Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify ... | \begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2} & = \frac{2\sqrt{2}-4}{4-\sqrt{2}}\cdot \frac{4+\sqrt{2}}{4+\sqrt{2}} - \sqrt{2} \\
& = \frac{4\sqrt{2}-12}{14} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6}{7} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6 -7 \sqrt{2}}{7}\\
& = \frac{-5\sqrt{2} -6 }{7}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
| Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $\geq X$? We are generating random numbers (integers) in the range $[a, b]$. All values are equally likely.
We will continue to generate random numbers in this range, and add up successive values until their combined s... | The expected number is dominated by the term based on the mean $\frac{2X}{a+b}$ - but we can do better than that. As $X\to\infty$,
$$E(n) = \frac{2X}{a+b}+\frac{2(a^2+ab+b^2)}{3(a^2+2ab+b^2)}+o(1)$$
Why? Consider the excess. In the limit $X\to\infty$, the frequency of hitting a particular point approaches a uniform den... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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If $ f(x) = x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}.$ Then $ \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx$
If $\displaystyle f(x) = x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}.$ Then $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx$
Try: Given $\displaystyle f(x) =x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}... | Hint
Note that $$f(x)={(x-{1\over 2})^3}+1$$and $$\int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(f(u+{1\over 2}))du=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(u^3+1)du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction proof for stirling of first kind. I have to show that for every stirling number of the first kind $\forall n \geq 2 : s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $\frac{1... | Induction step:
$\begin{aligned}s\left(n+1,n-1\right) & =ns\left(n,n-1\right)+s\left(n,n-2\right)\\
& =\frac{1}{2}n^{2}\left(n-1\right)+\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(3n-1\right)\\
& =\frac{1}{24}n\left(n-1\right)\left[12n+\left(n-2\right)\left(3n-1\right)\right]\\
& =\frac{1}{24}n\left(n-1\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Calculate $\lim_{x\to 0}\left(\frac{3\sin(x)-3x \cos(x)}{x^3}\right)^\frac{1}{x}$ This is an exercise from today's exam and I think I did something wrong:
Calculate the limit:
$$\lim_{x\to 0}\left(\frac{3\sin(x)-3x\cos(x)}{x^3}\right)^{1/x}.$$
To tackle this problem I used equivalent infinitesimals: $1-\cos(x) \sim \fr... | $$3\sin{x}-3x\cos{x}=3\left(x-\frac{x^3}{6}+\frac{x^5}{120}-...\right)-3x\left(1-\frac{x^2}{2}+\frac{x^4}{24}-...\right)=$$
$$=x^3-\frac{1}{10}x^5+...,$$
which says that your limit is equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3077567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers. Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.
WLOG, let a$\leqq b\leqq c$,
| If $a \le b$ but $b|a+1$ then $b \le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c \le a+2$ but $a < b=a+1 \le c\le a+2 $ so either $c = b = a+1$ or $c= a+2$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
| Maybe works, $$3=\sqrt{3^{2}}=\sqrt{5+4}=\sqrt{5+\sqrt{16}}=\sqrt{5+\sqrt{11+5}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{25}}}=\sqrt{5+\sqrt{11+\sqrt{19+6}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Why is $(x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1)/(x-1)^3$ irreducible? Consider the following polynomial
$$
f_n(x)=x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1
$$
Try the first $n$, I find that the $(x-1)^3$ is its factor:
$$
f_1(x)=(x-1)^3,\\
f_2(x)=(x-1)^3 (x^2+3x+1),\\
f_3(x)=(x-1)^3 (x^4+3x^3+6x^2+3x+1),\\
\cdots
$$
... | Here is a proof in the special case that $2n+1$ is prime (and $n>1$).
Let $g_n(x)=f_n(x)/(x-1)^3$ and let $h_n(x)=g_n(x+1)$. By dividing $f_n(x)$ by $x-1$ three times using long division, we find $$g_n(x)=\binom{2}{2}x^{2n-2}+\binom{3}{2}x^{2n-3}+\dots+\binom{n}{2}x^n+\binom{n+1}{2}x^{n-1}+\binom{n}{2}x^{n-2}+\dots+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Probability of $2$ pairs on $5$ dice We throw $5$ dice. What is the probability of getting $2$ pairs ?
My solution says it is $$\frac{6\cdot 5\cdot 4\cdot 5! }{6^5\cdot 2\cdot 2!\cdot 2!},$$ where as for me it's $$\frac{6\cdot 5\cdot 4\cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
I do as follows: Throwing $5$ dice is the same thin... | Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-\dfrac16=\dfrac56$
Taking your approach, you would say the probability should be $\dfrac{6\cdot 5 \cdot 2!}{6^2\cdot 1!\cdot 1!} = \dfrac{10}{6}$, which is clearly wrong. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I ha... | HINT:
The Cauchy-Schwarz Inequality
\begin{eqnarray*}
(x_1^2+ \cdots +x_n^2) (y_1^2+ \cdots +y_n^2) \geq (x_1 y_1+ \cdots +x_n y_n)^2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the area of parallelogram and its missing vertex Given three radius-vectors: $OA(5; 1; 4), OB(6;2;3), OC(4;2;4)$, find the missing vertex $D$ and calculate the area of obtained parallelogram.
My attempt:
Firstly, we are to find the vectors which form the parallelogram. $$AB = OA - OB = (-1; -1; 1)$$
$$AC=OA-OC=(1... | A simpler way to obtain $D$ is $D = C+AB$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\righ... | try it with $$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2dt}{1+t^2}$$
You will get this integral $$\int \frac{4 t
\left(t^2-1\right)^2}{\left(t^2+1
\right)^3 \left(t^2+2 t-1\right)}dt$$
and this is $$\int \left(1/2\,{\frac {-t+1}{{t}^{2}+1}}+{\frac {-4\,t+4}{ \left( {t}^{2}+1
\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Cauchy Schwarz - Finding minimum of a function For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$. This is a weekend brain teaser in the 2nd week of Calc 3.
| This problem can be solved by using elementary math only (no derivatives, no Lagrange multipliers, no C-S ineqaulity etc).
We have to minimize the expression:
$$f=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$$
Introduce substitution:
$$u=x-1, \ \ v=y+2, \ \ w=z$$
The expression that has to be minimized now becomes:
$$f=u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. Wh... | As $s:=a+b$ and $a-b$ have the same parity, we can write
$$\left\lfloor\frac{s-2b}2\right\rfloor+\left\lceil\frac s2\right\rceil=\left\lfloor\frac s2\right\rfloor+\left\lceil\frac s2\right\rceil-b=s-b=a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| Let $\cos x = c, \sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $\sin x = 0$ or $\cos x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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What is the value of $\operatorname{arctan} \left(-\frac{12}{5}\right)$? I am trying to find the value of $\sin(\operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $\operatorname{tan}(\theta) = -12/5$. But this is not one of the easy... | Let $\theta = \arctan \frac {-12}{5}$
$\tan \theta = \frac {-12}{5} = \frac {\sin \theta}{\cos \theta}$ which means there there is a right triangle with angle $\theta$ and the opposite side = $r\sin \theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r \cos \theta = 5$ where $r$ is the hypote... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then ... | So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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On a multisection of a series Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $r$ be a positive integer. Let $a$ and $n$ be nonnegative integers such that $n-a+1\geq r$. Prove that
$$ \sum_{k=a}^{n} f(k) = \sum_{j=0}^{r-1} \sum_{k=\lfloor{\frac{a+r-1-j}{r}}\rf... | This is an answer based on series manipulation. In order to ease things we make a two-step approach. At first we show the identity is valid in case $a=0$ and then the generalisation for integral $a\geq 0$ follows easily.
Step 1: $a=0$
The following is valid for integral $1\leq r\leq n$:
\begin{align*}
\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from her... | Your integral $$ \int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-\gamma+\int_{0}^{1}\frac{1-t^s}{1-t}dt=\psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
How to find the value $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ converges to? We have been finding the value of sums related to the Harmonic series by using that
$\sum_{n=1}^{N}\frac{1}{n}= ln(N)+\gamma + \varepsilon_N$.
When I try to calculate $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ using that method there... | Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=\sum_{k=0}^n{\frac{(-1)^n}{2n+1}}=\int_0^1{\sum_{k=0}^n{(-x^2)^n}\,dx},$$ and then prove that $$\left|S_n-\int_0^1{\frac{1}{1+x^2}\,dx}\right| \leq n^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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perfect square of summation of odd numbers This is not homework , and I am old enough to be your father :-).We know the summation of odd numbers results in perfect squares , like $1 + 3 = 4 $, $1+ 3 + 5 = 9$ and so on.
My question is , if we start from a random odd number (say $315$) , how many odd numbers after $31... | Lets suppose that the random odd number in which we stop is $2k+1$. We known that the sum:
$$1+3+5+\cdots +(2k+1) = k^2$$
Consider two cases:
Case 1: k is even
If this is the case then que can write $k = 2n$. Follows to see that $k^2 = (2n)^2$ and,
$$(2n)^2+(n^2-1)^2 = (n^2+1)^2$$
We know that:
$$1+3+4+\cdots+(2(n^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}... | I would substitute $$a=\sqrt[3]{8-x}$$ then you term is given by $$\frac{2\sqrt{a^3-7}-a}{8-a^3}$$ and then multiply numerator and denominator by $$2\sqrt{a^3-7}+a$$ and then you will get
$$\frac{(a-2)(14+7a+4a^2)}{-(a-2)(a^2+2a+4)(2\sqrt{a^3-7}+a)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$.
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$
*
*make exponent postive
$2x\frac{1}{\s... | This step is wrong
square everything
$4x^2\frac{1}{\sqrt{4-x}} - 9(4-x) =0$
Once you square everything you get (i.e. $a=b$, then $a^2=b^2$, then $a^2-b^2=0$)
$$4x^2\frac{1}{4-x} - 9(4-x) =0$$
Now multiply by $(4-x)$, we get
$$4x^2 - 9(4-x)^2 =0$$
or
$$(2x)^2 - [3(4-x)]^2 =0$$
which gives
$$(2x -12 + 3x)(2x +12 -... | {
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"timestamp": "2023-03-29T00:00:00",
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Factorise polynomial with real and complex roots How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$
I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard ... | I like the following way.
For all real value of $k$ we have:
$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(10k-6)x+k^2+4\right).$$
Now, we'll choose a value of $k$, for which $2k+\frac{9}{4}>0$ and $$(5k-3)^2-\left(2k+\frac{9}{4}\right)(k^2+4)=0.$$
Easy to see that $k=0$ i... | {
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"source": "stackexchange",
"question_score": "7",
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How many ways can 26 letters be divided into 13 pairs? The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I... | Another approach is: the first pair can be selected in ${26\choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13\choose 1}$:
$$\frac{{26\choose 2}}{{13\choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$\frac{{26\choose 2}}{{13\choose 1}}\cd... | {
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"source": "stackexchange",
"question_score": "1",
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Inequality $\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$ Suppose $a_i$ are dinstinct positive integers $\forall1\le i\le n$. Prove that
$$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$$
My approach:
I will procede by proo... | I think your proof is correct, but I think it's better to get it without contradiction.
$(1,2,...,n)$ and $\left(\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}\right)$ have an opposite ordering.
Thus, by Rearrangement
$$\frac{a_1}{1^1}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\geq\frac{a'_1}{1^1}+\frac{a'_2}{2^2}+...+\frac{a... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...
$\displaystyle|f(x,y)-L|=\left |\frac... | You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):
*
*$\sqrt{ab}\leq \sqrt{\frac{a^2+b^2}{2}}$
So, you get
\begin{eqnarray*} \left| \frac{x^2+xy+y^2}{\sqrt{x^2+y^2}}\right|
& = & \left| \frac{x^2+y^2}{\sqrt{x^2+y^2}} + \frac{xy}{\sqrt{x^2+y^2}}\right| \\
& \leq & \s... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How to show $\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}$? By numerical evaluation, it seems that the following identity holds. For any odd positive integer $L$,
$$\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}.$$
I was t... | When looking at OPs identity the left-hand side smells like a telescoping sum where we have something like the first and last entry at the right hand side, and indeed we can transform it accordingly. We start with some preliminary arrangements. Shifting the index $i$ by one to start with $i=0$ we want to show
\begi... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Formula for partial sum of binoms: $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom nk \binom mk$ I am dealing with some expressions containing combinatoric numbers. Does anybody know a formula for this?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$
| Not simple and we need to consider three cases since we do not know the parity of $n$ and what $\left[ \dfrac{n}{2} \right]$ means (floor or ceiling function).
*
*Suppose that
$n$ is odd $(n=2p+1)$ and that $\left[ \dfrac{n}{2} \right]=\left\lfloor \frac{n}{2}\right\rfloor=p$, the expression is
$$S_p=\sum_{k=0}^{p}... | {
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"source": "stackexchange",
"question_score": "1",
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recurrence relation limit $x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$ Let $0<b<a$ and $(x_{n})_{n\in \mathbb{N}}$ with $x_{0}=1, \ x_{1}=a+b$
$$x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $0<b<a$ and $L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$ then $L= ?$ ($L=a$ right answer )
b) If $0<b<a<1$ and $L=\lim_{... | Hint: Try rewriting your recurrence as follows:
$$
x_{n+2} = (a+b)x_{n+1} - abx_n \\
x_{n+2} = ax_{n+1}+bx_{n+1} - abx_n \\
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n)
$$
So you get:
$$
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n) = a^2(x_{n} - bx_{n-1}) = \cdots
$$
Can you proceed from here?
| {
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What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles?
Circles of radii 5, 5, 8 and $\frac{m}{n} $(the smallest circle) are
mutually externally tangent to all circles, where $m$ and $n$ are
relatively prime positive integers. Find $m + n$.
Source: Bangladesh Math Olympia... | The triangle has a base side of length $10$ and two sides of length $13$ each. The hight this trinagle is $h = \sqrt {13^2-5^2} = 12$. This height consists of the radius of the large circle, the radius $r=\frac mn$ of the smallest circle and the hight of another isoceles triangle with the same base side and two other s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Show that the equation represents a sphere, find center and radius $2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1$
So here's my guess:
$2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1$
Get all variables on one side
$2x^2 - 8x + 16 + 2y^2 + 2z^2 + 24z + 144 = 1 + 16 + 144$ Complete square
$(2x - 8)^2 + 2y^2 + (2z^2 + 24)^2 = 161$ Stopped h... | Here's a way to do it: divide through by $2$ to get
$$ x^2+y^2+z^2=4x-12z+\frac{1}{2}$$
$$ x^2-4x+y^2+z^2 +12 z=\frac{1}{2}.$$
Now, to complete the square for $x^2-4x$ we need to add $4$ to both sides. To complete the square for $z^2+12z$ we need to add $36$ to both sides. In total we add $40$ to both sides to get
$$ x... | {
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"url": "https://math.stackexchange.com/questions/3104569",
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"source": "stackexchange",
"question_score": "1",
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How do we use the picture to get the polynomial? Calculate the solution $p\in \mathbb{P}^4$ off the following interpolatin exercise:
\begin{align*}&p(0)=2 , \ p'(0)=3, \ p''(0)=1 \\ &p(1)=2, \ p'(1)=0\end{align*}
$$$$
I have done the following:
The polynomial that we are looking for is of the form \begin{equation*}... | You may solve this directly with Taylor:
$$p(x) =\sum_{k=0}^4\frac{p^{(k)}(0)}{k!}x^k$$
Pluggin in the given values you get
$$p(x) = p(0) + p'(0)x + \frac{p''(0)}{2}x^2 + \frac{p^{(3)}(0)}{6}x^3 + \frac{p^{(4)}(0)}{24}x^4 $$ $$= 2 + 3x + \frac{1}{2}x^2 + ax^3+bx^4$$
Now, use the other two conditions to find $a$ and $b$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int^{\pi/2}_0 \frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$
Finding
$$I=\int_0^{\pi/2}\frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$$
What I tried:
\begin{align*}
I &= \int^{\pi/2}_0 \left(\frac{\sin x}{1+\cos x}\right)^n\csc^2(x)\mathrm dx \\
&= \int^{\pi/2}_0 \tan^n(x/2)\csc^2(x)\mathrm dx
\end{align*}
... | Note that the $\csc^2(x)$ term can be expressed in terms of $\tan\left(\frac x2\right)$ aswell. To be precise we got that
\begin{align*}
\csc^2(x)=\left(\frac1{\sin(x)}\right)^2=\left(\frac1{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}\right)^2=\left(\frac{\frac1{\cos^2\left(\frac x2\right)}}{2\frac{\sin\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107807",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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A sum of series problem with alternating sign of terms I came across a problem that requires me to find the sum of a series. The term of the series $T_n$ is given by
$$T_n = (-1)^{\frac{n(n+1)}2}n^2$$
Sum till $4n$ terms is to be found.
Writing down the first few terms:
$$-(1)^2-(2)^2+3^2+4^2-(5)^2-(6)^2+7^2+8^2+ \ldot... | $$\begin{align}
&\;\;\;\ \sum_{r=1}^{4n}(-1)^{\frac {r(r+1)}2}r^2\\
&=\sum_{r=1}^n (4r)^2+(4r-1)^2-(4r-2)^2-(4r-3)^2\\
&=\sum_{r=1}^n \underbrace{(4r)^2-(4r-2)^2}+\underbrace{(4r-1)^2-(4r-3)^2}\\
&=\sum_{r=1}^n \qquad 2(8r-2)\;\;\;+\qquad 2(8r-4)\\
&=\sum_{r=1}^n4(8r-3)\\
&=4\sum_{r=1}^n (8r-4+1)\\
&=4\sum_{r=1}^n 4(2r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110394",
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"source": "stackexchange",
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How to compute remainder of division of $P(x)$ by $x^2 -3x+2$?
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.
I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+... | You want to find $r(x)$ where the degree of $r$ is less than 2 in the following equation:
$$P(x) = (x^2-3x+2)S(x) + r(x) = (x-1)(x-2)S(x) + r(x)$$
You also have the two formulas you wrote above (representing the polynomial remainders you are given). If you plug in $x=1$ and $x=2$, you have two different values of $x$ t... | {
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Find out the value of the integral $\int_{-2}^{2} \lfloor x^2-1\rfloor dx$ Find out the value of the integral $$\int_{-2}^{2} \lfloor x^2-1\rfloor dx$$ where $[x]$ denotes the floor function (i.e., $[x]$ is the greatest integer $\le x$.)
My attempt ..... $$\int_{-2}^2 \lfloor x^2 – 1\rfloor dx = 2\int_0^2 \lfloor x^2-1... | Since $f$ is even we have $$ I/2 =\int_0^2[x^2-1]\;dx=$$ $$=\int_0^1-1\;dx+\int_1^\sqrt{2}0\;dx+\int_\sqrt{2}^\sqrt{3}1\;dx+\int_\sqrt{3}^22\;dx=...$$
$$=(-1)+0+(\sqrt{3}-\sqrt{2})+(4-2\sqrt{3})=3-\sqrt{3}-\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $. Decide all solutions to the following trigonometric equation:
$$
\sin \left( 4\,x \right) =\cos \left( 3\,x \right).
$$
in the interval $[0, 2π[$.
I start by expanding $\sin \... | It'd be much easier to use the identities
$\sin \theta = \sin(2k \pi + \theta)$
$\cos \theta = \cos (\pm \theta +2k\pi)$
$\sin \theta = \cos (\frac \pi 2 - \theta); \cos(\theta) = \sin (\frac \pi 2-\theta)$.
So solutions occur when
$4x = \pm (\frac \pi 2 - 3x) + 2k\pi$
$\pm 3x = \frac \pi 2 - 4x + 2k \pi$ so
$7x = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\... | It happens that your sequence is a cyclic sequence. More precisely, its first six terms are$$-2+\frac{\sqrt3}2,2+\frac{\sqrt3}2,4,2-\frac{\sqrt3}2,-2-\frac{\sqrt3}2,\text{ and }-4$$and then it repeats itself again and again. Therefore:
*
*$\max\{X_n\,|\,n\in\mathbb N\}=\sup\{X_n\,|\,n\in\mathbb N\}=\limsup_nX_n=4$;
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114035",
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"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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Why is $(5\sqrt{5p}-3\sqrt{5q})(5\sqrt{5p}+3\sqrt{5q}) \equiv 5(5p-3q)(5p+3q)$? I was working on the difference of two squares, $125p^2-45q^2$
Writing my answer, $$(5\sqrt{5}p-3\sqrt{5}q)(5\sqrt{5}p+3\sqrt{5}q),$$ onto Pearson, I got a popup that said my answer was equivalent to the correct answer but in incorrect for... | $$125p^2-45q^2=5(25p^2-9q^2)=5(5p+3q)(5p-3q)$$
Your answer was incorrect because it should have been
$$125p^2-45q^2=(\sqrt{125}p^2+\sqrt{45}q^2)(\sqrt{125}p^2-\sqrt{45}q^2)=(5\sqrt{5}p+3\sqrt{5}q)(5\sqrt{5}p-3\sqrt{5}q)=5(5p+3q)(5p-3q)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Evaluate $\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx$
Evaluate $$\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx=\int_1^{\infty} \frac {\arcsin(\sin x)}{x^3}dx$$
I tried to use Feynman's trick in the second one as
$$I(a)=\int_1^{\infty} \frac {\arcsin(\sin ax)}{x^3}dx$$
We have $I(0)=0$
No... | Write $I_n = \left[\left(n-\frac{1}{2}\right)\pi, \left(n+\frac{1}{2}\right)\pi\right]$ and notice that the behavior of $\arcsin\sin x$ on $I_n$ depends on the value of $n$. In particular, $\arcsin\sin x$ is linear on each interval $I_n$ with the slope $(-1)^n$.
So, we can compute the integral as
\begin{align*}
\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Convergence of the Arithmetic mean of a sequence taking values in $\{0, 1\}$ Let $(a_i)_{i \in \mathbb{N}}$ be a sequence taking values in $\{0, 1\}$. Now define $x_n$ as the Arithmetic mean of the first $n$ elements of $(a_i)_{i \in \mathbb{N}}$. So let
$$x_n = \frac{1}{n}\sum_{i = 1}^{n}a_i.$$
Can we find such a sequ... | This is not the case. Take for example $a_n=1$ for certain consecutive $n$ and otherwise $a_n=0$ such that the mean of the sequence wiggles around between a third and two third. To let this happen, the sequence needs to contain longer and longer sequences of consecutive $1$s and $0$s.
To be precise one could take the f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculation of $\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$
Calculation of $$\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$$
My Try: Let $$I = \int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x+1)(x+2)}dx$$
Put $x=\sin^2 \theta$ and $dx = 2\sin \theta \cos \theta$ and changing limits
So $$I = \int^{\frac{\pi}{2}}_{0}\frac{2}{(... | Hint. Note that
$$\frac{2}{(1+\sin^2 \theta)(2+\sin^2 \theta)}
=\frac{2}{1+\sin^2 \theta}-\frac{2}{2+\sin^2 \theta}.$$
Moreover, for $k>0$,
$$\frac{1}{k+\sin^2 \theta}=\frac{\frac{1}{k}d(\tan \theta)}{1+\frac{(k+1)\tan^2 \theta}{k}}.$$
Finally let $t=\frac{\sqrt{k+1}\tan \theta}{\sqrt{k}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $Tr(B)+Tr(C)$ If $B,C$ are $2 \times 2 $ matrices with integer entries such that:
$$\begin{bmatrix}
-1 &1 \\
0& 2
\end{bmatrix}=B^3+C^3$$
Find value of $Tr(B)+Tr(C)$
My try:
Taking trace on both sides we get:
$$Tr(B^3)+Tr(C^3)=1$$
Any clue from here?
| Let $\displaystyle A=\begin{bmatrix}
-1 &1 \\
0& -2
\end{bmatrix}$ and $\displaystyle I =\begin{bmatrix}1&0 \\0&1\end{bmatrix}$.
Then Using Characteristic Equation(Caly -Hamilton Theorem)
$|A-\lambda I|=0\Rightarrow \begin{vmatrix}
-1-\lambda && 1& \\
0& & -2-\lambda\\
\end{vmatrix}=0$
$\Rightarrow \lambda^2+3\... | {
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"source": "stackexchange",
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List of asymptotic expressions for $\sum_r r^\beta \ln r$? Expression for $\beta =1$ in $\sum_r r^\beta \ln r$
I recently had the following idea to use the below identity:
$$ (1!2! 3! \dots n!) (12^2 3^3 4^4 \dots n^n) = n!^{n+1}$$
Dividing both sides by $n^{n(n+1)/2}$ and using $1+2+3 \dots + n = \frac{n(n+1)}{2}$:
$$... | By the Euler-Maclaurin summation formula,
$$\sum_{k=1}^nk^\beta\ln k\approx\frac{n^{\beta+1}}{\beta+1}\log n-\frac{n^{\beta+1}-1}{(\beta+1)^2}+\frac{n^\beta\ln n}2+\frac{\beta n^{\beta-1}\ln n-n^{\beta-1}+1}{12}+\cdots$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that any number that is not a power of 2 can be expressed as a sum of two or more consecutive positive integers 1) Prove that any number that is not a power of $2$ can be expressed as a sum
of two or more consecutive positive integers, but that this is not possible for powers of $2$.
2) Which numbers can be expre... | 1. Assume $2^m;\ m\ge 2$ can be expressed as the sum of consecutive integers $n+1,n+2,\dots n+k$. Then it is the difference between the sum of the first $k$ integers and the sum of the first $n$ integers. $$2^m=\frac{k(k+1)}{2}-\frac{n(n+1)}{2}=\frac{k^2-n^2+k-n}{2}\Rightarrow 2^{m+1}=(k+n)(k-n)+(k-n)=(k+n+1)(k-n)$$ No... | {
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Existence of closed form solution of the below DE I have a problem finding an analytical closed form solution of the following DE.
$ \displaystyle \left( \frac{y(x)+2}{3} \right) \frac{dy}{dx} + y(x) +\frac{1}{x} \left( a y(x)^2 + b y(x) -c \right) =0. \\
\{a,b,c\} \in \mathbb{R}, \qquad \{a,b,c\}>0.
$
I encountered t... | Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $\dfrac{du}{dx}=\dfrac{dy}{dx}$
$\therefore\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(a(u-2)^2+b(u-2)-c)=0$
$\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(au^2-(4a-b)u+4a-2b-c)=0$
$u\dfrac{du}{dx}=-\dfrac{3au^2}{x}-3\left(1-\dfra... | {
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Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\l... | You can write
$$x^2 \ln \left( \frac{x^2+1}{x^2+3} \right) = x^2 \ln \left( 1- \frac{2}{x^2+3} \right) \sim x^2 \times \frac{-2}{x^2+3}$$
So the limit is $-2$.
| {
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"source": "stackexchange",
"question_score": "3",
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Geometric center of convex set of positive semidefinite matrices Consider the set of $d\times d$ matrices that are positive semidefinite and have unit trace. This is a convex set, $S$. Is it possible to think of a geometric center of this set? The criterion for geometric center is an element $\rho$ such that if $\rho +... | When $d\ge3$, there cannot be any geometric centre. Since the trace of a matrix and the property of being a centre are preserved by unitary similarity, we may consider only the case where $\rho$ is a nonnegative diagonal matrix.
If $\rho=\frac1nI$, then a similar example to yours shows that $\rho$ is not a centre.
If $... | {
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Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$. $$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$
Using $\lfloor y ... | Your proof is correct. Here is another way to compute this limit.
Note that $\frac{i}{x}-1 < \left \lfloor \frac{i}{x} \right \rfloor \le \frac{i}{x}$,$\forall i=\overline{1,k}$.
Therefore, $\frac{k(k+1)}{2x} -k < \left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \f... | {
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"question_score": "7",
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Factor $2x^3 + 3x^2 - 2x$ I am having the most difficult time factoring this equation out. I should have paid more attention in precalculus in high school :(
$2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2) = x(2x-1)(x+2)$
How do you get the $(2x-1)(x+2)$ out of $(2x^2+3x-2)$?
Quadratic equation didn't work, I can't take out a com... | You can always use the quadratic equation to factor a quadratic polynomial $Q(x)$.
If the solutions to $Q(x)=0$ are $x=r$ and $x=s$, then the polynomial can be factored as
$$Q(x)=k(x-r)(x-s)$$
where $k$ is the leading coefficient of $Q(x)$.
In your case, $Q(x)=2x^2+3x-2$ and the quadratic formula says the solutions to ... | {
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}... | Note that $$\ln\frac{x+2-2\sqrt{x+1}}{x^2}=\ln\frac{(\sqrt{x+1}-1)^2}{x^2}=2\ln\frac{\sqrt{x+1}-1}{x}$$
if we further simplify,
$$2\ln(\frac{\sqrt{x+1}-1}{x}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1})
=2\ln\frac{1}{\sqrt{x+1}+1}=-2\ln(\sqrt{x+1}+1)$$
You did nothing wrong.
| {
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"source": "stackexchange",
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
... | There are two cases here:
*
*There's an even number of heads: 0, 2, 4, 6 or 8 heads
*There's an odd number of heads: 1, 3, 5, 7 or 9 heads
But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:
*
*There's an even numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is co... | Your method is fine, but you dropped the inequalities prematurely. Indeed, by comparing the extreme members, we must fulfill
$$\max\left(\dfrac m3,\dfrac{5-m}4\right)<\min\left(\dfrac{m+1}3,\dfrac{6-m}4\right)$$
which is only possible with $m=2$.
Then
$$\max\left(\dfrac 23,\frac34\right)\le\frac1x<\min\left(1,1\right)$... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the ... | You have $3$ intervals to consider: $(-\infty,-\sqrt3), (-\sqrt3,\sqrt3)$ and $(\sqrt3,\infty)$. The sign of the product $(x-\sqrt3)(x+\sqrt3)$ is constant on each interval. (That's by continuity, since the function $f(x)=x^2-3$ must pass through zero to change sign.)
Use a test point in each interval: \begin{align... | {
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"source": "stackexchange",
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Proof that $a\nabla^2 u = bu$ is the only homogenous second order 2D PDE unchanged/invariant by rotation Looking for feedback and maybe simpler intuition for my proof of the theorem, shown below
The statement of the theorem:
Theorem
Among all second-order homogeneous PDEs in two dimensions with constant coefficients,... | We say that a linear operator $L$ is rotationally invariant if and only if $L$ commutes with the orthogonal group, i.e.
$[L, O] = 0$ for every $O \in \text{O}(n)$.
Hence what you are proving is that if $L$ is a second order linear operator then
\begin{align}
LO[f](x) = L[f(O x)] = [Lf](O x) = OL[f](x)
\end{align}
if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere $ x^2 + y^2 + z^2 = 4 $. calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere
$ x^2 + y^2 + z^2 = 4 $.
my trial :
The Domain of integration on the YZ plane is :
solving ... | Let $\gamma : [0, 2\pi] \to \mathbb{R}^2, \gamma(t) = (\cos t, 1+\sin t)$ be the parameterization of the circle $x^2+(y-1)^2 = 1$ in the $xy$-plane.
The height of the cylinder over a point $(x,y)$ on this circle is given by $z = \sqrt{4-x^2-y^2}$ so the area is
$$A = 2\int_\gamma z\,d\gamma = 2\int_\gamma \sqrt{4-x^2-y... | {
"language": "en",
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Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + ... | It must be $$\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic... | \begin{align}
\frac{x+4}{x^2+2x+5} &= \frac{x+4}{(x+1)^2 +4} \\
&=\frac{\color{blue}{x+1}}{\color{blue}{(x+1)}^2 +4} +\frac{3}{\color{blue}{(x+1)}^2 +4} \\
&=\frac{\color{blue}{u}}{\color{blue}{u}^2 +4} +\frac{3}{\color{blue}{u}^2 +4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A triangle has sides x, x+4, and 3x−5. What is the possible range of x?
A triangle has sides $x, x+4,$ $and$ $3x−5$. What is the possible range of x?
The answer to this is not:
$2x-9$ $<$ $x$ $<$ $4x-1$
Which is gotten from:
--> $(3x-5)-(x+4)<x<3x-5+x+4$
--> $3x-5-x-4 < x < 4x-1$
--> $5x-9 < x < 4x-1$
So what is th... | Hint Consider using the triangle inequality
$$a+b>c$$
Thus you have that $$x+(x+4)>3x-5\iff 9>x\tag{1}$$
$$x+(3x-5)>x+4\iff 3x>9\iff x>3$$ Hence
$$x\in(3, 9)\text{ or equivalently }\color{red}{3<x<9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144286",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If the equation $\sin^2x-a\sin x+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?
If the equation $\sin^2(x)-a\sin(x)+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?
What I try:
$$\displaystyle \sin x=\frac{a\pm \sqrt{a^2-4b}}{2}\in\bigg(0,1\bigg)$$
for one real solutio... | To $x$ to have only one solution in the range $(0,\pi)$, then $x$ must be $\pi\over 2$, according to the unit circle for trigonometry.
Since, $\sin{\frac{\pi}{2}} = 1$,
$$\sin^2 x - a\sin x +b = 1-a+b =0$$
$$\implies a-b=1 \to\text{eq.1} $$
From what you have tried we get,
$$a^2 =4b \to \text{eq.2} $$
Thus, to satisfy ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3... | I would consider this a coincidence.
$$\frac{-4x^4-2x^3-26x^2-8x-4\mathbf{5}}{(x+1)(x^2 +3)^2},$$
for example, requires the linear term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} ... | Note that
$$\ln\left| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| = \ln\left|x+\sqrt{x^2-9}\right|-\ln 3$$
and $\ln3$ is a constant. Since $C$ is an arbitrary constant, you can define a new constant $C_1 = C - \ln3$ and the answer is
$$\ln\left|x+\sqrt{x^2-9}\right| + C_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\f... | For each term involving both $x$ and $y$, differentiate the $x$ and $y$ parts separately and combine the two using the product rule:
$$(x^4y)'=x^4(y)'+(x^4)'y=x^4\frac{dy}{dx}+4x^3y$$
$$(xy^5)'=x(y^5)'+(x)'y^5=5xy^4\frac{dy}{dx}+y^5$$
Thus
$$55x^4+x^4\frac{dy}{dx}+4x^3y+5xy^4\frac{dy}{dx}+y^5=0$$
$$(x^4+5xy^4)\frac{dy}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x... | It seems that you know why $R=\sqrt{26}$,
Hence $5\cos x - \sin x$ can be expressed as $\sqrt{26}(\frac{5}{\sqrt{26}}\cos x - \frac{1}{\sqrt{26}}\sin x)$
$$\text{Let, } \cos\alpha = \frac{5}{\sqrt{26}} \text{ and } \sin\alpha = \frac{1}{\sqrt{26}}$$
Hence, $5\cos x - \sin x$ can now be expressed as $\sqrt{26}\cos (x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3148923",
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"source": "stackexchange",
"question_score": "2",
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$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{... | Let $t=1/x$, then $e^{1/x} < 1 + \frac{1}{x-1}$ for $x>1$ is equivalent to
$e^t -1< +\frac{t}{1-t}$ for $0<t<1$.
We have $e^t-1=t+ \frac{t^2}{2!}+\frac{t^3}{3!}+.....$ and $\frac{t}{1-t}=t+t^2+t^3+...$for $|t|<1.$
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Hint: Showing $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$ converges in $L^{p}$ to $x^{-\frac{3}{2}}\sin{(x)}$ Let $f_{n}:]0, \infty[\to \mathbb R$ and $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$
Show that $f_{n}$ converges in $L^{p}$ to $f$ where $f(x):=x^{-\frac{3}{2}}\sin{(x)}$ and $p \in [1,2[$
My idea:
\... | First notice that
$$\frac{n\sqrt{x}\sin x}{1+nx^2} = \frac{\sqrt{x}\sin x}{\frac1n+x^2} \xrightarrow{n\to\infty} x^{-3/2}\sin x$$
pointwise.
Then notice that
$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2}\right|^p \le \frac1{x^{p/2}}\chi_{(0,1]}(x) + \frac1{x^{3p/2}}\chi_{[1,\infty)}(x) =: g(x) \in L^1(0,\infty)$$
and also... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ prov... | Putting $t = 2\pi - x$ we get $$C(a, b) = \int_{0}^{2\pi}\frac{2\pi - t}{a+b\cos^2t}\,dt= 2\pi\int_{0}^{2\pi}\frac{dt}{a + b\cos^2t} - C(a, b)$$ so that $$C(a, b) = \pi\int_{0}^{2\pi}\frac{dx}{a+b\cos^2x}$$ Since the integrand satisfies equation $f(2\pi - x) = f(x)$ it follows that $$C(a, b)=2\pi\int_{0}^{\pi}\frac{dx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the area of $y = 12-x^2, y=x^2-6$ I found the x-intercepts via my graphing calculator to be $(-3,3)$ and $(3,3)$ and from it I formed the Area equation,
$$A = \int_{-3}^{3} (12-x^2 - x^2 - 6)dx = \int_{-3}^{3}(-2x^2+6)dx \\
= -\frac{2}{3}x^3 + 6x \bigg]_{-3}^{3}$$
$$A = \bigg(3\frac{2}{3}\cdot 3^3 + 6(3) \bigg) ... | $$
12-x^2-\color{red}(x^2-6\color{red})=\color{red}{18}-2x^2\implies A=\left.18x-2\frac{x^3}3\right]_{-3}^3=72.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$.
Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$\frac{1}{6}$. If three marbles are drawn, then the probability all... | Let's look at the second equation:
$$\underbrace{\dfrac{x}{y}\cdot \dfrac{x-1}{y-1}}_{\dfrac{1}{6}}\cdot \dfrac{x-2}{y-2} = \dfrac{1}{21}$$
The first two terms are the same as the LHS of your first equation. So, we can substitute.
$$\dfrac{x-2}{y-2} = \dfrac{6}{21}$$
Reducing the RHS to lowest terms gives:
$$\dfrac{x-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How can I simplify my boolean expression further I have the following boolean expression that I want to simplify
$$B\cdot D+ \overline{A\cdot B\cdot D} + \overline{B}\cdot C\cdot \overline{D}$$
Here is what I have been able to due so far
$$B\cdot D+ \overline{A} + \overline{B} + \overline{D} + \overline{B} \cdot \ov... | Factorising
$$\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot (1+\overline{B}\cdot C)$$
$$=\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot 1$$
$$=\overline{A}+B\cdot D + \overline{B}+\overline{D}$$
Then by the absorption law,
$$B\cdot D + \overline{B}=D + \overline{B}$$
So this simplifies to
$$=\overlin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How can I block diagonalise this matrix? I have this matrix:
$$A = \left(
\begin{array}{cccc}
0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & -1 \\
\end{array}
\right)$$
and I would like to show that it can be block diagonalised into :
$$ B = \left(
\begin{array}{cccc}
\cos 2\pi/5 & -\sin 2\pi/5 ... | When you diagonalise a matrix over the complex numbers, you start by finding its eigenvalues, which may be complex. If the matrix is diagonalisable, then we can expect a basis of eigenvectors corresponding to these eigenvalues. If we compute the transformation $x \mapsto Ax$ in terms of this basis of eigenvectors, we g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Existence of 3 natural numbers that divide each other when squared and have 1 taken away from them Does there exist natural numbers, $a,b,c > 1$, such that;
$a^2 - 1$ is divisible by $b$ and $c$,
$b^2 - 1$ is divisible by $a$ and $c$ and
$c^2 - 1$ is divisible by $a$ and $b$.
| No such numbers exist.
First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.
Since $b,c$ are coprime ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Shell integration to find volume Find the volume generated by rotating the region bounded by the given curves about the y-axis.
$$y = \sqrt[3]{x}, y = 0, x = 1$$
Here is my work:
$$\begin{align}
V &= \int_{0}^{1}2\pi(x)\sqrt[3]{x}dx = 2\pi\int_{0}^{1}x^\frac{4}{3}dx \\
&= 2\pi \bigg[\frac{3}{6}x^\frac{6}{3} \bigg]_{0}^... | The setup is correct. The only thing that you got wrong is the antiderivative of the function $x^{\frac{4}{3}}$:
$$\int x^{\frac{4}{3}}\,dx=\frac{1}{\frac{4}{3}+1}x^{\frac{4}{3}+1}+C=
\frac{1}{\frac{7}{3}}x^{\frac{7}{3}}+C=
\frac{3x^{\frac{7}{3}}}{7}+C.$$
$$
V=2\pi\int_{0}^{1}x\sqrt[3]{x}\,dx=
2\pi\int_{0}^{1}x^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solving $\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$ This is the last indefinite integral I am attempting before proceeding to definite integrals study. Any ideas how to solve it?
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$$
My attempt: (not the longest one... :P)
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx... | So you have brought the ques down to solving integral of u^(3/2)/(u+3)
Now simply put u= z^2 so after taking constants out it becomes
z^4/(z^2+3)= (z^4 - 9 + 9)/(z^2+3)
=(z^2 - 3)(z^2+3)/(z^2+3) + 9/(z^2+3)
= z^2 - 3 + 9/(z^2+3)
Here, z^2 - 3 is very easy to solve. For 9/(z^2+3) simply put z= sqrt(3).tan(u) and the int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6... | In your way you proved that the minimal value is greater than $48$.
It's true, but the equality does not occur, which says that $48$ is not a minimal value.
The right solution can be the following, for example.
Let $\sin^2x\cos^2x=t.$
Thus, by AM-GM
$$t\leq\left(\frac{\sin^2x+\cos^2x}{2}\right)^2=\frac{1}{4}.$$
The eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Sum of combinations - convergence I am trying to evaluate the sum of combinations for a specific probability problem, but I am kind of stuck.
Let A have toss a coin $n+1$ times and B toss another coin $n$ times. What is the probability that A has more heads than B. (the probability of heads is $P = 1/2$). All experimen... |
On the one hand we obtain
\begin{align*}
\color{blue}{\sum_{y=0}^n\sum_{x=y+1}^{n+1}\binom{n+1}{x}\binom{n}{y}}
&=\sum_{y=0}^n\left(2^{n+1}-\sum_{x=0}^y\binom{n+1}{x}\right)\binom{n}{y}\tag{1}\\
&\,\,\color{blue}{=2^{2n+1}-\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}}\tag{2}
\end{align*}
on the other hand we ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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coefficient $x ^ n$ in development The advisor asks to verify that the coefficient of
$$x^n$$
in the development of:
$$(1+x)^{2n}+x(1+x)^{2n−1}+x2(1+x)^{2n−2}+......+x^n(1+x)^n$$
is equal to
$$\binom{2n+1}{n}$$
I tried for summations but not.
I did with the denominator change too, but I can not even,how can i match the... | You can explicitly evaluate your sum and extract the coefficient afterward. Recall that $$(y-x) (y^n + xy^{n-1} + \cdots + x^{n-1}y + x^n) = y^{n+1} - x^{n+1}$$ Letting $y = 1+x$ (so $y-x = 1$) gives
$$(1+x)^n + x(1+x)^{n-1} + \cdots + x^{n-1}(1+x) + x^n = (1+x)^{n+1} - x^{n+1}.$$
Then multiplying by $(1+x)^n$ gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my ... | Hint: $x^2+y^2=1-xy$, so that $x^3y+xy^3+4=xy(x^2+y^2)+4=xy(1-xy)+4$. Substitute $z=xy$. What do you end up with?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Recurrence Relation: $B(n) = 5\cdot B(n/3) + c(n^2)$ I'm working on deriving this recurrence relation in the form $O(n^d)$ for some value of d:
$B(n) = 5\cdot B(\frac{n}{3}) + cn^2$
The initial condition is:
$B(1) = c$
I'm having trouble incorporating the n^2 into my derivation. This is what I have so far:
We set $n =... | \begin{equation}
\begin{split}
B(n) &= 5B(\frac{n}{3}) + cn^2 \\
&= 5(5B(\frac{n}{3^2}) + c(\frac{n}{3})^2) + cn^2 \\
&= 5^2B(\frac{n}{3^2}) + c[5(\frac{n}{3})^2 + n^2] \\
&= 5^2(5B(\frac{n}{3^3}) + c(\frac{n}{3^2})^2) + c[5(\frac{n}{3})^2 + n^2] \\
&= 5^3B(\frac{n}{3^3}) + c[5^2(\frac{n}{3^2})^2 +5(\frac{n}{3})^2 + n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Why partial fraction decomposition of $\frac{1}{s^2(s+2)}$ is $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$? Can someone please explain why: $$\frac{1}{s^2(s+2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$$
And not:$$\frac{1}{s^2(s+2)}=\frac{A}{s^2}+\frac{B}{(s+2)}$$
I'm a bit confused where the extra s term comes from in... | You can do it step-by-step:
$$\begin{align}\frac{1}{s^2(s+2)}&=\frac1s\cdot \frac{1}{s\cdot (s+2)}=\\
&=\frac1s\cdot \left(\frac As+\frac{B}{s+2}\right)=\\
&=\frac A{s^2}+\frac1s\cdot \frac B{s+2}=\\
&=\frac{A}{s^2}+\frac{C}{s}+\frac{D}{s+2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What are the integer coeffcients of a cubic polynomial having two particular properties? Let $f(x) = x^3 + a x^2 + b x + c$ and $g(x) = x^3 + b x^2 + c x + a\,$ where $a, b, c$ are integers and $c\neq 0\,$. Suppose that the following conditions hold:
*
*$f(1)=0$
*The roots of $g(x)$ are squares of the roots of $f... | The conditions 1 and 2 imply that
*
*$\;c\,$ must be $\,0\,$ or $\,-1$,
*$\;a= -c^2$, and $\:b= c^2-c-1\,$.
Condition 1 gives $0=f(1)=1+a+b+c=g(1)\,.\,$ Hence both $f$ and $g$ have a zero at $1$ and factor as
$$f(x) \,=\, (x-1)\big(x^2 + (a+1)x -c\big)\\[1.5ex]
g(x) \,=\, (x-1)\big(x^2 -(a+c)x -a\big)\,.$$
Denot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$h^2 = x^2 + (x+1)^4$ This is a question I had of finding an exact value of $h$ as I was interested in the process and techniques used. The question was one to do with Pythagoras and the math shown is the question in $a^2 + b^2 = c^2$ form.
The question was what is the size of the hypotenuse of a right-angled triangle... | \begin{align}
c^2 &= a^2 + b^2 \\[1em]
h^2 &= x^2 + \left[(x+1)^2\right]^2 \\
h^2 &= x^2 + (x+1)^4 \\
h^2 &= x^2 + x^4 + 4x^3 + 6x^2 + 4x +1\\
h^2 &= x^4 + 4x^3 + 7x^2 + 4x +1\\
h &= \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}
\end{align}
You probably put into Wolfram Alpha some other task - you wanted the square root, and you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Implicit differentiation of an equation of a hyperbola Prove that an equation of the tangent line to the graph of the hyperbola :
$(x^2/a^2) - (y^2/b^2) = 1$
at the point ($x_0$, $y_0$) is
$x x_0/a^2 - y y_0/b^2 = 1$ (1)
I implicitly differentiated the equation and then found the gradient by substituting in the ... | \begin{align}
y-y_0 & = \frac{b^2x_0}{a^2y_0}(x-x_0)
\\ a^2y_0(y-y_0)& = (b^2x_0)(x-x_0)
\\ a^2y_0y-a^2y_0^2& = b^2x_0x-b^2x_0^2
\\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y
\end{align}
$(x_0,y_0)$ satisfaces that $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$ doing the operations, $b^2x_0^2 -a^2y_0^2=a^2b^2$, then:
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Unitriangular matrix group over primary finite field I have unitriangular matrix group over a field $F_q$.
It consists of the following elements:
$$E + A = \left[\begin{matrix}
1 & a12 & ... & a1n\\
0 & 1 & a23 & ...\\
... & ... & ... & ... \\
0 & 0 & ... & 1\\
\end{matrix}\right]$$
I'm trying to p... | Try to use induction on $n$. Note that you can think $A = A_{n}$ as a block matrix
$$
A = A_{n} = \begin{pmatrix} A_{n-1} & v_{n-1} \\ \mathbf{0}_{1, n-1} & 0 \end{pmatrix}
$$
where $v_{n-1} \in \mathbb{F}_{p}^{n-1}$. Then it is not hard to check that
$$
A^{m} = \begin{pmatrix} A_{n-1}^{m} & A_{n-1}^{m-1}v_{n-1} \\ \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the minimum of $(1+a^2)(1+b^2)(1+c^2)$ where $a,b,c\geq 0$
Find the minimum of:
$$(1+a^2)(1+b^2)(1+c^2) \ \ \ a,b,c\geq 0$$
Knowing that $$ab+bc+ac=27$$
I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)... | We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$\begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\\
&=26^2+(a+b+c-abc)^2\geq 26^2=676
\end{align}$$
where the cond... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\fr... | By definition, $\arctan x=\theta$ means
$$\tan\theta=x\quad\hbox{and}\quad -\frac\pi2<\theta<\frac\pi2\ .$$
Let $\theta$ be the RHS of your equation. We have
$$-1<\frac{x}{\sqrt{1+x^2}}<1\quad
\Rightarrow\quad 0<\arccos\frac{x}{\sqrt{1+x^2}}<\pi\quad
\Rightarrow\quad -\frac\pi2<\theta<\frac\pi2\ ,$$
so the second ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty? In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty?
What if the boxes can be empty?
| Using the Polya Enumeration Theorem we get for the two cases using the
cycle index of the symmetric group with empty boxes
$$Q_1 = [R^3 B^4] Z\left(S_3;
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right)$$
and without
$$Q_2 = [R^3 B^4] Z\left(S_3; -1 +
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right).$$
Now the cycle index is
$$Z(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $
Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $
My try:
$$f(2)=53$$
$$f^{(1)}(x)=5x^4+4x^3+2 \Rightarrow f^{(1)}(2)=114$$
$$f^{(2)}(x)=20x^3+12x^2 \Rightarrow f^{(2)}(2)=208$$
$$f^{(3)}(x)=60x^2+24x \Rightarrow f^{... | You have polynomials on both sides of $$x^5+x^4+2x+1=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$$
Thus the two sides match on the entire range of $(-\infty, \infty).$
You do not have to worry about the interval of convergence of either side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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maximum value of $\sum (a-b)^2$
If $a^2+b^2+c^2=5$ and $a,b,c \in \mathbb{R},$ find the maximum
value of $(a-b)^2+(b-c)^2+(c-a)^2$.
My Try: $(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)$
$$=10-2(ab+bc+ac)$$
Now this implies $ab+bc+ac\le 5$. So
$$(a-b)^2+(b-c)^2+(c-a)^2 = 10-2(ab+bc+ac) \geq 20.$$
Could some hel... | $$\sum_{cyc}(a-b)^2=3(a^2+b^2+c^2)-(a+b+c)^2\leq15.$$
The equality occurs for $a+b+c=0,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving $a^{ab}+b^{ab}\leq a^{a^2}+b^{b^2}$ and an identity . I was working on this Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$ when I have discovered the following identity :
$$\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{8}}\pm\Big(\frac{1}{4}\Big)^{\frac{x}{8}}\Bigg|=\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{4}}\pm\Big(\... | The identity is trivial, but the inequality is not trivial.
I cannot see any important connection between these two statements.
The following section is just a hint.
Let $~f(x) := x^{x(\frac{3}{4}-x)} + (\frac{3}{4}-x)^{x(\frac{3}{4}-x)}~$ and $~g(x) := x^{x^2} + (\frac{3}{4}-x)^{ (\frac{3}{4}-x)^2}~$
with $~\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How can this equation be true? I'm currently stuck on an equation that no matter what numbers I insert to it is true, but I can't prove that is true. The equation is:
$$\frac{1}{\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=(1+\frac{vu}{c^2})*\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}*\frac{1}{\sqrt{1-\frac{v^2}{... | $$\frac{1}{\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=(1+\frac{vu}{c^2})*\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}*\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \iff$$
$${\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=\frac{1}{(1+\frac{vu}{c^2})}*{\sqrt{1-\frac{u^2}{c^2}}}*{\sqrt{1-\frac{v^2}{c^2}}} \iff$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving or disproving $12\mid x$ given $x^2+2\mid y^2-2$
Let $x$, $y$ be positive integers such that $x^2+2\mid y^2-2$. Prove or disprove that $12\mid x$.
This conclusion comes when I was dealing with another problem, and I feel it is right because when $x=12$ and $y=32$,
$$\frac{32^2-2}{12^2+2}=7.$$
Let me explain t... | If $3\nmid x$ then $$ x^2\equiv 1\pmod 3\implies x^2+2\equiv 0\pmod 3$$ so $$ 3\mid y^2-2$$ which means $2$ is a square modulo 3. But this does not hold so $3\mid x$.
If $x$ is odd, then $x^2+2 = 4k(k+1)+3$, so there exists prime $p= 8l\pm 3$ such that $p\mid x^2+2$, but then $p\mid y^2-2$ and this is not true since b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Primality test for numbers of the form $N=4 \cdot 3^n-1$ Can you provide proof or counterexample for the claim given below?
Inspired by Lucas-Lehmer primality test I have formulated the following claim:
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 4 \cdot 3^{n}-1 $ where $n\ge3$ . Let $S_i... | This is a partial answer.
This answer proves that if $N$ is prime, then $S_{n-2}\equiv 0\pmod N$.
Proof :
First of all, let us prove by induction on $i$ that
$$S_i=s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}}\tag1$$
where $s=\sqrt 2-1,t=\sqrt 2+1$ with $st=1$.
We see that $(1)$ holds for $i=0$ since
$$S_0=P_9(6)=(3-2\sqrt 2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral $\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx$
Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar.
Anyway I thought it's quite a trivial integral so I'm gonna s... | One more by Feynman’s Technique
Parameterising our integral
$$I=\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x$$
by the integral
$$
I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,
$$
where $a\in [-\frac{\pi}{2}, \frac{\pi}{2}]. $
Differentiating $I(a)$ w.r.t. $a$ yields... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
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Solve the equation $\frac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$.
Solve the equation $\dfrac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$ over the reals.
Remember when I posted a question almost identical to this one?
Well, turned out my teacher ... | Using the Bunyakovsky and AM - GM inequalities for $(\sqrt{3x^2 - 1}, \sqrt{x^2 - x}, \sqrt{x^2 + 1})$, $(1, 1, -x)$ and $(2x^2 + 4)$, $(5x^2 - x)$, we have that $$(\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1})^2 \le (1 + 1 + x^2)(3x^2 - 1 + x^2 - x + x^2 + 1)$$
$$ = (x^2 + 2)(5x^2 - x) = \dfrac{(2x^2 + 4)(5x^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
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"url": "https://math.stackexchange.com/questions/3206701",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Relation defined on the set of real numbers by xRy when $x^2 + y^2 = 1$. Show whether or not R is reflexive, symmetric, antisymmetric or transitive.
Let R be the relation defined on the set of real numbers by $xRy$
whenever $x^2 + y^2 = 1.$ Show whether or not $R$ is reflexive,
symmetric, antisymmetric or transiti... | You misunderstood. Instead of using the $R$ notation, say that $x \sim y$ if and only if $x^2+y^2=1$. So
*
*Not reflexive, as we have elements $x$ such that $x \not\sim x$. For example, $1 \not\sim 1$, as $1^2+1^2 = 2 \neq 1$.
*It is symmetric, because if $x \sim y$, then $x^2+y^2 = 1$. So $y^2+x^2=1$ says that $y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be? $a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be?
I tried using $(a^2 + b^2)^2$ but I don't know what to do after.
| You have $a^2-ba^2+ba-b^2=0$ then $a=\dfrac{-b\pm\sqrt{b^2-4(1-b)(-b^2)}}{2(1-b)}$ so that $5-4b$ must be a square. The only possible value is $b=1$ thus $a=1$ and $a^4+b^4=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.