Composition of Permutations Suppose that

$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 7 & 4 & 1 & 6 & 5 & 3   \end{pmatrix}$$

$$\tau = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 2 & 5 & 4 & 3 & 7 & 1 \end{pmatrix} $$

$$\mu = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 4 & 1 & 3 & 2 & 6 & 8 & 5 & 7 \end{pmatrix}$$

Compute $\tau\mu\sigma^{-1}$

Workings:

$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 7 & 4 & 1 & 6 & 5 & 3   \end{pmatrix}$

$\sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 1 & 7 & 3 & 6 & 5 & 3  \end{pmatrix}$

Now I only know how to do the composition of permutations if they are of the same size. Not different sizes.

So I don't know how $\mu\sigma^{-1}$ is supposed to be computed.

If I were to guess

$\mu\sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 &5 & 6 & 7 \\ 2 & 4 & 5 & 3 & 8 & 6 & 3\end{pmatrix}$

Any help will be appreciated

Before multiplying, convert the smaller permutations to the size of the larger:
$$\sigma = \pmatrix{1&2&3&4&5&6&7&\color{blue}8\\2&7&4&1&6&5&3&\color{blue}8}$$
$$\sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 4 & 1 & 7 & 3 & 6 & 5 & \color{red}2 & 8\end{pmatrix}$$

Then proceed as usual.
$$\mu\sigma^{-1} = \pmatrix{1&2&3&4&5&6&7&\color{red}8\\2&4&5&3&8&6&\color{red}1&\color{red}7}$$
(red here is an error in your post).

Writing $\sigma$, $\tau$, $\mu$ as products of disjoint cycles, you have:

$\sigma=(1,2,7,3,4)(5,6)$
$\tau=(1,6,7)(3,5)$
$\mu=(1,4,2)(5,6,8,7)$

Now proceed as usual for compositions of permutations....


Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\frac{au -b}{a - bu}  $$
$$\sin x=\sqrt{1-\left(\frac{au -b}{a - bu}\right)^2}$$
 It is becoming messy with this .I also tried it using half angle formula but didn't find it good.

I think the tangent half-angle substitution should work. Here's what I have:

\begin{align*}
\int \frac{\mathrm{d}x}{(a+b\cos{x})^2} &= \int \frac{2\,\mathrm{d}t}{(1+t^2)(a+b\frac{1-t^2}{1+t^2})^2}, \qquad t = \tan{(x/2)} \\
&= 2 \int \frac{(1+t^2)dt}{a^2(1+t^2)^2+2ab(1-t^2)(1+t^2)+b^2(1-t^2)^2}\\
&= 2\int \frac{(1+t^2)\mathrm{d}t}{(a^2-2ab+b^2)t^4 + 2(a^2-b^2)t^2+a^2+2ab+b^2}\\
&=2\int \frac{(1+t^2)\mathrm{d}t}{((a-b)t^2 + (a+b))^2} \\
&=2\int \frac{(1+\frac{a+b}{a-b}\tan^2{\theta})\sqrt{\frac{a+b}{a-b}}\sec^2{\theta} \, \mathrm{d}\theta}{(a+b)^2\sec^4{\theta}} \qquad t=\sqrt{\frac{a+b}{a-b}}\tan{\theta} \\
&= 2\sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\int(\cos^2{\theta}+\frac{a+b}{a-b}\sin^2{\theta})\mathrm{d}\theta \\
&= \sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\left[\frac{2a}{a-b}\theta - \frac{2b}{a-b}\sin{\theta}\cos{\theta}\right]
\end{align*}

HINT
Put $$A=\frac{\ sin x}{a+b\cos x} $$now differentiate both sides and after simplification integrate it up to get the desired result.


Intuition of a Smooth Curve I'm currently taking an intro to Differential Geometry course, and am having trouble with the definition of a smooth curve. If you consider the curve $\lambda(t) = \left (\cos^3(t),\sin^3(t)\right )$  then clearly each of $x(t)$ and $y(t)$ are infinitely differentiable, and so by definition $\lambda(t)$ should be smooth. However plotting the curve shows that its graph has 4 cusps, and doesn't seem to be 'smooth' in any way that makes sense to me. Could somebody clarify for me the intuition behind what makes a curve smooth?

The curve that you describe is indeed defined by smooth, differentiable functions in both coordinates.

Anyway, the tangent vector has a direction depending on the two derivatives at the point considered. And if these derivatives turn out to be zero simultaneously, the direction is undetermined and the curve is free to form an angular point or a cusp.

As T. Bongers points out in the comments, the cusps you're getting reflect that the derivative is 0 at those points. I figured I'd expand on this and give a bit more intuition.

This is one of those times where it's important to distinguish between a manifold and its parametrization.

As you've already said, $\lambda(t)$ is a smooth map, $\Bbb{R}\to \Bbb{R}^2$, and we can call its image a curve (it is after all still a one dimensional manifold). However, its image is not a smooth curve.

What do I mean by this?

Well, we need to say precisely what we mean by a curve in $\Bbb{R}^2$. That is, we need to define the term submanifold. It turns out that there are some subtleties here, see the wikipedia page for the discussion. However, I suspect what would agree with your intuition here is the notion of an embedded (or regular) submanifold.

  An embedded submanifold, $N$, of a smooth manifold $M$ is a subset $N\subseteq M$ which can be given a smooth atlas compatible with the subspace topology such that the inclusion $N\hookrightarrow M$ is a smooth immersion.

  There is an equivalent local definition, which is that an embedded $k$-submanifold of an $n$-manifold $M$ is a subset $N$ of $M$ such that every point $p\in M$ has a coordinate neighborhood in $M$, $\phi: U\to \Bbb{R}^n$, such that $\phi(N\cap U)$ is the intersection of a $k$-plane with $\phi(U)$. These coordinate neighborhoods give the charts for the submanifold.

Now we can see that $\lambda$ fails to be a smooth immersion, since $d\lambda$ is zero at the cusps, which means that $d\lambda$ doesn't have full rank everywhere.

To see why the image can't be a smooth submanifold though, it's perhaps easier to look at the local definition. If we take $p$ to be a cusp, then there would need to be a neighborhood $U$ of $p$, and a smooth chart $\phi: U\to \Bbb{R}^2$ such that $\phi$ took the neighborhood of $p$ in the curve to a line in $\Bbb{R}^2$, which means that $\phi$ would have to "unfold" the cusp, which is impossible for a smooth map to do.


Division theorem for polynomials with integer coefficients I can see that the Division Theorem holds for polynomials in $\mathbb{Q}[x]$, but does not necessarily hold for polynomials in  $\mathbb{Z}[x]$, e.g.

Let $f=x^2+3x$ and $g=5x+2$.

Then the Division Theorem yields unique polynomials $q$ and $r$:

$f=gq+r$, in essence,

$x^2+3x=(\frac{1}{5}x+\frac{13}{25})\cdot(5x+2)-\frac{26}{25}$

where $q=\frac{1}{5}x+\frac{13}{25}$ and $r=-\frac{26}{25}$, but $q,r \notin \mathbb{Z}[x]$.

  The question is, how does one show that any $q,r$ fails to be in $\mathbb{Z}[x]$ to satisfy $f=gq+r$?

More concretely, for the existence of some $f$ and $g$ in $\mathbb{Z}[x]$, how does one show that there does not exist $q,r$ in $\mathbb{Z}[x]$ such that $f=gq+r$ with the property that $r=0$ or $\deg(r)<\deg(g)$?

If the leading coefficient of $g$ does not divide the leading coefficient of $f$ (in $\mathbb Z$) then there are no $q,r\in\mathbb Z[X]$ such that $f=gq+r$ with $r=0$ or $\deg r<\deg g$.

It's kind of silly honestly, but if you can't divide f by g for some reason like if g has a higher degree what we do we just set q to 0 and r to f and the statement holds.

but that's not super satisfying so what about if the degree of g is less than or equal well I can come up with something that still works. So let's refine what we mean by f. I'm gonna say f is a sum of variables and coefficients of this form

f = (a0)+(a1)x+(a2)x^2+...(an)*x^n (i don't know how to add subscripts)

basically I'm saying f is a polynomial and I'm listing out the x's and their coefficients because I'm gonna use the big (an)*x^n soon. I'm also gonna do the same thing for g. so g= (b0)+(b1)x+(b2)x^2+...(bk)*x^k.

so I'm gonna make a q now for f.
so I wanna call f1 = f -[a(n-k)/b(k)]  *x^(n-k)*g

: note (a(n-k)) is a coefficient in g not a* (n-k) again I don't know how to add subscripts yet

this is where I think the crux of your problem is, we need the biggest coefficient in f do be divisible by the biggest coefficient in g if we don't have it, then it doesn't divide any further we're stuck. Luckily that theorem assumes that (bk) is a unit

here's a quoted definition
"Let R be a commutative ring. Let f , g be two polynomials in R[x] with [g] ̸= 0, and suppose that the leading co- efficient of [g] is a unit of R..."

i.e.

that all this time an has to have a defined inverse in our assumptions.

now this gets easier we're nearly done.

f= f1+ [a(n-k)/b(k)]  *x^(n-k)*g
Now what?we induct. What do I mean we induct? well we have a base case let degree of f be 0, then if we remember at the top we can just let the whole expression be f=0*g+r where r=f. So now we can induct upwards using complete induction.

so
f= f1+ [a(n-k)/b(k)]*x^(n-k)g
 = gq1+r1+[a(n-k)/b(k)]*x^(n-k)g
 = [ gq1+[a(n-k)/b(k)]*x^(n-k)*g ]  +r1

= g*(q1+[a(n-k)/b(k)]*x^(n-k)*g)       +r1
hey wait this is what we wanted to prove.
done.


Find a real function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = -x$? I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:

  Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.

I am also wondering, can we find $f$ so that is continuous?

I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.

Source: http://www.halfaya.org/Casti/CalculusTheory2/challenge.pdf

An important piece of information is:

Theorem: $f$ is not continuous.

Proof: Observe that $f$ is invertible, because

$$f(f(f(f(x)))) = f(f(-x)) = x$$

and so $f \circ f \circ f = f^{-1}$. Any continuous invertible function on $\mathbb{R}$ is either strictly increasing or strictly decreasing.

If $f$ is strictly increasing, then:

$1 < 2$
$f(1) < f(2)$
$f(f(1)) < f(f(2))$
$-1 < -2$

contradiction! Similarly, if $f$ is strictly decreasing, then:

$1 < 2$
$f(1) > f(2)$
$f(f(1)) < f(f(2))$
$-1 < -2$

contradiction! Therefore, we conclude $f$ is not continuous. $\square$

For the sake of completeness, the entire solution space for $f$ consists of functions defined as follows:

Partition the set of all positive real numbers into ordered pairs $(a,b)$
Define $f$ by, whenever $(a,b)$ is one of our chosen pairs,

$f(0) = 0$
$f(a) = b$
$f(b) = -a$
$f(-a) = -b$
$f(-b) = a$

To see that every solution is of this form, let $f$ be a solution. Then we must have $f(0) = 0$ because:

Let $f(0) = a$. Then $f(a) = f(f(0)) = 0$ but $-a = f(f(a)) = f(0) = a$, and so $f(0) = 0$

If $a \neq 0$, then let $f(a) = b$. We have:

$f(b) = f(f(a)) = -a$
$f(-a) = f(f(b)) = -b$
$f(-b) = f(f(-a)) = a$

From here it's easy to see the set $\{ (a,f(a)) \mid a>0, f(a)>0 \}$ partitions the positive real numbers and so is of the form I describe above.

One particular solution is

$$ f(x) = \begin{cases}
   0   & x = 0
\\ x+1 & x > 0 \wedge \lceil x \rceil \text{ is odd}
\\ 1-x & x > 0 \wedge \lceil x \rceil \text{ is even}
\\ x-1 & x < 0 \wedge \lfloor x \rfloor \text{ is odd}
\\ -1-x & x < 0 \wedge \lfloor x \rfloor \text{ is even}
\end{cases}$$

e.g. $f(1/2) = 3/2$, $f(3/2) = -1/2$, $f(-1/2) = -3/2$, and $f(-3/2) = 1/2$.

(This works out to be Jyrki Lahtonen's example)

Closed form and generalised solutions
The piecewise function referred to in user14972's, Jyrki's and Alex's answers can be given in the closed form $$f(x)=\operatorname{sign}\left(x\right)-\left(-1\right)^{\operatorname{ceil}\left|x\right|}x$$
where $\operatorname{ceil}$ is the ceiling function. Let $n\in\mathbb{N}_0,\ \alpha\in(0,1]$, such that all $x\in\mathbb{R}^+$ can be represented by either $(2n+\alpha)$ or $(2n+1+\alpha)$. By substitution, we have $f(2n+\alpha)=2n+1+\alpha$ and $f(2n+1+\alpha)=-2n-\alpha$. Since $f$ is odd, we also have $f(-(2n+\alpha))=-(2n+1+\alpha)$ and $f(-(2n+1+\alpha))=2n+\alpha$, and therefore have the following cycle of values under $f$. Along with the trivial cycle $f(0)=0$, this proves that $f(f(x))=-x$ for all $x\in\mathbb{R}$.
$$\require{AMScd}
\begin{CD}
    2n+\alpha @>>> 2n+1+\alpha\\
    @AAA @VVV\\
    -2n-1-\alpha @<<< -2n-\alpha
\end{CD}$$
We may generalise this solution to $f^*(x)=c\cdot f\left(\frac{x}c\right)$, for all $c\in\mathbb{R}/\{0\}$. Even more general permissible functions may be defined by defining each cycle separately, as in the figure below, where $a,b$ are arbitrary real values.
$$\require{AMScd}
\begin{CD}
    a @>g>> b\\
    @AAA @VVhV\\
    -b @<<< -a
\end{CD}$$
The function, $g$, that transitions through the cycle can be defined quite broadly, and may not necessarily be even locally continuous. However it must not map $x$ to itself, nor $-x$, and cannot overlap with another cycle. The cycle is completed by defining $h$ as the negative inverse of $g$ and then defining both $g$ and $h$ as odd functions. For instance, if we define $g(x)=ax$ with $a>1$, we would have $h(x)=-\frac{x}{a}$ and $$f(x)=\frac{x}{2a}\left(\left(a^{2}+1\right)\left(-1\right)^{\operatorname{ceil}\left(\log_{a}\left|x\right|\right)}+a^{2}-1\right)$$
for nonzero $x$, which is admits another solution if we take $f(0)=0$. We can methodically generate solutions with this method by picking $g(x)$ as a continuous, strictly increasing function from $\mathbb{R}^+$ to itself, that is greater than $x$. If we then take a seed value $c\in\mathbb{R}^+$, we may partition $\mathbb{R}^+$ into the two disjoint sets,
$$
S_1=\bigcup_{n\in\mathbb{Z}}\ (g^{2n},g^{2n+1}],\quad
S_2=\bigcup_{n\in\mathbb{Z}}\ (g^{2n-1},g^{2n}]
$$
where $g^n$ denotes the $n$th iteration of $g$ on $c$ and where negative superscripts denote iterated inverses, e.g. $g^{-2}=g^{-1}\circ g^{-1}(c)$. We then have that whenever $x\in S_1$, $g(x)\in S_2$ and vice versa that $x\in S_2$ implies $g(x)\in S_1$. We can then define the solution $$f(x)=\begin{cases}0,&x=0\\g(x),&x\in S_1\\-g^{-1}(x),&x\in S_2\\-f(-x),&\text{else}\end{cases}$$


A linear transformation is an isomorphism if and only if it maps basis to basis A linear transformation $\tau\in L(V,W)$ is an isomorphism if and
only if there is a basis $B$ for $V$ for which $\tau B$ is a basis for $W$. In this case, maps any basis of $V$ to a basis of $W$?

Any linear transformation of Linear Space is isomorphism, if the V, W have the same dimensionality <-> transformation is not nullable <-> core is rank 0 <-> det L != 0 <->  there exist basis of V that maps in basis of W. if its true for 1 case its true for all cases, by definition of basis.

Because of that the only real thing that defines "face" of LS is number of its dimensions. All Linear spaces of same dimension are isomorph to each other.


largest number that leaves same remainder while dividing 5958, 5430 and 5814? Largest number that leaves same remainder while dividing 5958, 5430 and 5814 ?

$$5958 \equiv 5430 \equiv 5814 \pmod x$$
 $$3\times 17\times 19 \equiv 5\times 181\equiv 3\times 331\pmod x$$
$$969 \equiv 905\equiv 993\pmod x$$

After a bit of playing with the calculator,
I think the answer is $48$ but I don't know how to prove it.

Sorry if the answer is too obvious, I am still trying to wrap my head modular arithmetic and not very successful yet.It would be great help if anybody would give me some hints on how to proceed ahead. Thanks. $\ddot \smile$

Okay, first we set it up even though we have no idea where we are going.

$5958 \equiv n \mod x$

$5430 \equiv n \mod x$

$5814 \equiv n \mod x$.

Then we noodle a bit to make it smaller and more pallatable.

$5958 - 5814 = 144 \equiv 0 \mod x$

$5814 - 5430 = 384 \equiv 0 \mod x$

and

$5958 - 5430 = 528 \equiv 0 \mod x$

Well, now it's very clear that $x$ is a common divisor and all we have to do is find the $x = \gcd(144,384, 528)$

$x = \gcd(144 = 2^43^2, 384 = 2^7*3, 528 = 2^4*3*11) = 2^4*3 = 48$.

And we're done.

The only "small" solutions (below 10000000) seem to be ${1,2,3,4,6,8,12,16,24,48}$.
Of course, you can add $lcm({5958, 5430, 5814}) = 1741612770 $ to a solution and get another, bigger one.

The set above was found by letting do the computer what a computer can do best: Computing stuff. Here is the Haskell Code I used:

[x |  x <- [1..870806385 + 49], 5958 `mod` x == 5430 `mod` x, 5430 `mod` x == 5814 `mod` x]


Intuitive explanation of a positive semidefinite matrix What is an intuitive explanation of a positive-semidefinite matrix? Or a simple example which gives more intuition for it rather than the bare definition. Say $x$ is some vector in space and $M$ is some operation on vectors.

The definition is:

A $n$ × $n$ Hermitian matrix M is called positive-semidefinite if

$$x^{*} M x \geq 0$$

for all $x \in \mathbb{C}^n$ (or, all $x \in \mathbb{R}^n$ for the real matrix), where $x^*$ is the conjugate transpose of $x$.

One intuitive definition is as follows. Multiply any vector with a positive semi-definite matrix. The angle between the original vector and the resultant vector will always be less than or equal $\frac{\pi}{2}$. The positive definite matrix tries to keep the vector within a certain half space containing the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.

Let us take an example of a vector V = i + 2j which in matrix notation is

It will be represented as

POSITIVE DEFINITE MATRIX

A positive definite matrix is

How do we know it is positive definite matrix

The altered vector is

POSITIVE SEMI-DEFINITE MATRIX

The altered vector is [0 , 0]

NEGATIVE MATRIX

The altered vector in this case is

We can see the quadrant has changed


Determine eigenvalue pairs. Assume that $A$ is an $n\times n$ matrix. Assume that $u$ and $v$ are eigenvectors corresponding
to eigenvalues $5$ and $9$. Determine all pairs of real numbers $α$ and $β$ such that $αu+βv$ is also an eigenvector of $A$.

Could someone please advise if this works!
Since u and v are eigenvectors corresponding to eigenvalues 5 and 9,

We have Au = 5u …(1)
      Av = 9v  …(2)
Now we want to find all pairs of α and β  lets call them  α_i  and β_i  such that (α_i u+β_i v)
 is eigen vector of A i.e.
          $$      A(α_i u+β_i v)=e_i (α_i u+β_i v)
   < => Aα_{i} u+Aβ_{i} v=e_{i} (α_{i} u+β_{i} v)
   < => α_{i} Au+β_{i} Av=e_{i} (α_{i} u+β_{i} v)
    < => α_{i} 5u+β_{i} 9v=e_{i} (α_{i} u+β_{i} v)
    < => e_{i}=((α_{i} 5u+β_{i} 9v))/((α_{i} u+β_{i} v) )   …(3)

 e_{i} $$ will be=k when
      $$ α_{i}$$  happens to be the multiple of 9 i.e.$$α_{i}=9*k,
      β_{i} $$ happens to be multiple of 5 i.e.5*k
 except when k = 0

 Thus for$$ (α_{i} u+β_{i} v) $$ to be eigenvector of A $$ (α_{i},β_{i} )=[(9*k,5*k)]  ∀ k = ±1,±2,±3$$…and k!=0

So my question is different from the possible duplicate highlighted below - as it has a specific linear combination to deal with

Suppose $\alpha u+\beta v$ is an eigenvector with respect to the eigenvalue $\lambda$:
$$
A(\alpha u+\beta v)=\lambda(\alpha u+\beta v)
$$
By assumption, this can be rewritten as
$$
5\alpha u+9\beta v=\lambda\alpha u+\lambda\beta v
$$
and so
$$
(5-\lambda)\alpha u+(9-\lambda)\beta v=0
$$
Now recall that eigenvectors relative to distinct eigenvalues are linearly independent.

you have $$Av=5v$$ and $$Au=9u$$ now you have to find all a nad b such that $$ A(au+bv)=e(au+bv)$$ where e is new eigenvalue.

you have $$A(au+bv)=(aAu+bAv)=(a*9u+b*5v)=((9a)u+(5b)v)=e(au+bv)$$
but 5 and 9 are prime related to each other, you never can find such a and b,


Asymptotic expansion of a sum containing binomial coefficients I'm trying to find an asymptotic expansion for the following sum:

$$\sum_{k=2}^{n}\frac{n!}{k\left(n-k\right)!}=\sum_{k=2}^{n}\left(\begin{array}{c}
n\\
k
\end{array}\right)\left(k-1\right)!$$

for large $n$. According to Maple, this sum is equivalent to:

$$\frac{n^{2}-n}{2}\,_{3}F_{1}\left(1,2,2-n;3;-1\right)$$

where $_{3}F_{1}$ is a special case of the generalized hypergeometric function. But on the internet I cannot find any asymptotic expansion of $_{3}F_{1}$, because most work focused on other cases, e.g. $_{2}F_{1}$. I think it could be easier to work directly on the original sum. But by searching in previous posts I cannot find the specific sum I'm interested in. Moreover, asymptotic approximations of $\left(\begin{array}{c}
n\\
k
\end{array}\right)$ do not look very useful since they do not work for $k\approx n$. Help would be very appreciated!

Can be done in a similar way to this. Let $k = n - k$ for convenience:
$$S = \sum_{k = 2}^n \frac {n!} {k (n - k)!} =
\sum_{k = 0}^{n - 2} \frac {n!} {(n - k) k!} =
(n - 1)! \sum_{k = 0}^{n - 2} \frac 1 {k!} \frac 1 {1 - \frac k n}.$$
The main contribution comes from small values of $k$, and it is sufficient to use an approximation of the summand that holds for large $n$ and fixed $k$. The $i$th order term in the approximation will be simply $(k/n)^i$. Then the summation range in $S$ can be extended to infinity, and we get a complete asymptotic series for $S$:
$$S \approx e (n - 1)! \sum_{i = 0}^\infty \frac {B_i} {n^i},$$
where $B_0 = 1, B_1 = 1,  .\!.\!.$ are the Bell numbers.

You must be more specific. Do you want an asymptotic expansion for large $n$? if so, are you aware of Sterling formula to approximate $n!$ for large $n$? This should help you finding your answer.


The functional equation $f(xy)=f(x)f(y)$ Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.

With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,

$f(x)=x$
$f(x)=1$
$f(x)=0$

However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$.
So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.

Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?

If you take the equation
$$ \tag{*} f(x) = x^n $$
and "set $x=0$", what you get is not $f(x)=0$, but
$$ f(0) = 0^n $$
which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).

As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $\text{(*)}$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).

We may prove that the non-identical function $f(x)$ which is continuous over $(0,+\infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^\alpha$$ only.


Find $\sup \delta$ such that $f(x)$ is $\epsilon-$close to $L$ Let $f(x) = x^3 + 3x^2 - x +1$. Let $c=2$. Fine the least upper bound of $\delta$ so that $f(x)$ is bounded within $\epsilon$ of $f(2)$ for all sufficiently small $\epsilon>0$.

The solution says that

  In general, the bound is given by $\frac{1}{f'(c)}$ when $f$ is differentiable and $f'(c)\ne 0$.

Why $\frac{1}{f'(c)}$?

The derivative tells how fast a function in increasing at a point.

Intuitively (but not completely rigorously), if $|x-c| < \delta$ for small enough $\delta$,

$$|f'(c)| \approx \frac{|f(x) - f(c)|}{|x-c|},$$

which implies

$$|f(x) - f(c)| \approx |x-c| \cdot |f'(c)| < \delta |f'(c)|.$$

So in general, $\delta = \frac{1}{|f'(c)|}\epsilon$ will be your least upper bound of $\delta$.


Why is $0^0$ also known as indeterminate? I've seen on Maths Is Fun that $0^0$ is also know as indeterminate.  Seriously, when I wanted to see the value for $0^0$, it just told me it's indeterminate, but when I entered this into the exponent calculator, it was misspelled as "indeterminant".  Let's cut to the chase now.  I've seen this:$$5^0=1$$$$4^0=1$$$$3^0=1$$$$2^0=1$$$$1^0=1$$$$0^0=1$$and this:$$0^5=0$$$$0^4=0$$$$0^3=0$$$$0^2=0$$$$0^1=0$$$$0^0=0$$Right here, it seems like $0^0$ can be equal to either $0$ or $1$ as proven here.  This must be why $0^0$ is indeterminate.  Do you agree with me?  I can't wait to hear your "great answers" (these answers have to go with all of your questions (great answers).  What I mean is that you have to have great answers for all of your questions)!

The underlying problem here is that there are several related but differently defined concepts that are all notated $a^b$:

Exponentiation $\mathbb R\times \mathbb N_0 \to \mathbb R$ defined as repeated multiplication, which actually works as $R\times\mathbb N_0\to R$ for any ring-with-identity $R$. According to this concept $0^0$ is unambiguously $1$.
Exponentiation $\mathbb R^\times \times \mathbb Z \to \mathbb R^\times$ defined by extending the above to negative powers as reciprocals. This works for any field (or for the units of any ring). This doesn't define $0^0$ because it only works for numbers that have multiplicative inverses.
Exponentiation $\mathbb R_{\ge 0}\times \mathbb Q_{\ge 0}\to \mathbb R_{\ge 0}$ defined by roots of integral powers (themselves defined as above). If one bothers to include $a=0$ in this definition we again get $0^0=1$.
Exponentiation $\mathbb R_+\times \mathbb Q\to \mathbb R_+$, as above plus reciprocals and $a^0=1$.
Exponentiation $\mathbb R_{\ge 0}\times \mathbb R_+ \to \mathbb R_{\ge 0}$ and/or $\mathbb R_+\times \mathbb R\to\mathbb R_+$ by limits of the two above choices. This cannot be extended to give a value for $0^0$, because it has no limit for $(a,b)\to(0,0)$.
Exponentiation $\{e\}\times \mathbb R \to \mathbb R_+$ defined as a special case by the exponential function, which is again defined for example by power series. This extends to $\{e\}\times \mathbb C \to \mathbb C^\times$ for complex numbers.
Exponentiation $\mathbb R_+\times\mathbb R\to \mathbb R_+$ defined by $a^b = e^{b\log a}$. This doesn't extend to $a=0$ at all, and only with difficulty to complex $a$; we get multiple values due to the non-injectivity of the complex exponential.

Fortunately these definitions agree for those $(a,b)$ pairs where more than one of the concepts is defined. That's why it doesn't usually create problems to use the same notations for all of them. Note that all the definitions that do give meaning to $0^0$ result in $0^0=1$. One way to express this would be that in practice $a^b$ denotes the union of all these functions.

In practice some disagreement can arise, however, when someone claims that $a^b$ is "not defined" for some particular $(a,b)$ just because one of the definitions doesn't work for that $a$ and $b$. This is especially the case for $a=b=0$.

When $0^0$ is said to be an indeterminate form, what that means is neither more nor less than the fact that the limit
$$ \lim_{x\to a} f(x)^{g(x)} $$
cannot be evaluated by taking limits of $f(x)$ and $g(x)$ separately if $f(x)\to 0$ and $g(x)\to 0$.

In particular, the word "indeterminate" does not say anything about what happens if we do apply one of the exponentiation functions to $(0,0)$ and we haven't got those zeroes by "taking limits under the limit sign".

You're absolutely right. I would write it with limits, but the point is that there is no single value $0^0$ could be.


Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$. Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.

My try :

I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.

To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.

Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k>m$ ?

And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?

Thank you all in advance.

Suppose that $g$ has a zero at $a$. If $f$ is identically zero, we are done, as $0=0g$, so we may assume this is not the case, which implies that $g$ is not identically zero either, and therefore we may assume instead that it has a zero of order precisely $k$ at $a$, so $g(z)=(z-a)^kh(z)$, where $h$ is analytic in a neighborhood of $a$, and $h(a)\ne 0$.

Since $|f|\le|g|$, then $f(a)=0$, so $f(z)=(z-a)^mj(z)$ where $j(a)\ne0$ and $j$ is analytic in a neighborhood of $a$. For $z\ne a$ near $a$, $|f|\le|g|$ means that $$\displaystyle\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\le 1.$$

Since $j$ and $h$ are not zero near $a$, by continuity there is a constant $K$ such that $|j/h|\ge 1/K$ in a neighborhood of $a$, so $$\left|\frac{(z-a)^{m-k}j(z)}{h(z)}\right|\ge\frac{|z-a|^{m-k}}K,$$ or $|z-a|^{m-k}\le K$ for $z\ne a$ near $a$. But this is impossible if $m<k$.

Let $z_0$ be a zero of $g(z)$.Unless $g$ is identically zero, it can not be zero at points arbitrarily close to $z_0$. (By "identity principle".) And in that case $\displaystyle \lim_{z \to z_0} (z-z_0)\frac{f(z)}{g(z)} = 0$, (Since $|f(z)| \le |g(z)|$ ) which tells you that the singularity indeed is removable.


Stuck on proof and need help to solve it This is not homework
I do it for fun.

I am using the SD system.
Derive $L\supset T$

$L\supset\left ( C\vee T \right ).$ Assume
$\left ( \sim L\vee B \right )\wedge\left ( \sim B\vee\sim C \right ).$ Assume

As we see from the image it is valid in SL truth tree form

There does not seem to be an easy way to derive it. I figure the the easiest method is to derive a contradiction, but how to do it after the initial set up ?

$\sim\left ( L\supset T \right ).$ Assume

For "$Q$" and "$\sim Q$" from #2, we could use $B$ and $\sim B,$ but how to separate them so they could be used ?
Any help would greatly appreaciated.

A proof by contradiction for this one will be more work than doing a conditional proof, i.e. start a subproof, assume $L$, try to get to $T$, end the subproof, and perform a Conditional Introduction to get $L \supset T$
The rest of this proof is doing the Disjunctive Syllogism pattern over and over. So, if you have the Disjunctive Syllogism rule from SD+ this should actually be quite straightforward:
\begin{array}{lll}
1. & L \supset (C \lor T) & Assumption\\
2. & (\sim L \lor B) \& (\sim B \lor \sim C) & Assumption\\
3. & \quad L & Assumption\\
4. & \quad \sim L \lor B & \& \ Elimination \ Premise \ 2\\
5. & \quad \sim \sim L & Double \ Negation \ 3\\
6. & \quad B & Disjunctive \ Syllogism \ 4, 5\\
13. & \quad ...\\
? & \quad T\\
?+1 & L \supset T & \supset \ Introduction \ 3-?\\
\end{array}
In SD this will be a lot more tedious., but I'll show you how to do one of these. In particular, once you have assumed $L$ as part of your conditional proof, you should be able to get to $B$ using $\sim L \lor B$:
\begin{array}{lll}
1. & L \supset (C \lor T) & Assumption\\
2. & (\sim L \lor B) \& (\sim B \lor \sim C) & Assumption\\
3. & \quad L & Assumption\\
4. & \quad \sim L \lor B & \& \ Elimination \ Premise \ 2\\
5. & \quad \quad \sim L & Assumption\\
6. & \quad \quad \quad \sim B & Assumption\\
7. & \quad \quad \quad L & Reiteration \ 3\\
8. & \quad \quad \quad \sim L & Reiteration \ 5\\
9. & \quad \quad B & \sim \ Elimination \ 6-8\\
10. & \quad \quad B \ & Assumption\\
11. & \quad \quad B \ & Reiteration \ 10\\
12. & \quad B & \lor \ Elimination \ 4, 5-9,10-11\\
13. & \quad ...\\
? & \quad T\\
?+1 & L \supset T & \supset \ Introduction \ 3-?\\
\end{array}

Hint: Start with your given assumptions. Assume L. Use $P\to Q ~\equiv \neg P \lor Q$ to make a chain of implications to obtain T from L.


Two different results of calculation of the limit of $(2+4+\dots+2n)/n^2$ I would like to know where is the problem when i calculated those two limit with the following ways :
method (1):

$$\lim_{n\to \infty} \dfrac { 2+4+6+\cdots 2n} {n²} = \lim_{n\to \infty} \dfrac {n(n+1)}{n²}  =1 $$( using arithmitic sequence sum ).

Method (2) : now by using operation over limit we get this:
{n²}$$\lim_{n\to \infty} \dfrac { 2+4+6+\cdots 2n} {n²} = \lim_{n\to \infty} \dfrac {2}{n²}+\lim_{n\to\infty}\dfrac {4}{n²}+\cdots \lim_{n\to\infty}\dfrac {2n}{n²}  =0+0+\cdots 0=0 $$ .

Then is there some one who show me where is the problem in second method ?
Thank you for any help

You are trying
$$
\lim_{n\to\infty} \sum_{k=1}^n \frac{2k}{n^2}
"=" \sum_{k=1}^n \lim_{n\to\infty} \frac{2k}{n^2}
"=" \sum_{k=1}^n \lim_{m \to\infty} \frac{2k}{m^2}
$$
but $n$ has to be constant within the summation scope.
For the middle term it can not stay both fixed and be subject of a limit.
Only the first and last term are well defined (and different).

in method 2 you are summing an infinite number of zeros: the result is undefined, not 0.


Counting the number of surjections. How many functions from set $\{1,2,3,\ldots,n\}$ to $\{A,B,C\}$ are surjections? $n \geq 3$

Attempt

I was hoping to count the number of surjections by treating $A,B,C$ like bins, and counting the number of ways to fill them with $1,2,3,\ldots,n$ such that no bins are empty. This is equivalent to putting two "separator" bars in between the $n$ numbers or $n-1 \choose 2$.

However, I think I am missing a step here: prior to putting in the separator bars, I should permutate the $n$ elements. So, multiply by $n!$.

The order of elements in the same bins do not matter. So I should be divided by the number of permutations of the elements in each bins. This is where I am stuck.

Additional Info

I made the question up myself. Feel free to modify the question into a more solvable form if it is too difficult/complicated.

You’ve actually made a pretty good start: your first idea doesn’t work, but it’s a reasonable one to try, and you’ve spotted the difficulty with it. What you want now is an inclusion-exclusion argument. There are $3^n$ functions altogether. $2^n$ of them are functions from $\{1,\dots,n\}$ to $\{A,B\}$, missing $C$ altogether, so we need to throw them out. There are also $2^n$ functions that miss $B$ and $2^n$ that miss $A$, so our improved approximation to the number of surjections is $3^n-3\cdot2^n$.

Unfortunately, we’ve now gone overboard. The function that sends every $k\in\{1,\dots,n\}$ to $A$, for instance, has been thrown out twice, once for not hitting $B$ and once for not hitting $C$. The same goes for the other two constant functions. We need to throw each of them back in once, so that on net we’ve thrown each of them out just once, not twice. The revised number of surjections is then

$$3^n-3\cdot2^n+3=3\left(3^{n-1}-2^n+1\right)\;.\tag{1}$$

A little thought should convince you that no further adjustments are required and that $(1)$ is therefore the desired number.

formula for finding the total number of surjections (onto functions)


simplex algorithm - minimization So I get the basic concept of simplex algorithm but I am working on a project where I have to implement any linear programming algorithm (I chose simplex method) to minimize a function, but I don't have enough numbers (stay with me here). Basically, I tried by setting M_fe = 0 and M_sp = 0, but I still need a value for sigma in order to get the values of the 8 variables in eqn(1) and 3 variables in eqn(2). Variables in eqn(1) and eqn(2) hold the same value. Here's the following information:

Constraint equation:

4.534*F_bic + 3.167*F_bra + 2.930*F_brd + 0.174*F_pro - 1.997*F_tri + 0.02*F_cr + 0.248*F_ecu + 0.726*F_ecrl = M_fe (moment of flexion) ----------------- eqn(1)

1.28*F_bic - 0.1954*F_brd - 0.1736*F_pro = M_sp (moment of supination) ---- eqn(2)

Inequality constraint:

F_i/PCSA_i <= sigma ----------------- eqn(3)

Objective function:

Minimizing sigma ---------- eqn(4)

Some conditions:
M_fe can be equal to zero if that makes the problem easier to solve.
PCSA_i are variables and their values are given as follows:

PCSA_bic = 4.6
PCSA_bra = 7.0
PCSA_brd = 1.5
PCSA_pro = 3.4
PCSA_tri = 18.8
PCSA_fcr = 2.0
PCSA_ecu = 3.4
PCSA_ecrl = 2.4

Basically, my question is that is there any way to solve this problem without having a value for sigma? Funny thing is sigma is the same in eqn(3) for all 8 variables, which made me plugin the value of sigma for each corresponding variable in eqn(1) and eqn(2). Doing that gave me

13.094*sigma = M_fe ---------- eqn(5)

5.005*sigma = M_sp ----------- eqn(6)

But now how would i solve for M_fe and M_sp. This problem is like a merry-go-round. Any help is greatly appreciated. Also, this is my first post and technically my first time so please forgive me if I didn't follow any specific protocols in this post.

Thanks.

So I actually figured out how to solve this problem in May but never had a chance to post my answer. So sigma is minimized and the inequality constraints can then be solved. Using a simplex algorithm, we can set the equation as follows:

Ax = B

Aeq = [4.534 3.167 2.930 0.248 0.0201 0.716]

Beq = [20]

x = [FBIC; FBRA; FBRD; FECU; FFCR; FECRL]

The values from x can then easily give us the F_i value, which I was trying to solve for in equation 3.

Although this problem was very specific and extremely difficult, similar simpler problems can also be solved using a simplex algorithm and the following equation set up.

I can also share my MATLAB code if anyone is further interested in looking at this problem.

Cheers!


Sequence Convergence proof check I am trying to solve the following sequence question:

  Show that $\{a_n\}_{n=1}^\infty$ converges to $A$ if and only if $\{a_n - A\}_{n=1}^\infty$ converges to $0$.

Proof: Assume $\{a_n\}_{n=1}^\infty$ converges to $A$. Then for each epsilon > 0, there exist N such that for all n>=N, abs($\{a_n\}$- $A$) < epsilon. Now consider the sequence $\{a_n - A\}_{n=1}^\infty$. For this sequence to converge, it must be that for each epsilon > 0, there exist P such that for each n>=P, abs($\{a_n - A\}$ - $B$) < epsilon. Let $B$ = 0, so we have abs($\{a_n\}$- $A$) < epsilon. Since we know that abs($\{a_n\}$- $A$) < epsilon when n>=N, it follows that N=P. Therefore, since there exist this P where for all n>=P, abs($\{a_n - A\}$ - $B$) < epsilon, $\{a_n - A\}_{n=1}^\infty$ converges to $0$.

This is only the proof for the first assumption.

(1) $a_n \to A \Rightarrow \forall \epsilon>0, \exists N$ such that $|a_n-A|< \epsilon, \forall n >n$. To have $(a_n-A) \to 0$ then we need that for a given $\epsilon, \exists N_0$ s.t $|a_n - A - 0| = |a_n - A|< \epsilon$ for all $n >N_0$. Just take $N_0 = N$ from above.

(2) If $(a_n-A) \to 0$ then given $\epsilon>0$ there exists $N_1$ s.t $|a_n - A - 0| = |a_n-A|< \epsilon$ for all $n>N_1$ but again if you choose $N = N_1$ in the definition of $a_n \to A$ then you are done.

Looks good. You can actually prove both directions at once as follows:

$\{a_n\}_{n=1}^{\infty}$ converges to $A$ if and only if for every $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $\vert a_n - A \vert < \epsilon$, which is true if and only $\{a_n - A\}_{n=1}^{\infty}$ converges to $0$.


Absolute value laws - $|a \cdot b| = |a| \cdot |b|$ - Is my reasoning correct? I have the expression \begin{vmatrix}{ \dfrac{ (-1)\sqrt{n}(x+3) }{ \sqrt{n+1} } }\end{vmatrix}

We know that the absolute value laws state that $|a \cdot b| = |a| \cdot |b|$.

Therefore, would I be correct in saying that $ \begin{vmatrix}{ \dfrac{ (-1)\sqrt{n}(x+3) }{ \sqrt{n+1} } }\end{vmatrix} = \dfrac{ |x+3||-1|\sqrt{n}}{\sqrt{n+1} }$ ?

Thank you.

No. We have

$\begin{vmatrix}{ \dfrac{ (-1)\sqrt{n}(x+3) }{ \sqrt{n+1} } }\end{vmatrix}= \dfrac{ \sqrt{n}|(x+3)| }{ \sqrt{n+1} } $

Edit: You are right. My mistake, I have read $ \dfrac{ |x+3|-1\sqrt{n}}{\sqrt{n+1} }$


Complex solutions of $e^{e^z}=1$ I know that the solution of $e^z = 1$ is $2\pi k$, for integers $k.$  Note $z$ is complex.  I just don't really know how to start tackling $e^{e^z} = 1$

It seems that this means we want $e^z = 2\pi k$, i.e $z = \log(2\pi k)$, but this somehow seems like it's too simple to be right.

First off, there is an error in your question: if $\omega \in \mathbb{C}$, then
$$ \mathrm{e}^{\omega} = 1 \implies \omega = 2\pi k\color{red}{i}, $$
where $k$ is any integer.  Note that this is a purely imaginary number, rather than the real value $2\pi k$ in your question.  Taking $\omega = \mathrm{e}^{z}$, we have
$$ \mathrm{e}^{\mathrm{e}^z} = 1
\implies \mathrm{e}^z = 2\pi k i, $$
where $k$ is any integer.

Next, it might be helpful to (rather pedantically) expand out the right-hand side of the equation in terms of exponentials:
$$ \mathrm{e}^z
= 2\pi k i
= (2\pi k) (i)
= \mathrm{e}^{\log(2\pi k)} \mathrm{e}^{i\left(\frac{\pi}{2} + 2\pi\ell\right)}
= \mathrm{e}^{\log(2\pi k) + i\left(\frac{\pi}{2} + 2\pi\ell\right)},
$$
where $\ell$ is an arbitrary integer and $\log$ is the the natural logarithm defined on $\mathbb{R}$; we are using Euler's formula to write out $i$ as an exponential expression.  Now, equating the real and imaginary parts on the left- and right-hand sides of the equation, we have
$$ \Re(z) = \log(2\pi k)
\qquad\text{and}\qquad
\Im(z) = \frac{\pi}{2} + 2\pi\ell, $$
where $k \in \mathbb{Z}\setminus\{0\}$ (note that the logarithm of zero is undefined, hence $k$ cannot be zero) and $\ell \in \mathbb{Z}$.  In other words, with $k$ and $\ell$ as above, we conclude that
$$ z = \log(2\pi k) + i \left( \frac{\pi}{2} + 2\pi\ell \right). $$

The solution should be for $k\neq 0$

$$e^z=2ik\pi\implies z=2ik_1\pi+\log 2ik\pi=2ik_1\pi+\log (2k\pi\cdot e^{\frac{\pi}2i+2ik\pi})=\\=2ik_1\pi+\log 2k\pi+\frac{\pi}2i+2ik\pi=\\=\log 2k\pi+\frac{\pi}2i+2in\pi$$

Refer here for Complex logarithm.


How to prove $\sum_{n=1}^{\infty} \frac{3^n +7n}{2^n (n^2+1)} $ diverges? $$\sum_{n=1}^{\infty} \frac{3^n +7n}{2^n (n^2+1)} $$

It seems clear to me that this seires diverges since the dominant term is $(3/2)^n$, a geometric series with $r > 1$

However I am required to prove this using convergence tests, presumably the comparison test to prove divergence.

I cannot work out a suitable comparison to make to prove divergence, suggestions?

you can prove that the term doesn't tend to 0, so the necessary condition of convergence is not satisfied, and since all terms are positive, the sum has to be $\infty$.

If you want to compare:
clearly $\frac{3^n +7n}{2^n (n^2+1)}>\frac{3^n}{2^n (n^2+1)}\ge \frac{3^n }{2^n (n^2+n^2)}$ and now you can easily use Cauchy's test to prove that $\sum\frac{3^n }{2^n \cdot 2n^2}$ diverges.

Note that

$$ \frac{3^n +7n}{2^n (n^2+1)} \ge \frac{3^n }{2^n (n^2+1)}, $$

However, $$\sum_{n=1}^{\infty}\frac{3^n}{2^n (n^2+1)} $$ diverges, so the original serries cannot converges.


Financial Mathematics, Simple interest question. Help. Laurie deposits $\$60,000$
in a bank at $5\%$ interest per annum.

Andrew deposits $\$40,000$ in bank at $8\%$ per annum.

How long wil it take, by simple interest, for Andrew to have more money than Laurie?

What I have done so far:

I calculated the simple interest earned by both in $5$ years:

Laurie:
$\$60,000 \cdot 5 \cdot 5/ 100 =\$15000$

Andrew:
$\$40,000 \cdot 8 \cdot 5 / 100 = \$16000$

Hint: The simple interest per year for Laruie is $0.05\cdot 60000=3000$, while the simple interest for Andrew is $0.08\cdot 40000=3200$. So the total balance of Andrew has to be greater than Laurie's, i.e. solve
$$40000+3200x>60000+3000x$$

Let $L(n)$ and $A(n)$ be the amount of money Laurie and Andrew have after $n$ years, respectively. Then

$$\begin{align*}
L(n) &= 60000\left(1+\frac1{20}\right)^n=60000\left(\frac{21}{20}\right)^n\\
A(n) &= 40000\left(1+\frac2{25}\right)^n=40000\left(\frac{27}{25}\right)^n\\
\end{align*}$$
We want to find $$n^* = \operatorname{argmin}_n\{A(n) > L(n) \}.$$

Since $L$ and $A$ are positive, increasing functions, this is equivalent to $\frac{A(n)}{L(n)}>1$. We have

$$ \frac{A(n)}{L(n)} = \frac{40000}{60000}\left(\frac{27}{25}\right)^n\left(\frac{21}{20}\right)^{-n}= \frac23\left(\frac{36}{35}\right)^n$$
Since $\frac{A(n)}{L(n)}>1$ if and only if $\log\left(\frac{A(n)}{L(n)}\right) >0$,
$$ \log\left(\frac{A(n)}{L(n)}\right) = \log\frac23 + n\log\frac{36}{35}>0\iff n>\frac{\log\frac23}{\log\frac{36}{35}}\approx 14.393.$$
So Andrew will have more money after $$\lceil 14.393\rceil = 15$$ years.


I need to do math from ground up, so what is a good workbook? Can you guys recommend me a workbook that begins with arithmetic and ends with calculus.
Or from pre-algebra to calculus.
Like all "Master Math Series" books but in one complete book.

It would really help me if you knew a few.

Thanks.

Technically these are many books, but check out the Easy Way series. It has algebra, geometry, trig, pre-calculus, and calculus, (maybe some other books that I did not know before) with plenty of practice problems. There are also other books in this series if you are interested.


How do we know that the number $1$ is not equal to the number $-1$? How do we know that the number $1$ is not equal to the number $-1$? (I am not talking about the multiplicative inverse of an arbitrary field, but the integer/rational, real or complex number $1$.)
Is that an axiom?
Since someone in the comments asked for more background/motivation: I'm just a naive bachelor student and I wanted to prove that for any number $x$, $x=-x$ implies $x=0$ (which is true if and only if $1\neq -1$, see my last question).
Why is this question different to the question in my last post?
In my last post, I wanted to know if you can prove that $x=-x$ implies $x=0$ for an element $x$ of an arbitary field. It turned out that you can not (there are fields where this is not the case). In addition, it turned out that this is true if and only if $1\neq -1$. ($1$ stands for the multiplicative identity, not the number one.)
Since we normally assume that $x=-x$ implies $x=0$ if $x$ for some number $x$ (at least in my lectures), I wanted to know wheter you can prove that $1\neq -1$ or not. (To be honest, I didn't think that you have to distinguish e.g. between the integer $1$ and the natural number $1$...)
To put it into a nutshell: My first question was wheter you can prove $1\neq-1$ where $1$ stands for the multiplicative identity of an arbitrary field (the answer is no.) But we know that the answer is yes if $1$ stands for the integer/real number/complex number one, which led to this post.

That depends on the context, e.g. $1=-1$ in $\mathbb{Z}/2\mathbb{Z}$. Mathematics is a very strict discipline, you should always be precise about the meaning of symbols.
However judging by your tags you are asking about integers. To answer that question we first need to know what those symbols are and what integers are. One way to construct the set of integers $\mathbb{Z}$ is via $(\mathbb{N}\times\mathbb{N})/\sim$ where $(a,b)\sim (a',b')$ iff $a+b'=a'+b$. In that situation our $1$ is formally $[1,0]_\sim$ while $-1$ is $[0,1]_\sim$. For them to be equal we would have to answer the question whether $1+1=0+0$ over naturals?
Now Peano arithmetic jumps in. First of all $1$ is defined as the successor of zero: $1=S(0)$ by definition. The successor function $S$ and the zero $0$ are primitives in the Peano axioms. The addition is then defined recursively by $a+0=a$ and $a+S(b)=S(a+b)$.
Thus $0+0=0$. On the other hand $1+1=0$ has no chance of happening because
$$1+1=S(1+0)=S(1)$$
by definition, while $0$ is not a successor of anything by a Peano axiom.
Note that similar reasoning can be applied to any non-zero integer $x\in\mathbb{Z}$ since either $x$ or $-x$ can be represented by $[n,0]_\sim$ for some non-zero $n\in\mathbb{N}$. Non-zero is essential because the only natural that is not a successor is zero.

In any ring we have $1=-1\iff 1+1=0$ (just add/subtract $1$ from both sides).
Though, in $\Bbb Q$ [or $\Bbb R$ or $\Bbb C$] we have $1+1=2\ne 0$.


$\lvert\operatorname{Aut}(F/K)\rvert$ divides $[F:K]$ Let $F/K$ be a finite extension.
I want to prove $\lvert\operatorname{Aut}(F/K)\rvert$ divides  $[F:K]$.
I know  $\lvert\operatorname{Aut}(F/K)\rvert\leq [F:K]$ but I cannot show the former.

Take $\alpha$ from $K$, then $\lvert\operatorname{Aut}(F/K)\rvert＝\lvert\operatorname{Root}P_{\alpha}(K)$.
$[L:K]＝\deg P_{\alpha}$.
But I cannot go further. Thank you in advance for helping me.

Denote $G=\mathrm{Aut}_K(F)$. Then $F/F^G$ is a Galois extension with Galois group $G$. Thus, we have
$$|G|=[F\colon F^G]=\frac{[F\colon K]}{[F^G\colon K]}\Big\vert[F\colon K].$$

Every automorphism of $F$ over $K$ is determined by its action on a basis, which has $[F\colon K]$ elements. Thus, if the basis is mapped into itself by the automorphism group, then Aut$(F/K)$ can be represented as a subgroup of the permutation group on $[F\colon K]$ elements. Thus, the number of elements of Aut$(F/K)$ divides the size of the permutation group, which is $[F\colon K]$ factorial.

Your question asks about a much smaller bound, without the factorials. And for this we need a more refined approach. The way that the inequality $|\textrm{Aut}(F/K)|\leq [F\colon K]$ is proven here is to decompose the finite extension into a tower of simple extensions. Both sides of the inequality are multiplicative, and for simple extensions it is straightforward to get the bound.

Now if you follow along in this proof, each time you obtain an inequality between two integers, it is actually showing that one divides the other. Hence your result follows.

An earlier version of this answer was incorrect. The example I was considering was the following: a cubic extension with automorphism group $S_3$, which I believed could be obtained from an explicit construction in the comments below. That construction is incorrect, and no such extension exists.


Regarding order of integration I have following integration problem. If i change the order of integration I get different answer. Can anybody which order of integration is correct and why?
The problem is
$$\int_0^x\int_0^\infty\frac{e^{-a/b}}{b}e^{-b}dbda $$
Now if I integrate with respect to $b$ first then
$$\int_0^x 2K_0(2\sqrt{a})da$$
$$-2\sqrt{a}K_1(2\sqrt{a})|_0^x$$
$$-[2\sqrt{x}K_1(2\sqrt{x})-2\sqrt{0}K_1(2\sqrt{0})]$$

Now if I integrate with respect to $a$ first then the problem becomes
$$\int_0^\infty\int_0^x\frac{e^{-a/b}}{b}e^{-b}dadb$$
$$\int_0^\infty[1-e^{-x/b}]e^{-b}db$$
$$1-\int_0^\infty e^{-(x/b)-b}db$$
Now using 3.371-9 from Ryzhik and Gradshteyn i can find it to be
$$1-2\sqrt{x}K_1(2\sqrt{x})$$

These answer can become equal if $$2\sqrt{0}K_1(2\sqrt{0})=1$$
So can we say them to be equal?

The relevant theorem is Fubini's theorem

Here your function is non-negative, so it apply and both order gives the same result.

Now, I didn't check your calculations, so you may have a mistake somewhere in them.


How to show if $\sum_{n=1}^\infty 7X_{n+1} + X_n$ converges and $X_n \to 0$ when $n \to \infty$, then $\sum_{n=1}^\infty X_n$ converges? Note that $X_n$ is just a sequence of real numbers.

I'm a bit out of ideas because I'm used to having the extra hypothesis of $X_n$ being of positive terms. Any hints or suggestions will be greatly appreciated!

Let $L:=\sum_{n=1}^\infty (7X_{n+1} + X_n)$ and $S_N=\sum_{n=1}^N X_n$. Note that $$\sum_{n=1}^N( 7X_{n+1} + X_n) = 8S_N + 7X_{N+1}-7X_1$$
thus $\lim_{N\to \infty} (8S_N + 7X_{N+1}-7X_1) = L$. From the hypothesis $\lim_N X_N=0$, we can infer that $S_N$ converges and $\lim_{N\to \infty} S_N = \frac 18(7X_1+L)$.

$$\sum _{n=1}^\infty 7X_{n+1}+X_n=X_1+8\sum _{n=2}^\infty X_n$$
The second hypothesis is unneccessary. Indeed, it is implied by the first.


Show that the function $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$. Show that $\phi (x,y)=\arctan{\frac{2x}{x^2+y^2-1}}$ is harmonic by considering $w(z)=\frac{i+z}{i-z}$.

I know that if $\phi$ is harmonic then it satisfies Laplace's equation but I don't see how considering $w(z)$ will help me.. I've tried looking at the real and imaginary parts of $z$ but they don't seem to match up. I'm guessing the solutions will have something to do with the Cauchy Riemann equations but I've tried differentiating $w(z)$ with respect to $x$ and $y$, where $z=x+iy$, and that hasn't shed any light either.

Hint:

  $w$ is a fractional linear transformation, hence a conformal mapping (of what domains?). Therefore its inverse is also conformal. Composing a harmonic function with a conformal map gives you another harmonic function. So it remains to see what function you get by composing $\phi$ and $w$.

suppose $f(z) = Re^{i\Theta}$ and $f$ is differentiable. then, it can be shown that $\Theta$ is harmonic.

now, $$w(z)=\frac{i+z}{i-z}=\frac{x+i(y+1)}{-x - i(y-1)}= \frac{[x+i(y+1)][-x+i(y-1)]}{[-x - i(y-1)][-x+i(y-1)]}=\frac{(1-x^2-y^2)-2ix}{x^2+(y-1)^2}=Re^{i\Theta}$$

where $\Theta = \tan^{-1}\left(\frac{2x}{x^2+y^2-1}\right)$ and therefore is harmonic.


Probability of bit strings Suppose you pick a bit string of length $10$. Find the probability that the bit string has exactly two $1$'s, given that the string begins with a $1$.
Can someone please explain to me how to do it?

Since you know that the first bit is $1$, you just want to find the probability that exactly one of the remaining $9$ digits is $1$. The total number of possible $9$-bit sequences is $2^9$. Of these only $9$ have exactly a single $1$, one for each possible position of the $1$. So the probability is $\frac{9}{2^9}$.

X~The number of 1s in a bit string of length 10
X~B(10,0.5)
We just split it by cases since we know first one is already 1.
Y~The number of 1s in a bit string of length 9
Y~B(9,0.5)
P(Y=1)=0.01757
P(given first is 1 and we want 2 1s)=(0.01757×0.5)/P(X=2)
=0.2


Showing $\{X_n\}$ is uniformly integrable so I want to prove that $\{X_n\}$ is uniformly integrable, given that E[$|X_n|^2$] < C for every n
where C is a constant.
so using the markov inequality we get P($|X_n|^2$ $\geq$ N) $\leq$ $\frac{(E[|X_n|^2])}{N}$  $\leq$  $\frac{C}{N}$ , $\lim \limits_{N \to \infty}$$\frac{C}{N}$ = 0 therefore we can conclude
that:$\lim _{N\rightarrow \infty}\int_N^\infty P(|X_n|^2\geq N)dx \rightarrow 0$
but then I get that $\{X_n^2\}$ is uniformly integrable
how can I continiue from here or did I missed something?

Note that
$$1\{|X_n| \geq N\} N |X_n| \leq 1\{|X_n| \geq N\} |X_n|^2,$$
hence
$$\int_{|X_n| \geq N} |X_n| dP\leq \frac{1}{N} \int_{|X_n| \geq N} |X_n|^2 dP.$$
Can you conclude?


Are there any proofs that use $0/0$ is indeterminate? I'd imagine that "$n/0$ is undefined $\forall n\neq 0$" is very useful in finding contradictions, but are there any proofs that somehow use "$0/0$ is indeterminate"?

Yes. You can use that $0/0$ is indeterminate to conclude that certain expressions are in fact $0$ (and this is pretty much the only thing you can use $0/0$ for). However, nobody writes down such arguments because:

Writing such an argument requires you to be very careful, hence requires a lot of effort.
Any such argument can be replaced by a much easier argument of the form "Either the expression is $0$, or I get a contradiction: therefore the expression is $0$".

For an example of how to use that $\dfrac 00$ is indeterminate (and for a more general explanation of what it means to "divide by $0$", see this answer). On the flip-side, the fact that no number corresponds to the faction $\dfrac n0$ for $n\neq0$ is incredibly useful: for example, if you have an expression $\lim\dfrac {f(x)}{g(x)}$ and you (carefully) reduce it to an expression of the form $\dfrac n0$ where $n\neq0$, you can conclude that the limit does not exist, without mucking about with $\epsilon$'s and $\delta$'s.

Hope I understood you correctly, but here's a very simple example.

consider $\lim_{x \to 0} \frac{x}{x}$. it tends to one.

Now consider $\lim_{x \to 0} \frac{2x}{x}$. it tends to two.

We have 2 different limits to something that is close to $\frac{0}{0}$.

Was this helpful?


Characteristic Function of Geometric Brownian motion - PDE with terminal condition approach I want to compute the characteristic function of the standard geometric brownian motion. Note that I know that it is easy when you exploit the distributional properties of the process, but I'm trying to come up with some exercises by myself in order to apply the same approach to broader classes of stochastic processes.
Consider the usual equation of the GBM:
$$
dX_t = \mu X_t dt + \sigma X_t dW_t
$$

Then, by Feynman-Kac, we have to solve the PDE:
$$
\frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2 x \frac{\partial^2 f}{\partial x^2} = 0
$$
In order to find the characteristic function, we have to take into account the terminal condition
$$
f(x,T) = \exp(i \omega x)
$$
Substituting $f(x,T)$ inside the PDE yields
$$
\mu x i \omega \exp(i \omega x) - \frac{1}{2}\sigma^2 x \omega^2 \exp(i \omega x)=0
$$
But then, how do I proceed? The above doesn't look at all like the characteristic function of a random variable with lognormal distribution).
Obviously considering the logarithm of $\frac{dX_t}{X_t}$ simplifies things a lot, however I really want to came up with the characteristic function of $X_t$, not its logarithm (see p. 41-42 in Pricing Bermudan and American Options Using the FFT Method).

I hope that somebody can help me, or even discuss things a little bit. By the way, excuse me for my poor way of handling PDEs. :)

First of all there is a typo in the partial differential equation(PDE) that you state. Let us firstly recall that PDE describes the time evolution of the one-point probability density function of the stochastic process in question. That PDE is termed the backward Chapman-Kolmogorov equation; you can look up the Wikipedia page to learn how to get from the Langevin equation (the stochastic differential equation) to the Chapman-Kolmogorov equation.
Now the correct version is :
\begin{equation}
\frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2x^2 \frac{\partial^2 f}{\partial x^2} = 0
\end{equation}
Now solving PDEs is in general a hard task but in this particular case the  substitution $y=\log(x)$ turns the PDE into one with constant coefficients. Indeed we have:
\begin{eqnarray}
\frac{\partial f}{\partial y} &=& x \frac{\partial f}{\partial x}\\
\frac{\partial^2 f}{\partial y^2} &=& x \frac{\partial f}{\partial x} + x^2 \frac{\partial ^2 f}{\partial x^2}
\end{eqnarray}
and therefore our PDE takes the form:
\begin{equation}
\frac{\partial f}{\partial t} + \left(\mu-\frac{1}{2} \sigma^2\right) \frac{\partial f}{\partial y} + \frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2} = 0
\end{equation}
which is basically saying that the logarithm of a geometrical Brownian motion with parameters $(\mu,\sigma^2)$ is a Gaussian stochastic process with drift $\mu - 1/2 \sigma^2$ and variance $\sigma^2$. Interestingly enough we would have arrived at the same conclusion by using the original Langevin equation and Ito's lemma only.
Now going back to the PDE in question the standard way of solving it is to take Fourier transforms which is actually something you seek to obtain.
We define:
\begin{equation}
\tilde{f}(k,t) := \frac{1}{2\pi} \int\limits_{\mathbb R} e^{\imath k y} f(y,t) d y
\end{equation}
and we transform the PDE above into a ODE:
\begin{equation}
\frac{d \tilde{f}}{d t} + \left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right] \cdot \tilde{f} = 0
\end{equation}
which, given an initial condition at time $t=T$, is solved by:
\begin{equation}
\tilde{f}(k,t) = \tilde{f}(k,T) e^{\left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right](T-t)}
\end{equation}
This is the characteristic function that you were seeking. In applications like option pricing the initial condition $\tilde{f}(k,T)$ is equal to a Fourier transform of the option's payoff at maturity and the fair price of an option at some time before maturity is obtained by taking an inverse Fourier transform of the characteristic function.

Now, the really interesting question is to generalize that approach to more generic stochastic processes for example to L\'{e}vy stable processes. This leads to a fractional PDE. Those are much more difficult to handle yet it is worthwhile to pursue this approach because in here the Ito's lemma does not work.

Disclaimer: I took only one class of undergrad PDE and no grad PDE courses. In grad, our use of FK was just to solve a PDE. We didn't prove anything (well, there were some proofs but not required to study for exam hehe). There are most likely errors here assuming the whole thing isn't severely wrong, but I hope to gain better understanding through others' feedback.

Summary:(I assume you mean $\sigma^2 x^2$)

I think what you did wrong was that you got some $f$ and then plugged it into the PDE. I'm not sure that gets us anywhere. I think we are supposed to solve the PDE for $f$ (To clarify, this is $f = f(x,t)$, a function of two variables; I think you plugged in the '$f(x,T)$' in the initial condition, which is a function of only one variable) and then get the characteristic function from there.

The difference between this and, for example, that seems to be that in that, we come up with a stochastic process to solve the PDE while in this, we are given a stochastic process and then we need to solve the PDE.

If I'm not mistaken, FK seems to be implying two things:

Given a certain PDE, this conditional expectation solves it.
Any solution to a certain PDE can be expressed as a conditional expectation.

(Is that existence and uniqueness? Idk)

So...

In that: we use (1) in considering what stochastic process (Ansatz ) to guess to try to solve the PDE i.e. the PDE dictates the stochastic process (Ansatz) we guess.

In this: we solve the PDE through non-probabilistic methods and then we would know that by (2), our solution is a conditional expectation s.t. if we set $t=0$, we get the characteristic function i.e. the stochastic process dictates the PDE to solve wherein we use an Ansatz that does not explicitly involve a stochastic process (of course FK is saying that there is an implicit stochastic process).

Based on Arithmetic Brownian Motion (see 21.1,21.4):

Let T > 0, and suppose we have a stochastic processes $\{X_t\}_{t \in [0,T]}$ given by $$dX_t = \sigma(X_t,t)dW_t + \mu(X_t,t)dt$$

Then, the characteristic function of $X_T$ is $E[u(X_T)|X_0 = x]$ with $u(x) = \exp(i\omega x)$.

In the article, the author is given ABM

$$dX_t = \pm\sigma dW_t + \mu dt$$

Here, we are given GBM

$$dX_t = \pm\sigma xdW_t + \mu xdt \tag{*}$$

Now for the same stochastic processes but for any $u \in C^{2}(\mathbb R)$ (so this part doesn't have to do with characteristic functions), we can compute $E[u(X_T)|X_0 = x]$ by computing $E[u(X_T)|X_t = x]$ and then setting $t=0$ (duh!).
But how do we compute $E[u(X_T)|X_t = x]$?

(Come to think of it, I'm guessing you could compute it directly, but it would be difficult or something which is kind of why we're using FK.)

FK tells us that for any $u \in C^{2}(\mathbb R)$ (so this part too doesn't have to do with characteristic functions),  $f(x,t) := E[u(X_T)|X_t = x]$ is a (the?) solution of

$$\frac{\partial f}{\partial t} + \mu(x,t) \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2(x,t) \frac{\partial^2 f}{\partial x^2} = 0$$

Whose parameters are dictated by the stochastic process $(*)$. Hence, we must solve the ff PDE:

$$\frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2}\sigma^2x^2 \frac{\partial^2 f}{\partial x^2} = 0 \tag{**}$$

In Arithmetic Brownian Motion, the author used separation of variables i.e. used Ansatz $f(x,t) = \chi(x)\tau(t)$ and then plugged it into the PDE there which is simpler than ours because the coefficients are all constant.

Ours however has nonconstant coefficients for partial derivative of $f$ so it's not as simple. I'm not even sure it's possible: if we use the same Ansatz for GBM as with ABM namely $f(x,t) = \chi(x)\tau(t)$ to $(**)$, I think we will see that $f(x,t)$ is actually not separable, as in the case of the Ornstein-Uhlenbeck SDE (see 21.5).
Perhaps we may try the Ansatz in Ornstein-Uhlenbeck

$$f(x,t) = \exp(i \omega \alpha(t)x + \beta(t))$$

to find $\alpha(t), \beta(t)$ which are supposed to be deterministic functions s.t. $\alpha(T) = 1$ and $\beta(T) = 0$, but I think we will end up with

$$i\theta\alpha'(t) x + \beta'(t) + \mu x [i \theta \alpha(t)] + \frac12 \sigma^2 x^2 [i \theta \alpha(t)]^2 = 0$$

and so we cannot group the coefficients like in Ornstein-Uhlenbeck.

So I guess it all comes to down to finding the right Ansatz (or perhaps there's none?)

Alternatively:

Let $\{Y_t\}_{t \in [0,T]}$ be a stochastic process s.t. $dY_t := \frac{dX_t}{X_t}$. If we can find the cf of $X_t$ given the cf of $Y_t$ then note that $Y_t$ is Arithmetic Brownian Motion, whose cf we know.


Absolute Minima and Maxima on Triangles at the boundary Find absolute minimum and maxima for $f(x, y)= 11-3x+7y$

on closed triangular region with vertices $(0, 0), (7, 0)$ and $(7, 11)$.

How would I approach this type of question. I understand how to find the critical points on the interior which is just the partial derivative with respect to $x$ and $y$ which is $-3$ and $7$. But how do I find the critical points on the boundary with the vertices. I do not know what the end points would be to aid me with this question.

There are two ways we can approach this problem. One is the traditional calculus way. You would first have to check the interior of the region and look for critical points. It's straightforward to see that there isn't one. We now must test the boundaries, which are lines. For each side of the triangle, we can treat it as a constrained optimization problem. For example, when looking for optimal solution along the side with endpoints $(0,0)$ and $(7,0)$, we can treat it as the following problem:
$$\text{optimize } f(x,y)=11-3x+7y \text{, subject to } g(x,y) = y = 0$$
You can use the method of Lagrange multipliers to solve this equation:
$$\nabla f=\lambda \nabla g \text{, at optimal points} \implies (-3, 7) = \lambda\cdot(0,1)$$
which clearly has no solution. Repeating this method on each side will yield a similar result. The last candidates to be checked are the vertices themselves (they can be seen as "boundaries" of the sides). Checking their values will show that $f(7,11)$ is the absolute maximum, and $f(7,0)$ is the absolute minimum.

The second way makes use of principles in linear programming. We know that this is equivalent to the problem of finding the absolute minima and maxima of the linear function $f(x,y)=-3x+7y$ over the triangle. If you've studied the problem of linear programming, you know that optimal solutions (max and min) all occur at extreme points of a convex set. The triangular region you're considering is convex, so really, you only need to evaluate the function at the vertices to find the absolute min and max.

So, if you notice the given function, you can see that in 3- Dimensional space, the given equation is just a plane. And this plane, to find the maximum value of the height from the x,y plane, you can just assume that it lies on one of the boundary points and so to get the
actual value just set x,y of each of the boundary points and manually check which one is greater.

Now, in theory you can prove that if you have this kind of setting where you want to maximise/minimise the value of the height of the plane from the x,y plane, you can just consider the endpoints of the polygonal region of the input space that you want to optimise in.

Also, refer to https://en.wikipedia.org/wiki/Linear_programming.


Implementing sets in $\lambda$-calculus Both Set theory and $\lambda$-calculus are considered to be valid foundations for mathematics. Since these are both equivalent (in the sense that any structure that can be implemented in set theory can be implemented in $\lambda$-calculus, and vice-versa), is there a way to implement sets in $\lambda$-calculus?

I am sorry to inform you that the lambda calculus (Church, 1932) as a foundation of mathematics was shown to be logically inconsistent (Kleene and Rosser, 1935). Therefore your question is moot.

Sets can be implemented as lists in $\lambda$-calculus. The order of the elements in the list can be ignored by allowing only one type of function on the list - a membership test. All lists that produce the same results on performing membership tests are equivalent.


Inverse of a given function Let the function $f: R \to R$ be defined by $f(x)=x^2-2$. Find $f^{-1} \{- \infty<x \leq2\}$

What is $f^{-1} \{- \infty<x \leq2\}$ is supposed to mean here? If I take $ - \infty<x \leq2\ $ as domain, the $f$ is not injective in that interval, hence $f^{-1}$ does not exist. Could someone help me with this notation?

Suppose $f$ is a function from $X$ to $Y$, and suppose $V$ is a subset of $Y$. Then $f^{-1}(V)$ is defined to be the preimage of $V$, i.e. it is the set of all $x$ whose image under $f$ is contained in $V$:
$$ f^{-1}(V) = \{ x \in X : f(x) \in V \}.$$

Note that this definition works for arbitrary $f$, bijective or not! There doesn't need to be an inverse function $f^{-1}$. We write it as $f^{-1}$(V)$ purely because it is a convention.

[If $f$ is bijective, then the inverse $f^{-1}$ does exist, and you can show that the $f^{-1}(V)$ that I defined is the same as the image of $V$ under the function $f^{-1}$, i.e $f^{-1}(V) = \{ f^{-1}(y) : y \in V \}$.]

It doesn't seem like standard notation.  I have two guesses:

It's a typo and they meant the domain of $f$ to be $-\infty <x\leq \sqrt{2}.$
$f^{-1}$ may not be a function, but it is a relation.  And maybe they
want the set of all $x$ such that $f(x) \in (-\infty,2]$.


Apparent contradiction for probability density functions? Consider a probability density function $\it{pdf}$, $f\left(x\right)$, which can be expanded as:

$$ f\left(x\right) = \sum_{k=1}^{\infty} \alpha_k \delta\left(x-x_k\right)$$

It is easy to verify, by integrating both sides that, $\int_{-\infty}^{\infty}f\left(x\right) = 1$, implies $\sum_{k=1}^{\infty}\alpha_k=1$. Also the orthogonality of dirac delta functions centered on distinct points $x_k$ and $x_m$ leads to $\alpha_k=f\left(x_k\right)$. Here's the problem, the only bound on a $\it{pdf}$ is non-negativity, so since $f\left(x\right) \in \left[0,\infty\right)$, we should have $\alpha_k \in \left[0,\infty\right)$. If $\alpha_k$ has no upper bound, how can they sum to $1$?

Your implication $\sum_{k=1}^{\infty}\alpha_k=1$ is correct. The problem is your assumption $\alpha_k=f(x_k)$. This is not the case, since the delta function is defined by its integral; it is a distribution and does not have a function value in the traditional sense. So evaluating $\delta(x-x_k)$ at $x=x_k$ is meaningless.

EDIT: This is a reaction to your example in the comment below.
$$f(x_m)=\int_{-\infty}^{\infty}f(x)\delta(x-x_m)\;dx =
\int_{-\infty}^{\infty}\sum_{k=1}^{\infty}\alpha_k\delta(x-x_k)\delta(x-x_m)\;dx$$
Assuming that we can interchange integration and summation, we get
$$f(x_m) = \sum_{k=1}^{\infty}\alpha_k\int_{-\infty}^{\infty}\delta(x-x_k)\delta(x-x_m)\;dx$$
Now we evaluate the integral in a way that is common in physics and engineering; I believe mathematicians usually frown upon such manipulations (please correct me if I'm mistaken):
$$f(x_m) = \sum_{k=1}^{\infty}\alpha_k\delta(x_m-x_k)\tag{1}$$
which is not the same expression that you derived in your comment. Note that (1) is exactly what you get when you evaluate $f(x)$ at $x=x_m$ in your original expression for $f(x)$. So $f(x_m)\neq \alpha_m$, because here, $\delta(x)$ is not the Kronecker delta, but the delta function.

But it does!! Since
$$\int_{-\infty}^{\infty} f(x)dx =1 \Rightarrow \sum_{k=1}^{\infty} \alpha_k = 1$$

Now we have $\alpha_k = f(x_k) \geq 0$, and since $\sum_{k=1}^{\infty} \alpha_k$ is a convergent series, we also have $\alpha_k \to 0$. Moreover

$$\alpha_n \leq \sum_{k=1}^{\infty} \alpha_k = 1$$

Thus $0 \leq \alpha_n \leq 1$ for all n.

By chance I have misunderstood your question this might help:

$$\alpha_k \delta{\{x-x_k\}} \leq \sum_{k=1}^{\infty} \alpha_k\delta{\{x-x_k\}} = f(x)$$
Integrate both sides to get (Noting all terms are nonnegative) that
$$\alpha_k \leq 1$$


The map $f:\mathbb{Z}_3 \to \mathbb{Z}_6$ given by $f(x + 3\mathbb{Z}) = x + 6\mathbb{Z}$ is not well-defined By naming an equivalence class in the domain that is assigned at least two different values prove that the following is not a well defined function.

  $$f : \Bbb Z_{3} \to \Bbb Z_{6} \;\;\;\text{ given by } f(\overline x) = [x]  $$

  In this case we represent an element of the domain as an $\bar x$ and use the notation $[x]$ for equivalence classes in the co-domain.

$f(\overline0) =  [0] \;,$  $ \Bbb Z_{3}  \quad (3x+0)\;\; \overline 0 = \{ ...-6,-3,0,3,6...  \}, \; \Bbb Z_{6}\;  (6x+0)\; \overline0 =\{ ...-12,-6,0,6,12...\}$

$f(\overline1) = [1], $ $\qquad   \; (3x+1) \; \;\;\;\overline   1 = \{ ...-5,-2,1,4,7 ... \},\; \;  (6x+1)\;\overline1 =\{...-11,-5,1,7,13.. \}$

$f(\overline2) = [2], $ $\qquad  \qquad \qquad \;\overline 2 = \{ ...-4,-3,2,5,8 ... \},\;\;\overline 2 = \{ ...-10,-4,2,8,14  ...\},\;$

$f(\overline3) = [3] ,$  $\qquad \qquad \qquad \qquad \qquad \qquad,\; \quad \quad \quad \; \; \; \;\overline 3 = \{ ...-9,-3,3,9,15 ... \},$

$f(\overline4) =  [4],\qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \; \;\overline 4 = \{ ...-8,-2,4,10,16...  \}, $

$f(\overline5) = [5], \qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \;\;\overline 5 = \{ ...-7,-1,5,11,17...  \},  $

$f(\overline6) = [6] ,$

So my main question for this problem is how to find out if this question is not a function. From the information I have gathered here I still cannot see why this is not a function any help on showing how this is not function would be much appreciated.

  The set of equivalence classes for the relation $\cong_{m}$ is denoted $\Bbb Z_{m}$

The $ 3x+0 \text{ and } 6x+0$ are just showing how I got $\overline 0 $

If a function $f \colon \mathbb{Z}_3 \to \mathbb{Z}_6$ with $f(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ would exist, then
$$
    [0]
  = f(\overline{0})
  = f(\overline{3})
  = [3],
$$
where we used for the second equality that $\overline{0} = \overline{3}$.
By the definition of $\mathbb{Z}_6$ we have for all $x, y \in \mathbb{Z}$ that $[x] = [y]$ if and only if $x - y$ is divisible by $6$, i.e. if $x - y \in 6\mathbb{Z} = \{6n \mid n \in \mathbb{Z}\}$.
So $[0] = [3]$ is equivalent to $3 - 0 = 3$ being divisible by $6$, which doesn’t hold.

This contradiction shows that no such function $f$ exists.

More generally one can show that there exists a function $g \colon \mathbb{Z}_n \to \mathbb{Z}_m$ with $g(\overline{x}) = [x]$ for every $x \in \mathbb{Z}$ if and only if $m$ divides $n$.

It is an onto relation from a set of lesser cardinality to a set of larger cardinality. Thus it cannot be a function (by the definition of cardinality).


Probability of drawing balls from two urns (Bayes) An urn contains 5 white and 5 black balls. 4 balls are drawn from this urn and put into another urn. From this second urn a ball is drawn and is found to be white. What's the probability of drawing a white ball again at the next draw. (The first white ball drawn is not replaced.)

What I did was:-

E0= Drawing 0 white and 4 black balls

E1= Drawing 1 white and 3 black balls

E2= Drawing 2 white and 2 black balls

E3= Drawing 3 white and 1 black balls

E4= Drawing 4 white and 0 black balls

We ignore E0 because if there were 0 white balls transferred, the first one drawn couldn't be white. Out of the 4 remaining events, each of them have the probability of occurring at 1/4.

In E1, if the first ball drawn is white, there are 0 white balls left.(Total left 3)

In E2, if the first ball drawn is white, there is 1 white ball left. (Total left 3)

In E3, if the first ball drawn is white, there are 2 white balls left. (Total left 3.)

In E4, if the first ball drawn is white, there are still 3 more white balls left. (Total left 3)

So, the final answer should be [(1/4*0/3) + (1/4*1/3) + (1/4*2/3) + (1/4*3/3)]

Which equals to 1/12 + 2/12 + 3/12 = 1/2

I believe this to be the correct answer however the book has got 4/9 for the answer. Where exactly did I go wrong?

(Thanks for your time and sorry if there are any formatting errors in question.)

Hint: We can model this by arranging the ten balls in a line, so that the first four are those moved into the second urn then removed in that order.   Given that the first ball is white, what is the probability that the second ball is too?

That is the short method.

The long method is:

Let $E_x$ be the event of $x$ white balls among the four balls drawn from five white and five black, for $x\in\{0,1,2,3,4\}$.   This corresponds to a hypergeometric distribution.

$$\mathsf P(E_x) ~=~ \dfrac{\dbinom 5 x\dbinom 5 {4-x}}{\dbinom {10}4}\cdot\mathbf 1_{x\in\{0,1,2,3,4\}}$$

The events $E_0,E_1,E_2,E_3,E_4$ are thus not equally probable.

Let $A, B$ be the event of drawing white balls in two consecutive draws from among those four.

$$\mathsf P(B\mid A)=\dfrac{\sum_{x=2}^4 \mathsf P(E_x)~\mathsf P(A\cap B\mid E_x)}{\sum_{x=1}^4 \mathsf P(E_x)~\mathsf P(A\mid E_x)}=\dfrac{\sum_{x=2}^4 \binom 5x\binom 5{4-x}\binom x 2/\binom 4 2}{\sum_{x=1}^4 \binom 5x\binom 5{4-x}\binom x 1/\binom 4 1}$$

I would try drawing a tree diagram click here for examples of tree diagrams.

After you are done drawing the tree diagram you can then draw another tree diagram of all the combinations that have white balls in them. This is a rather long method but works as well.


Evaluating $\lim_{x\to 0}\frac{e^x+e^{-x}-2}{1-\cos x}$ without using L'Hospital's Rule Compute without using L'Hospital's Rule $$\lim_{x\to 0}\dfrac{e^x+e^{-x}-2}{1-\cos x}.$$

I thought of simplifying the limit as shown below.
\begin{align}
\lim_{x\to 0} \dfrac{e^x+e^{-x}-2}{1-\cos x}&=\lim_{x\to 0}\dfrac{2\sinh^2 x(1+\cos x)}{\sin^2 x(1+\cosh x)}\\
&=\space2\lim_{x\to 0}\dfrac{\sinh^2 x}{\sin^2 x}\cdot \lim_{x\to 0}\dfrac{1+\cos x}{1+\cosh x}
\end{align} But I'm a little stuck as to how to show that
$$\lim_{x\to 0}\dfrac{\sinh^2 x}{\sin^2 x}=1$$
so that the original limit is $2$. I can't just use a simple substitution.

Any help would be nice.

Edit: No use of L'Hospital's Rule is allowed. If I were allowed to use that, I would simply do the following:

$$\lim\limits_{x\to 0}\dfrac{e^x+e^{-x}-2}{1-\cos x} = \lim\limits_{x\to 0}\dfrac{e^x-e^{-x}}{\sin x}=\lim\limits_{x\to 0}\dfrac{e^x+e^{-x}}{\cos x}=2$$
That is literally trivial.

Divide numerator and denominator by $x^2$

And use the limit $\lim_{x \to 0} \frac{sinh (x)}{x}= 1$ and $\lim_{x \to 0} \frac{sin (x)}{x}=1$

Here you can find the proof of limit of sinh
Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.

Directly with the original limit and L'Hospital's Rule:

$$\lim_{x\to0}\frac{e^x+e^{-x}-2}{1-\cos x}\stackrel{L'H}=\lim_{x\to0}\frac{e^x-e^{-x}}{\sin x}\stackrel{L'H}=\lim_{x\to0}\frac{e^x+e^{-x}}{\cos x}=2$$


Mathematical term for "opposite" in this context? A friend wanted to play a game where we would roll a die, and if it landed even then the wish would come true, but if it landed odd then the opposite of the wish would come true. So let's say I have five dollars and my wish is to gain five dollars.

I know the negation of gaining five dollars is not gaining five dollars. But that is not the 'opposite' of my wish in this context. The opposite would be losing five dollars. What is the operation called that's being applied to "gaining five dollars" such that it becomes "losing five dollars"?

Change the rules of the game. The game costs \$5 to play and you can gain $10 by winning.


Every power series expansion for an entire function converges everywhere I would like to show that every power series expansion for an entire function converges everywhere.

Using Cauchy's Theorem and integration by parts yields
$$
\begin{align}
\left|\frac{f^{(n)}(w)}{n!}\right|
&=\left|\frac{1}{2\pi i}\oint\frac{f^{(n)}(w+z)}{n!\,z}\mathrm{d}z\right|\\
&=\left|\frac{1}{2\pi i}\oint\frac{f(w+z)}{z^{n+1}}\mathrm{d}z\right|\\
&\le\frac{1}{r^n}\max_{B(w,r)} |f|\tag{1}
\end{align}
$$
where the integration is around the circle $z=r\,e^{it}$ for $t$ from $0$ to $2\pi$.

Estimate $(1)$, called Cauchy's
 Estimates, says that the radius of convergence of the Taylor series for $f$ is at least $r$. Since $f$ is entire, we can set $r$ as large as we want.

Therefore, the Taylor series for $f$ at $w$ converges for all $z$.

Maybe something is wrong with this answer, but it seems to be pretty simple.

First, we know that the power series of an analytic function is unique.  So if a function is entire (analytic in the whole complex plane), then its power series is unique on the whole plane, and by definition is convergent.


Radius of convergence of complex series around 0+i Let $f(z):=\frac{\sin z}{z(z-\pi)(z+\pi)}$.  What is the radius of convergence of the series
$$
\sum_{n=0}^{\infty} \frac{f^{(n)}(i)}{n!}(z-i)^n\quad\text{?}
$$Explain.

My Attempt :

Note that the series in question is just the Taylor series of the function $f(z)$ about the point $i$.  So the solution reduces to determining the distance between the closest non-removable isolated singularity and the point $0+i$.  All of the isolated singularities are removable and we must conclude that the radius of convergence is $\infty$.  I verify that the singularities are all removable below.

$$
\lim_{z\to 0} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/\pi^2
$$
$$
\lim_{z\to \pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2
$$
$$
\lim_{z\to -\pi} \frac{\sin z}{z(z-\pi)(z+\pi)} =-1/2\pi^2
$$

Note that

  The radius of convergence is the distance to the nearest singularity for the function $f(z)$.

After editing my attempt to change the RoC to $\infty$ from $\mathbb{C}$ I believe my attempt is correct.


How many odd three-digits numbers are there whose all three digits are different I faced this problem on one test. I wrote my solution but then I found out that my solution is wrong, I still cannot find where my mistake is.

The problem says: How many three-digits numbers are there such that they are odd and their digits are all different.

Here is my approach:

We have three digits. Since the number should be odd, the last digit should be one of those numbers $1, 3, 5, 7, 9$. Now the second digits can be one of the digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ There are 10 different digits for the second digits, but since the digits should be different we cannot place 10 digits, but we can place 9 digits. And for the first digits we can place digits in the range $1...9$ but we cannot place the digits that are used in the two other digits and we can place only 7 digits.

So my result is $7\cdot9\cdot5 = 315$

However the result is not correct, because there are $320$ odd three-digits numbers with different digits.

Can you point me where is my mistake, thanks in advance.

Problems of this type are often solved more simply if we place the digits in a suitable order, instead of necessarily left to right.

Here we'll place the digits in the order last-first-second, i.e. in the order $Units-Hundreds-Tens$

$5$ choices for the last digit (to make the number odd)
$8$ choices left for the first digit (since $0$ can't be used)
$8$ choices still left for the second digit (since $0$ is now available for use)

Thus, without breaking up into cases, we get $5\cdot8\cdot8 = 320$

Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9

With repetition of digits we would have had 5 * 5 * 5 = 125

But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)

And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)

5 * 4 * 3 = 60 is my answer


Which is more correct, $\| \vec{a} \|$ or $|\vec{a}|$? The question is pretty self explanatory, but I’ve encountered situations where, for the length of some vector $\vec{a}$, to denote the length (or magnitude, which ever you prefer) as either $\| \vec{a}\|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ or $|\vec{a}|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and I was wondering which notation is more widely accepted, per say? I’ve tried researching this and different websites actually use different notation.

Any help is appreciated, thank you.

Definitions are not judged "correct" or "incorrect".  Both of your notations are used in various contexts in mathematics.  There is a problem only if the writer and the reader do not understand each other.

The first one is the more correct, in the sense that is preferable when you are dealing both with vectors and numbers and it is necessary to avoid misunderstanding.

The notation $|\cdot|$ indicates the absolute value for numbers, but it is also frequently and widely used for vectors when it is clear from the context.


On Ph.D. Qualifying Exams Where can I find Ph.D. qualifying exams questions.Is there any website that keeps a collection of such problems?
I need it for doing some revision of the basic topics. I know of a book but that do not have the full collection.

These are the sets of qualifying/preliminary examinations of US universities that I collected some time ago for the same purposes as you. (Dave L. Renfro points out in a commentary below that he compiled a similar list a decade ago, the following includes new departments, updated old broken links and removes unavailable sources). These exams are of much help and some even include solutions!:

University of Arizona Ph. D. Qualifying Exams
Arizona State University Qualifying Examinations
University of Alabama Qualifying Exams.
University of Alabama at Birmingham Qualifying Exams
SUNY Albany Ph.D. Preliminary Examinations.
University of California-Berkeley Preliminary Examinations.
Boston University Preliminary Exams.
University of Colorado-Boulder Prelim Exams.
Columbia University Qualifying Exams.
University of Connecticut Prelim Exams.
University of California-Davis Exams.
University of California-Irvine Exams
University of California-Riverside Exams
University of Florida Graduate Exams.
University of Georgia-Athens Qualifying Exams.
Harvard University Qualifying Exams.
University of Illinois at Urbana-Champaign Comprehensive Exams.
Indiana University-Bloomington Tier I Examinations.
University of Iowa Qualifying Exams.
Iowa State University Qualifying Exams
Johns Hopkins University Qualifying Exams.
Kansas State University Qualifying Exams.
University of California-Los Angeles Qualifying Exams.
Louisiana State University Comprehensive Exams.
University of Maryland Qualifying Exams.
University of Massachussetts-Amherst Qualifying Exams.
University of Michigan-Ann Arbor Qualifying Review Examinations.
University of Minnesota Prelim Exams.
University of Missouri-Columbia Qualifying Exams.
University of Nebraska-Lincoln Qualifying Exams.
University of New Mexico Qualifying Exams.
University of North Carolina at Chapel Hill Comprehensive Ph.D. Exams (requires one to log in).
Northeastern University Qualifying Examinations.
Northwestern University Preliminary Examinations.
Ohio State University Qualifying Exams.
Oklahoma State University Comprehensive Exams.
University of Oklahoma Qualifying Exams.
Pennsylvania State University Qualifying Exams.
University of Pittsburgh Preliminary Exams.
Purdue University Qualifying Exams.
University of Rochester Preliminary Exams.
Rutgers University-New Brunswick Qualifying Exams and oral exams.
Stanford University Qualifying Exams.
Syracuse University Prelim and Qualifying Exams.
Temple University PhD. Exams.
University of Texas at Austin Preliminary Exams.
Texas A & M University Qualifying Examinations.
University of Utah Qualifying Examinations.
University of Washington Preliminary Examinations.
University of Wisconsin-Madison Qualifying Exams.

The Written Qual Book by Daryl DeFord and David Freund.


The initial value problem $y'=\sqrt {y}, \:\: y(0)=\alpha$. The initial value problem

$$\begin{cases}y'=\sqrt {y}\\y(0)=\alpha\end{cases}$$ has

$A.$ At least two solutions if $\alpha =0$

$B.$ No solution if $\alpha > 0$

$C.$ At least one solution if $\alpha > 0$

$D.$ A unique solution if $\alpha = 0$

ATTEMPT

I solved the differential equation as

$y(x) = \frac{x^2}{4} + \alpha +x\sqrt {\alpha}$

How do i choose correct options?

Answer. A true, B false, C true, D false.

If $a=0$, uniqueness is violated.

Solution I. $\varphi(x)\equiv 0$.

Solution II. $\varphi(x)=\frac{x^2}{4}$

If $a\ne 0$, solution is unique (locally).

The general solution is
$$
y^{-1/2}dy=dx\rightarrow 2\sqrt{y}=x+c
$$
Considering initial value we have
$$
2\sqrt{y}=x+2\sqrt{\alpha}.
$$
Hence: i) If $\alpha>0$ the i.v.p has a unique solution in a neighborhood of $0$ (by existence-uniqueness theorem). ii) The i.v.p has not solution if $\alpha<0$. iii) If  $\alpha=0$ then it has at least two solutions: 1) $y=0$
, 2) $\sqrt{y}=x/2$. You can now verify which of the above options is correct or not.


All natural numbers are equal. I saw the following "theorem" and its "proof".

I can't explain well why the argument is wrong. Could you give me clear explanation so that kids can understand.

Theorem: All natural numbers are equal.
Let $a, b \in \mathbb{N}$, then a=b.

Proof by induction. Let $m=\max\{a, b\}$. We will prove that the theorem holds for all $m\in \mathbb{N}$.
If $m=1$, then $\max\{a,b\}=1$, so $a=b=1$.
Now assume that it holds for $m=k$ for some number $k$.
Now let $\max\{a, b\}=k+1$. Then $\max\{a-1, b-1\}=k$ and thus by assumption $a-1=b-1$, so $a=b$. Therefore, the proof is complete.

The error lies in the fact that when decreasing $1$ you may get out of the set $\mathbb{N}$ - and indeed, $\max\{1,2\}=\max\{0,1\}+1$, but $0 \ne 1$, and $0\notin \mathbb{N}$ as you defined it.

the first assumption was $m=k$ or $\max(a,b)=k$, but the theory is letting $\max(a,b)=k+1$ meaning "$\max(a,b)=\max(a,b)+1$" which drives the proof wrong!


Necessary but not sufficient in logic I am working through sample questions and am having a bit of trouble understanding the solution.

Write using logical connectives:

p : Grizzly bears have been seen in the area.

q : Hiking is safe on the trail.

r : Berries are ripe along the trail.

For hiking on the trail to be safe, it is necessary but not
sufficient that berries not be ripe along the trail and
for grizzly bears not to have been seen in the area.

I came up with

$$q \rightarrow(\lnot r \land \lnot p)$$

However, the solution is

$$q \rightarrow (\lnot r \land \lnot p) \land \lnot((\lnot r\land \lnot p)\rightarrow q)$$

I am wondering why the attached part is required. I guess my problem would be with understanding the question at hand. Would the last two propositions be sufficient enough for the first proposition to be true, are they both together still insufficient?

Any help is appreciated! Thank you.

For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the trail and for grizzly bears not to have been seen in the area.

$$(\underbrace{(\lnot r \land \lnot p)\leftarrow q}_{\text{it is neccessary}}) \underbrace{\land}_{\text{and/but}} (\underbrace{\lnot((\lnot r\land \lnot p)\rightarrow q)}_{\text{it is not sufficient}})$$

For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the trail and for grizzly bears not to have been seen in the area.

This is a bit vague. Do ripe berries on their own really pose a threat to hikers? Or is it the combination of bears and berries that is dangerous?

Unless this is some kind of trick question, I think the author meant: If the berries are ripe along the trail ($r$) and grizzly bears have been seen in the area ($p$), then hiking is not safe on the trail ($\neg q$). (Bears eat berries. Berries alone are seldom dangerous.)

If this is the case, the answer should be:$$r\land p\implies \neg q$$or equivalently  $$q\implies \neg r \lor \neg p$$


Looking to understand the rationale for money denomination Money is typically denominated in a way that allows for a greedy algorithm when computing a given amount $s$ as a sum of denominations $d_i$ of coins or bills:

$$
s = \sum_{i=1}^k n_i d_i\quad\text{with } n_i > 0, d_i > d_{i-1}
$$

We start with the highest denomination $d$ not exceeding $s$ and subtract it.  We repeat this process until $s$ is completely represented. As long as all denominations are available, this simple algorithm guarantees the minimum number of coins or bills used. (This is a little sketchy but I guess we are all familiar with it and it's not central to my question.)

My question is: what property of $d_i$ makes this work? I once heard that the value of any coin must be bigger than the sum of all coins of lower denomination for the greedy algorithm to work. This holds for the typical denomination:

$$
1,2,5,10,20,50,100
$$

However, I wasn't able to come up with a counter example or even a proof. Surely this problem has been studied and I would be glad to know more about it. In fact, I'm not even sure about the right tags -- feel free to edit them. I assume the algorithm to find the sum with the minimum number of coins becomes tricky when we no longer can assume that all denominations are available.

The greedy solution is not optimal in general, even for the case when the value of any coin is bigger than the sum of all coins of lower denomination.  For example, suppose we have coins of denominations 25, 9, 4, and 1.  For 37 cents, the greedy solution uses five coins: one 25, one 9, and three 1s.  However, the optimal solution is four coins: one 25 and three 4s.

The problem of finding the minimum number of coins given a set of denominations is called the change-making problem.  It's a variant of the famous knapsack problem and is known to be difficult in the general case.

An integer-programming formulation for the change-making problem was given in the answer to this math.SE question:  On problems of coins totaling to a given amount.

Added: While the Wikipedia article on the change-making problem claims that the greedy algorithm is optimal for U.S. coins, it provides no proof.  The article "Canonical Coin Systems for Change-Making Problems" discusses characterizations for when the greedy algorithm is optimal for coin systems of up to five denominations.  It also implies that for systems of larger than five denominations determining when the greedy algorithm is optimal is still an open problem.

I recently came up with 1 solution that seemed to show if the given 2 conditions are satisfied, the greedy algorithm would yield the optimal solution.

a) The G.C.D (All the denominations except 1) = 2nd Smallest denomination.

b) The sum of any 2 consecutive denominations must be lesser than the 3rd consecutive denomination.

For eg. c2 + c3 < c4.

(Where c1 = 1; c2, c3, c4 are coin denominations in ascending order).

I understand this is not a complete solution. However, I believe that if these 2 conditions are met, the greedy algorithm will yield the optimal solution.


Find the distribution of $Y = -\log (1-X)$ given that $X\sim U(0,1)$. If $X \sim U (0,1)$ then if we define a new random variable $Y=-\log (1-X)$ then what will be distribution of $Y$.
Please explain.

Let $y>0$.

$P(Y\leq y)=P(-\log(1-X)\leq y)=P(\log(1-X)\geq -y)=P(1-X\geq e^{-y})$

Now if $X\sim U(0,1)$, $1-X\sim U(0,1)$ too. So $P(1-X\geq e^{-y})=1-e^{-y}$.

We get the cdf of a standard exponential distribution.

The answers above are right, you find the PDF of a positively-truncated exponential.

In fact, this is one of the most common way to generate (on a computer) numbers following an exponential distribution, assuming that you have a PRNG that produces number $X \sim \mathcal{U}([0,1])$.
See here for more details, and here about the general method (Inverse transform sampling).


Why is cross product defined in the way that it is? $\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector?  Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?

So why is cross product defined in the way that it is?
I am mainly interested in the right hand rule defintion too as it is out of reach?

The cross product originally came from the quaternions, which extend the complex numbers with two other 'imaginary units' $j$ and $k$, that have noncommutative multiplication (i.e. you can have $uv \neq vu$), but satisfy the relations

$$ i^2 = j^2 = k^2 = ijk = -1 $$

AFAIK, this is the exact form that Hamilton originally conceived them. Presumably the choice that $ijk = -1$ is simply due to the convenience in writing this formula compactly, although it could have just as easily been an artifact of how he arrived at them.

Vector algebra comes from separating the quaternions into scalars (the real multiples of $1$) and vectors (the real linear combinations of $i$, $j$, and $k$). The cross product is literally just the vector component of the ordinary product of two vector quaternions. (the scalar component is the negative of the dot product)

The association of $i$, $j$, and $k$ to the unit vectors along the $x$, $y$, and $z$ axes is just lexicographic convenience; you're just associating them in alphabetic order.

The real reason is because it is convention to use a right-handed coordinate system.  If you switched to a left-handed coordinate system for some reason (I'm not sure why), then you would need to do your cross-products using a left-hand rule to get the same answers.  That's really as simple as it gets.  It's a result of our conventions in defining our coordinates the way we do.