Determine $\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$ Determine $\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$ if it exists.

I know that it exists, but I do not know how to show it. I tried to use the series definition $$\arctan (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1},$$ but I realised that we have only defined it that way for $x \in [-1, \, 1]$ shortly after.

Thanks in advance!

Note,

$$\arctan{\sqrt{n+1}} - \arctan{\sqrt{n}}
=\arctan \frac{\sqrt{n+1}-\sqrt{n}}{1+\sqrt{n+1}\cdot\sqrt{n}} $$
$$=\arctan  \frac{1}{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})} $$

Thus,
$$\lim_{n \rightarrow \infty}\arctan({\sqrt{n+1}}) - \arctan({\sqrt{n}})$$
$$=\lim_{n \rightarrow \infty}\arctan\frac1{(1+\sqrt{n^2+n})(\sqrt{n+1}+\sqrt{n})}=\arctan(0)=0$$

Why did you go in for a round about series option?

When $n$ is large $ \sqrt {n+1}\rightarrow \sqrt{ n} $ and limit is directly seen zero as difference of two equal terms. Arctan tends asymptotically to $\pi/4$ for large arguments without monotonic increase.


Can a logical formula that is a contradiction be represented as Disjunctive Normal Form? Hi I was trying to solve this math problem in mathematics.

  Obtain the D.N.F, C.N.F, P.D.N.F, P.C.N.F of the following expression : $(Q \to P) \land (\lnot P \land Q)$

I understand that there is no Particular Disjunctive Normal Form for this statement, as the the statement is a contradiction.

However, I'm not sure if it's suffice to say that there is no D.N.F for this statement as well.

I understand that the definition of a Disjunctive Normal Form is "a disjunction of one or more conjunctions of one or more literals"(from wikipedia). And that "All logical formulas can be converted into an equivalent disjunctive normal form"(again from wikipedia).

But I was wondering if $(P \land \lnot P) \lor (Q \land \lnot Q)$ could be an answer. As each of the conjunctions hold more than one literals can I say that this is the D.N.F of the given contradiction?

P.S. Pls correct my errors in the argument above. It would be more than welcome. Thank you in advance!!

$P \land \neg P$ is in DNF

It may not look like it, but it really is.

Here's the thing: a statement is in DNF iff it is a generalized disjunction of generalized conjunctions of literals. And by generalized we mean that the disjunction/conjunction can have any number of disjuncts/conjuncts. So the above statement is in fact a disjunction ... of exactly one disjunct.

Indeed, even a statement like $P$ is in DNF: it is a disjunction of one disjunct $P$,  which is itself a conjunction of one conjunct $P$, which is a literal. By similar reasoning, $P$ is in CNF as well.

In fact, $P \land \neg P$ is both a DNF and a CNF of your original formula. It is in CNF since it is a conjunction of two conjuncts, each of which is a disjunction of one literal.

$\begin{align}(Q \to P) \land (\lnot P \land Q) &\equiv (\lnot Q \lor P) \land(\lnot P \land Q)\tag{def. $\to$}\\ \\
&\equiv (\lnot Q \land (\lnot P \land Q)) \lor (P\land (\lnot P \land Q))\tag{distribution}\\ \\
&\equiv (\lnot Q \land \lnot P \land Q) \lor (P \land \lnot P \land Q)\tag {assoc.}\\ \\
&\equiv ((\lnot Q \land Q)\land \lnot P) \lor ((P\land \lnot P) \land Q))\tag{comm.+assoc.}\\ \\
&\equiv (\bot \land \lnot P) \lor (\bot\land Q)\\ \\
&\equiv \bot \lor \bot\\ \\
&\equiv \bot\end{align}$

The third line above is in Disjunctive Normal Form.


Linear Transformations given vector I'm having trouble understanding how to solve linear transformations given a vector transforming into another vector. Any suggestions for the example below? This is not merely someone trying to get a homework answer, but rather a simple explanation to someone taking Linear Algebra for the first time.

  Let $T:\mathbb{R}^2\to \mathbb{R}^2$ be a linear transformation that sends the vector $u=(5,2)$ into $(2,1)$ and maps $v =(1,3)$ into $(−1,3)$. Calculate $T(-3u), T(7v),$ and $T(-3u+7v)$.

Think of the definition of a linear transformation. Suppose $T$ is a linear transformation from $\mathbb{R}^n \mapsto \mathbb{R}^n$. Then $T$ satisfies the followings:

(1) For all $x,y \in \mathbb{R}^n$, $T(x+y) = T(x)+T(y)$ (Addition)

(2) For all $\lambda \in \mathbb{R}$ and $x \in \mathbb{R}^n$, $T(\lambda x) = \lambda T(x)$ (Scalar multiplication)

These are things that we need for solving the problem. In other words, there is no need for calculating the exact value of the matrix representation of $T$. In the problem, we have
$$ T(u) = \begin{pmatrix}2\\ 1\end{pmatrix}, \quad T(v) = \begin{pmatrix}-1 \\ 3\end{pmatrix}$$
for $u^T = (5,2)$ and $v^T = (1,3)$. And the calculation for $T(-3u),T(7v),$ and $T(-3u+7v)$ can be derived from the definition above.
$$\begin{align}T(-3u) &= -3T(u) \quad (\text{by the definition (2)})\\
&= -3\begin{pmatrix}2 \\ 1\end{pmatrix}\\
&= \begin{pmatrix} -6 \\ -3\end{pmatrix}\end{align}$$
Likewise, $T(7v)$ becomes $\begin{pmatrix}-7\\ 21\end{pmatrix}$. By using the definition (1), the calculation of $T(-3u + 7v)$ becomes $T(-3u) + T(7v)$ and we can easily get the result by simply adding up the previous calculations.

You can pose this as a linear least squares problem if you first remake your matrix into a vector with a process called vectorization, then we can write multiplication with another matrix or a vector as a matrix multiplication with a vectorized matrix. You can find a quite good description how to do this on This wikipedia site under "Matrix equations".

You will get a matrix equation system of the type $\bf Mx =b$ which you can solve for example with linear least squares.


How does $p_0(t)\mid k(t)$? Let  $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$. Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$. After this, it is written in a text that

Then, we have $p_0(t)\mid k(t)$. So, $\alpha_1$ is also a root of $k(t)$.

My question is how we know  $p_0(t)\mid k(t)$?
EDIT
Let me rephrase my problem.

the first statement:  $\alpha_i$ is a root of $k(t) \in \mathbb Z[t]$, this is according to a previous hypothesis.

the second statement:  Let $p_0(t)$ be the minimal polynomial of $\alpha_1$ over $\mathbb Z$, this is true because for $\alpha_1$, there exists a minimal polynomial.

the third statement:   We have $p_0(t)\mid k(t)$.

Now every algebraic number $\alpha_1$ has a minimum polynomial, say $p_0$, and  if polynomial $k(t)$ has root $\alpha_1$ then  $p_0$ divides $k(t)$ (in general every polynomial that$\alpha_1$ satisfies). But at this point we don't know $\alpha_i=\alpha_1$, then how we know $p_0(t)\mid k(t)$?

the fourth statement: "So", $\alpha_1$ is also a root of $k(t)$

Now if we already used the fact that  $\alpha_1$ is also a root of $k(t)$ imply $p_0(t)\mid k(t)$ then how can we say $p_0(t)\mid k(t)$ implies  $\alpha_1$ is also a root of $k(t)$?

He has constructed $k(t)$ so that $\alpha_1$ is a root of $k(t)$ by showing that $h(g(\alpha_1))=0$ and naming $k(t)=h(g(t))$.  Then if $p_0(t)$ is the minimal polynomial of $\alpha_1$, the theorem you quote in the comments "(for) every algebraic number α the minimum polynomial that α satisfies divides every polynomial that α satisfies." gives us $p_0(t)|k(t)$ because $k(\alpha_1)=0$


Solve the equation $y=2xy'+y^2y'^3$ Solve the equation $y=2xy'+y^2y'^3$.

I have tried to solve it but I can't recognize what type of equation I have learnt.

$$y=2xy'+y^2y'^3$$
Multiply by $y$
$$y^2=2xyy'+y^3y'^3$$
Substitute $y^2=u$
$$u=xu'+\frac 1 8(u')^3$$
It's Clairaut 's equation
$$y=xy'+g(y')$$
And the general solution is given by
$$u(x)=Cx+\frac {C^3}{8}$$
$$y^2(x)=Cx+\frac {C^3}{8}$$


are the two polynomials the same? If you have three polynomials $f, g$ and $h$. And the following is true:$\int^{1}_{-1}f\cdot g=\int^{1}_{-1}f\cdot h$. Does this directly imply that $h=g$ (assuming $f$ is not the zero polynomial)? I think it does imply, but I am not sure. Thanks.

Two diff polynomials can yield same integral value(we can also say it as grazed same area)
in same given limits(a,b) in this case (-1,1)

it doesnt mean both polynomials are same.

let f.g=k,f.h=l

and integral k in -1 to 1 =integral l in -1 to 1.

it doesnt mean k and l are same.


Moebius band not homeomorphic to Cylinder. I have been trying to think of a rather basic way of proving this, but it seems a bit elusive. In the case with boundary (looking at them as quotients in $ [0,1] \times [0,1]  $), they can be distinguished from the connectedness of the boundary (thanks Stefan), but I'm interested in the open case (looking at them as quotients in $ [0,1] \times (0,1)  $, i.e. as manifolds without boundary).

If there is not such basic way to do this, it would be interesting to read about some advanced methods you guys know about.

Here is the most elementary argument I can come up with.  Let $C$ denote the open cylinder and $M$ denote the open Moebius band.  Note that there exists a compact subset $K\subset C$ such that for any compact $K'$ such that $K\subseteq K'\subset C$, $C\setminus K'$ is disconnected (for instance, you can take $K$ to be the "equator" of $C$).  On the other hand, I claim that no compact subset $K\subset M$ has this property.  Indeed, let $K\subset M$ be compact; identifying $M$ as a quotient of $[0,1]\times(0,1)$ in the standard way, there is then some $\epsilon>0$ such that $K$ is contained in the image of $[0,1]\times[\epsilon,1-\epsilon]$ under the quotient map (for instance, because the images of sets of the form $[0, 1] \times (\epsilon, 1-\epsilon)$ are open and cover $K$). Let $K'$ be this image; then $K'$ is compact, $K\subseteq K'\subset M$, and $M\setminus K'$ is connected.

Let's be clear we have $X = [0,1]\times (0,1)$ with two different identification maps:

$f:(0,x) \in X  \mapsto (1,x) \in X$ this gives a cylinder $Y$
$f:(0,x) \in X  \mapsto (1,1-x) \in X$ this gives a Möbius band $Z$

Even though $Y$ and $Z$ are open, we can take their closure and $\partial \overline{Y} = S^1 \cup S^1$ while $\partial \overline{Z} = S^1 $.  See Chapter 4 of the notes of Allen Hatcher on Point Set Topology for a discussion on quotient spaces.

#1 Consider the subspace $X_0= (0,1) \times [\frac{1}{4}, \frac{3}{4}]$ common to both spaces.  Let $f:Y\to Z$ be a homeomorphism and the closure $\overline{X_0}\subset Y$ is a cylinder while $\overline{f(X_0)} \subset Z$ is a Mobius band.

#2 Cylinder is orientable.  Mobius band isn't.

#3 Remove the meridian circle $ [0,1] \times \frac{1}{2} $ from a cylinder - it is not connected.  Remove a meridian circle from a Möbius band, it remains connected.  Jordan curve is not necessary, merely that if a continuous map $\phi:[0,1] \to \mathbb{R}$ has positive and negative values, then $\phi(x) = 0$ somewhere.


solving bessel equation numerically. Assuming there's an equation (bessel) and I'm told to solve numerically. This means, to solve this type of equation, we must convert the equation to a system of first order ODE's by letting $z=y'$ and $z'=y''$. Now, assuming this is the bessel equation: $$x^2y''+xy'+(x^2-(3^2))y=0$$ But I need the initial conditions to do the conversion. How can I find the initial condition so as to be able to solve it numerically?

The Initial Conditions is not needed to do the conversion unless you want to use Euler's method. In the case of writing a program, any initial conditions can be used if not given and after , you compare it with the actual or theoretical solution


Competition Diophantine Equation I am looking for the number of integer solutions to this system of equations

$$a^2+b^2+c^2+d^2=2500$$ and

$$(a+50)(b+50)=cd$$

I tried moving terms around in the first equation and using the difference of two squares to try and gain some information, but came up empty. I was wondering what the best way to attack this problem is. Should I take the equations mod something, or should I try and place some bounds on the variables and try to derive the number of solutions from that? I ask that y'all only give me hints. Thanks

Hint: Use the fact $$2cd\leq c^2+d^2$$

and replace $cd$ and $c^2+d^2$ with a given constrains.


Need some advice to solve this integral $\int\frac{\sin^2x}{1+\sin^2x}\mathrm dx$ I'm trying to use this subtitution $t=\tan(x/2)$. But I don´t get anywhere. I've tried $t=\tan(x)$ too.  Appreciate your help.

$$\int\dfrac{\sin^2x}{1+\sin^2x}\mathrm dx$$

Note that $$\frac{\sin^2{x}}{1 + \sin^2{x}} = \frac{1 + \sin^2{x}}{1 + \sin^2{x}} - \frac{1}{1 + \sin^2{x}}$$

The first factor integrates to $x$. For the second integral, note that $$\frac{1}{1 + \sin^2{x}} = \frac{\sec^2{x}}{\tan^2{x} + \sec^2{x}} = \frac{\sec^2{x}}{2\tan^2{x} + 1}$$

Proceed using the substitution $u = \tan{x}$.

Write the ratio as

$$1- \frac{\cos^2 x}{1+\sin^2 x},$$

and recall that the derivative of $\arctan x$ is $1/(1+x^2)$ to obtain

$$x- 2^{-1/2} \arctan\bigl( \sqrt{2} \tan x\bigr)$$

(plus a constant)


Conditional Expectation Problem With Noise I was given the following problem:
let $X,N\sim \mathcal{N}(0,1)$ and let $A$ equal $1$ w.p $p$ and $0$ w.p $1-p$. Also, let $X,N,A$ be independent.
Define $Y=AX+N$. Find $\mathbb{E}(X\mid Y)$.

My idea was to use the 'nested conditioning' property $\mathbb{E}(X\mid Y)\overset{\text{a.s}}{=}\mathbb{E}[\mathbb{E}(X\mid Y,A)\mid Y]$.
I defined $Z=Y\mid A$ (which is likely at the very least a horrible abuse of notation).
$(X,Z)$ is undoubtedly a Gaussian vector (i.e jointly Gaussian) as a linear transformation of $(X,Y)$. Noting $Z\sim\mathcal{N}(0,A^2+1)$ we have $\mathbb{E}(X\mid Z)=\frac{\text{cov}(X,Z)}{\text{var}(Z)}Z=\frac{A}{A^2+1}Z$ , where I found $\text{cov}(X,Z)$ via the covariance matrix of $(X,Z)$.

I figure $\mathbb{E}(X\mid Y,A)=\mathbb{E}(X\mid Z)$, so substituting into the nesting property yields $$\mathbb{E}(X\mid Y)=\mathbb{E}[\mathbb{E}(X\mid Z)\mid Y]=\mathbb{E}\left( \frac{A}{A^2+1}Y\mid Y\right) =Y\mathbb{E}\left( \frac{A}{A^2+1}\mid Y\right)$$
And now I'm lost.

I'd like to know whether I've made any forbidden manipulations, and of course, I'd like a hint to help me out!

PS written some months later: This answer has some issues, which I will soon address in some edits.  See the comments below. End of PS.

$\newcommand{\E}{\mathbb E}$
$$
\E(X\mid Y)=\E(\E(X\mid Y,A)) = \E\left.\begin{cases} \E(X\mid X+N) & \text{if }A=1 \\ 0 & \text{if }A=0 \end{cases}\right\} = p\E(X\mid X+N).
$$
Whether all of these equalities are correct I leave for the moment as an exercise, which I haven't yet completely done myself.  The last conditional expected value can be found with formulas found in all the books, which it appears you probably know.


Why is pi/2 the angle associated with orthogonality in euclidean geometry? This is sort of a weird question, but I am trying to understand why 90 degrees or pi/2 radians is the angle that corresponds to orthogonality in 3-d (or really any dimension of) Euclidean space.  Said another way, what's so special about 1/4 of a complete rotation? Why is that the amount of rotation needed to end up in an orthogonal direction to where you started?

I think this must have a connection to the symmetries of the space but I can't put my finger on how to express it.

In 2D we can use the definition of a circle $\langle X,\,X\rangle = 1$, whence if a point with position vector $X$ moves smoothly on the circle, we get $\langle X,\,\dot{X}\rangle = 0$. With a little bit of work from this beginning, one can now prove that the linear mapping keeping the circle invariant is of the form:

$$R = \left(\begin{array}{cc}c&-s\\s&c\end{array}\right)$$

with $c^s+s^2=1$. Which mapping of this kind maps a vector into an orthogonal vector? In other words, we seek solutions to:

$$(x,\,y)\left(\begin{array}{cc}c&-s\\s&c\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right) = 0$$

and there are precisely two of them:

$$i=\pm\left(\begin{array}{cc}0&-1\\1&0\end{array}\right)$$

For either of these solutions, we can readily show that $\mathrm{id},\,i,\,i^2,\,i^3$ are all distinct and that $i^4=\mathrm{id}$. So four applications of our operator $i$ must impart exactly one full rotation.

Therefore, the isometry that maps a vector to an orthogonal one is the same as the isometry that corresponds to a quarter turn.

So, in summary, we can see that the answer to your question comes from both Danu's AND WillO's comments:

From Danu:

  It's a consequence of the definition of the inner product on $\mathbb{R}^N$.

And from WillO:

  It's because $i^4=1$

Further Questions from OP

  ...The derivation you give shows explicitly why a quarter turn maps a vector to an orthogonal one, but I think you probably couldn't even write down a rotation matrix in that form if the space had a structure in which that was false. ... was thinking about the meaning of orthogonality in Minkowski space and in non-Euclidean spaces that might be relevant for actual physical space.

Think of it this way: the matrix's structure encodes orthogonality: witness that its rows/ columns are orthogonal and it conserves inner products. Indeed in 2D it is the only linear mapping that both does so and preserves sense (reflexions also conserve inner products, but they multiply the angle between the two vectors concerned by -1). So it goes hand in hand with the definition of orthogonality. To understand this, witness that a matrix $M$ conserves inner products in Euclidean space iff

$$X^T\,M\,Y= X^T\,Y =\langle X,\,Y\rangle\,\forall\,X,\,Y\in\mathbb{R}^N\tag{1}$$

and write down the general matrix that does this: you'll find that its a rotation composed with a reflexion (indeed one can begin with the seemingly weaker notion of norm conservation $X^T\,M\,X= X^T\,X\,\forall\,X\in\mathbb{R}^N$, write down this condition for vectors $X$, $Y$ and $X+Y$, combine the resulting equations by dint of billinearity and you can derive (1) above). So the matrix's structure follows from the definition of orthogonality in Euclidean space.

A gross difference in Minkowski space is that the intersection of a vector space and its orthogonal complement is not trivial: there are nonzero null vectors ("lightlike vectors") s.t. $\langle X,\,X\rangle=0$, so you can't split the vector spce into a direct sum of subspace $U$ its orthogonal complement $U^\perp$ since $U\cap U^\perp\neq \{0\}$: the intersection of the two in general holds vectors other than 0.

  ...It seems like something important about physics would break if rotations worked some other way - that is, if it took 6 rotations by 90 degrees to get back to where you started....

Have you heard of spinors, covers of Lie groups and/or representations of groups? There are representations of rotations that take 8 lots of π/2 to get back to the identity. This is not quite the same thing, but it's intriguing nonetheless. $SU(2)$ is the double cover of $SO(3)$ and so if you represent multiplication by quaternions, quaternions act on vectors in a way that the unit $i$ corresponds to a half turn. You might like to see my answer here for further details. The scale factor here is 2, and $SU(2)$ is the universal cover of $SO(3)$ so all covers of $SU(3)$ must be contained in $SU(2)$ and so you get objects that take two full turns to return to the identity, objects that take one full turn to do so and nothing else: there are no scale factors of 3 or 6 for example.

This may be more simplistic than you require.
You said " ... any dimension..." so start with 2 dimensions.
It seems obvious to me that it you turn through 1/4 of a full rotation in 2 dimensions that you have turned through the angle of orthogonality.
If that is no  "obvious" in the context of your question (and it may not be) then the following simplistic answer does not satisfy. Otherwise ...

It's related to the fact that a circle has circumference 2 x Pi x radius (= R)  and that a radian is defined as the angle which encompasses an arc of length R. Given that definition the result "just drops out".

ie a full circle = 360 degrees = 2 x Pi radians
So 90 degrees = angle of "orthogonality"
= 360 degrees / 4 = 2 x Pi / 4 = Pi/2


An equality of sets involving two directions LEt $A,B$ be sets. Prove that $A \subseteq B \iff P(A) \subseteq P(B) $.

Attempt:

First, take any element of $P(A)$, say $Y$, we know by definition that $Y \subseteq A$ and so $Y \subseteq B$ (given). Therefore, by definition $Y \in P(B)$. and we have the result.

Conversely, suppose $x \in A$, then $\{x\} \in P(A)$ and so $\{ x \} \in P(B)$. (Can we conclude that $x \in B$) ?

Yes, that looks right.

For the single remaining step, once you know that $\{x\}\in\mathcal P(B)$, this is the same as $\{x\}\subseteq B$, which again means that $x\in B$, as needed.

I am afraid you seem to be getting a little confused on notation $\in$ is used to denote an element belonging to a set and $\subset$ is when saying a set is a subset of another set.

Anyway to show the converse assume the right hand side of the implication and let $x\in A$ then $x \in P(A)$ indeed taking the subset with just the element $x$ then $x \in P(B)$ and since $x$ is just a single element if $x \in P(B)$ then it follow $x \in B$ and we have proved the converse.

The first direction is fine one you sort out the notation. I think just need to make sure you are clear on notation.


For Which General Distributions Does This Inequality Hold? Let $X$ be a  random variable with mean $\mu$, where $0 < \mu < 1$.

Let  $X(n)$ be the sum of $n$ independent ,identically distributed, $X$ variables.

Under what conditions on $X$ , possibly dependent on $n$, is the following result true:

$$P[ X(n) \ge  (n+1)]  <  P[ X(n-1)\ge n]$$

For a simple illustration , if $X$ is discrete taking the values $0$ and $2$ with probabilities $0.7$ and $0.3$ respectively and $n=2$, the result is true as $0.09 < 0.3$.

The general question  arose from considering strategies for the players in the weaker of two  chess teams in a match.

X having a Normal distribution with mu < (3-2^0.5)/(2-2^0.5) , implicitly n>=2 will do: consider z scores.

Alternatively X (negative)  exponential with Erlangian sums will do for appropriate mu.

Other examples may well be possible


What is the element of this set? If K is any field. What is the elements of the sets
$$K[x,y]$$ and $$K[x,y]/(x^2,y^2)$$ where $(x^2,y^2)$ is the ideal?

The elements of $K[x,y]$ are polynomials in the variables $x$ and $y$ with coefficients from $K$. For example, $x + xy + y^2$.

The elements of $K[x,y]/(x^2,y^2)$ are cosets of the ideal $I$ generated by $x^2$ and $y^2$. For example, $x + xy + y^2 + I$. In this ring, the elements $x + xy + y^2 + I$, $x + xy + I$, and $x - x^2y + xy + I$ are all equal, since $y^2$ and $-x^2y$ are in $I$.

Now every element $p(x) + I$ in $K[x,y]/(x^2,y^2)$ is equal to a unique element of the form $ax + by + cxy + d + I$, where $a,b,c,d$ are in $K$. It's sometimes useful to identity the elements of $K[x,y]$ with $\{ax+by+cxy+d\mid a,b,c,d\in K\}$, remembering that multiplication is carried out modulo $I$ ("$x^2$ and $y^2$ are $0$"). What we've done here is picked a canonical representative for each equivalence class. This is analogous to how it's sometimes useful to identify the elements of $\mathbb{Z}/n\mathbb{Z}$ with $\{0,1,\dots,n-1\}$, remembering that addition and multiplication are carried out modulo $n$.

$K[x,y]$ are all polynomials in $x$ and $y$ with coefficients in $K$.

$K[x,y]/(x^2,y^2)$ is just a quotient ring. Roughly speaking, a property of that quotient ring is that all terms in the polynomial that have $x^2$ or $y^2$ become zero in the quotient ring.


Left Equivalence Relation I need to prove that if $G$ is a group, $H$ is a subgroup of $G$ and $\equiv_{L}$ is the equivalence relation from the partition $\{xH\:|\:x\in G\}$, then for $x_{1},x_{2}\in G$, $x_{1}\equiv_{L}x_{2}\Leftrightarrow x_{1}^{-1}x_{2}\in H$.

Attempt at Proof: Assume $x_{1}\equiv_{L} x_{2}$. By definition,

$x_{1}\equiv_{L} x_{2} \iff x_{1}H=x_{2}H$

$\iff H=x_{1}^{-1}x_{2}H$.

The last "if and only if" statement happens only when $\overline{x_{1}^{-1}x_{2}}=\overline{1_{G}}$, where the overline denotes an equivalence class. But $\overline{x_{1}^{-1}x_{2}}=\overline{1_{G}} \iff x_{1}^{-1}x_{2}\in H$. Hence, $x_{1}\equiv_{L}x_{2}\Leftrightarrow x_{1}^{-1}x_{2}\in H$.

Does this proof look okay? I think it is logical, but I don't want to make any mistakes on such a trivial problem. Thanks in advance for any help!

If you are using the partition $\{xH\mid x\in G\}$ to define your equivalence relation, then by definition $x_1\equiv_L x_2$ if and only if $x_2\in x_1H$. Therefore, the proof would be:
\begin{align}
x_1\equiv_L x_2 &\iff x_2\in x_1H\\
&\iff x_2=x_1h\mbox{ for some }h\in H\\
&\iff x_1^{-1}x_2=h\in H.
\end{align}

If you're doing a step-by-step

A $\Leftrightarrow$ B

B $\Leftrightarrow$ C

etc.

proof, you don't need to assume A at the beginning.

Your proof is essentially structured as claiming A $\Leftrightarrow$ B and B $\Leftrightarrow$ C and then going back and saying D $\rightarrow$ (B $\Leftrightarrow$ C) and D $\Leftrightarrow$ E, which a confusing structure.
It might be good to just work out why $x_{1}H=x_{2}H\iff H=x_{1}^{-1}x_{2}H$ from the definition of $x_{1}H$. On high level proofs, one can assume simpler properties that have already been established in the course, but if you're doing a basic proof involving X, you should at some point be relying on the definition of X.


Interpreting the statement Just started quantifiers, and I'm having some trouble with interpreting this.

Here's what I understand:

For every Epsilon $> 0$, there exists a Delta $> 0$ for all $x$ in $\mathbb{R}$.

The antecedent of the implication is false, because there is no Delta that is always greater than $x$.

The precedent of the implication is also false, because not every Epsilon greater than $0$ is also greater than $x^2 -1$ for any $x$.

So the statement is false. Is my logic sound? I'm not sure if I'm reading this right.

I believe your difficulty arises from a shortage of parentheses in the formula.  I think the formula was intended to be
$$
(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in\mathbb R)\Big([0<|x-1|<\delta]\implies [|x^2-1|<\varepsilon]\Big).
$$
That is, the quantifiers at the beginning apply to the whole implication, not just to its antecedent.  This is also how the two answers preceding mine (by Ahmed and Michael Cotton) interpreted the formula.

You could read the implication as:

$|x^2 -1|<\varepsilon\:$ whenever $\:
|x-1|<\delta$

which seems just fine.

With the quantifiers, this whole thing just says the function $x^2$ is continuous at $x=1$.

Added note: See Andreas Blass' answer. I was assuming the quantifiers applied to the entire implication. Depending on the text it's from, that's probably what is intended.


Subtraction of inequalities Suppose I have two inequalities:
$$k_1 \leq a$$
$$k_2 \leq b$$

where $k_1,k_2,a,b$ are all positive numbers

I know that that summation of them can be written as:
$$k_1+k_2 \leq a+b$$

But I want to find $$k_1-k_2 \leq ?$$ or $$? \leq k_1-k_2$$

Is there any way to solve this problem?

First thing you can do is invoke the triangle inequality, i.e. $|x + y| \leq |x| + |y|$. Note that this equality works for negative $x$, $y$ as well. This gives you $$
\begin{eqnarray}
|k_1 - k_2| &\leq& |k_1| + |k_2| &\leq |a| + |b| \text{ and since }k_1-k_2 \leq |k_1-k_2| \\
k_1 - k_2 &\leq& |k_1| + |k_2|  &\leq |a| + |b|
\end{eqnarray}
$$

For arbitrary $k_1 \leq a$, $k_2 \leq b$, you can't do better than that, but since you stated that $k_1,k_2 \geq 0$, you can. $k_1,k_2$ being positive means that $$
\begin{eqnarray}
-b &\leq& -k_2 &\leq& k_1 - k_2 &\leq& k_1 &\leq& a\\
-a &\leq& -k_1 &\leq& k_2 - k_1 &\leq& k_2 &\leq& b
\end{eqnarray}
$$
because subtracting a positive number can only make a number smaller, never bigger.

Finding the inequality with k(1) - K(2) as the subject

The simplest solution is usually the best solution----Albert Einstein

Given:

k(1) ≤ a

k(2) ≤ b

Required: To determine if  k(1) - K(2) can be made the subject of an inequality

       derived from the above inequalities

Possible solution:

One will operate on the given system of inequalities.

Generally, subtracting c < d from a < b yields a - d < b - c; but below, one

will apply the permissible operations on a system of inequalities. To add a

system of two inequalities, both inequalities must have the same sense (or

direction). To subtract an inequality, multiply this "subtrahend"  inequality by

-1 while reversing the sense of the inequality, and then add to the "minuend

inequality", making sure the two inequalities have the same sense.

Now,  one will subtract K(2) ≤ b from k(1) ≤ a).

Step 1: Multiply the "subtrahend" inequality,  K(2) ≤ b  by -1 and reverse the

     sense to obtain - K(2) ≥ - b......(1)

Step 2: Rewrite (1) so that it has the same sense as the "minuend" inequality,

    k(1) ≤ a. Then one obtains -b ≤ -k(2)

Step 3 Add the left sides and add the right sides below

   k(1) ≤ a
     -b ≤ -K(2)
 ---------------

k(1) - b ≤ a - k(2)  <-----the "difference" inequality,

From the above result, k(1) - k(2) cannot be made the subject of the above

inequality.

Conclusion:

Therefore, explicitly, k(1) - k(2) does not exist.

However, implicitly, k(1) - k(2) ≤ a + b - 2k(2)

Repeating the above procedure, using a, b, c, d

Given

a ≤ b...........(1).

c ≤ d............(2);

determine if a - c can be made the subject of the inequality derived from
the above system of inequalities.

Solution

From (2)  -c ≥ -d         (multiplying by -1 in order to subtract)

     -d ≤ -c....(3) (rewriting in the same sense as (1))

Adding the left sides of (1) and (3); and adding the right sides of (1) and (3),

one obtains

a - d ≤ b - c...........(4)

Conclusion

Explicitly, in (4), a - c cannot be made the subject of (4),

However, implicitly, a - c ≤ b + d - 2c


If a property in $\mathbb{N}$ is true up to $10^{47}$ are there reasons to think it is probably true in all $\mathbb{N}$? You have probably heard at some point statements like that the twin prime conjecture (namely that $2$ is an infinitely ocurring prime gap) is "probably" or "almost certainly" true. Same goes for a number of other problems in number theory asserting properties that we believe are satisfied by all natural numbers. An argument that seems to give ground to such beliefs is the observation that said conjecture holds up to a certain large number.

  My question would be if there some solid argument to think that a given conjecture is "probably" true, given that we know it is true up to some large number (as big as you wish but fixed).

The answer upfront seems to be no, and still the thought is so unavoidable I wonder if someone has thought of an argument to justify it.

EDIT: Do not take the question rigurously. I was looking for a broad take on this. Think for example of the case of the twin prime conjecture. Of course I know there are properties satisfied by numbers only above a given number (for example the property of being bigger than said number). Be imaginative. Tell me results or arguments that you think might relate. Tell me about conjectures where, while not knowing if they hold, this line of thought is somehow justified and why.

EDIT $2$: Case where this question would apply. Let's say we conjecture some property of the natural numbers. About the size of a possible minimum counterexample we are not able to find anything besides that it must be greater than $10^{47}$. At this point the thought might reach our minds that the conjecture is "probably" true. Now, is there any convincing argument that justifies this thought?

No.

In fact, most positive integers are larger than $10^{47}$. An example of a property that is true for integers below $10^{47}$ but not true for integers larger than $10^{47}$ is being smaller than $10^{47}$.

A more apt but common example can be found in number theory. The number of primes up to $x$ is denoted by $\pi(x)$, and is commonly approximated by the logarithmic integral
$$\text{Li}(x) = \int_2^x \frac{1}{\ln t}dt$$ Asymptotically, they are equal. It was believed for a long time that $\pi(n) < \text{Li}(n)$ for all $n$ (both Gauss and Riemann conjectured this). But it is not true. The first counterexample is thought to be somewhere around $10^{300}$.

$10^{10^{47}}$ have all digits at zero starting from unity up to the $10^{47th}$. Still, I advice you to verify one more step ;-)


Prove that $n\log(n) = O(2^{\log^2(n)})$ I have to prove that $n\log(n) = O(2^{\log^2(n)})$, where $\log(x) = \log_2x$. I tried saying that: $2^{\log^2(n)} = 2^ {\log(n)\log(n)}= 2^{\log(n)^{\log(n)}} = n^{log(n)}$. I also know that $n\log(n) = \log(n^{n})$.I thought of just saying that it's true because the first function is logarithmic, but I am not sure this is correct. I also tried with the fact that $n\log(n) = O(2^n)$, but proving that $2^{\log(n)} = O(2^n)$ doesn't help. Any recommendations would be appreciated.

Note that $\log n + \log \log n \le 2 \log n \le (\log n)^2$ provided that $n \ge 4$ in which case $$n \log n = 2^{\log n + \log \log n} \le 2^{(\log n)^2}.$$

Actually, $n\log n=o(2^{\log^2 n})$ (little oh), which is a stronger statement.
To see the, consider the functions $f(x)=x\log x$ and $g(x)=2^{\log^2 x}$.
Clearly both functions tend to infinity as $x\to\infty$. Using L'hopital's rule,
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\frac{
\frac{1+\ln x}{\ln 2}
}
{\frac{2\exp(\frac{\ln^2 x}{\ln 2})\ln x}{x\ln 2}}=\frac{x\ln 2(1+\ln x)}{2\ln 2\exp(\frac{\ln^2 x}{\ln 2})\ln x}$$
Since for sufficiently large $x$ we have $\ln x>2\ln 2$, we deduce that
for sufficiently large $x$ $\exp(\frac{\ln^2 x}{\ln 2})>\exp(2\ln x)=x^2$, and so the r.h.s of the displayed equation above is bounded above by
$$\frac{x\ln 2(1+\ln x)}{2\ln 2\cdot x^2\ln x}=O(\frac{1}{x})$$
Therefore $f(x)=o(g(x))$ as $x\to\infty$, and in particular, $n\log n=o(2^{\log^2 n})$ as $n\to\infty$.
Remark. The argument above shows that actually a much stronger statement is true, namely for every positive integer $k$,
$$\frac{f(x)}{g(x)}=o(x^{-k})\quad (x\to\infty)$$
Therefore, the ratio
$$\frac{n\log n}{2^{\log^2 n}}$$
tends to zero faster than any power of $\frac{1}{n}$ as $n\to\infty$.
In other words, for every positive integer $k$,
$$\lim_{n\to\infty}\frac{n^k\log n}{2^{\log^2 n}}$$
This of course answers the OP's question, although its  primary purpose is to show that the OP's question, is in a sense, "not the right question".


$12$ Prisoner Hat Problem with $5$ different Hats There are $12$ prisoners named $A,B,C,D,E,F,G,H,I,J,K$ and $L$ and each of them was given a hat with the number $0,1,2,3$ or $4$ on it. They cannot see what is on their own hat but can see what is on the other $11$ prisoners' hats.

The guard calls them forward in alphabetical order and asks them to whisper in his ear what number they think is on their hat. If they are correct they are allowed to leave otherwise they get executed.

Before the prisoners are given hats they are allowed to devise an optimum strategy so that the most prisoners leave with their life. Once hats are given no communication is allowed between the prisoners (including any sort of secret code).

What is the optimum strategy and how many prisoners can survive?

According to the OP's comments, later prisoners can observe the fate of earlier prisoners. But now it's easy to ensure that at least eight prisoners survive:

$A,B,C,$ and $D$ assume that the sum of all the hats modulo $5$ is equal to $0,1,2,$ and $3$ respectively, and base their answer on this assumption. As soon as one of them goes free, everybody else knows this sum, so they can answer correctly.

If none of them go free, the sum must be $4$. And of course if $A,B,$ or $C$ goes free, then $B,C,$ and/or $D$ are freed from their obligations, and will go free too. So the number of survivors can be anything from $8$ to $12$.

The probability of a prisoner guessing his hat number is $\frac{1}{5}$

The probability of a prisoner not guessing his hat number is $\frac{4}{5}$.

In other words, the prisoner knowing other prisoner's hat number has no bearing on his guessing his hat number correctly.  Each prisoner guesses independently of other prisoners.

P ( all guessed correctly) $={12\choose12} \frac{4^0}{5^{12}}$

P ( 11 guessed correctly) $={12\choose11} \frac{4^1}{5^{12}}$

...

P ( 0 guessed correctly) $={12\choose0} \frac{4^{12}}{5^{12}}$

This looks like binomial distribution and thus the expected number of prisoners who will survive $$\sum_{i=0}^{12} i{12\choose 12-i} \left(\frac{4}{5}\right)^{12-i} \left(\frac{1}{5}\right)^{i}$$

$$=\boxed{2.4}$$


What function to use to show one to one correspondence? This problem is from Discrete Mathematics and its Applications

Here's an example problem that the author gave

I'am working on problem 2e. I first recognized the set as countably infinite. If you do the Cartesian product of A with some zs say a portion of it, {1,2,3}, you get coordinates {(2,1),(2,2),(2,3)...} and so on which are countable and keep going(because the set of integers keeps going on. My question is what function to use to show the one to one correspondence between the set of positive integers and this set(Cartesian product of A and z+. set of positive integers)? I have a hard time with coming up with some function that will map a positive integer to a member of this set. The example problem's function was pretty intuitive. If you have a positive integer, to map to a odd positive integer, have a function like f(n) = 2n -1, or multiply it by 2 and subtract 1. What about this case?

Let $f:\mathbb{N}\to \mathbb{Z} \setminus \{0\}$ be

$$f(n) = \begin{cases}\dfrac{n+1}{2} & n \ \text{ is odd}\\ -\dfrac{n}{2} & n \ \text{ is even}\end{cases}$$

Then define $g:\mathbb{Z}\setminus \{0\}\to A\times \mathbb Z^+$ as

$$g(n) = \begin{cases}\{2\}\times \{-n\} &: n <0\\ \{3\}\times \{n\} &: n >0\end{cases}$$

$f$ and $g$ are both bijections. So $g\circ f$ is also a bijection.

Here's a simple function that maps the tuples in $\{2, 3\}\times \mathbb Z^+$ to the positive integers.

Let $(t_0, t_1)$ be a member of $\{2, 3\}\times \mathbb Z^+$

Then $(t_0 - 2)(2t_1 - 1) - (t_0 - 3)(2t_1)$ maps each $(t_0, t_1)$ to a unique member of $\mathbb Z^+$

If $t_0 = 2$ the 1st term vanishes and we get $-(-1)(2t_1)$ = $2t_1$

If $t_1 = 3$ the 2nd term vanishes and we get $(1)(2t_1 - 1)$ = $2t_1 - 1$

So we've mapped tuples of the form $(2, t_1)$ to even numbers and tuples of the form $(3, t_1)$ to odd numbers.

Inverting this function will be left as an exercise for the reader. :)


Efficient way to solve $31^6 \mod 189$ without using any external devices. Assuming basic knowledge of modulus arithmetic, is there a way, a trick or formulas that allow you to solve modulus question with very large numbers without using calculator? For example,

$$76^7 \mod 187=32$$ $$7^{23} \mod 143=2$$

This question ties into RSA Encryption where I'm trying to figure out how to solve these questions. So if I have $31^6 \mod 189$, I can simply use Wolfram Alpha but how can I work out that $31^6 \mod 189$?

Chinese remainder theorem int tandem with Eulers theorem will reduce is as far as possible.  As to whether the calculations are easy or not is another issue.

So $31^6 \mod 189$.  $189 = 7*3^3$ so by chinese remainder theorem we can solve this by solving $31^6 \mod 7$ and $31^6 \mod 27$.

So $7$ is prime and $31$ is not a multiple of $7$ we know $31^{7-1}\equiv 1 \mod 7$ so ... that's that.  And $\phi(27) = 3^2*2 = 18$ so $31^6 \equiv .... \mod 37$.  Well, it's a cube root of one but ... $31^6 \equiv 4^6 \equiv (3+1)^6 \equiv {6 \choose 2}3^2 + 6*3 + 1 \equiv \frac {6*5}2*3^2 + 18 + 1\equiv 19\mod 27$ ... or $4^6 \equiv 64^2 \equiv 10^2 \equiv 100 \equiv 19 \mod 27$.

So $31^6 \equiv 1 + 7k \equiv 19 + 27j \mod 189$.  By CRT there is one solution.

So solve $1 + 7k = 19 + 27j$

$27 = 4*7 - 1$

$1 = 4*7 - 27$.

$1 + 7k = 19 + 27j$

$7k - 27j = 18 = 21 - 3 = 3*7 - 3(4*7 - 27) = -9*7 +3*27$.

So if $k = -9$ and $j = -3$ we get $1-7*9 = -62$ and $19 - 3*27 =-62$

And so $31^6 \equiv -62 \equiv 127 \mod 189$

$31^2=961$ and $5\times 189=945$ so that $31^6\equiv 16^3$ and if you know your powers of $2$ this is efficiently sorted.


Probability All Players Receive a Spade 52 cards were distributed among 4 players, 13 cards for each. Calculate the probability that all players have received at least one spade.

I do not know what to do with this question, i have the answer in my textbook however i do not understand where does it come from. How to solve this question?

EDIT: I am still looking for a good explanation.

Consider 4 compartments, each having $13$ slots, and although cards are not distributed in this manner, it will not matter if we first distribute all the spades.

If one compartment is barred, all the spades have to go into the remaining $39$ slots, if two compartments are barred, they have to go into the remaining 26 slots, and so on.

Then applying the principle of inclusion and exclusion, $Pr = \frac{\dbinom{52}{13} - \dbinom41\dbinom{39}{13} + \dbinom42\dbinom{26}{13} - \dbinom43\dbinom{13}{13}}{\dbinom{52}{13}} \approx 0.9489$

I would say

Favorable event: Each player has one spade + 12 additional cards of, selected from the remaining 48 cards.

The number of favorable events: $\quad \omega ={48\choose12}$

the number of possible events: $\quad \Omega ={52\choose13}$

$\displaystyle P = \frac{\omega}{\Omega}=\frac{{48\choose12}}{{52\choose13}}=\frac{9139}{83300} \doteq 0.110$

Edit: $\color{green}{ \text{Apparently the wrong solution.}}$ georg


Minimum number of points for a good exponential curve fit We are currently trying to fit data from a process that generates data that decays exponentially.  We employ several techniques for fitting this exponential decay in including FFT analysis and a weighted least-squares algorithm.  We are attempting to fit a lot of these decays every second (in some cases 8000 decays/s) on varied computing platforms.

We want to minimize our computational time so we now have the question as to how do we determine the minimum number of points required for a "good" fit (I know the term is ambiguous).  Just to be clear, we aren't interested in how long the sample required for a good fit is; we want to be able to determine how many points for a given portion of a curve would be required for a good fit.  Is there an explicit approach to determining this or will this require modeling?

If the data is clean, you only need two points because there are only two degrees of freedom-the original rate and the decay time.  You use more points when there is noise in the data to get a better estimate, so the number of points needed depends on how noisy the data is.  You can linearize your fit by taking the logairithm, getting $\log(\text{counts}(t))=\log (\text{counts at t=0})-\text{(decay rate*time)}$  Now a linear least squares fit will give you estimates of the errors in the parameters.  I would take a few curves and fit them each with lots of points.  You can then take a random sample of the points and see how much the fit changes.

As stated in the original post, the function is a single exponential.

Also, as stated in the original post, we are trying to DETERMINE the number of points needed.  As a rough guideline, we sample 100s to 1000s of points for each curve currently.  But maybe we only need 10 or 50 or 100.


Why can't we add a non-square matrix $A$ to its transpose $A^T$? The addition operation is commonly defined as follows:

  Two matrices must have an equal number of rows and columns to be added

But this is a very shallow definition/interpretation.

A deeper interpretation of a $n \times n$ matrix, I would believe, would be that it is an element of a vector space, e.g. $A \in \mathbb{R}^{n \times n}$ as you need $n^2$ linear equations to compute each entry in the matrix (a dot product for each entry)

Now if you use that interpretation of a matrix, then let's assume we have a column vector $B \in \mathbb{R}^{n \times 1}$ and it's transpose, a row vector, $B^T \in \mathbb{R}^{1 \times n}$

$$\begin{align}B = \begin{bmatrix}
b_1 \\
b_2\\
. \\
.\\
.\\
b_n\end{bmatrix}
&&\text{and}&&B^T = \begin{bmatrix}
b_1 &
b_2&
. &
.&
.&
b_n\end{bmatrix}
\end{align}$$

But both $B$ and $B^T$, are both elements of the same vector space $B, B^T \in \mathbb{R}^n$. So why is their addition undefined?

Why can't you add a matrix to its transpose, just like you could two vectors that are elements of the same vector space?

Is it wrong to interpret matrices as I have done? Is there a more rigorous definition of the addition operation for matrices? Are there better ways to interpret matrices?

$B$ and $B^T$ are not elements of the same vector space: $B$ is an element of $\Bbb R^{n\times 1}$, and $B^T$ is an element of $\Bbb R^{1\times n}$. Both of these vector spaces are isomorphic to $\Bbb R^n$, but no two  of $\Bbb R^n$, $\Bbb R^{n\times 1}$, and $\Bbb R^{1\times n}$ are actually equal to each other.

There is no one deciding what you can and can't do in mathematics. You can do whatever you want to do. The only catch is that you have to convince people that what you do makes sense and is interesting. If you want to define the addition of your $B$ and your $B^T$, go ahead. I'm sure it can be useful in some cases.


Homework: What is the probability that I have picked Box A? Suppose I have two boxes. They look identical to me. However, I am told:
“Box A contains 1 red ball and three white balls; Box B has 2 red balls and 2
white balls.”

I randomly pick a box and select a ball from the box without looking into the
box.

It is a red ball.
What is the probability that I have picked Box A?
Choose one of the following options.

  A. Less than 30%.
  B. 30% ~ 50% (30 % is included; 50% is not included).
  C 50%.
  D. 50% ~ 60% (50% is not included; 60% is included).
  E. More than 60%.

    Solution

    P(Picking Red ball)$=(1+2)/(1+3+2+2)=3/8$

    P(Picking Red Ball From Box A)$=1/2*1/4=1/8$

    We want P(Picking Box A|Red Ball Picked)=P(Picking a Red ball from Box A)\P(Red Ball Picked)

    Therefore, probability is $=(1/8)/(3/8)=1/3$ Answer is A.

Am I doing this correctly? My tutor is known for giving not-so straightforward questions, so I'm wondering if I need to consider another way, or I could be wrong. Any alternatives welcome too!

$P\text{(picking a red ball)} =\frac 12 (\text {pick A}) \frac 14 (\text{\pick red ball (from A))}\\+ \frac 12  (\text {pick B}) \frac 12 (\text{\pick red ball (from B))}=\frac 18 + \frac 14=\frac 38$

This agrees with your value, but only because there were the same number of balls in each box, so each ball is equally likely to be picked.    If box B had 2 red and 4 white the probability of a red ball would be $\frac 12\frac 14 + \frac 12 \frac 13=\frac 18 + \frac 16=\frac 7{24}$ , but you would have $\frac {1+2}{4+6}=\frac 3{10}$

When you randomly pick a box so it is A or either B so probability would be 0.5


Finding Basic Limits For the function given by $f(x) = |x − 2| + 42$ and $x → 2$, make a guess of the limit, and then prove your
  assertion using the formal definition of a limit.

This was confusing; please help.

My guess is $f(x) = 42$ as $x$ approaches $2$.

Claim: $\displaystyle \lim_{x \to 2} f(x) = 42$.

Proof: Let $\epsilon > 0$ be given, choose $\delta = \epsilon$. If $0 < |x-2| < \delta \Rightarrow |f(x)-42| = ||x-2|+42-42| = |x-2| < \delta = \epsilon$, hence the result.

A function can have a limit so long as for $f(x)=L$ such that there is a distance $\delta$ to the left and right of $a$ such that there correspond to $x$ values that produce $y$ outputs that are $\epsilon$ values on the top and bottom of the limit.

$|x-a|>\delta$ such that $|f(x)-L| > \epsilon$

$f(2)=42$ as $x$ approaches
So....
$|x-2|>\delta$
say $\epsilon=0.01$, and we have values $y=42.01$ and $y=41.99$ casing $42$.

If $42.01=|x-2|+42$, then $2.01=x$.

If $41.99=|x-2|+42$, then $-0.99+2=1.99$.

If $2+\delta=2.01$, $2-\delta=1.99$, then $\delta=0.01$.

Hence there exists these two $\delta$ values such that there is also an $\epsilon$ distance between $42$.


taking the power of upper triangular matrices When a matrix $A$ is diagonalizble, i.e. $A = P \Lambda P^{-1}$, then it is very easy to compute $A^i$: it is just $P \Lambda^i P^{-1}$ where $\Lambda^i$ is the diagonal matrices with the eigenvalues exponentiated.

I have an upper triangular matrix where the diagonal is all 0s (this means that all the eigenvalues are 0). I would expect somehow to be able to do something similar to the exponentiation of $A$ above, but that's not possible because such upper triangular matrix is not diagonalizible. Is there another efficient way of computing the power of an upper triangular matrix with all 0s on the diagonal?

Every square matrix has a unique Jordan decomposition up to permutations of nonsingular matrix P. Let T be your upper triangular matrix where diagonal elements are all zero. Then $T=P^{-1}JP$. $T^n=P^{-1}J^n P$.


Why $\|u\|_{L^r}\leq C\|\nabla u\|_{L^p}^{a}\|u\|_{L^q}^{1-a}$? Let $u\in \mathcal C_c^1(\mathbb R^n)$ and $p>n$. Let $r>0$ s.t. $$\frac{1}{r}=a\left(\frac{1}{p}-\frac{1}{n}\right)+\frac{1-a}{q},$$
with $a\in [0,1]$, $p>1$ and $q\geq 1$. I'm trying to understand why $$\|u\|_{L^r(\mathbb R^n)}\leq C\|u\|_{W^{1,p}(\mathbb R^n)}^a\|u\|_{L^q(\mathbb R^n)}^{1-a},$$
with $C$ independent of $u$.

I thought first using the fact that $W^{1,p}(\mathbb R^n) \subset L^\infty (\mathbb R^n)$, but it's not conclusive.
But I know using Morrey's theorem that there is $C$ independent of $u$ s.t. $$\frac{|u(x)-u(y)|}{|x-y|^{1-\frac{n}{p}}}\leq C\|\nabla u\|_{L^p(\mathbb R^n)},$$
but even with this information, I can't conclude.
By a scaling argument, I can consider $\|u\|_{L^q}=\|u\|_{W^{1,p}}=1$, and thus, I have to prove that $u$ is uniformly bounded for the $r$ norm. I know that the injection $W^{1,p}(\mathbb R^n)\subset L^\infty (\mathbb R^n)$ is continuous and thus there is $C$ independent of $u$ s.t. $$\|u\|_{L^\infty }\leq C.$$
Now, I know that on a compact $K$, $L^{\infty }(K)$ is continuously injected in $L^{r}(K)$. So if I denote $K$ the support of $u$, I indeed have $\|u\|_{L^r(\mathbb R^n)}=\|u\|_{L^r}(K)\leq C$, and thus I can conclude, but since $K$ depend on $u$, the constant will depend on $u$, and thus, it doesn't solve my problem.

Any help is welcome :-)

$$\int_\Omega |u|^r=\int_\Omega |u|^{r-q}|u|^q\leq \|u^{r-q}\|_{L^\infty }\|u\|_{L^q}^q\leq C,$$
et voilà !

Please see the proof below, which is from Adams' book "Sobolev space".  Also, in the Theorem 5.9 in Adams' book, the restriction $p\leq q$ is removed.


Find all roots of $2x^3+16$ in $\mathbb C$ Find all roots of $p(x) = 2 x^3 + 16$ in $\mathbb C$.

I found my answers to be x = 2, -1+i$\sqrt{3}$, -1+i$\sqrt{3}$.

But when I put the expression into Symbolab, it gives me -2, 1+i$\sqrt{3}$, 1+i$\sqrt{3}$ as roots of p(x) in C.

Can someone explain where I went wrong?

This is how I did it;

First, I rewrite p(x) and get

$2x^3+16=0$

$2x^3=-16$

$x^3=-8$ ------ (1)

Express (1) in Euler form

$x=re^{\theta i}$

$x^3=r^3e^{3\theta i}$

Let w = -8

$ |w| = \sqrt{(-8)^2} = 8$

$\theta = tan^{-1}(0) = 0$

$w = 8e^{0\theta}$

Now I have

$r^3e^{3\theta i} = 8e^{0\theta}$

Equate the modulus and argument

$r^3 = 8$

$r = 2$

$3\theta = 0+2\pi k$ ------- for k $\in Z$

$\theta = {2\pi k \over 3}$

Now I have

$ x = 2(cos {2\pi k \over 3} + i sin {2\pi k \over 3})$

Calculating x by substituting k = 0,1,2

k = 0, x = 2

k = 1, x = 2(${-1 \over 2} + i {\sqrt{3} \over 2}$) = -1+i${\sqrt{3}}$

k = 2, x = 2(${-1 \over 2} - i {\sqrt{3} \over 2}$) = -1-i${\sqrt{3}}$

So all roots of p(x) in C are 2, -1-i${\sqrt{3}}$, -1+i${\sqrt{3}}$

.

.

.

I will be very appreciated if someone can help me out.

I will give three answers to three different questions.

  Can someone explain where I went wrong?

The complete rule for the argument of a complex number $z=x+i\,y$ is not simply $\theta=\tan^{-1} \frac{y}{x}$ but:
$$
\theta=
\begin{cases}
\arctan\frac{y}{x} & x>0 \\
\arctan\frac{y}{x}+\pi & x<0 \text{ and } y\ge 0 \\
\arctan\frac{y}{x}-\pi & x<0 \text{ and } y  < 0 \\
\frac{\pi}{2} & x=0 \text{ and } y  > 0 \\
-\frac{\pi}{2} & x=0 \text{ and } y  < 0 \\
\text{undefined} & x=0 \text{ and } y  = 0 \\
\end{cases}
$$
where $\arctan$ provides a value in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

  Which is the deep mathematical issue going on here?

The definition of inverse function. Some functions do not have a globally defined inverse. In this case, $\tan 0=\tan \pi=0$ so $\tan^{-1} 0$ is ambiguous. However, we can define $\tan:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\mathbb{R}$ and then this function is monotonically increasing thus invertible. The problem is then that we cannot obtain any possible argument, so we must consider all the cases above.

  3.a. In general, how can I do the exercises about complex numbers?

Drawing the unit circle. When $x>0$-i.e. the complex number is in the first or fourth quadrant with argument $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$-then the formula $\arctan\frac{y}{x}$ gives the correct result. But imagine that you change the sign of a complex number, i.e. $x\gets -x$ and $y\gets -y$. Then:
$$\arctan\frac{-y}{-x}=\arctan\frac{y}{x}$$
and the $\arctan$ is unable to detect the change of sign. But if you draw the situation, you realize that you are in the second or third quadrants, so you get the value of the corresponding argument in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ by the formula  $\arctan\frac{y}{x}$ and then "change sign"... but taking the opposite complex corresponds to half a turn of the circle, i.e. adding an angle of $\pi$ radians.

Now, a final detail: you can add multiples of $2\pi$ to the argument and you obtain the same complex number, because it corresponds to full turns of the circle. Then you have to fix a criterion, by defining a principal argument. Some people prefer $\theta\in[0,2\pi)$ whereas other use $\theta\in(-\pi,\pi]$. Whatever is your convention, in the computation above, you may end up with an argument outside your chosen interval, so you can freely add or subtract $2\pi$ as needed. For instance, in the formula above, when you are in the third quadrant, you should add $\pi$ to do the "half turn" corresponding to changing sign, but then you go out of the conventional interval, so you subtract $2\pi$ to go back. This is the explanation of the $-\pi$ of the third line of the formula, but mind that there are two processes going on here: adding $\pi$ is mandatory to go to the correct quadrant; subtracting $2\pi$ is somewhat optional, since it is only a convention to give the result in a unique way.

In other words,

  3.b. How can I learn by heart the formula above?

Don't even try. Nobody can. I strongly discourage it. It is much more instructive and less prone to errors to draw and understand really what is going on.

Another way you can calculate this is to use De Moivre's theorem, which states that if $z = r(\cos \theta + i\sin \theta)$, the $n$th roots of $z$ are $$r^{1/n}\left(\cos\frac{\theta+2\pi k}{n} + i \sin \frac{\theta+2\pi k}{n}\right)$$

From $(1)$, which is $x^3=-8$, we find that $z = 8(\cos \pi + i \sin \pi)$, so $r=8$, $\theta=\pi$ (because $8$ and $-8$ are opposite to each other), and $n=3$ (we are finding the $3$rd roots of $z$). Substituting these values into the formula, we have:

First root: $8^{1/3}(\cos \frac{\pi+2\pi *0}{3} + i \sin \frac{\pi+2\pi *0}{n}) = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}) = 1+i\sqrt{3}$

Second root: $8^{1/3}(\cos \frac{\pi+2\pi *1}{3} + i \sin \frac{\pi+2\pi *1}{3}) = 2(\cos\pi+i \sin \pi) = -2$

Third root: $8^{1/3}(\cos \frac{\pi+2\pi *2}{3} + i \sin \frac{\pi+2\pi *2}{3}) = 2(\cos \frac{5\pi}{3} + i\sin\frac{5\pi}{3}) = 1-i\sqrt{3}$

Use the fact that $n$th roots of $z$ have $n$ complex roots. Have we have found all the complex $3$rd roots of $8$ yet?


What am I doing wrong ? $\int \csc(x) \cot^2 (x) dx$ $$
\int \csc(x)\cot^2(x) dx= \int \csc(x) (\csc^2 (x) - 1) dx = \int (\csc^3(x) - \csc(x))dx
$$

Integrating by parts:

let u = $\csc(x)$, then du = $-\csc(x)\cot(x)dx$.

let dv = $\csc^2 (x)dx$, then v = $-\cot(x)$.

The above integral becomes

$$\int \csc(x)\cot^2(x) dx=-\csc(x)\cot(x) - \int \csc(x)\cot^2 (x)dx - \int \csc(x) dx.$$

Using $\cot^2 (x) = \csc^2 (x) - 1$ identity:

$$\int \csc(x)\cot^2(x) dx= -\csc(x)\cot(x) - \int \csc^3(x)dx + \int \csc(x) dx - \int \csc(x) dx\\
=-\frac{1}{2} \csc(x)\cot(x) - \frac{1}{2}\ln |\csc(x) + \cot(x)| + \ln |\csc(x) + \cot(x)|\\
=-\frac{1}{2} \csc(x)\cot(x) + \frac{1}{2}\ln |\csc(x) + \cot(x)| + C$$

just the last, as your try, you need to find $\int \csc^3(x)dx$, and $$\int \csc^3(x)dx \neq -\frac{1}{2}\csc x\cot x$$
so the last part is wrong.


Multiple-choice question about the probability of a random answer to itself being correct I found this math "problem" on the internet, and I'm wondering if it has an answer:

  Question: If you choose an answer to this question at random, what is the probability that you will be correct?

  a. $25\%$

  b. $50\%$

  c. $0\%$

  d. $25\%$

Does this question have a correct answer?

No, it is not meaningful. 25% is correct iff 50% is correct, and 50% is correct iff 25% is correct, so it can be neither of those two (because if both are correct, the only correct answer could be 75% which is not even an option). But it cannot be 0% either, because then the correct answer would be 25%. So none of the answers are correct, so the answer must be 0%. But then it is 25%. And so forth.

It's a multiple-choice variant (with bells and whistles) of the classical liar paradox, which asks whether the statement

  This statement is false.

is true or false. There are various more or less contrived "philosophical" attempts to resolve it, but by far the most common resolution is to deny that the statement means anything in the first place; therefore it is also meaningless to ask for its truth value.

Edited much later to add: There's a variant of this puzzle that's very popular on the internet at the moment, in which answer option (c) is 60% rather than 0%. In this variant it is at least internally consistent to claim that all of the answers are wrong, and so the possibility of getting a right one by choosing randomly is 0%.

Whether this actually resolves the variant puzzle is more a matter of taste and temperament than an objective mathematical question. It is not in general true for self-referencing questions that simply being internally consistent is enough for an answer to be unambiguously right; otherwise the question

  Is the correct answer to this question "yes"?

would have two different "right" answers, because "yes" and "no" are both internally consistent. In the 60% variant of the puzzle it is happens that the only internally consistent answer is "0%", but even so one might, as a matter of caution, still deny that such reasoning by elimination is valid for self-referential statements at all. If one adopts this stance, one would still consider the 60% variant meaningless.

One rationale for taking this strict position would be that we don't want to accept reasoning by elimination on

  True or false?

  The Great Pumpkin exists.
  Both of these statements are false.

where the only internally consistent resolution is that the first statement is true and the second one is false. However, it appears to be unsound to conclude that the Great Pumpkin exists on the basis simply that the puzzle was posed.

On the other hand, it is difficult to argue that there is no possible principle that will cordon off the Great Pumpkin example as meaningless while still allowing the 60% variant to be meaningful.

In the end, though, these things are more matters of taste and philosophy than they are mathematics. In mathematics we generally prefer to play it safe and completely refuse to work with explicitly self-referential statements. This avoids the risk of paradox, and does not seem to hinder mathematical arguments about the things mathematicians are ordinarily interested in. So whatever one decides to do with the question-about-itself, what one does is not really mathematics.

Well, I must be pretty insane if I start competing with these already heavily upvoted answers from high rep users. But although the following solution may sound a little creative or even frivolous, it could easily be the right one. You could say that this solution is reverse engineered, as follows:

The question instructs to only choose one single answer out of four. And assume a uniform distribution, since that is most likely intended, then each answer has a chance of 25% to become chosen.

So the correct answer should be: 25%.

This computes to answer A being correct, as well as answer D. Could that be? Yes, it can. The question does not reveal how many of the four given answers are correct, but since there is one to be picked, assume that at least one of the four answers is correct.

Let's call answer A + answer D the correct answer pair.

Now, there are two possible choices (A or D) that result in 50% of the correct answer (A and D). Secondly, there is 50% chance of picking one (A or D) of two (A and D) out of four (A to D). So whether answer A or answer D is chosen, in either case the probability of being correct (50% × 50%) is 25%, which evaluates true.

Thus, yes, the question has 2 correct answers.

And now I realize that this post is the long version of the by joriki ages ago given comment. ;)

Ok, to be clear about what I mean, I believe the question is a special variant of the following trick question:

  What is the color of the car?

  Black
  Blue
  Gray
  Metallic

As owner of the car I know the correct answer is metallic black. But this would render the question unfair, because it is never possible to give this answer by only selecting one. The difference with the question in question is the equality of both answers to give, which makes it slightly more fair. But since you can select only half of the full solution, the probability is still 25%.


Let $D$ be a distribution. Show that every point of $p \in M$ has a submanifold $N$ such that $p \in M$ and $T_p N = D_p$. Let $D$ be a distribution. Show that every point of $p \in M$ has an embedded submanifold $N$ such that $p \in M$ and $T_p N = D_p$.

Attempt: Assume $M$ is a smooth manifold of dimension $m$ and $D$ is a rank $k$-subbundle of $\pi: TM \to M$.

Let $p \in M$. Somehow, we will have to construct an embedded submanifold $N$ of dimension $k$ such that $D_p = T_p N$.

How can I construct such an embedded manifold? Any hints will be appreciated!

The problem is to find a submanifold that takes a certain tangent space at a certain point. You can do this when your manifold is an open subset of $\mathbb{R}^n$ trivially; just take a little piece of an affine space. Luckily, every manifold is locally diffeomorphic to an open subset of $\mathbb{R}^n$.

More specifically: let $\varphi\colon U \to \mathbb{R}^n$ be a chart of the manifold at $p$. Let $N$ be an embedded submanifold in $\varphi(U) \subseteq \mathbb{R}^n$ that takes the tangent space $(d \varphi)_p(D_p)$ at $\varphi(p)$. (For example, let $N$ be a restriction of an affine subspace.) Then, $\varphi^{-1}(N)$ is an embedded submanifold of $U$, and hence of $M$, that takes the tangent space $D_p$ at $p$. (To verify this, it is most convenient to regard an embedded submanifold as an inclusion map, and the tangent space "taken at a point" as the image of the differential of this inclusion.)

This is not always true there exist  distributions which are not integrable.

https://mathoverflow.net/questions/294276/a-strongly-non-integrable-distribution


Constructing a divergent series Please would you help me with this question? I've been thinking about it for ages but I've made very little headway, so if possible a hint would be ideal.

Let  $\sum_{n=1}^∞{x_n}$ be a divergent series, where $x_n > 0$ for all $n$. Show that there is a divergent series  $\sum_{n=1}^∞{y_n}$ with $y_n > 0$ for all $n$, such that $(\frac{y_n}{x_n}) → 0.$

I have not been taught analysis formally, hence my lack of progress. I know to consider the series as a sequence of partial sums, and I tried to take the contrapositive of the statement but that just overcomplicated matters. I know I don't have many ideas to present but I have been trying this for days.

Thank you in advance.

You can break up your divergent series $\sum x_n$ into bunches of
consecutive terms, each of which are $>1$. In detail there are
$n_1$, $n_2,\ldots$ with
$$x_1+x_2+\cdots +x_{n_1}>1,$$
$$x_{n_1+1}+x_{n_1+2}+\cdots +x_{n_2}>1,$$
etc. Pick suitable constants $u_1,u_2,\ldots$ and set $y_k=u_ix_k$
whenever $n_{i-1}<k\le n_i$.


Invertibility of Gram Matrix - Proof I'm an undergrad that just got introduced to ordinary least squares and am trying to understand why a Gram matrix is invertible only if the column vectors are linearly independent. The thread below (3rd post ) uses the exact same proof as my lecturer. However, there is still something that is not clear for me which probably is something silly.

Gram matrix invertible iff set of vectors linearly independent

  JasonMond's "only if" part is not as general as it should, because
  s/he assumes that AA is square. In the following, I complete the proof
  that holds whether AA is square or not.

  Let $G=A^TA$. If the column vectors of $A$ are linearly dependent,
  there exists a vector $u≠0$ such that

  $Au = 0$

  It follows that

  $0=A^TAu=Gu$

So i understand that if the column vectors of A are linearly dependent, then I can find a non-zero vector u such that $Au=0$. But how does this tell me that G is not invertible ? Because there are other vectors such that $Au\neq0$

For any square matrix G to be invertible, does it require the result of $Gu$ to be non-zero for all non-zero vectors $u$ ? If so, where can I read about this ?

Your question is related to the null space of a matrix. Suppose $A$ is a real $n\times n$ matrix, we define the null space of $A$, as
$$Null(A)=\left\{ x \in R^n : Ax=0\right\}$$
If $A$ is invertible, it follows that $Null(A)=\{0\}$. This fact can be proven as follows. Suppose $x \in Null(A)$, then $Ax=0$. Multiply both sides by $A^{-1}$, we obtain
$$A^{-1}Ax=0 \Rightarrow Ix=0 \Rightarrow x=0$$

So we know that $Gu=0$ for some non-zero vector $u$. Now assume for contradiction that $G$ is invertible. Then $G^{-1}Gu=G^{-1}0$ which simplifies to $u=0$. But this is a contradiction because we assumed $u$ is non-zero.

Therefore, if $Gu=0$ for some non-zero vector $u$, then $G$ is not invertible.


Are elements of categories sets? (Subtlety in Hungerford's Algebra.) Hungerford's book Algebra defines class, set, and category informally. Class  and $\in$ are left as primitive notions. The definitions are, mostly verbatim:

  A class $A$ is defined to be a set if and only if there exists some
  class $B$ such that $A\in B$.

And:

  A category is a class $\mathcal{C}$ of objects (denoted
  $A,B,C,\ldots$), together with (morphisms, composition, associativity,
  etc.)

I want to write $A\in\mathcal{C}$, which by the definitions in Hungerford imply that $A$ is a set. However, every document I see uses the  terminology "Let $A$ be an object of $\mathcal{C}$" rather than "$A\in\mathcal{C}$". Is this just convention, or is $A\in \mathcal{C}$ incorrect notation?

Secondly, does this cause any problem with the notion of a concretizable category? I can imagine that even though all objects of $\mathcal{C}$ are sets, the morphisms may only be abstract and may not be able to be written as actual functions between the sets. That way, there would be no contradiction between the existence of non-concretizable categories.

Hungerford says that a category is a class $\mathcal C$ of objects, together with a set of morphisms for every pair of objects. ...So what's the category? You're tempted to say the category is a pair $(\mathcal C,\mathcal M)$ of objects and morphisms. But as has been discussed, $\mathcal C$ is generally not a set, so there exists no such class in Hungerford's foundations.

In a sufficiently strong foundation, a category is indeed an ordered pair $\mathcal C=(\mathrm{ob}\mathcal C,\mathrm{mor}\mathcal C)$ as suggested above. Then your notation becomes confusing: if you say $c\in \mathcal{C}$, do you mean $c$ is an object or a morphism? So, as a matter of practice, many authors avoid your usage, in favor of $c\in\mathrm{ob}\mathcal C$.

However, many authors are happy to say $c\in \mathcal C$ when $c$ is an object, by an abuse of notation familiar from group theory etc. In Hungerford's rough foundations, this isn't even an abuse! And finally we return to the question of whether an object of a category is a set. The answer is that it depends on the founda. The most flexible foundation is usually yes, depending on foundations: in particular this is true in ZFC with inaccessible cardinals, probably the best foundation for category theory, and in NBG set theory, which is what Hungerford is approximating.

Categories are classes rather then sets.  This is because you can have the category $\mathbb{set}$ which is the category where the objects are sets and the morphisms are functions between these sets.

Since you can't have the set of all sets (at least in ZFC) you have to deal with classes.  The operator $\in$ is a relation between sets so saying $A \in \mathcal{C} $ is incorrect as $ \mathcal{C} $ is not a set.

A class is collection of sets for which a certain predicate is true.  While a set is a collection of sets created via the axioms of set theory.


characterisation of direct sum of abelian groups The direct sum of abelian groups can be defined in several equivalent ways, but I have some problem proving the equivalence.

Definition 1:
Given abelian groups $A$ and $B$, the direct sum is defined as the castesian product $A \times B$, equipped with the operation $(a,b)(a',b') = (aa',bb')$. Then $A$ can be identified as the subgroup $A \times \lbrace e_B \rbrace$ in $A \times B$. The case for $B$ similar.

Definition 2:
Let $A$ and $B$ be subgroups of an abelian group $G$. $G$ is said to be the direct product of $A$ and $B$ iff every element $g \in G$  can be expressed uniquely as $g = a + b$, with $a \in A$ and $b \in B$.

However, I have difficulty proving the following: i.e. if [EDIT: additional assumption: A and B are subgroups of an abelian group G and] $G \cong A \times B$  as defined in definition 1, then every element $g \in G$ can be expressed uniquely as $g = a + b$, with $a \in A$ and $b \in B$. Given only an isomorphism $\phi: A \times B \rightarrow G$, I can only conclude that $g = \phi(a,e_B) + \phi(e_A,b)$, but I do not know whether $\phi(a,e_B) = a$ and $\phi(e_A,b) = b$. I know they almost look the same, and should not matter much in the spirit of 'up to isomorphism', but I still cannot see how I can proceed to show $g = a + b$.

Can anyone help me proceed? Thank you very much.

Related to last comment:

But then there is no much problem imo, either: in the cartesian product every pair $\;(a,b)\;$ is unique, meaning: $\;(a,b)=(a',b')\iff a=a'\,,\,\,b=b'\;$ , so if $\;\phi:A\times B\to G\;$ is a group isomorphism, then if $\;\phi(a,1_B)=g_A\;,\;\;\phi(1_A,b)=g_B\;$ , we have then $\;g=g_Ag_B\in G\;$, and this expression is unique, no matter where $\;A\times 1_B\,,\,\,1_A\times B\;$ are mapped to by $\;\phi\;$ .

Also, observe that $\;A\times1_B\cong A\;,\;\;1_A\times B\cong B\;$ and this is as much as we can say in this respect.


$E[\Pr(X<Y)]$ where $X$ and $Y$ are normally distributed? Sorry if this is badly worded, not sure of the best way to express this.
Suppose we have $X \sim N(0,1)$ and $Y \sim  N(0,\sigma)$ and the CDF of $X: F(z) = \Pr(X < z)$. If we take the expectation of $F(X)$ we get 0.5, as discussed here:
Expectation of CDF of continuous random variable $X$, evaluated at $X$
But what if we take the expectation of $F(Y)$? In this case the PDF in question is no longer the derivative of the CDF, so the integral seems less obvious. Any help would be appreciated!

This actually works for any distributions that are symmetric about $0$, i.e. $X$ and $-X$ have the same distribution and $Y$ and $-Y$ have the same distribution, where the distribution of $X$ is continuous.
Then the CDF of $X$ satisfies
$$F(y) = \mathbb P(X \le y) = \mathbb P(X \ge -y) = 1 - F(-y)$$
so
$$ \mathbb E[F(Y)] = 1 - \mathbb E[F(-Y)] = 1 - \mathbb E[F(Y)]$$
and therefore $\mathbb E[F(Y)] = 1/2$.

Use this expression for  expectation, $\mathbb{E}(X)= \int_{\mathbb{R}} xdF(x)$   $$\mathbb{E}(F(Y))=\int_{\mathbb{R}} F(y) dF(y) = \int_0^1 x dx = \frac{1}{2}$$


Prove that $\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right)\mathrm e^x\,\mathrm dx\le\frac{b^4-a^4}{4} \;?$ If $1<a\leq b$, how to prove that

$$\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right)\mathrm e^x\,\mathrm dx\le\frac{b^4-a^4}{4} \;?$$

We have
$$ e^x = 1+x+\frac{x^2}{2}+\sum_{k=3}^\infty \frac{x^k}{k!}$$
and thus $$e^x \geq 1+x\qquad \text{ and } \qquad e^x\geq 1+x+\frac{x^2}{2}.$$
Now if $0\leq x\leq 1$ then $x^2\leq x$ and thus $$e^x \geq 1+x \geq 1+x^2.$$ Finally, if $x\geq 1$, let $f(x)=e^x-1-x^2$. We have $f(1)=e-2>0$, let us show that $f$ is increasing for $x\geq 1$. We have $f'(x)=e^x-2x$ and thus $f'(1)=e-2>0$. Moreover, $f''(x)=e^x-2>0$ for every $x\geq 1$. Thus $f'$ is increasing for $x\geq 1$. It follows that $f'(x)\geq f(1)>0$ for every $x\geq 1$. Thus $f$ is increasing for every $x\geq 1$. It follows that for $x\geq 1$ we have
$$0 < f(1)\leq f(x) = e^x-1-x^2 \iff 1+x^2\leq e^x.$$
Thus $1+x^2\leq e^x$ for every $x \geq 0$. We finally get
$$\int_{\ln{a}}^{\ln{b}}(1+x)(1+x^2)\left(1+x+\frac{x^2}{2}\right) e^xdx \leq\int_{\ln{a}}^{\ln{b}}e^{3x}e^x dx=\frac{b^4-a^4}{4}$$

Here  x > ln(a) > 0  so each of the first three terms in the integrand is less than e^x.
So the integrand is less than e^(4x) and this can be quickly integrated.


How can I show that $\sup(AB)\geq\sup A\sup B$ for $A,B\subset\mathbb{R}$ where $A\cup B$ is positive and bounded? The question is based on the following exercise in real analysis:

  Assume that $A,B\subset{\Bbb R}$ are both bounded and $x>0$ for all $x\in A\cup B$. Show that $$
\sup(AB)=\sup A\sup B
$$
  where
  $$
AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}.
$$

Since $0<a\leq\sup A$ and $0<b\leq\sup B$ for all $a\in A$ and $b\in B$, we have
$$
ab\leq\sup A\sup B
$$
for all $ab\in AB$ which implies that $\sup AB\leq\sup A\sup B$. I have trouble with another direction:
$$
\sup AB\geq\sup A\sup B
$$
I was trying to show that for every $\epsilon >0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon>0$,
$$
\sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b
$$
for some $a\in A,\ b\in B$. It follows that
$$
\sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2
$$
which seems quite close to what I want. How can I go on?

There's a simple way to prove the result. For all $0 < a\in A$, $0 < b\in B$ we have:
$$\sup(AB)\geq ab\iff\frac{1}{a}\sup(AB)\geq b$$
hence $B$ is bounded above by $\frac{1}{a}\sup(AB)$ so
$$\frac{1}{a}\sup(AB)\geq \sup B\iff \frac{1}{\sup B}\sup(AB)\geq a$$
hence $A$ is bounded above by $\frac{1}{\sup B}\sup(AB)$ so
$$\frac{1}{\sup B}\sup(AB)\geq \sup A\iff \sup(AB)\geq \sup(A)\sup(B)$$

I think the key here is that all x are positive, and the sets are bounded so the required $\mathrm{sup}$'s are finite. Let $a \in A$. Then $a \leq \mathrm{sup}A$. Let $b \in B$. Then $ab \leq  (\mathrm{sup}A)b \leq (\mathrm{sup}A) (\mathrm{sup}B)$, using the fact that $b \leq \mathrm{sup}B$.


Rearranging a Staircase Grid into a Square Is there any way to rearrange the above "staircase" grid into three pieces that can be rearranged into the 6x6 square grid below it? I have tried this problem for over six hours and have not arrived at any solution. There has to be some general strategy in approaching this problem other than brute force, right??

I would appreciate any and all help--hints, suggestions, anything... I've gone mad trying to solve it

:(

You first the 6 by 6 on the staircase. Even thought it is not total, you can take the left over pieces and do it.


Approximating commuting matrices by commuting diagonalizable matrices Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that

$A_n \rightarrow A$, $B_n \rightarrow B$.
Each $A_n$ is diagonalizable  and the same for each $B_n$.
For every $n$, $A_n$ commutes with $B_n$.

Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well.

To get this off the Unanswered list:

The question was subsequently posted to MathOverflow, where it was answered in the affirmative. Some useful references from Mark Wildon’s answer:

  I'm fairly sure M. Gerstenhaber was the first to prove that the irreducibility result. His paper is: On dominance and varieties of commuting matrices, Annals Math. 73 (1961), 324-348. However the result asked for in the question was already known from Theorem 5 in T. S. Motzkin, Olga Taussky, Pairs of matrices with property L II, Trans. Amer. Math. Soc. 80 (1955), 387-401. There is a short account of this work after Remark 3.4 in the paper by Meara and Vinsonhaler mentioned in SJ's answer.

The last reference appears to be to K. C. O'Meara and C. I. Vinsonhaler, On approximately simultaneously diagonalizable matrices, Linear Algebra Appl., 412 (2006), 39 - 74.

Part $1:$ I know that if $M$ is the algebra generated by $A,~ B,~ AB = BA,$ then the dimension of $M$ as a subspace of all commuting matrices, whose dimension is of course $~n^2$, is at most $n$.

I read a note somewhere which suggests that if the number of commuting matrices is $~k > 2$, that it is not known that $\dim(M)$ is less than or equal to $~n~$.

The same note put a bound on it less than $~n^2~$, but did not do any better.  In particular, the note I read says that even for $~k = 3~$, $~A,~ B,~ C~$ commute, that it has not known that $~\dim(M)~$ is bounded by $~n~$.

Part $2:$  Why is this question of interest?  There may be many reasons why it is of interest, but it was presented to me as an abstract problem in Algebra.

Thanks in advance for any useful additions to this thread.


complex result in numerical integral I'm in a contradiction that I open the topic is here because it is also related to programming. I try to calculate definite integral using Simpson's Rule. However, after some tries, I have encountered a problem referred as nan meaning not a number. How can I know ? I checked some equations. I get that my program shows nan when result of the equation is complex number. The interesting point is that I discussed the result with my lecturer, he said if you find complex number from numerical integral calculation I will give you two prizes(no pun intended). At the first glance, I acknowledged him to be right. However, I found complex number. How is it possible? For example, when I try to calculate $$ \sin^{t}(2t) + \cos^{t}(\frac{t}{9})  = pow(sin(2*t), t ) + pow( cos(t / 9), t ) $$

.
.
29)  1.066912
30)  pow(sin(2*1.525000), 1.525000 ) + pow( cos(1.525000 / 9), 1.525000 )
30)  1.004298
31)  pow(sin(2*1.575000), 1.575000 ) + pow( cos(1.575000 / 9), 1.575000 )
31)  nan
32)  pow(sin(2*1.625000), 1.625000 ) + pow( cos(1.625000 / 9), 1.625000 )
32)  nan
.
.
.

This is not an anwer but an enlargement.

I was at first incredulous.

I tried a routine calculation with Mathematica, which is known to be rigorous in its treatment of real vs. complex numbers, and I was very surprised to find a similar (even slightly worse) situation.

Mathematica (an old version), at the request :

Cos[2.629]^2.629 gives the result : -0.27461 + 0.640097 I  !

Sin[2*(2.477)]^(2.477)  gives 0.06711 + 0.92716*I ...

and the same result if you separate the computation into 2 steps...

a=Sin[2*(2.477)] ; a^(2.477)

Of course, we know that the trigonometric functions can take complex values, but for complex arguments ! Why this behaviour ? It's valuable to ask it in a numerical analysis forum...


Number formation from given digits I was asked this question in a recent interview. You are given two digits $0$ and $9$. Using only these two digits you must form the least positive number that is a multiple of a given '$n$' . The value of $n$ will be provided before hand. For example if we are given $n$ as $7$ then the answer will be $9009$ .

I tried using some basic divisibility principles but was not able to come up with a correct solution and code it up. Probably I need some help in number theory

The simplest solution is to use a recurrence:

Let $f(k,x) \Leftrightarrow \text{There is a number with k digits made of 0s and 9s that is equal to x modulo n}$

Then it is not hard to compute $f(k,x)$ using $\{ f(k-1,y) : y \in [0..n-1] \}$

This algorithm takes $O(cn)$ time where $c$ is the number of digits in the answer. From intuition I would guess that $c$ is generally $O(\log(n))$.

Start with $9$. Divide this by $n$ and take the remainder. Call the remainder $R$. If $n$ divides $10R+9$ then add the digit 9 and you are done. If not add the digit $0$ and divide $10 R$ by $n$ and take the remainder. Repeat the process.

The proof of this follows from the observation: If we have a number $X$ and add an extra digit $Y$ then the new number is $10X + Y$ and the new remainder is obtained by dividing $10 R+ Y$.

In your example:

$$
 9 = 1 \times 7 + 2 ~~~ R=2, ~~\text {7 does not divide $2\times 10  +9$}\\
 2\times 10 = 2 \times 7 + 6 ~~ R = 6, ~~\text {7 does not divide $6\times 10 +9$}\\
 6 \times 10 = 8 \times 7 + 4 ~~ R = 4, ~~ \text {7 divides $4\times 10 + 9$}\\
$$
So the answer is $9009$

Added as after thought

This is the smallest number of the form $900\cdots 09$. Not sure if there is a simple way to find the answer. For example, if $n$ is 367, then the answer is $9909$ but there is no simple way to find it other than some form of a $\alpha-\beta$ pruning.


Complex Analysis Branch cuts Take the branch of $\log(z)$ to lie in $(-\pi, \pi]$. With complex numbers when does $\sqrt{z^2} = z$ hold and when doesn't it? If we take $z=-1$, this equality holds but then $1=-1$. I have a feeling that it is because $-1$ lies on the branch cut but I am still confused as to why this occurs?

The function is considered to be undefined at the branch cut.


Picking balls from a bag until there are three of same color A bag contains $20$ balls:

$3$ Blue
$4$ Purple
$5$ Orange
$8$ Green

You are allowed to pick balls until you have three the same colour  eg (3 - 9 picks ) No balls are replaced until a new round is begun. What is the probability of getting each colour on each round.

$3$ Blue,
$3$ Purple,
$3$ Orange,
$3$ Green.

Clarification: Each "round" you pick until you have three the same, this can take of course between 3 - 9 picks. From picking three the same in three picks to picking 2 the same during your first 8 picks in any order and the 9th would give you a 3 of a kind. I want to work out the total expected probability for each round eg 5% chance of blue 10% purple 25% orange 60% green If someone can help work out one colour I should be able to do the other 3.

I seem to be struggling  with the fact that, until the $3$rd ball, the order isn't important.

What I did for 1 of the 27 ways of getting blue for example was
$$\frac{\binom{3}{3}\binom{4}{2}\binom{5}{2}\binom{8}{2}}{\binom{20}{9} \binom{9}{1}} = 0.001111376$$

There may be a more elegant way than this, but here's one way.

First, write down all of the color distributions that are not terminating, but could terminate on the next draw.  This means that you need to have drawn at least one color twice.  I'll designate the colors $1-4$ and not worry about order:

$2$ balls:  $11$
$3$ balls:  $112$
$4$ balls:  $1122, 1123$
$5$ balls:  $11223, 11234$
$6$ balls:  $112233, 112234$
$7$ balls:  $1122334$
$8$ balls:  $11223344$

Next, for each of these color distributions, calculate the number of ways to arrange the digits.  This can be done with multinomials.  For example, the number of ways to arrange $11223$ is
$$\frac{5!}{2!2!1!} = 30.$$

Next, observe that, for a given set of colors, the probability of drawing those colors is independent of the order.  For example,

$$BBGGP = \frac{3}{20} \frac{2}{19} \frac{8}{18} \frac{7}{17} \frac{4}{16},$$

but

$$PGBGB = \frac{4}{20} \frac{8}{19} \frac{3}{18} \frac{7}{17} \frac{2}{16},$$

which is exactly the same value.  All that happened was that the numerators got rearranged.

Now comes the somewhat tedious part, which is to calculate the individual probabilities for each substitution of colors.  We can at least be a little smart about it, and calculate the numerator part for each color:

$$B = 3, BB = 6, P = 4, PP = 12, O = 5, OO=20, G = 8, GG = 56.$$

Let's do the case for drawing three blue balls, and drawing the blue on the fifth ball.  We assign $BB$ to $11$ in the combination, and then write down all the others:  $$BBPP, BBOO, BBGG, BBPO, BBPG, BBOG.$$

(The first three are $1122$ types, and the last three are $1123$ types.)

To calculate the probability of getting each one of these, we multiply the probability of getting them in the stated order with the number of ways to rearrange them.  So,

$$P(BBPP) = \frac{6 \cdot 12}{20 \cdot 19 \cdot 18 \cdot 17}\cdot \frac{4!}{2!2!}.$$

Now, to find the probability of ending on the next turn with blue, given that you have two blues and two purples (drawn in any order), multiply by $1/16$.

Then it's just a lot of rinse/repeat. Do the same calculation for the rest of the four-ball colors.  Then do all of the other numbers of balls, for all of the applicable color combinations each time.

And that's just the first part of a four-part problem!

You can possibly reuse some of the calculations you did before if you keep track.

Hint:

Since you are drawing without replacement, you can imagine picking all the balls in order. You want the probability that the last three are all a particular colour.

By symmetry, this is the same as the probability that the first three are all that particular colour, an easier calculation I will leave for you.


Find a and b of $x^3+ax^2+bx−26=0$ I am doing practise papers and one of the questions is:

  The cubic equation $x^3+ax^2+bx−26=0$ has $3$ positive, distinct,
  integer roots.

  Find the values of $a$ and $b$

The mark scheme says:

  $3$ roots are $1, 2, 13$. Equation is $x^3-16x^2+41x-26=0$

I'm guessing this has something to do with the factor theorem thing where $x-a$ is a factor if $f(a)=0$ but with 3 unknowns I don't get how you find any of these.

I tried just putting $x$ as $1$:

$$
1^3+a+b-26=0
$$
$$
a+b=25
$$
then doing the same with $2$:
$$
2^3+4a+2b-26=0
$$
$$
4a+2b=18
$$
and after subsituting the two equations I got the answer of a $a=-16$ and $b=41$. However, I realised this was just by chance that these numbers were factors (it didn't work with $x=3$ for example).

What is the proper way to solve this?

If $p$, $q$, $r$ are distinct positive integer roots of $P(x) = x^3 + ax^2 + bx - 26$, then we must have $$(x-p)(x-q)(x-r) =  x^3 - (p+q+r)x^2 + (pq+qr+rp)x - pqr = P(x),$$ hence equating the constant coefficient, $$pqr = 26 = 2 \cdot 13.$$  Since the prime factorization of $26$ contains only $2$ and $13$, it immediately follows that there is only one unique solution (up to permutation) for this equation that satisfies the given conditions, namely $\{p, q, r\} = \{1, 2, 13\}$ in some order.  The rest follows easily.

Here's a hint: every cubic can be factored in the following way:
$$p(x) = (x-r_1)(x-r_2)(x-r_3)$$
You know, from the given, that $r_1$, $r_2$, $r_3$ are distinct positive integers, and you know that $r_1r_2r_3 = 26$.  It turns out that there is only one possible unordered triple $(r_1, r_2, r_3)$ that satisfies these properties – can you show this?

Once you find these roots, you can use Vieta's, as others have mentioned, to find the values of $a$ and $b$.


find the limit to infinity $a_n= 3^n ~\sin(4^{-n})~$ I am having problems solving limits when its infinity multiplied by $~0~$

$$\lim_{n\to \infty} a_n= 3^n \sin (4^{-n})~$$

I can't use L'Hôpital's rule here so how do I solve this

We have $a_n=3^n \sin(\frac{1}{4^n}).$

Hint: $| \sin(x)| \le |x|$ for all $x$.

Can you proceed ?

Alternatively, consider the series:
$$\sum_{n=1}^\infty 3^n \sin (4^{-n})$$
By the ratio test:
$$\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty} \frac{3^{n+1}\sin (4^{-(n+1)})}{3^n\sin (4^{-n})}=
\lim_{n\to \infty} \frac{3\sin(4^{-(n+1)})}{\sin(4\cdot 4^{-(n+1)})}=\\
\lim_{n\to \infty} \frac{3\require{cancel}\cancel{\sin(4^{-(n+1)})}}{4\require{cancel}\cancel{\sin (4^{-(n+1)})}\cos{(4^{-(n+1)})}\cos(2(4^{-(n+1)}))}=\frac34<1.$$
The convergence of the series implies the limit of the general term is $0$.


Finding the Probability function of the remainder. Let X have a geometric distribution with $f(x) = p(1−p)x$, $x = 0,1,2,\ldots$ Find the probability function of $R$, the remainder when $X$ is divided by $4$.

I am stuck on the above practice question: If I say $R= 0, 1/4, 2/4, \ldots$ and then use the continuous case equivalent of geometric distribution to find the probability function of $R$, which is exponential distribution, will I be correct?

I'm not sure what you mean by "using the continuous case equivalent of geometric distribution to find the probability function of R".  However, here's how I would approach the problem:

When you divide $X$ by $R$, there are three possible remainders: $0,1,2,$ and $3$.  You will get a remainder of $0$ for multiples of $4$, that is, when $X = 0,4,8,\dots$.  You will get a remainder of $1$ for numbers one more than the multiples of $4$, such as $1,5,9,\dots$.  Hopefully, you get the idea by now.

With that in mind, the probabilities are as follows: clearly, $P(R=r) = 0$ for all numbers $r$ besides $0,1,2,3$.  For $0,1,2,3,$ we have:
$$
\begin{align}
P(R = 0) &=
p(1-p)^0 + p(1-p)^4+p(1-p)^8+\cdots\\
&= \sum_{k=0}^\infty p(1-p)^{4k}\\
P(R = 1) &=
p(1-p)^1 + p(1-p)^5+p(1-p)^9+\cdots\\
&= \sum_{k=0}^\infty p(1-p)^{4k+1}\\
P(R = 2) &=
p(1-p)^2 + p(1-p)^6+p(1-p)^10+\cdots\\
&= \sum_{k=0}^\infty p(1-p)^{4k+2}\\
P(R = 3) &=
p(1-p)^3 + p(1-p)^7+p(1-p)^11+\cdots\\
&= \sum_{k=0}^\infty p(1-p)^{4k+3}\\
\end{align}
$$
From there, you could use the formula for the sum of a geometric series.  Alternatively, you could simply make the following observations:
$$
P(R = 3) = (1-p)P(R = 2)\\
P(R = 2) = (1 -p)P(R = 1)\\
P(R = 1) = (1 - p)P(R = 0)\\
P(R = 0) + P(R = 1) + P(R = 2) + P(R = 3) = 1
$$
And solve the above system of equations to find the required probabilities.


How to construct a partial ordering from Peano's 5 Axioms? I am trying to formally construct the usual partial ordering LTE from Peano's 5 Axioms. Would the following construction work?
$$\forall a,b: [(a,b)\in LTE \iff(a,b)\in N^2$$
$$\land ~ \forall c\subset N^2:[\forall d\in N: [(d,d) \in c]$$
$$ \land ~ \forall d,e:[(d,e)\in c \implies(d, S(e))\in c]$$
$$\implies (a,b) \in c]]$$
where $N$ is the set of natural numbers and $S$ is the usual successor function on $N$.

EDIT:
Informally, for all $x,y\in N$, we can recursively define $\le$ on $N$ as the "smallest" set of ordered pairs of natural numbers that satisfies:
$$x \le x$$
$$x\le y \implies x\le S(y)$$

(Answering my own question)
Thanks all for your help, but I didn't find my construction here to be very workable. I was trying to avoid using an addition function to "simplify things." That was probably a mistake.
Using Peano's Axioms, some basic set theory and a form of natural deduction, I formally constructed (in order) the function $+$ and the relations $\leq$ and $\lt$ as sets of ordered n-tuples such that, for all $x, y \in N$ we have:
$x+0 = x$
$x+S(y) = S(x+y)$
$x\leq y \iff \exists z\in N: x+z=y$
$x \lt y \iff x\leq y ~\land ~ x\neq y$
Everything then fell into place. I was just now able to derive the usual fundamental properties of $+$, $\leq$ and $\lt$ for a larger project that I am working on.


Why do engineers use derivatives in discontinuous functions? Is it correct? I am a Software Engineering student and this year I learned about how CPUs work, it turns out that electronic engineers and I also see it a lot in my field, we do use derivatives with discontinuous functions. For instance in order to calculate the optimal amount of ripple adders so as to minimise the execution time of the addition process:
$$\text{ExecutionTime}(n, k) = \Delta(4k+\frac{2n}{k}-4)$$
$$\frac{d\,\text{ExecutionTime}(n, k)}{dk}=4\Delta-\frac{2n\Delta}{k^2}=0$$
$$k= \sqrt{\frac{n}{2}}$$
where $n$ is the number of bits in the numbers to add, $k$ is the amount of adders in ripple and $\Delta$ is the "delta gate" (the time that takes to a gate to operate).
Clearly you can see that the execution time function is not continuous at all because $k$ is a natural number and so is $n$.
This is driving me crazy because on the one hand I understand that I can analyse the function as a continuous one and get results in that way, and indeed I think that's what we do ("I think", that's why I am asking), but my intuition and knowledge about mathematical analysis tells me that this is completely wrong, because the truth is that the function is not continuous and will never be and because of that, the derivative with respect to $k$ or $n$ does not exist because there is no rate of change.
If someone could explain me if my first guess is correct or not and why, I'd appreciate it a lot, thanks for reading and helping!

In general, computing the extrema of a continuous function and rounding them to integers does not yield the extrema of the restriction of that function to the integers. It is not hard to construct examples.
However, your particular function is convex on the domain $k>0$. In this case the  extremum is at one or both of the two integers nearest to the unique extremum of the continuous function.
It would have been nice to explicitly state this fact when determining the minimum by this method, as it is really not obvious, but unfortunately such subtleties are often forgotten (or never known in the first place) in such applied fields. So I commend you for noticing the problem and asking!

If you consider that k and n are continuous variables, you obtain a continuous function that may be derivated and is entangled to the initial discontinuous function. The real problem is if the extreme point of this function can approximate the extreme point of the discontinuous one. The error may be evaluated with the Cauchy reminder formula for Newton or Lagrange interpolation which has a k! in the denominator and may be tolerable if k is enough big.


Some digit summation problems What is the sum of the digits of all numbers from 1 to 1000000?
  In general, what is the sum of all digits between 1 and N?
  f(n) is a function counting all the ones that show up in 1, 2, 3, ..., n. so f(1)=1, f(10)=2, f(11)=4 etc. When is the first time
  f(n)=n.

So for the first question, I tried thinking about it such that between 000000 and 999999, each digit will appear the same number of times, so if I find out how many times one digit appears I can just apply that to the other 9 digits (then add 1 for the last number 1000000):

(the number of times 1 digit appears)*(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = ...

1 Appears once between 1 and 9
1 appears 11 times between 1 and 99
1 appears 11 * 10 + 10 = 120 times between 1 and 999
...I'm not sure how to find the pattern

But firstly I'm not so sure of my approach, secondly I'm not sure about how to find how many times one particular number appears, and third if this method worked it doesn't seem very good for solving the second part of the question.

Lastly, I had a similar question to the first 2 (question 3) so I just grouped it with those. I hope they are related, and if not I can make a seperate question for that one.

Thanks.

For question three, we can simply compute.  Besides n=1, the next solution is 199981, as generated by this PARI/GP code:

Define $r(n)$ to be the number of ones in the digits of $n$:

r(n) = nn=n;cc=0;while(nn>0,if((nn%10)==1,cc=cc+1);nn=floor(nn/10));return(cc)

Then run a loop and output $i$ if the sum of $r(i)$ from 1 up to $i$ is equal to $i$:

yy=0;for(i=1,199981,yy=yy+r(i);if(yy==i,print(i)))

The output is this:

1
199981

This shows $f(1)=1$ and $f(199981)=199981$ and there are no other solutions less than 199981.

Similarly for question one, we define $g(n)$ to be the sum of the digits of $n$:

g(n) = nn=n; cc=0; while(nn>0,cc=cc+(nn % 10);nn=floor(nn/10));return(cc)

then calculate:

sum(i=1,1000000,g(i))

which yields the sum 27000001.


A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that the balls drawn are blue and then subtracting it from 1. But I want to solve it straight forward and directly. Is it possible? If yes, how?

Assumption: The first ball drawn is not put back in the bag.

Let E be the event that none of the two picked balls are blue. Further let $E_1$ and $E_2$ be the events that the first ball picked is not blue and the second ball picked is not blue respectively.

$P(E) = P(E_1 \cap E_2) = P(E_1)P(E_2 | E_1)$

$\implies P(E) = \left( {5 \choose 1}*1/7 \right) \left( {4 \choose 1}*1/6 \right)$

$\implies P(E) =  10/21 $

In a bag there are 4 red balls ,2green balls, and 3yellow balls .if two balls are drawn at random ,what is the probability that at least one of the balls is green in color.