discrete mathematics - Find All Solutions to System of Congruence

$$
\begin{cases}
x\equiv 2 \pmod{3}\\
x\equiv 1 \pmod{4}\\
x\equiv 3 \pmod{5}
\end{cases}
$$

$
n_1=3\\

n_2=4\\
n_3=5\\
N=n_1 * n_2 * n_3 =60\\
m_1 = 60/3 = 20\\
m_2 = 60/4 = 15\\
m_3 = 60/5 = 12\\
gcd(20,3)=1=-20*1+3*7\\
gcd(15,4)=1=-15*1+4*4\\
gcd(12,5)=1=12*3-5*7\\
x=-20*2-15*1+12*3\equiv -19 \equiv 41 \pmod{60}\\

$

The above is what I've tried so far. Can someone tell me where I'm going wrong? It's my first time doing this and looking at the explanation and examples for the Chinese Remainder Theorem makes my head want to explode.

Answer

I think you're a bit confused about the recipe used to solve the system of congruences using the Chinese Remainder Theorem. Using your notation, the actual solution is given by
\begin{align*}
x = 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) + 1 \cdot m_2 \cdot ({m_2}^{-1} \text{ mod } 4) + 3 \cdot m_3 \cdot ({m_3}^{-1} \text{ mod } 5) \, .
\end{align*}
Let's see why this answer works. Since $m_1$ is divisible by both $4$ and $5$, the first term is "invisible" when we consider $x$ mod $4$ and mod $5$: it is congruent to $0$ mod $4$ and mod $5$, so it doesn't contribute to the answer of the last two congruences. Thus, we can focus on making this first term satisfy the first congruence. Okay, so far we know the first term should have as factors $2$ (the congruence we want) and $m_1$ (to get rid of the effect mod $4$ and $5$). But now we don't satisfy the first congruence; the $m_1$ throws things off. To fix this, we multiply by the ${m_1}^{-1}$ mod $3$. Then mod $3$ we have
\begin{align*}

x &= 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) + 1 \cdot m_2 \cdot ({m_2}^{-1} \text{ mod } 4) + 3 \cdot m_3 \cdot ({m_3}^{-1} \text{ mod } 5)\\
&\equiv 2 \cdot m_1 \cdot ({m_1}^{-1} \text{ mod } 3) \equiv 2 \pmod{3}
\end{align*}
as desired. Similar reasoning shows that $x$ satisfies the given congruences mod $4$ and $5$.

You have the right idea with your solution, but you're missing some factors. Since you've shown $1 = 20 \cdot (-1) + 3 \cdot 7 \equiv 20 \cdot (-1) \pmod{3}$, then
$$
{m_1}^{-1} \text{ mod } 3 = 20^{-1} \text{ mod } 3 = -1 \equiv 2 \pmod{3} \, .
$$
So the first term should be $2 \cdot 20 \cdot 2$. Can you figure out the other two terms now?

calculus - Find $limlimits_{n to infty} sumlimits_{k = 0}^{n} dfrac{binom{n}{k}}{n2^n+k}$.

I have to find the following limit:

$$\lim\limits_{n \to \infty} \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k}$$

I thought I can use something from this other, seemingly similar question, but I don't see any way of manipulating this sum into something easier to work with. So how should I approach this limit?

Answer

When $k \in \{0, 1, \dotsc, n\}$
$$ \binom{n}{k} \frac{1}{n2^n+n} \leq \binom{n}{k} \frac{1}{n2^n+k} \leq \binom{n}{k} \frac{1}{n2^n}, $$
whence
$$ \frac{2^n}{n(2^n + 1)} \leq \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} \leq \frac{2^n}{n2^n}. $$

By squeezing,
$$ \lim_{n \to \infty} \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} = 0.$$

inequality - How to solve $|z^2-1|

How to solve $|z^2-1|<|z|^2$ where $z$ is a complex number? I have tried it both with cartesian and polar coordinates but did not get a solution.

I got that far: $z=x+yi$ and then I got: $$\pm x >(\frac{y^2+0.5}{1+4y^2})^{0.5}$$ but I don't know how to visualise that in the coordinate system.

Answer

That is equivalent to
$$|z^2-1|<|z^2|$$
This means that $z^2$ is at a shorter distance from $1$ than from $0$. Then $Re(z^2)>1/2$.

Now, write $z=x+iy$, thus, $z^2=(x^2-y^2)+2xyi$. The former inequality becomes
$$x^2-y^2>\frac12$$

calculus - Prove that $int_0^infty frac{sin nx}{x}dx=frac{pi}{2}$

There was a question on multiple integrals which our professor gave us on our assignment.

QUESTION: Changing order of integration, show that $$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy=\int_0^\infty \frac{\sin nx}{x}dx$$
and hence prove that $$\int_0^\infty \frac{\sin nx}{x}dx=\frac{\pi}{2}$$

---

MY ATTEMPT: I was successful in proving the first part.

Firstly, I can state that the function $e^{-xy}\sin nx$ is continuous over the region $\mathbf{R}=\{(x,y): 0

$$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy$$
$$=\int_0^\infty \sin nx \left\{\int_0^\infty e^{-xy}\,dy\right\} \,dx$$
$$=\int_0^\infty \sin nx \left[\frac{e^{-xy}}{-x}\right]_0^\infty \,dx$$

$$ =\int_0^\infty \frac{\sin nx}{x}dx$$

However, the second part of the question yielded a different answer.

$$\int_0^\infty \int_0^\infty e^{-xy}\sin nx \,dx \,dy$$
$$=\int_0^\infty \left\{\int_0^\infty e^{-xy} \sin nx \,dx\right\} \,dy$$
$$=\int_0^\infty \frac{ndy}{\sqrt{n^2+y^2}}$$

which gives an indeterminate result, not the desired one.

Where did I go wrong? Can anyone help?

Answer

You should have obtained $$\int_{x=0}^\infty e^{-yx} \sin nx \, dx = \frac{n}{n^2 + y^2}.$$ There are a number of ways to show this, such as integration by parts. If you would like a full computation, it can be provided upon request.

---

Let $$I = \int e^{-xy} \sin nx \, dx.$$ Then with the choice $$u = \sin nx, \quad du = n \cos nx \, dx, \\ dv = e^{-xy} \, dx, \quad v = -\frac{1}{y} e^{-xy},$$ we obtain $$I = -\frac{1}{y} e^{-xy} \sin nx + \frac{n}{y} \int e^{-xy} \cos nx \, dx.$$ Repeating the process a second time with the choice $$u = \cos nx \, \quad du = -n \sin nx \, dx, \\ dv = e^{-xy} \, dx, \quad v = -\frac{1}{y} e^{-xy},$$ we find $$I = -\frac{1}{y}e^{-xy} \sin nx - \frac{n}{y^2} e^{-xy} \cos nx - \frac{n^2}{y^2} \int e^{-xy} \sin nx \, dx.$$ Consequently $$\left(1 + \frac{n^2}{y^2}\right) I = -\frac{e^{-xy}}{y^2} \left(y \sin nx + n \cos nx\right),$$ hence $$I = -\frac{e^{-xy}}{n^2 + y^2} (y \sin nx + n \cos nx) + C.$$ Evaluating the definite integral, for $y, n > 0$, we observe $$\lim_{x \to \infty} I(x) = 0, \quad I(0) = -\frac{n}{n^2 + y^2},$$ and the result follows.

calculus - Calculate the limit: $lim_{nrightarrowinfty}left(frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}right)$

Calculate the limit: $$\lim_{n\rightarrow\infty}\left(\frac{1^{2}+2^{2}+...+n^{2}}{n^{3}}\right)$$

I'm planning to change the numerator to something else.

I know that $1+2+3+...n = \frac{n(n+1)}{2}$

And now similar just with $2$ as exponent but I did many tries on paper and always failed..

The closest I had is this but it still seems wrong:

$1^{2}+2^{2}+...+n^{2} = \frac{n(n^{2}+1)}{2}$

Well the idea is replacing numerator and then forming it, then easily calculate limit.. But I cannot find the correct thing for numerator..

Any ideas?

Answer

For variety,

$$\begin{align}
\lim_{n \to \infty} \frac{1^2 + 2^2 + \ldots + n^2}{n^3}
&=
\lim_{n \to \infty} \frac{1}{n} \left( \left(\frac{1}{n}\right)^2
+ \left(\frac{2}{n}\right)^2 + \ldots + \left(\frac{n}{n}\right)^2
\right)
\\&= \int_0^1 x^2 \mathrm{d}x = \frac{1}{3}
\end{align}
$$

probability theory - If $Z=X$ on $A$ and $Z=Y$ on $A^c$ then $Z$ is a random variable

Let $X$ and $Y$ be random random variables and let $A \in \mathcal{B}$. Prove that the function $Z$ defined by
$$Z(\omega)=\begin{cases}
X(\omega),& \text{ if } \omega \in A \\

Y(\omega),& \text{ if } \omega \in A^{c}
\end{cases}$$
is a random variable

Proof so far:
$$Z^{-1}((-\infty,a])=\{\omega:Z(\omega)\geq a\}=\{\omega: Z(\omega)\geq a, \omega \in A\}\cup\{\omega: Z(\omega)\leq a, \omega \in A^{c}\}=Y^{-1}[a,\infty) \cup X^{-1}([a,\infty))$$
So $Z$ is measurable

Answer

Let $X, Y$ be random variables in $(\Omega, \mathcal B, \mathbb P)$.

If $A \in \mathcal B$, then $1_A$ and $1_{A^C}$ are random variables.

Note that

$$Z = X1_A + Y1_{A^C}$$

Since sums or products of random variables in $(\Omega, \mathcal B, \mathbb P)$ are random variables in $(\Omega, \mathcal B, \mathbb P)$, $Z$ is a random variable in $(\Omega, \mathcal B, \mathbb P)$.

---

As for your proof, I think you should say:

1. $\forall a \in \mathbb R$




2. have $Z \ge a$ instead of $Z \le a$




3. $Z$ is $\mathcal B$-measurable

Probability that any outcome of a dice roll happens more than X times out of Y trials

I'm trying to determine the probability that a person experiences a "lucky number" when rolling a single, fair, 6-sided dice over a set of rolls in a single trial. A "lucky number" in this case is any face of the die that occurs visibly more common than one would normally expect. If you roll a six-sided die 100 times, you expect the outcome to occur with ~16.6 results of 1, 2, 3, 4, 5, and 6 ea, on average.

For example, you roll a six-sided dice in 100 independent trials, what is the probability that the occurrence of rolling any side of the dice happens at least 33 times over the course of the 100 independent trials? It doesn't matter if the roll was 1, 2, 3, 4, 5, or 6, just that the same result happened at least 33 times out of the 100 trials.

How would I calculate this?

Thanks.

Answer

The chance that $1$ comes up exactly $33$ times in $100$ comes from the binomial distribution. The chance of success is $\frac 16$ and failure is $\frac 56$ so it is ${100 \choose 33}(\frac 16)^{33}(\frac 56)^{67}\approx 0.00003$ If we sum from $33$ to $100$ we get the chance of at least $33\ 1$s, which is about $0.0005$ per Alpha. You can multiply these by $6$ to get the chance for any number, as it is very unlikely we doublecount by having at least $33$ of two different numbers. So the chance of a "lucky number" happening by chance is about $0.0003$ or one in $3300$. Pretty unlikely, but rarer things happen all the time.