Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the least positive value of alpha for the following trignometric equation Given question is : $$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)$$
We are to find the least positive value of $\alpha$ for which above equation holds.
The way I did is as, $$\sin5\alpha \cos\theta + \cos5\alpha \sin\theta = \cos\theta \... | My suggestion would be:
$$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)=\sin(\pi/2-\theta+3\alpha)$$
so,
$$\sin(5\alpha + \theta) -\sin(\pi/2-\theta+3\alpha)=0$$
$$2\sin\left(\frac{(5\alpha + \theta)-(\pi/2-\theta+3\alpha)}{2}\right)\cdot \cos\left(\frac{(5\alpha + \theta)+(\pi/2-\theta+3\alpha)}{2}\right)=0$$
$$2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Calculating permutations for a specific password policy The security researcher Troy Hunt posted an example of an obscure password policy and I've been trying to figure out how to calculate the possible permutations
(Source: https://twitter.com/troyhunt/status/885243624933400577)
The rules are:
*
*The password must... | A brute force solution:
Nine numbers: Each number appears exactly once. We choose nine of the ten digits, then arrange the selected digits.
$$\binom{10}{9}9!$$
Eight numbers: We must use of the ten digits twice. We choose that digit, choose two of the nine locations for that number, choose seven of the other nine ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How many $6$ digit numbers can you make with the numbers $\{1, 2, 3, 4, 5\}$ so that the digit $2$ appears at least 3 times? I can't seem to understand.
I thought about it this way. We take a first example:
$222aaa$ for each blank space we have $5$ positions, so in this case the answer would be $125$ ($5 \times 5 \... |
How many six-digit numbers can be formed using numbers from the set $\{1, 2, 3, 4, 5\}$ with replacement if the digit $2$ must appear at least three times?
There are $\binom{6}{k}$ ways of choosing exactly $k$ positions for the $2$'s and $4^{6 - k}$ ways to fill the remaining $6 - k$ positions with a number different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequal... | You may use Lagrange multipliers method to show that we need $a=b=c$. The optimisation problem would be to minimise $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, having the constraint $a+b+c=t , t\leq \frac{3}{2}$.
$$F=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\lambda(a+b+c-t)$$
Then, we need to find the right $t$, for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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How to prove that a length is equal to the inradius of a triangle $D$ is the midpoint of the side $BC$ of the triangle $ABC$. The line joining $D$ and the incentre $I$ of the triangle intersects altitude $AA'$ at the point $P$.
Prove that the length of $AP$ is equal to the radius of the incircle of the triangle..
| Let $E$ be a tangency point with side $BC$ and let $c>b$.
Hence,
$$DE=BE-BD=\frac{a+c-b}{2}-\frac{a}{2}=\frac{c-b}{2},$$
$$BA_1=c\cos\beta=c\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2a}.$$
Thus, $$DA_1=BA_1-BD=\frac{a^2+c^2-b^2}{2a}-\frac{a}{2}=\frac{c^2-b^2}{2a}.$$
Now, since $\Delta DIE\sim\Delta DPA_1,$ we o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^7+y^7+z^7$ is divisible by $7^3$, then $x+y+z$ is divisible by $7^2$
Let $x, y, z$ be positive integers, and $7 \nmid xyz$. If $7^3|x^7+y^7+z^7$, show that $7^2|x+y+z$.
by Fermat's little theorem, $x^7 \equiv x \pmod7$, then $x^7+y^7+z^7\equiv x+y+z \equiv 0 $ (mod 7)
so we have $7 | (x+y+z)$.
what should I do nex... | HINT:
If $x+y\equiv0\pmod7,$ we are done
Otherwise
We can write $z=7a-x-y$
$$x^7+y^7+z^7=x^7+y^7+(7a-x-y)^7$$
$$(7a-x-y)^7\equiv-(x+y)^7+7(7a)(x+y)^6\pmod{7^3}$$
$$\implies-(x+y)^7+7(7a)(x+y)^6\equiv0\pmod{p^3}$$
As $7\nmid(x+y),$ $$x+y\equiv-49a\pmod{7^3}$$
which is impossible as $7\nmid(x+y)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Express $355$ as a sum of three squares. PROBLEM
Find all the ways to express $355$ as a sum of three squares.
MY ATTEMPT
By Legendre's three-square theorem, since $355$ is not of the form $n = {4^a}(8b+7)$ (for $a, b \in \mathbb{Z}$), then $355$ can be expressed as a sum of three squares.
WLOG, we can restrict to po... | I do not consider any order between $x,y,z$ here.
$x^2+y^2+z^2\equiv 1\pmod 3$ and since $(0,1,2)\xrightarrow{x^2}(0,1,1)$ then we can only be in the $0+0+1$ configuration, so two numbers are divisible by $3$.
$x=3a,\ y=3b$ gives $9(a^2+b^2)+z^2=355\iff z^2\equiv 4\pmod 9\iff z\equiv 2,7\pmod 9$
Since $19^2=361>355$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int (x^2+1)/(x^4+1)\ dx$ I have Divided the numerator and Denominator by $x^2$
to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:
$$
\int\frac{x^2+1}{... | Since $x^4+1=x^4+2x^2+1-2x^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$, we obtain
$$\frac{x^2+1}{x^4+1}=\frac{1}{2}\left(\frac{1}{x^2+\sqrt2x+1}+\frac{1}{x^2-\sqrt2x+1}\right).$$
Now, use $\int\frac{1}{1+x^2}dx=\arctan{x}+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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$\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$
Prove that \begin{align}\begin{vmatrix}
1 & a &bc \\
1& b & ac\\
1&c & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\
1& b&b^2 \\
1& b & c^2
\end{vmatrix}\\&=(c-a)(b-a)(c-b)\... | From the last column, we remove the common factor $abc$;
Then we multiply the first line by $a$, the second line by $b$ and the last line by $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis o... | Rotating the parabola $y=x^2$ by $\theta$ clockwise gives $v=u^2$, where
$$\left(u\atop v\right)=\left(\cos\theta\quad-\sin\theta\atop\sin\theta\quad\;\;\;\cos\theta\right)\left(x\atop y\right)$$
i.e.
$$x\sin \theta+y\cos\theta=(x\cos\theta-y\sin\theta)^2$$
Putting $\theta=\frac\pi 4$ gives
$$\frac 1{\sqrt2}(x+y)=\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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$33=\left\{a^3+b^3+c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$ I was just wondering, if I had an equation like:
$$33 = \left\{a^3 + b^3 + c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$$
What are the values of $(a, b \land c)$. Is there a way of proving or disproving that such integers of $a$, $b$, and $c$ exist to satisfy t... | Note in your case, the max value of $a,b,c$ is achieved when both others are zero, and hence is $33^{1/3} \approx 3.2$, so since they must be integers, you have $a,b,c \in \{0,1,2,3\}$.
Now it is easy to find all combinations of these in 4 variables. Without loss of generality, assume $a \ge b \ge c$, and you get the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Improper Integral Involving A Probable Partial Fraction Conversion I was having trouble evaluating the following integral:$$\int^{+\infty}_{-\infty} \frac{dx}{(x^2\pm ax+a^2)(x^2\pm bx+b^2)}$$
I have tried approaches that involve converting into partial fractions using thumb rule and without using it.I also tried the o... | We have
$$\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} = \frac{cx+d}{x^2+ax+a^2} + \frac{ex+f}{x^2+bx+b^2}.$$
Solving for $c,d,e,f$ one should find
\begin{align*}
\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} &= \frac{1}{(a-b)(a^2+ab+b^2)}\left(
\frac{x-b}{x^2+ax+a^2} + \textrm{something similar}\right).
\end{align*}
Now use the hint in t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Committee selection:Where is my mistake?
There are $5$ boys and $6$ girls. A committee of $4$ is to be selected so that it must consist at least one boy and at least one girl?
I consider $1$ boy can be chosen in $5$ ways.
Also, consider $1$ girl can be chosen in $6$ ways.
So, others $2$ can be chosen in $9C2$ ways.
S... | Let's denote the girls $\{g_1,g_2,\ldots,g_6\}$ and the boys $\{b_1,b_2,\ldots,b_5\}$.
You count some configurations more than once. The selection $g_1,g_2,b_1,b_2$ is counted four times namely as
\begin{align*}
(g_1,b_1),(g_2,b_2)\\
(g_1,b_2),(g_2,b_1)\\
(g_2,b_1),(g_1,b_2)\\
(g_2,b_2),(g_1,b_1)
\end{align*}
One way ... | {
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"url": "https://math.stackexchange.com/questions/2365017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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GRE Practice Question I need a calculus refresher! This question was one of the lowest percent correct on the practice (27%).
A curve in the xy-plane is given by
\begin{align*}
x &= t^2+2t \\
y &= 3t^4+4t^3
\end{align*}
for all $t > 0.$ The value of $\frac{d^2y}{dx^2}$ at the point $(8,80)$ is
(a) 4, (b) 24, (c) 32, (... | Solve the quadratic equation $t^2+2t-x=0$
to find $t=-1+\sqrt{1+x}$ and substitute into $y$ to get:
$$y=3(-1+\sqrt{1+x})^4+4(-1+\sqrt{1+x})^3.$$
Thus:
$$y'=12(-1+\sqrt{1+x})^3\cdot \frac{1}{2\sqrt{1+x}}+12(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}.$$
$$y''=-\frac{6}{2(1+x)^{3/2}}\cdot((-1+\sqrt{1+x})^3+(-1+\sqrt{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2366189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$
I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^... | Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2(A+60^\circ)+\cos^2A=1+\cos(A+60^\circ+A)\cos(A+60^\circ-A)=1+\dfrac{\cos(2A+60^\circ)}2$$
Using Werner's formula,
$$\cos(A+60^\circ)\cos A=\dfrac{\cos(A+60^\circ+A)+\cos(A+60^\circ-A)}2=?$$
Can you recognize $A$ here?
| {
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"url": "https://math.stackexchange.com/questions/2368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Quartic with 3 distinct roots Consider a quartic equation
$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$
The value of $k$ so that given equation has three real and distinct solutions can be equal to - i think that for any four degree to have 3 roots, its $f'(... | Hint
Observe that $0$ is not a root of this equation and since three real roots exist, therefore fourth must be real as well and there must be a repeated root.
Start with:
\begin{align*}
x^4 – kx^3 + 11x^2 – kx + 1 & = 0\\
x^2 – kx + 11 – \frac{k}{x} + \frac{1}{x^2} & = 0\\
\left(x+\frac{1}{x}\right)^2-k\left(x+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Understanding a Symmetry Argument
A rectangle has dimensions $a$ units by $b$ units with $a > b$. A diagonal divides the rectangle into two triangles. A square, with sides parallel to those of the rectangle, is inscribed in each triangle. Find the distance between the vertices (of the squares) that lie in the interior... | We can draw a picture of this situation as follows:
We need to find $c,d,e,f$ to know the interior vertices of the squares.
Can you see that this is rotationally symmetric, if we spin it around by $180^\circ$, it will be exactly the same problem?
This means that the distance between $(0,b)$ and $(c,f)$ is equal to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Limit $\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$ $$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$
I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fract... | Let $x-2=t$.
Hence,
$$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )=$$
$$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t}}\left(\frac{1}{\sqrt[3]{t+1}}+\frac{1}{\sqrt[3]{t-1}}\right)=$$
$$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t+1}\sqrt[3]{t-1}}\lim_{t\rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find derivative of $u = \sin(y^2 + u)$ using implicit differentiation Find $\frac{du}{dy}$, when $u = \sin(y^2 + u)$, using chain rule and expanding I get:
$1 = (2y\frac{du}{dy} + 1) \cos(y^2 + u)$
$2y \cos(y^2 + u)\frac{du}{dy} + \cos(y^2 + u) = 1$
$2y \cos(y^2 + u)\frac{du}{dy} = 1- \cos(y^2 + u)$
$\frac{du}{dy} = \... | Taking derivative towards $u$ w.r.t.$y$ is not $1$, but should be $\frac{d u}{d y}$
Thus, it instead should be:
$$\frac{du}{dy}=\cos(y^2 + u)(2y + \frac{du}{dy})$$
$$\frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2 + u)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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A game concerning Sudoku We (me and you) start with a blank $9\times 9$ blank square (as an empty Sudoku) and I fill the first three rows legally according to Sudoku rules. Is it always possible that you complete the Sudoku with this given three rows?
| Suppose you filled the first three rows according to Sudoku rules, let a1, a2, a3 be the 3x1 vectors in the top-left block; b1, b2, b3 be the next three 3x1 vectors in the top-middle block; and c1, c2, c3 be the last three 3x1 vectors in the top-right block.
Then consider:
\begin{array}{|ccc|ccc|ccc|}
\hline | & | & | ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Guessing a number Find the number $n$ which has exactly $3$ prime divisors: $3$, $5$, and $7$, and has $4$ divisors that are powers of $5$. Moreover, $n$ has as many divisors that are powers of $3$ as there are divisors that are powers of $7$. Finally, $24$ divisors of $n$ are multiples of $3$.
What I understood is tha... | "$24$ divisors of $n$ are multiples of $3$". So, $n/3$ itself has $24$ divisors which are each one third of a divisor of $n$.
To have four power of $5$ divisors and $m$ divisors apiece that are powers of $3$ and powers of $7$, the number must have the form $n=3^{m-1}×5^3×7^{m-1}$ counting $1=p^0$ as a power of each pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring limits Couldn't find this limit someone help me?
$$\lim_{x\rightarrow0} \frac{ (1+x)^{1/3} - (1-x)^{1/3}}{x}$$
I tried to take $x^{1/3}$ common from above expression
| Let $A=(1+x)^{1/3}$ and $B=(1-x)^{1/3}$ then
$$2x=A^3-B^3=(A-B)(A^2+AB+B^2)$$
Hence, as $x\to 0$, we have that $A\to 1$, $B\to 1$ and
$$\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x=\frac{A-B}{x}=\frac{2}{A^2+AB+B^2}\to\frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer. Question:
Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer.
My attempt:
Proof by contradiction:
Now, assume $ \sqrt{2k}$ is rational. Then, $ \sqrt{2k} = \frac{a}{b}$ where $a,b \in Z$, $b$ not equal $0$ and $a,b$ have no common factors.
$ \s... | You are almost there, suppose that $a$ and $b$ are relatively prime,
write $a^2=b^2(4m+2)=2b^2(2m+1)$, you deduce that $2$ divides $a^2$, and $a$, write $a=2a_1$, you have $2b^2(2m+1)=4a_1^2$, you deduce that $b^2(2m+1)=2a_1^2$. This implies that $2$ divides $b$, contradiction since $a$ and $b$ are relatively prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\sqrt{\phi }\right)\right|$ a radical? In this post, Reshetnikov gave the enormous even $80$-deg equation satisfied by,
$$x=\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{... | The scaled Bring quintic
$$x^5-\frac5nx+\frac4n=0\tag1$$
is solved by,
$$x =\frac45\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};n\right)$$
while the decic,
$$y^4(n y^2-5) (n y^2+5)^2 + \frac{16^2}n = 16 y^3 (2 n y^2+5)\tag2$$
is solved by,
$$y=\left(\frac45\right)^2\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$ Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$
I am not allowed to use induction. I do not know h... | Another proof
Let $f(x) = \frac{\log(x)}{x}$, then for $x, y > 0$
$$f(x y) = \frac{\log(x)+ \log(y)}{x y} = \frac{f(x)}{y} + \frac{f(y)}{x}$$
Derivating $n$ times with respect to $y$ yields
$$x^n f^{(n)}(x y) = f(x) \left(\frac{1}{y}\right)^{(n)} + \frac{f^{(n)}(y)}{x}$$
Now replacing $y$ by $1$ yields
$$x^n f^{(n)}(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM a... | If $a=3$ and $b=-1$ and $c=-2$ we get a value $1$.
But
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=-\frac{abc(a+b+c)\sum\limits_{cyc}(a^2-ab)}{\prod\limits_{cyc}(2a^2+bc)}=0.$$
Thus, $P=1$ for all $a+b+c=0$ and $\prod\limits_{cyc}(2a^2+bc)\neq0$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find all solutions to $x^3+(x+1)^3+ \dots + (x+15)^3=y^3$
Find all pairs of integers $(x, y)$ such that
$$x^3+(x+1)^3+ \dots + (x+15)^3=y^3$$
What I have tried so far:
The coefficient of $x^3$ is $16$ in the left hand side. It is not useful then to trying bound LHS between, for example, $(ax+b)^3$ and $(ax+c)^3$ a... | You are looking for $(x, y) \in \mathbb{Z}^2$ for which
$$ \sum_{i= 0}^{15} (x+i)^3 = y^3, \tag{0}$$
that is,
$$ \sum_{i= 0}^{15} \left( x^3 + 3 i x^2 + 3i^2 x + i^3 \right) = y^3, $$
that is,
$$ 16 x^3 + 3 \frac{15 (15+1)}{2} x^2 + 3 \frac{15 (15+1)(2 \times 15 + 1)}{6} + \left( \frac{15 (15+1)}{2} \right)^2 = y^3, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $... | Let this matrix be denoted as:
\begin{equation}
\mathbf{Q}=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
x & y & z
\end{array}\right]
\end{equation}
From the definition of the orthogonal matrix $\mathbf{Q}\,\mathbf{Q}^T = \mathbf{I}$:
\begin{equation}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \f... | There is a shortcut way to evaluate $$p(x):=(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4)\\ - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3).$$
It is clear that this is a polynomial of at most second degree, as all cubic term will cancel out*.
Then we perform the easy evaluations
$$p(1)=-6,\ p(2)=-2,\ p(3)=2,\ p(4)=6$$ and clearly $$p(x)=4x-10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\fra... | One might recall (from the proof of Kepler's first law, for instance) that
$$ \rho(\theta) = \frac{\frac{b^2}{a}}{1+\frac{c}{a}\cos\theta} $$
is the polar equation of an ellipse (with semi-axis $b<a$ and $c=\sqrt{a^2-b^2}$) with respect to a focus. The associated cartesian equation clearly is $\frac{(x+c)^2}{a^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\fr... | $$|2[1-(-\frac{1}2)^n]-2|<0.001\\
|-2(\frac{-1}2)^n|<0.001\\
+2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\
+2.\frac{1}{2^n}<\frac{1}{1000}\\
\frac{2^n}{2}>1000\\
2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), ... | We perform an orthogonal projection to map the ellipse to the unit circle.
Let the maximal area of our isosceles triangle be $\mathcal{A}$ which we wish to find, and let $\triangle ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\triangl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Maclaurin Series $ \int \frac{\sin(x)}{5x} $ I am supposed to evaluate the indefinite integral as an infinite series centered at $ x=0 $ and give the first five non-zero terms of the series.
$ \int \frac{\sin(x)}{5x} $
Here is what I have done so far:
$ g(x) = \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)... | Your result is perfectly good (before the expansion of the factorials, what I cannot check).
Forgetting the signs, the denominators are (as you wrote)
$$5 \qquad 5\times 3!\times 3=90\qquad 5\times 5!\times 5=3000 \qquad 5\times 7!\times 7=176400$$ $$\qquad 5\times 9!\times 9=16329600$$ Making then $$\frac{x}{5}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1... | $$\cot\alpha = -\frac{7}{24} \implies \tan\alpha = -\frac{24}{7}$$
Since $1 + \tan^2\alpha = \sec^2\alpha$,
$$\sec^2\alpha = 1 + \left(-\frac{24}{7}\right)^2 = 1 + \frac{576}{49} = \frac{625}{49} \implies |\sec\alpha| = \frac{25}{7}$$
Observe that $450^\circ < \alpha < 540^\circ \implies \cos\alpha < 0$. Hence,
$$\se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2393005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for ... | All of them fit in this infinite sequence of inequalities:
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots \ge 0 \ \forall x \in \mathbb{R} \\
\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} -\cdots > 0 \ \forall x>0 \\
\frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots > 0 \ \forall x \in \mathbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2394867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\la... | I get the following equations
$$2x-y-z=\lambda(2x-2) \\-x+2y-z=\lambda(2y+2) \\-x-y+2z=\lambda(2z+6) $$
or equivalently
$$\begin{bmatrix} 2-2\lambda & -1 & -1 \\-1 & 2-2\lambda & -1 \\ -1 & -1 & 2-2\lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}-2 \\ 2 \\ 6 \end{bmatrix} $$
If th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m}... | $(\sqrt{3}-3i)^6=27(1-\sqrt{3}i)^6=27\times 2^6\Big(\dfrac{1}{2}-\dfrac{\sqrt3}{2}i\Big)^6=27\times64\space\omega^6=27\times 64$
$\omega$ is a non-real solution of the equation $x^6=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
| Consider three cases:.
i) $\bf x\leq -1$. Then $x\leq -1$, $2^x< 1$, and the equation becomes
$\frac{1}{2}2^{-x} - 2^x = 1- 2^x + 1$, that is $2^{-x}=4$ which implies that $x=-2\leq -1$.
ii) $\bf -1<x<0$. Then $x>-1$, $2^x< 1$, and the equation becomes $2^x=2\cdot 2^{x} - 2^x = 1- 2^x + 1=2-2^x$, that is $2^x=1$ which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How many positive integer cubes are divisors of $1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!$? What I have tried:
$$N= 1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$
| Now, for each prime factor, how many possible exponents are there?
For example, for the prime factor $2$, the exponent can be $0,3,6,9,12,15,18$ or $21$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determining if a vector is in the row space I am just trying to determine if the vector $[0, 7, 4]$ belongs in the row space of the matrix $$A = \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}
$$
What I have done so far is created an augmented matrix like so
$$ \l... | Your working is fine.
Consider the matrix$$ \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}$$
Minus $2$ times of row $1$ plus row $2$ gives us row $3$, the row space is spanned by the first two rows.
Suppose $$a\begin{bmatrix} 1 & 2 & 0 \end{bmatrix} + b\begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\si... | Result: $\sin x\geq x-\dfrac{x^2}{2}\dots(*)$
Now $\sin \dfrac{2\pi}{9}=\dfrac{1}{\sqrt2}(\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})$
To show $\sin \dfrac{2\pi}{9}<\sqrt{\dfrac{3}{7}}\Leftrightarrow (\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})<\sqrt{\dfrac{6}{7}}\Leftrightarrow\sin\dfrac{\pi}{18}>\dfrac{1}{7}$
Usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Is this the only solution to the problem? Find all mutually non-equivalent $A(p,q)$ so that $(Aq\rightarrow \overline{p}) = (p+A)$ is true regardless of $p$ and $q$, where $\overline{x}$ is negation, $xy$ is conjunction, $x+y$ disjunction, $x\rightarrow y$ implication and $x=y$ equivalence.
\begin{matrix}p&q&A&(Aq\righ... | It seems to me you are correct up to here:
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
= 1.$$
It does not follow that
$A=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)}
= \overline p + \overline q,$ however.
Yes, it is true that
$$
(A = \overline p + \overline q)\rightarrow
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Smallest diameter of a balanced subset of the Hamming cube Let $\{0,1\}^n$ be the Hamming cube with the Hamming metric. It's a metric space of diameter $n$.
Let's call a set $B\subset \{0,1\}^n$ balanced if its center of mass is the center of the cube; that is, the average of all vectors contained in $B$ is
$(1/2,\do... | (This would probably be better as a comment than an answer but it looks like I can't comment yet.)
I think I can slightly improve on your partial results, with $d_7=4$ and (hence) $d_8=5$. (Since whenever $d_{2k-1} = k$ your bounds $(1)$-$(2)$ give $d_{2k}=k+1$.)
For the first claim, it's enough to find a balanced... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a so... | Like my answers in
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$
In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
let $S=\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
$2S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Find polynomials : $ xP(x-1)=(x-11)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ xP(x-1)=(x-11)P(x)$$
(Attempted work has been moved to answer)
| Substitute $x=0$, we have $P(0)=0$ and substitute $x=11$, we have $P(10)=0$,
so $P(x)$ has $0$ and $10$ as its roots, i.e., $\exists Q(x)$, $P(x)=x(x-10)Q(x)$
Since $ xP(x-1)=(x-11)P(x)$, so $x(x-1)(x-11)Q(x-1)=(x-11)x(x-10)Q(x)$
thus, $(x-1)Q(x-1) = (x-10)Q(x)$ ---[2]
Similarly, substitute $x=1$ and $x=10$ in [2], $P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
| Let $K>1$ be a constant. Let $y = K^x$. Then $\ln y = x \ln K$. So
$\dfrac{y'}{y} = \ln K$. Hence $$\frac{d}{dx}K^x = \ln(K) K^x \tag 1$$.
As a reality check, note that this gives you $\frac{d}{dx}e^x = e^x$.
Using L'Hospital's rule, we see that
$\displaystyle \lim_{x \to \infty}\dfrac{x}{K^x} =
\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
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What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be ... | Rather than give a new example, I'd like to present a way to expand on your own example. You mention that
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix},\quad
A^n =
\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}.$$
Though, as @Henrik mentions, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 26,
"answer_id": 25
} |
Show $f_{2n} = f_{n}^2 + 2f_{n-1}f_{n}$ where $f_n$ is the $n^{th}$ fibonacci number Note that for this problem $f_0 = 0, f_1 = 1, f_n = f_{n-1} + f_{n-2}$
My work so far is to write down the base case and the induction hypothesis. Then the problem becomes showing:
$f_{2n} = f_{n}^2 + 2f_{n-1}f_{n} \implies f_{2(n+1)} ... | Alternatively, using:
$$f_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right),$$
we get:
$$f_{2n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n}-\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\right),$$
$$f_n^2=\frac15 \left(\left(\frac{1+\sqrt{5}}{2}\right)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Logarithmic inequality with substitution I have this problem that I can solve halfway, but I'm struggling to find the interval for the solution.The inequality is this: $$\log_3(4^x+1)+\log_{4^x+1}(3)>2.5.$$
Now here is the method how I tried to solve this inequality:
$\log_3(4^x+1)+ \frac{1}{\log_{3}(4^x+1)}>2.5$
Subst... | $log_3(4^x+1)+log_{4^x+1}(3)>2.5$
$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$
Before I go any further, let me establish something. Assuming x is finite a real number, $4^x$>0. This means $4^x+1$>1. ln($4^x$+1)>0 if $4^x$+1 is greater than 1, which I have just established it is. Now, continuing from where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$ I was reading a solution for an analysis problem and they argued that $$\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$$ for $x\geq 1$ and $n\geq 1$, $x$ a real number and $n$ a natural number. Why is that?
| We have $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(x)+\ln(2)}{x^2}$ such that $n \in \mathbb{N} \geq 1$ and $x \in \mathbb{R} \geq 1$,
Using the fact that $\ln x+\ln y = \ln(x y)$ we get that $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(2x)}{x^2}$.
We will assume that $n\geq 3$ so $1+x^2 \ln n$ and $x^2$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}$
Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}.$
I'm currently trying to finish a task that requires me to use mathematical induction to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$. This is the first time that t... | You should use Wilsons theorem: It says that if $(m,n)=1$ then $m^{\phi(n)}\equiv$ 1 (mod $n$ ), where $\phi(n)$ is the number of natural numbers less than $n$ co-prime to $n$
Using this it follows that $m^{2\cdot 3^n} \equiv 1$ ( mod $3^{n+1}$ ) if $(m,3) = 1$
Now whats left is that $5$ is not a square mod 3 (one can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$a^2=ab+b^2+b+5$ has no integer solutions Prove that the equation $a^2=ab+b^2+b+5$ has no integer solutions.
My attempt:
$4a^2=4ab+4b^2+4b+20$
$4a^2-4ab+b^2 = 5b^2+4b+20$
$5(4a^2-4ab+b^2) = 5(5b^2+4b+20)$
$5(2a-b)^2=(5b+2)^2+96$
Let $2a-b = x$, $\;5b+2=y$
we get $\;5x^2-y^2=96$.
Please suggest on how to proceed.
| $$5x^2-y^2 = 84 \implies 2x^2-y^2 \equiv 0 \pmod{3}$$
which implies $3\mid x$ and $3\mid y$. But $9\nmid 84$.
Your original equation should be equivalent to $$5x^2-y^2 = 96$$
where $x=2a-b, y=5b+2$. But the above method works fine for $96$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the ... | Part 1:
$$\begin{align}
(1+x)^{2n}&=\sum_{m=0}^{2n}\binom {2n}mx^m\tag{1}\\
(1+x)^{2n}&=(1+x)^n(1+x)^n\\
&=\sum_{r=0}^n\binom nr x^r\sum_{s=0}^n\binom ns x^s\\
&=\sum_{r=0}^n\sum_{s=0}^n\binom nr\binom nsx^{r+s}\\
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\tag{2}\\
[x^m](2)=[x^m](1):\qquad
\sum_{r=0}^m\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solving a System of Two Differential Equations and getting the wrong answer Problem:
Solve the following system of differential equations.
\begin{eqnarray*}
5x' + y' - 3x + y &=& 0 \\
4x' + y' - 3x &=& -3t \\
\end{eqnarray*}
Answer:
To solve this, we use the operator method.
\begin{eqnarray*}
(5D-3)x + (D+1)y &=& 0 \\
... | This is the screen copy of your answer with the mistake pointed out :
NOTE :
I got :
\begin{eqnarray*}
x &=& c_1 e^{t} + c_2 e ^{3t} + t + \frac{7}{3} \\
y &=& - c_1 e^{t} - 3c_2e^{3t} + 3t - 1 \\
\end{eqnarray*}
which isn't exactly the expected result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | Because $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n=$$
$$=20(50^n-12^n)+38\cdot12^n$$ and since
$$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1}),$$
we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
} |
What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am hav... | $$a_0=1, \quad \text{and }a_n=\frac{x^{2n}}{2}\left(\prod_{k=0}^{n-1} \frac{2k+1}{2k+4}\right) \text{ for } n\ge1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find $\tan^{-1} (i\sqrt{2})$. Problem: Find $\tan^{-1} (i\sqrt{2})$.
My attempt: We must find $z \in \mathbb{C}$ such that $\tan z = i\sqrt{2}$.
$$\tan z = \frac{\sin z}{\cos z} = i\frac{e^{-iz} - e^{iz}}{e^{iz} + e^{-iz}}$$
Let $u = e^{iz}$. Then $$i\frac{u^{-1}-u}{u+u^{-1}} = i\sqrt{2}$$ so $$1-u^2 = \sqrt{2}(u^2 + 1... | A much easier solution is as follows:
$$
\tan^{-1}(z)=-i\tanh^{-1}(iz)\\
\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}\\
\begin{align}
\tan^{-1}(i\sqrt{2})
&=-i\tanh^{-1}(-\sqrt{2})=i\tanh^{-1}(\sqrt{2})\\
&=\frac{i}{2}[\ln(1+\sqrt{2})-\ln(1-\sqrt{2})]\\
&=\frac{i}{2}[\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-i(\pi+2n\pi)]\\
&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \g... | Start with
\begin{eqnarray*}
(2y^2z+2z^2x+2x^2y-yz^2-zx^2-xy^2-3xyz)^2 \geq 0
\end{eqnarray*}
This can be expanded out to give
\begin{eqnarray*}
4 \sum_{cyc} y^2 z^4 -4 \sum_{cyc} y^3 z^3 +\sum_{cyc} y^4 z^2 -4 \sum_{cyc} x y z^4 +\\14 \sum_{cyc} xy^2 z^3 -10 \sum_{cyc} x y^3 z^2 -3x^2 y^2 z^2 \geq 0
\end{eqnarray*}
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Evaluate $\int \frac{dx}{(x^2-x+1)(\sqrt{x^2+x+1})}$ Evaluate $$I=\int \frac{dx}{(x^2-x+1)(\sqrt{x^2+x+1})}$$
My Try:
we have $x^2-x+1=(x+w)(x+w^2)$ where $w$ is complex cube root of unity
I have splitted $I$ as
$$I=AI_1+BI_2$$ where $A,B$ are some constants
$$I_1=\int \frac{dx}{(x+w)\sqrt{x^2+x+1}}$$
By taylor's seri... | Substituting
$$ x = \frac{\alpha + \beta t}{1 +t}$$
we have:
$$ x^2- x+1 = \frac{(\alpha +\beta t)^2-(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$
$$ x^2+x +1 = \frac{(\alpha +\beta t)^2+(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$
Numbers $ \alpha, \beta $ we define like, that coefficients at $ t $ are zero.
Hence
$$ 2\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Minimise a sum of unknown function.
Let $f(x)$ be a real valued function such that $f(x) > 0\ \ \forall x
\in \Bbb R$. $f(x)$ is symmetrical about $x = 2$ and $x = 4$. If
$\displaystyle S = \sum^{50}_{r= 1} f(r + 4)$ and
$\displaystyle\prod^{50}_{r = 1}f(r) = 2^{50}$. Find the minimum value
of $S$.
$$\begin{... | From the symmetry at $x=4$, $f(5) = f(3)$ and from the symmetry at $x=2$, $f(3)=f(1)$. Therefore, $f(1)=f(3)=f(5)$. Similarly, we can show that for $\forall r:$
$$f(r+2)=f(r)$$
Consequently,
$$S=25 (f(1)+f(2))$$
Also, note that $\prod_{r=1}^{50}f(r)=(f(1)f(2))^{25}=2^{50}$. Thus,
$$f(1)f(2)=4$$
Therefore, $S$ is minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with a proofs question.. Q : If $x_1$ and $x_2$ are two solutions to quadratic equation $ax^2+bx+c = 0$, then show that
$$x_1 + x_2 = -\frac{b}{a}\qquad\text{and}\qquad x_1x_2 = \frac{c}{a}.$$
From this question I gathered that the discriminant $b^2-4ac > 0$
I also tried to use the AGM and I got the inequalit... | when you try to prove any thing try to think in the simple way first and start with what you already know.
$$x_1 = \frac{- b + \sqrt{b^2 - 4ac}}{2 a}$$
$$x_2 = \frac{- b - \sqrt{b^2 - 4ac}}{2 a}$$
Thus
$$x_1 + x_2 = \frac{-2b}{2a} = -\frac{b}{a};$$
the square root part cancel out because the had different signs.
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Integrate $\int \frac{1}{\sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}}dx$ For a physics assignment we are analyzing an anharmonic oscillator whose force is given by: $$ F =-kx + \frac{kx^3}{a^2},$$ where $k$ and $a$ are the spring constant and an arbitrary positive constant respectively. One of the quest... | Just observe $$\int \frac{1}{\sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}}dx = \int \frac{1}{\sqrt{\frac{k}{2ma^2}(x^2-a^2)^2}} dx$$
and then using substitution like $x = a \sin \theta$ to solve the integral ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Eigenvalues of $A^n$, $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$
Let $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$ and $\alpha_n$ and $\beta_n$ denote the two eigenvalues of $A^n$ such that $|\alpha_n|\geq |\beta_n|$. Then
*
*$\alpha_n\rightarrow \infty$ as $n\to\infty$
*$\beta_n\to 0$ as $n\to\infty$
*$\beta_n$ is pos... | It can be easily noticed that for all $n\geq 3$, $A^n=A^{n-1}+A^{n-2}$ and so $\alpha_n=\alpha_{n-1}+\alpha_{n-2}$ and $\beta_n=\beta_{n-1}+\beta_{n-2}$. Also note that
$$\alpha_1=\dfrac{1+\sqrt{5}}{2} \quad \alpha_2=\dfrac{3+\sqrt{5}}{2} \qquad \beta_1=\dfrac{1-\sqrt{5}}{2} \quad \beta_2=\dfrac{3-\sqrt{5}}{2} $$ Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Number of permutations of $9$ people of three nationalities in which no two people of the same nationality are adjacent $9$ different people must be put in a row.
Three of them are of nationality $X$, three are of $Y$, and the remaining three are of $Z$.
In how many combinations there will be no two people of the sam... | There are $9!$ ways to arrange nine individuals. From these, we must exclude those seating arrangements in which two people from the same country are seated in adjacent seats.
One pair of adjacent people: We have three ways to choose the nationality of the pair. We have $\binom{3}{2}$ ways to choose the people of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$ Find the $x$ :
$$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$$
$$x>0 , \to \lfloor \sqrt[3]{8x}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200 \\ \lfloor \sqrt[3]{8x}\rfloor \in \mathbb{Z}+\l... | We can write
$$
\left\{ \matrix{
0 < x\quad \Rightarrow \quad \left\lfloor {\root 3 \of {8x} } \right\rfloor + \left\lfloor {\root 3 \of {\left\lfloor {8x} \right\rfloor } } \right\rfloor = 200 \hfill \cr
x = - y < 0\quad \Rightarrow \quad \left\lfloor {\root 3 \of {8y} } \right\rfloor + \left\lfloor {\root ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls?
1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?
I am getting different answer from this solution manual. My argument is the followin... | The manual is right. The difference between your solution and the books solution is a factor of $2.$ That is ${7\choose1}{6\choose1} = 2{7\choose 2}$
You have placed 3 balls in one of 7 cells and 3 balls in one of 6 remaining. But you have double counted at this step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Turkevicius inequality From Zdravko Cvetkovski's Inequalities — Theorems, Techniques and Selected:
Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that:
$$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$
Please suggest how to prove this inequality using basic methods. No logarit... | Let $a\geq b\geq c\geq d$.
Thus, $$d(d-a)(d-b)(d+2c)+d(d-a)(d+2b)(d-c)+d(d+2a)(d-b)(d-c)\geq0$$ or
$$d^4-(ab+ac+bc)d^2+2abcd\geq0.$$
Thus, it's enough to prove that
$$a^4+b^4+c^4\geq(a^2+b^2+c^2-ab-ac-bc)d^2+a^2b^2+a^2c^2+b^2c^2$$ or
$$\sum_{cyc}(a^2-b^2)^2\geq\sum_{cyc}(a-b)^2d^2$$ or
$$\sum_{cyc}(a-b)^2((a+b)^2-d^2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | For odd $n$, $(-1)^{n-1}=1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=1^2+(-2^2+3^2)+(-4^2+5^2)+\cdots+(-(n-1)^2+n^2)\\
=1+\color{blue}{(-2+3)}(2+3)+\color{blue}{(-4+5)}(4+5)+\cdots+\color{blue}{(-(n-1)+n)}((n-1)+n)\\
=\color{blue}{1\cdot}\left(1+2+3+4+5+\cdots+(n-1)+n\right)\\
=\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gi... | Hint:
$$\sin x =\sqrt{1-\cos^2 x} =\sqrt{1-\frac{1}{1+\tan^2 x}} $$
$$\cos x =\sqrt{\frac{1}{1+\tan^2 x}} $$
Since
$$ 1+\tan^2 x =1+\frac{\sin x^2}{\cos^2 x} = \frac{\cos^2x+\sin x^2}{\cos^2 x} =\frac{1}{\cos^2 x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
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How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3... | We show that for any positive integer $n$
$$
S_n:=\sum_{j=1}^n\left(\sum_{k=j}^n\frac{1}{k}\right)^2+\sum_{k=1}^n\frac{1}{k}=\sum_{j=1}^{n} (H_n-H_{j-1})^2 +H_n=2n.$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$.
We have that
\begin{align*}
S_n&=nH_n^2-2H_n\sum_{j=1}^{n}H_{j-1}+\sum_{j=1}^{n}H_{j-1}^2+H_n\\
&=
nH_n^2-2H_n((n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$... | The RHS should be $$1^{ 2 }+3^{ 2 }+...+\left( 2n-1 \right) ^{ 2 }+\left( 2n+1 \right) ^{ 2 }=\underset { \left( n\left( 2n-1 \right) \left( 2n+1 \right) \right) /3 }{ \underbrace { 1^{ 2 }+3^{ 2 }+...+(2n-1)^{ 2 } } } +{ \left( 2n+1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } +{ \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $3^n$ > $n^4$ if $n\geq 8$ As the title states, I am tasked with proving $3^n$ > $n^4$ if $n\geq 8$
The base case is trivial to prove. It is obvious that $3^8 > 8^4$ since
$3^8 = 6561$ and $8^4 = 4096$, and $6561 > 4096$, thus the base case $n = 8$ is true.
Now for the Induction Hypothesis (IH) we let $n = m$ whi... | Pulling it together into a compact induction proof:
Base case: $3^8 = 9^4>8^4$
Inductive hypothesis: $3^k >k^4$ for some $k\ge 8$
$\begin{align}
\text{Observe }(k+1)^4 &= k^4 + 4k^3 + 6k^2 + 4k + 1 \\
&< k^4 + (4+6+4+1)k^3 \\
&< k^4 + 16k^3 \\
& \le 3k^4
\end{align}$
Then $3^{k+1}=3\cdot3^k>3\cdot k^4> (k+1)^4$ as requ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
better upper bound of Cauchy - Schwarz inequality for 4 variables Given $a,b,c,d,x,y,z,w \geq 0$.
By the Cauchy - Schwarz inequality we have that:
\begin{equation}
\label{1}\tag{1}
\left( ax + by + cz + dw \right) ^{2} \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) .
\en... | Both your inequalities are wrong.
$$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right)
\leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2}
$$ is wrong for $a=y=c=w\rightarrow0^+$.
$$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a m... | Let me try. Note that $2(ab+bc+ca) = a^2+b^2+c^2-(a+b+c)^2 = 2$, then $ab+bc+ca=1$.
We have $$(a-3)^3 + (b-2)^3+(c-2)^3 = 0$$
$$(a-3)^3+(b+c-4)(b^2+c^2+4-2b-2c-bc) = 0$$
$$(a-3)^3 + (-2-a)(6-a^2+4-2(2-a)-(1-a(2-a)))=0$$
$$(a-3)^3-(2+a)(-2a^2+4a+5)=0$$
$$3a^3-9a^2+14a-37=0$$
$$3(a-1)^3+5(a-1)-29=0$$
Solve this equation,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
A simple factoring problem $3x^3 - x^2 -12x + 4$ I'm stuck on a simple factoring problem of
$$3x^3 - x^2 -12x + 4$$
I keep coming up with
$$x^2(3x - 1) -4(3x - 1) = (3x - 1) (x^2 - 4) = 3x^3 -12x - x^2 + 4 =
3x^3 - x^2 - 12x + 4$$
but it doesn't seem to be right, can anyone point out my mistake?
| From $x^2(3x - 1) -4(3x - 1)$, factor out $(3x-1)$ to get $(x^2-4)(3x-1)$.
Knowing that $a^2-b^2=(a+b)(a-b)$, we get $(x-2)(x+2)(3x-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that there exist no natural number k such that $3^k+5^k$ is a square of an integer number? How to prove that there exist no natural number $k$ such that $3^k+5^k$ is a square of an integer number
| An elementary solution without modular arithmetic. :)
If $k$ is odd, then write
$$a^2 = 3^k+5^k= 8(3^{k-1}-3^{k-2}5+...+5^{k-1})$$
Since expression in bracket is odd (we have odd odd numbers) we have $8|a^2$ but $16\not|a^2$. A contradiction.
Say $k$ is even. Then $k=2n$ so $(a-3^n)(a+3^n)= 5^{2n}$, so $a-3^n= 5^{x}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Why does the angle between a Pair of Straight Lines depend only on the Homogeneous Part? The most general form of a quadratic is: $ax^2 + by^2 + 2gx + 2fy + c + 2hxy= 0 $ and that for a homogeneous second degree equation is : $ax^2 + by^2 + 2hxy=0$
I derived the formula $$\tan \theta = \left|\dfrac{2\sqrt{h^2 - ab}}{a... | For $2^\circ$ homogeneous equation $ax^2+2hxy+by^2=0$, we say $m_1+m_2=-\dfrac{2h}{b}$ where $m_1$ is the slope of first line and $m_2$ is the slope of second line.
Proof:
$$ax^2+2hxy+by^2=0$$
$$a+2h\dfrac{y}{x}+b\dfrac{y^2}{x^2}=0$$
$$bm^2+2hm+a=0$$
$$m_1+m_2=-\dfrac{2h}{b}$$
But if we talk about general $2^\circ$ equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\... | Hint :$$a=\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=b+c \\a^3=b^3+c^3+3bc(b+c)\\
a^3=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}}.\sqrt[3]{18-5\sqrt{13}}(a)\\a^3=36+3\sqrt[3]{324-325}a\\a^3=36-3a$$solve for a
$$a^3+3a-36=(a-3)(a^2+3a+12)=0 \to a=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Limit as $x\to -\infty$ for function having square root
Evaluate
$$\lim_{x\to -\infty} \frac{\sqrt{3x^2+2}}{x-2}.$$
My work: when I solved
$$\lim_{x\to -\infty} \frac{\sqrt{3+2/x^2}}{1-2/x}$$
I got answer $\sqrt{3}$
but if we take $|x|$ then I got $-\sqrt{3}$
what should be answer ?
$\sqrt{3}$ or $-\sqrt{3}$ ?
| Note that $\sqrt{x^2}=|x|$ and, for $x<0$, $|x|=-x$. Therefore, as $x\to -\infty$,
$$\frac{\sqrt{3x^2+2}}{x-2}=\frac{\sqrt{x^2}\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=\frac{|x|\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=-\frac{\sqrt{3+\frac{2}{x^2}}}{1-\frac{2}{x}}\to -\sqrt{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the number of solutions of x+y+z=17? Find the number of solutions of $x+y+z=17$ where $2\le x\le 5, 3\le y \le 6, 4\le z\le7$.
My approach:
The number of solutions with the indicated constraints is the coefficient of $x^{17}$ in the expansion of
($x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$
I have changed... | If $x,y,z \in \mathbb{Z}$ and then the constraints you had; then consider the following case where $x,y,z$ are the most they can be, i.e. $5$, $6$ and $7$:
$$5 + 6 + 7 = 18$$
$18$ is one more than $17$, which means that our solutions will be when only $x$ or only $y$ or only $z$ is one less than their maximum. We have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to get the same answer for integral $\int\frac{dz}{\sqrt{1+z^2}}$ as in textbook I have some answer:
$$\int\frac{dz}{\sqrt{1+z^2}}=\ln ( z + \sqrt {1 + z^2})+C$$
But I can't get the same expression. So here what I've done and got stuck. How can I get the same answer?
\begin{align}
& \int\frac{dz}{\sqrt{1+z^2}}=\le... | I'm not sure about how obvious this seems, but you can just make the direct substitution $z=x+\sqrt{1+x^2}$ in$$\int\frac {\mathrm dx}{\sqrt{1+x^2}}$$
Because$$\frac {\mathrm dz}{\mathrm dx}=1+\frac {x}{\sqrt{1+x^2}}=\frac {z}{\sqrt{1+x^2}}$$
This can be noted because the derivative of $\sqrt{1+x^2}$ is $x/\sqrt{1+x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2445250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
... | Define $\vec e_i = (0, \dots, 1, 0, \dots)$ where the 1 is in the $i^\text{th}$ position. Then note that your question is
$$ T(-5\vec e_1 + 4 \vec e_2 + 4 \vec e_3) = -5\pmatrix{3 \\ 1 \\ 4} + 4\pmatrix{-1 \\ -1 \\ 3} + 4\pmatrix{4 \\ -3 \\ -1} = \pmatrix{-3 \\ -21 \\ -12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$ and these numbers are roots of $x^3 - 8x + b = 0$ Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$, that they are roots of $x^3 - 8x + b = 0$
I find that $a+b=5x(-x+3) $
Also I tried to solve second equation to get roots, depending on b, but m... | Let $P(x) = x^3-5x^2+7x-a$ and $Q(x) = x^3-8x+b$ and $\gamma, \delta$ be their common roots.
We can decompose them as
$$\begin{cases}
P(x) &= (x-\alpha)M(x),\\
Q(x) &= (x-\beta)M(x)
\end{cases}\quad\text{ where }\quad M(x) = (x-\gamma)(x-\delta)
$$
Subtract them, we get
$$5x^2-15x+b+a = Q(x)-P(x) = (\alpha-\beta)M(x)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because o... | Reduce the rational degrees with successive substitutions as follows
$$\begin{align}
\int_{-\infty}^{\infty}\frac{(x^{4}+1)^{2}}{x^{12}+1}dx=&\ 2\int_{0}^{\infty}\frac{x^{4}+1}{x^{8}-x^{4}+1}\overset{x\to\frac1x}{dx}=\int_{0}^{\infty}\frac{(x^{4}+1)(x^2+1)}{x^{8}-x^{4}+1}
\overset{ x-\frac1x\to x}{ dx} \\
=& \ 2\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What is the minimum value of? On positive reals if $3x+4y+7z=1$ what is the minimum value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$ I have tried using the arithmetic mean and harmonic mean inequality but I failed. Not good at inequalities though. Please help.
| From $3x+4y+7z=1$ I got $z= \frac{1}{7} (-3 x-4 y+1)$ and plugged in
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
gives $$f(x,y)=\frac{7}{1-3 x-4 y}+\frac{1}{x}+\frac{1}{y}$$
$$f'_x=\frac{21}{(1-3 x-4 y)^2}-\frac{1}{x^2};\;f'_y=\frac{28}{(1-3 x-4 y)^2}-\frac{1}{y^2}$$
$f'_x=0$ gives $21x^2=(1-3x-4y)^2$ and $f_y=0\to 28y^2=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\... | $$\lim_{x\rightarrow-\infty}\left(\sqrt{x^2+3x}+x\right)$$
Multiply by the conjugate
\begin{aligned}
\sqrt{x^2+3x}+x &= \left(\sqrt{x^2+3x}+x\right) \cdot \left(\frac{\sqrt{x^2+3x}-x}{\sqrt{x^2+3x}-x}\right)\\
&= \frac{\left(\sqrt{x^2+3x}\right)^2-x^2}{\sqrt{x^2+3x}-x}\\
&= \frac{x^2+3x-x^2}{\sqrt{x^2+3x}-x}\\
&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $32x^2 -y^2 = 448$ I am trying to find all integer solutions to the following equation:
$$32x^2 - y^2 = 448$$
This is what I have tried so far:
The equation describes a hyperbola, and so I try the usual trick of intersecting the curve with a line of rational slope to find rational solutions first.
Knowing the poi... | Note that you can simplify and substitute in the following way:
$$\begin{array}{lll}
y^2 = 2^5(x^2-14)&&\text{substitute } y = 2^2z \\
z^2 = 2(x^2-14)&&\text{substitute } z = 2a\\
2a^2 = x^2-14&& \\
2(a^2+7) = x^2&&\text{substitute } x = 2b\\
a^2+7 = 2b^2&&\text{substitute } a = 2c+1\\
2(c^2+c + 2) = b^2&&\text{substit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm... | note that $$\left(\frac{-1}{2}y^{-5/2}+\frac{1}{2}y^{5/2}\right)^2+1=1/4\,{\frac { \left( y+1 \right) ^{2} \left( {y}^{4}-{y}^{3}+{y}^{2}-y
+1 \right) ^{2}}{{y}^{5}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
find the equation of the cone $z=\sqrt{x^2+y^2}$ in spherical coordinates I have the following...
$$z=\sqrt{x^2+y^2}$$
I need to write this as an equation in spherical coordinates.
I know that $p^2 = x^2+y^2+z^2$
and that...
$$x = p\sin\phi \cos\theta$$
$$y=p\sin\phi \sin\theta$$
$$z = p\cos\theta$$
The answer is $\phi... | The idea is to plug in the values of $x$, $y$ and $z$ in $$z = \sqrt{x^2+y^2}.$$
Specifically, by using the given expressions, we get
$$p \cos \phi = \sqrt{p^2\sin^2\phi \cos^2 \theta + p^2\sin^2\theta \sin^2 \phi}$$ $$p \cos\phi = \sqrt{p^2\sin^2 \phi \ (\sin^2 \theta + \cos^2 \theta)} $$
$$p \cos\phi = p \sin \phi$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt... | Try $x = \sin u$, so that $dx = \cos(u)\, du$.
Then use the fact that: $$\sqrt{1+\sqrt{1-\sin^2 u}} = \sqrt{1+\cos u} =\sqrt{2} |\cos\tfrac{u}{2}|$$
The integral converts to:
$$\begin{align}
I &= \int_{0}^{\tfrac{\pi}{2}} \frac{\sqrt{2} \cos\tfrac{u}{2} \, \cos u}{1+\sin^2 u} du \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Does $A,B>0$ imply that the projection onto the positive subspace of $A-B$ is smaller than $A+B$ for positive semidefinite matrices? I am trying to find a counterexample for the following matrix inequality which I suspect not to hold:
$$P^+(A-B)\leq A+B$$
where $A,B$ are positive semidefinite matrices and $P^+(X)$ the ... | Let $A=\frac{1}{10}\begin{bmatrix} 2&4\\4&8\end{bmatrix},$ $B=\frac{1}{10}\begin{bmatrix}9&-3\\-3&1\end{bmatrix}.$ Then $A-B=\frac{1}{10}\begin{bmatrix}-7&7\\7&7\end{bmatrix},$ which has eigenvalues $\pm \sqrt{\frac{49}{50}},$ with corresponding eigenvectors $\begin{bmatrix}1\\1\pm\sqrt{2}\end{bmatrix}.$ Thus $P^{+}(A-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} ... | if $\sqrt{a+b} = \sqrt a + \sqrt b$ for some $a,b \in \mathbb{R^{\geq 0}}$, then $a+b = a + 2\sqrt{a}\sqrt{b} + b$, then $\sqrt{a}\sqrt{b} = 0$. So $a =0$ or $b=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Evaluate The Arc Length
Let $$\alpha(t)=(\alpha \cos^3t,\alpha \sin^3 t)$$ Evaluate the arc length when $\alpha>0$ and $t\in[0,2\pi]$
$$\alpha'(t)=(-3\alpha \cos^2{t} \sin{t} ,3\alpha \sin^2t\ \cos t)$$
$$||\alpha'(t)||=\sqrt{(-3\alpha \cos^2{t} \sin{t})^2+(3\alpha \sin^2t\ \cos t)^2}=\sqrt{9\alpha^2\cos^4{t} \sin^2{... | Here another way to do this, this time in the complex plane. Let the astroid be given by
$$z=a(\cos^2 t+i~b\sin^3 t),\quad t\in[0,2\pi]$$
The arc length is given by
$$
\begin{align}
s
&=\int_0^{2\pi}|\dot z|~dt\\
&=4\int_0^{\pi/2}|\dot z|~dt,\quad \text{by symmetry}\\
&=4a\int_0^{\pi/2}|-3\cos^2 t~\sin t+3\sin^2 t~\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
not understanding identity theorem for polynomials
Suppose $f(x)$ is a polynomial such that $f(x) = a_nx^n +
a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots +a_1 x + a_0$ and there are
at least $n+1$ different values of $x$ for which $f(x) =0$.
(a) If $a_n\neq 0$, then what does the Fundamental Theorem of Algebra
tell us ab... | A nonzero polynomial of degree $d$ has at most $d$ roots. And the expression
$$\sum_{k=0}^n a_kx^k$$
is a polynomial of degree at most $n$. Hence if it has $n+1$ roots, it must be identically $0$.
Then if there are two polynomials of degree at most $n$, let $p(x)$ and $q(x)$, achieving the same value at $n+1$ points, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimizing $\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$ for nonnegative parameters with $a+b+c+d=4$
Minimize
$$\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$$
where $a$, $b$, $c$, $d$ are nonnegative and $a+b+c+d = 4$.
I know the minimum is $2/3$ when, say, $a=b=2$. I found thi... | For $(a,b,c,d)=(2,2,0,0)$ we get a value $\frac{2}{3}$.
We'll prove that it's a minimal value.
Indeed, by AM-GM
$$\sum_{cyc}\frac{a}{b^3+4}=\frac{1}{4}\sum_{cyc}\left(a-\frac{ab^3}{b^3+4}\right)=\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{2b^3+8}\right)\geq$$
$$\geq\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{3\sqrt[3]{b^6\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solu... | Using $ \sin^2 \alpha + \cos^2 \alpha =1$
\begin{eqnarray*}
\sin^4 \alpha + \cos^4 \alpha =(\sin^2 \alpha + \cos^2 \alpha)^2 -2\sin^2 \alpha \cos^2 \alpha = 1-2\sin^2 \alpha \cos^2 \alpha \\
\sin^6 \alpha + \cos^6 \alpha =(\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha -\sin^2 \alpha \cos^2 \alpha+ \cos^4 \alpha) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 5
} |
Prove identity in a triangle I want to show that if $ABC$ is a triangle then
$$\sin^2(A/2)+ \sin^2(B/2) + \sin^2(C/2) =1-2\sin(A/2) \sin(B/2) \sin(C/2)$$
Well I eventually got it after much algebra, but I am looking for a shorter solution, or maybe even a geometric one?
| As $\dfrac A2+\dfrac B2=\dfrac\pi2-\dfrac C2,\sin\dfrac C2=\cos\dfrac{A+B}2$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2$$
$$=1-\left(\cos^2\dfrac A2-\sin^2\dfrac B2\right)+\sin^2\dfrac C2$$
$$=1-\cos\dfrac{A+B}2\cos\dfrac{A-B}2+\cos^2\dfrac{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac... | Write $(x+1)^2 = (x-1)(ax+b)+c$ and plug in $x=1$, you get $4=0+c$ so $c=4$. Now
$$ (x+1)^2-4 = x^2+2x-3 = (x-1)(x+3)$$ so $a= 1$ and $b=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would ... | It's $$5(x^3-1)^2=(\sqrt3(2x^3-x)-1)^2$$ and from here
$$(5(x^3-1)^2-3(2x^3-x)^2-1)^2=12(2x^3-x)^2,$$
which is
$$(7x^6-12x^4+10x^3+3x^2-4)^2-12x^2(2x^2-1)^2=0$$ or
$$49x^{12}-168x^{10}+140x^9+186x^8-240x^7-76x^6+60x^5+153x^4-80x^3-36x^2+16=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How can I find the points of intersection of $f(x)=\sin(\sqrt{x^2+4})$ and $g(x)=4-x^2$ analytically How can I find the points of intersection of $f(x)=\sin(\sqrt{x^2+4})$ and $g(x)=4-x^2$ analytically without having to resort to a graph? Thank you very much
I know that $-1\leq\sin(\sqrt{x^2+4})\leq 1$, with which $-1\... | Just to add a few things after Jam's answer.
Sooner or later, you will learn than, better than with Taylor expansions, functions can be approximated using Padé approximants which, built around $u=u_0$, are in the form of $$f(u)=\frac{\sum_{i=0}^m a_i (u-u_0)^i } {1+\sum_{i=1}^n b_i (u-u_0)^i}$$ Using, for simplicity, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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