TITLE: Prove $\lim_{N\to\infty} \int_{a}^{b} f_N (x) dx = \int_{a}^{b} \lim_{N\to\infty} f_N(x) dx$
QUESTION [0 upvotes]: If $f_N $ is a function sequence that is integrable on $(a; b)$ for all $N $ and $\lim_{N\to\infty} f_N (x)$ exists for all $x$ in $(a; b) $
Does the following necessarily hold?:
$$\lim_{N\to\infty} \int_{a}^{b} f_N (x) dx = \int_{a}^{b} \lim_{N\to\infty} f_N(x) dx$$

REPLY [1 votes]: The statement is correct only if the functions $f_N(x)$ tend to $f(x)$ uniformly.
Proof: Let $f(x)=\lim_{n\to\infty}f_n(x)$. Uniform convergence implies that:$$\forall\epsilon>0\quad,\quad \exists N\quad,\quad\forall x\in(a,b), n>N\quad,\quad |f_n(x)-f(x)|<\epsilon$$which leads to $$f(x)-\epsilon<f_n(x)<f(x)+\epsilon$$since both $f_n(x)$ and $f(x)$ are integrable we can argue that:$$\int_{a}^{b}f(x)-\epsilon dx<\int_{a}^{b}f_n(x)dx<\int_{a}^{b}f(x)+\epsilon dx$$or$$\int_{a}^{b}f(x)dx-(b-a)\epsilon <\int_{a}^{b}f_n(x)dx<\int_{a}^{b}f(x) dx+(b-a)\epsilon$$which means that$$|\int_{a}^{b}f_n(x)dx-\int_{a}^{b}f(x) dx|<(b-a)\epsilon$$and implies $$\lim_{n\to\infty}\int_{a}^{b}f_n(x)dx=\int_{a}^{b}f(x) dx=\int_{a}^{b} \lim_{n\to\infty}f_n(x)dx$$


text:
\begin{definition}[Definition:Canonical Homomorphism from Ring to Unital Algebra]
Let $A$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
Let $B$ be an [[Definition:Unital Algebra|unital algebra]] over $A$ with [[Definition:Unit of Algebra|unit]] $1$ whose [[Definition:Underlying Module of Algebra|underlying module]] is [[Definition:Unitary Module|unitary]].
The '''canonical homomorphism''' $A \to B$ is the [[Definition:Algebra Homomorphism|algebra homomorphism]] that sends $a \in A$ to $a \cdot 1$.
\end{definition}

config: wiki


TITLE: Elevator stop, other approach
QUESTION [1 upvotes]: There is a $10$-floor building, $10$ people get in the elevator in the ground floor and each gets off at one of the floors randomly and independently. What is the probability that the elevator stops at floor $5$?
$$1 - (9/10)^{10}$$
I understand the approach of using the probability that no one gets out on the $3$rd floor, which is $(9/10)^{10}$.
But, my question is how do I approach this problem using the probability that people gets out on the $3$rd floor?
I thought if one person chooses the $5$th floor the probability is $1/10$, and if $2$ people choose the $5$th floor then the probability is $1/9$, and so on.
Thank you.

REPLY [1 votes]: Yes, there is an alternative way to solve this, although it's more difficult than the method you mentioned.
Start by considering the case where there are 2 floors and 2 people.  The probability of at least one person exiting on our chosen floor, say floor 2, is
$$P(\geq\text{1 person on floor; 2 people and 2 floors}) = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}\right) = \frac{3}{4}$$
To see where this came from, we first consider where person 1 gets off.  There is a $1/2$ chance the person will exit on our chosen floor, so we add that to our probability.  In the event that she doesn't get off on floor 2 (an event with probability $1/2$), we consider the other person.  Person 2 has a $1/2$ probability of exiting on floor 2, so we multiply that by the probability that the second person even matters (ie. the probability of the first person not getting off on floor 2).
For 3 people and 3 floors, we'd have
$$P(\geq\text{1 person on floor; 3 people and 3 floors}) = \frac{1}{3} + \frac{2}{3}\left(\frac{1}{3}+\frac{2}{3}\left(\frac{1}{3}\right)\right) = \frac{19}{27}$$
since there's a $1/3$ chance of the first person getting off on our chosen floor, and a $2/3$ probability that they don't and that we have to look at the second person.  The same logic is applied to the second person and so on.
For 4 people and 4 floors, we'd have
\begin{align}P(\geq\text{1 person on floor; 4 people and 4 floors}) &= \frac{1}{4} + \frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}+\frac{3}{4}\left(\frac{1}{4}\right)\right)\right) \\
 &= \frac{175}{256}\end{align}
For $N$ people and $N$ floors, we'd have
\begin{align}&P(\geq\text{1 person on floor; N people and N floors}) \\ =& \frac{1}{N} + \frac{N-1}{N}\left(\frac{1}{N}+\frac{N-1}{N}\left(\;\;\;\;\left.\ldots+\frac{N-1}{N}\left(\frac{1}{N}\right)\right)\right)\ldots\right)
\end{align}
where the pattern continues until there are $N$ of the fractions $1/N$ present in the expression.
By multiplying this out, we have
\begin{align} P =& \frac{1}{N}+\frac{\left(N-1\right)}{N^2}+\frac{\left(N-1\right)^2}{N^3}+\ldots+\frac{\left(N-1\right)^{N-1}}{N^N} \\
=& \sum_{k=0}^{N-1} \frac{1}{N} \left(\frac{N-1}{N}\right)^k
\end{align}
and using the formula for the sum of a finite geometric series, ie
$$\sum_{k=0}^{n-1} a r^k = \frac{a\left(1-r^n\right)}{1-r}$$
we get,
\begin{align}P =& \frac{\left(\frac{1}{N}\right) \left(1-\left(\frac{N-1}{N}\right)^N\right)}{1-\left(\frac{N-1}{N}\right)} \\
=& 1 - \left(\frac{N-1}{N}\right)^N
\end{align}
which is the same expression as the one given by the more simple method.


TITLE: Pressure Equation Clarification
QUESTION [1 upvotes]: $$\text{pressure} = \frac{\text{force}}{\text{area}}$$
Does pressure in this equation refer to the gauge pressure or absolute pressure?

REPLY [1 votes]: This is how we define Pressure,
Pressure = Force/Area.

That's the basic definition. Now you need to understand what is absolute pressure and gauge pressure. They refer to the value of pressure at two different points. Assume a point at depth h below the surface of the water*[simplest of example]* now the total pressure experienced by a body is P+(density)(g)(Depth[h]) i.e. the pressure applied by the liquid above the body and the atmospheric pressure.
On the contrary, absolute pressure simply means that if you calculate the pressure experienced by the body in an Isolated environment , or the pressure applied just by the liquid column above it which would be P(absolute) = (density)(g)(depth).
Hope this helps. Feel free to ask questions if any.
Vishwas


TITLE: How to evaluate an integral involving the derivative of the delta function?
QUESTION [0 upvotes]: $$\int^1_{-1}e^{-5x}\delta'(x)dx$$
I know how to solve these integrals involving delta functions generally, but how do you do it when it is a derivative of the delta function?
Appreciate the help!

REPLY [1 votes]: Integrate by parts $$
\int_{-1}^1 e^{-5x}\delta^{'}(x) \ dx = - \int_{-1}^1 (e^{-5x})^{'}\delta(x) \ dx
$$


TITLE: Is $5^\pi$ an integer?
QUESTION [8 upvotes]: I saw this very short math problem on Twitter:

Is $5^\pi$ an integer?

It isn't (it's 156.992545309…), but is there some technique to prove this without a calculator?
My first and only idea so far is to prove $156<5^\pi<157$ by taking logarithms, but it's annoyingly close to 157, and anyway I used a calculator to find those bounds.

REPLY [1 votes]: This proof is intended be a more elementary than the existing proofs, relying more on "brute force". In exchange, it's more tedious, as you need to manually compute a bunch of square roots and multiplications.
As input, I assume that one knows that: $\pi$ has an expansion which is about 3.14159265...
We will try to prove that $5^\pi$ is not an integer
by multiplying $5^3 \times 5^\frac{1}{10} \times 5^\frac{4}{100} \times \ldots$
and using bounds.
Immediately, the problem one runs into is:
how do you compute powers like $5^\frac{4}{100}$ by hand? That's not usually covered in school math, but one does learn how to compute
square roots by hand.
So let's work with the expansion in base 2 instead of base 10.
$$
0.14159265... (\textrm{base 10}) \rightarrow 0.00100100001... (\textrm{base 2})
$$
We should first compute a bunch of values for $5^\frac{1}{2^N}$.
A good "checkpoint" value for $N$ that we will need (roughly) is such that
$5^\frac{1}{2^N} < 1.008$, because we will end up wanting
to bound:
$$
5^{3 + \frac{1}{8} + \ldots} < 5^\pi < 5^{3 + \frac{1}{8} + \ldots + \frac{1}{2^N}}
$$
and $5^3 \times 0.008 = 1$.
Tediously computing a bunch of square roots by hand:
$$
5^\frac{1}{2} = 2.23606\ldots \\
5^\frac{1}{4} = 1.49534\ldots \\
5^\frac{1}{8} = 1.22284\ldots \\
5^\frac{1}{16} = 1.10582\ldots \\
5^\frac{1}{32} = 1.05158\ldots \\
5^\frac{1}{64} = 1.02546\ldots \\
5^\frac{1}{128} = 1.01265\ldots \\
5^\frac{1}{256} = 1.00630\ldots
$$
Let's try the following:
$$
5^{3 + \frac{1}{8} + \frac{1}{64}} < 5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{256}}
$$
For the left side:
$$
125 \times 1.2228 \times 1.0254 < 5^{3 + \frac{1}{8} + \frac{1}{64}} < 5^\pi \\
156.73\ldots < 5^\pi
$$
For the right side:
$$
5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{256}} < 125 \times 1.2229 \times 1.0255 \times 1.0063 \\
5^\pi < 157.74\ldots
$$
Drat! We still have one integer $157$ within the bounds. But this was expected, because the interval size we're working with
is roughly 1, so when we got a lower bound of 156.73, the upper
bound should be around 157.73.
At this, point, the next 1 in the binary expansion of $\pi$ is 3 more digits
away, so the value must be relatively close to the lower bound. Let's try constraining the upper bound more.
$$
5^\frac{1}{512} = 1.0031\ldots \\
5^\frac{1}{1024} = 1.00157\ldots
$$
Retrying:
$$
5^\pi < 5^{3 + \frac{1}{8} + \frac{1}{64} + \frac{1}{1024}} < 125 \times 1.22285 \times 1.02547 \times 1.00158 \\
5^\pi < 156.99...
$$
So $156 < 5^\pi < 157$.


TITLE: Table of planar connected graphs
QUESTION [6 upvotes]: For the other day I need to use the table of planar connected graph with few vertex. In the wolfram's mathworld , they listed only the graph with $4$ vertex.
Does anyone knows webpages or pdf on the web which carries the table of planar connected graph with up to $10$ edges?

REPLY [11 votes]: A lot depends on what the question means, in particular are the graphs simple, and is equivalence defined by embedding-preserving isomorphism or all graph isomorphisms.
For simple graphs up to general isomorphism, see my page for files up to 11 vertices.
For simple graphs up to embedding-preserving isomorphism, use plantri to make them.  The command for $n$ vertices is "plantri -pc1m1 $n$", or "plantri -opc1m1 $n$" if mirror-image doesn't count as isomorphism. (At first, add "-u" to just count rather than output.) Tables up to 14 vertices are in this paper, Table 22.