The relation between the convexity and the monotonicity of a function Assume the function $ F:\; ] 0,+\infty)^{n} \rightarrow] 0,+\infty) $
is  convex and $ F(x_{1},x_{2},\cdots, x_{n}) = 0 $ if and only if $ x_{i} = 0 \text{ for } i=1, 2 , \ldots , n. $ I would like to know if it is true that the function $\mu$ defined by:
$$ \mu ( x_ { 1 },x_ { 2 },\ldots, x_ { n} ) = F \left( \mu _ { 1 } \left(x_ { 1 } \right) , \mu _ { 2 } \left(x_ { 2}\right) , \ldots , \mu _ { n } \left( x_ { n} \right) \right)\quad \forall ( x_ { 1 },x_ { 2 },\ldots, x_ { n})\in X = ] 0,+\infty)^{n}  ,$$
is a nondecreasing function, where each $\mu _ { i}:\; ] 0,+\infty)\rightarrow] 0,+\infty)\text{ for } i=1, 2 , \ldots , n $ is nondecreasing and convex.

If $n\ge2$ then $F$ has to be zero, if I understand the assumptions right.

Suppose $F(x)\ge0$ for all $x\ge0$, and $F(x)=0$ if $x_i=0$ for one $i\in \{1\dots n\}$. Then $F(x)=0$ for all $x$:

Take two points $\hat x$ and $\tilde x$, both have one zero entry, so $F(\hat x)=F(\tilde x)=0$. Take a point $x$ on the line connecting $\hat x$ and $\tilde x$. Then by convexity, $F(x)\le0$. Since $F(x)\ge0$, $F(x)=0$ follows. Since every point $x\ne 0$ is on at least one such line,
$F\equiv0$ follows.

Professor A.Aghajani answered me as follow:

The reason is that the function "$F$" satisfies the assumptions of Thm 1.3 should be increasing with respect to each component. To see this, for simplicity assume "$F$" is one variable, and remember that "$F$" is convex, nonnegative and $F(0)=0.$ Now assume $x<y$, then there is a positive $"t<1"$ such that $x=ty.$ Now we can write, using the convexity:
$$F(x)= F(ty)=F( (1-t).0 + ty )< (1-t)F(0)+t F(y)=0+tF(y)<F(y).$$
So we proved "$F$" is increasing, now if "$\mu$" is a measure then it should be increasing (as a set function) thus $F(\mu)$ is a composition of two increasing function which is also increasing...


Right identity and Right inverse implies a group Let $(G, *)$ be a semi-group. Suppose

$ \exists e \in G$ such that $\forall a \in G,\  ae = a$;
$\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.

How can we prove that $(G,*)$ is a group?

It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:

1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:

$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$

This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.

2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse.  We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:

$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$

3) It is now clear that the right identity is also a left identity. For any $a$:

$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$

4) To show the uniqueness of the inverse:

Given any elements $a$ and $b$ such that $ab=e$, then

$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$

Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.

See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.

Most proofs I've seen contains many ingenious equations. The motivation behind these equations is not very clear. Here I present a more intuitive proof.

For any element $a$ in $G$, we can map it to a function $f_a:G \to G$, simply define $f_a(x) = xa$. We can see that every right identity element $e$ is mapped to the identity function $id$,
$$f_e=id,$$
and the operation on $G$ amounts to function composition:
$$ f_{ab} = f_b \circ f_a, $$
and the existence of right inverses means for any $a$ there is an $a^{-1}$ such that
$$f_{a^{-1}} \circ f_a = id.$$
That will imply that every function $f_a$ is injective, otherwise the composition of it, which is $id$, can't be injective either. Actually, it's bijective, but we don't have to use that strong conclusion here.

We want to show $a^{-1}a=e$. Because $f_{a^{-1}a} = f_a \circ f_{a^{-1}}$, we wonder whether $f_a \circ f_{a^{-1}} =id$. Intuitively that should be true since $f_{a^{-1}} \circ f_a = id$ and both $f_a$ and $f_{a^{-1}}$ are bijective. For a proof that only assumes that one of them is injective, see Lemma below.

Now that $f_{a^{-1}a} = f_e$, to prove $a^{-1}$ is a left inverse, we only need to show that for any $c \in G$ and $c' \in G$, $f_c = f_{c'}$ implies $c=c'$. We try to calculate $f_c(e)$:
$$f_c(e) = ec = (cc^{-1})c = c(c^{-1}c)=f_{c^{-1}c}(c)=id(c)=c.$$
For the same reason $f_{c'}(e) = c'$, so $c=c'$. Note here $ec=c$, so it has already proved every right identity is a left identity.

The uniqueness of the identity element is induced by the uniqueness of the identity function, for $f_e=f_{e'}$ implies $e=e'$, and also the uniqueness of inverse elements is induced by the uniqueness of inverse functions.

So we've proved $G$ is a group.

  Lemma.  Let $X$ and $Y$ be two arbitrary sets. Let $f$ be a function $f: X \to Y$ and $g$ be another function $g: Y \to X$.
  Suppose $g$ is injective, and $g \circ f = id$, then $f \circ g = id$.

  Proof. Let $y \in Y$, we want to show that $(f \circ g)(y) = y$. Let $y' = (f \circ g)(y)$. Apply $g$ to both side, $$g(y')=(g \circ f \circ g)(y) = (id \circ g)(y) = g(y).$$ So $y'=y$.


Significance of Rank of Jacobian I have been struggling with this question. What is the significance of finding the rank of a Jacobian matrix of a function?

I understand that the Rank of a matrix signifies the number of linearly independent rows / columns. How does this idea extend to finding the rank of a Jacobian matrix?

Thanks !! :)

If the Matrix has not a zero determinant it is called the full rank Jacobian. It means if J is Jacobian matrix. det(J) must not be zero to have the full rank.


In each of the following cases determine if it is a linear map or not? In each of the following justify why it is a linear map or not :

Let $f:V\rightarrow W$ be a linear map.

(a)

$f\left( u,v\right) =(u-v,vu), W=V=\mathbb{R}^{2}$

(b)

$f\left(u,v,w\right)=(u+w,3w), V=\mathbb{R}^{3}, W=\mathbb{R}^{2}$

(c)

$f\left( u,v\right)=(u-v,u+v,1), V=\mathbb{R}^{2}, W=\mathbb{R}^{3}$

I understand that in order for them to be linear maps they need to respect scalar multiplication and vector addition. However how do we sow that they do not hold ?

If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear map, then for any $\vec{v},\vec{w}$ in $\mathbb{R}^n$, $f(\vec{v}+\vec{w})=f(\vec{v})+f(\vec{w})$, and $f(\lambda \vec{v})=\lambda f(\vec{v})$.

part(a)

Let $\vec{u}=(r,s), \vec{v}=(a,b)$.

$f(\vec{u}+\vec{v})=f(r+a,s+b)=(r+a-s-b,(r+a)(s+b))$

$f(\vec{u})+f(\vec{v})=f(r,s)+f(a,b)=(r-a+t-b,rs+ab)$

$f(\vec{u}+\vec{v})\neq f(\vec{u})+f(\vec{v})$
Therefore this is not a linear map.

part(b)

Let $\vec{u}=(r,s,t), \vec{v}=(a,b,c),\vec{w}=(o,p,q)$.

$f(\vec{u}+\vec{v}+\vec{w})=f(r+a+o,s+b+p,t+c+q)=(r+a+o+t+c+q,3(t+c+q))$

$f(\vec{u})+f(\vec{v})+f(\vec{w})=f(r,s,t)+f(a,b,c)+f(o,p,q)=(r+t,3t)+(a+c,3c)+(o+q,3q)=(r+a+o+t+c+q,3t+3c+3q)$

$f(\vec{u}+\vec{v}+\vec{w})=f(\vec{u})+f(\vec{v})+f(\vec{w})$

$f(\lambda\vec{u})=f(\lambda r,\lambda s,\lambda t)=(\lambda r+\lambda t,3\lambda t)$

$\lambda f(\vec{u})=\lambda f(r,s,t)=\lambda (r+t,3t)=(\lambda (r+t),3\lambda t)$

$f(\lambda \vec{u})=\lambda f(\vec{u})$

Therefore this is a linear map.

part(c)

Let $\vec{u}=(r,s),\vec{v}=(a,b)$

$f(\vec{u}+\vec{v})=f(r+a,s+b)=(r+a-s-b,r+a+s+b,1)$

$f(\vec{u})+f(\vec{v})=f(r,s)+f(a,b)=(r-s,r+s,1)+(a-b,a+b,1)=(r-s+a-b,r+s+a+b,2)$

$f(\vec{u}+\vec{v})\neq f(\vec{u})+f(\vec{v})$.

Therefore this is not a linear map.

In each case we check the criteria

$a(u,v)=(au,av) to (au-av,a^2uv)=a(u-v,auv)$ not equal $ a(u-v,uv)$ So this is not a linear transform.

$a(u,v,w)=(au,av,aw) to (au+aw,3aw)=a(u+w,3w)$ so the first condition holds. $(x,y,z)+(u,v,w)=(x+u,y+v,w+z) to (x+z,3z)+(u+w,3w)=(x+u+w+z,3w+3z)$ so the second condition holds.

$(au,av)to (au-av,au+av,1)$ not equal $a(u-v,u+v,1)$ so this is not a linear mapping. There's a shortcut for this one namely that any linear mapping maps the zero vector to the zero vector because of scalar multiplication by 0. $(u-v,u+v,1)$ is never equal to the zero vector so this is not a linear mapping.


How to Calculate the 2 balance points "Swinging Sticks" physics toy/desk sculpture?? Please, this is something I have thought of a long time, but don't have unlimited supplies to keep doing it wrong. I want to make one of these for my nephew, he's interested in science!

There is a Physics Toy, called "Swinging Sticks", used in the movie "Iron Man 2", on Pepper Pott's desk. How can I calculate the 2 balance points for the "sticks" in this toy,  to make it keep spinning a long time? - Here is a great Youtube video of the toy:

https://www.youtube.com/watch?v=DhFHAEQ5x4c

I'm really brainless when it comes to calculating, but I think this can be determined easily without any calculation.
Let's call the bar attached to the fixed stand the "primary bar", the other one the "secondary bar".

The balancing point on the secondary bar is slightly offset from the secondary's center (gravity) point. As a ballpark figure, I would assume that an offset of $\frac{1}{20}$th of the secondary's length should do. This causes the secondary to always assume the upright position when in rest. This point is attached to the primary bar somewhere close to its end.

The balancing point of the primary again is offset slightly from its gravity center point, only this time the secondary's mass is part of the account. I would try to find the primary's center of mass by balancing it horizontally while the secondary dangles down from its end. When the sweet spot is found, shift it $\frac{1}{20}$th of the primary's length towards the secondary bar. This will be the pivot point for the primary on the stand, causing the long end of the primary to outweigh its short end plus the secondary.

As a rule of thumb, it can be assumed that the larger the offsets, the more rapid the sticks will act. Smaller offsets will result in calmer movements. It might be an interesting project to build a model with variable pivot points to experiment with.


Probability of 20 consecutive success in 100 runs. Suppose a chess player have a win rate equal 90%, what is the chance to have 20 consecutive wins (successes) playing 100 games? Consider that lose/draw = fail.

I've studied basic statistics in college and it seems like a binomial distribution problem (right?), but honestly I can't figure out a way to solve this problem considering "consecutive" successes.

Is there a statistical distribution for this kinda problem?

Thanks very much! I really appreciate any thoughts!

Feller has this all worked out on p. 325 of An Introduction to Probability Theory and Its Applications, 3rd Edition, equation 7.11:
$$q_n \sim \frac{1-px}{(r+1-rx)q} \cdot \frac{1}{x^{n+1}}$$
where $q_n$ is the probability of no success run of length $r$ in $n$ trials, $p$ is the probability of success, $q=1-p$, and $x$ is the root near 1 of

$$ 1-x + q p^r x^{r+1} = 0 $$

With your data, we find $x \approx 1.017502$ and $q_{100} \approx 0.2247$.

So the probability that the chess player will have at least one run of 20 successes is $0.7753$, approximately.

Edit: Actually "$x$ is the root near 1 ..." is slightly misleading. The equation has two positive roots, and we must peak the one that is not $1/p$. Details here.

The probability of getting $20$ consecutive wins is:

$$0.9^{20}$$

The first win of these consecutive wins can be at any trial from $1$ to $80$, and the probability of it being any on any of these trials is evenly distributed, so the probability of getting $20$ consecutive wins out of $100$ is:

$$\frac{80}{100}*0.9^{20}$$

A generalization for this formula would be:

$$\frac{n-k}{n}p^k$$

Where $p$ is the probability, $n$ is the number of trials, and $k$ is the number of consecutive wins.


Linear Algebra Standard Basis of $\mathbb{R}^3$ I have no idea what to do in this question, other then the fact that the formula $T$ is suppose to have some kind of $x$, $y$, $z$.

The standard basis is $E_1 = (1,0,0)$, $E_2 = (0,1,0)$, and $E_3 = (0,0,1)$. So if $X = (x,y,z) \in \mathbb{R}^3$, it has the form $$X = (x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1) =  x E_1 + y E_2 + z E_3.$$
Now use linearity of $T$ (which says $T(aX + bY) = aT(X) + bT(Y)$ for any numbers $a,b \in \mathbb{R}$ and vectors $X,Y \in \mathbb{R}^3$, and this extends to any finite sum of vectors) to get
$$
T(X) = T(xE_1 + y E_2 + z E_3) = xT(E_1) + yT(E_2) + zT(E_3).
$$
You are given $T(E_1)$, $T(E_2)$, and $T(E_3)$, so substitute in to get the answer.

Use matrix, this application would not be more than the multiplication of the transpose of the image, ie, the matrix of T in the canonical bases. Where E is the canonical base, TE = Im (T). Note that the transpose of the canonical is herself. It is relatively simple, just imagine what their eyes are two dimensions and the third touch, movement, ie move your body is a linear application from R3 to R3, if you cut the arm of R3 to R2. The first thing is to understand what is the linear algebra. They are rotations of a abelian body relative origin. Body parallel to that form the space, which is always a abelian because infinite body is always finite. Poincare you recommend reading her books disclosure abstracion to understand mathematics. We facilitate life. Translations of google


Probability of having exactly 1 pair from drawing 5 cards I have an exercise as follows: There is a collection of cards consisting of 52 cards (13 types and 4 colours each type). We draw 5 cards from the collection. Then what is the probability of having exactly 1 pair (pair means same colour or same type)?
Thanks for any indication.

We have $u$ different colors $C=\{c_i, i=1..u: k,l=1..u, k\ne l: c_k\ne c_l\}$ and $v$ different types $T=\{t_i, i=1..v: k,l=1..v,k\ne l: c_k\ne c_l\}$
Next we have $k$($=2$) cards with the properties that we are interested in(the same colors or the same type), then we have these cards add a $m-k$ cards that do not form any other with the properties that we are interested(question is "having exactly 1 pair").

First, assume the option of the same color:
Here we deck of cards imagine as subsets of types:
$\{c_1,c_1,...\}\cup\{c_2,c_2,...\}...\cup\{c_u,c_u,...\} $, where each subset contains $v$ elements with the same colour. Now we'll take from the first subset $k$ elements($k$ the same colours). This we can carry out $\binom{v}{k}$ ways(here, naturally, each has different type). Now we have selected a group of appropriate elements(pairs with the same colour) and next we need to add $m-k$ elements with different color and different type. Elements with different colour we can add by $\binom{u-1}{m-k}(m-k)!$ while any two cards must have the same type. This can be done in such a way that the from first unused colors subset($c_2$) and continue with next color($c_2$). Thus we can take $(v-k-1)(v-k-2)...(v-m)=\frac{(v-k-1)!}{(v-m-1)!}$.
So total $m$ options with the same color:$\binom{v}{k}k!\binom{u-1}{m-k}(m-k)!\frac{(v-k-1)!}{(v-m-1)!}m!$
Next, assume the option of the same type:
Here we deck of cards imagine as subset of types:
$\{t_1,t_1,...\}\cup\{t_2,t_2,...\}...\cup\{t_v,t_v,...\} $, where each subset contains $u$ elements with the same type. And we're similarly as in the first case.
So total $m$ options with the same type:$\binom{u}{k}k!\binom{v-1}{m-k}(m-k)!\frac{(u-k-1)!}{(u-m-1)!}m!$

The total number of permutations of $m$ elements from $n$ elements equals to $\binom{n}{m}m!$.

Probability of having exactly 1 pair (pair means same colour or same type) equals to:
$$p(n,u,v,m,k)=\frac{\left(\binom{v}{k}\binom{u-1}{m-k}\frac{(v-k-1)!}{(v-m-1)!}+\binom{u}{k}\binom{v-1}{m-k}\frac{(u-k-1)!}{(u-m-1)!}\right)k!(m-k)!}{\binom{n}{m}}$$

Probability for $u=4$ colors, $v=13$ types, $n=u\cdot v=52$, $k=2$ and $m=5$:
$$p(52,4,13,5,2)=\frac{\left(\binom{13}{2}\binom{3}{3}\frac{10!}{7!}+\binom{4}{2}\binom{12}{3}\frac{1!}{1}\right)2!3!}{\binom{52}{5}}\doteq 27\%$$

That's quite a lot.
In addition, if any of these cases occurs, it will be much more likely a pair with the same color than a pair with the same type (about 42 times more).

I assume that three of a kind are forbidden.

For the pair there are $13$ different types, and you may select any $2$ out of the $4$ cards. so there are
$$13\binom{4}{2} = 13\cdot 6 = 78$$
ways to select the pair.
The third card must be of a different type, so $52-4 = 48$ possibilities.
Since no second pair is allowed, for the fourth card there are $44$ possibilities, and for the fifth card $40$. The order in which those three single cards are drawn does not matter, so we have to divide by $3!$. We have found that
$$ 13\binom{4}{2} \cdot \frac{48\cdot 44\cdot 40}{3!} = 1098240$$
hands out of the $\binom{52}{5} = 2598960$ hands contain exactly $1$ pair. So the probability is $$1098240/2598960 \approx 42.26\%.$$


What is arcsec (-2)? The question asks to solve $\operatorname{arcsec}(-2) $.

Options are

a) π/3

b) −(π/3)

c) 5π/3

d) 4π/3

e) 2π/3

So my thoughts are: this is quadrant 2 or 3; on the unit circle, $ x = -1/2 $.

$\cos(t) = 1/2$ where $t$ is 60°. In quadrant 2 that is 2π/3. In quadrant 3 it's 4π/pi.

I though that the answer could be either d or e.

The answer key says that the only right answer is d.

The graphing calculator shows that the answer must be e.

Who is right and why?

It depends on how you define $\operatorname{arcsec} x$. Its domain is $(-\infty,-1]\cup[1,\infty)$, and one can define it so that
\begin{cases}
0\le \operatorname{arcsec}x<\frac{\pi}{2} & \text{if $x\ge1$}\\
\frac{\pi}{2}<\operatorname{arcsec}x\le\pi & \text{if $x\le-1$}
\end{cases}
You're in the second case, so (e) should be the correct answer, because $\cos(2\pi/3)=-1/2$.

However a different definition of the function is conceivable, although answer (d) would sound quite strange.

The cos -.5 is either 120 degrees in Q II or 240 degrees in Q III.  both answers are correct!!


Dominoes and induction, or how does induction work? I've never really understood why math induction is supposed to work.

You have these 3 steps:

Prove true for base case (n=0 or 1 or whatever)
Assume true for n=k.  Call this the induction hypothesis.
Prove true for n=k+1, somewhere using the induction hypothesis in your proof.

In my experience the proof is usually algebraic, and you just manipulate the problem until you get the induction hypothesis to appear.  If you can do that and it works out, then you say the proof holds.

Here's one I just worked out,

Show $\displaystyle\lim_{x\to\infty} \frac{(\ln x)^n}{x} = 0$

So you go:

Use L'Hospital's rule.
$\displaystyle\lim_{x\to\infty} \frac{\ln x}{x} = 0$.
Since that's
$\displaystyle\lim_{x\to\infty} \frac{1}{x} = 0$.
Assume true for $n=k$.
$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^k}{x} = 0$.
Prove true for $n=k+1$.  You get
$\displaystyle\lim_{x\to\infty} \frac{(\ln x)^{k+1}}{x} = 0.$

Use L'Hospital again:
$\displaystyle\lim_{x\to\infty} \frac{(k+1)(\ln x)^{k}}{x} = 0$.

Then you see the induction hypothesis appear, and you can say this is equal to $0$.

What I'm not comfortable with is this idea that you can just assume something to be true ($n=k$), then based on that assumption, form a proof for $n=k+1$ case.

I don't see how you can use something you've assumed to be true to prove something else to be true.

The inductive step is a proof of an implication: you are proving that if the property you want holds for $k$, then it holds for $k+1$.

It is a result of formal logic that if you can prove $P\rightarrow Q$ (that $P$ implies $Q$), then from $P$ you can prove $Q$; and conversely, that if from assuming that $P$ is true you can prove $Q$, then you can in fact prove $P\rightarrow Q$.

We do this pretty much every time we prove something. For example, suppose you want to prove that if $n$ is a natural number, then $n^2$ is a natural number. How do we start? "Let $n$ be a natural number." Wait! Why are you allowed to just assume that you already have a natural number? Shouldn't you have to start by proving it's a natural number? The answer is no, we don't have to, because we are not trying to prove an absolute, we are trying to prove a conditional statement: that if $n$ is a natural number, then something happens. So we may begin by assuming we are already in the case where the antecedent is true. (Intuitively, this is because if the antecedent is false, then the implication is necessarily true and there is nothing to be done; formally, it is because the Deduction Theorem, which is what I described above, tells you that if you manage to find a formal proof that ends with "$n^2$ is a natural number" by assuming that "$n$ is a natural number" is true, then you can use that proof to produce a formal proof that establishes the implication "if $n$ is a natural number then $n^2$ is a natural number"; we don't have to go through the exercise of actually producing the latter proof, we know it's "out there").

We do that in Calculus: "if $\lim\limits_{x\to x_0}f(x) = a$ and $\lim\limits_{x\to x_0}g(x) = b$, then $\lim\limits_{x\to x_0}(f(x)+g(x)) = a+b$." How do we prove this? We begin by assuming that the limit of $f(x)$ as $x\to x_0$ is $a$, and that the limit of $g(x)$ as $x\to x_0$ is $b$. We assume the premise/antecedent, and proceed to try to prove the consequent.

What this means in the case of induction is that, since the "Inductive Step" is actually a statement that says that an implication holds:
$$\mbox{"It" holds for $k$}\rightarrow \mbox{"it" holds for $k+1$},$$
then in order to prove this implication we can begin by assuming that the antecedent is already true, and then proceed to prove the consequent. Assuming that the antecedent is true is precisely the "Induction Hypothesis".

When you are done with the inductive step, you have in fact not proven that it holds for any particular number, you have only shown that if it holds for a particular number $k$, then it must hold for the next number $k+1$. It is a conditional statement, not an absolute one.

It is only when you combine that conditional statement with the base, which is an absolute statement that says "it" holds for a specific number, that you can conclude that the original statement holds for all natural numbers (greater than or equal to the base).

Since you mention dominoes in your title, I assume you are familiar with the standard metaphor of induction like dominoes that are standing all in a row falling. The inductive step is like arguing that all the dominoes will fall if you topple the first one (without actually toppling it): first, you argue that each domino is sufficiently close to the next domino so that if one falls, then the next one falls. You are not tumbling every domino. And when you argue this, you argue along the lines of "suppose this one falls; since it's length is ...", that is, you assume it falls in order to argue the next one will then fall. This is the same with the inductive step.

In a sense you are right that it feels like "cheating" to assume what you want; but the point is that you aren't really assuming what you want. Again, the inductive step does not in fact establish that the result holds for any number, it only establishes a conditional statement. If the result happens to hold for some $k$, then it would necessarily have to also hold for $k+1$. But we are completely silent on whether it actually holds for $k$ or not. We are not saying anything about that at the inductive-step stage.

Added: Here's an example to emphasize that the "inductive step" does not make any absolute statement, but only a conditional statement: Suppose you want to prove that for all natural numbers $n$, $n+1 = n$.

Inductive step. Induction Hypothesis: The statement holds for $k$; that is, I'm assuming that $k+1 = k$.

To be proven: The statement holds for $k+1$. Indeed: notice that since $k+1= k$, then adding one to both sides of the equation we have $(k+1)+1 = k+1$; this proves the statement holds for $k+1$. QED

This is a perfectly valid proof! It says that if $k+1=k$, then $(k+1)+1=k+1$.  This is true! Of course, the antecedent is never true, but the implication is. The reason this is not a full proof by induction of a false statement is that there is no "base"; the inductive step only proves the conditional, nothing more.

By the way: Yes, most proofs by induction that one encounters early on involve algebraic manipulations, but not all proofs by induction are of that kind. Consider the following simplified game of Nim: there are a certain number of matchsticks, and players alternate taking $1$, $2$, or $3$ matchsticks every turn. The person who takes the last matchstick wins.

Proposition. In the simplified game above, the first player has a winning strategy if the number of matchsticks is not divisible by $4$, and the second player has a winning strategy if the number of matchsticks is divisible by 4.

The proof is by (strong) induction, and it involves no algebraic manipulations whatsoever.

In explanations of the principle of induction, I think the 'domino' example is often not explored enough for it to be intuitive. Let me endeavour a more understandable explanation using the domino analogy.

As always, you first prove that P(1) is true (or P(2) or P(3) or P(4 septillion), depending from where you want your hypothesis to be valid. If you don't care that P(1) be true, you can begin from P(2) or wherever.).

Now, since P(k) implies P(k+1) (Which you prove by algebraic manipulation), and P(1) is true, P(2) is true. Since P(2) is true, P(2+1) is true. Since P(3) is true, P(4) is true. Since P(100,000) is true, P(100,001) is true.

This way, as you see, you can say P(n) is true for any n is the natural n.


Uniformly integrable sequence tending to 0 a.s. but with $\mathbb{E}(\sup_n|X_n|) = \infty$ I am trying to find a uniformly integrable sequence of random variables $(X_n: n \in \mathbb{N})$ such that both $X_n \to 0$ almost surely and $\mathbb{E}\left(\sup_n|X_n|\right) = \infty$. I think this is probably a well known counterintuitive result in the theory, but I haven't come across it before, so any help would be great.

It is possible to construct for any $p>1$ a sequence $(X_n)_{n\geqslant 1}$ of random variables such that

the sequence $(X_n)_{n\geqslant 0}$ converges to $0$ almost surely;
the sequence $\left(\mathbb E\left|X_n\right|^p\right)_{n\geqslant 1}$ is bounded and
the random variable $\sup_n\left|X_n\right|$ is not integrable.

Indeed, consider the unit interval endowed with the Borel $\sigma$-algebra and Lebesgue measure. Define for $n\geqslant 1$:
$$A_n:=\left(\left(n+1\right)^{-p/(p-1)-1},n^{-p/(p-1)-1}\right),\quad
X_n:=n^{1/(p-1)} \mathbf 1(A_n).$$
In this way, we have that
$$\tag{*} cn^{-p/(p-1)} \leqslant \lambda(A_n)\leqslant Cn^{-p/(p-1)}$$
where $c$ and $C$ are independent of $n$.

holds because the $A_n$ are pairwise disjoint.
By (*), $\mathbb E\left|X_n\right|^p\leqslant Cn^{-p/(p-1)}n^{p/(p-1)}=C$.
Since the $A_n$ are disjoint, $\sup_n\left|X_n\right|=\sum_{n=1}^{+\infty}X_n$ and by (*), $\mathbb EX_n\geqslant cn^{-1}$.

Added: This only works for functions without the restriction that they are defined on a probability space. The requirement for random variables makes my example fail, because I'm relying on the fact that the support of the functions gets smaller as you move toward $\infty$. Examples relying on the support of the function probably won't work for random variables, and it seems quite a bit harder to think of examples for this that have support on the whole real line or the positive reals and satisfy the needed integrability constraints by virtue of decaying quickly.

I'm really rusty with this stuff and I might misunderstand the question, but here is an idea.

Define

$$ \displaystyle END(X):= \sup \textrm{ }\{x \textrm{ s.t. } X(x) > 0  \}$$

Then make the sequence $X_{n}$ as:

$$ \displaystyle X_{n} = \begin{cases} 1 \textrm{ if } x \in \biggl[END(X_{n-1}), \frac{1}{n} + END(X_{n-1})\biggr] \\ 0 \textrm{ otherwise.}\end{cases} $$

for $n > 0$ with $END(X_{0}) := 0$.

I think the above definition is easier to read, but another way to write it is:

$$ X_{n} = \displaystyle \begin{cases} 1 \textrm{ if } x \in \biggl[\displaystyle\sum_{j=1}^{n-1}\frac{1}{j}, \displaystyle\sum_{j=1}^{n}\frac{1}{j}\biggr] \\ 0 \textrm{ otherwise.}\end{cases}$$

With the understood definition that for $n=1$ we start off with the indicator function of $[0,1]$.

The pointwise limit for a fixed $x$, $\displaystyle \lim_{n\to\infty} X_{n}(x) = 0 \quad\forall x $, but $\sup_{n} X_{n}(x)=1$ for $x \ge 0$, so

$$\int\sup_{n}X_{n}(x) = \int_{x=0}^{\infty}1dx = \infty.$$

To see that it's uniformly integrable, consider that

$$ \int X_{n}dx = \int_{END(X_{n-1})}^{END(X_{n-1})+\frac{1}{n}}dx = \frac{1}{n}.$$

For a given $\epsilon > 0$, choose $n$ such that $1/n < \epsilon$ and set $K = n$. The integral over $[K, \infty]$ for all $X_{m}$ with $m < n$ is 0 (since those functions are zero at or before they hit $x=K$). We just showed that the integral is smaller than $\epsilon$ for $n=K$ and such an integral only becomes smaller as
$n$ gets larger after that point. So the definition for uniform integrability should be satisfied.


Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$

Approach:

$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$

So the equation will have $(15,6)$ solutions. Now we divide by 3

$$5y \equiv 1\pmod 2$$.

Solving the Diophantine equation we get $y \equiv1\pmod 2 $, so $y=1+2m$

$$6x \equiv 9-15y\pmod {18}$$
$$6x \equiv 9-15(1+2m)\pmod {18}$$
$$6x \equiv -6-30m\pmod {18}$$

Divide by 6

$$x \equiv -1-5m\pmod 3$$

The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$
$$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$

Wolfram alpha says $y=1+2t$ and $x=t+3d+2$

Dividing everything by $3$, including the modulus, we get the equivalent congruence
$$2x+5y\equiv 3\pmod{6}.$$
It is convenient to rewrite this as $2x-y\equiv 3\pmod{6}$, or equivalently $y\equiv 2x-3\equiv 2x+3\pmod{6}$.

Now we have a parametric solution: $x\equiv t\pmod{6}$, $y\equiv 2t+3\pmod{6}$. To write it out at length, set $t=0,1,2,3,4,5$. The first two solutions are then $x\equiv 0\pmod{6}$, $y\equiv 3\pmod{6}$ and $x\equiv 1\pmod{6}$, $y\equiv 5\pmod{6}$. Four more to go.

Wolfram answer:
$y = 2c + 1$
$x = c + 3d + 2$


Finite Difference Spacing of Points for PDE's for Convergence of Explicit Forward-Stepping Scheme I realize that this question could be pretty broad, but I'm wondering at least what the conditions are for my simulation.

I'm developing an Explicit Forward-Stepping Finite Difference scheme to solve a nonlinear, time-dependent, heat flow (thermal diffusion) equation in cylindrical symmetry.  As such, I am interested in the temperature as a function of time t, radius R, and axial position Z (no $\theta$ dependence).

In my case, the radius is larger than the thickness (R >> Z), roughly 10 to 1.

My question is this:  when deciding the spacing (or stencil?) between points/nodes for my simulation, is it better to:

(1)  Use the same number of points in each spatial dimension (say, 100 in R, and 100 in Z)?

or,

(2)  Use the same ratio of points at the ratio of the spatial dimensions (say, 20 in R and 2 in Z, i.e. 10 to 1)?

As I said, I'd like to know the proper way to do it for general cases, but knowing for my case would be helpful by itself.

Also, does the time step also need to obey some equality in comparison to the spacing in R and Z at the same time?  I've seen conditions for convergence, but those appear to be some ratio between time and one spatial dimension, not time and all spatial dimensions.  Thanks!

My understanding is (IMHO), that it is more important that the spacing (or in your case, or ratio of points) be equal in each dimension.  That is to say, for your problem, that dr = dz, otherwise, diffusion will occur at different rates in different directions, leading to "false diffusion".  I don't have a source or reference for this.  So if R>>Z (R:Z = 10:1), then for dr to equal dz, you would need 10 times as many points in the r-direction than the z-direction.  For Cartesian this is more straight forward because the size of your cells remains constant if you take this approach, however with cylindrical (and spherical ?), it is a little more difficult to see.

If someone else wants to weigh in on this and/or has a proof or reference, I'd like to know!


$x + y = 7 $and $x^3 - y^3 = 37$, find $xy$ $x+y=7$
  $x^3-y^3 = 37$

  Find $xy$

I have tackled this one but I am stuck, when I get to this point:
$$(x-y)((x+2)^2 -xy) = 37$$
I would be glad if you could give me any suggestions.

Setting $y=7-x$, we obtain
$$
0=x^3-y^3-37=(2x^2 - 13x + 95)(x - 4).
$$
For integer solutions, the quadratic term has no integer root, so that $x=4$ and $y=3$, hence $xy=12$. Over the real numbers, this is also the unique solution, since $2x^2-13x+95=0$ has two complex, non-real solutions.


How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now. The answer is suppose to be $x = -by$, but I have no clue how to get there (have I made a calculation error??)
EDIT: Thanks everyone!!

The expression means that z subtends an angle $\frac{\pi}{2}$ at the points $2i$ and $6$
Ponder upon the following visual

The green arrows show some more positions which are possible for $z$. Clearly, as it subtends $90^o$ at those points, so the locus of $z$ is a semicircle, the end point of hose diameter is $(0,2)$ and $(6,0)$

You have not made a mistake. Observe that $x$ is a multiple of $y$:
$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$
which is what the answer states.


Real life applications of general vector spaces Students familiar with Euclidean space find the introduction of general vectors spaces pretty boring and abstract particularly when describing vector spaces such as set of polynomials or set of continuous functions. Is there a tangible way to introduce this? Are there examples which will have a real impact? I would like to introduce this in an engaging manner to introductory students. Are there any real life applications of general vector spaces?

Many years ago I was having a beer with a couple of fellow math grad students at some place around Harvard Square, and we overheard some guy at the next table trying to impress a girl telling her that he was taking a Linear Algebra course which was "so difficult" having to deal with spaces of "many dimensions".

I am not sure that this Linear Algebra technique of picking up girls at a bar may be listed among "real life applications", but, if it works, sure offers an important motivation.

In many Mathematical problems practical or theoretical we have a Set which may be sequence of numbers, continuous Functions etc. In which addition, subtraction, multiplication and division of two number belongs to that set. This suggests the concept of vector space


How to find the crossing number of a hypercube Q4? I'm struggling to find the crossing number of Q4, I think I have trouble visualizing the cube and finding the crossing number. Any idea what theorem or lemma I can use?

The Crossing number of Hypercube Q4 is 8.
Q4 can be constructed using two disjoint Q3 which is having a crossing number of 0, by adding an edge from each vertex in one copy of Q3 to the corresponding vertex in the other copy.
The lower bound for the crossing number of Qn is 4n/20 + O(4n/20).
The upper bound for the crossing number of Qn is (165/1024)*(4^n).


Prove that $\sqrt2-\sqrt5$ is irrational I've attempted this proof through contradiction and just wanted to check whether I did this correctly.

Assume that the claim is false, that is

$\sqrt2 - \sqrt5$ = $\frac {a}{b}$

Now square both sides

$2 - 5$ = $\frac {a^2}{b^2}$

$-3$ = $\frac {a^2}{b^2}$

$-3b^2$ = $a^2$

This is impossible as $a^2$ cannot be negative, so this is a contradiction.

Is this proof valid?

Hint: Your squaring must be: $$(a-b)^2=a^2+b^2-2ab$$

you square it wrong

if you wish to get 2-5 multiple by square root of 2 + square root of 5

but now be could have written the original question as (sqrt(2)-sqrt(5))/1

now when we multiple we get -3/sqrt(2)+sqrt(5)

now square this equation you get 4.5/sqrt(10) as sqrt(10) is irrational so division is also irrational so the square root of the equation is also irrational


can not find root of the equation $(0,0)$ with semi-major axis a and semi-minor axis $b$">

I want find out for each point $(x,y)$ in the white region, the nearest point $(a \cos \theta, b \sin \theta )$ on the ellipse curve.I tried with the following approach.
$$
\text {distance} = \sqrt{(x-a \cos \theta)^2+(y-b \sin \theta)^2}
$$
I find out the first derivative of the $$F(\theta)=(x-a \cos \theta)^2+(y-b \sin \theta)^2.$$ The equation after first derivative is:
$$
ax \sec \theta- by \csc \theta=a^2-b^2.
$$
How to calculate theta for this equation?

You can get it by solving a constrained optimization problem. You want the point $(u,v)$ that minimizes $f(u,v)=(u-x)^2+(v-y)^2$ subject to the constraint $\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1$. The point $(x,y)$ doesn't need to be inside the ellipse.

The Lagrangian is given by
$$
L(u,v,\lambda) = (u-x)^2+(v-y)^2-\lambda\left(\frac{u^2}{a^2} + \frac{v^2}{b^2}-1\right)
$$

Now you just need to compute the critical points of $L$ and choose the ones that yield the smallest distance to $(x,y)$.

You can solve this by hand but, out of laziness, I give you Wolfram's solution... Two critical points

WARNING: Wolfram is not giving the full set of solutions. After computing $u,v$ in terms of $\lambda$ from the first two equations of the system $\nabla L = 0$ and substituting them in the last equation we get a fourth degree polynomial equation in $\lambda$. The solution must be conmputed in an alternative way if this polynomial has four real roots.

$$
\left( -\frac{a \sqrt{b^2-y^2}}{b}, \frac{-a^2 b^4 y+a^2 b^2 y^3+a b^3 x y \sqrt{b^2-y^2}+b^6 y-b^4
   y^3}{a^4 b^2-a^4 y^2-2 a^2 b^4-a^2 b^2 x^2+2 a^2 b^2 y^2+b^6-b^4
   y^2}\right)
$$

and

$$
\left( \frac{a \sqrt{b^2-y^2}}{b}, \frac{-a^2 b^4 y+a^2 b^2 y^3-a b^3 x y \sqrt{b^2-y^2}+b^6 y-b^4
   y^3}{a^4 b^2-a^4 y^2-2 a^2 b^4-a^2 b^2 x^2+2 a^2 b^2 y^2+b^6-b^4
   y^2}\right)
$$

One will correspond to the minimum distance and the other will correspond to the maximum distance.

This is not relevant to your question, but the Lagrange multipliers are
$$
\frac{a^2 b^2-a^2 y^2+a b x \sqrt{b^2-y^2}}{b^2-y^2}
$$

and

$$
\frac{a^2 b^2-a^2 y^2-a b x \sqrt{b^2-y^2}}{b^2-y^2},
$$

respectively.

You have $axsec(\theta)- bycsc(\theta)= a^2- b^2$. Let $z= sec(\theta)$.  Since $csc(\theta)= \frac{1}{sec(\theta)}$, the equation becomes $axz- \frac{by}{z}= a^2- b^2$.  Muliply on both sides to get $axz^2- by= (a^2- b^2)z$.
We can write that as $(ax)z^3+ (b^2- a^2)z- by= 0$ and solve it using the quadratic formula:  $z= sec(\theta)= \frac{a^2- b^2\pm\sqrt{(b^2- a^2)^2+ 4abxy}}{2ax}$.


Determinant of a matrix equation Let $A=\begin{bmatrix}
1 &2  & 4\\
 0& 0 &3 \\
0 &0  &-1
\end{bmatrix}$, then we are asked to find the determinant of $A^{17}+A^{10}-I$
My approach is that the characteristic equation of this matrix gives $x^{3}=x$ so $A^{17}=A$ and $A^{10}=A^{2}$ hence $A^{17}+A^{10}-I$=$A^{2}+A-I$
Now $A^{2}$=$\begin{bmatrix}
1 &2  & 6\\
 0& 0 &-3 \\
0 &0  &1
\end{bmatrix}$ and $A^{2}+A$=$\begin{bmatrix}
2 &4  & 10\\
 0& 0 &0 \\
0 &0  &0
\end{bmatrix}$ and $A^{2}+A-I$=$\begin{bmatrix}
1 &4  & 10\\
 0& -1 &0 \\
0 &0  &-1
\end{bmatrix}$ which, because its lower triangular has a determinant of $1(-1)(-1)=1$;
But apparently the answer in my textbook for this is 0, could it be an error in the textbook or am i wrong somewhere in the reasoning, any help would be much appreciated

Another way to get the determinant is to see that
$$A^{17}=\begin{bmatrix}
1 &*  & *\\
 0& 0 & * \\
0 &0  &-1
\end{bmatrix} \quad \quad \text{and} \quad \quad A^{10}=\begin{bmatrix}
1 &*  & *\\
 0& 0 & * \\
0 &0  &1
\end{bmatrix}$$
so $$A^{17} + A^{10} - I =\begin{bmatrix}
1 &*  & *\\
 0& -1 & * \\
0 &0  & -1
\end{bmatrix}$$
Therefore the determinant is, indeed, equal to $1$.


Integral how to choose the good substitution? I have a problem with this integral. I don't know what kind of substitution to use.

$$\int t\sqrt{1+9t^4}dt$$

$$
\Big( t\,dt\Big)
$$
What you see above suggests at least considering $u=t^2$ so that $du=2t\,dt$, and then what you see above is $\frac 1 2 \, du$.

Since you have $9t^4 = (3t^2)^2$, using $u=3t^2$ works; then $du = 6t\,dt$ and $\frac 1 6 \, du = t\,dt$.

And then you have $9t^4 = (3t^2)^2 = u^2$, so $\displaystyle\int \sqrt{1+9t^4}\ t\, dt = \frac 1 6 \int\sqrt{1+u^2}\  du$.


New Question: Proving Divergence of a Sequence After I negated the definition of a convergent sequence, I ended up with the following mathematical statement:

$$\exists\ \epsilon > 0,\ \forall\ N \in \mathbb R\ \exists\ \mathbb N \ni n > N : |x_n - l| \ge \epsilon$$

Is this correct? I'd like clarification.

Anyway, I'm now asked to use my negation of the definition of convergence to prove that $(a_n)=((-1)^nn)$ is divergent. Can someone show me how?

Your negation is right, except for the little mistake that $N\in \mathbb{N}$. For your example, take $\varepsilon=1$. $\forall N \in \mathbb{N}$, take $n=\max\{2N,2l\}$. So we have examples that your statement is satisfied $\forall l\in \mathbb{R}$.


How to solve $y''' = y$ I'm trying to solve the following differential equation

  $ y''' = y$

and given conditions:

  $  y(1) = 3$, $y'(1) = 2$ and $y''(1) = 1 $

I began by making it:

  $y''' - y = 0$

But I'm uncertain of what to do after, or if I'm even headed in the right direction.
Please explain step by step as I'm taking an introductory course to Differential Equations!

The other solution is good.  Here is another method.

First, set $y' = x$.  Then we have $x'' = y$.

From here, set $x' = z$.  This gives us $z' = y$.

Notice that we have now expressed the original ODE as a system of equations:

$$x' = z$$
$$y' = x$$
$$z' = y$$

Therefore, we can rewrite our system in the form $A \mathbf{x'} = \mathbf{x}$, where $A$ is:

$$\left( \begin{array}{ccc}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right)$$

And I'll let you proceed with the usual eigenvalues method.

What remains the same through so many derivations? Well the first thing that one thins is always $Ae^x$

We can prove it properly aswell :): $$y=Ae^{mx}$$
$$y'=Ame^{mx}$$
$$y''=Am^2e^{mx}$$
$$y'''=Am^3e^{mx}$$

$$Am^3e^{mx}-Ae^{mx}=0$$
$$m^3-1=0,m=1$$
$$y=Ae^x$$


Evaluating $\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$ I am trying to solve the following integral:
$$\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$$
However the polynomial have imaginary roots:
$$-(-i A + B^2)^{1/2},\; (-i A + B^2)^{1/2},\; -(i A + B^2)^{1/2},\;(i A + B^2)^{1/2}$$
I am only interested on the solution from $0$ to $B$. How should I proceed?

Let $p=\sqrt{1+\frac{A^2}{B^4}}$ and substitute  $x=Bt$
$$I=\int_0^B \dfrac{1}{A^2 + (B^2-x^2)^2} dx
= \frac1{B^3} \int_0^1 \dfrac{1}{t^4-2t^2+p^2} dt
$$
Decompose  the integrand as
\begin{align}
\frac{1}{t^4-2t^2+p^2} = \frac1{2ps}\left( \frac{t+s}{t^2+st+p}- \frac{t-s}{t^2-st+p}\right)
\end{align}
with $s=\sqrt{2(p+1)}$. Then, integrate to obtain
$$I= \frac{1}{4B^3p}\left(\frac1{\sqrt{2(p+1)}}
\ln\frac {\sqrt{1+p}+\sqrt{2}}{\sqrt{1+p}-\sqrt{2}}
+ \sqrt{\frac2{p-1}}\tan^{-1}\sqrt{\frac2{p-1}}\right)
$$

I don't think those roots are correct.  Any complex roots of a polynomial with real coefficients must come in complex conjugate pairs.  If a+ bi and a- bi are roots then two factors are (x- a- bi) and (x- a+ bi) and their product is $(x-a)^2- (bi)^2= x^2- 2ax+ a^2+b^2$ and that will give a "partial fraction" of the form $\frac{Ax+ B}{x^2- 2ax+ a^2+ b^2}$.


How to solve the trigonometric inequality $\csc(2x) \leq \sec(x + \frac{\pi}{6})$ in $\mathbb{R}$? I have difficulty in solving the inequality

  $\csc(2x) \leq \sec(x + \frac{\pi}{6})$ in $\mathbb{R}$.

First, I need to solve the equation $\csc(2x) - \sec(x + \frac{\pi}{6}) = 0$. I see that the period is $2\pi$. My problem is that I have no idea how to factor $\csc(2x) - \sec(x + \frac{\pi}{6})$. Which method can I use  to transform it into a product of basic trig equations? Or is there another way to solve this equation ?

Thanks for your help.

$\cos x = \sin \frac \pi2 - x \implies \sec (x + \frac{\pi}{6} = \csc(\frac {\pi}{3} -x)\\
\csc 2x \le \csc (\frac \pi3 - x)$

When the LHS and RHS have the same sign: (careful, it is more involved than it appears on the face of it to find these regions.)

$2x \ge \frac \pi3 - x\\
x \ge \frac {\pi}{9} + k\frac {2\pi}{3}$

Plus the regions were the $RHS >0$ and $LHS < 0$

And keep in mind that $\csc x$ is undefined at certain points.

Actually it is a little trickier than that.  Here is a picture.

your inequality can be simplified to
$$\frac{1}{2\sin(x)\cos(x)}\le \frac{1}{\frac{1}{2}\sqrt{3}\cos(x)-\frac{1}{2}\sin(x)}$$
substituting $$a=\sin(x),b=\cos(x)$$ we get
$$\frac{1}{ab}\le \frac{2}{\sqrt{3}b-a}$$ with $a^2+b^2=1$
this gives $$-\frac{a-\sqrt{3}b+4ab}{2ab(a-\sqrt{3}b)}\geq 0$$


Is vector $w$ in the span of $\{v_1,v_2,v_3\}$? Let $v_1 = (1,0,-1)^T$, $v_2 = (2, 1, 3)^T$, $v_3 = (4,2,6)^T$ and $w = (7,4,7)^T$

From what I understand, if I set up an augment matrix in the form

\begin{pmatrix}1&2&4&7\\
0&1&2&4\\
-1&3&6&7\\
\end{pmatrix}

And reduce it using Gaussian elimination, if it is consistent, $w$ is in fact in the span of $\{v_1,v_2,v_3\}$. Else, if it is inconsistent, $w$ does not span $\{v_1,v_2,v_3\}$

This, $w$ does not span $\{v_1,v_2,v_3\}$. Is this true?

You are correct. If $w$ were an element of $\operatorname{span}(v_1, v_2, v_3)$, then we could represent $w$ as $w=xv_1 + yv_2 + zv_3$.  Thus there'd be a solution to the matrix equation $$\pmatrix{1&2&4\\0&1&2\\-1&3&6}\pmatrix{x \\ y \\ z} = \pmatrix{7 \\ 4 \\ 7}$$

Solving this, we see that
$$RREF\left(\begin{array}{ccc|c}1&2&4&7\\
0&1&2&4\\
-1&3&6&7
\end{array}\right) = \left(\begin{array}{ccc|c}1 & 0 & 0 & 0\\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1\end{array}\right) \\ \iff \begin{cases} x=0 \\ y+2z=0 \\ 0=1\end{cases}$$ but that last equation is never true so this system of equations has no solution.

Thus $w \not\in \operatorname{span}(v_1, v_2, v_3)$.

In other words, @tommy is wrong: $w$ is NOT a linear combination of those $3$ vectors.

yes as it is a linear combination of those 3 vectors


Linear Permutation of Grouped Things I can not search this over the internet so I thought some of you guys can help me out. The question goes like this.

There are 15 books. Of these are 5 Mathematics, 7 Physics, and 3 books on Chemistry. In how many ways can they be arranged in a shelf if the three subjects are to be grouped together?

P.S. I tried the using the permutation formula involving identical elements. But it's not arriving to the right answer in my book. Thanks in advance.

It doesn't seem like the permutation formula involving identical elements should apply here -- after all, the different Mathematics books are not identical. None of the fifteen books are identical.

I think the best way to think about this question is as follows: since each group must remain together, first choose the order of the groups, then choose the order within each group.

The number of ways of each books can be arranged is: $5!\times7!\times3!$ = $3628800$. The number of ways the subjects (group) can be arranged is: $3! = 6$.

Multiply both results = $21772800$.


Prove a lower bound of the smooth convex function with Lipschitz continuous gradient how to prove this conclusion:

Suppose that $f$ is a smooth convex function with Lipschitz continuous gradient on $X$, then there exists a self-adjoint and positive semi-definite linear operators $\sum_{f}$ such that for any $x, x^{'}\in X$,
$$f(x)\geq f(x') + \langle x -x', \nabla f(x')\rangle + 0.5 *\|x-x'\|_{\sum_{f}}^2.$$

Thanks for your attention and response.


Evaluate: $\lim_{x\to y} \frac{x^y-y^x}{x^x-y^y},$ without using L'Hospital's rule. I found this question in my younger brother's mathematics text book and they are not allowed to use L'Hospitals rule in exams. They evaluate limits using some fundamental formulas such as

$$\lim_{x\to0}\frac{\sin x}x=1,\;\lim_{x\to 0}\frac{e^x-1}{x}=1,\;\lim_{x\to0}\frac{a^x-1}{x}=\ln a,\; \lim_{x\to0}(1+x)^{\frac 1x}=e\text{ or }\lim_{x\to\infty}\left(1+\frac 1x\right)^x=e.$$

Though he knows about L'Hospital's rule but for exam purposes he asked me to solve it without using the same. But I'm not able to solve it using those formulas. So I posted it here to know if there's a way to solve it without the L'Hospital's rule. Thank you in advance.

Write $x^y=e^{(y\ln x)}$, then factor the numerator and denominator.


Where will they meet? A and B are running along the wall of a square park. The corners of the park are facing North, South, East and West and are named N, S, E and W respectively. They start at E and run towards S. If the speed of A is 6 times than of B, where do they meet for the 27th time?

a) between S and W
b) at S
c) between W and N
d) at N

Since $A$ runs faster, every time he meets $B$, he will have outruned him by an extra lap. Every lap measures $4$ times the length of the square, so if $x$ is the distance $B$ runs between two encounters with $A$ then:

$6x = 4+x \Rightarrow x = \frac{4}{5}$. So when they meet, $B$ would have ran $4/5$ times the side of the square.

So the $27$th time, $B$ would have run $27·\frac{4}{5} = 21 + \frac{3}{5}$ times the length of the side of the square.

Since every $4$ times, he's back on $E$, He'll be $1+\frac{3}{5}$ lenghts after $E$, i.e. between $S$ and $W$.

For $1$ complete round by $B$, $A$ completes 6 rounds of the square park.

Now for $B$ to meet $A$ for $27th$ time, $B$ must be completing $\frac {27} {6}$
rounds that is $4{\frac{1}{2}}$ rounds or
$\text{starting from E towards S, he will be at a position -}\\$

$\text{$a)$ between $S$ and $W$ side of the park}\\$


How do you find the center of a circle with a pencil and a book? Given a circle on a paper, and a pencil and a book. Can you find the center of the circle with the pencil and the book?

Let the corner of the book just touch the circle's edge. You can draw two perpendicular lines along book's edge which intersect with the circle at two different points. Connect the two points and it's a diameter.

Repeat this and get another diameter. The intersect of these two diameters is the center of the circle.

Hang the circle by its top edge on a wall close to a door or corner, draw a line straight down using the book's edge parallel to the door/corner edge. Rotate the circle f.ex 90 deg., draw a new line and you have the center and radius. This technique can also find the center of more complicated shapes. Gravity is the keyword here.


Total number of possibilities I have 3 buckets, 1st bucket has 5 red balls, 2nd bucket has 3 green balls and 3rd one has 2 blue balls. so I have total 10 balls in 3 buckets.

I need to know, what are the possible combination available if I need to pick up 4 balls( irrespective of colors) from 3 buckets. i.e I can pick up 4 from bucket 1 , or I can pick them up from different buckets as well.i.e 2 red, 1 green, 1 blue or 4 red or 2 green and 2 blue or 2 blue and 2 red...etc..what will be the mathematical formulation

We assume that balls of the same colour are identical. Let $r$ be the number of red we pick, $g$ the number of green, and $b$ the number of blue. We want to find the number of solutions of the equation $r+g+b=4$ in non-negative integers.

This would be taken care of by a general formula ("Stars and Bars") if there were $4$ or more balls in each bucket. However, there are only $2$ blue balls, so for example the solution $(0,1,3)$ is not possible.

I do not know whether you are expected to use a "general" procedure, or just plain count. We will do the latter, in a reasonably organized manner.

Divide into cases.

4 red: There is only $1$ way to do this.

3 red: We can pick a green, or a blue, $2$ ways.

2 red: We pick $2$ green, or $2$ blue, or $1$ of each, a total of $3$ ways.

1 red: We pick $3$ green, or $2$, or $1$, and the rest blue, a total of $3$ ways.

0 red: We pick $3$ green, or $2$, and the rest blue, a total of $2$ ways.

Add up.

Remark: Since the blues are fewest, it is obviously more efficient to let the cases be $2$ blue, $1$ blue, $0$ blue. You should do it that way, in order to make the solution "your own."

Since you don't care about colors, you will need to find possible combinations of choosing 4 balls from 10 balls.
$10!/6!4!=210$ combinations


simple geometry question- equation of cylinder A cylinder is $(x-a)^2+(y-a)^2=r^2$ with axis at $z$. I don't see where the '$z$' is in the equation. The book (calc 3) I'm using mentions the equation works for any $z$, but I don't see where the $z$ output is in the equation

Here is the excerpt from the book, the only part referencing to the basic equation of the cylinder as I've read 1-2 chapters beyond this point (it is a 4 chapter vector calc text)

The equation of an object is a way of telling whether a point is part of an object -- if you substitute the coordinates of the point into the equation and the equation is true, then the point is on the object; if the equation is not true for that point, then the point is not on the object. There is no $z$ because the z-coordinate is not part of the decision of whether a point is on the cylinder.

The fact that there is no z tells you that all points where the x- and y-coordinates satisfy the equation are part of the cylinder, regardless of the value of z. For example, if the equation is $(x-1)^2 + (y+2)^2 = 4$, then one of the points on the cylinder is (1,0,0), but so is (1,0,1) and (1,0,-1) and (1,0,5) and (1,0, -789) and so on. This means that the cylinder goes on forever both up and down.

An equation for a smoothed cylinder can be :

$$ x^2+y^2+z^{2n} = 1 $$

with $n\in\mathbb{N}$, $n\neq0$.

The greater the $n$ the smoother the top and bottom edges resulting from the "cut" of the infinite cylinder.

See for example https://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2Bz%5E20+%3D+1

smoothed cylinder


Finding the radius of a circle inside of a triangle What is the measure of the radius of the circle inscribed in a triangle whose sides measure $8$, $15$ and $17$ units?

I can easily  understand that it is a  right angle triangle  because of the given edges.  but I don't find any  easy formula to find the radius of the circle.

Hint:

use the fact that the area $A$ (of the triangle) is given by: $A=\frac{pr}{2}$ where $p$ is the perimeter and $r$ the incircle radius. This formula can easily be proved ( divide the triangle in three triangle with a common vertex at $O$) and is valid for a convex polygon..

The nswer is $3$, because one formula for the inradius is given by:
$$\frac{\text{area of triangle}}{\text{semi-perimeter of triangle}}$$


How many strings of length 12 can we compose using letters A, B, C, and D if every letter should appear at least once? How many strings of length 12 can we compose using letters A,B,C, and D if every letter
should appear at least once?

can someone walk me through this? I believe using the concept of the sieve formula is what i am supposed to use but I can't figure out where to start, my first thought is that 12! is the total number of permutations but then I know i have to take into account that each letter must be used once but I can't get any farther

EDIT: total number of strings with no restrictions is 4^12

please if someone can walk me through that would be great

Hint: you need the inclusion-exclusion principle.  How many total strings are there?  How many strings using only A, B, C are there?  It is easier to approach it this way, by removing the ones that have only three letters.  There are four times as many missing one letter, but the ones that have only A, B are overcounted in the subtraction.

I think the answer is 56(24 + 32).
since every letter should appear once so break the 12 letters into 8 + 4,
where 4 is compulsory (a combination of A,B,C,D so 4*3*2*1 = 24 arrangements).
And the rest 8 can be all A, all B , all C,all D so 4 choices for each position so 4*8 = 32.

So, total is 24 + 32 = 56.
Wat Say ??


Prove that $f_n(x) = n / (nx + 1) $ is pointwise convergent on $(0,1)$ Prove that $f_n(x) = n / (nx + 1) $ is pointwise convergent on $(0,1)$.

I know that the pointwise limit will be $1/x$ but I want to do it using the $\epsilon$ definition of convergence of a sequence of functions.

I think its pointwise limit is $1/x$ for every $x \in ]0,1[$.
To justify this, taking any $x \in ]0,1[$ and any $\varepsilon > 0$. We have
$$
\big| \frac{n}{nx+1} - \frac{1}{x} \big| = \frac{1}{nx^{2}+x} < \frac{1}{nx^{2}} < \varepsilon
$$
if $n > 1/\varepsilon x^{2}$,
so taking $N := \lceil \frac{1}{\varepsilon x^{2}} \rceil$ suffices.

Given $\epsilon>0$ and $x\in (0,1)$ choose $N\in \mathbb{N}$ as $N = $ (some function of $\epsilon$ and $x$).  Now for a $n>N$ we have
\begin{align*}
 |f_n(x) - 1/x| = \left| \frac{n}{nx+1} - 1/x \right| = \frac{1}{x(1+nx)}\le \frac{1}{x(1+N x)} \le \cdots \le \epsilon
\end{align*}

So you need to find the function and fill in the ellipses.


Difference between the dimensions $years^{-1}$ and, for instance, $\dfrac{animals}{years}$ in mathematical modelling using differential equations? I am told that an animal population has a natural growth rate of $k \ years^{-1}$ and is harvested at a rate of $a \ \dfrac{animals}{years}$.

Note that the constants $k$ and $a$ are both given as rates but, from a dimensional analysis perspective, they are different. I do not understand why one rate is given in the units $years^{-1}$ and the other is given in $\dfrac{animals}{years}$; indeed, I would have expected the growth rate to also be given in $\dfrac{animals}{years}$, since it is, after all, the growth rate of animals per year?

Obviously, this difference matters, since our units must make sense when modelling using differential equations. But I'm having trouble reasoning about this difference; it seems to make no sense when attempting to reason about it intuitively.

Why are the units $years^{-1}$ instead of $\dfrac{animals}{years}$, as they are for the other rate? Doesn't it make sense to have them both as $\dfrac{animals}{years}$, since they are rates?

I would greatly appreciate it if people could please take the time to help me understand this.

EDIT:

If readers are interested in the specific problem, it is from this MIT differential equations problem set:

One way to see the difference is to think about the differential equation involved.  If $N$ is the number of animals, then the differential equation will be in terms of $\frac{dN}{dt}$, which has units of animals per year.  This means that every term has to have units of animals per year.  The first term will have $k$, but $k$ by itself has the incorrect units; it has to be multiplied by $N$ to have the correct units.  The second term will simply be $a$ since $a$ already has the same units as the differential term.  For the first term, the rate depends on the population; for the second, the rate is a constant.  So the final differential equation is
$$\frac{dN}{dt}=kN-a$$

I think it is the same reason that angular speed is sometimes measured in $\mathrm{time}^{-1}$ (such as $\mathrm{s}^{-1}$) rather than $\frac{\mathrm{angle}}{\mathrm{time}}$ (such as $\frac{\mathrm{rad}}{\mathrm{s}}$).

It is because you consider some variables (angles, animals...) as dimensionless variables. This is because they are not physical quantities and they can be reduced to a proportionality constant (you measure the quantity 'animals' in the unit 'number of animals').
$$\frac{\mathrm{animals}}{\mathrm{years}}=\frac{1}{\text{total number of animals}}\frac{1 \text{ animal}}{\mathrm{years}}\simeq k \,\mathrm{years}^{-1}$$

It may seem reasonable to show the results in $\frac{\mathrm{animals}}{\mathrm{years}}$, specially is you want to talk about several ratios (like $\frac{\mathrm{predators}}{\mathrm{years}}$ or $\frac{\mathrm{preys}}{\mathrm{years}}$), but notice that all these ratios are comparable between each other (for example, 'predator population grow at a slower rate than prey population') because they have the same physical unit: $\mathrm{years}^{-1}$.


Fast way to check if two curves intersect? Say I had $y = 2x^2 + 4x - 3$ and $4 = x^2 + y^2$ and I wanted to know whether they intersected. I could simultaneously solve them, but that takes time. Is there a faster way of checking if two curves intersected, if I don't care about where the intersection is itself?

Generally speaking, there is no easy way to check that two curves intersect (even algebraic ones).

Here possible approach is to eliminate one of the variables to obtain a univariate polynomial. Then there are theorems about the approximate location of real roots.

In the case at hand, plugging $y$ in the second equation yields:

$$4x^4+16x^3+5x^2-24x+5=0.$$

The number of real roots can be settled by means of the Sturm Sequences.

The absolute fastest way? Suppose you are working over the complex field with points added at infinity. That is we stipulate that parallel lines actually intersect far out into the horizon (we say the line at infinity) similar to the way train track appear to the human eye. We call this projective geometry (Richard Souther has a visually epic account on youtube). Then by Bezout's theorem any two polynomials always intersect in either infinitely many points or in the product of their degrees counting multiplicity of roots. In this case they intersect in 4 points. Intersection becomes almost tautological (true by definition). In fact a form of Bezout holds for real space. It states that 2 polynomials intersect in at most the product of their degree points given they aren't the same polynomial. Here your parabola has degree 2 and your circle degree 2 so we know there can only be a maximum of 2x2=4 intersections. One sees there are less.

If you wanna be ${\rm I\!R}$eal though the fastest way (as C Marley said) is certainly graphically/intuitively. In this case your polynomials have exactly 2 intersections. 2nd degree equations in 2 variables are completely classified (they're all just intersections of a plane with a cone the so called conic sections (parabola, hyperbola, ellipse, circle)) so that you can always graph them relatively easily by completing the square to put them into standard forms. Higher degrees are much trickier not only geometrically but algebraically. Even this case results in a degree 4 polynomial which can seem impossible to solve exactly if it isn't in a nice easily factorable form although it still is plausible. There in fact is a cubic and a quartic equation similar to the quadratic equation! Although they are awfully long... Anything degree 5 or higher however doesn't have an equation similar to 2nd,3rd, and 4th degree polynomials and you'll need symmetric polynomials, permutations groups, automorphisms of field extensions, and much more a la Galois theory to determine if you can even theoretically solve it using the algebraic tools taught in HS (addition, subtraction, multiplication, division, and radicals).

The reality is the question you are asking is something that has been studied for millennium. It has greatly evolved over time and still grows to this day exponentially (if not factorially!) in the field of Algebraic Geometry. From humble Greek Euclidean beginnings on cones, to Rene Descarte's introduction of coordinates into geometry (the reason you can even write a geometric figure using algebra) to the modern day abstract algebra allowing us to study infinitely many different algebraic structures all at once, it is clear that the answer to your question is one that is highly non trivial. Algebraic Geometers were obsessed with precisely the question you asked, hence why they work in certain kinds of geometries to make polynomials behave nicely with such theorems as Bezout. What's so beautiful about the field is that the study of zeros of polynomial leads us to an understanding of all geometries as one further abstracts the very ideas that lead one to represent geometry through algebra. The breathtaking unity, elegance, and truth of Algebraic geometry and mathematics in general lead one to believe they are exploring the very thoughts of God that we mere humans spend lifetimes struggling to mine so we can grasp an infinitesimal fraction of infinity.


Find an upper bound for the modulus of a complex function For real $x$ let $r(x)$ be defined (implicitly) through

$$
e^{i\,x} = \left(1+i\,x\right)\,e^{\left(-(1/2)\,x^2+r(x)\right)},
$$
where $i$ is the complex unit. I need to prove that
$$
\left|r(x)\right|\leq \left|x\right|^3,\quad \text{for} ~~\left|x\right|\leq 1.
$$
Unfortunately I do not have any clue on this.

===Some computations====

Using expansion I get
$$
r(x) = i\,x+\frac{1}{2}\,x^2-\ln(1+i\,x)=i\,\frac{x^3}{3}+\frac{x^4}{4}-i\,\frac{x^5}{5}-\frac{x^6}{6}+i\,\frac{x^7}{7}+\frac{x^8}{8}+...
$$
that is
$$
r(x)=\sum_{k=2}^{\infty}(-1)^k\frac{x^{2\,k}}{2\,k}+i\,\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{2\,k+1}}{2\,k+1} = \frac{1}{2} \left(x^2-\log \left(x^2+1\right)\right)+i\,\left(x-\tan ^{-1}(x)\right)
$$

so that

$$
|r(x)| = \sqrt{\frac{1}{4} \left(x^2-\log \left(x^2+1\right)\right)^2+\left(x-\tan ^{-1}(x)\right)^2}
$$

I think I did it, if someone disagrees please let me know.

I think that the inequality does not hold. Let's find $r(x)$ explicitly. Let $p(x)$ and $q(x)$ be real functions such that $r(x)=p(x)+iq(x)$. To find $p(x)$, we take modulus on given equation:
$$
1=\sqrt{1+x^2}e^{-\frac{1}{2}x^2+p(x)}
$$
Then we get
$$
p(x)=\frac{1}{2}x^2 - \frac{1}{2}\ln(1+x^2).
$$
Substitute $p(x)$ to the original equation, then
$$
e^{ix}=(1+ix)e^{-\frac{1}{2}\ln(1+x^2)+iq(x)}=\frac{1+ix}{\sqrt{1+x^2}}e^{iq(x)}=e^{i(\theta(x)+q(x))},
$$
where $\theta(x)$ is a real number such that $\cos\theta(x)=\dfrac{1}{\sqrt{1+x^2}}$ and $\sin\theta(x)=\dfrac{x}{\sqrt{1+x^2}}$. Thus
$$
q(x)=x-\theta(x)+2n\pi
$$
for some $n\in \mathbb{Z}$. Define a complex-valued function $f:\mathbb{R}\to\mathbb{C}$ by
$$
f(x)=\frac{1}{2}x^2-\frac{1}{2}\ln(1+x^2)+i\left(x-\arccos\frac{1}{\sqrt{1+x^2}}+2018\pi\right),
$$
then $f$ satisfies the original equation. However, $f(0)\ne 0$.


Period of $f(2x+3)+f(2x+7)=2$ I have a problem in finding the period of functions given in the form of functional equations.

  Q. If $f(x)$ is periodic with period $t$ such that $f(2x+3)+f(2x+7)=2$. Find $t$. ($x\in \mathbb R$)

What I did:

$$f(2x+3)+f(2x+7)=2.........(1)$$

Replacing $x$ with $x-1$ in $(1)$,

$$f(2x+1)+f(2x+5)=2.........(2)$$

And replacing $x$ with $x+1$ in $(1)$,

$$f(2x+5)+f(2x+9)=2.........(3)$$

Subtracting $(2)$ from $(3)$, I get $$f(2x+1)=f(2x+9)$$

Since $x \in \mathbb R \iff 2x \in \mathbb R$, replace $2x$ with $x$ to get$$f(x)=f(x+8)\implies t=8$$

But sadly, my textbook's answer is $t=4$.

Is my method correct? How can I be sure that the $t$ so found is the least?

I found this tricky and I may have got it wrong, but I think that you are correct and the book is not.

If $f(2x+3)+f(2x+7)=2$ for all real $x$, then we can take $x=(y-3)/2$ to give
$$f(y)+f(y+4)=2\tag{$*$}$$
and then, essentially following your argument,
$$f(y)=f(y+8)$$
for all real $y$.  Now suppose that $f$ has period $t$: recall that this means $t>0$ and $f(y)=f(y+t)$ for all real $y$, and that no smaller positive $t$ has the same property.  Suppose that there is an integer $k$ such that
$$kt<8<(k+1)t\ ;$$
this can be rewritten as
$$8=kt+t'\quad\hbox{with}\quad 0<t'<t\ .$$
Then we have
$$f(y)=f(y+8)=f(y+t'+kt)=f(y+t')\quad\hbox{for all $y$},$$
which contradicts the fact that $f$ has period $t$.  Therefore we must have $8=kt$ for some positive integer $k$, that is, $t=8/k$.

Now $k$ cannot be even, for if $k=2m$ then for all $y$ we have $f(y)=f(y+mt)=f(y+4)$; in conjunction with $(*)$ this shows that $f(y)=1$ for all $y$, which is not (strictly speaking) periodic.

However it is possible that the period could be $t=8/k$ with $k$ odd.  For example, take
$$f(y)=1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)\ .$$
Then, as is well known, $f$ has period $t$; also
$$f(y)+f(y+4)
  =1+\sin\Bigl(\frac{2\pi y}{t}\Bigr)+1+\sin\Bigl(\frac{2\pi y}{t}+k\pi\Bigr)=2$$
for all $y$, as required.

The period is $4$. Actually $f(x)$ and $f(2x)$ are different functions. You see, $\cos(2x)$ and $\cos(x)$ are different. The period of $\cos(x)$ is $2\pi$ but the period of $\cos(2x)$ is $\pi$. Similarly, the period of $f(x)$ is the the period of $f(2x)/2$.


Spectrum of an upper triangular infinite matrix Is there any theory about spectra of triangular infinite marices? If so, where could I read something about it?

Any results about the spectra of triangular infinite matrices will be part of the spectral theory of linear operators. You can find it in any respectable functional analysis book. I found a book by Kreyszieg very approachable as I learned and my go-to reference is Methods of Modern Mathematical Physics by Reed and Simon.


Vector space as direct sum of kernel and image Let $T: V\to V$ be a linear map on an infinite-dimensional vector space $V$ over a field $F$. Suppose that $T(V)$ is finite dimensional, and $T^2(V)=T(V)$.

Show that $V=\ker T\oplus T(V)$.

(Note that this is not a duplicate of Show that the direct sum of a kernel of a projection and its image create the originating vector space., it seems to be more difficult since we only have $T^2(V)=T(V)$ rather than $T^2=T$.)

What I tried: I can prove $V=\ker T+T(V)$, what is difficult seems to be showing the direct sum, i.e. the intersection is 0.

Let $v\in V$. Since $T(v)\in T(V)=T^2(V)$, $T(v)=T^2(x)$ for some $x\in V$. Then $T(v-T(x))=0$, so that $v-T(x)=y\in\ker T$. Thus $v=y+T(x)\in\ker T+T(V)$. Hence $V\subseteq\ker T+T(V)$. $\ker T+T(V)\subseteq V$ is clear since $V$ is a vector space, so $V=\ker T+T(V)$.

I also have a previous result Linear Transformation from Infinite dimensional to Finite dimensional Space which may or may not be useful.

Thanks for any help!

The restriction of $T$ to $T(V)$ is an automorphism $T(V)\rightarrow T(V)$, in particular its kernel is $0$.

Proof. Since $v = \pi (v) + (v - \pi(v))$ and π(v – π(v)) = 0, V = im V + ker V. To show that the sum is direct, let v = π(w) ∈ ker π. Then 0 = π(v) = π^2(w) = π(w) = v, so im π∩ker π = 0.

Ran across this in a book I am reading.


When $\frac {a^3-b^3}{a^2-b^2}$ is an integer? If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer.

  For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?

Assuming $a\neq b$, we want that $\frac{a^2+ab+b^2}{a+b}=(a+b)-\frac{ab}{a+b}$ is an integer. The solutions of
$$ \frac{ab}{a+b} = k, $$
assuming $a+b\neq 0$, are the solutions of
$$ ab-ka-kb = 0, $$
i.e. the solutions of
$$ (a-k)(b-k) = k^2, $$
which depend on the couples of divisor/complementary divisor of $k^2$. In general, for any $d\mid k^2$ we have the solution $$a=d+k,\qquad b=\frac{k^2}{d}+k.$$

A trivial example is for $a,b \in \mathbb{Z}$ such that $a+b=1$


Is the series $\sum \frac{n+7^n}{n+5^5}$ converges or diverges I am asked to show whether the series

    $$\sum_{i=1}^{\infty}\frac{n+7^n}{n+5^n}$$

is convergent or divergent. I tried using Comparison Test but I was unsure of what to compare it to since there is addition in both the numerator and denominator. I also tried factoring out n from both parts of the fraction but this was not helpful either. I could use Ratio Test but we did not learn it yet. At this point I am stuck and have no idea what other methods to try.

Apply limit comparison  with $$\sum _1^\infty (7/5)^n$$ which is a geometric series with  $r=7/5$.

Note that $$\lim _{n\to \infty} \frac {(\frac {n+7^n}{n+5^n})}{ (\frac {7}{5})^n} =1$$

Thus the series $$\sum_{i=1}^{\infty}\frac{n+7^n}{n+5^n}$$ diverges as well.


Taylor series for tetration I would like to know what is known about Taylor series for tetration (and other hyper-exponentiations).

Surprisingly, such information is rare on internet. Numerical values for expansion in hyperexponent for tetration with basis $e$ can be found here:

http://en.citizendium.org/wiki/Tetration#Taylor_expansion_at_zero

However I am interested in expansion in the basis of tetration. I look for

$$ x \uparrow \uparrow m = \sum_{n=0}^{\infty} c_n(m)x^n $$

Wolfram mathematica gives me wired result (obviously a bug) at $x=0$ for $x \uparrow 3$:

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)

where in the output $log(x)$ appears (not a polynomial).

It is better at $x=1$

https://www.wolframalpha.com/input/?i=taylor+series+x%5E(x%5Ex)+at+x%3D1

I imagine the point $x=0$ may be "peculiar", but at least expansion at $x=1$ should be possible.

List of my questions:

1) Is tetration analytic at $x=0$? At $x=1$?

2) If yes, is an explicit closed-form formula for $c_n(m)$ known? (at any of these points)

3) If not, is an explicit closed-form formula for $c_n(m)$ know for some specific values of $m$?

4) Same questions for extension to higher hyper-exponentiations...

Thank you.

$x \uparrow \uparrow m$ is never analytic at $x = 0$ for $m > 1$. For example, consider $m = 2$, then you have the function $x \uparrow \uparrow 2 = x^x = e^{x \ln x}$ which clearly has a branch point at $x = 0$ and hence is not analytic. You can see this if we substitute $\ln x = 2\pi i n + \log x$ where $\log$ represents the principal branch of the logarithm, in which case we have $x^x = e^{x (2\pi i n + \log x)} = e^{x 2 \pi i n} e^{x\log x}$, which certainly depends on $n$ as long as $x$ is not an integer. You have similar problems for all $m > 1$.
The function $x \uparrow \uparrow m$ is always analytic at $1$, which can be seen inductively: Clearly for $m=1$ it is analytic at $1$. If $x \uparrow \uparrow m$ is analytic at 1, then $x \uparrow \uparrow{(m+1)} = e^{(\ln x)(x\uparrow \uparrow m)}$ is analytic at $1$ because $\exp$ is analytic everwhere and $\ln$ is analytic at 1. The Taylor series of $x \uparrow \uparrow m$ at $1$ is not very nice, but the terms can be computed explicitly. Wolfram Mathworld tells us that $$ (e^x) \uparrow\uparrow m = \sum_{n=0}^m \frac{(n+1)^n}{(n+1)!} x^n + \sum_{n = m+1}^\infty a_{m n} x^n$$ where $a_{m n}$ have a rather complicated recursive formula. To compute the Taylor series of $x \uparrow\uparrow m$ at $1$, you could compose the above series with the series for $\ln(x)$ at $1$ using Faa di Bruno's formula. Not very simple, but it probably as close to a closed form as you will get.

I think really the extension of the progression of operations by introducing a new one that is the $n$-fold iteration of the thus-far highest one is a process that on the face of it kindof looks like it ought to be susceptible of being plied indefinitely ... but at the end of the day isn't! (Except in this case to the natural numbers only.) You get it with complex-quaternion-octonion-etc also - that progression seems also to fizzle-out rather disappointingly; and a similar principle seems to be abroad in very many areas of mathematics. Sometimes I so want this or that progression to be extensible without limit ... but then I find eventually that I need to resign myself to the first two or three items of the progression being beautiful special cases. I must admit, though ... sometimes it's actually quite a relief!

However - I do think that business of relating the tetration by $n$ of the exponential function to the Lambert w function as $n\rightarrow\infty$, and the way there is a taylor series for the function at finite $n$ with coefficients that can be found by a recursion algorithm - and the way they 'peel-away' term-by-term to leave the coefficients of the lambert w function: I do find that to be an extraordinarily sweet little oasis in the midst of a parched desert ... and it is certainly an item that I treasure amongst my crown jewels!


Sequence of Events - Basic understanding I have an idea of the meaning of a sample space, and the events included in a sigma algebra. However, I am stuck in the definition of a sequence of events.

My difficulty is in the fact that in a sequence order matters, whereas in a set it does not. An event is a set of outcomes, and as such, it is not supposed to be ordered.

One way to look at it is to just to think of the real line, and Borel sets defined as open intervals, such that the intervals $(1,2), (2,3), (3,4)$, for example, would correspond to the sequence of events $\{(1,2)\},\{(2,3)\},\{(3,4)\}$. But I don't think this is the idea, and not all sets in a sigma algebra are defined on the real line.

Perhaps the idea is of nested "Russian dolls", whereby $E_1\subseteq E_2 \subseteq E_3 \subseteq E_4,\cdots$

So what is the definition and some valid examples of sequences of events (or sets)?

If you have any sample space, label it $\Omega$, then the Cartesian product $\Omega{\times}\Omega$ is also a sample space; and so in any $n$-order Cartesian exponential, $\Omega^n$ (for any positive integer $n$).

Any event drawn from the sample space $\Omega^n$ is also a sequence of $n$ events drawn from the sample space $\Omega$.

For instance, take the sample space of a fair coin toss, $\Omega=\{{\small\rm H},{\small\rm T}\}$.   A sequence of the events resulting from two fair coin tosses, such as $({\small\rm T},{\small\rm H})$ , is a event from the sample space of $\{({\small\rm H},{\small\rm H}),({\small\rm H},{\small\rm T}),({\small\rm T},{\small\rm H}),({\small\rm T},{\small\rm T})\}$; which is $\{{\small\rm H},{\small\rm T}\}^2$


Linear algebra terminology: unique, trivial, non-trivial, inconsistent and consistent Let me get this straight:

Unique solution- has an exact solution (such as a POI of 2 intersecting lines)
Trivial solution- when 0 needs to equal the zero vector in $Ax=0 vector$
Non-Trivial-? Perhaps when 0 does not equal the zero vector in $Ax=0$. This is due to the presence of free variables in Matrix A.
Consistent- Where there is a unique solution?
Inconsistent? When there is an inequality such as 4 can't equal 3 meaning the solution can not be solved. This is due to parallel lines.

Am I correct on these  technical terms?

I put inequality in italics because I'm not sure if I used it correctly.

Your formulations/phrasings are not very precise and should be modified:

Unique solution: Say you are given a $\mathbf{b}$ for which $A\mathbf{x}=\mathbf{b}$; then there is only one $\mathbf{x}$ (i.e., $\mathbf{x}$ is unique) for which the system is consistent. In the case of two lines in $\mathbb{R^2}$, this may be thought of as one and only one point of intersection.
Trivial solution: The only solution to $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.
Non-trivial solution: There exists $\mathbf{x}$ for which $A\mathbf{x}=\mathbf{0}$ where $\mathbf{x}\neq\mathbf{0}$.
Consistent: A system of linear equations is said to be consistent when there exists one or more solutions that makes this system true. For example, the simple system $x+y=2$ is consistent when $x=y=1$, when $x=0$ and $y=2$, etc.
Inconsistent: This is the opposite of a consistent system and is simply when a system of linear equations has no solution for which the system is true. A simple example $x-x=5$. This is the same as saying $0=5$, and we know this is not true regardless of the value for $x$. Thus, the simple system $x-x=5$ is inconsistent.

Inconsistent is when the system has NO solution. Consistent is the negation of this, e.g., we might have just one solution or infinitely many solutions. And a nontrivial solution might be a solution that's not the zero vector. So say $Ax = \begin{bmatrix}1 & 2 & 3\end{bmatrix}^T$, where $x=(1, 2, 3)^T$ and $A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$.


Is there a polynomial (or series) expression for summing $S_d(a,N)=\sum_{k=0}^{N-1} \log(1+{1\over a+k \cdot d})$? (perhaps Bernoulli-type) I need a quickly evaluatable expression for sums of consecutive logarithms of the type
$$ S_{d}(a,N) = \log(1+ {1\over a})+\log(1+ {1\over a+d})+\log(1+ {1\over a+2d})+ \cdots + \log(1+ {1\over a+(N-1)d})
$$
Here $N$ is typically very large (as well as $a$ while $d$ may be a small integer) why it is not feasible to evaluate the direct sum. The goal is to find some $a$ (not necessarily integer) with given fixed $d$ and $N$ which gives the sum a pre-determined value. For instance by a binary search inside a range of $a_\min$ and $a_\max$, and such a search needs to evaluate the sum multiple times.
I thought of some idea like assuming a Hurwitz-zeta-like ansatz of an infinite series and take the difference of two "exemplars" of them
$$ \begin{array}{lll}
 S^{^\star}_{d}(a) &= \log(1+ {1\over a})+\log(1+ {1\over a+d})+\log(1+ {1\over a+2d})+ \cdots  \\
 S_d(a,N) & \overset?= S^{^\star}_{d}(a/d) -  S^{^\star}_{d}(a/d+N)
\end{array}$$

Fumbling with transposing the Bernoulli-polynomials as a matrix of coefficients I seem to have got a possible solution, but perhaps there is a well known expression, perhaps like $\psi()$ function (in Pari/GP) or the like, or possibly even some more obvious expression.

  Q: Is there a polynomial or powerseries-expression for the finite sum of $\log(1+1/a_k)$ with equally spaced $a_k$ (having integer difference)?

Update: I've now an exposition of that method which employs the Bernoulli-numbers  /Zeta-values for an asymptotic series. It was too difficult to start this explanation in an answerbox here. I began with a draft in a pdf-file (see here) which I shall convert soon in a valid answer here.

I do not know how much this could help.

$$\sum_{k=0}^{N-1} \log(1+{1\over a+k \, d})=\sum_{k=0}^{N-1} \log(1+ a+k \, d)-\sum_{k=0}^{N-1} \log( a+k \, d)$$
$$\sum_{k=0}^{N-1} \log(b+k \, d)=\log\left(\prod_{k=0}^{N-1} (b+k \, d) \right)=\log \left(b d^{N-1} \left(\frac{b+d}{d}\right)_{N-1}\right)=\log \left(\frac{b d^{N-1} \Gamma \left(\frac{b}{d}+N\right)}{\Gamma
   \left(\frac{b+d}{d}\right)}\right)$$
$$S_{d}(a,N) =\log \left(\frac{\Gamma \left(\frac{a}{d}\right) \Gamma
   \left(\frac{a+1}{d}+N\right)}{\Gamma \left(\frac{a+1}{d}\right) \Gamma
   \left(\frac{a}{d}+N\right)}\right)$$ Using the  log gamma function could make the problem "simple" from a numerical point of view.

The derivative is
$$\frac{\partial} {\partial a} S_{d}(a,N)=\frac{\psi
   \left(\frac{a+1}{d}+N\right)-\psi \left(\frac{a}{d}+N\right)+\psi \left(\frac{a}{d}\right)-\psi\left(\frac{a+1}{d}\right)}{d}$$

I tried using $d=3$, $N=10^6$ and $S_{3}(a,10^6)=4.567$ using Newton method with $a_0=1$. The iterates are
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 1.00000 \\
 1 & 2.19160 \\
 2 & 3.54105 \\
 3 & 4.19248 \\
 4 & 4.27054 \\
 5 & 4.27141
\end{array}
\right)$$

It seems that the function looks like an hyperbola with an infinite branch when $a \to 0^+$. Trying polynomials does not seem (at least to me) to be a good idea.

Edit

If $N$ is really large, using Stirling approximations
$$S_{d}(a,N)\sim \log \left(\frac{\Gamma \left(\frac{a}{d}\right)}{\Gamma
   \left(\frac{a+1}{d}\right)}\right)+\frac{\log \left({N}\right)}{d}$$ Now, using an expansion around $a=1$ would give as a crude estimate
$$a=1+\frac {d \left(S_{d}(a,N)-\log \left(\frac{\Gamma \left(\frac{1}{d}\right)}{\Gamma
   \left(\frac{2}{d}\right)}\right)\right)-\log (N) } {\psi \left(\frac{1}{d}\right)-\psi \left(\frac{2}{d}\right) }$$

For the worked example, this gives almost exactly the first iterate of Newton method : $2.191599091$ instead of $2.191599310$. So, this is not of any interest.

For the generation of the starting guess, we can avoid the use of the $\Gamma(.)$ and $\psi(.)$ functions using the integral over $k$ instead of the sum. This would then require to solve (even approximately) for $a$ the equation
$$d\,S_{d}(a,N)=-(a+d (N-1)) \log (a+d (N-1))+(a+d (N-1)+1) \log (a+d (N-1)+1)+a \log (a)-(a+1)
   \log (a+1)$$ Considering that $N$ is large, this could reduce to
$$(a+1) \log (a+1)-a \log (a)=C\qquad \text{where} \qquad C=\log (edN)-d\,S_{d}(a,N)$$
Using a quick and dirty regression (data were generated for the range $0 \leq a \leq 10^4$) revealed as a good approximation
$$\log(a)=C-1$$

Applied to the worked example, this would give $a=3.364$


Difference between a deterministic and a nondeterministic finite automaton I am currently working on the automata theory and just passed through the definition of a nondeterministic and a deterministic finite automaton. The definition are quite similar and I don't understand well the difference between them. Can someone maybe passe again through them and also underline the difference with an example, if it is possible? Many thanks!

Both automata start from a single starting state.
A deterministic automaton, if it is in some state $q$ and it reads an input character, it will transit to a single state $q'$.
A non-deterministic automaton can transit from a state $q$ to many states $q_1, q_2, \ldots, q_k$ at the same time by reading a single input character. Also, a non-deterministic automaton can transit from one state to another even without reading input (these are $\varepsilon$-transitions). Intuitively, the non-deterministic automaton can make several transitions in parallel (and you have to take into account all of them), while the deterministic one has only one way to move at every step.
This difference is encoded in the signature of the transition functions for DFAs and NFAs. If $Q$ is the set of states and $\Gamma$ the set of characters, then the transition functions have the following signatures:
$$\delta_{D}: Q\times\Gamma \to Q\quad\quad\quad\quad\quad \delta_N:Q\times(\Gamma\cup\{\varepsilon\})\to \mathcal{P}(Q)$$
where $\delta_D$ is the transition function of a DFA and $\delta_N$ of an NFA.
The signature on the left means that from a single state and by reading a single input character, the automaton transits to a single state.
The signature on the right means that from a single state and by reading a single input character or by reading no character at all, the automaton can transit to a set of states.

In deterministic finite automata there is a unique final state but in non deterministic finite automata,the machine can be in many state at once called non determinism and have the capability to process epsilon transitions.


Probability of winning tic tac toe game. You and your friend like to play tic tac toe, but are not very good at it. On each turn, the player picks a random empty square to place their mark. The game ends when the first person gets three in a row, as usual. If you make the first move, what is the probability that you win the game?  If you make the first move, what is the probability that you win the game? If the answer can be expressed as $\frac{a}{b}$, a and b are coprime, find a+b.

My attempt:

No of probable ways of arranging tic tac toe symbols so as to win the game

=3horizontal ways+3 vertical ways+2 criss-cross ways=8

The total number of ways in which 3 spaces can be filled out of 9 spaces=9C3.

Therefore, Total number of outcomes=The total number of ways in which 3 spaces can be filled out of 9 spaces (and) One extra space where either of the two tic tac toe symbols can be filled up=9C3*2

Probability=8/(2*(9C3))=1/21

This is Problem E1324 from the American Mathematical Monthly, June-July 1958, p. 447. According to the solution the first player wins with
probability
$$\frac{1}{126}\left(62 + 36\cdot \frac{13}{40}\right)=\frac{737}{1260}.$$
See the details HERE.

Everybody takes this question wrong. For if you truly play, all games end in a draw. But the question begs, what are the possibilities? You have $9$ places, $X$ moves 1st. there are $3$ moves for $X$ corner, side, middle. $O$ moves 2nd, off an $X$ corner move there are $5$ possibilities for $O$, an $X$ side move there are $5$ possibilities for $O$, and off an $X$ middle move there are $2$ possibilities for $O$. So at the end of $2$ turns there are only $12$ possible games so for. So the question begs, how many possibilities are there at the end of $9$ moves?  How many does $X$ win? How many does $O$ win? How many end in a draw?


A property of the ring of integers of cyclotomic fields Let $t$ be a positive integer and consider $\mathbb Q(\zeta_t)$ a cyclotomic field. It is known that its ring of integer is $R=\mathbb Z[\zeta_t]$. I would like to know why there exists a nonzero prime ideal $P$ of $R$ that does not contain $t$.

Probably it is not so difficult to prove it, so I would appreciate some hints about it.

Correct me if I'm wrong, but try considering a prime $p$ that doesn't divide $t$. $(p)$ might not be prime but there are primes above it.


Find the smallest number which leaves remainder $8,12$ when divided by $28$ and $32$. The question is-

  Find the smallest number which leaves remainder $8,12$ when divided by $28$ and $32$.

My book gives directly a formula:

  Required number $= \mathrm{lcm}(\text{the two numbers; here}\ 28\ \text{and}\ 32) - \text{Sum of their remainders}$

without any proof.

I am not able to derive the proof or find a generalized form of this. I am also wondering what would happen if the numbers remain the same but the remainders are interchanged. Will the same method continue?

Any help is appreciated.

This formula isn't true, actually it works just in some cases, yours for example. This is because if $x$ is the required number then working modulo $28$, we have: $x \equiv -20 \pmod{28}$, which coincidentally is equal to $8$ modulo $28$. Try this formula for $22$ and $18$ and respecitve remainders of $5$ and $3$ and you will notice that it doesn't work

To find such a number (which sometimes might not exist) you need to solve the following congurence relations

$$x \equiv 12 \pmod{32} \implies x = 32t + 12$$
$$x \equiv 8 \pmod{28} \implies x = 28s + 8$$

Equating them and solving them you get:

$$32t + 12 = 28s + 8$$
$$28s \equiv 4 \pmod {32}$$
$$7s \equiv 1 \pmod 8 \implies s \equiv 7 \pmod 8 \implies s = 8k + 7$$

Substituting you will get $x = 28(8k + 7) + 8 = 224k + 204$. So the smallest two numbers satisfying the condition are $204$ and $428$.

The  formula given in your book is wrong.  I will explain but let me tell about the general types of these questions in detail.
 The questions are of the form  Find the number which when divided by x,y,z ...last divisor leaves the remainders p,q,r...last remainder of corresponding divisor. THESE QUESTIONS ARE ONLY ASKED WHEN x-p= y-q=z-r =...=k (In the given question 28-8=32-12 = 20) . The required number is (LCM OF x,y,z ...) -k
The proof of this formula is explained below.
Let the required number be L. .
L = xa+p = yb+q= zc-r and so on( a,  b, c are  variables here and are to be written as variables in all questions)(x y z...;p,q,r... will be numbers given in the question)
as p = x-k, q = y-k, r = z-k.
The above equations can be rewritten as
L = xa + x -k= yb +y -k=zc+z -k and so on [L =28a + 28 -20= 32b+32 -20]
L= x(a+1) -k=y(b+1)-k=z(c+1) -k and so on [ L=28(a+1)-20=32(b+1)-20]
These numbers x(a+1),  y(b+1) ,z(c+1) are equal and therefore equal to (m)(LCM OF x, y,z...).(m is any natural number)[LCM =224]
L=m( LCM OF x,y,z...)-k [224m -20]
and when the smalest value of L is asked m =1[224-20 = 204]


Logic in Philosophy vs. Mathematical Logic At my university, students majoring in philosophy take a course called "Logic in Philosophy" and there is also a course offered in the Math Department called "Mathematical Logic". There are also different books published in each field (Logic for Philosophy by Sider vs. Mathematical Logic by Enderton and Ebbinghaus).

Are these two distinct fields? If so, do they share common elements, concepts, and terminology/definitions?

Students majoring in philosophy take a course called "Logic in
  Philosophy" and there is also a course offered in the Math Department
  called "Mathematical Logic". Are these two distinct fields? If so, do
  they share common elements, concepts, and terminology/definitions?

Three points.

(A) Most philosophy students (at least those "majoring" in philosophy) are expected to do at least a Baby Formal Logic course, often in their very first year. Coverage will vary, but they should pick up an understanding of what makes a formally valid argument, of the truth-functional connectives, of the logic of quantifiers and identity, maybe one or two other things too like  familiarity with set notation. This will usually be done very slowly, remembering that most philosophy students have no maths background and many will be symbol-phobic. There may be more formal logic taught later in later years in optional courses, and eventually (though less and less these days) they might be offered a mathematical logic course.

Beginning university mathematics students will at various points, in e.g. courses on discrete mathematics, pick up an understanding of the connectives, of quantifiers, and of course of set notation, as they go along. But they won't usually have a stand-alone course on Baby Formal Logic (which hasn't much mathematical excitement!). So usually the first stand-alone logic course they might be offered is at senior undergraduate/beginning postgraduate course, in which Baby Logic gets covered (again) very fast, but this time with added completeness proofs etc. etc., before going on to other stuff.

So one main difference between an intro course in formal logic for philosophers and the early lectures in a course in mathematical logic for mathematicians is simply speed and level of mathematical sophistication. As noted, by mathematical standards, the combinatorial ideas involved in the philosophers' Baby Formal Logic course are extremely easy and elementary (even if it doesn't always look that way to the students having to take the course!). The same ideas can be covered at a cracking pace at the beginning of a  mathematical logic course.

(B) As I said, there may then be more formal logic offered to philosophers in later years in optional courses -- some of these courses covering various things of particular interest to philosophers, but not so much to (most) mathematicians, like modal logic. The second half of Sider's book indicates the sort of coverage typical of such courses. So here the focus of the coverage of courses in philosophy and mathematics department can indeed diverge.

(C) To take up some remarks in other answers, you need to sharply distinguish "Logic for Philosophers", meaning a treatment of some topics in formal logic that are of interest/use to philosophers, from "Philosophical Logic". The latter is a standard label for an area which is really a department of the broader philosophy of language, interested in questions about e.g. the nature of reference, the nature of logical necessity, the nature of a valid argument, the meaning of ordinary language conditionals, and so on. One technique for addressing such questions is to compare conditionals (to run with that example) in ordinary language with various constructed connectives in artificial languages of logic. So languages of formal logic get into the story as useful objects of comparison. And perhaps that's what unites the ragbag of topics in philosophical logic. But still, the real focus of courses on so-called philosophical logic is in issues that arise informally or pre-formally, not centrally in issues of formal logic.

As a mathematician (not a philosopher), I think that the biggest difference between philosophical logic and mathematical logic is that in mathematics we assume the existence of the most fundamental objects that we're trying to study.

The most common mathematical assumption is the existence of the integers.  They have bizarre properties.  For example, you can always add one to an integer to get a larger integer, i.e, there's an infinite number of them.  Estimates of the number of elementary particles in the visible universe are in the $10^{80}$ to $10^{97}$ range [1], so there might not be $10^{100}$ particles in the universe, yet $10^{100}$ is a well defined integer. $10^{10^{100}}$ is an integer with more digits than there are particles in the universe, yet it "exists", and what's more, I can  construct an integer that's larger than it: $10^{10^{100}} + 1$.  I can even prove that there are infinitely more integers that are larger than $10^{10^{100}}$.

I went to college to learn these proofs.  Aristophanes would approve.  [2]

Of course, integers do not really exist as physical objects.  If we accept the mathematician's claim that they do "exist", then we could do Anselm [3] one better:

Assume the existence of God.
Therefore, God exists.

I call it the "Mathematician's Proof of the Existence of God".

I know of no field of philosophy that so cavalierly assumes things into existence, although all philosophical arguments seem to be based on some claims assumed to be true.  Once these assumptions are clearly stated, then the same process of logical deduction (or induction) can be applied.

The great irony here is that by assuming the existence of physically implausible objects like integers and complex numbers, we can construct theories that explain the behavior of the actual physical universe to previously unimaginable precision.


If there exists a continuous non-constant map to the integers, then the space is not connected Let X be a topological space. show that if there exists a continuous, non constant map from X to the integers with the discrete topology, then X is not connected.

So I know that connected subspaces of integers with the discrete topology are just points. Also the image of a connected space under a continuous map is connected.

Here is where my reasoining for the proof eludes me. If I take the inverse image of those points is it that I haven now created a separation in the inverse image thus showing that X is not connected? Or is it that since the image of a connected space under a continuous function is connected, but since this maps to a point, the function is therefore constant?

If you already know that the image of a connected set is connected, then you're done: if $f:X\to \mathbb{Z}$ is connected, then $f(X)$ is connected.  The only connected subsets of $\mathbb{Z}$ are points, so $f(X)$ is a point, i.e. $f$ is constant.

Let $f$ be a non constant continuous function. It should suffice to prove it for two different points (Why?). Let $\{0,1\} = Im(f)$. Now $0$ and $1$ are open sets, and because $f$ is continuous we have that $f^{-1}(0)$ and $f^{-1}(1)$ are open and $f^{-1}(0) \cup f^{-1}(1) = X$, but $f^{-1}(0)\cap f^{-1}(1) = \emptyset$!