Which one result in mathematics has surprised you the most? A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

If a function of a complex variable is once differentiable, it's infinitely differentiable.

Computational instability of the Quadratic Formula. Who would have thought?

Due to this computational stability an alternative formula is also employed. Here is the relevant quote from the Wikipedia article:

“The alternative formula can reduce loss of precision in the numerical evaluation of the roots, which may be a problem if one of the roots is much smaller than the other in absolute magnitude.”

And here is the link to the full article:

http://en.wikipedia.org/wiki/Quadratic_equation


Easy way of tackling $\sqrt{x^2}+x=\sqrt{5}$ Solve for x
$$\sqrt{x^2}+x=\sqrt{5}\tag1$$

$$\sin(\sqrt{x^2}+x)=\sin({\sqrt{5}})\tag2$$

$$\sin(\sqrt{x^2})\cos(x)+\cos(\sqrt{x^2})\sin(x)=\sin(\sqrt{5})\tag3$$

$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\tan(x)=\sin(\sqrt{5})\sec(x)\tag4$$

$$\sin(\sqrt{x^2})+\cos(\sqrt{x^2})\sqrt{\sec^2(x)-1}=\sin(\sqrt{5})\sec(x)\tag5$$

finally after all the hard work $(5)$ reveals

$$x+x=\sqrt{5}\tag6$$

$$x=\frac{\sqrt{5}}{2}\tag7$$

Is there another easy way solving this $(1)?$

There is another way to solve this.

Using the fact that $\sqrt{x^2} = |x|$. Your equation turns out to be $$|x|+x=\sqrt{5}$$

First look for solutions where $x\geq 0$ in this case you get $x+x=\sqrt{5}$ and so $x=\frac{\sqrt{5}}{2}$.

Now if $x<0$ then $|x|=-x$ and so you get $-x+x=\sqrt{5}$ and clearly no $x$ satisfies this condition.

Use cases. Certainly $x \neq 0.$

Case I: $x > 0.$ Then $\sqrt{x^2} + x = 2x.$ Thus $x = \frac{\sqrt{5}}{2}.$

Case II: $x < 0.$. Then $\sqrt{x^2} = -x$, but then this says $ 0 = \sqrt{5},$ contradiction. Hence we must have the only solution $x = \frac{\sqrt{5}}{2}.$


What is difference between “true in ZFC” and “true”? Let $P$ be a claim. There are two questions.

  Is it true in ZFC that $P$?

And the other is:

  Is it true that $P$?

What is difference between these two questions?

I suspect that "true in ZFC" is intended to mean "provable in ZFC."  The relationship between that and just plain "true" is more a philosophical question than a mathematical one.  Let me explain two of the many possible philosophical positions, in the hope that they give an idea of the range of possibilities.

(1), a version of formalism: In mathematics, we have no way of knowing things except by proving them.  We accept, more or less arbitrarily, ZFC as the standard basis for our proofs.  So when something is proved in ZFC, we are willing to say that it is true.  It makes no sense to speak of truth apart from provability.  Therefore, "true" simply means provable in ZFC.

(2), a version of Platonism: There is a world of abstract entities, including the sets and related entities studied in mathematics.  To say that a mathematical statement is true is to say that it corresponds to the actual situation in this ideal world.  We have only a partial understanding and intuition of the ideal world; much of that understanding is summarized in the ZFC axioms, so anything provable from those axioms is known to be true.  But there are other truths that we may not be able to prove and, indeed, may not have any other way of knowing.  Therefore, "true" is a considerably broader notion than provability in ZFC.

To see how these viewpoints work out in a nontrivial situation, consider the fact (provable in ZFC) that, if ZFC is consistent, then it neither proves nor refultes the continuum hypothesis (CH).  A formalist (or, more precisely, someone who subscribes to the version of formalism that I've described) would conclude that, if ZFC is consistent, then CH is neither true nor false.  (He might reconcile this conclusion with the law "p or not p" of classical logic by declaring CH to be meaningless.) And he might have a problem with the issue of consistency of ZFC, because, if ZFC is consistent then this fact is not provable in ZFC.

A Platonist would begin by brushing aside the clause "if ZFC is consistent"; it's obviously consistent because it's true.  (So the consistency of ZFC is one example of truth beyond what is provable in ZFC). He knows that CH either holds or doesn't hold in the world of sets, but he probably doesn't know which way it goes. So he would be happy to find new axioms that are clearly true (as the ZFC axioms are) and decide CH.

I think I've stated both the formalist and the Platonist positions in fairly extreme forms, so I wouldn't be surprised to get comments from actual formalists and actual Platonists objecting that I've unfairly distorted their positions.  I hope such comments would include corrections of the distortions, so as to build a more accurate picture of the positions that people actually take.

The first type of assertion is well-defined and the second is undefinable.


Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Intuition? Why is $\operatorname{Div}\big(\operatorname{Curl} F\big) = 0$? Is there an intuitive explanation to what this means as well as an algebraic proof?

Also I understand that $\operatorname{Curl} \operatorname{Grad}F =0$, I interpet this as the gradient not having any rotation since its the direction of steepest ascent and if this formed some sort of loop this would not make sense? Is this somewhat correct? Any explanations would be appreciated.

Algebraically, both of these are simply based on the equality of mixed partials:
$$
\begin{align}
\nabla\cdot\nabla\times u
&=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(u_1,u_2,u_3\right)\\
&=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\cdot\left(\frac{\partial u_3}{\partial y}-\frac{\partial u_2}{\partial z},\frac{\partial u_1}{\partial z}-\frac{\partial u_3}{\partial x},\frac{\partial u_2}{\partial x}-\frac{\partial u_1}{\partial y}\right)\\
&=\color{#0000FF}{\frac{\partial^2u_3}{\partial x\partial y}}-\color{#00A000}{\frac{\partial^2u_2}{\partial x\partial z}}+\color{#C00000}{\frac{\partial^2u_1}{\partial y\partial z}}-\color{#0000FF}{\frac{\partial^2u_3}{\partial y\partial x}}+\color{#00A000}{\frac{\partial^2u_2}{\partial z\partial x}}-\color{#C00000}{\frac{\partial^2u_1}{\partial z\partial y}}\\[9pt]
&=0\tag{1}\\[20pt]
\nabla\times\nabla u
&=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)\times\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)u\\
&=\left(\frac{\partial^2}{\partial y\partial z}-\frac{\partial^2}{\partial z\partial y},\frac{\partial^2}{\partial z\partial x}-\frac{\partial^2}{\partial x\partial z},\frac{\partial^2}{\partial x\partial y}-\frac{\partial^2}{\partial y\partial x}\right)u\\[6pt]
&=(0,0,0)\,u\tag{2}\\[20pt]
\end{align}
$$
Intuitively(?), $(1)$ says that a flow induced perpendicular to the rotation of a smooth vector field has no source, and $(2)$ says that the flow induced by the gradient of a smooth scalar field has no rotation.

Here's an informal way of interpreting it. Taking the curl of a function is a cross product operation: curl. Now let's try to take the divergence of this curl, however, recall that cross products produce orthogonal vectors. Since the divergence is a dot product operation divergence, and the dot product of orthogonal vectors is 0, the divergence of a curl is 0.


Verify the following identity by a committee-selection model $${n \choose 0} + {n+1 \choose 1} + {n+2 \choose 2} +\cdots + {n + r \choose r} = {n+r+1 \choose r}$$

The "committee selection model" is a model used to view combinations as selecting a committee from a group of people. E.g., ${15 \choose 4}$ is the number of 4-person committees that can be formed from a group of 15 people.

What I'm tasked with doing is a combinatorial proof using this model--showing that the left counts the same thing as the right using the committee model.

On the RHS, you count the number of committees of $r$ people that can be chosen from a group on $n+r+1$ people.

Now select one particular person of the group, say John. Then you have two mutually exclusive cases:

John is not chosen in the committee, so the number of committees that ca be formed without John is ${n+r\choose r}$
John is chosen in the committee, so then you have to choose $r-1$ people out of a group of $n+r$. Now consider another person from the group, say Susie

2.1 John is chosen in the comitee, but Susie is not, which can be done in ${n+r-1\choose r-1}$ ways

2.2 Both John and Susie are chosen, in which case...

... and so on and so forth...

So if you label the people as $1,2,\ldots,n+r+1$, then the number of ways to choose $r$ people equals the sum of the ways to form the committee by mandatory choosing the labels $0,1,\ldots k-1$ and not to choose label $k$, for $k=1,2\ldots,r+1$

You have that $\binom x y$ counts the ways to select $y$ persons from a list of $x$ persons.

For instance $\binom 1 1$ counts the $1$ way to select one person from a list of one person.

Hint: Notice that $\tbinom x y = \tbinom x {x-y}$ for all integers $x,y$ where $0\leq y\leq x$, and so likewise $\tbinom {n+k} k=\tbinom {n+k}n$ for all $0\leq n\leq n+k$.

Thus the LHS, $\sum_{k=0}^r \binom {n+k}k$ equals $\sum_{k=0}^r \binom {n+k}n $ and counts ...

Likewise the RHS, $\binom {n+r+1}{r}$ is $\binom {n+r+1}{n+1}$, thus counts...

Show what they count can be the same task.


Proof of unique coordinatization of tensor space once bases are chosen I am trying to prove the statement:

If $K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \xrightarrow{G} K^{n_1 \times n_2 \times \dots n_d}$ is isomorphic as a tensor space to $K^{n_1} \times K^{n_2} \times \dots \times K^{n_d} \xrightarrow{\otimes} K^{n_1} \otimes K^{n_2} \otimes \dots \otimes K^{n_d}$, then f = G up to some permutation of the indices ${i_1}{i_2}\ldots{i_d}$.

Let me define everything precisely.

Let $V^1$, $V^2$, $\ldots$ , $V^d$, $W$ be vector spaces over $K$ and let $\otimes$ be the multilinear mapping $\otimes :V^1 \times V^2 \times \dots \times V^d \rightarrow W$ such that $W$ equals the space spanned by the image of $\otimes$, and for any multilinear mapping $f: V^1 \times V^2 \times \dots \times V^d \rightarrow H$ for any vector space $H$, there exists a unique linear mapping $F: T \rightarrow H$ s.t. $F(\otimes({v^1 \times v^2 \times \dots \times v^d})) = f(v^1 \times v^2 \times \dots \times v^d)$.

Let $V^1 \times V^2 \times \dots \times V^d \xrightarrow{\otimes} W$ and $V^1 \times V^2 \times \dots \times V^d \xrightarrow{\otimes '} W'$ be two tensor products of ${V^1 \times V^2 \times \dots \times V^d}$. $\eta: W \rightarrow W'$ is a tensor space morphism if $\eta$ is a vector space morphism and ${\eta (v^1 \otimes v^2 \otimes \ldots \otimes v^d) = \eta (v^1 \otimes ' v^2 \otimes ' \ldots \otimes ' v^d)}$ for all $v^1 \times v^2 \times \dots \times v^d \ \in \ V^1 \times V^2 \times \dots \times V^d$, where  $\otimes(v^1 \times v^2 \times \dots \times v^d)$ is denoted $v^1 \otimes v^2 \otimes \dots \otimes v^d$.

The $d$-fold tensor space is isomorphic as a tensor space to the space of $d$-fold hypermatrices. A d-fold hypermatrix over K is a function from $(1, 2, \ldots, n_1) \times (1, 2, \ldots, n_2) \times \ldots \times (1, 2, \ldots, n_d)$ to K, and the space of all such hypermatrices is denoted $K^{n_1 \times n_2 \times \ldots \times n_d}$. Hypermatrices in $K^{n_1 \times n_2 \times \ldots \times n_d}$ are written as $A = (a_{{i_1}{i_2}\ldots{i_d}})$ where each $a_{{i_1}{i_2}\ldots{i_d}} \in K$.

To see that these spaces are isomorphic as tensor spaces, choose ordered bases $(e^k_i)_{i=1}^{n_k}$ of $V^{k}$ for ${k = 1, \ldots, d}$, and consider the mapping
\begin{align*}
V^1 \times V^2 \times \ldots \times V^d \ &\xrightarrow{f} \quad  K^{n_1 \times n_2 \times \ldots n_d} \\
v^1 \times v^2 \times \ldots \times v^d \ &\longmapsto \quad  (a_{{i_1}{i_2}\ldots{i_d}} = a^1_{i_1}a^2_{i_3}\ldots a^d_{i_d} )
\end{align*}
where $v^k = \sum_{i = 1}^{n_k} a^k_i \ e^k_i  $ for k = 1, 2, $\dots$ d. Note that f satisfies the universal property of the tensor product, and thus the map F below must be an isomorphism of tensor spaces.
\begin{align*}
V^1 \otimes V^2 \otimes \ldots \otimes V^d \ &\xrightarrow{F} \quad  K^{n_1 \times n_2 \times \ldots n_d} \\
v^1 \otimes v^2 \otimes \ldots \otimes v^d \ &\longmapsto \quad  (a_{{i_1}{i_2}\ldots{i_d}})
\end{align*}
where $v^1 \otimes v^2 \otimes \ldots \otimes v^d = \sum_{i_1, i_2, \ldots, i_d = 1}^{n_1, n_2, \ldots, n_d} \  a_{{i_1}{i_2}\ldots{i_d}} \ e_{i_1}^1 \otimes e_{i_2}^2 \otimes \ldots \otimes e_{i_d}^d  $.

I did not read your post too closely, but at the end you are listing products of components. That is exactly what the coefficients are for a basis of a tensor product space built up with simple tensors from a basis of the original two spaces. Do you know how to make a basis of a tensor product space?


Similar Triangles Problem For Geometry I In quadrilateral QRTS, we have QR = 11, QS = 9, and ST=2. Sides RQ and ST are extended past Q and S, respectively, to meet at point P. If PS = 8 and PQ = 5, then what is RT?

Try drawing it out. You should end up with triangle PTR, with point S lying on PT and Q lying on PR. You also know the length of PT and PR, as well as that of SQ.

In triangle PSQ, we have all the side lengths so you can find the angle between TP and PR using the law of cosines.

Now you have, in PTR, two side lengths and one angle, so you can then use the law of cosines once again to find TR.

you can solve it without cosines by flipping the PQS triangle, so that you can use SAS similarity to prove they are similar.
PQ =5  PT = 10
PS = 8 PR = 16
and the angle between them is the same angle, so the triangles are similar and have a 2:1 ratio
so 9*2 = 18 and youre done.


Can every indefinite integral of a discontinuous function be written in a way that "proves" something false? I just saw the following fake proof.

$$\int \frac1x dx =\int 1\cdot \frac1x dx=x\frac1x+\int x \frac1{x^2} dx = 1+ \int \frac1x dx$$

Which would imply $1=0$, hence the fake proof tag.

The explanation given was that there was a discontinuity in the function that we were in essence "integrating over", however the integral is indefinite so we are not explicitly integrating over anything (or maybe integrating over everything?).

Will any function with a discontinuity be exploitable in this way? Could we actually solve the problem by converting the integral to a definite integral in a region there is no discontinuity? I think the trick used would still work, or am not sure why it would provide a result that make sense.

I disagree with alexqwx. His answer is misleading. There is no constant of integration generated in integration by parts until integration actually occurs. The above answer also does not mention that the fallacy occurs when subtracting indefinite integrals not because of a lack of a constant of integration.

There is no reason to write $\int \frac{1}{x}dx=1+\int \frac{1}{x}dx +C$. The error is in assuming $\int \! f\, dx - \int\! f\, dx = 0$ when, in reality, $\int \! f\, dx - \int\! f\, dx = \int \! 0 \, dx = C.$

There is nothing wrong with the statement of $$\int \frac1x dx =1+ \int \frac1x dx,$$

only the conclusion obtained from it. It is just a proof that $1$ is constant. If you decide at this point to add bounds to your integral you find

$$\int_a^b \frac1x dx =1_a^b+ \int_a^b \frac1x dx$$

$$\int_a^b \frac1x dx = \int_a^b \frac1x dx$$

Because $1_a^b = 1 - 1 = 0$.

Again:

$$\int \frac1x dx =1+ \int \frac1x dx$$

$$\int \frac1x dx -  \int \frac1x dx =1$$

$$\int \frac1x - \frac1x dx = 1$$

$$\int 0\, dx = 1 $$

$$C = 1$$

The fallacy occurs when subtracting the indefinite integrals, not with anything mentioned by alexqwx.

$\int \frac{1}{x}dx=1+\int \frac{1}{x}dx \color{red}{+C}$ (you forgot the constant of integration).

Then we have that $C=-1$, which is fine.


Don't understand the Fundamental Theorem of Calculus If $f$ is continuous on $[a, b]$ and defining
  $$
F(x) = \int_a^x \! f(t) \, dt
$$
  for $x \in [a, b]$, then $F'(x) = f(x)$ for $x \in (a,b)$.

I don't understand what function the variable $t$ is for.  What is the definition of $f(t)$?  And is $F(x)$ a set?  Because it seems like $\int_a^x$ can have varying answers depending on what $x$ is.

Remember that a function is like a machine where you for each input there is exactly one output. We usually write $y = f(x)$ when $x$ denotes the input and $y$ denotes the output. $f$ is the function (rule) that tells you how you can find $y$ given $x$.

Ok, so now let's say that you have a function $f$. Let's call the input variable $t$. So for each number $t$ you get the (output) number $f(t)$. Now, given two numbers $a$ and $b$, you can find the integral
$$
\int_a^b f(t)\; dt.
$$
This integral gives you a number. Here we use $t$ in $f(t)$ and $dt$ to say that $t$ is the (input) variable for the function $f$.

Now if you choose another $b$ then you will get another number for the integral. So, for example, you might have
$$
\int_a^5 f(t)\;dt, \quad
\int_a^8 f(t)\;dt, \quad
\int_a^{11.5} f(t)\;dt.
$$
So, you actually have a function. Let's call the input variable for this function $x$ (we don't want to use the same variable as the one for $f$). So for each (input) number $x$ we get a number (output):
$$
\int_a^x f(t)\;dt
$$
If $x=5$ then you get $\int_a^5 f(t)\;dt$ (a numbers).

We haven't given this function a "name", so let's call it $F$ (again we don't want to call it $f$ because we already have a different function with that name).

So for each input $x$ you get the output $F(x)$ and this is defined by
$$
F(x) = \int_a^x f(t)\; dt.
$$
This is then a new function - a function who's output depends on the input $x$. What the Fundamental Theorem of Calculus tells you is that (under the given assumptions) this function is differentiable (that is you can find the derivative) with derivative
$$
F'(x) = f(x).
$$
So, for example
$$
F'(5) = f(5).
$$
Remember that the derivative of a function is just a new function. So again we have an output ($F'(x)$) for each input $x$. The way you calculate the output of the derivative of $F$ is by evaluating $f$ at the input.

A simple and concise answer is as follows:

$$F(x)=∫_a^b f(x)\,dx$$

Therefore,

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

So $F(x)$ is the integral of $f(x)dx$ from $a$ to $b$ , right? All the fundamental theorem of Calculus is saying is that along this set (interval $a,b$) exists the inverse operation:

$$F'(x)=f(x) $$


A basic doubt on joint distribution How to calculate the following probability $P(X \leq x, Y=y)$ where $X$ is a continuous random variable and $Y$ is a discrete random variable. I have been given the distribution of $X$ and distribution of $P(Y=y|X=x)$.

Mixing discrete and continuous distributions makes for interesting problems since one cannot rely anymore on fully-cooked formulas without understanding them. Here the first notion to understand is the conditional distribution of $Y$ conditionally on $X$. Assume that $X$ has density $f_X$. When we declare that
$$
P[Y=y\mid X=x]=p_Y(y\mid x),
$$
what we are actually imposing is that, for every $y$ and every measurable $B$,
$$
P[Y=y,X\in B]=\int_Bp_Y(y\mid x)f_X(x)\mathrm dx.
$$
This is the strict analogue of the fact that, when $(X,Z)$ has density $f_{X,Z}$, then, with obvious notations,
$$
f_{X,Z}(x,z)=f_{Z\mid X}(z\mid x)f_X(x),
$$
and of the integrated version, stating that, for every measurable $B$ and $C$,
$$
P[X\in B,Z\in C]=\int_C\int_Bf_{Z\mid X}(z\mid x)f_X(x)\mathrm dx\mathrm dz.
$$
Note that, since $X$ has a density, $P[Y=y,X=x]=0$ for every $y$ and $x$ hence the integrated version above is necessary. As a consequence,
$$
P[Y=y,X\leqslant x]=\int_{-\infty}^xp_Y(y\mid u)f_X(u)\mathrm du.
$$

Well P(X=t,Y=y)=P(Y=y)P(X=t|Y=y). Of course since X is a continuous random variable the probability that X=t is zero but we can sum up all of these values by taking in integral of the right hand side as t goes from negative infinity to x to get the probability that X≤x,Y=y


Is there a difference between $y=\frac{\sqrt{1-x}}{\sqrt{1+x}}$ and $y=\sqrt{\frac{1-x}{1+x}}$ Is there a difference between $$y=\frac{\sqrt{1-x}}{\sqrt{1+x}}$$ and $$y=\sqrt{\frac{1-x}{1+x}}$$
If there is a difference, why when I give the square for both equations, they will be equal.

Yes. There are differences. In the region where both of $x-1$ and $x+1$ are negative, the upper one will be wrong and lower one is correct. This region is $(-\infty, -1).$

For complex numbers there are no difference because any number is acceptable under the sqrt (positive or negative)

In general for real numbers, $ sqrt(\dfrac{a}{b}) = \dfrac{sqrt(|a|)}{sqrt(
|b|)} $ for $ (a \wedge b) \epsilon \mathbb{R}$


Relationship "finite mean"  "absolutely integrable What is the relationship between the property of a random variable (i.e. a measurable function defined on some probability space) being absolutely integrable, i.e. $$\mathbb{E}|X|<\infty$$and having finite mean, i.e.$$\mathbb{E}X<\infty\quad ?$$Obviously the first implies the second, but does the converse hold too ?

Exact equivalence, since E(X) is only defined when E(|X|) is finite. In Lebesgue theory of integration, the integral of a real-valued measurable function $f$ exists if and only if $|f|$ has finite integral.

nope. in general it does not hold. take the sin for example..


For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?

For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13

if x = 8, y = 325 which is divisible by 13

if x = 16, y = 1157 which is divisible by 13

if x = 21, y = 1937 which is divisible by 13

I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.

How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?

Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?

Let us try with this approach:

$\begin{align} 4x^2+8x+5 &= 4(x+1)^2+1 &\equiv 0 &\quad(\text{mod} 13)\\
&\Rightarrow 4(x+1)^2 &\equiv 12 &\quad(\text{mod} 13)\\
&\Rightarrow (x+1)^2 &\equiv 3 &\quad(\text{mod} 13)\end{align}\\$

Substituting for $x+1 = f$ we are looking to take the square-root of 3 modulo 13. The following holds true:
$$n^2 \equiv (13-n)^2 \quad (\text{mod} 13)$$

It suffices to look at numbers from 0 to 6:

$\begin{align}
0^2 &\equiv 0 (\text{mod} 13),\\
1^2 &\equiv 1 (\text{mod} 13),\\
2^2 &\equiv 4 (\text{mod} 13),\\
3^2 &\equiv 9 (\text{mod} 13),\\
4^2 &\equiv 3 (\text{mod} 13),\\
5^2 &\equiv 12 (\text{mod} 13),\\
6^2 &\equiv 10 (\text{mod} 13).
\end{align}$

Thus numbers of the form $f = 13m + 4$ solve the problem. Then also $f = 13m + 9$ will solve the problem. Since $x = f-1$ our solutions are numbers of the form: $x \in \{13m + 3, 13m +8, \text{where } m\geq 0\}$.

Solution

Check that image, I have tried solving for x = 13i + 3. You can try similarly for x = 13i + 8


Whats the result of double integral $\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$ Whats the result of double integral
 $$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$

I was trying to get this $$\int_{0}^{1}\sqrt{1+x^2}\arctan {\frac{1}{\sqrt{1+x^2}}}dx=\int_{0}^{1}(1+x^2)\frac{1}{\sqrt{1+x^2}}\arctan {\frac{1}{\sqrt{1+x^2}}}dx$$
$$=\int_{0}^{1}(1+x^2)\int_{0}^{1}\frac{1}{1+x^2+y^2}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
So far I don't know what to do next.any helps are to be grateful.

By symmetry one has
$$Q:=\int_0^1\int_0^1{1+x^2\over1+x^2+y^2}\>dy\>dx={1\over2}\int_0^1\int_0^1{2+x^2+y^2\over1+x^2+y^2}\>dy\>dx$$
and therefore
$$Q={1\over2}\int_0^1\int_0^1\left(1+{1\over 1+x^2+y^2}\right)\>dy\>dx={1\over2}+\int_0^{\pi/4}\int_0^{\sec\phi}{r\over 1+r^2}\>dr\>d\phi\ .$$
The last inner integral evaluates to $${1\over2}\log(1+r^2)\biggr|_0^{\sec\phi}={1\over2}\log(1+\sec^2\phi)\ ,$$
so that we finally obtain
$$Q={1\over2}+{1\over2}\int_0^{\pi/4}\log(1+\sec^2\phi)\>d\phi\doteq0.819755\ .$$
(Mathematica can evaluate the last integral in terms of arcsines and polylogs.)

Changing the variable $u=1+x^2$, integrating for y, and reversing the change, you can get:
$$I = \int(\sqrt{(x^2+1)}\arctan(\frac 1 {\sqrt{(x^2+1)}}), x = 0..1 ) = 0.819755$$


Prove that a number composed of only $2n$ "1"s, minus another number composed of only $n$ "2"s, is a perfect square. Prove that a number composed of only $2n$ "1"s, minus another number composed of only $n$ "2"s, is a perfect square.

Examples:

$$11-2 = 9 = 3^2$$

$$1111 - 22 = 1089 = 33^2$$

My try

I saw that I can rearrange some of the "equations" like this

Examples:
$$1111-22 = 101(11)-(11)2=11(101-2)=11\cdot99 = 11^2 \cdot 9 = 11^2 \cdot3^2 = (11\cdot 3)^2$$

$$111111-222 = 1001(111)-(111)2 = 111(1001-2) = 111 \cdot 999 = 111^2 \cdot9 = (111\cdot 3)^2$$

I think this is enough to prove that statement but I want a more "elegant" or general way (maybe an algebraic way?), and also to know if my solution is correct.

The number composed of $n$ ones is $(10^n-1)/9$. So your number is
$$\frac{10^{2n}-1}9-2\frac{10^n-1}9=\frac{a^2-1-2(a-1)}9$$
with $a=10^n$. I'll leave you to carry on...

An elegant way is by saying:

Without loss of generality, $n=3$. Then
$$111111-222 = 1001\cdot111-2\cdot111 =  999\cdot111 = 9\cdot111^2 = (3\cdot111)^2.$$

The educated reader has enough information to understand the pattern and generalize it to other $n$. That makes it a perfectly rigorous and concrete proof.


Establishing the upper and lower bounds of normal using standard deviation I understand the concept of standard deviations and z-values, but I'm trying to figure out if standard deviations alone are good for establishing the upper and lower bounds for normal. For example, if I have the following dataset:

$x = 1,4,1,10,112,6,22,7,18,113,1,4,1,10,112,6,22,7,18,113,1,4,1,10,112,6,22,7$

$\mu = 26.82$

$std(x) = 41.16$

I have been establishing the range for normal values like this:

$lowerNormal = 26.82 - 41.16 = -14.78$

$upperNormal = 26.82 + 41.16 = 67.98$

So, when a new value comes along:

$a = 72$

I consider that value to be abnormal, since it is above the $upperNormal$ value.

My question is whether or not this is a statistically sound way for determining whether a value is normal. The real issue I'm seeing is that none of the values in $x$ will ever be negative, so it seems that at least the lower bound is somewhat arbitrary.

Anyways, I apologize if I am missing something simple. I'm learning statistics on my own and it's a little puzzling sometimes. Thanks for your help

Your problem is that you are subtracting the full standard deviation from the mean. When you do that, it means that you are multiplying the standard deviation by 2, since you are subtracting $1*std(x)$ and adding $1*std(x)$.

Instead, add/subtract $\frac{1}{2}*std(x)$.

Thus, your new normals are:

$$lowerNormal=26.82−\frac{41.16}{2}=6.24$$
$$upperNormal=26.82+\frac{41.16}{2}=47.4$$

If a negative number is not a possible/feasible data point than your lower limit is zero in this particular example. It will stay zero until you get more data that shifts your mean higher.


Proving independence of events Let $A$, $B$ and $C$ be mutually independent indicator variable (also called dummy variable, >Boolean indicator, binary variable).
Prove that the events $(A = B)$, $(B = C)$ are independent if and only if:
a) $A$ is uniform (i.e. $P(A=1)=P(A=0)=0.5$)
OR
b) $C$ is uniform
OR
c) $B$ is constant(i.e. $P(B=1) = 0$ OR $1$).

I can't even figure out even where to start. Any help would be really appreciated.
Thanks a lot!

For convenience, let $a = P(A=1)$, $b=P(B=1)$, $c = P(C=1)$.  Then $P(A=B) = ab + (1-a)(1-b)$,
$P(B=C) = bc + (1-b)(1-c)$, and $P(A=B=C) = abc + (1-a)(1-b)(1-c)$.
You want to solve $P(A=B=C) = P(A=B) P(B=C)$.
Try expanding out $P(A=B+C) - P(A=B) P(B=C)$ and factoring.  If the problem is correct, $a - 1/2$, $c - 1/2$, $b - 1$ and $b$ should be factors.

I'm the OP. Could not comment above. So I'm writing here.

What am I to solve exactly here?

$$P(A=B=C)=P(A=B)P(B=C)$$ or
$$P(A=B+C)−P(A=B)P(B=C)$$

Also, what is $$P(A=B+C)$$ equal to?

One last thing: To show that the events [A=B] and [B=C], we need to show that $$P( A=B \cap B = C) = P(A=B) \cdot P(B = C)$$, right? I'm not sure how this relates to solving the above equations.

Many thanks!

-Nick!


Is vector space a field? Or more than that? As you all know, vector space is closed under scalar multiplication,  scalar product, vector product and addition. If I take scalar product, vector space is a field, but if i take vector product, vector space is a group. Is there any term designating this kind of set? Plus scalar multiplication is a binary operation from two same sets to different set. What kind of operation is this?

Let $K$ be a field. A vector space over $K$ is a set $V$, together with an operation $+\colon V\times V\to V$, and a function $\cdot\colon K\times V\to V$ (or alternatively, a family of functions $\lambda_c\colon V\to V$, indexed by elements $c\in K$) subject to certain identities and conditions.

The set $V$ does not normally contain $K$, so it makes no sense to talk about the vector space $V$ as being "closed under scalar multiplication". In addition, there is in general no "product" of vectors, so it does not make sense to talk about "vector multiplication."

Even in the case where we have a cross product of vectors (e.g., $\mathbb{R}^3$), this operation does not make $V$ into a group (the product is not associative).

Scalar multiplication is not an operation; an operation is always a function from a cartesian power of a set to the set. However, you can use currying to view the scalar multiplication as a family of unary operations on $V$, as indicated above, where for each $c\in K$ and each $\mathbf{v}\in V$, we define $\lambda_c(\mathbf{v})=c\mathbf{v}$.

(Added. Once you take into account the vector space axioms that relate to the scalar multiplication, it turns out that the scalar multiplication $K\times V\to V$ is actually an action of the field $K$ on the additive group $(V,+)$. But to truly call it an "action", you need the function to satisfy certain properties; just having a function $K\times V\to V$ does not make it an action, whereas any function $S\times S\to S$ is a binary operation on $S$, regardless of its properties.)

So pretty much all of your questions are based on misstatements of fact. Seems hard to answer them accurately.

Now, there is a situation in which some of what you say might make sense: if $F$ and $K$ are fields, and $F\subseteq K$, then we can view $K$ as a vector space over $F$ by "forgetting" about multiplication inside of $K$ when neither factor is in $F$. But here you really are going the other way: you already have a field, and you are obtaining a vector space by restricting the multiplication $\cdot\colon K\times K\to K$ to $F\times K$.

There is another situation in which you may have a product between "vectors": when you have an algebra. If $F$ is a field, a $K$ algebra is a ring with (a copy of) $F$ in the center of $K$. But again, what you have is a richer structure that, by "forgetting" part of the structure, yields a vector space. Much like you can have a ring and, by forgetting the product, obtain a group.

A vector space over a field is a set with its own definition

A field is a set with its own definition

But later on you will find that the field axioms agree with vector space axioms

Therefore, a field over itself is a vector space but a vector space does not imply a field
For example, polynomials are vector space but also a ring, not a field


symplectic lie algebra is simple The symplectic lie algebra defined by $sp\left(n\right)=\left\{ X\in gl_{2n}\,|\, X^{t}J+JX=0\right\}$  when $J=\begin{pmatrix}0 & I\\
-I & 0\end{pmatrix}$. So $X\in sp\left(n\right)$ is of the sort $X=\begin{pmatrix}A & B\\
C & -A^{t}\end{pmatrix}$ when $B=B^{t},\, C=C^{t}$.

This far I was able to get, but how can I prove that it is simple?

Thanks!

Let ${\mathfrak g}=sp(n)$.

Short answer, from an advanced viewpoint : $\mathfrak g$ is semisimple (because $K(x,y)=tr(xy)$ defines a nondegenerate invariant bilinear form on $\mathfrak g$), so it will be simple iff its Dynkin diagram is connected. But we know that the diagram is $C_n$ and this is clearly connected.

Longer answer, from a more pedestrian viewpoint : a standard basis $\cal B$ for $\mathfrak g$ is $\lbrace h_i,a_i,b_i\rbrace_{1 \leq i \leq n} \cup \lbrace c_{i,j},d_{i,j},f_{i,j},g_{i,j}\rbrace_{1 \leq i<j \leq n}$ where

$$
h_i=e_{ii}-e_{n+i,n+i}, \ \ \ a_i=e_{i,i+n}, \ \ \  b_i=e_{i+n,i},
$$

$$
c_{i,j}=e_{ij}-e_{i+n,j+n}, \ \ \ d_{i,j}=e_{ji}-e_{j+n,i+n},
$$

$$
f_{i,j}=e_{i,j+n}-e_{j,i+n}, \ \ \  g_{i,j}=e_{i+n,j}-e_{j+n,i}
$$

(and $e_{ij}$ is the matrix all of whose coefficients are zero except for the $(i,j)$-th which is $1$).

Now let $\mathfrak i$ be an ideal of $\mathfrak g$. Then $\mathfrak i$ is stable by $ad(x)$ for any $x\in {\mathfrak g}$. In particular, if $H$ is the vector space spanned by
$h_1,h_2, \ldots ,h_n$, then $\mathfrak i$ is stable by all the $ad(h)$, for
$h \in H$. Now those $ad(h)$ are simultaneously diagonalized by the basis we have just described above. We deduce that $\mathfrak i=H' \oplus {\sf Vect}(F)$, where $H'$ is a linear subspace of $H$ and $F$ is a subset of ${\cal B}'={\cal B} \setminus \lbrace h_1,h_2, \ldots ,h_n \rbrace$. Let us now consider the root spaces
$$
AB_i={\sf Vect}(h_i,a_i,b_i), \ CD_{ij}={\sf Vect}(h_i-h_j,c_{ij},d_{ij}), \
FG_{ij}={\sf Vect}(h_i+h_j,f_{ij},g_{ij})
$$
Each root space $R$ is a Lie subalgebra of $\mathfrak g$ isomorphic to ${\mathfrak sl}_2$. It easily follows that either ${\mathfrak i} \cap R=\lbrace 0 \rbrace$ or
$R \subseteq {\mathfrak i}$ for any $R$. Combining this with the decomposition of $\mathfrak i$  above, we see that $\mathfrak i$ is the direct sum of some root spaces, plus some subspace  of $H$.

Suppose $F$ contains $a_i$ for some $i$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i,a_i,b_i)$. For $j>i$, $\mathfrak i$ also contains
$[a_i,g_{ij}]=c_{ij}$, $h_i-[c_{ij},d_{ij}]=h_j, [h_j,\frac{a_j}{2}]=a_j$. Similarly, for $k<i$,
$\mathfrak i$ also contains $[a_i,-g_{ki}]=d_{ki}$, $h_i+[c_{ki},d_{ki}]=h_k, [h_k,\frac{a_k}{2}]=a_k$. So $\mathfrak i$ contains all the $a_i$, and we easily deduce that $\mathfrak i=A$.

Suppose $F$ contains $c_{ij}$ for some $i<j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i-h_j,c_{ij},d_{ij})$. And $\mathfrak i$ also contains $h_i+h_j=[c_{ij},f_{ij}]$. So $\mathfrak i$ contains both $h_i$ and $h_j$, it also contains the root space
${\sf Vect}(h_i,a_i,b_i)$, so $\mathfrak i=A$ by the preceding paragraph.

Suppose $F$ contains $f_{ij}$ for some $i<j$. Then $\mathfrak i$ contains the whole root space ${\sf Vect}(h_i+h_j,f_{ij},g_{ij})$. And $\mathfrak i$ also contains $c_{ij}=[f_{ij},b_{j}]$, so $\mathfrak i=A$ by the preceding paragraph.

Finally we see that $\mathfrak i=A$ unless $F$ is empty. In that case, $\mathfrak i$ is a linear subspace of $H$. For any $i$, $ad (a_i)$ must be zero on $\mathfrak i$, otherwise
$\mathfrak i$ would contain $a_i$. It is easily deduced that ${\mathfrak i}=\lbrace 0 \rbrace$,
so $\mathfrak g$ is indeed simple.

$sp(n)$ is a Lie subalgebra of the more familiar lie algebra $sl(n)$ because $Sp(n)$ is a Lie subgroup of $SL(n)$ (determinant $= 1$ condition). Since $sl(n)$ is semisimple, and therefore has non-degenerate Killing form, $tr(xy)$ would also be non-degenerate on its lie subalgebras.


Prove that $ 4^{n} + 6n - 10$ is divisible by $18$, $ \ \forall n \in \mathbb{N}$ Question:

Prove that $ 4^{n} + 6n - 10$ is divisible by $18$, $ \ \forall n \in \mathbb{N}$

My attempt:

Proof by Induction:

Base Case:

$ n = 1,  4^1 + 6(1) - 10 = 0 = 18(0)$. So base case holds.

I.H: Assume $ \ 4^{k} + 6k -10 = 18m$ for some $ m \in \mathbb{Z},\  k \in \mathbb{N}$.

WTP $ \ 4^{k+1} + 6(k+1) - 10 = 18m$.

$ 4^{k+1} + 6(k+1) - 10 = 4\cdot4^k +6k + 6 -10 = 4(18m + 10 -6k) + 6k + 6 -10$ , by I.H.

$ 4(18m + 10 -6k) + 6k + 6 -10 = 72m +40 - 18k -4 = 72m -18k +36 = 18(4m -k +2)$.

Hence by induction, the result holds for all natural numbers.

I have seen similar proofs in which instead of substituting an expression using the induction hypotheses, the $ k+1$ statement is manipulated until its brought in the desired form. Can I use the induction hypothesis to sub in $ \ 4^{k}$ like this? Is this approach correct?

Your proof looks okay.

Here is an alternate approach:

mod $18$, we have
$$
\begin{array}{c|c}
n&1&2&3&4&\cdots\\
\hline
4^n&4&16&10&4&\cdots\\
\hline
6n&6&12&0&6&\cdots
\end{array}
$$
Since each column is dependent only on the previous one, the sequences for $4^n$ and $6n$ repeat every $3$ columns. Also from the table above,
$$
4^n+6n-10\equiv0\pmod{18}
$$
will also repeat every $3$ columns.

Nit-pick: Your WTP should be $4^{k+1}+6(k+1)-10=18\cdot m'$ for some $m' \in \mathbb{Z}$, rather than using the specific $m$ from the IH


Dummit and Foote, Abstract Algebra, Question 7, Cyclic Groups I'm trying to solve the following question Dummit and Foote's Abstract Algebra.

  Let $Z_{48} = \langle x \rangle$. For which integers a does the map $\phi_a : \bar{1} \mapsto x^a$ extend to an ismorphism from $\mathbb{Z}/48 \mathbb{Z}$ onto $Z_{48}$

Proof: Assume that $\phi_a$ is an isomorphism. We then must have that $| \bar{1} | = | x^a |$. Since $| \bar{1} | = 48$, then $| x^a| = 48$. Using the formula $|x^a| = 48/(48,a)$, we then must have that $(48, a) = 1$. That is, $a$ is the set of all integers such that $2,3 \nmid a$.

I think this is a sufficient condition to put on $a$. For if $2, 3 \mid a$, then $\phi_a$ will not be an isomorphism, as given by the use of the previous isomorphism theorem.

But I get a contradiction (I suppose) when I try and individually impose the conditions of homomorphism and injectivity (bijectivity) to derive a condition on $a$.

$\mathbf{\phi_a}$ is a homomorphism

We can then derive the following recrusion relation.

\begin{align}
\phi_a(\overline{n-1} \; * \; \bar{1}) = \phi_a(\bar{n}) = \phi_a(\overline{n-1})\phi_a(\bar{1}) = \phi_a(\overline{n-1})x^a
\end{align}

Using this, we can show that $\phi(\bar{n}) = x^{na}$. So, a possible map is given by this equation. It is easily shown that if we start from this map, then it is a homomorophism.

$\mathbf{\phi_a}$ is a injective

To prove that $\phi_a$ is a bijection, it suffices to check when is $\phi_a$ an injection. We have, with $1 \leq d < c \leq 48,$

\begin{align}
\phi_a (\bar{c}) &  = \phi_a (\bar{d}) \\
\Longrightarrow  x^{ca} & = x^{da} \\
\Longrightarrow  x^{a(c - d)} & = 1 \\
\end{align}

This implies that $48 \mid a(c-d)$. Since $48 \nmid (c - d)$, we must have that $48 \mid a$. But this is in contradiction with the conditions derived above. If $a = 48$, then $2, 3 \mid 48$. Also, under the mapping $\bar{1} \mapsto x^{48}$, where $x^{48}$ is the identity in $Z_{48}$. Since a homormophism maps the idenintity of one group to the identity of the group this, this is again a contradiction.

Where am I going wrong?

Edit:

Here's a correct proof of injectivity. with $1 \leq d < c \leq 48,$

\begin{align}
\phi_a (\bar{c}) &  = \phi_a (\bar{d}) \\
\Longrightarrow  x^{a(c - d)} & = 1 \\
\end{align}

This implies that $48 \mid a(c-d)$. Since $(48, a) = 1$, we must have that $48 \mid c - d$. But this is a contradiction, sine $c - d < 48$. So we must have that $c = d$.

It's not true for general $r$ that $r \mid st$ implies $r \mid s$ or $r \mid t$ for all $s, t$; this is only true when $r$ is prime.

At any rate, to check injectivity of a group homomorphism, it's sufficient to show that its kernel is trivial. So, it's enough to show that $\phi_a(x^n) = \bar 0$ (that is, $an \equiv 0 \pmod {48}$) implies $x^n = e$, (that is, that $n \equiv 0 \pmod {48}$). But this condition on $a$ is essentially by definition coprimality of $a, n$.


Showing that multilinear maps factor through tensor products Let $R$ be a commutative ring and let $M_{1},...M_{n},L$ be $R$-modules.  Let $\iota:M_{1},...M_{n}\rightarrow M_{1}\bigotimes...\bigotimes M_{n}$ be defined by $\iota(m_{1},...,m_{n})\mapsto m_{1}\otimes...\otimes m_{n}$ and let $\phi:M_{1}\times...\times M_{n}\rightarrow L$ be an $n$-multilinear map.  Claim: there is a unique $R$-module homomorphism $\Phi:M_{1}\bigotimes...\bigotimes M_{n}\mapsto L$  such that $\phi =\Phi \circ \iota$.

Assume this holds for $n=2$ (in other words, assume the universal property for the tensor product of pairs of modules).  I want to show that this case implies the remaining cases, for example, $n=3$.  Apparently this is immediate, but I have absolutely no idea why it is true.  I have tried every way of pairing the modules but nothing has worked.

Edit: I am using the construction from free groups, etc. definition of tensor products.

Existence

$\Phi: m_1 \otimes \cdots \otimes m_n \mapsto \phi(m_1, \cdots, m_n)$ satisfies the condition.

Uniqueness

Let $\Phi$ and $\Phi'$ be maps satisfying the condition.

Then, I will prove that $\Phi = \Phi'$.

Let $m_1 \otimes \cdots \otimes m_n \in M_1 \otimes \cdots \otimes M_n$. Then:

$$\begin{array}{rcl}
\Phi(m_1 \otimes \cdots \otimes m_n)
&=& \Phi(\iota(m_1,\cdots,m_n)) \\
&=& \phi(m_1,\cdots,m_n) \\
&=& \Phi'(\iota(m_1,\cdots,m_n)) \\
&=& \Phi'(m_1 \otimes \cdots \otimes m_n)
\end{array}$$


Proof that ${\left(\pi^\pi\right)}^{\pi^\pi}$ (and now $\pi^{\left(\pi^{\pi^\pi}\right)}$) is a noninteger. Conor McBride asks for a fast proof that $$x = {\left(\pi^\pi\right)}^{\pi^\pi}$$ is not an integer.  It would be sufficient to calculate a very rough approximation, to a precision of less than $1,$ and show that $n < x < n+1$ for some integer $n$.  But $x$ is large enough ($56$ digits) that that method is not obviously practical without electronic computation tools.

Is there a purely theoretic proof that works with no calculation at all, or is there a calculational argument that would be practical prior to $1950$?

Edit: Let me phrase the question a little differently, as a spur to your mathematical imagination. I go back in time to $1847$ and have lunch with Carl Friedrich Gauss.  We are chatting about the developments in $20$th and $21$st-century mathematics, and because I am a horrible person, I mention that $(\pi^\pi)^{\pi^\pi}$ has been proven to be an integer. Gauss is incredulous, but I insist. After I leave, can Gauss convince himself that I am nothing but a troll?

Edit: After I looked back at McBride's original question, I saw to my dismay that he had not asked about $x= (\pi^\pi)^{\pi^\pi}$, but $$y=\pi^{\left(\pi^{\pi^\pi}\right)}$$ which is a rather different animal. The value of $x$ has $57$ decimal digits; the value of $y$ has $10^{57}$ digits.  So Alexander Walker's impressive display of direct numerical calculation will avail nothing for deciding that $y$ is not an integer, neither in $1873$ nor in $2013,$ and perhaps not in $2153$ either. ("We should attempt to destroy the aliens.") More powerful theoretical methods will be required.  I am posting an additional bounty to see if anyone can suggest anything of that sort; it need not be theory that was available in the $19$th century.

I think this calculation would have been doable in 1873, as follows:

(1) Compute $\log \pi$ to $60$ digits.  For this, we begin with the expansion

$$\log \pi = \log \left(\frac{\pi}{22/7}\right)+\log 2 + \log 11-\log 7$$

and take full advantage of the fact that the logarithm tables of small integers were known to great precision.  As for the logarithm of $\pi/(22/7) \approx 1.00041$, the Taylor series for the logarithm spits out $60$ digits after only $17$ terms.  (Surprisingly, this exact analysis has been done on MSE before.)

(2) Compute $(\pi+1)\log \pi$ to $60$ digits.  This is no big deal, given our value for $\log \pi$ from (1), since $\pi$ was known (to Gauss, no less!) to at least $100$ digits back in 1844.  For reference, this value is

$$ \approx 4.7410048855785583722294291029994190930164741026691888020108672.$$

(The multiplication will of course be a pain, as it requires around $1800$ flops.  Nevertheless, this computation would likely be delegated to a lesser mathematician.  The Wikipedia article on Zacharias Dase (a temporary assistant to Gauss) puts these computations into perspective:

  At age 15 [Zacharias Dase] began to travel extensively, giving exhibitions in Germany, Austria and England. Among his most impressive feats, he multiplied 79532853 × 93758479 in 54 seconds. He multiplied two 20-digit numbers in 6 minutes; two 40-digit numbers in 40 minutes; and two 100-digit numbers in 8 hours 45 minutes. The famous mathematician Carl Friedrich Gauss commented that someone skilled in calculation could have done the 100-digit calculation in about half that time with pencil and paper.

(3) Exponentiate the product, again to $60$ places. Using just the series
$$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$
such a calculation would require $77$ terms.  For this reason, we instead calculate
$$e^{4.741} \approx 114.54869311801681310751748724665811195370661075419665168411647;$$
$$e^{0.0000048\cdots} \approx 1.0000048855904928305900123833767696556988185632721564706179420.$$
The latter approximation requires a mere $10$ terms of the exponential Taylor series to achieve $60$ digits, a commitment of around 18000 flops.  By Gauss's metric, we might expect a skilled mathematician to complete this task in just over seven hours.

The prior calculation could be done in one of two ways: directly, at a cost of another $18000$ flops (e.g. $77$ consecutive multiplications of a $4$-digit and a $60$-digit number); or by calculating $\mathrm{exp}(4)$ and $\mathrm{exp}(.741)$ independently (a slight time savings, I believe, even after the final multiplication).  Of course, now it just takes another $1800$ flops to multiply these numbers to $60$ places.

Note: In hindsight, it appears that calculating the triple product
$$e^4 e^{3/4}e^{-9/1000}$$
may expedite this most recent step.

In case you've lost track, we now know the value of
$$e^{(\pi +1)\log \pi}=\pi^{\pi+1}$$
to $60$ digits, all within a day or two of starting our calculations.

(4) Multiply $\log \pi$ and $\pi^{\pi+1}$ to $60$ digits.  This step is easy, given steps (1) and (3).  This value is
$$\approx 131.12795303153615589452803943707399841542170349230159549341360.$$
Of course, this value is also the logarithm of $(\pi^\pi)^{\pi^\pi}$, so it remains to:

(5) Exponentiate the term from (4).  Since it worked out so well in (3), we'll again split our exponential into a product of simpler terms.  Here, we luck out - since the binary expansion of $131.127953\ldots$ begins as
$$10000011.001000001100000\cdots_2,$$
the partial approximation $131+1/8$ is surprisingly accurate: to within $\approx 0.002953$.  The exponential of this remainder can be made accurate to over $60$ digits with a mere $18$ terms (i.e. $32000$ flops).

Secondly, we compute $e^{131}$ to $60$ digits, using iterated multiplication and an approximation of $e$ to $62$ digits ($205$ digits were known to William Shanks in 1871).  Since it suffices here to successively compute
$$e^2,e^4,e^8,e^{16},e^{32},e^{64},e^{128},$$
this step can be done in less than $15000$ flops.  Thirdly, we compute $e^{1/8}$ to $60$ digits using three applications of Newton's method for the square root (another $6000$ flops).  We find a value of
$$ \approx 887455172183124295874631455225434602688412866765466125005\color{red}{.16},$$
a non-integer.

All said and done, I would be surprised if this calculation took much longer than a week.  At the same time, I stress that this problem would have been barely doable in that era.  If twice the number of digits were required, for example, this work may very well have taken the better part of a year.

The idea is to try to use this theorem, but in essence it will be just a leading force into simplifying it all considerably.

Theorem

$e^x$ is an integer iff for every $n>ex$

$$1-\{\sum_{k=0}^{n}\{\frac{x^k}{k!}\}\} \leq \frac{2x^{n+1}}{(n+1)!}$$

This is easy to derive and 19th century mathematicians had tools to find that it is sufficient to take $ex$ terms in order to achieve the precision less than $1$ and the right part is a crude estimation of the error term.

However in order to make this more precise, we find an integer as close to $(\pi^\pi)^{\pi^\pi}$ as we can. Since we know $\pi^3 \approx 31$, the candidate is simply $31^{31}$ which is not that difficult to calculate after all:

$$31^{31}=\frac{(((((31)^2)^2)^2)^2)^2}{31}$$

Using that we are reducing the calculation from $300$ terms to about $80$ terms of working with smaller numbers. Yet we can do better:

$$\pi^{\pi+1}\ln(\pi)-31\ln(31) \approx 25$$

Going into this direction we can try to find max integer solution of

$$x^x < e^{24}$$
$$x = 10$$

That leaves

$$\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10) \approx 1.64$$

making

$$(\pi^\pi)^{\pi^\pi}=31^{31}10^{10}e^{\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10)}$$

Further

$$(\pi^\pi)^{\pi^\pi}=31^{31} 10^{10} 2^{2} e^{{\pi^{\pi+1}}\ln(\pi)-31\ln(31)-10\ln(10)-2\ln(2)}$$

And finally

$$1<e^{\pi^{\pi+1}\ln(\pi)-31\ln(31)-10\ln(10)-2\ln(2)} <2$$

As we are dealing with small numbers the idea is very obvious. Express

$$\pi^{\pi+1}\ln(\pi) = r+\sum_{k=1}^{m}a_k\ln(b_k), a_k \in \mathbb{N},b_k \in \mathbb{N}, 0 \leq r < \ln(2)$$

Once this is done it is only $\{e^r\}$ that decides if the expression is an integer or not which we can calculate to almost any desirable precision or limited but sufficient one manually. Still we need to do the final multiplication.

Notice that even without the final multiplication, just by calculating $\{e^r\}$ we may confirm beyond doubt that the number cannot be rational number (which is the only way of making the above expression an integer), or that it is not a correct rational number if it happens to be rational, for which we would not have to multiply with the large number $31^{31} 2^{2}$ at all. A nice thing to have is $10^{10}$ if we work in decimal notation, but any integer pair $p\ln(q)$ would do equally well.

Simplifications can be revolving around these two ideas, of course, starting from instantly trying to find the best matching $10^d$ to still using the series with a couple of terms, or trying out only small primes in order to get the simplest possible proof that $e^r$ is not rational and so on.

For example

$$\pi^{\pi+1}\ln(\pi) = 189\ln(2)+r$$

Now you calculate $e^r$ to about 60-digit precision. The worst case scenario is that it is of the form $\frac{p}{2^{189}}$ which would be obvious that it is not from its form. (Not that mathematicians were not capable of using binary form I doubt dealing constantly just with $0$ and $1$ would be beneficial. Still it is a very nice feature to have that $r$ has to stop in binary representation.)

Finally we pick

$e^{\pi^{\pi+1}\ln(\pi)- 56\ln(10)} = 8.874551721831242958746314552254346026884128667654661250051588548428...$

proving it is not an integer. The precision is about 65 digits and it does not end with the stream of $0$'s or possibly $9$'s.

19th century mathematicians were much smarter than this sample of ideas for sure.

Not to forget then about the "and now" part.

All we need is to prove that

$$e^{\pi^{\pi^\pi}\ln(\pi)-2213275961092431622\ln(2)}$$

is irrational. Piece of cake for 22nd century mathematician. :)

The most likely future proof is going to claim that

$$ \{ 1,^{n}\pi | n \in \mathbb{N} \} $$

are all linearly independent over $\mathbb{Q}$, where $^{n}x$ is tetration. Meaning none of exponents is an integer.


What is the theory of distribution that makes possible to calculate Fourier transform of the Sine function? I am an Engineering student.
Sine wave function is a power signal present from $-\infty$ to $+\infty$.
I have have read that because of distribution theory,the Fourier transform of the Sine function  is possible. But what is distribution theory behind it?

Note : Please try to understand that I am talking about function that tends from $-\infty$ to $+\infty$ , not 0 to $+\infty$

Ok let me see if I can't answer your question from a mathematical perspective in a way that might be helpful for an engineer. The first thing you need to understand is that $\varphi_{n}(x)=\frac{1}{\sqrt{2\pi}}e^{ikx} , \, k \in \mathbb{Z}$ forms an orthonormal sequence in $L^{2}([-\pi,\pi]))$. Now, anytime you take the inner product of two orthonormal functions such as
$$
\langle \varphi_{n},\varphi_{m}\rangle=\int^{\pi}_{-\pi} e^{i(m-n)x} \, dx=\delta_{mn}=\begin{cases} 0 & \text{if } m \neq n \\1 & \text{if }m=n \end{cases}
$$
for example with $n \neq m$
$$
\langle\varphi_{n},\varphi_{m}\rangle=\frac{1}{2\pi}\int^{\pi}_{-\pi}e^{i(m-n)x} \, dx
=0$$
and
$$
\langle \varphi_{n},\varphi_{n}\rangle=\frac{1}{2\pi}\int^{\pi}_{-\pi}e^{i(n-n)x}\, dx=1
$$
Now IMO, the Fourier transform is best understood as a linear operator $F:L^{2}(\mathbb{R}) \to L^{2}(\mathbb{R})$ given by
$$
Ff=\int^{\infty}_{-\infty} f(x)e^{-i 2 \pi \omega x} dx
$$
which more generally belongs to the class of integral transforms of the form
$$
Tf=\int^{\infty}_{-\infty}K(x,\omega)f(x) dx
$$
Hopefully by now you can see that the Fourier transform looks a lot like an inner product of an orthonormal basis function in $L^{2}(\mathbb{R})$ and a function $f \in L^{2}(\mathbb{R})$. For an engineer, you might interpret the inner product as a sort of correlation between the given function and the basis function. However, note that the Fourier transform gives does not give us a number like the inner product but it does give us a function $\hat{f}(\omega)$ and for any $\omega_{0} \in \mathbb{R}$, we have
$$
\hat{f}(\omega_{0})=\lambda_{0}
$$
and this value $\lambda_{0}$ scales our basis function $e^{i 2\pi \omega_{0} x}$ so that we have $\lambda_{0} e^{-i 2 \pi \omega_{0} x}$ so that we have
$$
\lambda_{0}e^{ik \omega_{0}x}=\lambda_{0} \cos(\omega_{0} x)+i\lambda_{0} \sin(\omega_{0}x)
$$
which tells us the amplitude (which can be interpreted as the contribution) of our basis function at a given frequency. Now, if we only had countably many values of $\lambda_{0}$, we could write our original function $f$ as
$$
f=\sum^{\infty}_{k=-\infty} \lambda_{k}e^{-i 2 \pi \omega_{k}x}
$$
where
$$\lambda_{k}=\int^{\infty}_{-\infty} f(x)e^{i 2\pi \omega_{k}} \, dx$$
but since $\omega_{0} \in \mathbb{R}$ for most functions (think about how many terms you'd need to write a constant function or any non-periodic function in terms of sines and cosines) then we have that
$$
f(x)=\int^{\infty}_{-\infty} \hat{f}(\omega) e^{i 2 \pi \omega x} \, dx
$$

So now to finally answer your question, the connection between distribution theory and the Fourier transform of a sine function has to do with something called the uncertainty principle or Gabor limit. This just means that there is a tradeoff between time localization and frequency localization. This is to say that the more that one zooms in to a specific value of time $f(t_{0})$, the more spread out the the values of $\hat{f}(\omega)$. This is formalized in the following statement where $D[f]=\int^{\infty}_{\infty} x^{2} f(x) \, dx$
$$
D[f]D[\hat{f}]=\int^{\infty}_{\infty} \lvert x^{2} f(x)\rvert^{2} \, dx \int^{\infty}_{\infty} \lvert \omega^{2} \hat{f}(\omega) \rvert^{2} \, d\omega \geq \frac{\lvert \lvert f \rvert \rvert^{4}_{2}}{16 \pi^{2}}$$
where it should be fairly obvious that $D[f]$ is a linear functional that measures the "spread" or "variance" of $f$. Now, imagine that one wants to take the Fourier transform of a distribution $\delta: C^{\infty}_{c} \to \mathbb{R}$ which is clearly a linear functional defined by
$$
\delta[\varphi]=\varphi[0]
$$
for every $\varphi \in C^{\infty}_{c}$. Now we often define $\delta$ as a distributional derivative of the heaviside function
$$
H(x)=\begin{cases}0 & \text{if } x<0 \\ 1 & \text{if } x\geq 0 \end{cases}
$$
such that
$$
\delta[\varphi]=-\int^{\infty}_{-\infty} \varphi'(x) H(x) \, dx
$$
for every $\varphi \in C^{\infty}_{c}$
and then you end up with a Lebesgue-Steiltjes integral
$$
-\int^{\infty}_{-\infty} \varphi'(x) H(x) \, dx=\int^{\infty}_{-\infty} \varphi(x) \, dH(x)
$$
and I personally prefer this definition so we write
$$
\delta[\varphi]=\int^{\infty}_{-\infty} \varphi(x)\, dH(x)
$$
for all $\varphi \in C^{\infty}_{c}$ which then gives rise to the heuristic definition
$$
\delta(x)=\begin{cases} 0 & \text{if } x \neq 0 \\ \infty & \text{if } x=0 \end{cases}
$$
where $\delta(x)$ is also subject to the constraint that
$$
\int^{\infty}_{-\infty} \delta(x) \, dx=1
$$
and if we try to take the Fourier transform of the tempered distribution $\delta[\varphi]$, we get
$$
\hat{\delta}(\omega)=\int^{\infty}_{-\infty} \delta(x) e^{-i 2 \pi \omega x} \, dx=1
$$
Now note that that
$$
\delta(x)=\int^{\infty}_{-\infty} 1(e^{i 2 \pi \omega x}) \, dx
$$
Now, consider that for $f(x)=\sin(x)=\frac{e^{-ix}-e^{ix}}{2i}$, we have
$$
\int^{\infty}_{-\infty} \frac{e^{-ix}-e^{ix}}{2i} e^{-i 2 \pi \omega x} \, dx\\
=\int_{- \infty}^{\infty} - \frac{\left(\mathrm{e}^{\mathrm{ix}}\!\left(\mathrm{e}^{- 2 \pi i \mathrm{\omega} x}\right) - \mathrm{e}^{- \mathrm{ix}}\!\left(\mathrm{e}^{- 2 \pi i \mathrm{\omega} x}\right)\right)}{2 i} \,d x \\ = \int^{\infty}_{-\infty} \frac{e^{2 \pi i \omega x+ix}-e^{-2 \pi i \omega x-ix}}{2i} \\ = \frac{1}{2 i}\int^{\infty}_{-\infty} e^{ix(2 \pi \omega +1)}-e^{ix(- 2 \pi \omega -1)} \, dx \\ = \frac{1}{2i} \left(\delta(2 \pi \omega +1)-\delta(2 \pi \omega -1)\right) $$
which intuitively makes sense as a function $f(x)=sin(x)$ can be written in terms of a single frequency $f_{0}=\frac{1}{2 \pi}$


Three equilateral triangles form a hexagon As I posted yesterday, I was learning about vectors yesterday. I know how to add and subtract them, but I can’t multiply yet. So here is an extra problem from my teacher I need help with:

Given a hexagon $ABCDEF$ and a point $P$ on the plane sarisfying that the triangles $ABP, CDP, EFP$ are equilateral.

Prove that the midpoints of segments $AF, BC, DE$ form an equilateral triangle!

To rotate a vector around any point in the plane it's the same that to rotate this vector around his tail.

Let $R^{\alpha}$ be rotation on the plane by angle $\alpha$.

Also, let $K$, $L$ and $M$ be midpoints of $AF$, $BC$ and $DE$ respectively.

Thus, $$R^{60^{\circ}}\left(\vec{KM}\right)=R^{60^{\circ}}\left(\frac{1}{2}\left(\vec{AD}+\vec{FE}\right)\right)=\frac{1}{2}R^{60^{\circ}}\left(\vec{AP}+\vec{PD}+\vec{FE}\right)=$$
$$=\frac{1}{2}R^{60^{\circ}}\left(\vec{AB}+\vec{PC}+\vec{FP}\right)=\frac{1}{2}R^{60^{\circ}}\left(\vec{AB}+\vec{FC}\right)=\vec{KL}$$
and we are done!

This extra problem? It's built for the tools you don't have yet.

Hints: Put the origin at $P$. For your choice of vector systems, use complex numbers. Rotation by $60^\circ$ or $120^\circ$ (which is given by multiplication by a particular complex number) will feature prominently.

The problem also makes assumptions about the configuration; it's possible to build a hexagon for which it fails. A picture of such a hexagon is included below; the blue segments from the six labeled vertices intersect at $P$, and the triangle that we hope to be equilateral is drawn in red. Clearly, it isn't equilateral in this case.

You might object - that hexagon isn't convex. True, it isn't - but neither is $ABDCFE$ using the same vertices, and that one (with the midpoints of $BD$, $CF$, and $AE$) works.


Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.

step 1 $\frac {2x^2}{x+2} < x-2$

step 2 $\frac {2x^2}{x+2} - (x-2) < 0$

step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$

step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$

step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$

step 5 $\frac {x^2 + 4}{x+2} < 0$

step 6 I used character study to get the result x > 2. But this is incorrect.

Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.

Thanks!

HINT : $\frac{x^2+4}{x+2}\lt 0$ is correct. Then, note that $x^2+4\gt 0$ for any $x\in\mathbb R$.

Your inequality holds good if $x>-2$, not $2$.


Is $A$ a linear transformation on $\mathbb{C}^n$, where $A(u) = \| u\| u$? Is  $A$  a  linear transformation  or  not  on $\mathbb{C^n}$, where
$A(u) = \|u\| u$ and $\|u\| $ is the  length of  the  vector $u$.

I know  the definition  of linear  transformations: $T$ is linear if it satisfies

$$
T(\mathbf x+\mathbf y) = T(\mathbf x) + T(\mathbf y)\\
T(c\mathbf x) = cT(\mathbf x)
$$

But  here  how can I apply it? I am confused. . .

Since there is a factor of $\| u \|$ in the definition of $A(u)$, it means that $A$ is not scaling linearly. The magnitude of the scaling increases as the magnitude of the vector $u$ increases. But this contradicts the second property of a linear transformation, which essentially says that scaling is uniform.

So, we expect that $A$ is not linear. A good choice of $u$ to verify this is to take a nonzero vector $u$ and multiply it by a factor $c$ which is not equal to $1$, and then check. So, let's fix a nonzero $u \in \mathbb{C}^n$, say $u = (1,\dots,1)$, and $c = 2$. Then, we get
$$
A(2u) = \| 2u \| \cdot 2u = 4 \| u \| u.
$$
But if $A$ were linear, then we would have had $$A(2u) = 2 A(u) = 2 \|u \| u.$$ Since this is not the case, we have shown that $A$ is not linear.

Also note that the other answer by @GinoCHJ is incomplete. Merely knowing that $$A(u+v) \leq A(u) + A(v) + \| v\| u + \|u\|v$$ does not imply that $A$ is not a linear transformation. Perhaps by some wicked magic it turns out that $\|v\| u + \|u\|v = 0$ for all $u,v \in \mathbb{C}^n$? Their answer would be satisfactorily completed if specific vectors $u$ and $v$ are produced such that the inequality is strict.

Let $u, v \in \mathbb{C}^n$, then
\begin{align*}
  &A(u + v) = \lVert u + v \rVert(u + v) \\
\Rightarrow &A(u + v) = \lVert u + v \rVert u + \lVert u + v \rVert v \leq \lVert u \rVert u + \lVert v \rVert u + \lVert u \rVert v + \lVert v \rVert v \\
\Rightarrow &A(u + v) \leq A(u) + \lVert v \rVert u + \lVert u \rVert v + A(v)
\end{align*}
Hence, $A$ is not a linear transformation.


Find UMVUE of $P(X_1+X_2+X_3 =2)$ Where $X_1, ... X_n \sim$ Poisson$(\lambda)$ (independent) Let $X_1, ... , X_n \sim $ Poisson$(\lambda)$ (independence is assumed) Find the UMVUE $T(X_1,...,X_n) $ of $g(\lambda) := P(X_1+X_2+X_3=2)$
So we have $g(\lambda) := P(X_1+X_2+X_3=2) = e^{-3\lambda}\frac{(3\lambda)^2}{2!}$
$S(\underline{X}) = \sum_{i=1}^n X_i$ is complete and sufficient. and $S(\underline{X}) \sim $ Poisson$(n\lambda)$
So I want to find $h$ such that  $T(\underline{X}) = h(S(\underline{X}))$.
Setting up we have $\displaystyle E[h(S)] = e^{-3\lambda}\frac{(3\lambda)^2}{2!} \implies \sum_{k=0}^\infty \frac{e^{-n\lambda(n\lambda)^k}}{k!}h(S) = e^{-3\lambda}\frac{(3\lambda)^2}{2!}$
The next step I made was to get a sum on the RHS so I have and moved the $e^{-n\lambda} $ over:
$\displaystyle \sum_{k=0}^\infty \frac{(n\lambda)^k}{k!}h(k) = \sum_{k=0}^n\frac{((n-3)\lambda)^k}{k!}\frac{(3\lambda)^2}{2!} = \sum_{k=0}^\infty \frac{(n\lambda)^k}{k!} \bigg( \sum_{j=0}^k {k \choose j}(n\lambda)^{-j}(-3\lambda)^{j}\frac{(3\lambda)^2}{2!} \bigg)$
However, I don't think I have computed this correctly, from what I have above I should be able match to get this $T(\underline{X})$
Looking for input if I did this correctly.
EDIT: I have $h(S)$ but I should have $h(k)$ so I will change this

You want to solve for $h(\cdot)$ where for every $\lambda>0$,
$$E[h(S)]=\sum_{k=0}^\infty h(k)\frac{e^{-n\lambda}(n\lambda)^k}{k!}=c\,e^{-3\lambda}\lambda^2$$
for some positive constant $c$.
That is,
$$
\sum_{k=0}^\infty \frac{h(k)n^k}{k!}\lambda^k=c\,e^{(n-3)\lambda}\lambda^2
=c\sum_{j=0}^\infty \frac{(n-3)^j}{j!}\lambda^{j+2}
=\sum_{k=2}^\infty \frac{c(n-3)^{k-2}}{(k-2)!}\lambda^{k}
$$
Now compare coefficients of $\lambda^k$.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
By definition, the answer is given by
\begin{align}
&\bbox[10px,#ffd]{\sum_{k_{1} = 0}^{\infty}{\lambda^{k_{1}}\expo{-\lambda} \over k_{1}!}
\sum_{k_{2} = 0}^{\infty}{\lambda^{k_{2}}\expo{-\lambda} \over k_{2}!}
\sum_{k_{3} = 0}^{\infty}{\lambda^{k_{3}}\expo{-\lambda} \over k_{3}!}\
\overbrace{\bracks{z^{2}}z^{k_{1} + k_{2} + k_{3}}}^{\ds{\large \delta_{k_{1} + k_{2} + k_{3}, 2}}}}
\\[5mm] = &\
\expo{-3\lambda}\bracks{z^{2}}\bracks{\sum_{k = 0}^{\infty}{\pars{\lambda z}^{k} \over k!}}^{3}
\\[5mm] = &\
\expo{-3\lambda}\
\underbrace{\bracks{z^{2}}\expo{3\lambda z}}
_{\ds{{\pars{3\lambda}^{2} \over 2!}}} =
\bbx{\large{9 \over 2}\,\lambda^{2}\expo{-3\lambda}} \\ &
\end{align}


What was the book that opened your mind to the beauty of mathematics? Of course, I am generalising here. It may have been a teacher, a theorem, self pursuit, discussions with family / friends / colleagues, etc. that opened your mind to the beauty of mathematics. But this question is specifically about which books inspired you.

For me, Euler, master of us all is right up there. I am interested in which books have inspired other people.

Gödel, Escher, Bach by Douglas R. Hofstadter

Dungeons & Dragons, Basic edition player's manual.


Prove that $\prod_{i=1}^n (1+x_i/n) \sim \exp (\sum_{i=1}^n x_i/n)$ as $n\rightarrow\infty$? Let $x_1,x_2,\dots$ be an infinite sequence of real numbers. Assume that they are bounded, $|x_i| \le C < \infty$ for all $i$ for some $C$.

Is it true that, for any such sequence

$$\lim_{n \rightarrow \infty} \prod_{i = 1}^n \left( 1 + \frac{x_i}{n} \right)
= \lim_{n \rightarrow \infty}  \exp \left(\frac{1}{n} \sum_{i = 1}^n x_i
\right)?\qquad\qquad (1)$$

So far I only have the heuristic argument

$$\prod_{i = 1}^n \left[ \left( 1 + \frac{x_i}{n} \right)^n \right]^{1 / n}
\rightarrow \prod_{i = 1}^n \mathrm e^{x_i / n} = \exp \left( \frac{1}{n} \sum_{i= 1}^n x_i \right)$$

but I can't be sure that it is correct.

Update: The answer to my original question, that the lims in (1) are equal, is NO. In fact the limits might not exist. However their ratio goes to 1, so the two expressions are asymptotically equivalent. I've updated the title to reflect the actual true statement that was proved in the answers here to make this easier to find.

(I). $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ need not exist, even if $(x_j)_j$ is a bounded sequence.

(2). Let $A_n=\prod_{j=1}^n(1+x_j/n) .$ Let $B_n=\exp (n^{-1}\sum_{j=1}^n x_j).$ Let  $n>C.$

We have $\forall j\leq n\,( |x_j|<n)$ so $$\log A_n=\sum_{j=1}^n\log (1+x_j/n)=$$ $$=\sum_{j=1}^n\sum_{s=1}^{\infty }(-1)^{s-1}(x_j/n)^s/s=$$ $$=(\sum_{j=1}^nx_j/n) + R_n$$ $$\text { where } \quad  |R_n|=|\sum_{j=1}^n\sum_{s=2}^{\infty}(-1)^{s-1}(x_j/n)^s/s|\leq$$ $$\leq  \sum_{j=1}^n \sum_{s=2}^{\infty}(C/n)^s/s=n\sum_{s=2}^{\infty}(C/n)^s/s=$$ $$=(C^2/n)\sum_{t=0}^{\infty} (C/n)^t/(t+2)\leq$$ $$\leq (C^2/n)\sum_{t=0}^{\infty}(C/n)^t=(C^2/n)\cdot 1/(1-C/n).$$ So $\lim_{n\to \infty}R_n=0.$

But we also have $R_n=\log A_n-\log B_n=\log(A_n/B_n).$ So $\lim_{n\to \infty}\log (A_n/B_n)=0.$ So $\lim_{n\to \infty}(A_n/B_n)=1.$

So if the Cesaro mean $\lim_{n\to \infty}n^{-1}\sum_{j=1}^n x_j$ exists then the LHS and RHS in your Q are equal. If the Cesaro mean does not exist, then the ratio of the LHS to the RHS still converges to $1.$

It follows quite simply from the one of the definitions of $e^x$ (saying "one of" the definitions a bit loosely to indicate that any one of them could be taken to be the definition) - $$e^x\equiv\lim_{n\rightarrow\infty}\left(1+{x\over n}\right)^n .$$

The product of $n$ factors is a generalisation of the raising to the power of $n$ in the given (by me) limit definition, and that the argument of your exponential term is the arithmetic mean over all $x$ is the generalisation in that place corresponding to it.

One way we have of proving it in the simple case (let's call it) when it's just a single value of $x$, is to show that the binomial expansion morphs into the Taylor series for $e^x$ with $(n)_k$ (Pochhamner notation desending) becoming ever more nearly $n^k$ for a given value of $k$ (index along the series - the binomial or the Taylor) as $n$ increases. This will work also for the compound case (let's call it - infinite product of $n$ factors), if we can show that the sum of the products of the $x_i$ taken $k$ at a time becomes more nearly like $${\left(\sum_{j=1}^n x_j\right)^k\over k!}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\operatorname{¶}$$ for a given $k$ as $n$ increases in the same manner as, or a similar manner to that in which, $(n)_k$ becomes more nearly like $n^k$ ... or even in a very different manner, as long as it becomes more nearly like it: but that is not really to be expected. The difference consists in this last sum having $k$-fold products in it that are not fully hererogeneous - whereas in $$\prod_{j=1}^n\left(1+{x_j \over n}\right)\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\operatorname{§}$$ all the $k$-fold products occuring in a sum of them are fully heterogeneous. The total number of $k$-fold products including the non-fully-heterogeneous ones in ¶ is $n^k$, so it becomes the more nearly completely dominated by the fully-heterogeneous ones in the same proportion as $(n)_k$ becomes the more like $n^k$. This is an exact analogy of the similutude - ever increasing with $n$ - of the $(n)_k$ in the binomial expansion to the $n^k$ in the Taylor series in the simple case - and differs in that rather than being a pure number, it is the number of terms in a sum.

I cannot prove at this present time that there is absolutely no way a pathological set of $x_j$ could be devised such that the fully heterogeneous terms in the sum fail to dominate the non-fully heterogeneous ones despite their ever-increasing preponderance - constituting a proportion $${(n)_k\over n^k}$$ of the total termage. But it certainly would be an extremely pathological set, the devising of which would demand surpassing ingenuity.

Ah! but I've just noticed that you said "assume they are bounded"! That prettymuch precludes anysuch "pathological set", I should think.


A proof of $\sum{\mu(n)/n}=0$ I am looking for a proof (or references) of the following statement
$$\sum_{n=1}^{\infty}{\frac{\mu(n)}{n}}=0$$
where $\mu$ is the Möbius function.

Many thanks !

(I remember being utterly astonished by this result when I first came across it, in Apostol's introductory book on analytic number theory, and I mean to study it properly some time.)

According to Apostol, there's a proof in Ayoub's book (see BibTeX items below) that it is equivalent to the Prime Number Theorem.

From section 5.6 of Edwards' book one learns
that even before Landau, de la Vallee Poussin proved it
in 1899, obtaining the estimate $|\sum \mu(n)/n| < K/\log x$
for all sufficiently large $x$.

Indeed, the result was already stated by Euler!  On reflection,
that's not surprising, because he probably merely assumed that
you could expand and rearrange the product.

According to Ivic (pp. 50-51), the result
was first proved by H. von Mangoldt in 1897.
He refers the reader to Titchmarsh
for details.

@BOOK{apostol:analytic,
   AUTHOR = "Tom M. Apostol",
   TITLE = "Introduction to Analytic Number Theory",
   PUBLISHER = "Springer",
   ADDRESS = "New York",
   YEAR = "1976"}

@book{ayoub:analytic,
   author    = "Raymond G. Ayoub",
   title     = "An Introduction to the Analytic Theory of Numbers",
   publisher = "American Mathematical Society",
   year      = "1963"}

@book{edwards:riemann,
   author    = "H. M. Edwards",
   title     = "Riemann's Zeta Function",
   publisher = "Dover",
   address   = "New York",
   year      = "2001",
   note      = "Republication of the edition published by
      Academic Press in 1974."}

@book{ivic:riemann,
   author    = "Aleksandar Ivi\'c",
   title     = "The Riemann Zeta-Function",
   publisher = "Dover",
   address   = "New York",
   year      = "2003",
   note      = "Republication of the edition published by Wiley in 1985."}

@article{mangoldt:moebius,
   author    = "H. von Mangoldt",
   title     = "Beweis der {Gleichung} $\sum_{k = 1}^\infty \mu(k)/k = 0$",
   journal   = "Sitz. Preuss. Akad. Wiss., Berlin",
   year      = "1897",
   pages     = "835-852"}

@book{titchmarsh:riemann,
   author    = "E. C. Titchmarsh",
   title     = "The Theory of the Riemann Zeta-function",
   edition   = "Second",
   publisher = "Oxford University Press",
   year      = "1987",
   note      = "Revised by D. R. Heath-Brown."}

For a "proof" you can consider this:

Expand the product

$$ \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right) \cdots $$ for all primes, which yields
$$ \sum_{n=1}^\infty \frac{\mu(n)}{n} $$ your desired sum. To prove that the product diverges to zero, you can use the zeta function, since the product is the reciprocal of $$\left(\frac{2}{2-1}\right)\left(\frac{3}{3-1}\right)\left(\frac{5}{5-1}\right)\left(\frac{7}{7-1}\right)\cdots $$ which can be shown to be equal to
$$ \frac{1}{1} + \frac{1}{2}+\frac{1}{3}+ \frac{1}{4} + \cdots$$ which can be shown to diverge by the most basic analysis.

If you wish to make the entire argument more rigorous and formal you can consider the limits of most of these expressions and work form there.


Definition of manifolds with finite atlas It is known that any manifold of finite dimension $n$ has an finite atlas (with at most $n+1$ charts).
My question is then the following :
why don't we define an atlas of a manifold (in the finite dimensional case) as a finite set of charts that covers the manifold,
for teaching purposes.
Wouldn't that simplify the presentation and the theory (partition of unity, etc.) ?
Or in other words : do we encounter naturally manifolds with an infinite atlas (as part of their definition) ?

There's a relevant discussion on pp. 12-13 of Introduction to Smooth Manifolds by Lee:

"Our plan is to define a "smooth structure" on $M$ by giving a smooth atlas, and to define a function $f:M \to \mathbb R$ to be smooth if and only if $f \circ \phi^{-1}$ is smooth in the sense of ordinary calculus for each coordinate chart $(U,\phi)$ in the atlas.  There is one minor technical problem with this approach: In general, there will be many possible choices of atlas that give the "same" smooth structure, in that they all determine the same collection of smooth functions on $M$.  For example, consider the following pair of atlases on $\mathbb R^n$:
\begin{align}
\mathcal A_1 &= \{ (\mathbb R^n, \text{Id}_{\mathbb R^n}) \} \\
\mathcal A_2 &= \{(B_1(x),\text{Id}_{B_1(x)}) : x \in \mathbb R^n \}.
\end{align}
Although these are different smooth atlases, clearly a function $f:\mathbb R^n \to \mathbb R$ is smooth with respect to either atlas if and only if it is smooth in the sense of ordinary calculus.

"We could choose to define a smooth structure as an equivalence class of smooth atlases under an appropriate equivalence relation.  However, it is more straightforward to make the following definition: A smooth atlas $\mathcal A$ on $M$ is maximal if it is not contained in any strictly larger smooth atlas.  This just means that any chart that is smoothly compatible with every chart in $\mathcal A$ is already in $\mathcal A$.  (Such a smooth atlas is also said to be complete.)"

The oriented surface with infinite genus does not have a finite atlas.


Lagrange multipliers and critical points (differential form description). On $M \times V^*$, where $M$ is a differentiable manifold (not necessarily equipped with a metric) and $V^*$ is dual to a vector space $V$, one can define a Lagrange function $F = f +v^*h(x)$ using functions $f: M \to \mathbb{R}$ and $h: M \to V$, and with $v^* \in V^*$. In this set up what is the standard argument for showing that the critical point set of $F$ is in bijection with the critical points of $f$ restricted to $h^{-1}(0)$?

In general case, let $N=h^{-1}(0)$ and $x\in N$ is a critical point of $f\big|_N$, that is, $df_x(\alpha)=0, \forall \alpha\in T_xN\subset T_xM$. Since $h\equiv 0$ on $N$, $dh_x(\alpha)=0$ too. Choose any $\beta\in T_xM-T_xN$ such that $df_x(\beta)=1$, denote $\gamma=dh_x(\beta)\in V$ and choose $v^*$ such that $v^*(\gamma)=-1$ then we have $df_x=-v^*\circ dh_x$, that is $df_x+v^*\circ dh_x=0$, implying that $(x, v^*)$ is a critical point of $F$.

In David Bressoud's text on calculus (from celestial mechanics to special relativity), there is an elegant and annotated proof (section 9.7).


Question operation on indicator function [uniform distribution] I do not understand a particular step in deriving the likelihood function for a continuous uniform distribution $U[0,\theta]$. This has to do with a particular operation on the indicator function.

My problem arises in Step 4. What is the rationale in this step? What is the performed operation and why is it allowed? My tutorial teacher said the inverse is taken of all terms but how do we in that case come up with the interval $[0,1]$. She said to use 1/0 = $\infty$ but if I recall correctly this is not true (division by zero is undefined).

It is allowed because if $y>0, \theta>0$ then
$\frac{\theta}{y_{(n)}}\geq1\Rightarrow \theta\geq y_{(n)}\Rightarrow 1\geq\frac{y_{(n)}}{\theta}>0$

So
$I_{[1,\infty]}\bigg(\frac{\theta}{y_{(n)}}\bigg)=I_{(0,1]}\bigg(\frac{y_{(n)}}{\theta}\bigg)$
or
$I_{[1,\infty]}\bigg(\frac{\theta}{y_{(n)}}\bigg)=I_{[0,1]}\bigg(\frac{y_{(n)}}{\theta}\bigg)$ a.e. (almost everywhere).

Because for positive $n$, if $n< x\leqslant 1$ then $1\leqslant x^{-1}<n^{-1}$ so therefore taking the limits:
$$\lim_{n\to 0^+}\mathbf I_{[n;1]}(x)=\lim_{m\to\infty}\mathbf I_{[1;m]}(x^{-1})\qquad (m=n^{-1})$$

Which we can abbreviate as: $$\mathbf I_{[0;1]}(x)=\mathbf I_{[1;\infty)}(x^{-1})$$

That is all.


Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible. Let $P(x)$ be a polynomial in $\mathbb{Z}[x]$ of degree 5 such that $P(1)=3$ and $P=(x-1)^5\bmod 3$. Show that as a polynomial in $\mathbb{Q}[x]$, $P(x)$ is irreducible.

So here I find that $$P(x)=(x-1)^5+3=x^5-5x^4+10x^3-10x^2+5x+2.$$ How can I show this irreducible? Can I use Eisenstein's Criterion? How would I do that? For instance, if my prime is 5 then it divides 4 of the 6 terms but what about those other 2?

$\!\bmod 3\!:\ p \equiv (x\!-\!1)^5\ $ so Eisenstein applies to $ f(x) := p(x\!+\!1) \equiv x^5\,$ by $\,3^2\nmid f(0)\! =\! p(1)\!=\! 3$

Remark $ $ This is the standard shifted Eisenstein criterion, e.g. see here for motivation.

It suffices to show it is irreducible over $\Bbb Z$, by Gauß' lemma.  Go mod $3$ and use Eisenstein.


$\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ converges uniformly to a $1-$ periodic meromorphic function. Given a sequence of meromorphic function $(f_n)_n$ on a open set $\Omega\subset \Bbb{C}.$

How does it mean that $\sum_{n=-\infty}^{+\infty}f_n$ converges uniformly on every compact of $D?$

Does it mean that $\sum_{n=1}^\infty (f_n+f_{-n})$ converges uniformly on every compact?

I ask because I find an exercise when it's asking to prove the convergence (on every compact) of $\sum_{n=-\infty}^{+\infty}\frac{1}{z-n}$ to a $1-$periodic meromorphic function. If I guess correctly the definition we have $$\frac1{z-n}+\frac1{z+n}=\frac{2z}{z^2-n^2}.$$

If $A=\sup\{\vert z\vert:z\in K\}$ then $\vert\frac{2z}{z^2-n^2}\vert\le\frac{2A}{n^2-A^2}$ for $n>A.$ So that the serie converges uniformly and it's meromorphic by a theorem.

But why it's $1-periodic?$

$\displaystyle\sum_{n=-\infty}^\infty \frac{1}{z-n}$ diverges but $\displaystyle\frac{1}{2}\sum_{n=-\infty}^\infty (\frac{1}{z-n}+\frac{1}{z+n})$ converges uniformly away from its poles. It is obviously $1$-periodic.

is the sine function remember that $ \frac{1}{z-n}+\frac{1}{z+n} $ is just $ \frac{2z}{z^{2}-n^{2}}$ for every n , now integrate over z and you get

$$ \sum_{n=0}^{\infty}ln(z^{2}-n^{2} $$

which is the logarithm of the sine functio divided by x $ ln(sinx) -ln(x) $


Trigonometric bearing problem I have two trigonometric problems that I solved, however it does not match the answer in the book:

  1) A yacht crosses the start line of a race on a bearing of $31$ degrees. After $4.3$ km, it rounds a buoy and sails on a bearing of $346$ degrees. When it is due north of its start, how far has it sailed altogether.

I calculated the answer to be $6.353$ km. However the book's answer is $13$ m. The units are themselves different.

  2) The bearing of $B$ and $A$ is $65$ degrees. The bearing of $C$ from $B$ is $150$ degrees, and the bearing of $A$ from $C$ is $305$ degrees. If $AC=300$m, find $BC$.

I calculated the answer to be $-1333.838$. However the answer in the book was $261$ km.  The units are themselves different as well.

  3) The bearing of $Y$ from $X$ is $205$ degrees. The bearing of $Z$ from $Y$ is $315$ degrees, and the bearing of $X$ from $Z$ is $85$ degrees. If $XY = 4$ km, find the distance $XY$.

I calculated the answer to be $3.26$ km.

The questions are from Edexcel IGCSE Ex-181*
Please check if my answers are correct? If not please show the steps to get the correct answer. I used the sine rule to solve these sums.

The boat leaves at a bearing of 31 degrees and ends up north of it's start point. Therefore first corner of the triangle is 31 degrees.

At the buoy it has a bearing of 346 which is NW - ish. This angle is equal to 135 degrees.
1st part of angle is 90-31= 59degrees using construction lines and the alternate angles theorem. The second part of the angle is 346-270=76 degrees. (the bearing minus the East-West construction line) so 76+59=135 degrees.

The last angle is 180-135-31=14 degrees.

I now use Sine rule: x/sin135  =  4.3/sin 14. x = 12.57 - which rounds to 13km.

This is the answer for the question no2

this is the answer for the question no3


Understanding why single-variable expansion of modular arithmetic is valid. I was reading: https://arxiv.org/pdf/1206.5114.pdf page 10 and came across an odd theorem.

It states, that the system:

$$ \begin{matrix} 2x_1+ 3x_2  \equiv 1 \mod 5  \\ 3x_1 + 5x_2 \equiv 1 \mod 7 \end{matrix} $$

Can be "re-written" as:

$$ \begin{matrix} 2x_1+ 3x_2 +5x_3 =  1   \\ 3x_1 + 5x_2 + 7x_3 = 1  \end{matrix} $$

Now what exactly is meant by "re-written here?" is it that for every solution in the first system there exists a solution in the second system? Or that for every solution in the second system there exists a solution in the first system? Or both?

I find at least the notion of "BOTH" to be impossible. For the simple reason that if we let $X = 2x_1 + 3x_2$ and let $Y = 3x_1 + 5x_2$

Then we are claiming that if

$$ X \equiv 1 \mod 5$$
$$ Y \equiv 1 \mod 7$$

That there exists an $x_3$ such that

$$ X + 5x_3 = 1 $$
$$ Y + 7x_3 = 1$$

Now to me this seems very odd, since if we analyze

$$ X \equiv 1 \mod 5$$

To me all that says is there is SOME integer $s$ such that $X = 5s + 1$ and similarly there is SOME integer $r$ such that $Y = 7r + 1$. The idea that $s = r$ is necessary seems blatantly false (let X = 11, Y = 8 as a counter example). With BOTH ruled out, (And our counter-example ruling out that every solution the first system implies a solution to the second system) it's pretty clear that at best every solution to the second system is a solution to first.

Now certainly I would call that a severe loss of information and not a simple "re-write", so does this paper have a mistake and they meant to communicate something else? or am I missing a major detail? or is the analysis up to here correct and the term "re-written" has a hidden loss of information not made clear by the paper.

The solutions to $$2x_1+3x_2+5x_3=1, 3x_1+5x_2+7x_3=1$$ satisfy $$2x_1+3x_2≡1, \mod5 $$

$$3x_1+5x_2≡1,  \mod7$$

But the solutions to the second system are not necessarily solutions to the first system because you may have  $$2x_1+3x_2+5x_3=1, 3x_1+5x_2+7x_4=1$$ where $x_3\ne x_4$

Now what exactly is meant by "re-written here?"

The expressions are equivalent. For a simpler example, consider:

$$4x_1 + 3x_2 \equiv 5 \pmod 6 \tag 1$$

I will write this expression in a similar way to that PDF file. Consider the definition of congruence in this context:

$$a \equiv b \pmod m \iff \frac{a-b}{m} = k \in \mathbb Z$$

Then $(1)$ becomes

$$4x_1 + 3x_2 \equiv 5 \pmod 6 \iff \frac{4x_1 + 3x_2 - 5}{6} = k \in \mathbb{Z} \tag 2$$

Multiply by $6$ on each side in $(3)$, add $5$ to each side, and subtract the integer $k$ from each to obtain

$$\frac{4x_1 + 3x_2 - 5}{6} = x_3 \in \mathbb{Z} \iff 4x_1 + 3x_2 -6k = 5 \tag 4$$

Since $k \in \mathbb{Z}$, we know an integer $-k$ exists. We can take $k = -x_3$ for some integers $x_3$ and thus $(4)$ becomes

$$4x_1 + 3x_2 -6k = 5 \iff 4x_1 + 3x_2 -6(-x_3) = 5 \iff 4x_1 + 3x_2 + 6x_3 = 5$$

Now, of course, the prior is something I think you're already aware of in some form, which you touch on. So I'll address another point:

  Now to me this seems very odd, since if we analyze
  $$ X \equiv 1 \mod 5$$
  To me all that says is there is SOME integer $s$ such that $X = 5s + 1$ [...]

While this is sort of true, solutions $X$ to a modular congruence as above are really just an equivalence class of integers. The "some integer" you're referring to is probably the "least residue" (at least, that's what I often find people misinterpret it as), but in this above equation there are multiple solutions because, per our definition of modular congruence,

$$X \equiv 1 \pmod 5 \iff \frac{X-1}{5} = k \in \mathbb{Z}$$

Or, less formally, solutions $X$ have a remainder $1$ when divided by $5$ (not the best definition, mind you, it's sometimes misleading to look at it that way). So which $X$ are a solution?

$$X = \{ ..., -9, -4, 1, 6, 9, ... \} = \{ k \in \mathbb{Z} | k = 5n, n \in \mathbb{Z} \} $$

This is an equivalence class of solutions, which we denote by $[1]$ (since $1$ is the least residue). All are solutions to the congruence.

In extrapolating this logic to a system of congruences, think of it like an ordinary system of equations. Many inputs could satisfy any of the equations (not necessarily one), and the system seeks only those that satisfy all of the equations (provided they exist). Similar idea at play here.


Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$

Thanks in advance for any help.

Heres a nice combinatorial proof: Lets say you have $n+1$ kids, and want to form a committee of three. Order the kids $a_1, a_2, \cdots, a_{n+1}$. There are $\dbinom{n+1}{3}$ ways to form the committee. On the other hand, if $a_1$ is the first person on the committee, we need to choose two more, in $\dbinom{n}{2}$ ways. If $a_2$ is the first person on the committee, we can choose two more in $\dbinom{n-1}{2}$ ways. In general, if $a_n$ is the first person on the committee, we can choosse two more in $\dbinom{n-k+1}{2}$ ways. Therefore, we have $$ \dbinom{n+1}{3} = \sum_{i=1}^{n+1} \dbinom{n-k+1}{2} = \dbinom{n}{2} + \dbinom{n-1}{2} + \cdots + \dbinom22$$

If the identity is true, we can rewrite this specific case of the well-known addition formula
$$\binom m3+\binom m2=\binom{m+1}3$$
as the trivial recurrence
$$\sum_{n=2}^{m-1} \binom{n}{2}+\binom m2=\sum_{n=2}^m \binom{n}{2}.$$


Prove that $x^{2} \equiv -1 \pmod p$ has no solutions if prime $p \equiv 3 \pmod 4$. Assume: $p$ is a prime that satisfies $p \equiv 3 \pmod 4$

Show: $x^{2} \equiv -1 \pmod p$ has no solutions $\forall x \in \mathbb{Z}$.

I know this problem has something to do with Fermat's Little Theorem, that $a^{p-1} \equiv 1\pmod p$. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated.

Suppose $x^2\equiv -1\pmod{p}$. Then $x^4\equiv 1\pmod{p}$. Since $p = 4k+3$, we have
$$x^{p-1} = x^{4k+2} = x^2x^{4k} \equiv -1(x^4)^k\equiv -1\pmod{p},$$
which contradicts Fermat's Little Theorem.

A solution to the given congruence would imply there exists an element of multiplicative order 4 in the finite field of size p which is false because (p-1)/2 is odd and therefore p-1 is not divisible by 4.


How does Horner method evaluate the derivative of a function From my understanding, Horner method is mainly used to evaluate polynomial functions by altering the equation into a simpler recursive relation with lesser number of operations.
Say for example, I was given $f (x) = 4x^4 + 3x^3 +2x^2+x+5$
This can be rewritten as $5 +x (1+x (2+x (3+x (4)))$

Were we can evaluate the function as a recurrent relation of simpler terms starting from:

$b_n=4 $

$b_{n-1} = 3 + b_n* x$

And $b_0$ would be the whole term evaluated and therefore the image of the function.
What I want to understand how is running horner method to the $b_n$ values result in the derivative?

If you apply the Horner scheme, or perhaps the Ruffini method (both originally published papers on variants of the Newton method that use a trick for fast polynomial evaluation that was previously well-known), then you perform a polynomial division with remainder by a linear term. From the Horner table, you can read off the coefficients for a polynomial $q$ so that
$$
p(x)=p(a)+(x-a)q(x)\iff q(x)=\frac{p(x)-p(a)}{x-a}
$$
which tells you that $q(a)=p'(a)$. A second Horner evaluation below the first table will thus evaluate the derivative value.

You can also see this as algorithmic differentiation, if the original algorithm is

val = 0
for k=0 to deg
    val = val*x + a[deg-k]
end for

then the derivative by using the chain rule in every step gives

dval = 0; val = 0
for k=0 to deg
    dval = dval*x + val
    val = val*x + a[deg-k]
end for

The derivative steps needs to be computed first as it uses the last value of val.

The key is to make the indices the same and use original Horner's rule in the same for loop to evaluate both polynomial and its derivative.

Let's say the original problem is to compute f(x)=sum[ c[n] x^n, for {n,0 ,k }].

The derivative is of the form f'(x)=sum[ m c[m] x^(m-1), for {m,1 ,k }].
A change of index n=m-1 make the base of iterator the same, i.e. f'(x)=sum[ (n+1) c[n+1] x^(n), for {n,0 ,k-1 }]

Now we have same iterator which can be used except the case n=k which only applies to evaluation of function and not the derivative. This case can be computed outside the loop easily.

Here is the way to do it in Python:

def horner_eval(x, a):

'''Uses Horner's rule to return the polynomial
   a[n] + a[n+1] x^1 + a[n+2] x^2 + ... + a[0] x^(n)
AND its derivative evaluated at x.
NOTICE: coefficient vector is given backward for ease of indexing
'''

dist, speed = 0, 0

for i in range(len(a) - 1):
    dist = a[i] + (x * dist)
    speed = i * a[i] + (x * speed)

dist = a[-1] + (x * dist) # since function has one term more that the derivative
return dist, speed

dist stores the values of the polynomial at x.

speed stores the values of derivative of the polynomial at x.


Another limit related to pi number Find the value of the limit:

$$\lim_{n\to\infty} \sum_{k=0}^n \frac{{k!}^{2} {2}^{k}}{(2k+1)!}$$

I'm trying to find out if this limit can be computed only by using high school
knowledge for solving limits. Thanks.

Mimicking robjohn's solution to the series, and after proving convergence, we may proceed as follows:

$$\sum\limits_{k = 0}^\infty  {\frac{{k!^2{2^k}}}{{\left( {2k + 1} \right)!}}}  = \sum\limits_{k = 1}^\infty  {\frac{{\left( {k - 1} \right)!^2{2^{k - 1}}}}{{\left( {2k - 1} \right)!}}}  = \sum\limits_{k = 1}^\infty  {\frac{{{\Gamma ^2}\left( k \right)}}{{\Gamma \left( {2k} \right)}}{2^{k - 1}}}  = \sum\limits_{k = 1}^\infty  {\operatorname{B} \left( {k,k} \right){2^{k - 1}}} $$

$$\sum\limits_{k = 1}^\infty  {\operatorname{B} \left( {k,k} \right){2^{k - 1}}}  = \sum\limits_{k = 1}^\infty  {\int\limits_0^1 {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}dt} }  = \int\limits_0^1 {\sum\limits_{k = 1}^\infty  {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}} dt} $$

Then

$$=\int\limits_0^1 {\frac{{dt}}{{1 - 2t\left( {1 - t} \right)}}}  = \int\limits_0^1 {\frac{{dt}}{{1 - 2t + 2{t^2}}}}  = \frac{1}{2}\int\limits_0^1 {\frac{{dt}}{{{{\left( {t - \frac{1}{2}} \right)}^2} + \frac{1}{4}}}} $$

Now let $t-1/2=u$.

$$\frac{1}{2}\int\limits_{ - 1/2}^{1/2} {\frac{{du}}{{{u^2} + {{\left( {1/2} \right)}^2}}}}  = \arctan 2\frac{1}{2} - \arctan 2\left( { - \frac{1}{2}} \right)$$

$$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$$

If students don't know about the Gauss error function -- which is defined in terms of a non-elementary integral -- then no! Because the exact value of this infinite sum is $$\sqrt\frac{e \pi}2  \operatorname{erf}\left(\frac{1}{\sqrt 2}\right)$$.


No cont function $f\colon\mathbb{R}\to\mathbb{R}$ with $f(x)$ rational $\iff f(x+1)$ irrational. Prove that there are no continuous functions $f\colon \mathbb{R} \to \mathbb{R}$ with the property:

  For any $x \in \mathbb{R}$, $f(x)$ is a rational number if and only if $f(x+1)$ is an irrational number.

Source: 6th University of Rochester Math Olympiad.

Suppose $f$ were such a beast. Then

$$g(x) = f(x+1) - f(x)$$

defines a continuous function on $\mathbb{R}$ that takes only irrational values. By the intermediate value theorem and the denseness of $\mathbb{Q}$ in $\mathbb{R}$, $g$ must be constant, say $g \equiv c$.

Let $x_0 \in \mathbb{R}$ with $f(x_0) \in \mathbb{Q}$. Then $f(x_0+2) - f(x_0) = 2c\in \mathbb{Q}$, contradicting the irrationality of $c$.

$\Bbb Q$ is countable but $\Bbb R$ is uncountable, so we can easily show that $f$ is constant by using the fact that $f$ is continuous.


Low Level Well-Ordering Principle Proof Suppose that I have k envelopes, numbered $0,1,...,k−1$, such that envelope i contains $2^i$ dollars. Using the well-ordering principle, prove the following claim.

Claim: for any integer $0 ≤ n < 2^k$, there is a set of envelopes that contain exactly n dollars between them.

How would I tackle this problem? I am stumped. I know that the well-ordering principle states that a non-empty set will always have a smallest element in the set, but I don't exactly know how to apply it to write the proof.

Call the positive integer $k$ bad if there is a set of $k$ envelopes, numbered $0$ to $k-1$, with contents as described in the question, and an integer $n\lt 2^k$, such that no subset of the envelopes contains exactly $n$ dollars between them.

We want to prove there are no bad $k$.  Suppose to the contrary that there are. Then the set $B$ of bad $k$ is non-empty. It follows that there is a smallest bad $k$.  Call it $m$.

$m$ is greater than $0$, because $m=0$ holds obviously.

So $m$ is bad, and all positive integers $\lt m$ are good.

We arrive at a contradiction by showing that $m$ is good, that is, that any number $n\lt 2^m$ of dollars can be obtained by choosing an appropriate collection of the envelopes.

Case (i), $n\lt 2^{m-1}$: Since $m-1$ is good, there is a subset of the first $m-1$ envelopes that gives sum $n$.

Case (ii), $2^{m-1}\le n\lt 2^m$. Take the contents of the biggest envelope, that is, envelope with label $m-1$, which has $2^{m-1}$ dollars.

Then we still need an amount $a$ of dollars, where $0=2^{m-1}-2^{m-1}\le a\lt 2^m-2^{m-1}=2^{m-1}$.  Because $m-1$ is good, we can produce this amount $a$ from the first $m-1$ envelopes. Hence by adding the contents of the envelope with label $m-1$, we get our $n$ dollars.

Remark: This is just an ordinary proof by induction, rewritten to use "well-ordering principle" language.

Let S denote the values in $\{0,1,2,…,2k−1\}$ that cannot be expressed with the envelopes. $0$ is not in $S$, because "no envelopes" has zero dollars. $S$ is well-ordered because it is a subset of the naturals; if it is nonempty it has a smallest element $s$. Hence $0,1,2,…,s−1$ can all be expressed with combinations of envelopes, but s cannot be. You need to use the combination of these two facts to get a contradiction; hence $S$ is empty and your theorem is proved.

If you need a hint for how to get that contradiction, let me know.

How do I get to a contradiction?


Local rings characterization According to Wikipedia, a (non-commutative) ring $R$ is local if and only if there do not exist two proper (principal) (left) ideals $I_1, I_2$ such that $R = I_1 + I_2$. It is easy to see that local rings have this property, but I'm having trouble showing the other way: if there do not exist two proper co-maximal (principal) (left) ideals then the ring is local.

HINT $\ $ If $\rm\ R \ne 0\ $ is nonlocal then there exists an $\rm\: r\:$ such that both $\rm\: r\:$ and $\rm\: 1-r\:$ are nonunits. Choose maximal ideals containing them.


Expected value of number comprising randomly selected digits and a decimal point You are given a bag with eleven tiles, one each marked with a different digit $0$ to $9$ or a decimal point.  The tiles will be pulled from the bag one at a time and placed in sequence on the table to produce a single decimal number.  This resulting number will be your prize.  Compute the expected value of your prize.

Here is my solution to the problem:

The average value considering only the digits is $4999999999.5$ (i.e., if the decimal point is the last chosen).  Each candidate position for the chosen decimal point tile has an equal $\dfrac{1}{11}$ chance of being the case.  Therefore, the answer is $\dfrac{1}{11} \big(4999999999.5(10^{0}+10^{-1}+\cdots+10^{-11}\big)\big) = 505050505$.

The last equation is incorrect. 10 digits have 11 positions to put the decimal point. So it should be sum of series (10^0, 10^-1, 10^-2....10^-10).

10^-11 should not appear in the equation