Apostol proof divides by zero? The following is a proof from Apostol's linear algebra book that $\{1,t,t^{2},...\}$ is independent. To my eye, he's dividing by zero repeatedly. Is this really as huge an error as it seems, or are there missing details that would make this rigorous?

It suffices to show that for each $n$ the $n+1$ polynomials $1,t,...,t^{n}$
are independent. If these polynomials span the zero polynomial we
have $$\sum_{k=0}^{n}c_{k}t^{k}=0$$ for all real $t$. Putting $t=0$ we see that $c_{0}=0$. Now divide
by $t$ and put $t=0$ again to find $c_{1}=0$. Repeating the process
we find every $c_{k}$ is $0$, so $1,t,...,t^{n}$ are independent.

Edit: I'm treating "polynomial" as a garden-variety function from $\mathbb{R}$ to $\mathbb{R}$, and it seems like that may be part of my problem.

As David mentioned in his comment, the issue here is that continuity is implicitly used. When one deduces that $c_0 = 0$, you are left with
$$
t(c_1 + c_2 t + c_3 t^2 + \dots + c_n t^{n-1}) = 0
$$
which means that for all $t \neq 0$, $c_1 + c_2 t + \dots + c_n t^{n-1} = 0$. But since polynomials are continuous functions over the reals, we can use the continuity of polynomials to see that $c_1 + c_2 t + \dots + c_n t^{n-1}$ = 0 also when $t =0$, where we find that $c_1 = 0$ and the process can continue.

This is where we use a very nice property of polynomial functions over the reals ; the only polynomial with real coefficients that vanishes everywhere over the reals is the zero polynomial.

I'm saying this because I've studied for a summer polynomials with this "vanishing property", i.e. that when evaluated in their ground ring, they're worth zero everywhere. Here the "ground ring" would be $\mathbb R$ (I wouldn't say "ground ring" is a standard term, but it works for me here.). Replace $\mathbb R$ by say, $\mathbb Z / p^n \mathbb Z$, and things get all messed up. Here's a nice result (just for show =D ) that I've managed to find out :

Theorem. Let $f(x) \in \mathbb Z / p^n \mathbb Z$. Then the associated polynomial function $f : \mathbb Z / p^n \mathbb Z \to \mathbb Z / p^n \mathbb Z$ is identically zero if and only if when we write
$$
f(x) = \sum_{i=0}^n c_i (x)_i
$$
where $(x)_i = x(x-1)\dots(x-(i-1))$ with the convention $(x)_0 = 1$, we obtain that $c_i \cdot i! \equiv 0 \mod p^n$.

Proof. ($\Longleftarrow$) Suppose $c_i \cdot i! \equiv 0 \mod p^n$ for $0 \le i \le n$. Then
$$
f(0) = c_0 \equiv c_0 \cdot 0! \equiv 0 \mod p^n.
$$
Assuming this is true for $k-1$, then
$$
f(k) = \sum_{i=0}^k c_i (k)_i
$$
and since $(k)_i = i! \begin{pmatrix} k \\\ i \end{pmatrix}$, we have
$$
\sum_{i=0}^k c_i (k)_i = \sum_{i=0}^k c_i i! \begin{pmatrix} k \\\ i \end{pmatrix} \equiv \sum_{i=0}^k 0 \equiv 0 \mod p^n,
$$
because binomial coefficients are integers.

($\Longrightarrow$) Suppose $f$ is always zero $\mod p^n$. Then since for $k \ge 0$, we have
$$
f(k) = \sum_{i=0}^n c_i i! \begin{pmatrix} k \\\ i \end{pmatrix}.
$$
Computing $f(0)$ gives $0 \equiv f(0) = c_0 \mod p^n$. By induction,
$$
0 \equiv f(k) = \sum_{i=0}^k c_i i! \begin{pmatrix} k \\\ i \end{pmatrix} \equiv c_k k! \begin{pmatrix} k \\\ k \end{pmatrix} = c_k k! \mod p^n,
$$
so that $c_k k! \equiv 0 \mod p^n$.

This gives you a wide class of rings over which there is a bunch of polynomials that are non-zero and can vanish everywhere. Note that these rings are "discrete" in some sense, i.e. there is no notion of continuity that works in a similar fashion than with the reals, which is why we had hope that those polynomials might exist. One familiar example is of course $x^{p-1} - 1$ in $\mathbb Z / p \mathbb Z$ but there are many other such polynomials, for instance $(x^{p-1} - 1)^n q(x) \mod p^n$ where $q(x)$ is an arbitrary non-zero polynomial in $\mathbb Z / p^n \mathbb Z$. Note that those polynomials also form an ideal, which can be interesting. =)

Hope that was fun to read! And that it helps.

You mighty  use induction for  $ n= 1$ $ p (t)$ is constant on $\mathbb{R}\0$ iff $ c_{0}, c_{1}$ are zero . suppose true for $ n$ then consider
$$ c_{0}+...+ c_{n +1} t^{n +1}=0$$  didn't substituting for $ t =0$  we get $ c_{0}=0$ and
$$ c_{1}+...+ c_{n +1} t^ n =0$$ and hypothesis of induction we get the result.


Mean of a squared random variable if I have a random variable $x$ ~ $N(0,1)$, what is the expected value of $x^2$? Is there a general rule to calculate expected values of squared variables? Thanks.

Hint: There is a connection between the variance of $x$ and the expected value of $x^{2}$. If you write both down separately you'll see it.

If $X$ follows Standard Normal.then  $X^2$ follows Chi Square with 1 degrees of freedom..
And the Expected value of a  Chi Square is the degrees of freedom itself So $$E(X^2)=1$$
As far as a general rule, there is none you have to figure it out for yourself by practicing


The Linear Independence of Powers of x over finite fields My professor asked us to prove that $P =\{1,x,x^2,\dots, x^n \dots \}$ is a basis for polynomials. This is pretty simple if you are working over $\mathbb{R}$, however he did not specify which field (though it was clear he meant $\mathbb{R}$).

In order for $P$ to be a basis, its elements must be linearly independent.
This is easy to prove over an infinite field, by finitely many roots. However, $P$ is not necessarily linearly independent over finite fields. For instance, in $\mathbb{Z}_2$, $x^n + x = 0$ for all $x$.

My question is if there is a basis for all polynomials over a finite field? My abstract algebra is weak and this was just something I noticed in my introductory linear algebra course.

This is a frequent point of confusion.

It is true that, say, $x^3$ and $x$ take the same values $\pmod 3$ as $x$ runs through the residues $\pmod 3$.  However, they are not the same polynomial.  Two polynomials $\sum_{i=0}^na_ix^i$ and $\sum_{i=0}^{m}b_ix^i$ are said to be the same if and only if $a_i=b_i$ for all $i$.  Note that this implies that the degrees are the same.

With this definition, the linear independence of the monomials $\{x^i\}$ is clear.

As a partial justification of this definition, note that, while $x^3$ and $x$ take the same values in the finite field $\mathbb F_3$, they take different values in extension fields, $\mathbb F_9$ for instance.  Indeed, by the usual results from field theory, the polynomial $x^3-x$ could only have at most three roots in any field.

The definition of an $n$-dimensional vector space, $V$ is that any vector $u \in V$ can be written as a linear combination of the space's basis vectors and that no linear combination of these basis vectors is in the nullspace of $V$.

Let the set of vectors $S =  \{u_1, ..., u_n \}$, by definition of linear independence then $S \subset  span\{U\}$ iff $U$ is at least $n$ dimensional.


Is there a function with the property$ f(n)=f^{(n)}(a)$ Is there a not identically zero, real-analytic function $f\colon\mathbb R\to\mathbb R$, which satisfies

$$f(n)=f^{(n)}(a),n\in\mathbb N \text{ or }\mathbb N^+?$$

and $a\in \mathbb R$

I saw a special case when $a=0$

I try to solve it by :

$$f(x)=e^{cx}$$
$$f(n)=e^{nc}$$
$$f^{(n)}(x)=c^ne^{cx}$$
$$f^{(n)}(a)=c^ne^{ca}$$
so $$e^{nc}=c^ne^{ca}$$
so $$c=\frac{nW(\frac{a-n}{n})}{a-n}$$

the problem is we always see  n with c
but the special case when a=0 give

$$c=\frac{nW(\frac{0-n}{n})}{0-n}$$
$$c=\frac{W(\frac{-1}{1})}{-1}=-W(-1)$$

I think there is no solution when $a\neq 0$

may be there is another function can solve it

Is there any solution in general?

thanks for all

I can't explicitly find an example, so perhaps turning to an existence proof, that can also be used to construct an example. To do so, consider the operator $\mathcal L_a: C^\infty \times \mathbb N \to \mathbb R$, where $\mathcal L_a\{f,n\} = f(n) - f^{(n)}(a)$. Then inspect Banach fixed-point theorem, if you can choose a norm in $C^\infty \times \mathbb N$ where $\mathcal L_a$ is a contraction, then jack-pot. Or, if it is not a contraction, for many reasonable norms, then you can say there is probably no such function in Banach space.
Hope this helps.

I don't know about a general solution. But if we assume the function is invertible, then:
$$ f(n) = f^n(a)$$
$$f^{-1}f(n) = n= f^{n-1}(a)$$

Then, $$f^0(a) = a = 1$$
$$f(1) = f(a)= 2$$
 $$f(2) = f^2(1) = 3$$ and soon. In this case $f$ is an increment function.


Prove or disprove: $A-(B-C) = (A-B)-C$ This is what I have so far:

First, we will attempt to show $A-(B-C) \subseteq (A-C) - C.$
    Let $x \in A -(B-C)$.
    Then $x\in A$ and $x \notin (B-C)$. By DeMorgan's we have that
$$x \notin (B-C) = x \notin B \vee x \in C.$$
    We have that $x \in A \wedge (x \notin B \vee x \in C)$.
    Then we have that $(x \in A \wedge x \notin B) \vee (x \in A \wedge x \in C)$.

So by definition of $\cup$ and $\cap$, we have $x \in (A-B) \cup (x\in (A \cap C))$. I'm kind of stuck here.

I can see a contradiction here in that LHS says $x \in C$ but the RHS says $x \notin C$.

Can anyone tell me what to do from here? I'm not quite sure how to formally state that the statement is false. I feel like I'm not being complete here.

Why would you start trying to prove some random identity without first testing it in some easily computed cases to see if it's right or wrong??? Because I like to look for easy solutions before hard ones, I always try plugging in zero for one of the variables, or in this case because we're dealing with sets, the empty set.

If $A=\emptyset$ then $A-(B-C)=(A-B)-C$ turns into $\emptyset=\emptyset,$ check.

If $C=\emptyset$ it turns into $A-B=A-B,$ check.

If $B=\emptyset$ it turns into $A=A-C.$ Now that looks fishy, doesn't it? What if $C=A,$ then we get $A=\emptyset.$ That sure isn't an identity . . .

Counterexample: Let $B=\emptyset$ and $A=C=\mathbb R$ (or any nonempty set). Then
$$A-(B-C)=\mathbb R-(\emptyset-\mathbb R)=\mathbb R-\emptyset=\mathbb R$$
while
$$(A-B)-C=(\mathbb R-\emptyset)-\mathbb R=\mathbb R-\mathbb R=\emptyset.$$


The Golden Ratio and the Secant Method The Secant Method is basically a way of replacing the second derivative of a function with an approximation. So $$f''(x) \approx \frac{f'(x_2)-f'(x_1)}{x_{2}-x_{1}}$$

We then use this in Newton's formula $$x_{k+1} = x_{k}-\frac{f'(x_k)}{f''(x_k)}$$

It turns out that the secant method converges with rate equal to the golden ratio. Is this just a coincidence?

It is not a coincidence.  It comes from the similarity from one stage of the root finding to the next and the fact that $\frac 1{\phi}=\phi-1$.  A good discussion is in section 9.2 of Numerical Recipes

The secant method replaces the first derivative, not the second.

Also, the Newton's method iteration is
$x_{n+1} = x_n - f(x_n)/f'(x_n)$.


How to determine probability of an outcome when the number of tries is variable. Given a known probability of an event being successful, how do you calculate the odds of at least one successful outcome when the number of times you can attempt is determined by a separate dice roll.

For example:  I know the probability of a favorable outcome on dice X is "P".      I first roll a six sided dice. That dice roll determines the number times I can roll dice X.  What is the probability that I will have at least one successful roll of dice X as a function of "P"?

P(desired outcome from N rolls) will be P(desired outcome from 1 roll)^N if you want each roll to satisfy a given condition.

P(at least 1 * desired outcome from N rolls) will be 1-P(not desired outcome)^N.

P(at least M * desired outcome from N rolls, $N \geq M$) will be
$1-\sum_{b=0}^{M-1} {N \choose b} \left(\frac {1} {P(desired outcome)}\right)^N$

If you roll the X die $k$ times, the probability of failing on all $k$ times  is $(1-p)^k$.

Hence the probability of succeeding at least once in $k$ tries is $1-(1-p)^k$. (because succeeding at least once and failing every time are complementary events.)

Because the initial die roll is independent of the X die roll, their probabilities multiply. The probability of succeeding when you roll a six-sided die then roll your X die that many times is:
$$\sum_{k=1}^6 \frac{1}{6}\times \left[1-(1-p)^k\right] = 1-\frac{1}{6}\sum_{k=1}^6(1-p)^k$$

The geometric sum formula states that $\sum_{k=1}^n q^k = \frac{q(1-q^n)}{1-q}$.  In our case, this means that the overall probability of success with a $n=6$ sided die and $q=(1-p)$ geometric probability is:
$$1-\frac{1}{6}\frac{(1-p)(1-(1-p^6))}{p},$$ which, if you like, is just the polynomial
$$\Pr(p) = -\frac{1}{6}\left(p^6 - 7p^5 + 21 p^4 -35 p^3 + 35 p^2 - 21 p^1 \right)$$

As an example, when $p=\frac{1}{2}$ (X is a two-sided coin), the probability is around 0.83. When $p=\frac{1}{6}$ (X is another six-sided die), the probability is around 0.45.


Real world applications of Pythagoras' Theorem I have a school assignment, and it requires me to list a few of the real world applications of Pythagoras Theorem. However, most of the ones I found are rather generic, and not special at all.

What are some of the real world applications of Pythagoras' Theorem?

There was an older person nearby who used to use it in his job making window frames.  He needed them square or rectangular with 90 degree angles.  So he'd measure the two sides and a diagonal and if Pythagoras worked he'd got the 90 degree right.

Say you're playing frisbee with a few of your friends, and the frisbee gets stuck in a tree. You want to get a ladder to reach it, but you don't know how long the ladder needs to be. You can make a point where you want the ladder to touch the ground, and then measure from there, whereas $ds^2=dx^2+dy^2$ is true, so we can say, as an example, that $dx=3 \text{m}$ long and $dy=4\text{m}$ long. Now, we must solve for $ds$. Here's my calculations:
$ds^2=3^2+4^2=9+16=25$
$\sqrt{ds^2}=ds$
$\sqrt{25}=5$
$ds=5\text{m}$
You'd need a ladder that is $5\text{m}$ long to reach the frisbee.


Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral:
$$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$
I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.

Any suggestions are welcome. Thanks.

Let $I:=\int\frac{\cos x}{\cos x+\sin x}dx$ and $J:=\int\frac{\sin x}{\cos x+\sin x}dx$. Then $I+J=x + C$, and
$$I-J=\int\frac{\cos x-\sin x}{\cos x+\sin x}dx=\int\frac{u'(x)}{u(x)}dx,$$
where $u(x)=\cos x+\sin x$. Now we can conclude.

Just for some new ideas!
I would reccomend a completely different method. This method uses the Gudermannian $\text{gd}$ function. So you would substitute $x=\text{gd}(a);\text{d}x=\text{sech}\space a\text{d}a$ That transforms the integral into:

$$\int \frac{\tanh a}{\tanh a+\text{sech}\space a}(\text{sech}\space a)\mathrm da$$

Through some hyperbolic trig properties, we get (correct me if I'm wrong)

$$\int {\frac{1}{\cosh a+\coth a}}\text{d}a$$

You could probably take it from here


There are 100 people in a queue waiting to enter a hall. The hall has exactly 100 seats numbered from 1 to 100. The first person in the queue... There are $100$ people in a queue waiting to enter a hall. The hall has exactly $100$ seats numbered from $1$ to $100$. The first person in the queue enters the hall, chooses any seat and sits there. The $n$-th person in the queue where $n$ can be $2,\ldots,100$, enters the hall after $(n-1)$-th person is seated. He sists in seat number $n$ if he finds it vacant; otherwise he takes any unoccupied seat. Find the total number of ways in which $100$ seats can be filled up, provided the $100$-th person occupies seat number $100$.

I could not realise how this chaotic behaviour will end. I think the solution lies in finding that. Please help.

Let $s_r$ be the number of the seat taken by the $r^{th}$ person.

Suppose $s_1=k_1$, then person $k_1$ is the first to find their seat already filled. Either person $k_1$ takes seat $1$, when the remainder of the passengers take their own seat, or $s_{k_1}=k_2\gt k_1$. Person $k_2$ is the second person to find their seat occupied (repeat the argument).

So you need to count the number of increasing sequences $2\leq k_1\lt \dots \lt k_r\le 99$ where $k_r$ indexes the people who find their seat occupied and have to occupy a seat other than their own.

   To compute the count note that any number between $2$ and $99$ inclusive can either be in the sequence or out of it. So that comes to $2^{98}$.

The normal question about this situation is to find the probability that the $100^{th}$ person sits in their own seat. If this is part of your method for solving that, it is not the easiest way of getting an answer.

The correct answer is 99!
Check it out for 4 people.
This is an application of funtions. i.e No.  of one-one functions when co-domain(No. of seats)  and domain(passengers)  are equal.


elements of order 15 in $Z_5 \times Z_{15}$ I know that for an element $(a,b) \in \mathbb{Z}_j \times \mathbb{Z}_k$, the element will have order $lcm(m,n)$ where m is the order of a in $\mathbb{Z}_j$ and n is the order of b in $\mathbb{Z}_k$. So if I am supposed to instead find such a,b do I just go through all possibilities for m and n that give 15 then check if a and b can have such an order in their respective quotient groups?

I am ultimately trying to find the number of elements of order 15 in $\mathbb{Z}_5 \times \mathbb{Z}_{15}$

$$\mathbb{Z}_5\times \mathbb{Z}_{15}\cong \mathbb{Z}_5\times \mathbb{Z}_5\times \mathbb{Z}_3$$

Hence, we need to select an element of order $5$ from $\mathbb{Z}_5\times \mathbb{Z}_5$, and combine with an element of order $3$ from $\mathbb{Z}_3$.  There are $24$ ways to do the former (anything but the identity), and $2$ ways to do the latter.  Hence the answer you seek is $48$.


Does $\sum\limits_n \log\left(1+{1\over n}\right)$ diverge or converge? How do I find out if $\sum\limits_n\log(1+{1\over n})$ diverges or converges? Wolfram recommends me to use comparison test, but I do not know series which diverges and less than this.

$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)=\log\prod_{n=1}^{N}\frac{n+1}{n}=\log(N+1).$$

If you want to use a comparison, notice that since $f(t)=\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, we have:
$$\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\geq\frac{1}{n+1/2}$$
by Jensen's inequality, hence:
$$\sum_{n=1}^{N}\log\left(1+\frac{1}{n}\right)\geq 2\sum_{n=1}^{N}\frac{1}{2n+1} = 2H_{2n+1}-H_n.$$

Another way to think of this problem is that $\sum log(1+1/n)$ = $\log \prod(1+1/n)$ = $\log \prod((n+1)/n)$, which goes to infinity. Thus the sum diverges.


What are imaginary numbers? At school, I really struggled to understand the concept of imaginary numbers.  My teacher told us that an imaginary number is a number that has something to do with the square root of $-1$.  When I tried to calculate the square root of $-1$ on my calculator, it gave me an error.  To this day I still do not understand imaginary numbers.  It makes no sense to me at all.  Is there someone here who totally gets it and can explain it?

Why is the concept even useful?

Let's go through some questions in order and see where it takes us. [Or skip to the bit about complex numbers below if you can't be bothered.]

What are natural numbers?

It took quite some evolution, but humans are blessed by their ability to notice that there is a similarity between the situations of having three apples in your hand and having three eggs in your hand. Or, indeed, three twigs or three babies or three spots. Or even three knocks at the door. And we generalise all of these situations by calling it 'three'; same goes for the other natural numbers. This is not the construction we usually take in maths, but it's how we learn what numbers are.

  Natural numbers are what allow us to count a finite collection of things. We call this set of numbers $\mathbb{N}$.

What are integers?

Once we've learnt how to measure quantity, it doesn't take us long before we need to measure change, or relative quantity. If I'm holding three apples and you take away two, I now have 'two fewer' apples than I had before; but if you gave me two apples I'd have 'two more'. We want to measure these changes on the same scale (rather than the separate scales of 'more' and 'less'), and we do this by introducing negative natural numbers: the net increase in apples is $-2$.

  We get the integers from the naturals by allowing ourselves to take numbers away: $\mathbb{Z}$ is the closure of $\mathbb{N}$ under the operation $-$.

What are rational numbers?

My friend and I are pretty hungry at this point but since you came along and stole two of my apples I only have one left. Out of mutual respect we decide we should each have the same quantity of apple, and so we cut it down the middle. We call the quantity of apple we each get 'a half', or $\frac{1}{2}$. The net change in apple after I give my friend his half is $-\frac{1}{2}$.

  We get the rationals from the integers by allowing ourselves to divide integers by positive integers [or, equivalently, by nonzero integers]: $\mathbb{Q}$ is (sort of) the closure of $\mathbb{Z}$ under the operation $\div$.

What are real numbers?

I find some more apples and put them in a pie, which I cook in a circular dish. One of my friends decides to get smart, and asks for a slice of the pie whose curved edge has the same length as its straight edges (i.e. arc length of the circular segment is equal to its radius). I decide to honour his request, and using our newfangled rational numbers I try to work out how many such slices I could cut. But I can't quite get there: it's somewhere between $6$ and $7$; somewhere between $\frac{43}{7}$ and $\frac{44}{7}$; somewhere between $\frac{709}{113}$ and $\frac{710}{113}$; and so on, but no matter how accurate I try and make the fractions, I never quite get there. So I decide to call this number $2\pi$ (or $\tau$?) and move on with my life.

  The reals turn the rationals into a continuum, filling the holes which can be approximated to arbitrary degrees of accuracy but never actually reached: $\mathbb{R}$ is the completion of $\mathbb{Q}$.

What are complex numbers? [Finally!]

Our real numbers prove to be quite useful. If I want to make a pie which is twice as big as my last one but still circular then I'll use a dish whose radius is $\sqrt{2}$ times bigger. If I decide this isn't enough and I want to make it thrice as big again then I'll use a dish whose radius is $\sqrt{3}$ times as big as the last. But it turns out that to get this dish I could have made the original one thrice as big and then that one twice as big; the order in which I increase the size of the dish has no effect on what I end up with. And I could have done it in one go, making it six times as big by using a dish whose radius is $\sqrt{6}$ times as big. This leads to my discovery of the fact that multiplication corresponds to scaling $-$ they obey the same rules. (Multiplication by negative numbers responds to scaling and then flipping.)

But I can also spin a pie around. Rotating it by one angle and then another has the same effect as rotating it by the second angle and then the first $-$ the order in which I carry out the rotations has no effect on what I end up with, just like with scaling. Does this mean we can model rotation with some kind of multiplication, where multiplication of these new numbers corresponds to addition of the angles? If I could, then I'd be able to rotate a point on the pie by performing a sequence of multiplications. I notice that if I rotate my pie by $90^{\circ}$ four times then it ends up how it was, so I'll declare this $90^{\circ}$ rotation to be multiplication by '$i$' and see what happens. We've seen that $i^4=1$, and with our funky real numbers we know that $i^4=(i^2)^2$ and so $i^2 = \pm 1$. But $i^2 \ne 1$ since rotating twice doesn't leave the pie how it was $-$ it's facing the wrong way; so in fact $i^2=-1$. This then also obeys the rules for multiplication by negative real numbers.

Upon further experimentation with spinning pies around we discover that defining $i$ in this way leads to numbers (formed by adding and multiplying real numbers with this new '$i$' beast) which, under multiplication, do indeed correspond to combined scalings and rotations in a 'number plane', which contains our previously held 'number line'. What's more, they can be multiplied, divided and rooted as we please. It then has the fun consequence that any polynomial with coefficients of this kind has as many roots as its degree; what fun!

  The complex numbers allow us to consider scalings and rotations as two instances of the same thing; and by ensuring that negative reals have square roots, we get something where every (non-constant) polynomial equation can be solved: $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$.

[Final edit ever: It occurs to me that I never mentioned anything to do with anything 'imaginary', since I presumed that Sachin really wanted to know about the complex numbers as a whole. But for the sake of completeness: the imaginary numbers are precisely the real multiples of $i$ $-$ you scale the pie and rotate it by $90^{\circ}$ in either direction. They are the rotations/scalings which, when performed twice, leave the pie facing backwards; that is, they are the numbers which square to give negative real numbers.]

What next?

I've been asked in the comments to mention quaternions and octonions. These go (even further) beyond what the question is asking, so I won't dwell on them, but the idea is: my friends and I are actually aliens from a multi-dimensional world and simply aren't satisfied with a measly $2$-dimensional number system. By extending the principles from our so-called complex numbers we get systems which include copies of $\mathbb{C}$ and act in many ways like numbers, but now (unless we restrict ourselves to one of the copies of $\mathbb{C}$) the order in which we carry out our weird multi-dimensional symmetries does matter. But, with them, we can do lots of science.

I have also completely omitted any mention of ordinal numbers, because they fork off in a different direction straight after the naturals. We get some very exciting stuff out of these, but we don't find $\mathbb{C}$ because it doesn't have any natural order relation on it.

Historical note

The above succession of stages is not a historical account of how numbers of different types are discovered. I don't claim to know an awful lot about the history of mathematics, but I know enough to know that the concept of a number evolved in different ways in different cultures, likely due to practical implications. In particular, it is very unlikely that complex numbers were devised geometrically as rotations-and-scalings $-$ the needs of the time were algebraic and people were throwing away (perfectly valid) equations because they didn't think $\sqrt{-1}$ could exist. Their geometric properties were discovered soon after.

However, this is roughly the sequence in which these number sets are (usually) constructed in ZF set theory and we have a nice sequence of inclusions
$$1 \hookrightarrow \mathbb{N} \hookrightarrow \mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{R} \hookrightarrow \mathbb{C}$$

Stuff to read

The other answers to this question give very insightful ways of getting $\mathbb{C}$ from $\mathbb{R}$ in different ways, and discussing how and why complex numbers are useful $-$ there's only so much use to spinning pies around.
A Visual, Intuitive Guide to Imaginary Numbers $-$ thanks go to Joe, in the comments, for pointing this out to me.
Some older questions, e.g. here and here, have some brilliant answers.

I'd be glad to know of more such resources; feel free to post any in the comments.

As no one has mentioned about this, here is an excellent intuitive video series (that shouldn't be missed) by Welch Labs as Imaginary Numbers are Real.

These are a set of 13 videos that beautifully help with the visualization of complex numbers from A Simple Complex Plane, all the way up to Riemann Surfaces. Here is something from the videos to get you interested.

Consider a Function $f(x)=x^{2}+1$.

Here is the plot of the function in the real x-y plane:

(Source: WolframAlpha)

Now according to the Fundamental Theorem of Algebra we should have n-roots for n-th degree polynomial but if you consider the graph for the given function it doesn't appear to intersect the x-axis right ?

Well, the thing is, we are not seeing it correctly and have not included a fundamental set of numbers : Complex Numbers which have both real and imaginary part but don't get confused yet as both the parts are quite real.The below GIF from the above videos beautifully plots the the function in the Complex plane (The vertical axis that comes out of the paper being the imaginary axis).

(Source: Welch Labs)

So you can see that the function actually intersects the x-axis but in a different dimension and that is the imaginary plane. This gives an excellent intuitive sense of why imaginary numbers are very real! I would recommend the users to view all the videos as there's a lot of graphic animations for better visualization and understanding.


Laplace transformations for dummies Is there a simple explanation of what the Laplace transformations do exactly and how they work?  Reading my math book has left me in a foggy haze of proofs that I don't completely understand. I'm looking for an explanation in layman's terms so that I understand what it is doing as I make these seemingly magical transformations.

I searched the site and closest to an answer was this. However, it is too complicated for me.

There are beautiful video lessons at MIT Opencourseware. I'm particularly in love with this presentation of the Laplace transform.

Think that the laplace transformation is a kind of a machine, the machine eats function of t f(t) out comes F(s). you do a transformation from time to frequency. Inside the machine you have this integral expression that you already know. it is similar when you transform from one vector space to another. for instance you go from R to R^2


if $$ is a finite abelian group $()$ divides the exponent of $$, for each $\in $ Define the exponent of a group as $\sup o(a)$ ,where $a \in G$. Show that if $G$ is a finite abelian group $o(a)$ divides the exponent of $G$, for each $\in $
Here $o(a)$ denotes the order of an element $a$?
How can I show this? Can someone please show the right approach to attempt this question.
Thanks.

By the fundamental theorem of finitely generated groups, we can write $G = \bigoplus\limits_{i = 1}^n \bigoplus\limits_{j = 1}^{m_i} \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z}$ with inclusion maps $k_{ij} : \mathbb{Z} / p_i^{e_{ij}} \mathbb{Z} \to G$, such that the $p_i$ are distinct primes and for each $i$, the $e_{ij}$ are sorted in increasing order.
Let $M = \prod\limits_{i = 1}^n p_i^{e_{i m_j}}$. Consider the map $f: G \to G$ defined by multiplication by $M$. It's easy to see that $f = 0$. Thus, we see that for all $a \in A$, $o(a) | M$. And in particular, for all $a \in A$, $o(a) \leq M$.
Now note that for $x = \sum\limits_{i = 1}^n k_{i m_j}(1)$, we have $o(x) = M$.
Then we see that $M$ is the exponent of $G$ and that for all $a \in G$, $o(a)$ is a factor of $M$.


Prove that any set in (X,d) is either closed or open. Let $X = \mathbb{R^2}$, $d(x,y) = |x| + |y|$ for $ x \neq y$. Prove that any set in (X,d) is either closed or open.

For the solution:

$ A \subset (\mathbb{R^2},d)$ for A to be open $ \forall a\in A\ \exists\epsilon\ B(a,\epsilon) \subset A$

By choosing $\epsilon$ as $|a|/2 $ $\ B(a,|a|/2) = \{a\}$ Thus A is open.
I can do the same logic for X\A but I do not know how to do either step or if A is such that the open... step.

Note that every $x \neq 0$ is an isolated point: $B(x, |x|)=\{x\}$.

So if $0 \notin A$, $A$ is open. Otherwise the complement is open, so $A$ is closed.

First note that the sets $X$ and $\emptyset$ are both open and closed in a topological space $X$. Depending on your topology, every set in the topology might also be an open set, but in a metric space, this cannot happen.

Consider a subset $A\subset X$. Let $A$ be open. We can show $A$ is not closed.
Note since $A$ is open, this means $X\setminus A$ is closed. Since $A$ is an open set in a metric space, we can write it as an open ball $B_{\epsilon}(x)\subset X$ (since the topology is induced by the open balls). But then we have $X\setminus
 B_{\epsilon}(x)=\{\forall x\in X: d(x,p)\geq \epsilon\}$ which is a closed set. We can apply a similar argument to show that $A$ cannot be open if $A\subset X$ is closed.


How many possible invalid AIS message body combinations are there for a specific checksum? According to the CATB AIS Specification, there is a 2 digit checksum to check an incoming AIS message for validity:

Here is a sample AIS message:

  AIVDM,1,1,,B,177KQJ5000G?tO`K>RA1wUbN0TKH,0*5C

The actual AIS message body is the payload which is:

  177KQJ5000G?tO`K>RA1wUbN0TKH,0

And the checksum is on the end which is:

  5C

Each character placeholder can be a total of 64 different characters. The checksum, however, is tested using a weird XOR calculation which may skew the calculation using finite math. My question is, how many invalid message payload combinations exist for each checksum? That is, if the checksum only has 64^2 or 4,096 combinations, then there are X chunks of payload combinations that match each checksum, and since only 1 of those are valid per AIS message the others would constitute invalid messages (that would still pass the checksum).

You can view the payload as being 8 bit strings of length N where N is the length of the playload.  The first bit string being all the most significant digits, the next the second most significant digits etc.

A parity check will eliminate half of these possibilities. So the number of
possible passing strings will be $8(2^{n-1})$. Note many of the passing payloads will contain strings with characters outside the given alphabet.


Solve the equation $e^x = yx^n$ for $x$? I've been looking at this equation for some time now:

$$e^x = yx^n$$

where $n \in \mathbb{N}_+$.

I found I can transform it to:

$$e^x = yx^n$$
$$xe^x = yx^{n+1}$$
$$x=W(yx^{n+1})$$

Or to this:

$$e^x = yx^n$$
$$x = \ln(yx^n)$$

If $yx^n$ is positive:

$$x = \ln(x^n)+\ln(y)$$

And if both $x^n$ and $y$ are positive:

$$x = n\ln(x)+\ln(y)$$

But I didn't find anyway to solve these for $x$ from there.

$$
yx^n=e^x
$$
First, divide by $ye^x$
$$
x^n\frac{1}{e^x}=\frac 1y
$$
Then, take the $n$-th root, as suggested by Wojowu in comments:
$$xe^{-\frac xn}=y^{-\frac 1n}$$

Now, multiply by $-\dfrac 1n$ in order to use $W(x)e^{W(x)}=x$:
\begin{align*}-\frac xne^{-\frac xn}&=-\frac{y^{-\frac 1n}}{n} \\
-\frac{x}{n}&=\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right) \\
x&=-n\mathop W\left(-\frac{y^{-\frac 1n}}{n}\right)
\end{align*}
and the equation is solved.

You can not solve the above equation explicit in the form $x=f(y)$


Sets such that inner measure equals outer measure is a sigma algebra Let $\mu$ be a measure on a $\sigma$-algebra $\mathcal{A}$ in $\mathcal{P}(\Omega)$ such that $\mu(\Omega)<\infty$. Define inner and outer measures as follows.

$\mu^*(E)=\inf\{\sum_{n=1}^{\infty}\mu(A_n): A_n\in\mathcal{A}, E\subset\bigcup_{n=1}^{\infty}A_n\}$

and

$\mu_*(E)=\sup\{\mu(A): A\in\mathcal{A}, A\subset E\}$.
I need to prove that the set $\bar{\mathcal{A}}=\{E\subset\Omega: \mu_*(E)=\mu^*(E)\}$ is a $\sigma$-algebra.

First, I tried to prove it in a direct manner. But I got stuck.

After trying some other exercises I thought, $\bar{\mathcal{A}}=\{E\subset\Omega: \mu_*(E)=\mu^*(E)\}=\{E\subset\Omega: \exists A,B\in\mathcal{A}, A\subset E\subset B, \mu(B\setminus A)=0\}$. Is this true?

Should I prove the first one direct, or can I prove that the second one is a $\sigma$-algebra?

If $A_1,A_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n$ and $B_n=A_n-\bigcup_{i=1}^{n-1}A_i$ then $B_1,B_2,\dots$ is a sequence in $\mathcal A$ with $E\subseteq\bigcup_{n=1}^{\infty}A_n=\bigcup_{n=1}^{\infty}B_n$.

Setting $B=\bigcup_{n=1}^{\infty}B_n$ we find that $B\in\mathcal A$ with $$\mu^*(E)\leq\mu(B)=\sum_{n=1}^{\infty}\mu(B_n)\leq\sum_{n=1}^{\infty}\mu(A_n)$$

Then $\mu^*(E)<\infty$ and this also makes clear that for every integer $k$ we can find a set $C_k\in\mathcal A$ with: $$\mu^*(E)\leq\mu(C_k)\leq\mu^*(E)+\frac1{k}$$

Then $C:=\bigcap_{k=1}^{\infty}C_k\in\mathcal A$ with $E\subseteq C$ and $\mu^*(E)=\mu(C)$.

Also we can find a sequence $D_1,D_2,\dots$ in $\mathcal A$ with $D_1\subseteq D_2\subseteq\cdots\subseteq E$ and $\mu(D_n)\to\mu_*(E)$.

Then $D:=\bigcup_{n=1}^{\infty}D_n\in\mathcal A$ with $D\subseteq E\subseteq C$ and $\mu(D)=\mu_*(E)$.

Then also $\mu(C-D)=0$.

Proved is now that $\bar{\mathcal{A}}\subseteq\{E\subset\Omega: \exists C,D\in\mathcal{A}, D\subseteq E\subseteq C, \mu(C-D)=0\}$.

The side $\supseteq$ is quite evident.


If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$.
Also... $a,b$ and $n$ are natural numbers.
I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from there...

Because $n$ is coprime to $a$ and $b$, we know there are linear combinations that equal one. Let the coefficients be $x,y,z,w$ such that $ax + ny = 1$ and $bz + nw = 1$. Multiply these together:
$$abxz + axnw + bzny + n^2yw = 1 \cdot 1$$
$$ab(xz) + n(axw + bzy + nyw) = 1$$

So there's a linear combination of $ab$ and $n$ that is equal to $1$. Thus they are coprime.

There is a useful theorem for these kind of problems that says:

For $a,b \in \Bbb Z,a\neq 0, b\neq 0$ then $gcd(a,b)=1 \Leftrightarrow \exists m,n \in \Bbb Z : am+bn=1$

wich is not difficult to prove.

It is a useful exercise to prove or disprove this theorem when $gcd(a,b)=d$ and $am+bn=d$ where $d\ge 1$


Exercise: Evaluating integration $\int_{|z|=r} \frac{1}{(z-a)(z-b)}dz$, $|a|<r<|b|$ This is an exercise from Stein-Shakarchi's Complex Analysis: evaluate integration $$\int_{|z|=r} \frac{1}{(z-a)(z-b)}dz, \,\,\,\, |a|<r<|b|. $$

The problem I am facing is the following. It is sufficient to find $\int_{|z|=r} \frac{1}{z-a}dz$ and $\int_{|z|=r} \frac{1}{z-b}dz$ (and use partial fraction methd).

This exercise is in first chapter, where the author introduces the integration of $f$ over a parametrized smooth curve $\gamma$. However, I didn't find any theorem in first chapter applicble to evaluate this integration. I tried to evaluate it through parametrization $\gamma(t)=re^{it}$ for $0\leq t\leq 2\pi$. Then $$\int_{|z|=r} \frac{1}{z-a}dz=\int_0^{2\pi} \frac{rie^{it}}{re^{it}-a}dt$$. But I couldn't solve this last integration. Can you help me?

I have seen that this can be solved using some Cauchy's integration formua; BUT, this is taken in second chapter of the book, whereas this exercise is in first chapter.

Assigned

$$f(z) ≔ \frac{1}{(z-a)(z-b)},\space\space\space\space\space\space\space z\in\mathbb{C}\setminus\{ a,b\}$$

and given the parametrization

$$γ(t) ≔ re^{it},\space\space\space\space\space\space\space0≤t≤2π,$$

observe that $a$ and $b$ are single poles of $f$, holomorphic over its domain, with $a$ lying within the circle of boundary $γ$, oriented in counterclockwise direction.
Hence

$$∮_γ f = 2πi \space \mathrm{res}_a f = 2πi \space \lim_{z→a}(z-a)f(z) = \frac {2π}{a-b}i.$$


Distribution of $(XY)^Z$ if $(X,Y,Z)$ is i.i.d. uniform on $[0,1]$ $X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$

How is random variable $(XY)^Z$ distributed?

I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.

Hints:

The random variable $X$ is uniform on $(0,1)$ if and only if $-\log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $w\mapsto w\mathrm e^{-w}\mathbf 1_{w\gt0}$.
If $W$  is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.

Conclusion:

If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.

Using the definition of weak convergence, it is so easy. First, for any positive integer $k\ge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(\cdot)$, we can find a polynomial function $f(\cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $W\sim U(0,1)$.


Linear algebra and multivariable calculus practice exercises for khan-academy videos? I have completed linear algebra in khan-academy and I am going to watch the multivariable calculus lectures from there .There is no exercise in the linear algebra section. And I heard that linear algebra and multivariable calculus is the different way of looking at the same thing ,

so I am seeking for some practice exercises(possibly worked out so that I can check my soln) so that both my linear algebra skill and intuition behind the relation multivariable calculus become strengthen. Thanks in advance

There are problems and solutions in this course:

http://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/index.htm

See under the different units.

There is a website named Lemma run by one Pavel Grinfeld that has a course in linear algebra that is similar to Khan Academy's more developed material, in that it has short lecture videos interspersed with exercises that are graded automatically by the website.
I think Dr Grinfeld's linear algebra material is certainly good for its price ($0.00), although my notes from going through the first eight or so chapters note the following small number of issues:

§2.12, “Linear Subspaces of the Plane”, is confusing enough that I think it ought to be redone — for now, read the comment under the video.

Dr Grinfeld habitually abuses notation for null spaces, writing a linear combination with unknown coefficients to represent the set of all linear coefficients of its form. In §7.5, he attempts to address the confusion this caused one student who complained, but I would think it better to bother to use set-builder notation and thus to avoid the problem altogether, as do three of the four freely-available linear algebra textbooks recommended by the American Institute of Mathematics (the one that doesn’t omits null spaces entirely).

Dr Grinfeld, and perhaps his students, likely would make fewer mistakes in Gaussian elimination were he to use augmented matrix notation, as is used in the said three textbooks.


How do I find the probability of these independent events The following is taken from the ETS math review for the GRE:

Let A, B, C, and D be events for which P(A or B)=0.6, P(A)=0.2, P(C or D)=0.6, and P(C)=0.5 The events A and B are mutually exclusive, and the events C and D are independent.
Find P(D).

I'd appreciate it if someone could show me how to solve this. The answer is given as 0.2, but I don't know how it's arrived at. Thanks!

Using the independence of $C$ and $D$ we have that
$$
P(C)P(D)=P(C\cap D)=P(C)+P(D)-P(C\cup D).
$$
Use this to find $P(D)$. You don't need to involve $A$ and $B$ to find $P(D)$.

P(c∩d)= 0 as they are independent so

p (c∪d)= p(c)+p(d)-P(c∩d)

so p(d)= 0.6-0.5

p(d)=0.1


"Slow" and "fast" rates of convergence I have recently read about convergence and divergence. However, I am having trouble understanding how something can converge/diverge "slowly" or "fast". If you sum up two series (that converge to the same number) infinitely, they will converge, not at any particular rate, but just to the number- this is how I see it.

So what does it mean for a series to converge (or diverge) slowly?

Suppose $\sum x_n$ converges. Given a positive $\varepsilon$; let $S:(0,\infty)\to \Bbb N$, $S(\varepsilon)$ give the smallest $N$ such that $|x-\sum^N x_n|<\varepsilon$. This measures convergence, in some sense.

Suppose $\sum x_n$ diverges to positive infinity. Given a positive $M$, let $S:(0,\infty)\to \Bbb N$, $S(M)$ give the smallest $N$ such that $|\sum^N x_n|>M$. This measures divergence, in some sense.

As an example, if $x_n=1/n$, then $S(N)\sim e^N$, this says that the divergence is very slow. That is, the larger $S(N)$ is, the slower the series diverges.

For convergence, we're worried at what happens with $S(\varepsilon)$ as $\varepsilon\to 0$. For $S(M)$; we're interested at what happens as $M\to\infty$.

Think about those two series (we'll call them S1 and S2) you mentioned that converge to the same number. Let say that both series get to that number after at least 50 terms in each series. So, at 50 terms, the sum you have at that point in S1 will be different than the sum of S2 at 50 terms. The series with the sum at 50 terms that is closer to the convergence point is the series that is approaching it faster at 50 terms.


Fourier transform of even/odd function How can I show that the Fourier transform of an even integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is even real-valued function? And the Fourier transform of an odd integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is odd and purely imaginary function?

Let $f: \mathbb{R} \to \mathbb{R}$ be an integrable function and let $\hat{f}$ denote its Fourier transform, i.e.
$$
\hat{f}(\xi)=\int_\mathbb{R}e^{ix\xi}f(x)dx.
$$
We have
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{-ix\xi}f(x)dx=\int_\mathbb{R}e^{iy\xi}f(-y)dy.
$$
If $f$ is even then
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{iy\xi}f(y)dy=\hat{f}(\xi),
$$
i.e. $\hat{f}$ is an even real-valued function.

If $f$ is odd then
$$
\overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}-e^{iy\xi}f(y)dy=-\hat{f}(\xi),
$$
i.e. $\hat{f}$ is an odd purely imaginary function.

The FT of even functions are also even; The FT of odd functions are also odd.
The real part of the FT of a real function is even; The imaginary part of the FT of a real function is odd.

So the Fourier Transform $F(\omega)$ of a real and even function $f(x)$ must satisfy both:

Because $f(x)$ is even: $F(\omega)$ is even (for both real and imaginary parts)
Because $f(x)$ is real: the real part of $F(\omega)$ is even, and the imaginary part is odd

Now for the imaginary part of $F(\omega)$ to be both even and odd, it must be zero, thus $F(\omega)$ is real-only.


line integral hard function to differentiate $$\int_\gamma \frac{(x^2+y^2-2)\,dx+(4y-x^2-y^2-2) \, dy}{x^2+y^2-2x-2y+2}$$

where $\gamma = 2\sin\left(\frac{\pi x}{2}\right)$ from $(2,0)$ to $(0,0)$.

I think it should be a shortcut to this problem that I cannot see , if that is not the case I will keep trying to simplify it .

Thnaks in advance

Complete the squares in the denominator $(x-1)^2+(y-1)^2$ and change the variables $x-1\mapsto x$ and $y-1\mapsto y$. You get the vector field
\begin{align}
&\left[\frac{x^2+2x+y^2+2y}{x^2+y^2},\frac{4y-x^2-2x-y^2-2y}{x^2+y^2}\right]=\\
&=\left[1+\frac{2x}{x^2+y^2},-1+\frac{2y}{x^2+y^2}\right]-2\left[\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right].
\end{align}
The first term is conservative (with an easy potential), the second term has zero curl (easy to check).

P.S. Actually the second term has a potential too if you restrict you domain e.g. to the plain without the negative $y$-axis, but the potential is less obvious (easier to get in the polar coordinates).

Hint: Let $I$ denote the integral. Note that the curve can be parameterized using curve using

$$x(t) = t$$
$$y(t) = \sin\left(\frac{\pi x}{2} \right)$$

for $0 \leq t \leq 2$

Thus, separate your integral into two different integrals:

$$\int_\gamma \frac{x^2 + y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dx \qquad \mathrm{and} \qquad\int_\gamma \frac{4y - x^2 - y^2 - 2}{x^2 + y^2 - 2x - 2y + 2} dy$$

whose sum is equal to $I$.


When the primitive of a function is the function itself, then $f(x)=0$ (using the FTC)? Let f: R → R be a continuous function such that:
$f(x)=\int_{0}^{x}f(t)\,dt$
Show that $f(x)=0$; for all $x$.
I did it like this:
According to the statement, $ f (x) $ is already the primitive of $ f (x) $, so $ f (x) $ is equal to $ F (x) $. By the Fundamental Theorem of Calculus we already know that $ F´ (x) = f (x) $ just like $ f (x) = F (x) $ we have $ F´ (x) = f´ (x) $ and therefore $ f (x) = f´ (x) $. I don't know how to show that it's zero.

As $f$ is continuous, all its antiderivatives are differentiable and the differential equation
$$f'(x)=f(x)$$ holds.
We know that the general solution of this ODE is $f(x)=ce^x$ (including $c=0$, from $f'(x)=f(x)=0$).
Now as $f(0)=0$, only the solution with $c=0$ is possible.

If $f(x)=f'(x)$ then $\frac{f'(x)}{f(x)}=1$. Integrating, you get that
$\ln f(x)=x+K$ whence $f(x)=ce^x$ for some $c\in\Bbb R$.
But you also know that $f(x)=\int_0^xf(t)dt$, so $f(0)=\int_0^0f(t)dt=0$.
Putting altogether $0=f(0)=ce^0=c$ and you'll get that $f(x)=0$.
EDIT: Thh above argument works whenever $f\ne0$. Observe that, in fact, $c\ne0$ but $f(x)=0$ is a trivial solution of $f'(x)=f(x)$. So we can write that all solutions of $f'(x)=f(x)$ are $f(x)=ce^x$ with $c\in \Bbb R$.


Area swept by a moving line Consider a line segment on an $\Bbb R^2$ plane with one end (call it $p_1$) fixed along x-axis and another end (call it $p_2$) fixed along y-axis. The point $p_1$ starts at $(0,0)$ and the point $p_2$ starts at $(0,1)$. I allow the line segment to move with one of the following constraints:

$1$. The point $p_1$ moves to the right with the constraint that the length of the line segment is fixed. The line segment stops moving when $p_1$ reaches $(1,0)$.

$2$. The point $p_1$ moves to the right and the point $p_2$ moves downward with the constraint that the distances travelled by each of the points are equal at any moment, i.e. when $p_1$ moves by $(t,0)$, $p_2$ shall move by $(0,-t)$. The line segment stops moving when $p_1$ reaches $(1,0)$.

In each of both cases, there is a region swept by the line segment. What are their areas respectively?

It may look like a simple integration problem, but I have absolutely no idea how to find the function for the curve above the region. Do I miss out something obvious, or is it really a non-trivial problem?

It's not trivial, but it's not that hard either.  Each family of line segments crosses each vertical $x=x_0$ (for $x_0\in[0,1]$) at a range of heights $\{ y(x_0) \}$; you want the function $f(x)$ such that $f(x_0)=\sup\{y(x_0)\}$.

In the first case, you have line segments from $(0,\cos\theta)$ to $(\sin\theta,0)$, crossing $x_0$ at $\cos\theta - x_0 \cot\theta$.  This is maximized when
$$
-\sin\theta + x_0\csc^2\theta=0,
$$
or $\sin\theta=x_0^{1/3}$; so $f(x)=\cos\theta - \sin^3\theta\cot\theta=\cos^3\theta=(1-x^{2/3})^{3/2}$.

In the second case, you have line segments from $(0,t)$ to $(1-t,0)$, crossing $x_0$ at $t - x_0 t/(1-t)$.  This is maximized when
$$
1 - \frac{x_0}{1-t} - \frac{x_0 t}{(1-t)^2}=0,
$$
or $t=1-\sqrt{x_0}$; so $f(x)=(1-\sqrt{x})- x(1-\sqrt{x})/\sqrt{x}=(1-\sqrt{x})^2$ in this case.

Isn't this $1 - \frac{\pi}{4}$ ?

The line describes a quarter-circle of radius 1 centred at $(1,1)$, which sits in inside a unit square.

So the area underneath the 'slipping ladder' is a unit square minus a quarter of a unit circle.


What are $S$ and $T$ in $C(T) = \{S \in L(V,V)\mid ST=TS\}$? Let $C(T) \subseteq L(V, V)$ be the subset of all linear operators which commute with $T$, i.e.:
  $$C(T) = \{S \in L(V, V ) \mid ST = TS\}$$
  Prove that: $C(T)$ is a subspace of $L(V, V)$, $C(T)$ contains the identity operator $I_V$, $C(T)$ is closed under composition of operators, and $T$ is an isomorphism if the linear map is a nonzero linear map.

Am I supposed to assume that $T$ is also in $L(V, V)$ so that $S: V \to V$ and $S = T$ or that $S$ is an inverse of $T$? I feel like if I know what $S$ and $T$ are, I can solve the question. Please help.

You need to show that $C(T)$ is closed under addition and scalar multiplication and also contains the identity. Clearly, $C(T)$ contains the identity since the identity commutes with all operators.  To show that $C(T)$ is closed under addition, let $U, V \in C(T)$.  Then, $(U + V)T = UT + VT = TU + TV = T(U + V)$.


Combinatorial proof involving sum of factorials I need help with this combinatorial proof:
$1\cdot1!+2\cdot2!+\cdots+n\cdot n!=(n+1)!-1$
So far I came up with this:
Let S be a set of numbers $1, 2, \ldots, n+1$
So LHS could be: How many permutations of length $k$ there is over set $S$ such that number $1$ is not the first in permutation. $k$ can be $2, 3,\ldots, n+1$. But the RHS gives me trouble...
Thanks.
(Sorry for the poor formating, I am not good with HTML)

$n\cdot n!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ without $n+1$ being first.

$(n-1)\cdot (n-1)!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ with $n+1$ being first and $n$ not second.

$(n-2)\cdot (n-2)!$ counts the number of ways to order $\{1,2,3,\ldots,n+1\}$ with $n+1$ being first and $n$ second and $n-1$ not third.

Etc...

The RHS is all the ways to order $\{1,2,3,\ldots,n+1\}$ without all the elements in descending order.

hint: $n\cdot n! = (n+1)!-n!$. Can you continue?


How to plot $\frac{2x-4}{1+x^2}$ Can someone suggest me how to plot this function (without any software)?

$$\frac{2x-4}{1+x^2}$$

I have found out such points as:

$(0,-4)$ and $(2,0)$
I know that $y = 0$ is horizontal asymptote (when $x \rightarrow\pm\infty$)
I have found out also extrema points and those are $(x=2\pm\sqrt(5))$
And some other stuff, but not sure how to draw a plot

Here I gave a method without Calculus.

Let $$y=\frac{2(x-2)}{(x^2+1)}.$$ Note that $$\lim_{x\to\pm\infty}y=0$$ $y$ has only one zero $x=2$ and $$y\ge  0\iff x\ge 2.$$ Also $$yx^2-2x+(y+4)=0$$ Since we consider only real $x$ values discriminant of this equation is non-negative. $$\Delta_x=2^2-4y(y+4)\ge 0$$ $$y^2+4y-1\le 0$$ $$(y+2)^2-5\le 0.$$ Therefore $y$ is always in the range $$-2-\sqrt5\le y\le -2+\sqrt5.$$ Find $x$ values corresponding to $y=-2\pm\sqrt5.$
Now you can draw the graph of your function using these details. your graph will looks like

Find inflection points $ y^{''} =0 $  with tangent slopes $y^{'}$ , maximum/minimum points and the asymptotes by standard procedure ( divide by $ x^2, x\rightarrow  \infty $  etc. ), find coordinates of points for $ y=0, x=0, $ joining them with a smooth curve. Connect the curve at a few points between these points calculated by hand.


How to evaluate the following limit? $\lim\limits_{x\to \infty}\big(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\big)$ How to evaluate the following limit? $$\lim _{x\to \infty }\left(\sqrt[3]{x^3+x^2}-\sqrt[3]{x^3-x^2}\right).$$

What I first did is multiply by the conjugate, but having trouble finishing the problem. I believe the final answer is $\tfrac23$.

Edit: Here's what I got so far

$$\lim _{x\to \infty }\left(\frac{2x^2}{\sqrt[3]{x^3+x^2}+\sqrt[3]{x^3-x^2}}\right)$$


Equation of plane given a point and contain line of intersection of 2 plane What is the equation of the plane that passes through the point $(−1, 2, 1)$ and contains the line of intersection of the planes $x + y − z = 2$ and $2x − y + 3z = 1$?

How can i find the equation of the plane?

Hint:

Consider the pencil of planes that have this line in common:

$$\lambda(x+y-z-2)+(1-\lambda)(2x-y+3z-1)=0.$$

By plugging the coordinates of the given point, you will determine $\lambda$.

First, find the line of intersection.  You know that the line of intersection is parallel to the plane.  Then, find the vector orthogonal to this line.  Use this orthogonal vector and the point given to construct the equation of the plane.


For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m<x<n$ For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m<x<n$.

Assume that $m < n$. I apply first Archimedian property to get $x < n$. But I need to show that $m < x$ to complete proof?

Could you provide some hints?

From the wording, it looks as if we can assume that the reals have the Archimedean property. So there is a natural number $n$ such that $x\lt n$. It remains to deal with the "$m$" part of the assertion.

If $x\ge 0$, let $m=-1$. Then $m\lt x\lt n$.

Now suppose that $x\lt 0$. There  is a natural number $w$ such that $|x|=-x\lt w$. Let $m=-w$. Show that $m\lt x$.

Look at $[x]-1$ and $[x]+1$ where $[x]$ denotes the integer part of $x$.


If $T: R^2 \rightarrow R$ is a linear transformation then $T: R^2 \rightarrow R$ is a linear transformation such that $T(v) \neq 0$ for some $v$  in $R^2$ then T is

one-to-one
onto
$nullity(T) = 0$
none of the above

My first thought was that since $dim(R^2) > dim(R)$ then T cannot be one-to-one. I was inclined to put none of the above, but I feel $T(v) \neq 0$ is a crucial piece of information. Is this function onto? If so why?

$T$ is onto. Indeed, let $v\in R^2$ be such that $T(v)\ne 0$.
Let $\alpha$ be any real number. Then
$$
T\left(\frac{\alpha}{T(v)}v\right)=\frac{\alpha}{T(v)}T(v)=\alpha.
$$

Nullspace is not empty so the dimension of Nullspace is not zero.
By rank and nullity theorem $\text{dim}(N(T)) + \text{dim}(R(T)) = \text{dim}(R^2) = 2$.
But $dim(N(T))$ is also not $2$ because not for all $v$, $T(v) = 0$.
Thus $\text{dim} (N(T)) = 1$ which implies $\text{dim}(R(T)) = 1$.
since R(T) is subspace of $R^2$ and $\text{dim}(R(T)) = \text{dim}(R)$ then $R(T) = R$ which means that $T$ is onto.


Does $〈Ax,y⟩=〈x,A^*y⟩$ hold for any inner product? Consider $\Bbb C^n$. Let $A^*$ be the conjugate transpose of matrix $A$, I see in many materials $A^*$ is also called the adjoint of $A$. I am now confused by that does $〈Ax,y⟩=〈x,A^*y⟩$ hold just for the standard inner product (whose proof is simple), or any inner product? Thanks!

It does not hold for every inner product. The analogous situation in the real plane illustrates this. Consider
$$
H = \pmatrix{2 & 0 \\ 0 & 1}
$$
and the inner product $x \cdot y = x^t H y$.
Let
$$
A = \pmatrix{0 & 1 \\ 1 & 0}
$$
$x =  \pmatrix{1\\0}$ and $y =  \pmatrix{0\\1}$.

Then $Ax \cdot y = 1$, but $x \cdot A^t y = 2$.

Since the real plane is a subspace of the complex plane, this suffices as a counterexample in $\Bbb C^2$ as well (at least if you extend the definition of the dot product by writing $x \cdot y = x^t H \bar{y}$).

$〈Ax,y⟩=〈x,A^*y⟩ $ this holds for any inner product ! when you assume $A^*$ is adjoint of $A$ in space X equipped with inner product of $〈.,.⟩$

This is how adjoint operators  defined in finite dimension.


Exponential simplification help Why is it true that
$$
\frac{exp(x)}{1+exp(x)} = \frac{1}{exp(-x) + 1}
$$
Why did we negate $x$?
I tried to divide the left-hand side with $exp(x)$ and got
$$
\frac{exp(x)}{exp(x) + (exp(x))^2}
$$
but I cannot get forward.

Perhaps you should tell us what $\exp(x)$ is defined as.
The simplest explanation is that you know there is a number $e$ and you know what raising a base to a power is and $\exp(x)$ is simply defined as $\exp(x)= e^x$.  And therefore $\exp(-x) = e^{-x}= \frac 1{e^x} = \frac 1{exp(x)}$ (and we know that $e^{-x}= \frac 1{e^x}$ because $e^{-x}e^{x} = e^{-x+x} = e^0 = 1$).
But that assume you know what raising numbers to a power means.  We usually use the concept of $\exp(x)$ to define what raising a number to a power is by $b^x = \exp(x\cdot \ln b)$ where $\ln b$ is defined as $\int_1^b \frac 1t dt$ and $\exp(x) = \ln^{inverse}(x)$ (i.e. $\exp(x)$ is the value $k$ so that $\ln (k) = x$.
Then its a matter of proving that $\exp(x+y) = \exp x\cdot \exp y$.  If we know that then we have to know that $\exp(x)\exp(-x) = \exp(x-x)=\exp(0) = 1$ and therefore $\exp(-x) =\frac 1{\exp(x)}$.
I may provide a proof of that in a few hours.


Function to apply to a linearly increasing positive real number to reach an arbitrary limit I've got a friend who is making a browser game and he's trying to figure out how to make a function that acts like a logarithm in that it returns higher values quickly but eventually mellows out and reaches a limit.

For example, say someone invests "points" into a characters speed. This person should get less and less speed boost for every point they put in. 1 point gives maybe .10 back as a direct modifier on speed things (this character would now be 10 percent faster than other characters with no points).  2 points might give back .19 total(.1 + .09), 3 would be .26 total(.1 + .09 + .07), etc (only an example, not the expected results).

Is there a way someone can help us understand a good way to come about this so that we can play with the numbers a bit and tweak bonuses for balance?

I have a feeling someone will have to help me with tags and whether this is the right place to ask this question :)

You mention log but I think a reflected and shifted exp function might work better for what you describe. For example

$$f(x)=1-e^{-x}$$

More generally you may vary this idea introducing and adjusting some parameters, say $a,b,c$ and consider the function

$$f(x)=a-b e^{-c x}$$

A logarithmic function will do. In fact, any increasing, concave-down function will do. That is, the function must have a positive but decreasing derivative. See Diminishing Returns.

I'm not here to give you an exact function. You'll have to tweak it yourself to get the numbers right. However if you start with $f(x) = \ln(x)$ and provide necessary transformations (translations, dilations, etc.) then you should be able to tweak it as necessary.

Go to Desmos.com, type in
$$ y= a \ln (bx - h) + k $$
and add sliders for $a$, $b$, $h$, and $k$. Play around with them and see what happens.

I might add that if you're looking for a linear decrease in returns, your function must be quadratic. For example, if your returns are
$$ 1.0, 0.9, 0.8, 0.7, \dots $$
then this means your function's derivative is decreasing at a constant rate (it is linear) so your original function must be of the form
$$ y = ax^2 + bx + c $$
where $a < 0$.


Apparent paradox with power series convergence I'm having problem figuring out how these three facts are mutually consistent:

Call a set $S \subset \mathbb{C}$ to be concirclic if for some real number $R$, if you let  $A$ be the (open) ball centered at zero with radios $R$ and $\overline{A}$ be the closure of $A$ then $A \subset S \subset \overline{A}$.  For any power series $f(z) = \sum a_i z^i$, let $S_f := \{ z \in \mathbb{C} | f(z) \text{ is finite } \}$. Then $S$ is concirclic
If you let $f(z) = \sum (z-.5)^i$ then $S_f = \{ z \in \mathbb{C}, |z - .5| < 1 \}$. In particular, $S_f$ is not concirclic.
If you define $b_i$ as the sum of coefficients of $z^i$ in $\sum_{k \geq 0} (z-.5)^k$, then $b_i$ is well defined (i.e is not divergent). Then $f(z) = \sum b_i z^i$, and hence $S_f$ is concirclic.

I'm making some very stupid mistake but can't figure it lol

The short answer is you cannot arbitrarily rearrange/regroup a series and always expect it to stay convergent/divergent. For example, $0+0+0+\dots$ converges, but $(1-1)+(1-1)+\dots=1-1+1-1+\dots$ diverges. You start with $\sum_{k\ge 0}(z-1/2)^k$, which is absolutely convergent on an open circle of radius $1$ centered at $1/2$. After some rearrangements, you end at $\sum_i b_i z^i$. This is still convergent on the original set, and equal to the original series there, but is also convergent on the larger circle which is centered at the origin with radius $3/2$.

In detail, you have
$$
\sum_{k\ge 0}(z-1/2)^k=\frac1{1-(z-\frac12)}=\frac23\cdot \frac1{1-\frac23z}=\sum_{i\ge 0}\underbrace{(2/3)^{i+1}}_{b_i}z^i
$$

The fact that your rearrangements lead to a correct answer in the original region of convergence (the circle of radius $1$ centered at $1/2$) suggests that they can be justified in this region, but for now I cannot see how.

You defined $S_f$ as the domain of convergence of Taylor's series centered at $0$, and such domain is indeed concirclic (upd: if radius of convergence is finite). But in 2 you take Taylor series centered at $\frac{1}{2}$ - and it's domain of convergence in general differs from domain of convergence of series centered at $0$. So $S_f \neq \{z: |z - 0.5| < 1\}$

If you calculate $b_i$ you will note that $b_i = \frac{2}{3} \cdot \left(\frac{2z}{3}\right)^i$, and $\sum b_i z_i$ converges at, say $z = -1$, while original series diverges there.


Gluing axiom of a TQFT In the book, Lectures on tensor categories and modular functors by Bakalov and Kirillov they construct a TQFT.

When they come to prove the gluing axiom, they just mention that "...This statement is of purely topological nature, and we omit its proof." And I don't see how to prove it.

Basically, they claim that if we cancel two coupons of a special link we get the same manifold  via surgery. (See the picture from page 87 of the book)

Could someone provide me a proof of the gluing axiom or give me a reference containing a complete proof? I read a Turaev's book but that proof is also not easy to understand.

Given topological spaces $\rm X, Y$ and $\rm Z$ with two continuous functions $f : \rm Z \to X$, $g : \rm Z \to Y$, it is always possible to build the push-out  $$ \rm X \coprod_{Z} Y$$ which is obtained by gluing $\rm X$ and $\rm Y$ along $\rm Z$.

It is defined as the quotient of the disjoint union by the equivalence relation $f(z) = g(z)$.

In your examples, all the gluings are just with one point i.e $\rm Z = \{\ast\}$.


Dividing 100% by 3 without any left In mathematics, as far as I know, you can't divide 100% by 3 without having 0,1...% left.

Imagine an apple which was cloned two times, so the other 2 are completely equal in 'quality'. The totality of the 3 apples is 100%. Now, you can divide those 3 apples for 3 persons and you will get 100% divided by 3 and none left.

Is this because                                    1: mathematics is not real                   2: there is no 1 or 2, and it's in fact just an invention for measurements? So,  in the division of 100% by 3 WITHOUT any left, is NOT  accurate?

So in mathematics, as far as I know, you can't divide 100% by 3, without having that 0,1..% left....

No! we can in Math and also in real life. This is similar to ask can we divide $1$ into 3 parts? And the answer is again yes. $$1÷3=\frac{1}{3}$$ because adding $\frac{1}{3}$ three times give $1$.

Consider 3 sticks of same length. Align these three sticks and call the total length as 1 unit. Now the length of any of the individual stick is exactly $\frac{1}{3}$ unit.

Moreover you can use the number system having base 3 to remove the (apparent) incompleteness of the base-ten expression $1÷3=0.333333333...$ In the number system having base 3, the number '3' itself will be written as $10$ and the number $1$ as it is.

The division $1÷3$ is now $1÷10$ which is equal to $0.1$. so you see writing (in base ten) $100÷3=33.333333$ does not mean that we cannot divide $100$ into three equal parts. What it means is that we are using a number system having $10$ base so we cannot write $\frac{100}{3}$ in decimals.

Here, in a practical way – I understand all your concepts, sirs – I was using a simple calculator on my computer, and looks like mine's got a 10 decimal digit precision. Which make this possible:

9 decimal "3s"
33.333333333 * 3
= 99.999999999

10 decimal "3s"
33.3333333333 * 3
= 100

If I divide 100 by 3, it'll give me 10 decimal "3s". And this expression was valid:
100/3 * 3 = 100


Gradient descent vs. Newton's method -- which one requires more computation? Broadly speaking, when numerically minimizing a d-dimensional objective function:

Gradient descent generally requires more iterations, but each iteration is fast
(we only need to compute 1st derivatives)

Newton's method generally requires fewer iterations, but each iteration is slow
(we need to compute 2nd dervatives too)

My question is: in terms of the total amount of computation required, which one generally ends up being faster -- Newton's method or gradient-descent? Does this depend on $d$? How?
If this is a better question for another site please let me know.
Minor update
If it matters for the sake of comparison, let's assume the function is convex and "typical" (i.e. I'm not going to explicitly choose a function that exhibits the worst-case behavior of either algorithm). I'm just trying to understand what the rule of thumb is regarding the performance of each method.

I think this depends a lot on the structure of the function you are optimizing (duh). In general non-convex cases, both algorithms have the same worst-case complexity for the number of iterations taken to drive the norm of the gradient below some given tolerance. Not sure what this means in terms of actual computation time for instances because the constant factors come into play.

You can look at this paper by Gould et al and references therein for more details.

I think a good thumbrule is - if your problem is convex and you have a reasonably good initial guess, Newton's (or Quasi-Newton) is usually much faster in practice.

In general a function one is trying to find a zero of is given by an explicit formula so the calculation of the derivative is a single-occurrence event and need not to be incorporated into the iteration. For the advantages of Newton's method see http://en.wikipedia.org/wiki/Newton%27s_method


Calculating probability of an 11-card straight My friends and I were playing a card game where you're dealt 11 cards (it was one of those "family games", so I'm not sure what we were playing. It was kind of like Canasta).

Anyway, on one of my deals my friend was dealt an 11-card straight, 3-K. Before dealing, I shuffled 7 times.

What are the odds of this happening? I never took probability courses, but it seems exceedingly rare from a layman's perspective.

I'm interested in the result, but I'm also very interested to see how the result is produced! I'm just starting to get back into math after a series of awful high school teachers.

Thank you!

For just a 3-K straight, the probability is
$$\frac{4^{11}}{\binom{52}{11}}=6.944×10^{-5}$$
The numerator arises because there are 4 possible suits for each card in the straight; the denominator is just the number of ways to deal an 11-card hand from the full deck.

If any 11-straight was allowed, the above probability would be multiplied by four because there are four such straights (A-J, 2-Q, 3-K, 4-A), leading to a probability of $2.778×10^{-4}$.

Number of ways you can choose $11$ cards from a deck of $52$ cards is $\binom{52}{11}=\frac{52!}{11!41!}$. Number of ways those a set of $11$ cards contains $3$ K's, is (the number of ways of choosing $3$ K's from the $4$ K's)$\times$(the number of ways of choosing $8$ cards from the remaining $(52-4)$ cards)$=\binom{4}{3}\times\binom{48}{8}$. Thus the required probability is $$\frac{\binom{4}{3}\binom{48}{8}}{\binom{52}{11}}$$


Find the inverse Laplace transform of $\frac{1}{1+3^\sqrt{s}}$ How can I find the inverse Laplace transform of $1/(1+3^\sqrt{s})$?
or does it exist?

The inverse transform does not exist.  Consider an integration of the complex integral

$$\oint_C dz \frac{e^{t z}}{1+e^{a \sqrt{z}}} $$

where $a \gt 0$ and $C$ being a Bromwich contour deformed to go around the branch cut along the negative real axis and around the branch point at $z=0$.  By Cauchy's theorem, we find that the inverse transform may be written as

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{1+e^{a \sqrt{s}}} = \frac1{2 \pi} \int_0^{\infty} dx \, e^{-x t} \tan{\left ( \frac{a}{2} \sqrt{x} \right )}$$

The integral on the RHS does not converge in any sense due to the poles at $x=(2 k+1)^2 \pi^2/a^2$ for nonnegative integer $k$.  Whether the inverse transform exists as a Cauchy principal value, I leave as an exercise for the reader.


Lipschitz function and incomplete range Find a Lipschitz function $f:\mathbb{R} \rightarrow \mathbb{R}$ that has incomplete range.

What is incomplete range of a function? Does it has to do with completeness of a metric space?

Any help would be appreciated.

If by incomplete range we mean that the image $f(\Bbb R)$ of $f:\Bbb R \to \Bbb R$ is not a complete subset of $\Bbb R$, then the function

$f(x) = \tanh x \tag 1$

provides an example.  It is everywhere continuously differentiable, which is sufficient for it to be
Lipschitz; but

$f(\Bbb R) = \tanh(\Bbb R) = (-1, 1), \tag 2$

which is incomplete as there are sequences $\alpha_i \in (-1, 1)$ such that

$\alpha_i \to -1, \; \text{or} \; \alpha_i \to 1 \; \text{as} \; i \to \infty, \tag 3$

but

$-1, \; 1 \notin (-1,  1). \tag 4$


Is $x^{\frac{3}{2}}$ a differentiable function? When we speak of differentiable function, do we mean $C^1$ function?

I know $C^1$ means continuously differentiable.

Is $x^{\frac{3}{2}}$ a $C^1$ function or $C^2$ function? Its second derivative blow up at $x = 0$. Is it $C^1$?

Does Taylor series exist for $x^{\frac{3}{2}}$?

Yes. The domain of your function is $x\ge 0$ and the derivative $\frac 32\sqrt x$ is well defined in this domain.

Assume $x>0$. Then
$$
f(x)=x^{3/2},\quad f'(x)=\frac32 \sqrt{x},\quad f''(x)=\frac3{4\sqrt{x}}.
$$ Observe that all these functions are continuous over $(0,\infty)$, that is $f$ is $C^1(0,\infty)$. In fact, $f$ is $C^\infty(0,\infty)$.


Method of Variation of Parameters - Assigning zero works? I have yet to find a decent answer on this, and so I don't think this question is inappropriate. Also, this question is mainly meant for people that are very familiar with this method.

In the method of variation of parameters method for solving the non-homogeneous (heterogeneous?) part of a second order linear differential equation - that is, one in the form $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$I'm still boggled with the exact purpose of setting parts of the first derivative of the particular solution equal to solution.

I will explain what I'm talking about here.

Suppose we have a $2^{nd}$ order linear non-homogeneous ODE,  $$y''(t)+a(t)y'(t)+b(t)y(t)=g(t)$$
Also, suppose we've already obtained a general solution for the homogeneous version of the ODE, $y''(t)+a(t)y'(t)+b(t)y(t)=0$, that's in the nice form $$y_h=c_1y_1+c_2y_2$$ To get the particular solution for the non-homogeneous part, we start off with the weird (but successful) method of variation of parameters - change the $c_1$ and $c_2$ to $u_1(t)$ and $u_2(t)$ respectively (hence, variation of parameters), and call that $y_p$.

$$y_p=u_1(t)y_1+u_2(t)y_2$$

The goal is to find out $u_1(t)$ and $u_2(t)$, and so we now get the $1^{st}$ and $2^{nd}$ derivatives of $y_p$ so we can slap them into the original ODE and hopefully solve for $u_1(t)$ and $u_2(t)$. The first derivative follows by chain rule:

$$y'_p = u'_1(t)y_1 + u_1(t)y'_1 + u'_2(t)y_2 + u_2(t)y'_2$$

Here is the part I don't understand.

In every textbook I've seen this in, the next step goes basically as follows:

  For no reason and no motivation (as far as we're concerned right now), we now set the following:
  $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$
  This will turn out to work in the end very nicely, and it will also make our computations much more simple.

I've seen full well that this does work out nicely, and that it does make the computations much more simple, but my question is why is this done, and why does it work out so nicely? Why does that equation $$u'_1(t)y_1 + u'_2(t)y_2 = 0$$ always have to be satisfied for particular solutions to $2^{nd}$ order linear non-homogeneous ODE's of this form?.

The explanation that you are looking for relies on the fact that $y_1$ and $y_2$, being linearly independent solutions to the homogeneous equation, form a basis for the solution space of the inhomogeneous equation. This means that they span the space, so that the solution vector $\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix}$ can be written as a linear combination of the vectors $\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix}$ and $\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$. Hence, for some functions $u_1(t)$ and $u_2(t)$: $$\begin{pmatrix}y_p \\ y_p^{\prime}\end{pmatrix} = u_1\begin{pmatrix}y_1 \\ y_1^{\prime}\end{pmatrix} + u_2\begin{pmatrix}y_2 \\ y_2^{\prime}\end{pmatrix}$$
We consider the two components:
$$y_p = u_1y_1 + u_2y_2$$
$$y_p^{\prime} = u_1y_1^{\prime} + u_2y_2^{\prime}$$
By differentiating the first equation and subtracting it from the second we get that
$$0 = u_1^{\prime}y_1 + u_2^{\prime}y_2$$
which is what you were looking for. In my opinion, the subtlety lies in the observation that the above vectors form a basis for the solution space. Perhaps someone has a good explanation of why that is the case.

the whosleon's answer is pretty good but I just wanna add something.
the particular solution is outside of homogeneous solution space, that means you cant write yp = c1y1 + c2y2. so you have to jump into another dimension by chaning either y1 or y2's dimension using u1 and u2. you could call u1 and u2 the dimension booster. (for example, consider x+y=0 the homogeneous solution is a line pass the (0,0) but if x+y = c , the line will move up and down. you can never jump when you only moving along the line. so you need boost the dimension (again since the homogeneous equal 0, it is impossible to get something other than 0 within solution space))
since y is function of t(or x or whatever you call it) and as well as u1 and u2. you can achieve any dimension you want. so yp can be expressed as yp = u1y1 + u2y2
same to the prime of everything. since the second order differential equation need y and y' to determine the whole thing(its like if you know postion and velocity of a object attached by spring you will determine the whole process) so y and y' are independent basis. it is nonsense to say you have to have 5 m/s when you at the origin of the spring, you can have whatever speed you want thats why they are independent.
Since y and y' are independent and form the basis of the solution, the differential equation doesn't care product rule anymore, thats why you can write
yp = u1y1 + u2y2
yp'= u1y1' + u2y2'
again the reason y and y' use the same u is they form a basis.
but in reality there is a product rule (ab)' = a'b + ab' the only way to obtain above is to set any u' = 0 to achieve the mapping process. so ui'yi = 0
I know its kinda painful to understand, the main point is to have independent basis combination you have to get ride of those term. maybe the spring example will help alot. if you know the physics, homogeneous mean no external force where non homo mean there is external force. when you have external force, you need a "dimension booster" to get to the "process"  that original physics process can never achieve.
im a physics guy so my explanation is bit physical but I think that will cover your question. Sorry for the formatting, im lazy.


Countability of removable discontinuities I saw a problem of proving that number of removable discontinuities in a function is countable. I was not able to prove it and tried many things. Can anyone do it?
Thanks in advance.

For $n\in \{1,2,\dots,\}$, let $E_n$ be the set of removable discontinuities $x_0$ where the difference between $f(x_0)$ and $\lim_{x\to x_0}f(x)$ is at least $1/n$. If we can prove $E_n$ is countable for all $n$, we will be done, since the set of removable discontinuities is the union of the sets $E_n$.
Given $x_0\in E_n$, let $\lim_{x\to x_0}=L$. There is a neighborhood around $x_0$ where points in the neighborhood are all with $\frac1{4n}$ of $L$. Therefore, they are all within $\frac1{2n}$ of each other. This means there cannot be any removable discontinuities with a gap of $\frac1n$ in that neighborhood, so this neighborhood is disjoint from the rest of $E_n$. Therefore, the points in $E_n$ are all isolated from each other, and therefore countable.

Let $x_{0}$ be removable discontinuity, then $$\lim_{x\to x_{0}^{-}}=\lim_{x\to x_{0}^{+}}=L$$then for $\epsilon=0.5$  $\exists \delta >0$ s.t $\forall x\in [x_{0}-\delta,x_{0}+\delta] \neq x_{0}$ we have:$$L-0.5 \leq f(x) \leq L+0.5$$ now from the density of the rational number there exists a rational number $q\in [x_{0}-\delta,x_{0}+\delta]$ and we can show that $x$ is the only discontinuity in this interval, thus we assign a unique rational number for every removable discontinuity, but since the rationals are countable so are the removable discontinuities


"A Queer Coincidence," riddle from Dudeney's book I'm stuck on a mathematical riddle from Dudeney's Amusement in Mathematics.

I will write here the text of the riddle and my attempt. I just don't get how to proceed further.

  Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards, Francis, and Gudgeon, were recently engaged in play. The name of the particular game is of no consequence. They had agreed that whenever a player won a game he should double the money of each of the other players—that is, he was to give the players just as much money as they had already in their pockets. They played seven games, and, strange to say, each won a game in turn, in the order in which their names are given. But a more curious coincidence is this—that when they had finished play each of the seven men had exactly the same amount—two shillings and eightpence—in his pocket. The puzzle is to find out how much money each man had with him before he sat down to play.

Attempt

I started to mind like this: first of all $2$ shillings and $8$ pence are $32$ pence ($1$s = $12$p).

Hence at the end they all have $32$ p, which means in total there are $224$ p since there are seven players.

Now, Assume the players are called $A B C D E F G$.

$A$ is the first and he possesses $a$ pence, hence the other ones, in total, have $R$ pence.

The first useful equation is trivial and its

$$a + R = 224$$

Now: $A$ wins first, hence he has to pay everyone else twice what they have.

But $R$ is what they all have, hence he pays $R$: after the first turn $A$ remains with $a - R$ pence.

But then he always loses (other wins), and each turn until the end $A$ gets his sum doubled. In this way the turn are described by:

$a - R$
$2(a-R)$
$4(a-R)$

$\cdots$

$64(a.R)$

After seven rounds, $A$ has $64(a-R)$ pence, which must be equal to $32$ since the text says it's what they all have at the end.

Trivially we conclude

$$
\begin{cases}
a + R = 224 \\
a - R = 1/2
\end{cases}
$$

Which means $a = 112.25$ which in farthings is $449$ f.

Now my problem is to understand how to get the sum of the other players. I tried with a similar method but I failed.

Any help?

It is essentially bookkeepping (or literally bookkeeping, since they are playing for money). Working backwards, each player has half as much as the subsequent round, except the "winner" of the next round additionally has half the total of $7\times 32=224$, so $112$ more than half.

Hint:  After $G$ wins they each have $32$, so before the last game the first six had $16$ and $G$ had $128$.  Then before $F$ wins they had ????


Radical İdeal-Nullstellensatz $f=y{^2}+2xy-1$ and $g=x^2+1$.Prove $<f,g>$ is not radical ideal.
Hint :What is $f+g$ $?$

I have an idea: neither $x$ nor $y$ belong the ideal.but $(x\cdot y)^{2}\in I$.Since
$x^{2}\in I$ ; thus $x\cdot y\in\sqrt{I}$.So $x+y\in\sqrt{I}$

To show that the ideal is not radical, you can exhibit some polynomial $p(x)$ so that $p(x)^n \in I$, while $p(x)$ is not.

hint: Can you show that $(x+y)^2 \in I$?

As for your work: how did you show that $x^2 \in I$?. If $x^2 \in I$, why would this imply that $xy \in I$? If $xy \in I$, how do you know that $x+y \in I$? Where is the contradiction?

you say $p(x)^n$ is radical but $p(x)$ is not.I say $x^2$.I have a theorem : An ideal $I$ is $radical$ if $f^m\in I$ for some integer $m>=1$ implies that $f\in I$


Hilbert class field application If $K$ is an imaginary quadratic field and $M$ is an unramified Abelian extension of $K$, the prove that $M$ is Galois over $\mathbb{Q}$

Let see... If $L$ is the Hilbert class field of $K$, then $L$ is the maximal extension unramified of $K$, then $\mathbb{Q} \subset K \subset M \subset L$ and $L$ is Galois over $K$...

Thanks!

Partial answer:

We have $K \subset L$, $M \subset L$ and $\mathbb{Q} \subset K$, Galois extensions. Also, $K \subset M$ Galois, because $Gal(L/M)$ is normal subgroup of $Gal(L/K)$.

What can we say about  $Aut(M/\mathbb{Q})$?

$|Aut(M/\mathbb{Q})|= 2|Gal(M/K)|$?

The following simple argument seems to work for any $K$ that is Galois over $\mathbb{Q}$ (could it be that $K$ is specified to be imaginary quadratic in order to ensure one understands that $M/K$ is unramified not only at the finite but also at the infinite primes?): since $M$ is unramified, abelian, it must be contained in $CF(K)$. As $CF(K)/K$ is Galois and abelian, all the intermediate fields must be Galois and abelian. So $M/K$ is Galois. Since $K/\mathbb{Q}$ is also Galois, $M/\mathbb{Q}$ must be Galois as well.


Another "can't use Alt. Series Conv. test" (How do I determine if it converges or diverges?) Stumped on yet another absolute convergence problem.
(Converge conditionally, absolutely, or diverges)

$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} $$

First, I tested for absolute convergence, and the series reduced to $\frac{\ln(n)}{n}$   .....Using direct comparison, $\frac{\ln(n)}{n} > \frac{1}{n}$ I determined this diverges.  So, it does not converge absolutely.

Next, to use the alternating series test on the original series, you need to make sure the magnitude of the terms are decreasing.  Well, they are not:

$$\sum_{n=1}^\infty \frac{(-1)^n \ln(n)}{n} = -0 + .35 - .37 + .35.... $$

So, without using the alternating series test, how do I determine if the orig. series converges or diverges?

How can I use the divergence test (Nth term test)?  I'm not sure what the nth term tends to with that (-1) oscillation.  Just focus on $\frac{\ln(n)}{n}$  But we established that diverges, right?   But, I am told this series converges conditionally.

Do I use integral test?

Suppose you break the series into its first 100 terms (which has a finite sum) and the remaining infinite series. You can then use the alternating series test on that remaining series, in which the terms DO decrease (although you'll need to prove this).

This is a general trick for infinite series: what happens in the first few terms never affects convergence, so you're welcome to apply any of the tests starting at the $k$th term (for any finite value of $k$ that you like). This can be pretty useful.


What is the goal of harmonic analysis? I am taking a basic course in harmonic analysis right now. Going in, I thought it was about generalizing Fourier transform / series: finding an alternative representation of some function where something works out nicer than it did before.

Now, having taken the first few weeks of this, it is not at all about Fourier analysis but about the Hardy-Littlewood-Maximal-operator, interpolation theorems, Stein's theorem/lemma and a lot of constants which we try to improve constantly in some bounds. We are following Stein's book on singular integrals, I guess.

Can anyone tell me where this is leading? Why are we concerned with this kind of operators and in which other areas are the results helping?

Harmonic analysis is a very powerful tool to deal with many mathematical problems: i restrict myself on non-linearities in Partial differential equations (since it is my domain of research and you can adapt mutatis mutandis for any other domain because only technicalities will be discussed) for three important reasons:

It is based on divide and conquer philosophy: it uses particular ways and methods (Fourier series/transform, Littlewood Paley decomposition) of decomposing the phase plane which takes a single function and writes it as a superposition of a countably infinite family of functions of varying frequencies.
Other  advantage of harmonic analysis methods is that a partial differential equation (those who involves more than one variable $t$: time and $x$: space for example)  will be seen as an ordinary differential equation.
In fact, derivatives in phase space became multiplications in frequency space $$\mathcal{F}(\nabla f)(\xi)=2\pi i \xi\hat{f}(\xi)$$ where by $\hat{f}$ and $\mathcal{F}(\nabla f)$    we refer to Fourier transform of $f$ and $\nabla f$ respectively.
Then the answer to your question is: the goal of harmonic analysis is to serve as a generous bag of tools and powerful tricks.
A final modest information that i allow myself to add and it may contribute as a part of the answer: The idea of fractional derivative also becomes clear in the context of harmonic analysis. In fact, by applying formally the inverse  Fourier transform on $(2\pi i \xi)^{\gamma}\hat{f}(\xi)$ to obtain the derivative of order $\gamma$ of  $f(x)$ for all  $\gamma\in \Bbb{R}.$
Some operations can be performed only in Frequeny space, namely when we apply the Fourier transform and then take the inverse to go back to phase space.
As for operators and the whole theory, we must always remember that harmonic analysis is a "huge bag" that contains the fruit of contributions of many mathematicians developped according to necessity.

With that set-up I would expect to see boundedness of operators such as Hilbert transform, Reisz transform, etc. Expect to see fractional powers of Laplacian $\Delta$. Of course it ultimately depends on the instructor. There are an array of topics she/he could choose from.


Stability of dynamical system with purely imaginary eigenvalues I have a 4 dimensional dynamical system and I want to know the stability of one of its fixed points, the problem is that all the eigenvalues of the Jacobian there are purely imaginary:
$$
0 - 5.64799 i, 0 + 5.64799 i, 0 - 5.71808 i, 0 + 5.71808 i
$$
The eigenvectors of the Jacobian at the fix point are linearly independent and are the following:
$$
\left(-5 \sqrt{\frac{5}{192 \sqrt{5}-125}},1,0,0\right) \\
\left(5 \sqrt{\frac{5}{192 \sqrt{5}-125}},1,0,0\right) \\
\left(0,0,-5 i \sqrt{\frac{5}{125+64 \sqrt{5}}},1\right) \\
\left(0,0,5 i \sqrt{\frac{5}{125+64 \sqrt{5}}},1\right)
$$
What is the stability of this fix point and how can one figure it out?
Extra info:
My system is a continuous system of ODEs describing a physical system.
(I found a similar question here but I did not fully understand how it would apply to my case since he is talking about an eigenvalue with multiplicity 2.)

In $\mathbb{R}^2$ the modulus of the eigenvalues of the Jacobian gives you the criteria for stability:
\begin{gather}
0 < |\lambda_1| \leq |\lambda_2|<1 \; \text{ stable} \\
1 < |\lambda_1|\leq  |\lambda_2| \; \text{ unstable} \\
0 < |\lambda_1| < 1 < |\lambda_2| \; \text{ saddle point}
\end{gather}
But since you're working in $\mathbb{R}^4$, I don't think it should be too different. Whichever eigenvalues have modulus less than one will correspond to a local stable manifold of the fixed point (given by the corresponding eigenvector), whereas the ones with modulus more than one are unstable.
So in summary I guess it would be sort of weird to just call the fixed point "stable" or "unstable" since there are many more possibilities, but if all the eigenvalues' norms are less than one the fixed point is stable, and if all of them are more than one then it is unstable.
Good Luck! Hope this helps!


Random binary array shows patterns around prime numbers First post, so please let me know if I'm doing something wrong or if this question does not belong here.

I have been toying with java to visualize an interesting 2D binary array I thought of today in my math class. My thought was to start with one origin cell, and set all cells in an infinite grid equal to 1. Then, starting with the 8 cells surrounding the origin and moving outward, each cell would be evaluated by the following rule:

-If the cell at (x,y) == 1, then any cell at (nx,ny) = 0, for all positive integers n > 1

Basically I was thinking that if (1,0) was on, then all of the other cells behind it - (2,0), (3,0), etc... would be "in it's shadow" and therefore equal to zero. I figured the result would have some significance with prime numbers, and it did.

I wrote a program to plot all the cells in the 4th quadrant, up to about x=2000, y=1000 and took screenshots at various zoom levels. Here are the results:

The top left cell (0,0) is arbitrarily set to 1. Here's the algorithm:

    boolean [][] grid = new boolean[height][width];

    for(int x=0; x<width; x++)
    {
      for(int y=0; y<height; y++)
      {
          grid[x][y] = true;
      }
    }

    for(int x=0; x<width; x++)
    {
      for(int y=0; y<height; y++)
      {
          if(grid[x][y])
          {
              int xt = 2*x;
              int yt = 2*y;

              while((xt < width && yt < height) && (xt != 0 || yt != 0))
              {
                  grid[xt][yt] = false;

                  xt += x;
                  yt += y;
              }
          }
      }
    }

On a large scale, it looks somewhat uniform, but on a small scale it looks random (or at least I can't see the pattern).

Here's my questions:

1 - Has this "function" been researched before? It seems interesting to me, but I can't find anything online about it.

2 - Where is this randomness coming from? The code is so simple, but the result looks very complex.

3 - Is there any mathematical way for me to search this grid for patterns more efficiently than the naked eye? I can see that the vertical and horizontal lines are most prominent along the prime numbers, which is interesting.

Thank you, and sorry if this is a dumb question. I couldn't think of any other place to ask it.

This pattern is usually described in the following way: Suppose you stand at the origin, and at each lattice point $(x,y)\in\mathbb{Z}^2$ there is a thin pole stuck into the ground. Which lattice points are visible, i.e. which poles are not hidden behind other poles? The answer is that $(x,y)$ is visible if and only if $x$ and $y$ have no common factor $>1$, since if $d$ is a common factor, then $(x,y)$ is hidden by $(x/d, y/d)$.

If you search for visible lattice points, you will find articles ranging from introduction to research papers.

I've also made this graph sometime in the early 90's and noticed the diagonals. I note that they are formed by the multiples of primes. The multiples of 1: rise/run = 1/1; multiples of 2: rise/run = 2/1; ... etc. (I graphed in the 1st quadrant.)

Rather than focusing on only the red dots, it's interesting to focus on the white solid lines as well. Primes occur at the intersection of solid lines (no divisors of lesser value other than 1). So assigning a color to those lines can be of interest, as can be focusing on the patterns inside those squares, and the on-off pattern of the next higher line and the squares being formed by those unique on-off patterned lines.

Primes are definitely deterministic in where they appear, but the exact nature of the determinism is unknown and may even be unknowable. (I have a feeling it's recursive.) I came up with an algorithm that can output all primes in order if you seed it with a given prime and the primes that come before it that are higher than it's square root (you can start it at a very low prime in the list of all primes), but it doesn't appear to be good for finding huge primes because the computing time becomes outrageous.

I have an idea that the task could potentially be simplified, but I'm one of those "amateur mathematicians" and uncertain of that particular statement. The other statements, I'm confident about.


Differential Equations: Calculating Barometric Pressure at an Altitude I'm trying to solve a differential equation problem that has to do with barometric pressure at a certain altitude.

The rate at which barometric pressure decreases with altitude is proportional to the pressure at that altitude. If the altitude in feet, then the constant of proportionality is 3.7 x 105. The barometric pressure at sea level is 29.92 inches of mercury.

How would I calculate the barometric pressure, in inches of mercury, at any altitude using the information above?

The differential equation that I came up with looks like this:
Let $P=$ Barometric Pressure
Let $a=$ Altitude
$\dfrac{dP}{da}=-k(29.92-P)$
But when I solve it, it doesn't look like it's the right answer.

Thanks for your help!

Pressure decreases exponentially with altitude and thats what you will get provided you fix your signs in the equation. Your equation will look like:

$\frac{dP}{da} = -kP$

where $k$ is some positive constant.

Solve this to get $P(a) = P(0) e^{-ka}$.

and at $a = 0$, $P(0) = 29.92$. Make sure to keep track of all your units.

If your constant k doesn't allow the organic factors to coexist within this specific equation
Just use a new one with your original formula you should get the right answer accurate to 6% of the correct answer because it was never officially empirically determined.


Solve the quadratic congrunce Solve the following quadratic congreunce

$x^2+ 7x + 10 \equiv 0$ (mod $11$).

I want to know a general and easy method how to solve this kind of questions.

For $ax^2+bx+c\equiv 0$ (mod $p$), if $p$ is an odd prime not dividing $a$, then this is the same as solving a simpler question about quadratic residues $y^2\equiv b^2-4ac$ (mod $p$), and $y\equiv 2ax+b$ (mod $p$).

Use the quadratic formula $$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

so $b^2-4ac = 7^2 - 40 = 9 = 3^2$

so $\pm \sqrt{b^2-4ac} = \pm 3$

and $-b \pm 3 = -10,-4$

and so the answers are -5 and -2.


Show that a subset of $C[0,1]$ is closed I have a question I my metric spaces course book I cannot solve:

Show that the following subset is closed in $C[0,1]$: $\{ f \in C[0,1] \mid f(a)=0 \textrm{ for all }a \in A \}$, where $C[0,1]$ is the space of continuous real-valued functions on $[0,1]$ with the sup metric and $A \subseteq [0,1]$.

If $A$ was just one point, could you see why it's true? Can you see how this proves it for general $A$?

First proof it taking $A$ as a singleton which is shown by someone else above then it is obvious because intersection of closed sets is closed.


If $f$ and $f\circ h$ are nonconstant and real-analytic, with $h\in C^\infty$, does it follow that $h$ is also real-analytic? Let $f,g:I\to\mathbb{R}$ be two real, non-constant, analytic functions on an open interval $I\subset\mathbb{R}$. If there is a smooth function $h:I\to\mathbb{R}$ (i.e. $h\in C^{\infty}(I)$) such that: $(f\circ h)(t)=g(t),\ \forall\ t\in I$, is it true that $h$ is also analytic on $I$?

If we add that $f'$ does not vanish nowhere, then the answer is yes.

It can be checked locally. As $f$ is real-analytic, it extends analytically in an neighbourhood of $h(I)$ in $\mathbb C$ - We assume that $f$ is real analytic on an interval $J$, with $h(I)\subset J$.

Let now $h(t_0)=w_0\in h(I)$.

As $f'(w_0)\ne 0$, then $f$ possesses an analytic inverse $F$ in a neighbourhood $D(w_0,r)\subset \mathbb C$ of $t_0$,  and hence $h=F\circ g$, for $t\in I$, such that $h(t)\in (w_0-r,w_0+r)$, and therefore $h$ is analytic in a neighboorhood of $t_0$.

However, if $f'$ vanishes, then it DOES NOT hold. For example $h(x)=\lvert x\rvert$ and $f(x)=x^2$.