Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2... | Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = 8$$
Now we can write $$x-y = (x-2)-(y-2) = \left[(x-2)+(2-y)\right]$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
$$\displaystyle \left[(x-2)^2+(2-y)^2\right]\cdot [1^2+1^2]\geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
| Let $x = n + \epsilon$ where $n\in \mathbb Z$ and $0 \le \epsilon \lt 1$.
Then
$x^2 = (n + \epsilon)^2 = 19n - 88$
Since $0 \le \epsilon \lt 1$, then $n^2 \le 19n - 88 \lt (n+1)^2$
The line $y = 19x - 88$ intersects the parabola $y = x^2$ at $x = 8$ and at $x = 11$. Since the parabola $y = x^2$ is concave up, this imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attem... | Edit: with some reasoning, and without controversial part
I see no way of calculating this primitive using human tools like integration by parts, substitution and so on. However, inspired by the example here, we could try a function in the form
$$
a\log\bigl[p(x)\sqrt{x^4+4x^3-6x^2+4x+1}+q(x)\bigr],
$$
where $a$ is a c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that the sequence $(a_n)$ where $a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$ is of Cauchy. Show that the sequence $(a_n)$ where
$$a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$$
is of Cauchy.
Comments:
I will use the definition
$$\forall \epsilon > 0, \exists n_0 \in \mathbb{N}; n,m > n_0 \Rightarrow |a_n - a... | If $m>n>1$, then
\begin{align}
|a_m-a_n|&=\left|\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{m!}\right|\\&<\left|\frac{1}{(n+1)!}+\frac{1}{2}\frac{1}{(n+1)!}+\ldots+\frac{1}{2^{m-n-1}}\frac{1}{(n+1)!}\right|\\
&=\frac{1}{(n+1)!}\sum_{k=0}^{m-n-1}\frac{1}{2^k}\\
&<\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{2^k}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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a conjectured continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{... | The proposed continued fraction
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}
\end{equation}
can be written as
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
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Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have non-integer solutions? Does $\lfloor x\lceil y\rceil\rfloor = \lceil x\lfloor y\rfloor\rceil$ have any non-integer solutions? If so, how do you find them?
| I'm assuming your $x$ and $y$ are real numbers. Now, we can write $x= \lfloor x \rfloor + \{x \} $ and $ y=\lfloor y \rfloor + \{y\} $ where the $\{ \}$ here denotes the fractional part. We have that $\{x\} \neq 0$ and $\{y\} \neq 0$ since $x$ and $y$ are not integers. Your equation then reads
$$ \lfloor x \lceil y \rc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $e^{x+y}=e^xe^y$ using $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$. I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$
So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1... | \begin{align*}
e^x e^y &= \lim_{n\rightarrow \infty}(1+\frac{x}{n})^n\lim_{n\rightarrow \infty}(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x}{n})^n(1+\frac{y}{n})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n}+\frac{xy}{n^2})^n\\
&=\lim_{n\rightarrow \infty}(1+\frac{x+y}{n})^n \frac{(1+\frac{x+y}{n}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
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higher college level combinatorics Show the following equality:
$$\sum_{n_1,\ldots,n_k\geq 0} \min(n_1,\ldots,n_k)x_1^{n_1} \cdots x_k^{n_k}=\frac{x_1\cdots x_k}{(1-x_1)\dots(1-x_k)(1-x_1x_2\cdots x_k)}.$$
This is a problem from Stanley's Enumerative Combinatorics, and I am trying to prove the equality. I tried to plu... | Looking at the RHS, you should know that $\frac{1}{(1-x_i)} = (1+x_i+x_i^2+x_i^3+\dots)$
Let us simplify notation a little by letting $x_1x_2x_3\dots x_k=X$
and since the RHS is a product of such terms you have:
$=X(1+x_1+x_1^2+x_1^3+\dots)(1+x_2+x_2^2+x_2^3+\dots)\dots(1+x_k+x_k^2+x_k^3+\dots)(1+X+X^2+X^3+\dots)$
Dist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Infinitely rational points in $y^2 = x^3 - 4$? If the $x$-coordinate of a rational point $P$ of $y^2 = x^3 - 4$ is given by $m/n$, the $x$-coordinate of $2P$ is given by$${{(m^3 + 32n^3)m}\over{4(m^3 - 4n^3)n}}.$$Using this fact, how do I show that$$144 \cdot H(x\text{-coordinate of }2P) \ge H(x\text{-coordinate of }P)... | Let $m$ and $n$ be relatively prime integers and let$$A = |(m^3 + 32n^3)m|,\text{ }B = |4(m^3 - 4n^3)n|.$$Denote by $D$ the greatest common divisor of $A$ and $B$. In order to show the inequality in question, it suffices to show that $D$ is a divisor of $144$. For, if that is the case and if the $x$-coordinate of $P$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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What is the coefficient of $x^3 y^4$ in the expansion of $ (2x-y+5)^8$ I was thinking of doing $\binom{8}{4}$ but not sure if right.
| Pre-expansion, there are $8$ factors of $2x - y + 5$.
From those $8$ factors, choose the $3$ that contribute to the $x^3$, from the remaining $5$ factors, choose the $4$ that will contribute to $y^4$. There is only 1 factor left so it chooses itself.
Post-expansion, that's ${8 \choose 3}{5 \choose 4}{1 \choose 1}$ ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$ Show that $5\cos^2x - 2\sqrt{3}\sin x\cos x + 3\sin^2x = \cos 2x - \sqrt{3}\sin2x + 4$
Attempt.
I'm preparing for a trigonometry examination tomorrow. So I'm solving as many questions as possible. This is confusing. I've tried solvin... | The first thing to remember is
$$
\sin2x=2\sin x\cos x
$$
that settles the $-\sqrt{3}\sin2x$ part. Thus you just need to massage
$$
5\cos^2x+3\sin^2x
$$
recalling $\cos2x=\cos^2x-\sin^2x$; so rewrite it as
$$
\cos^2x+4\cos^2x-\sin^2x+4\sin^2x=\cos2x+4(\cos^2x+\sin^2x)
$$
You can also equivalently prove that
$$
5\cos^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finishing my $\epsilon - \delta$ proof Hi fellas I have this limit: $$lim_{x\to-1}\dfrac{x^2-1}{x^2+x}=2$$
$Dem:$
Let $\epsilon >0$ random but fixed then:
$$|\dfrac{x^2-1}{x^2+x}-2|=|\dfrac{x^2-1-2x^2-x}{x^2+x}|=|\dfrac{-x^2-2x-1}{x^2+x}|=|\dfrac{(x+1)^2}{x(x+1)}|=|\dfrac{x+1}{x}|<M|x-1|<M(\dfrac{\epsilon}{M})=\epsilon... | What you want to show is that,
for any $\delta > 0$
there is an $\epsilon > 0$
such that
if
$|x+1| < \epsilon$
then
$|\dfrac{x^2-1}{x^2+x}
-2| < \delta
$.
You have worked out
very nicely that
$|\dfrac{x^2-1}{x^2+x}
-2|
=|\frac{x+1}{x}|
$.
Suppose
$|x+1| < \epsilon$.
To make sure that
the $x$ in the denominator
does no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proof that $3^c + 7^c - 2$ by induction I'm trying to prove the for every $c \in \mathbb{N}$, $3^c + 7^c - 2$ is a multiple of $8$. $\mathbb{N} = \{1,2,3,\ldots\}$
Base case: $c = 1$
$(3^1 + 7^1 - 2) = 8$ Base case is true.
Now assume this is true for $c=k$.
Now I prove this holds for $c=k+1$ $(3^{k+1}+7^{k+1}-2)$.
$(... | Notice
\begin{align}
3^k\cdot 3+7^k\cdot 7-2&=3^k\cdot(3+4)-3^k\cdot 4+7^k\cdot 7-2\cdot 7 +2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+2\cdot 7-2\\
&=7(3^k+7^k-2)-3^k\cdot 4+12\\
&=7(3^k+7^k-2)-12\cdot(3^{k-1}-1)\\
\end{align}
by hypothesis $8$ divides $3^k+7^k-2$, and for $k\ge 1$ the number $3^{k-1}-1$ is even, then $8$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving that $\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} =\tan \left ( \frac{\alpha+\beta}{2} \right )$ Using double angle identities a total of four times, one for each expression in the left hand side, I acquired this.
$$\frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{\sin \left ( \f... | Assume that $$P=(\cos\ a,\sin\ b),\ Q=(\cos\ b,\sin\ b)$$ Then what is
slope of line passing through $-Q,\ P$ :
For convenience $0<a<b<\pi/2$, let $R=(0,1)$. Then $$\angle\
ROQ=\frac{\pi}{2}-b,\ O=(0,0)$$
and $$\angle\ Q(-Q)P = \frac{b-a}{2} $$
When $\angle\ PP'(-Q) = \pi/2$, then $\angle\ P(-Q)P'=
\frac{a+b}{2}$
Hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Min of $\frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^{7}+ c^{7}} +\frac{c^{10}+ a^{10}}{c^{7}+ a^{7}} $ I got this problem I tried several time to solve it by many inequalities but I got stuk. My question is how I get the minimum value of $$ \frac{a^{10}+ b^{10}}{a^{7}+ b^{7}} +\frac{b^{10}+ c^{10}}{b^... | First note that
$$(a^3-b^3)(a^7-b^7)\ge 0\tag{1}$$
so
$$2(a^{10}+b^{10})\ge (a^7+b^7)(a^3+b^3).\tag{2}$$
It follows that
$$\frac{a^{10}+b^{10}}{a^7+b^7}+\frac{b^{10}+c^{10}}{b^7+c^7}+\frac{c^{10}+a^{10}}{c^7+a^7}\ge a^3+b^3+c^3.$$
Note that the equality happens when $a=b=c$. Next, we show that $a^3+b^3+c^3$ attains it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Finding the Roots of Unity I have the following equation, $$z^5 = -16 + (16\sqrt 3)i$$I am asked to write down the 5th roots of unity and find all the roots for the above equation expressing each root in the form $re^{i\theta}$. I am just wondering if my solutions are correct. Here are my solutions,
5th roots of unity,... | For the fifth roots of unity, the roots are $\frac{2\pi}{5}$ radian apart from one another. Since the principal value for arg(z) is (-$\pi$,$\pi$], hence, the fifth roots of unity are $$z = 1, e^{\frac{2\pi}{5}i}, e^{\frac{4\pi}{5}i}, e^{-\frac{2\pi}{5}i}, e^{-\frac{4\pi}{5}i}$$
For the all the roots of the above equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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A trigonometry question involving medians and a sum of cotangents The medians of a triangle $ABC$ make angles $\alpha$, $\beta$ and $\gamma$ with each other. I have to prove that:
$$\cot\alpha+ \cot\beta+ \cot\gamma +\cot(A)+\cot(B)+\cot(C)= 0.$$
Any ideas?
|
By the sine and cosine theorems,
$$\cot(\widehat{BGC}) = \frac{\cos\widehat{BGC}}{\sin\widehat{BGC}} = \frac{\frac{BG^2+CG^2-BC^2}{2\, BG\cdot CG}}{\frac{2[BGC]}{BG\cdot CG}}=\frac{1}{4}\cdot\frac{BG^2+CG^2-BC^2}{[BGC]}\tag{1}$$
but $BG=\frac{2}{3}m_a$ and Stewart's theorem gives $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$, hen... | {
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"url": "https://math.stackexchange.com/questions/1473203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rationalise the denominator and simplify $\frac {3\sqrt 2-4}{3\sqrt2+4}$ Does someone have an idea how to work $\dfrac {3 \sqrt 2 - 4} {3 \sqrt 2 + 4}$ by rationalising the denominator method and simplifying?
| For problems like this, you need to multiply numerator and denominator by the denominator's conjugate. A conjugate is found by flipping the sign from positive to negative or negative to positive in a binomial. So, the denominator's conjugate in the above's equation is $ 3\sqrt{2} -4 $. Multiplying an irrational binomia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $1+x^2=\sqrt{3}x$ then $\sum _{n=1}^{24}\left(x^n+\frac{1}{x^n}\right)^2$ is equal to I tried this problem by different methods but i am not able to get the answer in easy way . First i found the roots of equation and then represented it in polar form of complex number . I got $\cos \left(30^{\circ}\right)\pm i\sin ... | For $\quad x=\cos 30^{\circ}+i\sin 30^{\circ}\quad$ from De Moivrè's formula follows
\begin{align}
x^n&=\cos (30^{\circ}n)+i\sin (30^{\circ}n)\\
\frac{1}{x^n}=x^{-n}&=\cos (30^{\circ}n)-i\sin (30^{\circ}n)\\
\end{align}
Hence,
$$x^n+\frac{1}{x^n}=2\cos(30^{\circ}n)\tag{1}$$
On the other hand, $\cos (x+360^{\circ})=\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first ine... | You have to apply it repeatedly on different levels.
First, $(a^4+b^4)/2\ge a^2b^2$. Also, $(a^4+c^4)/2\ge a^2c^2$. Now apply the AM-GM to these two.
$${(a^4+b^4)/2+(a^4+c^4)/2\over2}\ge\sqrt{a^2b^2\cdot a^2c^2}=a^2bc=abc\cdot a$$
Now write two other similar inequalities with $a,b,c$ in different order, and add them al... | {
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"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
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Positive integer solutions to $5a^2 + 8a + 4 = 4b^2$ I'm convinced $5a^2 + 8a + 4 = 4b^2$ can be somehow turned into Pell's equation.
My first steps: Rewrite as $5(a + 4/5)^2 + 4/5 = 4b^2$. Rewrite $B = 2b, A = 5a + 4$ to get $A^2 + 4 = 5B^2$. I'm almost there, I just need to know how to solve this Pell-like equation.... | user236182's answer: $a$ must be even, $a = 2k$. Rewrite as $5(4k^2) + 16k +4 = 4b^2$, then multiply by $5/4$ to get $25k^2 + 20k + 5 = 5 b^2 \iff (5k + 2)^2 - 5b^2 = -1$. This is Pell's equation in the form $x^2 - 5y^2 = -1$. This is sufficient for all positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Nonlinear first order ODE In trying to compute an integral curve for a certain fiber of the Hopf fibration in $\mathbb R^3$ under stereographic projection, I've come across the following IVP:
$$
2y' - y^2 - 1=0, \quad y(0) = 0.
$$
I'm at a loss on how to solve this simple looking thing. Any help would be appreciated!
| This is Riccati equation
Rewrite $$y'-\frac{1}{2}y^2=\frac{1}{2}$$
Let $y=-2\frac{w'}{w}$
then $$y'=-2\frac{w''(x)}{w(x)}+2\frac{w'(x)^2}{w(x)^2}=-2\frac{w''(x)}{w(x)}+\frac{1}{2}y^2$$
or
$$y'-\frac{1}{2}y^2=-2\frac{w''(x)}{w(x)}$$
thus we got
$$-2\frac{w''(x)}{w(x)}=\frac{1}{2}$$ or
$${w''(x)}+\frac{1}{4}w(x)=0$$
whi... | {
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"url": "https://math.stackexchange.com/questions/1487071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim _{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$ Find the limit of $f(x)=\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$ when x approaches 4.
I have no idea where to start with this problem.
| Without using L'Hospital's rule
Notice,
Rationalizing the numerator & denominator as follows
$$\lim_{x\to 4}\frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$
$$=\lim_{x\to 4}\frac{(\sqrt{5-x}-1)(\sqrt{5-x}+1)(2+\sqrt{x})}{(2-\sqrt{x})(2+\sqrt{x})(\sqrt{5-x}+1)}$$
$$=\lim_{x\to 4}\frac{(5-x-1)(2+\sqrt{x})}{(4-x)(\sqrt{5-x}+1)}$$
$$... | {
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"question_score": "1",
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rank of product of full rank matrices Please help me with the following question:
Let $A$ be $m \times n$, $B$ be $n \times p$ matrices with $\text{rank}(A)=m$, $\text{rank}(B)=p$, where $p < m < n$.
What are conditions such that $\text{rank}(AB)=\text{rank}(B)$???
Thanks!
| No. In general that is wrong, consider the following example: $p= 1$, $m=2$, $n=3$, $\def\M#1#2{\operatorname{Mat}_{#1,#2}(\mathbf R)}$
$$ B = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \in \M 31,
D = \begin{pmatrix} 2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1 \end{pmatrix} \in \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Continuity of $\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ at (0, 0) I am having trouble proving that $\dfrac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$ is continuous at $(0, 0)$ if we set the value at $(0, 0)$ to be $0$.
I don't see a way to prove this as I cannot factor this into partial fractions.
| Hint. By considering polar coordinates,
$$
x:=r\cos\theta,\quad y:=r\sin\theta,
$$ you get
$$
\begin{align}
f(r\cos\theta,r\sin\theta)&=\frac{r^5(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)}{r^4}\\\\
&=r(\cos^5\theta-4\cos^3\theta \sin^2\theta-\cos\theta\sin^4\theta)
\end{align}
$$ then observe that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
how to find this limit : lim x to infinity how can I find this limit which is become infinite
$\lim _{x\to \infty }\left(x(\sqrt{x^2+1}-x)\right)$
can I use conjugate method ?
that what I'm doing until now
$= x\left(\frac{\sqrt{x^2+1}-x}{1}\cdot \:\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\right)\:$
$= x\left(\frac{1}{\sqrt... | Yes Using Conjugate.
$$\displaystyle \lim_{x\rightarrow \infty}x\left(\sqrt{x^2+1}-x\right) = \lim_{x\rightarrow \infty}x\left[\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x}\times \sqrt{x^2+1}+x\right]$$
So $$\displaystyle \lim_{x\rightarrow \infty}x\left[\frac{1}{\sqrt{x^2+1}+x}\right] = \lim_{x\rightarrow \infty}\frac{x}{x\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
| $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
$\frac{(x+1)(x-3)}{(x-1)(x+2)(x-3)}-\frac{(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{(x+1)(x-3)-(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$
the inequations are:
*
*$x>-2$
*$4x+3>0$ or $x>-\frac{4}{3}$
*$x>1$
*$x>3$
We need to have one or three inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Solve using AM GM Inequality if possible Let x, y, z be non-zero real numbers such that $\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x} = 7$ and $\dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} = 9$, then $\dfrac{x^3}{y^3} + \dfrac{y^3}{z^3} + \dfrac{z^3}{x^3}$ is equal to?
I don't really know how to solve this. any methods wou... | let $a= \frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then some manipulations would show that
$(a+b+c)^3= 3(a+b+c)((a^2+b^2+c^2)) - 2(a^3+b^3+c^3) + 6abc$
we already have $a+b+c = 7, abc=1$, now its left to find what $a^2+b^2+c^2$ is.
$(\frac{x}{y} + \frac{y}{z}+ \frac{z}{x})^2 = (\frac{x}{y})^2 + (\frac{y}{z})^2+ (\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Distributing multiplication of rational functions I am having trouble distributing with fractions. This
$$ \left(\frac{1}{(x + 3)} + \frac{(x + 3)}{(x - 3)}\right)\, (9 - x^2) = -\frac{(x^2-3)}{9-x^2}(9-x^2) $$
has the answer $\left\{\left[x = - \frac{9}{7}\right]\right\}$.
I start solving by cancelling $(9-x^2)$ on th... | Cancelling $(9-x^2)$ on the RHS leaves you with $-(x^2-3)$. Next I'd recommend finding a common denominator and combining the fractions inside the parentheses on the LHS. You should get $$ \frac{1}{\left(x + 3\right)} + \frac{\left(x + 3\right)}{\left(x - 3\right)} = \frac{x-3+(x+3)^2}{x^2 - 9}$$ which means the LHS w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
showing global min point let's $F(x)= x'Ax + a'x $
where $x'Ax$ is a quadratic form and $a'$ is defined as a vector.
$$A:= \left[ \begin{matrix} 6 &1&1 \\ 1&2&0 \\ 1&0&4\end{matrix}\right] $$
Does there exist a global minimum point (absolute min) for this function $F(x)$?
note: I know that if the function is given as ... | Isn't minimum or maximum were the gradient is zero (vector)?
$$ \frac{{\rm d}F(x)}{{\rm d}x} = 2 A x + a' =0 $$
So the unique solution
$$ x = - \frac{1}{2} A^{-1} a' $$
if it exists must be a minimum or maximum.
Edit 1
So for your example suppose $a'=(a_1,a_2,a_3)$ and $x=(x_1,x_2,x_3)$ then
$$ \frac{{\rm d}F(x)}{{\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the smallest possible value of $\lfloor (a+b+c)/d\rfloor+\lfloor (a+b+d)/c\rfloor+\lfloor (a+d+c)/b\rfloor+\lfloor (d+b+c)/a\rfloor$? What is the smallest possible value of
$$\left\lfloor\frac{a+b+c}{d}\right\rfloor+\left\lfloor\frac{a+b+d}{c}\right\rfloor+\left\lfloor\frac{a+d+c}{b}\right\rfloor+\left\lfloor\f... | Let $$g(a,b,c,d)=\dfrac{a+b+c}{d}+\dfrac{b+c+d}{a}+\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}$$
use Cauchy-Schwarz inequality we have
$$g(a,b,c,d)=(a+b+c+d)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)-4\ge 16-4=12$$
so
$$\lfloor\dfrac{a+b+c}{d}\rfloor+\lfloor\dfrac{b+c+d}{a}\rfloor+\lfloor\dfrac{c+d+a}{b}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the lim f(x) = 1 using the $\epsilon-\delta$ definition
Show directly (from the $\epsilon-\delta$ definition ) that
$$ \lim_{x\to-1}\frac{x^{2}-3}{x-1} = 1$$
Attempt
Using
$$
\left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert
= \left\lvert \frac{(x-2)(x+1)... | I find it easier to make simplifying assumptions first. You have an
expression $| { (x-2)(x+1) \over x-1 } |$ which you would like to bound near
$x=-1$. Start by bounding $\delta$ by, say, $1$ (that is $|x+1| < 1$), in which case you have
$|x-1| > 1$, and $|x-2| < 4$. Then you have
$| { (x-2)(x+1) \over x-1 } | \le 4 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the angle of complex number Let
$$
z = \frac{a-jw}{a+jw}.
$$
Then the angle of $z$ is
$$
-\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right).
$$
How is that so?
| Assuming $a,w\in\mathbb{R}$:
$$\arg\left(\frac{a-wi}{a+wi}\right)=$$
$$\arg\left(\frac{a-wi}{a+wi}\cdot\frac{a-wi}{a-wi}\right)=$$
$$\arg\left(\frac{(a-wi)^2}{a^2+w^2}\right)=$$
$$\arg\left(\frac{a^2-w^2-2awi}{a^2+w^2}\right)=$$
$$\arg\left(a^2-w^2-2awi\right)-\arg\left(a^2+w^2\right)=$$
$$\arg\left(a^2-w^2-2awi\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$ I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And th... | You seem to have some very unfortunate ideas about algebra! As David Mitra said, When you are adding powers of z you do not add the powers themselves. That is a property of multiplication: $z^n\cdot z^m= z^{n+ m}$.
To multiply $(z- 1)(1+ z+ z^2+ z^3)$ use the "distributive law" $a(b+ c)= ab+ ac$ and $(b+ c)a= ab+ ac$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ How can one solve the equation :
$[\frac{x^2-x+1}{2}]=\frac{x-1}{3}$ ?
Such that $[x]$ is the integer part of $x$.
By definition :
$[\frac{x^2-x+1}{2}]<=\frac{x^2-x+1}{2}<[\frac{x^2-x+1}{2}]+1$.
So :
$\frac{x-1}{3}<=\frac{x^2-x+1}{2}<\frac{x-1}{3}+1$.
1)
$\frac{x-1}{3}<=\frac{... | The key is to recognize that the left hand side is, by definition of the floor function, an integer, so the right hand side must also be an integer, and hence we must have $x=3k+1$ for some $k\in\mathbb{Z}$. This gives (on doing a little algebra)
$${x^2-x+1\over2}={3k(3k+1)\over2}+{1\over2}$$
Note that $2$ divides eit... | {
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$ x+y+z = 3, \; \sum\limits_{cyc} \frac{x}{2x^2+x+1} \leq \frac{3}{4} $
For positive variables $ x+y+z=3 $, show that $ \displaystyle \sum_{cyc} \dfrac{x}{2x^2+x+1} \leq \dfrac{3}{4} $.
Apart from $ (n-1)$ EV, I could not prove this inequality. I've tried transforming it into a more generic problem - it looks fairly ... | Also, the Tangent Line method helps.
Indeed, we need to prove that:
$$\sum_{cyc}\left(\frac{1}{4}-\frac{x}{2x^2+x+1}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x-1)}{2x^2+x+1}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x-1)}{2x^2+x+1}-\frac{x-1}{4}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(5-2x)}{2x^2+x+1}\geq0,$$
which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the remainder of $31^{2008}$ divided by $36$? What is the remainder of $31^{2008}$ divided by $36$?
Using Euler's theorem, we have:
$$
\begin{align*}
\gcd(31,36) = 1 &\implies 31^{35} \equiv 1 \pmod{36} \\
&\implies 31^{2008} \equiv 31^{35(57) + 13} \equiv (31^{35})^{57+13} \equiv 1^{57}*31^{13} \\
&\implies 31... | The line that goes $31^{35}\equiv 1 \pmod {36}$ is false. Euler's theorem says that $31^{\phi(n)}\equiv 1 \pmod {36}$, but $\phi(36) = 12$. So we would say $31^{2008} \equiv 31^{4} \pmod{36}$ using the same reduction method (taking away multiples of 12) to get $13 \pmod{36}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ Task
Calculate the volume of the region trapped by $z^2=x^2+y^2, z=2(x^2+y^2)$ using a triple integral.
I'm kind of lost on this one, here's my (probably wrong) attempt:
Calculate the intersection, I get $z=0 \wedge z=\frac 12$. That means the ... | Your definition of the region is incorrect (though your bounds for $z$ are correct). When $0\le z\le\frac{1}{2}$, the cone lies above the paraboloid; the circle of intersection at $z=\frac{1}{2}$ is $x^2+y^2 = \frac{1}{4}$. So one definition of the region is
\begin{equation*}
-\frac{1}{2}\le x\le \frac{1}{2},\quad
... | {
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"source": "stackexchange",
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Prove inequality with binomial coefficient: $6 + \frac{4^n}{2 \sqrt{n}} \le \binom{2n}n$ I have to prove inequality, where $n \in N$
$$6 + \frac{4^n}{2 \sqrt{n}} \le {2n \choose n}$$
I have checked and it is true when $n\ge4$, however I have no idea how I should start. Can anyone give a hint?
| $$\frac{1}{4^n}\binom{2n}{n} = \frac{(2n-1)!!}{(2n)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)$$
and by squaring both sides:
$$\left(\frac{1}{4^n}\binom{2n}{n}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$ If $A$ is a symmetric and $B$ is a skew symmetric matrix and $A+B$ is non singular and $C=(A+B)^{-1}(A-B)$,then prove that $C^TAC=A$.
My Attempt:
$C^T=((A+B)^{-1}(A-B))^T=(A-B)^T((A+B)^{-1})^T=(A^T-B^T)((A+B)^T)^{-1}$
$C^T=(A+B)(A-B)^{-1}$
$C^TAC=... | First of all, we show that $C^T(A+B)C=A+B.$ To do that:
$$\begin{align}(A+B)(A-B)^{-1}(A+B)(A+B)^{-1}(A-B) & \\ & \underbrace{=}_{(A+B)(A+B)^{-1}=I}(A+B)(A-B)^{-1}(A-B)\\ & \underbrace{=}_{(A-B)(A-B)^{-1}=I}A+B.\end{align}$$
In a similar way we can show that $C^T(A-B)C=A-B.$ Indeed:
$$\begin{align}(A+B)(A-B)^{-1}(A-B)... | {
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"url": "https://math.stackexchange.com/questions/1506308",
"timestamp": "2023-03-29T00:00:00",
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To determine if a polynomial has real solution I have the following polynomial : $x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$
I must determine if this polynomial has at least 1 real solution and justify why. We have a theorem which says that all polynomials with real coefficients can be decomposed in a product of polynomi... | Since
$$x^7+x^6+x^5+x^4+x^3+x^2+x+1=\frac{x^8-1}{x-1}$$
we can use the fact that $x^8-1$ has roots at $-1$ and $1$ to get the root of $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How many solutions exist for a non-linear system How many solutions exist to the following system:
$$
\begin{eqnarray}
xy+xz &=& 54+x^2 \\
yx+yz &=& 64+y^2 \\
xz+yz &=& 70+z^2
\end{eqnarray}
$$
I have guessed that the solutions, if exist, have to be integer, then we get:
$$
\begin{eqnarray}
x(y+z-x) &=& 2\cdot 3\cdot... | Firstly as you had factorized:
$x(y+z-x) = 54$.
$y(z+x-y) = 64$.
$z(x+y-z) = 70$.
Also the differences between equations are:
$(x-y)z = (x-y)(x+y) - 10$.
$(y-z)x = (y-z)(y+z) - 6$.
$(x-z)y = (x-z)(x+z) - 16$.
Which are equivalent to:
$(x-y)(x+y-z) = 10$.
$(y-z)(y+z-x) = 6$.
$(x-z)(z+x-y) = 16$.
And co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Sum of $(a+\frac{1}{a})^2$ and $(b+\frac{1}{b})^2$ Prove that:
$$
\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge\frac{25}{2}
$$
if $a,b$ are positive real numbers such that $a+b=1$.
I have tried expanding the squares and rewriting them such that $a+b$ is a term/part of a term but what I get is completely... | For your revised question, another way is to note that $(x + \frac1x)^2$ is convex, so by Jensen's inequality:
$$\left(a + \frac1a\right)^2 + \left(b + \frac1b\right)^2 \ge 2\left(\frac{a+b}2 + \frac2{a+b}\right)^2=2\left(\frac12 + 2\right)^2=\frac{25}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How do I complete solving this homogeneous system of linear equations using Gaussian elimination? $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=0\\ { x }_{ 1 }+{ x }_{ 2 }+2{ x }_{ 3 }-{ x }_{ 4 }=0\\ 2{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=0\\ -{ x }_{ 1 }-{ x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }=0\\ $$
Steps I took:
$$... | You end up with
$$\left[\begin{array}{rrrr|r}
1 & 1 & 0& 1& 0 \\
0 & 0 & 1 & -1& 0 \\
\end{array}\right].$$
All you have to do is paint in just straightening the tableau, i.e., create an upper triangle matrix whose diagonal elements are $\pm1$ in the following way. The first row of your reduced matrix has ... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Show that $\sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $
Show that $\displaystyle \sum_{n=1} ^{\infty} \frac {1}{(x+n)^2} \leq \frac{2}{x} $
For any real number x $\geq 1$, I want to show the above
Very rusty on my analysis, I think I need to do a comparison test to show the series converges, but then no... | Let $n\ge 1$, for $x> 0$ we have $$\frac{1}{(x+n)^2}\le\frac{1}{(x+t)^2}\le\frac{1}{(x+n-1)^2}\qquad\text{for }n-1\le t\le n$$
Then
\begin{align}
\int_{n-1}^{n}\frac{1}{(x+n)^2}\,dt&\le \int_{n-1}^{n}\frac{1}{(x+t)^2}\,dt\\[6pt]
\frac{1}{(x+n)^2}&\le\frac{1}{x+n-1}-\frac{1}{x+n}\\[6pt]
\sum_{n=1}^N\frac{1}{(x+n)^2}&\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516295",
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"source": "stackexchange",
"question_score": "4",
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For which $n$ is $2^{2x+2} \equiv 2\pmod n \quad\text{and}\quad 2^{2x+2} \equiv 4\pmod {n-1} $ We begin with
\begin{align*}
2(2^p+n^2-1-2n) \equiv 0\pmod n \quad&\text{and}\quad 2(2^p+n^2-1-2n) \equiv 0\pmod {n-1} \\
\end{align*}
Which we reduce to
\begin{align*}
2^p\equiv1\pmod{\frac n{2}} \quad&\text{and}\quad 2^{p... | EDIT: For a counterexample, try $p=11$ and $n=23$ We have $2(2^p+n^2−1−2n)=5060=10 \cdot 23 \cdot 22$.
INCORRECT ANSWER:
I think I found a counterexample. Set $p = 3$ and $n = 7$. Then, $2(2^p+n^2−1−2n) = 2(8+49-1-14) = 2(42) = 84$. This is divisible by both $n = 7$ and $n-1 = 6$, but $p = 3$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$ How can one evaluate the below sum? Any help would be greatly appreciated.
$$\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$$
| We can built the formula, bottom up.
Let $s_n = \sum_{k=1}^\infty k^n r^k $
$n-(n-1)=1 \Rightarrow (1-r)s_1 = s_0$
$n^2-(n-1)^2=2n-1 \Rightarrow (1-r)s_2 = 2 s_1 - s_0$
$s_0 = r + r^2 + r^3 + \cdots = \frac{r}{1-r}$
$s_1 = \frac{s_0}{1-r} = \frac{r}{(1-r)^2}$
$s_2 = \frac{2 s_1 - s_0}{1-r} = \frac{r(r+1)}{(1-r)^3}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?
Steps I took:
*
*Let $A= \left[\begin{array}{rr}
a & b \\
c & d \\
\end{array}\right]$ and let $B= \left[\begin{array}{rr}
w ... | Since $AB=A$ and $BA=B$, to show that $A$ is an idempotent we have to show that $A=A^2$
Therefore $A^2=(AB)^2 = A^2B^{ } = AB = A(BA) = AA = A^2 = A$. Proven
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | Your proof is correct, because each inequality you write is equivalent to the previous one (it should be noted, probably).
Changing all $\ne$ into $=$ would make it a proof by contradiction, that's however unnecessary.
In a different way, you could just swap terms and square, again changing inequalities into equivalent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
find the solution to recurrent relation Solving some math problem, I have faced this recurrent equation: $$S(n) = 3 S(n - 3) + 2 \sum\limits_{k = 2}^{n / 3} S(n - 3 k) \times k.$$
Here $n = 3 \alpha$, means, $n$ can be divided by $3$ ($\alpha$ is integer), and $S(0) = 1$.
Could anyone please help me to find $S(n)$ in a... | Call $S(3 n) = s_n$, write the recurrence with no subtractions in indices:
$\begin{align}
s_{n + 2}
&= 3 s_{n + 1} + 2 \sum_{2 \le k \le n + 2} k s_{n + 2 - k} \\
&= 3 s_{n + 1} + 2 \sum_{0 \le k \le n} (k + 2) s_{n - k} \\
&= 3 s_{n + 1} + 2 \sum_{0 \le k \le n} (n - k + 2) s_k
\end{align}$
Defin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$ using the definition
Prove $$\lim_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$$ using the definition.
Let there be $\varepsilon>0$ we need to find $N<n$ such that $\Big|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}$$
\Big|<\varepsilon$
$$\left|\frac{n^... | Hint: $\frac{n-3}{3n^2+1}\leq \frac{1}{3n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
System of recurrence relations with Taylor series expansion Find $a_n,b_n$ where $a_0=1,b_0=0$ for the following relations:
$a_{n+1}=2a_n+b_n$
$b_{n+1}=a_n+b_n$
Using generating functions, the system is:
$f(x)-a_0=2xf(x)+xg(x)$
$g(x)-b_0=xf(x)+xg(x)$
Solving for $f(x)$ and $g(x)$ gives:
$$f(x)=\frac{1-x}{x^2-3x+1}$$
$$... | You seem to have done all the necessary work, but confused a couple things. The Taylor series should be evaluated about the point $x=0$, not about the first terms of the sequences $a_0=1$, $b_0=0$. In other words, the $n^{\rm th}$ derivatives of $f$ and $g$ that you computed with partial fractions should be evaluated a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Helping Im stuck with Question of Integral of $\frac{e^{2x}}{6e^{x}+2}$ So for this question I set $u= e^x$ and $du= e^x dx$, if put it back in it will be integral of $\frac{u}{6u+2}$. How would I continue on? Do I have to break it down into parts if so how?
P.S: the answer should come out to $(\frac{e^x}{6})-(\frac{1}... | If you do what Andre recommended. Let $u = 6e^x + 2$. In order to write the top, we can square $u$.
\begin{align}
u^2 = (6e^x + 2)^2 &= 36e^{2x} + 24e^x +4
\end{align}
So
\begin{equation}
e^{2x} = u^2 - 35e^{2x} - 24e^{x} - 4
\end{equation}
Also $\frac{du}{6e^x} = dx$. Now we can solve the integral(remember $6e^x = u -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding $\int _0^a\sqrt{1+\frac{1}{4x}}dx$ to calculate arclength So I'm trying to find the arclength of $x^{0.5}$ and its tougher than I thought. Tried substitutions like $\dfrac{\cot^2x}{4}$ and some other trig subs but they got me nowhere. Any tips?
$$\int _0^a\sqrt{1+\frac{1}{4x}}dx$$
Edit:
This is what I got so fa... | If $u^2=1+\frac1{4x}$ then $x=\frac1{4(u^2-1)}$ and $\mathrm{d}x=-\frac{u}{2(u^2-1)^2}\,\mathrm{d}u$
$$
\begin{align}
\int_0^a\sqrt{1+\frac1{4x}}\,\mathrm{d}x
&=\frac12\int_{\sqrt{1+\frac1{4a}}}^\infty\frac{u^2}{(u^2-1)^2}\,\mathrm{d}u\\
&=\frac18\int_{\sqrt{1+\frac1{4a}}}^\infty\left(\frac1{(u-1)^2}+\frac1{u-1}+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Rationality of the Lemniscate. This question is exercise 2 of Chapter 4 in Kunz' textbook of algebraic curves.
Let $f$ be the lemniscate with equation
$$(X^2 + Y^2 )^2 = α(X^2 − Y^2) \;\; (\alpha \in K^\times )$$
and let $x, y \in K[f]$ be the associated coordinate functions. Prove the
rationality of $f$ by showin... | In the case $K=\mathbb{R}$ (where we can graph the lemniscate) the way to parametrize the lemniscate is by using the circles $x^2+y^2=t(x-y)$. You can check that:
$$
\begin{split} &&x^2+y^2=t(x-y)\\ &\implies& \left(x-\frac{t}{2}\right)^2+\left(y+\frac{t}{2}\right)^2=\frac{t^2}{2} \end{split}
$$
so that the parametriza... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding a sequence satisfying this recurrence relation? I just don't even know where to start with this,
Find a sequence $(x_n)$ satisfying the recurrence relation:
$2x_n$$_+$$_2$ = $3x_n$$_+$$_1$ + $8x_n$ + $3x_n$$_-$$_1$ Where n is a natural number and
$x_0$ = -1,
$x_1$ = 3 and
$x_2$ = 3
Thanks in advance!
| A general way to solve such is using generating functions. Define:
$\begin{align*}
g(z)
&= \sum_{n \ge 0} x_n z^n
\end{align*}$
Write your recurrence shifted (subtraction in indices gets messy), multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
2 x_{n + 3}
&= 3 x_{n + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
computing $\lim\limits_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$ without Taylor series how can i compute this limit without using Taylor series?
$\lim\limits_{(x,y) \to(0,0)} \frac{e^{x^2-y^2} -1}{ x - y}$, while $x\neq y$.
i tried to put $r^2=x^2-y^2$ and transfer to polar but it didn't work for me.
Thank you fo... | By making the substitution $z = e^{x^2-y^2}-1$, we obtain that
\begin{align*}
\lim_{(x, y)\rightarrow (0, 0)}\frac{e^{x^2-y^2}-1}{x^2-y^2} &= \lim_{z\rightarrow 0}\frac{z}{\ln(1+z)}\\
&= \lim_{z\rightarrow 0}\frac{1}{\ln\left((1+z)^{1/z}\right)}\\
&=1.
\end{align*}
Then,
\begin{align*}
\lim_{(x, y)\rightarrow (0, 0)}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Determinant of an $n\times n$ Toeplitz matrix Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:
$$a_{ij}= |i-j|$$
So we have the symmetric matrix
\begin{bmatrix}
0 & 1 & 2 & 3 & 4 & \dots & n-1 \\
1 & 0 & 1 & 2 & 3 & \dots & n-2 \\
2 & 1 & 0 & 1 & 2 & \dots & n-3 \\
\vdots & \vdots & \vdots & \vdots & \vdo... | If we do row reduction, i.e. $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_2-R_1$, and then again do $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_3-R_2$ and finally $R_2+R_1$, then we get
\begin{align}
&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-2 \\
2 & 1 & 0 & 1 & 2 & \cdots & n-3 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
if $a ≡ b\pmod {2n}$ then prove $a^2 ≡ b^2 \pmod {4n}$
Let $n$ be positive number, if $a \equiv b \pmod{2n}$, prove that
$a^2 \equiv b^2 \pmod{4n}$.
By the congruence in hypothesis, we have $a-b = 2nk$ where $k$ is an integer.
Then $a = b+2nk$ and $a^2 = b^2+4n^2k^2+4knb$. From this we get $a^2-b^2 = 4kn(kn+b)$.
No... | Here's an alternative proof. Assume that $2n|(a-b)$. Then $a$ and $b$ are either both even or both odd, so $2|(a+b)$. Since $a^2-b^2 = (a-b)(a+b)$, we have that $4n|(a^2-b^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Determining matrix $A$ and $B$, rectangular matrix Let $A$ be a $3\times 2$ matrix and $B$ be a $2\times 3$ be matrices satisfying $$AB=\begin{pmatrix} 8 & 2 & -2\\ 2 & 5 & 4\\ -2 & 4 & 5\end{pmatrix}$$
Calculate $BA$. How would you go for this problem? Do we start by noticing the matrix is symmetric? Any hints/ideas? ... | Using the suggestion of @WillJagy and the direct approach suggested by @Dac0, we have
$$
A =
\begin{bmatrix}
a & b\\
c & d\\
e & f\\
\end{bmatrix}\qquad
B =
\begin{bmatrix}
a & c & e\\
b & d & f\\
\end{bmatrix}
$$
and from $AB$ we get the following set of equations
\begin{align}
a^2 + b^2 &= 8\\
c^2 + d^2 &= 5\\
e^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.
My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce t... | We have
$$x \equiv 2 \pmod {3}$$
$$x \equiv 3 \pmod {4}$$
$$x \equiv 4 \pmod {5}$$
in the form $x\equiv a_i\pmod{m_i}$ and $M=m_1m_2m_3=3\cdot4\cdot5 = 60$
Then using chinese remainder theorem we have to find $b_i$ such that $b_i\dfrac{M}{m_i} = 1\pmod{m_i}$
Then
$$b_1 \cdot\frac{60}{3}\equiv 1 \pmod {3} \implies b_1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Improper integral $\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ I have to determine the convergence of
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$
Can I say that
$(1-x^{2}\sin(\frac{1}{x^{2}}))\leq1+x^2$
and since $\int_0^3 1+x^2 dx$
is not even improper, the integral
$\int_0^3 (1-x^{2}\sin(\frac{1}{x^{2}})) dx$ is... | $$\int_0^3 \left(1-x^{2}\sin\left(\frac{1}{x^2}\right)\right)\ dx$$
$$=\int_0^3\ dx-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
$$=3-\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx$$
Since
$$\int_0^3 x^{2}\sin\left(\frac{1}{x^2}\right)\ dx \leq \int_0^3 x^2\ dx$$
And
$$\lim\limits_{t\to 0^+}\int_t^3 x^2\ dx=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to Evaluate $\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$? How to find this limit without using L'Hospital rule
$$\lim_{x\to0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
| Use, binomial expansion of $(1-x)^{1/3}$ & Taylor's series expansion of $4^x$ & $3^x$ as follows $$\lim_{x\to 0}\frac{(1-x)^{1/3}-1}{4^x-3^x}$$
$$=\lim_{x\to 0}\frac{\left(1+\frac{\frac{1}{3}}{1!}(-x)+\frac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}(-x)^2+\ldots\right)-1}{\left(1+x\ln 4+\frac{x^2}{2!}(\ln 4)^2+\ldots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Another way to express $\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$? I believe that the sum $$\lim\limits_{m\to\infty}\sum_{n=1}^m\frac{\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)}{n}$$ converges and it is about $1.85193$. Is there another way that this ... | $\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\sin\left(\frac{2\pi n}{2m+2}\right)
=\sin\left(\frac{\pi n}{m+1}\right)
$
so
$\frac1{n}\sin\left(2\pi n\left(1+\frac{1}{2m+2}\right)\right)
=\frac1{n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{m+1}{\pi n}\sin\left(\frac{\pi n}{m+1}\right)
=\frac{\pi}{m+1}\frac{m+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Coordinate of $S(s,t)$ for Which Area of Quadrilateral is Maximum.
Let $P(-2,3)\;\;,Q(-1,1)\;\;,R(s,t)$ and $S(2,7)$ be $4$ points in order on the parabola
$y=ax^2+bx+c$. Then the coordinates of $R(s,t)$ such that the area of Quadrilateral
$PQRS$ is maximum.
$\bf{My\; Try::}$ Here $P(-2,3)\;,Q(-1,1)\;\;,S(2,7)$ be th... | All that calculation was really unnecessary. Here's a fast way:
The area of the quadrilateral PQRS is equal to the sum of the areas of the triangles $\triangle RQS$ and $\triangle PQS$. And since the area of the triangle $\triangle PQS$ is fixed, we need only to maximise the area of the $\triangle RQS$
Additionally sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral with substitution For my engineering math course I got a couple of exercises about indefinite integrals. I ran trought all of them but stumbled upon the following problem.
$$\int \frac{1-x}{\sqrt{1+x-2x^2}}\,dx $$
We can write $1+x-2x^2$ as $(1-x)(2x+1)$
So I got:
$$
\int \frac{1-x}{\sqrt{1+x-2x^2... | Let $u^2=\frac{1-x}{1+2x}$. Then $x=\frac{1-u^2}{1+2u^2}$ and $\mathrm{d}x=-\frac{6u}{(1+2u^2)^2}\,\mathrm{d}u$.
Let $\sqrt2u=\tan(\theta)$, then $\sqrt2\,\mathrm{d}u=\sec^2(\theta)\,\mathrm{d}\theta$.
$$
\begin{align}
\int\frac{1-x}{\sqrt{1+x-2x^2}}\,\mathrm{d}x
&=\int\frac{1-x}{\sqrt{(1-x)(1+2x)}}\,\mathrm{d}x\\
&=\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
System of linear and quadratic congruences Solve the system:
$x^2+x+1\equiv 0\pmod {7}$
$2x-4\equiv 0\pmod {6}$
Completing the square for first equation gives:
$$(2x+1)^2\equiv 4\pmod {7}$$
$$y^2\equiv 4\pmod{7}\Rightarrow y_1=2,y_2=5$$
$$2x+1\equiv 2\pmod{7}\Rightarrow (2,7)=1|7\Rightarrow x=4$$
$$2x+1\equiv 5\pmod{7}... | You solved the quadratic congruence correctly, but the solutions of $2x-4\equiv 0\pmod{6}$ are $x\equiv \{2,5\}\pmod{6}$.
Divide everything by $\gcd(2,4,6)=2$: $$\iff x-2\equiv 0\pmod{3}\iff x\equiv 2\pmod{3}$$
$$\iff x\equiv \{2,5\}\pmod{6}$$
The solutions of $x^2+x+1\equiv 0\pmod{7}$ are $x\equiv \{2,4\}\pmod{7}$ (as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further... | Let $x=2 u^2$:
$$2 \sqrt{2} \int du \, \frac{u}{\sqrt{1+u^2}} \arctan{u} = 2\sqrt{2} \sqrt{1+u^2} \arctan{u} - 2 \sqrt{2} \int \frac{du}{\sqrt{1+u^2}}$$
The latter integral is easily done using the sub $u=\sinh{v}$, so we have as the integral
$$2 \sqrt{x+2} \arctan{\sqrt{\frac{x}{2}}} - 2 \sqrt{2} \log{\left (\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
$c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Find the value of $a+b+c$ $a=\sqrt{57+40\sqrt2}-\sqrt{57-40\sqrt2}$ and $b=\sqrt{25^{\frac{1}{\log_85}}+49^{\frac{1}{\log_67}}}$ and $c$ is the value of $x^3+3x-14$ where $x=\sqrt[3]{7+5\sqrt2}-\frac{1}{\sqrt[3]{7+5\sqrt2}}$.Fin... | $$a^2=57+40\sqrt2+57-40\sqrt2-2\sqrt{57^2-40^2\cdot2}=2\cdot57-2\sqrt49=100$$
$$a=10$$
$$b=\sqrt{5^{2\log_5 8}+7^{2\log_7 6}}=\sqrt{8^2+6^2}=\sqrt100=10$$
$$x=a-\frac1a,\quad a=(7+5\sqrt2)^{\frac13}$$
$$x^3+3x-14=\left(a^3-3a+\frac3a-\frac1{a^3}\right)+3\left(a-\frac1a\right)-14\\
=a^3-\frac1{a^3}-14=7+5\sqrt2-\frac1{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | Actually there are solid arguments why $0/0$ can be defined as $0$.
Here is a graphic of the function $f(x,y)=x/y$:
The function is odd against both $x$ and $y$ variables.
Along the axis $x=0$ it is constant zero.
Along the axis $y=0$ it is unsigned infinity, but its Cauchy principal value is constant zero.
Along the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 11
} |
Find the area of the region that lies inside the first curve and outside the second curve. $r = 10 \cos\theta,\ r = 5$ I am not sure of my answer. In the figure, $r=10 \cos\theta$ is a circle that doesn't look like a circle.
The area of $r=5$ is $\pi r^2 = 25 \pi$.
You remove the area from $-\pi/3$ to $\pi/3$ of $10 \... | Let $S$ be the area of the red part in the following figure :
$\qquad\qquad\qquad$
You want to find $2S$.
$S$ can be obtained by
$$S=\frac 12\times 5^2\pi-[\text{sector $OAB$}]-[\text{the gray part}]\tag1$$
Since $\triangle{OAB}$ is an equilateral triangle, we have $\angle{AOB}=\frac{\pi}{3}$. So, $$[\text{sector $OAB... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
2 problems related to the number 2015
*
*Let $p=\underbrace{11\cdots1}_\text{2015}\underbrace{22\cdots2}_\text{2015}$. Find $n$, where $n(n+1) = p$
*Prove that $\frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{2015^2} < \frac{2014}{2015}$
For 1, I tried dividing in various ways until I got a simpler expression, bu... | For 1 you can use the quadratic equation:
$n^2+n-p=0 \rightarrow n=\frac{-1 \pm \sqrt{4 \cdots 4 8 \cdots 89}}{2} $.
Now show $\sqrt{4 \cdots 4 8 \cdots 89}=\underbrace{66\cdots67}_\text{2015}$ and take the positive root to get $n=\underbrace{33\cdots3}_\text{2015}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How can i solve something like that?
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
How should I start? Should I try rewrite it in partial fractions?
| HINT:
As $x^2+7x-3=\dfrac{(2x+7)^2-61}4$ start with $2x+7=\sqrt{61}\sec y$
Alternatively,
as the highest power of $x$ in $(x^2+7x-3)^2$ is $4,$
differentiate $\dfrac{ax^3+bx^2+cx+d}{(x^2+7x-3)^2}$ wrt $x$
and compare with $\dfrac{x+1}{(x^2+7x-3)^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving of this trigonometric identity $$\frac{\cot \beta}{\csc \beta - 1} + \frac{\cot \beta}{\csc \beta + 1} = 2 \sec \beta$$
What I've done:
$$\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} +1} +
\frac{\frac{\cos \beta}{\sin \beta}} {\frac{1}{\sin \beta} -1}=\\
=\frac{\frac{\cos \beta}{\sin \beta}} {\fra... | You first mistake was listening to your teacher. This has turned you into a $\sin$ / $\cos$ robot, causing you go against your better intuition and find a common denominator. So let's do that now
$$\begin{array}{lll}
\frac{\cot\beta}{\csc\beta-1}+\frac{\cot\beta}{\csc\beta+1}&=&\frac{\cot\beta}{\csc\beta-1}\cdot\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b... | with $\cos(\beta/2)=\sqrt{\frac{s(s-b)}{ac}}$ and $\cos(\gamma/2)=\sqrt{\frac{s(s-b)}{ab}}$ and $s=(a+b+c)/2$ we get after squaring
$$-1/2\,bc \left( b-c \right) \left( {a}^{3}+{a}^{2}b+{a}^{2}c+3\,abc+{
b}^{2}c+b{c}^{2} \right)
=0$$ thus $b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $y^3-3y-\sqrt{2}=0$ using trigonometry This is a part of a larger question.
I had to show that for $4x^3-3x-\cos 3\alpha=0$ one of the solutions is $\cos \alpha$ and then find the other two solutions. Here they are:
$$4x^3-3x-\cos 3\alpha = (x-\cos \alpha)(2x+\cos \alpha + \sqrt{3} \sin \alpha)(2x+\cos \alpha - \... | Thee trigonometric method relies on the trigonometric identity:
$$\cos 3\theta=4\cos^3\theta -3\cos\theta.$$
Set $\;y=A\cos\theta,\enspace A>0$. The equation becomes
$$A^3\cos^3\theta -3A\cos\theta=\sqrt2$$
We choose $A>0$ such that the left-hand side of the equation is proportional to the development of $\cos3\theta $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find all incongruent solutions of $x^8\equiv3\pmod{13}$. Find all incongruent solutions of $x^8\equiv3\pmod{13}$.
I know that $2$ is a primitive root of $13$ and that $2^4\equiv3\pmod{13}$, so we want to solve $x^8\equiv2^4\pmod{13}$.
Now, $\gcd(8,\phi(13))=\gcd(8,12)=4$ divides the exponent of $2$, which is $4$, so $x... | We can just list all the values $x^8$ takes for $x\in\Bbb{Z}/13\Bbb{Z}$:
$$\begin{array}{c|cccccccccccccc}
x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10& 11& 12\\
\hline
x^8 & 0 & 1 & 9 & 9 & 3 & 1 & 3 & 3 & 1 & 3 & 9 & 9 & 1 \\
\end{array}$$
If you use the fact that $(-x)^8=x^8$ and $(xy)^8=x^8y^8$, the only 'big'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$? Find all solutions of $$\sin{14x} - \sin{12x} + 8 \sin x - \cos{13x} = 4$$ on an interval $(0^\circ, 360^\circ)$
| Using the trigonometric identity $$\sin A-\sin B\equiv2\cos\frac{A+B}2\sin\frac{A-B}2,$$with $A=14x$ and $B=12x$, we can write the equation as $$(2\sin x-1)(\cos13x+4)=0.$$The general solution is therefore $$x=n\pi+(-1)^n\frac\pi6\quad(n\in\Bbb Z).$$For the range you specify, $x$ is $30^\circ$ or $150^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\binom{n}{k}}{n^k} < \frac{\binom{n+1}{k}}{(n+1)^k}$ I have this math question that I'm kind of stuck on.
Prove that for all integers $1 < k \le n$, $$\frac{\binom{n}{k}}{n^k} <
\frac{\binom{n+1}{k}}{(n+1)^k}$$
I have to use mathematical induction on $k$ to prove this. So, I first check the base ca... | First,
I would just take a look
at their ratio.
$\begin{array}\\
\frac{\frac{\binom{n+1}{k}}{(n+1)^k}}
{\frac{\binom{n}{k}}{n^k}}
&=\frac{\frac{\frac{(n+1)!}{k!(n+1-k)!}}{(n+1)^k}}
{\frac{\frac{n!}{k!(n-k)!}}{n^k}}\\
&=\frac{\frac{(n+1)}{(n+1-k)}}{(1+1/n)^k}\\
&=\frac{\frac{(1+1/n)}{1+(1-k)/n}}{(1+1/n)^k}\\
&=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number Theory: If $d\mid(4^n+1)$, show that $d$ is a sum of two squares I have a proof for the following problem, but I'm not sure if it's correct:
If $d\mid(4^n+1)$, show that $d$ is a sum of two squares.
Proof
$d\mid(4^n+1)\implies dm=4^n+1$, some $m\in\mathbb{Z}$.
Suppose that one of $d,m\equiv3\pmod4$. Then, $dm\eq... | If $p$ is a prime divisor of $4^n+1$, then $p$ is odd and $(2^n)^2 \equiv -1\pmod{p}$. Therefore, $p \equiv 1\pmod{4}$. So if $d \mid (4^n+1)$ and $d>1$, then $d$ is a product of (not necessarily distinct) primes of the form $4k+1$. Each of these primes is a sum of two squares, and so $d$ is also a sum of two squares. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Supremum and Infimum of set I just got the set and Ι tried to find the supremum and infimum and prove it. $$(x-2)\sqrt{\frac{x+1}{x-1}} \quad \text{ for } \quad 2< x\leq 54$$
I succeed to get to this set $\frac{x-2}{x-1}\sqrt{x^{2}-1}$ but I'm stuck.
What can I do now ?
Thanks.
| On one hand, since $x>2$ we have
$$
(x-2)\sqrt{\frac{x+1}{x-1}}>0.
$$
On the other hand
$$
\lim_{x\rightarrow 2^+}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Hence
$$
\inf_{2<x\leq 54}(x-2)\sqrt{\frac{x+1}{x-1}}=0.
$$
Since $2<x\leq 54$
$$
(x-2)\sqrt{\frac{x+1}{x-1}}=\frac{x-2}{x-1}\sqrt{x^2-1}=\left(1-\frac{1}{x-1}\right)\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculating square roots using the recurrence $x_{n+1} = \frac12 \left(x_n + \frac2{x_n}\right)$
Let $x_1 = 2$, and define $$x_{n+1} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right).$$ Show that $x_n^2$ is always greater than or equal to $2$, and then use this to prove that $x_n − x_{n+1} ≥ 0$. Conclude that $\lim x_n =... | From
$x_{n+1}
= \frac{1}{2} \left(x_n + \frac{2}{x_n}\right)
$
we get, squaring,
$x_{n+1}^2
= \frac{1}{4} \left(x_n^2 +4+ \frac{4}{x_n^2}\right)
$
so that
$\begin{array}\\
x_{n+1}^2 -2
&= \frac{1}{4} \left(x_n^2 -4+ \frac{4}{x_n^2}\right)\\
&= \frac{1}{4} \left(x_n - \frac{2}{x_n}\right)^2\\
&= \frac{1}{4} \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
| Let's grind it out. We want all ordered quadruples, but to get them it is enough to find all ordered triples $(a,b,c)$ where $a\le b\le c$. Then any triple of this kind we find can be permuted arbitrarily.
There cannot be triples $(a,b,c)$ with $a\le b\le c$ and $a\gt 2$, since for these we get that $3$ divides $a!+b!+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question on Divergence Theorem Evaluate $\iint F.ds$
where $F= \frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k$, and $S$ is the surface:
$S= \{(x,y,z) | x^2 + y^2 +z^2 =9\}$
Please explain in detail how to get the answer.
| Simple application of the Divergence theorem:
$$\iint F\cdot ds = \iiint \nabla \cdot F dV$$
$$=\iiint \nabla \cdot (\frac{59}{3} x^3 \hat i + \frac{59}{3} y^3 \hat j + \frac{59}{3} z^3 \hat k) dV$$
$$=\iiint (59 x^2 \hat i + 59 y^2 \hat j + 59 z^2 \hat k) dV$$
$$=\iiint 59(x^2+y^2+z^2) dV$$
$$=59 \cdot \iiint r^2 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find all pairs $(x,y)$ of integers such that $y^2 = x(x+1)(x+2)$? Here $y^2$ is divisible by $12.$ And satisfying all those conditions I think $y=0$ is the only solution. But I can't show it mathematically.
| Setting $z=x+1$, we need $y^2 = z^3-z = z(z^2-1)$. Since $\gcd(z,z^2-1)=1$, we need $z=m^2$ and $z^2 - 1 = n^2$. This forces $z^2-n^2=1 \implies (z+n)(z-n) = 1 \implies z = \pm 1 \text{ and }n=0$. Apart from this, clearly $z=0$ is also a solution. Hence, the only solutions are $$(x,y) = (0,0);(-1,0);(-2,0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The functional equation $ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 $ I came across the functional equation
$$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \text . $$
So far I tried plugging $ x = f ( y ) $ and got $ f ( x ) = \frac { f ( 0 ) - x ^ 2 + 1 ... | You can show that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \big( x - f ( y ) \big) = f \big( f ( y ) \big) + x f ( y ) + f ( x ) - 1 \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = 1 - \frac { x ^ 2 } 2 $. It's straightforward to check that this function indeed satisfies \eqre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove $\sin (x)+ \sin(y) = 2 \sin \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)$ using addition theorems? I am stuck at one task, it's to proof the following equation using addition theorems.
From a draft I understood that the equation is correct or true. But that's it.
What's a good way to explain i... | The relevant addition formulae tell you that:
$$
\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\\
\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)
$$
Applying these to $2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ yields:
$$
\begin{align*}
2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right) & = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$ Find the value of $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})$
Let $\sin(\frac{1}{4}\arcsin\frac{\sqrt{63}}{8})=x$
$\arcsin\frac{\sqrt{63}}{8}=4\arcsin x$
$\arcsin\frac{\sqrt{63}}{8}=\arcsin(4x\sqrt{1-x^2})(2x^2-1)$
$\frac{\sqrt{63}}{8}=(4x\sqrt{1-x^2})(... | Let
$$x = \arcsin{\frac{\sqrt{63}}{8}} = \arccos{\frac18} $$
Use succession of half-angle formulae:
$$\sin{\frac{x}{4}} = \sqrt{\frac{1-\cos{(x/2)}}{2}} = \sqrt{\frac{1-\sqrt{\frac{1+\cos{(x)}}{2}} }{2}} $$
In this case, $\cos{x} = 1/8$ so that
$$\sin{\left (\frac14 \arcsin{\frac{\sqrt{63}}{8}} \right )} = \sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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I would like to calculate $\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$ I want to calculate the following limit: $$\lim_{x \to \frac{\pi}{6}} \frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}$$
or prove that it does not exist. Now I know the result is $-3$, but I am having troub... | You may just observe that, $\color{#3366cc}{\sin (\pi/6) =1/2}$ and that, as $x \to \pi/6$,
$$
\require{cancel}
\frac{2 \sin^2{x}+\sin{x}-1}{2 \sin^2{x}-3 \sin{x}+1}=\frac{\cancel{\color{#3366cc}{(2 \sin{x}-1)}}(\sin{x}+1)}{\cancel{\color{#3366cc}{(2 \sin{x}-1)}}(\sin{x}-1)}=\frac{\sin{x}+1}{\sin{x}-1} \color{#cc0066}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\... | Use equivalents:
*
*$\ln(1+4^x)\sim_\infty\ln 4^x=x\ln 4$,
*similarly $\ln(1+3^x)\sim_\infty\ln 3^x=x\ln 3$,
hence $$\frac{\ln(1+4^x)}{\ln(1+3^x)}\sim_\infty\frac{x\ln 4}{x\ln 3}=\frac{\ln 4}{\ln 3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
If two integer triples have the same sum of 6th powers, then their sums of squares agree $\bmod 9$ Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$
prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$
I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really ... | Write
$$
a^6+b^6+c^6-x^6-y^6-z^6 = (a^2-x^2)(a^4+a^2x^2+x^4) + (b^2-y^2)(b^4+b^2y^2+y^4) + (c^2-z^2)(c^4+c^2z^2+z^4) ;
$$
by Thomas's comment, we can also pair off in such a way that $a,x$ are either both $\equiv 0\bmod 3$ or both $\not\equiv 0\bmod 3$, and similarly for $b,y$ and $c,z$.
The non-zero squares $\bmod 9$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
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Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series... | Proceed inductively. Verify that for $k=1$, $1 = 1$.
Now suppose the result holds for $n - 1$. Then
$$ 1^2 + \dots + (-1)^{n-1} n^2 = 1 + \dots + (-1)^{n-2} (n-1)^2 + (-1)^{n-1} n^2 = (-1)^n \frac{n(n-1)}{2} + (-1)^{n-1} n^2 = (-1)^n \left(\frac{n(n-1)-2n^2}{2} \right) = (-1)^n \left(\frac{-n^2-n}{2} \right) = (-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Divisibility set of problems It may seem as a begginer question, but i was wondering if there exist a certain method to solve problems like the following :
$$1) \text{ if } n ∈ \Bbb N \\ n(n^2+5)\,⋮\,6$$
$$2) \text{ if } n ∈ \Bbb N \\n^4+6n^3+11n^2+6n\,⋮\,24$$
$$3)\text{ if } n=2k \text{ then } \frac{n}{12}+\frac{n^2}... | Another possible method is to write these numbers in terms of binomial coefficients which are always integers :
1) $$\frac{n(n^2+5)}{6}=\frac{n^3+5n}{6}=n+\frac{n^3-n}{6}=n+\frac{n(n+1)(n-1)}{3!}=n+\binom{n+1}{3}$$ which is an integer so $6 \mid n(n^2+5n)$
2) $$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$ so $$\frac{n^4+6n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Remainder of the numerator of a harmonic sum modulo 13 Let $a$ be the integer determined by
$$\frac{1}{1}+\frac{1}{2}+...+\frac{1}{23}=\frac{a}{23!}.$$
Determine the remainder of $a$ when divided by 13.
Can anyone help me with this, or just give me any hint?
| Note that
$$a = \frac{23!}{1} + \frac{23!}{2} + \dots + \frac{23!}{23}$$
Since each term is an integer, $a$ is an integer.
To find $a \pmod{13}$ (if that's what you're asking), note that due to the factorial term, each of these fraction has at least a factor of $13$, except $\frac{23!}{13}$.
As such, you'll have to d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of so... | More easier and obvious answer:
$$\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\geq\frac{3}{2}$$
$$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}\geq3$$
$$\frac{2a+b+c}{b+c}+\frac{b+2c+a}{a+b}+\frac{c+2b+a}{a+c}\geq6$$
$$\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right... | we have $x^4-x^2+1=(x^2-\frac 12)^2+\frac {3}{4}≥0$$$\frac{1}{2}-f(x)=1/2*{\frac { \left( {x}^{2}-x-1 \right) ^{2}}{{x}^{4}-{x}^{2}+1}}\geq 0$$
and $$f(x)+\frac{1}{2}=1/2*{\frac { \left( {x}^{2}+x-1 \right) ^{2}}{{x}^{4}-{x}^{2}+1}}\geq 0$$
thus $$|f(x)| \le \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the number of solutions of the equation $G=G'$ in $[0,2\pi]$ Let $G$ be the sum of infinite geometric series whose first term is $\sin\theta$ and common ratio is $\cos\theta$,while $G'$ be the sum of a different infinite geometric series whose first term is $\cos\theta$ and common ratio is $\sin\theta$.Find the nu... | So we have $G=\frac{\sin\theta}{1-\cos\theta}$ and $G'=\frac{\cos\theta}{1-\sin\theta}$,
an important condition for the sum of the geometric series is $-1<r<1$ where r is the common ratio, thus
$$\frac{\sin\theta}{1-\cos\theta}=\frac{\cos\theta}{1-\sin\theta}$$
$${\sin\theta}*(1-\sin\theta)={\cos\theta}*(1-\cos\theta)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$
I have no idea how to calculate it. Please help.
| Notice,
$$\int \frac{x-1}{\sqrt{x^2-2x}}\ dx=\frac 12\int \frac{2(x-1)}{\sqrt{x^2-2x}}\ dx $$
$$=\frac 12\int \frac{d(x^2-2x)}{\sqrt{x^2-2x}}$$
$$=\frac 12\int (x^2-2x)^{-1/2}d(x^2-2x)$$
$$=\frac 12\frac{(x^2-2x)^{-\frac 12+1}}{-\frac 12+1}+C$$
$$=\frac 12\frac{(x^2-2x)^{1/2}}{1/2}+C$$
$$=\color{red}{\sqrt{x^2-2x}+C}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Integrate expression with $x^6$ So I'm trying to integrate this expression, but I'm not figuring out what's the best substitution to do...
$ \int \frac {1}{x^6+1} dx $
I tried to take $x^6 +1 $ and write $ (x^2 + 1) (x^4 -x^2 + 1) $ and then do partial functions, so I reach to the sum of two expressions.
One of them i... | HINT: the partialfrac decomposition is given by $$1/3\, \left( {x}^{2}+1 \right) ^{-1}+1/3\,{\frac {-{x}^{2}+2}{{x}^{4}-
{x}^{2}+1}}
$$
and $$x^4-x^2+1=(x^2-1/2)^2+\frac{3}{4}$$
$$x^4-x^2+1=(x^2+1)^2-3x^2$$ is better (see above) since we get a product finally we obtain
$$1/3\,\arctan \left( x \right) -1/12\,\sqrt {3}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show $\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}$. Let $p$ be prime and $d \ge 2$. I want to show that
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}.
$$
I have a proof, but I think it is complicated, and the statement appears in a book as if it is very easy to see. So is there... | There are even easier proofs than the answers others have supplied. If you look at the numerator, one of the terms is guaranteed to be divisible by $p^2-1$, whichever has an even exponent. This is because $(x^2-1)|(x^{2n}-1)$. The division clearly results in a polynomial with constant term $1$, as does the division of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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.