mathbf{z  =  1:  } \quad 4.118·(x^2 \ - \ 3.433x \ + \ 2.947) \ + \ 1.882·(y^2 \ - \ 2.866y \ + \ 2.053) $$ $$ = \ \ 6 \ - \ c \ + \ 12.136 \ + \ 3.863 \ \ = \ \ 22 \ - \ c \ \ . $$
So indeed, all of the cross-sections are ellipses for $ \ c \ < \ 22 \ $ and "collapse" to the point $ \ (2,1,1) \ $ for $ \ c \ = \ 22 \ . $  Hence this is the global maximum for our function.

When there is only one critical point and the critical point is a local maximum, that critical point is global maximum as well.


A question in Probability, aces drawn from two halves of a shuffled deck "A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled and a card is drawn from it. Compute the probability that this drawn card is an ace."

Source : A First Course in Probability, Sheldon Ross, Chapter 3, Exercise 37

(My intention was not to be lazy and let the community do all the work for me, but I think that is what has been misunderstood seeing the 2 dislikes this question received. I just thought I should not clutter the question with more text than necessary)

What I've tried : Probability that the second half already contained 0,1,2 or 3 aces before the ace from the first half was added to it. Then, when the new ace was added to it from the first half, we could calculate the probabilities of drawing an ace considering each of these cases, and add them to get the answer. But I am not sure how do I calculate the P(second half contained 0/1/2/3 aces)..

Here are two solutions:
Let $A$ be the event an ace is drawn from the second half
Let $I$ be the event that the interchanged card is selected

$P(A) = P(A|I)\cdot P(I) + P(A|I^c)\cdot P(I^c)$
$P(A) = 1\cdot\dfrac{1}{27} + \dfrac{3}{51}\cdot\dfrac{26}{27}$
$P(A) = \dfrac{43}{459} = 0.094$

Note that $P(A|I^c) = \dfrac{3}{51}$ because there are 3 aces to choose from and all 51 other cards are equally likely to be in the second half

Second Solution:
Let $A$ be the event an ace is drawn from the second half
Let $N_i$ be the event that $i$ aces are in the second half with $0 \leq i \leq 3$

$P(A) = \sum_{i=0}^3 P(A|N_i)\cdot P(N_i)$
$P(A) = P(A|N_0)\cdot P(N_0) + P(A|N_1)\cdot P(N_1) + P(A|N_2)\cdot P(N_2) + P(A|N_3)\cdot P(N_3)$
$P(A) = \dfrac{1}{27}\cdot P(N_0) + \dfrac{2}{27}\cdot P(N_1) + \dfrac{3}{27}\cdot P(N_2) + \dfrac{4}{27}\cdot P(N_3)$

Note that there are only 3 aces to choose from and 51 cards total, since one ace is in the first half

$P(N_0) = \dfrac{\dbinom{3}{0}\dbinom{48}{26}}{\dbinom{51}{26}}$
$P(N_1) = \dfrac{\dbinom{3}{1}\dbinom{48}{25}}{\dbinom{51}{26}}$
$P(N_2) = \dfrac{\dbinom{3}{2}\dbinom{48}{24}}{\dbinom{51}{26}}$
$P(N_3) = \dfrac{\dbinom{3}{3}\dbinom{48}{23}}{\dbinom{51}{26}}$

$P(A) = \dfrac{43}{459} = 0.094$

Let $A$ be the event that an ace is drawn from the first half of the deck.

Let $X$ be the count of aces in the second half-deck before the additional ace is added from the first deck.

$$\mathsf P(X=x\mid A) ~=~ \dfrac{\dbinom{4}{x}\dbinom{48}{26-x}\mathbf 1_{x\in\{0,1,2,3\}}}{\dbinom{52}{26}-\dbinom{48}{26}}$$

Let $B$ be the event of drawing an ace after adding one and shuffling.   There are now $X+1$ aces among $27$ cards.

$$\mathsf P(B\mid X=x, A)~=~\dfrac{x+1}{27}$$

By the Law of Total Probability.

$$\begin{align}\mathsf P(B\mid A)~=~& \sum_{x=0}^3 \mathsf P(B\mid X=x, A)\mathsf P(X=x\mid A)
\\[1ex] ~=~& \dfrac{1\cdot \dbinom{3}{0}\dbinom{48}{25}+2\cdot \dbinom{3}{1}\dbinom{48}{24}+3\cdot\dbinom{3}{2}\dbinom{48}{24}+4\cdot \dbinom{3}{3}\dbinom{48}{23}}{27~\left(\dbinom{52}{26}-\dbinom{48}{26}\right)}\end{align}$$

Remark: Since we are given that event $A$ has happened, the probability that we seek is indeed conditional on this event: $\mathsf P(B\mid A)$


How do editors of mathematical journals know that a submited paper is original, that is unpublished? I wondered many times how the editors of mathematical journals know that the content of a paper or article that was submited by an author is original, this is, that the main content of such calculations and reasonings is unpublished.

  Question. How do editors of mathematical journals know that a submited paper is original, that is unpublished? Many thanks.

I would like to know it as curiosity, because I imagine that professors working in mathematical journals are experts in a field of mathematics. Then a day they receive a writing, a paper or a remark, with perhaps several results, and they have to evaluate if this was not published or known previously.
Additionally I imagine that these professors or editors have tools to search information for this purpose, and maybe they have contact with other colleagues who have historical memory about what results, and of what kind, have been published. Am I wrong in these believes? How do they know and they note the originality of such mathematical content?

When submitting a paper for publication, it is not only the editor who looks at it. The editor will (usually) take a first look to determine if the paper looks to be suitable for the journal. If they see that the main results are already well-known this will likely make them reject the paper (especially if the authors have not made any remark on this themselves).

If this is not the case and the paper generally looks fine (this also includes the results actually seeming to be interesting), then the paper will be sent to one or two experts in the field for review. These will (hopefully) take a much closer look at the paper and determine if they think it is suitable for the journal. Once again, this will include trying to determine whether the results are actually new, which is why it is important for the editor to choose reviewers who really are experts, as otherwise they will not be familiar enough with the field to determine this.

Finally, the reviewers will make a recommendation to the editor based on their reading, and using the editor will make a final decision on whether to accept the paper, possibly contingent on certain improvements.

Usually they have payed staff, some Prof. who are working in this field. So they send your paper to them, they review it. Based on their review it will be published or not. It depends on the quality of the journal, how well they check your work.


Find the value of $\alpha$ Let $f(x)=\frac{\alpha x}{x+1},x\neq-1$. Then for what value of $\alpha$ is $f(f(x))=x$?

My approach:
Given that,
 $f(f(x))=x$
Applying $f^{-1}$ on both sides, we get
$f(x)=f^{-1}(x)$
The above condition is only true when the curve $y=f(x)$ intersects with $y=x$.

$\therefore f(x)=x\implies \frac{\alpha x}{x+1}=x$
 On solving this, we get $\alpha $ as a function of $x$.

I know that there is another method to solve this in which we find the expression for $f(f(x))$ and equate it to $x$.

I don't need that solution. I just want to know what is wrong with this approach as this was the first thing that came to my mind on seeing the question and the method was much shorter compared to the above-mentioned method.

Anybody with alternative methods are welcome to share.

Thanks in advance.

Note that when $\alpha = -1$, the vertical and horizontal asymptotes will be equal: $x=-1, y=-1$ and $f(x)=f^{-1}(x)$.

In fact, that is the case for any hyperbola with equal asymptotes:

$$f(x)=\frac{ax+b}{c x+d}, \left(-\frac{d}{c}=\frac{a}{c}\right) \Rightarrow \\
y=\frac{ax+b}{c x+d} \Rightarrow x=\frac{b-yd}{yc-a} \Rightarrow \\
f^{-1}(x)=\frac{b-xd}{xc-a}=\frac{ax+b}{cx+d}.$$
You can equate the original function to its inverse to see the solution $\alpha=-1$ clearly:
$$f(x)=f^{-1}(x) \Rightarrow \frac{\alpha x}{x+1}=\frac{x}{\alpha-x} \Rightarrow \\
\frac{\alpha^2x-\alpha x^2-x^2-x}{(x+1)(\alpha -x)}=0 \Rightarrow \\
\frac{(\alpha +1)(\alpha-x-1)x}{(x+1)(\alpha -x)}=0$$

$f(f(x))=\alpha^2x/((x+1)(\alpha x/(x+1)+1))$
The equation you have to solve is:
$$f(f(x))=x$$
So you get:
$$\alpha=-1$$
and
$$\alpha=x+1$$


Convergence of a series with alternating denominator - Real Analysis Decide if the series converges absolutely, conditionally, or not at all.
\begin{equation}
\sum_{n=1}^{\infty}\frac{(-1)^n}{(2+(-1)^n)n}
\end{equation}

I'm having a lot of trouble with this one. I know the even terms will be of the form $\frac{1}{3n}$ and the odd terms of the form $-\frac{1}{n}$ but I'm not really sure how I can use that to prove convergence. Any hints would be greatly appreciated.

I agree with your analysis of the problem:
even terms will be of the form 1/3n and the odd terms of the form −1/n.  Then each of them will converge to the same number 0.  so the whole thing will converge to 0.  Afterward, you can prove by induction.


Why is $\tan(x)$ a function? A function $f:X\rightarrow Y$ maps each $x\in X$ to some $y \in Y$. So consider $\tan{\frac{\pi}{2}}$ for which $\tan(x)$ is undefined, so in this case, $\tan(x)$ does not map to an element of its range. This conflicts with my understanding of what a function is. Why do we still consider $\tan(x)$ a function?

The set $X$ in your definition is the domain of the function.  The domain of $\tan(x)$ is typically taken to be

$$
X=\bigcup_{k\in\Bbb{Z}} \left(-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi\right)
$$

Thus $\pi/2\notin X$, and so don't need to assign a value to $\tan(\pi/2)$ (or for any $\pi/2+k\pi,k\in\Bbb{Z}$ for that matter).

Actually that is the definition of a continuous function. The definition for a function is that for all inputs there is exactly one output. In a non-continuous function, when the input doesn't map to anything, it is not in the domain. So if f(pi/2) is undefined, pi/2 is not in the domain and therefore not an input.


For a compact covering space, the fibres of the covering map are finite. I am stuck on the following exercise:

  Let $Y$ be a compact topological space, and $p:\ Y\ \longrightarrow\ X$ a covering map. Show that for every $x\in X$ the fibre $p^{-1}(x)$ is finite.

Any help would be very welcome.

The space $X$ has a finite open cover $(U_i)_i$ of evenly covered neighborhoods. We can assume that the cover is minimal, that means none of these sets can be removed. The preimage of each $U_i$ is a disjoint union of open sets $\left(V_i^j\right)_{j\in J_i}$ each of which is mapped homeomorphically onto $U_i$. Since the $U_i$ cover $X$, the $V_i^j$ cover $Y$, so there must be a finite subcover.

By minimality of the cover of $X$, every $U_i$ has an element $x_i$ not contained in any other $U_j$. The finitely many $V_k^l$ cover $Y$, thus cover $p^{-1}(x_i)$. But this fiber is disjoint from any $V_k^l$ with $k\ne i$, so $p^{-1}(x_i)$ must be covered by sets $V_i^1,...,V_i^m$. Since each of them contains only one point of the fiber, it must be finite.

Now since the fibers of all points in $U_i$ have the same cardinality, we conclude that all fibers are finite.

Maybe these results will be helpful:
I'm assuming: Top space is Hausdorff, and singletons are closed in bottom space.

1)The fibers of a singleton are discrete.

2) If a singleton y in your space is closed; $U_y$ is a covering neighborhood of $y$,
with restricition homeomorphism $P_y: V_y \rightarrow U_y$ , then $P^{-y}$ is a closed
set. A closed subset of a compact Hausdorff space is compact.

3)The only compact discrete spaces are finite spaces.

Can you take it from here?


Is it possible to write a sum as an integral to solve it? I was wondering, for example,

Can:

$$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$

Be written as an Integral? To solve it.  I am NOT talking about a method for using tricks with integrals.

But actually writing an integral form. Like

$$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)} = \int_{a}^{b} g(x) \space dx$$

What are some general tricks in finding infinite sum series.

A General Trick

A General Trick for summing this series is to use Telescoping Series:
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+2)}
&=\frac13\lim_{N\to\infty}\sum_{n=1}^N\left(\frac1{3n-1}-\frac1{3n+2}\right)\\
&=\frac13\lim_{N\to\infty}\left[\sum_{n=1}^N\frac1{3n-1}-\sum_{n=1}^N\frac1{3n+2}\right]\\
&=\frac13\lim_{N\to\infty}\left[\sum_{n=0}^{N-1}\frac1{3n+2}-\sum_{n=1}^N\frac1{3n+2}\right]\\
&=\frac13\lim_{N\to\infty}\left[\frac12-\frac1{3N+2}\right]\\
&=\frac16
\end{align}
$$

An Integral Trick

Since
$$
\int_0^\infty e^{-nt}\,\mathrm{d}t=\frac1n
$$
for $n\gt0$, we can write
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(3n-1)(3n+2)}
&=\sum_{n=1}^\infty\frac13\int_0^\infty\left(e^{-(3n-1)t}-e^{-(3n+2)t}\right)\mathrm{d}t\\
&=\frac13\int_0^\infty\frac{e^{-2t}-e^{-5t}}{1-e^{-3t}}\mathrm{d}t\\
&=\frac13\int_0^\infty e^{-2t}\,\mathrm{d}t\\
&=\frac16
\end{align}
$$

This particular sum could be solved because you had two terms $ax+b$ and $ax+c$ and the difference between c and b is equal to a (I think it would work in a slightly more complicated way if it was a not-too-large multiple of a).

If you want numerical values in general cases, and the sum doesn't converge quickly for your taste, or you want just a partial sum, you can use that

$$\displaystyle f (k) = \int_{k-1/2}^{k+1/2} f(k) dx ≈ \int_{k-1/2}^{k+1/2} f(x) dx$$

and therefore

$$\displaystyle \sum_{k=n}^{m} f(k) ≈ \int_{n-1/2}^{m+1/2} f(x) dx$$

Assuming that you can solve the integral in closed form, if you let

$$\displaystyle g (k) = f(k) - \int_{k-1/2}^{k+1/2} f(x) dx$$

then

$$\displaystyle \sum_{k=n}^{m} f(k) = \int_{n-1/2}^{m+1/2} f(x) dx + \sum_{k=n}^{m} g(k)$$

$g (k)$ will usually converge much faster than $f (k)$.


Which of the following conditions are necessary for the positive integer $n$ to be divisible by 6? Which of them are sufficient? Question:
Are my answers correct? I am particularly concerned with (ii) and (vii).

Which of the following conditions are necessary for the positive integer $n$ to be divisible by 6? Which of them are sufficient?
(i) 3 divides $n$.
(ii) 9 divides $n$.
(iii) 12 divides $n$.
(iv) $n =$ 12.
(v) 6 divides $n^2$.
(vi) 2 divides $n$ and 3 divides $n$.
(vii) 2 divides $n$ or 3 divides $n$.

Answers:
(i) necessary,
(ii) none,
(iii) sufficient,
(iv) sufficient,
(v) sufficient and necessary,
(vi) sufficient and necessary,
(vii) necessary.

I think that your answer to numbers II and VII are correct.

$9$ is not a factor of $6$ so is not necessary.

Since $n$ might be divisible by $2$ but may not be divisible by $3$ (e.g. $4$) and since $n$ might be divisible by $3$ but may not be divisible by $2$ (e.g. $9$) VII is insufficient but must be necessary.


Let a function f be continuous. Assume $\int_a^b f=0$ Prove $f(x)=0$ for all $x\in [a, b]$ Let $f:[a, b]\to \mathbb{R}$ be continuous. Assume $f(x)\geq 0$ for all $x\in[a, b]$, and that $$\int_a^b f\text{ }\mathrm{ d}x=0$$ Prove that $f(x)=0$ for all $x\in[a, b]$.

Would you recommend using a direct proof or a proof by contradiction by assuming there exists $y\in[a, b]$ such that $f(y)\geq 0$?

Sketch of proof:

Assume the opposite, so $f(x)>0$ for some $x$. use continuity to prove $f$ is larger than a positive $l$ in a closed interval $[c,d]$ around $x$.

Use $\int_a^b\geq L\{a,c,d,b\}\geq (d-c)l>0$ to conclude if $f$ is strictly positive in a point the integral is greater than zero.

Therefore if $\int_a^bf=0$ we have $f$ is never strictly positive.

Let $F'(x) = f(x)$. Then $F(b) - F(a) = 0$. The differential is defined in $b$ and $a$, and thus $(F(b) - F(a))' = 0'$, so $f(b) = f(a)$.

Since the function $f$ is continuous, there is $c$ between $[a,b]$ such that $f(c) = f(a)$. We also have $F(b) - F(a) = F(b) - F(c) + F(c) - F(a)$, and as $f(x) \ge 0$, $F(c) = F(a)$ (because the integral is crescent) for every $c$ between $a and b$, so $f(c) = f(b) = f(a)= k$ for every $c$ in the interval and the integral reduces itself to $k(b-a)=0$, so $f(c)=k=0$.


Partial Derivatives of Cartesian and Polar Coordinate Systems Plane polar coordinates, $r$ and $\phi$, and Cartesian coordinates, $x$ and $y$, are related by
$$x=r\cos\phi,\ y=r\sin\phi,$$
and
$$r^2=x^2+y^2,\ \phi=\tan^{-1}({y \over x}).$$

To try to find the partial derivative $\partial r \over \partial x$, I used two approaches:

Using $x=r \cos \phi$,
$${\partial x \over \partial r}=\cos \phi,$$
and using ${\partial x \over \partial r}=({\partial r \over \partial x})^{-1},$
$${\partial r \over \partial x}=\frac{1}{\cos \phi}.$$
Using $r^2=x^2+y^2$,
$$r=(x^2+y^2)^{1/2},$$
$$\frac{\partial r}{\partial x}=\frac{x}{(x^2+y^2)^{1/2}}=\cos\phi.$$

Why are the two answers different? I think the second approach gives the correct answer but I do not understand what is wrong with the first approach.

Division in partial derivatives is just a notation so $\frac{\partial r}{\partial x} \neq \frac{\partial x}{\partial r}^{-1}$.

What you are doing using the "inverse" of the derivate, is a property that has the derivation when proceding with bijective functions, this means that they have and inverse function so the derivative of the inverse function is the inverse of the derivative of the initial function. Let me suggest you to read these pages:

Bjective function: https://en.wikipedia.org/wiki/Bijection
Inverse function theorem: https://en.wikipedia.org/wiki/Inverse_function_theorem

This is, the function $r^2 = x^2 + y^2$ is not bijective, so you can not do the derivative as you proposed in the second case.


A Physics Problem in Calculus. In our class, we encountered a problem that is something like this: "A ball is thrown vertically upward with ...". Since the motion of the object is rectilinear and is a free fall, we all convene with the idea that the acceleration $a(t)$ is 32 feet per second square. However, we are confused about the sign of $a(t)$ as if it positive or negative.

Now, various references stated that if we let the upward direction to be positive then $a$ is negative and if we let downward to be the positive direction, then $a$ is positive. The problem in their claim is that they did not explain well how they arrived with that conclusion.

My question now is that, why is the acceleration $a$ negative if we choose the upward direction to be positive. Note: I need a simple but comprehensive answer. Thanks in advance.

The gravity force is downward and so is the acceleration (by $F=ma$).

So if you choose a downward axis, the acceleration is positive.

And if you choose an upward axis, the acceleration is negative.

You can model this on a linear one dimensional space, the "heigth" $x$ of the particle. Different forces are acting on the ball, on the one hand the gravitation, which is directed to the "ground", on the other hand in the beginning a force is applied towards the "top" as the ball is thrown in that direction. You can derive the equations of motion through Newtons formular $F(t)=m\ddot{x}(t)$ and the superposition of forces, the resulting force is the sum of each force.


Find the value of $k$ if $\det(A^{–1}) = (\det A)^k$ where $A$ is a $3 \times 3$ matrix Now I'm getting two answers to this question i.e. $k = -1$ and $k = 1$

Reason i'm getting $k = 1$ :

$$A(A^{-1}) = I$$

Taking determinant on both sides we get

$$\det[A(A^{-1})] = \det(I)$$

$$\det(A)(\det(A^{-1})) = 1$$

Dividing both sides with $\det(A)$ we get

$$\det(A^{-1}) = \det(A)^{-1}$$

Hence, $k = -1$

Reason I'm getting $k = 1$ :

$$A^{-1} = \frac1{\det(A)}Adj(A)$$

Applying determinant on both sides we get

$$\det(A^{-1}) = \det{[\frac1{\det(A)}]Adj(A)}$$

$$\det(A^{-1}) = \frac1{\det(A)}\det[Adj(A)]$$

because $\det[Adj(A)] = \det(A)^{n-1}$

i.e. $\det[Adj(A)] = \det(A)^{3-1}$

Or, $$\det[Adj(A)] = \det(A)^2$$

Thus, $$\det(A^{-1}) = \frac1{\det(A)}[\det(A)^2]$$

Or, $$\det(A^{-1}) = \det(A)^1$$

Hence $k = 1$

Now my Question is which one is the right answer
And why am I getting two answers, have i neglected a condition or what
Any kind of help will be much appreciated.

P.S. $k = -1$ is the answer given in all books I've come across.

First the mistake:

$$\det(A^{-1}) = \frac{1}{\det(A)^\color{red}{3}}{\det[Adj(A)]}$$

because $\det[Adj(A)] = \det(A)^{n-1}$

i.e. $$\det[Adj(A)] = \det(A)^{3-1}$$

Or, $$\det[Adj(A)] = \det(A)^2$$

Thus, $$\det(A^{-1}) = \frac1{\det(A)^3}[\det(A)^2]=\frac{1}{\det(A)}$$

which is a known result.

Next, this is how I would approach the problem:

$$\det(A^{-1})= \det(A)^k$$

Hence we have $$1=\det(A)^{k+1}$$

If we want the equality to hold as long as $\det(A) \neq 0$, we require $k+1=0$.

Notice that I did not use the propery of the size of $A$.

Suppose that $A$ has the property $det(A)=1$, then $det(A^{-1})=1$, hence

$det(A^{-1})= ( \det(A))^k$  for all(!) $k \in \mathbb Z$.


Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$ I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$
My attempt: Letting
$$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$
$$
\begin{aligned}
I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta)
\end{aligned}
$$
$$
I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha
$$
But I can't go further.

Taking integration by parts,

$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$

Now

$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx
= \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$

and the first integral is easily computed by integration by parts:

$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx
= -\log 2 + \frac{\pi}{2}. $$

The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion

\begin{align*}
\log \sec\theta
 = -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right|
&= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\
&= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta),
\end{align*}

we have

\begin{align*}
\int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx
&= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\
&= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,}
\end{align*}

where $G$ is Catalan's constant. Combining altogether, we obtain

$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$

Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields

The function does not have a elementary anti derivative.

Approximately (by numerical integration) the answer is $0.779611255707666$.


Limit of integral is zero implies limit of measure of a sequence of sets is zero. Let $f$ be a measurable function on $[0,1]$, $f$ is positive a.e on $[0,1]$. If ${E_k}$ is a sequence of measurable sets in $[0,1]$ such that $\lim_{k\to\infty} \int_{E_k} f(x) dx =0$, prove that $\lim_{k\to\infty}  m(E_k)=0$.

I tried to apply the limit theorem but I have neither bounded functions or monotone functions. I can't find an integrable function to dominate $1_{E_k} f$.
Can someone help me?
Thank you

Sketch: Suppose $m(E_k)$ does not converge to $0.$ Then there exists $\epsilon>0$ and a subsequence $k_j$ such that $m(E_{k_j})>\epsilon$ for all $j.$

Verify that $\lim_{n\to \infty} m(\{f>1/n\}) = 1.$ Thus there exists $n_0$ such that $ m(\{f>1/n_0\}) > 1-\epsilon/2.$ This implies

$$m\left (E_{k_j}\cap \{f>1/n_0\}\right )>\epsilon/2\,\,\text { for all } j.$$

Thus

$$\int_{E_{k_j}} f \ge \int_{E_{k_j}\cap \{f>1/n_0\}} f > \frac{1}{n_0}m(E_{k_j}) > \frac{1}{n_0}\cdot \frac{\epsilon}{2}\,\,\text { for all } j.$$

That contradicts the hypothesis that $\int_{E_k} f \to 0.$

Since $f$ is positive almost everywhere, the increasing sequence of sets $A_{n} = \{x \in [0,1] \, \mid \, f(x) > n^{-1}\}$ satisfies $m(A_{n}) \to 1$ by continuity of measure.  Thus, for each $\delta \in (0,1)$, we can pick a $N \in \mathbb{N}$ such that $m(A_{N}) > 1 - \delta$.

Observe that $\int_{E_{k} \cap A_{N}} f(x) \, dx \leq \int_{E_{k}} f(x) \, dx$ since $f$ is positive almost everywhere.  Thus, $\lim_{n \to \infty} \int_{E_{k} \cap A_{N}} f(x) \, dx = 0$.  On the other hand,
$$0 \leq N^{-1} m(E_{k} \cap A_{N}) \leq \int_{E_{k} \cap A_{N}} f(x) \, dx$$
so $m(E_{k} \cap A_{N}) \to 0$ as $k \to \infty$ (since $N$ is fixed).

How big can $m(E_{k} \setminus A_{N})$ be?  No bigger than $m([0,1] \setminus A_{N})$.  Thus,
$$m(E_{k}) \leq m(E_{k} \cap A_{N}) + \delta.$$
Now take $k \to \infty$ to obtain
$$\limsup_{k \to \infty} m(E_{k}) \leq \delta,$$
which implies $\limsup_{k \to \infty} m(E_{k}) \leq \delta$.
Since $\delta$ was arbitrary and $m(E_{k}) \geq 0$ for each $k$, we deduce
$$\lim_{n \to \infty} m(E_{k}) = 0.$$


What's the negation of “At least three of the sentence are false”? Following with the question I asked before

What's the negation of "One of the sentence is false"?

The negation of “At least three of the sentence are false” would be "any, one or two of the sentence is/are false"?

Call $f$ the number of false sentences.
Your "At least $3$ of these sentences are false" is
$$
f \ge 3
$$
The negation of this is clearly
$$ f < 3$$
Thus, "the number of false sentences is smaller than $3$", i.e., "there are $0$, $1$ or $2$ false sentences", i.e., "there are at most $2$ false sentences."

No! The negation of the sentence "At least one of the three sentences is false" is "All three of the sentences are true" or, equivalently "None of the sentences are false".


How to prove that $\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$? Let $L$ be any lattice in $\mathbb{C},$ and $L'$ a lattice containing $L$ with index $n$ (i.e $n=\sharp L'/L$)  I found this statement

"The lattice $L'$ must  be contained in  $\frac{1}{n}L = \{\frac{1}{n}L\; ; \; l\in L\},$ because each element of $L'/L$ has order dividing $n=\sharp L'/L.$ Thus, $L'$  corresponds to a subgroup of order $n$ in $\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$
"
in  Koblitz Introduction to Elliptic Curves and modular forms  p 152

I don't understand why

$L'\subset \frac{1}{n}L$
$L'$ corresponds to a subgroup of order $n$ in
$\frac{1}{n}L$
$\frac{1}{n}L/L\simeq (\mathbb{Z}/n\mathbb{Z})^2$

Could someone please explain to me this, Thanks in advance.

When I was first looking at questions like this I recall needing to take my time to fully absorb the fact that $\frac1n L$ is BIGGER than $L$. Why would anyone think that multiplication by $1/n$ makes something smaller? Beats me:-) Anyway, with that off my chest, let's roll.

1. Let $x\in L'$. Consider the coset $x+L$ in the quotient group $L'/L$. Because $[L':L]=n$ we have, by Lagrange's theorem familiar from elementary group theory, that $$nx+L=n(x+L)=0_{L'/L}=L.$$ So $nx\in L$ and therefore $x\in\frac1n L$.

2. By part 1 we have the inclusions
$$
L\subseteq L'\subseteq \frac1nL.
$$
Moding out the smallest subgroup $L$ tells us that
$$
L'/L\le \frac1nL/L.
$$
We were given the fact that $\#L'/L=n$ so the subgroup $L'/L$ has, indeed, order $n$. In other words, we are using the fact that if $K$ is a normal subgroup of $G$, then there is a bijective correspondence between the subgroups of $G/K$ and the intermediate groups $H$ such that $K\le H\le G$. The correspondence relates $H$ and $H/K\le G/K$.

3. $L$ is a free abelian group of rank two generated by, say, $\alpha$ and $\beta$. So $\frac1n L$ is the free abelian group generated by $u=\alpha/n$ and $v=\beta/n$. We then switch points of view and think of $L$ as the subgroup of $\frac1nL$ generated by $nu$ and $nv$. The isomorphism is given by
$$
f:\frac1nL/L\to(\Bbb{Z}/n\Bbb{Z})^2, (au+bv)+L\mapsto (\overline{a},\overline{b}).
$$


I assume that by $\phi(t)$ you mean the area in the left tail of a normal distribution. If so, then the two below is incorrect and should be removed:
$$P(X > 1) = 1 - P(X \leq 1) = 1 - \color{red}{2}\phi(1) = \dots$$

thus $$P(X\le1)=\phi(1)$$

$$your answer =1-\phi(1)$$


What is the value of $i^i$? I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!

$$i^i = e^{i\log i} = e^{i(\log |i|+i\arg i)} = e^{i(i\arg i)} = e^{-\arg i} = e^{-\frac{\pi}{2}+2 \pi k} \qquad k \in \mathbb{Z}$$


Why do both sine and cosine exist? Cosine is just a change in the argument of sine, and vice versa.

$$\sin(x+\pi/2)=\cos(x)$$
$$\cos(x-\pi/2)=\sin(x)$$

So why do we have both of them? Do they both exist simply for convenience in defining the other trig functions?

You seem to be asking why we name them both, rather than why they exist, since the very relationships you've written shows that if one exists, the other does, too.
Essentially, all mathematical notation and names, except for a very small subset, are for convenience/clarity/human communication. Math does not require that we name any function consistently across separate proofs, but it becomes much easier to communicate and think about things when we do.
I prefer the relationship:
$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$ since this is symmetric and obviously geometric when $0\leq x\leq \pi/2$, and because this is also the relationship between "tangent" and "cotangent" and "secant" and "cosecant." It indicates a duality in these functions.
It is certainly something of a paradox that adding more names often simplifies our understanding. In particular, if you defined only one, it would give you a sense that one of these functions was "primary." There is a hint of that error even in the names "sine" and "cosine," which vaguely implies that "sine" is primary, but it would be particularly strong if we only defined "sine" and never defined "cosine." We would have a harder time grasping the duality that happens in trig functions.

If you actually must start with one function, most mathematicians wouldn't start with $\cos x$ or $\sin x,$ they'd start with the complex-valued function: $\operatorname{cis}(x)=\cos(x)+i\sin(x).$ This has the property $\operatorname{cis}(x)\operatorname{cis}(y)=\operatorname{cis}(x+y),$ and can also be written as $e^{ix}.$ You can define $\cos x$ and $\sin x$ in terms of $\operatorname{cis} x:$
$$\cos x=\frac{1}{2}\left(\operatorname{cis}(x)+\operatorname{cis}(-x)\right)\\
\sin x=\frac{1}{2i}\left(\operatorname{cis}(x)-\operatorname{cis}(-x)\right)$$

Addition: I recently heard, in a lecture that covered some of the accomplishments of the early Muslim world, that they acquired knowledge of the sine function from India, and developed the other trigonometric functions. So if you want someone to blame, we can blame them.

Just as every individual has a name , different from others ; so should every function have a name different from others . This enables us to remember different functions in short, without having to write the detailed mathematical expressions , representting these functions.


Lagrange to find the maximum. I do not know how to get the following:
the answer is 1/n or 1/n^n But I can't get that result.
We have to use Lagrange theorem $\triangledown f = \lambda \triangledown g$ and I got the gradiants:

$\triangledown f = 2x{_{k}}\prod x{_{j}}^{2}$

and

$\triangledown g = 2x{_{k}}$

but than I can't get the correct result. Please help me.

With

$f(x_1, x_2, \ldots, x_n) = \prod_1^n x_i^2 \tag{1}$

and

$g(x_1, x_2, \ldots, x_n) = \sum_1^n x_i^2 = 1, \tag{2}$

we have, as pointed out by our OP SoHCahToha,

$(\nabla f)_j = 2x_j \prod_{1, i \ne j}^n x_i^2, \tag{3}$

and

$(\nabla g)_j = 2x_j.  \tag{4}$

With Lagrange multiplier $\lambda$, we conclude that

$(\nabla f)_j = \lambda (\nabla g)_j \tag{5}$

for all $j$, which by (3)-(4) becomes

$2x_j \prod_{1, i \ne j}^n x_i^2 = \lambda (2x_j). \tag{6}$

We note that

$f(x_1, x_2, \ldots, x_n) \ge 0 \tag{7}$

for all vectors $(x_1, x_2, \ldots, x_n)$, and

$f(x_1, x_2, \ldots, x_n) = 0 \tag{8}$

only when $x_i = 0$ for at least one index $i$.  Therefore, the maximum of $f$ occurs at a point where $x_i \ne 0$ for all $i$, so division of (6) by $2x_j$ is permissible, leading to

$\prod_{1, i \ne j}^n x_i^2 = \lambda \tag{9}$

for every $j$.  Furthermore, from (6) we see that $\lambda \ne 0$ since no $x_j = 0$ at the maximum, so we can form the quotients

$\dfrac{\prod_{1, i \ne j}^n x_i^2}{\prod_{1, i \ne k}^n x_i^2} = 1 \tag{10}$

for every pair of indices $j, k$.  (10) is trivial when $j = k$, but for $j \ne k$ becomes

$\dfrac{x_k^2}{x_j^2} = 1 \tag{11}$

or

$x_k^2 = x_j^2; \tag{12}$

using (2) and (12), we see that

$n x_j^2 = 1 \tag{13}$

or

$x_j^2 = \dfrac{1}{n}; \tag{14}$

thus, for every $j$,

$x_j = \pm \dfrac{1}{\sqrt n}. \tag{15}$

It then follows that

$f(x_1, x_2, \ldots, x_n) = \prod_1^n \dfrac{1}{n} = \dfrac{1}{n^n} = n^{-n} \tag{16}$

is the maximum possible value of $f(x_1, x_2, \ldots, x_n)$ subject to (2).

Technically, we need to prove the points whose coordinates are given by (15) are in fact true maxima and not some other form of extrema such as saddle points.  But since $f(x_1, x_2, \ldots, x_n)$ is a smooth function on the compact sphere given by (2), we know it takes its maximal value, where the derivatives of $f$ vanish.  But we have seen that at every point where $\nabla f = 0$, $f$ takes the value $n^{-n}$; thus this must in fact be the greatest possible value of $f$ subject to (2).

This is very simple, and equivalent to:

$$
\mathcal{P}: \max f=y^{n}\\
 ny=1, y\ge 0\\
y=x_i^2 \ \forall i
$$

The solution is then
$$
x_i=\sqrt{1/n} \ \forall i
$$


What does it mean for rational numbers to be "dense in the reals?" What does it mean for rational numbers to be "dense in the reals?" I can't seem to find a decent explanation online...

It means that between any two reals there is a rational number.  The integers, for example, are not dense in the reals because one can find two reals with no integers between them.

That definition works well when the set is linearly ordered, but one may also say that the set of rational points, i.e. points with rational coordinates, in the plane is dense in the plane.  Then it must be defined differently: it means that every open set in the plane intersects the set of all rational points.  No matter how small you make an open disk in the plane, it cannot avoid containing some rational points; so the set of all rational points is dense in the plane.  In the case of the line, saying that every open interval contains some rationals amounts to the same thing as saying that between any two reals there is some rational number between them.

Between any two rational numbers there exist another rational number. For example 1/2 and 1/4 are two rational numbers, but there exist another rational number 1/3 between the two above.In the case of other subsets of numbers in real numbers for instance,integers,there cannot exist another integers between any two.


Evaluate a differential equation at a given point I have a second order differential equation $x^2\cdot y''(x)+x\cdot y'(x)+(x^2-10)=0$ with the initial condition $y(100)=1$ and $y'(100)=0$. I want to evaluate $y(x)$ when $x=103$ with 8 digit accuracy.

The first step is rewrite this equation to a system of first order equation.
$\begin{cases} y'=w\\ w' = -\frac{w}{x}-\frac{10}{x^2}+1\end{cases}$ with the initial condition $\begin{cases} y(100)=1\\ w(100)=0\end{cases}$. Then I can use fourth-order Runge-Kutta methods to solve the system. In order to achieve 8 digit accuracy, I need to decide the step size $h$. Solving $h^4\le 10^{-8}$, I get $h=0.01$.

Is this seems a reasonable approach? Is there anything special with such a differential equation with the initial condition so that I need use other method to evaluate it?

It is reasonable to begin with $h=0.01$. Since the theoretical bounds for RK4 are very conservative and also difficult to use, the step halving method may be preferable. It is based on running the computations with two or several step sizes and  comparing the results.

For illustration purposes, I used Euler's method (rather than RK4) for this equation. The results were:

with $h=0.02$, $y(103)\approx 5.4225$
with $h=0.01$, $y(103)\approx 5.4371$
with $h=0.005$, $y(103)\approx 5.4443$

Since Euler's method is of 1st order, each result is off by about $Ch$ from the true value. We can estimate $C$ as follows
$$
\frac{5.4225-5.4371}{0.02-0.01} \approx -1.46
$$
$$
\frac{5.4371-5.4443}{0.01-0.005} \approx -1.44
$$
The fact that two values of $C$ are close allows us to conclude confidently that the error is bounded by, say, $2h$ when $h$ is small.

For RK4 the error behaves as $Ch^4$, so you would have the differences like $0.02^4-0.01^4$ in the denominator. Try it (and report the results, if you don't mind).

I think that the method of Frobenius is applicable to this problem.


Is $f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$ bounded? $$f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$$Is this function bounded?

So obviously this converges because $|\frac1k\sin\frac x{2^k}|<|\frac x{2^k}|$ and $\sum\frac x{2^k}$ converges by the integral test.

Now I need to show that there exists a $N$ for all $y\in\text{range}\,f$ such that $|y|<N$.

So I think:

Let $a_k=\frac1k$ and $b_k=\sin\frac x{2^k}$. From the Schwarz inequality we get $|\sum a_kb_k|\leq\sqrt{\sum|a_k|^2\sum|b_k|^2}$. Since $a_k>0$ for all $k$ we have $a_k=|a_k|$ and thus $|a_k|^2=\frac1{k^2}$. $\sum\frac1{k^2}=\frac{\pi^2}6$ from the Reimann-Zeta. Now all we have left to prove is that $\sum|b_k|^2$ is bounded as well.

The function is unbounded. In fact, take $x_m = \frac{2^{3m}}7 2\pi$; then the sequence $f(x_m)$ tends to infinity. To see this, note that
$$
\sin\Big( \frac n72\pi \Big) \approx \begin{cases}
0.781831, &\text{if }n\equiv1\pmod 7, \\
0.974928, &\text{if }n\equiv2\pmod 7, \\
-0.433884, &\text{if }n\equiv4\pmod 7.
\end{cases}
$$
Therefore
\begin{align*}
f(x_m) &= \sum_{j=1}^m \bigg( \frac1{3j-2}\sin\Big(\frac{2^{3m-(3j-2)}}7 2\pi\Big) + \frac1{3j-1}\sin\Big(\frac{2^{3m-(3j-1)}}7 2\pi\Big) + \frac1{3j}\sin\Big(\frac{2^{3m-3j}}7 2\pi\Big) \bigg) \\
&\qquad{}+ \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&\approx \sum_{j=1}^m \bigg( \frac{0.974928}{3j-2} - \frac{0.433884}{3j-1} + \frac{0.781831}{3j} \bigg) + \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&> \sum_{j=1}^m \frac{0.44}j + \sum_{k=3m+1}^\infty 0 > 0.44 \ln m.
\end{align*}

Note that for $x\in[0,2\pi]$
$$
\left\lvert\,\sin\frac x{2^k}\right\rvert\le \frac{\lvert x\rvert}{2^k}\le \frac{2\pi}{2^k},
$$
and hence
$$\lvert\, f(x)\rvert\le \sum_{k\in\mathbb N}\left|\frac1k\sin\frac x{2^k}\right|\le
\sum_{k=1}^\infty \frac{2\pi}{k2^k}<2\pi.
$$


How to solve $3 + 0.075x = 1.075^x$ for x? I tried various methods... Could only solve it numerically not analytically. Any suggestions?

The equation
$$a+b x=c^x$$ has solutions in terms of Lambert function. It is given by
$$x=-\frac{1}{\log (c)}W\left(-\frac{\log (c) }{b}c^{-\frac{a}{b}}\right)-\frac{a}{b}$$ but this function has multiple branches.

Consider that you look for the zero's of $$f(x)=c^x-bx-a$$ which has as derivatives
$$f'(x)=c^x \log (c)-b\qquad \text{and} \qquad f''(x)=c^x \log ^2(c)\, \, >0 \, \forall x$$
The first derivative cancels at
$$x_*=\frac{\log \left(\frac{b}{\log (c)}\right)}{\log (c)}$$ which is aminimum. So, if $f(x_*)<0$ tow possible roots. If you want to approximate them, develop as Taylor series aound this point
$$f(x)=f(x_*)+\frac{1}{2} (x-x_*)^2 f''(x_*)+O\left((x-x_*)^3\right)$$ and solve the quadratic to get as estimates
$$x_\pm=x_* \pm \sqrt{-2\frac{f(x_*)}{f''(x_*)}}$$

Applied to your case, this would give $x_-=-26.6577$ and $x_+=27.6638$; using them as starting values, Newton method would converge very fast to the solutions.

You can solve it with MatDeck as well, the equations solutions are -39.218 and 21.033

Solving of nonlinear equations in MatDeck


Convergent sequences are closed subspace of $l^\infty$ Here is the exercise:

Let's consider the Banach space $l^{\infty}$ with the sup norm.

We consider the following subspace of $l^{\infty}$ :

$C = ${ $x \in l^{\infty} : \lim_{n\to \infty} x_n\quad exists $}.

We have to show that $C$ is a closed subspace of $l^{\infty}$ and is thus also a Banach space with respect to the sup norm.

Let $y$ denote $\lim_{n\to \infty}$ $x_n$.

The key there is to show that every convergent sequence in $C$, say every sequence in $C$ converges within $C$, hence we have to show that $y$ belongs to $C$.
I've successfully shown that $y$ is a Cauchy sequence in $l^{\infty}$, but I don't know what to do with that. When I'm looking at the correction, it says that { $y$ Cauchy } $\Rightarrow$ {$y$ convergent} $\Rightarrow$ {y $\in$ $C$}.

That's where I'm lost. I don't understand why we can deduce $y$ is convergent by having shown $y$ is Cauchy. Also, I don't understand why we have to show that $y$ is Cauchy.

I hope someone can explain me those two problems I have.

I do understand though that $C$ being closed implies that $C$ is a Banach space.

Consider $l^{\infty}\setminus C$
Let $y_{n} \in l^{\infty}\setminus C\,,$ then $\forall y \in\mathbb{R}, \exists \varepsilon_{y} > 0$ such that $\|y-y_{n}\|_{\infty} > 2\varepsilon_{y}$
Now consider $\mathbb{B}(y_{n},\varepsilon_{y})$
Let $z_{n} \in \mathbb{B}(y_{n},\varepsilon_{y})\Longrightarrow \|y_{n}-z_{n}\|_{\infty}<\varepsilon_{y}$
So $\forall y \in \mathbb{R}$;
$\|y-z_{n}\|_{\infty} = \|y-y_{n}-(z_{n}-y_{n})\|_{\infty} \geq \|y-y_{n}\|_{\infty} - \|z-y_{n}\|_{\infty} > 2\varepsilon_{y} - \varepsilon_{y} = \varepsilon_{y} > 0$
$\Longrightarrow \mathbb{B}(y_{n},\varepsilon_{y}) \subseteq l^{\infty}\setminus C\Longrightarrow l^{\infty}\setminus C$ is open $\Longrightarrow C$ is closed.

Let $X=\{1,2,3,\cdots\}\cup\{\infty\}$ and define a topology on $X$ that is discrete, but neighborhoods of $\infty$ include all large integers. Then $X$ is compact, and your sequences are the same as the continuous functions $C(X)$. This is the one-point compactification. $C(X)$ is a Banach space, and the elements of $C(X)$ are determined by their values on the integers, with a natural isometric inclusion in $\ell^{\infty}$.


Probability to win between 7 and 10 games of the next 15 games The probability that the Mets win is .8.  What is the probability they win between 7 and 10 games of the next 15 games?

Please help. Thank you.

Hint: Look at the binomial distribution.

look at the table of binomial probabilities.
N=15
P(710)= .003+.014+.043+.103=.163


A Man Invested Some Money at 10% and 12%............. A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of Rs.130. If he had interchanged the amounts, he would have received Rs. 40 more as yearly interest. How much did he invest at different rates?

Let the two invested rates be x and y.
Therefore, 12/100*x+10/100*y=130    (Dividing by 100)
          =0.12x+0.1y=130...(i)*0.12
0.1x+0.12y=170...(ii)*0.1
Now solve by yourself to get the answer.


Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$

My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$

then what I will do??

Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have
$$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$

The general solution to a cubic:

$$\begin{align}
r & =  \sqrt[3]{
\underbrace{  \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) }_{p}
+ \sqrt[2]{
    \left(\underbrace{  \frac{c}{3a} - \frac{b^2}{9a^2} }_{q} \right)^3   + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
  }
} \\
& +  \sqrt[3]{
\left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)
- \sqrt[2]{
     \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3  + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
  }
} \\
& - \underbrace{ \frac{b}{3a} }_{r}
\end{align}$$

The problem gives $p = 2$, $q^3 + p^2= 3$, $r = 0$.

$$\begin{cases}
0 = \frac{b}{3a} \\
2 = \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \\
3 = \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3  + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2  \\
\end{cases}$$

Since polynomials have the same roots up to a scaling factor (that is, $f(x)$ has the same roots as $k~f(x)$), we can choose $a = 1$.  And solve and get $b = 0$.

$$\begin{cases}
a = 1 \\
b = 0 \\
2 = -\frac{d}{2} \\
3 = \left( \frac{c}{3} \right)^3  + \left( \frac{d}{2} \right)^2  \\
\end{cases}$$

$$\begin{cases}
a = 1 \\
b = 0 \\
3 = \left( \frac{c}{3} \right)^3  + 4  \\
d = -4 \\
\end{cases}$$

$$\begin{cases}
a = 1 \\
b = 0 \\
c = -3 \\
d = -4 \\
\end{cases}$$

$$x^3 - 3x - 4 = 0$$


How to convince a math teacher of this simple and obvious fact? I have in my presence a mathematics teacher, who asserts that

$$ \frac{a}{b} = \frac{c}{d} $$

Implies:

$$ a = c, \space b=d $$

She has been shown in multiple ways why this is not true:

$$ \frac{1}{2} = \frac{4}{8} $$

$$ \frac{0}{5} = \frac{0}{657} $$

For me, these seem like valid (dis)proofs by contradiction, but she isn't satisfied. She wants a 'more mathematical' proof, and I can't think of any.

I'm worried that if she isn't convinced, it may be detrimental to some students. Is there another way to systematically demonstrate the untruth of her conjecture?

EDIT: Since the answer which worked was from a comment, but each answer is also very good, I'm upvoting all of them instead of accepting a specific one. Feel free to close this question for being too open if so a moderator desires.

You can prove that all the numbers are equal ;-)

Let's assume that for all $a,b,c,d \in \mathbb{R}$, $b \neq 0$, $d \neq 0$ we have

$$ \frac{a}{b} = \frac{c}{d}\quad \text{ implies }\quad a = c\ \text{ and }\ b = d. \tag{$\spadesuit$}$$

Now take any two numbers, say $p$ and $q$, and write

$$\frac{p}{p} = \frac{q}{q}.$$

Using claim $(\spadesuit)$ we have $p = q$. For the special case, where one of them equals zero (e.g. $q$), use $$\frac{2p}{2p} = \frac{p+q}{p+q}.$$

I hope this helps ;-)

Alternatively, $\frac{a}{b}=\frac{c}{d}$ if and only if $ad=cb$ by the definition of fraction equality. So, according to your teacher, $ad=cb\iff a=c\text{ and }d=b$, i.e., there is only one way of writing the integer $ad=cb$ as the product of two integers. However, in general we can write an integer as the product of two integers in many different ways. For example, $24$ can be written as:

$1\times 24$

$2\times 12$

$3\times 8$

$4\times 6$

$6\times 4$

$8\times 3$

$12\times 2$

$24\times 1$

I hope this helps!


Difference between Slope and Gradient It has been a few years since studying contour maps.

Often I hear slope and gradient interchangeably in describing steepness.

Does anyone know any good definitions and analogies of slope and gradient.

Thanks,

Amanda

Best Answer - Chosen by Voters

A gradient is a vector, and slope is a scalar. Gradients really become meaningful in multivarible functions, where the gradient is a vector of partial derivatives. With single variable functions, the gradient is a one dimensional vector with the slope as its single coordinate (so, not very different to the slope at all).

Source(s):
Currently studying multivariable calculus

As gradient is a vector and the gradient of a scaler is a vector ie it drives a scaler in a direction; slope is a scaler ( scaler can be positive and negative) yet it drives a line ( y= mx+ c ) with unique steepness in upward or downward in a direction ( deviation depends upon change in perpendicular per unit base ie tan(theta) which is proportional to theta and up and down  direction by + and - signs) unlike a vector which has finite length (magnitude)