Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Show that $\sqrt[3]{1+\sqrt{3}}$ isn't an element of the field $\mathbb{Q}(\sqrt{3} ,\sqrt[3]{2})$ Setting $\alpha = \sqrt[3]{2}$ and $a+b\alpha+c\alpha^2=\sqrt[3]{1+\sqrt{3}}$ for $a,b,c$ in $\mathbb{Q}(\sqrt{3})$ (The minimal polynomial for $\sqrt[3]{2}$ in $\mathbb{Q}(\sqrt{3})$ is $x^3-2=0$ since if it were reducib... | Here is an alternative solution, which is essentially the same as Lubin's but with a more elementary presentation.
Lemma. Let $K$ be a subfield of $\mathbb C$ with $\sqrt[3]{2}\not\in K$ (whence it easily follows that $[K(\sqrt[3]{2}):K]=3$). Let $k\in K$. Then $k$
is a cube in $K(\sqrt[3]{2})$ iff one of $k,\frac{k}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi... | A different formulation:
Let
$\displaystyle \alpha = \frac{\pi - A}{4}$
$\displaystyle \beta = \frac{\pi - B}{4}$
$\displaystyle \gamma = \frac{\pi - C}{4}$
We need to find the minimum value of
$\displaystyle \tan^2 \alpha + \tan^2 \beta + \tan^2 \gamma$
subject to
$\displaystyle \alpha + \beta + \gamma = \frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How to quickly compute the determinant of given matrix I need to find the determinant of given matrix :
$\begin{bmatrix}
1&0&0&0&0&2\\
0&1&0&0&2&0\\
0&0&1&2&0&0\\
0&0&2&1&0&0\\
0&2&0&0&1&0\\
2&0&0&0&0&1\\
\end{bmatrix}$
I know that It can be computed with the help of row operations;by applying
R1 $\to$ R1 + R2+R3+R4+R5... | By reordering both rows and columns you can see that the determinant is equal to that of a block diagonal matrix:
$$
\begin{vmatrix}
1&2&0&0&0&0\\
2&1&0&0&0&0\\
0&0&1&2&0&0\\
0&0&2&1&0&0\\
0&0&0&0&1&2\\
0&0&0&0&2&1\\
\end{vmatrix}
= \left(\begin{vmatrix}
1&2 \\ 2&1
\end{vmatrix}\right)^3
$$
Or with Schur complements: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so... | Making
$$
y = \frac 12\left(e^{i\beta}+e^{-i \beta}\right)-\frac{1}{2i}\left(e^{i\beta}-e^{-i\beta}\right)
$$
we have after powering and collecting
$$
y^3 = \frac 32\left(\cos\beta-\sin\beta\right)-\frac 12\left(\cos(3\beta)+\sin(3\beta)\right)
$$
then follows
$$
y^3 = \frac 32 y - \frac 12 x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Please explain the formula for the sum of the cubes and the difference: $a^3 - b^3$ and $a^3 + b^3$? I have not yet fully mastered the formula for accelerated multiplication. I was able to understand everything except the last two. I would like to deal with them. Why have : $$ a^3 + b^3 = (a+b)(a^2 -ab + b^2)^* $$ and ... | There is a very general factorisation formula, once taught in high school:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+a^{n-k}b^k+\dots+ab^{n-2}+b^{n-1}).$$
(To memorise it, the second factor is the sum of all monomials in $a$ and $b$ of total degree $n-1$.)
The simplest proof consists in showing first by induction the part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simple sum with $i = 2$ step I need to calculate the following sum:
$$S_{n} = \sum_{2 \leq i \leq n } (3i - 2)$$
(two more conditions for the above sum: $n$ is even and $i$ with step $2$ (not sure how to do multi-line))
Adding image of the task with multi-line visible:
I wrote down the first few terms and it looks li... |
We consider the original task
\begin{align*}
S_N=\sum_{{2\leq i\leq N}\atop{i\ \mathrm{with\ step\ } 2}}\left(3i-2\right)\qquad\qquad N\mathrm{\ even}
\end{align*}
We calculate for small $N=2,4,6$ (considering even $N$ only)
\begin{align*}
S_2&=\sum_{{2\leq i\leq 2}\atop{i\ \mathrm{with\ step\ } 2}}(3i-2)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integrating the real part of $(1-ix)/\sqrt{a^2-(1-ix)^2}$ What is the solution to the following integral (if there is any)?
$$I = \operatorname{Re}\int_0^{\infty}dx \frac{1-ix}{\sqrt{a^2-(1-ix)^2}},$$
where it is known that $a>1$.
My approach:
I'm not getting any progress with taking the integral before the real part, ... | Let $f(x) = (1 - i x)/\sqrt {a^2 - (1 - i x)^2}$ and assume that $\sqrt z$ is the principal value of the square root. Since $\lim_{x \to \infty} f(x) = -i$, $\int_0^\infty f(x) dx$ diverges. However,
$$\int_0^A \operatorname {Re}(f(x)) dx =
\operatorname {Re} \int_0^A f(x) dx$$
for $A \in \mathbb R$, because the integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the method to factor $x^3 + 1$? In the solution to a problem, it's stated that
We see that $x^3+1=(x+1)(x^2-x+1)$.
Why is this, and what method can I use for similar problems with different coefficients?
The full problem is
Find the remainder when $x^{81}+x^{48}+2x^{27}+x^6+3$ is divided by $x^3+1$.
| Since $(-1)^3+1=0,$ we know that $x^3+1$ is divisible by $x+1$, which gives the following way:
$$x^3+1=x^3+x^2-x^2-x+x+1=x^2(x+1)-x(x+1)+x+1=(x+1)(x^2-x+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How would I go about computing this finite sum? How would I go about computing the sum
$$
\sum_{k=1}^{n} \dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}.
$$
I have tried partial fractions but have gotten stuck trying to find the coefficients. I decomposed it like this:
$$
\dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a... | Observe that
$$\dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}= \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}.$$
Then take sum as $k$ varies from $1$ to $n$.
$$\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\dfrac{2^k}{k^2}=\sum_{k=1}^n \dfrac{-2^{k+1}}{(k+1)^2}+\sum_{k=1}^n \dfrac{2^k}{k^2}$$
This being an alternate sum of elements of same t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Reducing fractions How can I get [if it's possible]:
$$ 2- \frac{1}{n+1} $$
from this:
$$2- \frac{1}{n} + \frac{1}{n(n+1)} $$
EDIT:
I started like this:
$$2- \frac{1}{n} + \frac{1}{n(n+1)}= 2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$
How do I proceed from here?
|
I started like this: $$2- \frac{1}{n} + \frac{1}{n(n+1)}=
2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)} $$
The mistake happened here:
$$2-\frac{n+1}{n(n+1)}+\frac{1}{n(n+1)} =2-\frac{n+2}{n(n+1)}$$
because you forgot the minus sign. You can rewrite the left hand side as
$$2 +\frac{-n-1}{n(n+1)} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\be... | $90<x,y<180^\circ$
$180<x+y<360\implies x+y=360-\arccos(1/3)$
If $x-y>0,x-y=\arccos(1/5)$
$\sin2y=\sin(360-\arccos(1/3)-\arccos(1/5))=-\sin(\arccos(1/3)+\arccos(1/5))$
Now $\arccos(1/3)+\arccos(1/5)=\arcsin(2\sqrt2/3)+\arccos(2\sqrt6/5)$
Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{... | \begin{align}
\int\limits_{0}^{1} \dfrac{x^{2}}{\sqrt{1+x^{4}}}\,\mathrm{d}x &= \dfrac{1}{4}\int\limits_{0}^{1} \dfrac{1}{x^{1/4}\sqrt{1+x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1}^{2} \dfrac{1}{\left(x-1\right)^{1/4}\sqrt{x}}\,\mathrm{d}x \\
&= \dfrac{1}{4}\int\limits_{1/2}^{1} \dfrac{1 -\color{red}{x} + \color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we re... | We know the GP progression formula for when common ratio less than $1$ which is pretty simple to derive and is $\displaystyle \sum_{k=1}^\infty r^{k} = \dfrac{1}{1-r} $ from there we can write the following,
\begin{align} \int \sum\limits_{k=2}^\infty r^{k-1} dr = \int \frac{1}{1-r} dr \end{align}
After integration we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Algebraic Manipulations to solve for q If $p-q=\sqrt3$ and $\sqrt{\sqrt{3}-\sqrt{p}}=q$, find $q$.
Some work
$$\bigg(\sqrt{\sqrt{3}+\sqrt{p}}
\bigg)\bigg(\sqrt{\sqrt{3}-\sqrt{p}}\bigg)=\sqrt{3-p}$$
$$\sqrt{\sqrt{3}+\sqrt{p}}=\frac{\sqrt{3-p}}{q}$$
$$q^2\sqrt3+q^2\sqrt{p}=3-p$$
$$(\sqrt3+\sqrt{p})q^2+p-3=0$$
I can't fin... | Square the second equation and rearrange
\begin{eqnarray*}
\sqrt{3}-q^2 = \sqrt{p}.
\end{eqnarray*}
Square this and subtract $q$ and we can use the second equation to get
\begin{eqnarray*}
q^4 -2 q^2 \sqrt{3}-q +3 -\sqrt{3}=0.
\end{eqnarray*}
Now this can be factorised !
\begin{eqnarray*}
(q^2-q - \sqrt{3})(q^2+q +1 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$... | Hint $\ \overbrace{x = \color{#0a0}{-1\!+\!9}(a\!+\!5(b\!+\!4c))}^{\text{by iterated division}}\,$ so $\overbrace{\color{#90f}{\bmod 5}\Rightarrow a\equiv 3}^{\large\color{#90f}{x\ \equiv\ 1}};\ $ $\overbrace{\color{#c00}{\bmod 4}\Rightarrow b\equiv 0}^{\large\color{#c00}{x\ \equiv\ 2}},\,$ so $\ \bbox[5px,border:1px s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to apply Lagrange Multipliers with modulus ??
I had the question $$x^2+y^2+xy=1$$ and I had to find maximum value of $$\vert xy(x^2+y^2) \vert$$
I tried using Lagrange multipliers by forming
$x^3+3xy^2 = \lambda (2y+x)$
And $y^3+3yx^2 = \lambda (2x+y)$
Now I subtract it to get $(x-y)^3 = \lambda (y-x)$
And as $\l... | Hint.
Following the fact
$$
\max |f| = \max(-f,f) \ \ \text{and}\ \ \min |f| = \min(-f,f)
$$
With $f = x y(x^2+y^2)$ the lagrangian can be established as
$$
L(x,y,\lambda) = s f+\lambda(x^2+y^2+x y -1)
$$
where $s=\pm$
The stationary points are given by
$$
\nabla L = 0 = \left\{
\begin{array}{rcl}
2 s y x^2+\lambda (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$ for $x_i > 0$ If $ x_1, x_2 , x_3........x_n $ are n positive reals prove that
$$ \frac{x_1}{1+x_1^2} + \frac{x_2}{1+x_1^2+x_2^2} +...+\frac{x_n}{1+x_1^2+x_2^2+...x_n^2} \le \sqrt{n}$$
So this is an IMO 2001 propose... | Let us call the individual terms of the inequality as $A_i$, then by AM-RMS inequality we have
$$\frac{\sum A_i}{n} \le \sqrt{\frac{\sum A_i^2}{n}},~\mbox{where} ~A_i=\frac{x_i}{1+x_1^2+x_2^2+x_3^2+...x_i^2}.~~~~(1)$$
So it would suffice to prove that $\sum A_i^2 \le 1.$
Note that for $i\ge 2$, $$ A_i^2=\frac{x_1^2}{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now ... | Note that $n^5-n=n(n^2-1)(n^2+1)$ is divisible by $5$ by little Fermat. $n^3-n$ is likewise divisible by $3$ and either $n^2$ or $n^2-1$ is divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number.
I tried some bas... | $$x^3-x^2+px-1=0 \Rightarrow (x^3-1)^3-(x^2-px)^3=0 \Rightarrow x^9+(3p-.4)x^6+(p^3-3p+3)-1=0.$$
Let us transform this equation by $y=x^3+1 \rightarrow x=(y-1)^{1/3}$. Then
we get a cubic Eq. for $y$ as
$$y^3+y^2(3p-7)+(p^3-9p+14)y-p^3+6p-9=0,$$ $y_1, y_2, y_3$ are its roots.
Then $$y_1 y_2 y_3=p^3-6p+9=2019 \Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
Find $\displaystyle\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx$
My approach is as follow
$\cos^2x=t$;
$\sin2x\ dx=-dt$
Therefore,
\begin{align}
\int \frac{\sin2x+2\tan x}{\cos^6x+6\cos^2x+4}dx&=\int \frac{\sin2x+\frac{2\tan x\cos^2x}{\cos^2x}}{\cos^6x+6\co... | You are almost there.
Note that\begin{align}\frac{1+\frac1t}{t^3+6t+4}&=\frac{t+1}{t(t^3+6t+4)}\\&=\frac14\left(\frac1t-\frac{t^2+2}{t^3+6t+4}\right).\end{align}Now, use the fact that $(t^3+6t+4)'=3(t^2+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding the Taylor Series Expansion using Binomial Series, then obtaining a subsequent Expansion. Hi all doing some prep work for a course and really struggling to wrap my head around some of the revision questions. Some help would be really appreciated. There are two halfs, part one is as follows:
Consider the functi... |
We obtain from $f(x)=\frac{1}{1-x}=\sum_{n=0}^\infty x^n$
\begin{align*}
\color{blue}{x^2f(-x^2)}&=x^2\sum_{n=0}^\infty(-x^2)^n\\
&=x^2\sum_{n=0}^\infty(-1)^nx^{2n}\\
&=\sum_{n=0}^\infty (-1)^nx^{2n+2}\\
&\,\,\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}x^{2n}}
\end{align*}
In the last line we shift the index by $1$ in or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-... | When the example has
$$ n + 5 = n-2 + 7 $$
it might have been more clearly written
$$ n + 5 = 1\cdot(n-2) + 7 $$
to highlight the quotient, $1$, and remainder, $7$, of the division $\frac{n+5}{n-2}$.
For your problem,
$$ 3n + 11 = 3 \cdot (n+1) + 8 \text{.} $$
That is, the quotient is $3$ and the remainder is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What is $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$ Let $p_{n}$ a prime number and $p_{n+1}$ is the next prime.
How to calculate $\liminf_{n\to\infty} ( \frac{p_{n}}{p_{n}+p_{n+1}})$
Edited :
here is my attempt :
$\frac{p_{n}}{p_{n}+p_{n+1}}\approx \frac {n ln(n)}{p_{n}+p_{n+1}} \frac{p_{n}}{n ln(n)} $
| In fact, your lim inf is a lim. First, observe that
$$
(n+1)\ln(n+1) = n \ln(n+1) + \ln(n+1) = \\
n[\ln(n+1) - \ln(n) + \ln(n)] + \ln(n+1) = \\
n \ln(n) + n \ln\left(\frac{n+1}{n}\right) + \ln(n+1)
$$
Thus (by the prime number theorem), we have
$$
\lim_{n \to \infty} \frac{p_n}{p_{n+1}} =
\lim_{n \to \infty} \frac{p... | {
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"answer_id": 0
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Prove inequalities $\frac 34 \le I(a,b) \le 1$ Given the expression,
$$ I(a,b) = \frac{a^2}{(1+a)(a+b)} + \frac{b^2}{(1+b)(a+b)} + \frac{1}{(1+a)(1+b)}$$
where $a\ge 0$ and $b\ge 0$, prove the following inequalities:
$$\frac 34 \le I(a,b) \le 1$$
I had trouble figuring it out. Tried some inequity techniques I am aware ... | Putting everything over a common denominator shows that
$$I(a,b) = \frac{a^2+b^2+a^2b+ab^2+a+b}{a^2+b^2+a^2b+ab^2+a+b+2ab} = \frac{(1+a)(1+b)(a+b)-2ab}{(1+a)(1+b)(a+b)} \leq 1$$
since $a,b\geq 0$. Simplifying this expression, we get
$$I(a,b) = 1 - \frac{2ab}{(1+a)(1+b)(a+b)}$$
Now take a look at the second term and con... | {
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"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Determinant of matrix $[a_{ij}]$ where $a_{ij}=ij$ if $i\ne j$ and $a_{ij}=1+ij$ if $i=j$ Let $A=[a_{ij}]$ be a square matrix of order $n$ whose entries are given as follows.
For $1\leq i,j\leq n$ we have $a_{ij}=ij$ if $i\neq j$ and $a_{ij}=1+ij$ if $i=j$.
I have to evaluate the determinant.
I just wrote the matrix b... | Let the determinant be $D_n$. Then
$$
D_n = \det
\begin{bmatrix}
1 + 1^2 & 1 \times 2 & 1\times 3 & \cdots & 1\times n\\
2\times 1 & 1 + 2^2 & 2\times 3 & \cdots & 2 \times n \\
3\times 1 & 3 \times 2 & 1 + 3^2 & \cdots & 3 \times n \\
\vdots & \vdots &\vdots & \ddots & \vdots\\
n\times 1 & n \times 2 & n\times 3 &... | {
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Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$. It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le
8\sqrt{30}$ using numeric methods. For example by multiplying
$(1+\sqrt{2}) \le 3 $
$(1+\sqrt{3}) \le 3 $
$(1+\sqrt{5}) \le 4 $
We get:
$(1+\sqrt{2})(1+\sqr... | We use the inequality $$\frac{a+b}{2}\le \sqrt{\frac{a^2+b^2}{2}}$$ so we get
$$\frac{1+\sqrt{2}}{2}\times\frac{1+\sqrt{3}}{2}\times\frac{1+\sqrt{5}}{2}\le \sqrt{30}$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$.
In $\triangle ABC$, $AD$ and $BE$ are medians, and $AD \perp BE$. If $AC=14$ and $BC=22$, find $AB$.
I'm using Apollonius's theorem to find the medians. Instead, I found
$$BE^2-AD^2=72$$
Not sure how to proceed.
| Let $G$ be the centroid of $\triangle ABC$. Suppose that $3x=|AD|$ and $3y=|BE|$. Then by Pythagoras
$$7^2=\left(\frac{1}{2}|AC|\right)^2=|AD|^2=|AG|^2+|GD|^2=(2x)^2+y^2$$
and
$$11^2=\left(\frac{1}{2}|BC|\right)^2=|BE|^2=|BG|^2+|GE|^2=(2y)^2+x^2.$$
Thus
$$170=7^2+11^2=(4x^2+y^2)+(4y^2+x^2)=5(x^2+y^2).$$
Therefore,
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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First order linear differential equation (possible error in book) Consider following differential equation in $$\frac{dx}{(3x^2+y^2)/x}=\frac{dy}{(3y^2+x^2)/y}=\frac{dz}{-(x^2+y^2)/z}$$
$\require{enclose}$
Taking multipliers $x, y, 4z$ we get $$xdx+ydy+4zdz=0 \\ \enclose{box}{x^2+y^2+4z^2=C_1}$$
I have this function co... | You are right, two charecteristic equations are :
$$x^2+y^2+4z^2=C_1$$
$$\frac{(x^2-y^2)^2}{x^2+y^2}=C_2$$
From that, the general solution of the PDE expressed on the form of implicit equation $C_1=F(C_2)$ with arbitrary function $F$ :
$$x^2+y^2+4z^2=F\left(\frac{(x^2-y^2)^2}{x^2+y^2}\right)$$
On explicit form :
$$z(x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$ using Mathematical induction $x_1=1$ $x_2=2$
$x_n=\frac{1}{2}(x_{n-2}+x_{n-1})$ for n $\gt$ 2. We have to prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$
What I tried :
For n=1, $|x_1-x_{2}|=1 =\frac{1}{2^{0}}$
Let for n=k assumption be true. Hence $|x_k-x_{k+1}|=\frac{1}{2^{k-... | Using triangle inequality will probably not help you, as you're supposed to prove an exact equality, not an inequality. Instead, just substitute $x_{k+2} = \frac{1}{2}(x_k + x_{k+1})$. Then,
\begin{align*}
|x_{k+1} - x_{k+2}| &= \left|x_{k+1} - \frac{1}{2}(x_k + x_{k+1})\right| \\
&= \left|\frac{1}{2}x_k - x_{k+1}\righ... | {
"language": "en",
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"source": "stackexchange",
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Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{s... | Need to prove$:$ $$\displaystyle \left( {\dfrac {a+b}{c}}+{\dfrac {b+c}{a}}+{\dfrac {a+c}{b}} \right) ^{2
}\geqslant 4 \left( a+b+c \right) \left( {\dfrac {a}{bc}}+{\dfrac {b}{ac}}+{
\dfrac {c}{ab}} \right) $$
Or $$\displaystyle \,{\frac { 2\left( ab-2\,ac+bc \right) ^{2}}{ \left( {a}^{2}+{c}^{2} \right) {b}^{2}}}+{\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Line integral in parameterized curve $a,b\gt 0$ and $n \in \mathbb Z$ \ {$0$}. Parameterized curve $C: x=a\cos nt, y= b\sin nt$
where $0 \le t \le2\pi$ and n is how many times that goes around origo.
I want to calculate line integral:
$\omega(C) = \frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2}$
I have calculated:
$... | So you were able to find
$$
\frac{1}{2\pi}\oint_C \frac{x\,dy-y\, dx}{x^2+y^2}
= \dfrac{2\tan^{-1}\left(\frac{b\tan(2n\pi)}{a}\right)\vert a b\vert - ab(\mod(4n-1,2)-4n-1)\pi}{4\vert ab \vert\pi}
$$
That looks pretty complicated, but:
*
*$\tan(2n\pi) = 0$ for any integer $n$. Therefore
$$
\tan^{-1}\left(\fr... | {
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"source": "stackexchange",
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Let $ a \in \mathbb{N}^\ast $, prove that $ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $ I have the following problem to solve. It's about convergent sequences.
Let $ a \in \mathbb{N}^\ast $, prove that:
$$ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $$
To solve it, I first sol... | Note that $$\sqrt{a^2+1}<\sqrt{a^2+1+\frac{1}{4a^2}}=\sqrt{\left(a+\frac1{2a}\right)^2}=a+\frac1{2a}.$$
Thus
$$\sqrt{a^2+1}-a<\frac{1}{2a}.$$
On the other hand
$$\sqrt{a^2+1}+a<2a+\frac{1}{2a}.$$
Therefore
$$\sqrt{a^2+1}-a=\frac{1}{\sqrt{a^2+1}+a}>\frac{1}{2a+\frac{1}{2a}}.$$
However
$$\frac{1}{2a+\frac{1}{2a}}=\frac{1... | {
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What is probability to get two balls of different colors, if one of them is blue?
There are $12$ red, $8$ green and $10$ blue balls in a box. Two balls are taken out at random. What is probability to get two balls of different colors, if one of them is blue?
We can use the formula of Conditional probability. Let $B=... | You computed the probability of obtaining at least one blue ball incorrectly. For the favorable cases, you can have either two blue balls, which occurs with probability
$$\Pr(\text{two blue}) = \frac{\dbinom{10}{2}}{\dbinom{30}{2}}$$
or one blue ball and one ball of a different color, which you correctly calculated i... | {
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Demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ for the given conditions. I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.
The proof provides me with the information that $x^2\gt 2$ and $x>0$.
I've taken the following steps to simplify the algebra...to the point whe... | It is equivalent to $$\frac{(x^2-2)^2}{4x^2}>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of the periodic continued fraction with given terms
Find the value of the periodic continued fraction with the terms
$1, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, . . .$
We see that it starts to be periodic after $1$, i.e, $3,4,3,2$ then $3,4,3,2$ etc...
I know that $x= \frac{A_{k+1}}{B_{k+1}}$ = $\frac{A... | if we take $u=x+1,$ we find that u > 2 has purely periodic c.f. $(2,3,4,3).$
Next
$$
\left(
\begin{array}{cc}
2 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
3 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
4 & 1 \\
1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
3 & 1 \\
1 & 0 \\
\en... | {
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"url": "https://math.stackexchange.com/questions/3382551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Tricky epsilon-delta proof $$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$
What I've got so far is that:
$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\
\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\fr... | Take the case where $x\in [-2,0]$ (i.e. $\delta < 1$). Then we have the following inequality
$$\left|\frac{(x+1)(2x-1)}{3(x^2-x+1)}\right| < \frac{5}{3}|x+1|$$
by maximizing the numerator and minimizing the denominator. So set $\delta = \min(\frac{3}{5}\epsilon,1)$ and the proof step for the limit follows in both cases... | {
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but... | The quadratic formula alone won't help you obviate the $\pm$ sign. It's better to note $\sin x=\frac{2t}{1+t^2}$ ($t$ is more common notation than $B$) while $\cos x=\frac{1-t^2}{1+t^2}$, so $\frac{\sin x}{1+\cos x}=t$. Alternatively, $$\frac{\sin x}{1+\cos x}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}=... | {
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The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k \in \mathbb N$.
The polynomial $x^{2k}+1+(x+1)^{2k}$ is not divisible by $x^2+x+1$. Find the value of $k\in \mathbb{N}$.
I tried finding out the roots of $x^2+x+1$ which were $\dfrac{-1±\sqrt{3}i}{2}$ but in vain. I got no resul... | Using the remainder theoram, putting $x^2 + x + 1 = 0$ in the polynomial $P(x) =x^{2k} + 1 + (x + 1)^{2k}$ we will get the remaimder.
$$
P(x) = 1 + x^{2k} + (x^2 + 2x + 1)^k = 1 + x^{2k} + x^k
$$
We can also prove that if $x^2 + x + 1 = 0$, $x^3 = 1$.
We can now clearly see that for $k$ is not equal to $3q$, $P(x)=0$
B... | {
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Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has
1) exactly two distinct negative roots
2) at least two distinct negative roots
I tried to factorize it but didn't get any breakthrough.
| Hint: Try $x^4+2ax^3+x^2+2ax+1=(x^2+bx+1)(x^2+cx+1)$.
| {
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
M... | Let consider the reverse $\frac{n^3}{3} > 3n-3$ then we can prove the induction step as follows
$$\frac{(n+1)^3}{3} =\frac{n^3}{3}+\frac{3n^2+3n+1}{3}> 3n-3+\frac{3n^2+3n+1}{3}>3n-3+3=3(n+1)-3$$
we have used that for $n\ge 2$
$$\frac{3n^2+3n+1}{3}>3 \iff3n^2+3n-8>0$$
| {
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ ... | $y:=x+3$;
$f(y)=(y-1)^{1/3}+y^{1/3}+(y+1)^{1/3}=0$;
By inspection $y=0$ is a solution.
0) $y^{1/3}$ is an odd, increasing function.
1) $y \ge 1$ is ruled out, since $f(y)>0$
2) $y \le -1$ is ruled out, since $f(y)<0$.
Remains $1< y <1$;
3) $f(y)> 0$ for $0<y<1$; since $-1< (y-1) <0$, $1<(y+1) <2$:
4) Similarly $f(y)... | {
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"source": "stackexchange",
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Series using partial sums... Determine if the following series converge by studying the partial sums. If it converges, compute its value.
$\sum_{n=1}^{\infty} \frac{2n+3}{(n+1)(n+2)}$.
We use partial fractions so
\begin{equation*}
\begin{split}
\frac{2n+3}{(n+1)(n+2)} &= \frac{A}{n+1}+\frac{B}{n+2} \\
\Longleftrightarr... | Your method is correct, I just put a simpler approach if you are interested. We see that $$\frac{2n+3}{(n+1)(n+2)}>\frac{2n}{n^2+3n+2}>\frac{2n} {6n^2}=\frac{1}{3n} $$
But $\sum{\frac{1}{n}}$ diverges.....
| {
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How to get a closed-form for this sum? I'm trying to calculate the $[x^n]F(x)$ from a generating function $F(x) = \frac{1}{(1-2x^2)^2}$ and I came across with this expression involving a sum:
$$2^n\sum_{k=0}^n k^k (n-k)^{n-k}$$
What I did so far was calculate the $[x^n]\frac{1}{1-2x^2}$, witch is $(2n)^n$ if I'm not w... | It would seem that based on the series
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$
that
$$\frac{1}{1- 2 \, x^2} = \sum_{n=0}^{\infty} 2^{n} \, x^{2n} = \sum_{n=0}^{\infty} 2^{n/2} \, \left(\frac{1+(-1)^{n}}{2}\right) \, x^n.$$
From this then:
\begin{align}
\frac{1}{(1- 2 x^2)^2} &= \sum_{r=0}^{\infty} \sum_{s=0}^{\inf... | {
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Find all the value of $(x, y, z)$ with $x, y, z \in \Bbb R$ $$x^2 + 4 = y^3 + 4x - z^3$$
$$y^2 + 4 = z^3 + 4y - x^3$$
$$z^2 + 4 = x^3 + 4z - y^3$$
What's the proper method to solve this question? I've tried any method but still not get the answer.
| $$x^2 + 4 = y^3 + 4x - z^3 \implies x^2 -4x +4 = y^3-z^3 \implies (x-2)^2 = y^3-z^3 \quad\text{ (1.)}$$
$$y^2 + 4 = z^3 + 4y - x^3 \implies y^2 -4y +4 = z^3 -x^3 \implies (y-2)^2 = z^3-z^3 \quad\text{ (2.)}$$
$$z^2 + 4 = x^3 + 4z - y^3 \implies z^2 -4z +4 = x^3-y^3 \implies (z-2)^2 = x^3-y^3 \quad\text{ (3.)}$$
Adding ... | {
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$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(... | The equation is $ \frac{2\tan x}{\tan 2x} - 5 \tan x + 5 = 0$, as one sees from rearranging the first term. The double angle formula for $\tan$ gives $\frac{2 \tan x }{\tan 2x} = 1 - \tan^2 x$ and therefore $(1 - \tan^2 x) - 5 \tan x + 5 = 0$, which simplifies to $\tan^2 x + 5 \tan x - 6 = 0$. This finally gives $\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$ Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$
My attempt is as follows:
Taking LHS:
$$\left(\sin A-\cos A\right)(1+\sin A\cos A)$$
$$\left(\sin^2A-\cos^2A\right)\frac{\left(1+\sin A\cos A\right)}{\left(\sin A+\cos A\right)... | With $x=\pi$ the LHS is equal to $1$ while the RHS is equal to $-1$ therefore the identity is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse trigonometry proof How would I go about proving
$$\arccos x = \arctan \frac{\sqrt {c-x^2}}{x} $$
Where $c$ is a constant and $0< x ≤1$
| Suppose $\theta = \arccos x$
then $\cos \theta = x$
if $0<x\le 1$ then $0\le \theta < \frac {\pi}{2}$
$\sin \theta = \sqrt {1-x^2}$ and $\tan\theta = \frac {\sin\theta}{\cos\theta} = \frac {\sqrt {1-x^2}}{x}$
$\theta = \arccos x = \arctan \frac {\sqrt {1-x^2}}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a stat... | Observe that \begin{eqnarray*}
(\frac{3n-2}{3n+1})^{2n} & = & \left(\frac{3n+1-3}{3n+1}\right)^{2n}\\
& = & \left\{ \left(1+\frac{(-3)}{3n+1}\right)^{-1}\left(1+\frac{(-3)}{3n+1}\right)^{3n+1}\right\} ^{\frac{2}{3}}.
\end{eqnarray*}
We assume the fact without proof: For any $x\in\mathbb{R}$, $\lim_{n\rightarrow\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How can we attain 3 solutions when using Vieta's Substitution? Consider $x^3+px=q$. Substituting $x:=w-\frac{p}{3w}$ will reduce it to $w^3-\frac{p^3}{27w^3}-q=0$. Multiplying with $w^3$ will result in $(w^3)^2-qw^3-\frac{1}{27}p^3=0$. Using the quadratic formula, we get$$w^3=-\frac{q\pm \sqrt{q^2+\frac{4}{27}p^3}}{2}... | When $q^2<-\frac{4p^3}{27}$, $w^3$ becomes a complex number and there are three distinctive complex values for $w$ once the cubit root is taken on $w^3$.
The solution to the cubic equation $x^3+px+q$ is instead,
$$x=w-\frac{p}{3w}$$
As a result, there are three distinctive roots for $x$ evaluated from the three $w$’s.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
I tried the following but it doesn't seem to work...
$$= \lim_{x\to 0} \frac{x}{2} \cdot \frac{\sin(3x)}{3x}\cdot\frac{6x}{1-\cos(6x)}$$
$$= 0$$
But the result of this limit is $\frac{1}{6}$.
Am I missing something or did I make... | We can also make the substitution $x \mapsto \frac{u}{3}$:
$$\frac{1}{3} \frac{u \sin u}{1 - \cos 2u} = \frac{1}{3} \frac{u \sin u}{1 - \cos^2 u + \sin^2 u} = \frac{1}{3} \frac{u \sin u}{2 \sin^2 u} = \frac{1}{6}\frac{u}{\sin u}.$$
and since $\frac{u}{\sin u} \to 1$ as $u \to 0$:
$$\lim_{u \to 0} \frac{1}{3} \frac{u \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Why does $\sqrt{x\sqrt{x}}$ equal $x^{3/4}$ instead of $x^{1/4}$? When I try to solve the equation, I get $x^{1/4}$ since the two roots share the same index and exponent, which means that I ought to be able to multiply across.
Instead, the solution in the textbook is as follows.
$$\left (xx^{1/2} \right )^{1/2}=\left ... | The notation is $\sqrt{x\sqrt x}= m$
Then $m^2 = x\sqrt x$.
So $(m^2)^2 = (x\sqrt x)^2 = x^2*(\sqrt x)^2 = x^2*x = x^3$
So $m^4 = x^3$ and so
$m = \sqrt[4]{x^3} = x^{\frac 34}$
....
The book did it this way:
$\sqrt{x\sqrt x} = (x\cdot x^{\frac 12})^{\frac 12}=$
$(x^1\cdot x^{\frac 12})^{\frac 12} =$
$(x^{1+ \frac 12})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2... | Hint: Write your equation in the form $$- \left( \sin \left( x \right) \right) ^{2} \left( \cos \left( x
\right) \right) ^{2} \left( \left( \cos \left( x \right) \right) ^
{4}-3\, \left( \cos \left( x \right) \right) ^{2}+4 \right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2... | Play with linear combinations of $x^2+y^2$ and the expression from the given equation. This way we find for $a>1$:
$$a(x^2+y^2) - 6 = (a-1)x^2-2xy+(a+1)y^2=(\sqrt{a-1}x-\sqrt{a+1}y)^2+(2\sqrt{a^2-1}-2)xy$$
We get rid of the last term if we let $a=\sqrt 2$:
$$\tag1\sqrt 2(x^2+y^2) -6 = \left(\sqrt{\sqrt 2-1}\,x-\sqrt{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the nth derivative using the binomial theorem. Let n be an element of the set of natural numbers
Let $F(x)=x^2(1+x)^n$ and write $F^n$ for the nth derivative of the function $F$.
Compute $F^n$ by applying the Binomial Theorem to $(1+x)^n$.
I don't understand the step where I need to find the derivative of t... | Write it in decreasing powers of $x$ and use subscript $n$:
$$F_n(x)=x^2(1+x)^n=x^2(x+1)^n=\\
\color{red}{x^{n+2}}+\color{green}{nx^{n+1}}+\color{blue}{\frac{n(n-1)}{2}x^{n}}+\frac{n(n-1)(n-2)}{6}x^{n-1}+\cdots+nx^3+x^2$$
Note that for $f(x)=x^n$:
$$f^{(0)}(x)=f(x)=x^n\\
f^{(k)}(x)=\begin{cases}n(n-1)(n-2)\cdots(n-k+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proving an inequality using AM-GM
Let $ a,b,c > 0$ such that $ abc = 1$. Prove that
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1.$$
$$\frac{ab}{a^2+b^2+\sqrt{c}}=\frac{1}{\frac{a}{b}+\frac{b}{c}+\sqrt{c^3}}$$
(AM-GM): $$\frac{a}{b}+\frac{b}{a}+\sq... | By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{ab}{a^2+b^2+\sqrt{c}}\leq\sum_{cyc}\frac{ab}{2ab+\sqrt{c^2ab}}=\sum_{cyc}\frac{\sqrt{ab}}{2\sqrt{ab}+c}=$$
$$=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{\sqrt{ab}}{2\sqrt{ab}+c}\right)=\frac{3}{2}-\sum_{cyc}\frac{c}{2(2\sqrt{ab}+c)}\leq$$
$$\leq\frac{3}{2}-\frac{\left(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $x$ is an integer if $x^4-x$ and $x^3-x$ are integers. Given that $x^4-x$ and $x^3-x$ are integers.
Prove that $x$ is an integer.
Expanding, adding and subtracting the given expressions I can conclude that the followings are integers:$x(x-1)(x+1)$, $x^3(x-1)$, $x(x-1)(x^2+x+1)$ are integers.
Could you help ... | I will assume that $x$ is a priori living in some integral commutative ring, e.g. let's say $x$ is a complex number. It wouldn't be as easy (or as true) if $x$ were in some non-commutative or non-integral ring.
Write $a = x^4 - x$ and $b = x^3 - x$, which are integers by assumption.
We have
(1) $x^4 - x - a = 0$,
(2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Finding the value of $c$ to make $-\dfrac{1}{2}x^2+x+c$ a perfect square trinomial
I should find the value of $c$ to make: $$-\dfrac{1}{2}x^2+x+c$$ a
perfect square trinomial.
I really messed up: $-\dfrac{1}{2}x^2+x+c=x-\dfrac{1}{2}x^2+c$, but it seems like $-\dfrac{1}{2}x^2$ isn't $a^2\boldsymbol{-2ab}+b^2$. Can ... | We need $B^2-4AC=0$ that is
$$1+2c=0 \implies c=-\frac12$$
indeed we obtain
$$-\frac12(x^2-2x+1)=\left(\frac i{\sqrt2}(x-1)\right)^2$$
Therefore there is no $c$ to have a perfect square on reals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is it possible to represent the natural number "1" as the sum of p-series in this way? My argument: $$1=(\frac{1}{2})^2+(\frac{1}{3})^2+\cdots+(\frac{1}{2})^3+(\frac{1}{3})^3+\cdots=\sum_{k=2}^\infty (\frac{1}{k})^2+\sum_{k=2}^\infty (\frac{1}{k})^3+\cdots .$$
Explanation) First, for any natural number $n\geq2$, the fo... | So what you are saying is
$1
=\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
$.
Let's check.
$\begin{array}\\
\sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \dfrac1{k^m}
&=\sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \dfrac1{k^m}
\qquad\text{(reverse order of summation)}\\
&=\sum_{k=2}^{\infty}\dfrac{1/k^2}{1-1/k}
\qquad\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Simple continued fraction of $\sqrt{d}$ with period of shortest length $3$ This is the problem:
Does there exist positive integer $d$ ( which is not a perfect square ) such that the length of the least period in the simple continued fraction of $\sqrt{d}$ is $3$?
Consider the following theorem
Theorem : If the pos... | Just working numerically, $41$ is the least example, with $$\sqrt {41}=[6; \overline {2,2,12}]$$
here is a tabulated list of the periods of $\sqrt d$.
OEIS provides a list of $d$ for which the period is $3$, and that link provides a way of generating infinitely many examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Find the number of integers between 1 and 25 so that the expression is divided by 6 Find the number of integers so that the $n^2+3n+2$ is divided by 6 when $1 \le n \le 25$.
I start by $6\mid n^2+3n+2 6|(n+1)(n+2) \implies 6k=(n+1)(n+2) $ when $ k \in Z$
we can see that for $n=$ $1,2,4,5,7,8,10,11....$ will work, but ... | Being divisible by $6$ means being even and divisible by $3$.
*
*Modulo $2$, $n^2+3n+2\equiv n^2+n\equiv n+n \text{ (by Fermat) }\equiv 0$. So it is always even.
*Modulo $3$, $n^2+3n+2\equiv n^2+2$, and it is congruent to $0$ if and only if $n^2\equiv 1\mod 3$, which happens only if $n\equiv \pm 1$, in other words,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Complexity of Quicksort I am currently struggling with a problem on time complexity of the Quicksort algorithm in the average case.
I calculated the average time complexity of the algorithm to be
$T_n \quad = \quad \frac{1}{n} \cdot \sum_{k=1}^n \left( T_{k-1} + T_{n-k} + n - 1 \right)$
Which can also be written as
$T_... | I finally managed to get it down. I start from the above mentioned term
$n \cdot T_n \quad = \quad (n+1) \cdot T_{n-1} + 2(n-1)$
Since I only want to know magnitudes and asymptotic behavior I simply discard the $-2$ term, remaining with
$n \cdot T_n \quad = \quad (n+1) \cdot T_{n-1} + 2n$
which is equivalent to
$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$ for all $n \ge 2$? I'm trying to prove that $$n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$$ for all $n \ge 2$. I try induction on $n$ as follows:
My attempt:
The inequality holds for $n=2$. Let it holds for $n$. Our goal is to show that $$2 {... | Note that the statement itself tells you how to approximate $\sum_{k=1}^{n} \frac{1}{k}$, so with wishful thinking we hope it is sufficient (which need not be the case).
Change the statement to showing that $\sqrt{\frac{n}{2}} \geq \sum_{k=1}^{n} \frac{1}{k}$.
Then, the induction step requires us to show that $\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ Let $a$,$b$,$c$ and $d$ be non-zero, pairwise different real numbers such that $ \frac{a}{b} +\frac{b}{c} +\frac{c}{d} + \frac{d}{a}=4$ and $ac=bd$ .
Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ and that $-12$ is the max... | Let $u = \frac{a}{b},\,v= \frac{b}{c},$ then $\frac{c}{d}=\frac{1}{u},\,\frac{d}{a}=\frac{1}{v}$ we get $u+v+\frac{1}{u}+\frac{1}{v}=4,$ and
$$P = \frac{a}{c} + \frac{b}{d} + \frac{c}{a} + \frac{d}{b} = \left(u+\frac{1}{u}\right)\left(v+\frac{1}{v}\right).$$
Now see $u,v$ can't satisfied $uv > 0.$ Indeed if $u > 0,\,v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
r... | Your equation $$\left(x-\frac{r}{1+r}\right)^2+y^2-\frac{1}{(1+r)^2}=0$$ is equivalent to
$$rx^2+ry^2-2xr+x^2+y^2+r-1$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find $g$(3) if $g(x)g(y)=g(x)+g(y)+g(xy)-2$ and $g(2)=5$
If $g(x)$ is a polynomial function satisfying $g(x)g(y)=g(x)+g(y)+g(xy)-2$ for all real $x$ and $y$ and $g(2)=5$, then find $g$(3)
Putting $x=1,y=2$
$$
5g(1)=g(1)+5+5-2\implies \boxed{g(1)=2}
$$
And I think I can evaluate for $g(1/2)$ also.
Solution given in my... | The point is that $g$ is a polynomial.
Thus write $g(x) = a_n x^n + \dotsc + a_0$, and you have an equality:
$$\sum_{i, j = 0}^n a_i a_j x^i y^j = (\sum_{i = 0}^n a_i (x^iy^i + x^i + y^i)) - 2.$$
Since this is true for all $x, y$, it means that it is an equality as polynomials in $x, y$.
By comparing coefficients, we s... | {
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"url": "https://math.stackexchange.com/questions/3439843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow ... | You are re-stating the Fibonacci sequence, and by the theory of linear recurrences, the terms are quasi-proportional to the powers of the largest root of the characteristic equation
$$\phi^2-\phi-1=0.$$
Hence, the ratio of successive terms quickly tends to $\phi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 4
} |
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt... | Since
$(n+1)^3
=n^3+3n^2+$lower order terms,
$(n+2/3)^3
=n^3+2n^2+$lower order terms.
Therefore the left cube root
is about $n+2/3$.
Since the right cube root
is just about $n$,
the difference is about $2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove formula from Pascal's triangle I found this in my Norwegian mathematics book. I have solved a)-d) but am stuck at e). Nice if you have feedback on what I have attempted to do.
a) Where are the numbers of the form $\binom{k}{l}$ in Pascal's triangle? Where are the numbers of the form $\binom{k}{2}$? In general, if... | I have the solution for $\sum_{k=1}^nk^2$
$k^2=k(k-1)+k$ so that $\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\sum_{k=1}^nk$
$\sum_{k=1}^n k(k-1)(k-2)\dots (k-i+1)=\frac{(n+1)n(n-1)\dots(n-i+1)}{i+1}$
$\sum_{k=1}^n k(k-1)=\frac{(n+1)n}{2+1}=\frac{(n+1)n(n-1)}{3}$
$\sum_{k=1}^nk^2=\sum_{k=1}^nk(k-1)+\frac{n(n+1)}{2}=\frac{(n+1)n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statem... | As you suggested, it's notationally simpler to suppose $4^k+5^k+6^k$ is divisible by $15$ and consider
$$4^{k+2}+5^{k+2}+6^{k+2} = 16\cdot 4^k + 25\cdot 5^k + 36\cdot 6^k.$$
Subtracting the original expression, we get $15\cdot 4^k + 24\cdot 5^k + 35\cdot 6^k$. The first term is divisible by $15$. Now note that
$$24\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Solve for $x,y$ $\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
Designate $x,y\in\left(0,\frac{\pi}{2}\right)$ which fulfills
$\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
My proof:
$\cos2x=1-2\sin^2x\Rightarrow \sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x$
$\sin^2x+\s... | I think this is a question about mental arihmetic, which $x=\pi/6, y= \pi/4$.
Or proceed from third step, $\cos2x+\cos2y=2\cos(x+y)(\cos(x-y))=1/2$
hence $2\cos(5\pi/12)(\cos(x-y))=1/2$
then using calculator, or as follows, you compute $\cos(x-y)=\dfrac{\sqrt{2}(\sqrt{3}+1)}{4}$ (note that $\cos(5\pi/12)$=...
Either ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing ... | As $x>0$
WLOG $x=3\sec t, 0< t<\dfrac\pi2\implies \sin t,\cos t>0$
$$\dfrac{35}4=3\sec t+\dfrac{3\sec t}{3\tan t} \iff\sec t+\csc t=\dfrac{35}{12}$$
Method$\#1:$
Let $\sin t+\cos t=u$
$$\dfrac{35}{12}=\dfrac{2u}{u^2-1}$$
$$\iff0=35u^2-24u-35=7u(5u-7)+5(5u-7)=(5u-7)(7u+5)$$
As $u>0, u=\sin t+\cos t=\dfrac75$
Now use $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x... | As $x\to 0$, $x\sin x\sim x^2$. Also
$$\cos x=1-\frac{x^2}2+O(x^4),$$
$$\cos2x=1-2x^2+O(x^4),$$
$$\sqrt{\cos2x}=1-x^2+O(x^4),$$
$$(\cos x)(\sqrt{\cos2x})=1-\frac{3x^2}2+O(x^4),$$
$$1-(\cos x)(\sqrt{\cos2x})\sim\frac{3x^2}2.$$
Therefore
$$\lim_{x\to0}\frac{1-(\cos x)(\sqrt{\cos2x})}{x\sin x}=\frac32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Limit of this sequence $\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$ I am trying to calculate the limit of this sequence :
$$\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$$
I tried two methods and the two methods leaded me to infinity or 4/0.
Anything woul... | We can use that
$$ \frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}=$$
$$=\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}
\cdot \frac{\sqrt{2n+2} + \sqrt{2n-2}}{ \sqrt{2n+2} + \sqrt{2n-2}}
\cdot\frac{\sqrt{3n+1} + \sqrt{3n}}{ \sqrt{3n+1} + \sqrt{3n}}=$$
$$=4\cdot \frac{\sqrt{3n+1} + \sqrt{3n}}{ \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{... | For all $x \in \mathbb{R}$ holds
$$1 - \frac{x^2}2\le \cos x \le 1 - \frac{x^2}2 + \frac{x^4}{24}$$
so
$$\left(1 - \frac{x^2}2\right)^{1/x^2}\le (\cos x)^{1/x^2} \le \left(1 - \frac{x^2}2 + \frac{x^4}{24}\right)^{1/x^2}$$
Both bounds are easily seen to converge to $e^{-\frac12}$ when $x \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ Let $F_{a}(n)$ be the digit sum of $n$ in base $a$,
define $W(a,b)=F_{a}(a^{\lceil\frac{\log{b}}{\log{a}} \rceil}-b)$,
prove that $\displaystyle\ W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ if $n−1 \in 6\mathbb{N} \pm 1$.
| Let
\begin{align}
&a=n&
&b=\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6
\end{align}
We have
$$\frac{a^3}3\leq b\leq\frac{a^3}3+a^2$$
from which
$$2<\frac{\log(b)}{\log(a)}<3$$ hence $\lceil\log(b)/\log(a)\rceil=3$ and $W(a,b)=F_a(a^3-b)$.
Moreover, if $b=d_0+d_1a+d_2a^2$ with $0\leq d_i\leq a-1$ then
$$a^3-b=(a-d_0)+(a-d_1-1)a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many seating arrangements of $m$ people in a row of $n$ seats are there if $k$ of the people must sit together? Suppose there are five people $A, B, C, D, E$ to be seated in a row of eight seats $S_1, \ldots, S_8$.
(1) How many possibilities are there if $A$ and $B$ are to sit next to each other?
(2) How many possi... | Both of your solutions to the first problem are correct. However, your general formula is not. Notice that in the first problem, $n = 8$, $m = 5$, and $k = 2$, so your formula gives
$$2!P(8 - 2 + 1, 8 - 5) = 2!P(7, 3) = 2! \cdot 7 \cdot 6 \cdot 5 = 2 \cdot 210 = 420$$
Let's see what went wrong.
We wish to seat $m$ p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$ I have this simple limit
$$\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}$$
I have solved this with these steps:
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}=\frac{\ln(1+x^3\cos 3x)(1+\sqrt{1+\c... | The fact is that, $\ln(1+x^{3}\cos 3x)\approx x^{3}\cos 3x$, without the additional $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\... | You need to make use of the following standard limits $$\lim_{t\to 0}\frac{e^t-1}{t}=1,\lim_{t\to 0}\frac{\tan t} {t} =1,\lim_{t\to 0}\frac{\sinh t} {t} =1,\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}$$ Using the second limit above the limit in question is equal to the limit of $$\frac{\sqrt{\cosh(3x^2)}e^{4x^3}-1}{2x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve intial value problem using power series $xy''+y'+2y=0$ with y(1) =2, y'(1) =4.
What I tried:
\begin{equation}
xy'' +y'+2y = 0
\end{equation}
Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
x\sum_{k=2}... | $$
\begin {cases}
xy''+y'+2y=0, \\
y(1) =2, y'(1) =4
\end{cases}
$$
Hint
There are some mistakes in your calculations. And you should use this serie instead since you are given initial conditions at $x=1$ and not at $x=0$:
$$y=\sum_{k=0}^{\infty}c_k(x-1)^k,$$
$$ y'(x)=\sum_{k=0}^{\infty}(k+1)c_{k+1}(x-1)^{k}$$
$$ y''(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sum_{k=0}^\infty \frac{e^{-x}(k-x)^2x^k}{k!} = x$ Show that $$\sum_{k=0}^\infty \frac{e^{-x}(k-x)^2x^k}{k!} = x$$
I guess I can use Maclaurin expansion of $e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$.
| Consider instead
$$\begin{align*}
\sum_{k=0}^\infty (k(k-1)-2kx+x^2)\frac{x^k}{k!}
&= \sum_{k=2}^\infty k(k-1)\frac{x^k}{k!} - 2\sum_{k=1}^\infty k\frac{x^{k+1}}{k!} + \sum_{k=0}^\infty \frac{x^{k+2}}{k!}
\\ &= \sum_{k=2}^\infty \frac{x^k}{(k-2)!} - 2\sum_{k=1}^\infty \frac{x^{k+1}}{(k-1)!} + \sum_{k=0}^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^... | Hint: This is pretty straightforward if you accept that $\sqrt[n]n\to1$. For, $\sqrt[n]{n^2+n}=\sqrt[n]n\cdot \sqrt[n]{n+1}$. Note $\sqrt[n]{n+1}\to1$ easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Polynomial with integer coefficients: given values at point find minimal value for a point A polynomial $P(x)$ with integer coefficients satisfies the following: $P(5) = 25, P(7) = 49, P(9) = 81$. How do I find the minimal possible value of $|P(10)|$.
| We can write $p(x)=q(x)(x-5)(x-7)(x-9)+x^2$.
Now $p(10)=q(10)(10-5)(10-7)(10-9)+100=q(10) \cdot 5 \cdot 3 \cdot 1 +100=15 \cdot q(10)+100$
So we want $q(10)$ to be "as close as possible" to $-\frac{100}{15}=-6,67..$, i.e. we want $q(10)=-7$ and then $|p(10)|=|-15 \cdot 7 +100|=|-105+100|=5$. So for example take $q(x)=x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$ This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify ... | $$I=\int_{0}^{\pi/4}\frac{\sin x +\cos x}{9+16 \sin 2x} dx.$$
Use $\sin 2x=1-(\sin x- \cos x)^2$ and re-write
$$I=\int_{0}^{\pi/4} \frac{\sin x +\cos x}{25-16(\sin x -\cos x)^2} dx,$$
Now use $\sin x -\cos x=t \implies (\cos x+ \sin x ) dx=dt$, then
$$I=\frac{1}{16}\int_{-1}^{0} \frac{dt}{25/16-t^2}=\left .\frac{1}{16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Proving bijectivity of $f:[0,\infty)\to[0,2)$, where $f(x)=\frac{1}{x}+1$ for $x>1$, and $f(x)=x$ for $x\le 1$
Let $f: [0, \infty) \rightarrow [0,2)$ be defined by $$f(x) = \left\{
\begin{array}{ll}
\frac{1}{x} + 1, & x > 1 \\
x, & x\le 1 \\
\end{array}
\right.$$
The question asks to prove that the functi... | Injectivity: Suppose $f(x) = f(y)$.
There are four cases: $0 \le x \le 1$ and $0\le y \le 1$ and $f(x) = x = f(y) = y$. Thus $x = y$.
$0 \le x \le 1$ and $y > 1$ and $f(x) = x = f(y) = \frac 1y +1$. So $x = \frac 1y + 1$. And $y= \frac 1{x-1}$. But if $y > 1$ then $0 < \frac 1y < 1$ and $1 < \frac 1y + 1< 2$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof explanation of $\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$ I don't understand the proof of
$$\sum_{k=0}^{n-1}(n+k)(n-k) = \frac{1}{6}n(n+1)(4n-1), n \geq 1$$
via induction.
$$\text{Base case}: n=1 \Rightarrow \sum_{k=0}^0(1+k)(1-k)=1=\frac{1}{6}(1+1)(4-1) = \frac{6}{6}$$
$$\text{Inductive s... | Here
$$\sum_{k=0}^n \left[(n+1+k)(n+1-k)\right] = \sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]$$
we are using that
$$(n+1+k)(n+1-k)=((n+k)+1)((n-k)-1)=$$
$$=(n+k)(n-k)+(n-k)+(n+k)+1$$
and then since $(n+k)+(n-k)=2n$
$$\sum_{k=0}^n\left[(n+k)(n-k)+(n-k)+(n+k)+1\right]=$$
$$=\sum_{k=0}^n \left[(n+k)(n-k)\right]+$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | You can use the Squeeze Theorem, with
$$ n\cdot \frac{n}{n^2 + n} \leq a_n \leq n\cdot \frac{n}{n^2 + 1} $$
And as $ n \to \infty$, you get $ 1 \leq \lim_{n \to \infty} a_n \leq 1 $, So, using the Squeeze Theorem, $a_n \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
value of $k$ in binomial sum
If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is
what i try
$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}... | The binomial expansion is
$$\sqrt{1-x}=\sum_{n=0}^\infty (-1)^n\binom{\frac{1}{2}}{n}x^n$$ and so you need to solve for $k$
$$\sum_{n=0}^{10} (-1)^n\binom{\frac{1}{2}}{n}=\frac{\binom{20}{10}}{k^{10}}$$ The result should be a rather small integer. Just use Excel and inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding how many numbers in a given set contain a given binary pattern I came across a weird question recently in a competition, and now that the competition is complete I'm wondering how to solve similar problems (the actual competition was non-calculator and had values in place of $n_{10}$ and $k_{10}$.
Given $S = \{... | A practical method for tests
I had a go at solving your problem by recurrence relations in 'test conditions'. This did work but I found it much easier to write out the solution by splitting up the interval into simple chunks. So for your example the working would be as follows
$\begin{vmatrix}1&1&0&0&0&1&1&1 \\1&1&0&0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$ABC$ rectangle in $A$ or $C$ iff $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$ The triangle $ABC$ is rectangle in $A$ or $C$ if and only if $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$, where $\alpha$ is the angle in $A$, $\beta$ is the angle in $B... | Rewrite
$$\frac{\sin\alpha+\sin\gamma}{\sin\beta}
=\frac{2\sin\frac{\alpha+\gamma}{2}\cos\frac{\alpha-\gamma}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}
=\frac{\cos\frac{\alpha-\gamma}{2}}{\sin\frac{\beta}{2}}
=\cot\frac{\beta}{2}$$
or
$$\cos\frac{\alpha-\gamma}{2}=\cot\frac{\beta}{2}\sin\frac{\beta}{2}=\cos\frac{\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 ... | A somewhat recursive form of the integral can be derived as follows:
$$\int_0^\infty\frac{\sin^n(x)}{x^m}dx = \int_0^\infty\sin^n(x)\frac{1}{x^m}dx =$$
integrating per partes and assuming $n > m-1$ leads to
$$=\frac{n}{m-1}\int_0^\infty\frac{\sin^{n-1}(x)}{x^{m-1}}\cos(x)dx=\frac{n}{m-1}\int_0^\infty\frac{\sin^{n-2}(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\t... | Note,
$$\sin x\sin^{-1}x=(x-\frac{x^3}6 +\frac{x^5}{120}+O(x^7)) (x+\frac{x^3}6 +\frac{3x^5}{40}+O(x^7))=x^2+\frac{x^6}{18}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}
=\lim\limits_{x\to 0}\dfrac1{x^6} \left(\frac{x^6}{18}+O(x^8)\right)=\dfrac{1}{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that $\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}$ Show that $$\lim_{x\to 0} \frac{\tan x\tan^{-1}x-x^2}{x^6}=\frac{2}{9}.$$
Proceed:
$$\tan x\tan^{-1}x=(x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+O(x^7)) (x-\frac{1}{3}x^3+\frac{1}{5}x^5+O(x^7))=x^2+\frac{2x^6}{9}+O(x^8)$$
Thus,
$$\lim\limits_{x\to 0} \dfr... | $$\tan^{-1} x=x-x/^3/3-x^5/5+..., \tan x=x+x^3/3+2x^5/15+...$$
Then $$L=\lim_{x \rightarrow 0}\frac{\tan x \tan^{-1} x-x^2}{x^{6}} =\lim_{x \rightarrow 0} \frac{2x^6/9+x^8/45+...}{x^6}=\frac{2}{9.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of $\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$ as $x\to\infty$ I wish to find the limit of $$\sin\left(\sqrt {x+2}\right)-\sin\left(\sqrt{x+4}\right)$$ as $x\to\infty$. I think that this limit does not exist.
| Hint: $\sin A -\sin B =2\cos (\frac {A+B} 2) \sin (\frac {A-B} 2)$ and hence $|\sin (\sqrt {x+2} -\sin (\sqrt {x+4}) |\leq 2 |\sin (\sqrt {x+2} - \sqrt {x+4})/2| \to 0$ since $\sqrt {x+2} - \sqrt {x+4} =\frac {-2} {\sqrt {x+2} +\sqrt {x+4}} \to 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Multiple of a number and sum of a set Lets say there is a number 30, and this sum of numbers
3 + 5 + 7 + 9 ...
Now I want to find where this sum and a multiple of 30 gets equal.
for 30 its 4 = 30 * 4 = 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
I don't have any math experience, just wanted to know if this could be ... | We know that the sum of first $n$ odd numbers is $n^2$ . So the sum $$1+3+5+\cdots+(n+1) = n^2 \\ \implies 3+5+7+\cdots+(n+1)=(n+1)^2-1$$Since we want it to be a multiple of $30$ , We have the following equation:
$$(n+1)^2- 1= 30k \implies n^2+2n-30k = 0$$
Using quadratic formula , we get:
$$n = \dfrac{-2\pm\sqrt{4+120... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For which value of $k$ the equations $y=2x-5$, $y=x+2$ and $y=kx-12$ have common solution For which value of $k$ the equations $y=2x-5$, $y=x+2$ and $y=kx-12$ have common solution.
Let $G_{2x-5}$ intersects $G_{x+2}$ at point $N(x_N;y_N)$. We can get that $x_N=7$ and $y_N=9$. From here $N$ must lie on $G_{kx-12}$. Ther... | The problem is that $k = 3$ is the proper value since $3(7) - 12 = 9$. Your $k$ value of $\frac{20}{7}$ is $\frac{1}{7}$ too small, which is why the red line is slightly to the right & below where it should be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find maximum of $C=2(x+y+z)-xy-yz-xz$
Let $x,y,z\ge 0$ such that $x^2+y^2+z^2=3$. Find the maximum of $$C=2(x+y+z)-xy-yz-xz.$$
I tried Schur and AM-GM inequality but I really have no idea about this problem. It is not homogeneous so it's hard for me.
| Let $(x+y+z) = a$ . Then $$(x+y+z)^2 = a^2 \implies (xy+yz+zx)=\dfrac{a^2-3}{2}$$
So $$2(x+y+z)-xy-yz-xz = 2a -\dfrac{a^2-3}2$$
$$ C = \dfrac{-a^2+4a+3}{2}$$
The maximum of this quadratic is at $a = 2$ , for which the max becomes :
$$2(x+y+z)-xy-yz-xz = \dfrac 72$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+... | \begin{align}
7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc
&\ge0
\tag{1}\label{1}
\end{align}
As @DeepSea suggested, we can replace
the expressions in terms of side lengths $a,b,c$
with equivalent in terms of semiperimeter $\rho=\tfrac12(a+b+c)$,
inradius $r$ and
circumradius $R$ of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
$\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$
Solve the system:
$$\begin{array}{|l} \dfrac{5}{x}-\dfrac{3}{y}=-4 \\ \dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3} \end{array}$$
First, we have $x,y \ne 0$. If we multiply the first equation by $xy$ and the second by $3x^2y^2$, we get $5y-3x=-4xy$ and $9y... | HINT:
Let $\dfrac{1}{x}=a$, $\dfrac{1}{y}=b$ and solve the following system:
$$5a-3b=-4$$ and $$3a^2+6b^2=\frac{11}{3}.$$
$b=\dfrac{5a+4}{3}$ gives $$3a^2+\frac{2(5a+4)^2}{3}=\frac{11}{3}$$ or
$$59a^2+80a+21=0$$ or
$$(a+1)(59a+21)=0.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding a formula for a triangle similar to Pascal's. I need to find a formula expressed in binomial coefficients of the following triangle:
$$D_n^k=\begin{cases}
n \quad\quad\quad\quad\quad\text{ if } n=k \text{ or } k=0 \\
D_{n-1}^k+D_{n-1}^{k-1} \text{ otherwise}
\end{cases}$$
This triangle is the same as the Pascal... | We derive a formula with the help of generating functions for
\begin{align*}
D_n^k&=D_{n-1}^{k}+D_{n-1}^{k-1}\qquad\ \ n\geq 1,k\geq 1\tag{1}\\
D_0^k&=k\qquad\qquad\qquad\qquad k\geq 1\\
D_n^0&=n\qquad\qquad\qquad\qquad n\geq 1
\end{align*}
We obtain
\begin{align*}
D(x,y)&=\sum_{n=1}^\infty\sum_{k=0}^nD_n^kx^ny^k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
what $a$ that makes $(a - \sqrt 2) x^2 + (a - \sqrt 2)x + a - 1$ always below $x$-axis it means$(a - \sqrt 2) < 0$ also $D < 0$
so we get $a < \sqrt 2$ and also ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
Solve ${(a - \sqrt 2)}^2 - 4(a - \sqrt 2)(a - 1) < 0$
$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$
i get $\f... | Just be careful here:
$$(a - \sqrt 2)[ (a - \sqrt 2) - 4(a - 1)] < 0$$
$$(a - \sqrt 2)[ -3a +4- \sqrt 2] < 0$$
and then:
$$a<\frac{4-\sqrt{2}}{3} \quad \text{or} \quad a>\sqrt{2}$$
and after that you have to take the intesection with $a<\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.