Show that there is a continuous function $g$ over $\mathbb{R}$ such that $0 < g(x) \leq f(x)$ Suppose $f$ is a function over $\mathbb{R}$ such that $f(x) > 0$ for all $x \in \mathbb{R}$ and that $f$ is strictly decreasing. Show that there is a continuous function $g$ over $\mathbb{R}$ such that $0 < g(x) \leq f(x)$ and that $\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)} = 0$.

Attempt:

We have $f(x) < f(y)$ if $x < y$ and $f(x) > 0$ for all $x \in \mathbb{R}$. I am not sure how to take into account the condition that $0 < g(x) \leq f(x)$, but the condition that $\displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)}$ means that $f(x)$ grows faster than $g(x)$, so that intuitively fits the previous definition. Maybe proof by contradiction will work here?

For $n\in\Bbb N$ let $g(n)=\frac1nf(n+1)$ and linearly interpolate between these points. That is, let
$$ g(x)=\begin{cases}f(2)&x\le 1\\
(x-\lfloor x\rfloor)\frac1{\lceil x\rceil}f(\lceil x\rceil+1)+(\lfloor x\rfloor +1-x)\frac1{\lfloor x\rfloor}f(\lfloor x\rfloor +1)&x> 1\end{cases}$$
and verify that is is indeed continuous (the only "problematic" points being $x\in\Bbb N$).
Also, $g$ is clearly positive.
For $n\le x<n+1$ we have $g(x)\le \frac 1nf(n+1)$ and thus $0<\frac{g(x)}{f(x)}\le \frac 1n$. We conclude that $\lim_{x\to\infty}\frac{g(x)}{f(x)}=0$.

Alternatively: The function $h(x)=\frac{f(x)}{\max\{1,x\}}$ is also positive and strictly decreasing. It suffices to find continuous $g$ with $0<g(x)<h(x)$. Such $g$ can be found by "mollification", e.g. simply
$$g(x)=\int_0^1 h(x+t)\,\mathrm dt $$
The key point is that the monotonicity of $h$ implies integrability and that the integral is continuous as a function of $x$.

For simplicity, only consider $x \ge 0$. Since $f(x)$ is strictly decreasing, we know there are countable jump points $x_0 = 0 < x_1 < x_2 < ..<x_n < ...$ for $f(x)$. $f(x)$ is continnous in $(x_i, x_{x+1})$, and f(x) has finite left limit $f(x_i^+)$ at $x_i$ and the finite right limit $f(x_{i+1}^-)$ at $x_{i + 1}$.

We notice that $f(x_i^-) > f(x_i^+)$

Now we can construst a continuouds function $G(x)$ from f(x) by connecting $f(x_i - \epsilon_i)$ to $f(x_i^+)$ with the straight line, otherwise, G(x) = f(x), where $0 < \epsilon_i < (x_i - x_{i - 1})$.

The final $g(x) = G(x)/( x + 1)$.  Clearly,
$$0 <= g(x) \le f(x),$$ and $$limit_{x \to \infty} \frac{g(x)}{f(x)} \le limit_{x->\infty} \frac{1}{1 + x} = 0 $$


How can we show that $2^3+1^3+3^3+4^3+\cdots+L_n^3={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$ In my recent post I was intended to asked this question

$...,3,-1,2,1,3,4,7,...$ for $n=...,-2,-1,0,1,2,3,4,...$; It is the Lucas numbers

Let the sum of the cube of Lucas series be $S_L$

$S_L=2^3+1^3+3^3+4^3+\cdots+L_n^3$,

show that it has a closed form $$S_L={L_nL_{n+1}^2+5(-1)^n[L_{n-1}+2(-1)^n]+9\over 2}$$

I try: I can't think  of any simple identities to use.

This is the only one might have some sort of link to it,

$L_{n+1}^3+L_{n+2}^3={L_{n+3}\over 2}(L_n^2+L_{n+1}^2+L_{n+2}^2)$

$2^2+1^2+3^2+\cdots+L_n^2=L_nL_{n+1}+2$

Any further hints?

Some relations about Fibonacci and Lucas numbers:


Does $h \circ f = I = f \circ g \implies h = g$? Suppose that the identity function $I$ is defined as $I:X \to X$ such that $\forall x \in X$, $I(x) = x$.
I was hoping that $f(g(x)) = h(f(x)) = x \implies g(x) = h(x)$ for a proof I am constructing. I am stuck as to where to begin. Any help would be greatly appreciated. Cheers.

$$
h = h\circ\operatorname{id}_X = h\circ (f\circ g) = (h\circ f)\circ g = \operatorname{id}_X\circ g = g
$$

If $h(f(x))=x$ then it is also true that $f(h(x))=x$ now if $f$ is inyective we are done. If not there is no such inverse of $f$ for all the reals because then if $a$ and $b$ has the same image in $f$, $h$ will have to give different values for those images, which by definition is not a function.


Is this shape definition an ellipse? I want to define an ellipse-like shape with two radii. The radius at 0 degrees is 3. The radius at 90 degrees is 2. But in order to close the loop, we have to specify some smooth transition between them.

If the transition is linear (radius at 45 degrees is 2.5), does that make it an ellipse?

EDIT: "Linear" means the radius decreases linearly from 3 to 2 as the angle increases linearly from 0 to 90 degrees. So the equation is r = 2 + t/90 for quadrant 1.

"The angle" means the angle made by the positive x axis and whatever radius we're looking at, centered at the origin. The loop I'm describing is also centered at the origin.

A common way to define an ellipse centered at origin is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a$ is the horizontal semi-minor axis, and $b$ is the vertical semi-minor axis.

You often see this written parametrically as
$$\begin{cases}
x(t) = a \cos(t) \\
y(t) = b \sin(t) \end{cases}$$
with $0 \le t \le 2 \pi$, but it is important to notice that $t$ is not an angle here. If we use $P = (x(t), y(t))$, $O = (0, 0)$, and $A = (1,0)$, the angle $\angle AOP$ is actually $\theta$,
$$\theta = \arctan\left(\frac{y(t)}{x(t)}\right) = \arctan\left(\frac{b \sin(t)}{a \cos(t)}\right) = \arctan\left(\frac{b}{a}\tan(t)\right)$$
i.e. $\theta$ is the angle from positive $x$-axis counterclockwise to the point. Conversely,
$$t = \arctan\left(\frac{a}{b}\tan(\theta)\right)$$

In the circle case, $a = b = r$, and we have $t = \theta$, so for circles,
$$\begin{cases}
x(\theta) = r \cos(\theta) \\
y(\theta) = r \sin(\theta) \end{cases}$$
is true.

If we assume the angle is measured counterclockwise from the positive $x$ axis as usual, then you could use
$$\frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \tag{1}\label{1}$$

In general, we can also describe the same ellipse as a function of $x$, by solving equation $\eqref{1}$ for $y$:
$$y(x) = \pm \frac{b}{a}\sqrt{a^2 - x^2} = \frac{2}{3}\sqrt{3^2 - x^2}$$

At 45° from origin (center of the ellipse) we have $y = x$, and the distance to the center is $$r(45°) = \sqrt{x^2 + (y(x))^2} = \sqrt{\frac{2 a^2 b^2}{a^2 + b^2}} = \sqrt{\frac{72}{13}} \approx 2.3534$$

So, if your shape has $r(45°) = 2.5$, it is not an ellipse.

The principal radii of curvature are

$$ \frac{a^2}{b}, \frac{b^2}{a} $$

By looking at the ellipse evolute center and radii of curvature positions close to that of evolute can be chosen for a smooth transition tangential junction find two profiled ellipse approximations of any eccentricity.

Sketches of  EvoluteBasedEllipses


If $f$ is continuously differentiable in $[a,b]$, $f(a)=f(b)$, and $f'(a)=f'(b)$, then there exist $a<x_1<x_2<b$ such that $f'(x_1) = f'(x_2)$. This problem is from Apostol's Mathematical Analysis:

  Let $f$ be continuously differentiable in $[a, b]$. If $ f(a) = f(b)$ and if
  $f^{'}(a) = f^{'}(b)$, then prove that there exists $x_1$ and $x_2$ in $(a, b)$ such that $x_1\neq x_2$ but $f^{'}(x_1) = f^{'}(x_2)$.

My try:

By Rolle's Theorem $\exists x_0\in (a,b) $ such that $f^{'}(x_0)=0$.
How to guarantee existence of $x_1,x_2 $ from here?

Can it be solved from a geometrical point of view?

As you said, by Rolle's Theorem, we are guaranteed existence of $c \in (a,b)$ such that $f'(c) = 0$.  WLOG, let $f'(a) = f'(b) = k > 0$.  By the continuity of $f'$ and the intermediate value theorem, we get existence of $c_1 \in (a,c)$ such that $f'(c_1) = k_1$ where $0<k_1<k$.  Similarly, we get $c_2 \in (c,b)$ such that $f'(c_2) = k_1$.

Now, if $f'(a) = 0$, then let's look at some cases.  If $f(c) \neq f(a)$, then the mean value theorem gives existence of $d \in (a,c)$ such that $\alpha = f'(d) = \frac{f(c) - f(a)}{c - a}\neq 0.$  Now, we can apply the above argument again.

If $f(c) = f(a) = f(b)$, then if $f'$ is nonzero at some point in the interval, the above argument works.  Otherwise, $f'$ is $0$ and the result follows.

(i) Let's suppose that $f'(a)>0$ (then we'll see the case $f'(a)=0$).
Because of (i), you can find a $\eta > 0$ (as small as it is) that we have $f(a + \eta) > f(a)$ and $f(b - \eta) < f(b)$.

$f$ is continuous on $[a + \eta, b-\eta]$, therefore there exists $c \in [a + \eta; b - \eta]$ so that $f(c) = f(a) = f(b)$. ($f(c)$ in $[f(a+\eta], f(b-\eta)])$.

$f$ is continuous on both intervals $[a, c] $ and $[c, b]$ and reach its maximums in a $x_1$ and $x_2$ ($x_1$ in $(a, c)$ and $x_2$ in $(c, b)$ because $a, c$ and $b$ are not the maximum since $a+\eta$ and $b-\eta$).

We have $f'(x_1) = f'(x_2) = 0$

Opposite reasonning for $f'(a)<0$ leading to the same conclusion

(ii) if $f'(a) = 0$ because of Rolle's theorem you know that you have a $c$ in $(a,b)$ that $f'(c) = 0$. Assume that $f(c) = f(a)$ (then we reapply this reasonning on $a$ and $c$ instead of $a$ and $b$ until finding a $c_n$ that $f(c_n)$ is different from $f(a)$, if we can not find one then $f$ is constant and any $c_1<c_2$ in $(a, b)$ will do)

We have a $f(a) < f(c)$ and $f'$ continuous on $[a, c]$ and $f'(a) = f'(c) = 0$ but $f'$ not equal to $0$ and it exists a $c_1$ in $(a, c)$ so that $f'(c_1)>0$ because $f(a) < f(c)$.

Because of that and $f'$ continuous on this interval, $f'$ reaches its maximum in $d$ in $(a;c)$ and because of its continuity $f'$ reaches $f'(d)/2$ in two points $x_1,x_2$, where $a<x_1<d<x_2<c<b$.


Differential equation : $y' = (x+1)/(xy+x)$ So, I have the following differential equation to solve : $$y' = \frac{x+1}{xy+x}$$
After several steps, I get here :

$t^2 + 2t = 2x + 2ln(x) + c$

How do I isolate $t ?$

thank you!

By the way, $ t=f(x)=y$

let $W$ denote Lambert's W-function. after taking exponentials of both sides:
$$
Ae^{\frac12(t+1)^2} = xe^x =W^{-1}(x)
$$
so we have
$$
x = W(Ae^{\frac12(t+1)^2})
$$
if this is correct, then we have:
$$
t = \sqrt(2\log( A^{-1}W^{-1}(x)))-1
$$


What are the interesting applications of hyperbolic geometry? I am aware that, historically, hyperbolic geometry was useful in showing that there can be consistent geometries that satisfy the first 4 axioms of Euclid's elements but not the fifth, the infamous parallel lines postulate, putting an end to centuries of unsuccesfull attempts to deduce the last axiom from the first ones.

It seems to be, apart from this fact, of genuine interest since it was part of the usual curriculum of all mathematicians at the begining of the century and also because there are so many books on the subject.

However, I have not found mention of applications of hyperbolic geometry to other branches of mathematics in the few books I have sampled. Do you know any or where I could find them?

Maybe this isn't the sort of answer you were looking for, but I find it striking how often hyperbolic geometry shows up in nature. For instance, you can see some characteristically hyperbolic "crinkling" on lettuce leaves and jellyfish tentacles:![

My guess as to why this shows up again and again (and I am certainly not a biologist here, so this is only speculation) is that hyperbolic space manages to pack in more surface area within a given radius than flat or positively curved geometries; perhaps this allows lettuce leaves or jellyfish tentacles to absorb nutrients more effectively or something.

EDIT: In response to the OP's comment, I'll say a little bit more about how these relate to hyperbolic geometry.

One way to detect the curvature of your surface is to look at what the surface area of a circle of a given radius is. In flat (Euclidean) space, we all know that the formula is given by $A(r) = \pi r^2$, so that there is a quadratic relationship between the radius of your circle and the area enclosed. Off the top of my head, I don't know what the formula is for a circle inscribed on the sphere (a positively-curved surface) is, but we can get an indication that circles in positive curvature enclose less area than in flat space as follows: the upper hemisphere on a sphere of radius 1 is a spherical circle of radius $\pi/2$, since the distance from the north pole to the equator, walking along the surface of the sphere, is $\pi/2$. In flat space, this circle would enclose an area of $\pi^3/4 \approx 7.75$. But the upper hemisphere has a surface area of $2 \pi \approx 6.28$.

By contrast, in hyperbolic space, a circle of a fixed radius packs in more surface area than its flat or positively-curved counterpart; you can see this explicitly, for example, by putting a hyperbolic metric on the unit disk or the upper half-plane, where you will compute that a hyperbolic circle has area that grows exponentially with the radius.

So what happens when you have a hyperbolic surface sitting inside three-dimensional space? Well, all that extra surface area has to go somewhere, and things naturally "crinkle up". If you are at all interested, you can crochet hyperbolic planes (see, for instance, this article of David Henderson and Daina Taimina), and you'll see how this happens in practice.

I am not a mathematician, but my "Barron's dictionary of Mathematics Terms'  Second Edition
  1995 copyright edited by Douglas Dowing says one example of  hyperbolic function use is
"catenary";  catenary: study of the curve formed by a flexible rope hanging between two points.


Explain why perpendicular lines have negative reciprocal slopes I am not sure how to explain this. I just know they have negative reciprocals because one one line will have a positive slope while the other negative.

Translate two lines so that their intersection is the origin and then take two vectors along each line, say $u=(1,k_1), v=(1,k_2)$. The two lines are perpendicular if and only if $u\perp v$, viz
$$u\cdot v=1+k_1k_2=0$$
This explains why $k_1$ is the negative reciprocal of $k_2$.

Two 90 degree rotations will snap back to how it was originally:
Suppose you have line A, line A' which is perpendicular to A, and line A'' which is perpendicular to A' (all on the Cartesian plane). A and A'' are parallel (since A'' is A rotated ±90±90 degrees, and even the ± is a silly distinction as rotating 90 degrees clockwise is the same as rotating 90 degrees counterclockwise), which means that if perpendicular slopes have a hard and fast formula, they had better be reciprocals or negative reciprocals (or the same slope, but that's garbage), as otherwise p(p(m)), the perpendicular of the perpendicular of the slope, will not equal the original slope. By inspection, in a pair of perpendicular lines, one has a positive slope and one has a negative slope (I'm not account for the student who asks "What about zero?"). Then the only valid option is negative reciprocals.

In my opinion, this is the best way to show a middle school student that will allows them to teach it to other middle school students (but geometry/trigonometry should eventually show it formally). Rotating a piece of paper 90 degrees and describing how it transforms rise and run is my second favorite.


Another method of finding area of hypocycloids I was finding the are the of hypocycloids. Then it struck me that apart from integration, there could be another method of finding the area of the hypocycloid with different curves. But the problem is I am not getting my answer right. So could somebody please help me tell if my logic is wrong altogether or if I am doing some other mistake.

Here is the another method I am talking about:-

Take case of a deltoid - we can make a deltoid by taking an equilateral triangle, and on all three of its vertices drawing circles whose radius is half the side of the triangle. Now the figure left in the middle is a deltoid. Similarly, we could use n number of sides in the regular polygon to draw hypocycloids with n cusps. (see the pictures below - the red coloured drawing in the middle is the hypocycloid)

Is this idea wrong?
Thank you so much :)

I think you assume that hypocycloids consist of circle arcs, but this is not the case:

Consider e.g. the Astroid:

It is an algebraic curve of degree 6, defined by $$ (x^2+y^2-a^2)^3+27a^2x^2y^2=0 $$ which is nowhere close to circle.

To come back to your example: The Deltoid is defined by this equation of degree 4:

$$(x^2+y^2)^2+18a^2(x^2+y^2)-27a^4 = 8a(x^3-3xy^2)$$ which also is certainly not composed of circles.

I used a software to produce the hypocycloid and then I tried this with an Astroid. Here is what happened:

Hypocycloid been drawn

As you can see, the circumference of the circle doesn't go over the shape of the hypocycloid. It doesn't work like that unfortunately.


Asymptotics of the maximum of quantized standard normals This is a problem from my measure theoretic probability class.

Problem:
Given independent standard normals $Z_1,...Z_n$, let $X_i$ be the nearest integer to $Z_i$. Let $M_n$ be the maximum of $\{X_i,..., X_n\}$. Show that there exists an integer sequence $\{a_n\}$ and a sequence of probabilities $\{p_n\}$ such that $P(M_n = a_n) \sim  p_n$ and $P(M_n = a_n + 1) \sim 1 - p_n$. The symbol '$\sim$' means that the ratio of the left and right tends to 1 as $n$ goes to infinity.

Show that $p_n$ does not converge as $n$ goes to infinity.

Context:
The professor assigned this problem after we discussed how the CDF of the maximum of standard normals can asymptotically be written as a double exponential. He took $x = \sqrt{ 2 \log(n) - \log(\log(n)) + c }$ in order to show that the asymptotic distribution of the max of gaussians is $\exp(- \exp(-c/2) / (2 \sqrt{2 \pi}))$. He then went on to say that for discrete random variables "everything breaks."

My ideas:
I'm not really sure where to start, especially considering that the examples from class were all for continuous random variables. I think I understand the gist of the proposition: asymptotically the maximum of the $X_i$ will be in a window of two integers wp 1. The choice of $x$ for the continuous case is exotic and I think that the discrete case will require similar "magic."

Any hints or ideas?

Thanks!

One looks for an integer sequence $(a_n)$ such that, when $n\to\infty$, $P[M_n\leqslant a_n-1]\to0$ and $P[M_n\leqslant a_n+1]\to1$.
For every $x$, $P[M_n\leqslant x]=\Phi(x+\frac12)^n$. When $x\to+\infty$, $\Phi(x)\to1$ hence
$$
\log\Phi(x)\sim-(1-\Phi(x))\sim-1/(\sqrt{2\pi}x\mathrm e^{x^2/2}).
$$
Thus, one looks for some integer sequence $(a_n)$ such that $x\mathrm e^{x^2/2}\ll n$ for $x=a_n-\frac12$ and $x\mathrm e^{x^2/2}\gg n$ for $x=a_n+\frac32$. Equivalently, one asks that
$$
a_n\mathrm e^{a_n^2/2-a_n/2}\ll n\ll a_n\mathrm e^{a_n^2/2+3a_n/2},
$$
that is,
$$
\tfrac12a_n^2+\tfrac12a_n\pm a_n+\log a_n-\log n\to\pm\infty.
$$
Assume that
$$
a_n=\lfloor\sqrt{2\log n}\rfloor,
$$
then $\sqrt{2\log n}-1\leqslant a_n\leqslant\sqrt{2\log n}$ hence
$$
\tfrac12a_n^2-\tfrac12a_n\leqslant \log n-\tfrac12\sqrt{2\log n}+O(1),
$$
and
$$
\tfrac12a_n^2+\tfrac32a_n\geqslant \log n+\tfrac12\sqrt{2\log n}+O(1),
$$
hence the sequence $(a_n)$ yields the result.

The fact that $a_n$ stays constant during longer and longer intervals, then jumps to $a_n+1$, and similar estimates to those above, probably entail that $\liminf\limits_{n\to\infty}p_n=0$ and $\limsup\limits_{n\to\infty} p_n=1$.

Since "everything breaks" for discrete random variables, do this in terms of the continuous random variables $Z_j$.  Start by relating $M_n$ to the maximum of the $Z_j$.


There exists strictly increasing $\{x_n\}$ that converges to $\sup E$ I need to prove that, if $E \subseteq \mathbb{R}$ is a non-empty bounded set and $\sup E \not\in E$ then there exists a strictly increasing sequence $\{x_n\}$ that converges to $\sup E$ such that $x_n \in E$ for all $n \in \mathbb{N}$.

I've been trying to find a clue in the textbook, but couldn't. I don't even know how to start the proof. Could someone please give a clue?

Hint

Firstly, choose any element $x_1 \in E$. As stated, $\sup E \notin E \implies x_1 < \sup E$. Using the approximation property for suprema, for each $n \geq 2 \exists x_n \in E$ such that $\max(\sup E -\frac{1}{n},x_{n-1})<x_n<\sup E$ so that a sequence $x_1<x_2<x_3<...$ is strictly increasing.

Then go on to apply the Squeeze theorem and you should be home and dry.

Edit Show that the sequence above is increasing

Start at $n=1$ there exists an $x \in E$ such that $\sup  E  − 1 < x$. Denote this $x = x_1$. Now, for $n=2$ there exists an $x \in E$ such that $\sup E − \frac{1}{2}  < x_2$. Clearly $x_2 > x_1$. This works for any $n$ so there exists a sequence $x_n \in E$ with the property that $\sup E - \frac{1}{n}<x_{n}$

I would phrase a hint as this:

If there were no sequence approaching $\sup E$, how could $\sup E$ be the supremum?


Exponential of a Quartic I know there is a related post regarding this, but does anyone know a 'closed' form solution for the integral
\begin{equation}I=\int_{-\infty}^{\infty} dx\,e^{ax-bx^{4}}\end{equation}
I know you can do a series expansion of the quartic term, but I'd like to find a way that avoids that. The integral should converge as $e^{-bx^{4}}$ dominates at large $x$ and so the integrand quickly goes to zero (for $b>0$ obviously). Any help/ references would be greatly appreciated.

Integrating by parts leads to $$I=\left.x\cdot e^{ax-bx^4}\right\vert_{-\infty}^\infty-\int_{-\infty}^\infty x(a-4bx^3)e^{ax-bx^4}\,dx=-a\frac{\partial I}{\partial a}-4b\frac{\partial I}{\partial b},$$but $\,\partial I/\partial b=-\int_{-\infty}^\infty x^4e^{ax-bx^4}\,dx\;$ also equals $-\partial^4I/\partial a^4$, so $I(a)+aI'(a)=(aI(a))'=4bI^{(4)}(a)$, or $\;aI(a)=C+4bI'''(a)$. Since $I(a)$ is even (and entire, of course), $I'(0)$ and $I'''(0)=0$ and so $C=0$; and using the integral formula for the gamma function, $$I(0)=2\int_0^\infty e^{-bx^4}\,dx=2\int_0^\infty e^{-y}b^{-1/4}\frac{dy}{4y^{3/4}}=2^{-1}b^{-1/4}\Gamma(1/4)$$ and $$I''(0)=2\int_0^\infty x^2e^{-bx^4}\,dx=2\int_0^\infty b^{-1/2}y^{1/2}e^{-y}\frac{b^{-1/4}dy}{4y^{3/4}}=2^{-1}b^{-3/4}\Gamma(3/4)$$I plugged all of this into Wolfram|Alpha and got the same result as @GEdgar (with $x,a$ replacing, respectively, $a,b$).

This is called Gaussian, it is not integrable by elementary means. First you need to make the substitution $a \rightarrow ax$ so that the integral is easier. Now the integral is $$\int _{-\infty} ^{\infty} e^{ax^{2} -b x^{4}} dx$$
Now this is an easier integral if you make the substitution $x \rightarrow \sqrt{x}$ because the integral is now
$$\int _{-\infty} ^{\infty} e^{ax^ - b x^{2}} dx$$
Ok, now we are really on track, because you can complete the square. Rewrite the integral as
$$\int _{-\infty} ^{\infty} e^{-b(x-\frac{a}{2b})^{2}}e^{-(ax)^2} dx$$
Now we are almost done, just pull out the last factor to get
$$e^{-(ax)^2} \int _{-\infty} ^{\infty} e^{-b(x-\frac{a}{2b})^{2}} dx = e^{-(ax)^2} \sqrt{\pi}$$
Now we have to evaluate somewhere because this is a function of x whereas the original integral is not. I picked 0, but any other value should work. So, the answer is $\sqrt{\pi}$.


Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice A topological space is called quasi-compact if every open cover of it has a finite subcover.
Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$.
The usual proof that $Z$ is quasi-compact uses a maximal filter, hence Axiom of Choice. Can we prove it without using Axiom of Choice?

Edit(Apr. 14, 2014)
If I am not mistaken, I have come up with a proof without using Axiom of Choice.
I would like to post it as an answer.

Edit(Apr. 15, 2014)
I was mistaken. As Andres Caicedo pointed out, I used AC without noticing it in my "proof".

Here is a proof which is choice free. I should also mention that the proof using ultrafilters appears in Herrlich's The Axiom of Choice, where later he says that it can be modified to work without the axiom of choice using some ideas from other proofs. I did not check that claim in details, though.

First, let us prove the following lemma:

Lemma. Let $X$ be a topological space and $\cal B$ a  basis for the topology. $X$ is quasi-compact if and only if every cover with elements of $\cal B$ has a finite subcover.
Proof. One direction is trivial, if $X$ is quasi-compact, certainly every cover with elements of $\cal B$ has a finite subcover. In the other direction, suppose that $\mathcal U=\{U_i\mid i\in I\}$ is an open cover, consider $\cal U'$ to be the refined cover, $\{V\in\mathcal B\mid\exists i\in I: V\subseteq U_i\}$. Then $\cal U'$ is an open cover as well, since every point in $U_i$ lies within some element of $\cal B$. Let $V_1,\ldots,V_n\in\cal U'$ be a finite subcover, then we can choose $U_1,\ldots,U_n\in\cal U$ such that $V_i\subseteq U_i$ for all $i\leq n$, and this is a finite subcover as wanted. $\ \square$

Now we can prove our theorem.
Let $X,Y$ be two quasi-compact spaces, and let $\cal U$ be an open covering of $X\times Y$ using rectangles (that is, sets of the form $U\times V$ where $U$ is open in $X$ and $V$ open in $Y$). We say that $A\subseteq X$ is adequate [for $\cal U$] if $A\times Y$ has a finite subcover in $\cal U$. Our goal is to show that $X$ is adequate, then $X\times Y$ can be covered by a finite subcover of $\cal U$.
First we show that if $x\in X$, then there is some $U\subseteq X$ which is open and $x\in U$ such that $U$ is adequate. Let $U_1\times V_1,\ldots,U_n\times V_n$ be a finite subcover such that $\{x\}\times Y\subseteq U_1\times V_1\cup\ldots\cup U_n\times V_n$ (such finite cover exists since $\{x\}\times Y$ is quasi-compact). Now consider $U=\bigcap_{i=1}^n U_i$, then $U$ is open as a finite intersection of open sets, and non-empty since $\{x\}\in U$, as wanted. $U$ is adequate since given any $(u,y)\in U\times Y$ we have that for some $i\leq n$ it is true that $(x,y)\in U_i\times V_i$, so $(u,y)\in U_i\times V_i$ as well (since $u\in U_i$). Therefore $U_1\times V_1,\ldots,U_n\times V_n$ is a cover of $U\times Y$.
Next we note that the finite union of adequate sets is adequate (as it is covered by the [finite] union of the [finite] subcovers of the adequate sets).
Now $\{U\subseteq X\mid U\text{ is open and adequate}\}$ is an open cover of $X$, by the above fact that every $x\in X$ has an adequate neighborhood, and by quasi-compactness of $X$ it has a finite subcover. And therefore $X$ is the finite union of adequate sets and it is adequate as well.
Finally, since rectangles form a basis for the product topology, from the lemma above we have that $X\times Y$ is indeed quasi-compact, as wanted. $\quad\square$
(I found the proof on ProofWiki sometime in the past.)

Let $X, Y$ be quasi-compact spaces, $Z = X\times Y$.
We will prove that $Z$ is quasi-compact without using Axiom of Choice.
Suppose $(W_\alpha)_{\alpha\in A}$ is an open cover of $Z$.
Then $W_\alpha = \bigcup_{\beta\in B_\alpha} U_{\alpha\beta} \times V_{\alpha\beta}$, where $U_{\alpha\beta}$ is an open subset of $X$ and $V_{\alpha\beta}$ is an open subset of $Y$.
Let $x \in X$. Then $\{x\}\times Y$ is quasi-compact.
Since $(U_{\alpha\beta} \times V_{\alpha\beta})_{(\alpha,\beta)\in A\times B_\alpha}$ is an open cover of $\{x\}\times Y$, there exists a finite subcover $U_i(x) \times V_i(x), i = 1, \cdots , n_x$ of $\{x\}\times Y$, where each $U_i(x) \times V_i(x)$ is of the form $U_{\alpha\beta} \times V_{\alpha\beta}$ for some $(\alpha, \beta) \in A\times B_\alpha$.
Let $U(x) = U_1(x) \cap \cdots \cap U_{n_x}$.
Since $(U(x))_{x\in X}$ is an open cover of $X$, there exists a finite subcover $U(x_1),\cdots, U(x_m)$ of $X$.
Then $U(x_i) \times V_k(x_i)$, $i = 1, \cdots, m, k = 1,\cdots, n_{x_i}$ is a finite cover of $X\times Y$.
Then for every pair $(i, k)$, there exist $\alpha \in A$ and $\beta \in B_\alpha$ such that
$U(x_i) \times V_k(x_i) \subset U_{\alpha\beta} \times V_{\alpha\beta} \subset W_\alpha$.
Hence there exists a finite subcover of $(W_\alpha)_{\alpha\in A}$.
QED


Number of vectors in a set & span of a set I needed clarification on a linear algebra question that I had:

Given the matrices $v_1 = \begin{bmatrix}
    1 \\
    1 \\
    1 \\
\end{bmatrix}, $ $v_2 = \begin{bmatrix}
    1 \\
    -1 \\
    1 \\
\end{bmatrix}$ and $v_3 = \begin{bmatrix}
    1 \\
    1 \\
    -1 \\
\end{bmatrix}$,

1) How many vectors does the set {${v_1, v_2, v_3}$} have?

2) How many vectors are in Span{$v_1, v_2, v_3$}?

I think the answer to #1 is 3, simply because there are three matrices, and the answer to #2 is infinite, since there are an infinite number of linear combinations that can be made using these vectors.

I am uncertain of these answers, though.

Construct the matrix
$$
\mathbf{A} = \left[ \begin{array}{ccc}
  v_{1} & v_{2} & v_{3}
\end{array}\right]
 =
\left[ \begin{array}{rrr}
    \phantom{-}1 &  1 &  1 \\
    1 & -1 &  1 \\
    1 &  1 & -1
\end{array}\right] .
$$
Because $\det \mathbf{A} = 4 \ne 0$, the vector set is linearly independent.

The span of the vector set is
$$
\text{span}
  \left\{ v_{1}, v_{2}, v_{3} \right\}
= \mathbb{R}^{3}.
$$
These vectors are a linearly independent span, also called a minimal spanning set for $\mathbb{R}^{3}$.


Does there exist an integer $s$ such that every integer $> 1$ can be written as a sum of at most $s$ primes? Does there exist an integer $s$ such that every integer $> 1$ can be written as a sum of at most $s$ primes ?

Terrence Tao has proven that every odd number greater than one is the sum of at most five primes. Hence, using $3$ as a prime that we can substract to any even number greater than $5$, and noticing that $2$ is prime and $4=2+2$, every odd number is the sum of at most six primes.

As a conclusion, every number greater than one is the sum of at most six primes.

Of course, this is a very loose approximation. If it is proven, Goldbach's conjecture would allow us to lower the number of needed primes to $3$.


A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$ I couldn't relate this question to any of the topics specifically , I found this in a miscellaneous math problems book(non-calculus) . Here's how it goes,  A book has a few pages on which page numbers are written. Someone has torn one page out of it and now average of all page numbers is $\frac{105}{4}$. Answer the following:

(i) If the total numbers of pages in book is n then find the value of

$$\sum_{r=1}^{10} \biggl\lfloor{\frac{n+r}{r+1}}\biggr\rfloor\,.$$
 OPTIONS:
(A)100 (B)107 (C)105 (D)82

(ii) If the line $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is drawn ,then the total number of points with integral co-ordinates enclosed within the region bounded by $x=0,y=0$ and $x+y=\bigl\lfloor\frac{n}{3}\bigr\rfloor$ is -----?
(A)105 (B)153 (C)59 (D)78

STATUS: No clue how to start.

Any help is welcome.

OK, so here is a solution to (i) in case we know that when a page is torn out and odd and the following even page number is removed.

Given the book has $n$ pages the sum of the page numbers will be
$$
T_n=1+2+...+n=\frac{n(n+1)}{2}
$$
known as the $n$'th Triangular Number. So assume that page numbers $x$ and $x+1$ have been torn out where $x$ is odd. Then we can write
$$
\frac{105}{4}=\tfrac{1}{n-2}\left(T_n-(2x+1)\right)
$$
Plugging in the formula for $T_n$ from above and solving for $x$ then yields
$$
x=\frac{1}{4}n^2-\frac{103}{8}n+\frac{103}{4}
$$
which is a quadratic expression in $n$ with zeros $n=\frac{103\pm\sqrt{8961}}{4}$ which approximately is $n=2.08$ and $n=49.42$. On the other hand it is obvious that $x<n$ which then in turn yields a quadratic inequality in $n$ that can be solved to see that $n<\frac{111+\sqrt{10673}}{4}\approx 53.58$.

Having a slightly closer look at the expression for $x$ one realizes that $n-2$ must be divisible by $8$ for $x$ to be an integer. So unless we take $n=2$ (which actually works) we must have $49.42<n=50<53.58$ for $n$ to be an integer satisfying all requirements in that interval. Plugging $n=50$ into the expression for $x$ then yields $x=7,x+1=8$, and you can check as I did in the comments that the average of the remaining pages is really $\frac{105}{4}$. With $n=50$ one gets
$$
\sum_{i=1}^{10}\left\lfloor\frac{50+r}{r+1}\right\rfloor=105
$$
So option (C) answers the first question correctly.

Part (ii)

Still using $n=50$ we get $\lfloor\frac{50}{3}\rfloor=16$ so that the line is $x+y=16$ or $y=16-x$. Together with the axes $y=0$ and $x=0$ this encloses a triangle with $17$ lattice points on the line $y=16-x$ since you can start from $(0,16)$ and move down right step-by-step to 'visit' 17 lattice points before you hit the x-axis. Similarly you hit $16$ lattice points following the same procedure from $(0,15)$. With this we get the answer to (ii) which is:
$$
T_{17}=17+16+...+1=\frac{17\cdot 18}{2}=153
$$
So option (B) answers the second question correctly.

For 1, we have to assume the pages are  consecutively numbered from $1$ to $n$ before a single page is removed (which removes two page numbers).  The average of all the page numbers is $\frac {n+1}2$.  The remaining number of pages must be a multiple of $4$ so we can get the denominator in the average. Clearly it will not move the average much, so the average is about $26$ and $n$ is about twice that.  So we guess $n=54$ will work.  The sum of all the pages is $\frac {54 \cdot 55}2=1485$.  The sum of the remaining pages if $52 \cdot \frac {105}4=1365$, so the sum of the removed page numbers is $120$.  Unfortunately there is no page with that sum.  Trying $n=50$ gives a starting sum of $1275$ and an ending sum of $1260$, leading to removal of the page that includes numbers $12$ and $13$.  It is then arithmetic to evaluate the sum, getting $105$

An approach without guessing, but with more work is as follows:  If we let $m$ be the page number removed, we have $$\frac {\frac {n(n+1)}2-m}{n-1}=\frac {105}4\\
\frac {n(n+1)}2-m=\frac {105(n-1)}4\\2n^2+2n-4m=105n-105\\2n^2-103n+105-4m=0\\n=\frac 14\left(103 +\sqrt{103^2+32m-620}\right)$$ and searching for $m$ that will make the square root an integer will get there.


How does the classification using the 0-1 loss matrix method work? In this machine learning lecture the professor says:

  Suppose $\mathbf{X}\in\Bbb R^p$ and $g\in G$ where $G$ is a discrete
  space. We have a joint probability distribution $\Pr(\mathbf{X},g)$.

  Our training data has some points like:

  $(\mathbf{x_1},g_1)$, $(\mathbf{x_2},g_2)$, $(\mathbf{x_3},g_3)$ ...
  $(\mathbf{x_n},g_n)$

  We now define a function $f(\mathbf{X}):\Bbb R^p \to G$.

  The loss $L$ is defined as a $K\times K$ matrix where $K$ is the
  cardinality of $G$. It has only zeroes along the main diagonal.

  $L(k,l)$ is basically the cost of classifying $k$ as $l$.

  An example of $0-1$ loss function:

  \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}

  $\text{EPE}(\hat{f}) = \text{E} [L(G,\hat{f})]$ (where $\text{EPE=
 Expected Prediction Error}$)

  $=E_\mathbf{X} E_{G/\mathbf{X}} \{L[G,\hat{f}]|\mathbf{X}\}$

  $\hat{f}(\mathbf{x})=\text{argmin}_g\sum_{k=1}^{k}L(k,g)\text{Pr}(k|\mathbf{X}=\mathbf{x})=\text{argmax}_g\text{Pr}(g|\mathbf{X=x})$

  $\hat{f}(\mathbf{x})$ is the Bayesian Optimal Classifier.

I couldn't really follow what the professor was trying to say in some of the steps.

My questions are:

Suppose our loss matrix is indeed: \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} What it the use of this matrix? What does classifying $k$ as $l$ even mean? Then how do we read off the loss for (say) a certain input $\mathbf{x_i}$ from the matrix?
I couldn't understand what $\hat{f}$ and $\text{EPE}(\hat{f}\mathbf{(x)})$ stand for. Could someone please explain it with a simple example?

Consider a random variable $X$ and a radom variable $g $. $X $ is uniformly distributed over $[0,1] $ if $g=1$ and is uniformly distributed over  $[0.5,1.5] $ if $g=2$.  $g$ takes the value $1$ with probabity $0.1$. By this the joint distribution of $(X,g) $ is given.

Asssume that we have to guess $g $ after having observed $X $. If we miss then we have to pay $1$ rupy. If we don't then the penalty is $0$.

It is clear that if $X $ is below $0.5$ then our guess for $g $ is $1$. If $X $ is above $1$ then our guess for $g $ is $2$. But what do we do if $X $ falls between $0.5$ and $1$?

Now, define $\hat f $ the following way: let it be $1$ if $X <0.7$ and $2$ otherwise.

Could we do that any better?

Can you identify the current loss matrix?

Can you compute the expected loss belonging to the method given above?

Can you create a similar problem so that the loss matrix is the one given in the OP?

I think the loss matrix, for example in the ESL book page 20, the zero-one loss matrix, can be treated as a square look-up table. The number of rows and the number of columns are both the number of all possible classes. Assume it is $K \times K$.

In your example, $K = 3$. So there are three possible levels, let's say they are lvl-1, lvl-2, and lvl-3. If you have an observation of which the level is lvl-2 (truth), but your estimator is lvl-1. Then we got a penalty since this is a misclassification, and (lvl-2, lvl-1) corrsponds to the value that sits at the 2$_{nd}$ row and the 1$_{st}$ column, which is 1 in your table. You asked what's $k$ was classified as $l$. Here $k$ is the truth, which is lvl-2, and $l$ is an error, which is lvl-1 in my example.

In the ESL book, the EPE (Expected Prediction Error) is $$EPE = E[L(G,\hat{G}(X))]$$
, where $G$ and $\hat{G}(X)$ within $L(G, \hat{G}(X))$ can be treated as the row and column indices. Therefore $L(G, \hat{G}(X))$ can be considered as a random variable. The sources of its randomness are G and X.

Let's assume that we observed a $X$ and corresponding observed class $G$, then an estimator of $G$, that is the $\hat{G}(X)$, can be calculated by certain technique. Then we go to the 'look-up' table, check the row index by known G, and check the column index by estimator $\hat{G}(X)$. If $G = \hat{G}(X)$ then we have 0 'penalty', 1 otherwise. This is determined because we have already observed both $X$ and $G$, no randomness thereafter.

If we only observed a $X$, not $G$, you are still able to find a $\hat{G}(X)$, which is the column index, since you can just apply the technique on any $X$ as long as it is observed. But we don't know the 'penalty' because you need both row and column indices. Therefore we calculate the expectation of this random 'penalty'. The expectation is conditioning on the observed $X$. It is the weighted average of all possible 'penalties' (or ones, which means miscatches) that this 'penalty' can be, each 'penalty' being weighted according to the conditional probability of its occurence. For example, here is the weight for $k^{th}$ class: $Pr(g_{k}|X)$, of which $k \in \{1, 2, ..., K\}$. Now we have:

$$\tag{*} E[L(G, \hat{G}(X))|X]=\sum\limits_{k=1}^K [L(g_{k}, \hat{G}(X))Pr(g_{k}|X)]$$

The above "$| X$" means $X$ is known. But it is also random, which means it is unknown until we observe one. And we are trying to find the EPE, not a conditional expectation like (*). Therefore we need to find the expecation of the above conditional expectation, which is:
$$EPE = E_{X}\left(\sum\limits_{k=1}^K [L(g_{k}, \hat{G}(X))Pr(g_{k}|X)]\right)$$

So we should pick the estimator $\hat{G}(X)$ which minimizes the above EPE.

This is my understanding. I think there must be answer more mathematically solid.


find Number of connected components of $Y$? Let $Q,Q^c$ denotes  be the set of all rational and irrational numbers in $\mathbb {R}$. $Y=\mathbb {R^2}\setminus(Q×Q^C )$ with the usual subspace topology of $\mathbb {R^2}$.find  Number of connected components of $Y$?

My answer   :  i thinks  $Y = \mathbb{R} \times \mathbb{R} \cup  \mathbb{R} \times \mathbb{R}$. Since $Y$ is covered by lines in both coordinate directions,  it is path connected. so  $Y$ has one connected component.

is  its correct/incorrect ??

Pliz tell me

Notice that

$Y =\mathbb{R}^2\backslash(\mathbb{Q}\times\mathbb{Q}^c) = (\mathbb{R}\times (\mathbb{R}\backslash \mathbb{Q}^c))\cup((\mathbb{R}\backslash \mathbb{Q})\times \mathbb{Q}^c) = (\mathbb{R}\times\mathbb{Q})\cup(\mathbb{Q}^c\times\mathbb{Q}^c).$

So, if $y=(a,b)\in Y$ and $b\in \mathbb{Q}^c$, then $a\in\mathbb{Q}^c$ and $\{y\}$ is the connected component of $y$.

If, $y'= (a,b)\in Y$ and $b\in\mathbb{Q}$, then $a$ can be any real number, so $\{(t,b),\ t\in \mathbb{R}\}$ is the connected componet of $y'$.


Do you need real analysis to understand complex analysis? I'm debating whether I should take a course, in complex analysis (using Bak as a text).  I've already taken Munkres level topology and "very light" real analysis (proving the basic theorems about calculus) using the text Wade.

The complex analysis course is supposedly difficult and will even cover the Prime Number Theorem in the end.  Do you think it's better to take Rudin level real analysis first?

If the course teaches complex analysis from a geometric perspective-emphasizing the properties of analytic maps of the plane as a "calculus of oriented angles", as I did in my undergraduate complex analysis course-then believe it or not, you'll need very little if any real analysis except for certain results (like Cauchy's theorem and convergent series). For example, a good way to think of the derivative in the complex plane as a sequence of "infinitesimal" rotations of a tangent line to a circle centered at a point in the Argand plane-whereas the sequence of rotated tangent lines converges to the point by contracting in length along increasingly smaller subcircles. Also, most of the standard transformational geometry of the Euclidean plane has very elegant reformulations in terms of the standard analytic functions of the plane, such as the complex exponential in plane polar coordinates. If the course focuses on these aspects of elementary complex analysis, you'd be better off brushing up on your basic geometry then real analysis! However, if the course develops complex analysis via a rigorous development of the complex plane as a metric or normed space and focuses on infinite series, then that's a different story and you'll need a lot more rigorous real analysis to get comfortable with it.

'Real analysis' may refer to either 'elementary analysis' such as in Trench Real Analysis (RIP) or, well, real analysis such as in Royden Fitzpatrick Real Analysis. The former is necessary, but the latter is not except possibly a few topics such as uniform convergence, but I guess first courses in complex analysis will teach uniform convergence as if for the first time just as in real analysis. A few parts may depend on the school/course/textbook, but overall I believe elementary analysis is sufficient.


Independence in Coupon Collecting Problem If $T_i$ is the number of cards that we draw from a deck before seeing the $(i+1)$th new card (after seeing the ith new card), how can I show that Then $T_i$ and $T_j$ are independent (for i≠j)? I know the definition of independence, but I don't know how to get a workable second definition to compare to $P(T_j)P(T_i)$.

In both cases you have a geometric distribution.  If there are $n$ cards in total, the chance of success after having seen $i$ cards is $\frac {n-i}n$.  This clearly is independent of $j$


Geometrical Proof of a Rotation I wanna prove geometrically ( and not by linear algebra, doing transformations in the  bases ) the result of the rotation of a point. The proof should only include geometrical steps like using  similarity between triangles, pythagorean theorem and definition of cos and sen on a triangle, for example.

This picture clears my doubt    :

http://postimg.org/image/z60zv5d83/

Thanks in advance.

Let $(x,y)$ be the original coordinates, and $(x',y')$ the later one. Then $$x'=ax+by,~y'=cx+dy$$as they are linear combination. Notice that $$x^2+y^2=(x')^2+(y')^2$$Replacing $x',y'$ by $x,y$ in the above equation, we get $$a^2+c^2=b^2+d^2=1,~ab+cd=0$$Let $a=\cos\theta$ and do some work gives the result.

As required by OP, I will give an alternative method using vector decomposition:

Let $\vec{OA}=(x,y)$ be the original vector, $\vec{OB}=(x',y')$ the vector after rotation of angle $\theta$. Then
$$(x',y')=\vec{OB}=\vec{OC}+\vec{OD}=(x,y)\cos\theta+(y,-x)\sin\theta.~~\mbox{(DONE!)}$$

No transformation of bases is needed. Let $R:=R_\theta$ be the rotation about the origin by angle $\theta$.

All is needed, that this is a linear transformation. This can be viewed as a purely geometrical statement:

$0$ goes to $0$.
Rotation preserves line segments and takes triangles to triangles (corresponding to $R(a+b)=R(a)+R(b)$ for vectors $a,b$).
Rotation is exchangable with all zooms from the origin, i.e. $R(\lambda a)=\lambda R(a)$.

Let $e_1=(1,0)$ and $e_2=(0,1)$. Now, by definition of sine and cosine, we have that $R(e_1)=(\cos\theta,\sin\theta)$, and please check (geometrically, i.e. by drawing) that $R(e_2)=(-\sin\theta,\cos\theta)$.

Then, for our arbitrary point $P=(d,f)$, rotating it about $0$ is the same as rotating the $\vec{OP}$ vector, so by linearity, we have
$$R(P)=R(d\cdot e_1+f\cdot e_2)=d\cdot R(e_1)+f\cdot R(e_2)\,,$$
and we're there.


Deleting any digit yields a prime... is there a name for this? My son likes his grilled cheese sandwich cut into various numbers, the number depends on his mood.  His mother won't indulge his requests, but I often will.  Here is the day he wanted 100:

But today he wanted the prime 719, which I obliged.  When deciding which digit to eat first, he went through the choices, trying to make a composite with the digits left behind.  But he quickly realized that eating any digit would leave a prime: 71, 79, 19 are all prime.  Pleased with his discovery of this prime 719, he tried to find a larger one, but couldn't.

My questions:

Do these primes have a name?
Can you think of any more of them (clearly 23 is the smallest)?
Are there an infinite number of them?
Is there likely to be a way to find them short of using a computer?

Thanks to everyone for the interesting answers. Here's an improved heuristic that makes use of various parts of them -- the basic approach from Matthew, Fixee's observation of the repeated digits, and Numth' observations about restrictions on the digits. It's still a bit off, but I think it may be further improved taking into account Numth' observation 2; I've only taken into account  observation 1 (mod $2$ and $5$) so far; I think the mod $3$ part will need to be a bit more subtle; more on that later.

Consider numbers with $m$ digits consisting of $k$ strings of repeated digits. For instance, $38800999$ would have $m=8$ and $k=4$. The number of such numbers is $9^k\left(m-1\atop k-1\right)$: We have $9$ choices for the first repeated digit (we can't use $0$) and $9$ choices for the remaining ones (we can't use the previous one), and there are $\left(m-1 \atop k-1\right)$ different way to divide the repetitions. To check that this is right, we can calculate the total number of numbers with $m$ digits:

$$\sum_{k=1}^m 9^k\left({m-1 \atop k-1}\right)=9\sum_{k=0}^{m-1} 9^k\left({m-1 \atop k}\right)=9(1+9)^{m-1}=9\cdot10^{m-1}\;,$$

which is correct.

Like Matthew, I'll use $1/\log x$ as the "probability" of a number to be prime. This isn't quite right, since this is the total density below $x$, whereas we need the marginal density, $(x/\log x)'=1/\log x - 1/\log^2 x$. I'll do the analytic calculations using just $1/\log x$, which gives an upper bound and is thus good enough to show that the sum converges, but I'll give some numerical results using the more precise formula in the end.

So the probability of one of these numbers being prime is $1/\log x$, which we can bound from above by $1/\log 10^m=1/(m\log 10)$. Now we need the probability that the $k$ different numbers that we can get by deleting one of the digits are also prime. Note that for all but the last digit, the deletion doesn't change the last digit.

One of the main reasons why Matthew's heuristic significantly underestimates the abundance of these numbers is that we need to include a factor of $5/2$ for each digit except the last: For each of these digits, given that the original number is prime and the modified number has the same last digit, we already know it isn't divisible by $2$ or $5$, which raises its chances of being prime by a factor of $10/|\{1,3,7,9\}|=5/2$.

So multiplying all these probabilities, we have one factor of $1/(m\log 10)$ for the original number to be prime, $k$ factors of $1/(m\log 10)$ for all the modified numbers to be prime, and $k-1$ factors of $5/2$ to account for the last digit. (We could bound the probability for the modified numbers by $1/((m-1)\log 10)$, but I want to keep things simple first to derive the convergence; I'll get back to that in the numerical estimates).

Now we have all we need to write an upper bound for the sum of the "probabilities" over all numbers:

$$
\begin{eqnarray}
&&
\sum_{m=2}^{\infty}\sum_{k=1}^{m}9^k\left({m-1 \atop k-1}\right)\left(\frac{1}{m\log10}\right)^{k+1}\left(\frac{10}{4}\right)^{k-1}
\\
&=&
\frac{9}{\log^2 10}\sum_{m=2}^{\infty}\frac{1}{m^2}\sum_{k=0}^{m-1}\left({m-1 \atop k}\right)\left(\frac{45}{2m\log10}\right)^k
\\
&=&
\frac{9}{\log^2 10}\sum_{m=2}^{\infty}\frac{1}{m^2}\left(1+\frac{45}{2m\log10}\right)^{m-1}
\\
&<&
\frac{9}{\log^2 10}\sum_{m=2}^{\infty}\frac{1}{m^2}\mathrm{e}^{45/(2\log10)}
\\
&=&
\frac{9}{\log^2 10}\left(\frac{\pi^2}{6}-1\right)\mathrm{e}^{45/(2\log10)}
\\
&\approx&
19191
\;.
\end{eqnarray}
$$

Although this is quite a lot bigger than Matthew's result, it's still finite.

The bound is not very tight, for three reasons: I didn't use the fact that the modified numbers have one fewer digit; I dropped the $1/\log^2x$ term; and the exponential bound isn't tight for small $m$. To get a better estimate, let's keep the $1/\log^2x$ term and the power instead of the exponential, and let's roughly estimate the probabilities of the original and modified numbers being prime by using $1/((m-1)\log10)$ for all of them -- that overestimates the probability for the original number and underestimates it for the modified numbers, so should give a better estimate. Putting this all together yields the following estimate for the number of $m$-digit numbers, Fixee's $c(m)$, with $j:=m-1$:

$$9\left(\frac{1}{j\log10}-\frac{1}{(j\log 10)^2}\right)^2\left(1+\frac{45}{2}\left(\frac{1}{j\log10}-\frac{1}{(j\log 10)^2}\right)\right)^j\;.$$

Here's a plot. The numbers are reasonably close to the actual ones, but still too low, even at $m=9$, i.e. $j=8$, where the approximations should be reasonable and the maximum is almost attained, but the value is around 11 whereas the actual values are all around 16.

P.S.: I think I figured out how to correctly take into account the restrictions mod $3$; I'll be posting that later in the day.

P.P.S.: I just realized that part of the remaining underestimation stems from the fact that the last digit should also get a factor of $5/2$ if it's repeated. I think that together with the mod $3$ part might get the result roughly in line with the actual numbers.

[Update:]

I've produced some more numerical data and improved the estimate as described, and the two are now in quite satisfactory agreement.

For taking into account the restrictions mod $2$, $3$ and $5$, the general approach is to take the original number as given, with the generic probability of being prime, and then to analyze the modified numbers under the condition that the original number is prime. We need to distinguish two cases depending on whether the last digit is repeated. The repeated case is easier to analyze, so I'll treat that first.

So assume that the last digit is repeated. That excludes one place at which to divide the repetitions, so there are $9^k\left(m-2 \atop k-1\right)$ such numbers. In this case, each digit, including the last one, gets a factor of $5/2$ to account for the fact that the corresponding modified number is not divisible by $2$ or $5$. To help in getting the more complicated case of a non-repeated last digit right, it's worthwhile stating more explicitly how this factor of $5/2$ arises from a conditional probability. We can consider the probabilities that a number is coprime to all primes in a set $S$ and that it is coprime to all primes not in $S$ as independent, so that the probability of the number being prime is the product of these two probabilities. Then the probability of a number being coprime to $2$ and $5$ is $\lvert\{1,3,7,9\}/10\rvert=2/5$, and the probability of the number being prime is $p=2/5q$, where $q$ is the probability of it being coprime to all primes other than $2$ and $5$. If we estimate $p$ using the overall density of primes but we know that the number is coprime to $2$ or $5$, then the conditional probability of it being prime is $q=(5/2)p$. This is relatively obvious in the present case, but this way of looking at it will be helpful in dealing with the case of a non-repeated last digit.

Now let's look at the mod $3$ restrictions. The last digit is known to be $1$, $3$, $7$ or $9$. If we delete a $3$ or a $9$, the number will still not be divisible by $3$. If we delete a $1$ or a $7$, there is a 50% chance of the number becoming divisible by $3$. Thus, the conditional probability of the modified number being coprime to $3$ is $3/4$, and this has to be divided by the unconditional probability of it being coprime to $3$, which is $2/3$, to obtain the factor $(3/4)/(2/3)=9/8$ by which we need to multiply the unconditional probability estimate.

Now consider the remaining digits, starting from the end. Each digit is different from the one after it, and the one after it is known to be one of the $7$ digits allowed mod $3$. That leaves $6$ allowed digits and $3$ forbidden ones, for a conditional probability of $2/3$. This is equal to the unconditional probability, so we don't need to include any factors to account for the mod $3$ restrictions on the remaining digits.

That completes the considerations for the case of a repeated last digit. Let's call the estimate for the unconditional probability of a number with $m$ digits to be prime $p_m$ (more on that below); then we can write the sum of the probabilities for all $m$-digit numbers with repeated last digit as

$$
\begin{eqnarray}
&&
\frac{9}{8}p_m\sum_{k=1}^{m-1}9^k{\left(m-2 \atop k-1\right)} \left(\frac{5}{2}\right)^kp_{m-1}^k
\\
&=&
\frac{9}{8}\frac{5}{2}9p_mp_{m-1}\sum_{k=0}^{m-2}9^k{\left(m-2 \atop k\right)} \left(\frac{5}{2}\right)^kp_{m-1}^k
\\
&=&
\frac{405}{16}p_mp_{m-1}\left(1+\frac{45}{2}p_{m-1}\right)^{m-2}\;,
\end{eqnarray}
$$

where the orginal sum now only runs up to $m-1$ since the last digit cannot be repeated for $k=m$.

Now let's turn to the slightly more complicated case where the last digit isn't repeated. That reduces both the number of digits and the number of places at which to divide the repetitions, so there are $9^k\left(m-2 \atop k-2\right)$ such numbers. The sum works out as it should, since $\left(m-2 \atop k-1\right)+\left(m-2 \atop k-2\right)=\left(m-1 \atop k-1\right)$.

If the last digit doesn't repeat, we don't know whether deleting the last digit makes the number divisible by $2$ or $5$, and there is correlation between the events of the last deletion leaving the number coprime to $2$ and $5$ and the last two deletions leaving it coprime to $3$. Thus we need to determine the conditional probability that all three of these events occur given that the original number is prime.

Again, the last digit can be $1$, $3$, $7$ or $9$. But now the penultimate digit also has to be one of these, since otherwise the deletion of the last digit would render the number divisible by $2$. Thus we have $16$ combinations for the last two digits, $4$ of which are excluded because the digits cannot be the same. We have two different cases to consider, depending on whether $1$ and $7$ (which have the same remainder mod $3$) are allowed or not. Each of these cases occurs with probability $1/2$, depending on whether the remainder of the original number mod $3$ is $1$ or $2$. (Strictly speaking, the probabilities for these two cases are also slightly correlated with the probabilities being determined, but I believe this correlation should decay quickly as $m$ increases.)

So with probability $1/2$, only $3$ and $9$ are allowed, so the only combinations for the last two digits are $39$ and $93$. In the other case, all $4$ digits, and hence all $12$ combinations, are allowed. Thus, on average $7$ combinations of the last two digits lead to the last two deletions leaving the number coprime to $2$, $3$ and $5$, out of the $4\cdot9=36$ that are possible given that the original number is prime. That yields a conditional probability $7/36$, which we need to divide by the unconditional probability $(\lvert\{1,7,11,13,17,19,23,29\}\rvert/30)^2=(4/15)^2$ of two numbers being coprime to $2$, $3$ and $5$ to get the factor $(7/36)(15/4)^2=175/64$ by which we need to multiply the unconditional probability. All remaining digits contribute a factor of $5/2$ because deleting them doesn't make the number divisible by $2$, and we don't need any correcting factors for the mod $3$ restrictions for the remaining digits, for the same reasons as above, so we're done. Putting it all together, we have for the sum of the probabilities for all $m$-digit numbers with non-repeated last digit:

$$
\begin{eqnarray}
&&
\frac{175}{64}p_m\sum_{k=2}^{m}9^k{\left(m-2 \atop k-2\right)} \left(\frac{5}{2}\right)^{k-2}p_{m-1}^k
\\
&=&
9^2\frac{175}{64}p_mp_{m-1}^2\sum_{k=0}^{m-2}9^k{\left(m-2 \atop k\right)} \left(\frac{5}{2}\right)^kp_{m-1}^k
\\
&=&
\frac{14175}{64}p_mp_{m-1}^2\left(1+\frac{45}{2}p_{m-1}\right)^{m-2}\;,
\end{eqnarray}
$$

where the original sum starts at $2$ since the last digit necessarily repeats for $k=1$.

The ratio between the two results is $(35/4)p_{m-1}\approx(35/4)/((m-1)\log 10)\approx 3.8/(m-1)$, so for small $m$ there are more numbers with the last digit not repeated and for large $m$ there are more with the last digit repeated, in agreement with the numerical data.

We can bound $p_m$ by $1/\log m$ from above as before, and then show as before that for both cases the sum over the probabilities for all $m$ converges. To get a better estimate, we can take $p_m$ to be the average density of primes for all $m$-digit numbers, which is

$$p_m=\frac{\frac{10^m}{\log10^m}-\frac{10^{m-1}}{\log10^{m-1}}}{10^m-10^{m-1}}=\frac{1}{9\log 10}\left(\frac{10}{m}-\frac{1}{m-1}\right)\;.$$

Here are plots of the repeated case, the non-repeated case and the total. The maxima are at $m=18$, $7$ and $15$ with maximal values of and $21$, $8$ and $26$, respectively. The expected total counts (summed numerically) are $1794$, $209$ and $2003$, respectively. Thus, we can expect there to be around $2000$ of these numbers in total; about half of these have $74$ digits or more.

Here's a table comparing the actual counts for $3$ to $11$ digits (using the numerical data I report further down) to the above estimates:

$$
\begin{array}{|c|c|c|c|}
m&\text{last digit repeated}&\text{last digit not repeated}&\text{total}\\\hline\\
\begin{array}{c}
\mathrm{\vphantom{actual}\vphantom{estimate}}\\
\\
2\\
3\\
4\\
5\\
6\\
7\\
8\\
9\\
10\\
11\\
\end{array}
&
\begin{array}{c|c}
\mathrm{actual}&\mathrm{estimate}\\
\hline\\
0&\\
1&4\\
3&6\\
7&8\\
12&11\\
5&13\\
8&15\\
11&16\\
21&17\\
16&19\\
\end{array}
&
\begin{array}{c|c}
\mathrm{actual}&\mathrm{estimate}\\
\hline\\
4&\\
10&6\\
11&7\\
9&8\\
6&8\\
8&8\\
6&8\\
7&8\\
1&7\\
7&7\\
\end{array}
&
\begin{array}{c|c}
\mathrm{actual}&\mathrm{estimate}\\
\hline\\
4&\\
11&10\\
14&13\\
16&16\\
18&19\\
13&21\\
14&22\\
18&24\\
22&25\\
23&25\\
\end{array}
\end{array}
$$

The agreement turns out to be quite good, though there seems to be a slight overestimation now. I have no explanation for this, since the only systematic effect I can think of that I haven't taken into account is the correlation between the size of the original number and the size of the modified numbers, which should increase rather than decrease the probabilities. (In case you're wondering whether this is a case of changing the theory until it fits the data, it was actually the other way around: I got several incorrect results for the non-repeated case that would have removed the overestimation, but I believe the above analysis is the correct one.)

Here are all the numbers with up to $11$ digits (up to the 4,239,555,920th prime); I checked that they coincide with the ones already reported, but I'm putting them all here to have them together in one place:

23,
37,
53,
73,
113,
131,
137,
173,
179,
197,
311,
317,
431,
617,
719,
1013,
1031,
1097,
1499,
1997,
2239,
2293,
3137,
4019,
4919,
6173,
7019,
7433,
9677,
10193,
10613,
11093,
19973,
23833,
26833,
30011,
37019,
40013,
47933,
73331,
74177,
90011,
91733,
93491,
94397,
111731,
166931,
333911,
355933,
477797,
477977,
633317,
633377,
665293,
700199,
719333,
746099,
779699,
901499,
901997,
944777,
962233,
991733,
1367777,
1440731,
1799999,
2668999,
3304331,
3716633,
4437011,
5600239,
6666437,
6913337,
7333331,
7364471,
7391117,
13334117,
22255999,
33771191,
38800999,
40011197,
40097777,
44333339,
49473377,
79994177,
86000899,
93361493,
94400477,
99396617,
99917711,
110499911,
144170699,
199033997,
222559399,
333904433,
461133713,
469946111,
640774499,
679774391,
680006893,
711110111,
716664317,
743444477,
889309999,
900117773,
982669999,
999371099,
999444431,
1113399311,
1133333777,
1176991733,
1466664677,
1667144477,
1716336911,
2350000999,
3336133337,
3355522333,
3443339111,
3973337999,
4111116011,
4900001111,
6446999477,
6666116411,
6689899999,
6914333711,
7463333477,
8555555599,
8888333599,
8936093833,
9746666477,
10000000097,
10666610333,
11100077711,
11793334733,
19000019333,
23525555899,
30001114937,
33008299999,
33110666399,
33399911777,
37796941199,
40470660977,
44434133339,
46661333333,
46666133333,
55888856893,
61077333377,
66664441373,
66700447613,
66993413393,
71111164499,
77443733111,
99444901133

Some interesting specimens are 10000000097 (which is the only one with lots of $0$s after the first digit, showing that there's no significant effect from the overestimation of the size of the modified numbers in such a case) and the pair 46661333333 and 46666133333.

Here's the Java program I used to find these (using a bit set as Fixee suggested to cram as many primes as possible into my 8GB). The computation took half an hour on a MacBook Pro i7.

public class EdiblePrimes {
    static class LongBitSet {
        int [] bits;

        public LongBitSet (long nbits) {
            bits = new int [(int) ((nbits + 0x1f) >> 5)];
        }

        public void clear (long bit) {
            bits [index (bit)] &= ~mask (bit);
        }
        public void set (long bit) {
            bits [index (bit)] |= mask (bit);
        }
        public boolean get (long bit) {
            return (bits [index (bit)] & mask (bit)) != 0;
        }
        private static int mask (long bit) {
            return 1 << (bit & 0x1f);
        }
        private static int index (long bit) {
            return (int) (bit >> 5);
        }
    }

    final static long max = 0x1800000000L;
    final static long maxIndex = max >> 1;

    public static void main (String [] args) {
        LongBitSet composite = new LongBitSet (maxIndex); // bit at index n says whether 2n+1 is composite
        composite.set (0); // 1 isn't prime

        int maxDivisor = (int) (Math.sqrt (max) + 1);

        for (int divisor = 3;divisor < maxDivisor;divisor += 2)
            if (!composite.get (divisor >> 1))
                for (long multiple = (3 * divisor) >> 1;multiple < maxIndex;multiple += divisor)
                    composite.set (multiple);

    outer:
        for (long n = 11;n < max;n += 2) {
            if (!composite.get (n >> 1)) {
                long power = 1;
                do {
                    long nextPower = 10 * power;
                    long modified = n % power + (n / nextPower) * power;
                    if (modified != 2 && ((modified & 1) == 0 || composite.get (modified >> 1)))
                        continue outer;
                    power = nextPower;
                } while (power < n);
                System.out.println (n);
            }
    }
}

[Update:]

Since the probability has a simple $k$ dependence, we can calculate the expectation value for $k$, i.e. the expected number of strings of repeated digits. For the shifted sums over $k$, we have

$$
\begin{eqnarray}
&&
\frac{
\sum\left(m-2 \atop k\right) k q^k
}{
\sum\left(m-2 \atop k\right) q^k
}
\\
&=&
\frac{
q\frac{\partial}{\partial q}\sum\left(m-2 \atop k\right)q^k
}{
\sum\left(m-2 \atop k\right) q^k
}
\\
&=&
\frac{
q\frac{\partial}{\partial q}(1+q)^{m-2}
}{
(1+q)^{m-2}
}
\\
&=&
\frac{
(m-2)q(1+q)^{m-3}
}{
(1+q)^{m-2}
}
\\
&=&
\frac{m-2}{1+q^{-1}}\;.
\end{eqnarray}
$$

With $q\approx 45/(2(m-1)\log10)$, this becomes

$$
\frac{m-2}{1+\frac{2(m-1)\log10}{45}}\to_{m\to\infty}\frac{45}{2\log{10}}\approx 10\;.
$$

We shifted $k$ down by $1$ and $2$ for numbers with with last digit repeated and not repeated, respectively. Thus, if we don't count a non-repeated last digit as a separate string, we get the same expectation value in both cases, namely $1$ more than the above value. Thus, in the limit of large $n$, there will be about $11$ strings of repeated digits on average (not counting non-repeated last digits), but the values for small $m$ are not near the limit because of the large factor of $45$. Here's a comparison of the actual averages with the estimated expectation values (using the above better estimate for $p_m$):

$$
\begin{array}{|c|c|c|}
\hline\\
m&\text{actual}&\text{estimated}\\
\hline\\
2&1.0&\\
3&1.9&1.8\\
4&2.8&2.5\\
5&3.5&3.1\\
6&3.8&3.6\\
7&4.1&4.1\\
8&4.4&4.5\\
9&5.3&4.8\\
10&5.4&5.1\\
11&5.7&5.4
\end{array}
$$

Here, too, the agreement is quite good, but there seems to be a slight systematic error left.


How to prove that $j^\mu=\cos(\frac{\mu\pi}{2})+j\sin(\frac{\mu\pi}{2})$? I am trying to read a journal paper and I need to understand some math operations.

At this time, I feel it hard to understand this equation:

$j^\mu=\cos(\frac{\mu\pi}{2})+j\sin(\frac{\mu\pi}{2}), j=sqrt(-1), \mu\in R,$

Is this something from the complex number angular formation definition or some kind of Euler equation? Last year, I have asked my professor to help me go through this part. But now I still can't feel familiar to this part.

Based on some answers:

$j= e^{j \frac{\pi}{2}}$ is the part that I feel hard to understand.

I know some basic complex number operations and I often use MATLAB to solve math problem.

Thanks,

Bo

Updating my answer, as OP has said that $\mu\in \mathbb R$, so the assumption I made in my previous answers below that $\mu \in \mathbb Z$ no longer hold.

The first question to ask is, "What does $j^{\mu}$ even mean if $\mu$ is not an integer?"  There are three numbers that satisfy the equation, $x^3=j$, and while you could argue that one of them should be the value $j^{1/3}$, which of the three values should you take?

One way around this is by appealing to the exponential function, $\operatorname{exp}(x)=e^x=\sum \frac{x^n}{n!},$ where the sum is our actual definition, and $e^x$ is suggestive shorthand.  One can establish various properties like $e^{a+b}=e^a e^b$, which mean that $e^x$ behaves like normal exponentiation in cases where things are simple, i.e., if $e^x=\alpha$, then $e^{nx}=\alpha^n$, and $\left(e^{x/n}\right)^n=\alpha$.  Since $e^x>0$ whenever $x\in \mathbb R$, this allows us to define $x^{1/n}$ as $e^{\log x/n}$ for positive $x$, which always spits out the real $n$th root instead of one of the other $n-1$ complex roots we could have taken.  It also allows us to define $x^{\mu}=e^{\mu\log x}$ for positive $x$ even when $\mu$ is not rational.  Because of continuity of $e^x$, and we have that if $(\r_i)$ is a sequence of rational numbers approaching $\mu$, then by taking the exponential function to define $x^{r_i}$, we have $x^{r_i}\to x^{\mu}$, and so this definition fits in with just saying "define things as roots of polynomials, then extend using continuity."

However, we run into problems defining $x^{\mu}$ when $x$ is not a positive real number.  I don't want to get into all the ways that things can go wrong, but what I am about to do is not wholly sufficient.

By using the sum formula, we can define $e^x$ not just for $x\in \mathbb R$, but actually for $x\in \mathbb C$. By looking at the Taylor series for $\sin$ and $\cos$ (or otherwise), we can establish that $e^{jx}=\cos x + j\sin x$, and we can check that $e^{a+b}=e^a e^b$ still holds when $a$ and $b$ are complex numbers.

Now, we can answer your question.  Since $e^{j*\pi/2}=j$, we can define $j^{\mu}=e^{j\pi \mu/2}=\cos(\pi\mu/2)+j\sin(\pi \mu/2)$, and this will agree with the values you would get for $j^{\mu}$ when $\mu$ is an integer.  However, this is not the only way to go...

Since $e^{2j\pi}=1$, we also have that $e^{j*\pi/2+2jk\pi}=j$ whenever $k$ is an integer, and so we could define $j^{\mu}=e^{j\mu(\pi/2+2k\pi)}$ for some integer $k$.  This will give us the same values when $\mu$ is an integer, but will give different values when $\mu$ is not.  While one can argue that $k=0$ gives the simplest possibility, there isn't a compelling argument (that I can think of) to say that it is the "right" definition in any real sense.

Just to throw something interesting in, the above argument shows that there are an infinite number of possibilities for what the value $j^j$ should be.  However, all of them are real numbers!

Old answers below.

You can prove the formula by induction, but for the induction step you are going to need the following result, which is easy enough to prove by expanding things out, regrouping terms, and making an appeal to standard trig identities:

$$ \left(\cos(\theta)+j\sin(\theta)\right)\left(\cos(\varphi)+j\sin(\varphi) \right)=\cos(\theta+\varphi)+j\sin(\theta+\varphi)$$

A second approach that you could take (because you are only asking about $j^{\mu}$ and not powers of a different complex number) is to verify that both the left and right hand sides of your equation only depend on the remainder of $\mu$ when divided by $4$, and then break things up into four cases, depending on if that remainder is $0, 1, 2$, or $3$.


What was the notation for functions before Euler? According to the Wikipedia article,

  [Euler] introduced much of the modern mathematical terminology and notation, particularly for mathematical analysis, such as the notion of a mathematical function.
  — Leonhard Euler, Wikipedia

What was the notation for functions before him?

I scanned through parts of Newton's Pricipia found online, and was surprised that a search for the word "function" did not yield any results at all. There do appear to be equations acting as what we would call functions, such as when he describes force, we see things such as
$$F=\frac {2h^2}{SP^2}\cdot \frac {QR}{QT^2}$$
and $$R=\frac {\frac 1 2 L}{1+e\cos ASP}$$
(on page 223) but he refers to these as equations, not functions, and admittedly (written the way they are) that is exactly what they are. It seems anything that we would today write as a function, Newton described in words, such as:

  If a hyperbolic orbit be described under the action of a repulsive
  force tending from the center, the force varies as the distance and
  the velocity at any point as the diameter of the conjugate hyperbola
  parallel to the tangent at the point.

Or he used words within his equation:

$$\text{Velocity at P}=\frac {h.VA}{SP^2}$$

This last one almost assuredly would be written as a function if presented in a modern textbook. Newton is certainly not the only source one should consider, but it does give an idea of what was going on right before Euler began publishing.

The information on this website, which unfortunately does not include specific sources, indicates that Bernoulli proposed that $\phi$ or $\phi x$ be used as the notation for a function, and Euler introduced $f(x)$.

Edit: Reference #11 from David Renfro's answer gives references for the statements made about Bernoulli and Euler on the website, as described in the last paragraph above. In my brief skim of Newton's Principia, I also found exactly what was described in reference #11 to be true, specifically that the arguments were motivated almost exclusively from analytic geometry, and that what we consider a "function" was really only considered a variable, as is indicated in the few examples above. I would recommend reading #11, it explains in good detail what you would like to know, I think.

Let's observe an example :

$a)$ formal description of function (two-part notation)

$f : \mathbf{N} \rightarrow \mathbf{R}$

$n \mapsto \sqrt{n}$

$b)$ Euler's notation :

$f(n)=\sqrt{n}$

I don't know who introduced two-part notation but I think that this notation must be older than Euler's notation since it gives more information about function and therefore two-part-notation is closer to correct definition of the function than Euler's notation.

There is also good wikipedia article about notation for differentiation.


Show that the power series is solution to f ''(x) + f(x) = 0 Show that

$$f(x)=\sum_{n=0}^∞ \frac{(-1)^nx^{2n}}{(2n)!}$$

is a solution of:
$$f''(x)+f(x)=0$$

It seems that $f''x$ is just $-f(x)$ $=>$ $-f(x)+f(x)=0$, but when I try to get the solution  I get a weird power series that is not $-f(x)$. Can anyone tell me what went wrong? See below:

$$f'(x)=\sum_{n=1}^∞ \frac{(-1)^nx^{2n-1}*(2n)}{(2n)!}$$
 $$f''(x)=\sum_{n=2}^∞ \frac{(-1)^nx^{2n-2}*(2n)*(2n-1)}{(2n)!}$$

$$f''(x)=\sum_{n=0}^∞ \frac{(-1)^{n+2}x^{2n+2}*(2n+4)*(2n+3)}{(2n+4)!}$$
 $$=\sum_{n=0}^∞ \frac{(-1)^{n+2}x^{2n+2}}{(2n+2)!}$$

Anyone know what mistake I made? I need $f''(x)$ to be $-f(x)$

You have made mistakes in changing $n$ to $n+2$. In $f''(x)$, $x^{2n-2}$ the becomes $x^{2n+2}$ and $(2n)(2n-1)$ becomes $(2n+4)(2n+3)$. Also $(2n)!$ becomes $(2n+4)!$

Answer for the edited version. To compare your series for $f''(x)$ with the series for $f(x)$ you have to change $n$ to $n-1$ (so that $x^{2n+2}$ becomes $x^{2n}$).

‘’ has to star with n=1 not n=2. And be more careful when reindexing to 0, alternating power will switch by 1, x power by two.


$x_1$ such that given sequence $(x_n)$ converges Let $(x_n)_{n \in \mathbb N}$ be a sequence in $\mathbb R$ such that $|4-x_{n+1}|<q|4-x_n|^2$, $\ q>0$

Under what circumstances for the initial value $x_1$ can we guarantee that $(x_n)_{n\in \mathbb N}$ converges?

One can show by induction that
$$ \forall n\in\mathbb{N},\,|4-x_n|\leqslant q^n|4-x_0|^{2^n} $$
If $x_0\in]3,5[$, then $|4-x_0|<1$ and $\lim\limits_{n\rightarrow +\infty}x_n=4$, if $x_0\in\{3,5\}$, then $|4-x_0|=1$ and $(x_n)$ converges if $q<1$, in this case the limit is $4$. If $x_0\notin[3,5]$, then $|x_0-4|>1$ and $\lim\limits_{n\rightarrow +\infty}{|x_0-4|^{2^n}}=+\infty$, we can't say anything.


How can a transformation be linear transformation without linearity? My teacher at the University gave me a question I could not understand completely. Here is the question:

  Let $T: \mathbb R^3 \to P[x]$ be a linear transformation with $$T([1, 0,
0])=x+1, \quad T([0, 1, 0])=x^2-x, \quad T([0, 0, 1])=x^2,$$ find
  $T([a, b, c])$, also find the standard matrix $A$ for the
  transformation.

The part that I did not understand is that how can $T([0, 1, 0])$ and $T([0, 0, 1])$ be linear since they have $x^2$. Also the $T([1, 0, 0])$ term has a constant. Those violate the linear transformation rules. Don't they?

Linearity is nothing more or less than requiring  $T(u+v)=T(u) +(v), T(au)=aT(u)$. This does not rule out
$T(u)=x+1$.

In fact consider the function $M\colon P[x]\to P[x]$ which has the effect of multiplying anything by $x^2+3x+1$ (or any random, fixed polynomial).

$M(f(x)+g(x) )= M(f(x) ) + M(g(x))$. ANd M(af(x) ) = aM(f(x)).

Can see  $M(1)= x^2+3x+1$, a quadratic function! $M$ satisfies the requirement of linear transformation, Perfectly alright.


What is the negation of this statement? Let $(K_n)$ be a sequence of sets.

What is the negation of the following statement?

For all $U$ open containing $x$, $U \cap K_n \neq \emptyset$ for all but finitely many $n$.

Let us write the statement formally:
$$\forall U(x\in U\rightarrow\exists n\forall k(k>n\rightarrow U\cap K_k\neq\varnothing))$$
For every $U$ (open of course), if $x\in U$ then there is some $n$ that for all $k>n$ we have $K_k\cap U\neq\varnothing$.

Now negation flips quantifiers and $\lnot(\alpha\rightarrow\beta)$ is the same as $\lnot\beta\land\alpha$. So we have:

$$\exists U(x\in U\land\forall n\exists k(k>n\land U\cap K_k=\varnothing))$$
Or in words, there exists an open set $U$ such that $x\in U$ but for every $n$ there is some $k>n$ such that $U\cap K_k=\varnothing$. However in the natural numbers to say that something happens unboundedly often is the same as saying it happens infinitely often. So finally we can say:

  There exists an open set $U$ such that $x\in U$ and for infinitely many $n$ we have $U\cap K_n=\varnothing$.

$\neg$(for all $U$ open containing $x$, $U \cap K_n \ne \emptyset$ for all but finitely many $n$).


$\sum_{p} \chi(p)/p$ is conditionally convergent for non-principal character Let $\chi$ is a non-principal character. Show that the sum $\sum_{p}\frac{\chi(p)}{p}$ is conditionally convergent. Then show that the product $\prod_{p}(1-\frac{\chi(p)}{p})^{-1}$ is conditionally convergent to $L(1,\chi)$.

Using summation by parts, I can show that $\sum_{n}\frac{\chi(n)}{n}$ is conditionally convergent. But it seems that the method cannot be applied to $\sum_{p}\frac{\chi(p)}{p}$ since $\frac{1}{n}1_{\mathcal{P}}(n)$ is not monotone.

Use partial summation to show that
\[\sum_{p \leq x} \frac{\chi(p)}{p} = \frac{1}{x \log x} \sum_{p \leq x} \chi(p) \log p - \int_{2}^{x} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt.\]
Now
\[\sum_{p \leq x} \chi(p) \log p = \sum_{n \leq x} \chi(n) \Lambda(n) + O\left(\sqrt{x} \log x\right).\]
Via the usual methods as for proving the prime number theorem, one can then show that
\[\sum_{n \leq x} \chi(n) \Lambda(n) = O\left(x e^{-c\sqrt{\log x}}\right)\]
if $\chi$ is nonprincipal. (Note that this is by far the hardest part; it's essentially equivalent to the prime number theorem in arithmetic progressions, which in turn is essentially equivalent to the nonvanishing of $L(s,\chi)$ along the line $\Re(s) = 1$. A standard reference for this estimate is chapter 11 of Montgomery and Vaughan.)

This implies the result, because then the first term in the earlier expression for $\sum_{p \leq x} \frac{\chi(p)}{p}$ is $o(1)$ and the second term is
\[\int_{2}^{\infty} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt - \int_{x}^{\infty} \frac{\log t + 1}{t^2 (\log t)^2} \sum_{p \leq t} \chi(p) \log p \, dt,\]
and the first integral is some finite number and the second is $o(1)$ due to the bound $\sum_{p \leq x} \chi(p) \log p = O\left(x e^{-c\sqrt{\log x}}\right)$.

EDIT: Note that for $\Re(s) > 1$,
\[\log L(s,\chi) = \log \prod_p \frac{1}{1 - \chi(p) p^{-s}} = \sum_p \log \frac{1}{1 - \chi(p) p^{-s}} = \sum_p \sum_{k = 1}^{\infty} \frac{\chi(p)^k}{k p^{ks}}\]
via the Euler product for $L(s,\chi)$ and the Taylor series expansion for $\log$. All of this is justified because $\Re(s) > 1$, so everything is absolutely convergent. It follows that for $\Re(s) > 1$,
\[\sum_p \frac{\chi(p)}{p^s} = \log L(s,\chi) - \sum_p \sum_{k = 2}^{\infty} \frac{\chi(p)^k}{k p^{ks}}.\]
The sum on the right-hand side is absolutely convergent for $\Re(s) > 1/2$, and so defines a holomorphic function on this right-half plane. Moreover, by the zero-free region for $L(s,\chi)$, $\log L(s,\chi)$ defines a holomorphic function in the same zero-free region. It follows by analytic continuation that the left-hand side extends to a holomorphic function in this same zero-free region.

In spite of all of this, one cannot conclude immediately this holomorphic function is equal to $\sum_p \frac{\chi(p)}{p}$ when $s = 1$, and so this sum converges conditionally. It does show however that the limit $\lim_{s \to 1^+} \sum_p \frac{\chi(p)}{p^s}$ exists.

complicated. there is a proof sketch when $\chi$ is a non-principal non-quadratic character (i.e. when it is a complex valued character) :

the main trick is to use that $\frac{1}{\chi(q)}\sum_{\chi \pmod q} \chi(n) =  \delta_{\ n \ \equiv\  1 \pmod q}$ (the discrete Fourier transform of size $\phi(q)$)

and that $L(s,\chi)$ is holomorphic (on $ Re(s) > 0$ is enough and easy to prove) whenever $\chi$ is not the principal character $\chi_0$ (i.e. $\chi_0(n)=1$ whenever $gcd(n,q)$).

hence if $\chi \ne \chi_0$ and $\lim_{s \to 1^+} \sum_p \chi(p)/p^s$ diverges then $L(s,\chi)$ has a $0$ at $s=1$,

but if it is a non real character then the character comes with its complex conjugate $\overline{\chi}$ which in that case also has a zero at $s=1$, and hence two terms of the product
$$F_q(s) = \prod_{\chi \pmod q} L(s,\chi)$$
would have a zero at s=1, but only $L(s,\chi_0)$ has a pole at $s=1$, of order $1$, hence $F_q(s)$ has a zero at $s=1$, a contradiction since :
$$\ln F_q(s) = \phi(q) \sum_{p^k \equiv 1 \pmod q} \frac{1}{k p^{k s}}$$
which diverges to $+\infty$ when $s \to 1^+$.
then prove that $L(s,\chi)$ doesn't have any zero on $Re(s) =1$, using the same argument as for  $\zeta(s)$ : https://en.wikipedia.org/wiki/Prime_number_theorem#Proof_sketch

which is based on the same kind of trick : that $\zeta(s)$ is meromorphic, that it doesn't have any pole on $Re(s) = 1, s \ne 1$, and that $\ln \zeta(s)$ is a Dirichlet series with non-negative coefficients, hence using $$\zeta(x)^3\zeta(x+iy)^4\zeta(x+2iy)|=\exp\sum_{n,p}\frac{3+4\cos(ny\log p) +\cos (2ny\log p)}{np^{nx}}\ge 1$$ if it had a zero at $1+it$ it would have a pole at $1+2it$, a contradiction.
and you will  be ready for proving the prime number theorem for
Dirichlet L-functions when $\chi$ is a complex character, i.e. that $\ln L(s,\chi)$ is holomoprhic on $Re(s) \ge 1$, and hence that $$\sum_p \frac{\chi(p)}{p}$$ converges.
when $\chi$ is the real non-principal (quadratic) character, I don't remember how we prove that it doesn't have a zero at $s=1$, see for example Terrence tao blog  https://terrytao.wordpress.com/2014/11/23/254a-notes-1-elementary-multiplicative-number-theory/ just after the corollary 77.


Existence of biholomorphic function for two fixed points with $f(z_1) = z_2$ Let $U \subset C$ be a simply connected open subspace and $z_1, z_2 \in U$. I want to show that then there exists a biholomorphic function $f: U \rightarrow U$ with $f(z_1) = z_2$, but I have no idea how to do so.

If $f$ is biholomorphic, $f$ is injective. Hence, from $f(z_1) = f(z_2)$ we get $z_1=z_2$.


How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$

How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$?

Are they using some simple identity or is there some other observation happening?
Thanks!

HINT: $$\sin{3x}=3\sin {x} -4\sin^3 x \\ \sin {(A+B)}=\sin A \cos B + \cos A \sin B$$

Hint:
$$\sin x =\frac{opposite}{hypotenuse}$$


Prove that $A$ is closed I am asked to prove the following question:

Let "If $a \in \mathbb{R}$, $a_n \in A$ for all $n \in \mathbb{N}^+$, and $a_n \to a$, then $a \in A$." be a statement $P$. Show that if $P$ is true then $A$ is closed.

I was given a hint that if $a \notin \text{int}(\mathbb{R} \setminus A)$, then for all $n \in \mathbb{N}^+$, there exists $a_n \in (a - 1/n, a+1/n) \cap A$.

But I can't get my head around what is the hint trying to tell me and I don't know how to proceed. Can anyone give a hint? Thanks a lot!

I finally understood the question so here is an attempt at the proof.

Suppose $a \in \textbf{cl}A$, we have that $a \notin \textbf{int}(\mathbb{R} \setminus A)$. Thus, $a$ is not an interior point of $(\mathbb{R}\setminus A)$ and then we have for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \nsubseteq \mathbb{R}\setminus A$. This is equivalent to for all $n \in \mathbb{N}^+$, $(a - 1/n, a + 1/n) \cap A \neq \emptyset$. Hence, we let $a_n \in (a - 1/n, a + 1/n) \cap A$. Thus, $a_n \in A$ for all $n \in \mathbb{N}^+$. Moreoever, $a_n \to a$ since $\cap_{n = 1}^\infty(a- 1/n, a+ 1/n) = \{a\}$. Therefore, $a \in A$ by $P$. Hence, $\textbf{cl}A \subset A$ and $\textbf{cl}A = A$ since $A \subset \textbf{cl}A$. Therefore, $A$ is closed since $\textbf{cl}A$ is closed.

Hint: $a\not\in\operatorname{int}({\Bbb R\setminus A})$ is another way of saying $a\in\operatorname{cl}(A)$.  So, the hint is that every element of the closure of $A$ has a sequence in $A$ converging to it.  Now use $P$.


Complex power series and radius of convergence Let $c$ be a non-zero complex number, and consider the power series
\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots.
\end{equation}
By using the Ratio Test, or otherwise, show that the series has radius of convergence $|c|$. By differentiating term by term, show that $S'(z)= \frac{1}{z}$.

I've never done power series in complex analysis, so this is what I've attempted so far:

\begin{equation}
S(z)=\frac{z-c}{c}-\frac{(z-c)^2}{2c^2}+\frac{(z-c)^3}{3c^3}-\ldots\\
=\sum_{n=0}^{\infty} \frac{(-1)^n (z-c)^n}{nc}.
\end{equation}
Let $x_{n}=\frac{(-1)^n}{nc}$. Using the Ratio test, we have:
\begin{equation}
lim_{n \rightarrow \infty} \lvert \frac{x_{n+1}(z-c)^{n+1}}{x_{n}(z-c)^n} \rvert = \lvert z-c \rvert lim_{n \rightarrow \infty} \lvert - \frac{n}{n+1} \rvert =  -\lvert z-c \rvert <1.
\end{equation}

I'm pretty sure this is wrong somewhere, but I have no idea how to continue to show that the radius of convergence is $|c|$.

Notice (hint):

First of all:

When $n$ 'starts' with $0$ than we've got a problem, because we get (dividing by $0$):

$$\frac{(-1)^0(z-c)^0}{0c}$$

Use the ratio test, to proof that this series converges, when $|c-z|<1$.
So:

$$\sum_{n=1}^{\infty}\frac{(-1)^n(z-c)^n}{cn}=-\frac{\ln(1+z-c)}{c}\space\space\space\space\space\space\text{when}\space|c-z|<1$$


Probability of draw 1000 balls of the same color from 2000 Without replacement. We have 2000 balls. 1000 red and 1000 green.
What is the probability of draw the 1st 1000 balls to be the same color (red or green).
 This is how I worked on it, not sure.
My work

Let's begin with a numerically simpler problem. Suppose you have 8 balls
in the urn: 4 red and 4 green. You draw 4 balls from the urn without
replacement. Then there are ${8 \choose 4} = 70$ possible outcomes. Two
of the outcomes have all balls the same color: one all 4 red and the other
all 4 green. So the probability of getting 4 balls of the same color is
$2/{8 \choose 4} = 2/70 = 0.02857.$

In the same way, the answer to the original question is $2/{2000 \choose 10000},$
which is a very small number. Maybe you are allowed to leave your answer
in this 'combinatorial' form. And maybe you are expected to use
Stirling's Approximation (as suggested by @ThomasAndrews and @callciulus).
As you can see by the Wikipedia reference, there are several forms of the
Approximation, all of which give 'about' the same answer.

The answer you linked in your Question is correct for the probability
of getting 'all red balls'. Multiplying by 2, you'd get the probability
for 'all the same color'. which matches $2/{2000 \choose 10000}.$
 (Your first factor is 2.)

Edited: I removed the second answer. Thanks for letting me know where I went wrong. ;)
This answer applies if you assume that the ball is returned to the bag every time you draw. The probability of drawing 1 ball a certain color is 1/2. Since the first 1000 balls all need to be the same color, you take probability and raise it to the power of the number of draws: so your answer is (1/2)^1000


Question on time speed and distance A and B, who are separated by a distance 90 m, are approaching towards each other. The initial speed of A is 5 m/s and that of B is 10 m/s. If A increases his speed by 5 m/s every second, when will they meet each other?

HINT:

Let the distance $d$ between the two be expressed a a function of time $t$. Then you have that the rate of change of the distance is
$$d'(t)=-5-10-5t$$
$$d'(t)=-15-5t$$
The two will meet when $d(t)=0$. Can you integrate, find the constant that is added when you integrate, and then find when $d(t)=0$?

FULL ANSWER:
Integrate both sides to get
$$d(t)=-15t-\frac{5}{2}t^2+C$$
And since the starting distance is $90$, then $C=90$, so
$$d(t)=-15t-\frac{5}{2}t^2+90$$
And you must find when this is zero:
$$-\frac{5}{2}t^2-15t+90=0$$
$$\frac{5}{2}t^2+15t-90=0$$
This can be solved like a quadratic:
$$t^2+6t-36=0$$
Can you use the quadratic formula to solve this?

$A$ & $B$ are $90$ m distance apart.

Their relative speed is $(10+5)$ m/sec

Say, after $t$ second, they meet each other at a distance $x$ from the point from where $A$ starts.

We can say that both of them in combination have covered $90$ m distance within $t$ secs at a speed of $15$ m/s

so $t = \frac{90}{15} = 6$ secs


How to find a function using domain and range in math? I want to find a function, given domain and range.
Example:

Domain Range
2      6
4      8
6      20
7      24

  This is just an example. I want to know the method of creating functions regardless of the values in Domain and Range

Then how can I find a function using Domain and Range?

Any help will be appreciated :)

The table can be used to define the function and it is a completely correct definition.

If you are looking for a closed form, we can use for example polynomial interpolation.

Refer also to the following

Explanation of Lagrange Interpolating Polynomial

你能通过你已知的函数，例如指数函数，对数函数，幂函数这些你很熟悉的函数进行配凑，因为你知道它的性质，在哪个区间增长或减少，然后合理添加常数进行配凑。我建议你先画出x -y坐标系,标记出domain and range 所在的矩形区域，再考虑用哪一个函数将它放入这个区域，使得这个函数在矩形区域内并且可以取到最大最小值。these  are  answered by chinese you are suggested to use google  translator.

  Translated From Google Translator:
  You can use a function you
  know, such as an exponential function, a logarithmic function, a power
  function, a function that you are familiar with, because you know its
  nature, in which interval it grows or decreases, and then add a
  constant to match . I suggest that you first draw the x-y coordinate
  system, mark the rectangular area where the domain and range are
  located, and then consider which function to put it into this area, so
  that this function is in the rectangular area and can get the maximum
  and minimum


Problems with double integration I'm trying to integrate $\int_{-a}^a\int_b^cy^{2m+1}e^{xy^{2n}}dxdy$.But I have never seen an integral with so many parts to it and I am little overwhelmed. How do I solve this?

$$\newcommand{\i}[2]{\int_{#1}{#2}}$$
I believe you can split it up like
$$\int_{-a}^{a}\Biggr(\int_{b}^{c} \cdots dx \Biggr)dy$$

To learn more about double integrals, click on,

https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-topic/v/double-integral-1

$$\i{-a}{a} y^{2m+1-2n}(\i{b}{c} y^{2n} e^{y^{2n}x}dx)dy$$
$$\i{-a}{a} y^{2m+1-2n} e^{y^{2n}x} dy$$
Putting $e^{y^{2n}x}=t$
$$dt =y^{2n-1} e^{y^{2n}x}dx$$
$$\i{-a}{a} \frac{y^{2m+1-2n-2n+1}}{x} y^{2n-1}x e^{y^{2n}x} dy$$
For further steps go to,
https://www.integral-calculator.com/#expr=y%5E%7B2%28m-n%29%2B1%7D%20e%5E%7Bxy%5E%282n%29%7D&intvar=y
See the steps there


If $f(x)<g(x)$ prove that $\lim f(x)<\lim g(x)$ I have this question:

  Let $f(x)→A$ and $g(x)→B$ as $x→x_0$. Prove that if $f(x) < g(x)$ for all $x∈(x_0−η, x_0+η)$ (for some $η > 0$) then $A\leq B$. In this case is it always true that $A < B$?

I've tried playing around with the definition for limits but I'm not getting anywhere. Can someone give me a hint on where to start?

HINT: Suppose that $A>B$, and let $\epsilon=\frac12(A-B)>0$. Show that there is a $\delta>0$ such that $f(x)>A-\epsilon=B+\epsilon>g(x)$ for all $x\in(x_0-\delta,x_0+\delta)$; this contradicts the hypothesis that $f(x)<g(x)$ on an open interval around $x_0$. (Why?)

It’s easy to find examples in which $A=B$. You can do it with $f(x)=0$, in fact.

Quite obvious from the definition of continuity. By continuity,$\forall \epsilon$ there exist $\delta$ s.t $x\in B_{\delta}(x_0) \implies|f(x)-A|<\epsilon$ and $|g(x)-B|<\epsilon$ and so we have $ B-\epsilon<g(x) \ \text{and} f(x)<\epsilon +A$ so $ B<2\epsilon +A \  \forall \epsilon >0$. Since $\epsilon$ is arbitrary so we conclude that $B<A$. But i am not sure what is missing, i am not sure why i couldn't prove that $A=B$


$X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. $\phi$ is continuous $\iff$ its graph is closed. $X \subset \mathbb{R}^m$. Let $\phi:x \to \mathbb{R}^n$ be bounded. Then $\phi$ is continuous $\iff$ its graph is closed.

I was asked to prove this, but I believe it's false.

Let $Gra(\phi) = \{(x, \phi(x))|x \in X\}$ the Graph of $\phi$.

Suppose X is not closed. Then exists $a \in \bar{X}-{X}$ (being $\bar{X}$ the closure of X).

Therefore exists $x_n \in X$ s.t. $x_n \to a$. The sequence $\phi(x_n)$ has a convergent subsequence $\phi(x_{k_n})$ since it's bounded. Let $\lim \phi(x_{k_n})=b$

Then $(a,b) \in \overline{Gra(\phi)}$  (the closure of $Gra(\phi)$) but $(a,b)\not \in Gra(\phi) \implies Gra(\phi)$ is not closed.

Is this correct?

Of course, when $X$ itself is not closed in $\mathbb{R}^m$, the graph cannot be closed. We henceforth assume that $X$ is closed.

Let $\mathcal{C}(X)$ denote the set of convergent sequences $\mathbf{x}=(x_n)_{n\in\mathbb{N}}\subset X$. For any such sequence we will denote its limit by $x$. Notice that $x\in X$ by the assumption that $X$ is closed.

On the one hand, we have:

\begin{align}
\phi \text{ is continuous}&\iff \phi \text{ is sequentially continuous}\\
&\iff \forall \mathbf{x}\in\mathcal{C}(X), \,\phi(x)=\lim_{n\to\infty}\phi(x_n)\tag{1}\end{align}

On the other, we have

\begin{align}
Gra(\phi) \text{ is closed}&\iff Gra(\phi) \text{ is sequentially closed}\\
&\iff \forall \text{ convergent } \big(x_n,\phi(x_n)\big),\, \lim_{n\to\infty}\big(x_n,\phi(x_n)\big) \in Gra(\phi)\tag{2}
\end{align}

It is clear that $(1)\implies (2)$.
Indeed, if $\big(x_n,\phi(x_n)\big)$ is convergent, then $\mathbf{x}=(x_n)$ is convergent and hence by $(1)$ so too is $\big(\phi(x_n)\big)$,
with $\lim_{n\to\infty}\phi(x_n)=\phi(x)$.
Hence,

$$\lim_{n\to\infty}\big(x_n,\phi(x_n)\big)=\big(x,\phi(x)\big),$$

which belongs to $Gra(\phi)$ by definition.

We will show that when $\phi$ is bounded, $(2)\implies(1)$, which conludes the proof.

Suppose $\phi$ is bounded and let $\mathbf{x}\in\mathcal{C}(X)$.
Since $\phi$ is bounded and $(x_n)$ is convergent (and hence bounded), we have that $\big(\phi(x_n)\big)$ is a bounded sequence.
By the Bolzano-Weierstrass Theorem, there is a convergent subsequence $\big(\phi(x_{n_k})\big)$ of $\big(\phi(x_n)\big)$.

Of course, $(x_{n_k})$ it itself convergent and converges to $x$, the same limit of $(x_n)$.
It follows that $\big(x_{n_k},\phi(x_{n_k})\big)$ is convergent, and by $(2)$ we have that $\lim_{k\to\infty}\big(x_{n_k},\phi(x_{n_k})\big)=\big(x,\phi(x)\big)$.
In particular, $\lim_{k\to\infty}\phi(x_{n_k})=\phi(x)$.

It then suffices to show that $\big(\phi(x_n)\big)$ is convergent;
in this case, its limit must coincide with that of $\big(x_{n_k},\phi(x_{n_k})\big)$, that is, must equal $\phi(x)$. We show this by contradiction.

Indeed, if that were not the case, then there would be some $\epsilon>0$ and a subsequence $(x_{m_k})$ of $(x_n)$ with $d\big(\phi(x_{m_k}),\phi(x)\big)\geq \epsilon$ for all $k$.
Now, $\big(\phi(x_{m_k})\big)$ is of course bounded, so by the Bolzano-Weierstrass Theorem it must have a convergent subsequence, say $\left(\phi\left(x_{m_{k_j}}\right)\right)$.
By construction, we have that $$y=\lim_{j\to\infty}\phi\left(x_{m_{k_j}}\right)\neq \phi(x)$$

But $\left(x_{m_{k_j}}\right)$ is a subsequence of $(x_n)$, and hence converges to $x$.
It follows that $$\lim_{j\to\infty}\left(x_{m_{k_j}},\phi\left(x_{m_{k_j}}\right)\right)=(x,y)$$
and hence does not belong to $Gra(\phi)$, in contradiction with $(2)$. $\square$.

  Notice that in the proof above we didn't actually need the hypothesis that $\phi$ be bounded, but rather that $\phi$ 'preserves boundedness', ie, $\phi$ takes bounded sets to bounded sets.

This is wrong (for unbounded function). take $\phi : [0, \infty] \rightarrow R$ with $\phi(x) = \frac{1}{x}$ for all $x \in (0, \infty)$ and $\phi (0)=0$. its graph is closed but $\phi$ is not continuous!

EDIT: The above example is for unbounded function on limited domain (before editing the original question). If we assume function is bounded then the claim  is correct.

Proof: For Left to Right: the function $F(x) = (x, \phi(x))$  is continuous (now to show graph is closed in $X \times R^n$, use the sequential definition of continuity of $F$).
For Right to Left: similar to your argument, take $x_n \rightarrow x \in X$. want to show that $\phi(x_n) \rightarrow \phi(x),$ if actually $\phi(x_n) \nrightarrow \phi(x)$ then (since $\phi$  is bounded)  $\phi (x_n)$ has a convergent subsequence, say $\phi (x_{n_{k}}) \rightarrow y \neq \phi(x)$, i.e., $$(x_{n_{k}},  \phi (x_{n_{k}}) ) \rightarrow (x,y)\notin \text{Graph}.$$

Which is contradicting with Graph being Closed.


Question regarding basis and dimension So I was reading Linear Algebra: A geometrical approach by S.Kumaresan and there  is a problem saying prove that a vector space with finite number of elements in basis will be finite dimensional. Though it is easy to prove, in the very next part they said although in later part you will see that converse is not necessarily true. And here is my problem. Considering what they are saying then a finite dimensional vector space may have infinite number of elements in basis. Consider a vector space $V$ which is finite dimensional(let it be $n$ dimensional) have an infinite set as basis. Now there is a definition that statesA vector space is $k$ dimensional if it has a set of $k$ elements as basis. so the vector space $V$ which is finite dimensional(let it be $n$ dimensional) will have a finite element($n$) set as basis. But it is not true. Since in a finite dimensional vector space's any two bases have same number of elements. hence all of vector space$V$'s basis will have n element which is contradictory by our first assumption. Then we can't say the vector space is finite dimensional. But that seems contradictory too. I mean a vector space can't be finite dimensional and infinite dimensional at the same time because according to definition of finite dimensional: A vector space$V$ is finite dimensional if it has a finite elements set $S$ such that $L(S)=V$. Where $L(S)$ is span of the set $S$. Please tell me where am I wrong in this whole arguement. Thanks in advance.

OK. So these are the pictures of the definitions and theorems I have used on question.