Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to evaluate $\iiint_{\mathbb{R}^3} \exp \left(-x^2 -y^2-z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z$ How to evaluate
$$
\iiint_{\mathbb{R}^3} \exp \left( -x^2 -y^2 -z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z,
$$
My Attempt:
If we use the transformation to spherical polar coordinates $R, \theta, \phi$, th... | The integral $\int_{0}^{\infty}e^{-R^2}dR$ is the Gaussian Integral.
It is very well known, the wikipedia link I have given above also contains a derivation. It equals-
$\int_{0}^{\infty}e^{-R^2}dR=\frac{\sqrt{\pi}}{2}$
So for your original integral,
$\begin{align}&\int\int\int_{\mathbb{R}^3}e^{-(x^2+y^2+z^2)}dxdydz\\&... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the substitution be different in this proof of a Fermat theorem? I was reading one of the Fermat theorem proof from the wikipedia
Specifically the theorem: " If $2^{k}+1$ is an odd prime, then $k$ is a power of 2."
In the proof it says:
Substituting $a=2^{r}$,$b=-1$,$b=-1$, $m = s$ and using that $s$ is
odd
I ori... | If $k\in \Bbb Z^+$ and if $k$ is not a power of $2$ then $k\ge 3$ and $k$ has an odd divisor $p>1.$ So $k=k'p$ with $p$ odd and $p\ge 3$ and $k'\in \Bbb Z^+.$ Then, because $p$ is odd, we have $$2^k+1=2^{pk'}+1=(2^{k'})^p+1\equiv (-1)^p+1\equiv (-1)+1\equiv 0 \pmod {2^{k'}+1}$$ so $2^{k'}+1$ is a divisor of $2^k+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$ Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
My Attempt:
On rationalizing both fractions separately and then adding (since denominator becomes $\... | Hint
$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$
$$\implies\sqrt2-\sqrt{2-\sqrt3}=\dfrac{2-(\sqrt3-1)}{\sqrt2}$$
$$\implies\dfrac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}=\sqrt{\dfrac23}\cdot\dfrac{2-\sqrt3}{\sqrt3-1}$$
$$\dfrac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}=?$$
Now take the sum
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my method correct to answering this question? Is there a quicker method to solve this question?
What is the smallest number that can be written as the sum of three, four and five consecutive numbers?
I encountered this question while doing my Math summer homework. I have tried to make progress on this question.
Su... | Your reasoning contains a subtle (essentially notational) error, which does not greatly affect the result you obtain. You say:
The number must be the lowest common multiple of $3x+3$, $4x+6$ and $5x+10$, which is $60x + 30$.
But the value of $x$ in each of $3x+3$, $4x+6$ and $5x+10$ is not the same; by your own resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relationship between Chebyshev polynomials and square roots ($\sqrt{3}+\sqrt{2}=\frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}}$ etc.) (If my English is strange, I would appreciate it if you could correct it.)
There seems to be a property about the sum of square roots. (This is almost self-explanatory.)
let $ a,\ b,\ t \in \... | According to wiki, $T_n(x)$ can be expressed by the following equation too.
$$ T_n(x) = \frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{2} $$
Hence,
$$ T_{2^{t-1}}(a+b) = \frac{(a+b+\sqrt{(a+b)^2-1})^{2^{t-1}}+(a+b-\sqrt{(a+b)^2-1})^{2^{t-1}}}{2} $$
On the other hands,
\begin{eqnarray*}
B &=& \frac... | {
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"url": "https://math.stackexchange.com/questions/4216274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Decomposing a large block $7\times 4 \times 5$ into small ones $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$ Consider a block of size $7\times 4 \times 5$. We want to make this with cubes of size $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$. If $x,\ y,\ z$ cubes are needed respectivel... | As noted in the comments, there is a valid configuration with $(x,y,z)=(22,8,2)$.
Claim:$\;$Every valid configuration $(x,y,z)$ has $x\ge 22$.
Proof:
Suppose $(x,y,z)$ is a valid configuration.
Summing the volumes, we get $x+8y+27z=140$, hence $x\equiv 4-3z\;(\text{mod}\;8)$.
As you noted, we must have $z\in\{0,1,... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqr... |
You have done most of the work by coming up with the nice construction. I will just mark point $Q$, draw chord $CQ$ and use it to find the answer.
As $BP$ and $AQ$ are parallel, $\angle AFB = \angle FBP = 45^0$
So $\angle CFQ = 45^0$ and as $\angle CQF = 90^0$, $CQ = \cfrac{a}{\sqrt2}$
But also note that, $\angle CAQ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$ For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$
My solution :
let $$p = \alpha + \alpha^2 + \alpha^4$$ and $$q = \alpha^3 + \alpha^5 + \alpha^6.$$
We know $$1 + \... | Yet another way (albeit not too different from OP's), using that $\,|\alpha| = 1\,$, $\,\overline \alpha=\dfrac{1}{\alpha}\,$:
*
*$\require{cancel}\displaystyle \;p+\overline p = \alpha + \alpha^2+\alpha^4 + \overline \alpha + \overline \alpha^2 + \overline \alpha^4 =\alpha + \alpha^2+\alpha^4 + \frac{1}{\alpha}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin... | Here's an alternative that take advantage of convexity of sine in $(0,\pi)$
$$
\begin{align}
\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)\cdot 2\cdot\sin{\frac{A/2+B/2+C/2}{3}}&\geq\frac{2}{3}\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)^{2}\\
&\geq 2\left(\sin{\frac{A}{2}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write a linear system in a new basis thanks for spending some time helping :)
How exactly do you represent a linear system in a different basis?
Let's say the new basis B = {(1,1),(-1,1)}
How do I write the system
\begin{cases} x+y=7\\ 2x–4y=-16 \end{cases}
in the new basis?
| In terms of matrices your system is
$$\left[\begin{array}{cc}1&1\\2&-4\end{array}\right]
\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}7\\-16\end{array}\right],$$
but if your basis change is
$$v_1=e_1+e_2,$$
$$v_2=-e_1+e_2,$$
then the old basis in terms of the new one is expressed as
$$e_1=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the figure, the polygons shown are regular. Find $"x"$. For reference:
my progress:
I marked the angles I could find but couldn't finish
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle... | Lengthen $BA$ to the left and find the angles those it makes with $AM$ and $AL$. Then using them find $\angle MAL$ and $\angle LAN$. Then you can find $\angle ANL$ and eventually $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \... | First, reformulate the problem as the following:
$$
y^2 - 9x^2 = p \Rightarrow(y-3x)(y+3x)= p
$$
Now, for any given $p$, find its prime factors. Then, for any 2-partitions of them, solve a simple equation system.
To simplify some cases, suppose $p$ is factorized to $p_1 \times p_2$. Now, solve the following system:
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y... | It is not surprizing that Mathematica does not produce explicit expressions for $(x,y,z)$.
Using it with the restriction $x>y>z$, it gives that $y$ and $z$ are respectively the fourth and the first roots of the equation
$$56 t^6-336 t^5+378 t^4+819 t^3-2079 t^2+1512 t-351=0$$
Using @River Li's comment, $(x,y,z)$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}... | Rewrite the limmand as
$$\frac{n}{\log n}\left(n^{\frac{1}{n}}-1\right) = \frac{n}{\log n}\left(e^{\frac{\log n}{n}}-1\right)$$
Then with $x = \frac{\log n}{n}$ we have
$$\lim_{x \to 0^+} \frac{e^x-1}{x} = 1 $$
the classic limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Differential equation $y' = (2y^2 + x)/(3y^2 + 5)$ (Apostol, section 8.28, ex. 30) Problem
This is from Apostol's Calculus book, section 8.28, exercise 30.
Let $y = f(x)$ be that solution of the differential equation
$$y' = \dfrac{2y^2 + x}{3y^2 + 5}$$
which satisfies the initial condition $f(0) = 0$.
(a) The differen... | Based on Paul Sinclair's suggestion in the comments to the question, here is an attempt at a new argument for part (d).
We found that $\lim_{x \to +\infty} y' = 2/3$. This means that, given $\epsilon > 0$, there is a $N > 0$ such that, if $x > N$, then
$$\left|y' - \dfrac{2}{3}\right| < \epsilon$$
This implies
$$-\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of $\left(1+\frac1n\right)^n<3-\frac3n$ for $n\geq7$? We have the following inequality:
$$e_n:=\left(1+\frac1n\right)^n<3-\frac3n,$$
for $n\in\mathbb{N}$ and $n\geq7$.
My proof as follows: $e_1=2<2.75,e_2=2.25<2.75$, and for $\forall n\in\mathbb{N},n\geq2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&=2+\sum_{k=... | This isn't exactly elementary but does the job. We wish to show that $f(x)<3$ for all real $x\ge7$ where $f(x)=(1+1/x)^x/(1-1/x)$. Its derivative is given by $$f'(x)=g(x)\left(x(x^2-1)\log\left(1+1/x\right)-x^2-1\right)$$ where $g(x)=(1+1/x)^x/((x-1)^2(x+1))$ is positive. Since $$x(x^2-1)\log(1+1/x)-x^2-1<(x^2-1)(x\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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6-dimensional integral with strange result I am working through a physics paper where I encounter the following 6-dimensional integral:
$$ G(\mathbf{q}) = \frac{2}{(2\pi)^6 n^2} \int_{k \leq q_F} \int_{k' \leq q_F} \frac{\mathbf{q} \cdot (\mathbf{q} + \mathbf{k} -\mathbf{k}')}{|\mathbf{q} + \mathbf{k} -\mathbf{k}'|^2} ... | So for anyone still interested or coming across this later, here is the solution. A colleague and I finally managed to solve it. It is however quite long and tedious, so I will only present the main ideas and the end results.
We start with a transformation to $\mathbf{r} = \frac{\mathbf{k} + \mathbf{k}'}{2}$ and $\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute $\frac 17$ in $\Bbb{Z}_3$
Compute $\frac 17$ in $\Bbb{Z}_3.$
We will have to solve $7x\equiv 1\pmod p,~~p=3.$
*
*We get $x\equiv 1\pmod 3.$
*Then $x\equiv 1+3a_1\pmod 9,$ so $7(1+3a_1)\equiv 1 \pmod 9$ basically lifting the exponent of $p=3,$ we get $1+3a_1\equiv 4\pmod 9\implies a_1\equiv 1\pmod 3.$
*So ... | You are doing correct.
For a slightly more efficient approach, note that $7\mid 3^6-1$ by Fermat's little theorem. Then by noting that $3^6-1=7\cdot104$, we have
\begin{align*}
\frac{1}{7}
&= -\frac{104}{1 - 3^6} \\
&= -\sum_{n=0}^{\infty}104 \cdot 3^{6n} \\
&= 1 + \sum_{n=0}^{\infty} (3^{6} - 1)3^{6n} - \sum_{n=0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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What is the value of the PM segment in the figure below? For reference:
Given a rectangle, ABCD (AB > BC) by B draws a perpendicular to AC which
intercepts "P" to CD, and an "M" the perpendicular to BD drawn by B.
$\space$ The prolongation of DM and BC intersect at Q.
Calculate PM if DQ =$ 17$ and BP = $9$. (answer: $4... |
Hints: If you show that In figure M is midpoint of QT and DM=PM, then we have:
In right angled triangle TBQ, BM=MQ so:
$17-PM=9+PM\Rightarrow PM=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A method to derive the quadratic formula I wonder if this is a valid method to derive the quadratic formula.
It is not an optimal method at all, but I am intrigued if it is a genuine way or has any gap.
The method starts from the equation whose coefficients are such that it is not necessary to use the quadratic formula... | [Too long for a comment]
There is another way to derive the quadratic formula. But it is so complicated, it is almost barely usable.
Part 1
We begin with: $$x^2-px+q=0.$$ Let's assume this equation is of the form $A(x-\alpha)(x-\beta)$ for some constant $A$ and roots $\alpha$, $\beta$. So $$x^2-px+q=A(x-\alpha)(x-\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Transform ODE $(u-x)u_x + u + x = 0$ to polar coordinates According to Peter Olver in his book “Applications of Lie Groups to Differential Equations”, p. 104, Example 2.32, the ODE:
$$(u-x)u_x + u + x = 0 \tag{1}$$
is transformed in polar coordinates with $x = r\cos \theta$ and $u = r\sin\theta$ to:
$$\frac{dr}{d\theta... | $$(u-x)u_x + u + x = 0 \tag{1}$$
$$-xu_x + u =- x -uu_x$$
$$\left(\dfrac u x \right)' =\dfrac 1 {x^2}(x+uu_x)$$
$$\left(\dfrac u x\right)' =\dfrac 1 {2x^2}(x^2+u^2)'$$
Multiply both sides by $ \dfrac {dx}{d\theta}$:
$$\dfrac {d}{d \theta}(\tan \theta )=\dfrac 1 {2r^2 \cos^2 \theta}\dfrac {dr^2}{d\theta }$$
$$\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for the value of $a,b$ in $f(x)=(ax+b)(x^5+1)-(5x+1)$ s.t. $(x^2+1)|f(x)$
$Q.$ If $f(x)=(ax+b)(x^5+1)-5x-1$ is divisible by $x^2+1$ . Then the value of $2a+3b$ $?$
MY APPROACH : We have , $(x^2+1)|f(x)$ then $(x-i)|f(x)$ and $(x+i)|f(x)$ .
So by Factor Theorem we have $f(-i)=0$ and $f(i)=0$ $$f(i)=-a+bi+(a-5... | We have
$$\begin{align}&ax^6+bx^5+x(a-5)+(b-1)=P(x)(x^2+1)+mx+n \\\\
\implies &\begin{cases} mi+n=-a+bi+i(a-5)+(b-1)\\-mi+n=-a-bi-i(a-5)+(b-1)\end{cases}\\\\
\implies &\begin{cases} n=\frac{2(b-1)-2a}{2}=b-a-1\\m=\frac{2bi+2i(a-5)}{2i}=a+b-5\\
\end{cases}\\\\
\implies&\begin{cases}n=0,\thinspace b-a=1\\m=0,\thinspace a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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If $(f+g)(x)=4$ and $(f\circ g)(x)=7-4x$ What is the sum of the possible values for $g(2)$?
If $f$ and $g$ are polynomial functions so that $(f+g)(x)=4$ and
$(f\circ g)(x)=7-4x$, what is the sum of the possible values for
$g(2)$?
$1)2\qquad\qquad\qquad2)3\qquad\qquad\qquad3)-3\qquad\qquad\qquad4)-2$
To solve this pro... | Using the condition that $f(x) + g(x) = 4$, $f(x) = ax + b, \ g(x) = -ax + (4-b)$, which results in:
$$f(g(x)) = f(-ax + 4 - b) = a(-ax+4-b) + b = 7 - 4x$$
hence $-a^2x = -4x \implies a = \pm 2$ and $a(4-b)+b = 7 \implies b = 1, 5$. Thus $g(x) = -2x+3$ or $2x-1$, so the sum of the possible values of $g(2)$ is $-1 + 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But i... | Hint. From $\cos (A+B)=\cos A\cos B-\sin A\sin B$ with $B=A$ we have $\cos 2A=\cos^2A-\sin^2A=(1-\sin^2A)-\sin^2 A=1-2\sin^2 A.$ Hence $2\sin^2 A=1-\cos 2A.$ With $A=2x$ therefore $$2\sin^2(x/2)=1-\cos x.$$ So if $\cos x\ne 1$ then $$\frac {x\tan^{-1}x}{1-\cos x}=\frac {x\tan^{-1}x}{2\sin^2(x/2)}=$$ $$=\left(\frac {(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Suppose $a_1=1$ and $a_n=\sqrt{n-a_{n-1}}$ $\forall n \in\mathbb{Z_{\ge 2}}$. What is the least integer $n$ for which $a_n>4$? ⌊$a_{100}$⌋? Suppose $a_1=1$ and $a_n=\sqrt{n-a_{n-1}}$ $\forall n \in\mathbb{Z_{\ge 2}}$. What is the least integer $n$ for which $a_n>4$? What is $\left \lfloor a_{100} \right \rfloor$, where... | Your second question
*
*It's easy to see that $\forall n \geqslant 2 \ , \ a_n \leqslant \sqrt{n}\ $ . So $a_{100} < \sqrt{100} = 10$ and $a_{99} < \sqrt{99} < 10 $.
*Then $\ \ a_{100} = \sqrt{100 -a_{99}} > \sqrt{100-10} > 9 \ $ .
*So $\ \fbox{$\ \left\lfloor a_{100} \right\rfloor = 9\ $}\ $ .
Your first questio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Writing $ \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}} =\frac{m}{n}\left( H_n-H_{m-n} \right) $ Given the following series, where $m,n$ are integers such that $n<m$:
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\left( H_n-H_{m-n} \right)
\end{align}
We know that harmonic number $H_n = 1 + \frac... |
We obtain for positive integers $m>n$
\begin{align*}
\color{blue}{\frac{1}{n}\sum_{i=0}^{n-1}\frac{m}{m-i}}
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-i}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-(n-1-i)}\tag{1}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-n+1+i}\\
&=\frac{m}{n}\sum_{i=m-n+1}^m\frac{1}{i}\tag{2}\\
&=\frac{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Improper integral $\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$ I am having trouble solving
$$\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$$
let $\ln(x)=u$ then $x du= dx $
$$\Rightarrow \int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx=\int_0^{\infty} \frac{x+1-1}{(x+1)(\pi^2+u^2)}du=\int_0^{\infty} \frac... | Enforcing the substitution $x\mapsto e^{-x}$ reveals
$$\int_0^\infty \frac1{(x+1)(\pi^2+\log^2(x))}\,dx=\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx$$
Then, note that
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx&=\int_{-\infty}^0 \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx+ \int_0^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, ... | $\dfrac{n}{2n+1} = \dfrac{1}{2} \left( \dfrac{2n}{2n+1} \right) = \dfrac{1}{2} \left( \dfrac{2n+1 - 1}{2n+1} \right) = \dfrac{1}{2} \left(1 - \dfrac{1}{2n+1} \right)$
Since $ \dfrac{1}{2n+1} $ decreases with increasing $n$, then $ \left( 1 - \dfrac{1}{2n+1} \right)$ is increasing with $n$, thus $\dfrac{n}{2n+1} $ incre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
formular for sum of column in a Matrix I have a trivial problem. I need to create a vector, which has the sum of the columns in a AxB matrix. What would be the mathematical formula for that?
Thank you
| Hint: The following examples might be helpful.
\begin{align*}
\begin{pmatrix}a&b&c\\d&e&f\end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
&=\begin{pmatrix}a+b+c\\d+e+f\end{pmatrix}\\
\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\end{pmatrix}
&=\begin{pmatrix}a+d&b+e&c+f\end{pmatrix}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem in normalization inequalities Let $a,b,c>0$. Prove that: $$\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64}{27}(a+b+c)^3$$
First solution:
$\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=3$, we need to prove:
$$(a+b)(b+c)(c+a)\ge8\sqrt{abc}$$
$\bullet$ Notice that, $a+b+c=3$, we may hav... |
*Yes, of course.
For example, after full expanding we need to prove that:
$$\sum_{sym}(27a^4b^2-5a^4bc+27a^3b^3-30a^3b^2c-19a^2b^2c^2)\geq0,$$ which is true by Muirhead.
Also, we can prove that the inequality $$(a+b)^2(a+c)^2(b+c)^2\geq\frac{64}{27}abc(a+b+c)^3$$ is true for any reals $a$, $b$ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How can I find the Laurent series expansion How can I find the Laurent series expansion for
$$\frac{z^2-1}{z^2+1} $$ with $|z|>2$.
$$\frac{z^2-1}{z^2+1} =\frac{z^2+1-2}{z^2+1}=1-\frac2{z^2+1}$$
The fraction on the right-hand-side can be expressed as $$\frac1{z^2+1}=\frac i{2(z+i)}+\frac{-i}{2(z-i)}$$
also
$$\frac 1{z+i... | Many exercises concerning the computation of Laurent expansions can simply be solved by making use of the geometric series
$$\sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$
However, the important point is that the above identity is valid only for $x \in \mathbb{C}$ satisfying $|x| <1$. Since you are working under the conditio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
I managed to show this with a 3 instead of a 5 :
The inequality is symmetric in $a$, $b$, $c$ so... | Hint: Show that for non-negative real $ a \leq b$,
$$ \min \{ a^2, (b-a)^2 \} \leq \frac{ a^2 + b^2 } { 5 },$$
by considering cases of $ b \geq 2a$ and $ a \leq b < 2a$.
Corollary: The original statement holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the measure of $\overset{\LARGE{\frown}}{AB}$ in the figure below? For reference:
In the figure calculate $\overset{\LARGE{\frown}}{AB}$ if $ \overset{\LARGE{\frown}}{BC} = 90^o$
My progress:
Relationships I found:
$FO$ is angle bissector
$\triangle OBC(isosceles):\measuredangle OCJ=\measuredangle OBJ=45^o\\
... | $Let BD = a ~and~ r = radius\\
AODJ~ is~ a ~square\\
BC = r\sqrt2\\
Power ~of~ Point~ Theorem :a(a+\sqrt2) = r^2 \rightarrow r^2-ar\sqrt{2}-a^2=0\\
\therefore r=\frac{a\sqrt{2}+-a\sqrt{6}}{2}\implies \frac{r}{a}=\frac{\sqrt{6}+\sqrt{2}}{2}\\
law ~of~ sines:\frac{r}{a}=\frac{sen(45+\frac{x}{2})}{sen(\frac{x}{2})}\\
\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the mistake in my reasoning in calculating limit where $x \to -\infty$ I had this question in an exam today:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x}
$$
My question is: I have tried to solve it and I also found two other methods to solve it, one in my textbook and one from somebody el... | Exact explanations were made about the mistake you made in the comments. You missed that $x$ is negative.
You can observe this simple fact:
$$\begin{align}x=\begin{cases}-\sqrt{x^2},~\text{if}~ x<0\\\sqrt{x^2}, ~\text{if}~x>0\end{cases}\end{align}$$
This immediately reveals your mistake.
I would calculate this limit as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the area of a region in an isosceles triangle that contains squares of area $256$ and $49$ There was a quiz posted on F*cebook by someone. Here's the problem.
And here's my attempt:
First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green squar... | Let $\alpha$ is the base angle. Then the area of one (bigger) blue triangle is $\frac{16^2}{2} \tan (2\pi-2\alpha)$, the area of the second blue triangle is $\frac{7^2}{2} \tan \alpha$. On the other hand, $\tan \alpha=\frac{16\tan (2\pi-2 \alpha)+9}{7}$ or (using double-angle formula and considering that $\tan (2\pi-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
finding the characteristic polynomial of $A$ given the characteristic polynomial of $A^2$ $A$ is a $3\times3$ matrix, the Jordan form of $A^2$ is $$
\begin{pmatrix}
25 & 1 & 0\\
0 & 25 & 0\\
0 & 0 & 1\\
\end{pmatrix}
$$
Find the characteristic polynomial, minimal polynomial and Jordan form of $A$ if $A$ has only negati... | Let $\lambda_1, \lambda_2, \lambda_3$ the eigen values of $A$. $$P_A(X) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$$
Since the eigen values of $A$ are $25$, $25$, $1$. So you can assume that $\lambda_1^2=\lambda_2^2=25$ and $\lambda_3^2=1$. You can find the values of $\lambda_i$ since they are negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Expected value of the area of an annulus Consider the following problem:
Let $X,Y,Z,T$ be independent random variables with standard normal
distribution ($N(0,1)$). Consider the circles with center $(0,0)$ and radius
$(X,Y)$ and $(Z,T)$, respectively. Let $A$ be equal to the area of the
annulus defined by these circle... | The length of the radius is $U := \sqrt{X^2+Y^2}$ and $U^2=X^2+Y^2\sim\chi^2(2)$ therefore the random area is $A_{X,Y}=\pi U^2\sim \Gamma(1,2\pi)$. Same for $A_{Z,T}=\pi H^2$. The characteristic function of $\pi(U^2-H^2)$ is
$$E[e^{i\pi\theta(U^2-H^2)}]=E[e^{i\pi\theta U^2}]E[e^{-i\pi\theta H^2}]=(1-2i\pi\theta)^{-1}(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$ Solve $$\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$$
I tried to:
$$\lfloor x \rfloor ^2 + 9 (x-\lfloor x \rfloor)^2=6x-10$$
$$ 18 x \lfloor x \rfloor -10 \lfloor x \rfloor^2=(3x-1)^2 +9 $$
but I got stuck from here, or I star... | Notice that
$$x\ge \lfloor x\rfloor>x-1$$
$$x^2\ge \lfloor x\rfloor ^2>x^2-2x+1$$
$$x^2+9>9\{x\}^2+\lfloor x\rfloor^2>x^2-2x+1$$
$$x^2+9>6x-10>x^2-2x+1$$
$$4+\sqrt5\approx 6.2>x>4-\sqrt5\approx 1.7$$
Hence $\lfloor x\rfloor\in\{2,3,4,5,6\}$. Now do casework based on each value of $\lfloor x\rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Some intuition behind why we visit $\frac{2}{3}$ of Natural Numbers? Consider stating at $0$ on the number line of integers. Toss a fair coin and if heads advance $1$ space and if tails advance $2$ spaces.
*
*What number has the highest probability of being visited?
*What is the probability of visiting very large nu... | The intuition behind it is very simple: the average step length is $\frac32$. Therefore you would expect to traverse a stretch of length $n$ in about $\frac{n}{3/2}=\frac23 n$ steps. So about $\frac23$ of the integers in that stretch will be visited.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$
If inverse of the function $y=\sqrt{x+\sqrt{2x-1}}$ be equal to
$y=ax^2+bx+c$, then what is the value of $a^2+b^2+c^2$?
$A) 3$
$B) 4$
$C) 5$
$D) 6$
I tried to isolate $x$ from the equation $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy... | Since we are told that the inverse function is quadratic, we know that the curve function $ \ y \ = \ \sqrt{x+\sqrt{2x-1}} \ \ $ is the "upper half" of a "horizontal" parabola (which is not difficult to verify). We can find one point on the curve by considering that its domain is $ \ x \ \ge \ \frac12 \ \ ; $ the coor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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The sum of $\frac{999}{1999} + \frac{999}{1999} \frac{998}{1998} + ...+ \frac{999}{1999} \frac{998}{1998}...\frac{1}{1001}$? Ive tried to represent $\frac{999}{1999}$ as $1 - \frac{1000}{1999}$ and so on, but it didnt lead me anywhere. I also tried to represent it as $\frac{999!}{1999!} \cdot \frac{1998!}{998!} + \frac... | It becomes easier if the terms are added from right to left.
The sum of the last two terms is
$$
\frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003}\frac{2}{1002}\left(1 + \frac{1}{1001}\right) = \frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003} \cdot \frac{2}{1001} \, .
$$
The sum of the last three terms is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the value of $ abc \cdot (a + b + c) $ Let $ a, b $ and $ c $ be numbers such that $$ \begin {cases} a ^ 2-ab = 1 \\ b ^ 2-bc = 1 \\ c ^ 2-ac = 1 \end {cases} $$Calculate the value of $ abc \cdot (a + b + c) $
Attempt: Can I solve the problem by symmetry?
| Another way.
We have $$a^2b-ab^2=b,$$ $$b^2c-c^2b=c$$ and $$c^2a-a^2c=a,$$ which gives $$\sum_{cyc}(a^2b-ab^2)=a+b+c$$ or $$(a-b)(a-c)(b-c)=a+b+c.$$
In another hand, $$\prod_{cyc}(a^2-ab)=1$$ or $$abc(a-b)(b-c)(c-a)=1,$$ which gives $$abc(a+b+c)=-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $(I-\mathbf x\mathbf y^T)^{-1} = I-\frac{1}{\mathbf x^T\mathbf y- 1}\mathbf x\mathbf y^T$ Let $x,y\in\mathbb R^n$ and suppose that $x^Ty \neq 1$. Show that $(I-\mathbf x\mathbf y^T)^{-1} = I-\frac{1}{\mathbf x^T\mathbf y- 1}\mathbf x\mathbf y^T$. Note, I need to compute this directly not as some special case... | For (1), we will use the fact that $\frac{x^Ty-1}{x^Ty-1} = 1$ and $x^Ty = y^Tx$. We also use the fact that $xy^Txy^T = (x(y^Tx)y^T) = (y^Tx)xy^T =y^Txxy^T =x^Tyxy^T $. Obser for (1) we have that
$$\begin{equation}\begin{split}\left(I-x y^T\right)\left(I-\frac{1}{x^T y- 1} x y^T\right) &= I - \frac{1}{x^T y- 1} x y^T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}... | Applying Cauchy-Schwarz inequality: $LHS = \sum_{\text{cyc}} \dfrac{a^2+b^2}{b}=\sum_{\text{cyc}}\dfrac{\left(\sqrt{a^2+b^2}\right)^2}{b}\ge\dfrac{\left(\sum_{\text{cyc}}\sqrt{a^2+b^2}\right)^2}{\sum_{\text{cyc}} a}\ge RHS\iff \sum_{\text{cyc}}\sqrt{a^2+b^2}\ge \sum_{\text{cyc}} \sqrt{2}a$, but this is true since $\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt{3}\ge2\left(\sqrt{\frac{d}{a+b+c}}+\frac{\sqrt{ab+bc+ca}}{a+b+c}\right)$
Problem: Let $a,b,c,d>0.$ Prove that: $$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt{3}\ge2\left(\sqrt{\frac{d}{a+b+c}}+\frac{\sqrt{ab+bc+ca}}{a+b+c}\right)$$
It is my old teacher's problem.
My attempt: I guess a=... | We need to prove that:
$$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d}{a+b+c}}$$ and since $$\sqrt3(a+b+c)\geq2\sqrt{ab+ac+bc},$$ by AM-GM we obtain:
$$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d\left(\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$ I want to evaluate the series
$$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$$
I can rewrite this as
$$1+\sum_{n\geq 1} (-3)^{-3n}\left(\frac{3}{6n-1}+\frac{1}{6n+1}... | For context and inspiration, let's remember that the linear factors in the power series for $\log(1+x)$ come from integrating a geometric series:
$$
\log(1+x) = \int_0^x \frac1{1+t}\,dt = \int_0^x \sum_{n=0}^\infty (-1)^n t^n \,dt = \sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n+1}.
$$
So if we recognize our series as a spe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Cantor function integral. I'm trying to solve the following integral for a problem about fractals involving Cantor set:
$$\mathcal{I}=\int_{0}^{1}C\left(\sqrt{1-x^2}\right)dx$$
Where $C(x)$ denotes the Cantor ternary function.
I don't know a lot about measure theory and Lebesgue integrals, so I don't know what to do he... | Partial solution.
Put $t = \sqrt{1-x^2}$. Hence $x = \sqrt{1-t^2}$ and
$\mathcal{I} = \int_0^1 \frac{C(t) tdt}{\sqrt{1-t^2}}$.
Open Cantor set $A = \bigsqcup_{n \ge1, 1 \le k \le 2^{n-1}} (a_{nk}, b_{nk})$, where $(a_{11}, b_{11}) = (\frac13, \frac23)$, $(a_{21}, b_{21}) = (\frac19, \frac29)$, $(a_{22}, b_{22}) = (\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\int_{0}^{2\pi}\frac{d\phi}{2+\sin(\phi)}$ with complex exponentials I’m working through the textbook A First Course in Complex Analysis and I’m trying to solve problem 4.30:
Compute the real integral
$$\int_{0}^{2\pi}\frac{d\varphi}{2+\sin(\varphi)}$$
by writing the sine function in terms of the complex expo... | Here is another approach that relies on Calculus II.
We have
$$
I=\int_0^{2\pi}\frac{dx}{2+\sin x}=\int_{0}^{\pi}\frac{dx}{2+\sin x}+\int_{\pi}^{2\pi}\frac{dx}{2+\sin x}
$$
Setting $y=x-\pi/2, z=x-3\pi/2$ in the first and second integral, respectively, we get
\begin{eqnarray}
I&=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\angle CAD$ if $\triangle ABC$ is right angled at $B$, $\angle BAD = 30^\circ, \angle ADB = \angle ADC = 15^\circ$ Find angle $\theta$ in the below diagram.
This is a question that was brought to me by a high school student.
While I came up with a trigonometric solution and a synthetic solution, I am posting her... | Sine rule for $\Delta ABD$:
$$\frac{AB}{\sin 15^\circ}=\frac{BD}{\sin 30^\circ} \Rightarrow BD=\frac{AB}{2\sin 15^\circ}$$
Sine rule for $\Delta BCD$:
$$\frac{BC}{\sin 30^\circ}=\frac{BD}{\sin 105^\circ} \Rightarrow BC=\frac{BD}{2\sin 105^\circ}=\frac{BD}{2\cos 15^\circ}=\frac{AB}{4\sin 15^\circ\cos 15^\circ}=AB$$
Henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Discrete calculus: is my proof about the difference of two consecutive powers correct? Theorem statement: $\displaystyle \Delta^dF_n(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^{d-1}F_{n-1-k}(x)(-1)^k$
Proof:
$$\begin{align} F_n(x) &= x^n =\Delta^0F(x) \\[2ex] \Delta^1F_n(x) &= x^n -(x-1)^n \\[2ex] &= x^n - \sum_{k=0}... | There is a mistake early on:
\begin{align*}
x^n - (x - 1)^n &= x^n - \left(x^n - nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} - \cdots \right) \\
&= nx^{n-1} - \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \\
&= \sum_{i = 0}^{n-1} \binom{n}{i}(-1)^{n-i+1}x^i \\
\end{align*}
Whereas you said this is $x^{n-1} -(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Rationalizing $\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ in two ways gives different answers I have a doubt see when we rationalize denominator of expression $$\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}- \sqrt{1-\cos x}}$$
we get answer $$\frac{1+\sin x}{\cos x}$$ but when we ... | Putting $a$ for $\cos x$,
you have
(notice the MathJax)
$\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}
$
I now do the rationalizing
(using
$(u+v)(u-v)=u^2-v^2)$
and later substitute
$a = \cos(x)$
and use
$\sin^2(x)+\cos^2(x) = 1$.
I am doing this
in excruciating detail
so you can see
all the steps involved.
Once... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label... | Calculating the given limit is standard, when you prove that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $. Let's focus on the nontrivial part of the problem. Using $ g ( 1 ) = 0 $,
$$ g ( x + y ) = g ( x ) + g ( y ) \text , \tag 0 \label 0 $$
and
$$ g \left ( \frac 1 x \right ) = \frac { g ( x ) } { x ^ 4 } \text , ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4298129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Permutations of the set $\{1,\ldots,n\}$ are there such that, no two of the numbers $1,2,3$ are consecutive
How many permutations of the set $\{1,\ldots,n\}$ are there such that no two of the numbers $1,2,3$ are consecutive?
My first thought was the inclusion-exclusion formula, like here, but it got messy, so I decid... | I will solve the problem in two ways.
Method 1: We use the Inclusion-Exclusion Principle.
There are $n!$ permutations of the elements of the set $\{1, 2, 3, \ldots, n\}$. From these, we must exclude those permutations in which at least two of the numbers $1, 2, 3$ are consecutive.
Since there are only three numbers i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determinant of matrix with $2$'s and the pattern $3\ 1\ 3$
Let for $n\ge 1, D_n$ be the determinant of the $n\times n$ matrix $A_n$ where the entries along the main diagonal (i.e. of the form $(i,i)$ for $1\leq i\leq n$) are all $3$, entries of the form $(i,i+1)$ are equal to $1$, entries of the form $(i, i+2)$ are eq... | Let $e$ be the vector of ones and $N$ be the full-sized nilpotent Jordan block. Since $A_n$ is a rank-one update of $B_n=I-N+N^2$, by Sylvester's determinant lemma we get
$$
\det(A_n)=\det(B_n+2ee^T)=\det(B_n)(1+2e^TB_n^{-1}e)=1+2e^TB_n^{-1}e.
$$
So, we need to find the sum of all entries in $B_n^{-1}$. Since $B_n$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from bo... | Be careful to square both side,
Make sure both positive...
.
Here my solution.,
Hope this could be referenced....
${x}+\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}=\mathrm{0} \\ $
$\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{4}{x}+\mathrm{1}}+\mathrm{3}=\mathrm{0} \\ $
$\left(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using su... | Rather than trying to do a reduction formula on the original integrand, first perform the substitution $$y = x^3, \quad dy = 3x^2 \, dx.$$ Then for an integrand of the form $$f_n(x) = x^{3n+2} e^{-x^3}$$ we have
$$\int f_n(x) \, dx = \frac{1}{3}\int y^n e^{-y} \, dy.$$ This integral is now amenable to integration by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
prove that $\int_0^\infty \frac{\sin^2 x-x\sin x}{x^3} \, dx= \frac{1}{2} - \ln 2$
Prove that
$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$
Integration by parts gives
\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl(... | $$I=\int \frac{\sin^2 (x)-x\sin (x)}{x^3}\, dx=\int \frac{\sin^2 (x)}{x^3}\, dx-\int \frac{\sin (x)}{x^2}\, dx$$
$$\int \frac{\sin^2 (x)}{x^3}\, dx=\frac 1 2\int \frac{1-\cos(2x)}{x^3}\,dx$$
$$\int \frac{1-\cos(2x)}{x^3}\,dx=-\frac 1{2x^2}-\int \frac{\cos(2x)}{x^3}\,dx$$
$$\int \frac{\cos(2x)}{x^3}\,dx=-\frac{\cos(2x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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What's the measure of the $\angle BAC$ in the triangle below? For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.
My progress:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: L... | Just a note: This is a method for reference, it approximates your angle answer:
We know that:
$\tan(\frac{c}{2}) = \frac{BF}{x}$
$\tan(90-c) = \frac{x+BF}{x} = 1 + \frac{BF}{x} = 1+\tan(\frac{c}{2})$
Using the relation between sine, cosine and tangent:
$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sin(\frac{c}{2})}{\cos(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to evaluate $I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$ Let's assume that $r>a>0$ and $0<\alpha<\pi$, to evaluate the following integral:
$$I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$$
I used a change of variable $u=\sqrt... | For $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha$
$$
I = \cos\alpha \int_{0}^{\pi} \frac{\cos\theta\sin\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta - \sin\alpha \int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta
$$
The first integral is trivial, here I fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{ali... | Let $y=\sqrt{1-x^2}$, then $ydy=-xdx$.
Noticing that
$$d(x-y)= dx-\left(-\frac{x}{y} d x\right)=\frac{(x+y)dx}{y} \Rightarrow \frac{d(x-y)}{x+y}=\frac{d x}{\sqrt{1-x^2}}, $$
we split the integral into 2 parts as below.
$$
\begin{aligned}
I &=\int \frac{d x}{x+y} \\
&\left.=\frac{1}{2} \int \frac{(d x+d y)+(d x-d y)}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$ Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) ... | You are correct. Let $$f(x,y) = g + \frac{1}{g} + h + \frac{1}{h}$$ where $$g(x,y) = \frac{\sin x}{\cos y}, \quad h(x,y) = \frac{\cos x}{\sin y}.$$ Then since $0 \le x, y \le \pi/2$, we must have $0 \le \sin x, \cos x, \sin y, \cos y \ge 1$, hence $g, h \ge 0$. Then AM-GM proves $g + 1/g \ge 2$, as well as $h + 1/h ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What´s the value of the area of the triangle below? For reference: The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of the triangle (Answer:$9$)
My progress...
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipul... |
If $a = \sqrt{18}, b = \sqrt{20}, c = \sqrt{26}$, first note that $a^2 + b^2 \gt c^2$ where $c$ is the longest side. That means it is an acute-angled triangle and the orthocenter is inside the triangle. So if we drop a perp from vertex $A$, then using Pythagoras,
$AD^2 = 20 - x^2 = 26 - (\sqrt{18} - x)^2$
$ \implies x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Global extreme points of function wrt. two variables Hi I am currently working on a problem in theoretical chemistry and am struggling a bit. While discussing Born-Oppenheimer energy functions, I was asked to find all the extreme points and saddle points of the follwing function:
I honestly didn't encounter such a pro... | We will use these extrema in two varaibles notes as a guide.
We have
$$U(x, y) = (x\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2+3\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right)$$
Finding critical points
$$f_x = \left(x^3-x^2-2 x\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right) = 0\\ f_y = \left(\frac{x^4}{4}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality in cyclic order : $\sum\frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum\frac{8}{a+3}$ Prove:$$\sum_{cyc} \frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum_{cyc} \frac{8}{a+3}$$ where $a,~b,~c$ are positive real numbers.
My thought: I think Holder's inequality will be used. But can't understand how to start? Please give any... | Hint: Using A.M-G.M inequality,
$$(a+b)^2\le2(a^2+b^2)~~\text{and}~~ 4abc\le2c(a^2+b^2) $$
$$\therefore\; (a+b)^2+4abc\le 2(a^2+b^2)(c+1) $$
It remains to prove,
$$\sum_{cyc}\left( \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\right)\ge \sum_{cyc} \frac{8}{a+3}$$
$$ \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\ge {\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the generating function of $f(n) = \sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k}$ I want to find the generating function of $f(n) = \sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k}$, where $C_k$ is the $k$-th Catalan number. So, using the definition of an ordinary generating function:
$$F(x)=\sum_{n \ge 0} \biggl(\sum_{k... | \begin{align}
F(x) &= \sum_{n \ge 0} \left(\sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k} \right)x^n \\
&= \sum_{k \ge 0} (-1)^k C_{k} \sum_{n \ge k} \binom{n}{k} (-x)^n \\
&= \sum_{k \ge 0} (-1)^k C_{k} \frac{(-x)^k}{(1+x)^{k+1}} \\
&= \frac{1}{1+x}\sum_{k \ge 0} C_{k} \left(\frac{x}{1+x}\right)^k \\
&= \frac{1}{1+x}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate some integrals with hypergeometric function There are some interesting integral problems:
$$\int_{0}^{\infty} \cos\left (\frac{x}{2} \right ) \left ( \cosh(x)-\frac{4x^{3/2}{}_1F_2\left ( 1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4} \right ) }{3\sqrt{\pi} } \right )\text{d}x
=\frac{4}{5}.$$
$$\int_{0}^{\infty}... | I present an approach using a well-known property of the Laplace Transform, which is typically a good approach for such integrals.
For the first integral:
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A composite function problem The question is:
Suppose $$f(x) = x^2+1,$$ $$g(x) = 3-x.$$
Find the values for $x$ for such that $$(g\circ f)(x) = (f \circ g)(x).$$
I tried banging my head for one hour but my answer doesn't match the one given by the book which is $1/\sqrt{2}$ and $-1/\sqrt{2}$.
I think the answer given i... | Given that $f(x) = x^2 + 1$ and $g(x) = 3 -x$, the composite function should look as follows for the LHS of the equation, hence,
$$3-(x^2 + 1)$$
RHS,
$$(3-x)^2 + 1$$
Thus,
$$ 3-(x^2 + 1) = (3-x)^2 + 1$$.
Simplifying we get,
$$2x^2-6x+8 = 0\\ x^2 - 3x + 4 = 0.$$
In order to find the value's of $x$ satisfying the equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
limit, quotient of roots How can I find
$$
\lim_{n \to \infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \sqrt[6]{n - 3}?
$$
If I multiply by $\sqrt{n + 2} + \sqrt{n + 1}$ I could get no divisor, but I cannot get any result on this way. On the other hand I could also transform this expressio... | Your idea of multiplying and dividing by $\sqrt{n+2}+\sqrt{n+1}$ (or by the conjugate) is a good one, and we can do the same thing for the conjugate of the numerator. Since the denominator had square roots, its conjugate came from the difference of squares formula
$$a-b = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$$
Simila... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Ways to find basis of a span of matrices
let $W= $$\left(\begin{matrix}
b & -a & 3a+b \\
c & 0 & a+2c \\
-3c & c & b \\
\end{matrix} \right)
$ $ \in M_3( \Bbb R)$ and $a,b,c \in \Bbb R$ I would like to know ways to find the basis of W using basic methods so I can have more options while practicing.
for this case I f... | From the matrices
$$\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
,\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
,
\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$$
you ought to choose a linear combination
$$K\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ball drawing problem with a condition (with replacing and without replacing) You have $9$ balls in a bag: $4$ blue, $2$ purple, $3$ green. If you draw $3$ balls and I tell you that you've drawn at least $1$ blue, what is the probability that you've drawn $3$ blue balls?
*
*solve for both, replacing and not replacing ... | As an alternative approach without replacement,
*
*there in a sense ${9 \choose 3} = 84$ equally likely ways of choosing three balls from nine
*${4 \choose 3}=4$ are all blue
*${5 \choose 3}=10$ are none blue
*$84-10=74$ have at least one blue
so the conditional probability you want has probability $\frac{4}{74}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you solve the equation $2^{|x+1|}-|2^x-1| = 2^x+1$? | ... | is absolute value
$2^{|x+1|}-|2^x-1| = 2^x+1$, move $|2^x-1|$ to RHS: $2^{|x+1|} = 2^x+1+|2^x-1|$
Log base 2 each side:
$|x+1| = log_2 (2^x+1+|2^x-1|)$
Here we quickly observe: $-1$ is not a solution but $0$ is a solution. If you try negative numbers $-... | If $|x+1|\ge 0$ the equation becomes
$$
2\cdot 2^x-|2^x-1|=2^x+1
$$
that is equivalent to
$$
\begin{cases}
2^x-1\ge0\\
2\cdot 2^x-2^x+1=2^x+1
\end{cases}
\quad \lor\quad
\begin{cases}
2^x-1<0\\
2\cdot 2^x+2^x-1=2^x+1
\end{cases}
$$
can you solve these two systems? and solve also the other case: for $|x+1|< 0$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate Integral of intersection between two surfaces
Calculate
$$\int_{\partial F} 2xy\, dx + x^2 \, dy + (1+x-z) \, dz$$
for the intersection of $z=x^2+y^2$ and $2x+2y+z=7$. Go clockwise with respect to the origin.
My attempt:
I first calculate the intersection, which is pretty easy
$$z=x^2+y^2 \wedge 2x+2y+z=7 \... | You should consider the surface $S$ given by the intersection of the cylinder $(x+1)^2+(y+1)^2\leq 7+2=9$ with the plane $2x+2y+z=7$. Then $\mathbf{n}=-(2,2,1)/3$ (the ellipse $\partial S$ is clockwise oriented) and by Stokes' theorem:
$$
\begin{align*}\int_{\partial S} \mathbf{F} \, d\mathbf{s} &= \iint_S \text{curl }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derive an upper bound for $\log(x)$ A paper has given that for $x_0 > 0$,
$$
\log x \leq \frac{x}{x_0} + \log x_0 - 1.
$$
The way I tried to prove this is:
For $|t| < 1$,
\begin{align*}
\log(1 - t) = -\sum_{k=1}^n \frac{t^k}{k}
\end{align*}
We can replace $t = 1 - \frac{x}{x_0}$,
\begin{align*}
\log x - \log x_0 & = - ... | Hint: $\log x$ is concave, so it remains below its tangent at $x_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{... | As already commented by @DatBoi that the proof is incorrect, I would suggest you a different approach.
$ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$
a) Substitute $x= \tan\theta$
$\implies f(x)= \sin^{-1}(\sin2\theta)$
$\implies f(x) =2\theta\quad -\pi/2\leq2\theta\leq\pi/2$
$\implies f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Hard Integral without Partial Fraction Decomposition So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.
$\int{\frac{3x+4}{(x^2+5)^2}dx}=$
Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^... | Note that
\begin{align}
\left( \frac1{x^2+5}\right)’&= -\frac{2x}{(x^2+5)^2},\>\>\>\>\>
\left( \frac x{x^2+5}\right)’= \frac{10}{(x^2+5)^2}
- \frac{1}{x^2+5}
\end{align}
Their linear combination below yields
$$
\left( \frac {4x-15}{x^2+5}\right)’= \frac{10(3x+4)}{(x^2+5)^2}
-\frac{4}{x^2+5}
$$
Integrate both sides to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\fr... | This fact is useful in thinking about your problem: If $U,V\sim N(0,1)$ are independent standard normals, their ratio $X=U/V$ has the standard Cauchy distribution. Then $Y=(1-X)/(1+X) = (V-U)/(V+U) = S/T$ where $S=(V-U)/\sqrt 2$ and $T=(V+U)/\sqrt 2$. But $S$ and $T$ are also independent standard normals, so $Y$ has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Maximum value of $\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cdots + \sqrt{1-\sqrt{a_n}}$ for $a_i \in [0,1], a_1+a_2+ \cdots +a_n = 2$
Given $n \ge 3$ real numbers $a_1, a_2, \cdots, a_n \in [0,1]$ such that $a_1 + a_2+ \cdots +a_n =2$, find the maximum value of:
$$P =\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cd... | Hint : use the EV method with :
$$f(x)=\sqrt{1-\sqrt{1-\frac{x}{x+1}}}$$
We have on $(0,\infty)$:
$$f'''(x)>0$$
With the constraint (for $0<x_i\leq 1$):
$$\sum_{i=1}^{n}\frac{1-x_i}{x_i}=2\,,\sum_{i=1}^{n}\left(\frac{1-x_i}{x_i}\right)^2=\operatorname{constant}$$
Now use corollary 1.4 (see http://emis.maths.adelaide.ed... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^... |
It was an easy question, just one important result and a bit of number theory was needed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Equalities with 2 given substitions Let $\alpha$ and $\beta$ be real numbers such that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ Which of the following equalities hold?
A. $\alpha^2 + \beta^2 = 2$
B. $\alpha^3 + \beta^3 = 3$
C. $\alpha^4 + \beta^4 = 6$
D. $\alpha^5 + \beta^5 = 12$
E. $\alpha^6 + \beta^6 = 18... | Choice E. Use $\alpha^6+\beta^6 = (\alpha^3+\beta^3)^2 -2(\alpha\beta)^3 = 4^2 - 2(-1) = 18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Can $\mathbb{R}^3$ with Hadamard product be represented as matrices? It is known that $\mathbb{R}^2$ with Hadamard product, represented as pairs or numbers $(a,b)$ with element-wise operations is isomorphic to real matrices of the form $\left(
\begin{array}{cc}
\frac{a+b}{2} & \frac{a-b}{2} \\
\frac{a-b}{2} & \frac{a... | The following works for complex matrices. Let $\omega = \exp(i 2\pi/3)$ be a primitive cube root of $1$. Define
$$
H(a,b,c) :=
\begin{bmatrix}
\frac{a+b+c}{3} & \frac{a+\omega b+\omega^2 c}{3} &
\frac{a+\omega^2 b+\omega c}{3}
\\
\frac{a+\omega^2 b+\omega c}{3} & \frac{a+b+c}{3} &
\frac{a+\omega b+\omega^2 c}{3}
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Under these conditions, determine the polynomial in two variables. Let $f(x,y)=(x^2+y^2+2)^{xy}$ defined on $\mathbb R^2$ and let $g(x,y)$ be a degree $2$ polynomial in two variables.
Suppose $\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{g(x,y)-f(x,y)}{x^2+y^2}=0$. Then, determine $g(x,y)$.
Here is what I did so far.
Let... | Recall that for $t\to 0$, $e^t=1+t+O(t^2)$ and $\ln(1+t)=t+o(t)$.
Therefore,for $(x,y)\to (0,0)$,
\begin{align}(x^2+y^2+2)^{xy}&=\exp(xy\ln(x^2+y^2+2))\\
&=1+xy\ln(x^2+y^2+2)+O(x^2y^2)\\
&=1+xy\left(\ln(2)+\ln\left(1+\frac{x^2+y^2}{2}\right)\right)+o(x^2+y^2)\\
&=1+xy\left(\ln(2)+\frac{x^2+y^2}{2}+o(x^2+y^2)\right)+o(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4383309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do we find the exact value of $\int_{0}^{\infty} \frac{\ln ^{n}\left(1+x^{2}\right)}{1+x^{2}} d x$, where $n\in \mathbb N?$ Latest Edit
By @KStarGamer’s help, I can finally find a reduction formula for $I_n$ as below: $$\boxed{I_n= 2 \ln 2 I_{n-1}+ (n-1)!\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k!}\zeta(n-k) I_k}
$$
wher... | With differentiation of $$\frac2\pi\int_{0}^{\frac\pi2} (2\cos t)^adt = \frac{\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}$$
in $a$, it can be shown that $J_k=\frac2\pi \int_{0}^{\frac\pi2}\ln^k(2\cos t)dt$ are the power-series coefficients of the exponential function below
$$
\exp\bigg( \sum_{k=1}^\infty \frac{(-1)^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = ... | $$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
$$I = \int \frac{1}{x^{3} \sqrt{\frac{2}{x}-1}} dx$$
Substitute $\displaystyle u=\frac{1}{x}$ to get
$$I=-\int\frac{u}{\sqrt{2u-1}}du$$
$$-2I=\int\frac{2u-1+1}{\sqrt{2u-1}}du$$
Can you finish it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Sum of consecutive numbers with a rounded multiplier I know full well that the sum of the integers from $1$ to $n$ is $\frac{n * \left (n+1\right )}{2}$ but I need a formula that works for the rounded multiples of $\frac{9}{7}$ the integers from $1$ to $n$.
e.g.
$n$
$\lfloor\frac{9}{7}n\rceil$
cumulative sum
1
... | You’re adding all the numbers up to $\left\lfloor\frac97n\right\rceil$ except those with remainders $2$ and $7$ modulo $9$. So you have $\frac{\left\lfloor\frac97n\right\rceil\left(\left\lfloor\frac97n\right\rceil+1\right)}2$, but you need to subtract the missing numbers. There are $\left\lfloor\frac{n+5}7\right\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $\prod_{i=1}^{120} a_i$ Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $$\prod_{i=1}^{120} a_i$$.
My Attempt:
I am aware how to find the sum of the... | Let's order the $a_{1}, \ldots, a_{120}$ so that $a_{i} \cdot a_{121-i} = M$ for all $1 \leq i \leq 120.$ We can do this because an integer $d$ is a positive divisor of $M$ if and only if $\frac{M}{d}$ is a positive integer divisor of $M$ (and since $M$ is not a square, $d$ and $\frac{M}{d}$ are never equal).
Now, note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} ... | $y=f(x). y^3-y^2-x^2y+x^2=0$. Max $y$ is $1$. Min $y$ is $0$. Find $f(-4/3)+f(0)+f(1/2)$
$g(x,y)= (y-x)(y+x)(y-1)=0$.
The constraint requires only certain $y$ values for a given $x$.
$x=0\implies y\in\{0,1\}$
$x=1/2\implies y^3-y^2-y/4+1/4=0.\implies (2y-1)(2y+1)(y-1)=0\implies y\in\{1/2,-1/2,1\}$
$x=-4/3\implies y^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Closed form of $\sum_{i=0}^{\infty} x^i (i!(i+1)!)^{-1/2}$ I am looking for a closed form of this series
$$
\sum\limits_{i=0}^{\infty} \frac{x^i}{\sqrt{i!(i+1)!}},\:x \geq 0.
$$
The main problem is that this series contains roots of factorials. But many special functions just contain factorials, not their roots. Is it ... | You will not find closed form, but using
$$n^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-nt} \, {\rm d}t$$
for $s=1/2$ (it should work for general $s$) and substituting $e^{-t} \rightarrow t$, you can bound your sum above $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!\sqrt{n+1}} = \frac{1}{\sqrt{\pi}}\int_0^1 \frac{e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$ and find it arc length? How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$?
Should I make a new equation $x^{2} + y^{2} + z^{2} = x + y + z$?
O... | Yet another answer.
The unit sphere in $\mathbb{R}^n$ can be described by $\|x\|^2 = 1$, and the equivalent plane by $e^T x = 1$ where $e=(1,...,1)$. Let $Q$ be an orthogonal rotation such that $Qe = \sqrt{n}e_1$, where $e_1 = (1,0,...,0)$.
Using the transformation $y=Qx$ we get the equivalent equations $\|y\|^2 = 1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1... | Using the hint given by @GEdgar, notice that
$$\frac{1}{(1+x)(1-x)^{2}}=\frac{A}{1-x}+\frac{B}{(1-x)^{2}} +\frac{C}{1+x}.$$
Solving, we get
$$1=A(x^{2}-1)+B(x+1)+C(x-1)^{2}$$
Equality coefficients, we have
$$\begin{cases}-A+B+C=1,\\ 0A+B-2C=0,\\A+0B+C=0\end{cases}$$
Solving the system equations, we get
$$\left(A,B,C\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\i... | Since
$$
\frac{\sin(x)}{\cosh(x)+\cos(x)}=2\sum_{n\ge 1}(-1)^{n-1}e^{-nx}\sin(nx
)$$
(shown here), we can combine the above with $\sin^2(x)\sin(nx) = \frac14 \left[2 \sin(n x) + \sin((2 - n) x) - \sin((2 + n) x)\right]$ and $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}\mathrm{d}x = \arctan\left(\frac{b}{a}\right)$ to get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find the error in my workings for the proof of $\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\ge 1$ Here are my workings...
By Cauchy-Schwarz,
$$\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\left( \frac{x+2y}{x} + \frac{y+2z}{y} + \frac{z+2x}{z}\right) ≥ (1+1+1)^2$$
$$\left(\frac{x}{x+2y... | Observe,$$\sum_{\text{cyclic}}\frac{x}{x+2y}=\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}$$
Using Titu's Lemma,
$$\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}=1$$
$$\therefore\;\; \frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I fou... | Note that you get $$A^2=A+2I$$
Therefore $A^n$ is a linear combination of $A$ and $I\ $, i.e. $$ A^n=a_nA+i_nI$$
You get $A^{n+1}=a_nA^2+i_nA=a_n(A+2I)+i_nA=(a_n+i_n)A+2a_nI$
The system to solve is $\begin{cases}a_{n+1}=a_n+i_n\\i_{n+1}=2a_n\end{cases}\ $ and reporting for $a_{n+2}$ gives $\begin{cases}a_{n+2}=a_{n+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and the... | I advise you to apply the identity
\begin{align*}
\tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - \tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)}
\end{align*}
so that you arrive at the following equation:
\begin{align*}
\frac{3x - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3 & \Longle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the maximum value of:$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$ Let $a,b,c \in R$, find the maximum value of:$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$$
I have the solution:
$$5a^2+2b^2+c^2-2(ab+bc+ca)=5(a-\frac{b+c}{5})^2+\frac{1}{5}(3b-2c)^2\geqslant 0... | Alternate approach:
$a^2+b^2+c^2-ab-bc-ca$ takes its max value when $b =\frac{a+c}2$. So $a^2+b^2+c^2-ab-bc-ca=\frac 34(a-c)^2$, so we aim to maximize
$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}- |c-a|$$
Let $c-a=t$, so $c^2-3a^2=(a+t)^2-3a^2=-2a^2+2at+t^2$. This takes maximum if $a=t/2$, and $c^2-3a^2=3/2 t^2$. Therefore, we aim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of an expression using the roots of a given polynomial
Let $a,$ $b,$ $c,$ $d,$ and $e$ be the distinct roots of the equation $x^5 + 7x^4 - 2 = 0.$ Find
\begin{align*} &\frac{a^4}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^4}{(b - a)(b - c)(b - d)(b - e)} \\ &\quad + \frac{c^4}{(c - a)(c - b)(c - d)(c - ... | Hint: Turns out Vieta isn't needed. Irrespective of which polynomial they are roots of, as long as $a,b,c,d,e$ are distinct,
$$\sum_{cyc} \frac{a^4}{(a-b)(a-c)(a-d)(a-e)}=1$$
Will expand on the hint in case you can't complete the proof...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to pro... | You also need $a >1$.
There is another less direct way to do this. Let us write $A=2^a-1$, $B=2^b-1$, and $C=2^{ab}-1$. Then both $A$ and $B$ divide $C$ so write $C=BD$ for some integer $D$. Also, the inequality $a <b$ gives the inequality $A<B$. Now, $a \ge 2$ also gives
$$B^2 < 2^{2b} -1$$ $$\le 2^{ab}-1 = C,$$ so in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k... | As demonstrated in the other answers, it suffices to prove the statement for $k=1$, that is
$$
I = \int_{0}^{2\pi}f(x) \cos (x) \, dx \ge 0 \, .
$$
We can split the integral into four parts over the intervals $[0, \pi/2]$, $\pi/2, \pi]$, $[\pi , 3\pi/2]$ and $[3\pi/2, 2 \pi|$ and substitute $x = \pi - y$, $x=\pi + y$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Probability of spelling IDAHO correctly I'm working with a problem stated as follows:
A sign reads IDAHO. Two letters are removed and put back at random, each
equally likely to be put upside down as in the correct orientation. What is the
probability that the sign still reads IDAHO?
So, we form our events as:
*
*$A... | This answer identifies the errors in the question’s reasoning, mainly by restating the advice in Aaron Montgomery’s answer and Acccumulation’s comment, then uses all of Acccumulation’s advice to recalculate $P(A)$.
$P(A|B)$ is indeed $\frac{1}{2}$, but let us write it as $\frac{4}{8}$ instead, for consistency with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Compute $\sum_{k=0}^{n}{\frac{\binom{n}{k}}{k+1}}$ Compute the sum:
$$\sum_{k=0}^{n}{\frac{\binom{n}{k}}{k+1}}$$
I tried to solve it:
$$\sum_{k=0}^{n}{\frac{(k+1)(k+2)\binom{n+2}{k+2}}{(n+1)(n+2)(k+2)}}=\sum_{k=0}^{n}{\frac{(k+1)\binom{n+2}{k+2}}{(n+1)(n+2)}}=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}(k+1)\binom{n+2}{k+2}$$
Bu... | You should start by showing that:
$$ \frac{n\choose k}{k+1} = \frac{1}{n+1} {n+1\choose k+1} $$
Then notice you're summing over $k$, so you're free to take this $\frac{1}{n+1}$ out of the sum as a constant factor:
$$ \sum_{k=0}^n \frac{n\choose k}{k+1} = \frac{1}{n+1}\sum_{k=0}^n {n+1\choose k+1} $$
Finally,
$$ \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Cuboid with inner and outer Cubes A cuboid with integer side lengths consists of cubes with side length 1. Such a cube is called an inner cube if none of its sides are visible from the outside, otherwise, i.e. if at least one side is visible from the outside, the cube is called an outer cube. The $3 \times 4 \times 7$ ... | We will show that the set of all solutions are finite (solutions with length greater than $2$), and we give some bounds to find all of the solutions.
Without lose of generality we can assume $2 < a \leq b \leq c$ (Note that if $a=1, 2$ then every cube is a visible cube). We will show that there are only finitely many ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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.