Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Value of $\lim\limits_{n\rightarrow\infty}\prod\limits_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$ Find the value of m for the following $$\lim\limits_{n\rightarrow\infty}\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$$
| It is possible to find a closed form for your product, if fact we have for any $ k\geq 3 $ :
\begin{aligned}1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}&=\frac{1-\tan^{2}{\left(\frac{\pi}{2^{k}}\right)}}{\cos^{2}{\left(\frac{\pi}{2^{k}}\right)}}\\ &=\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos^{4}{\left(\frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$... | The quantity $T(x)\geq0$ is a ${\pi\over2}$-periodic function of $x$. It is extremal, i.e., maximal or minimal, iff $$T^2=7+2\sqrt{(5\cos^2 x+1)(5\sin^2 x+1)}$$ is extremal, and this is the case iff the radicand on the RHS
is extremal. Letting $\sin^2 x=:t$ this means that
$$\phi(t):=\bigl(5(1-t)+1\bigr)(5t+1)=6+25(t-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theor... | Introduce the dummy variable $w$. Then we want to minimize
$$
F(x,y,z,w)=x^4+y^4+z^4+w^4-1-4xyzw
$$subject to $x^2+y^2+z^2\leq 9, w=1$. By AM-GM
$$
(x^4+y^4+z^4+w^4)\geq4 \sqrt[4]{x^4y^4z^4w^4}=4xyzw,
$$with equality iff $x=y=z=w$. Then since we must have $w=1$, the minimum occurs at $(x,y,z)=(1,1,1)$, giving $f(1,1,1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I ... | From Inequality on AoPS:
Without loss of generality $x \ge y \ge z > 0$, so that $z \le 1$. We have
$$
2(2-x)(2-y) = (x+y-2)^2 +4 - x^2-y^2 \ge 4 - x^2 - y^2 = 1+z^2
$$
and therefore
$$
(2-x)(2-y)(2-z) \ge \frac 1 2 (1+z^2)(2-z) =: f(z) \, .
$$
An elementary calculation shows that the minimum of $f$ on $[0, 1]$ is $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx $ by differentiating parameter $a$ Can anyone help me calculate the integral using differentiation with respect to a parameter:
$$I(a) = \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx, |a|<1$$
UPD: I try to solve like this:
\begin{align*}
I'(a)&=\int_{0}... | Note
$$\lim_{x\to 1}{\frac{\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}= \frac{\arctan^{-1}(\infty)}{\sqrt{1+a^2}}=
\frac{\frac\pi2}{\sqrt{1+a^2}} $$
Then, you have
$$I’(a)=\frac\pi a\left(1-\frac1{\sqrt{1+a^2}} \right)
$$
and
$$ \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx
= \int_0^a I’(t) dt =\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pierre runs a game at a fair, where each player is guaranteed to win $10 - Die rolling question Pierre runs a game at a fair, where each player is guaranteed to win $10.
Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs.
When a ‘6’ occurs, Pierre gives the play... | The probability of not rolling $6$ for $k$ times is $\left(\frac{5}{6}\right)^k$.
The probability of not rolling $6$ for $k-1$ first times and then rolling $6$ on the $k$th roll is $\frac16\left(\frac{5}{6}\right)^{k-1}$.
So to find the expected number of rolls we are just to find the sum
$\sum\limits_{k=1}^\infty k\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}... | Now, $$\frac{(ad+bc)^2}{abcd}-4=\frac{a^2d^2-2abcd+b^2c^2}{abcd}=\frac{(ad-bc)^2}{abcd}\geq0.$$Also, by C-S
$$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq\left(\sqrt{\frac{b}{a}\cdot\frac{a}{b}}+\sqrt{\frac{d}{c}\cdot\frac{c}{d}}\right)^2`=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any... | $\big(x+y+z\big)^{3}=27xyz\tag{1}$
Let $s=x+y+z$ and substitute $z=s-x-y$ to equation $(1).$
$s^3-27xys+27x^2y+27xy^2=0\tag{2}$
For $x$ to be rational solution, the discriminant of the equation $(2)$ must be square.
Hence $-27y(4s-3y)(s-3y)^2$ must be square, then
$v^2 = (9y-6s)^2-36s^2$
We solve simultaneous equations... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Complex number - arctan $$z=-2+2\sqrt{3}i\implies x=-2, y=2\sqrt3$$
$$r=\sqrt{x^2+y^2}=\sqrt{4+12}=4$$
$$\text{Angle}=\arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\frac{2\pi}{3}=120^\circ$$
1) May I know how $\arctan\left(\dfrac{2\sqrt3}{-2}\right)+\pi$ turns into $\dfrac{2\pi}{3}$?
2) Can I use calculator to do the cal... | Notice, $\tan(-\theta)=-\tan\theta$
$$\therefore \arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\arctan\left(-\sqrt{3}\right)+\pi=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}=120^\circ$$
You can use calculator to find amplitude & argument. Do remember $\tan^{-1}\sqrt3=\frac{\pi}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is there an easier way to solve the given problem? If $x + 2 + \sqrt{3}i=0$ then find the value of $2x^4+3x^3-x^2-15x+36$
If you try to find the values of $x^2, x^3 $and $ x^4$, and then put the values in, we can find the solution to be 1. But is there a better way to solve this problem? Where you can probably break th... | Real coefficients in the quartic inspired me to multiply $x + 2 + \sqrt{3}i\;$ by the conjugate, which gives
$$(x + 2 + \sqrt{3}i)(x + 2 - \sqrt{3}i)=x^2+4x+7.$$
Dividing the given expression by $x^2+4x+7$ (just to see if we get a simple form) leads to $$2x^4+3x^3-x^2-15x+36=\underbrace{(x^2+4x+7)}_{0}\cdot(2x^2-5x+5)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Spivak Calculus chapter 1 problem 13 proof critique Question paraphrased: Prove that the maximum between two numbers $x$ and $y$ is given by:
$$\max(x,y)=\frac{x+y+|y-x|}{2}$$
Proof: Let $x$ and $y$ be two arbitrary numbers. Then, one and only one holds true: $x \ge y$ or $x \le y$
If $x \ge y$, $x=x+(y-y)=(x-y)+y=|x-... | Your proof is ok.
Rephrasing:
1) Let $y\ge x$:
Then $\max(x,y)=y$ and
$\dfrac{y+x+|y-x|}{2}=2y/2=y$, since $|y-x|=y-x$.
2) Let $y<x$ :
Then $\max(x,y)=x$ and
$\dfrac{y+x+|y-x|}{2}=x$ since $|y-x|=x-y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one... | Writing $u = x - 2$ and $v = y + 5$ shows that the problem is equivalent to proving that if $u^2 + v^2 = 9,$ then $|u - v| \leqslant 3\sqrt2.$ We have $$
(u - v)^2 + (u + v)^2 = (u^2 - 2uv + v^2) + (u^2 + 2uv + v^2) = 2(u^2 + v^2) = 18,
$$
therefore $(u - v)^2 \leqslant 18,$ and the result follows by taking square root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Uniform convergence of a sequence of functions which is integral of another sequence I was going through some questions on pointwise and uniform convergence. Got stuck in one of those which says:
Let $g_n(x) = \sin^2(x+\frac{1}{n})$ be defined on $[0,\infty).$
and $f_n(x) = \int_0^xg_n(t)\,dt.$
I am supposed to discuss... | You have
$$\begin{aligned}g_n(x)&=\sin^2\left(x + \frac{1}{n}\right) = \frac{1}{2}\left(1- \cos\left(2(x + \frac{1}{n})\right)\right)\\
&=\frac{1}{2}\left(1 - \cos 2x \cos\frac{1}{n} + \sin 2x \sin \frac{1}{n}\right).
\end{aligned}$$
Therefore
$$f_n(x)= \frac{1}{2}\left(x - \frac{1}{2}\cos\frac{1}{n}\sin 2x-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$ I want to calculate the following improper integral:
$$\int_A \frac{z}{y^2}\\
A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$$
First I noticed that the conditions imply $x^2<2x\rightarrow x<2;1<x^2<y^2\rightarrow y>1$, and thus $... | We can partition the domain $A$ into regions $A_1,A_2,A_3$ . . .
The diagram below shows the projections of $A_1,A_2,A_3$ onto the $xy$-plane.
Region $A_1$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{y^2-x^2}\\[4pt]
x\le\,&y\le\sqrt{2x}\\[4pt]
1\le\,&x\le 2\\[4pt]
\end{align*}
\right.
$$
so letti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Every positive integer can be written as the sum of four squares. But not all necessarily non-... | Assume there are $4$ such non-zero squares that add up to $2^{2n+1}$ for any $n \ge 1$, i.e., you have
$$2^{2n+1} = a^2 + b^2 + c^2 + d^2 \tag{1}\label{eq1A}$$
However, note all perfect squares are congruent to $0$, $1$ or $4$ modulo $8$. Since you just asked for a hint, the rest of the answer is in the spoiler below.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$? The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.
| First note that $\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{\frac{1}{2}}{2^{n-1}+1}-\frac{1}{2^{n+1}}$ (an alternative form). Also the series $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+1}<\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}=2<\infty$ and hence it converges by the comparison test. Since $\sum_{n=1}^{N}\frac{1}{2^{n+1}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so th... | You made mistake while finding residue
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{z \to \dfrac{-i}{3}} \frac{1}{3(z+3i)}=\frac{1}{3\left(-\frac i3+3i\right)}=\frac{1}{8i}
$$
$$
\therefore \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{1}{8i} = \frac{\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r... | Consider the path along the x-axis, y-axis and $y=x^{\frac{3}{2}}$.
Along the x-axis:
$$\lim_{(x,0) \to (0,0)} \frac{x^3 \cdot 0}{x^6 + 0}=0
$$
Along the y-axis:$$\lim_{(0,x) \to (0,0)} \frac{0 \cdot y^2}{0 + y^4}=0
$$
Along the curve $y=x^{\frac{3}{2}}$:
$$ \lim_{(x,x^{\frac{3}{2}}) \to (0,0)} \frac{x^3{(x^{\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I ... | It's wrong.
Try, $b>1$ and $bc\rightarrow-1^-$.
It's true for positive variables.
Indeed, let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and we need to prove that
$$\sum_{cyc}\frac{\frac{y}{x}-1}{\frac{y}{x}\cdot\frac{z}{y}+1}\geq0$$ or
$$\sum_{cyc}\frac{y-x}{x+z}\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve for $a_{n}$ where $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$ . $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$.
So I am trying to find $a_n$ by using the generating function let's call it $A(x)$.
the equation then is written as (if I doing this correctly) :
$A(x)-ao = 4xA(x) + \sum_{k=1}2^{2n-1}*x^k$
n... | Your initial equation for $A(x)$ is incorrect. You need to work the boundary condition $a_0 = 1$ into the equation. (The condition $a_1=6$ is redundant, since it can be computed from $a_0$ and the recurrence relation.)
There is more than one way to do this, but one way is to write the recurrence equation so it is true... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{2} \lt \sum_{r=1}^{n} \frac{1}{n+r} \lt \frac{3}{4} , n>1$ I have proven the left hand side inequality. Can someone give me a hint for the right hand side ?
Proof for the left hand side:
$\sum_{r=1}^{n} \frac{1}{n+r} >\sum_{r=1}^{n} \frac{1}{2n}$.
| By C-S $$\sum_{r=1}^n\frac{1}{n+r}=1+\sum_{r=1}^n\left(\frac{1}{n+r}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{r=1}^n\frac{r}{n+r}=$$
$$=1-\frac{1}{n}\sum_{r=1}^n\frac{r^2}{nr+r^2}\leq1-\frac{\left(\sum\limits_{r=1}^nr\right)^2}{n\sum\limits_{r=1}^n(nr+r^2}=1-\frac{\frac{n^2(n+1)^2}{4}}{n\left(\frac{n^2(n+1)}{2}+\frac{n(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find intersection between line and ellipsoid I want to find points $\space P(x,y,z) \space$ where a line intersect an ellipsoid with $$P = P_{1}+t(P_{2}-P_{1})$$
Here is where I stuck:
The ellipsoid can be described as:
$$\frac{(x-x_{3})^{2}}{a^{2}} + \frac{(y-y_{3})^{2}}{b^{2}} + \frac{(z-z_{3})^{2}}{c^{2}} = 1$$
afte... | Translate the ellipsoid to origin, by subtracting $(x_3, y_3, z_3)$ from $P1$ and $P2$, so the line is parametrised as
$$\vec{p} = \vec{v}_0 + t \vec{v}_1$$
where
$$\begin{aligned}
\vec{v}_0 = (\chi_0, \gamma_0, \zeta_0) &= (x_1 - x_3, y_1 - y_3, z_1 - z_3) \\
\vec{v}_1 = (\chi_1, \gamma_1, \zeta_1) &= (x_2 - x_1, y_2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What went wrong in the evaluation of $\int \frac{1}{3-2\sin(x)}dx$? I tried to evaluate the following integral:
$$\int \frac{1}{3-2\sin(x)}dx\,\, $$ with universal substitution, using the fact that
$t(x):=\tan\left(\frac{x}{2}\right)$
$\sin(x)=\frac{2t(x)}{1+t(x)^2}$ and $t'(x)=\frac{1+t(x)^2}{2}$
$\displaystyle\int \... | You made a mistake when applying
$$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right].$$
Define $t := (\sqrt{3}u + \frac{2}{\sqrt{3}})$ as you did implicitly, but make sure to rework $du$ into $c \cdot dt$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$? Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original question; namely,
$$\int_0^1 x^{n-1}\... | Mathematica choked when I gave it the general expression (at least the way I tried). Here's how the sum works out for some small values of $m$.
\begin{gather}
m = 0 : \tfrac{1}{2}\log^2(2) \\
m = 1 : -1 + 2 \log(2) - \tfrac{1}{2}\log^2(2) \\
m = 2 : \tfrac{5}{4} - 2 \log(2) + \tfrac{1}{2}\log^2(2) \\
m = 3 : - \tfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sin^2x + 3\sin x\cos x + 2\cos^2x=0$, for $0\leq x\leq 2\pi$
Solve
$$\sin^2x + 3\sin x\cos x + 2\cos^2x=0$$
for $0\leq x\leq 2\pi$.
My answers are
$$x=2.03, 5.18 \qquad\text{or}\qquad
x=\frac{3\pi}{4},\frac{7\pi}{4} \qquad\text{or}\qquad
x=\frac{\pi}{2}, \frac{3\pi}{2},$$
but the answer states $x=2.03, 5.18... | Look at this $$x^2+3xy+2y^2=(x+y)(x+2y)$$ can you see??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\f... | Now I am going to prove that:
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$
for all $x,y\in\left]0,+\infty\right[$ such that $xy>1$.
Proof:
For all $x,y\in\left]0,+\infty\right[$ such that $xy>1$, it results that:
$x>\frac{1}{y}=\cot\left(\tan^{-1}y\right)=\tan\left(\frac{\pi}{2}-\tan^{-1}y\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding a formula for a function using the 2nd fundamental theorem of calculus $$f(t) =\begin{cases} t &\text{if }0 \leq t \leq 1, \\
2-t &\text{if }1 < t \leq 2, \\
t^2 &\text{if }t > 2.
\end{cases}$$
Find a formula for the function $F: [0, 2] \rightarrow \mathbb{R}$ defined by $F(x) = \int_0^xf(t)\mathrm dt$ for x... | Your given function is given piecewise.
$$ f(t) = \begin{cases}
t ,& 0 \leq t \leq 1 \\
2-t ,& 1 < t \leq 2 \\
t^2 ,& 2 < t
\end{cases} $$
You should expect your accumulation function to also be given piecewise.
\begin{align*}
F(x) &= \int_0^x f(t) \,\mathrm{d}t \\
&= \begin{cases}
\int_0^x f(t) \,\mathrm{d}t &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate hypergeometric $_6F_5\left(\{\frac12\}_6;1,\{\frac32\}_4;1\right)$ Background: I'm searching for $_pF_q$ representations for MZVs. In related article On the interplay between hypergeometric series, Fourier-Legendre expansions and Euler sums by M. Cantarini and J. D’Aurizio, the series $_5F_4\left(\frac{1}{2},\... | All right, I am going to re-do my computations from scratch. This will probably take some time, so please do not downvote this answer in the meanwhile. For any $n\geq 3$ we have
$$ (-1)^{n+1}x^{n+1/2}\cdot \frac{d^n}{dx^n} \frac{\log^3(x)}{\sqrt{x}} = A_n+B_n\log(x)+C_n\log^2(x)+D_n\log^3(x)=S_n\tag{S}$$
with $A_n,B_n,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $a^4+2a^3 b+2ab^3+b^4 ≥ 6a^2b^2$ Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.
| You can easily prove it using AM-GM Inequality or Lagrange method. Using AM-GM Inequality,
$$ \frac {a^4 + b^4}{2} \ge a^2.b^2 $$
$$ \frac {a^2 + b^2}{2} \ge a.b$$
$$ \text {Also, you can write }a^3.b+a.b^3 = ab.(a^2+b^2)$$
$$ \text {So, } a^4 + b^4 +2a^3.b+2a.b^3 \ge 6.a^2.b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
| Clearly, $\lim_{n\to \infty} x_n = \infty$. We have
\begin{align}
\sqrt{x_{n+1}} - \sqrt{x_n} &= \sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} - \sqrt{x_n}\\[6pt]
&= \frac{a + \sqrt{b^2 + 4ax_n}}{\sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} + \sqrt{x_n}}\\[6pt]
&\to \sqrt{a} \quad \mathrm{as}\quad n \to \infty.
\end{align}
By the Stolz-C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{... | Also, $$a^2+b^2+c^2=\frac{1}{14}(3^2+2^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{14}(3a+2b+c)^2=3.5.$$
The equality occurs for $$(3,2,1)||(a,b,c),$$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+... | Because each expression is homogenous, we may assume that $ z = 1$ (by using the substitution $ x' = \frac{x}{z}$). The question becomes:
If $x^3 + y^3 = 1$, show that $x^2 - 6xy + y^2 + 6x + 6 y - 7 > 0$.
Note: This 2-variable inequality type is common, and there are several ways of dealing with it by exploiting $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Number of Real Solutions of $\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1$
Find the number of real solutions of the equation $$\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1\,.$$
By hit and trial i got the solution at $x=\pm 1$ but i am not able to solve it as it involves power of 7
| Note that $\frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}$ is an even function. So it's enough to consider $x\ge 0$.
Let $$f(x) = \frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-x}$$
If $x \gt \sqrt{2} \ $ then $3^{x^2 - 2}\gt 1$ and also $\frac{7^{1+cos\pi x}}{3} \gt 0 \ \ , \ 9^{\frac{1}{2}-x} \gt 0$. There... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find whether the sequence is convergent . Find whether the sequence $(a_n)$ given by $a_{n+1}= \sqrt{a_n}+\sqrt{a_{n-1}}$, where $a_1=1$ and $a_2=2$, is convergent.
So , $a_{n+1}-a_{n}= \sqrt{a_n} + \sqrt{a_{n-1}} -a_n \implies \sqrt{a_n}(1-\sqrt{a_n})+ \sqrt{a_{n-1}}.$
Now I assumed the sequence is $>1$ and I ... | We show that the sequence is convergent by showing that it's bounded and monotone.
Claim 1. $2 \le a_n \le 4$ for all $n \ge 2.$
Proof. We prove this via induction. For $n = 2, 3$, it is manually verified.
Let $P(n)$ denote the statement "$2 \le a_n \le 4$".
Assume that $n \ge 4$ and that $P(k)$ is true for all $2 \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold? By generalizing this (1) and this (2) questions and performing some research
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \... | Suppose $abc=1,$ and let $b=c=t,\,a=\frac{1}{t^2},$ the inequality become
$$k \leqslant \frac{6(t+1)(t^3+t^2+1)}{t(2t+1)} = F(t).$$
Easy to find
$$k \leqslant k_0 = F(t_0) = \frac{9\sqrt{665}}{8}\sin{\left(\frac{\pi}{6}+\frac{1}{3}\arccos{\frac{13117\sqrt{665}}{442225}}\right)}-\frac{141}{16} = 11.1086,$$
for $ \displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find matrix $B$ such that $B^2 = I - A$
$A$ is a squared matrix such that $A^4 = \mathbf{0}$. Find $B$ in form of $A$ such that $B^2 = I - A$.
I tried in this way:
\begin{gather*}
A = I - B^2 \\
A ^ 4 = ( I - B)^4 (I + B)^4
\end{gather*}
So $I = B$ or $- I = B$
But the question wants $B$ as a form of $A$.
any help?
| Assume $B$ can be written as a polynomial in $A$. That is,
\begin{align*}
B = c_0I+c_1A+c_2A^2+c_3A^3
\end{align*}
for unknown constants $c_0,c_1,c_2$, and $c_3$. Higher powers vanish since $A^4=0$. The relation $B^2=I-A$ gives
\begin{align*}
I-A = B^2 = c_0^2I+2c_0c_1A+(2c_0c_2+c_1^2)A^2+2(c_0c_3+c_1c_2)A^3.
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the values of $\theta$ for which the tangent line to the given curve is parallel to $x$ ,$y$ axis Given the curve $$r(\theta):=\sec\left(\theta\right)+a\cos\left(\theta\right) \tag{$a \in \mathbb R$}$$
Find the values of $\theta$ for which the tangent line to the curve is parallel to the $x$ and $y$ axis.
*
*Th... |
So where was I wrong?
It is wrong that "$\frac{a\pm\sqrt{a^{2}-8a}}{4a}$ is never between $0$ and $1$".
This answer proves the following two claims :
Claim 1 : $$0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1\iff a\ge 8$$
Claim 2 : $$0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1\iff a\le -1\quad\text{or}\quad a\ge 8$$
Claim 1 : $$0\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}$ Evaluate $$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
My attempt: $\cos(\frac{1}{1-x})=\cos(\frac{1}{x-1})$, so if we let $t=x-1$ we get
$$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{x-1})}{\sqrt{x-1}}=\lim... | For $x\rightarrow1^+$ we obtain: $$\frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}=\frac{\sin(x^3-1)}{x^3-1}\cdot\sqrt{x-1}(x^2+x+1)\cos\frac{1}{x-1}\rightarrow0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and ... | For any two functions $f$ and $g$, we have
$$\begin{align}
|f(x)|-|g(x)|\gt0
&\iff|f(x)|\gt|g(x)|\\
&\iff(f(x))^2\gt(g(x))^2\\
&\iff(f(x))^2-(g(x))^2\gt0\\
&\iff(f(x)-g(x))(f(x)+g(x))\gt0
\end{align}$$
For $f(x)=3x-5$ and $g(x)=2x+3$, we see that $f(x)-g(x)=x-8$ and $f(x)+g(x)=5x-2$, and so
$$|3x-5|-|2x+3|\gt0\iff(x-8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Why does $\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$? I really don't understand the expression.
$$\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$$
I generally have a hard time understanding non-intuitive things in math and this is one of them. Normally when I don't understand something I use an app , photomath, to help explai... | $\frac{1}{a}-\frac{1}{b} = \frac{1}{a} * \frac{1}{1} - \frac{1}{b} * \frac{1}{1}$
= $\frac{1}{a} * \frac{b}{b} - \frac{1}{b} *\frac{a}{a}$ = $\frac{b} {ab} - \frac{a}{ab} = \frac{b-a}{ab}$
using $ab=ba$ and $\frac{x}{x} = 1$ for any $x \neq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Need help with Alternative Factoring method I was working on some factoring, as I have always been terrible at it, when I found 3B1B's video on an easier method. There's a TL:DR at the bottom if you're familiar. The basics are as follows:
Imagine the graph of a quadratic. $x^2 - 1$ for example. It's got 2 roots $r$ and... | Given
$$x^2+bx+c=x^2-\frac{5}{2}x-\frac{3}{2}$$
We get $m=\frac{-b}{2}=\frac{5}{4}$ and $$d=\sqrt{m^2-c}=\sqrt{\frac{25}{16}+\frac{3}{2}}=\sqrt{\frac{49}{16}}=\frac{7}{4}$$
Therefore our roots are $\frac{5}{4}\pm\frac{7}{4}=\{\frac{-1}{2},3\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}... |
Evaluate $$L=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}$$
If there are more terms to apply the limit to, you should separate the terms of which limits are apparent to reduce the probability of such failure.
\begin{align}
L&=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ How can I prove the identity $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ for $|x|<1$? I am preferably looking for a derivation rather than using the RHS. I have tried using binomial expansion, but it only seems to give the LHS back. I also tried taking the logarith... | We can repeatedly double the power in the denominator as follows: (similar to rationalizing the denominator, if you are familiar wtih that)
$$
\begin{align}
&\frac1{1-x}=\frac1{1-x}\cdot\frac{1+x}{1+x}\\
=&\frac{1+x}{1-x^2}=\frac{1+x}{1-x^2}\cdot\frac{1+x^2}{1+x^2}\\
=&\frac{(1+x)(1+x^2)}{1-x^4}=\frac{(1+x)(1+x^2)}{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $a_{n+1}=2a_n −n^2+n$ Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$ If
$$a_{n+1}=2a_n −n^2+n$$
Define a sequence $a_n$
that satisfy the recurrence relation as described above, with $a_1 = 3$
Find the value of $$\dfrac{ |a_{20} - a_{15} | }{18133} $$
Attempt
First ev... | Let $a_m=b_m+c_0+c_1m+c_2m^2+\cdots+c_rm^r$
$b_{m+1}+c_0+c_1(m+1)+c_2(m+1)^2+\cdots=2(b_n+c_0+c_1m+c_2n^2+\cdots)+n-n^2$
If $r\ge3,$ compare the coefficients of $m^r$
$$c_r=2c_r\iff c_r=0$$
$\implies a_m=b_m+c_0+c_1m+c_2m^2$
Compare the coefficients of $n^2, c_2=2c_1-1\iff c_2=1$
Similarly, comparing the coefficients ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Differentiate $\left(x^6-2x^2\right) \ln\left(x\right) \sin\left(x\right)$ Differentiate
$$\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)$$
with respect to $x$
My work so far
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)\... | Let $f(x)=\left(x^6-2x^2\right)\ln x\sin x$. Then,
$$\ln f(x)=\ln(x^6-2x^2) +\ln (\ln x ) + \ln(\sin x)
$$
and
$$f’(x)=f(x)\left(\frac{6x^5-4x}{x^6-2x^2}+\frac{1}{x\ln x}+ \cot x\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $ \lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right) $ This question comes to my mind immediately after asking this question.
I was earlier unknown that limit of sum equal sum of limits only when there are finite terms. Now the problem is then how do I evalu... | Observe that, for $x\ne 0$,
$$ -\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots=\frac{1}{x^3} \left(-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots\right)
=\frac{1}{x^3} \left(\frac{x}{1!}-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
| $$-48=\dfrac p{d^3}+\dfrac pd+pd+pd^3=\dfrac p{d^3}(1+d^2+d^4+d^6)=\dfrac{p(1+d^2)(1+d^4)}{d^3}$$
Similarly,
$$3280=\dfrac{p^2(1+d^4)(1+d^8)}{d^6}$$
$$\implies\dfrac{(-48)^2}{3280}=\dfrac{(1+d^2)^2(1+d^4)}{1+d^8}=\dfrac{\left(d+\dfrac1d\right)^2\left(d^2+\dfrac1{d^2}\right)}{d^4+\dfrac1{d^4}}$$
Let $d^2+\dfrac1{d^2}=u$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \t... | You can easily integrate with suitable substitution as follows
Let $2x=3\tan\theta\implies dx=\frac{3}{2}\sec^2\theta\ d\theta$
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac{(\frac32\tan\theta)^3}{(9\tan^2\theta+9)^{3/2}}\ \frac{3}{2}\sec^2\theta\ d\theta$$
$$=\left(\frac{3}{2}\right)^4\frac{1}{3^3}\int \frac{\tan^3\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is c... | $${\sin x\over x}\approx1-{1\over6}x^2$$
so
$${\sin^2x\over x^2}\approx\left(1-{1\over6}x^2\right)^2\approx1-{1\over3}x^2$$
not just $1$. We get
$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the... | Multiplying by the conjugate and applying a couple of substitutions does work.
\begin{align*}
\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{d}x &=\int \underbrace{\frac{\sqrt{2x}}{x-4}}_{\sqrt{x} \to u} + \underbrace{\frac{\sqrt{x+4}}{x-4}}_{\sqrt{x+4} \to t} \; \mathrm{d}x\\
&=2\sqrt{2} \int \frac{u^2}{u^2-4} \;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$. Let $a,b$ be coprime integers. I am trying to find all integers $x,y$ such that:
$$x^3+5y^3=(a^3+5b^3)^3$$
What I have tried:
$$5y^3=(a^3+5b^3-x)[(a^3+5b^3)^2+(a^2+5b^3)x+x^2 ]$$
There are 2 Cases depending which factor of $5y^3$ is divisible by $5$:
Case 1: $5$... | The only solutions are the ones with $y=0$.
This is one of those problems where generalizing makes it easier, so I'm going to substitute the RHS with the cube of a general integer
$$x^3+5y^3=z^3$$
If $z=0$ you get the $(0,0,0)$ solution and otherwise $\frac x y =-5^{1/3}$ which is impossible. Now I'm going to let $x_1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to justify these approximations about prime numbers? A recently closed question asked for a possible closed form of the infinite summation
$$f(a)=\sum _{i=1}^{\infty } a^{-p_i}$$ for which I already proposed a first simple but totally empirical approximation.
Since we quickly face very small numbers, I t... | I found this estimate for $g$: $\ g(a)\sim \dfrac{a^2(a^2-1)}{a^2+a-1}$.
First inequality
$f(a) = \displaystyle \sum_{i=1}^{+\infty} \dfrac{1}{a^{p_i}} \leqslant \sum_{i=2}^{+\infty} \dfrac{1}{a^i} -\sum_{i=2}^{+\infty} \dfrac{1}{a^{2i}} = \dfrac{1}{a^2}\dfrac{1}{1-\dfrac{1}{a}} -\dfrac{1}{a^4}\dfrac{1}{1-\dfrac{1}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
} |
What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{... | First, the integral:
$$
I := \int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)\left(1+9\,x^2\right)\left(9+\,x^2\right)+\left(9+x^2\right)\ln\left(4+x^2\right)\left(10+10\,x^2\right)}{\left(9+x^2\right)^2\left(1+9\,x^2\right)}\,\text{d}x
$$
thanks to the linearity of the integrals it can be written as the algebraic sum of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$ I need help simplifying the radical $$y=\sqrt{x-\sqrt{x+\sqrt{x-...}}}$$
The above expression can be rewritten as $$y=\sqrt{x-\sqrt{x+y}}$$
Squaring on both sides, I get $$y^2=x-\sqrt{x+y}$$
Rearranging terms and squaring again yields $$x^2+y^4-2xy^2=x+y$$
At this p... | Perhaps it is more instructive to consider instead the following: let $$y = \sqrt{x - \sqrt{x + \sqrt{x - \sqrt{x + \cdots}}}}, \\
z = \sqrt{x + \sqrt{x - \sqrt{x + \sqrt{x - \cdots}}}},$$ so that if $y$ and $z$ exist, they satisfy the system $$y = \sqrt{x - z}, \\ z = \sqrt{x + y},$$ or $$y^2 = x - z, \\ z^2 = x + y.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x... | Denote $L$ the existing limit. Then, express it as
\begin{align}
L=\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}
&= \lim _{x \to 0}\frac {x (2\cos^2\frac x2 -1) -2 \sin \frac x2 \cos\frac x2} {2x^2 \sin \frac x2 \cos\frac x2}\\
&= \lim _{x \to 0} \frac {x (\cos^2\frac x2-1) +2 \cos\frac x2(\frac x2\cos \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in t... | Say $c$ and $d$ are zeroes of $x^2-x-1$ then they are zeros of $ax^9+bx^8+1$ too.
Since $c^2 = c+1$ we have $$c^4=c^2+2c+1=3c+2$$ and $$c^8 = 9c^2+12c+4 = 21c+13$$
and finnaly $c^9 = 34c+21$.
So we have $$a(34c+21)+b(21c+13)+1=0$$ or $$\boxed{(34a+21b)c+ (21a+13b+1)=0}$$
Simmilay we have for $d$: $$\boxed{(34a+21b)d+ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)... | We have $(1-z)(x^2-xy+y^2)=1-z^2.$
If $z=1$, so $x+y=0$ and we obtain $(t,-t,1)$, where $t$ is an integer.
Let $z\neq1$.
Thus, $$x^2-xy+y^2=z+1$$ and $$x+y=1-z,$$ which gives $$(1-z)^2-3xy=z+1$$ or
$$3xy=z^2-3z.$$
Thus, $z$ is divisible by $3$ and $$(1-z)^2-\frac{4}{3}(z^2-3z)\geq0$$ or
$$z^2-6z-3\leq0$$ or
$$3-\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Contest math application for Wilson's theorem
$$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{23} = \frac{a}{23!}$$ Find the remainder when $a$ is divided by $13.$
I found this online and got stuck a bit. I approached the problem as such:
From the expression we get $$a=\frac{23!}{1}+\frac{23!}{2}+\dots+\frac{23!}{23!}$$
s... | Since $12!=10!\times 11\times12\equiv-1\bmod13$, it follows that $10!\times-2\times-1\equiv-1\bmod13$,
so $10!\times2\equiv-1\equiv12\bmod13$, so $10!\equiv6\bmod 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Elementary Arithmetic Problem In 1988, in a French math competition for middle school grades, the following problem was given:
Complete this multiplication so that all the digits from 0 to 9 appear:
$... × .. = ....1$
I am stumped. Of course the last digits of the two numbers are 3 and 7 but this is as far as I can go!... | The rule of nines: $jkl \equiv j+ k + l \pmod 9$ so
So if we have $abc\times de = fghi1$ and $a,b,....,f,g,h,i,1$ are the digits from $0,....,9$ then $abc + de +fghi1 \equiv 0 \pmod 9$
And if $abc \equiv j\pmod 9$ and $de \equiv k \pmod 9$ we have $fghi1 \equiv jk \equiv -(j+k)$.
Or $(j+1)k \equiv -j$ and $(k+1)j \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Series Expansion by differentiation We know that 1/(1-x) = 1+x+x^2+x^3+....x^n
Say we need to find a suitable function for the expansion x+x^2+x^3+x^4+....x^(n+1)
We would differentiate 1/(1-x) = 1+x+x^2+x^3+....x^n
This would yield 1/(1-x)^2 = 1 + 2x+3x^2+4x^3 +...+nx^(n-1)**
Multiplying it by x would yield x/(1-x)^2... | Your work is fine, now we just have to subtract $x$ and add $1$ that is
$$g(x)=\frac{x}{(1-x)^2}-x+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A question based on quadratic forms in linear algebra ( rank, representation) This particular question was asked in masters of mathematics exam of a university and I am unable to solve it. So I am asking it here.
Consider the quadratic form $Q(v)=v^{t} A v$, where
$$A=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0... | The key to solving these questions (i.e. the classification of quadratic forms up to a change of basis) amounts to using Sylvester's law of intertia.
In particular, we find that $A$ has $3$ positive eigenvalues and $1$ negative eigenvalue, so its indices of innertia are $n_+ = 3, n_- = 1, n_0 = 0$.
With that, we can el... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | Another way.
By C-S and AM-GM we obtain: $$\prod_{cyc}(1+a^2)\geq(a+b)^2(1+c^2)=(a+b)^2+(ac+bc)^2\geq$$
$$\geq4ab+(3-ab)^2=a^2b^2-2ab+9=(ab-1)^2+8\geq8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\... | I shall be continuing from the last two integrals of your work $$\int_0^{\sqrt 2-1} \frac{\tanh^{-1}x}{x} dx =\frac{1}{2}\int_0^{\sqrt 2-1} \left(\frac{\log(1+x)}{x}-\frac{\log(1-x)}{x}\right)dx$$ These two integrals posses non-elementary antiderivatives in terms of special function Dilogarithm function so by definiti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\... | We have
$$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!}$$
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x)^n(\ln(2))^n}{n!}$$
$$2^{-x}-1 = -1+\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n(\ln(2))^n}{n!}=\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n+1}(\ln(2))^{n+1}}{(n+1)!}$$
$$\frac{2^{-x}-1}x =\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum positive integer $n$ and the matrix $A^{2020}$ Let $A=a\begin{pmatrix} 1 & -1\\
1 & 1 \end{pmatrix}$ $(a>0)$ and $I=\begin{pmatrix} 1 & 0\\
0 & 1 \end{pmatrix}$ satisfy $A^4+I=\begin{pmatrix} 0 & 0\\
0 & 0 \end{pmatrix}$
*
*Find $a$ For this I can do. I saw $a=\frac{1}{\sqrt{2}}$
*Find the minimum p... | For 2: $A^{2}\left(\begin{array}{c}
0\\
1
\end{array}\right)=\left(\begin{array}{c}
-1\\
0
\end{array}\right)$, so$$A^6\left(\begin{array}{c}
0\\
1
\end{array}\right)=-A^2\left(\begin{array}{c}
0\\
1
\end{array}\right)=\left(\begin{array}{c}
1\\
0
\end{array}\right).$$Hint for 3: $2020$ is an odd multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$ Here is a nice problem proposed by Cornel Valean
$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\... | Update: the problem and solution will be part of a new paper soon.
A solution by Cornel Ioan Valean
Let's denote the main integral by $\mathcal{I}$, and then we have
$$\mathcal{I=}\int_0^1\frac{(\pi/4-\arctan((1-x)/(1+x)))\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x$$
$$=\underbrace{\frac{\pi}{4}\int_0^1\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I ... | Let ${\rm R}_3$ be the $3 \times 3$ reversal matrix. Hence,
$$\det \left[\begin{array}{ccc|ccc} 1&0&0&0&0&2\\ 0&1&0&0&2&0\\ 0&0&1&2&0&0\\ \hline 0&0&2&1&0&0\\0&2&0&0&1&0\\ 2&0&0&0&0&1\end{array}\right] = \det \begin{bmatrix} {\rm I}_3 & 2{\rm R}_3\\ 2{\rm R}_3 & {\rm I}_3\end{bmatrix} = \det \left( {\rm I}_3 - 4 {\rm R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. Th... | You are almost there, check that
$$f'(x)=\lim_{h \to 0} \frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}$$
Divide by $h$ up and down:
$$f'(x)=\lim_{h \to 0}\frac{1-x^2-xh}{(x^4+2x^2+1)+x^2h^2+2x^3h+h^2+2xh}$$ $$\implies f'(x)=\frac{1-x^2}{(1+x^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Correct result for this integral $\int \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}}\, dx$ Wolfram|Alpha and its CAS, Wolfram Mathematica are, as far as I know, the only website and software that give the correct solution to this integral,
$$ f(x) = \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 ... | This is a partial solution, equivalent to the process used in the video, which is only valid if
$$\cos\frac t2$$
is non-negative.
Start with this identity:
$$\sqrt{2+2\cos t} = \sqrt{4\cos^2\frac t2} = 2\cos\frac t2$$
For applying this to the integrand, first make the substitution $t = \sqrt x$, then successively apply... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Extraneous solution from substituting in equations I came across this example of how you could end up with an extraneous solution but I was wondering how it arose.
We have the equation: $$x^2+x+1=0 $$ Since x=0 does not satisfy the equation, you can divide by x on both sides which yields: $$x+1+\frac{1}{x}=0$$ which is... | This substitution ($x+1=-x^2$) expands a set of roots of the equation
because $-x^2$ also depends on $x$.
You can substitute $x+1=y$, for example.
More example, when a similar substitution gives similar problems.
Let we need to solve $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1}.$$
We obtain:
$$\left(\sqrt[3]{2x+1}+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Closed form of hypergeometric $\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)$ How can we prove $$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\frac{\sqrt{k(z)}}{2}\right) K\left(\frac{1}{2}-\frac{1}{2 ... | This is not an answer.
For small values of $z$, we have
$$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=1+\frac{9 z}{128}+\frac{3675 z^2}{131072}+\frac{266805
z^3}{16777216}+O\left(z^4\right)$$ while, with $ k(z)=-2 \sqrt{z^2-z}-2 z+1$,$$\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{... | Given $\sqrt{xy}\left(x-y\right)=x+y$
let $yx=c$ , where $c>0$.
$$\sqrt{c}\left(x^{2}-c\right)=x^{2}+c$$
$$x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -[1]$$
Let a function $$F(x,c)=x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c$$ be defined. Then $$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$
Let $0\leq\alpha\leq \pi $. $V_n (\alpha) $ denote the number of sign changes in the sequence $\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha $. Then prove that $$\lim... | *
*I will assume that the case where $\alpha\in \pi \mathbb{Q}$ is easy, because the sequence $\big(\cos(k\alpha)\big)_{k\ge1}$ is periodic in this case, and if we consider $0$ as positive number then $V_{2q}(p\pi/q)=2p\pm 1$ and the result holds in this case.
*Now we assume that $\alpha\notin \pi\mathbb{Q}$. This im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displa... | Let
$$ a = \displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}}. $$
Notice that the first term is the multiplicative inverse of the second term. Thus
$$ a^3 = \displaystyle \frac{x+1}{x-1} + \frac{x-1}{x+1} + 3 \left(\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,... | Taking @user's hint a little further (and also assuming $x,y,z,\neq 0$), writing
\begin{align}
u &=\frac{1}{x}\\
v&=\frac{1}{y}\\
w&=\frac{1}{z}
\end{align}
Yields
\begin{align}
u+v &= 5\\
v+w &= 6\\
z+u&=7
\end{align}
thus
$$\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
u\\
v\\
w\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is a quick way (without calculator) to determine that $(2^9 + 1)^2 + 2^9 + 2$ is not prime? I came across the following expression $(2^9 + 1)^2 + 2^9 + 2$, which is divisible by 7 and thus not prime. Without this information and a calculator, how could I easily determine that this number is not prime?
Here is a wa... | $$A=(2^9 + 1)^2 + 2^9 + 2$$
$$\begin{align}\color{red}{A-7}=\left(2^9-1+2\right)^2+2^9+2-7=(a+2)^2+2^9-5=a^2+4a+2^9-1=a^2+4a+a=\color{red}{a^2+5a}\end{align}$$
,where $a=2^9-1=\left(2^3\right)^3-1=\left(2^3-1\right)\left(2^6+2^3+1\right)=7×\left(2^6+2^3+1\right).$
So, $(A-7) \mod 7=0 \Longrightarrow A\mod 7=0.$
The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=... | Without loss of generality suppose that $a\geq b\geq c$. Then,
$$
6\cdot\left(\frac{a^3+b^3+c^3}{3}-abc\right)=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
$$
Note that
$$
a+b+c\geq (a-c)+(b-c)=2(a-b)+(b-c)
$$
and
$$
(a-b)^2+(b-c)^2+(c-a)^2=2(a-b)^2+2(b-c)^2+2(a-b)(b-c).
$$
Denote $x=a-b$ and $y=b-c$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $a$ such that the complex solutions of $z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$ form a parallelogram in the complex plane Find all values of the real number $a$ so that the four complex roots of
$$z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$$ form the vertices of a parallelogram in the complex pl... | Let $x$ be the parallelogram center. Then the roots are $x+y,\,x-y,\,x+z,\,x-z$ for some $y,z$. Then, by Vieta's formulas:
$$\begin{cases}
6=4x\\
6x^2-y^2-z^2=11a\\
4 x^3 - 2 x y^2 - 2 x z^2 = 3(2a^2+3a−3)\\
(x^2-y^2)(x^2-z^2)=1
\end{cases}$$
Although it is doable by hand, too lazy to perform it here, feed it to WA to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\int {\cos ^4 x} \ dx$ using $\cos 2x$? How would I go about solving $$\int \cos{^4} x \ dx $$ given that $$\cos 2x = \cos^2 x − \sin^2 x = 1 − 2 \sin ^2 x = 2 \cos^2 x − 1?$$ I know that I could break it down to $$(\cos{^2}x){^2}$$ but how do I proceed from here?
| Another way:
$$ \cos^4 x = \left( \frac{e^{ix}+e^{-ix}}{2} \right)^4=\frac{1}{16}
\left( e^{4ix} + 4 e^{2ix} + 6 + 4 e^{-2ix} + e^{-4ix} \right)$$
$$\begin{aligned}
\int \cos^4 x \, dx & = \frac{1}{16} \left(-\frac{i}{4} e^{4ix} -2i e^{2ix} + 6 x+ 2i e^{-2ix} +\frac{i}{4} e^{-4ix} \right)+\text{const.}\\
&= \frac{3}{8}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all pairs of integers $(x, y)$ such that $x^3+y^3=(x+y)^2.$
Find all pairs of integers $(x, y)$ such that $$x^3+y^3=(x+y)^2.$$
Since $x^3+y^3 = (x+y)(x^2-xy+y^2)$ we get that $$x^2-xy+y^2=x+y$$
this can be expressed as $$x^2-(y-1)x+y^2-y=0.$$
Since we want integers we should probably look at when the discriminan... | You are almost there. Substitute $y = 0, 1, 2$ and solve for $x$ in each case.
When $y=0$, the equation is $x^3 = x^2$. The two solutions for $x$ are $0, 1$.
When $y = 1$, the equation is $x^3+1 = (x+1)^2$. Expanding and rearranging gets $x^3-x^2-2x=0$, and the solutions are $x = -1, 0, 2$.
When $y = 2$, the equation i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Maximum of the function $f_n(x) = (1 - (1 - x^2)^n) / x$
Consider the following function $f_n$ defined on the interval $(0,1]$:
$$f_n(x) = \frac{1 - (1 - x^2)^n}{x}$$
Determine the maximum of this function for any natural number $n$.
I have computed the derivative of this function: $f_n(x)' = \frac{2n x^2(1 - x^2)^{n... | I will add on to dezdichado's answer to find the exact constant on the rate of growth of the maximum. Let $x = \frac{c}{\sqrt{n}} + O\left(n^{-3/2}\right)$. Since it must be true that $$\lim_{n \to \infty}(1-2n)\left( 1-x^2 \right)^{n} + 2n\left(1-x^2 \right)^{n-1}-1=0$$
$x = \frac{c}{\sqrt{n}}$ can be plugged into thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the smallest power of $A$ such that $A^n = I$ Let $A=\begin{bmatrix}0 & 1\\-1 & 1\end{bmatrix}$ then the smallest positive positive integer $n$ such that $A^n = I$ is :
(a) $1$ (b) $2$ (c) $4$ (d) $6$
proof: option (d) 6.
The characteristic polynomial of $A$ is $\lambda^2 - \lambda + 1$. So the eigenvalues of $... | Since $\lambda^2-\lambda +1$ is characteristic, it follows that $A^2 - A + I = 0$, i.e. \begin{align}
A^2 &= A - I\\
A^3 &= A^2 - A = -I\\
A^4 &= -A \\
A^5 &= A^2\cdot A^3 = I - A\\
A^6 &= A^3\cdot A^3 = I.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I c... | $$\sin ^4(x)+\cos ^4(x)=\left(\sin ^2(x)+\cos ^2(x)\right)^2-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^2(2x)=\\=1-\frac{1}{2}\cdot\frac{1-\cos(4x)}{2}=\frac{1}{4}\left(3+\cos(4x)\right)$$
$$\cos(4x)=\frac{1-\tan^2(2x)}{1+\tan^2(2x)}=\frac{1-t^2}{1+t^2}$$
where $\tan(2x)=t$, $x=\frac{1}{2}\arctan t$, $dx=\frac{dt}{2(1+t^2)}$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 7
} |
If $\tan(\theta$) = $\frac{x}{7}$ for -$\frac{\pi}{2} < \theta < \frac{\pi}{2}$, find an expression for $\sin(2\theta$) in terms of $x$. I'm trying to solve this and am getting a very different answer from the book. Can someone please help.
I know that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. So I have to take $\... | Your solution is not full.
For example, you can not say immediately that $\sin\theta=\frac{x}{\sqrt{49+x^2}}$ because $|\sin\theta|=\frac{|x|}{\sqrt{49+x^2}}.$
We can make the following.
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{\frac{2x}{7}}{1+\frac{x^2}{49}}=\frac{14x}{49+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the inequality $ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2}$ have a name? This is somewhat embarrassing but if $a,b$ are nonnegative real numbers the following seems to hold
$$
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}.
$$
Does this inequality have a name?
| From AM-GM
$$\frac{a+b}{2}\geq \sqrt{ab} \iff\\
a+b\geq \frac{a+b}{2}+2\sqrt{\frac{a}{2}\cdot\frac{b}{2}} \iff\\
a+b\geq \left(\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}\right)^2 \overset{a,b\geq0}{\iff}\\
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}$$
Which makes it a special case or corollary of it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the ... | As you did not exactly clarify what you mean by abc (their product?) I can only give you halve an answer. I assume you do not want a integer solution rather than a continous one and also real solutions.
With some calculations you will find:
$$b_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
and
$$c_{1/2}=\frac{1}{2}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Continuity of piecewise $f$ and its partials, $f(x,y)= \frac{xy(x^2-y^2)}{x^2+y^2}$ for $x,y\neq 0$
Let $f: \mathbb{R}^2 \to \mathbb{R}$ and $$f(x,y) = \left\{
\begin{array}{ll}
0, & (x,y)=0 \\
\frac{xy(x^2-y^2)}{x^2+y^2}, & (x,y) \neq0 \\
\end{array}
\right.$$ Determine if $f, \partial_xf$ and $\partial_... | Hint:
Transform the partial derivatives via polar coordinates. Also there's an error in your partial derivative w.r.t. $x$:
$$\frac{\partial f}{\partial x}=\frac{y(x^4+4x^2y^2-y^{\color{red}4})}{(x^2+y^2)^2}=\frac{r^5(\cos^4\theta+4\cos^2\theta\sin^2\theta-\sin^4\theta)}{r^4}=r(\cos2\theta+\sin^22\theta),$$
therefor i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability of HTHT in n coin flips What is the probability of finding the sequence HTHT in $n$ flips of a coin? By brute force I get
*
*$n=4: \frac{1}{2^4}$
*$n=5: \frac{2(2)}{2^5}$
*$n=6: \frac{2^2(3)-1}{2^6}$
*$n=7: \frac{2^3(4)-2}{2^7}$
*$n=8: \frac{2^4(5)-6}{2^8}$
I hoped to extract a pattern but counting ge... | Let $A_n$ be the number of sequences that don't include the pattern $HTHT$. Then
$$A_n = \pmatrix{1 & 1 & 1 &1} T^n {\pmatrix{1 & 0 & 0 &0}}^{t} \tag1$$
where
$$T= \pmatrix{
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 \\
0 & 1 & 0 &0 \\
0 & 0 & 1 &0
} \tag2$$
Eq $(1)$ is obtained by counting recursively the "allowed" sequences t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Complete $\int \frac{x^2}{\sqrt{9x^2-1}}dx$ I am trying to solve the integral $$\int \frac{x^2}{\sqrt{9x^2-1}}dx,$$ but I am not sure how to solve it.
I have thought substitute $x$ by $\frac{\sec(u)}{3}$. Then $dx = \frac{1}{3} \tan(u)\sec(u)$,
$\sqrt{9x^2 -1} = \sqrt{\sec^2(u) - 1} = \tan(u)$ and $u = \sec^{-1}$(3x)
$... | There is a specific formula that helps in your case that helps "remove" the powers on a secant integral.
$$\int \sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx$$
Also I think you missed a 3 somewhere, during the substitution the dx should ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Diophantine equation:$x^5+x^4+1=p^y$
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.
My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tri... | Continuing your approach note that from the equality
$$
(x^3-x+1)(x^2+x+1)=p^y
$$
we obtain that $x^3-x+1=p^n$ and $x^2+x+1=p^m$ for some nonnegative integers $m$ and $n$.
For $x=1$ and $x=2$ we have solutions $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
Now note that for $x\geq 3$ we have
$$
x^3-x+1=(x^3-1)-(x-2)=(x-1)(x^2+x+1)-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to decide whether to use Bayes' theorem or not?
A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in next two drawn , exactly one white ball is drawn. ( Leave answer in $ \binom{n}{r}$)
What I did:
$$ P... | First a formula: Suppose we have $R$ red balls & $W$ white balls, then the probability of selecting a specific sequence of $r $ red balls & $w $ white balls (with $r \le R, w \le W$) is
$ { (R+W - (r+w))! \over (R+W)!} { R! \over (R-r)! } {W! \over (W-w)! }$. Since these can
be rearranged in $\binom{w+r}{w}$ ways, we s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl... | Using partial fraction decmposition
$$\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{8 (2 r+1)}-\frac{3}{16 (2 r+3)}+\frac{1}{16 (2 r-1)}$$ that you can rewrite
$$\frac 18 \left(\frac{1}{2 r+1}-\frac{1}{2 r+3} \right)+\frac 1{16} \left(\frac{1}{2 r-1}-\frac{1}{2 r+3} \right)$$ There would be a lot of telescoping
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve this integral $\int \frac{dx} {\sin^5{x}}$ Let us denote
$$I_{m,n} = \int \sin^m{x} \cos^n{x}\ dx$$
where $m,n$ are integers (possibly negative or zeros).
There are some well-known recurrent formulas for $I_{m,n}$
So... as an example I was trying to solve this particular integral
$$\int \frac{dx} {\sin^5{x}}$$
us... | You made a mistake somewhere. From WA and after simplifications you should have (what I get):
$$\frac{3}{8}\ln|\tan(\frac{x}{2})|-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}$$
Let me know!
I think the $24\ln|\tan(\frac{x}{2})|$ is clear and we multiplied it by $\frac{1}{64}$ to give t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3825698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve this integral $\int \frac{dx}{\sin^6x \cdot \cos^6x}$ I was solving this integral manually
$$\int \frac{dx}{\sin^6x \cdot \cos^6x}$$
It's not easy but it wasn't that hard too having all good formulas/recurrences in front of me and not rushing my calculations.
So I got this answer by hand:
$$(1/5) \cdot \sin(x)^... | Since your result has been confirmed.
It would have been very fast to use $x=\tan^{-1}(t)$ which makes
$$I=\int \frac{dx}{\sin^6(x) \, \cos^6(x)}=\int\frac{\left(t^2+1\right)^5}{t^6}\,dt $$ that is to say
$$I=\frac{t^5}{5}-\frac{1}{5 t^5}+\frac{5 t^3}{3}-\frac{5}{3 t^3}+10 t-\frac{10}{t}+C$$ Back to $x$
$$I=\frac{\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Help Finding Closed Form for Function of Decreasing Half Steps I would like to know if there is a closed form for this function:
$
f(x) =
\begin{cases}
\frac{1}{2}, & \text{if}\ 0 \leq x < \frac{1}{2} \\
\frac{1}{4}, & \text{if}\ \frac{1}{2} \leq x < \frac{1}{2} + \frac{1}{4} \\
\frac{1}{8}, & \... |
Fig. 1: Graphical representation of function $f$ (formula (0)). The red vertical segments are a plotting artefact.
Here is a solution:
$$y=f(x)=2^{\lfloor\log_2(1-x)\rfloor}\tag{0}$$
where $\lfloor \cdots \rfloor$ is the floor function and $\log_2$ is the logarithm function with base $2$.
Remark: Consideration of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I prove that $y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$ when $x>0$ and $1I would like to prove that
$$y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$$
for all real numbers $x > 0$ and $1 < y < 1.5$. This seems true when plotted on WolframAlpha, but I don't know how to pro... | We have that
$$g(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =1-\frac{2xy^3}{(1+x^2)^2}+\frac{2x^3y}{(1+y^2)^2}$$
with
$$g'(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =-\frac{6xy^2}{(1+x^2)^2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
how to find $\frac{a}{b}$ given that $2a^2 + 2007a + 3 = 0$ and $3b^2 + 2007b + 2 = 0$ Given that
$$\begin{cases}
2a^2 + 2007a + 3 = 0 \\
3b^2 + 2007b + 2 = 0
\end{cases}
$$
and $ab \ne 1$, how to solve for $\frac{a}{b}$?
My try:
\begin{align}
2007a = -3 - 2a^2 \\
2007b = -2 - 3b^2 \\
\frac{a}{b} = \frac{-3 - 2a^2}{-2... | $\mathcal{Hint}$
Consider $2x^2+2007x+3=0$
Note that $a$ is one the roots.
Now, replacing $x$ by $\frac{1}{x}$
How does that affect the equation and its roots?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$ I am asked to prove by induction that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$.
I wonder whether there is a more direct method, for example factorizing by $10$.
If an expression is divisible by $10$, does this mean that I can factoriz... | Best method? Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$ and as $3\cdot 7 \equiv 1 \pmod {10}$ then $3^{-1}\equiv 7 \pmod {10}$ and $7-7\equiv 0\pmod {10}$ so $10|7\cdot 11^{2n+1} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3832496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 5
} |
Evaluate $\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$
$$\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$$
I figured out that the functions are inverses of each other (kinda). If find the inverse of 1st function, we get
$y=(1+x^3)^{1/2}$
$y^2=1+x^3$
$(y^2-1)^{1/3} = x$
But the function given is, $((x+1)^2-1)^{1/3}$
Substitut... | By Integral of Inverse Functions we have:
$$\int_c^d f^{-1}(y)\,dy+\int_a^b f(x)\,dx=bd-ac$$
given that $f(a)=c, f(b)=d$.
Let $f(x)= (1+x^3)^{1/2}$. As you have observed, $f^{-1}(y) = (y^2-1)^{1/3}$.
Also we have $f(0)=1$ and $f(2)=3$. Thus:
$$\int_1^3 (y^2-1)^{1/3}\,dy + \int_0^2(1+x^3)^{1/2}\,dx = 3\times2-1\times0=6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ an... | Actually, the polynomial is irreducible over $\Bbb F_{19}$ and hence irreducible over $\Bbb Z$ and $\Bbb Q$. It has no root over $\Bbb F_{19}$ and no quadratic or cubic factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.