Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to determine the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$ I have a question
Find the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$
There ARE questions like this on stack exchange already I know, but I'm not able to formulate a pattern or know how to apply t... | Try:
$\begin{align*}
1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\
&= \frac{1 - x^{11}}{1- x} \\
(1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^4 \\
&= \frac{(1 - x^{11})^4}{(1- x)^4} \\
&= (1 - 4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
\cdot \sum_{k \ge 0} (-1)^k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Using ring R=$\mathbb{Z}[X]/(X^n+1)$ Currently studying Ring-LWE,here, and having trouble grasping a concrete definition of what this means, as a ring. My understanding is that everything in this ring is an integer polynomial, and that it must be reduced $X^n+1$, is this correct? Essentially working modulo $X^n+1$.
It ... | The polynomials are just multiplied as usual but any term $X^N$ is replaced by $-1$, $X^{n+1} \equiv -X$ etc. , as we consider $X^n+1$ to be equivalent to $0$.
All coefficients of those polynomials lie in $\mathbb{Z}_q$, so whenever they go above $q$, reduce them.
So a simple example: $n=3$, $q=3$, then $$(1+X+X^2)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$ I want to prove that
$$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$$
for positive numbers $x,y,z$.
I don't know how to even begin. I must say I'm not 100% certain the inequality always holds.
I tried the sort of ... | There is no loss of generality in assuming $0 < x \leq y \leq z$. Rewrite the inequality as
$$\frac{(x-y)^2+(y-z)(x-y)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-y)^2+(y-x)(z-y)}{z^2} \geq 0$$
and rearrange the terms as follows:
$$\frac{(x-y)^2}{x^2}+\frac{(z-y)^2}{z^2}+(z-y)(y-x)\left(\frac{1}{x^2}+\frac{1}{z^2}-\frac{1}{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
In how many ways can $5$ boys and $5$ girls stand in a queue such that exactly four of the girls stand consecutively in the queue? I have two combinatorics questions.
Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m... |
In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue?
You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue.
Method 1: We treat the block of five girls as a single object. We then have six objects to arrange, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
| Let $x=t^2,$ where $t>0$.
Thus, $$t^3-9t-\sqrt8=0$$ or
$$t^3+\sqrt8t^2-\sqrt8t^2-8t-t-\sqrt8=0$$ or
$$(t+\sqrt8)(t^2-\sqrt8t-1)=0$$ or
$$x-\sqrt{8x}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving non-trivial integer solutions don't exist for $z^2=10x^2-5y^2$ Given $z^2=10x^2-5y^2$.
I am wanting to show this equation has integer solutions only $0=x=y=z$.
I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated.
Solution:
Consider modulo... | As Hoseyn Heydari's comment stated, you should consider checking modulo $5$ instead for your equation of
$$z^2 = 10x^2 - 5y^2 \tag{1}\label{eq1}$$
As the RHS is a multiple of $5$, this means $z = 5z_1$ for some integer $z_1$. Thus, \eqref{eq1} becomes
$$5z^2_1 = 2x^2 - y^2 \tag{2}\label{eq2}$$
Since the possible congru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Define the domain of $f(x)=\sqrt{x^2+4}$ I am to define the domain of $f(x)=\sqrt{x^2+4}$.
I arrived at $[-2,\infty)$ whereas the textbook solution is $(-\infty,\infty)$.
To arrive at my solution I set the radicand to be greater than or equal to zero:
$x^2+4\ge0$
$x+2\ge0$ # square root of each side
$x\ge-2$
Thus I get... | As you said, the radicand must be greater than or equal to zero. Since the square of any real number is at least zero, $x^2 + 4 \geq 4 > 0$ for every real number $x$, which implies that the domain is indeed $(-\infty, \infty)$.
In your calculation, you should have had
\begin{align*}
x^2 + 4 & \geq 0\\
x^2 & \geq -4
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
An exercise from Apostol's book I am trying to solve following problem from Apostol's Mathematical Analysis. The problem could be very trivial, but I am not getting clue for it.
Let $\{a_n\}$ be a sequence of real numbers in $[-2,2]$ such that
$$|a_{n+2}-a_{n+1}|\le \frac{1}{8} |a_{n+1}^2-a_n^2| \,\,\,\, \mbox{ for... | Let $A_{m+1}^-:=\left|a_{m+1}-a_m\right|$ and $A_{m+1}^+:=\left|a_{m+1}+a_m\right|$ for convenience. Then for a positive integer $k$, \begin{align}A_{n+k+1}^-&\le\frac{1}{8}\left|a_{n+k}^2-a_{n+k-1}^2\right|=\frac{1}{8}A_{n+k}^+A_{n+k}^-\\&\le\frac{1}{8}A_{n+k}^+\cdot\frac18\left|a_{n+k-1}^2-a_{n+k-2}^2\right|=\frac1{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\... | Set
$$
w=\frac{1+iz}{1-iz}
$$
so you can solve for $z$, getting
$$
z=i\frac{1-w}{1+w}
$$
This is real if and only if
$$
i\frac{1-w}{1+w}=-i\frac{1-\bar{w}}{1+\bar{w}}
$$
that becomes
$$
1+\bar{w}-w-w\bar{w}=-1+\bar{w}-w+w\bar{w}
$$
that is, $w\bar{w}=1$.
Therefore $|a+bi|=1$ is a necessary condition. Now if we write $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Exercises to see if certain series converge.
*
*Does the series $\sum \frac{2^k+1}{3^k}$ converge?
*Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?
*Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?
*Does the series $\sum \frac{k!}{(2k)!}$ converge?
For the first question, do we use the comparis... | For 1) use the fact that $\sum (\frac 2 3)^{k}$ and $\sum \frac1 {3^{k}}$ both converge.
Your answer for 2) and 3) are OK.
In 4) $a_{k+1}$ is not $\frac {(k+1)!} {(2k+1)!}$. It is $\frac {(k+1)!} {(2k+2)!}$. [You have to replace $k$ by $k+1$ in the formula for $a_k$]. But your approach is good and you will be able to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | Let $a = 1$. Then $4^b = 3 \Rightarrow b = \log_4 3 = \frac{\log 3}{\log 4}$.
Therefore:
$$9^{a/b} + 16^{b/a}$$
$$=9^{\log 4 / \log 3} + 16^{\log3 / \log 4}$$
$$=3^{2 \log 4 / \log 3} + 4^{2 \log 4 / \log 3}$$
$$=3^{2 \log_3 {4} } + 4^{2 \log_4 {3}}$$
$$= \left( 3^{\log_3 4} \right )^2 + \left( 4^{\log_4 3} \right)^2 $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Generating function on number of ways to purchase a bouquet with exactly 50 dollars There are two kinds of flowers in a shop. Roses cost 3 dollars each while carnations cost 2 dollars each. How many different kinds of bouquets can be bought with exactly 50 dollars?
My solution:
We can purchase $1,2,3,...$ roses, each c... | Use trusty partial fractions...
$\begin{align*}
[x^{50}] \frac{1}{(1 - x^2) (1 - x^3)}
&= [x^{50}] \left(
\frac{1}{3 (1 + x + x^2)}
+ \frac{1}{4 (1 + x)}
+ \frac{1}{4 (1 - x)}
+ \frac{1}{6 (1 - x)^2}
\right) \\
&= [... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Result from $X^2+Y^2=p^{n-1}$ with $p^{n-1} \equiv 1 \pmod{8}$ I'm reading an article and they have the equation $X^2+Y^2=p^{n-1}$ where $p$ is a prime and $p^{n-1} \equiv 1 \pmod{8}$ and $p \equiv 1 \pmod{4}$.
They conclude that since $p^{n-1} \equiv 1 \pmod{8}$ it must follow that
$$\begin{cases}
X \equiv 0 \pmod{4... | Note $p^{n-1} \equiv 1 \pmod 8$ means both sides of $X^2 + Y^2 = p^{n-1}$ is odd. Thus, one of $X$ and one of $Y$ must be even and the other odd. Consider $X$ to be even and $Y$ to be odd. As $Y$ is odd (so $Y \equiv \pm 1 \pmod 4$), when squared, it's $Y^2 \equiv 1 \pmod 8$. With $X$ being even, if it's $X \equiv 2 \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find function $f(x)$ knowing that $[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$ is satisfied for $\forall x \in \mathbb R$.
Find function $f(x)$ knowing that $$\large [f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$$ is satisfied for $\forall x \in \mathbb R$ $(f(x) \ne c$ with $c$ being ... | If we assume a solution of the form $f(x)=\dfrac{x}{a}$ then
$$ \left[\frac{1}{a}(x^2-x)+1\right]\cdot\left[\frac{1}{a}(x^3+1)\right]
=\left[\frac{1}{a}(x^4-2x^3+x)+1\right]\cdot\left[\frac{1}{a}(x+1)\right] $$
$$ (x^2-x+a)\cdot(x^3+1)=(x^4-2x^3+x+a)\cdot(x+1) $$
So either $x=-1$ or
$$ (x^2-x+a)\cdot(x^2-x+1)=x^4-2x^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Here is my attempt:
$x = 0$ and $y = 0$, these both line go through the $x$ and $y$ axes. And also the circle touches those two lines. So the ce... | If the simple error in your calculation [coefficient in middle $2xy$ term in expansion of $(x+y)^2$ ] is removed the correct equation is
$$ (x^2+y^2)-a (x+y)+ \frac{a^2}{4}=0 $$
The origin/corner touching circle is in the first or fourth quadrant according as odd term $y$ is positive or negative.
$$ (x^2+y^2)-a (x-y)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{n\rightarrow \infty}\int_0^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} dx$ I feel it goes to zero I separate the integral $\int_0^1 \frac{x^{n-2}\cos(n\pi x)}{1+x^n}$ the sequence goes to zero and it bounded by $0.5$ so by bounded convergence theorem the limit is zero. Now for the $\int_1^\infty \frac{x^{n-2}\... | Your solution seems fine. For a bit of further simplification, we may do as follows. Write $I_n$ for the integral. Then
\begin{align*}
&\left| I_n - \int_{1}^{\infty} \frac{\cos(n\pi x)}{x^{2}} \, \mathrm{d}x \right| \\
&\leq \int_{0}^{1} \left| \frac{x^{n-2}\cos(n\pi x)}{1+x^n}\right| \, \mathrm{d}x + \int_{1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
From a set of N elements how many reflexive relations are out of the anti symmetric relations As I know in a set of N elements there are $2^{n^2}$ (two to the power of $n$ squared) relations, in which there are $3^{\frac12(n^2-n)}$ anti-symmetric relations.
How can I find out of those anti-symmetric relations the amoun... | A binary relation on a set $S$ with $N$ elements can be represented by an $(N \times N)$-table in which the cell $(a, b)$ is marked if and only if $(a, b)$ belongs to the relation (i.e., $a$ is related to $b$).
For example, if $S = \{ 1, 2, 3, 4 \}$, a simple example of binary relation on $S$ is:
\begin{array}{c||c|c|c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum angle $X$ in the range $0^\circ \leq x < 360^\circ$ which satisfies the equation $\cos^2(2x)+\sqrt{3}\sin(2x)-\frac{7}{4}=0$
Find the maximum angle $X$ in the range $0^\circ \leq x < 360^\circ$ which satisfies the equation
$\cos^2(2x)+\sqrt{3}\sin(2x)-\frac{7}{4}=0$
$\cos^2(2x)=\sin^2(2x)-1$, so we c... | Note you have $\cos^22x=1-\sin^22x$; with $\sin2x=t$ you get
$$
1-t^2+t\sqrt{3}-\frac{7}{4}=0
$$
so the equation can be rewritten
$$
4t^2-4t\sqrt{3}+3=0
$$
This is actually $(2t-\sqrt{3})^2=0$, so the only root is $t=\sqrt{3}/2$. Hence $2x=\pi/3+2k\pi$ or $2x=2\pi/3+2k\pi$. Hence
$$
x=\frac{\pi}{6}+k\pi\qquad\text{or}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all $(x,y)$ of positive integers s.t. $x^3-y^3=xy+61$ Given
$x^3-y^3=xy+61$.
Find all pairs (x,y) of positive integers which satisfies the given equation.
I tried solving this by following method-
$(x-y)(x^2+y^2+xy)-xy=61$
$(x-y)[(x-y)^2+3xy]-xy=61$
$(x-y)^3+3xy(x-y)-xy=61$
I don't know if doing this was in any m... | $x\geq y+1$ so:
$61 = x^3-y^3-xy=x(x^2-y)-y^3\geq (y+1)(y^2+y+1)-y^3=2y^2+2y+1$
$y^2+y\leq 30$, so $1\leq y\leq 5$ - just check this numbers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Number of particular paths of even length Let $W_n=\{ w=a_1a_2\cdots a_{2n}: a_i\in\{1,-1\},\ a_1+\cdots+a_j\geq 0,\ \forall\ 1\leq j\leq 2n\}$. Let $W_n^k=\{w\in W_n:w\ \text{does not cross the line}\ y=k \}$. Is there any way to find $\# W_n^k$?
Example: if $n=2$, then $\# W_2=6$ and $\# W_2^1=1$ and $\#W_2^2=4$.
| Fix $k$ and let $a_{n,i}$ be the number of paths of length $n$ with step sizes $\pm 1$ that never leave the strip $\{0, 1, \ldots, k\}$ and end in state $i$, $0 \leq i \leq k$. If $0 < i < k$, so $i$ is not on the boundary, there are two ways that we could have arrived at the state $i$: either from $i+1$ or $i-1$. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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I need to prove that $\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$ I need to prove the following limit
$$\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$$
Using the Squeeze Theorem
$$0\le|\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}|\leq\frac{|x|\log(1+y)}{y}\to0$$
For $(x,y)\to(0,0)$
Here I have used... | $\begin{aligned}
& \lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}} \xrightarrow{\begin{cases}x=r \cdot \cos\theta
\\ y=r \cdot \sin\theta
\end{cases}}\lim_{r\to 0}{\frac{|r \cdot \cos\theta| \ln{(1+r \cdot \sin\theta)} }{ \sqrt{(r \cdot \cos\theta)^2+(r \cdot \sin\theta)^2} } }
\\& \le \lim_{r\to 0}\frac{|r \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$
*
*Show that the sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ is monotone decreasing and bounded below.
Let $a_n = \frac{n^2}{9^n}$. For $n\geq 1$ we have
\begin{equation*}
a_n-a_{n+1} = \frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}.
\end{equation*}
T... | Notice that
\begin{align*}
a_n-a_{n+1}&=\frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}\\
&=\frac{1}{9^n}\left(\frac{9n^2-n^2-1-2n}{9}\right)\\
&=\frac{8n^2-1-2n}{9^{n+1}}.
\end{align*}
Now, $8n^2-1-2n=(4n+1)(2n-1)\gt 0$ because $n\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The proof of van der Corput inequality $\newcommand{\lrp}[1]{\left(#1\right)}$
$\newcommand{\lrmod}[1]{\left|#1\right|}$
I am trying to understand the proof of the van der Corput inequality given in Lemma 1 of this blog entry due to Tao. We will use the Big-O notation, whose definition can be found here.
Lemma. (Van d... | You are mistaking the inequality. Tao is saying that there is a constant $C > 0$ with $|\frac{1}{N}\sum_{n \le N} a_n| \le C(\frac{1}{H}\sum_{h \le H} |\frac{1}{N}\sum_{n \le N} a_{n+h}\overline{a_n}|)^{1/2} + C\frac{H}{N}$.
Actually, I think there should be a $C\frac{1}{H^{1/2}}$ on the right hand side as well (see t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can s... | Induction will work. Although, the observation made by José shows why induction works.
Suppose that $n\in\mathbb N$. For the base case when $n=1$ we have
$$\sum\limits_{k=1}^{2^2} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{2}$$
then for the induction hypothesis we will assume that
$$\sum\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula. So by easy means, I derived
$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
But then I checked for the actual formula it is this;
$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2... | You made a mistake in your derivation. You should have $(x^2+y^2)^\color{red}2=a^2+b^2$.
This follows from $a=x^2-y^2$ and $b=2xy$ (or from known properties of complex modulus).
Thus, $a=x^2-\dfrac {b^2}{4x^2}$;
solving this quadratic equation in $x^2$ yields $x^2=\dfrac{a+\sqrt{a^2+b^2}}2$ as the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
PMF for the maximum of two four-sided dices
Suppose we roll two four-sided dice, that is, each die has four sides, numbered $1, 2, 3, 4$. Let $X_1$ and $X_2$ be the numbers that appear on the first and second die respectively, and let $Z = \max\{X_1, X_2\}$, that is $Z$ is the larger of the two numbers rolled. Find th... | Assuming the die is fair, there are $4^2 = 16$ equally likely outcomes.
The maximum of a set, if it exists, is the largest element of the set.
\begin{array}{c c}
\text{outcome} & \text{maximum}\\ \hline
(1, 1) & 1\\
(1, 2) & 2\\
(1, 3) & 3\\
(1, 4) & 4\\
(2, 1) & 2\\
(2, 2) & 2\\
(2, 3) & 3\\
(2, 4) & 4\\
(3, 1) & 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I derive the following expression for the sum of orthogonal matrices? In Johansen's book 'Likelihood-based inference in cointegrated vector autoregressive models', in order to get the expression for the Granger's representation theorem he claims that:
$$\beta_\bot(\alpha'_\bot \beta_\bot )^{-1} \alpha'_\bot + \a... | The stated identity need not be valid under the assumptions provided. Consider for example $\alpha=\beta_{\bot}=\binom{1}{0}$ and $\beta=\alpha_{\bot}=\binom{0}{1}$ corresponding to $N=2$, $R=1$. These vectors all have rank $R=N-R=1$, and they satisfy $\alpha'\alpha_\bot = \beta'\beta_\bot = 0$. But $\alpha'_\bot \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means ... | Also, we can use C-S here:
$$19.5=\sqrt{6.5^2\cdot9}=\sqrt{(2.5^2+6^2)((2x-1)^2+(y+1)^2)}\geq$$
$$\geq2.5(2x-1)+6(y+1)=5x+6y+3.5,$$
which gives $$5x+6y\leq16.$$
The equality occurs for $$(2.5,6)||(2x-1,y+1),$$ which says that $16$ is a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Solve for $\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$ It is known that $$\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x=\pi \ln \left|2 \cos \frac{a}{2}\right|+a \ln \left|\tan \frac{a}{2}\right|+2 \sum_{k=0}^{+\infty} \frac{\sin (2 k+1) a}{(2 k+1... | Let's follow the advice by @Sonal_sqrt. For $-\pi/2<a<3\pi/2$,
$$\int_0^\infty\frac{\cos a\ dx}{1+2x\sin a+x^2}=\tan^{-1}\left.\frac{x+\sin a}{\cos a}\right|_{x=0}^{x=\infty}=\frac{\pi}{2}-a,$$
so, if $I(a)$ is the given integral, then (for the same range of $a$)
\begin{align}
I'(a)&=\int_0^\infty\frac{2x\cos a\ dx}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _... | Hint. Note that
$${1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}=\frac{(x^3+x^{-3})^2}{4}.$$
so it remains to evaluate
$$L=\frac{1}{2}\int ^2 _1(x^3+x^{-3})\,dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the following integral: $\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$ How can we compute this integral for all $\operatorname{Re}(a)>0$ and $\operatorname{Re}(b)>0$?
$$\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$$
Is there a way to compute it using methods from real analysis?
| **my attempt **
$$I=\int_{0}^{\infty }\frac{x\ cos(ax)}{e^{bx}-1}dx=\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}cos(ax)dx\\
\\
\\
=\frac{1}{2}\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}\ (e^{iax}-e^{-iax})dx=\frac{1}{2}\sum_{n=1}^{\infty }[\int_{0}^{\infty }x\ e^{-(bn-ia)}dx+\int_{0}^{\infty }x\ e^{-(bn+ia)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding polynomial functions that satisfy a given condition
Find all polynomial functions $P(x)$ which satisfy $P(x)^2=P(P(x))$.
Attempt at reaching answer:
Let $d$ be the degree of $p(x) $.
We have $d+d=d*d \implies d=0 \vee d=2$.
If $d=0$ we have that $p(x)$ is constant, and this is indeed a solution.
Otherwise, w... | You may proceed as follows:
Differentiating gives
$$2P(x)P'(x) = P'(P(x))P'(x)$$
Obviously the identity is satisfied, if $P'(x) = 0 \Rightarrow P(x) = c \stackrel{c^2=c}{\Longrightarrow} c=0$ or $c=1$ .
Now, if $P'(x) \neq 0$, then setting $y=P(x)$ gives us on a non-empty open interval the polynomial equation
$$2y = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Fully simplifying $\sqrt{13+2\left(\sqrt{2}-\sqrt{10}-\sqrt{20}\right)}$ I have this statement:
Simplify this: $\sqrt{13+2\left(\sqrt{2}-\sqrt{10}-\sqrt{20}\right)}$
I know how to simplify roots like $\sqrt{a \pm k\sqrt{b}}$, But i don't know how to simplify this.
Any hint for an elegant solution is appreciated.
| Notice that $$(-\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z-2\sqrt{xy}+2\sqrt{yz}-2\sqrt{xz}$$
So $$\sqrt{x+y+z-2\sqrt{xy}+2\sqrt{yz}-2\sqrt{xz}}=-\sqrt{x}+\sqrt{y}+\sqrt{z}$$
In your case on comparision you have $x+y+z=13$, $xy=20, zx=10, yz=2.$ So you get $xyz=20$, then $x=10,y=2,z=1$, Hence the answer is $\sqrt{10} -\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\... | As ganeshie8 suggested, your matrix formula is not working because the line does not pass through the origin.
When you translate everything up by $6$ units, the line now passes through the origin and you can continue as follows:
$$\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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An interesting identity involving the abundancy index of divisors of odd perfect numbers Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
A number $y$ is said to be perfect if $\sigma(y)=2y$.
Denote the abundancy index of $z$ by $I(z)=\sigma(z)/z$.
Euler proved that an odd perfect number $N$, if ... | There is nothing too special about the identities, as we shall soon see.
Hereinafter, we let $q^k n^2$ be an odd perfect number with special prime $q$, and we assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds true. Also, we denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show $13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0$ for $x\in(0,1)$ I am doing a problem that can be reduced to proving the following inequality:
$$13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0, x \in(0,1)
$$
B... | Here's my sketch proof:
For the following assume $x\in(0,1)$. First use the following lemma:
$$\left(x^2+1\right)+\sqrt{2}\sqrt{x^4+1}>2x+2\left(\frac{x^4+1}{x^2+1}\right)$$
Proof:
$${\frac{d}{dx}\left(2(x^4+1)-\left\{2x+2\left(\frac{x^4+1}{x^2+1}\right)-(x^2+1)\right\}^2\right)\\=\frac{4(x-1)^5(x^4+2x^3+4x^2+4x+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\d... | Take the step below,
$$(x_1-x_2)^2= x_1^2+x_2^2 - 2x_1x_2=\dfrac{7}{4}-2(-\frac 12) = \frac{11}{4}
$$
Thus,
$$|x_1-x_2|=\frac{\sqrt{11}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
computing an integral without making any substitution We are trying to evaluate
$$ \int\limits_0^1 ( \sqrt{2-x^2} - \sqrt{2x-x^2} ~) ~ dx $$
without any substitution (well, this is how this problem is supposed to be solved)
Idea:
We notice that if $y=f(x)$ is the integrand, then $f(1) = \sqrt{2}$ and $f(1)=0$ as is e... | One possible approach to solve this problem consists in interpreting integrals as areas. Indeed, consider the functions
\begin{align*}
\begin{cases}
f(x) = \sqrt{2 - x^{2}}\\
g(x) = \sqrt{2x - x^{2}} = \sqrt{1 - (x-1)^{2}}
\end{cases}
\end{align*}
The graph of the function $f$ is the upper semicircle centered at the or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | You may use
*
*$(x^2-1)|(x^{2k} -1)$ by applying $(x^2)^k - 1 = (x^2-1)(x^{2(k-1)} + \cdots + x^2 + 1)$
\begin{eqnarray*} x^{81} + x^{49} + x^{25} + x^{9} + x
& = & x\left((x^2)^{40} -1 + (x^2)^{24} -1 + (x^2)^{12} -1 + (x^2)^{4} -1 + 4+1\right) \\
& = & x((x^2-1)p(x) + 5)\\
& = & x(x^2-1)p(x) + 5x
\end{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Cant get partial fraction decomposition I would like to prove that: $$ L = (\frac{1+z^{-1}}{1+0.5z^{-1}} ) \cdot (\frac{1}{1-z^{-1}}) = \frac{0.166}{z + 0.5} + \frac{1.33}{z - 1} + 1$$
How do I get from left to right? The same problem have other solution that I succeed to get $L = \frac43\frac{z}{z-1} - \frac13\frac{z}... | First, write the expression as a ratio of expanded polynomials in lowest terms, that is,
$$
\frac{2z^2+2z}{2z^2-z-1}.
$$
Next, use long division to find the quotient and remainder of $\frac{2z^2+2z}{2z^2-z-1}$ and then rewrite it as the quotient plus the remainder over the denominator, as follows
$$
1+\frac{3z+1}{2z^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} $ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$.
Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $a, b, c$ are positives satisfying $abc = 1$.
We... | Let $a=b=c=1$.
Thus, we obtain a value $\frac{3}{16}.$
We'll prove that it's a maximal value.
Indeed, by AM-GM $$\sum_{cyc}\frac{bc}{(b+c)^3(a^2+1)}\leq\sum_{cyc}\frac{bc}{8\sqrt{b^3c^3}(a^2+1)}=\sum_{cyc}\frac{\sqrt{a}}{8(a^2+1)}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{3}{2},$$ where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$ Let $a_n = {1 \ \over n+1} +{1 \ \over n+2}+ \ldots\ +{1 \ \over 2n}. $
Prove that for $ n \ge\ 3 $ one has $ a_{n+1}\ge\ a_n $. and based on this conclude that $ {a_{2019}}>{3 \over 5}$
I try
$ a_{n+1}- a_n = {1 \ \over n+2}+{1 \ \over n+3} +{1 \ \... | Supppose $n \geq 3$.\begin{align*} a_{n+1} &= \frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n+1} + \frac{1}{2n+2} \\ & =\left( \frac{1}{n+1} + \frac{1}{n+2}+ \cdots + \frac{1}{2n-1} + \frac{1}{2n} \right) + \left( \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} \right) \\ & = a_n + \left( \frac{1}{2n+1} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Is the series $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$ convergent or divergent $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$
$\begin{align}
\sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\
&= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} ... | Yes. Whenever you have a convergent series $\sum_{n=0}^\infty a_n$ and a divergent series $\sum_{n=0}^\infty b_n$, the series $\sum_{n=0}^\infty(a_n+b_n)$ diverges.
Or you can apply the ratio test to reach the same conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/sma... | Observe:
$(x - \frac{1}{x})^2 = x^2 - 2 + \frac {1}{x^2}\\
(x + \frac{1}{x})^2 = x^2 + 2 + \frac {1}{x^2}\\
(x - \frac{1}{x})^2= (x + \frac 1x)^2 - 4$
We make this substitution in the original expression
$(x + \frac 1x)^4(x - \frac{1}{x})^2)\\
(x + \frac 1x)^4((x + \frac{1}{x})^2 - 4)\\
(x + \frac 1x)^6 - 4(x + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Summation or Algebra? I am trying to visualize the concept that should be involved with this problem: What is $x$ in the equation
$$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0\quad ?$$
What should I do? Express in summation (which I find very hard) or just algebra?
| $$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0$$
$$\implies x(1-2+3-4+\cdots-10)-(1-2^2+3^2-\cdots-10^2)=0$$
$$\implies x=\dfrac{1-2^2+3^2-\cdots-10^2}{1-2+3-4+\cdots-10}$$
$$\implies x=11$$
$$1-2^2+3^2-\cdots-10^2=\sum_{n=1}^{10}(-1)^{n-1}n^2=-55$$
and $$1-2+3-4+\cdots-10=\sum_{n=1}^{10}(-1)^{n-1}n=-5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Definite integral of rational expression involving quadratics I was given the following exercise on my Calculus class:
$$
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}dx
$$
I studied from a book (author, Stewart) various methods to solve integrals of the form $\int \frac{P(x)}{Q(x)}dx$; a basic idea common to all of this methods... | HINT
You may start by noticing that
\begin{align*}
\frac{8x+32}{x^{2}-4x+8} & = 4\times\frac{2x + 8}{x^{2}-4x+8} = 4\times\frac{2x - 4 + 4 + 8}{x^{2}-4x+8}\\\\
& = 4\times\frac{2x - 4}{x^{2}-4x+8} + 4\times\frac{12}{x^{2}-4x+8}
\end{align*}
In order to integrate the first expression, observe that $(x^{2}-4x+8)^{\prime}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + ... | As Steve Kass's question comment suggests, you have
$$\begin{equation}\begin{aligned}
(x + 1) (y + 2) (z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
((y + 1) + 1)\left((x + 1) (z + 1)\right) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1) (y + 1) (z + 1) + (x + 1)(z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1)(z + 1) & = 8
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Using integration to compute the area of a region bounded by a curve For the area bounded by curves
$$(x-1)^2 + y^2 =1 $$
and
$$x^2 + y^2 =1$$
I tried it by finding $y$ and integrating within the suitable limits through $y\,\mathrm{d}x$ method. But when I tried the $x\,\mathrm{d}y$ way the answer was different. For th... | Through geometry, we can see that the area bound by the curves is made up of $4$ equal areas -- namely the area bound by the upper red semicircle and the $x$-axis on the interval $x\in[0.5,1]$.
If the shaded area above is $A$, then the total area bound by the curves is $4A$.
We can then write the red semicircle as a f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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proving the identity $b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$ for $b \in \mathbb{N}$, $b > 2$, by counting a set in different ways I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$ without just using algebra. The idea I've thought about is that there are $b^{3}$ way... | $$b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$$
Method 1: We count ordered triples with entries drawn from the set $\{1, 2, 3, \ldots, b\}$ with replacement in two ways.
There are $b$ choices for each of the three entries, so there are $b^3$ such triples.
For the RHS, we consider cases, depending on how many different num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't? What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't?
I started by taking the prime factorization of $20^{19}$ to get : $2^{38}5^{19}$.
I then noticed, that t... | Well, If $x = 2^a5^c$ and $y = 2^b5^d$
thenn we have $a+b > 38$ or $c+d > 19$.
To have $a+b > 38$ while $a \le 38$ and $b \le 38$ if we have $a = k$ then $b$ can be as small as $39-k$ and can be as large as $38$. Those are $k$ options. $(38+1) - (39-k) = k$. So there are $\sum_{k=1}^{38}k = \frac {38*39}2 = 741$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find the maximal value of $c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$ in terms of $c^3 + a^3 = m$ where $a, b$ and $c$ are positives.
Let $a, b$ and $c$ be positive real numbers such that $c^3 + a^3 = m$ and $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0.$$ Find the maxi... |
Claim: The maximal value is $${\frak m}=\frac12\left(\frac m2\right)^{2/3}+\left(\frac m2\right)^{-1/3}+\frac32\tag1$$ attained at $a=b=c$.
Let $$d=c^2 - ca + a^2 - \dfrac{(b + 1)^2(b - 2)}{c + a}=\frac{m-b^3+3b+2}{a+c}\tag2$$ with $a^3+c^3=m$. Supposing that $$(b^3 - abc) + ab(a + b) - c(a^2 + b^2)=r\tag3$$$$(abc - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Oresme's proof of the harmonic series' divergence I've been trying to understand Oresme's proof that the harmonic series diverges since it's greater than the series of halves, which diverges.
I'm struggling to capture an aspect of the relationship which I think can be expressed as: "the series of halves is not surject... | Another way to write this.
Consider the first $m$
groups of $2^k$ terms
for $k = 0$ to $m-1$.
The $k$-th group is
$g(k)
=\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{j}
$.
Since $j$ for each term
is less than $2^{k+1}$,
and there are
$2^k$ terms,
the sum satisfies
$g(k)
\gt\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{2^{k+1}}
=\dfrac{2^k}{2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving $\cos \frac{165}{2}$ What are the steps to solve $$\frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4}$$ into $$\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
Please explain. I got them from derivation of $$\cos(82.5^\circ)$$ in 2 different ways.
$$\cos(60^\circ+22.5^\circ)$$ and $$\cos\frac{(90^\circ+75^\circ)}{2}$$
Thank... | Note that $\sqrt{2\sqrt{2}} = \sqrt{4/\sqrt{2}}$ so you have
$$
\begin{split}
f &= \frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4} \\
&= \frac12 \sqrt{\frac{2\sqrt{2}-1-\sqrt3}{\sqrt2}} \\
&= \frac12 \sqrt{2 - \frac{1+\sqrt3}{\sqrt2}}
\end{split}
$$
so remains to prove that $$ \frac{1 + \sqrt{3}}{\sqrt2} = \sqrt{2+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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"Altered" Alternating Series Diverges or Converges? Consider the series
$$
1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
which alternates between a block of positives and a block of negatives, with the block sizes,
$$
1, 1, 2, 3, 4, ... | Write $H_n = \sum_{k=1}^{n} \frac{1}{k} $ for the $n$-th harmonic number. It is well-known that $H_n$ has the asymptotic form
$$ H_n = \log n + \gamma + \mathcal{O}\left(\frac{1}{n}\right) \quad \text{as} \quad n\to\infty. $$
Now returning to OP's question, let $s_n$ denote the $n$-th partial sum. Also, write
$$m_k = \... | {
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"source": "stackexchange",
"question_score": "1",
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Recursive sequence. Value is resetting to 0
Solve
$$\begin{cases}
{a_n} = 5a_{n-1} - 6a_{n-2} + 6 \cdot 3^{n}\\
{a_0} = 1, a_1=50
\end{cases}$$
The problem I've got here is the C resetting.
It happens to me sometimes and I don't know why.
Hope someone can clear it up for me.
Thanks.
| $a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n$, with $a_0=1$, $a_1=50$.
First we find solution of homogenous recurrence relation
$$a_n-5a_{n-1}+6a_{n-2}=0.$$
$r^2-5r+6=0$, we have $r=2$ with multiplicity 1 and $r=3$ with multiplicity 1. So,
$$a_n^h=A\cdot 2^n+B\cdot 3^n.$$
Because in the right-hand side $$a_n-5a_{n-1}+6a_{n-2}=6\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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antidervative of $\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $ Question
How can the antiderivative of
$$
\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $$ be obtained?
| A hint from trigonometry
Consider the Figure below, where $\triangle ABC$ is right-angled, $\overline{AB} = |x|$ and $\overline{AC}=1$, so that, by definition
$$\alpha = \angle CAB = \arccos |x|.$$
Draw the bisector of $A$, that intersects $BC$ in $D$. From $D$ draw the perpendicular to $AD$. Let, on it, $E$ be a point... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Interesting sequence with a beautiful term : $U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$ Question :
Find the term of $U_{n}$ such that :
$U_{0}=2$ and $U_{1}=\frac{5}{2}$
$U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$
My try :
We have ; $U_{2}=\frac{5}{2}$ $U_{3}=2^{3}+2^{-3}$
After calculus ... I think the term is :
... | Given any $v > 1$, construct sequence $\displaystyle\;v_n = \begin{cases} 1, & n = 0\\ v, & n = 1\\ v_{n-1}v_{n-2}^2 & n > 1\end{cases}$
Notice
$$v_{n+1} = v_n v_{n-1}^2 \iff
\frac{v_{n+1}}{v_n^2} = \frac{v_{n-1}^2}{v_n}
\implies
\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}} =
\frac{v_{n-1}^2}{v_n} + \frac{v_n}{v_{n-1... | {
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"url": "https://math.stackexchange.com/questions/3383500",
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"source": "stackexchange",
"question_score": "2",
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Show that $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $z=-\frac12(1\pm i\cot(k\pi/8))$; and follow-up questions
*
*(a) Show that the equation $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $-\frac12\left(1\pm i\cot\frac{k\pi}{8}\right)$, where $k=1,2,3$.
*(b) Hence show that
$$(z+1)^8-z^8=\tfrac18(2z+1)(2z^2+2z+1)\left(4z^2+4z... | $$
(1+z)^8=z^8\tag1
$$
$(1)$ implies that
$$
1+z=e^{\pi ik/4}z\tag2
$$
That is
$$
\begin{align}
z
&=\frac1{e^{\pi ik/4}-1}\\
&=\frac1{\cos(\pi k/4)-1+i\sin(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{(\cos(\pi k/4)-1)^2+\sin^2(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{2-2\cos(\pi k/4)}\\
&=\frac{-\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Complex Numbers - Converting to Polar form I understand the basics of converting to polar form but I have just come across a question that I haven't seen before.
Usually the complex number is expressed as $z=a+bi$, but this time the complex number I was given is $z^3=-4+4{\sqrt3}i$.
Do I need to somehow remove the 3 p... | Now $z^3=8(\cos \frac{2\pi}{3}+i \sin \frac{2 \pi}{3})$: using DeMoivre's formula we get
$$
z_0=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 0 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 0 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{2\pi}{9}+i \sin \frac{2 \pi}{9}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Inverse element of $x^2 + x + 1$ in $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ I know that $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field since $x^3 + x + 1$ is irreducible in $\mathbb Z_2$. But I still can't seem to find any inverse element for $x^2 + x + 1$. I want to find an element $g(x) \in \mathbb Z_2[x]... | Write $(x^2+x+1)(ax^2+bx+c)=1\implies ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c=a(-x^2-x)+(a+b)(-x-1)+(a+b+c)x^2+(b+c)x+c=(b+c)x^2+cx+c=(b+c)x^2+cx+c-a=1\implies a=1,b=0,c=0$, where I have used the fact that $x^3+x+1=0$ to substitute for $x^4$ and $x^3$.
So $x^2$ is your inverse.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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$1^3+2^3+...+n^3= $? How I can find formula $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...
I tried like this
$1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$
after a long process i found it this way
$A... | You can expand
$$\left(\frac {n(n+1)}2\right)^2=\frac 14n^2(n^2+2n+1)=\frac 14n^4+\frac 12n^3+\frac 14n^2$$
and read off $A=\frac 14,B=\frac 12,C=\frac 14,D=0,E=0$
You can also solve simultaneous equations
$$1^3=1=A1^4+B^3+C1^2+D1+E=A+B+C+D+E\\
1^3+2^3=9=A2^4+B2^3+C2^2+D2+E=16A+8B+4C+2D+E$$
and three more similar. Yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
He... | Your argument is correct. However, you could come to the same conclusion through a different method. Let
$$a_n=\frac{2^n+5^n}{3^n+4^n}$$
and consider what happens as $n\to\infty$. In the numerator, $5^n$ will grow faster than $2^n$ and in the denominator $4^n$ will grow faster than $3^n$. Therefore
$$\lim_{n\to\infty}a... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Does the series $\sum_{n = 1}^{\infty}\frac{\cos n}{\sqrt{n}}(1 + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{n!})$ converge? I want to check if this series converge or diverge:
$$\sum_{n = 1}^{\infty}\frac{\cos n}{\sqrt{n}}(1 + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{n!})$$
Which technique can i use there ?
UPD:... | Recall that
$$
e= \lim_{n \to \infty} 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!}
$$
Now consider $S_N= \cos n (1+1/1!+\cdots+1/N!)$. We have $\sum_{n=1}^M S_n$ is bounded for every integer $M$; to see this, consider see this post and use the fact that you know the sum of the reciprocal f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find matrix A of linear transformation of two bases. If we have set E = {v1, v2, v3} with
$v1 = \begin{pmatrix}1\\1\\1\end{pmatrix}$ ,
$v2 = \begin{pmatrix}1\\0\\-2\end{pmatrix}$ ,
$v3 = \begin{pmatrix}1\\1\\0\end{pmatrix}$
and
$L(v1) = \begin{pmatrix}1\\2\\3\\0\end{pmatrix}$ ,
$L(v2) = \begin{pmatrix}3\\0\\1\\0\end... | You can write
$LV=U$, where $L_{4 \times 3}$ is the unknowm matrix $V_{3 \times 3}$ is the matricx made from the columns $v_i$ and $U_{4 \times 3}$ is made from the columns $u_i$
$V=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & -2 & 0 \end{bmatrix}$ $~~U=\begin{bmatrix} 1 & 3 & -1 \\ 2 & 0 &-2 \\ 3 & 1 & -3 \\ 0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Where is the error in this apparent contradiction arising from the function $f(\pm n)=\left(\frac{n}{n+1}\right)^{\pm 1}$? We define a function $ f \left( \pm n \right) =\left( \frac{ n }{ n + 1 } \right) ^ { \pm 1 } $.
That is, $$ f \left( m \right) = \begin{cases}\frac{ m }{ m + 1 } &\text{ if } m \ge 0 \\ \frac{ -m ... | you have , where you used alpha and beta
$\frac{a}{b}\div \frac{a + b}{b} = \frac{a}{a +b}$
but you've ended up with $b^2$ etc which is right, but you could cancel b
then you say 'since' and you don't cancel b out properly
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to solve this limit, and what did I do wrong? I have to solve:
$$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} $$
Here's my attempt:
$$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} = \lim_{n\to\infty}\frac{(1+\frac{2}{n+1})^{1/6}-1}{\frac{2}{n+1}}\cdot\frac{2}{(n+1)(\sqr... | From here
$$\ldots=\frac{1}{6}\lim_{n\to\infty}\frac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}=\frac{1}{6}\frac{2}{\ln 2+1-1}=\ldots$$
As an alternative
$$\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1}=\frac{\frac1n}{\sqrt[n]{2}-1}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\frac1n}\to \frac{1}{3\ln2}$$
indeed
*
*$\frac{\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because... | The limit of $\frac{x \log{x}}{(x^2+1)^2}$ has the same behavior as $\frac{\log{x}}{x^3}$.
You can use L'Hôpital's rule and see that it converges to $\frac{1}{3x^3}$, which goes to $0$ as $x$ approaches infinity.
Using integration by parts with $u=\log{x}$ and $\frac{dv}{dx} = \frac{x}{(x^2+1)^2}$. We know that $\frac{... | {
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"url": "https://math.stackexchange.com/questions/3409650",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Arrangement of letters with one pair of letters always between two letters
How many ways are there to arrange $11$ letters, $A,B,C,D,E,F,x,x,x,y,y$ so that every $y$ lies between two $x$ (not necessarily adjacent)$?
Attempt:I fixed the letters $ x \quad y \quad y \quad x$. Totally there are 5 gaps(all distinct) for t... | The capital letters can be arranged in $6!=720$ ways. The lowercase letters can be arranged in $\binom{3}{1}=3$ ways, since we are forced to have x's at either end.
With that done, the capital letters and lowercase letters can be merged in $\binom{11}{5}=462$ ways -- all we need to do is choose which five spots out of... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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using epsilon delta definition to proof limits
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
let $$f(x)=\frac{2x+3}{x-1}$$ then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(
0 < \left|x-1 \right|=x-1<\delta\Longrightarrow \large\left|\frac{2x+3}{x-1}\right|\right)>M$
$$M<\left|\frac{2x+3}{x-1}\right|... | FIRST:--- Choose $M>0$. Let $\delta>0$ be such that,$$0<\delta<\frac{5}{M+2}$$ $$\implies\frac{5}{\delta}-2=\frac{5-2\delta}{\delta}>M.$$ Hence, for any $x\in \Bbb R$ with $1<x<1+\delta$ we have, $$\frac{2x+3}{x-1}=\frac{2(x-1)+5}{x-1}>\frac{5-2(x-1)}{x-1}>\frac{5-2\delta}{\delta}>M.$$ So, $$\lim_{\large x \to 1+} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to find $\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$ how to find
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$$ without L'hopital or taylor/Laurent series
I tried bu... | Let $y=x-\pi/3$, so we are to compute
\begin{align*}
\lim_{y\rightarrow 0}\dfrac{\tan^{3}(y+\pi/3)-3\tan(y+\pi/3)}{\cos(y+\pi/2)}.
\end{align*}
Note that
\begin{align*}
\tan^{3}(y+\pi/3)-3\tan(y+\pi/3)&=\tan(3(y+\pi/3))(3\tan^{2}(y+\pi/3)-1)\\
&=(\tan(3y))(3\tan^{2}(y+\pi/3)-1),
\end{align*}
and the term reduces to
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int {\dfrac{x^2+\left(n-1\right)n}{\left(x\sin x+n\cos x\right)^2}}dx $
$$\int {\dfrac{x^2+\left(n-1\right)n}{\left(x\sin\left(x\right)+n\cos\left(x\right)\right)^2}}dx $$
My Try:
I multiple $x^{2n-2}$ to both N and D, then took D as $u$ and then solved to get $\dfrac{n\sin\left(x\right)-x\cos\left(x\righ... | Use shorthands $s= \sin x$ and $c=\cos x$ below. Decompose the integrand with $s^2+c^2 = 1$
\begin{align}
& {\dfrac{x^2+(n-1)n}{(x\sin x+n\cos x)^2}} \\
=&\frac{x^2+(n-1)n}{(xs+nc)^2}
= \frac{x^2(s^2+c^2)+(n-1)n(s^2+c^2)}{(xs+nc)^2} \\
=& \frac{[xs+(n-1)c](xs+nc)-[(1-n)xs+xc](ns-xc) }{(xs+nc)^2}\\
= & \frac{xs+(n-1)c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_{0}^{\pi/2}{\frac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$
$$\int_{0}^{\pi/2}{\dfrac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$$
I am unable to exploit the properties of definite integral, neither it seems that indefinite integration is possible.
| I think that power 2 of $\sin x$ is missed and solve the other one as below. At the first sight, the numerator of the integrand seems to be an antiderivative of $x\sin x$. However, the negative sign in between reminds me that it should be an antiderivative of $\dfrac{\sin x}{x}$. So I decided to divide both numerator a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the harmonic conjugate Let
$$u(x,y)=x^3+ax^2y+bxy^2+2y^3$$
Be a harmonic function and $v(x,y)$ be its harmonic conjugate .If $v(0,0)=1$ then find the
$$|a+b+v(1,1)|$$
The solution i tried-since $u$ is harmonic so by laplace equation we get
$$6x+2ay+2bx+12y=0$$
By comparing coefficients we get $a=-6 $ and $b=... | $$u(x,y) =x^3 -6x^2 y -3xy^2 +2y^3 =(x^3 -3xy^2 ) +2 (y^3 -3x^2 y) =Re( (x+iy)^3 ) - Im(2 (x+iy)^3 )= Re z^3 -2Im z^3 == Re [(1-2i)z^3 ]$$
Hence the conjugate is equal to
$$v(x,y)=Im [(1-2i) z^3 ]$$
$$|a+b +v(1,1) |=|-9 +Im [(1-2i) (1+i)^3]|=|-9 +6|=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate
\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}
We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{... | Hint:
$$ (1\pm f(x))^n \approx 1\pm nf(x) $$
$$ \ln(1\pm f(x)) \approx \pm f(x)$$
for $ f(x) \approx 0$ as $x\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1+2^N \bmod 2^{N+1}$ I want to prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1+2^N \bmod 2^{N+1}$ for $N\geq 4$.
First, let's prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1 \bmod 2^{N}$.
$$1+i-i^2\equiv1+j-j^2$$
$$\Leftrightarrow (i-j)(i+j-1)\equiv0$$
Since the parity of $... | Let $a_i=1+i-i^2$ and
$$p_n=\prod_{i=1}^{2^n}a_i$$
so that we have to show
$$p_{n-1}\equiv 1+2^n\pmod{2^{n+1}}\tag 1$$
Note that $(1)$ implies $p_{n-1}^2\equiv 1+2^{n+1}\pmod{2^{n+2}}$.
By induction on $n\geq 4$, we have:
\begin{align}
p_n
&=p_{n-1}\prod_{i=2^{n-1}+1}^{2^n}a_i\\
&=p_{n-1}\prod_{i=1}^{2^{n-1}}a_{2^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Parabola equation in form of quadratic $ax^2+bx+c=y$ where $a+b+c$ is an integer Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach
$(x-\frac{1}{4})^2=4a'(y+\frac{9}{8... | We know that the vertex x coordinates is $\frac{-b}{2a} = \frac{1}{4}$ solve for $b = \frac{-a}{2}$ now the point $\frac{1}{4} , \frac{-9}{8}$ satisfies the parabola so
$\frac{a}{16} - \frac{a}{8} + c = \frac{-9}{8}$
$a-2a + 16c = -18 $
$c = \frac{-18 +a}{16}$ finally
$a+b+c = a + \frac{-a}{2} +\frac{-18+a}{16} \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is my Proof for the following problem correct? Prove that $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$ If $a,b,c$ are positive real numbers prove that $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
I state the following:
Multiplying both sides by $abc$ yields $$(abc)(a+b+c)?{a^4+b^4+c^4}$$
By AM-G... | Your idea is correct, but the proof is a bit confusing because it is not always clear if you assume the desired inequality or not. I would write it as follows:
Using the AM-GM inequality $(1)$ and the power-mean inequality $(2)$ we have
$$
abc(a+b+c)\underset{(1)}{\leq}(a+b+c)\left(\frac{a+b+c}{3}\right)^3 = 3\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $364 \mid n^{91} - n^7$ [Generalization of Euler & Fermat Theorems] I am trying to prove this statement which is equivalent to show $n^{91} = n^7 \pmod{ 364}$. By splitting modulus theorem $n^{91} = n^7 \bmod 91$ and $\bmod 4$. Then I don’t know what to do next... Can anyone help me?
| Use Euler's theorem:
Mod $4$: either $n\equiv0 $ or $2$, so $n^{91}\equiv n^7\equiv0$, or $\color{blue}{n^2\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv (n^2)^{42}n^7\equiv n^7$
Mod $7$: either $n\equiv0$, so $n^{91}\equiv n^7\equiv 0$, or $\color{blue}{n^6\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv(n^6)^{14}n^7\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What 's wrong with this integral substitution? What's wrong with the following solution?
Let $t=\sin x,$ then $x=\arcsin t.$ By substitution method we have the following:
$\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^0 \frac{t^2}{\sqrt{1-t^2}}\rm{dt}=0$.
However, we know that $\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^{2\pi}\frac{1-... | The reason why the lower and upper limits turn into the same value $0$ after changing variable is due to the lack of bijectivity of the substitution, even though the calculation is incorrect. Bijectivity in changing the variable is not a strong requirement, but it needs to be careful in calculation when a non-bijective... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
given $ ∫ y dx$ between x=3 and x=1, show that it equals A + B√3 where A and B are integers to be found Where $$∫y dx =2x^{3/2} - 8x^{1/2}$$
I keep getting $6 + 6√3$, which is incorrect. The correct answer should be $6 - 2√3$
But I can't figure out how to correctly derive that answer.
Cheers
| It must be a calculation error. Indeed, given the antiderivative expression $2x^{\frac 32} - 8x^{\frac 12}$, there is no need to attempt to find $y$, because the integral between $1$ and $3$ is given by the value of this function at $3$ minus its value at $1$, by the fundamental theorem of calculus.
The value at $3$ :
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$ What is the value of $5x^2-5x-1?$ If $$x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$$ What is the value of $5x^2-5x-1?$
Efforts:
After rationalization, I got $x={\sqrt{5}+1\over 2}$ and $x^2={\sqrt{5}+1\over \sqrt{5}-1}$. Going by this method is very tedious and boring.
Is there a m... | Rationalizing the denominator of $x$, we get
$$x=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}}=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5+1}}=\sqrt{\frac{(\sqrt5+1)^2}{4}}=\frac{1+\sqrt5}2$$
For the expression to be evaluated, we have
\begin{align}
5x^2-5x-1&=5(x^2-x-\frac15)\\
&=5(x^2-x-\frac15-\frac45+\frac45)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Problem with equation in complex numbers
I am supposed to calculate:
$x^{2}=5+i$
I used formula:
$\left | \cos \frac{\alpha }{2} \right |=\sqrt{\frac{1+\cos \alpha }{2}}$
$\left | \sin \frac{\alpha }{2} \right |=\sqrt{\frac{1-\cos \alpha }{2}}$
and I came to the point where:
x= $\sqrt[4]{26}\left ( \frac{\sqrt{\sqr... | $5 + i = \rho (\cos \theta + i\sin theta)\\
\sqrt{\rho (\cos \theta + i\sin \theta)} = \sqrt {\rho}(\cos \frac \theta2 + i\sin \frac \theta2)\\
\rho = \sqrt {5^2 + 1} = \sqrt {26}\\
\theta = \arctan \frac {1}{5}\\
\cos \theta = \frac {5}{\sqrt {26}}\\
\sin\theta = \frac {1}{\sqrt {26}}\\
\tan \frac{\theta}{2} = \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determinant of a $3 \times 3$ Vandermonde matrix
Let $$A = \begin{bmatrix} p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{bmatrix}$$ Prove that $$\det(A) = (r-q)(r-p)(p-q)$$
| \begin{align}
\begin{vmatrix}p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{vmatrix}
&=\begin{vmatrix}q^2&q\\r^2&r\end{vmatrix}-\begin{vmatrix}p^2&p\\r^2&r\end{vmatrix}+\begin{vmatrix}p^2&p\\q^2&q\end{vmatrix}\\
&=(q^2r-qr^2)-(p^2r-pr^2)+(p^2q-pq^2)\\
&=q^2r-qr^2-p^2r+pr^2+p^2q-pq^2\\
&=\frac{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3439048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that the equation has only one real root. Prove that $(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only one real root.
It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root.
But i... | Let $y=\dfrac{x-1+x-2+x-3+x-4}4$
$x=y+2.5$
$$(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=(y+1.5)^3+(y+.5)^3+(y-.5)^3+(y-1.5)^3$$
Now use $(a-b)^3+(a+b)^3=2(a^3+3ab^2)$
$$0=4y^3+6y((1.5)^2+(.5)^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3439455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
} |
How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here pg.305 in Golan. The example is given below:
How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here?
I watched this video https://www.youtube.com/watch?v=msFp3vOYIoA and I followed the procedure mentioned... | If you're good with generalized eigenvectors as a concept, skip this. Recall for an eigenvector $Av=\lambda v$ so $(A-\lambda I)v=0$. For generalized eigenvectors we satisfy $(A-\lambda I)^kv=0$, so we can see that an eigenvector can be generalized by solving for $(A-\lambda I)u=v$ where $v$ is a generalized eigenvecto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Adjoint of Sum = Sum of Adjoints? Didnt find this anywhere, just verifying.
We know that: $$ (A+B)^T= A^T+B^T $$.
Does it follow that:$$ (A+B)^† = A^† +B
^† $$ for all A and B matrices (the dagger here representing the adjoint)?
If so what is the proof?
| This statement is not true in general.
Wikipedia tells us that for 3 by 3 matrices the first element of the adjoint obtains as,
$$
a_{1,1}^\dagger = a_{2,2}a_{3,3} - a_{2,3}a_{3,2} \quad \text{ and } \quad b_{1,1}^\dagger = b_{2,2}b_{3,3} - b_{2,3}b_{3,2},
$$
but
$$
\begin{align}
(a_{1,1} + b_{1,1})^\dagger &= (a_{2,2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate integral $\int_t^\infty (1-3x^{-4})e^{-x^2/2}dx$ I am trying to evaluate the integral
$$\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x.$$
This is as far as I could get before getting stuck.
$$
\begin{align}
\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x&=\int_t^\infty e^{-x^2/2}\,\mathrm{d}\left(x+\frac{1}... | $$I=I_1-I_2$$
$$I_1=\int_{t}^{\infty} e^{-x^2/2} dx= \sqrt{2} \int_{t/\sqrt{2}}^{\infty} e^{-z^2} dz=\sqrt{\frac{\pi}{2}}~Erfc(t/\sqrt{2})$$
$$I_2=3\int_{t}^{\infty} x^{-4}~ e^{-x^2/2} dx$$
Let $x^2=2u, xdx=du$, then
$$I_2=\frac{3}{4\sqrt{2}} \int_{t^2/2}^{\infty} u^{-5/2} e^{-u} du$$
Next, twice integration by parts l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\le... | Writing $A_2$ for $A/2$, etc, and noting that $A_2+B_2+C_2 = \pi/2$, we have
$$\begin{align}
\tan A_2-\tan B_2 &= \tan B_2 - \tan C_2 \\[6pt]
\frac{\sin A_2 \cos B_2 - \cos A_2 \sin B_2}{\cos A_2 \cos B_2}
&=
\frac{\sin B_2 \cos C_2 - \cos B_2 \sin C_2}{\cos B_2 \cos C_2} \\[6pt]
\sin(A_2-B_2)\cos C_2 &= \sin(B_2-C_2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
... | Instead of using L'Hospital, multiply the top and bottom by conjugate:
$$\frac{\sin^2{x}\Big((2\sin^2{x}+3\sin{x}+4)-(\sin^2{x}+6\sin{x}+2)\Big)}{\cos^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\Big)}=\\
\require{cancel}\frac{\sin^2{x}(\cancel{1-\sin x})(2-\sin x)}{(\cancel{1-\sin x})(1+\sin x)\Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Minimize $x + z$ subject to $x^2 + y^2 = 1$ and $y^2+z^2 = 4$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + z \\
\text{s.t} & \quad x^2 + y^2 = 1 \\
& \quad y^2+z^2 = 4
\end{align*}$$
One of the regularity conditions is linear independence constraint constrain... | Yes, your solution is correct, I just corrected a small typo in the notations of $f(x, y, z)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality... | I have a solution using Buffalo way, but it's ugly! I'm sorry about that!
Solution:
Without loss of generality, assume that $x=\min\{x,y,z\}$.
Let $x=a$, $y=a+u$, $z=a+v$ so $a>0$; $u,v \geq 0$
We need to prove: $$(x+y+z)^2 (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 - 81\frac{(x^2 +y^2 +z^2)}{xy+yz+zx} \geq 0$$
After redu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding the Maclaurin series for $f(x)=x\ln(x+1)$
Find the Maclaurin series for the function
$$f(x)=x\ln(x+1)$$
So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.
$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0... | You may use
$$\frac1{1-t} = 1 + t + t^2 + \>...$$
to obtain,
$$x\ln(1+x) = -x\int_0^{-x} \frac{dt}{1-t}=-x\int_0^{-x}\left(1+t+t^2+\>...\right)dt
= x^2 - \frac{x^3}2 + \frac{x^4}3+\>...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Give an asymptotic upper bound of this recurrence relation : $T(n) = 2\cdot T(n^{1/2}) + n$ I unrolled the recursion and got this:
$T(n) = 2^k\cdot T(n^{1/2^k}) + [2^0\cdot n^{1/2^0} + 2^1\cdot n^{1/2^1} + 2^2\cdot n^{1/2^2}+\cdots+2^{k-1}\cdot n^{1/2^{k-1}}]$
I considered every $n^{1/2^i}$ as $n$.
$\begin{align}
T(n)&... | We have
$$T(m^2)=2T(m)+m^2.$$
Then
$$T(4)=2T(2)+4,\\
T(16)=4T(2)+8+16,\\
T(256)=8T(2)+16+32+256,\\
T(65536)=16T(2)+32+64+512+65536,\\
\cdots\\
T(2^{2^k})=2^kT(2)+2^k2^2+2^{k-1}2^4+2^{k-2}2^8+\cdots2^{2^k},$$
which is
$$T(n)=\log_2n\ T(2)+\log_2n\,2^2+(\log_2n-1)2^4+(\log_2n-1)2^8+\cdots+n=O(n).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$ This question was asked in the Crux Mathematicorum , October edition , Pg -$5$ , which can found be Here.
The question states that :
Both $4$ and $52$ can be expressed as the sum of two squares as well as
exceeding another square by $3$... | Brainstorming from doing something similar with pythagorean triples.
If $b^2 = 2a + 1 + 3=2a+4$ then $a^2 + b^2 = a^2 + 2a + 1 + 3 = (a+1)^2 + 3$.
If $b$ is any even number, $2k; k > 1$ then if $a = \frac {(2k)^2-4}2= 2k^2-2$
And if $c = 2k^2 -1$ then
$a^2 + b^2 = $
$(2k^2 -2)^2 + 4k^2 =$
$4k^4 - 8k^2 + 4 + 4k^2 =$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If $2R+r=r_1$, then prove that $A=90$
R is the circumradius of the triangle ABC, $r$ is the inradius and $r_1$ is the ex-radius on side a which is opposite to angle A
In the expression
$$2R+4R\sin \frac A2 \sin \frac B2 \sin \frac C2=4R\sin \frac A2 \cos \frac B2 \cos \frac C2$$
R gets canceled and the expression be... | Hint
$$2R=\cdots=4R\sin A/2\cos(B+C)/2=4R\sin^2A/2=2R(1-\cos A)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\ti... | Take the upper right point $A$ of the red circle, the center $B$ of the yellow circle and the center $C$ of the big circle as vertices of a triangle. Call the radius of the smaller circles $r$. Then $AB=3r$, $BC=r\sqrt3$ and $\angle CBA=30^\circ$. Law of cosine gives $R=AC$, the radius of the big circle.
My results:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Limit with criteria $\lim_{n \to \infty}n \cdot \left [ \frac1e \left (1+\frac{1}{n+1} \right )^{n+1}-1 \right ]$ $$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]$$
I was trying to calculate a limit that drove me to this case of Raabe-Duhamel's test, but I don't know how t... | An overkilled method:
\begin{align*}
n\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right]&=\dfrac{n}{n+1}\cdot(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end{align*}
so we are to look at
\begin{align*}
(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculating definite integral with absolute value I need to evaluate, with $a,c>0$, the integral:
$$\int_{-a}^a\left(1-\frac{|x-y|}c\right)\,dy$$
This is what i tried
if $x>y:$
$$\int_{-a}^a\left(1+\frac{y-x}c\right)\,dy=2 a-\frac{2 a x}{c}$$
if $x<y:$
$$\int_{-a}^a\left(1+\frac{x-y}c\right)\,dy=2 a+\frac{2 a x}{c}$$
i... | For $x \in (-a,a) $
$$I \equiv \int_{-a}^a\left(1-\frac{|x-y|}c\right)\,dy \\ = \frac 1c\int_{-a}^x\left( c-x+y \right)\,dy
\\ + \frac 1c\int_{x}^a\left( c+x-y \right)\,dy
\\ = \frac 1c \left[ (c-x)(x+a)+(c+x)(a-x) \right]
\\ = \frac 1c (-2x^2+2ac) =2a-2\frac{x^2}{c} $$
for $x<-a$
$$I = 2a+2\frac{ax}c $$
for $x>a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$xy + ax^2 + bx + c = 0 , x+y$ has relative max or min if... Options :
$b^2-4ac > 0$
$b^2/4ac >0$
$b/(c-1) >0$
$c/(a-1)>0$
$a/(b-1)>0$
Relative max or min is local max or min, means max or min at certain open interval..
$xy + ax^2 + bx + c = 0 $
$ax^2 + (y+b)x + c = 0 $ if its a qudratic fuction.
Then the max or min ... | Get $y$ by itself so that $$y = \frac{-ax^{2}-bx-c}{x}$$
Then substitute this for $y$ in $f(x) = x+y$ so we have
$$f(x) = \frac{-ax^{2}-bx-c}{x} + x$$
and we want to know if $f(x)$ will have a relative minimum or maximum. The cases where it will not have a minimum or maximum are when the derivative has no real zeroe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did... | HINT:
$$cos\theta_1cos\theta_2=\frac{cos(\theta_1 - \theta_2) + cos(\theta_1 + \theta_2)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.