Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
If $f(x)\mod (x-1)=5$, $f(x)\mod (x+1)=3$, $f(x)\mod (x+2)=5$ then find the remainder when $f(x)$ is divided by $x^3+2x^2-x-2$
I don't have any idea as to how to proceed with this questio... | The final condition implies that the divisor is of order $3$, so the remainder is of order $2$. We can write $$f(x)=(x-1)(x+1)(x+2)Q(x) +ax^2+bx+c$$ We have $f(1) =5$, so $$a+b+c=5...(1)$$ and similarly $f(-1)=3$ and $f(-2)=5$, so , $$a-b+c=3...(2)$$ and $$4a-2b+c=5$$ Solving for $a,b,c$, we get $a=1, b=1, c=3$. Hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Mean and mode of a Beta random variable A continuous random variable is said to have a $Beta(a,b)$ distribution if its density is given by
$f(x) = (1 / \text{B}(a,b))x^{a-1} (1-x)^{b-1}$ if $0 < x < 1$
Find mean , var, mode if $a = 3, b = 5.$
This is throwing me off with the beta distribution. I'm not sure if it chang... | \begin{equation}
\begin{array}{rcl} f^\prime(x)=\dfrac{df(x)}{dx}& = & \dfrac{1}{B(a,b)}\dfrac{d}{dx}x^{a-1}(1-x)^{b-1}\\
& = & \dfrac{1}{B(a,b)}\left[(a-1)x^{a-2}(1-x)^{b-1}-x^{a-1}(b-1)(1-x)^{b-2}\right]\\
& = & \dfrac{1}{B(a,b)}x^{a-2}(1-x)^{b-2}\left[(a-1)(1-x)-(b-1)x\right]\\
\end{array}
\end{equation}
\begin{equa... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find two integer numbers given the LCM and the difference between them I started a Computer Science degree this year and I have a subject called 'Discrete Mathematics' which this unit (divisibility) is annoying me a lot, because I can't find the key to solve the problems as like as other units and subjects.
The proble... | In the search of a more elegant solution rather than brute force, we know that as $\text{lcm}(a,b)=2^2\cdot 3\cdot 5^2\cdot 13$ that if both $a$ and $b$ had a factor of $13$ that $a-b$ must also be divisible by $13$. Similarly if both $a$ and $b$ had a factor of $5^2$ that $a-b$ must be divisible by $5^2$.
Since $a-b=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Mathematical Induction with Exponents: $1 + \frac12 + \frac14 + \dots + \frac1{2^{n}} = 2 - \frac1{2^{n}}$ Prove $1 + \frac{1}{2} + \frac{1}{4} + ... + \frac{1}{2^{n}} = 2 - \frac{1}{2^{n}}$ for all positive integers $n$.
My approach was to add $\frac{1}{2^{n + 1}}$ to both sides for the induction step. However, I got ... | Let $S(n)$ be the statement: $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2^{n}}=2-\dfrac{1}{2^{n}}$
First do basis step:
$S(1):\hspace{5 mm}$ $1+\dfrac{1}{2^{1}}=\dfrac{3}{2}$
and$\hspace{9 mm}$ $2-\dfrac{1}{2^{1}}=\dfrac{3}{2}$
Then do inductive step:
Assume $S(k)$ is true, or $1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double Integrals - Finding the volume of a unit disk under a function For a Unit Disk
$x^2 + y^2 \ge 1$
And for a function of $x$ and $y$
$f(x, y) = 3 + y - x^2$
I want to find the volume underneath the function bound by the unit disk.
At first I integrated the function with respect to $y$, and made the upper bound $\s... | The volume is, by definition:
$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dydx\stackrel{(**)}=\int_{-1}^1(3-x^2)\left(2\sqrt{1-x^2}\right)dx=\int_{-1}^1\left(6-2x^2\right)\sqrt{1-x^2}dx$$
(**) Explanation: we have
$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(3+y-x^2)dydx=\int_{-1}^1\left(\left.(3-x^2)y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$
Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$
where $a,b,c>0$
My attempt:
I couldn't proceed after that. Please help me in this regard, thanks!
| If you want to use the weighted AM-GM inequality, considering
\begin{align*}
\frac{bc+ca+ab}{a+b+c} &=
\frac{2(bc+ca+ab)}{2(a+b+c)} \\
&= \frac{(b+c)\color{red}{\boldsymbol{a}}+
(c+a)\color{green}{\boldsymbol{b}}+
(a+b)\color{blue}{\boldsymbol{c}}}{(b+c)+(c+a)+(a+b)} \\
\left( \frac{bc+ca+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$
If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that
$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$
I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know ho... | replace $c$ with $a+b$ and it should simplify.
$a^4+b^4+c^4 = a^4+b^4+(a+b)^4 = 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$
$2(ab)^2+2(ac)^2+2(bc)^2 = 2a^2b^2 + 2a^2(a^2+2ab+b^2) + 2b^2(a^2+2ab+b^2) \\
= 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?
| Suppose there actually is such a triple, $(x,y,z)$, of real numbers. Let $u$ be a variable and let's force $x$, $y$, and $z$ to be the roots of a polynomial. We would consider \begin{align*}
0 &= (u-x)(u-y)(u-z) \\
&= u^3 - (x+y+z)u^2 + (xy + xz + yz) u - (x y z) \\
&= u^3 - u^2 + (xy + xz + yz) u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
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"answer_id": 14
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Show that $(1+x)^{1-x}(1-x)^{1+x}<1$ If $x$ is a positive proper fraction. Prove that $$(1+x)^{1-x}(1-x)^{1+x}<1$$
Actually this question has two parts I can't do the $1^{st}$ part but the $2^{nd}$ part was quite easy with respect to the $1^{st}$ one. The $2^{nd}$ was to show that $$a^bb^a<(\frac{a+b}{2})^{a+b} $$ I ob... | Since $\log$ is concave,
$$
\begin{align}
\frac{1-x}2\log(1+x)+\frac{1+x}2\log(1-x)
&\le\log\left(\frac{1-x}2(1+x)+\frac{1+x}2(1-x)\right)\\
&=\log\left(1-x^2\right)
\end{align}
$$
Therefore,
$$
\color{#090}{(1+x)^{1-x}(1-x)^{1+x}}\le\color{#C00}{\left(1-x^2\right)^2}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Simplifying the second derivative I need to simplify the below with respect to $y''$. I'm given: $$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}$$
The final result should look like:
$$-\frac{3x^2(y^4+x^4)}{y^7}$$
Here are my steps:
*
*Combine term: $-3x^3y^2\left(-\frac{x^3}{y^3}\right)$.
$$-3x^3y^2\... | $$y''=-\frac{3x^2y^3-3x^3y^2\left(-\frac{x^3}{y^3}\right)}{y^6}=-\frac{3x^2y^3+3x^3\left(\frac{x^3}{y}\right)}{y^6}=y''=-\frac{\left(\frac{3x^2y^4+3x^6}{y}\right)}{y^6}=-\frac{3x^2y^4+3x^6}{y^7}=-\frac{3x^2(y^4+x^4)}{y^7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Is $\int_0^\infty \vert \sin x \vert^{x^2} \ dx$ convergent? To study the question, I'm looking at the convergence of the series
$$u_k = \int_{k \pi}^{(k+1) \pi}\vert \sin x \vert^{x^2} \ dx,$$ using the inequalities
$$2 \int_0^{\frac{\pi}{2}} \sin^{(k+1)^2\pi^2} x \ dx \le u_k \le 2 \int_0^{\frac{\pi}{2}} \sin^{k^2 \p... | I finally found another solution based on more basic technics.
$$\begin{align}
v_k&= \int_0^{\frac{\pi}{2}} \sin^{k^2 \pi^2} x \ dx = \int_0^{\frac{\pi}{2}} \cos^{k^2 \pi^2} x \ dx\\
&\ge \int_0^{\frac{\sqrt{2}}{k \pi}} \cos^{k^2 \pi^2} x \ dx\\
&\ge \int_0^{\frac{\sqrt{2}}{k \pi}} (1-\frac{x^2}{2})^{k^2 \pi^2} \ dx\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26}=\frac{2}{3}(2^{27}+1)$ I need to prove the following identity
$$
\binom{28}{2}+\binom{28}{5}+\binom{28}{8}+\cdots+\binom{28}{26} = \frac{2}{3}(2^{27}+1).
$$
I have tried to use the fact that $\binom{m}{n}=\binom{m}{m-n}$ but it doesn't help.
| Hint. Let $w=\exp(2\pi i/3)$. Then $w^3=1$ and $1+w+w^2=0$, and hence
$$
(1+1)^{28}+w(1+w)^{28}+w^2(1+w^2)^{28}=3 \left(\binom{28}{2}+\binom{28}{5}+\cdots+\binom{28}{26}\right),
$$
since the sum above is a real number.
Next
$$
1+w=-w^2\quad\Rightarrow\quad(1+w)^{28}=w^{54}=w^2\\
1+w^2=-w\quad\Rightarrow\quad(1+w^2)^{28... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{... | For the maximum, note that:
$$9=7a^2-9ab+7b^2 = \frac{5}{2}(a^2+b^2) + \frac{9}{2}(a-b)^2 \ge \frac{5}{2}(a^2+b^2)$$
[ EDIT ] For the minimum, note that:
$$9=7a^2-9ab+7b^2 = \frac{23}{2}(a^2+b^2) - \frac{9}{2}(a+b)^2 \le \frac{23}{2}(a^2+b^2)$$
| {
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"url": "https://math.stackexchange.com/questions/2072169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | I don't think there can ever be a mixed fraction of the form $n\frac{n^2+5}{4}$ if $n$ $\in$ $\mathbb{N}$. Please note that if it were a mixed fraction then $n^2+5$ would denote the remainder while $4$ is the divisor and this would never be possible as for $n$ $\in$ $\mathbb{N}$, $n^2+5 \gt 4$ always.
Hence, this expr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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if $xP(x)=x^t+P(x-1)$ find the $t\equiv ?\pmod 3$ Let polynomials $P(x)\in Z[x]$,and such $\deg{(P(x))}=t-1$,and such for any real $x$ have
$$xP(x)=x^t+P(x-1)$$
Find the $t\equiv ?\pmod 3$?
I try Let $P(x)=x^{t-1}+a_{t-2}x^{t-2}+\cdots+a_{1}x+a_{0}$ where $a_{i}\in Z$
so we have
$$a_{t-2}x^{t-1}+a_{t-3}x^{t-2}+\cdot... | Let $P(x) = \displaystyle\sum_{k = 0}^{t-1}a_kx^k$ where $a_k \in \mathbb{Z}$. Also, for $r = 0,1,\ldots,5$ define $S_r = \displaystyle\sum_{\substack{k = 0\\k \equiv r \pmod{6}}}^{t-1}a_k$.
Note that for $x = e^{i\tfrac{\pi}{3}} = \tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i$ we have $x-1 = -\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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in a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is In a triangle $ABC,$ if $2c^2=a^2+b^2,$ then largest possible degree measure of angle $C$ is
cosine formula $\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{c^2}{2ab}<\frac{(a+b)^2}{2ab}=\frac{a^2+b^2}{2ab}+1$
wan,t be ... | We know that $$\cos C =\frac {a^2+b^2-c^2 }{2ab} =\frac {c^2}{2ab} =\frac {a^2+b^2}{4ab} $$ Now applying the AM-GM inequality, we get, $$\frac {a^2+b^2}{4ab}\geq \frac {1}{2} \Rightarrow C\leq \frac {\pi}{3} $$ Hope it helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A simpler proof of $(1-x)^n<\frac{1}{1+nx}$ I proved the following inequality:
Let $x\in\mathbb{R}, 0<x<1, n\in\mathbb{N}$, then $(1-x)^n<\frac{1}{1+nx}$
however, judging from the context in the exercise book, I feel like there is a much simpler way to prove it, but I can't see it. So I'm asking for that simpler alte... | First note:
$$\frac{1+nx}{1-x}=\frac{(1+nx)(1+x)}{(1-x)(1+x)}=\frac{1+(n+1)x+nx^2}{1-x^2}\ge\frac{1+(n+1)x}{1}$$
So:
$$\frac{1+nx}{1+(n+1)x}\ge1-x$$
Take a product of these expressions for $n=0,1,\cdots,N-1$ to see
$$\frac{1}{1+Nx}\ge(1-x)^N$$
(this can easily be recast as an induction)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots ... | As $y\ge 0$, $\;\mathrm e^y\ge 1$. Now $\mathrm e^y$ is a root of the quadratic polynomial $\;P(t)=t^2-2xt+1$. Just observe that, as $P(1)=2(1-x) < 0$ for $x>1$, $1$ separates the roots. Hence $\mathrm e^y$ is the greatest root — with a + sign.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Fix $\alpha > 1$. Take $x_1 > \sqrt{\alpha}$, and define $$x_{n+1} = \frac{\alpha + x_n}{1+x_n} = x_... | $$\text {Let } a_n=(\sqrt \alpha)\tan (d_n+\pi /4) \text { with } d_n\in (0,\pi /4).$$ Using $\tan (x +\pi /4)=(1+\tan x)/(1-\tan x)$ when $\tan x \ne 1,$ we arrive, after some calculation, that $$\tan d_{n+1}=k \tan d_n$$ $$ \text { where } k=\frac {1-\sqrt {\alpha} }{1+\sqrt {\alpha}}.$$ Observe that $0>k>-1$ becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why is $\lim\limits_{x\to -\infty} \frac{1}{\sqrt{x^2+2x}-x}$ equal to $0$? So I made one exercise, which was $\lim_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x}$ I solved this one by:
$\lim \limits_{x\to +\infty} \frac{1}{\sqrt{x^2-2x}-x} \frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x} $
$\lim \limits_{x\to +\infty} \frac{\sqrt{x^... | Let $-1/x=h$
$x^2+2x=\dfrac{1-2h}{h^2}\implies\sqrt{x^2+2x}=\dfrac{\sqrt{1-2h}}{\sqrt{h^2}}=\dfrac{\sqrt{1-2h}}{|h|}$
As $h\to0^+, h>0,|h|=+h$
$$\lim_{x\to-\infty}\dfrac1{\sqrt{x^2+2x}-x}=\lim_{h\to0^+}\dfrac h{\sqrt{1-2h}+1}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$ I want to solve the integral:$$\int x^2\sqrt{x^2+1}\;dx.$$
I did $x = \tan t$, then it is equal to:$$\int\frac{\tan^2 t}{\cos^3 t}\;dx.$$
Or:$$\int\frac{\sin^2 t}{\cos^5 t}\;dx.$$
I stuck there. Any help will be much appreciated...
| Going the hyperbolic instead of the trigonometric route, we can let $x:=\sinh t$, then $dx=\cosh tdt$.
\begin{align}
\int x^2\sqrt{x^2+1}dx &= \int\sinh^2t\sqrt{\sinh^2t+1}\cosh tdt\\
&=\int\sinh^2t\cosh^2tdt\\
&=\int\left(\sinh t\cosh t\right)^2dt\\
&=\frac{1}{4}\int\sinh^2(2t)dt\\
&=\frac{1}{8}\int\left(\cosh(4t)-1\r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that that $a>\frac{3}{4}$ Let $x^4+2ax^3+x^2+2ax+1=0$ .
If this equation has at least two real negative roots then show that that $a>\frac{3}{4}$.
I tried to solve the equation first by using the sub $t=x+ \frac{1}{x}$.
Then I got $t^2+2at-1=0$.
Thereafter how can I proceed to get the required inequality ?
| Another approach to determine the behaviour of $f(x)=x^4 + 2ax^3 + x^2 + 2ax + 1$ is by quartic equation analysis.
Define $a,b,c,d,e$ as the coefficients of $f(x)$.
$$a=1,b=2k,c=1,d=2k,e=1$$
where you want to prove that $f(x)$ has two real negative roots when $k>\frac{3}{4}$
Define the discriminant which will allow u... | {
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"source": "stackexchange",
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What is the value of $a+b+c+d$?
Four integers $a,b,c,d$ make all the statements below true. What is the value of $a+b+c+d$?
(i) $10 \leq a,b,c,d \leq 20$
(ii) $ab-cd = 58$
(iii) $ad-bc = 110$
Adding the two equations together gives $ab+ad-cd-bc = 168$ and so $$a(b+d)-c(b+d) = (a-c)(b+d) = 168 = 2^3 \cdot 3 \cdot 7.$$... | Consider $(iii)-(ii)$, we have
$$(a+c)(d-b) = (ad-bc) - (ab-cd) = 110 - 58 = 52$$
Since $10 \le a, c \le 20 \implies 20 \le a+c \le 40$ and the only divisor
of $52$ between $20$ and $40$ is $26$, we find
$$a+c = 26, d - b = 2$$
Consider $(ii)+(iii)$, we have
$$(a-c)(b+d) = (ab-cd) + (ad - bc) = 58+160 = 168$$
Since $10... | {
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Use the general formula $x^3+y^3=(x+y)\left((x+y)^2-3xy\right)$.
Here $x=\sqrt[3]{2+\sqrt{5}}$ and $y=\sqrt[3]{2-\sqrt{5}}$.
Let $a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ and
$$b=\sqrt[3]{2+\sqrt{5}}\sqrt[3]{2-\sqrt{5}}=\sqrt[3]{2^2-(\sqrt{5})^2}=-1$$
$$(2+\sqrt{5})+(2-\sqrt{5})=4$$
$$=a\left(a^2-3b\right)=a^3+3a$$... | {
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Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}... | Neither is correct.
It's not restrictive to assume $a>0$ (for $a=0$ the antiderivative has a different form; for $a<0$ change the intervals below accordingly).
The function $\frac{1}{x^2-a^2}$ is defined for $x\ne\pm a$ and is negative over $(-a,a)$, positive over $(-\infty,-a)$ and $(a,\infty)$.
The sign of $\frac{x-a... | {
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Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi... | Although this is same to answer of @schrodingersCat Let $x\geq y\geq0\geq z$
$$\cos(x+y)^2=1-\sin(x+y)^2$$
$$⇔1-(\sin{x}+\sin{y})^2=(\cos{x}+\cos{y})^2$$
$$⇔1+2(\cos{x}\cos{y}+\sin{x}\sin{y})=0$$
$$⇔\cos(x-y)=-1/2 $$
$$x-y=2\pi/3$$
and
$$\sin{x}+\sin{y}=-\sin{z}$$
$$⇔\sin(y+2\pi/3)+\sin{y}=\sqrt3/2$$
$$⇔(1/2)\sin{y}... | {
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"answer_id": 2
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Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\... | Notice that
$$x-\sqrt{x^2-x+1}=x-\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}\\\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}> (x-\frac{1}{2})
$$
Now consider if $x<\frac{1}{2}$ the inequality is trivial,consider then $x\geq \frac{1}{2}$ then you're allowed to square the inequality and since $\frac{3}{4}>0$ the inequality is always tru... | {
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The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms The sum of series $1\cdot 3\cdot 2^2+2\cdot 4\cdot 3^2+3\cdot 5\cdot 4^2+\cdots \cdots n$ terms
i have calculate $a_{k} = k(k+2)(k+1)^2$
so $\displaystyle \sum^{n}_{k=1}a_{k} = \sum^{n}_{k=1}k(k+1)^2(k+2)$
i wan,t be able g... | Hint:
Observe that
\begin{align*}
a_k&=k(k+2)(k+1)^2\\
&=(k+1-1)(k+1+1)(k+1)^2\\
&=\left[(k+1)^2-1\right](k+1)^2\\
&=(k+1)^4-(k+1)^2
\end{align*}
Now, here is something that can be useful in order to compute the sum of the fourth powers and squares
Geometric interpretation for sum of fourth powers
| {
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Find the inverse of a polynomial in a quotient ring Consider $\mathbb Z_5[x]/I$ with $I$ as ideal generated by $b=x^3+3x+2$. If $(x+2) + I$ is element of $\mathbb Z_5[x]/I$ that has an inverse. Find the inverse of $(x+2) + I$.
I stuck to get the inverse because the gcd of that is not $1$.
| Perform the Euclidean division by Horner's scheme (in $\mathbf F_5$):
$$\begin{matrix}\\ \\\times -2\quad\end{matrix}
\begin{matrix}
\hline
1&0&-2&2\\
\downarrow&-2&-1&1\\
\hline
\ 1&-2&2&-2
\end{matrix}
$$
Thus $0=x^3+3x+2\;(=x^3-2x+2)=(x+2)(x^2-2x+2)-2$, whence, multiplying both sides by $-2$,
$$(x+2)(x^2-2x+2)=2\... | {
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Help solve simultaneous equations I need to solve the system of equations
$$6x+y=3\tag 1 $$
$$x^2+y^2=16\tag 2$$
So first, I rearrange $(1)$ to $y=3-6x$ and substitute that into $(2)$.
I get $$x^2+(3-6x)^2 = 16$$
which is $$x^2 + 36x^2-36x+9-16=0$$
which is $$37x^2-36x-7=0$$.
This I need to solve by completing the squa... | $ 37 x^2 - 36 x - 7 = 0 $
Factor out the $37$ from the first two terms
$ 37 ( x^2 - \dfrac{36}{37} x ) - 7 = 0$
Take half of $\dfrac{36}{37}$ which is $\dfrac{18}{37} $ and add it with its sign to $ x $ then square both as follows
$ 37 ( x - \dfrac{18}{37} )^2 - 7 = 37 ( \dfrac{18}{37} )^2$
Multiply through by $37^2$
... | {
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Given the $ x+y+z =3$ Prove that $ x^3+y^3+z^3+6xyz \geq 27/4$ Also, $x$,$y$ and $z$ are positive.
I have figured out that it would be enough to prove that LHS is bigger than $(27-\text{LHS})/3$, then by substituting 27 with $ (x+y+z)^3 $, it's enough to prove that
$$x^3+y^3+z^3+6xyz \geq x^2(3-x)+y^2(3-y)+z^2(3-z).$... | Without loss of generality, we assume $z\ge \frac{x+y+z}{3}=1$. We temporarily fix $z$ and check how to arrange $x$ and $y$ to achieve a minimum. Now $y=3-x-z$.
So $f(x)=x^3+(3-x-z)^3+z^3+6x(3-x-z)z$, and the derivative of $f$ is
$$f'(x) = 3x^2-3(3-x-z)^2 + 6(3-x-z) z-6xz=x(-3*2*(z-3) -6z-6z) + (-3(z-3)^2+6(3-z)... | {
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"source": "stackexchange",
"question_score": "3",
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solving inhomogenous $2$ x $2$ ODE?
Let $y'=\begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ sin(x)\end{pmatrix}$ with $y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$
Now I want to solve it, but don't know how to continue:
$1)$ solve the homogeous ODE $y'=Ay$
$\Rightarrow y_{hom}(x)=e^{0x} \cdot c_1\begin{pma... | We are given
$$y'= Ay + g = \begin{pmatrix}1 & 2 \\ 3 & 6\end{pmatrix}y+\begin{pmatrix}x \\ \sin x\end{pmatrix}, y_0=\begin{pmatrix}0 \\ 0\end{pmatrix}$$
You correctly found the homogeneous solution
$$y_h(x) = c_1\begin{pmatrix}-2 \\ 1\end{pmatrix}+c_2 \cdot e^{7x}\begin{pmatrix}1 \\ 3\end{pmatrix}$$
Since you want to ... | {
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How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?
How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?
My Try:
$$\lim _{x\to \infty }\left(x^2\sin\left(\frac{1}{x}\right)\ln\left(\frac{x+3}{\sqrt{x^2-5x}}\right)\r... | Notice that
$$\frac{x+3}{\sqrt{x^2-5x}} = \left(\dfrac{x^2+6x+9}{x^2-5x}\right)^{1/2} = \left(1+ \dfrac{11x+9}{x^2-5x}\right)^{1/2}.$$
By setting $y = \dfrac{x^2-5x}{11x+9}$ and since $x \rightarrow \infty \implies y \rightarrow \infty$, we obtain
$$ x = \dfrac{5+11y + \sqrt{121y^2+146y+25}}{2}.$$
Thus, $$\lim_{x \rig... | {
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"source": "stackexchange",
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Show the recursive sequence is increasing
How do I show that the recursive sequence
$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad
n\geq 2, \phantom{x} a_1 = 3$$
is an increasing sequence?
1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.
\begin{align*}
a... | Suppose for all $m\le n-1$, $a_{m+1}\ge a_m$.
Case 1: $n=2k$. Note that
$$ \lfloor \frac{2k+1}2 \rfloor=k,\lceil\frac{2k+1}2 \rceil=k+1. $$
Then
\begin{eqnarray}
a_{n+1} - a_n & =&a_{2k+1}-a_{2k}\\
&=& a_{\left\lfloor \frac{2k+1}{2} \right\rfloor} + a_{\left\lceil \frac{2k+1}{2} \right\rceil} +3(2k+1)+1 - ( a_{\lfloo... | {
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Find the value after Remainder Theorem The expression $4x^3+21x-6$ has the same remainder when divided by $x-a$ or by $x+b$, where $a$ not equals to $b$. Find the value of $a^2+b^2-ab$. I have the answer but I don't know how to work through the question.
| Given same remainder, by Polynomial Remainder Theorem we have,
$\displaystyle\ \ \ \ \ \ f(a)=f(-b)$
$\displaystyle\Rightarrow 4a^3+21a-6=-4b^3-21b-6$
$\displaystyle\Rightarrow 4(a^3+b^3)+21(a+b)=0$
$\displaystyle\Rightarrow \frac{(a^3+b^3)}{(a+b)}=a^2+b^2-ab=\frac{-21}{4}$
| {
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What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$?
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.
I plotted the graph of $x^... | The cubic $x^3+y^3+xy+1$ is symmetric in $y=x$. So lets rotate it by $\frac{\pi}{4}$. Let $x=\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}$ and $y=\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}$. Note that the function to minimize becomes $x`^2+y`^2$.
$$\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}... | {
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$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
| Without loss of generality, suppose that $a\leq b\leq c$. Also, write $N:=1001$. Since $$(a+1)(b+1)(c+1)=N\,,$$ we have that $$|a+1|\,|b+1|\leq N\,.$$ This clearly means
$$|a+1|+|b+1|\leq N+1\,.$$
(If $x$ and $y$ are positive integers with $xy\leq N$, then $(x-1)(y-1)\geq 0$, which implies that $x+y\leq xy+1\leq N+1... | {
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Sum $\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$ What is the sum of the following series:
$\frac {1}{9} + \frac {1}{18} +\frac {1}{30} +\frac {1}{45} +\frac {1}{63} \ldots$
I tried to do it through telescope method but it didn't work.
| I believe we have,
$$\frac{2}{3} \sum_{n=2}^{\infty} \frac{1}{n(n+1)}$$
Clearly telescoping will work.
How I got it,
Let's look at denominators.
$$9,18,30,45,63,...$$
Take difference of consecutive two.
$$9,12,15,18,..$$
This is clearly given by $3(x-1)+9$ because the difference between every two consecutive numbers i... | {
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Area of the shaded part.
My Attempt,
Area of square $=10^2=100 cm^2$.
Area of circle $=\pi r^2=25\pi cm^2$.
What should I do further?
|
$\begin{align} \color{darkorange}{\textsf{Shaded}} &= (\color{red}{\textsf{Triangle}} + \color{green}{\textsf{Polygon}})-\color{blue}{\textsf{CircleSegment}}\\&= \color{red}{\left(\frac{1\times2}{2}\right)} + \color{green}{\left(3 + \frac{3\times 4}{2}\right)}-\color{blue}{\left(\frac{\arctan{(3/4)}}{2\pi}\pi \times 5... | {
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"answer_count": 2,
"answer_id": 1
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Area between geodesic and arc of constant lattitude A geodesic is a line representing the shortest route between two points on a sphere, for example on the Earth treated here as a perfect sphere.
Two points on Earth having the same latitude can be also connected with the line being a part of a circle for selected const... | Summary: If $A$ and $B$ lie on the latitude line at angle $0 < \alpha < \pi/2$ north of the equator on a sphere of unit radius, and at an angular separation $0 < \theta = \beta_{2} - \beta_{1} < \pi$, then the "digon" bounded by the latitude and the great circle arc $AB$ (in blue) has area
\begin{align*}
\pi - \theta... | {
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"question_score": "2",
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Series shown up in Quantum Mechanics Here is the series, by some concepts in quantum mechanics, it should be 1, however, I don't know how to prove it. Could someone show me the proof? Thanks.
$$\lim_{k\to \infty}\sum_{n=1}^{k} \frac{4}{n^2\pi^2}(1-\cos\frac{n\pi}{2})^2
$$
| Let $S$ be defined as
$$
S=\frac{4}{\pi^2}\sum_{k\geq 1}\frac{1}{k^2}\left(1-\cos\left(\frac{\pi k}{2}\right)\right)^2
$$
Decompose $S$ according to the remainder of $n$ modulo $4$.
Since $\cos(\pi n/2)=1$ for $n=0 \mod 4$ one part of the sum exactly cancels. Furthermore $\color{blue}{\cos(n\pi/2)=0}$ for $\color{blue... | {
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"question_score": "3",
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Find integer solution of a system of equations Find all such natural $n$ such that
$
\dfrac{12n-21143}{25}
$
and
$
\dfrac{2n-3403}{25}
$
both are squares of prime numbers.
So far I have found by computer only solution: $ 2014.$
How to prove taht there are no more solutions?
I was tried first to solve the syst... | Your equations are
\begin{cases}
12k= 839+ p^2\\
2k= 135+ q^2
\end{cases}
and we observe that $$p^2-q^2 = 10k - 704 = 5(2k-141)+1$$
so $\,p^2 - q^2 = 1 \pmod{5}$ which tells us that $p^2 = 1 \mod 5$ and $q = 0 \mod 5$, or $p=0 \mod 5$ and $q^2 = -1 \mod 5$. We consider the first case here, and the second later. Since ... | {
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"source": "stackexchange",
"question_score": "2",
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What is the coefficient of $x^2$ in the polynomial.. What is coefficient of $x^2$ in polynomial $(1-4x)^6(1+3x)^8$ ?
I know how to do it for $(1-4x)^6$ for example, but how to do it for the product of two polynomials?
| Let $f(x) = (1-4x)^6(1+3x)^8 = uv$, where $u = (1-4x)^6$ and $v=(1+3x)^8$
\begin{align*}
f''(x) &= u'' v + 2 u'v' + v''\\
&= 6 \cdot 5 \cdot (-4)^2 (1-4x)^4 (1+3x)^8 + 2 \cdot 6 \cdot (-4) \cdot 8 \cdot 3 (1-4x)^5 (1+3x)^7 \\
& \qquad + 8 \cdot 7 \cdot (3)^2 (1-4x)^6 (1+3x)^6
\end{align*}
Hence coefficient of... | {
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Solve $y'=\frac{2}{x^2}-y^{2}$
$$y'=\frac{2}{x^2}-y^{2}$$
$y=\frac{k}{x}$ is a solution, find the general solution
So we have riccati differential equation.
$y=z+\frac{k}{x}\iff z=y-\frac{k}{x}$
$y'=z'-\frac{k}{x^2}$
Plugin it to the ode we get:
$$z'-\frac{k}{x^2}=\frac{2}{x^2}-(z+\frac{k}{x})^2$$
$$z'=\frac{2-k^2... | Hint: if $y=\frac{k}{x}$ is a solution then pluging in the equation we get $$\\ -\frac { k }{ { x }^{ 2 } } =\frac { 2 }{ { x }^{ 2 } } -\frac { { k }^{ 2 } }{ { x }^{ 2 } } \\ { k }^{ 2 }-k-2=0\\ \left( k-2 \right) \left( k+1 \right) =0\\ $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt... | Since $y=x^3-x^2$, we get $x+(x^3-x^2)^2=(x^3-x^2)^3$ or
$$x(x^2-x-1)(x^6-2x^5+2x^4-2x^3+2x^2-x+1)=0$$
and since $x^6-2x^5+2x^4-2x^3+2x^2-x+1=(x^6-2x^5+x^4)+(x^4-2x^3+x^2)+(x^2-x+1)>0$,
we get your answer.
| {
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"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
Find the sum of the series $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+...=$ Given : $$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+ ...$$
What is the sum of this series, how can one rewrite it to look simpler ?
EDIT :
Actually I found how to rewrite i... | Let's find the sum of the geometric series: $$\sum_{k=0}^\infty x^{3k}=\frac{1}{1-x^{3}}$$
For $0<x<1$. Now let's integrate this series by term three times:
$$\sum_{k=0}^\infty \frac{x^{3k+1}}{3k+1}=\int_0^x \frac{dt}{1-t^{3}}$$
$$\sum_{k=0}^\infty \frac{x^{3k+2}}{(3k+1)(3k+2)}=\int_0^x \int_0^y \frac{dt~ dy}{1-t^{3}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \fra... | $$I=\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
thus
$$2I=1-\frac{2}{2^1}+\frac{3}{2^2}-\frac{4}{2^3}+\ldots-\frac{100}{2^{99}}$$
as a result
$$2I+I=1-\left(\frac{2}{2^1}-\frac{1}{2^1}\right)+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)-\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Fourier Coefficients for $\frac{3}{5-4\cos{2\theta}}$ I want to compute the (even) fourier coefficients for
$\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.
Namely, I want to compute the integral:
$$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$
Integrating this in mathem... | Hint: Observe
\begin{align}
\frac{3}{5-4\cos 2\theta} = \frac{3/5}{1-\frac{4}{5}\cos 2\theta} = \frac{3}{5}\sum^\infty_{n=0} \left(\frac{4}{5}\right)^n \cos^n 2\theta
\end{align}
and
\begin{align}
\cos^n 2\theta = \left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)^n = \frac{1}{2^n}\sum^n_{k=0}\binom{n}{k} e^{2ik\theta}e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding basis of subspace from homogenous equations (checking) The question;
$U = \{x |Ax = 0\}$ If $ A = \begin{bmatrix}1 & 2 & 1 & 0 & -2\\ 2 & 1 & 2 & 1 & 2\\1 & 1 & 0 & -1 & -2\\ 0 & 0 & 2 & 0 & 4\end{bmatrix}$
Find a basis for $U$.
To make it linearly independent, I reduce the rows of $A$;
$\begin{bmatrix}1 & 0 ... | Observe that row operations doesn't change the solution for the system of linear equations. From the last column of the reduced row echelon form (RREF) of $A$,
$$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating $\sum_{n \in \mathbb{N}} \frac{1}{n^2}$ with the Poisson summation formula So, there is the following exercise at the end of a lecture script
By using the function $f(x)=e^{-a|x|}$ ($x \in \mathbb{R}, a>0$) evaluate
$$\sum_{n \in \mathbb{N}} \frac{1}{n^2+a^2}$$
and by letting $a \to 0$ find $\sum_{n \i... | The Poisson summation formula leads to:
$$ \sum_{n\geq 0}\frac{1}{n^2+a^2}=\frac{1+\pi a \coth(\pi a)}{2a^2}\tag{1}$$
and $\lim_{a\to 0^+}\sum_{n\geq\color{red}{ 1}}\frac{1}{n^2+a^2}$ can be computed through de l'Hospital theorem or just by recalling that in a neighbourhood of the origin
$$ z\coth(z)=1 + \frac{z^2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find that solution $ϕ$ which satisfies $ϕ(0) = 0, ϕ(1) = 3$ for $y'' = 3x + 1$. Find the particular solution to the following second order linear ordinary differential equation that satisfies the $ϕ$
$y'' = 3x + 1$, $ϕ(0) = 0, ϕ(1) = 3$
My solution:
$y' = \frac{3x^2}{2} + x + c_1$
$y = \frac{x^3}{2} + \frac{x^2}{2} ... | You already have c2=0 substitute it and find c1.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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While I am solve $(4+t^2)\frac{dy}{dt} + 2ty = 4t$, I don't know why my answer is wrong.. $(4+t^2)\frac{dy}{dt} + 2ty = 4t$
$(4+t^2)\frac{dy}{dt}= 4t -2ty$
$(4+t^2)dy = 4t dt -2tydt$
$\int (4+t^2)dy = \int( 4t -2ty )dt$
$4y+t^2y = 2t^2 -t^2y+ c$
$y(4+2t^2)= 2t^2 +c $
$y = \frac{2t^2 + c}{4+2t^2}$
This is my answer, but... | The basic error was integrating before the variables were separated.
\begin{eqnarray}
(4+t^2)\frac{dy}{dt} + 2ty &=& 4t\\
(4+t^2)\,dy+2ty\,dt&=&4t\,dt\\
(4+t^2)\,dy+(2ty-4t)\,dt&=&0\\
(4+t^2)\,dy+2t(y-2)\,dt\\
\frac{1}{y-2}\,dy+\frac{2t}{4+t^2}\,dt&=&=0\\
\ln\vert y-2\vert+\ln\vert 4+t^2\vert&=&\ln\vert c\vert\\
\ln\ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove decimal number pattern for its n-bit 2's compliment If we let $B = b_{n−1}b_{n−2} · · · b_1b_0$ be an $n$-bit $2$’s complement integer.
How can we show that the decimal value of $B$ is $−b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2} + b_{n−3}\cdot 2^{n−3} + \cdots + b_1\cdot 2 + b_0$.
| Hint: let $\;A=-B\;$ then, by the definition of $2$'s complement, $A$ is obtained from $B$ by flipping all bits and adding $1\,$:
$$
A = (1-b_{n-1})\cdot 2^{n-1} + (1-b_{n-2})\cdot 2^{n-2} + \cdots + (1-b_1) \cdot 2 + (1-b_0) + 1
$$
Let $B\,'$ be the given expression $B\,' = −b_{n−1}\cdot2^{n−1} + b_{n−2} \cdot 2^{n−2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle
Now i remember a identity which was like if $a+b+c=0$,then $a^3+b^3+c^3=3abc$. So i have $\sum_{}^{} \tan(A)=0$. How do i procee... | Note that if $A,B,C$ are angles of a triangle, we have that $$\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ As seen here. If $$0=\tan A+\tan B +\tan C=\tan A \tan B \tan C$$
Then we have that for at least one angle among $A,B,C$ is $0$ or $\pi$. Thus, it is a contradiction. So $$\tan A+\tan B +\tan C \neq 0$$
Now exploi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Question on Linear Combinations and Vectors This is a very simple question I just wanted to make sure I was doing correctly.
Express the vector
$$
\underline{v} = \left(\matrix{2\\-1\\5\\-3\\6}\right)
$$as a linear combination of $\mathbf{e_1}, \mathbf{e_2}, \mathbf{e_3}, \mathbf{e_4}$, and $\mathbf{e_5}$ in $\mathbb... | An easy way to see this is to put the standard basis vectors into a matrix. This may be overkill, but this question is currently unanswered.
Note that taking each standard basis vector in $\mathbb{R}^5$ as columns into a matrix simply yields the identity matrix $I_5$ or some permutation of it.
$
\begin{align*}
I_5 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the minimum of $\overline{AP}+\overline{AQ}$. My niece asked me a math question: There is a point $A(-3,2)$ on the $x$-$y$ plane. A line $L$ is perpendicular to the line $x=y$. Line $L$ and $y=f(x)=2^x$ has an intersection point $P$. Line $L$ and $y=g(x)=\log_2 x$ has an intersection point $Q$. Find the minimum of... | Sorry, but AM-GM is unable to be applied in this case as $\overline{AP} \neq \overline {AQ}$ for all $P$, $Q$.
However, the triangle inequality can be applied nicely in this case. First, we must reformulate the points.
Let us name the points $A(-3,2)$, $B(2,-3)$, $P(a,2^a)$, $Q(2^a,a)$,
Note that $$ \overline{AP}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the prob... | $$f(x)=\dfrac{7}{\sqrt{x}}\\f'(x)=\lim\limits_{h \to0}\dfrac{f(x+h)-f(x)}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}}{h}\\f'(x)=\lim\limits_{h \to0}\dfrac{\bigg(\frac{7}{\sqrt{x+h}}-\frac{7}{\sqrt{x}}\bigg)\bigg(\frac{7}{\sqrt{x+h}}+\frac{7}{\sqrt{x}}\bigg)}{h\bigg(\frac{7}{\sqrt{x+h}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integrate $\int \frac{dx}{\sqrt {7-6x}-x^2}$ $$\int \frac{dx}{\sqrt {7-6x}-x^2}$$
Have no idea how to start with this. How do i make a perfect square of the variable?
| I assume you are trying to solve the integral
$$I:=\int \frac{\mathrm d x}{\sqrt{7-6x-x^2}}$$
because you mentioned to make a perfect square.
Since $$7-6x-x^2= 16-(x+3)^2= \frac{1}{16}\left(1- \left( \frac{x+3}{4}\right)^2\right)$$
the integral becomes
$$I= \int \frac{\mathrm d x}{\frac{1}{4}\sqrt{1- \left( \frac{x+3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
| Hint: Since $(a^2+b^2)c=c^4=(a+b)^2$ we have that $a^2+b^2$ divides $(a+b)^2$. This implies $a=b$ for positive integers. Then we have $2a^2=c^3$, so that $(a,b,c)=(2,2,2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternate methods to solve this equation The question is to find the roots of the equation:
$2(x^2+ \frac{1}{x^2}) - 3(x+ \frac{1}{x}) - 1 =0$
My attempt:
$2(x^2+ \frac{1}{x^2} + 2) - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - 4 = 0$
⇒$2(x+ \frac{1}{x})^2 - 3(x+ \frac{1}{x}) - 1 - ... | Hint -
$(x+ \frac{1}{x}) = t$
Also,
$(x^2+ \frac{1}{x^2}) = t^2 - 2$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
| Apply law |a+b| <= |a| + |b|
Let $L = \frac{f(x)-\sqrt{x^2+9}}{x}$
$$
|L| = \lvert\frac{f(x) - 3 + 3 - \sqrt{x^2+9}}{x}\rvert \le |\frac{f(x)-3}{x}| + |\frac{3 - \sqrt{x^2+9}}{x}| \\
\le |\frac{x^2}{x}| + |\frac{(3-\sqrt{x^2+9})(3+\sqrt{x^2+9})}{x(3+\sqrt{x^2+9})}| = |x| + |\frac{x}{3+\sqrt{x^2+9}}|
$$
Hence,
$$
\lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I d... | $$\lim_{x\to1}\frac{\ln x}{x-1}=1$$ is a basic limit from high school, and expresses the logarithm is differentiable at $x=1$ and has derivative equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Equality using floor function Let $n\in \mathbb{N}$. How we can show this :
$$\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor =\lfloor \sqrt{16n+20}\rfloor$$
by using the concavity of $x\longmapsto \sqrt{x}$.
I read an article a few years ago about this but I can not find it on the internet.
| Let's first show the $\leq$ direction: By concavity,
$$\frac{\sqrt{n} + \sqrt{n+1}}{2} < \sqrt{n + \frac{1}{2}}$$
so
$$\sqrt{n} + \sqrt{n+1} < \sqrt{4n + 2}\,\,\,\,\, (1)$$
Similarly
$$\sqrt{n+2} + \sqrt{n+3} < \sqrt{4n + 10} \,\,\,\,\, (2)$$
And using the same principle again:
$$\sqrt{4n + 2} + \sqrt{4n + 10} < \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151826",
"timestamp": "2023-03-29T00:00:00",
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How to solve $x - 3\sqrt{\frac{5}{x}} = 8$ for $x - \sqrt{5x}$? I have a problem from my textbook. By using $x - 3\sqrt{\frac{5}{x}} = 8$ how can we find the value of $x - \sqrt{5x}$. I have derived the equation that's given so much, but i couldn't find the answer. Solvings or hints are appreciated.
| we can write $$x-8=3\sqrt{\frac{5}{x}}$$ after squaring this equation we get
$$x^2-16x+64=9\cdot \frac{5}{x}$$ multiplying by $$x\ne 0$$ we obtain
$$x^3-16x^2+64x-45=0$$ factorizing this equation we get
$$(x-5)(x^2-11x+9)=0$$ can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Epsilon-Delta: Prove $\frac{1}{x} \rightarrow 7$ as $x \rightarrow \frac{1}{7}$
Prove that $\displaystyle\frac{1}{x} \rightarrow 7$ as $\displaystyle x \rightarrow \frac{1}{7}$.
I need to show this with an $\epsilon-\delta$ argument. Still figuring these types of proofs out though, so I could use some tips/critiques ... | for $\beta \gt 0$ write:
$$
x = \frac{1+\beta}7
$$
then
$$
\begin{align}
7-\frac1x\ & \le 7\bigg(1-\frac1{1+\beta}\bigg) \\[14pt]
& = \frac{7\beta}{\beta+1}\\[14pt]
& \le 7\beta
\end{align}
$$
thus:
$$
x-\frac1{7} \le \frac{\beta}7 \Rightarrow 7 -\frac1x\le 7\beta
$$
with this established, the $\epsilon-\delta$ argumen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If a matrix commutes with a set of other matrices, what conclusions can be drawn? I have a very specific example from a book on quantum mechanics by Schwabl, in which he states that an object which commutes with all four gamma matrices,
$$
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
... | Call your four matrices $A,B,C,D$ respectively. While they indeed don't span $M_4(\mathbb C)$, the point is that the algebra they generate is the whole matrix space. So, any matrix that commutes with $A,B,C,D$ must in turn commute with all members of $M_4(\mathbb C)$. In fact, if we put $X=\frac{B\,(AC-C)\,A}{2i}$ and ... | {
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"url": "https://math.stackexchange.com/questions/2154750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 0
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Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
... | Alternatively, by cosine law
\begin{align*}
b^2+c^2-a^2 &= 2bc\cos A \\
c^2+a^2-b^2 &= 2ca\cos B \\
a^2+b^2-c^2 &= 2ab\cos C \\
a^2+b^2+c^2 &= 2(bc\cos A+ca\cos B+ab\cos C) \\
& \le 2(bc+ca+ab)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Prove that $\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$ is always odd for any natural $n$.
Prove that $$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
is always odd for any natural $n$.
I attempted to write the binomial expansion and sum it so the root n... | HINT:
Say $\left(\frac{3+\sqrt{17}}{2}\right)=a$ and $\left(\frac{3-\sqrt{17}}{2}\right)=b$.
Now observe that:
$$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
$$=a^n+b^n$$
$$=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$$
$$=\color{red}{3\cdot\left[\left(\frac{3+\sqrt{17}}{2}\right)^{n-1}+... | {
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"answer_id": 0
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Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | For 9: if $n$ is divisible by 3, then the conclusion is obvious. If $n \equiv 1 \bmod 3$, then $n^3 \equiv 1 \bmod 9$ (proof: write $n^3 = (3k+1)^3 = 27k^3 + 27k^2 + 9k + 1$; all terms but the constant are killed modulo 9). If $n \equiv -1 \bmod 3$, then $n^3 \equiv -1 \bmod 9$ (proof is similar).
For 7, the conclusion... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
How can I find all the matrices that commute with this matrix? I would like to find all the matrices that commute with the following matrix
$$A = \begin{pmatrix}2&0&0\\ \:0&2&0\\ \:0&0&3\end{pmatrix}$$
I set $AX = XA$, but still can't find the solutions from the equations.
| If
$B:= \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)$,
then your matrix becomes $A = 2I + B$. Thus a matrix $C$ will commute with $A$ if and only if $C$ commutes with $B$. But note
$BC = \left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left( \begin{matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".
Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$
My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\... | One option is to notice the right side is a binomial expansion. But your solution should work. For the induction step, we have
$$\begin{align*}
& \sum_{k=0}^{r+1}(-1)^k\binom{r+1}{k}4^{r+1-k} \\
=& 4^{r+1}+\sum_{k=1}^{r+1} (-1)^k\binom{r+1}{k}4^{r+1-k} \\
=& 4^{r+1}+\sum_{k=0}^r (-1)^{k+1}\binom{r+1}{k+1} 4^{r-k} \\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac12+\frac16+...+\frac{1}{n(n+1)}=\frac{n}{n+1}$ for $n \in \mathbb{N}$ I am using Induction:
Base Case $n=1$ holds ; $\frac12$= $\frac{1}{(1)+1}$
Assume $\frac{n}{n+1}$ is true from some $n \in \mathbb{N}$.
Then $\frac12+\frac16+...+\frac{1}{n(n+1)}+ \frac{1}{(n+1)((n+1)+1)}=\frac{n}{n+1}+ \frac{1}{(n+1)((n+... | For completeness sake, we evaluate the sum without induction: Note that $$ \frac{1}{k(k+1)} = \frac{(k+1) - k}{k(k+1)} = \frac{k+1}{k(k+1)} - \frac{k}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.$$ Therefore, $$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^n \frac{1}{k} - \sum_{k=2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence m... | Hint
Write $a+b+c=k$, so
$$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 3
} |
Solving Linear Equation with non-invertible matrix? Find X when AX = B.
A= \begin{bmatrix}1&2&1\\2&5&4\end{bmatrix}
B = \begin{bmatrix}3&1\\5&4\end{bmatrix}
I know how to solve for x when A is invertible. I know it would just be $A^{-1}$B.
I'm confused with how to do this when A is non-invertible, I was trying to use t... | reducing, saving a little typing
$$
\left(
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & 1 \\
2 & 5 & 4 & 5 & 4
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & 1 \\
0 & 1 & 2 & -1 & 2
\end{array}
\right)
$$
$$
\left(
\begin{array}{rrr|rr}
1 & 0 & -3 & 5 & -3 \\
0 & 1 & 2 & -1 & 2
\end{array}
\right)
$$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that if LCMs are equal, then the numbers are equal too. For $a,b \in \mathbb{N}$, how do I prove:
$$lcm(a,a+5)=lcm(b,b+5) \implies a=b$$
| As Batominovski pointed out in his comment, $a$ and $b$ have to be positive integers, otherwise there are counterexamples to the statement.
We're going to use the following identities:
i) $ab=[a,b](a,b)$ where $[a,b]$ and $(a,b)$ denotes the $\text{lcm}$ and the $\gcd$ of $a$ and $b$, respectively.
ii) $[ca,cb]=c[a,b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the difference of two pronic numbers whose sum is 240. How do I do this without Excel/brute force, and how do I explain this to an 11 year old please? There's apparently supposed to be a trick to this because it's in a standardised exam.
Well I noticed that the difference between two consecutive numbers in the seq... | if $a(a+1)+b(b+1)=240$, then must one of them less than $120$ or equal $120$. we can let $a < b$. as $11 \cdot 12 > 120 > 10 \cdot 11$, let us try $A$ less than $10 \cdot 11$, B great than $11 \cdot 12$, here $A=a(a+1), B=b(b+1)$.
the minimum $A$ is $A = 1 \cdot 2$, if $A=1 \cdot 2$, $B$ is the maximum, but if $B = 15 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfra... | A shortcut that I found (Correct me if I am wrong in spotting):
Using AM-GM Inequality, we get
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
$$\frac{1}{a}+\frac{1}{b}\ge \frac{4}{1}$$
Using Cauchy-Schwarz Inequality,
$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Prove that if $x_{n+2}=\frac{2+x_{n+1}}{2+x_n},$ then $x_n$ converges
Let $x_0 > 0$, $x_1 > 0$ and $$x_{n+2} = \dfrac{2+x_{n+1}}{2+x_n}$$ for $n \in \{0,1,\dots\}$. Prove that $x_n$ converges.
My attempt:
If $x_n$ converges and $\lim\limits_{n\rightarrow\infty}x_n=a$ so $a=\frac{2+a}{2+a}$, which gives $a=1$. Now,
$... | Clearly every $x_n$ is positive. With this we know that $$x_{n+2} = \frac{2 + x_{n+1}}{2 + x_n} = \frac{2}{2+x_n} + \frac{x_{n+1}}{2+x_n} < 1 + \frac{x_{n+1}}{2}.$$
Applying this recursively we get $$x_n < 1 + \frac{x_{n-1}}{2} < 1 + \frac{1 + \frac{x_{n-2}}{2}}{2} = 1 + \frac{1}{2} + \frac{x_{n-2}}{4} < $$
$$< 1 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Abstract Algebra (Multiplicative Inverse using the Euclidean Algorithm) The question I am trying to solve is to find the multiplicative inverse of
$2-\sqrt[3]{2}+\sqrt[3]{4}$.
So far I am come up with the following.
Let $\alpha = 2-\sqrt[3]{2}+\sqrt[3]{4}$.
Then $s(x)=2-x+x^2$ and $x=\sqrt[3]{2}.$
Then $t(x)=x^3-2$.
T... | The following Maple command
gcdex(x^3-2,2-x+x^2,x,'s','t'); s; t;
will produce the gcd (1) and the factors $s(x),t(x)$ such that
$$
s(x)(x^3-2) +t(x) (2-x+x^2)=1.
$$
You'll see that
$$
s(x)=- 5/22 + 1/22 x
$$
and
$$
t(x)= 3/11 + 2/11 x - 1/22 x^2.
$$
Accordingly,
$$
t(\sqrt[3]2) \cdot (2-\sqrt[3]2+\sqrt[3]4)=1,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the following determinant is equal to $\sin(2(x+y))$
Prove that
$$\left|\begin {array} c
\cos(x+y) & \sin(x+y) & -\cos(x+y)\\
\sin(x-y) & \cos(x-y) & \sin(x-y) \\
\sin{2x} & 0 & \sin (2y)
\end {array}\right|
=\sin(2(x+y))$$
It is my question. I have tried several types of operations such as $C_1=C_1... | Posing $X=x+y$ and $Y=x-y$, the determinant becomes:
$$\Delta=\left|\begin {array} c
\cos(X) & \sin(X) & -\cos(X)\\
\sin(Y) & \cos(Y) & \sin(Y) \\
\sin(X+Y) & 0 & \sin (X-Y)
\end {array}\right|$$
Because of the zero at $(3,2)$, Sarrus's rule yields an expression of the determinant as a sum of four terms (instead o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Can the cosine function be written in "simpler" expressions? I recently found that the constants $\pi$ and $\phi=\frac{1+\sqrt{5}}{2} $ can be related by the identity
$$
2\cos \frac{\pi}{5} = \phi.
$$
Is there some way to write the cosine function by "simpler" mathematical expressions?
| Gauss introduced a method for cyclotomic polynomials that quickly gives such expressions. Let $n$ be odd and $\omega$ a primitive $n$-th root of unity. Then one of th values indicated by the expression $\omega + (1/\omega)$ is, indeed, $ x =2 \cos (2 \pi / n).$ These satisfy monic polynomials with integer coefficients.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Is $\frac1p(\sin^px+\cos^px)-\frac1q(\sin^qx+\cos^qx)$ constant for some reals $p$ and $q$. We know for
$$f(x)=\dfrac14(\sin^4x+\cos^4x)~~~;~~~g(x)=\dfrac16(\sin^6x+\cos^6x)$$
have $f(x)-g(x)=\dfrac{1}{12}$. My question is
Are there other real $p$ and $q$ such that
$$f(x)=\dfrac1p(\sin^px+\cos^px)~~~;~~~g(x)=\dfra... | Let $p,q$ be integers. If $p=q$, one has nothing to do. Suppose $p<q$. Let
$$F(x)=\dfrac1p(\sin^px+\cos^px)-\dfrac1q(\sin^qx+\cos^qx)$$
and then
\begin{eqnarray}
F'(x)&=&\sin^{p-1}x\cos x-\cos^{p-1}x\sin x-\sin^{q-1}x\cos x+\cos^{q-1}x\sin x\\
&=&\sin x\cos x(\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x).
\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^... | $x+y = 2\\
(x+y)^2 = 4\\
(x-y)^2 \ge 0$
add them together
$2x^2 + 2y^2 \ge 4\\
x^2 + y^2 \ge 2$
repeat:
$x^4 + y^4 \ge 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal I want to find the rank of this $n\times n$ matrix
\begin{pmatrix}
1 & a & a & \cdots & \cdots & a \\
a & 1 & a & \cdots & \cdots & a \\
a & a & 1 & a & \cdots & a \\
\vdots & \vdots & a& \ddots & & \vdots\\
\vdots & \vdots & \... | If you haven't figured it out yet here's the solution: if $a=1$ $\mathrm{Rank}(A)=1$ otherwise $\mathrm{Rank}(A)=n$ where
$$A:= \begin{pmatrix}
1&a&a&\cdots&a\\
a&1&a&\cdots&a\\
a&a&1&\cdots&a\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
a&a&1&\cdots&1
\end{pmatrix}$$
You've already shown that you know if $a\in\{0,1\}$. So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For $2x+2y-3z\geq0$ and similar prove that $\sum\limits_{cyc}(7z-3x-3y)(x-y)^2\geq0$
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(7z-3x-3y)(x-y)^2+(7y-3x-3z)(x-z)^2+(7x-3y-3z)(y-z)^2\geq0$$
I have a proof for the following weaker ineq... | Proof:
Because both the conditions and the conclusion are homogeneous, we may assume that $x+y+z=1$.
Geometrically,
$$\Delta=\{(x,y,z)\in \mathbb{R}^3 |\ 2x+2y−3z≥0, 2x+2z−3y≥0, \\
2y+2z−3x≥0, x+y+z=1\}$$
is a triangle . By solveing three linear equations
$$2x+2y−3z=0,\ 2x+2z−3y=0,\ x+y+z=1;$$
$$2x+2z−3y=0,\ 2y+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 ... | You have calculated det$(B) $ wrong - you just got the order of signs mixed up, when calculating the determinant by expanding into $ 2 \times 2 $ matrices.
It is det $(B) = 1 \begin{vmatrix} 1 & 2 \\ 2 & 1 \\ \end{vmatrix} - 0 \begin{vmatrix} 1 & 2 \\ 1 & 1 \\ \end{vmatrix} + 1 \begin{vmatrix} 1 & 1 \\ 1 & 2 \\ \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Using L'Hopital Rule, evaluate $\lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$ Using L'Hopital Rule, evaluate $$ \lim_{x \to 0} {\left( \frac {1} {x^2}-\frac {\cot x} {x} \right)}$$
I find this question weired.
If we just combine the two terms into one single fraction, we get$$\lim_{x \to 0} {\fra... | Note that
$$
\lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1
$$
so $\dfrac{1-x\cot x}{x^2}$ is an indeterminate form $0/0$ at $0$.
You can certainly use l’Hôpital:
$$
\lim_{x\to0}\dfrac{1-x\cot x}{x^2}
=
\lim_{x\to0}\frac{-\cot x+\frac{x}{\sin^2x}}{2x}=
\lim_{x\to0}\frac{x-\sin x\cos x}{2x\sin^2x}
$$
However, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Extrema under constraints
Find the critical points of the function $f(x_1, x_2)=x_1x_2$ under the constraint $2x_1+x_2=b$.
Using the method of Lagrange multipliers I got the following:
\begin{equation*}L(x_1,x_2,\lambda )=x_1x_2-\lambda \cdot \left (2x_1+x_2-b\right )\end{equation*}
\begin{align*}&L_{x_1}(x_1,x_2,... | $$\begin{array}{ll} \text{maximize} & x y\\ \text{subject to} & 2x + y = b\end{array}$$
From the equality constraint, we have $y = b - 2 x$. Let
$$g (x) := x (b - 2 x)$$
The derivative of $g$ vanishes at $\frac b4$. Hence, the maximizer is $(\bar x, \bar y) := \left(\frac b4, \frac b2\right)$ and the maximum is $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Cubic Root verification I'm working on a cubic root question and goes like this...
Let, r,s, and t be the roots of the equation $x^3+ax^2+bx+c = 0$. Re-write the expressions in terms of a,b, and c.
I'm given $r^2+s^2+t^2$. I know that, given 3 roots, I can factor the cubic equation as $(x-r)(x-s)(x-t)$, but don't know ... |
We haven't studied Vieta's formula in my class just yet.
Pretending to not know Vieta's formulas, note that if:
$$P(x)=x^3+ax^2+bx+c = (x-r)(x-s)(x-t)$$
then:
$$P(-x)=-x^3+ax^2-bx+c = -(x+r)(x+s)(x+t)$$
It follows that:
$$
\begin{align}
P(x)P(-x) & = (x^3+ax^2+bx+c)(-x^3+ax^2-bx+c) \\
& = (ax^2+c)^2-(x^3+bx)^2 \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2189679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $k$ such that $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}$ for $xyz=1$.
Find all $k\in\mathbb{R}^+$ such that for all $xyz=1$, $x,y,z\in\mathbb{R}^+$, we
have
$$\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}.$$
If we set $$x=\frac{a}{b},\,y=\frac{b}{c},\,z=... | I think the answer is $k_{min}=\log_23$.
Let $x=e^a$, $y=e^b$, $z=e^c$ and $f(x)=\frac{1}{(1+e^x)^k}$.
Hence, $a+b+c=0$ and we need to prove that
$$\sum_{cyc}f(a)\geq3f\left(\frac{a+b+c}{3}\right)$$
But $f''(x)=\frac{2ke^x(ke^x-1)}{(1+e^x)^{k+2}}\geq0$ for all $k\geq1$ and $x\geq0$.
Thus, by Vasc's RCF Theorem for all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a... | Assume the sequence converges i.e. $\displaystyle\lim_{n\to \infty} x_n = C< \infty.$
Then, as $n\to\infty$ we have $\,x_{n}\approx x_{n+1}$.
Taking limit of $x_{n+1}=x_n+\frac{1}{x_n}$ we get
$$
\lim_{n\to\infty} x_{n}= \lim_{n\to\infty} \left(x_n+\frac{1}{x_n} \right)\implies C = C + \frac{1}{C} \iff \frac{1}{C} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x... | I would do it this way:
Using
$\ln(1+z)
=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}x^k}{k}
$,
which converges for your range,
$\begin{array}\\
\int_0^{1/2}x\ln(1+x^3)dx
&=\int_0^{1/2}x\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{3k}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int_0^{1/2}x^{3k+1}dx\\
&=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $(a+b)(a+c)(b+c)=8$ then $\prod\limits_{cyc}(2a+bc)\leq27$
Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$(2a+bc)(2b+ac)(2c+ab)\leq27$$
My trying:
We need to prove that
$$8abc+a^2b^2c^2+\sum_{cyc}(4a^2b^2+2a^3bc)\leq27$$ or
$$abc\prod_{cyc}(a+b)+a^2b^2c^2+\sum_{cyc}a^2b... | let : $p= a+b+c, \ q= ab+bc+ca, \ r=abc$
then : $pq=8+r \ (\ p\ge 3, \ 0\le q\le 3)$
$$q^2\ge 3pr \Rightarrow q \le \dfrac{3p^2-\sqrt{9p^4-96p}}{2}$$
$$by \ Shur : q^3+9r \ge 4pq \Rightarrow q \ge \dfrac{72-p^3}{5p}$$
$$f(p,q) = (p-2)^2q^2+2(p^2-4)(p-4)q+16p(4-p) \le 27$$
So we only need to consider the inequality in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using the substitution method for a simple integral I have been playing with the substitution rule in order to test some ideas with computational graphs. One of the things I'm doing is applying the substitution to well known, and easy, integrals. For example, let's use that method to find the indefinite integral for
$... | $\int \frac{1}{2x}u du$ = $\int \frac{1}{2\sqrt{u}}\cdot u du$ = $ \int \frac{\sqrt{u}}{2}du$ = $u^{\frac{3}{2}}\cdot\frac{2}{3}\cdot \frac{1}{2} + C$
$= \frac{u^{\frac 3 2}}{3} + C = \frac{x^3}{3} + C$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
... | Let's use, from hypothesis, that $$3^{2k-1}+2^{2k-1}=5p\to 3^{2k-1}=5p-2^{2k-1}$$ so
$$3^{2k+1}+2^{2k+1}=9\cdot3^{2k-1}+4\cdot2^{2k-1}=9\cdot(5p-2^{2k-1})+4\cdot2^{2k-1}=\\
45p-5\cdot2^{2k-1}=5(9p-2^{2k-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solve $x^2 + 12 = y^4$ in the integers.
Solve $x^2 + 12 = y^4$ in the integers.
My thought was to factor $x^2 - y^4$ = $(x - y^2)(x + y^2) = -12$. I'm not sure if this is the right approach, and I'm not sure where to go from here.
| Rearrange as
$$y^4-x^2=12$$
Because they are squares, any positive value of $y$ or $x$ has a corresponding negative value, so it’s sufficient to find the positive values.
$$(y^2-x)(y^2+x)=12$$
So $[(y^2-x),(y^2+x)]$ is $(1,12)$ or $(2,6)$ or $(3,4)$
(Note: because of the symmetry, we can ignore $(12,1)$ or $(6,2)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the inequality $(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$
Solve the inequality $$(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$$
My work so far:
1) $a>0$
2) Let $f(x)=(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)$.
$f(x)=0$ if and only if $x^2-2\sqrt{a}\cdot x+1=0$ or $2^x+\lg a=0$
$$x^2-2\sqrt{a}\cdot x+1=0$$
$$x_{1,2}=\sqrt a \p... | you must do case work for $a$:
1) $$a>1$$ then we have $$2^x+\lg(a)>0$$ and we have to solve $$x^2-2\sqrt{a}x+1<0$$ this is equivalent to
$$(x-\sqrt{a})^2<a-1$$ and further
$$|x-\sqrt{a}|<\sqrt{a-1}$$
can you proceed?
2)$$a=1$$ then we have $$(x^2-2x+1)<0$$ which is impossible.
3) the case $$0<a<1$$ is for you!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+... |
I entered this into Wolfram alpha and this is what it gives me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How t... | You're almost there. You had
$$ \frac{a+2b+c}{(a+b)(b+c)} = \frac{3}{a+b+c} $$
which implies
$$ a^2 + ab + ac + 2ab + 2b^2 + 2bc + ac + bc + c^2 = 3ab + 3b^2 + 3ac + 3bc $$
so
$$ a^2 + c^2 - ac = b^2 $$
Now recall the cosine rule $b^2 = a^2 + c^2 - 2ac\cos(\angle B)$. This means that $\angle B = \arccos(\frac{1}{2}) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone woul... | First of all get rid of $x^2+y^2$
$$ \frac{12}{4x} +\frac{4}{5y} = 2,\quad 2x+6y = 5xy $$
It is a rectangular hyperbola passing through origin with asymptotes parallel to axes reminding one of $ x= t, y= 1/t $ when axes are asymptotes. So solve for $x,y$
$$ x= \frac{6y}{5y-2},\, y= \frac{2x}{5x-6}$$
The asymptotes ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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.