Thursday, May 19, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


study the convergence of the sequence of function $f_{n}(x)=(1+\frac{x}{n})^{\frac{1}{n}}$

Posted: 19 May 2022 12:29 AM PDT

study the convergence of the sequence of function $f_{n}(x)=(1+\frac{x}{n})^{\frac{1}{n}}$ I know that it converges to 1 for all x in $R$ but I want to prove it what could I write. I also want to study whether it converges uniformly or not also, What should I write in this type of question?

Digits tower power iterate

Posted: 19 May 2022 12:20 AM PDT

Stack the digits of a natural number into a power tower, iterate until only one digit remains.

Does this iteration always terminate for any positive integer?

Additionally specify $0^n = 0$, even when $n=0$.

For example

$$ \begin{aligned} f_0&=58\\ f_1&=5^8=390625\\ f_2&=3^{9^{0}}=3^1=3\\ \end{aligned} $$


The smallest uncalculable number is 28:

$$28 \to 256 \to 39235776294252497421590\cdots$$

Free Boolean algebra over a commutative idempotent monoid

Posted: 19 May 2022 12:04 AM PDT

On the nLab page for BoolAlg (the category of Boolean algebras), there is a construction of a free Boolean algebra over a set $X$ which first takes $X$ to its free commutative idempotent monoid (namely by taking $\mathcal P_\text{fin}(X)$ with the operation of intersection). Then, to get to a Boolean algebra, we take $\mathcal P_\text{fin} \mathcal P_\text{fin} (X)$ with symmetric difference as addition and intersection as multiplication (unless I am mistaken).

Is it possible to construct a free Boolean algebra over an arbitrary commutative idempotent monoid?

Taking a commutative idempotent monoid $M$, I can give $\mathcal P_\text{fin}(M)$ addition defined by the symmetric difference. However, I seem to run into problems trying to define multiplication on $\mathcal P_\text{fin}(M)$ using the multiplication from $M$ so that the new multiplication is also idempotent. It seems like I'd need to further restrict $\mathcal P_\text{fin}(M)$ to finite subsets of $M$ that are closed under multiplication. In particular, we could define $A \cdot B = \{ a \cdot b : a \in A, b \in B \}$ for $A,B \in \mathcal P_\text{fin}(M)$. For idempotence, need $A \cdot A = A$, that is, $a \cdot a' \in A$ for all $a,a' \in A$. Is this the right path? It seems like we'd then run into problems with addition (symmetric difference) being well-defined.

Compare a computer simulation (of a chemical reaction network) to an SDE

Posted: 19 May 2022 12:22 AM PDT

I am currently reading the book 'Stochastic Analysis of Biochemical Systems' of Anderson and Kurtz. In this book, one studies chemical reaction networks and also how to simulate these with MATLAB. I know, that there is also a package in R Project, called GillespieSSA, which can be used for those kind of simulations, too. So far, I was able to implement some of the codes. However, I am wondering, how I could compare my simulated graphs to an actual SDE. To be more precise: I have the assumption that one graph (or one species in terms of chemical reaction networks) should be following a specific SDE

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

How can I check, that my simulation is in fact 'solving' this SDE? Do I need to work with the expectation and variance of the simulation over time and compare it to some theoretical value? I have the feeling, that there must be a way to compare a computational value to a theoretical value, to verify, that my simulation is a possible candidate for a solution of this SDE, so to speak.

I'd be happy for any hint or help! Any book or paper recommendation are very welcome, too!

Thank you in advance!

Prove that a relation R on set A is antisymmetric if and only if $R \cap R^{-1} \subseteq \{(a,a):a \in A\}$.

Posted: 19 May 2022 12:26 AM PDT

Can someone check to see if my proof is correct? If it actually is correct, can someone tell me how to be less verbose and "make it mathy and less wordy" for my backwards implication $(\Leftarrow)$ portion of the proof? Here's the problem:

Prove that a relation $R$ on set $A$ is antisymmetric if and only if $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$.

My proof attempt:

$(\Rightarrow)$ Let $R$ be an antisymmetric relation on $A$. Suppose $(a,b),(b,a) \in R$ where $a, b \in A$.

Then by definition of antisymmetric it must the case that $a=b$, so $(a,a) \in R \Rightarrow (a,a) \in R^{-1}$

$\Rightarrow (a,a) \in R\cap R^{-1} \Rightarrow (a,a) \in \{(a,a) : a \in A\}$.

Since $a, b$ were arbitrary elements, $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$

$(\Leftarrow)$ Let $R\cap R^{-1} \subseteq \{(a,a) : a\in A\}$. To be a subset of $\{(a,a) : a \in A\}$, $R\cap R^{-1}$ must be a set of ordered pairs where the 1st and 2nd terms of the ordered pairs are equal to a single element from $A$. Consider then, any arbitrary element of A, say $a \in A$. Then $(a,a) \in A \times A$ and $(a,a) \in R \cap R^{-1}$. By definition of intersection, $(a,a) \in R$ as well. These results hold for all elements of $A$, so $R \subseteq A\times A$. Thus, $R$ is a relation on $A$ by definition.

To say $R$ is an antisymmetric relation on $A$ is vacuously true since there will never be an ordered pair of the form $(x, y)$ or $(y,x)$ in $R$, unless, of course, $x = y$.

Since $R\cap R^{-1} \subseteq \{(a,a) : a\in A\} \Rightarrow$ the relation $R$ on set $A$ is antisymmetric

And since the relation $R$ on set $A$ is antisymmetric $\Rightarrow R\cap R^{-1} \subseteq \{(a,a) : a\in A\}$

Therefore the relation $R$ on set $A$ is antisymmetric $\Leftrightarrow$ $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$.

Difference in definition of Reflexive Relation between Set Theory and Class Theory

Posted: 18 May 2022 11:58 PM PDT

The class-theoretical context of this comes from Smullyan and Fitting's Set Theory and the Continuum Problem (revised ed., 2010).

Context: self-study.

Consider a relation $R$ on a set: $R \subseteq S \times S$

Consider a relation $R'$ on a class: $R' \subseteq V \times V$, where $V$ is a basic universe.

Now in the set-theoretical context, $R$ is a reflexive relation iff: $$\forall x \in S: (x, x) \in R$$

However, in the field-theoretical context, $R'$ is a reflexive relation iff: $$\forall x \in \mathrm {Field} (R'): (x, x) \in R'$$

where $\mathrm {Field} (R')$ is defined as the (class) union of the domain and range of $R'$, which are themselves defined as the subclasses of $V$ containing (respectively) the first coordinates of $R'$ and the second coordinates of $R'$.

The latter definition is according to S&F Chapter $4$: Superinduction, Well-Ordering and Choice, $\S 1$: Introduction to Well-Ordering.

This makes the definition of a reflexive relation between classes different from that between sets. But a set is a class. Hence the definitions are inconsistent.

Note further that this has the knock-on effect that if a (class-theoretically-defined) relation is both symmetric and transitive, then it is as a result reflexive.

This is at odds with the standard set-theoretical result that it is not necessarily the case that a relation both symmetric and transitive is also reflexive.

How is this resolved?

  • Have Smullyan and Fitting made a mistake here?
  • Are there sources which define a reflexive relation for class theory consistent with that for set theory, in which a reflexive relation is defined on a general class $C$ in which, if there exist $x \in C$ for which $x$ is not actually in the field of $C$, then the relation is specifically not reflexive?
  • Is it glossed over or otherwise brushed under the carpet?
  • Is it genuinely unimportant in the general scheme of things?

Considering that class theory was specifically invented to address inconsistencies and loose thinking in the whole area of set theory, it would be disappointing to learn that the above is considered a minor enough matter to ignore.

Please comment with requests for clarification if I have explained this insufficiently clearly.

How to get an entire function $f(z)$

Posted: 19 May 2022 12:22 AM PDT

$f$ is an entire function such that $|zf(z)-1+e^z|\leq 1+|z| \ \forall z\in \mathbb C$ then which of the folowing are true?

(1) $f'(0)=1$,

(2) $f'(0)=-1/2$,

(3) $f'(0)=-1/3$,

(4) $f'(0)= -1/4 $

My approach:

Since $f$ is an entire function so it has a power series expansion then $f(z)=\sum_{n=0}^{\infty}a_nz^n$.

So, $|z\sum_{n=0}^{\infty}a_nz^n-1+e^z|=|z\sum_{n=0}^{\infty}a_nz^n+\sum_{n=1}^{\infty}\frac{z^n}{n!}|=|a_0z+\sum_{n=1}^{\infty}(a_nz+\frac{1}{n!})z^n|$

So, $|a_0z+\sum_{n=1}^{\infty}(a_nz+\frac{1}{n!})z^n|\leq 1+|z|$

Now I can not understand how to proceed. Can anyone please help me? Thank you in advance.

What does it mean : infinite norm over the distance between two function? [duplicate]

Posted: 18 May 2022 11:51 PM PDT

I am trying to understand the assumption proof of Theorem 3 in the paper "A Universal Law of Robustness via isoperimetry" by Bubeck and Sellke.

Theorem 3. Let $\mathcal{F}$ be a class of functions from $\mathbb{R}^{d} \rightarrow \mathbb{R}$ and let $\left(x_{i}, y_{i}\right)_{i=1}^{n}$ be i.i.d. input-output pairs in $\mathbb{R}^{d} \times[-1,1]$. Fix $\epsilon, \delta \in(0,1)$. Assume that:

  1. The function class can be written as $\mathcal{F}=\left\{f_{\boldsymbol{w}}, \boldsymbol{w} \in \mathcal{W}\right\}$ with $\mathcal{W} \subset \mathbb{R}^{p}$, $\operatorname{diam}(\mathcal{W}) \leq W$ and for any $\boldsymbol{w}_{1}, \boldsymbol{w}_{2} \in \mathcal{W}$, $$ \left\|f_{\boldsymbol{w}_{1}}-f_{\boldsymbol{w}_{2}}\right\|_{\infty} \leq J\left\|\boldsymbol{w}_{1}-\boldsymbol{w}_{2}\right\| $$

what does mean by the infinite norm over $(f_{w1}-f_{w2})$. Can anyone explain it with an example. I know infinite norm of a vector means max of the absolute value of the component like if v is a vector consist of (1,-3,2) $||v||_{\infty}$ = max(1,|3|,2) = 3.

But how does it work for function? , further how it works for this case i.e subtraction of the two function?

Find $2^{(3^{5782})} \pmod{20}$ [duplicate]

Posted: 19 May 2022 12:11 AM PDT

How to find $2^{(3^{5782})}\pmod{20}$?

Attempt: Using Fermat's little theorem with the numbers $2$ and $5$ we get $2^{4} \equiv 1 \pmod{5}$.

Therefore, $$ 2^{(3^{5782})} \equiv 2^{4t+x} \equiv (2^4)^{t}·2^x \equiv 2^x \pmod{5}. $$ So, $x \equiv 3^{5782} \pmod{4}$.

If we use Euler's theorem we get $3^{\phi(4)} \equiv 3^2 \equiv 1 \pmod{4}$. Thus, $$ x \equiv 3^{5782} \equiv (3^2)^{2891} \equiv 1 \pmod{4} $$ and the result is $2^{(3^{5782})} \equiv 2^x \equiv 2 \pmod{5}$. Now how does that help me, if at all?

Prove that $ z = \cos( \pi/2 + 2\pi k ) + i \sin( \pi/2 + 2\pi k )$ has only two values for $k \in Z $

Posted: 19 May 2022 12:14 AM PDT

So, I want to actually find and prove the number of distinct values of

1

Mary (M) is twice as old as Ann (A) was when M was half as old as A will be when A is $3$ times as old as M was when M was $3$ times as old as A was

Posted: 18 May 2022 11:59 PM PDT

The combined ages of Mary and Ann is 44 years, and Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann was.

How old is Mary?

Translating into equations seems to be the way to go with this but the difficulty is where to start after the second "and".

"The combined ages of Mary and Ann is 44" clearly translates to:

$$M + A = 44$$

Any suggestions?

Showing $\int _{0} ^{\pi/4} \frac{\cos^{2022}(x)}{\sin^{2022}(x) + \cos^{2022}(x) } dx \approx \frac{\pi}{4}$

Posted: 19 May 2022 12:27 AM PDT

Show that $$\int_{0} ^{\pi/4} \frac{\cos^{2022}(x)}{\sin^{2022}(x) + \cos^{2022}(x) } dx \approx \frac{\pi}{4}$$

My method was this: I tried using $x \to \pi/4-x$ conversion but that doesn't lead to common denominator. Next thing I tried was to take help of approximation as the answer too is an approximation, so I thought as $\cos x >\sin x$ in $(0,\pi/4)$, but didn't got a good reason to neglect the $\sin^{2022}(x)$ in comparison to $\cos^{2022}x$. Can anyone explain how we can do so?

Note: also I think at $x= \pi/4$ we cannot neglect at all since both would be equal, so what in that point, will the integral be not a bit more then?

Correct notation to illustrate the product of a weighted path

Posted: 19 May 2022 12:26 AM PDT

I am struggling to represent a situation using the correct mathematical notation. I have a graph that consists of a series of nodes (1-4) and edges which represent movement pathways between nodes. The edges have an attribute d, which represents the distance between the two adjacent nodes.

an example of a network

The probability of successfully moving between two adjacent nodes i and j is a function of distance (d):

$$P_{success} = e^{-d_{i,j}/10}$$

e.g. for each adjacent pair of nodes in the figure above the probability of success would be:

$$P_{success, 1,2} = e^{-0.5/10} = 0.95$$

$$P_{success, 2,3} = e^{-0.1/10} = 0.9$$

$$P_{success, 3,4} = e^{-1.5/10} = 0.86$$

Now I apply an additional toll to the probability of success ($P_{toll}$) which can range from 0 to 1. If $P_{toll}$ = 0.5 then the probability of moving from node 1 to node 4 would be:

$$(0.95 * 0.5) * (0.9 * 0.5) * (0.86 * 0.5) = 0.09$$

My Question:

How do I present a general form of the above situation mathematically?

This is my attempt (which I know is wrong), hopefully I am at least on the right track…

$$\prod^j_{i:j}(e^{-d_{i,j}/10} P_{toll})$$

Edit

To clarify, I am not only interested in the distance from node 1 to 4, but from any node to any other node.

Interchanging almost surely equal random variables in a conditional expectation

Posted: 19 May 2022 12:19 AM PDT

Suppose we have random variables $U, V, Y$, where $Y$ is real-valued, and some event $A$ such that, almost surely, $$ U \, \mathbb{1}_A = V \, \mathbb{1}_A $$ where $\mathbb{1}_A$ is an indicator function. Does it follow that $$ \mathbb{E}[Y \mid \mathbb{1}_A = 1, U] \, \mathbb{P}(A \mid U) = \mathbb{E}[Y \mid \mathbb{1}_A = 1, V] \, \mathbb{P}(A \mid U) $$ almost surely?

If everything is discrete, then, provided $\mathbb{P}(U = u) > 0$, we have $$ \mathbb{E}[Y \mid \mathbb{1}_A = 1, U = u] \, \mathbb{P}(\mathbb{1}_A = 1 \mid U = u) %= \left(\sum_{y} y \, \mathbb{P}(Y = y \mid \mathbb{1}_A=1, U=u)\right)\mathbb{P}(\mathbb{1}_A = 1 \mid U = u) \\ = \frac{\sum_y y \, \mathbb{P}(Y = y, \mathbb{1}_A=1, U=u)}{\mathbb{P}(\mathbb{1}_A=1, U=u)} \, \mathbb{P}(\mathbb{1}_A = 1 \mid U = u)\\ = \frac{\sum_y y \, \mathbb{P}(Y = y, \mathbb{1}_A=1, V=u)}{\mathbb{P}(\mathbb{1}_A=1, V=u)} \, \mathbb{P}(\mathbb{1}_A = 1 \mid U = u) \\ = \mathbb{E}[Y \mid A = a, V = u] \, \mathbb{P}(\mathbb{1}_A = 1 \mid U = u). $$ I am wondering if the same holds true in general, i.e. under the general definition of conditional expectation, with $Y$ an arbitrary (integrable) random variable, and $U$ and $V$ arbitrary measurable functions (subject to the condition above).

From algebraic point of view, what are the similarities and differences between these two hypercomplex systems?

Posted: 19 May 2022 12:22 AM PDT

I would like to know how compare from the algebraic point of view these two 3-dimensional hypercomplex number systems.

3-dimensional split-complex numbers

Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication.

Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$.

A number would be written in the form $a+bj+ck$. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$.

Here is a Mathematica code to experiment with:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];  $Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} ->        x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;  

Using this code one can see that

$j^2=k^2=1$

$jk=j+k-1$

$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$

$j^j=j^k=j$

$k^k=k^j=k$

$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$

$0^{j+k}=1-\frac{j}{2}-\frac{k}{2}$

The division formula would be:

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\right)+\frac{a_1+b_1-c_1}{2 \left(a_2+b_2-c_2\right)}+\frac{a_1-b_1+c_1}{2 \left(a_2-b_2+c_2\right)}$

If we add a complex unity $i$, we will get a 6-dimensional number system.

Particularly, we will see that

$i^{j+k}=1-j-k$

and

$\log (j k)=i\pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$

Triplex numbers

This is a realization of triplex numbers, described in this video.

Here,

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$

$j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$

$k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$

Mathematica code:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];  $Pre = (# /. ({j -> {1, E^(2 I \[Pi]/3)},             k -> {1, E^(-2 I \[Pi]/3)}}) /. {x_, y_} ->           FullSimplify[(x/3 + Im[y]/Sqrt[3] - Re[y]/3) j + (x/3 -                Im[y]/Sqrt[3] - Re[y]/3) k +             1/3 (x + y + Conjugate[y])] // FullSimplify // Expand) &;  

Particularly, we will see that

$j^2=k$, $k^2=j$, $jk=1$

$j^k=-\frac{1}{3} 2 e^{\frac{\pi }{\sqrt{3}}} j+\frac{j}{3}+\frac{1}{3} e^{\frac{\pi }{\sqrt{3}}} k+\frac{k}{3}+\frac{e^{\frac{\pi }{\sqrt{3}}}}{3}+\frac{1}{3}$

$0^{j + k + 1}=-\frac{j}{3}-\frac{k}{3}+\frac{2}{3}$

$\log j = \frac{2 \pi j}{3 \sqrt{3}}-\frac{2 \pi k}{3 \sqrt{3}}$

$\log(j+k)=\frac{\pi j}{\sqrt{3}}+\frac{1}{3} j \log (2)-\frac{\pi k}{\sqrt{3}}+\frac{1}{3} k \log (2)+\frac{\log (2)}{3}$

The division formula is

$\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{a_2^2 \left(a+b k+c j\right)-a_2 \left(b_2 \left(a k+b j+c\right)+c_2 \left(a j+b+c k\right)\right)+c_2^2 \left(a k+b j+c\right)-b_2 c_2 \left(a+b k+c j\right)+b_2^2 \left(a j+b+c k\right)}{a_2^3+b_2^3+c_2^3-3 a_2 b_2 c_2}$

and

$N(a+bj+ck)=\sqrt[3]{a^3+b^3+c^3-3abc}$

is the analog of modulus.

If we add the complex unity, we will see that

$i^{j+k}=-\frac{j}{3}-\frac{j}{\sqrt{3}}-\frac{k}{3}+\frac{k}{\sqrt{3}}-\frac{1}{3}$


It looks that the both systems are commutative associative algebras with zero divisors but without nilpotents. Can something be said about their differences from the algebraic point of view?

How do I find the partial sum of the Maclaurin series for $e^x$?

Posted: 18 May 2022 11:48 PM PDT

In one of the problems I am trying to solve, it basically narrowed down to finding the sum $$\sum^{n=c}_{n=0}\frac{x^n}{n!}$$ which is the partial sum of the Maclaurin series for $e^x$.

Wolfram | Alpha gave the output $$\frac{e^x\Gamma(c+1,x)}{\Gamma(c+1)}$$ But I am unsure how to derive this. All my usual summation evaluation tricks are not helpful and I am clueless.

Help is appreciated!

How to compute $\frac{63.5\times 0.5\times 10\times 60}{2\times 96500}$ without a calculator?

Posted: 18 May 2022 11:54 PM PDT

Thankfully, I have been provided 4 options:

(a) $0.0987$

(b) $0.0897$

(c) $0.0798$

(d) $0.0789$

My attempt:

$$\frac{63.5\times 0.5\times 10\times 60}{2\times 96500}$$

$$\frac{63.5\times 0.5\times 600}{2\times 96500}$$

$$\frac{63.5 \times 300}{2\times 96500}$$

$$\frac{63.5\times 150}{96500}$$

$$\frac{635\times 15}{96500}$$

$$\frac{6350+3000+150+25}{96500}$$

$$\frac{9500+25}{96500}$$

$$\frac{9525}{96500}$$

I'm still stuck with a long division.

$$0.09$$

$$\overline{96500)952500}\tag{1}$$

$$868500$$

[I couldn't display the long division nicely.]

We do not need to continue the long division further. We can understand the answer will be (a) by checking the options.

My question:

  1. This is actually a chemistry question from a competitive exam, which involves stoichiometric calculations. I had the most trouble doing the long division. I had to guess that the number is $9$ and had to multiply $9$ by $96500$. Is there a quicker and easier way?

Solving $|x-3|^{3x^2-10x+3}=1$ [closed]

Posted: 19 May 2022 12:24 AM PDT

Consider the equation $$|x-3|^{3x^2-10x+3}=1$$

Rewriting it in logarithmic form, $$\log_{|x-3|}1=3x^2-10x+3.$$

Since the base of a logarithm must be neither $0$ nor $1,$ $$3x^2-10x+3=0\\x=3\quad\text{or}\quad\frac13\\x=\frac13.$$

However, the given solution is $$x=\frac13\quad\text{or}\quad2\quad\text{or}\quad4.$$

Please tell me what I am missing here.

Linear function of independent uniform distributed random variables.

Posted: 18 May 2022 11:59 PM PDT

Given that two independent uniform random variables X and Y, i am trying to find the function of Z = aX + bY +c. The method which I used was jacobian transformation to find the pdf. Let W = Y. Let us assume X is uniform on interval $\left[ l_1, r_1 \right] $ and Y is uniform on interval $ \left[ l_2, r_2 \right] $

So,

\begin{align} g(z) = \int_{-\infty}^{\infty} \frac{f(\frac{Z - bW-C}{a}, W)}{\lvert a \rvert}dW \end{align}

The above can be verified by Jacobian.

since \begin{align} f(X, Y) = \frac{1}{\left( r_1 - l_1 \right) \left( r_2 - l_2 \right)} \end{align}

\begin{align} g(z) = \int_{-\infty}^{\infty} \frac{1}{\lvert a \rvert \left( r_1 - l_1 \right) \left( r_2 - l_2 \right)} dW \end{align}

Let us take a problem, where suppose $R_1$ is the rate of interest of the 1st stock and $R_2$ is the rate of interest of the 2nd stock, and $ R_1 \in \left[ -10, 20 \right] and R_2 \in \left[ -4.5, 10 \right] and R_1 , R_2 $ are independent, we want to find the pdf of portfolio, $ Z = 54R_1 + 110R_2 $

substuting in the above formula gives,

\begin{align} g(z) = k \left(z + 1035 \right) , -1035 < z < 560 \end{align}

\begin{align} g(z) = 14.5k, 560 < z < 585 \end{align}

\begin{align} g(z) = k \left(2180 - z \right) , 585 < z < 2180 \end{align}

where k is found to be $ 4.26 * 10^{-5} $

But if we integrate the above pdf for z, the value is much larger than 1.

The correct answer is they take $ P = 54R_1 $ and $ Q = 110R_2 $, take Z = P + Q and apply the above formula. where P is uniform on interval $\left[ -540, 1080 \right] $ and Q is uniform on interval $ \left[ -495 , 1110 \right] $

then the pdf is

\begin{align} g(z) = k \left(z + 1035 \right) , -1035 < z < 560 \end{align}

\begin{align} g(z) = \frac{k}{1620}, 560 < z < 585 \end{align}

\begin{align} g(z) = k \left(2180 - z \right) , 585 < z < 2180 \end{align}

where k is found to be $ 3.87 * 10^{-7} $

My question is why doesnt the initial method does not work ?

I found k to be \begin{align} \frac{1}{a \left( r_1 - l_1 \right) \left( r_2 - l_2 \right)} \end{align}

But the correct solution finds k to be \begin{align} \frac{1}{ab \left( r_1 - l_1 \right) \left( r_2 - l_2 \right)} \end{align}

The problem is in Probablity and Statistics 4th ed. Problem 12 on sec 4.2. Page 225

Kindly explain why this difference. Thankyou

Binary Sequences and Combinatorics

Posted: 19 May 2022 12:27 AM PDT

I have the following problem:

Consider binary sequences of length $n$. We know there are $2^{n}$ of them. Now I take $2^{n-2}$ of them (except the zero sequence). If we take only 2- sums can we generate the remaining sequences ?. Example: $n =4$. Take 4 of them. Lets take $(1000), (0010), (1001), (0100),$

The possible 2 - sums are

$(1000) + (0010) =(1010), (1000) + (1001)=(0011), (1100), (1011), (0110), (1101).$

Here I dont have enough sums to generate all the remaining sequences.

What about this in general. My guess says that it is not possible.

Any hints towards this will highly be appreciable.

Edit:The problem can be reformulated as taking any set with $n$ elements. Take $2^{n-2}$ non - empty subsets. Can we obtain all the remaining subsets by the 2 - symmetric differences of these elements. By 2 - symmetric I mean the symmetric difference of two subsets from our list.

Find the remainder of $3^{333}$ divided by $100$

Posted: 19 May 2022 12:31 AM PDT

Find the remainder of $3^{333}$ divided by $100$

So I can find that $100=2^2\cdot 5^2$

Then I want to find $3^{333}$ mod $4$ and mod $25$ and use chinese remainder theorem to find a solution mod $100$.

I can find that $3^{333}\equiv (-1)$ mod $4$

But then $3^{333}=((3^3)^3)^{37}\equiv (27^3)^{37}\equiv (2^3)^{37}\equiv 8^{37}$ mod $25$

But I cannot find $8^{37}$ mod $25$

Inclusion/exclusion, at least and exactly arrangements?

Posted: 19 May 2022 12:10 AM PDT

The question wants to count certain arrangements of the word "ARRANGEMENT":

a) find exactly 2 pairs of consecutive letters?

b) find at least 3 pairs of consecutive letters?

I have the answer given from the tutor but it doesn't make sense to me.

Let's start with the base case:

$S_2 = \frac{(11-2\times2+2)!}{(2!)^{(4-2)}}\binom{4}{2}$:All possible combinations for 2 pairs of consecutive letters are known.

$S_3 = \frac{(11-2\times3+3)!}{(2!)^{(4-3)}}{4\choose3}$:All possible combinations for 3 pairs of consecutive letters are known.

$S_4 = (11-2\times4+4)!\binom{4}{4}=7!$:All possible combinations for 4 pairs of consecutive letters are known.

The equation for exactly m conditions:$E_m = S_m - {m + 1\choose1}S_{m+1} + {m + 2\choose2}S_{m+2}$.

The equation for at least m conditions:$L_m = S_m - {m \choose1}S_{m+1} + {m + 1\choose2}S_{m+2}$.

Answer for (a):$E_2 = S_2 - {3\choose1}S_3 + {4\choose3}S_4$.

Answer for (b):$L_3 = S_3 - {3\choose1}S_4$.

For (a), I don't understand why we need to multiply $\binom{3}{1}$ with $S_3$ and $\binom{4}{3}$ with $S_4$?

If we have ${S_3}$ that satisfies the requirement for ${S_2}$ (as three pairs would include two pairs), then wouldn't ${E_2} = S_2 - S_3$?

For (b), wouldn't the answer just be ${S_3}$ since we it contains the combination for every triple pair?

I don't understand the given formula used for ${E_m}$ and ${L_m}$, mainly the combinatorics part because it looks to me that we already handled that combinations in the calculations of ${S_2}$, ${S_3}$, ${S_4}$.

Could someone please explain the formula and why the answers are as such?

Thanks!

EDIT: I got the question from https://www.youtube.com/watch?v=D1T3xy_vtxU&index=8&list=PLDDGPdw7e6Aj0amDsYInT_8p6xTSTGEi2 - start the video at (4.35) for the question

How can I calculate $24^{33} \bmod 80$

Posted: 19 May 2022 12:25 AM PDT

How can I calculate $24^{33} \mod 80$. I don't see how to start. I only know that $\phi (80) = 32$. A hint would suffice, thanks!

How to solve definite integral

Posted: 19 May 2022 12:24 AM PDT

Is there an easy way to compute the definite integral \begin{align*} \int_{-\pi}^{\pi} x^{n}e^{\mathrm{i}a x^2-\mathrm{i}b x}\,dx, \end{align*} where $ n\in \mathbb{N} $ and $ a,b\in \mathbb{R} ?$ I found some formulas for $\int_{-\infty}^{\infty}$ or $\int_{0}^{\infty}$ but I want to solve for finite limits.

What is the mathematical symbol for range?

Posted: 19 May 2022 12:23 AM PDT

An unsophisticated question from a neophyte:

Given the numbers: $1,2,3,4,5$

What is the symbol for the range of the numbers?

i.e. the lowest-highest number in the set. For example, the min max is $1-5$.

The ____ is $1-5$. (insert math symbol into blank).

Should such a beast exist, I'd be particularly interested in it's unicode character...

Tried searching a few online resources, such as: http://rapidtables.com/math/symbols/Basic_Math_Symbols.htm without success.

Proof that there are infinitely many primes congruent to 3 modulo 4

Posted: 18 May 2022 11:52 PM PDT

I'm having difficult proving this.

As a hint the exercise to prove first, that if $a\lneqq \pm 1$ satisfies $a \equiv 3 \pmod4$, then exist $p$ prime, $p \equiv 3 \pmod 4$ such $p\mid4$. But I'm not really getting for what purpose can this be used.

Finding volumes — when to use double integrals and triple integrals?

Posted: 19 May 2022 12:00 AM PDT

This is not a technical question at all, but I'm quite confused about what should I use to compute volumes in $\mathbb{R}^3$ with integration.

I've read somewhere that a double integral gets the volume swapping across the $x$ and $y$ axis while a triple integral just integrate the whole thing at once, how accurate is this? Can a volume expressed by a double integral be expressed by a triple integral?, And can a triple always be expressed by a double? This one doesn't seem true, but I don't have a good answer to why.

I also found this comments while reading openstudy.com made by someone named KingGeorge one year ago:

For a double integral you have to integrate some function, for a triple integral, you integrate 1.

Does this mean that using an integral to get a volume always should look like $\iiint dxdydz$ without any function?

Geometrically, there are a few things you can be looking at. One, you're finding a 4-volume. That is, the 4-dimensional equivalent of volume. Two, if the volume in the region you're integrating in has a changing density, you could be finding the total mass.

I'm not sure if I understand correctly, but this means that a triple integral does not compute exactly the volume I want but a 4-D equivalent?.

Break-even analysis with net profit

Posted: 19 May 2022 12:15 AM PDT

You have total fixed costs of $\$12,000$. Your manufacturing and shipping of the widgets costs $\$7$ per widget. You sell each widget for $\$22$. Whats the break-even point?

If you sell the widgets for $\$22$,

$$\frac{12000}{22-7}= 800$$

widgets. How many widgets do you have to sell to realize a net profit of $15,000?

Proof that every repeating decimal is rational

Posted: 19 May 2022 12:12 AM PDT

Wikipedia claims that every repeating decimal represents a rational number.

According to the following definition, how can we prove that fact?

Definition: A number is rational if it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.

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