Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I co... | First observe that
\begin{eqnarray*}
5^6 \equiv 5 \pmod{22}.
\end{eqnarray*}
So we only need to calculate powers of $5$ modulo $22$, they are $5,3,15,9,1$ (by tedious calculation), the minimal value of $k$ is $\color{red}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so fa... | Without DCT: In absolute value the integrand is bounded above by
$$\frac{n^2 (x/n)}{n^3x + x(1+x^3)} = \frac{n}{n^3 + (1+x^3)} \le \frac{n}{n^3 + x^3}.$$
Let $x = ny$ to see that
$$\int_0^\infty \frac{n}{n^3 + x^3}\, dx = \frac{1}{n}\int_0^\infty \frac{1}{1 + y^3}\, dy.$$
Since the last integral converges, the integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Explanation of this hint involving the summation of cosine and sine I have this question
Let $n \ge 2$ be an integer. Prove that $$\sum_{k=0}^{n-1}\cos(\frac{2k\pi}{n}) = 0 = \sum_{k=0}^{n-1}\sin(\frac{2k\pi}{n})$$
I was given the hint to
Set $z = \cos(\frac{2\pi}{n}) + i\sin(\frac{2\pi}{n})$, so $z^n=1$. Now writ... | When you set $z:=\text{cis}\left( \dfrac{2\pi}{n} \right)$ you win all the solutions to $z^n-1=0$, in factorization, using one of moivre law's ($\text{cis}^m(x)=\text{cis}(m.x)$),
$$ (z-1)\cdot \sum_{k=0}^{n-1} \text{cis}\left( \dfrac{2\pi}{n} \cdot k\right) = 0 + 0.\text{i} $$
If $z \neq 1$, we have that crazy sum a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Extrapolating $\pi$ using Taylor expansion Let $P_n$ be a polygon inscribed in a circle with diameter $1$. Each side of the polygon has length $l_n=\sin(\pi /n)$ and the circumference of $P_n=nl_n$.
With
$$P_n= \pi - \frac{\pi^3}{3!}\frac{1}{n^2}+ \frac{\pi^5}{5!}\frac{1}{n^4}- \frac{\pi^7}{7!}\frac{1}{n^6}+\dots$$
We ... | Denote by $a,b,c,d$ the coefficents for $P_2,P_3,P_4,P_6$.
The errors cancel if
$$\left(
\begin{array}{ccc}
\frac{1}{2^2} & \frac{1}{2^4} & \frac{1}{2^6} \\
\frac{1}{3^2} & \frac{1}{3^4} & \frac{1}{3^6} \\
\frac{1}{4^2} & \frac{1}{4^4} & \frac{1}{4^6} \\
\frac{1}{6^2} & \frac{1}{6^4} & \frac{1}{6^6} \\
\end{array}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Expand $(1+u)^{\tfrac12}$ up to order $3$:
$$(1+u)^{\tfrac12}=1+\frac12 u-\frac18u^2+\frac1{16}u^3+o(u^3),$$
and compose with $u=-2x+3x^2-4x^3$:
*
*$u^2=4x^2-12x^3+o(x^3)$,
*$u^3=u^2\cdot u=-8x^3+o(x^3)$.
One finally obtains$$1-x+x^2-x^3+o(x^3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac... | \begin{align*}
\sum_{n=1}\dfrac{1}{(2n-1)(2n+1)}&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2(n+1)-1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{f(n)}-\dfrac{1}{f(n+1)}\right)\\
&=\dfrac{1}{2}\dfrac{1}{f(1)},
\end{align*}
where $f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve the Integral: $\int{\sqrt{8y-x^2}dx}$ I am trying to solve the following integral:
$$\int{\sqrt{8y-x^2}}dx$$
For which I don't understand clearly how to work with root values.
What I've tried:
$$
\int{\sqrt{8y-x^2}dx} = \int{(8y-x^2)^{\frac{1}{2}}dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\int{(8y-x^2)dx}... | this isn't correct, Substitute $$x=\sqrt{8y}\sin(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions a... | You can do it with Taylor expansions:
$x\left(1+x^{2}\right)^{1/2}-x\left(1+x^{3}\right)^{1/3}=
x^{2}\left(1+\frac{1}{x^{2}}\right)^{1/2}-x^{2}\left(1+\frac{1}{x^{3}}\right)^{1/3}
=x^{2}\left(1+\frac{1}{2}\cdot\frac{1}{x^{2}}+o\left(\frac{1}{x^{2}}\right)\right)-x^{2}\left(1+\frac{1}{3}\cdot\frac{1}{x^{3}}+o\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Splitting the square root of complex function into real and imaginary parts I have these functions below: $$\sqrt{(x+iy)^2-a^2}$$
$$\frac{b(x+iy)}{\sqrt{(x+iy)^2-a^2}}$$
How do I split these to get the real and imaginary parts of these functions?
If anyone could help me out, that really would be helpful!!!!
It will hel... | Basically we need a complex number $A+ib$ such that its square equals $(x+iy)^2-a^2=(x^2-a^2-y^2)+i(2xy)$.
We have: $$(A+ib)^2=(A^2-b^2)+i(2Ab)=(x^2-a^2-y^2)+i(2xy)$$ giving us: $A^2-b^2=x^2-a^2-y^2$ and $Ab=xy$.
Now, we will get, $$A^2+b^2 = \sqrt{(A^2+b^2)^2} = \sqrt{(A^2-b^2)^2+4A^2b^2}=\sqrt{(x^2-a^2-y^2)^2+4x^2y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many distinct numbers between $10$ and $1000$ can be formed from the digits $2$, $3$, $4$, $5$, $6$, and $0$ without repetition of any digit? Here is my attempt:
If $0$ is not to be any digit of the number, then there are $5 \times 4 \times 3 = 60 = ^5P_2$ possibilities for a 3-digit number, whereas there are $5 \... | First we deal with all two digit numbers ($11 - 99$). There are two numbers to pick, and tens place cannot be $0$. Hence number of ways are $5\cdot 5$.
Now three digit numbers ($100-999$). Again we cannot have $0$ at hundreds place, so we have $5\cdot 5 \cdot 4$ ways.
In all we have $5^3$ numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$ ?
EDIT: So ... | In order to compute $\arcsin(2x\sqrt{1-x^2})$, you need that
$$
\bigl|2x\sqrt{1-x^2}\,\bigr|\le 1
$$
which becomes
$$
4x^2(1-x^2)\le 1
$$
and therefore
$$
4x^4-4x^2+1\ge0
$$
Since the left-hand side is a square, the inequality is satisfied for every $x$, provided $1-x^2\le1$. This is equivalent to $-1\le x\le 1$.
You ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& =... | This is my first attempt at answering my own question. I realise the question is of a low quality, but I'm focusing on the question-and-answer aspect.
There are a few issues with the OP's attempt. Firstly, the base step was skipped. Secondly, and crucially, $f(k+1)$ is no longer odd, so it cannot be used in the induct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.... | Notice $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. For the identity at hand,
$$\sum_{k=1}^n k^5 + \sum_{k=1}^n k^7 \stackrel{?}{=} 2 \left(\sum_{k=1}^n k\right)^4$$
If one compute the difference of successive terms in RHS, we find
$$\begin{align}{\rm RHS}_n - {\rm RHS}_{n-1}
&= 2\left(\frac{n(n+1)}{2}\right)^4-2\left(\frac{n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| Write
$$\lim_{x\to0}\dfrac{\cos x-1+1-\sqrt{\cos 2x}+\sqrt{\cos 2x}-\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}=$$
$$\lim_{x\to0}\dfrac{\cos x-1}{x^2}+\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}+\lim_{x\to0}\sqrt{\cos 2x}\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}$$
and we have
\begin{align}
&\lim_{x\to0}\dfrac{\cos x-1}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find $A^n$, if $n$ is Natural number Find eqation for:$\ A^n$ when:
$$A =\left( \begin{array}{cc}
a & 1 & 0\\
0 & a & 1\\
0 & 0 & a\end{array} \right)$$
I calculated $\ A^2$ $\ A^3$ and compared entries:
At the end I've got something like:
$$A^n =\left( \begin{array}{cc}
a^n & na^{n-1} & ???\\
0 & a^n & na^{n-1}\\
0 & ... | $$A^2 = \left(
\begin{array}{ccc}
a^2 & 2 a & 1 \\
0 & a^2 & 2 a \\
0 & 0 & a^2 \\
\end{array}
\right)$$
$$A^3 = \left(
\begin{array}{ccc}
a^3 & 3 a^2 & 3 a \\
0 & a^3 & 3 a^2 \\
0 & 0 & a^3 \\
\end{array}
\right)$$
$$A^4 = \left(
\begin{array}{ccc}
a^4 & 4 a^3 & 6 a^2 \\
0 & a^4 & 4 a^3 \\
0 & 0 & a^4 \\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the absolute value of the difference of two variables? The problem is as follows:
Let $x$ and $y$ integers which satisfy the following equations:
$$x+y-\sqrt{xy}=7$$
$$x^2+y^2+xy=133$$
Find the value of $\;|x-y|.$
I'm stuck on this problem due the fact that there appears a square root of $xy$ and t... | Let $a=x+y,b=\sqrt{xy}$. The two given equations can be rewritten as $a-b=7$ and $a^2-b^2=133$. As $a^2-b^2=(a+b)(a-b)$, $a+b=133/7=19$, so $a=(19+7)/2=13$ and $b=(19-7)/2=6$. Thus, $x+y=13$ and $xy=36$.
$$|x-y|^2=(x+y)^2-4xy=13^2-4(36)=13^2-12^2=13+12=25=5^2,$$
so $|x-y|=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Compare sum of radicals I am stuck in a difficult question:
Compare $18$ and
$$
A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}
$$
without using calculator.
Thank you for all solution.
| Lets be methodical about this. Consider finding a nice upper bound for
$\sqrt a + \sqrt b$.
$$(\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}$$
So maybe we should replace $ab$ with the smallest $n$ such that $ab \le n^2$
Then $(\sqrt a + \sqrt b)^2 < a + b + 2n$. So
$$\sqrt a + \sqrt b < \sqrt{a + b + 2n}$$
We note that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding a specific base for matrices space I need to find a base that includes only matrices with Rank=1 for the following space: $$V=\left \{ \right.(\begin{smallmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{smallmatrix}\bigr):x,y,z\in\mathbb{R}\left. \right \}$$
What I did is:
1) Extracted $x,y,x$ to find the... | A good start is to see which matrices out of
$$
e_1=\begin{pmatrix}
1&0\\0&0
\end{pmatrix},~
e_2=\begin{pmatrix}
0&1\\0&0
\end{pmatrix},~
e_3=\begin{pmatrix}
0&0\\
1&0
\end{pmatrix}
\text{ and }
e_4=\begin{pmatrix}
0&0\\0&1
\end{pmatrix}
$$
are contained in $V$.
You can directly check that $e_1,e_4\notin V$. But we sti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Triangular numbers is relating to Pythagoras triples Let
$$T_n:=1,3,6,10,15,...$$
are triangular numbers, $T_n={n(n+1)\over 2}.$
Pythagoras triples $(5,12,13),(7,24,25),(9,40,41),(11,60,61)$ and so on,...
Observe that $S_n=T_n+2T_{n+1}+T_{n+2}$ for $n\ge1$
$S_1=13$
$S_2=25$
$S_3=41$ and so on, ...
Can anyone explain h... | Pythagorean triples can be gnerated by $(m^2-n^2, 2nm , n^2+m^2)$ and if $m=n+1$ you will obtain your sequence.
Note that your sequence $S_n$ are the sum of two consecutive squares. $S_=n^2+(n+1)^2$. Also
\begin{eqnarray*}
S_n&=& \frac{n(n+1)}{2}+ 2\frac{(n+1)(n+2)}{2}+\frac{(n+2)(n+3)}{2}= 2n^2+6n+5 \\&=&(n+1)^2+(n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\l... |
I thought it might be instructive to present an approach that does not rely on calculus, but rather uses the squeeze theorem, a set of inequalities that can be obtained with pre-calculus tools only, and the values of the sums $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$. To that end we proceed.
TOOL $1$: Elementary Ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we wan... | To expound a bit on Catalin Zara's answer:
Your answer should give all possible antiderivatives; i.e., if you plug in any particular combination of constants for the generic constants $C_1, C_2, \dots$, you get a function which, when you differentiate, gives you the integrand. So the answer with three constants is more... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
If $f(x) $ is a continuous function with a given identity, how do I find $f(\sqrt{3})$?
If $f(x) $ Is a continuous function $ \forall\ x \in R$ and satisfies
$x^2+x \{f(x)\} - 3 = \sqrt{3} \ f(x) \ \forall\ x \in R$
Find $f(\sqrt{3})$.
$\{ t\}$ is the fractional part of $t$.
--
My attempt: I substituted $\sqr... | By substituting $\sqrt{3}$ in there, you get that $f(x) - \{f(x)\} = 0$, which implies that $0 \leq a < 1$ as you concluded.
Now suppose $f(\sqrt{3}) = a, 0 < a < 1$. Then, by continuity, $f$ satisfies the equation $x^2 + xf(x) - 3 = \sqrt{3} f(x)$ in a neighborhood of $x = \sqrt{3}$, so that $$f(x) = \frac{x^2 - 3}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mat... | (Alternative approach, without induction.) The following uses the notation $\,\mathcal{M}n\,$ to denote any integer multiple of $\,n\,$. The obvious rules apply $\,2 \cdot \mathcal{M}3 = \mathcal{M}6, \,\mathcal{M}3 \cdot \mathcal{M}3 = \mathcal{M}9\,$ etc.
*
*$\,4^k-1 = \mathcal{M}3\,$ for all $\,k \ge 0\,$, theref... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there an infinite number of numbers like $1600$? My reputation is at this moment at $1600$.
I did some experimenting with $1600$ and obtained the following:
Evidently, it is a perfect square $1600=40^2$
Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.
Also, it can be written ... | Yes. We know that $5^2=3^2+4^2$ and, if we consider $k\in\mathbb{N}$ and a semejant triangle for this with scale $2k$ then $(6k)^2+(8k)^2=(10k)^2$.
The number $(10k)^2$ is our candidate. In fact, its is a square number, its a hypothenuse of a pythagorean integer-triple.
Also, $$(10k)^2=(2\cdot 5k)^2=4\cdot(5k)^2=(5k)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Interesting Olympiad Style Problem about Invariance Problem: The following operations are permitted with the quadratic polynomial $ax^2 +bx +c:$ (a) switch $a$ and $c$, (b) replace $x$ by $x + t$ where $t$ is any real. By repeating these operations, can you transform $x^2 − x − 2$ into $x^2 − x − 1?$
My Attempt: Notice... | I don't understand. Is it $S = a+b+c\pmod{t}$ or $S= a+b+c$. Because I don't understand how you get $S\equiv a+b+c\pmod{t}$ in the second case.
Anyway, just calculate the discriminant and show that it doesn't change:
Mark new polynomial with $a'x^2+b'x+c'$
Case 1. If we change only we get from $ax^2+bx+c$ this $cx^2+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove that $\int_0^\infty\frac1{x^x}\, dx<2$
Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$
Note: This inequality is rather tight. The integral approximates to $1.9955$.
Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiat... | Remarks: Here is an alternative proof.
I used the same bounds for this question
Improper integral inequality including the golden ratio and the Sophomore's dream
We will use the following auxiliary results (Facts 1-2).
Fact 1: $x^{-x} \le \frac{3 - x}{x^2 - x + 2}$ for all $x \in [1, 2]$.
(RHS is the Pade $(1, 2)$ app... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
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Find the remainder when $\frac{13!}{7!}$ is divided by $17$. How to find the remainder when $\frac{13!}{7!}$ is divided by $17$?
I started with Wilson's Theorem which states that every prime $p$ divides $(p-1)!+1$. That is $(p-1)!=-1 mod(p)$.
$16!=-1 mod(17)$
Kindly help me how to get to $\frac{13!}{7!}$.
| By Wilson's theorem $16!\equiv -1\pmod{17}$ and since $17$ is a prime of the form $8k+1$ and the square roots of $-1$ in $\mathbb{F}_{17}^*$ are $4$ and $13$, $8!\equiv 13\pmod{17}$ and $7!\equiv 8\pmod{17}$. By using the notation $\frac{1}{a}$ for the inverse of $a$,
$$ \frac{13!}{7!}\equiv\frac{16!}{(-1)(-2)(-3)8}\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$ I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic.
$$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{... | Everything is fine till you write $du=\frac{1}{\sqrt{k}}dv$. It is instead, $du=\sqrt{k}dv$. Aside from that, your solution method is great.
Also, for the argument to the arctan function, note that $\frac{x+\frac{b}{2a}}{\sqrt{\frac{c}{a}-\left(\frac{b}{2a}\right)^2}}=\frac{2a\left(x+\frac{b}{2a}\right)}{2a\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\lim_{x \to 2} x^3 = 8$ by using epsilon-delta Prove that $$\lim_{x \to 2} x^3 = 8$$
My attempt,
Given $\epsilon>0$, $\exists \space \delta>0$ such that if $$|x^3-8|<\epsilon \space \text{if} \space 0<|x-2|<\delta$$
$$|(x-2)(x^2+2x+4)|<\epsilon$$
I'm stuck here. Hope someone could continue the solution a... | Note that
$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$
so
\begin{align}|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\\
&\le |x-2|(|x-2|^2 + 6|x-2| + 12)
< \delta(\delta^2 + 6 \delta + 12)
\end{align}
For $\varepsilon > 0$ you would take $\delta < \min\left\{\frac\varepsilon{19}, 1\right\}$ because then $|x-2| <... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How many positive integers from 1-1000 have 5 divisors?
How many positive integers from $1-1000$ have $5$ divisors?
Any answers would be greatly appreciated. If you have any questions, I will edit for clarification.
| (Filling in some details, on the off chance that other answers are unfamiliar to potential readers!)
Consider the prime factorization of $75 = 3^1 \cdot 5^2$. How many factors does $75$ have? Well, in forming a factor of $75$, we have to choose how many $3$s to include - there is one available, but we could also choos... | {
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"url": "https://math.stackexchange.com/questions/2593855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Righ... | Note that $x$ cannot be negative as $|x+1|$ is always nonnegative.
With that mind, observe that $|x+1| = x+1, \forall x > 0$. Therefore
$$ \frac{1}{x+1} < \frac{1}{2x} $$
which results in $x < 1$ as you have done.
So the final result is $x \in (0,1)$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cro... | at "Cross Multiplication" it should be $$ p(3p+4)(p+1) $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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For what positive numbers $a,b$ is the series $\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$ convergent? Last one for tonight. For what positive numbers $a,b$ is the series $$\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$$ convergent?
I'm defeated by this one. I don't even k... | Because$$\def\d{\mathrm{d}}
\sum_{j = 0}^k \frac{1}{b + j} > \int_0^{k + 1} \frac{\d x}{b + x} = \ln(b + k + 1) - \ln b,
$$
for $a \geqslant 1$,$$
a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} \geqslant a^{\ln(b + k + 1) - \ln b} \geqslant a^{\ln(b + 1) - \ln b},
$$
which implies $a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In the figure, how do I find the value of the variable?
According to the problem, I have to find the value of the variable in simplest radical form. I began with the equation 18√9 = 6 (1/3)x, so x = 9√3. Now, I know we have to find the area of the base in terms of x. But how do I find the height; do I use Pythagorean ... | \begin{gathered}
V\ =\ \frac{Bh}{3} \ \ \ where\ B\ is\ the\ base\ area\ of\ the\ equilateral\ triangle\ of\ length\ x\\
\\
B\ =\ \frac{1}{2} \ ( base\ triangle\ length)( base\ triangle\ height) \ =\ \ \frac{1}{2} \ ( x)\left(\frac{\sqrt{3} x}{2} \ \right) \ =\ \frac{\sqrt{3} x^{2}}{4}\\
\\
V\ =\ \frac{1}{3} \ *\ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to bui... | We have $$2\cos\frac{3A+3B}{2}\cos\frac{3A-3B}{2}-2\sin^2\frac{3C}{2}=0$$ or
$$\cos\left(\frac{3\pi}{2}-\frac{3C}{2}\right)\cos\frac{3A-3B}{2}-\sin^2\frac{3C}{2}=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}-\sin\frac{3C}{2}\right)=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}+\cos\frac{3A+3B}{2}\right)=0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Find the largest real $k$ such that: $(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$
Find the largest real $k$ such that for every non negative real numbers $a,b,c$ : $$(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$$
I expanded the LHS but the problem got more complicated and no progress...
| For $c=0$ and $a=b=1$ we obtain $4\geq k$.
We'll prove that $4$ is a maximal value.
Indeed, by Muirhead we obtain:
$$(a+b+c)^2(ab+ac+bc)=\sum_{cyc}(a^2+2ab)\sum_{cyc}ab=$$
$$=\sum_{cyc}(a^3b+a^3c+a^2bc+2a^2b^2+4a^2bc)=\sum_{cyc}(a^3b+a^3c+2a^2b^2+5a^2bc)\geq$$
$$\geq\sum_{cyc}(4a^2b^2+5a^2bc)\geq4(a^2b^2+a^2c^2+b^2c^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expre... | Simply note that
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}=(m^2 + 2)(m^2-1)^{2-\frac32}=(m^2 + 2)(m^2-1)^{-\frac12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting f... | $2^{x-3} + 15\cdot 2^{x-3} = 256;$
$2^{x-3}(1+15) =256;$
$2^{x-3} =256/(16)= 16;$
$x-3=4$ , by inspection.
$x=7.$
Note:
$\dfrac{1}{2^{3-x}}= 2^{x-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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limit with summation and product
Given $L=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\bigg(\frac{1}{r!}\prod^{r}_{i=1}\left(\frac{i}{2}+\frac{1}{3}\right)\bigg)$. then $\lfloor L \rfloor$ is
Try: $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1}{r!}\bigg[\left(\frac{1}{2}+\frac{1}{3}\right)\cdot \left(\... | For any $|a| < 1$,$$
\sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n (ak + b) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \prod_{k = 1}^n \left( -\frac{b}{a} - k \right) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n}.
$$
Note that $|a| < 1$, by the generalized binomial theorem,$$
(1 - a)^{-\frac{b}{a}... | {
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Summation. What does is evaluate to? What is $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$ if $a_{n+2}=a_{n+1}+a_{n}$ and $a_{1}=a_{2}=1$?
| I found a nice answer, you guys might be curious to see it as well:
Let S be the sum.
Now $S=\frac{1}{16}+\frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\frac{5}{4096}$...
Multiply S by 4 to get that
$4S = \frac{1}{4}+\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{5}{1024}$...
$3S = \frac{1}{4}+\frac{1}{64}+\frac{1}{256}+\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Calculate $\int \frac{1}{x^2+x+1} \, dx$ $$ \int \frac{1}{x^2+x+1}\, dx = \int \frac{1}{(x+\frac 1 2)^{2} + \frac 3 4}\, dx $$
Substitute $x+\frac 1 2 = u$, $dx = du$:
$$\int \frac 4 3 \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du = \frac 4 3 \int \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du$$
Substitute... | What you did is fine. The book, on the other hand…
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+... | From here
$$= \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2
=(-1)^{(n+2)}\left(\frac{-n(n+1)}{2} +\frac{2(n+1)^2)}2\right)
=(-1)^{(n+2)}\left(\frac{(n+1)(n+2)}{2}\right)$$
| {
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"url": "https://math.stackexchange.com/questions/2601559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/comm... | Define
$$
f(a,\lambda) = -\frac{a}{a^{11}+1} + \lambda \log(a) + \frac{1}{2}
$$
Then, for any choice of $\lambda$,
$$
f(a,\lambda) + f(b,\lambda) + f(c,\lambda) = -\frac{a}{a^{11}+1} -\frac{b}{b^{11}+1} -\frac{c}{c^{11}+1} + \frac{3}{2}
$$
and we need to show that this is $\ge 0$.
It suffices to show that, for some $\... | {
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"url": "https://math.stackexchange.com/questions/2602035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Complex integral involving Cauchy integral formula Where $C$ is the circle $|z|=\frac{3}{2}$, evaluate the following integral using the Cauchy integral formula.
$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz$$
Clearly the simple poles at $z=\pm i$ are the ones that are inside the circle $C$, and the simple poles at $z=\p... | Write the integral as a sum of integrals around $\{i,-i\}$. These integrals are easily evaluated using the Cauchy integral formula:
$$
\frac{1}{2\pi i}\oint_{|z-i|=1/2}\frac{1}{(z-i)}\frac{e^z}{(z+i)(z^2-4)}dz
= \frac{e^i}{(i+i)(i^2-4)} \\
\frac{1}{2\pi i}\oint_{|z+i|=1/2}\frac{1}{(z+i)}\frac{e^z}{(z-i)(z^2-4... | {
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"url": "https://math.stackexchange.com/questions/2606994",
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"source": "stackexchange",
"question_score": "2",
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Identify $\mathbb{Z}[x]/(x^2-3,2x+4)$. I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.
I've been solving such problems in an approach like:
$$
2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\
(2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\
(4x+8)-(4x+6)=2
$$
What shall I do next, please? Thank you.
Simon
| First divide $x^2-3$ by $x+2$, dividing by a linear polynomial is the same as evaluating at the root of the linear term, so $x^2-3=4-3=1$. Thus $x^2-3=q(x)(x+2)+1$ for some $q(x)$. Multiplying by 2, and rearranging, we get $2(x^2-3)-q(x)(2x+4)=2$. Hence $2\in (x^2-3,2x+4)$, so since $2\mid 2x+4$, we have
$$(x^2-3,2x+4... | {
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"timestamp": "2023-03-29T00:00:00",
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If $f\left(x\right)=$ $\sum_{n=0}^{\infty}$ $\frac{x^{3n}}{\left(3n\right)!}$ then prove $f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
QuestionIf $f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
then prove that
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
My Appro... | Note that we can write: $$f(x) = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \ldots $$ and $$f'(x) = \sum_{n=0}^{\infty} \frac{x^{3n+2}}{(3n+2)!}$$ $$ = \frac{x^2}{2!} + \frac{x^5}{5!} + \ldots$$ $$f''(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)!} $$ $$= x + \frac{x^4}{4!} +\ldots$$
Add them up to get $e^x$.
| {
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Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k... | by geometrics sum wfor any complex such that: $|z|<1$ we have
Therefore
$$\sum_{k=1}^{\infty} z^k= \lim_{n\to\infty}\sum_{k=1}^{n} z^k =\lim_{n\to\infty} \frac{z(1-z^{n+1})}{1-z} = \frac{z}{1-z} $$
since $\lim_{n\to\infty} z^n =0 $ since $|z|<1$.
Hence
\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$... | You wish to derive a formula for the sum $\sum_{k=1}^n k^6$ using "high-school methods".
Consider $$\frac{x^7}7-\frac{(x-1)^7}7=x^6-3x^5+5x^4+\cdots\\\frac{x^6}6-\frac{(x-1)^6}6=x^5-\cdots\\\cdots$$
So, defining $$I_n:=\frac{n^7}7+\alpha_1\frac{n^6}6+\alpha_2\frac{n^5}5+\cdots+\alpha_6 n+\alpha_7$$ for some constants $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$
Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$.
I have no idea how to approach this question. However, I managed to solve it using a rather childish approach. I change this equation by multiplying $2$, getting $2|x+1|+4|x-5|+4|x-3.5... | Case 1:$x\le -1\\f(x)=-x-1+10-2x+7-2x+5.5-0.5x=21.5-5.5x$
Case 2:$-1\le x\le 3.5\\f(x)=x+1+10-2x+7-2x+5.5-0.5x=23.5-3.5x$
Case 3:$3.5\le x\le 5\\f(x)=x+1+10-2x-7+2x+5.5-0.5x=9.5+0.5x$
Case 4:$5\le x\le 11\\f(x)=x+1-10+2x-7+2x+5.5-0.5x=-10.5+4.5x$
Case 5:$11\le x\\f(x)=x+1-10+2x-7+2x+0.5x-5.5=-21.5+5.5x$
so we have: $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Geometry Problem: Find the area of $ABCD$ square
My Problem is:
The $E-$ point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.
My way:
$$a=\frac{k+17}{\sqrt2}$$
$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow... | Let $F$ be the projection of $E$ on $BC$.
$CE=17$ implies $EF=FC=\frac{17}{\sqrt{2}}$. $BE=13$ and the Pythagorean theorem imply
$$ BF = \sqrt{13^2-\frac{17^2}{2}}=\frac{7}{\sqrt{2}} $$
hence
$$ BC=BF+FC = \frac{17+7}{\sqrt{2}} = 12\sqrt{2} $$
and the area of $ABCD$ is clearly $288$. You do not really need the cosine ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
F... | Let $u=1-2x-3x^2$, then $\mathrm{d}u=-\left(6x+2\right)\mathrm{d}x$
$$
\begin{align}
\int\frac{4+12x}{(1-2x-3x^2)^{1/3}}\,\mathrm{d}x
&=\int\frac{-2\,\mathrm{d}u}{u^{1/3}}\\
&=-3u^{2/3}+C\\[3pt]
&=-3\left(1-2x-3x^2\right)^{2/3}+C
\end{align}
$$
Looking at the answer in the question, this is pretty much exactly what wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
B... | When there are $3$ real roots, Cardano-Tartaglia's formulæ can't be applied, and we need a trigonometric solution:
Setting $x=A\cos \theta\;$ ($A>0,\; 0\le \theta <\pi$), the equation becomes
$$A^3\cos^3\theta -15A \cos\theta=10.$$
We'll try to write the l.h.s. as $\;B\cos 3\theta=10\;$ to obtain an easy-to-solve trig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{a+b+c}{d+e+f}$ Given real numbers $a, b, c, d, e, f$, such that:
$a^2 + b^2 + c^2 = 25$
$d^2 + e^2 + f^2 = 36$
$ad + be + cf = 30$
What is the value of $\frac{a+b+c}{d+e+f}$?
I've tried combining equations in several ways but haven't gotten very far. Any hints would be appreciated.
| Hint: by the Cauchy-Schwarz inequality:
$$
900 = 30^2 = (ad + be + cf )^2 \le (a^2 + b^2 + c^2 )(d^2 + e^2 + f^2) = 25 \cdot 36 = 900
$$
Equality occurs when $a,b,c$ and $d,e,f$ are proportional.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× ... | Note that $\forall x\in \mathbb{R}$
$$\sin x= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.
$$
which converges “very fast for small” x.
In this case $x=\frac{\pi}{10}$ so we have that
$$ \sin \left(\frac{\pi}{10}\right) \approx \frac{\pi}{10} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
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Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is... | by part method
$$\int(x^2 + 1)^7x^3dx=x^2*\frac{(x^2+1)^8}{2}-\int(x^2 + 1)^8xdx$$
$$x^2*\frac{(x^2+1)^8}{2}-\frac{1}{18}(x^2+1)^9+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find all possible values of $ d_{17}$ knowing that $ (d_7)^2 + (d_{15})^2= (d_{16})^2$
Let $ d_1,d_2 \ldots, d_r$ be the positive divisors of $ n$. Namely,
$$ 1=d_1<d_2< \ldots <d_r=n$$
Now if $ (d_7)^2 + (d_{15})^2= (d_{16})^2$ find all possible values of $ d_{17}$
I don't know which tools should I use here. I ... | Hint:
It is well-known, that if $a^2+b^2=c^2$ then $60|abc$ so $d_2=2,d_3=3,d_4=4,d_5=5,d_6=6$ and $d_7=7,8,9$ or $10$
Further Hint
If $d_7=7$ then $d_{16}-d_{15}=1$, hence $d_{16} + d_{15}=49 \implies d_{16}=25,d_{15}=24$ then $4200|n$ and $d_8=8$ and...
$10,12,14,15,20,21,24|n$ \to $d_{15} \leq 24$ but it means ,th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Minimum distance from the points of the function $\frac{1}{4xy}$ to the point $(0, 0, 0)$ I am trying to find the minimum distance from the points of the function $\large{\frac{1}{4xy}}$ to the point $(0, 0, 0)$.
This appears to be a problem of Lagrange in which my condition: $C(x,y,z) = z - \frac{1}{4xy} = 0$, and my ... | Note that by substitution of the constraint we need to minimize
$$f(x,y)=\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+\frac{1}{16x^2y^2}}$$
and we don't need Lagrange's method.
In this case AM-GM is the most effective method, indeed
$$\frac{x^2+y^2+\frac{1}{16x^2y^2} }{3}\ge\sqrt[3]{x^2\cdot y^2\cdot \frac{1}{16x^2y^2}}\iff x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Is there any trig identity which can be used for $\int \sin^4x$? What is $\int \sin^4x$?
I tried to use the identity $\sin^2x = \frac{1-\cos 2x}{2}$, but I'm stuck.
| Using the substitution above:
$\sin^4 x = (\frac {1-\cos 2x}2)^2 = \frac 14 - \frac 12 \cos 2x + \frac 14\cos^2 2x $
$\cos^2 2x = \frac {1+\cos 4x}{2}$
Or, using a little complex analysis
$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\sin^4 x = \left(\frac {e^{ix} - e^{-ix}}{2i}\right)^4\\
\frac {e^{4ix} - 4e^{3ix}e^{-ix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n... | You can’t do this step
$$\lim_{n\rightarrow \infty } \frac{ \frac{1^4}{n^4} + \frac{2^4}{n^4} +...+1 }{n}\color{red}{=\frac{0 + 0 +0 +...+1 }{n}= 0} $$
since you have infinitely many terms which tend to 0 and their sum not necessarily is $0$.
Think to $\sum_1^{\infty} \frac1k$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Limits connected to a right angled triangle Given a right angled triangle with sides $1,x$ and hypotenuse $y$. Let $\theta$ be the angle contained by side $1$ and hypotenuse. Then evaluate the following limits:
*$\lim_{\theta\to\pi/2}\sqrt{y}-\sqrt{x}$
*$\lim_{\theta\to\pi/2}y-x$
*$\lim_{\theta\to\pi/2}y^2-x^2$
*$... | First note that$$y^3-x^3=\dfrac{y^6-x^6}{x^3+y^3}=\dfrac{3y^4-3y^2+1}{x^3+y^3}\ge\dfrac{3y^4-3y^2+1}{2y^3}=1.5y-1.5\dfrac{1}{y}+\dfrac{1}{y^3}$$Also $\cos\theta=\dfrac{1}{y}\to 0^+$ leads to $y\to \infty$ and makes the limit $\infty$ either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Origin Triangle Tetrahedron Volume I have a problem that goes like this:
Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?
I really can... | Because $A$ is on the $x$-axis, there exists a number $a$ such that the coordinates of $A$ are $(a,0,0)$. Similarly, there exist numbers $b$ and $c$ such that the coordinates of $B$ and $C$ are $(0,b,0)$ and $(0,0,c)$, respectively.
We know that $AB,BC,$ and $CA$ are $5,6,7$ in some order. We may assume $AB=5,BC=6,$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Probability of being dealt a bridge hand with... Hey Guys I'd like to know if this question can be tackled this way.
What is the probability of being dealt a bridge hand with exactly 4 honour cards and exactly 4 cards from the 5 through 10?
I wrote that having in mind that the 10 is not included as it wasn't specified... | We split the bridge cards into $4$ mutually exclusive sets:
*
*$A$ is the set of $10$s; it consists of $4$ cards.
*$B$ is the set of honour cards which are not $10$s; it consists of $16$ cards.
*$C$ is the set of cards with number $5$ through $9$; it consists of $20$ cards.
*$D$ is the set of cards which are not ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\f... | Using $\cos2x=1-2\sin^2x=2\cos^2x-1,$
$$I=\dfrac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}{4\sqrt2|\cos2x|}=\dfrac{1+\cos^22x}{2\sqrt2|\cos2x|}$$
For $\cos2x>0,$
$$2\sqrt2I=\sec2x+\cos2x$$
Now use Integral of the secant function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Solve congruence with large exponents I'm trying to solve:
$$ x^{1477} \equiv 54 \mod 97 $$
Applying Euler-Fermat gives:
$$
x^{1477} = x^{15\cdot 96 + 37} = x^{37}\cdot x^{{96}^{15}} \equiv x^{37}\cdot 1^{15} = x^{37} \mod 97
$$
So instead of solving $ x^{1477} \equiv 54 \mod 97 $ one can solve $ x^{37} \equiv 54 \m... | If $x^{37} \equiv 54 \pmod {97}$ then $x^{37\cdot 24} \equiv 54^{24} \equiv 1 \pmod {97}$ because the order of $54 \pmod {97}$ is $24$.
This means the order of $x$ is a common divisor of $96$ and $37 \cdot 24$.
Therefore the order of $x$ is a divisor of $24$, and so $x^{24} \equiv 1 \pmod {97}$.
But if $k$ is a divisor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $x+\frac1{4x} ≥ 1$ for $x>0$
Let $x$ be a real number such that $x > 0$. Prove that x +$\frac {1} {4x} ≥ 1$.
Not really sure on the correct way to approach it/is valid and could use some help.
Answer:
Proof Strategy: Proof by cases:
*
*$x = 1$
*$x > 1$
*$x < 1$
--
Case 1: $x = 1$
$x+\frac1{4x}$
$= ... | For fun:
Let $x>0,$ real. Multiply by $4x$ :
$4x^2-4x +1 \ge 0.$
Need to show that above inequality is true for $x \gt 0$.
$4x^2 -4x +1 = 4(x^2 -x) +1 = 4(x-1/2)^2 \ge 0$
(why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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integral of $\cos (x+y+z)$ over a sphere Let $r>0$ and $S_r=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=r^2\}$.
Show that
$$\int_{S_r}\cos x\cos y\cos z\ dS=\int_{S_r}\cos (x+y+z)\ dS$$
and find the value of the integrals.
I know that we can parametrize $S_r$ and try to solve the integrals by hand or try to expand the integra... | As a partial answer I can show the first equation is valid:
On the upper surface of the sphere,
$$\vec r=\langle x,y,z\rangle=\langle x,y,\sqrt{r^2-x^2-y^2}\rangle$$
So
$$d\vec r=\langle 1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}}\rangle dx+\langle 0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}\rangle dx$$
And then
$$\begin{align}d\vec A&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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P.M.F and expected value/expected payout The question is:
Let a random variable X be the number of days that a certain patient needs to be in the hospital. Suppose that X has the p.m.f.
$$\displaystyle f(x) = \frac{5 - x}{10}, \quad x = 1, 2, 3, 4 $$
If the patient is to receive 166 dollars from an insurance compan... |
What am I doing wrong?
$$E(X) = 166( 1(\frac{5-1}{10}) + 2(\frac{5-2}{10})) + \underbrace{118( 3(\frac{5-3}{10}) + 4(\frac{5-4}{10}))}$$
If you spend three or four days in the hospital, you pay \$166 for each of the first two days and \$118 for the remainder. Not $118 for each of the days.
$$E(X) = 166( 1(\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding the distance between two parallel tangents of a rational function A Curve C has equation $y=\frac{2x-5}{x-1}$
a) A line y = mx+c is tangent to the curve. Find a condition for c in terms of m.
This can be solved relatively easily
$mx + c = \frac{2x-5}{x-1}$
$mx^2 + (c-m-2)x -c+5=0$
take discriminant as 0, as the... | Given $y=\frac{2x-5}{x-1}$, the two tangents must have same slope:
$$m=y'(p)=\frac{3}{(p-1)^2}=y'(r)=\frac{3}{(r-1)^2} \Rightarrow p=-\sqrt{\frac{3}{m}}+1; r=\sqrt{\frac{3}{m}}+1 \ \ \ (*)$$
The squared distance between the tangents is
$$d^2=(p-r)^2+(q-s)^2=(p-r)^2+\left(\frac{2p-5}{p-1}-\frac{2r-5}{r-1}\right)^2.$$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if div... | \begin{eqnarray*}
=\color{red}{27} \times 2^{5n+1}+5 \color{red}{(2^{5n+1}+5^{n+1})}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Find all $f$ that satisfies $f:\mathbb{R}\rightarrow\mathbb{R};f(x+y)+f(x)f(y)=(1+x)f(y)+(1+y)f(x)+f(xy)$ Find all $f$ that satisfies:
$1, ~f:\mathbb{R}\rightarrow\mathbb{R};\\
2,\forall x,y\in\mathbb{R},f(x+y)+f(x)f(y)=(1+x)f(y)+(1+y)f(x)+f(xy);
$
Maybe we can prove it's derivable or it's a linear function.
Any idea?
| The only solutions are $f(x)=0$, $f(x)=3x$, and $f(x)=x(x+1)$. It is easy to verify these all work; let me now prove there are no other solutions.
First note that setting $y=0$ gives $$f(x)f(0)=(2+x)f(0).$$ So either $f(0)=0$ or $f(x)=2+x$ for all $x$. Since $f(x)=2+x$ does not satisfy the functional equation, we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | The coefficient of $x^n$ is the number of integer compositions of $n$ into two parts where every part has length between $0$ and $5$. You can find the coefficients with a stars and bars approach. For every $n$, draw $n$ stars in a line, and see how many places it is possible to insert one bar such that there are betwee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Deriving the closed form of $M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$ I have the sequence, let $M_0=1$
$$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$$
Which I would like first to study the convergence and fine the closed form.
I failed to show that $M_n$ is bound... | Define $a_n = \dfrac{M_n}{2^n} \ (n \in \mathbb{N}_+)$, then$$
a_{n + 1} = \frac{\sqrt{1 + a_n^2} - 1}{a_n} > 0.
$$
Define $b_n = \sqrt{1 + a_n^2} \ (n \in \mathbb{N}_+)$, then$$
b_n > 1 \Longrightarrow a_n = \sqrt{b_n^2 - 1},
$$
and\begin{align*}
&\mathrel{\phantom{\Longrightarrow}}\sqrt{b_{n + 1}^2 - 1} = \frac{b_n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| Also, the Tangent Line method works: $$\sum_{cyc}\frac{1}{3x^2+1}=\sum_{cyc}\left(\frac{1}{3x^2+1}-\frac{1}{13}+\frac{12}{169}(x-2)\right)+\frac{3}{13}=$$
$$=\sum_{cyc}\frac{3(x-2)^2(12x+11)}{169(3x^2+1)}+\frac{3}{13}\geq\frac{3}{13}>\frac{3}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$
If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
Equality is achieved in some strange cases: For example, if $x = -\dfrac{1}{\sqrt{2}}$, $y = 0$ and $z = \dfrac{1}{\sqrt{2}}$, then $(x-y)(y-z)(z-x)=\dfrac{1}{\sqrt{2}}$... | If $\prod\limits_{cyc}(x-y)<0$ then it's obvious.
But for $\prod\limits_{cyc}(x-y)\geq0$ it's enough to prove that
$$(x^2+y^2+z^2)^3\geq2(x-y)^2(x-z)^2(y-z)^2.$$
Now, let $x\leq y\leq z$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that
$$(3x^2+2(u+v)x+u^2+v^2)^3\geq2(u-v)^2u^2v^2$$ or
$$3x^2+2(u+v)x+u^2+v^2-\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding the area of a region of circle Consider the following shape. It is requested to find the area of the shaded region. The radius of the circles and the overlapping half circles is 1 unit.
The region is irregular and I couldn't find a way to find it.
| Rotate the image, and place the origin of the Cartesian system at the center of the main circle. The other two circles will be centered at $(\pm 1/ \sqrt{2},-1/ \sqrt{2})$.
Obviously, it's enough to find half of the area for $x>0$.
For the green circle for $y>0$ we have:
$$y=\sqrt{1-x^2}$$
While for the right blue cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$ but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind ... | The other answers to this question use complex numbers. Here is a "real-only" answer.
Let us first work out
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx\qquad y\ge0$$
"But wait a minute, isn't this always infinite?" Yes, but the integral from $0$ to $y$ (the generally implied bounds of the indefinite integral) can... | {
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"url": "https://math.stackexchange.com/questions/2651254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Cosine of a matrix I came across this question, asked in a competitive exam. It is as follows.
Given a matrix $M = \begin{bmatrix}2&1\\1&2\end{bmatrix}$ what is the value of $cos(πM/6)$?
I've tried series expansion but I think there is an alternative way doing it, any help is appreciated.
Options given are
\begin{b... | $$M=A+B$$
where
$$A=2I$$
and
$$B=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$
We have
$$[A,B]=0$$
and
$$B^2=I$$
So
$$\exp\left(i\frac{\pi}{6}M\right)=\exp\left(i\frac{\pi}{6}2I\right)\exp\left(i\frac{\pi}{6}B\right)$$
$$=\exp\left(i\frac{\pi}{3}I\right)\left(I+i\frac{\pi}{6}B+\cdots\right)$$
$$=\exp\left(i\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x... | Explanation 1: If you have only $(x+1)$, your denominators are too weak:
$$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}+\frac{Cx}{x(x+1)}$$
Thus, you your denominator is quadratic whereas the desired denominator is $x(x+1)^2$, cubic. I mean, $\frac{C}{x+1}$ is redundant. Why? Just l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$? Is there a geometric interpretation for this?
How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$?
I know I can mod out the ideal one piece at a time and show $\mathbb C[x]/(x^2+1)$ is not a field since $(x^2+1)$ is not maximal in $\ma... | Let $I = (x^2+1,y^2+1)$, let $J=(x^2+1,x-y)$, and let $K=(x^2+1,x+y)$.
If $1 \in J$, then $a(x^2+1)+b(x-y)=1$, for some $a,b \in \mathbb{R}[x,y]$. But then, letting $y=x$, we get $a(x,x)(x^2+1)=1$, contradiction, since a nonzero multiple of $x^2+1$ must have degree at least $2$.
If $1 \in K$, then $c(x^2+1)+d(x+y)=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number ... | Yet another way:
Let's call rolling all five dice a "turn". Let $X$ be the number of the first turn on which all five dice have rolled at least one six. If $X>n$ then we have at least one die which has not rolled a six by turn $n$. So
$$\begin{align}
E(X) &= \sum_{n>0} P(X > n) \tag{1} \\
&= \sum_{n=0}^{\infty} \{1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{... | Hint:
Using $\cos2x=1-2\sin^2x,$
$$\sin3x=3\sin x-4\sin^3x=\cdots=\sin x(1+2\cos2x)$$
Put $2x=\dfrac\pi{20},\dfrac{3\pi}{20},\dfrac{9\pi}{20},\dfrac{27\pi}{20}$ one by one to recognize the Telescoping nature.
Finally use $\sin(2m\pi+y)=\sin y$ where $m$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Inequality with positive reals For each $x,y,z>0$, define
$$
f(x,y,z)=\frac{3x^2(y+z)+2xyz}{(x+y)(y+z)(z+x)}.
$$
Fix also $a,b,c,d>0$ with $a\le b$. How can we prove "reasonably" the following inequality?
$$
\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)} \le \frac{a+c}{b+c}.
$$
| The expression
$$\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}
\le \frac{a+c}{b+c}$$ is equivalent to
$$(b+c)(f(a,b,d)+f(c,b,d))\le(a+c)(f(b,a,d)+f(c,a,d))$$ The $LHS$ is equal to
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2bcd}{(b+d)(c+d)}$$
The $RHS$ is equal to
$$\frac{... | {
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"url": "https://math.stackexchange.com/questions/2659411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.
Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.
I tried the following solution, with no result:
Considering $z=x+yi$,
$$|x+yi-10|=3|x+yi-2|$$
$$|(x-10)+yi|=3|(x-2)+yi|$$
$$\sqrt{(x-10)^2+y... | You are almost at end of your exercise. Just notice that
$$0=x^2+y^2-2x-8=(x-1)^2+y^2-9=|z-1|^2-3^2\Leftrightarrow |z-1|=3$$
where $z=x+iy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1... | The result seems to me weird. You wrote $\displaystyle \left(2018n+1\right)^4=4\left(2018n+1\right)$ which is false. And you when an expression is at power $n$, you cannot say the argument tends to $0$ hence it tends to $0^n$.
I would rather write
$$
\left(1-\frac{5}{n^4}\right)^{\left(2018n+1\right)^4}=e^{\left(2018n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4... | Since $w+\overline{w}\in \mathbb{R}$ for each $w\in \mathbb{C}$
we have $z^2+ \overline{z}^2 =1$ and from $|z|=1$ we have $z\overline{z}=1$.
So $$(z+\overline{z})^2= z^2+2z\overline{z} +\overline{z}^2 =3$$
and thus we have $z+\overline{z} =\pm \sqrt{3}$
So, since $\overline{z} = {1\over z} $ we have $z^2\pm \sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2... | HINT
Use
$$\frac{1}{n^2-\sqrt{n^4+4n^2+n}}\frac{n^2+\sqrt{n^4+4n^2+n}}{n^2+\sqrt{n^4+4n^2+n}}=\frac{n^2+\sqrt{n^4+4n^2+n}}{-4n^2-n}=\frac{1+\sqrt{1+4/n^2+1/n^3}}{-4-1/n}$$
In your attempt the following
$$n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}=n^2\cdot \sqrt1$$
is uncorrect, indeed by binomial series expansion
$$\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)... | use that $$\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$$
$$\sin\left(\frac{3\pi}{10}\right)=\frac{\sqrt{5}+1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$. Find the Maximum and Minimum Value of the expression $A=3x-2y$. For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$.Find the Maximum and Minimum Value of the expression $A=3x-2y$.
| Using secondary school "tools", you can replace
$$
x = \left( {A + 2y} \right)/3
$$
into the quadric equation, to get, after some simple simplifications
$$
25y^{\,2} - \left( {8A + 60} \right)y + \left( {8A + 60} \right)^{\,2} = 0
$$
whose solution is
$$
\eqalign{
& y = {{2\left( {2A + 15} \right) \pm \sqrt {\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\fr... | \begin{align}\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}&=\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)
\\&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)
\\&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2} \\&= (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ Solving this system of equations :
$$4=a+c$$ $$5=d+ac+b$$ $$4=ad+bc$$ $$4=bd$$
Anyone knows how to solve this without a CAS?
| We obtain $$c=4-a$$ and $$d=\frac{4}{b},$$ which gives
$$\frac{4}{b}+b+a(4-a)=5$$ and
$$a\cdot\frac{4}{b}+b(4-a)=4.$$
From the last equation we obtain:
$$a=\frac{4b^2-4b}{b^2-4},$$ which gives
$$\frac{4}{b}+b+\frac{4b^2-4b}{b^2-4}\left(4-\frac{4b^2-4b}{b^2-4}\right)=5$$ or
$$\frac{b^2-5b+4}{b}+\frac{16b(b-4)(b-1)}{(b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge? For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?
Solution says $0<s<2$. Any ideas?
| Let us split the integral into two parts.
\begin{align}
\int^\infty_0 \frac{\sin x}{x^s}\ dx = \int^\pi_0 \frac{\sin x}{x^s}\ dx + \int^\infty_\pi \frac{\sin x}{x^s}\ dx
\end{align}
where
\begin{align}
\int^\pi_0 \frac{\sin x}{x^s}\ dx=\int^\pi_0 \frac{\operatorname{sinc} x}{x^{s-1}}\ dx< \infty \ \ \ \text{ if and on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\m... | The hint.
Use for all $n\geq2$ $$s_{n+1}-\sqrt3=\frac{(s_n-\sqrt3)^2}{2s_n}>0$$ and
$$s_{n+1}-s_n=\frac{3-s_n^2}{2s_n}<0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$? When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?
Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.
$(ab_5... | Continuing along your route, where:
$$4103=(10a+b)(10c+d)$$
(notice all values are represented in base $5$ here)
All values will be in base $5$ throughout.
Notice that $40^2=3100$, which is less than $4103$, so $a$ and $c$ are both $4$:
$$4103=(40+b)(40+d)=3100+40(b+d)+bd$$
Now, notice that $\left|{b-d}\right|=2$ in or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | For $0 \leqslant x \leqslant 1$,$$
x(x - 1)(x^2 + 1) \geqslant \left(-\frac{1}{4}\right)(x^2 + 1) > \left(-\frac{1}{4}\right) \times 2 = -\frac{1}{2}.
$$
For $x < 0$ or $x > 1$,$$
x(x - 1)(x^2 + 1) > 0 > -\frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Solving $4^{32^x} = 16^{8^x}$
Solve for $x$: $4^{32^x} = 16^{8^x}$.
So I have tried using log (so $x\log{4^{32}} = x\log{16^8}$), but that wasn't very helpful to me, and some random guessing gave me an answer of $1/2$, but I was wondering how I could be more mathematical…
| Since$$
4^{32^x} = 2^{2 \cdot 32^x} = 2^{2 \cdot 2^{5x}} = 2^{2^{5x + 1}}
$$
and$$
16^{8^x} = 2^{4 \cdot 8^x} = 2^{4 \cdot 2^{3x}} = 2^{2^{3x + 2}},
$$
then$$
4^{32^x} = 16^{8^x} \Longleftrightarrow 2^{2^{5x + 1}} = 2^{2^{3x + 2}} \Longleftrightarrow 2^{5x + 1} = 2^{3x + 2}\\\Longleftrightarrow 5x + 1 = 3x + 2 \Longlef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of a 5th-root expression. Simplify and find the value of the expression:
$$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}.$$
I tried to rationalise it.
It was of no use..
| $$\frac{123+\sqrt{15125}}2=\frac{123+55\sqrt5}2=\left(\frac{1+\sqrt5}2\right)^{10}=\left(\frac{3+\sqrt5}2\right)^5.$$
Likewise
$$\frac{123-\sqrt{15125}}2=\left(\frac{3-\sqrt5}2\right)^5.$$
So
$$\sqrt[5]{\frac{123+\sqrt{15125}}2}+\sqrt[5]{\frac{123-\sqrt{15125}}2}
=\frac{3+\sqrt5}2+\frac{3-\sqrt5}2=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Unable to use Chinese Remainder Theorem on a question. So I had recently learned about Chinese Remainder theorem, and I got a Question to solve:
Find X if-
$X\equiv 1\bmod 2\\X\equiv 2 \bmod 3\\X\equiv 3 \bmod 4\tag*{}$
So I went on to find the solution modulus, $N=24$ & $N_1=12\\ N_2 = 8\\ N_3 = 6\tag*{}$
And now I ju... | $x \equiv 1 (\mod 2) \Rightarrow x-1$ divisible by $2$ $\Rightarrow x-1+2=x+1$ is divisible by $2$.
$x \equiv 2 (\mod 3) \Rightarrow x-2$ divisible by $3$ $\Rightarrow x-2+3=x+1$ is divisible by $3$.
$x \equiv 3 (\mod 4) \Rightarrow x-3$ divisible by $4$ $\Rightarrow x-3+4=x+1$ is divisible by $4$.
Because $x+1$ is div... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove the identity $ \sum\limits_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0.$ Prove the identity
$$
\sum_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0
$$
I want to reduce
$$
\frac{(a+k-2)!}{(a+k-n)!}=(2-n)\binom{a+k-n}{a+k-2}^{-1}=(2-n) (a+k-2) \int_0^1 z^{a+k-2}(1-z)^{... |
We obtain for integral $n>0$ and $a\in\mathbb{C}\setminus\{n-1,n-2,n-3,\ldots\}$
\begin{align*}
\color{blue}{\sum_{k=0}^{n-1}}&\color{blue}{(-1)^k\binom{n}{k}(n-k)\frac{(a+k-2)!}{(a+k-n)!}}\\
&=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k\frac{(a+n-k-2)!}{(a-k)!}\tag{1}\\
&=(-1)^nn(n-2)!\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that perspective projection of circle is ellipse How to prove that perspective projection of a circle is an ellipse?
I start with the parametric equation of circle and ellipse:
Circle:
$x = r\cos t$
$y = r\sin t$
Ellipse:
$x = a\cos(t)$
$y = b\sin(t)$
Then I have perspective projection matrix:
$$
\begin{pma... | As noted in the answer of @rfabbri ( +1), the perspective projection of a circle is not always an ellipse, but it is in general a conic section: an ellipse, a parabola, or a hyperbola.
If you want a matrix that transform the circle in an ellipse by projection, than note that the general equation of a conic:
$$
Ax^2+Bx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) ... | $$\dfrac{n(n+5)+4}{4(n+4)(n+5)}=\dfrac{n^2+5n+4}{4(n+4)(n+5)}=\dfrac{(n+1)(n+4)}{4(n+4)(n+5)}$$
This simplifies to $\dfrac {n+1}{4(n+5)}$ and you want to substitute $n$ for $n+1$ for $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.