Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all integer values of $m$ such that the equation $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$ has exactly four distinct real roots.
Find all integer values of $m$ such that the equation $$\large \sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ has exactly four distinct real roots.
$\left(x \in [0, 9], m \in ... | We have $$\sqrt{x}+\sqrt{9-x}=\sqrt{3m-x^2+9x},$$ which is for $0\leq x\leq9$ and $3m-x^2+9x\geq0$ is equivalent to
$$9+2\sqrt{x(9-x)}-x(9-x)=3m$$ or
$$10-\left(1-\sqrt{x(9-x)}\right)^2=3m,$$ which gives $3m\leq10$ and $m\leq3.$
Also, since $$\sqrt{x}+\sqrt{9-x}=\sqrt{9+2\sqrt{x(9-x)}}\geq3,$$ we obtain:
$$3m-x^2+9x\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$.
I'm thinking to show that $n$ is divisible by each of the prime factors of $930$, is that right? I'm stuck
| $N=5 + .... + 5^{150} = 5(1+...... + 5^{149} = 5\cdot \frac {5^{150}-1}{4}$.
$930| N \iff $
$4*930| 4*N = 5(5^{150}-1) \iff $
$\frac {4*930}5=\frac{4*93*10}5=4*93*2|5^{150}-1\iff$
$8*3*31|5^{150}-1\iff$
$8|5^{150}-1$ and $3|5^{150}-1$ and $31|5^{150}-1\iff$
$5^{150}\equiv 1\pmod {8,3,31}$.
And we can's Euler's Th|Ferma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$
Question: Show that
$$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$
From Wolfram alpha, it seems that the equality above is indeed correct.
But I do not know how to prove it.
Any hint is appreciated.
| The Generalized Binomial Theorem says
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac12\frac{x}{1!}+\frac12\!\cdot\!\frac32\frac{x^2}{2!}+\frac12\!\cdot\!\frac32\!\cdot\!\frac52\frac{x^2}{3!}+\cdots\\[6pt]
&=\sum_{k=0}^\infty(2k-1)!!\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\infty\frac{(2k!)}{2^kk!}\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
What are the equations of the 3D transformation matrices? Please have a look at the image below. There is a transformation matrix and graphical representation of 2 coordinate systems. I understand what the numbers in the matrix represent, but I can't figure out how we could write this in terms of equations.
I've been... | The equation for this matrix comes from transformation rule:
$$
\begin{pmatrix}x_1\\y_1\\z_1\\1\end{pmatrix} =
\begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\
R_{21}&R_{22}&R_{23}&t_2\\
R_{31}&R_{32}&R_{33}&t_3\\
0&0&0&1\end{pmatrix}
\begin{pmatrix}x\\y\\z\\1\end{pmatrix} $$
$$
x_1=R_{11}x+R_{12}y+R_{13}z+t_1,\\
y_1=R_{21}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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At least one perfect square in $[S_n,S_{n+1}]$ Given $\{a_n\}$ a sequence of real numbers such that $a_1 > 1$ and $a_{n+1}-a_n\geq 2$, with $S_n=a_1+a_2+...+a_n$, prove that for any $n\geq 3$, there is at least one perfect square in the range $[S_n, S_{n+1}]$.
I was thinking that $a_{n+1}\ge 2+a_n \ge 4+a_{n-1}\geq ..... | Let $(a_n)$ be a sequence of real numbers with $a_1 \geq 0$ and $a_{n+1} - a_n \geq 2$. Define $S_n = a_1 + \cdots + a_n$. We will show that $[S_n, S_{n+1}]$ contains a perfect square for all $n \geq 1$.
Lemma. Let $0 \leq x < y$ be real numbers with $y -x \geq 2\sqrt x + 1$. Then $[x, y]$ contains a perfect square.
P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(... | Your error comes from the fact that you cannot say that if $M_1M_2=0$ that $M_1=0$ or $M_2=0$...
For example :
$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
AM GM inequality with constants I was trying to find the minimum value of $a + \frac{1}{a} + 3$ given that $a$ is a positive real number. I found $2$ ways to do it.
Minimum value of $a + \frac{1}{a}$ is $2$ using the AM-GM inequality and therefore the minimum value of $a + \frac{1}{a} + 3$ is $2 + 3 = 5$
Then, I tried ... | (1) You correctly used the AM-GM inequality when you showed that
$$a+\frac{1}{a}\ge2,$$
and of course, this implies that
$$a+\frac{1}{a}+3\ge5.$$
Also, you correctly used the AM-GM inequality when you showed that
$$a+\frac{1}{a}+3\ge3\cdot3^{\frac{1}{3}}.$$
Note that
$$5>3\cdot3^{\frac{1}{3}}\approx4.32,$$
so your firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\... | A solution without derivatives
Clearly, $t \ge 2$ or $t \le -2$.
Denote the minimum of $f(t)=\frac{(t-1)^3}{t^3-3t-1}$ on $|t|\ge 2$ by $m > 0$.
Since $f(3) = \frac{8}{17}$, we have $m \le \frac{8}{17}$.
Note that $f(t) - 1 = \frac{3t(2-t)}{t^3-3t-1} \ge 0$ for $t \le -2$.
Thus, the minimum occurs on $t \ge 2$.
$f(t) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
... | The second integral can be carried as follows,
$$\int\frac{1-x}{x^2+x+1}dx=\int\frac{-(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34}dx$$
$$=-\frac12\int\frac{d[(x+\frac12)^2]}{(x+\frac12)^2+\frac34}
+\frac32\int\frac{dx}{(x+\frac12)^2+\frac34}$$
$$=-\frac12\ln\left[(x+\frac12)^2+\frac34\right]+\frac{1}{\sqrt{3}}\arctan\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$ I'm trying to evaluate
$$\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$$
I've found that it is
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3+x^3\ln x +\text{o}(x^3)}$$... | What you do is right! Follow your method:
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\left(1+\ln x +\frac{\text{o}(x^3)}{x^3}\right)}=\lim_{x\to0^+}\frac{-\frac{1}{2}+\text{o}(1)}{1+\ln x +\text{o}(1)}=0.$$
Actually,
$$(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}\sim\frac{-1}{2}x^3.$$
So you also can evaluate this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which one is larger? I want to prove that for positive integer $n$,
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n}$$
but I am stuck on how to proceed. Can someone help me?
| Note that, for $n \gt 1$,
$$\frac{1}{1 - \frac{1}{n}} = \frac{n}{n - 1} = 1 + \frac{1}{n - 1} \tag{1}\label{eq1A}$$
Using \eqref{eq1A}, for $n \gt 1$, multiplying both sides of your proposed inequality by $\left(\frac{1}{1 - \frac{1}{n}}\right)^n$ gives
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice:
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$.
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sur... | By the given $$\frac{5}{3}<x<\frac{7}{3}$$ and we need to prove that:
$$4<x^2+x<8,$$ which is true because $x^2+x$ increases for $x>0$:
$$x^2+x<\left(\frac{7}{3}\right)^2+\frac{7}{3}=\frac{70}{9}<8$$ and $$x^2+x>\left(\frac{5}{3}\right)^2+\frac{5}{3}=\frac{40}{9}>4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(... | $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$
At least one pair from $(a+b) , (b+c) , (a+c)$ is even number.
Therefore, $6 | 3(a+b)(b+c)(a+c)$. Consequently,
$a^3+b^3+c^3 \equiv (a+b+c)^3 \mod{6}$
$a^3+b^3+c^3 \equiv a+b+c \mod{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$ Must I verify that Left hand side(LHS)= Right hand side(RHS)or can I prove that RHS= LHS?
I don’t know how to prove from LHS=RHS. How to separate the $4r^3+r$ into two terms, i.e. $\displaystyle\Bigl(r+\frac{1}{2}\Bigr)^4-\Bigl(r-\frac{1}{2}... | $(x+y)^{4}-(x-y)^{3}=8x^{3}y+8xy^3$
Just substitute $x=r$ and $y=\frac{1}{2}$ to obtain
$4r^{3}+r=(r+\frac{1}{2})^{4}-(r-\frac{1}{2})^{4}$
As for the infinite sum,
$\sum_{r=1}^{n}({4r^{3}+r})=-\left( \frac{1}{2} \right)^4+\left( \frac{3}{2} \right)^4 - \left( \frac{3}{2} \right)^4+...+\left( n + \frac{1}{2} \right)^4$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displays... | We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}\color{blue}{(-4)^k\binom{n+k}{2k}}&=\sum_{k=0}^n(-4)^k\binom{n+k}{n-k}\\
&=\sum_{k=0}^n(-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrles... | The limit is equal to $-\dfrac{1}{2}$.
A LOWER bound: for $0<x<1$, $\log(x)<0$ and
$$\begin{align}
\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}
&\geq
\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n+\binom{n}{3}x^2}\\
&\geq
\frac{1}{2\ln x}\int_{1/2}^{\infty}\frac{ds}{s+\frac{s^3x^2}{6}}\\
&=\frac{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$... | Let $x=a+b$ and $y=\frac{ab}{a+b}$.
\begin{align}
a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} &=(a+b)^2-2ab+\left(\frac{ab}{(a+b)}\right)^2-\frac{2((a+b)^2-ab)}{a+b}
\\&= x^2-2xy+y^2-2(x-y)=\underset{\text{quadratic in } x-y}{\underbrace{(x-y)(x-y-2)}}\\
&\geq -1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to solve a rotated ellipse equation for y? Given an x, I want to get the value of y on the Cartesian plane.
It's easy to solve it for a non-rotated ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{1}$$
$$y = \frac{b}{a} \sqrt{(a^2 - x^2)} \tag{2}$$
How do I solve it when the ellipse is rotated and the equation ... | Your third equation can be written as
\begin{eqnarray*}
\color{blue}{y^2} \left( \frac{ \sin^2 \theta}{a^2}+ \frac{ \cos^2 \theta}{b^2} \right) + 2x\color{blue}{y} \sin \theta \cos \theta \left( \frac{ 1}{a^2}- \frac{ 1}{b^2} \right)+x^2 \left( \frac{ \cos^2 \theta}{a^2}+ \frac{ \sin^2 \theta}{b^2} \right)-1=0.
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\... | Calculate the limit of inverse:
$g(x)=\frac{6\sin\left(\frac{\pi x}{18}\right)-x}{3-x}=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)+3-x}{3-x}$
$=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)}{3-x}+1$
However
$\sin\left(\frac{\pi x}{18}\right)-\sin\left(\frac{\pi}{6}\right)=2\sin\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summa... | We seek to evaluate
$$\sum_{k=0}^n \frac{(-1)^k}{k+1} {n\choose k}^2.$$
This is
$$\frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose k+1}
= \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose n-k}
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1}
\sum_{k=0}^n (-1)^{k} {n\choose k} z^k
\\ = [z^n] (1+z)^{n+1} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Showing that $1+2020^y$ ($y\in \mathbb{Z}$) is not a perfect square for even $y$ it is obvious that $1+2020^y$ is not a perfect square since $2020^y$ is a perfect square $+1$.+
for uneven $y$, i wanted to show that $z^2-1=(z-1)(z+1)=2020^y$. How can I show that I can't distribute the factors of $2020^y$ into two intege... | To answer the question you asked: "How can I show that I can't distribute the factors of $2020^y$ into two integers n and n+2?"
$2020^y = 101^y*2^{2y}*5^y$.
If $n= z-1, n+2 = z+1$ then $\gcd(z+1, z-1) = \gcd((z+1)-(z-1),(z-1))=\gcd(2,z-1)=\begin{cases}1\\2\end{cases}$
As $z-1, z+1$ are either both even or both odd, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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What's the sum of elements of this multi-set? Let S be a $\mathbf{multi}$ set of integers such that:
$$S = \left \{\min(\mathbf{C}, a + b - 1) | a \in \left [ \mathbf{A} \right ], b \in \left [ \mathbf{B} \right ] \right \}$$
*
*Note that $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are positive integers and we know ... | If $C \ge A+B-1$, then the min against $C$ will have no effect and the sum just reduces to $$\sum_{a=1}^A \sum_{b=1}^B (a + b - 1) = BA(A+1)/2 + AB(B+1)/2 - AB = AB(A+B)/2.$$
Let’s assume now that $A \le B \le C < A+B-1$. Then some of the values $a+b-1$ get replaced by the smaller $C$, so we just need to add up the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$ Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$
edit: the homework might be $5xy+\ln(x^2y)=2$, but this is close enough to get the point
$Solution:$
First lets use properties of $\ln$ to write our equation in a way that will be e... | $$5xy+ \ln(xy^2)=2$$
Differentiating w.r.t $x$
$$5y+5xy'+\frac{y^2+2xyy'}{xy^2}=0$$
$$y'=-\frac{(5xy+1)y}{x(2+5xy)}=\frac{dy}{dx}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving System of Symmetric Equations I am to solve the system below.
$$\begin{cases} (x^2+1)(y^2+1)=10 \\ (x+y)(xy-1)=3 \end{cases}$$
I have tried to factor the equations of the system but they do not factor. How can I approach the problem?
| We have $$x^2+y^2+x^2y^2=9$$ or
$$(x+y)^2-2xy+x^2y^2=9$$ and since $$x+y=\frac{3}{xy-1},$$ we obtain:
$$\frac{9}{(xy-1)^2}-2xy+x^2y^2=9$$ or
$$\frac{9(2-xy)xy}{(xy-1)^2}-xy(2-xy)=0$$ or
$$xy(2-xy)(4-xy)(2+xy)=0.$$
Now, $xy=0$ gives $(-3,0)$ and $(0,-3)$;
$xy=2$ gives $x+y=3$ and $(2,1)$ and $(1,2)$;
$xy=4$ does not gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arccos telescopic sum: $\sum_{k=1}^{n} \arccos \frac{2k^2}{4k^4+1}$ Compute the finite sum:
$\arccos \frac{2 \cdot 1^2}{4\cdot 1^4+1}+...+\arccos \frac{2 \cdot n^2}{4\cdot n^4+1}$
Please help. I have no idea what to do. There should be a telescopation but I can’t find it.
| This is not a complete answer, just a try. By partial fraction decomposition, you get
$$z_k:= \frac{2k^2}{4k^4+1} = \frac{k}{2(2k^2-2k+1)} - \frac{k}{2(2k^2+2k+1)} $$
Then you could use the fact that
$$\arccos(z) = \frac \pi 2 - \arcsin(z) $$
and the identity
$$\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)=\arcsin(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cd... | You can also use the root test together with the inequality between geometric and arithmetic mean (GM-AM):
$$\sqrt[k]{\frac{k!}{k^k}}= \frac{\sqrt[k]{k!}}{k}\stackrel{GM-AM}{\leq}\frac{\frac{k(k+1)}{2k}}{k}=\frac{k+1}{2k}\stackrel{k\to\infty}{\longrightarrow}\frac 12 <1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Easy way to determine the sign of Gaussian curvature without explicit computation. I know how to compute Gaussian curvature via the first and second fundamental form. i.e.,
\begin{align}
K = \frac{eg-f^2}{EG-F^2}
\end{align}
In this computation one usually set $n = \frac{X_u\times X_v}{||X_u\times X_v||}$
Without det... | I suppose you are considering surfaces (ie 2-dimensional) embedded into $\mathbb{R}^3$. You can try to formalize the intuition that montains and valleys have positive curvature and saddle points have negative curvature.
Compute the tangent plane of the surface at a point. If in a small neighbourhood of the point, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all positive integer pairs $(a, b)$ such that $(ab + a + b) \mid (a^2 + b^2 + 1)$.
Find all positive integer pairs $(a, b)$ such that $$(ab + a + b) \mid (a^2 + b^2 + 1)$$
Let $a^2 + b^2 + 1 = k(ab + a + b), k \in \mathbb N, k \ge 1$.
For $k = 1$, we have that $$a^2 + b^2 - ab - a - b + 1 = 0$$, where $(a, b) = ... | Reference is HURWITZ 1907
Caution: this is about POSITIVE $x,y$ values. For $k=5,$ there is a solution at $-5,-3.$
The main thing is that the jumping step leads to inequalities for what Hurwitz called ground solutions. We have
$$ x^2 - kxy + y^2 -kx-ky +1=0 $$
To jump, the coefficient of $x$ is $-ky-k.$ The two $x$ so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ I am trying to prove $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ for an acute triangle. There is:
$$\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt{3\sum\frac{b^2+c^2-a^2}{a^2+2bc}}$$
So we have to prove: $\sum\fra... | Let $x=\sqrt{b^2+c^2-a^2}, y=\sqrt{c^2+a^2-b^2}$ and $z=\sqrt{a^2+b^2-c^2}$ (they are positive real numbers due to acute-angled restriction). Then:
$$a=\sqrt{\frac{y^2+z^2}{2}},\ b=\sqrt{\frac{z^2+x^2}{2}},\ c=\sqrt{\frac{x^2+y^2}{2}}$$
and the inequality is equivalent with:
$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $ \min_{a,b,c \in \mathbb{R}} \int_{-1}^1 |e^x-a-bx-cx^2|^2dx $ The question is the following:
Compute
$$ \min_{a,b,c \in \mathbb{R}} \int_{-1}^1 |e^x-a-bx-cx^2|^2dx $$
I tried to expand the integral, and it gives me:
$$
\int_{-1}^1 |e^x-a-bx-cx^2|^2dx= 2a^2+\frac{2}{3}(2ac+b^2)+\frac{2}{e}(a-2b+5c)-2e(a+... | Hint.
You can compute the derivative of $f(a,b,c):=\int_{-1}^{1} | e^x - a - bx - cx^2 |^2 dx$ with respect to $a$, $b$ and $c$ and set them to zero to find the values of $a,b$ and $c$ that minimise that expression.
Values for comparison
We have $\frac{df(a,b,c)}{da} = 4 a + \frac{4c}{3} - 2( e - e^{-1})$ and $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$... | $$y-1=\dfrac{2t-3}{t^2-t+2}$$
Let $z=3-2t\implies5\ge z\ge1$
$$\dfrac4{1-y}+4=4+\dfrac {z^2-4z+11}z=z+\dfrac{11}z\ge2\sqrt{z\cdot\dfrac{11}z}$$ which is attained if $z^2=11$
$$y\ge?$$
Now for $5\ge a>b\ge1,$
$$a+11/a-(b+11/b)=\dfrac{(a+b-11)(a-b)}{ab}<0$$
So, $z+\dfrac{11}z$ will be maximum for the the minimum value of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\arctan{x}+\arctan{y}$ from integration I was trying to derive the property
$$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$
for $x,y>0$ and $xy<1$ from the integral representation
$$
\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.
$$
I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}... | Fixing $y$, define $f(x):=\arctan x+\arctan y-\arctan\frac{x+y}{1-xy}$ so $f(0)=0$ and$$\begin{align}f^\prime(x)&=\frac{1}{1+x^2}-\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\partial_x\frac{x+y}{1-xy}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1-xy-(x+y)(-y)}{(1-xy)^2}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of $\frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) -x }{\tan x - x}$ as $x \to 0$ $$\lim_{x \to 0} \frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) -x }{\tan x - x}$$
I tried to solve this using L'Hopital rule but the resulting differential got too messy
$$=\lim_{x \to 0} \frac{e^{\tan x}\sec^2x - e^x + \sec... | This can also be accomplished with power series. First, the denominator because it's easier:
\begin{multline}
\tan(x) - x =\left[x + \frac{x^3}{3} + O(x^4)\right] -x = \frac{x^3}{3} + O(x^4).
\end{multline}
Thus, we need to expand the numerator out to third order to determine the limit:
\begin{multline}
e^{\tan{x}}-e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove, by the process of Mathematical Induction, that: $\sqrt 1 + \sqrt 2 + ... + \sqrt{n} < \frac{4n+3}{6}\sqrt{n}$ for all integers n > 0. So what I am stuck on is proving the inequality for n=k+1. I have my teacher's solution, but it doesn't quite make sense to me.
Teacher's solution:
$\sqrt 1 + \sqrt 2 + ... + \sqr... | Hypothesis:
$$\sum_{j=1}^k \sqrt{j} < \frac{4k+3}{6}\sqrt{k}$$
We wanted to show that
$$\sum_{j=1}^{k+1} \sqrt{j} < \frac{4(k+1)+3}{6}\sqrt{k+1}$$
From the hypothesis, we have
$$\sum_{j=1}^{k+1} \sqrt{j} < \frac{4k+3}{6}\sqrt{k}+\sqrt{k +1}$$
Hence it suffices to verify that
$$\frac{4k+3}{6}\sqrt{k}+\sqrt{k +1} < \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Defi... | Hint:
\begin{align*}
\frac{\partial}{\partial x}\left(x\cos\left(y\right)\right)-\frac{\partial}{\partial y}\left(\sin\left(y\right)\right)&=\cos\left(y\right)-\cos\left(y\right)=0
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck ... | The author finds the identity
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}],$$
by applying the quadratic formula to the previous equation, which is a quadratic in $a$:
$$a^2+b(2-n)\cdot a+b^2=0.$$
Plugging the coefficients into the quadratic formula yields
\begin{eqnarray*}
a&=&\frac{-b(2-n)\pm\sqrt{(b(2-n))^2-4b^2}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that for every $n\in\mathbb{N}$, $n^2$ is divisible by 3 or has a form $3k+1$? I tried to do this by induction, but it doesn't make any sense: $n^2=3k$ or $n^2=3k+1$
*
*option: $(n+1)^2= 3k+2n+1$
*option: $(n+1)^2= 3k+2n+2$
Is there any other way on proving this problem?
| Because you wanted to proceed by induction ...
Base case: $0^3 = 0$ is divisible by 3
For any $n$, we have 2 cases.
First take $(n + 1)^2 = n^2 + 2n + 1$
Case 1: $n^2$ is divisible by 3
Let $n = 3k$. What can we conclude?
Case 2: $n^2$ is $3k + 1$.
$$ n^2 + 2n + 1 = n^2 - 1 + 2(n + 1) = (n - 1)(n + 1) + 2(n + 1)$$
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2... | You can try $$x^4+2x^3y+3x^2y^2+2xy^3+y^4= x^2y^2({x^2\over y^2}+2{x\over y}+3+2{y\over x}+ {y^2\over x^2})$$
Let $t= {x\over y}+{y\over x}$ then we have ${x^2\over y^2}+{y^2\over x^2}=t^2-2$, so $$...= x^2y^2(t^2-2+2t+3) =x^2y^2(t+1)^2= $$ $$=x^2y^2({x\over y}+{y\over x}+1)^2 =(x^2+y^2+xy)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How to evaluate : $\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)$ Thank you to &robjohn for proving this question, I got motivated by it, so I went on to investigate this sum $(1)$,
$$S_{x}=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\rig... | This seems to be a telescoping series:
$$
\begin{align}
S_{x}
&=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)\\
&=\sum_{n=0}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}-\sum_{n=2}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}\\
&=\frac{F_x}{F_{x+1}^2}+\frac{F_{x+1}}{F_{x+2}^2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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solution of the cubic equation $x^3 = 0$? it is clear that one solution of the above equation is $x = 0$, but since it is a cubic equation so it should have $3$ complex roots, so what will be its other roots? Also how can we determine that a cubic equation has $3$ distinct roots or $2$ same $1$ distinct or all distinct... | First,
$$ (x-0)(x-0)(x-0) = x^3 $$
exhibits the three roots of $x^3$. They're all $0$.
Much like the discriminant of the quadratic, the discriminant of the cubic indicates the types of the roots. The polynomial $x^3$ has discriminant $0$ so at least two roots are the same. Generally, if the coefficients of a cubi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate o... | $\frac{(2 x^2+x y-y^2+x z-2 y z-z^2)}{(2 y^2-x^2+x y-2 x z+y z-z^2)}=\frac{(2x-y-z)(x+y+z)}{(2y-x-z)(x+y+z)}=\frac{(2x-y-z)}{(2y-x-z)}.
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$x^{3}+ax^2+bx+c$ has all roots negative real numbers and a<3. Establish an inequality between only b and c
A cubic equation $x^{3}+ax^2+bx+c$ has all negative real roots and $a, b, c\in R$ with $a<3.$
Prove that $b+c<4.$
My attempt :
Let the cubic be $f(x)$
Plotting graph we see that ,
$f(x\geq 0)>0$.
So we can see ... | We can write our polynomial in the form $(x+r_1)(x+r_2)(x+r_3)$, where $r_i$ are positive real numbers. Note that $b=r_1r_2+r_2r_3+r_3r_1$ and $c=r_1r_2r_3$.
$\color{Green}{\text{Maybe you will find}}$ Maclaurin's inequality $\color{Green}{\text{very interesting}}$.
By Maclaurin's inequality, we know that $\sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integral involving elliptic integral functions Recently I came across this identity:
$$\int _0^1\:\frac{K\left(x\right)}{1+x}dx=\frac{\pi ^2}{8}$$
Where:
$$K\left(x\right)=\int _0^{\frac{\pi }{2}}\:\frac{1}{\sqrt{1-\left(x\sin \left(\theta \right)\right)^2}}d\theta $$
Any hints to how to prove this identity?
| For $a \in (0,1)$ we have
\begin{align}
\int \limits_0^1 \frac{\mathrm{d} x}{(1+x) \sqrt{1-a^2 x^2}} &\stackrel{x = \frac{2 t}{a (1+t^2)}}{=} \frac{2}{a} \int \limits_0^\frac{1 - \sqrt{1-a^2}}{a} \frac{\mathrm{d}t}{1 + \frac{2}{a} t + t^2} = \frac{2 a}{1-a^2} \int \limits_0^\frac{1 - \sqrt{1-a^2}}{a} \frac{\mathrm{d}t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$
If $x$ and $y$ are rational numbers such that : $$(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$$ then which of the following is true?
A) $\;x=1$, $y=1$
B) $\;x=2$, $y=1$
C) $\;x=5$, $y=1$
D) $\;x$ and $y$ can take infinitely ... | And why is this down voted?
$$2y-x = \sqrt{2}\Big(2y-x+\sqrt{3}(x-y-1)\Big)\;\;\;\;\;\;/^2$$
$${(2y-x)^2\over 2}+x-2y = \sqrt{3}(x-y-1)$$
If $x-y-1\ne 0$ we have
$$\sqrt {3}= \underbrace{{(2y-x)^2\over 2}+x-2y\over x-y-1}_{\in\mathbb{Q}}$$
which is impossible. So $y=x-1$ and $${(2y-x)^2\over 2}+x-2y=0$$
Solve this sys... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(... | You can do directly about the $x,y$ : $ 0 \le \dfrac{x^2}{\sqrt{x^2+y^2}} \le |x|$ and similarly for the other one. This means ...the expression in the $\cos \to 0$ and the answer is clearly $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality involving the angle bisectors of a triangle
Let $l_a,l_b,l_c$ denote the lengths of angle bisectors of a triangle with sides $a,b,c$ and semiperimeter $s$. I am looking for the best constant $K>0$ such that
$$l_a^2+l_b^2+l_c^2> K s^2.$$
I found that $K=2/3$ works, but I suspect that best constant is $K=8... | As pointed out by Michael Rozenberg, for $K=8/9$ we have to show that
$$\frac{bc(s-a)}{(b+c)^2}+
\frac{ca(s-b)}{(c+a)^2}+\frac{ab(s-c)}{(a+b)^2}>\frac{2s}{9}.$$
Let $a=x+y$, $b=y+z$, $c=z+x$ with $x,y,z>0$, then the inequality is equivalent to
$$\sigma_1^3(\sigma_1^2-4\sigma_2)^2+6\sigma_1\sigma_3(\sigma_1\sigma_2
-9\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Quadrilateral inscribed in semicircle. How to find BC using sin ABD? AB = 3, BD = 5 , tanABD = 0.75
BC is diameter
Question : How to find BC using sin ABD?
I can find sin cos ABD and lenght AD.
I can find BC using Ptolemy's theorem.
| Use $$\measuredangle ABD=\arccos\frac{3}{x}-\arccos\frac{5}{x},$$ where $BC=x=2R$.
Thus, $$\frac{3}{4}=\frac{\sqrt{\frac{x^2}{9}-1}-\sqrt{\frac{x^2}{25}-1}}{1+\sqrt{\frac{x^2}{9}-1}\cdot\sqrt{\frac{x^2}{25}-1}}$$
Can you end it now?
I got $BC=\frac{5}{3}\sqrt{10}.$
Ptolemy also helps.
Indeed, $AD=\sqrt{10},$ $AC=\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt... | Substitute $x = \sin 2t $ to have
$$\sqrt{1-x} = \cos t - \sin t, \>\>\>\>\>\sqrt{1+x} = \cos t + \sin t$$
and,
$$\begin{align}
& \int_0^1 \frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2} \\
& = \int_0^{\pi/4}\frac{\cos2t}{1+\cos t}dt \\
&= \int_0^{\pi/4}\frac{2(1+\cos t)^2 -4 (1+\cos t) +1}{1+\cos t}dt \\
& =\int_0^{\pi/4} \left(-2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Challenging integral with root in denominator I'm wondering if anyone has ideas to get started with this integral:
$$
I = \frac{b}{\pi}\int_{-\infty}^{\infty}dx\,\frac{1}{\sqrt{\left[(x-1)^2+b^2\right]\left[(x+1)^2+b^2\right]}}
$$
for $b>0$. Mathematica 12 can't get it done (except for $b=1$) and the most obvious metho... | We have
\begin{align}
I &= \frac{b}{\pi} \int \limits_{-\infty}^\infty \frac{\mathrm{d} x}{\sqrt{[(x-1)^2 + b^2][(x+1)^2+b^2]}} = \frac{2b}{\pi} \int \limits_0^\infty \frac{\mathrm{d} x}{\sqrt{[(x-1)^2 + b^2][(x+1)^2+b^2]}} \\
&= \frac{2b}{\pi} \int \limits_0^\infty \frac{\mathrm{d} x}{\sqrt{(1+b^2)^2 - 2 (1-b^2) x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive a,b,c,d, if $a^2+b^2+c^2+d^2+abcd=5$, show $a+b+c+d\leq 4$. One can use Lagrange multiplier, but I am looking for a more elementary proof.
I try to find the maximum of $a+b+c+d$ and follow the standard approach.
Construct $a+b+c+d+\lambda (a^2+b^2+c^2+d^2+abcd-5)$.
One can obtain $1+2a\lambda +bcd =0$ and... | Let $a+b+c+d>4$, $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $$x+y+z+t=4.$$
Thus, $$k(x+y+z+t)>4,$$ which gives $$k>1.$$
But, $$5=a^2+b^2+c^2+d^2 + abcd=k^2(x^2+y^2+z^2+t^2)+k^4xyzt>x^2+y^2+z^2+t^2+xyzt,$$
which is a contradiction because we'll prove now that $$x^2+y^2+z^2+t^2+xyzt\geq5$$ or
$$16(x^2+y^2+z^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Change of Variables in Multiple Integrations (rotated ellipse) Let $D$ be an ellipse rotated counterclockwise by $\frac{\pi}{3}$ radians, as pictured below. Before rotation, the horizontal radius was 2 and vertical radius was 6.
picture referenced
Convert $\iint\limits_{D} \, \bigl(8x + 4y\bigr)\, dA$ into an integral ... | To follow along, draw three copies of the plane. Label one set of axes $x,y$, another set of axes $u,v$ and the third set of axes $\xi, \eta$. In the $(x,y)$ plane, draw the rotated ellipse $D$, as in the figure. In the $(u,v)$ plane, draw the same ellipse, but with semi-major axes along the coordinate axes; i.e the eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^\frac{\pi}{2} \min (\sin x, \frac{1}{2}) dx$
Evaluate
$\int_0^\frac{\pi}{2} \min (\sin x, \frac{1}{2}) dx$
Now, what I've tried to do is to write $\min (\sin x, \frac{1}{2})$ : $\sin x \leq \frac{1}{2}$ and $\frac{1}{2} < \sin x$. but I don't quite understand how to choose the sets for each m... | Notice two things:
*
*On the interval $(0,\frac{\pi}{2})$, function $\sin{x}$ is always increasing.
*$\sin(0) =0$
That means that $\min \left(\sin x, \frac{1}{2} \right)$ will be equal to $\sin x$ when $x<\frac{\pi}{6}$ because $\sin{\frac{\pi}{6}}=\frac{1}{2}$. Now we have
$$
\int_0^{\frac{\pi}{2}} \min \left(\sin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
| Better version.
Notice that $3 \mid x!$ for $x \geq 3$ and $3 \not \mid 2^d$. Therefore, at least one of $a, b, c \leq 2$. WLOG let $a\leq b\leq c$.
If $c\leq 1$, $a\leq b\leq c\leq 1, a!=b!=c!=1$ which gives no solution.
If $c=2$, $a!+b!+2=2^d$. $a,b\in \{0,1\}$ gives $4$ solutions, while $(a,b)=(1,2)$ and $(2,2)$ doe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove thi... | We will prove a more general result.
Let
$$
f(x)=\frac{\sin nx}{\sin x}=\sum_{k=0}^\infty a_kx^{2k}\tag1
$$
and
$$
f_m(x)=\sum_{k=0}^m a_kx^{2k}.\tag2
$$
Then the following inequalities hold:
$$\begin{cases}
f(x)\le f_m(x),& m\text{ even}\\
f(x)\ge f_m(x),& m\text{ odd}\\
\end{cases}.\tag3
$$
Your inequality will the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Conditional probability of continuous throwing a dice A person will keep throwing a dice until he gets $6 .$ If he gets 6 then the experiment will be stopped.
The probability that experiment ends at 6 th trial, given that prime number does not appear is
(A) $\left(\frac{2}{5}\right)^{5}\left(\frac{1}{6}\right)$
(B) $\l... | In a single roll of the dice, the only way a prime number does not appear is, if the die lands $2,4,6$. Restricting our attention to these three outcomes, the probability that the game does not end, conditional on a prime not appearing is $2/3$, as you are interested only in $2,4$. All trials are independent. So, if we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is tr... | $2^{k} + 2 < 2^{k + 1}$
$2^{k} + 2< 2\cdot 2^{k}$
$2< 2^{k}$
This is trivially true for $k = 3$, and $2^{k}$ is an increasing function, so the inequality is true and the induction is complete. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Simplifying $\frac{ \cos^2 x - \sin^2 x }{\sin{2x}}$ Please consider the following problem and my answer to it:
Problem:
Simplify the following expression:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} $$
Answer:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{ \cos^2 x - \sin^2 x }{ 2 \sin x \cos x} $$
$$ \frac{ \cos^2 ... | Continue with
$$ \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x}
=\frac1{2\tan x} -\frac12 \tan x
= \frac{1-\tan^2x}{2\tan x}=\frac1{\tan 2x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy... | Solving for $p$ we get
$$
y' = \frac{3x\pm\sqrt{9x^2-4ay^2}}{2ay}
$$
or
$$
2ay y'=3x\pm\sqrt{9x^2-4ay^2}
$$
now calling $u = a y^2$ we have
$$
u' = 3x\pm\sqrt{(3x)^2-4u}
$$
and now following with $z = (3x)^2-4u$ we arrive at
$$
z'\pm 4\sqrt{z}=6x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Let T be linear map on $P_2(R)$. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix. Let T be linear map on V. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix.
$V = P_2(R), T(f(x)) = xf'(x)+f(2)x+f(3)$.
I was ab... | An eigenvalue of a linear operator $T$ consists of a scalar $\lambda$ such that $Tv = \lambda v$, for some $v\in V\backslash\{0\}$.
This is the same as to require that $\ker(T - \lambda I) \neq \{0\}$. This happens iff $\det([T]_{\mathcal{B}} - \lambda I) = 0$ for some (arbitrary) matricial representation of $T$.
At yo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this has real roots $(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$ Prove this has real roots
$(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$
My Work
\begin{align*}
\Delta&=(a+b)^2(a-c)^2-4a(b-c)(a^2-bc) \\
&=a^4+2a^3c+a^2c^2-2a^3b+b^2a^2-4a^2bc-2abc^2+2ab^2c+b^2c^2.
\end{align*}
How do I show that this is positive? Simplification doesn'... | Note that $-1$ is a root, so the other root is also real. It turns out to be $\dfrac{a(b-c)}{bc-a^2}$, but that's not really necessary for what you need to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $ \sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x} $ I am reviewing my high-school math (year 10-ish) but I am hitting a wall with this nasty inequality:
$$
\sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x}
$$
I find (steps below) the solution to be
$$
1 \le x \le 3 \text{ with } x \ne 2
$$
but the text-book and W... | First of all for real cases, $$3\ge x\ge1$$
Observe that $x+3-x=x-1+4-x$
We need $$\sqrt x-\sqrt{x-1}>\sqrt{4-x}-\sqrt{3-x}$$
$$\iff\sqrt x+\sqrt{3-x}>\sqrt{4-x}+\sqrt{x-1}$$
As both sides are $>0$
we can safely take square in both sides
$$x+3-x+2\sqrt{x(3-x)}>4-x+x-1+2\sqrt{(4-x)(x-1)}$$
Again as both sides are $>0$
w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If x,y,z is positive then find the minimum value of $x+y+z+\frac{1}{x} + \frac{1}{y} +\frac{1}{z}+1$ If x,y,z is positive then find the minimum value of $x+y+z+\frac{1}{x} + \frac{1}{y} +\frac{1}{z}+1$
I know this can be solved by the relation AM$\geq$GM$\geq$HM. But I do not how to apply here and proceed.
Please help ... | $x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1 \geq 6 \sqrt[6]{x \cdot y \cdot z \cdot \frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z} } + 1=7$, where the equality is obtained in $x = y = z = \frac{1}{x} = \frac{1}{y} = \frac{1}{z} =1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evalua... | You may continue with
$
\frac{q^k}{1+q^{2k}}= \frac{1}{2\cos k\alpha}
$, where $\alpha=\frac{2\pi j}7,\>j=1,2,...,6$, and write the expression as,
$$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
=\frac{\cos\alpha+ \cos2\alpha+ \cos3\alpha}{2\cos\alpha\cos2\alpha\cos3\alpha}=\frac ND\tag1\\$$
where we used $2\c... | {
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"url": "https://math.stackexchange.com/questions/3649434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Factorise $Σa^{2}(b^{4}-c^{4})$ I expanded this cyclic expression then found the three factors and assumed third to be constant but after that I got stuck at this question. The options are
(a)$(a-b)^{2}(b-c)^{2}(c-a)^{2}$
(b)$(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
(c)$(a+b)^{2}(b+c)^{2}(c+a)^{2}$
(d)None of these
The answer... | We can consider $\sum a(b^2-c^2)$ and then make the substitutions $a\rightarrow a^2,b\rightarrow b^2, c\rightarrow c^2$ at the end.
$$\sum a(b^2-c^2)$$
$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$
We can group by the squares:
$$=a^2(c-b)+ b^2(a-c)+c^2(b-a)$$
Now we force a factor of $(b-a)$ from the first two terms:
$$=a^2c-aab+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x)$ is a common factor of $g(x)$ and $h(x)$ find $f(x)$ Given that $f(x)$ is a common factor of $g(x)=x^4-3x^3+2x^2-3x+1$ and $h(x)=3x^4-9x^3+2x^2+3x-1$ find $f(x)$
I tried to factorised $g(x)$ but it doesn't have any rational roots as I've already tried 1 and -1.
So how do I solve this?
| $$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) $$
$$ \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) $$
$$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) = \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) \cdot \color{magenta}{ \left( 3 \right) } + \left( - 4 x^{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)... | $\!\bmod x^2\!+\!x\!+\!1\!:\,\ \color{#c00}{x^3\equiv 1}\,$ so $\, 0\equiv h(\color{#c00}{x^3})\!+\!xg(\color{#c00}{x^3})\equiv h(\color{#c00}1)+xg(\color{#c00}1)\iff h(1)=0=g(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Which integration is correct? Or are both correct? What's the correct method to integrate the function $\frac{1}{2(3x+1)}$?
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \int\frac{1}{6x+2}dx\\
&= \frac{1}{6}ln(6x+2)+c
\end{align}
$$or$$
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \frac{1}{2}\int\frac{1}{3x+1}dx \\
&= (\frac{... | \begin{align}
\frac{1}{6}\ln(6x+2)+c&= \frac{1}{6}\ln(2(3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6} \ln (3x+1)+c',
\end{align}
with $c'=c+\frac{1}{6}\ln2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trying to find a general equation for an ellipse given the foci and sum of focal distances I'm trying to find an equation for the ellipse in the form $$Ax^2 + Bxy + Cy^2 +Dx +Ey +F = 0$$ given the foci $(a,b)$ and $(c,d)$ and the sum of focal distances $r$. I started from the definition $$\sqrt{(x-a)^2+(y-b)^2} + \sqrt... | Doing it for $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ is fun.
$S_1(-ae,0(, S_2(ae,0)$, $P(x,y)$ given that
$$PS_1+PS_2=2a \implies \sqrt{(x+ae)^2+y^2}+\sqrt{(x-ae)^2+y^2}=2a$$ $$\implies U+V=2a~~~~~(1)$$
Then $$U^2-V^2=4aex ~~~ U-V=2ex~~~~~~(3)$$
From (1) and (2), we get
$$U=a+ex \implies (x+ae)^2+y^2=(a+ex)^2$$
$$\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$ Question: let P(x) is a polynomial function
Given $P(x^2)= x^2(x^2+1) P(x)$ and
$P(2)=3$, Find $P(3)$
If I put $x=2$, we get $P(4)$ which is not required.
I tried to find $P(x)$ to let it general polynomial to find its roots bu... | Elaborating on some details:
$P(0)=P(0^2)=0^2(0^2+1)P(0)=0$
$P(1^1)=P(1)=1^2(1^2+1)P(1)$ so $P(1)=2P(1)$ implying $P(1)=0$
$0=P(1)=P((-1)^2)=(-1)^2((-1)^2+1)P(-1)=2P(-1)$ implying $P(-1)=0$
So, we have found three roots so far.
As for the degree of $P$, if we were to suppose $P(x)$ were a degree $n$ polynomial, it fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximize $\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$ if $x+y+z\leq 1$ and $(y+z)^2+2x-x^2-2xy\leq 1-2\gamma$, $0.24 \leq \gamma \leq 0.25$ I am trying to maximize the function
$$f(x,y,z)=\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$$
with the following constraints:
$$x\geq 0, y\geq 0, z \geq 0,$$
$$x+y+z\leq 1,$$
$$x+y\geq 4/5,... | In this answer we solve a particular case of the problem, when $z=0$.
Now the last constraint becomes $g(x,y)=y^2+2x-x^2-2xy\le 1-2\gamma$. Since $\frac{\partial g}{\partial y}=2y(1-x)\le 0$, for $x$ fixed we can increase $y$ increasing $f$ by this, until $y$ will be bounded by a constraint $x+y\le 1$ or $g(x,y)= 1-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sequence $\left\{ a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. As title. Sequence $\left\{a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$.
Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$.
I tried the i... | If you define $b_n=a_n-3$, you want to show that $$|b_{n+1}|<\frac{|b_n|}{3}.$$
Now, the recursion is $$b_{n+1}=a_{n+1}-3=\sqrt{a_n+6}-3=\sqrt{b_n+9}-3=\frac{b_n}{\sqrt{b_n+9}+3}$$
Since $b_1=1$, this shows that $b_n>0$ for all $n$ by induction, so you can forgot about the absolute values. Moreover $\sqrt{b_n+9}>\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $f(x,y)=\cos^4y-(1-\tan^4x-\sin^2y)(\sin^2x\sin^2y+\cos^2y)^2\geq0$ in some open neighborhood of $(0,0)$ Time is ticking and I must prove the Hourglass (or Figure 8) Inequality, which states
$f(x,y)\geq 0$ in some open neighborhood of $(0,0)$, where
$$f(x,y)=\cos^4 y -(1-\tan^4 x - \sin^2 y)(\sin^2 x \sin^... | Let us prove that $f(x, y) \ge 0$ for $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$.
When $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$, we have $|\sin x| \le |x| \le |\tan x|$,
$\frac{3}{4} \le \cos^2 y \le 1$, and $1-\tan^4 x - \sin^2 y > 0$.
Denote $v = \cos^2 y \in [\frac{3}{4}, 1]$. We have
\begin{align}
f(x, y) &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to solve the following question involving floor/greatest integer functions.
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$ with the notations $\lfloor x \rfloor$ denoting the greatest integer less than or equal to $x$ and $\{x\}$ t... | Let $x = n + r$ where $n = [x]$ and $r = \{x\}$.
Then we have $3n - [n^2 + 2nr + r^2]=2r$
$3n - n^2 - [2nr + r^2] = 2r$
and.... oh, hey, the LHS is an integer the RHS being $2\{x\}$ means $\{x\} = 0$ or $0.5$.
Two options $x$ is an integer and $x = [x] = n$ and $r=\{x\} = 0$ and we have
$3n-n^2=0$ and $n^2 = 3n$ and $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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With a given matrix $A \in M_3(\mathbb{R})$ show that $A^{2009} + A^{2008} = 2 ^{2008} (A + I_3)$. I am given the matrix:
$$A = \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{pmatrix}$$
I have to show that the following is true:
$$A ^ {2009} + A ^ {2008} = 2 ^ {2008} (A + I_3)$$
I calcucalted $A^2, A^3, A^... | More generally, $A^{n+1} + A^{n} = 2 ^{n} (A + I)$ for all $n \in \mathbb N$. This is proved by induction using that the minimal polynomial of $A$ is $x^2- x - 2=(x-2)(x+1)$ and so $A^2=A+2I$:
$$
A^{n+2} + A^{n+1} = A(A^{n+1} + A^{n})= A(2^{n} (A + I))=2^n(A^2+A)= 2^n(2A+2I)=2^{n+1} (A + I)
$$
(The characteristic poly... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Value of $\lim\limits_{n\rightarrow\infty}\prod\limits_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$ Find the value of m for the following $$\lim\limits_{n\rightarrow\infty}\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$$
| Long, but detailed
$$\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=
\prod_{k=3}^n\left(1-\tan^2\frac{\pi}{2^k}\right)\left(1+\tan^2\frac{\pi}{2^k}\right)=\\
\prod_{k=3}^n\left(\frac{\cos^2{\frac{\pi}{2^k}}-\sin^2{\frac{\pi}{2^k}}}{\cos^2{\frac{\pi}{2^k}}}\right)\left(\frac{\cos^2{\frac{\pi}{2^k}}+\sin^2{\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$... | For minimum,
$$T=\sqrt{(\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1})^2}\\
=\sqrt{7+2\sqrt{25\cos^2x\sin^2x+6}}\\
\ge \sqrt{7+2\sqrt6}=1+\sqrt6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding Laurent series of $f(z) = \frac{1}{z^{2}+4}$ on two different domains. I need to find the Laurent series of $$f(z) = \frac{1}{z^{2}+4}.$$ First for $ z \in \mathbb{C}: |z|<2$ and then for $z \in \mathbb{C}: 1<|z-i|<3$.
Now for the first restriction I do the following:
\begin{align}
f(z) &= \frac{1}{4} \frac{1}{... | You can write
\begin{align*}
\frac{1}{z^2+4} & =\frac{1}{4i}\left[\frac{1}{z-2i}-\frac{1}{z+2i}\right]\\
& =\frac{1}{4i}\left[\frac{1}{(z-i)-i}-\frac{1}{(z-i)+3i}\right]\\
& =\frac{1}{4(z-i)}\left[\frac{1}{1-\left(\frac{i}{z-i}\right)}\right]+\frac{1}{12}\left[\frac{1}{1+\left(\frac{z-i}{3i}\right)}\right]\\
&=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I ... | Wlog $z\le 1$ because the variables cannot all be $\ge1$.
For fixed $z$, we want to minimize $(2-x)(2-y)$ under a constraint $x^2+y^2=3-z^2$, which is a constant between $2$ and $3$.
Now
$$ \begin{align}(2-x)(2-y)&=4-2(x+y)+xy\\&=4-2(x+y)+\frac12(x+y)^2-\frac{3-z^2}2\\
&=\frac12\left(x+y-2\right)^2+2-\frac{3-z^2}2
\\ &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I... | $a^2+b^2=2c^2\tag{1}$
We can get infinitely many integer solutions by Diophantus's method below.
Substitute $a=t+p, b=t+q, c=t+r$ to equation $(1)$, then we get
$$(2p+2q-4r)t+p^2-2r^2+q^2=0$$
Let $t = -1/2(p^2-2r^2+q^2)/(p+q-2r)$.
Then we get a parametric solution below.
$a = p^2+(-4r+2q)p+2r^2-q^2$
$b = p^2-2r^2-q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. Let (x,y) be a pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. If $|x|+|y|=\frac{p}{q}$ (where p and q are relatively prime), then find the value (6p – q).
I used the concept $\frac... | Equation (1) $\times x$ + Equation (2) $\times y$ when added together gives
\begin{align*}
56x^2+66xy+56y^2&=0\\
28x^2+33xy+28y^2&=0\\
28\left(x+\frac{33}{56}y\right)^2+\frac{2047}{112}y^2&=0.
\end{align*}
Since left side is always non-negative (for $x,y \in \mathbb{R}$), this can only happen when both
\begin{align*}
x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proving $S:=\left\{1\Big/\left(x^2-3\right)\mid x\in\Bbb Q\right\}$ is unbounded
Let $S:=\left\{\dfrac1{x^2 -3} :x\in\Bbb Q\right\}$. Prove
carefully that $S$ is unbounded
My proof
By contradiction, if we can find some $M$ so that $|s| \leqslant M\ \ \forall s \in S$, that is, if
$$ \dfrac{1}{|x^2-3|} \leqslant M\ \f... | One could show that $S$ is neither bounded from above nor bounded from below. However, we are only asked to sow that it is not bounded. This can be done ab ovo without any sophisticated converging sequences a la Heron formula.
So assume there exists $M\in \Bbb R$ with $|s|\le M$ for all $s\in S$.
Pick $n\in\Bbb N$ with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Derive a bound for a general Gauss sum (from Iwaniec and Kowalski pg 199) I am reading 'Analytic Number Theory' by Iwaniec and Kowalski and get stuck on this:
On page 199,they say
...we get
$$\left| S_f(N) \right|^2\leq N+\sum_{ 1\leq \mathcal{l} < N} \min(2N,\left \| 2\alpha \mathcal{l} \right \| ^{-1}).$$
Hence one ... | We can assume $\alpha < 1/4$ because otherwise we can apply the trivial bound $|S_f(N)| \leq N$.
Let's partition the sum as follows:
\begin{align*}
\sum_{ 1\leq \ell < N} \min(2N,\| 2\alpha \ell \| ^{-1}) &= \sum_{\substack{1\leq \ell < N \\ 2\alpha\ell < 1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})\\
&+ \sum_{1 \leq m < 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $23a^2$ is not the sum of 3 squares. I know that Legendre's theorem states that a number is expressible as a sum of 3 squares iff. it's not of the form $4^x (8m+7)$, so I need to show that $23a^2$ is of this form, how could I go about doing this?
| If a number $n$ is of this form, then:
$$n = 4^x \left(8m + 7\right) = 4^x \cdot 8m + 7 \cdot 4^x = 23a^2$$
Therefore
$$n \equiv 7 \cdot 4^x \equiv \begin{cases}0 & \text{if } x > 1 \\4 & \text{if } x = 1\\7 & \text{if } x = 0\end{cases}\pmod{8}$$
and
$$n \equiv 23 a^2 \equiv \begin{cases}0 & \text{if } a \equiv 0 \pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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The area of a triangle determined by two diagonals at a vertex of a regular heptagon
In a circle of diameter 7, a regular heptagon is drawn inside of it. Then, we shade a triangular region as shown:
What’s the exact value of the shaded region, without using trigonometric constants?
My attempt
I tried to solve it wit... | Expressing the sides $a,b,c$ via the sines of corresponding central angles one obtains:
$$
A=\frac{abc}{4R}=2R^2\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7.\tag1
$$
For a product of sines we have the following theorem:
$$
\prod_{0<m_i<n}2\sin\frac{\pi m_i}{n}=n.
\tag2
$$
Therefore: $$2^6\sin\frac\pi7\sin\frac{2\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Taylor Series of $f(x) =\frac{1}{x^2}$ I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.
$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$
I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$
So I got the expansion:
$$f(x)=1+2(x... | For $x$ such that $|x+1|<1$
$$f(x)=\frac{-1}{x}=\frac{1}{1-(x+1)}$$
$$=1+(x+1)+(x+1)^2+....$$
$$=\sum_{k=0}^\infty (x+1)^k$$
As a sum of a power series,
$f $ is differentiable at $ (-2,0)$ and
$$f'(x)=\frac{1}{x^2}=\sum_{k=1}^\infty k(x+1)^{k-1}$$
$$=\sum_{k=0}^\infty(k+1)(x+1)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}... | Taking a different approach entirely, note that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)=1+{ad\over bc}+{bc\over ad}+1$$
Thus, letting $ad/bc=x$ and noting that $x\gt0$, we see that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)\ge4\iff x+{1\over x}\ge2\iff x^2-2x+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Prove $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold?
Prove that $(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold?
My attempt:
By Cauchy-Schwarz inequality, we have:
$$\begin{align*}(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}... | They called it the Titu's lemma in the inequality world or the Angel form of the CS inequality. The inequality follows from this lemma: $\dfrac{1}{a}+\dfrac{4}{b} = \dfrac{1^2}{a}+\dfrac{2^2}{b} \ge \dfrac{(1+2)^2}{a+b} = \dfrac{9}{a+b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\co... | $$z+\frac 1z=2\cos \theta\Rightarrow z^2-2\cos\theta z+1=0$$
Solve this quadratic to obtain
$$z=cos\theta\pm i\sin\theta$$
Or,
$$z=e^{i\theta},e^{-i\theta}$$
Now,
$$\frac{z^{2n}-1}{z^{2n}+1}=\frac{z^n-1/z^n}{z^n+1/z^n}$$
In either of the above cases(for $z$), we have,
$$z^n+1/z^n=e^{in\theta}+e^{-in\theta}=2\cos n\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3694013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $a \in \mathbb N$ such that $x^2+ax-1 = y^2$ has a solution in positive integers Question: Find all the positive integers $a$ such that $x^2+ax-1 = y^2$ has a solution in positive integers $(x,y)$.
Comments: It's easy to see that this equation rarely has a solution (in the sense that for a fixed $a$, $x^2+ax-1$ is... | Actually, every $a$ that is not $2 \pmod 4$ yields a solution. These $a$ are impossible because then
$$x^2 + ax - 1 \equiv x^2 + 2x - 1 \equiv (x - 1)^2 - 2 \pmod 4,$$
which is either $2$ or $3$ mod $4$, but squares are only $0$ or $1$ mod $4$.
For the other $a$, I will construct values of $x$ that work.
For $a = 4k$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ I am trying to understand each step in order to get from $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the... | Alternatively, recall the general formula $$a^2-b^2=(a+b)(a-b).$$
This can be seen by expanding the RHS, or by writing
$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b).$$
Then, with $a=x$ and $b=2$, we have
$$x^2-4=x^2-2^2=(x+2)(x-2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\exists !c \in \mathbb{R} \exists ! x \in \mathbb{R} (x^2 + 3x + c = 0)$ This is an exercise from Velleman's "How To Prove It". I am struggling with how to finish the final part of the uniqueness proof, so any hints would be appreciated!
*a. Prove that there is a unique real number $c$ such that there ... | Let $P(X) = X^2 + 3 X + c$. It is a real polynomial of degree $2$, and its discriminant is $\Delta = 9 - 4c$. Then
\begin{align}
P \text{ has a unique root} &\iff \Delta=0 \\
&\iff c = \frac94
\end{align}
Thus, $c=\frac94$ is the unique number for which there is a unique solution $x$ for $P(x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find rectangular Cartesian coordinate system Find rectangular Cartesian coordinate system in $\mathbb{R}^3$ to bring quadratic surface $2y^2-3z^2+4xz-12y+15=0 $ to standard form.
After splittig off square the equation can be rewritten as follows $2(y-3)^2-3z^2+4xz=3$.
What to do with mixed term $4xz$?
| Just because it's kind of fun, here's a method using the Matrix Representations of Conic Sections. Essentially, for
$$ Q(x,z) = -3z^2 + 4xz$$
We can write
\begin{align}
Q(x,z) &= \begin{pmatrix}
x & z
\end{pmatrix}
\begin{pmatrix}
0 & 2\\
2 & -3
\end{pmatrix}
\begin{pmatrix}
x \\ z
\end{pmatrix}
\end{align}
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one... | Hint:
Let $x-y=c$
Use $x=y+c$ to form a Quadratic Equation in $y$
As $y$ is real, the discriminant must be $\ge0$
One may replace $y$ with $x-c$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do I find the sum of a power series $\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}$? I have found the area of convergence to be $ x \in (-\infty, \infty)$, and this is how far I had gotten before getting stuck:
$$
\begin{aligned}
\sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}} &=\sum_{k=0}^... | Hint
Let $x=3y$ $$A=\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}=9\sum_{n=3}^\infty \frac{ y^n}{n (n+1)!}=9y\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}$$
$$\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}=\left(\sum_{n=3}^\infty \frac{ y^{n}}{ (n+1)!}\right)'=\left(\frac 1y\sum_{n=3}^\infty \frac{ y^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
| When you squared you forget to put the coefficient of 2.
You should have $2\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$
Or $\sqrt{(x^2+8x+7)(x^2+3x+2)}=2(x+1)^2$
....
But anyway....
Square again. But it'll help to factor components.
$(x^2+8x+7)(x^2+3x+2)=4(x+1)^4$
$(x+7)(x+1)(x+2)(x+1)=4(x+1)^4$.
If $x = -1$ we get $0=0$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $ Solve $$(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $$
I kind of know the only integer solutions are $\{0, 0\}, \{-1, -1\}$ but I don't know how to prove that
| Rewrite it like this: $$xy(x+y)+4xy+x+y=0$$ and let $a=x+y$ so $$xy = -{a\over a+4}$$
then $a+4\mid a$ so $$a+4\mid (a+4)-a =4\implies a+4\in \{-4,-2,-1,1,2,4\}$$
So $a\in \{-8,-6,-5,-3,-2,0\}$
*
*If $a=0$ then $xy =0$ and $y=-x$ so $x=y=0$
*If $a=-2$ then $xy =1$ and $y=-x-2$ so $x=y=-1$
*If $a=-3$ then $xy =3$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Stokes's Theorem for the Cone Consider the vector field $$F = \biggl \langle \sin x-\frac{y^3}3, \cos y+\frac{x^3}3, xyz \biggr \rangle.$$ Let $S$ be the surface given by the cone $$z^2 = x^2 + y^2 \text{ for } 0 \leq z \leq1.$$
I would like to verify Stokes's Theorem for the surface of the cone over the region $S.$ I ... | Recall that Stokes's Theorem says that $$\oint_{\partial S} \mathbf F \cdot d \mathbf r = \iint_S (\nabla \times \mathbf F) \, dS,$$ where $S$ is a parametrized surface; $\partial S$ is the boundary of $S$ parametrized by the curve $\mathbf r(t)$ on some closed interval $a \leq t \leq b;$ and $\nabla \times \mathbf F$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3708556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$PA = LU$ decomposition Consider a matrix $A= \begin{pmatrix}
1 & 2 & 1\\
3 & 6 & 1\\
0 & 4 & 1
\end{pmatrix}$
I am applying the transformations on matrix $A$ to convert it to $U$ using the following matrices:
(The i,j in $E_{ij}$ denotes the element in matrix A which is fixed in order to transform it into $U$)
... | Basically the idea behind writing a Gaussian elimination outcome in the form $PA=LU$ is that you could have done all the swaps at the beginning and then just done the row scaling/addition afterward.
So here suppose you had swapped 2 and 3 initially. The scale/add step that you would have done to zero out the first col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $\frac{1}{16} \sum \frac{(b+c)(c+a)}{ab} +\frac{9}{4} \geq 4\sum \frac{ab}{(b+c)(c+a)}$ For $a,b,c>0$. Prove: $$\frac{1}{16} \sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+\frac{9}{4} \geq 4\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}$$
My SOS's proof is:
... | We need to prove that:
$$\sum_{cyc}\left(\frac{(a+c)(b+c)}{ab}-4\right)\geq16\sum_{cyc}\left(\frac{4ab}{(a+c)(b+c)}-1\right)$$ or
$$\sum_{cyc}(c^2+ac+bc-3ab)\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}((c-a)(3b+c)-(b-c)(2a+c))\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^{2\pi} \frac{x \cos x}{2 - \cos^2 x} dx$. I have to find the integral:
$$\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx$$
I rewrote it as:
$$\int_0^{2\pi} \frac{x \cos x}{1 + \sin^ 2 x} dx$$
But nothing further. I plugged it in a calculator and the result was $0$. I can see that the following relation hol... | We'll use a trick to get rid of the $x$.
As you've noted, $\sin(2\pi -x) = \sin x$, and $\cos(2\pi - x) = \cos x$.
Replacing $x$ with $2 \pi - x$, we get
$$I = \int_0^{2\pi} \frac{x \cos x}{1+\sin^2x} \, dx = \int_0^{2\pi} \frac{(2 \pi - x) \cos x}{1+\sin^2x} \, dx.$$
Adding the integral to itself, we get
$$2I = \int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.