Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Eigenfunction of the Fourier transform I want to show that $$ \frac{1}{ \sqrt{ 2 \pi}} \int_{-\infty}^{\infty} \frac{e^{-iwx}}{\cosh{ (x \sqrt{\frac{\pi}{2}}} ) } = \frac{1}{\cosh{ (w \sqrt{\frac{\pi}{2}}} ) } .$$
My attempt is to first make a substitution $y = x \sqrt{ \frac{\pi}{2} }$, and write $\cosh$ as exponen... | You are just about there. One trick that I find helps over this hump is to recognize that your sum may be rewritten as
$$ \frac{2}{\pi} \sum_{k=-\infty}^{\infty} (-1)^k \frac{ 2k+1 }{ (2k+1)^2 + (w \sqrt{\frac{2}{\pi}} )^2}$$
Now you can apply the residue theorem to this infinite sum. I will state the following resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Trigonometry - Finding the range of the function Problem :
$$f(\theta)=(2\sqrt{3}+4)\sin\theta +4\cos \theta $$
I have studied if the function is in the form : $f(\theta)=a\cos\theta + b\sin\theta$ then the range of this function can be given as $$[-\sqrt{a^2+b^2} ,\sqrt{a^2+b^2}]$$
So, here the range will be [-$44+1... | The answer is indeed $ [-\sqrt{a^2+b^2},\, \sqrt{a^2+b^2}]$.
You made a calculation mistake somewhere. The range is $[-\sqrt{44+16\sqrt{3}},\sqrt{44+16\sqrt{3}}]$
Here's a proof:
Let $a=2\sqrt{3}+4$ and $b= 4$
Choose $y \in [0,2\pi]$ such that $\sin y = \dfrac{b}{\sqrt{a^2+b^2}}$ and $\cos y = \dfrac{a}{\sqrt{a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \se... | I would like to handle the integral by integration by parts. $$
\begin{aligned}
\int \frac{d t}{\left(t^{2}+9\right)^{2}} &=-\int \frac{1}{2 t} d\left(\frac{1}{t^{2}+9}\right) \\
&=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1}{2} \int \frac{1}{t^{2}\left(t^{2}+9\right)} d t \\
&=-\frac{1}{2 t\left(t^{2}+9\right)}-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. Tha... | If you know that
$$\gamma_n=\frac{1}{1}+\frac{1}{2}+..+\frac{1}{n}- \ln(n) \,,$$ is convergent then your sequence is exactly
$$\gamma_{n^2}-\gamma_n+ \ln(n) \,.$$
As $\gamma_{n^2}-\gamma_n \to 0$ and $\ln(n) \to \infty$ it follows that your sequence diverges to $\infty$.
Edit With the change, your sequence is
$$\ga... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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Analytic solution of $ \int_t^{t+r} \frac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx $ Could anyone help me to solve this defined integral analytically?
$$
\int_t^{t+r} \dfrac{x^2}{(t+r-\sqrt{2rx-x^2})^4}dx
$$
| The antiderivative of the integrand of the integral under consideration can be found by the Euler changes $\sqrt{2rx-x^2}=tx$ or/and $\sqrt{2rx-x^2}=t(x-2r)$. The output done with Maple by the command $int(x^2/(r+t-2*sqrt(2*r*x-x^2))^4, x)$ is huge. It can be seen in the worksheet exported as a pdf file (20MB). The def... | {
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"source": "stackexchange",
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Sum of one, two, and three squares If a square $n^2$ can be written as the sum of two nonzero squares as well as the sum of three nonzero squares, then can we conclude that it can be written as the sum of any number of nonzero squares up to $n^2 - 14$ nonzero squares?
Example: $13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2$. Bu... | Yes, because all numbers can be written as the sum of four squares.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Summing Finitely Many Terms of Harmonic Series: $\sum_{k=a}^{b} \frac{1}{k}$ How do I calculate sum of a finite harmonic series of the following form?
$$\sum_{k=a}^{b} \frac{1}{k} = \frac{1}{a} + \frac{1}{a+1} + \frac{1}{a+2} + \cdots + \frac{1}{b}$$
Is there a general formula for this? How can we approach this if not?... | The formula for sum of H.P. remained unknown for many years, but now I have found the formula as an infinite polynomial.
The formula or the infinite polynomial which is
equal to the sum of th H.P. $\frac{1}{a} + \frac{1}{a+d} + \frac{1}{a+2d} + \frac{1}{a+3d} + ........$ is :-
$Sum of H.P. = (1/a) [1 + \frac{1}{1×(1+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 6
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Proving $\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$ For each $a,b,c > 0$ and $a^5+b^5+c^5=3 $ .How to prove that :
$$\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$$
| Since $a, b, c$ are positive, we can set
$$
a = x^2\quad b = y^2 \quad c = z^2
$$
The inequality becomes
$$
\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq 3
$$
with the condition $x^{10} + y^{10} + z^ {10} = 3$.
Applying Hölder's inequality we get:
$$
3 = (x^{10} + y^{10} + z^{10})^{9/10} \cdot (1 + 1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
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Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
Prove:$|x-1|+|x-2|+|x-3|+\cdots+|x-n|\geq n-1$
example1: $|x-1|+|x-2|\geq 1$
my solution：(substitution)
$x-1=t,x-2=t-1,|t|+|t-1|\geq 1,|t-1|\geq 1-|t|,$
square,
$t^2-2t+1\geq 1-2|t|+t^2,\text{Since} -t\leq -|t|,$
so proved.
question1 : Is my proof right? Alternatives?
on... | You want
$S(x)
=\sum_{i=1}^n |x-i|$.
I will show
$\begin{cases}
\text{if } x \le 1&S(x) \ge \dfrac{n(n-1)}{2}\\
\text{if } x \ge n &S(x) \ge \dfrac{n(n-1)}{2}\\
\text{otherwise} &S(x) \ge \dfrac{n^2-1}{4}
\end{cases}
$
Exact results are
$\begin{cases}
\text{if } x \le 1, & S(x) = n(1-x) + \dfrac{n(n-1)}{2}\\
\text{if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Calculating this Riemann sum limit
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$
How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
| Recall that if $f$ is integrable on $[a,b]$, then:
$$
\int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right)
$$
Notice that:
$$
\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}
= \sum_{k=1}^{n} {\left(\dfrac{k}{n^2} - \dfrac{1}{n^3}\right) \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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For a positive integer $n$, $p_n$ denotes the product of the digits of n, and $s_n$ denotes the sum of the digits of n. For a positive integer $n$, $p_n$ denotes the product of the digits of $n$, and $s_n$ denotes the sum of the digits of $n$. What is the number of integers between 10 and 1000 for which $p_n + s_n = n$... | Given the conditions, your last equation doesn't have any solution. Let's see.
As $x \neq 0$ let's divide the equation by $x$:
$$yz=99+ \frac{y}{x}$$
Therefore $x |y$.
But for $y=0$ there's no solution, and for $y \neq 0$:
$$yz >99 \Rightarrow y>10 \, \mathrm{or} \, z>10.$$
Which is impossible since $y$ and $z$ are dig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/440417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove inequality by induction Once again, I'm stuck in a demonstration by induction, this time, it's really proving that an inequality is valid. So, here is the inequality:
Prove that $\binom{2n}{n} \geq (n+5)^2 \ \forall n \geq 5, n \in \mathbb{N} $
Then, what I wanted to prove is that:
$\binom{2n+2}{n+1} \geq (n+6... | What you have is right, you want to show that if $n \geq 5$, then
$$2(2n+1)(n+5)^2 \geq (n+6)^2(n+1) $$
This can be shown in various ways. For example, if you just expand everything, you get that
$$2(2n+1)(n+5)^2 \geq (n+6)^2(n+1) \Leftrightarrow 3n^3 + 29n^2 + 72n + 14 \geq 0 $$
which is obviously true for the region... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find an equation of each tangent plane to a cone that is parallel to a given plane. Is the solution correct? Please check it and tell me my mistakes. Thank you.
Find an equation for each plane tangent to $K$ which is parallel to the plane $x-y+z=5$, where $K$ is the half-cone $z=\sqrt{x^2+y^2}$
Let $f(x,y,z) = z-\sqrt... | This solution looks correct. However, there is one careless error towards the very end. Where you have:
$$-\frac{y}{z} = 1$$
It should read:
$$-\frac{y}{z} = -1$$
But, it appears to be a careless error, because you ended up correcting for it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem :
If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.
Plea... | You can directly substitute the value P in the given term
We get:
$\large \large \frac {3\sinθ + \cosθ}{\sinθ - \cosθ} + \frac {3\cosθ + \sinθ}{\cosθ - \sinθ}$
which simplifies to $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| $$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
Put x-1 = t^2
so the equation becomes
$$\sqrt{t^2+4-4t} + \sqrt{t^2+9-6t} =1$$
complete the squares .
EDIT:
My answer was incorrect. Thanks to egreg and Samatix.
| {
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"url": "https://math.stackexchange.com/questions/449300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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$\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$? Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily verifies that
\begin{equation}
\det(I+A) = 1 + tr(A) + \det(A)
\end{equation}
or
\begin{equation}
\det(I+tA) = 1 + t\ tr(A) + t^2\det(A)
\end{equation}
for so... | $$
\left|
\begin{array}{ccc}
1+a_{11} & a_{12} & a_{13} \\
a_{21} & 1+a_{22} & a_{23} \\
a_{31} & a_{32} & 1+a_{33} \\
\end{array}
\right|
=
\left|
\begin{array}{ccc}
1 & a_{12} & a_{13} \\
0 & 1+a_{22} & a_{23} \\
0 & a_{32} & 1+a_{33} \\
\end{array}
\right|
+
\left|
\begin{array}{ccc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ Use the addition or subtraction formula for cosine to compute $\cos(-5\pi/12)$ (Leave your answer in exact form.)
I have $$\begin{align}\cos(-5\pi/12)&=\cos((\pi/4)-(5\pi/6))\\
&=\cos(\pi/4)\cos(5\pi/6)+\sin(\pi/4)\sin(5\p... | Alternative, you can use the half-angle formula (I am aware that this was not specified):
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \sqrt{\frac{1+\cos{(-5 \pi/6)}}{2}} = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Now, $2-\sqrt{3} = (\sqrt{3}-1)^2/2$
so that
$$\cos{\left(-\frac{5 \pi}{12}\right)} = \frac{\sqrt{3}-1}{2 \sqrt{2}} = \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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An inequality involving arctan of complex argument I have the following conjecture:
\begin{equation}
\text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0.
\end{equation}
Which seems to be true numerically.
Can anyone offer some advice on how to approach pr... | I start at the beginning, with a symmetric non-increasing density $\varphi$ on $\mathbb R$. The layer cake representation reduces the matter to $\varphi=\chi_{[-a,a]}$. Therefore, the goal is to show that
$$\int_{-a}^a \operatorname{Re}\frac{1+iy}{1+i(y-t)}\,dt > \int_{-a}^a \operatorname{Re}\frac{1 }{1-i t}\,dt \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Fractions in Ancient Egypt In ancient Egypt, fractions were written as sums of fractions with numerator 1. For instance,$ \frac{3}{5}=\frac{1}{2}+\frac{1}{10}$. Consider the following algorithm for writing a fraction $\frac{m}{n}$ in this form$(1\leq m < n)$: write the fraction $\frac{1}{\lceil n/m\rceil}$ , calculate ... | We proceed by induction on $m\in \Bbb{N}$.
Base Case: When $m=1$, $\dfrac{m}{n}$ is already in the desired form, so terminate.
Induction Hypothesis: Assume that for all $m \in \{1,...,k\}$ (where $k \geq 1$), the algorithm will terminate when its input is $\dfrac{m}{n}$.
It remains to prove the claim true for $m=k+1$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c$ are in Geometric Progression and$ a-b,c-a,b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to? If $a,b,c$ are in Geometric Progression and $a-b$,$c-a$,$b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to?
The statements I could get are
1)$b^2=ac$
2)$(c-a)^2=2(b-a)(b-... | Let $b=ra$ and $c=r^2a$. Then $c-a=\frac{2(a-b)(b-c)}{a-c}$, so $(c-a)^2=2(b-a)(b-c)$ and therefore $(r^2-1)^2 a^2=2(r-1)ar(1-r)a=-2r(r-1)^2a^2$. Dividing by
$(r-1)^2a^2$ gives $(r+1)^2=-2r$ and therefore $r^2+4r+1=0$. Thus $a+4b+c=a+4ra+r^2 a=a(r^2+4r+1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Indefinite integral of normal distribution How does one calculate the indefinite integral?
$$\int\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)dx$$
Where $\sigma$ is some constant.
Work so far:
Integrating from exp as rest is constant.
$$\begin{align}
\int\exp\left(-\frac{x^2}{2\sigma^2}\right)dx&=\s... | It's been a while since you've asked, but this is the indefinite integral:
[\begin{array}{l}
\int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}^2}}}}}{{\sigma \sqrt {2\pi } }}} \right)\;dx = } \frac{{\int {\left( {\frac{{{e^{ - \frac{1}{2}\;\cdot\;{{\left( {\frac{x}{\sigma }} \right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is wrong with my solution of finding remainder of $50^{(51^{52})}$ when divided by 11? I used the following method using remainder theorem. (I used method from here: Find the remainder of $128^{1000}/153$.)
$$\begin{align}
(50^{{51}^{52}})/11 & = (50^{2652})/11 \implies \\
(6^{2652})/11 & = (36^{1326})/11 \impl... | Since $50\equiv 6\pmod{11}$, you should start by looking at powers of $6$ modulo $11$. If you do, you'll find that $50^{10}\equiv6^{10}\equiv 1\pmod{11}$. That means that when computing $50^x\pmod{11}$ we can discard all multiples of $10$ in $x$ and just concentrate on the remainder mod $10$. For example, if we needed ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using the Legendre Symbol Which have solutions? $x^{2} \equiv7 \mod{53}$, $x^{2} \equiv53 \mod{7}$, $x^{2} \equiv 14 \mod{31}$, $x^{2} \equiv 25\mod{997}$?
I have all of these properties for this Legendre symbol and no idea what to do with it to find whether these have a soltion, let alone solve them if they do.
| They all have solutions. Since $53 \equiv 1 \pmod{4}$, we have
$$\left(\frac{7}{53}\right) = \left(\frac{53}{7}\right) = \left(\frac{4}{7}\right) = \left(\frac{2}{7}\right)^2 = 1,$$
which settles the first two. $31 \equiv 7 \pmod{8}$, so
$$\left(\frac{14}{31}\right) = \left( \frac{2}{31}\right) \left(\frac{7}{31}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an equation involving binomial coefficients and complex numbers Question:
Solve the following equation for $x$:
$$\sum_{k=0}^{n}\binom{n}{k}x^{k}\cos(k\theta )=0$$
Attempt:
I think this equation come from:
$$(x\cos\theta+ix\sin\theta)^{k}$$
Is that right?
I don't know what to do after that.
| $$\sum_{0\le r\le n}\binom nkx^ke^{ik\theta}=(1+xe^{i\theta})^n=(1+x\cos\theta+ix\sin\theta)^n$$
Let $1+x\cos\theta=R\cos\alpha, x\sin\theta=R\sin\alpha$
$$R^2=(1+x\cos\theta)^2+(x\sin\theta)^2=1+x^2+2x\cos\theta$$
$$\implies \cos\alpha=\frac{1+x\cos\theta}{\sqrt{1+x^2+2x\cos\theta}}$$
$$\implies (1+x\cos\theta+ix\sin... | {
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"url": "https://math.stackexchange.com/questions/463550",
"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
} |
If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) ... | As already found $\cos^2\theta-\sin^2\theta=x,\implies \cos2\theta=x$
Using $a^3-b^3=(a-b)^3+3ab(a-b),$
$$\cos^6\theta-\sin^6\theta=(\cos^2\theta)^3-(\sin^2\theta)^3$$
$$=(\cos^2\theta-\sin^2\theta)^3+3\cos^2\theta\sin^2\theta(\cos^2\theta-\sin^2\theta) $$
$$=x^3+3x\cos^2\theta\sin^2\theta$$
$$=x^3+\frac{3x}4\sin^22\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well ... | Let $\;$ $LHS=f(a,b,c,d)$. Note that $f(a,b,c,d)>f(a-k,b-k,c-k,d-k)$ for $0<k\le\min\{a,b,c,d\}$. So, WLOG we can take $d=0$ to prove the inequality. In that case we should show $ab^2(b-c)+bc^3\ge0$. If $b\ge c$ obviously we are done. For $c\ge b \ge a$ and $c\ge a \ge b$ cases arranging the inequality shows us $ab^3+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 3
} |
$7^{10}$ by $51$ what rest? How to find the remains of Division $7^{10}$ by $51$ using arithmetic debris?
$$$$
$$7\equiv51\;\text{mod}11$$
| $7^3=343=17\cdot20+3$, so $7^3\equiv3\pmod{17}$, and $7^{10}\equiv3^3\cdot7\equiv27\cdot7\equiv10\cdot7\equiv2\pmod{17}$, since $4\cdot17=68$. Of course $7^{10}\equiv1^{10}\equiv1\pmod3$. Thus, we want an integer $n\in\{0,1,\dots,50\}$ such that $n\equiv2\pmod{17}$ and $n\equiv1\pmod3$; $2+17=19$ clearly does the trick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/466796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
| Assuming the contrary, for some $0\le x \le 1$, and $y=1-x$,
$$\begin{align}
\sqrt{x^2+1}+\sqrt{y^2+1} <& \sqrt{5}\\
\left( \sqrt{x^2+1} + \sqrt{y^2+1} \right)^2 <& 5 &\text{(Square both sides)}\\
x^2 +1 + 2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + 1 \right)} + y^2 + 1 <& 5\\
2\sqrt{ \left( x^2 + 1 \right)\left( y^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/467183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Why is $\frac{6\sin(2\pi/3)}2 = \frac{3\sqrt3}{2}$ The source of my problem is that I have an integral:
$$\int_{\pi/3}^\pi{(6\cos 2x)dx}$$
For anyone having the same problem as me, to see the boundaries, They are $\pi/3$ and $\pi$
The primitive function of $\,6\cos 2x\,$ is, if I'm not mistaken, $\dfrac{6\sin2x}2$
I ha... | $$\dfrac {6\sin\left(\frac{2\pi}{3}\right)}{2} = 3 \sin \left(\frac {2\pi}{3}\right) = 3\sin \left(\frac{\pi}{3}\right) = 3\left(\frac{\sqrt 3}{2}\right)$$
Re: your question in the comment below your question: $\sin\left(\frac{\pi}{3}\right) = \sin\left(\pi - \frac {\pi}{3}\right) = \sin \left(\frac{2\pi}{3} \right)$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$ Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.
I've done the first two -
$$\lim_{x\to0}f(x)\\
=\lim_{x\to0}\sqrt{x^2-3x}=0$$
$$\lim_{x\to-\infty}\frac{f(x)}{g(x)}\\
=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\fra... | Your second one is incorrect, since $\sqrt{x^2}=-x$ for all $x<0$. Hence:
$$
\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1-\frac{3}{x}}=-1
$$
For the last one, we multiply by the conjugate:
\begin{align*}
\lim_{x \to \infty} [f(x)-g(x)]
&= \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
| Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient.
In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping:
\begin{align*}
2x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/471690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Congruence Equation $3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$ How to solve the following congruence equation?
$$3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$$
If $t_n$ be the $n$th triangular number, then
$$t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}$$
so if we solve this congruence equation we fin... | We solve it by looking at different primes one at a time:
modulo 2:
The equation becomes $0\equiv0\pmod2$, since $x^n\equiv x\pmod2$ for every integer $x\gt0$. Hence this is always true.
modulo 3:
We obtain now: $n+2\equiv0\pmod3$, i.e. $n\equiv1\pmod3$.
modulo 5:
The equation is seen to be $(1+n^2)(3n+2)\equiv0\pmod5,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is there any solution to the following system of equations? Is there any solution to the following system of diophantine equations?
$$
\left\{\begin{array}{l}
2.a^2 = b^2+c^2+d^2 \\
a^2 = e^2+f^2+g^2 , & \mbox{with }((a,b,c,d,e,f,g)>2)\in N\mbox{ and differents among them}
\end{array}
\right.
$$
| In the system of equations:
$\left\{\begin{aligned}&X^2+Y^2+Z^2=2R^2\\&F^2+G^2+T^2=R^2\end{aligned}\right.$
Solutions have the form:
$X=4(c^2+f^2+k^2)k^2-(2c^2-2d^2+f^2)f^2-(d^2-c^2)^2$
$Y=4kd(2k^2+d^2-c^2-f^2)$
$Z=4(f^2+c^2-k^2)k^2+(f^2+2c^2-2d^2)f^2+(d^2-c^2)^2$
$F=4kc(2k^2+d^2-c^2-f^2)$
$G=4kf(2k^2+d^2-c^2-f^2)$
$T=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is there another simpler method to solve this elementary school math problem? I am teaching an elementary student. He has a homework as follows.
There are $16$ students who use either bicycles or tricycles. The total
number of wheels is $38$. Find the number of students using bicycles.
I have $3$ solutions as follo... | Consider the total wheel function $f$ of two variables $x$ the number of bicyclists and $y$ the number of "tricyclists". Since every time we add a bicyclist we increase the number of wheels by two, we have
$$
f_x = 2.
$$
Similarly for case of tricylists,
$$
f_y = 3.
$$
So we have the solutions
$$
f(x,y) = 2x + A(y) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "158",
"answer_count": 19,
"answer_id": 6
} |
Modified Pell equation: $x^2-D y^2 = m$, $m\neq1$. How does one solve the Diophantine equation
$$ x^2-Dy^2=m, $$
where $m$ is some fixed arbitrary integer?
I understand that given the fundamental solution to $r^2-D s^2=1$, and any solution to the above I can generate an infinite sequence of solutions. But how do I fin... | Here I analyse the equation $x^2 -D y^2 = z^2$.Generally these type of equations can have several solutions if x, y, D and z are particular function of a certain parameter like m.
Suppose $x=2m^2 + p$; plugging in equation we get:
$(2m^2 +p)^2 - z^2 = D y^2 $ ⇒ $4 m^4 + 4p m^2+p^2 -z^2=Dy^2 $
If $4m^2 | p^2 - z^2$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| A probability approach.
Let $\{C_n\}_{n=1}^{\infty}$ be independent random bits such that $P(C_n=1)=\frac{1}{2(n+1)}$.
Then $$\prod_{k=1}^{n}P(C_k=0)=\prod_{k=1}^{n}\left(1-\frac{1}{2(k+1)}\right)\to 0\text{ as }n\to\infty$$
So we can define a random variable $X$ to be the least $n$ such that $C_n=1$, and you get tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 6
} |
Limit of $\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$ Find the limit of:
$$\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$$
| $$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{-2\sin^2 \frac{1}{2x}}{-2\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} = \lim_{x \to \infty} \frac14 \frac{\sin^2 \frac{1}{2x}}{\frac{1}{(2x)^2}} \frac{\frac{1}{x^2}}{\sin^2 \frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/481421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 3
} |
Solving for $x$ in an inequality $$\frac{x^2-3x-2}{x^2+5x+6}<\frac{2-x}{x^2-4}$$
First, I got the things it couldn't be, which were: -3, -2, and 2
I thought to factor everything I could, and I got
$$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{2-x}{(x+2)(x-2)}$$
I also cancelled out one value on the right side
$$\frac{x^2-3x-2}{(... | Starting from where you stopped, we have
$$\dfrac{x^2-3x-2}{(x+3)(x+2)}+\dfrac{1}{x+2}<0$$
Adding, we have
$$\dfrac{(x-1)^2}{(x+3)(x+2)}<0$$
As the numerator is non-negative, we can ignore it for now. So we need solutions to
$$(x+3)(x+2)<0$$
You should be able to do this now simply by splitting the number line into t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area. Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area.
I could find the area of this r... | If $y=b$,
then $y=x^2$ intersects that
at $x=\sqrt{b}$.
That whole area is a rectangle
with vertices
$
( \sqrt{b}, 0),
( \sqrt{b}, b)
( -\sqrt{b}, b),
( -\sqrt{b}, b)
$.
Its area is
$2b\sqrt{b}
=2b^{3/2}
$.
The area below the parabola is
$2\int_0^{\sqrt{b}} x^2 dx
=2 \frac{x^3}{3}\big|_0^{\sqrt{b}}
=\frac23 b^{3/2}
$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/482557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$ After some calculations with WolframAlfa, it seems that
$$
\frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}
$$
Where
$$
\eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}}
$$
is the Dirichlet Eta function.
Co... | If we can change the order of summation, we obtain
$$\begin{align}
1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1} \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 0
} |
Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
| $$
{\rm F} \equiv x^{2} + y^{2} -\mu\left(3x - 4y - 12\right)\,,
\quad
\left\vert
\begin{array}{rcl}
{\partial F \over \partial x} = 0
&\Longrightarrow&
2x - 3\mu = 0\ \Longrightarrow\ x = {3 \over 2}\,\mu
\\[1mm]
{\partial F \over \partial y} = 0
&\Longrightarrow&
2y + 4\mu = 0\ \Longrightarrow\ y = -2\mu
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
L... | Here's the idea. First note that your right hand side is actually $\frac{1}{4}N^2(N+1)^2$. Let's add $(N+1)^3$ to both sides of the inductive step to see if we can get $\frac{1}{4}(N+1)^2(N+2)^2$.
Doing this, we have
$$1^3 + 2^3 + \cdots + (N+1)^3 = \frac{1}{4}N^2(N+1)^2 + (N+1)^3 = (N+1)^2\left(\frac{1}{4}N^2 + N + 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/485806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not kno... | First
$$(a+1/a)^2 + (b+1/b)^2 \geq\frac{1}{2} (a+b+1/a+1/b)^2=\frac{1}{2}(1+1/(ab))^2.$$
Then note that
$$ab\le(a+b)^2/4=1/4.$$
Take it into the first one, you may get your inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 4
} |
Limit $\lim\limits_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$ Evaluate the following limit: $$\lim_{x\to\infty} \frac{5x^2}{\sqrt{7x^2-3}}$$
I'm not really sure what to do when there is a square root for an infinity limit.
Please Help!
| You can consider $$\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2}{\sqrt{7x^2-3}}\frac{\sqrt{7x^2-3}}{\sqrt{7x^2-3}}=\lim_{x\to\infty}\frac{5x^2\sqrt{7x^2-3}}{7x^2-3}=
\lim_{x\to\infty}\frac{5\sqrt{7x^2-3}}{7-\frac{3}{x^2}}$$
Now for $x\to\infty$ we have $\sqrt{7x^2-3}\to\infty,\frac{3}{x^2}\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/487752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
A difficult differential equation $ y(2x^4+y)\frac{dy}{dx} = (1-4xy^2)x^2$ How to solve the following differential equation?
$$ y(2x^4+y)\dfrac{dy}{dx} = (1-4xy^2)x^2$$
No clue as to how to even begin. Hints?
| $$y(2x^4+y)\frac{dy}{dx}=(1-4xy^2)x^2$$
$$2x^4y\frac{dy}{dx}+y^2\frac{dy}{dx}=x^2-4x^3 y^2$$
$$2x^4y\frac{dy}{dx}+4x^3y^2+y^2\frac{dy}{dx}=x^2$$
$$\frac{d}{dx}(x^4 y^2)+y^2\frac{dy}{dx}=x^2$$
$$x^4y^2+\frac{y^3}{3}=\frac{x^3}{3}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}} \mathrm dx$ I am having real trouble evaluating this indefinite integral
$$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$
Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ ... | We first split the integral into two.
$$
\begin{aligned}
I:=& \int \frac{\sin x+\cos x}{\sin ^{4} x+\cos ^{4} x} d x \\
=&-\int \frac{d(\cos x)}{\left(1-\cos ^{2} x\right)^{2}+\cos ^{4} x}+\int \frac{d(\sin x)}{\sin ^{4} x+\left(1-\sin ^{2} x\right)^{2}} \\
=& \frac{1}{2}\left[\underbrace{-\int \frac{d y}{y^{4}-y^{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How find this maximum and minimum of$|x+1|+|x-1|+\sqrt{4-x^2}$
show that
$$2+\sqrt{3}\le|x+1|+|x-1|+\sqrt{4-x^2}\le2\sqrt{5}$$
This problem have nice methods? Thank you
my ugly methods,
since $-2\le x\le 2$,and $f(x)=|x-1|+|x+1|+\sqrt{4-x^2}\Longrightarrow f(x)=f(-x)$
so we only find $x\in [0,2]$ $f(x)_{\max},f(x)_... | The function is defined only for $|x| \le 2$, and is even. Hence let $x = 2 \sin A$, for $A \in [0, \frac{\pi}{2}]$. Then:
$F = |x+1| + |x-1| + \sqrt{4-x^2} = 2 \sin A + 1 + |2\sin A - 1|+ 2\cos A$
For $A \in [0, \frac{\pi}{6}]$, $F = 2 + 2 \cos A $ has the obvious range $[2 + 2 \cos \frac{\pi}{6}, 2 + 2 \cos 0] = [2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Relationships between the elements $(a,b,c,d)$ of a solution to $A^2+B^2+4=C^2+D^2$ I have reduced a certain equation (in positive integers) to the equation
$$A^2 + B^2 + 4 = C^2 + D^2. \quad(\star)$$
Assume the positive integers $(a,b,c,d)$ are any solution to $(\star)$. Are there any algebraic restrictions on [i.e., ... | For the equation: $X^2+Y^2+4=Z^2+D^2$
You can draw a simple formula.
$X=(2c+b+2)c+b$
$Y=(2c+b+2)c$
$Z=2c^2+bc-2$
$D=(2c+b+4)c+b$
more:
$X=\frac{(a^2-q^2)b}{2}+(b-2)(q-1)$
$Y=ab$
$Z=bq+2-b$
$D=\frac{(a^2-q^2)b}{2}+(b-2)q+2$
more:
$X=\frac{(a^2-q^2)b}{2}-(b+2)(q+1)$
$Y=ab$
$Z=b+bq+2$
$D=\frac{(a^2-q^2)b}{2}-(b+2)q-2$
$c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/494534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Roots of $ x^3-3x+1$ How can I find the roots of $x^3-3x+1$ using Cardano's formula?
So far, I found that
$$x = \sqrt[3]{\dfrac{-1+\sqrt{-3}}{2}} + \sqrt[3]{\dfrac{-1-\sqrt{-3}}{2}}$$
$$x = \sqrt[3]{\dfrac{-1+i\sqrt{3}}{2}} + \sqrt[3]{\dfrac{-1-i\sqrt{3}}{2}}$$
$$x = \sqrt[3]{\frac{-1}{2}+\frac{i\sqrt{-3}}{2}} + \sqrt... | Use the Cardano approach:
Let
$$ x=u+v $$ then we will get
$$u^3+v^3+3(uv-1)(u+v)+1=0$$
by setting $uv-1=0$ we get $$uv=1$$ and $$u^3+v^3=-1$$ by cubing the first equation we obtain a simple quadratic system in terms of $u^3$ and $v^3$ which you can be reduced to a quadratic equation in terms of $u^3$ for example: $$u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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condition for a cubic polynomial to have a real root Let $a,b \in R$ and assume that $x=1$ is a root of the polynomial
$p(x)= x^4+ax^3+bx^2+ax+1$. Find the ranges of values of $a$ for which $p$ has a complex root which is not real.
Here first I factored out $x-1$ which left me with a cubic polynomial and then i thoug... | There is a well known method to solve equations of the type
$$x^4+ax^3+bx^2+ax+1=0 \tag{1}$$
If $x \ne 0$ we can transform it to
$$x^2(x^2+\frac{1}{x^2})+ax^2(x+\frac{1}{x})+bx^2=0 \tag{2}$$
Now we observe that
$$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$$
Using the substitution $y=x+\frac{1}{x}$ from $(2)$ a quadrati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Real solutions of $x$ in $\{x\} = \{x^2\} = \{x^3\}$
Calculate real values of $x$ in $\{x\} = \{x^2\} = \{x^3\}$ , where $\{x\}$ is the fractional part of $x$.
My Attempt:
Let $\{x\} = \{x^2\} = \{x^3\} = k$. Because the fractional part of $X$ is given by $\{X\} = X-\lfloor X \rfloor$, we know the following to be tru... | For any $x$ such $\{x\} = \{x^2\}$, we have
$$x^2 - x = \lfloor x^2\rfloor + \{x^2\} - \lfloor x\rfloor - \{x\} = \lfloor x^2\rfloor - \lfloor x\rfloor \in \mathbb{Z}.$$
Likewise $x^3 - x \in \mathbb{Z}$ and $x^3 - x^2 \in \mathbb{Z}$. If $x^2 - x \neq 0$ (that is, $x \neq 0, 1$) then
$$x = \frac{x^3 - x^2}{x^2 - x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Common tangent lines of two quadratic functions Find all such lines that are tangent to the following curves:
$$y=x^2$$ and $$y=-x^2+2x-2$$
I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the e... | If $y=x^2$ we have that $\frac {dy}{dx} = 2x$
This gives the gradient of the tangent at the point $(x, y)=(a, a^2)$
If the tangent line is $y=mx + c$ we therefore have $a^2=(2a)\cdot a+c$ whence $c=-a^2$ and the general tangent line to $y=x^2$ is $$y=2ax-a^2$$
If $y=-x^2+2x-2$ we have $\frac {dy}{dx} = -2x+2$, so that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/500632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Proof of $(a+b)^{n+1}$ I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.
I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.
I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
| Note that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, so
$$\begin{align}
(a+b)^{n+1}&=(a+b)^n(a+b)\\&=\left(\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k}\right)(a+b)\\
&=\sum_{k=0}^{n}\binom{n}{k}a^{k+1}b^{n-k}+\sum_{k=0}^{n}\binom{n}{k}a^kb^{n-k+1}\\
&=a^{n+1}+\sum_{k=1}^{n}\binom{n}{k-1}a^kb^{n-k+1}+\sum_{k=1}^{n}\binom{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int \frac{x^2}{\sqrt{2+3x}}$ using u-sub and integration by parts I'm supposed to integrate this problem by two methods and show that they are the same or that they differ by a constant.
Problem: $$\int \frac{x^2}{\sqrt{2+3x}} dx$$
First method: u-sub with $u = 2+3x$
Second method: integration by parts $u... | In the first case, you need to express $x^2$ in terms of $u$, by noting $$u = 2 + 3x \iff x =\frac 13(u -2).$$ So $$x^2 = \dfrac 19 (u - 2)^2$$
In the second case, $u = x^2 \implies du = 2x dx \iff \dfrac 12 u\,du = x\,dx$
Also, for $dv = \dfrac 1{\sqrt{2 + 3x}}$, and $w = 2 + 3x \implies \dfrac 13 dw = dx$, then $$dv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I calculate summated math questions without using some kind of computer script or program? For example, how can I find the sum of $\frac 1 {(2 + n) \cdot (3 + n)}$ + ... for $n = 0$ to $n = 97$
i.e.
$$\frac{1}{2 \cdot 3} + \frac 1 {3 \cdot 4} + ... + \frac 1 {99 + 100}$$
| Note that $$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$$
Hence, your second series can be rewritten as
$$\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ...$$
terminating, of course, at an appropriate point.
Similarly,
$$\frac{1}{(2 + n)(3 + n)} = \frac{1}{2 + n} - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How find this interesting sum $\sum_{n=1}^\infty\ln(f(n))+\ln(g(n))$ How find this following interesting sum
$$\sum\limits_{n = 1}^\infty {\left\{ {{{\ln }^2}\left( {n - \frac{1}{4}} \right) + {{\ln }^2}\left( {n{\rm{ + }}\frac{1}{4}} \right) - {{\ln }^2}\left( {n - \frac{1}{2}} \right) - {{\ln }^2}\left( {n + \frac... | Here is my solution to the problem. The idea behind it is not difficult. It's the same thing that one can do to evaluate series like
$$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = 6$$
using generating functions. Note that
$$\frac{d}{ds} \frac{1}{(n+a)^s} = \ln^2(n+a) \frac{1}{(n+a)^s}$$
We can view the series in question as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but w... | $$3\sin^4 x - 2\sin^6 x = 3 (\sin^2 x)^2 - 2(\sin^2x)^3 = (\sin^2 x)^2(3 - 2\sin^2 x)\cdots$$
Now we can use the Pythagorean Identity: $$\sin^2x + \cos^2x = 1 \iff \sin^2 x = 1 -\cos^2 x$$
$$\begin{align} 3\sin^4 x - 2\sin^6 x & = 3 (\sin^2 x)^2 - 2(\sin^2 x)^3 \\ \\
& = (\sin^2x)^2 (3 - 2\sin^2 x) \\ \\
& = (1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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solving trigonometric equation $3^{\sin^2(x)}+3^{\cos^2(x)}=4$ Please help me to solve this trigonometric equation.
$$3^{\sin^2(x)}+3^{\cos^2(x)}=4.$$
| As $\sin^2x+\cos^2x=1$
Let $3^{\cos^2x}=a$
$\displaystyle 3^{\sin^2x}=3^{1-\cos^2x}=\frac3{3^{\cos^2x}}=\frac3a$
So,we have $\displaystyle \frac3a+a=4\iff a^2-4a+3=0\iff a=1,3$
If $a=1,$ $3^{\cos^2x}=1=3^0\iff \cos^2x=0\iff \cos x=0,x=(2n+1)\frac\pi2$
If $a=3,$ $3^{\cos^2x}=3=3^1\iff \cos^2x=1\iff \sin x=0,x=n\pi=2n\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Prime trios of the form $p$,$p+2$,$p+4$ My question is about number theory about the prime numbers.
Here is the question:
Prove that there is only one prime trio of the form $p$,$p+2$,$p+4$.
I mean,we can only find the primes 3,5,7
That satisfies the condition.
| Let $n$ be an integer. Then one of $n$, $n+2$, $n+4$ is divisible by $3$.
For the remainder when $n$ is divided by $3$ is $0$, $1$, or $2$.
If it is $0$, then $n$ is divisible by $3$.
If it is $1$, then $n+2$ is divisible by $3$.
If it is $2$, then $n+4$ is divisible by $3$.
A number $\gt 3$ which is divisible by $3$ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$ let $a,b,c>0$,and such
$a+b+c=3$,
show that
$$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$
I think this inequality use this
$$ab\le\dfrac{(a+b)^2}{4}$$
| By C-S $$\sum_{cyc}\frac{1}{(a+b)(4-ab)}=\sum_{cyc}\frac{(c+6)^2}{(a+b)(4-ab)(c+6)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(c+6)\right)^2}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}=\frac{441}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}.$$
Thus, it remains to prove that
$$882\geq\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2.$$
Let $a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/510202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to decompose Partial Fraction. Is this the method? Decompose $\frac{(x^3+x+1)}{(x^2+1)^2}$
Based on my understanding so far:
The partials of the denominator are $(x^2+1)$ and $(x^2+1)^2$
$\frac{(x^3+x+1)}{(x^2+1)^2}$ can be decomposed into partial fractions as below:
$\frac{(x^3+x+1)}{(x^2+1)^2} = \frac{A}{x^2... | By inspection, the partial fractions decomposition is $\dfrac{x}{x^2+1}+\dfrac{1}{(x^2+1)^2}$. This is because $x^3+x=x(x^2+1)$.
Remark: If the top were less simple, we would look for constants $A, B, C, D$ such that our expression is identically equal to $\frac{Ax+B}{x^2+1}+\frac{Cx+D}{(x^2+1)^2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$. Show that $\frac{z}{(z-1)(z-3)} = -3\sum\limits_{n=0}^{\infty}\frac{(z-1)^n}{2^{n+2}} - \frac{1}{2(z-1)}$.
I tried breaking it up using partial fractions but it did not seem to get me any closer to what I want.
| $$\frac z{(z-1)(z-3)}=-\frac1{2(z-1)}+\frac3{2(z-3)}=-\frac1{2(z-1)}-\frac34\frac1{1-\frac{z-1}2}\;\;(**)$$
Now, we have that
$$|z-1|<2\implies \frac{|z-1|}2<1\;,\;\;\text{so}$$
$$(**)=-\frac1{2(z-1)}-\frac34\sum_{n=1}^\infty\frac{(z-1)^n}{2^n}=-3\sum_{n=0}^\infty\frac{(z-1)^n}{2^{n+2}}-\frac1{2(z-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove an equation in complex numbers Prove an equation ($n \in N$, $z \in C$, $a \in R$):
$$ z^n + \frac1{z^n} = 2 \cos\alpha n$$
if $$ z + \frac1{z} = 2 \cos\alpha$$
I tried math induction, but did not solve.
| Using de Moivre's formula $(\cos \alpha + i \sin \alpha)^n = \cos n\alpha + i \sin n\alpha$, we see that setting $z = r(\cos \alpha + i\sin \alpha)$, we have $z^{-1} = r^{-1}(\cos \alpha - i \sin \alpha)$; thus
$z + z^{-1} = r(\cos \alpha + i \sin \alpha) + r^{-1}(\cos \alpha - i \sin \alpha) = 2 \cos \alpha. \tag{1}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| suppose $gcd(a^2, b^2) = d$ and $d \ne 1$
Then $a^2 = dm, b^2 = dn \implies a^2b^2 = d^2mn$, where $gcd(m,n) = 1$
This implies both $m$ and $n$ are perfect squares. $m = p^2, n = q^2$
We have $a^2 = dp^2, b^2 = dq^2 \implies d$ is a perfect square. Let $d = x^2$
Now we have $a^2 = x^2p^2, b^2 = x^2q^2 \implies a = xp, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/512924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
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A basic doubt on a infinite series problem I see in Rudin the following statement is claimed for the following convergent series:
$$1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$
If $s$ is the sum of this series then $$s \lt 1 -\frac{1}{2} + \frac{1}{3}$$. How is that possible to tell ... | Notice that
$$ 1-\frac{1}{2}+\frac{1}{3}-\left( \frac{1}{4}-\dots \right)=1-\frac{1}{2}+\frac{1}{3}-a< 1-\frac{1}{2}+\frac{1}{3}$$
Since $a>0$.
Note:
$$a=\frac{1}{4}-\frac{1}{5}+\dots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/513336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Verify this identity: $\sin x/(1 - \cos x) = \csc x + \cot x$ Verify this identity $\frac{\sin {x}} {1 - \cos {x}}\ = \csc x + \cot x$ I got the left side to $\frac{1-\cos x} {\csc x}$, but I can't get any farther. Am I on the right path? Can I get some help?
| The right-hand side can be rewritten as
$$\csc{x} + \cot{x} = \frac{1}{\sin{x}} + \frac{\cos{x}}{\sin{x}} = \frac{1 + \cos{x}}{\sin{x}}$$
Multiplying the left side top and bottom by $1 + \cos{x}$, we find that
$$\frac{\sin{x}}{1 - \cos{x}} = \frac{\sin{x}(1 + \cos{x})}{1 - \cos^2{x}} = \frac{\sin{x}(1 + \cos{x})}{\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limits of integral: $\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$ , where $D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$ What are the limits of the integral:
$$\iiint_{D} \frac {\mathrm{d}x\mathrm{d}y\mathrm{d}z}{(x + y + z + 1)^3}$$ where
$$ D =\{ x > 0 , y > 0 , z > 0 , x + y + z < 2\}$$
... | Hint: $(x,y,z)\in D$ if and only if $0<x<2$, $0<y<2-x$, and $0<z<2-x-y$. Then, $$\underset{D}{\iiint}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{(x+y+z+1)^3}=\int_{x=0}^{x=2}\left[\int_{y=0}^{y=2-x}\left(\int_{z=0}^{z=2-x-y}\frac{1}{(x+y+z+1)^3}\mathrm{d}z\right)\mathrm{d}y\right]\mathrm{d}x.$$ Be careful: since the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc
| One way without MacLaurin or L'Hospital: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} = \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/516483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Complex Numbers.... Suppose a is a complex number such that:
$$a^2+a+\frac{1}{a}+\frac{1}{a^2}+1=0$$
If m is a positive integer, find the value of:
$$(a^2)^m+a^m+\frac{1}{a^m}+\frac{1}{(a^2)^m}$$
My Approach:
After I could not solve it using the usual methods I tried a bit crazier approach. I thought that as the ques... | If we multiply
$$a^2 + a + \frac1a + \frac{1}{a^2} + 1 = 0$$
with $a^2$, we obtain (since evidently $a \neq 1$)
$$0 = a^4 + a^3 + a^2 + a + 1 = \frac{a^5-1}{a-1},$$
so $a^5 = 1$,
$$a = e^{(2\pi ik)/5},\quad k \in \{1,2,3,4\}.$$
Then, if $m$ is a multiple of $5$, we have $$(a^2)^m + a^m + \frac1{a^m} + \frac{1}{(a^2)^m}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x... | The following method may be unnecessarily time intensive, but will definitely work. Call $f(x) = y$ and $f(y) = f(f(x)) = z$. The equality you need to solve is $f(z) = 1$. This gives a quadratic equation for $z$. Solve it for $z$; this gives you two answers. For these two answers $z_1, z_2$, solve $f(y) = z_1, z_2$, gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/518683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Taylor Series for $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$. The question reads as follows:
Let the function $f$ be given by $f(z)=\frac{z}{2}+\frac{z}{e^z-1}$ if
$z\neq0$ and $f(z)=1$ if $z=0$. Show that $f$ is analytic at $z=0$ and
that $f(z)=f(-z)$. Deduce that there is a Taylor series
$f(z)=a_0+a_2x^2+a_4x^4+...$ v... | To find the first terms of the Taylor series of $f$, there are various possibilities. One can simply differentiate the function and use $a_n = \frac{1}{n!}f^{(n)}(0)$ to obtain the coefficients, but that soon becomes unwieldy. Still, for the coefficients up to order $4$, it's doable.
Another standard way, when one has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/519963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| Let's use the arithmetic-geometric inequality.
That gives us $(xyz)^{\frac{1}{4}}≤\frac{1}{4}(x+y+z)\leq\frac{3}{4}$
thus $(xyz)^4\leq(\frac{3}{4})^{16}$
Now all you have to prove is that $x^3+y^3+z^3\leq3*(\frac{4}{3})^{16}$
But $x^3+y^3+z^3\leq 3^3+3^3+3^3=81$ because $x\leq3$ and $y\leq3$ and $z\leq3$.
Thus you have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Apollonius circle, its radius and center I've got the following set:
$\{|z-a|=k|z-b|\}$, where z is a complex number, a an b are fixed, and $k>0$,$k \ne 1$.
I need to prove that this is a circle (called Apollonius circle).
I also have to prove that this circle's radius is equal to $k|a-b||1-k^2|^{-1}$ and it's centre i... | This is an old post, but I wanted to see if the calculation could be illuminated by unpacking to real coordinates later, and by solving a simpler equation. After doing this I show how to do the calculation without unpacking to real coordinates.
We want to show that the set of $z$ satisfying $|z - a| = k |z - b|$ forms ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
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Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane.
Here is what I did.
Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\f... | I'm afraid you have make something wrong.Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. You should get $y^2=\frac{3}{5}-\frac{x^2}{3}$, rather than $y^2=9-5x^2$ this is the equation of the curve while $z=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find the $(x^2_{1}+x_{1}x_{2}+x^2_{2})(x^2_{2}+x_{2}x_{3}+x^3_{3})(x^2_{3}+x_{3}x_{1}+x^2_{1})$ Let $x_1$, $x_2$, and $x_3$ be the roots of the equation
$$4x^3-6x^2+7x-8=0.$$
Find this value:
$$(x_1^2+x_1x_2+x_2^2)(x_2^2+x_2x_3+x_3^3)(x_3^2+x_3x_1+x_1^2)$$
My try:
$$x_1+x_2+x_3=\frac 3 2$$
$$x_1 x_2+x_2 x_3+x_1 x_... | Assuming that we need to calculate $$(x^2_1+x_1x_2+x^2_2)(x^2_2+x_2x_3+x^2_3)(x^2_3+x_3x_1+x^2_1)$$
Let $\displaystyle x^2_1+x_1x_2+x^2_2=y_3,x^2_2+x_2x_3+x^2_3=y_1, x^2_3+x_3x_1+x^2_1=y_2$
So, $\displaystyle y_3=x^2_1+x_1x_2+x^2_2=(x_1+x_2)^2-x_1x_2$
$=\left(\frac32-x_3\right)^2-\frac2{x_3}$ as $x_1+x_2+x_3=\frac32$ a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generating functions word problem(balloons) Find the generating function for the number of ways to create a bunch of n balloons selected from white, gold, and blue balloons so that the bunch contains
at least one white balloon, at least one gold balloon, and at most two blue balloons. How many ways are there to create ... | You have $$(x^2-x^5)\frac{1}{(1-x)^3}$$
We have the generating function $$\frac{1}{(1-x)^3}=\sum_{n\ge 0} {n+2\choose 2}x^n$$
Multiply by $x^k$ gives $$x^k\frac{1}{(1-x)^3}=\sum_{n\ge 0}{n+2\choose 2}x^{n+k}$$
We reindex using $m=n+k$ to get $$x^k\frac{1}{(1-x)^3}=\sum_{m\ge k}{m-k+2\choose 2}x^m$$
We do this twice, fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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integrate $ \frac {7x^2 + 2x − 7}{ x^3 − x}$ dx I keep messing up with the integration part of this I think. Before that I have to factor out the $x$ on the bottom, and then set up the $A$ and $B$ right?
Evaluate the integral. (Remember to use $\ln|u|$ where appropriate. Use $C$ for the constant of integration.)
$$\int... | HINT:
As $x^3-x=x(x^2-1)=x(x+1)(x-1),$
Using Partial fraction decomposition
$$\frac{7x^2+2x-7}{x(x-1)(x+1)}=\frac A{x+1}+\frac Bx+\frac C{x-1}$$
Multiply either sides by $x^3-x=x(x^2-1)=x(x+1)(x-1)$
and compare the coefficients of the different power of $x$ to find $A,B,C$
Finally use $\displaystyle \int\frac{dx}{x+m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Con... | The following is a proof of the formula $$S= \sum_{k=1}^{\infty} \frac{H_{k}}{ (k+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
This formula is mentioned in a comment under Bennett Gardiner's answer.
(For $a=0$, the right side of the equation should be interpreted as a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 5
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Coordinate vectors for different size matrices Hi am having trouble with this question
Let $A = \left[\begin{array}{cc}-1&3&-4&-3\\2&-1&0&-3\end{array}\right]$ Find the coordinate vector for the matrix $A \cdot A^{T}$
with respect to the standard basis for $R^{2 \times 2}$
I figured out the $A \cdot A^{T} = \left[\beg... | $$AA^t=\begin{pmatrix}-1&\;\;3&-4&-3\\\;\;2&-1&\;\;0&-3\end{pmatrix}\begin{pmatrix}-1&\;\;2\\\;\;3&-1\\-4&\;\;0\\-3&-3\end{pmatrix}=\begin{pmatrix}35&4\\4&14\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the probability of $a>b+c$, where $a$, $b$, $c$ are $U(0,1)$ What is the probability that $a > b + c$?
$a, b, c$ are picked independently and uniformly at random from bounded interval [0,1] of $\mathbb{R}$.
| Let $d = b + c$.
$d$ has the Irwin-Hall distribution with $n=2$ independent variables.
Its CDF is equal to $F_d(x)=\frac{1}{n!}\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom{n}{k}(x-k)^n$.
Now,
$$P(a<d) = \int_{-\infty}^{\infty} F_a(x)\, F'_d(x)\,dx.$$
Explanation of this equality. $a$ and $d$ are independent and continuous... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.
| Other people have basically already said it, but the FOIL is only something that you can apply when you have expressions of exactly this form:
$$
(a + b)(c+d).
$$
In this case you do: (F)irst (O)utside (I)nside (L)ast: The First part of $(a + b)^2$ is $ac$. The Outside is $ad$, the Inside if $bc$ and the Last is $bd$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1)$ is true for any $n\in\mathbb N$? Question : Is the following true for any $n\in\mathbb N$?
$$\left\lfloor\left(\sum_{k=n}^{\infty}\frac 1{k^3}\right)^{-1}\right\rfloor=2n(n-1).$$
Note that $\lfloor x\rfloor$ is the largest integer not ... | Yes, it is true:
$$\sum_{k=n}^{\infty} \frac{1}{k^3} < \sum_{k=n}^{\infty} \frac{1}{k(k^2-1)} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k(k-1)} - \frac{1}{2k(k+1)}\Big) = \frac{1}{2n(n-1)}$$ and $$\sum_{k=n}^{\infty} \frac{1}{k^3} > \sum_{k=n}^{\infty} \frac{k}{k^4 + \frac{1}{4}} = \sum_{k=n}^{\infty} \Big( \frac{1}{2k^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/527214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Existence of $2^{1/2}$ I'm trying to prove, that the square root of 2 exists in $\mathbb{R}$.
I'm looking at the set $A:=\{y\in\mathbb{R}: y\geq 0, y^2\geq 2\}$. I have already proven that the greatest lower bound $b$ of $A$ exists. However, now I don't know how to proceed. I know that I should consider the cases $b^2 ... | Suppose that the infimum is $b$ and there is no real number $c$ such that $c^2=2$. Then if $b^2> 2$, then we have $0<a<1$ and moreover:
$$
b(1-a)< b\\
b^2(1-a)^2=b^2\left(\frac 12+\frac 1{b^2}\right)^2\geq b^2\left(\frac {\sqrt 2}{b}\right)^2=2
$$
where the last inequality comes from simple AM-GM inequality. But this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Calculate volumes using triple integrals I have to calculate these two volumes using triple integrals:
volume of $A = \{(x,y,z) \in \Bbb R^3 : {x^2\over a^2} + {y^2\over b^2} \leq z \leq 1 \}$
volume of $B = \{(x,y,z) \in \Bbb R^3 : x^2+y^2+z^2 \leq a^2, x^2+y^2-ax \geq 0,x^2+y^2+ax \geq 0 \}$
So I want to calculate th... | Volume A is an elliptic paraboloid with its "vertex" at the origin, truncated by the plane $ \ z \ = \ 1 \ $ . Because it has a symmetry axis along the $ \ z-$ axis, it is actually convenient to integrate using a single variable, since every "slice" parallel to the $ \ xy-$ plane is an ellipse, the area of which has a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A triangle has side lengths 4,6,8. A tangent is drawn to incircle parallel to side 4 cutting ..... Problem :
A triangle has side lengths $4,6,8$. A tangent is drawn to incircle parallel to side $4$ cutting other two sides at M and N, than length of MN is
(a) $\frac{10}{9}$
(b) $\frac{20}{ 9}$
(c) $\frac{5}{3}$
(d) $... | Let the triangle be $ABC$, such that $M$ is on $AC$, and $N$ is on $AB$ and $x=MN$.
As $ABC$ and $AMN$ are similar we want to derive the scaling factor $k$ by calculating their areas. Let the games begin!
Let $F_1$ be the area of $ABC$, $F_2$ the area of $AMN$ and $F_3$ the area of $NMBC$. From Heron we know $F_1=3\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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$\gcd(c^a + 1, c^b + 1)$ for even $a$ and $b$? Following on this question, what is the Greatest Common Denominator of $c^a + 1$ and $c^b + 1$, where $a, b, c \in N$. I know that for odd a and b, we have $\gcd(c^a + 1, c^b + 1) = c^{\gcd(a, b)} + 1$ Thanks, Aleks Vlasev
I also found that for a odd and b even (or vice ve... | If $b=a$, then we have the trivial case $c^b+1 = c^a+1$.
Now take $b>a$ (without loss of generality). Note that $\gcd(r,s) \mid \gcd(r+s,r-s)$ for all integers $r,s$. Hence here we need consider
\begin{align}
&\gcd(c^b+c^a+2,c^b-c^a)\\
&\qquad= \gcd(c^b+c^a+2, c^a(c^{b-a}-1)) \\
&\qquad= \gcd\left(c^b+c^a+2, c^a(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/533769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help. The question asks
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.
Here is what I have done so far. Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + ... | Your function can be expressed as a convergent infinite product
$$f(z) = (1-z)(1-z/2)(1-z/4)(1-z/8)\dots$$
It's easy to see directly from this expression that it satisfies the functional equation $f(2z) = (1-2z)f(z)$.
I don't think that it can be expressed in terms of elementary functions, but I may be wrong.
Its value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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About the inscribed sphere and the exspheres of a $n$-dimensional simplex Let us consider $n$-dimensional simplex $K$ in $n$-dimensional Euclidean space. Let $r_0$ be the radius of the inscribed sphere of $K$, and let be $r_1, r_2, \cdots, r_{n+1}$ be each radius of the exsphere of $K$. Then, here is my question.
Quest... | I'll prove the formula for a triangle, and it'll be "obvious" that the argument extends to any-dimensional space.
Consider $\triangle ABC$ with incenter $O$ and inradius $r$. Recall the fundamental observation that point $O$ helps dissect the triangle into sub-triangles of equal height ($r$), so that
$$\begin{align}
|\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/535692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Laurent series of $z^{-3}$ at $z_0 = i$. Is there a way to do this by hand or is the question just evil? I have to find the two Laurent series expansions of $\frac{1}{z^3}$ about $i$. The only approach I can think of is to do:
$$\frac{1}{z^3} = \frac{1}{(z-i)^3} \left( \frac{z-i}{z} \right) ^3 = \frac{1}{(z-i)^3} \left... | I'm sure you have been taught about this before. $\frac{1}{z^3}$ is regular at
$z = i$, so the Laurent series there is simply the ordinary Taylor series expansion (in the vicinity of $i$).
For any $\alpha > 0$ and $a \in \mathbb{C}$ not on the negative real axis, we have
$$\begin{align}
\frac{1}{z^\alpha}
= & \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/536121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| $$\begin{align}\int_0^1\frac{\ln(x)}{x^2-1}dx&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)\\
&=\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 5
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How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| We can show that there is no solution for $x,y,z>0$. Firstly we have:
$$
x^{2}+y^{2}+z^{2}=xyz \implies xyz\geq 3(xyz)^{\frac 23}\\
\implies xyz\geq 27.
$$
On the other hand we have:
$$
x^{2}+y^{2}+z^{2}=xyz\geq \frac 13(x+y+z)^{2}=\frac 13(x^3+y^3+z^3)^{2}\geq \frac 13(3xyz)^{2}=3(xyz)^2\\
xyz\leq \frac 13
$$
So we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proofs of the properties of Jacobi symbol The definition and properties of Jacobi symbol are stated in this article.
I don't have a textbook handy containing the proofs of the following properties of Jacobi symbol.
It seems to me that not many textbooks on elementary number theory contain them.
So I think it is nice no... | 1.
Let $n$ be a positive odd integer.
Let $a, b$ be integers such that $a \equiv b$ (mod $n$).
Then
$$\left(\frac{a}{n}\right) = \left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ with every prime factor repeated according to its multiplicity.
Since $a \equiv b$ (mod $p$) for every pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$ Can anybody give me a hint about how to find smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$?
I thought that I will find it piece by piece with help from my friend Fermat's Littl... | First of all, $p-1$ is not necessarily the Multiplicative Order ord$_pa$ for prime $p$ and $(a,p)=1$
For example, ord$_72=3\ne 7-1$
Again, $9=3^2$ is not prime, we need Euler's Totient Theorem here
Now as ord$_{11}2=10$ and ord$_{13}2=12,$ there can be no number $<$lcm
$(10,12)=60$ to satisfy $2^n\equiv1\pmod {5\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find the Minimum value of this expression Let $x,y,z>0$ such that: $x+y+z=1$. Find the minimum value of this expression.
$P=(\frac{x+1}{x})^3.\frac{z^2}{z^2+1}+(\frac{y+1}{y})^3.\frac{x^2}{x^2+1}+(\frac{z+1}{z})^3.\frac{y^2}{y^2+1}$
| Using rearrangement inequality:
$$P=\sum_{cyc} \left[\left(\frac{x+1}{x}\right)^3 \frac{z^2}{z^2+1}\right] \ge \sum \left[\left(\frac{x+1}{x}\right)^3 \frac{x^2}{x^2+1}\right] = \sum \frac {(1+x)^3}{x \, (1+x^2)}$$
Noting $f(x) = \dfrac{(1+x)^3}{x \, (1+x^2)}$ is convex in $(0, 1)$, by Jensen's inequality:
$$\frac{P}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/545892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $?
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $
| Look at $xy$ divisors and $yz$ divisors. (Because we have that $x,y,z \in \mathbb{N}$.)
$$
\begin{array}{|c|lcr|}
\hline xy \text{ and } yz & \text{} \\
\hline
22 & 1 & 2 & 11 & 22 \\
26 & 1 & 2 & 13 & 26 \\
\hline
\end{array}
$$
So if $xy=22=\color{red}{2}\cdot11 = \color{green}{1} \cdot 22$, and $yz=26=\color{red}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Pre Calculus Synthetic Division two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
| The cubic must factor as $2(x-r_1)(x-r_2)(x-r_3)$ and the coefficient of $x^2$ is $2 \cdot (-3/2).$ So the sum of the roots is $3/2$ and you know two roots are $3,-2$ with sum $1$, forcing the third root to be $1/2$.
Note that we do not have to find $p,q$ to use this method, since the sum of the roots of a monic cubic ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$\sin 2\theta +\sin \theta =1$ I tried Wolframalpha to solve this equation. The solution is $\theta\approx 0.355$. Since once a wise guy at MSE told me not to trust machines, I would like to know what methods can be used to solve this equation. I would appreciate for a brief explanation.
By the way $0\leq\theta\leq\pi... | Well, note that if we put $t=\tan\frac\theta 2,$ then we have $$\begin{align}\sin\theta &= \sin\left(2\cdot\frac\theta2\right)\\ &= 2\sin\frac\theta2\cos\frac\theta2\\ &= 2\tan\frac\theta2\cos^2\frac\theta2\\ &= \cfrac{2t}{\sec^2\frac\theta2}\\ &= \cfrac{2t}{\tan^2\frac\theta2+1}\\ &= \frac{2t}{t^2+1},\end{align}$$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/554933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\d... | By Holder $$3=x^n+y^n+z^n=\frac{1}{3^{n-1}}(1+1+1)^{n-1}(x^n+y^n+z^n)\geq\frac{1}{3^{n-1}}(x+y+z)^n,$$ which gives $$x+y+z\leq3.$$
Now, let $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and $a+b+c=3$.
Thus, $k(a+b+c)\leq3$ gives $$0<k\leq1$$ and
$$\sum_{cyc}\frac{x}{x+yz}=\sum_{cyc}\frac{ka}{ka+k^2bc}=\sum_{cyc}\frac{a}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to prove the following inequality of logarithm? Let $x,y,z\in\mathbb{C}.$ Suppose $$z=\frac{1}{2}(xy\pm\sqrt{x^2y^2-4(x^2+y^2)} ).$$
Show that $$log^+|z|\leq log^+|x|+log^+|y|+log 2.$$
Where $log^+\phi=max\{0,log\phi\}.$ Here we are also considering the complex root function with respect to the principle branch of ... | To prove what you want, we basically just need to apply the triangle inequality to your formula. I'll look at cases separately depending whether $|x|,|y|$ are greater than or smaller than unit magnitude.
*
*If $|x| \le 1$, $|y| \le 1$, then $|x^2y^2-4x^2-4y^2| \le |x^2y^2|+|4x^2|+|4y^2|\le 9$,so:
$$2z=xy\pm\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/561999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Third-degree cosine inequality for obtuse triangle Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$
My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get
$$a^4b^2+... | Using $\sin2x,\cos2x$ formula
$$a^3\cos A=(2R\sin A)^3\cos A= 2R^3(2\sin^2A)(2\sin A\cos A)$$
$$=2R^3(1-\cos2A)\sin2A=R^3(2\sin2A-2\sin2A\cos2A)=R^3(2\sin2A-\sin4A)$$
Using this, $\sum \sin2A=4\prod \sin A$
Now, $\sin4A+\sin4B+\sin4C=2\sin(2A+2B)\cos(2A-2B)+2\sin2C\cos2C$
Now, $\cos2C=\cos\{2\pi-2(A+B)\}=\cos2(A+B)$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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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.