Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| Because $x^2+3x+1=0$, we have $x^2=-3x-1$ and also $x^2+2x+1=-x$, for $x=a,b$. Hence $$\left(\frac{a}{b+1}\right)^2=\frac{a^2}{(b+1)^2}=\frac{-3a-1}{-b}=\frac{3a+1}{b}$$
By symmetry, the desired expression is $$\frac{3a+1}{b}+\frac{3b+1}{a}=\frac{3a^2+a}{ab}+\frac{3b^2+b}{ab}=\frac{3a^2+a+3b^2+b}{ab}=\frac{3(-3a-1)+a+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqr... | It's probably better to cube $S$:
$$S^3 = (5 + 2\sqrt{13}) + 3\sqrt[3] {5+2 \sqrt {13}}\sqrt[3] {5- 2 \sqrt {13}}\bigg(\sqrt[3] {5+2 \sqrt {13}} + \sqrt[3] {5-2 \sqrt {13}}\bigg) + (5 - 2\sqrt{13})$$
$$= 10 + 3\sqrt[3] {5^2-(2 \sqrt {13})^2}S$$
$$= 10 -9S$$
So $S$ satisfies the equation $S^3 + 9S -10 = 0$. The polynom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac{1}{2\pi}\frac{xdy-ydx}{x^2+y^2}$ is closed I would like to prove that $\alpha = \frac{1}{2\pi} \frac{xdy-ydx}{x^2+y^2}$ is a closed differential form on $\mathbb{R}^2-\{0\}$ . However when I apply the external derivative to this expression (and ignore the $\frac{1}{2\pi}\cdot\frac{1}{x^2+y^2}$ factor... | Well, let us write $\alpha=f\cdot \omega $ with $f(x,y)=\frac{1}{x^2+y^2}$ and $\omega=xdy-ydx$. Then
\begin{align}
d\alpha=df\wedge \omega+fd\omega&=-\frac{1}{(x^2+y^2)^2}\left(2xdx+2ydy\right)\wedge\omega+
\frac{1}{x^2+y^2}2dx\wedge dy=\\
&=-\frac{1}{(x^2+y^2)^2}\left(2x^2 dx\wedge dy-2y^2dy\wedge dx\right)+
\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Telescoping series of form $\sum (an+1) ..(an+k)$ If we are given a series say
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k)$$ we can find a telescoping series by noting that
$(n+1)(n+2)..(n+k)(n+k+1) - n(n+1)..(n+k) = (n+k+1-n) (n+1)...(n+k)$
and hence able to write
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k) = \frac{1}{k+1}\sum_{n=1}^... | The original identity comes from finding an antidifference for the function $n^{\underline{k}}$. Unfortunately, no such antidifference exists for $(2n)^{\underline{k}}$, due to lack of a "chain rule" for antidifferences. However here's an approach that you might like:
$$a_k=\sum_{n=1}^N(2n+1)(2n+2)\cdots(2n+k)$$
$$b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ I am trying to prove the following inequality
For all positive numbers $a$, $b$ and $c$ we have
$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$
I can probably solve this by reduc... | We need to prove that
$$\sum_{cyc}\left(\frac{a^3}{b^2-bc+c^2}-a\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^2+bc-b^2-c^2)}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}\frac{a((a-b)(a+2b-c)-(c-a)(a+2c-b))}{b^2-bc+c^2}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a(a+2b-c}{b^2-bc+c^2}-\frac{b(b+2a-c)}{a^2-ac+c^2}\right)\geq0$$ or
$$\sum_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sq... | Let $p=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$
Define :$$x_1=\sqrt{1+2\sqrt{2}}-1$$
$$x_2=\sqrt{1+2\sqrt{2+\sqrt{3}}}-\sqrt{1+2\sqrt{2}}$$
$$.$$
$$.$$
$$x_{n-1}=\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n}}}}-\sqrt{1+2\sqrt{2+\sqrt{3+...+\sqrt{n-1}}}}$$
From the summation , we can see that $$p-1=x_1+x_2+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 5,
"answer_id": 3
} |
Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\rig... | \begin{equation}
\begin{split}
\ & x^4+y^4+(x+y)^4\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2+2xy)^2\\
\ =& (x^2+y^2)^2-2x^2y^2+(x^2+y^2)^2+4xy(x^2+y^2)+4x^2y^2\\
\ =& 2((x^2+y^2)^2+x^2y^2+2xy(x^2+y^2))\\
\ =& 2(x^2+y^2+xy)^2
\end{split}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then ... | $\bf{My\; Solution::}$ Given $a\;,b$ are the roots of the equation $x^2-2cx-5d=0$ So
$\displaystyle a+b=2c............................(1)\;\;\;\;\;\; ab = -5d.....................(2)$
similarly $c\;,d$ are the roots of the equation $x^2-2ax-5b=0$ So
$\displaystyle c+d=2a............................(3)\;\;\;\;\;\; cd=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculating $\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$
Find $$\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor$$ where $\lfloor\cdot\rfloor$ is greatest integer function
I am a high school teacher. One of my students came up to ask this limit.
For $\lfloor\frac{\sin x}{x}\rfloo... | Let $[.]$ denote GIF.
If $x \to 0$, then $\sin x=x-x^3/6+O(x^5)$ and $\tan x =x+x^3/3+O(x^4)$
then $$f(x)=\left(\frac{x^2}{\sin x \tan x}\right)=\left(\frac{x^2}{x^2(1-x^2/6+...)(1+x^2/3+...)}\right)=\left(\frac{1}{1+x^2/6-x^4/18}\right)$$
Using $(1+z)^k =1+kz+O(z^2)$, we can write
$$1-x^2/6 \le f(x) \le 1-x^2/6+x^4/18... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 5
} |
Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$
I know the conventional way of solving it would be (That's what i showed her):
$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\l... | The key here is that you can only break up limits over addition/subtraction/etc when you know that those limits exist.
So, you cannot write
$$
\lim_{x\rightarrow2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)=\lim_{x\rightarrow2}\frac{1}{x-2}-\lim_{x\rightarrow2}\frac{4}{x^2-4},
$$
because these limits are both non-exist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Area enclosed by curves $\; y=x^2;\;\;y^2=2x-x^2$ Find the area enclosed by the curves: $$ y=x^2;\quad y^2=2x-x^2$$
I know how to set-up the problem. I am having difficulty figuring out the integral however. I first found the points of intersection by setting the two equations equal to one another. I got the lower limi... | You're set up is spot on.
Indeed, we're finding the area bounded below the curve $y = \sqrt{2x - x^2}\,$ and above the curve $y = x^2,\;$ between $x = 0$ and $x = 1$:
So indeed, we need to integrate:
$$\begin{align} I & = \int_0^1 \left(\sqrt{2x - x^2} - x^2\right)\,dx \\ \\
& = \int_0^1 \sqrt{1 - 1 + 2x - x^2} \,dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
tough algebric problem? I wanted to know how can i prove that if
$xy+yz+zx=1$, then
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}
= \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}}$$
I did let $x=\tan A$, $y=\tan B$, $z=\tan C$
given $xy+yz+zx =1$ we have $\tan A \tan B+ \tan B \tan C+\tan C \tan A=1$
$\tan C(\tan... | The accepted answer by `lab bhattacharjee' is perfectly fine. For what it's worth, there's a nice geometric view that might enrich this nearly-4-year-old post a tiny bit.
The identity to be proven is, given $\alpha + \beta + \gamma = \frac{\pi}2$, $$ \frac12 \big( \sin 2\alpha + \sin 2\beta + \sin 2\gamma \big) = 2 \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The indu... | Combinatorial argument:
Take $n$ pairs of people, that is $2n$ people total, and then you can pick:
*
*one person from each pair in $2^n$ ways,
*any $n$ individuals in $\binom{2n}{n}$ ways,
*any subset of those people in $2^{2n}$ ways.
I hope this helps ;-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
At least one member of a pythagorean triple is even
I am required to prove that if $a$, $b$, and $c$ are integers such that $a^2 + b^2 = c^2$, then at least one of $a$ and $b$ is even. A hint has been provided to use contradiction.
I reasoned as follows, but drew a blank in no time:
Let us instead assume that both $... | Let us try to write this using congruences, which is a very useful and compact notation.
We know that a square can only have remainder $0$ or $1$ modulo $4$. I.e., for any integer $x$ one of these two congruences must be true: $x^2\equiv 0 \pmod 4$ or $x^2\equiv1 \pmod4$. The first case happens if $x$ is even, the seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}... | $$\cos \alpha + \sin \alpha = \sqrt{2} (\frac{\cos \alpha}{\sqrt{2}} + \frac{\sin \alpha}{\sqrt{2}}) = \sqrt{2}(\sin \frac{\pi}{4} \cos\alpha + \cos \frac{\pi}{4} \sin \alpha)
= \sqrt{2}\sin(\frac{\pi}{4} + \alpha) = .2 $$
Taking the inverse sin of each side yields
$$ \alpha = 2.2143 + 2 \pi n_1 \mid n_1 \in \mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
How to factor $8xy^3+8x^2-8x^3y-8y^2$ How can I factor $8xy^3+8x^2-8x^3y-8y^2$ or the different form $2x(4y^3+4x)-2y(4x^3+4y)$
Is there any general methods that work?
A possible solution should be $8(x^2-y^2)(1-xy)$ But please do not start from here as in the general case I will not know the answer...
Thanks!
Alexander... | First take out $8$ as factor
$8xy^3+8x^2-8x^3y-8y^2=8(xy^3-y^2+x^2-x^3y)$
Now,
$xy^3-y^2+x^2-x^3y=y^2(xy-1)-x^2(xy-1)=(xy-1)(y^2-x^2)=(xy-1)(y+x)(y-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
| We have $$\frac{a^2+b^2}{2ab}=\frac31$$
Applying componendo and dividendo, $$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{3+1}{3-1}$$
$$\implies \left(\frac{a+b}{a-b}\right)^2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?
If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?
I tried to add $-16a^2$ to both sides and doing some algebra but it didnt help much.
| Here's how to find the factorization, it is very similar to completing the square, we have $$\begin{align*}a^4 + 64 &= a^4 + 16a^2 + 64 - 16a^2\\ &= (a^2 + 8)^2 - 16a^2 \\ &= (a^2 - 4a + 8)(a^2 + 4a + 8)\end{align*}$$
See also http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness.
Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$
So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 1
} |
Providing a closed formula for a linear recursive sequence I am studying for an exam in linear algebra and I have some trouble solving the following:
Let $(a_n)$ be a linear recursive sequence in $\mathbb{Z}_5$ with
\begin{align}
a_0 = 2, a_1 = 1, a_2 = 0 \text{ and } a_{n+3} = 2a_{n+2} + a_{n+1} + 3a_n \text{ for } ... | $$\begin{pmatrix}a_{n+3}\\a_{n+2}\\a_{n+1}\end{pmatrix}=\begin{pmatrix}2&1&3\\1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}a_{n+2}\\a_{n+1}\\a_n\end{pmatrix}=\ldots=\begin{pmatrix}2&1&3\\1&0&0\\0&1&0\end{pmatrix}^n\begin{pmatrix}2\\1\\0\end{pmatrix}$$
Now:
$$\det(tI-A)=\begin{vmatrix}t-2&-1&-3\\
-1&t&0\\
0&-1&t\end{vmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ Problem :
Find the sum of :
$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$
My approach :
Here the $n$'th term... | This is not an independent answer but a response to Lord Soth's speculation of a geometric proof.
Consider two right-handed triangles $ABC$ and $ABD$ with base $1$ and heights $\sqrt{n}$ and $\sqrt{n-1}$. Let $\theta$ be the angle $\measuredangle CBD$. The area of the triangle $BCD$ can be computed in two ways:
*
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | In the very first line, you assumed that $a=|a|=\sqrt{a^2}.$ But this is not true in general. In particular, it is false for negative $a.$ And in fact, $$4-\frac92<0,$$ so in this case we do not have that $$4-\frac92=\sqrt{\left(4-\frac92\right)^2}=\left|4-\frac92\right|$$ as assumed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 7
} |
Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.
$$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$
I just want to know if my proof is okay or there is a flaw, and if there is one I request to give som... | $x>-1\iff \underline{x+1}=\sqrt{(x+1)^2}=\sqrt{1+2x+x^2}=\sqrt{1+x\cdot(\underline{\underline{x+2}})}$
$\begin{align}x>-2\iff \underline{\underline{x+2}}=\sqrt{(x+2)^2}=\sqrt{[(x+1)+1]^2}&=\sqrt{1+2(x+1)+(x+1)^2}=\\&=\sqrt{1+(x+1)(\underline{\underline{\underline{x+3}}})}\end{align}$
$\to x+1=\sqrt{1+x\sqrt{1+(x+1)(\un... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ How to show that $\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2$ for coprime $a$ and $b$?
I know the fact that $\gcd(a,b)=1$ implies $\gcd(a,b^2)=1$ and $\gcd(a^2,b)=1$, but how do I apply this to that?
| Hint: Suppose gcd(a,b)=1 and let $d=gcd(a+b,a^2+b^2) \implies d|(a+b) $ and $ d|(a^2+b^2)$. Let $dr=a+b$ and $ds=a^2+b^2$ where $r,s \in\mathbb{Z}$. We see that by squaring $dr=a+b$ we get $d^2r^2=a^2+2ab+b^2$. Then $d^2r^2-d=a^2+2ab+b^2-a^2-b^2=2ab$. Thus $d(dr^2-1)=2ab\implies d|2ab$ From this we break the proof into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Geometry problem on circles from a competition Triangle $\triangle ABC$ is an equilateral triangle whose side is $16$. A circle meets the sides of the triangle at $6$ points:
*
*it intersects $AC$ at $G$ and $F$ and $|AG|=2$, $|GF|=13$, $|FC|=1$.
*it intersects $AB$ at $H$ and $J$ and $|AH|=3$, $|HJ|=7$, $|JB|=6$.
... | Suppose the triangle is drawn with $A$ as the topmost point, $B$ as the left base point and $C$ as the right base point.
The equation for the circle is $(x-x_0)^2+(y-y_0)^2=r^2$, where $(x_0, y_0)$ is the centre coordinates and $r$ is the radius. Assume a coordinate system with the origin at $B$. Now the coordinates f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descarte... | A way to do it is to write $(x+2)(x+3) = x^2 + 6x + 6 - x$, $(x+1)(x+6) = x^2 + 6x + 6 + x$, so
$$
(x+2)(x+3)(x+1)(x+6) = (x^2 + 6x + 6)^2 -x^2
$$
which gives that
$$
(x+2)(x+3)(x+1)(x+6) - 3x^2 = (x^2 + 6x + 6)^2 -4x^2 = (x^2+ 4x + 6)(x^2 +8x + 6).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 0
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Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3... | This is a little bit not fun, but will give your algebra a good workout. First note that
$$(x+h)^{4/3}-x^{4/3}=\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{2/3}-x^{2/3}\right)
.\tag{1}$$
and the right-hand side of (1) can be written as
$$\left((x+h)^{2/3}+x^{2/3}\right)\left((x+h)^{1/3}+x^{1/3}\right)\left((x+h)^{1/3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
| My solution:
I. $\sin x\sin2x\sin 3x<0=>$
$ 1 < 1 - \sin x\sin2x\sin 3x = \cos x\cos 2x\cos 3x\leq 1$ False!
II. $\sin x\sin2x\sin 3x = 0 => \cos x\cos 2x\cos 3x\ = 1 =>... x=n\pi, n$ integer
III. $\sin x\sin2x\sin 3x>0=>$
$ 1 = |\sin x\sin2x\sin 3x + \cos x\cos 2x\cos 3x|\leq$
$ |\sin x\sin2x\sin3x|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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Help with this inequality I am given four numbers $a,b,c,d$, such that $c>a>b,c>d>b$ and $0 \le a,b,c,d\le 1$ Can the following two inequalities hold strictly
$ad\le bc$ and $(1-a)(1-d)\le (1-b)(1-c)$.
| My interpretation of your question is that you want to find a set of 4 real numbers such that $0 \leq a, b, c, d \leq 1$, $c > d > b$, $c > a > b$, $ad < bc$, $(1-a)(1-d) < (1-b)(1-c)$
If so, then no set of numbers exist.
Consider $[\frac{1}{c^2} - \frac{1}{c} ] (bc) + [\frac{1}{c}] ( 1-b)(1-c) + [ 1 - \frac{1}{c}] \\=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can someone explain this trigonometric limit? I have
$$\lim \limits_{x\to 0} \frac {\tan(2x)}{\sin(x)}$$ and in my case the result is $\frac{2}{1}$ =2 not whether it is right.
This is my procedure.
$$\lim \limits_{x\to 0} \frac{\frac {\sin(2x)}{\cos(2x)}}{\frac{\sin(x)}{1}}= \lim \limits_{x\to 0} {\dfrac {\sin(2x)}{... | That does give you the correct answer, but it takes a bit more work than necessary.
An alternative method is to use the double-angle formula for $\sin$. That is:
$$\sin(2x) = 2\cos(x)\sin(x)$$
Thus:
$$\begin{align}\require{cancel}
\lim_{x\to0} \frac{\tan(2x)}{\sin(x)} &= \lim_{x\to0} \frac{\sin(2x)}{\cos(2x)\sin(x)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
| HINT:
$$ \lim_{x\to -\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} =\lim_{x\to +\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \lim_{x\to +\infty}\frac{\sqrt{(\frac{1}{x^2}+4)}}{\frac{8}{x^2}+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation for x, y and z: $\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$ I am having some trouble with this problem,
Solve for $x,y,$ and $z$.
$$\sqrt{x-y+z}=\sqrt x - \sqrt y + \sqrt z$$
Here is my work so far,
$$x - y +z = x+y+z-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy}$$
$$2y-2\sqrt{xy} + 2\sqrt{xz}- 2\sqrt{zy} =... | Your last equation can be written as
$$ \sqrt{zx} = \sqrt{y} \left[ \sqrt{x} - \sqrt{y} + \sqrt{z} \right] = \sqrt{y} \sqrt{ x-y+z} $$
Squaring both sides gives us
$$ zx = xy - y^2 + zy $$
Which simplifies to $$(y-x)(y-z) = 0 $$
Hence, we require $x=y$ or $y=z$. It is clear that in either case, the equation is satisfie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solving recurrence equation using generating function method I've been trying to solve the following equation intuitively (I only know the method if there are minuses in the equation - $a_{n-1}, a_{n-2}...$).
$$a_{n+2}=4a_{n+1}-4a_{n}$$
$$a_{0}=3$$
$$a_{1}=8$$
$$
\begin{align}
A(x)&=\sum\limits_{n>=0}a_{n}x^{n} \\
&= \... | Just another approach for verification.
Assume
$$
a_{n+2}=4a_{n+1}-4a_{n}
$$
Then
$$
\begin{align}
f(x)&=\sum_{k=0}^\infty a_kx^k\\
xf(x)&=\sum_{k=1}^\infty a_{k-1}x^k\\
x^2f(x)&=\sum_{k=2}^\infty a_{k-2}x^k\\
\end{align}
$$
Then we get
$$
\begin{align}
f(x)(1-4x+4x^2)&=a_0+a_1x-4a_0x+\sum_{k=2}^\infty(a_k-4a_{k-1}+4a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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question about applying a comparison test to a sequence In lecture, we were asked:
Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$ converge?
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2 }$
In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosin... | The easiest way to deal with this situation is to use the Limit Comparison Test, since then you don't have to worry about choosing appropriate constants. However, if you want to use the Comparison Test directly, trying a comparison with $\sum_{n=1}^\infty \frac{a}{n^2}$ and solving $\frac{2n^3+3n-8}{n^5-5n^3-n^2+2}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $ [Corrected question]
I'm struggling at proving the following combinatorical identity:
$$\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} $$
I would like to see a combinatorical (logical) solution, or an algebraic solution.
| Here is what might be the worst possible solution to this problem. We begin like ccorn does by noting
$$\begin{aligned}
\sum_{k=0}^n k\,\binom{n}{k}^2 &= \sum_{k=0}^n k\,\binom{n}{k}^2\,x^{k-1}\bigg|_{x=1} =
\left(\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n \binom{n}{k}^2\,x^k\right)\bigg|_{x=1}
\end{aligned}.$$
We exam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\inft... | This does work. Since your series is positive, the comparison test you've imployed ($2^{n-1}\leq 2^n-1$, hence $\frac{1}{2^n-1}\leq\frac{1}{2^{n-1}}$) implies that the sequence of partial sums of our series is increasing and bounded from above by the sequence $\sum_{n=1}^{N}\frac{1}{2^{n-1}}\to 2$, hence convergent and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Expected value sum of dots We throw $n$-times the die. Let $E_n$ be expected value sum of dots (got in all throws).
Compute
*
*$E_1$
*$E_2$
*$E_3$
*$E_4$
So i know how can I do it, for example in 1. I have:
*
*$E_1 = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \... | Let $X_n$ denote the number of dots thrown on the $n$th try. Then $$E_n = E[\sum_1^n X_i] = \sum_1^nE[X_i] = n E_1$$
do you see how to finish?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$4^\text{th}$ power of a $2\times 2$ matrix $$A = \left(\begin{array}{cc}\cos x & -\sin x \\ \sin x & \cos x\end{array}\right)$$ is given as a matrix. What is the result of $$ad + bc \text{ if } A^4=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$
Note that $A^4$ is the $4^\text{th}$ power of the matrix $A$.
I trie... | \begin{eqnarray*}
A
& = &
\left(%
\begin{array}{rr}
\cos\left(x\right) & -\sin\left(x\right)
\\
\sin\left(x\right) & \cos\left(x\right)
\end{array}
\right)
=
\cos\left(x\right) - {\rm i}\sin\left(x\right)\,\sigma_{y}
\\
A^{2}
& = &
\cos^{2}\left(x\right) - \sin^{2}\left(x\right)
-
2{\rm i}\sin\left(x\right)\cos\left(x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | From this and this, $\sin7x=7t-56t^3+112t^5-64t^7$ where $t=\sin x$
Now, if $\sin7x=0, 7x=n\pi, x=\frac{n\pi}7$ where $n=0,1,2,3,4,5,6$
Clearly, $\sin\frac{r\pi}7$ are the roots of $7-56t^2+112t^4-64t^6=0$ where $r=1,2,3,4,5,6$
As $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7,$
$\sin^2\frac{r\pi}7$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Initial value problem with $x^2u''-2xu'+2u=24/x^2$ I have some trouble finding the coefficients and powers of $x$ in the following problem.
Solve $x^2u''-2xu'+2u=\frac{24}{x^2}$ with $u(1)=2$ and $u'(1)=-5$.
I first set $$Lu_H=x^2u''-2xu'+2u=0$$ to find the homogeneous solution. After differentiating and substituting... | Easier way to see the solution:
$$\frac{d}{dx} \frac{u}{x} = \frac{u'}{x}-\frac{u}{x^2}$$
$$\frac{d^2}{dx^2} \frac{u}{x} = \frac{u''}{x}-2 \frac{u'}{x^2} + 2 \frac{u}{x^3}$$
So...divide both sides of the ODE by $x^3$ and get
$$\frac{u''}{x}-2 \frac{u'}{x^2} + 2 \frac{u}{x^3} = \frac{d^2}{dx^2} \frac{u}{x} = \frac{24}{x... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \sigma = \left( \begin{array}{ccccccccc} 1&2&3&4&5 \\ 2&3&4&1&5 \end{array} \right) \neq s_1 \circ s_2 \circ \ldots \circ s_k $ Prove that we can't write permutation $ \sigma = \left( \begin{array}{ccccccccc} 1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 1 & 5 \end{array} \right) $ as $s_1 \circ s_2 \circ \ldots \circ ... | Hint/Spoiler: Have you checked the parities of all the permutations here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Wallis Product (Long infinite Product) I'm almost finished proving Wallis product in a given question. But the last step is:
What am I doing wrong? Am I on the right track? Hints?
Show that $$\prod_{k=1}^{m} \left(\frac{(2k-1)(2k+1)}{(2k)^2}\right) = \frac{(2m+1)((2m!))^2}{2^{4m}(m!)^4}$$
Currently I have tried expand... | You are quite close. If we multiply out, then on top we get $2m+1$ times the product of the squares of the odd numbers up to $2m-1$. So we get
$$(2m+1) \left(1\cdot 3\cdot 5\cdots(2m-1)\right)^2.$$
"Improve" $1\cdot 3\cdot 5\cdot (2m-1)$ to $(2m)!$ by filling in the even numbers, that is, by multiplying by $2\cdot 4\c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
| If you multiply a number $n$ by any prime number, then its number of divisors is multiplied by $\frac{m+1}m$ where $p^{m-1}$ is the highest power of $p$ dividing$~n$. This is because the multiplicity of $p$ in divisors of $n$ can be any one of the $m$ values $0,1,\ldots,m-1$, but that multiplicity can be chosen among t... | {
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"url": "https://math.stackexchange.com/questions/475002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$
If $x \sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x^2)\frac{dy}{dx}+1=0.$
The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.
| the points $(0,0)$ and $(-1,-1)$ no longer lie on the graph.
Btw on actual differentiation it yield $\frac{dy}{dx} = \frac{y^2 - 2x(1+y)}{x^2 - 2y(1+x)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it always possible to factorize $(a+b)^p - a^p - b^p$ this way? I'm looking at the solution of an IMO problem and in the solution the author has written the factorization $(a+b)^7 - a^7 - b^7=7ab(a+b)(a^2+ab+b^2)^2$ to solve the problem. It seems like it's always possible to find a factorization like $(a+b)^p - a^p ... | It is true in general. These are often called Cauchy polynomials — see, for example, Fermat's Last Theorem for Amateurs, Chapter VII. The power of $a^2+ab+b^2$ is either $1$ or $2$ as $p \equiv -1$ or $+1$ modulo $6$, respectively. There are many proofs in the literature (Cayley, Glaisher, etc.).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
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Divide polynomials with exponents and simplify The expression is
$$\frac{p^2q^2}{m^2-n^2} \cdot \frac{m^2+2mn+n^2}{3p^2+2pq-q^2}.$$
How could I divide and simplify this?
| We can write the product of two rational fractions as one fraction, and use commutativity of multiplication to rearrange factors, as needed:
$$\frac{p^2q^2}{m^2-n^2} \cdot \frac{m^2+2mn+n^2}{3p^2+2pq-q^2} = \frac{p^2q^2(m^2 + 2mn+n^2)}{(3p^2 + 2pq - q^2)(m^2 - n^2)}$$
Here, you have quadratics in each of the numerator ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| Putting $\frac1x=h$
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
$$=\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}$$
Method $1:$
Using Binomial Expansion (1, 2) , $(1+x)^n=1+nx+O(x^2)$ and as $\lim_{h\to0}\frac... | {
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"url": "https://math.stackexchange.com/questions/480463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y... | $y'=-x^{n-1}y^{1-n}$.
$y''=-(n-1)x^{n-2}y^{1-n}-x^{n-1}(1-n)y^{-n}y'$
$y''=-(n-1)x^{n-2}y^{1-n}+x^{n-1}(1-n)y^{-n}x^{n-1}y^{1-n}$
$y''=-(n-1)x^{n-2}y^ny^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$
$y''=-(n-1)x^{n-2}(1-x^n)y^{1-2n}+x^{n-1}(1-n)x^{n-1}y^{1-2n}$
Should be easy from there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? ... | Since $(x+3)^2\geq 0$ for all real values of $x$, $x^2+6x+10=x^2+6x+9+1=(x+3)^2+1\geq 1$, and the inequality follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Effective method to solve $ \frac{x}{3x-5}\leq \frac{2}{x-1}$ I need to solve this inequality. How can I do so effectively?
$$ \frac{x}{3x-5}\leq \frac{2}{x-1}$$
|
$$\frac{x}{3x-5}\leq \frac{2}{x-1}$$
Subtracting the right-hand side from each side, then finding the common denominator and subtracting resulting numerators gives us:$$\begin{align}
\dfrac{x(x-1)-2(3x-5)}{(3x-5)(x-1)}\leq 0 & \iff \dfrac{x^2 - 7x + 10}{(x-1)(3x-5)} \leq 0 \\ \\
& \iff \dfrac{(x-2)(x-5)}{(x-1)(3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simpler solution to a geometry problem In a set of geometry problems, I got this one:
If in a triangle $ABC$ with segments $AB=8$, $BC=4$, and $3A+2B=180^{\circ}$, calculate the side $AC$
My solution was
Let $A=2\alpha$,$B=90^{\circ}-3\alpha$, where $\alpha<30$, then the second condition is always met.
So
$$tan(2\al... |
We might also "drop a perpendicular" from vertex $ \ B \ $ to the extended line $ \ AC \ \ , \ $ the length of which we will call $ \ y \ \ . \ $ Using your notation for angle measure, we have $ \ y \ = \ 8 \sin (2 \alpha) \ = \ 4 \sin (90º - \alpha) \ = \ 4 \cos \alpha \ \ . $ From this, we find
$$ 8 \ · \ 2 · \si... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$
| Without loss of generality
let $0\le a\le b\le c\le 1$,then
\begin{align*}
&\dfrac{a}{c+b+1}+\dfrac{b}{a+c+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&\le\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(a+b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(a+1)(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/488273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality With Recurrent Relation $a_{1}=1$, $~$ $a_{n+1}-a_{n}=\sqrt{\dfrac{a_{n+1}^{2}-1}{2}}+\sqrt{\dfrac{a_{n}^{2}-1}{2}}$ , $~$ $a_{n+1}>a_{n}$
Prove that $~$ $\displaystyle\sum_{k=1}^{\infty}\frac{1}{a_{n}}<e$
| Note that sequence $\{a_n, ~ n\in \mathbb{N}\}$ is non-decreasing (because RHS can't be negative, if $a_n\in \mathbb{R}$).
First, let's consider increasing sequence $\{a_n, ~ n\in \mathbb{N}\}$:
$$
a_{n+1}>a_n.\tag{1}
$$
1.
Let's show that there is other (easy) recurrent relation
$$
a_n = 6a_{n-1}-a_{n-2}, \qquad (n\ge... | {
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How prove this $x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$ show that
$$x^3<\sin^2{x}\tan{x},x\in\left(0,\dfrac{\pi}{2}\right)$$ have nice methods? Thank you
my try:
$$\Longleftrightarrow \cos{x}\cdot x^3<(\sin{x})^3$$
let
$$f(x)=\cos{x}\cdot x^3-(\sin{x})^3$$
$$\Longrightarrow f'(x)=-\sin{x}\cdot x^3+3\... | We have that
$$\sin^3x-x^3 \cos x=\frac14\left(3\sin x-\sin(3x)\right)-x^3 \cos x\ge0 \iff 3\sin x-\sin(3x)-4x^3\cos x \ge 0$$
and since by Taylor's series
*
*$\sin x\ge x-\frac16x^3+\frac1{120}x^5-\frac1{5040}x^7$
*$\sin (3 x)\le 3x-\frac 9 2 x^3+\frac{81}{40}x^5-\frac{243}{560}x^7+\frac{243}{4480}x^9$
*$\cos x\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}... | $$
\begin{align}
\log(2)&=\sum_{n=0}^\infty\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\\
\frac12\log(2)&=\sum_{n=0}^\infty\hphantom{-\frac1{4n+2}}\frac1{4n+2}\hphantom{\;+\frac1{4n+3}}-\frac1{4n+4}\\
\frac\pi4&=\sum_{n=0}^\infty\frac1{4n+1}\hphantom{-\frac1{4n+2}}-\frac1{4n+3}\\
\frac12\left(\log(2)-\frac12\log... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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Inequalities with radicals I have a question regarding the following inequality; It may sound petty but yet;
The following inequality $$ \sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$$ has the solution $\frac {-5}{4}\le x \lt 1 $
During the process of solving it we need to check the correctness of the solution to... | As a follow-up to my comment above, I'd like to walk through a solution of this inequality.
$$ \sqrt{-6x+10} + \sqrt{-x + 2} > \sqrt{4x+5}$$
*
*First determine the restrictions on $x$ implied by the radicals.
$$
-6x + 10 \geq 0 \;\Rightarrow\; x \leq 5/3.
$$
$$
-x + 2 \geq 0 \;\Rightarrow\; x \leq 2.
$$
$$
4x+5 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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CS problem, turned to mathematics I am trying to solve some of the projecteuler problems using a much of a programmers approach. However, I would like to get more into the math, and therefore would try to do some mathematical reasoning to this specific projecteuler exercise:
Surprisingly there are only three numbers th... | If a number $n$ has seven digits, the sum of fifth powers of digits is $\le 7\cdot 9^5<10^6\le n$ and with each additional digit the upper limit for the power-sum grows by at most $9^5$ whereas the lower lmit for $n$ grows by a factor of $10$. We conclude that $n$ has at most six digits.
Moreover, from $d^5\equiv d\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/495737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Taking Radicals Out of Denominator How does one take radicals out of the denominator of
$$\frac{1}{a + b(\sqrt{2} + \sqrt{3})}$$
assuming $a,b \in \mathbb{Q}$ not both equal to $0$. I know the trick is to multiply this expression by $1$ s.t. the radicals are removed, but I can't think of such $c/d = 1$ to try.
| Try
$$ \frac{ a + b \sqrt{2} - b \sqrt{3} } { a + b \sqrt{2} - b \sqrt{3} } \times \frac{ a - b \sqrt{2} + b \sqrt{3} } { a - b \sqrt{2} + b \sqrt{3} } \times \frac{ a - b \sqrt{2} - b \sqrt{3} } { a - b \sqrt{2} - b \sqrt{3} } .$$
The idea is that you have to multiply by all possible conjugates.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that $10^n - 1$ is divisible by $9$ How can I show that $10^n-1, 10^{n-1}-1,...., 10-1$ are all divisible by 9? I was considering using Euclid's algorithm, but I can't find a way to get that to work.
| $$ a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) $$
Proof of the identity by expansion:
$$\begin{align}
&(a-b)(a^{n-1} + a^{n-2}b + \ldots + b^{n-1}) \\\\
&=a^{n} + a^{n-1}b + a^{n-2}b^2 + \ldots + ab^{n-1}\\\\
&\;\;\;\;\;\;\;\;- ba^{n-1} - b^2a^{n-2}-\ldots -b^{n-1}a - b^n\\\\
&=a^n-b^n
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $n=7k$ or $7k+1$, if $n=x^2=y^3$. $n$ is both a square and a cube, I need to prove $n=7k$ or $n=7k+1$, for some integer $k$.
I need some guide lines.
| If $n$ is both a square and a cube, it is a perfect $6$-th power.
Let $z$ be a number. Then $z$ has shape $7k$, or $7k+1$, or $7k+2$, and so on up to $7k+6$.
Then $z^6$ has shape (respectively) $7k$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, and $7k+1$.
To do the calculations, let us take the example $z$ of the shape $7... | {
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"timestamp": "2023-03-29T00:00:00",
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On the roots of $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$ The Problem Prove that $\tan \frac{\pi}{15}$ is a root of the equation $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$, and find the other roots.
Source: Question 17 from the complex numbers chapter from Bostock's Further Pure Mathematics
Thoughts Using the identity $\tan 5\theta =... | First of all, $\tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
Set $\tan5\theta=\sqrt3=\tan\frac\pi3$
$\implies 5\theta=n\pi+\frac\pi3=\frac{(3n+1)\pi}5$ where $n$ is any integer
So, the values of $\theta$ can be set from $n=0,1,2,3,4$
Observe that for $n=3,\tan\theta=\tan\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$ equal zero? Limit $$\lim_{x\to 0+} \sqrt{\frac {1}{x}+2}-\sqrt{\frac {1}{x}}$$
equals zero.
Could you help me prove it?
| Let $y = \dfrac1x$. We then have
$$L = \lim_{x \to 0^+} \left(\sqrt{\dfrac1x+2} - \sqrt{\dfrac1x}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) = \lim_{y \to \infty} \left(\sqrt{y+2} - \sqrt{y}\right) \times \dfrac{\sqrt{y+2} + \sqrt{y}}{\sqrt{y+2} + \sqrt{y}}$$
This gives us
$$L = \lim_{y \to \infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/501675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| $$x^2+y^2\geq 2xy\implies 2(x^2+y^2)\geq (x+y)^2$$
Hence, $$(x^2+y^2)\geq \frac{6^2}{2}=18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
| Let's use degree measure (for writing convenience):
we need to prove identity
$$
\tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ = 1.\tag{1}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 54^\circ \sin 72^\circ =^?= \cos 12^\circ \cos 48^\circ \cos 54^\circ \cos 72^\circ;\tag{2}
$$
$$
\sin 12^\circ \sin 48^\circ \sin 5... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding basis for the space spanned by some vectors. Find a subset of vectors $\{v_1, v_2, v_3, v_4, v_5\}$ that forms the basis for the space spanned by these vectors: $$v1=\left ( \begin{array}{c} 1\\-2\\0\\3 \end{array}\right), v2= \left ( \begin{array}{c} 2\\-5\\-3\\6\end{array}\right)\, ,\, v3=\left ( \begin{array... | You don't need to guess; just write down the matrix having the vectors as columns:
$$\begin{bmatrix}
1 & 2 & 1 & 2 & 3 \\
-2 & -5 & -1 & -1 & 2 \\
0 & -3 & 3 & 4 & 14 \\
3 & 6 & 1 & -7 & -17
\end{bmatrix}$$
and proceed with Gaussian elimination; first do $R_2+2R_1$ (sum to the second row the first multiplied by $2$)... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extended euclidean algorithm So I am trying to figure this out.
And for one of the problem the question is x*41= 1 (mod 99)
And the answer lists
x | 41x mod 99
0 99
1 41
-2 17
5 7
-12 3
29 1
And conclude x=29
How did they get this value(can someone explain)? To better put, how do you calculate thi... | When two numbers $e,\varphi$ are given, $e<\varphi$, $GCD(e,\varphi)=1$, and we need to find $x$, such that
$$
x\cdot e = 1 (\bmod~\varphi),\tag{1}
$$
then denote
$$
r_0 = \varphi, \qquad v_0 = 0;
$$
$$
r_1 = e, \qquad\; v_1 = 1;
$$
then for each $n\geqslant 1$ to build values:
$$
s_n = \left\lfloor \frac{r_{n-1}}{r_n... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving the identity: $\sin3x + \sin x = 2\sin2x\cos x$ I need some help proving this identity:
$$\sin3x + \sin x = 2\sin2x\cos x$$
I don't know where to start. I thought about expanding $\sin 3x$ into $\sin (2x + x)$ but I don't think that does me any good. Any hints would be appreciated.
Thanks!
| $$\sin 3x+\sin x=2\sin 2x\cos x$$
$$LHS
=\sin(2x+x)+\sin x\\
=\sin2xcosx+\sin x\cos2x+\sin x\\
=2(\sin x \cos x)\cos x+\sin(cos^2x-\sin^2x)+\sin x\\
=2\sin x \cos^2x+\sin x \cos^2x-\sin^3x+\sin x\\
=\cos^2x(2\sin x+\sin x)-\sin^3x+\sin x\\
=(1-\sin^2x)3\sin x-\sin^3x+\sin x\\
=3\sin x-\sin^3x-\sin^3x+\sin x\\
=4\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/513349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How to prove or disprove $n$ is prime? Let $n$ be an odd number such that
$$2^{\frac{n-1}{2}}\equiv -1 \pmod{n}$$
How do I prove or disprove $n$ must is prime?
This problem is from when I solved another problem. Thank you
| The basic method for fast exponentiation is to first find the base 2 expansion of $1638$, i.e., notice that $1638=2+2^2+2^5+2^6+2^9+2^{10}$.
On the other hand, $2^2\equiv 4 \pmod{3277}$, $2^{2^2}\equiv 16 \pmod{3277}$, $2^{2^3}\equiv 256 \pmod{3277}$ and $2^{2^4}\equiv -4 \pmod{3277}$, $2^{2^5}\equiv 16 \pmod{3277}$, $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solve for $x: 2/\sqrt{2-x^2} = 4- 2x^2?$ How do you expand and solve for $x$?
It is $2 = 4\sqrt{2-x^2} - 2x^2\sqrt{2-x^2}$
Thank you!
How would I solve for x?
| Note that real solutions of $x$ occur for $|x|\lt\sqrt 2$. Then start by dividing by $2$:
$$1=(2-x^2)\sqrt{2-x^2}=(2-x^2)^{3/2}$$
$$\implies 1=2-x^2 \implies x^2=1$$
Which implies that $x=\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the included formula relating cos(nx) and cos(x) I'm struggling with the below problem. Can anyone shed some light on it?
Show that the below formula is a correct relation between $y = \cos n\theta$ and $x = cos \theta$ for all $n$:
$$ x = \frac 12 \sqrt[n]{y + \sqrt{y^2 - 1}} + \frac 12 \sqrt[n]{y - \sqrt{y^2 - ... | This may not be exactly the proof you want, but here is a proof (here, we will only use the fact that $i=\sqrt{-1}$ a couple of times, and nothing else about the complex number $i$...I will make note of when we use this).:
We proceed first to prove the identity
$\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\t... | {
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Computing an inverse modulo $25$ Supposed we wish to compute $11^{-1}$ mod $25$. Using the extended Euclid algorithm, we find that $15 \cdot 25 - 34 \cdot 11 =1$. Reducing both sides modulo $25$, we have $-34 \cdot 11 \equiv 1$ mod $25$. So $-34 \equiv 16 $ mod $25$ is the inverse of $11$ mod $25$.
im realy confused ab... | Note that $25 | 15 \cdot 25$, so $15 \cdot 25 \equiv 0 \pmod{25}$. On the other hand, we see that
$$1 \equiv -15 \cdot 25 - 34 \cdot 11 \equiv 0 + (-34) \cdot 11 \pmod{25}$$
Finally,
$$-34 \equiv -34 + 50 \equiv 16 \pmod{25}$$
so we can rewrite the above as
$$1 \equiv 16 \cdot 11 \pmod{25}$$
and $16$ is the desired inv... | {
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Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we... | We have $$\Delta x=\frac{b-a}{n}=\frac{4}{n}$$ and $$\xi_1=-2+\frac{4}{n};\;\;\xi_2=-2+2\frac{4}{n};\;\;...\;\;;\xi_n=-2+n\frac{4}{n}$$ then $$\xi_i=-2+\frac{4i}{n}=\frac{4i-2n}{n}$$ $$\int_{-2}^2 {4-x^2}\;dx=\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\frac{4}{n}\\\lim_{n\to+\infty} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Probability of rolling 6's on 3 dice - adjusted gambling odds? I'm struggling to work out odds on a game that were working on. It's probably best if I write an example as I'm really not a mathematician!
I'm working on a dice game where the player bets 1 coin and rolls 3 dice. If any of the dice are a 6 we payout based ... | The probabilities you have are wrong. Denote by $X$ the number of sixes rolled. Then the correct probabilities are:
$P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
$P(X = 1) = P$ (first dice six) $ + P$(second dice six) $ + P$(third dice six) $= \frac{1}{6} \frac{5}{6} \frac{5}{6} + \frac{5}{6} \frac{1}{6} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$
I think I got it, but is this proof correct?
We can write any integer x in the form: $x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4$, and $x = 6k + 5$.
If $x = 6k$, then $x^3 = 216k^3$. Then $x^3 - ... | Yes, that works just fine.
Another way is to observe that if $$x^3\equiv x\pmod 6$$ then also $$x^3-x\equiv 0\pmod 6$$
and so the original statement is equivalent to the assertion that $x^3-x$ is always a multiple of 6.
Since $x^3-x$ factors as $(x-1)\cdot x\cdot (x+1)$ it is a product of three consecutive numbers; on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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The probability that it will take at least $k$ tosses until the first head A fair coin is tossed until a head is obtained. So for example, the sample space can be modeled as {$H, TH, TTH, TTTH, ...$}. The probability measure is modeled so that $P(H)=\frac 12$, $P(TH)=\frac 14$, $P(TTH)=\frac 18$ etc.
Find the probabili... | Let $X$ be the number of moves. Then
$$P(X=k) = \frac{1}{2^k}$$
(a) $$P = 1 - P(X > k) = 1 - P(k \text{ tails}) = 1 - \frac{1}{2^k} = \frac{2^k - 1}{2^k}$$
(b) $$P = \sum_{k=0}^\infty P(X=2k+1) = \sum_{k=0}^\infty \frac{1}{2^{2k+1}} = \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k} = \frac{1}{2} \frac{1}{1- \frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/526964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $6$ divides $n(n + 1)(n + 2)$ I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows:
Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$.
Note: An integer $a$ divides an integer $b$, written $a|b$, if there exists $q ∈ Z... | If we let $m=n+1$, we need to prove that $6 \vert m(m^2-1)$ for all $m \geq 2$. Equivalently, we want to show that
$$m^3 \equiv m \bmod6$$
First note that $m \equiv 0,1,2,3,4,5 \pmod 6$. Now observe that following:
$$m \equiv 0 \bmod 6 \implies m^3 \equiv 0^3 \bmod 6 \implies m^3 \equiv 0 \bmod6 $$
$$m \equiv 1 \bmod 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/527300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 5
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Did I do this limit right? $\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1}$ This is how I did this limit:
$$\lim_{x \to 0} \frac{(\sqrt{x^2+2}-\sqrt{2})}{\sqrt{x^2+1}-1} =\lim_{x \to 0}\frac{(\sqrt{x^2+2}-\sqrt{2})(\sqrt{x^2+1}+1)}{(\sqrt{x^2+1}-1)(\sqrt{x^2+1}+1)} = \lim_{x \to 0} \frac{(\sqrt{x^2+2}-\s... | If you correctly simplify the very last line in your solution, you'll see you get the same answer: $\frac 2{2\sqrt 2}$
$$\frac{(\sqrt{0^2+1}+1)}{(\sqrt{0^2+2}+\sqrt{2})}=\frac{2}{2\sqrt{2}} = \frac 1{\sqrt 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$|3^a-2^b|\neq p$, from a contest I recently came across an old contest problem: (I did not find the solution anywhere)
Find the least prime number which cannot be written in the form $|3^a-2^b|$ where $a$ and $b$ are nonnegative integers.
At the beginning I thought that the answer was the number $2$ but the ''n... | One readily finds representations for all primes $<41$.
Assume $41=|3^a-2^b|$ that is $2^b\pm 41=3^a$.
By not finding powers of three among the numbers $2^b\pm41$ with $b\le 2$, we conclude $b\ge 3$.
Then $3^a\equiv \pm41\pmod{8}$, which is equivalent to $a\equiv 0\pmod 2$.
Especially $3^a\equiv1\pmod {8}$ and hence $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Elementary Algebra Inequality question $$
\frac{1}{x-1} < -\frac{1}{x+2}
$$
(see this page in wolframalpha)
Ok, so I think the main problem is that I don't really know how to do these questions. What I tried to do was move $-1/(x + 2)$ to the LHS and then tried to get a common denominator. I ended up with
$$
\frac{(x ... | $\frac{1}{x-1}<-\frac{1}{x+2} \rightarrow \frac{1}{x-1}+\frac{1}{x+2}<0 \rightarrow$
$\frac{(x+2)+(x-1)}{(x+2)(x-1)} <0 \rightarrow \frac{(x+2)+(x-1)}{(x+2)(x-1)}((x+2)(x-1)^2<0 \rightarrow (x+2+x-1)(x+2)(x+1)<0 $
for $x$ different to $-2,1$. (-2 and 1 arent soultions cause they are undefined.)
using the fact that rat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of $\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}$dx Question:
$$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$
What we did:
we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the ... | Alternatively, if we call our integral $I$ and perform a dummy variable substitution, it all falls out rather nicely! $$I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x \ \overset{x \mapsto \frac{\pi}{2} - x}= \ \int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
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How to obtain $y$
The question was written with dark-blue pen. And I tried to solve this question. I obtained $x$ as it is below. But I cannot obtain $y$ Please show me how to do this.
By the way, $\gamma (t)$ may not be clearly readable. So, I wrote again.
$$\gamma (t)=( \cos ^2 (t)-1/2, \sin(t)\cos (t), \sin (t)... | From your work,
$$x^2+y^2=\frac{1}{4} \Rightarrow y^2=\frac{1}{4}-x^2,$$
and
$$x=\cos^2 t -\frac{1}{2}.$$
Substituting the latter into the former produces
\begin{align*}
y^2 &=\frac{1}{4}-x^2 \\
&=\frac{1}{4}-\left( \cos^2 t - \frac{1}{2} \right)^2 \\
&=\frac{1}{4}-\left( \cos^4 t - \cos^2 t + \frac{1}{4} \right) \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Determinant of a General Expression Matrix I don't really know how to do this question:
" Let $A = $
$\begin{pmatrix}1 & 1& 1 & \cdots & 1 \\ 1 & 1-x & 1 & \cdots & 1 \\ 1 & 1& 2-x &\cdots & 1 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1& 1 & 1 &\cdots& n-1-x \end{pmatrix}$"
be an n×n matrix with n≥3. Solve the e... | Subtract the first column from the last, then $$\det A=\det\begin{pmatrix}1 & 1& 1 & \cdots & 1 & 0 \\ 1 & 1-x & 1 & \cdots & 1 & 0 \\ 1 & 1& 2-x &\cdots & 1 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1& 1 & 1 &\cdots& n-2-x & 0\\1& 1 & 1 &\cdots& 1 & n-2-x \end{pmatrix}=$$$$(n-2-x)\det\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/531756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that a system of Diophantine equations will have irrational solutions as well as integers Solve $\begin{cases} 3xy-2y^2=-2\\ 9x^2+4y^2=10 \end{cases}$
Rearranging the 2nd equation to $x^2=\dfrac{10-4y^2}{9} \Longrightarrow 0\leq x^2 \leq 1$ if $x^2=1$ than $y=\pm\dfrac{1}{2}$ and $x=\pm1$ but how do I show the... | The number of solutions has nothing to do with some of the intersection points being rational, which is a happy coincidence. The way to approach this problem is plain old algebra. Add twice the first equation to the second, and divide through by 3 to get
$$2x y + 3x^2 = 2.$$
Then manipulating the second equation gives
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Complex Number - Find all roots of the equation $$e^{i \frac{{\pi}}{3}}z^5+4e^{i\frac{(2+3){\pi}}{6}}z^3 + z^2 + 4i = 0.$$
By using Euler's formula, I got:
$$e^{i \frac{{\pi}}{3}} = \cos{\frac {\pi}{3}} + i\sin{\frac {\pi}{3}} = (\frac{1}{2} + i{\frac {\sqrt{3}}{2}}) $$
$$4e^{i \frac{(2+3){\pi}}{6}} = 4(\cos{\frac {5\... | Hint: The polynomial can be factored as
$$
e^{i\pi/3} (z^3+a)(z^2+b).
$$
More: Multiply it out:
$$
e^{i\pi/3} (z^3+a)(z^2+b) = e^{i\pi/3}z^5+be^{i\pi/3}z^3 + ae^{i\pi/3}z^2 + abe^{i\pi/3}.
$$
To make this match the original polynomial, $e^{i\pi/3}z^5 + 4e^{i5\pi/6}z^3 + z^2 + 4i$, we need
$$
be^{i\pi/3} = 4e^{i5\pi/6},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to prove $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$? If $m$ and $n$ are any two positive integers then how do we show that the inequality $$\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$$ always holds ?
| let
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}$$
then we easy to
$$f(1,1)=f(1,2)=f(2,1)=\dfrac{1}{12}\le\dfrac{4}{45}$$
so,when $m,n\ge 2,$,then we let $j=m+n+2\ge 6$
then
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}\le\dfrac{1}{j-1}-\dfrac{4}{j^2}\le\dfrac{4}{45}$$
It is clearly,
because $$f(x)=\dfrac{1}{x-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\... | let $a=x^2,b=y^2,c=z^2$
$$\Longleftrightarrow x^2+y^2+z^2=3\Longrightarrow \dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge 3$$
note
$$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2z^2x^2+2y^2z^2$$
use
AM-GM inequality,then we have
$$4\dfrac{x^2}{y}+2x^2y^2+x^4\ge (2\cdot 2+2+1)x^{\frac{2\cdot 2\times 2+2\cdot 2+2\cdot 2}{2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Intersection of two tangents on a parabola proof There are two tangent lines on a parabola $x^2$. The $x$ values of where the tangent lines intersect with the parabola are $a$ and $b$ respectively. The point where the two tangent lines intersect has an $x$ value of $c$.
Prove that $c=(a+b)/2$
I have tried taking the d... | Let's try to work through that algebra, then.
The derivative of $x^2$ is $2x$, so those tangent lines are given by
$$
y = 2a(x-a) + a^2\\
y = 2b(x-b) + b^2
$$
Thus, at the $x$-coordinate of the lines' intersection, we have
$$
2a(x-a) + a^2 = 2b(x-b) + b^2
$$
Expanding the above expressions, we have
$$
2ax - 2a^2 + a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multivaraible calculus,Global minima maxima How do I compute the global minimum or maximum of the function
$f(x,y)=-\sin x\cos y$.
Given it is on a square $(0\leq x\leq 2\pi)$ and $(0\leq y\leq2\pi)$
| The inequality chain already mentioned by user7530 establishes the range of the function rather quickly:
$$ -1 \ \le \ \sin \ x \ \le \ +1 \ \ \Rightarrow \ \ -1 \ \le \ -\sin \ x \ \le \ +1 $$
$$ \Rightarrow \ \ -1 \ \le \ -\cos \ y \ \le \ -\sin \ x \ \cos \ y \ \le \ \cos \ y \ \le \ +1 \ \ . $$
It's a little mor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/542298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? It seems like an expression from a simple induction proof problem that's missing its result.
| It can be transformed into
$$
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{1+2+3+\cdots +n}}=
\frac{(2-1)(2^2-1)(2^3-1)\cdots(2^{n-1}-1)}{2^{n(n+1)/2}}
$$
but whether this is a simplification I don't know.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/543616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| Note that (can be easily seen with rule of Sarrus)$$
\begin{vmatrix}
x & y & z \\
z & x & y \\
y & z & x \\
\end{vmatrix}=x^3+y^3+z^3-3xyz
$$
On the other hand, it is equal to (if we add to the first row 2 other rows)
$$
\begin{vmatrix}
x+y+z & x+y+z & x+y+z \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 1
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$Z_1$, $Z_2$, and $Z_3$ are independent standard normal variables. Obtain the correlation matrix Define
$X_1= 2Z_1 - Z_2 +3Z_3$
$X_2=Z_1 + Z_2 - Z_3$
$X_3 = 3Z_1 + 2Z_2 + Z_3$
So would the correlation matrix of $(X_1, X_2, X_3)$ just be the coefficients of each variable?
| Initially you have
$$
\mathbf{Z}=\begin{pmatrix}Z_1 \\ Z_2 \\ Z_3 \end{pmatrix}\sim N\left(\boldsymbol{0}, \mathbf{I}\right)
$$
Define $\mathbf{X}$:
$$
\mathbf{X}=\begin{pmatrix}X_1 \\ X_2 \\ X_3 \end{pmatrix}=\begin{pmatrix}2Z_1-Z_2+3Z_3 \\ Z_1+Z_2-Z_3 \\ 3Z_1+2Z_2+Z_3 \end{pmatrix}=\underbrace{\begin{pmatrix}2 & -1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/546969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is the minimum value of $a+b+\frac{1}{ab}$ if $a^2 + b^2 = 1$? For the case when $a,b>0,$ I used AM-GM Inequality as follows that:
$\frac{(a+b+\frac{1}{ab})}{3} \geq (ab\frac{1}{ab})^\frac{1}{3}$
This implies that $(a+b+\frac{1}{ab})\geq 3$. Hence, the minimum value of $(a+b+\frac{1}{ab})$ is 3
But the answer is $... | Constrant:
$$q=x^2 +y^2-1=0$$
Goal:
$$ c= x+ y + \frac{1}{xy}$$
Now, by LaGrange multipliers , we can say that:
$$ \nabla q = \lambda \nabla c$$
Hence,
$$2x = 1 - \frac{1}{x^2y} \tag{1}$$
$$ 2y = 1- \frac1{y^2 x} \tag{2}$$
By multiplication of $x$ in eqtn (1) and $y$ in eqtn (2):
$$ 2x^2 = x - \frac{1}{xy} \tag{3}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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How to get $\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}- \frac{1-z^n}{n}\bigg)$"? I asked a question here and got answer committing :
$$(1+yz^n)(1+y)^{n-1} - (1+yz)^n=\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}-
\frac{1-z^n}{n}\bigg) \tag{1}$$and$$\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}=
... | $1.$
\begin{align*}
(1+yz^n)(1+y)^{n-1} - (1+yz)^n &= (1 + yz^n) \sum_{k=0}^{n-1} \binom{n-1}{k}y^k - \sum_{k=0}^{n} \binom{n}{k}(zy)^k \\
&= \sum_{k=1}^{n-1} y^k \left(\binom{n-1}{k} + z^n \binom{n-1}{k-1} - \binom{n}{k} z^k \right )\\
&= \sum_{k=1}^{n-1} \binom{n-1}{k-1} y^k \left( \frac{n-k}{k } + z^n - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$ \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0$ implies that $ \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0$ $a ,b , c$ are real numbers such that $ \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0$ , then how do we prove
(without routine laborious manipulation) that $\dfrac a{(bc-a^2)^2} + \... | Multiply out denominators, and rearrange the first equation to $(ab+bc+ca)^2=(ab+bc+ca)(a^2+b^2+c^2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Non-Homogeneous System [Problem] "Find a general solution of the system and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system."
$\begin{bmatrix}3 & 4 & 1 & 2 \\ 6 & 8 & 2 & 5\\9 & 12 & 3 & 10 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ ... | Try row reducing the augmented matrix to reduced row echelon form:
$$
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
6 & 8 & 2 & 5 & 7\\
9 & 12 & 3 & 10 & 13
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}
3 & 4 & 1 & 2 & 3\\
0 & 0 & 0 & 1 & 1\\
0 & 0 & 0 & 4 & 1
\end{array}\right]
\sim
\left[\begin{array}{cccc|c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\frac{mi}{n}+\frac{1}{2}\right\rfloor$ is an even number Let $m$, $n$ be positive odd numbers such that $\gcd(m,n)=1$. Show that
$$\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\dfrac{mi}{n}+\dfrac{1}{2}\right\rfloor$$
is an even number, where $\lfloor{x}\rfloor$ is the largest... | Note $[x+\frac{1}{2}]=[2x]-[x]$
$$\sum_{i=1}^{\frac{n-1}{2}}\left\lfloor\dfrac{mi}{n}+\dfrac{1}{2}\right\rfloor=\sum_{i=1}^{\frac{n-1}{2}}(\left\lfloor2x\right\rfloor-\lfloor x\rfloor)\\=\sum_{i\ge \frac{n-1}{2},i\equiv0\pmod{2}}\left\lfloor x \right\rfloor-\sum_{i\le \frac{n-1}{2},i\equiv1\pmod{2}}\left\lfloor x \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/552860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Implicit derivative - $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Let $y$ be a function of $x$ determined by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Find $\Large\frac{dy}{dx}$ and $\Large\frac{d^2y}{dx^2}$
I've obtained $\Large\frac{dy}{dx} = \frac{-xb^2}{ya^2}$ and $\Large\frac{d^2y}{dx^2}=\frac{-b^2a^2y^2... | Your answer is correct, just unsimplified. Note:
$$
-b^2a^2y^2-x^2b^4=-b^2(a^2y^2+b^2x^2)=-b^2
$$
The second equality holds when you cross multiply the equation for the ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/553004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the complex $z$ such $\max{(|1+z|,|1+z^2|)}$ is minimum find the complex $z$,such
$$\max{(|1+z|,|1+z^2|)}$$ is minimum
My try: let $z=a+bi$,then
$$|1+z|=\sqrt{(a+1)^2+b^2}$$
$$|1+z^2|=|1+a^2+2abi-b^2|=\sqrt{(1+a^2-b^2)^2+4a^2b^2}$$
Then I can't,Thank you
| go further, there is two cases:
case I: $|1+z|\ge|1+z^2|$
$(a+1)^2+b^2\ge(1+a^2-b^2)^2+4a^2b^2 \iff a^4+2 a^2 b^2+a^2-2 a+b^4-3 b^2 \le 0 \iff (b^2-x_1)(b^2-x_2)\le 0, x_1=\dfrac{3-2a^2-\sqrt{D}}{2},x_2=\dfrac{3-2a^2+\sqrt{D}}{2},D=-16a^2+8a+9 \iff x_1\le b^2 \le x_2$
it is trivial that $D\ge 0$ for this case $\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
| By putting $x=\sinh(z)^2$ we have:
$$ I = \int_{0}^{+\infty}2\tanh z\operatorname{arccot}\left(\sinh z-2\cosh z\right)dz; $$
integrating by parts we have:
$$ I = \int_{0}^{+\infty} 2\frac{-1+2\tanh z}{5\cosh z-4\sinh x}\log\left(\cosh z\right)dz;$$
now putting $z=\log t$ we get:
$$ I = 4\int_{1}^{+\infty} \frac{t^2-3}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long wa... | Let me add a more general answer.
Note that
$$
{1 \over {\left( {x + 1} \right)\left( {x + 2} \right) \cdots \left( {x + m} \right)}}
= {1 \over {\left( {x + 1} \right)^{\,\overline {\,m\,} } }}
= {{\Gamma \left( {x + 1} \right)} \over {\Gamma \left( {x + 1 + m} \right)}}
= x^{\,\underline {\, - m\,} }
$$
where $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 10,
"answer_id": 9
} |
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.