Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Combinatorial proofs that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$. I am trying to prove that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.
The one I could come up with is that the equivalence relation where $A\sim B$ if $B$ is of the form $A+k$ for some $k$ with th... |
I am trying to prove hat $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.
Assumed that it is intended that $n,k \in \Bbb{Z}.$
Alternative (direct) approach that uses
Legendre's Formula.
While less elegant than other approaches, this approach provides a nice fail-safe, if elegance eludes you.
Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Differential equation. $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ Check my steps I have equation $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ and I need to solve it with separation method.
My try:
$-x^2\cdot dy-x\cdot dy=-\sqrt{3+y^2}\cdot dx$
$-x(x+1)\cdot dy=-\sqrt{3+y^2}\cdot dx $
$\frac{-x(x+1)}{dx}=-\frac{\sqrt{3+y^2}}{... | Just rewrite the equation as
$$
\dfrac{1}{x+x^2} dx = \frac{1}{\sqrt{3+y^2}}dy.
$$
By integration, you get
$$
\log \frac{|x|}{|1+x|} = \textrm{arcsinh} y + C
$$
So, the solution can be given in an explicit form:
$$
y = \sinh\left(\log \dfrac{|x|}{|x+1|}-C\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If... | If you write $z$ as $a+bi$, then you get the system$$\left\{\begin{array}{l}(a+1)^2+b^2=1\\(a^2-b^2+1)^2+(-2ab)^2=1,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}a^2+b^2+2a=0\\2 a^2 b^2+a^4+2 a^2+b^4-2 b^2=0.\end{array}\right.$$From the first equation, you get that $b^2=-2a-a^2$, and if you replace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y =... | We have,
$$\begin{align}&\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x≥0\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=2x\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=3x\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=\sqrt{3x}\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=2\sqrt{3x}\\
\if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove $\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1 $ Prove $$\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1,
$$
where $a,b,c \in (0,1)$.
I tried to solve it in this way:
$$\log_{a}{(\frac{b^2}{... | Your idea is good but I think it would be more rigorous in the following way:
Before proving
$$F(a,b,c) = \log_{a}\left(\frac{b^2}{ac}-b+ac\right)\cdot\log_{b}\left(\frac{c^2}{ab}-c+ab\right)\cdot\log_{c}\left(\frac{a^2}{bc}-a+bc\right)\geq1$$
one can prove that
*
*$\frac{b^2}{ac}-b+ac \ge b$
*$\frac{c^2}{ab}-c+ab ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4278468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integral$ \int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$ $$\int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$$
I tried to substitute $1+\cos^2x$ , tried to change cos in sin by complementary formula etc . But nothing seems to work out. I think it's not as simple as I thought ?
| $$ I = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1+\cos^2 x}} dx = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{2-\sin^2 x}} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx $$
Recall the elliptic integral of the first and second kind, respectively:
$$F(\phi,k) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=... | You can also appeal to the rational root theorem,
$$ \begin{align*} x &= \sqrt{3} + \sqrt{6}\\
x^2 &= 9 + 6\sqrt{2}\\
x^4-18x^2 &+9=0 \end{align*} $$
So if $x = p/q$ with $p,q$ coprime then $p |9, q|1$ or $p = 1,3,9$ and $q=1,-1$, but none of the options qualify as $x$ is not an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Probability of randomly assigning elements to groups
10 students are tested in an exam with 4 different versions. Each
student is randomly assigned to one of the versions. What is the
probability that there are exactly $i$ versions in which exactly > $3$ students were assigned. Answer separately for $i=2, i = 3$.
I... |
$10$ students are tested in an exam with $4$ different versions. Each student is randomly assigned to one of the versions. What is the probability that there are exactly $i$ versions in which exactly $3$ students were assigned. Answer separately for $i=2,i=3$.
$i = 2$
As you observed, the sample space has size $|\Ome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Factoring a third degree polynomial with a given root My professor gave us the following polynomial:
$f(x) = 3x^3-4x^2-x+2$
Given is that $x = 1$ is a root of this function. We are asked to find the other ones.
He then told us that, given $x=1$ is a root, we now know that we can factorize this polynomial into $(x-1)$, ... | Factoring a polynomial is basically the same as factoring an integer into primes. $34= 2(17), 36= 2^2(3^2$). Any polynomial can be factored, over the real numbers, into either linear terms, such as "$(x- a)$" where $a$ is a real number, or a irreducible quadratic term. (quadratics that cannot be factored further with r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculating the joint probability density $f(x,y)=\frac{1}{y}$ If the joint probability density of $X$ and $Y$ is given by:
$$f(x,y) = \frac{1}{y}$$
for $0<x<y, 0<y<1$
Find the probability that the sum of the values of $X$ and $Y$ will exceed 1/2.
What I have tried:
I have tried sketching out the region and got this:
... | According to nejimban's comment you have to integrate on the set
$$
\begin{align}
A&=\{(x,y):0<x<y<1,\ \frac{1}{2}< x+y\}\\
\end{align}
$$
Given this information you find
$$y>\frac{1}{2}-x>\frac{1}{2}-y$$
such that
$$1>y>\frac{1}{4}$$
Furthermore you find, that
$$y>x>\frac{1}{2}-y$$
However, note that for $y>\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of the inradius of triangle $PHQ$? For refrence: The right triangle $ABC$, right at $B$, the height $BH$ is drawn. Let $P$ and $Q$ be the intersections of triangles $AHB$ and $BHC$,
$PQ$ intersects at $E$ the $BH$, where $\frac{BE}{EH} = 5\sqrt2$ and the inradius of the triangle $ABC$ measures $10$.
C... | $S$ the point of touching incircle ABH with BH. $T$ is the point of intersection of straight lines $PS$ and $QN$. Triangles $PES$ and $PQT$ are right and similar. $PS=r_1$, $PT=r_1+r_2$, $QT=r_2-r_1$. From similarity:
$$ES=PS\cdot QT/PT=r_1\frac{r_2-r_1}{r_1+r_2}$$
$$EH=ES+SH=r_1\frac{r_2-r_1}{r_1+r_2}+r_1=\frac{2r_1r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y... | As shown in other answer, you can just form an equation for sum of distances to foci. Going by your method, the goal would be to write
$$\left( \frac{\text{Distance from Line perp to Axis, thro' center }}{\text{Semi-major axis}} \right)^2 + \left( \frac{\text{Distance from Axis}}{\text{Semi-minor axis}} \right)^2 = 1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | Or
$$
\begin{align}
4x^2-2xy-4x+3y-3 &= \\
4x^2-4x-3-y(2x-3) &= \\
(2x-3)(2x+1)-y(2x-3) &= \\
(2x-3)(2x+1-y)
\end{align}
$$
As far as factoring quadratics is concerned (your question about factoring $y^2-12y+12$) if the roots are $\alpha$ and $\beta$ the quadratic factors as
$$
(y-\alpha)(y-\beta)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 4
} |
Minimize the ratio involving the ellipse
Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,
and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and
$(1,1)$ respectively. Find the minimum value of
$\dfrac{|PA|^2}{|PB|}$.
Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then
$$\dfrac{|PA|^2}{|PB|}... | Doing what @Paul Sinclair already commented and using multiple angle formulae
$$\dfrac{|PA|^4}{|PB|^2}=\frac{-256 \cos (\theta )+92 \cos (2 \theta )-16 \cos (3 \theta )+2 \cos
(4 \theta )+259}{8 \left(-4 \sqrt{3} \sin (\theta )-8 \cos (\theta
)+\cos (2 \theta )+11\right)}$$ Computing the derivative, its numerator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$? $$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$
$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3... | After some truly horrific partial fraction decomposition (I'm trusting WolframAlpha (check near the bottom of the page) for now, until I can verify it), we get
\begin{align}
\frac{ t^3 }{3 (t^6 - t^3 + 1)} + \frac{t}{6 (t^6 - t^3 + 1)} - \frac{t}{2 (t^6 + t^3 + 1)} - \frac{1}{6 (t^6 - t^3 + 1)} + \frac{1}{2 (t^6 + t^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6... | An alternative method using some known identities .
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x.\end{aligned}$
Let $\displaystyle x=t^{\frac{1}{6}}$.
Then we get $$\int_{0}^{\infty}\frac{1}{6}\frac{t\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$ I was trying to solve the following question:
\begin{align*}
\sinh x &= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\
\cosh x &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
\end{align*}
Usi... | In order to multiply two power series, say
\begin{align*}
\def\bl#1{\color{blue}{#1}}
\def\gr#1{\color{green}{#1}}
\bl{A}(x) &= \bl{a_0} + \bl{a_1}x + \bl{a_2}x^2 + \cdots \\
\gr{B}(x) &= \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots,
\end{align*}
we have to imagine opening parentheses:
\begin{align*}
\bl{A}(x) \, \gr{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What's the measure of the segment $EF$ in the triangle below? For reference:
The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)
My progress:
Here are the relations I found:
$EF = R - OE \\
\trian... | Here is another approach -
Using Heron's formula, $\triangle_{ABC} = 12 \sqrt5$
$\triangle_{ABC} = \frac12 \cdot BH \cdot AC = 12 \sqrt5 \implies BH = \frac{8 \sqrt5}{3}$
Using Pythagoras, $ \displaystyle AH = \frac{11}{3} \implies HE = \frac92 - \frac{11}3 = \frac56$
As $BG$ is angle bisector, $ \displaystyle AG = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What's the the value of $x$ in the circumference below? For reference:
In the figure; calculate $x$, if $r =\sqrt2$.
(Answer: $x = \sqrt2$)
My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.
Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implie... |
$AB = 2R\\
r_1+r_2= R\\
AO = R$
Euclid's Th.: $\triangle OO_1O_3$ and $OO_2O_4$
$(R-r)^2=(r_1+r)^2+r_2^2-2(r_1-r)r_2\\R^2-2Rr+r^2 = r_1^2+2r_1r+{r^2}+r_2^2-2r_1r_2+2rr_2\implies\\\boxed{R^2-2Rr = (r_1-r_2)^2+2r(r_1+r_2)}(I)$
$(R-x)^2=(r_2+x)^2+r_1^2-2(r_2-x)r_1=\\
R^2-2Rx+{x^2} =r_2^2+2r_2x+{x^2}+r_1^2-2r_2r_1+2xr_1=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Linear transformation of a line Question:
Find the image of the line $y=5x+9$ by rotation with center $O(0,0)$ with rotation angle of $270^{\circ}$ followed by dilatation at the center $O(0,0)$ with a scale factor of $3$.
My Attempt:
Given the line $y=5x+9$, it passes through $(-1.8,0)$ and $(0,9)$. I used these two p... | This is correct. You can verify the answer by using linear algebra as well. The original line is given by
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0\\9 \end{pmatrix}+ t \begin{pmatrix} 1 \\ 5 \end{pmatrix}; (t\in\mathbb{R})$$
The rotation by $270$° represented by the matrix R, and the dilation by 3 repr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral of $1 / \sqrt x$ using Limits Actually the problem here is to find out the INTEGRAL of $\frac{1}{\sqrt x}$ using the limit definition. I am very well able to solve the question using POWER rule but that is not allowed in the question.
$$b-a=nh$$
where $h$ is very small.
Then
$$\int_a^b{\frac{1}{\sqrt x}} dx= \... | I'm posting another answer based on fixed-width partitions and a "harmonic mean approximation" for fun. I think this answer can hardly be generalized to other rational powers of $x$. The image and the arguments were copied from my former messages on Discord.
We would make a little trick of approximating $1/\sqrt{x}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x... | First multiply by $\sqrt{3x+2}$ to get
$$3x+2 + x^2 = 2x \sqrt{3x+2}$$
Then square both sides
\begin{align*}
(x^2+3x+2)^2 &= 4x^2(3x+2)\\
x^4+6x^3+13x^2+12x+4 &= 12x^3+8x^2 \\
x^4-6x^3+5x^2+12x+4 &=0
\end{align*}
Try to solve this polynomial equation, giving you the true solution and one false solution that you have to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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What's the rate of the stream's current? At his usual rate a man rows 15 miles downstream in five hours less than it takes him to return. If he doubles his usual rate, the time dowstream is only hour less than the time uqstream. What's the rate of the stream's current?
Let the man's rate in still water be $r$ and the r... | This is correct. I would have made things a little simpler, noting that I have $5r^2-5c^2=30c=4r^2-c^2$, and so $r^2=4c^2$. Substituting $r=2c$ in either of these equations then gives the result a little more easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Real function $f$ such that $f(f(x)) = x^3 - 3x^2 + 3x$ Does there exist a function $f:\mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x^3 - 3x^2 + 3x$?
Since $f(x) = f(y)$ implies $f(f(x)) = f(f(y))$ and so $(x-1)^3 - 1 = (y-1)^3 - 1$, i.e. $x=y$, we have that $f$ must be injective. In particular, $f(f(0)) = 0$ and $f... | Take the function
$$
f(x)=
\begin{cases}
(x-1)^{\sqrt 3}+1,\qquad x\geq 1,\\
1-(1-x)^{\sqrt 3},\qquad x< 1.
\end{cases}
$$
Then for $x\geq 1$ we have
$$
f(f(x))=\left((x-1)^{\sqrt 3}+1-1\right)^{\sqrt 3}+1=\left((x-1)^{\sqrt 3}\right)^{\sqrt 3}+1=(x-1)^3+1=x^3-3x^2+3x.
$$
If $x<1$ we get
$$
f(f(x))=1-\left(1-\left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4311717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)... | Let $z=x-y$ then $$f(x,y) = z^2+5y^2-12z-10y+41 $$ $$ = (z-6)^2+5(y-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integral representation of Bessel function $J_1(x)$ In "The Handbook of Mathematical Functions" by Abramovitz and Stegun, according to Eq. 9.1.24,
\begin{align}
J_0(x)=&\frac{2}{\pi}\int_{1}^\infty \frac{\sin(xt)}{\sqrt{t^2-1}}dt,\quad x>0.
\end{align}
Naively using $J_1(x)=-dJ_0(x)/dx$,
\begin{align}
J_1(x)=&-\frac{2}... | If $x>0$, then
$$ \begin{align}J_{0}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \frac{\sin (xt)}{\sqrt{t^{2}-1}} \, \mathrm dt \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt + \frac{2}{\pi} \int_{1}^{\infty}\frac{\sin (xt)}{t} \, \mathrm dt \\ &= \frac{2}{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\a... | Here is another variation. We show the identity
\begin{align*}
\color{blue}{[z^n]\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{1}
\end{align*}
is valid for integral $n>0$ and $\alpha \in\mathbb{C}\setminus\{-1,\ldots,-n\}$.
We start with the LHS of (1) and obtain
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Show that if $|z|=2$, $\text{Im}(1-\bar{z}+z^2)\le 7$. Show that if $|z|=2$, $|\text{Im}(1-\bar{z}+z^2)|\le 7$.
My attempt:
Let $z=x+yi$, $|z|=\sqrt{x^2+y^2}$, $|z|^2=(\sqrt{x^2+y^2})^2=x^2+y^2$.
Since $|z|=2$, then $|z|^2=2^2=4$. So $x^2+y^2=4$. Also, $(x-y)^2\ge0$ $\implies x^2+y^2\ge2xy$;
$$xy\le2 $$
So $|y|\le2$ an... | By the triangle inequality:
$$|1 + (-\bar{z}) + z^2 | ≤ |1| + |-\bar{z}| + |z^2| ≤ 1+2+2^2 = 7$$
(think about what happens to $z = 2e^{i \theta}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Amount of solutions to an equation Briliant.org asks for how many solutions there are to this equation:
$x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 = x + 2x^3 - 2x - 2x^3 + x$
Now if I put them in order:
$ 3x^3 + x^3 -8x^2 -7x^2 + 3x^2 = 2x^3 - 2x^3 - 2x + x + x$
and combine
$ 4x^3 -12x^2 = 0$
I could add $12x^2$ to both sides a... | A cubic equation has either $3$ real roots or $1$ real root and $2$ complex roots. Here, we have three real roots, two of which are identical.
\begin{align*}
x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 &= x + 2x^3 - 2x - 2x^3 + x\\
4 x^3 - 12 x^2 &= 0\\
(x-0)(x-0)(x-3) & =\frac{0}{4}
\implies x\in\big\{0,0,3\big\}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $12x^5+16x^4-17x^3-19x^2+5x+3=0$ Solve the equation $$12x^5+16x^4-17x^3-19x^2+5x+3=0$$ The divisors of $3$ are $\pm1;\pm3$ and the divisors of $12$ are $\pm1;\pm2;\pm3;\pm4;\pm6;\pm12$, so the possible rational roots are $$\pm1;\pm\dfrac12;\pm\dfrac13;\pm\dfrac14;\pm\dfrac16;\pm\dfrac{1}{12};\pm3;\pm... | Step 1:
Try $\pm1$, they work so we can reduce the problem to a cubic by dividing by $(x-1)(x+1)=x^2-1$:
$$
P(x)=(12x^5+16x^4-17x^3-19x^2+5x+3)/(x^2-1)\\=12 x^3 + 16 x^2 - 5 x - 3
$$
Step 2: To find a 3rd solution we find the extrema of $P(x)$ we solve
$$
P'(x)=36x^2-32x-5=0
$$
Finding that it has 2 solutions at $x=-1.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate this limit $(\infty/\infty)$ I want to prove that $1=0.999...$ in this way, we have
$$0.9...9=0.9 \left(10^{-n-1}\right)\left(\sum_{i=0}^n 10^i\right)$$
Hence $$0.999...=\lim_{n \to \infty} 0.9 \left(10^{-n-1}\right)\left(\sum_{i=1}^n 10^i\right) \\ =0.9 \lim_{n \to \infty} \frac{\sum_{i=0}^n 10^i}{10^{n+1}}$$... | When summing a geometric sequence, we want the common ratio to be less than $1$ in magnitude:
$$
\sum_{n=0}^\infty r^n = 1 + r + r^2 + \cdots = \frac{1}{1-r}
$$
as long as $\lvert r \rvert < 1$.
Thus, to evaluate $0.\overline{9} = 0.999\dots$, we write it in terms of a geometric series with common ratio $r=\tfrac{1}{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The constraints are not all considered
The solution says that domain of integration is delimited above by the sphere of equation $x ^ 2 + y ^ 2 + z ^ 2 = 2$ and below by the cone $z = \sqrt{x ^ 2 + y ^ 2}$. I have the impression that we do not consider the fact that $x$ is between $0$ and $1$ and that $y$ is between ... | You are right and the solution is incorrect. If we wished to consider the region bounded by the sphere $x^2 + y^2 + z^2 = 2$ and the cone $z = \sqrt{x^2 + y^2}$, then the correct bounds would be
$$\int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}} \frac{1}{\sqrt{x^2 + y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Upper bound: Given $L$-smooth convex $f$; $( y- x)^T \left( \nabla f(z)-\nabla f(x)\right)\leq(L/2) ( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2)$? Given $L$-smooth convex $f$, I would highly appreciate if you can confirm whether the following bound is correct or not.
\begin{align}
\left( y- x\right)^T \left( \nabla f(z)-\nabla ... | What about this simple estimate:
$$
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right)
\le \|y-x\| \cdot \|\nabla f(z)-\nabla f(x)\|
\le L \|y-x\| \cdot \|z-x\|
\le \frac L2(\|y-x\|^2 + \|z-x\|^2)
$$
??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4336013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
number of real roots of the equation $f(f(x)) = c$ The question is stated as follows:
Let $f(x) = x^4 − 4x + 1$ for all real number $x$.
$\ $Determine, for each real number $c$, the
number of real roots of the equation $f(f(x)) = c.$
By calculating $f(f(x))$ explicitly it is easy to see that $f(f(x))=u^4+4u^3+6u^2-2$... | drew some graphs, fiddled with that. I don't see how you would know what to do without some sketches.
$$f(x) = x^4 - 4x + 1$$
$$g(x) = f(f(x)) = (x^4-4x+1)^4-(x^4-4x+1) + 1 \; , \; \; $$
then
$$ 2+g(x) = x^2 \; (x^3-4)^2 \; \left( \; 2 + (x^4 -4x+2)^2 \; \right) $$
where
$$ 2 + (x^4 -4x+2)^2 \; > \; \; 0 $$
We see... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$ It's an integral which seems simple but I confess I cannot evaluate this :
$$\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$$
I can evaluate another integral where I start from :
$$\int_{\pi}^{\infty}\left(x^{2}-x-1\righ... | Here is an attempt at an antiderivative. You can consider it a comment.
Attempt 1:
Here is a series expansion for the antiderivative using geometric series which includes the $[\pi,\infty)$ interval of convergence:
$$\int \frac{dx}{x^2-\sin(x)-1}=-\int \sum_{n=0}^\infty \left(i\frac{e^{-ix}-e^{ix}}{2}\right)^n(x^2-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a mu... | Since $y$ is an integer,$\frac{-11x}{x+11}$ will be an integer. When $\frac{x}{x+11}$ is an integer, it refers that $\frac{11}{x+11}$ is an interger, or $x=0$. Then you can find all such $x$ in this situation. Otherwise $\frac{x}{x+11}$ is not an integer but $\frac{-11x}{x+11}$ is. Thus $x$ is a multiple of $11$, write... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the Maclaurin polynomial of order 6 of: $f(x)=x\ln(1+x^{3})\ln(1-x^{2})$ Find the Maclaurin polynomial of order 6 of:
$$f(x)=x\ln(1+x^{3})\ln(1-x^{2})$$
The result I get doesn't make sense.
my try:
because: $\ln(1+x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+1}}{n+1}$
$$\ln(1+x^{3})=x^{3}-\frac{x^{6}}{2}+\frac{x^{... | after writing out the first terms of the maclaurin expansion of $\ln(1+x^3)$ and $\ln(1-x^2)$ , observe that the only way to get an $x^6$ term is to multiply $x$ and the 2 first terms $x^3$ and $-x^2$. thus the order 6 maclaurin polynomial has only one term $-x^6$.
the maclaurin expansion does not have an $x^7$ term as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the CEN triangular region For reference: If ABCD is a square, AB = DE.
Calculate the area of the CEN triangular region.(Answer:$6m^2$)
My progress:
$S_{\triangle MCE}=X=\frac{MC⋅DC}{2}=\frac{MC⋅ℓ}{2}(I)\\
I+S_{\triangle MBA}=\frac{l^2}{2}\\
\therefore X+5 = \frac{l^2}{2}\implies 9 + S_{\triangle CE... | You can use a system of $4$ equations :
$\begin{cases} \frac{l \cdot BM}{2} = 5 \\ \frac{(l -BM)\cdot CN}{2} = 4 \\ \frac{(l -BM)\cdot l}{2} = A_{CNE}+4 \\ \frac{l\cdot CN}{2} = A_{CNE} \end{cases} $
and solve for $A_{CNE}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$. A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$.
Calculate $E(X|Y=2)$.
My attempt
A... | $\mathbb E(X\mid X\geq3)$ is the expectation of the number of rolls needed to obtain a $5$ under the condition that the rolls $1$ and $2$ do not provide a $5$.
We might say that the sequence of relevant rolls actually starts after $2$ rolls that are in vain, so that:$$\mathbb E(X\mid X\geq3)=\mathbb E(2+X)=2+\mathbb EX... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The Frobenius norm is submultiplicative
Let $M_{d \times d} (\Bbb R)$ be the space of real $d \times d$ matrices. For $A = (a_{ij} ) \in M^{d×d}$, define $$ \| A \|_2 := \left( {\sum_{i,j=1}^{d}|a_{ij}|^2} \right)^\frac{1}{2} $$ Show that $$\|AB\|_2 \leq \|A\|_2 \|B\|_2$$
I don't know where to start this step. How d... | In my humble opinion brute-force calculation should suffice.
Denote $a_{ij}$ to be the element on the $i$-th row and $j$-th column of matrix $\mathbf{A}$; similar for $b_{ij}$. Then
$$\mathbf{A B} =
\begin{bmatrix}
\sum a_{1j} b_{j1} & \sum a_{1j} b_{j2} & \cdots & \sum a_{1j} b_{jd} \\
\sum a_{2j} b_{j1} & \sum a_{2j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to determine the basis with given vectors? So I tried to solve this problem, but I can't seem to find an answer.
So the question is: "Using determinants, indicate the sets of values of $x$, $y$, and $z$ for which the vector sequence is a basis.
$$
S = ((0, z, -y), (-z, 0, x), (y, -x, 0))
$$
I tried to put these thr... | Have you considered the possibility that you have not made a mistake?
It seems like you already know that if the determinant of the matrix $\begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$ is $0,$ then the column vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} 0 \\ z \\ -y \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the quadrangular region $EBCD$ For reference:
Calculate the area of the quadrangular region $EBCD$, if $BC = 5$, and $AD = AC$. $P$ is a tangent point. (Answer: $64$)
My progress:
$\triangle OPD \sim \triangle ADC$
$AD=2OD \implies k=1:2$
Therefore $P$ is the midpoint $CD$ and $AP$ is perpendicu... | From the relations you've discovered, we find $AC=5\sqrt5$ and so is $AD$.
Considering the angle $26.5^\circ$ is approximately equal to the angle we get when bisecting the second largest angle in a $3:4:5$ right triangle, we have $\tan26.5^\circ\approx\frac12$. (See also)
Therefore we can easily get $$\sin(3\cdot26.5^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $n$ is a positive integer, $a$, $b$, and $\sqrt[n]{a}-\sqrt[n]{b}$ are rational, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational. Assume that $a$ and $b$ are rational numbers with $a\neq b$ such that $\sqrt{a}-\sqrt{b}$ is rational. The each of $\sqrt{a}$ and $\sqrt{b}$ must be rational. Because if $\s... | Easiest is to say the following. Assume $x^k-a$ and $x^l-b$ are irreducible over $\mathbb{Q}$ then $\sqrt[k]{a}-\sqrt[l]{b}$ is irrational, for otherwise if $k\leq l$,
$$a=(\sqrt[l]{b}+r)^k$$ is an $k$th degree equation for $\sqrt[l]{b}$, or $k-1$ degree if they are equal.
Note that if $$x^n-a=f(x)g(x)$$ is reducible w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_0^1 \frac{\ln x\ +\ \ln(\sqrt x\ +\sqrt {1+x})}{\sqrt {1-x^2}} dx=0$ If a simple way exists, I am looking to show that
$$\boxed{K=\int_0^1 \frac{\ln(x)+\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx=0}$$
say, with symmetry, a clever change of variables, or integrations by parts, without evaluating integral... | I know this is not exactly what it is asked for.
On the path of Zacky,
\begin{align*}
J&=\int_0^1\frac{\text{arcsinh}\left(\sqrt{x}\right)}{\sqrt{1-x^2}}dx\\
&\overset{\text{IBP}}=\Big[\arcsin (x)\text{arcsinh}\left(\sqrt{x}\right)\Big]_0^1-\frac{1}{2}\underbrace{\int_0^1 \frac{\arcsin x}{\sqrt{x}\sqrt{1+x}}dx}_{_{z=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
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Algebraic proof that $\sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} = \frac{1}{n+2}$ I tried evaluating the integral $\int\limits_{[0,1]^n} \min(x^1,x^2,\ldots,x^n) \lvert d^nx\rvert$.
In my first attempt, I used a recursive approach and managed to defined the integral as being $I_n^k = \int\limits... | This answer comes a bit late but I thought it is worth mentioning it because it shows a simple algebraic proof of the formula in question.
First note the following two facts:
*
*(1): $\frac 1{(k+1)(k+2)}\binom nk = \frac 1{(n+1)(n+2)}\binom{n+2}{k+2}$, Indeed
\begin{eqnarray*} \frac 1{(k+1)(k+2)}\binom nk
& = & \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
Integral $\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx$ I am trying to compute the integral
$$\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx \tag{1}$$
Context: Originally I was trying to prove the following result:
$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})\tag{2}$$
Where $\beta(2)$ is th... | Using integration by parts:
\begin{align*}
\int{\frac{x}{\cos(x)}dx} &= \int{x\sec(x)\ dx} \\
&= x\sec(x)\tan(x) - \int{\sec(x)\tan(x)dx} \\
&= x\sec(x)\tan(x) - \int{\cos(x)\sec^2(x)\tan(x)dx} && ...multiply \ by \cos(x)\sec(x) = 1 \\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\sin(x)\tan(x)dx}]\\
&= x\sec(x)\tan(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer... | In this post, I will attempt to improve on the upper bound
$$I(n^2) \leq \dfrac{2q}{q+1}.$$
(Although this attempt is unsuccessful, I am retaining this answer here, mainly for my own benefit.)
Following mathlove's lead in the comments, I tried to find $a, b, c, d, e, f \in \mathbb{R}$ such that
$$2 - 2\cdot\Bigg(\dfra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful tr... | Glad to see there are several nice answers in various perspectives. I am now going to give one more by inverse substitution followed by integration by parts.
Using $x \mapsto \frac{1}{x}$ yields
$$
I(m, n, r):=\int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+1\right)^{n}} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
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Approximating factorial using identity $\frac{1}{x}!\frac{2}{x}!\cdots\cdot\frac{x}{x}!=\frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$ I created a function that describes the product of the inverse multiples of a factorial
$$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\... | Here is the proof of how I got $m(x)$
and since we're till speaking on approximation i also want to share this to you (degrees) @claude leibovici
$$ \sin{5} \approx \frac{1}{6+\sqrt{3}+\sqrt{14}}$$
The factorial function above is actually extended to the gamma function, but i am using factorial notation for simplicity ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the vertical asymptotes of this equation, $f(x)=2\tan(4x-32)$? Could someone show how to use the vertical asymptote formula? I am having a hard time getting it into the right form. I thought maybe I had to put $(4x-32)$ equal to the vertical asymptote equation. I can't get the math to work out.
Asymptote ... | The tangent function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \tan x$ has vertical asymptotes at $x = \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z}$.
We can solve for the vertical asymptotes of $g(x) = A\tan(Bx - C)$, where $B \neq 0$, by setting
$$Bx - C = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
then solving for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral involving product of dilogarithm and an exponential I am interested in the integral
\begin{equation}
\int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast)
\end{equation}
where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to evaluate the double integral
\begin{equation}
I (a,b) = \int... | This is a partial answer that is too long for a comment.
Let
\begin{align*}
J(a) = \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u .
\end{align*}
Note that
\begin{align*}
\frac{d }{d u} \left( u \mathrm{Li}_2 (u) - u - (1-u)\ln{(1-u)} \right) = \mathrm{Li}_2 (u) .
\end{align*}
Then, we may integrate-by-parts to find
\begin{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternative Solution for Cubic Equation I was solving a problem and got to the following equation $x^3 + x^2 + x - 1 = 0 \; \; (1)$, numerically I found that the solution was:
$$x =\frac{1}{3} \left(-1 - \frac{2}{\left(17 + 3 \sqrt{33} \right)^{-1/3}} + \big(17 + 3 \sqrt{33} \big)^{-1/3} \right)$$
Which gives the same... | I shall follow the steps given here.
We have $\Delta=-44$ so only one real root.
Using $p=\frac 23$, $q=-\frac{34}{27}$ and the hyperbolic method
$$t_0=\frac{2}{3} \sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2
\sqrt{2}}\right)\right)$$
$$x=t_0-\frac b {3a}=\frac{1}{3} \left(-1+2 \sqrt{2} \sinh \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find integer solution for square equation
Does $a^2 + b^2 = 6 c^2$ have any integer solution?
My thought:
(0,0,0) is obviously a solution but I don't think there are any others.
Because if I take equation of modulo 6, it gives $a^2 + b^2 = 0 \pmod 6$
a and b can only be 3 mod 6 or 0 mod 6. However, I am not sure what... | Consider a $\pmod{3}$ argument.
Since $6c^2$ is a multiple of $3$, you must have that
$a^2 + b^2$ is a multiple of $3$.
However, for any integer $n$, either
$n^2 \equiv 0 \pmod{3}$ or $n^2 \equiv (+1) \pmod{3}$.
Therefore, $a$ and $b$ must each be a multiple of $3$.
Let $r$ denote the largest positive integer exponent ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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rolling $4$ dice probability question I thought I understood the $4$ dice rolling probabilities until considering the sum of all possible results.
Probability of getting different results while rolling 4 standard dices:
*
*all $4$ dice have one value - $1/216$
*having $2$ different values:
$\dfrac{4C2 \cdot 6 \cdo... | You made an error when considering the case with two different values. There are two possibilities:
*
*One value appears three times and another value appears once.
*Two values each appear twice.
One value appears three times and another value appears once: There are six ways to select the value that appears thre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations an... | First solution: We have
$$8(xy+yz+zx)^3 = [ \sum (xy+yz) ]^3 \leq 9( (xy+yz)^3 + (yz+zx)^3 + (zx+xy)^3 ), $$
so it remains to show that
$$ (xy+yz)^3 + (yz+zx)^3 + (zx+xy)^3 \leq 3 (x^2+yz)(y^2+zx)(z^2 + xy). $$
Let $ M (a, b, c) = \sum_{sym} x^a y^b z^c$. Then, by expanding everything, we get that
$$3(x^2+yz)(y^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\sqrt{2x+1}-\sqrt{2x-1}=2$ I solved $\sqrt{2x+1}-\sqrt{2x-1}=2$
by squaring both sides.
$$\sqrt{2x+1}=\sqrt{2x-1}+2$$
$$2x+1=(2x-1)+4\sqrt{2x-1}+4$$
$$-1=2\sqrt{2x-1}$$
$$1=4({2x-1})$$
$$x=\frac{5}{8}$$
Now when I put $x=\frac{5}{8}$ to the given equation,I get $\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=1 \neq... | Your re-arranged equation $ \ \sqrt{2x+1} \ = \ \sqrt{2x-1} \ + \ 2 \ \ $ would give the $ \ x-$coordinates for the intersection points between two square-root curves, if such points exist. However, "squaring" this equation reveals a difficulty. The curve equation $ \ y^2 \ = \ 2x + 1 \ \ $ implied by the left side r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluat... | Let $I\left(\lambda\right)=\int_0^{\infty} \frac{x^{2 \lambda-1}}{\left(a^2+x^2\right)^n} d x,$ then
$$I_n=\int_0^{\infty} \frac{\ln x}{\left(a^2 +x^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) .$$
Now we are going to express $I\left(\lambda\right)$ as a beta function by letting $x=a\tan \theta$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
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For a non-negative integer $k$, $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$ Suppose that $f$ defined on $(a,\infty)$ is bounded on each finite interval $(a,b),a>b$. For a non-negative integer $k$, it is to be shown that $\lim_{x\to \infty} \frac{f(x+1)-f(x)... | First we prove the statement for the case $l=0$, i.e.
$
\lim_{n \to \infty}\frac{f(x+1)-f(x)}{x^k} = 0
$.
Given $\epsilon > 0$ there is a $y > a$ such that for all $x \ge y$:
$$
\left | \frac{f(x+1)-f(x)}{x^k} \right | < \epsilon \, .
$$
Similarly as in If $\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell,$ then $\lim\limits_{x\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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To prove: $\bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2 $ How could one show that
$$ \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2
$$
I have tried the below approach:
\begin{align*}
LHS & = \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr)\\
& = 1-\sin(A-B)+\sin(A+B)-\... | \begin{aligned}
\text { LHS }=&(1+\sin (A+B))(1-\sin (A-B)) \\
=& 1-\sin (A-B)+\sin (A+B)-\sin (A+B) \sin (A-B) \\
=& 1- \sin A \cos B + \cos A \sin B + \sin A \cos B + \cos A \sin B -\sin (A+B) \sin (A-B)\\
=& 1 + 2\cos A \sin B - \sin (A+B) \sin (A-B) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (cos(A+B-A+B) - cos(A+B+A-B... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s.
A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s.
I know the standard way to calculate this, but wasn't sure why another approach - stars ... | Although a somewhat offbeat approach, Stars and Bars can be used to solve the problem.
You have $10$ slots, $4$ of which are going to be taken up by $2$ 2's and $2$ 5's. You can construe these as Stars. You will then have $6$ Bars that are divided into $5$ regions, by the $4$ Stars.
So, you compute the number of solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $... | Let $f(x) \in \Bbb Z[x]$. We have $m \mid f(n) \iff f(n)\equiv 0\pmod m$, but when written as this it's clear that $n$ may be replaced by any integer (not only natural numbers) congruent to $n$ modulo $m$, to get a equivalent congruence. The means that
"The largest positive integer that divide $f(n)$ for all positive i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability Problem with 10 players, Bob and a friend being on a team Please check my work.
Out of 10 players including Bob and his two best friends a team of 5 players will be formed. What is the likelihood of Bob making the team with at most one of his friends?
Total Possible Teams
$$\binom{10}{5} = 252$$
Bob with at... | There are indeed $$\binom{10}{5}$$ ways to form a team of five players from among the ten people.
If Bob is selected to be on the team, there are $$\binom{7}{4}$$ ways to select four teammates for Bob from among the seven people who are not his friends.
If Bob is selected to be on the team, there are $$\binom{2}{1}\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic inequality with negative roots Assume the following quadratic inequality: $$0\lt x^2+4x-100$$ The solutions are: $$
x\lt -2-\sqrt{104},\qquad x\gt-2+\sqrt{104}
$$
In this case, the positive root keeps the original direction of the inequality ($\gt$), but the negative root inverts it.
However, for the general ... | You can tackle this by completing the square. Notice that
$$
\begin{align}
0&<ax^2+bx+c \\
&=a\left(x^2+\frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right)+c \\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c
\end{align}
$$
which means that
$$a>0 \implies \left(x+\frac{b}{2a}\right)^2>\frac{b^2-4ac}{4a^2} \imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$ \sqrt a+\sqrt b +\sqrt c = 1$ Prove: $\frac{a^2+bc}{\sqrt{2a^2(b+c)}}+\frac{b^2 + ac}{\sqrt{2b^2(a + c)}}+\frac{c^2 +ab}{\sqrt{2c^2(a+b)}}\geq1$ I'm having some trouble with this problem:
a, b, c are positive real numbers where $ \sqrt a + \sqrt b + \sqrt c = 1 $ .
Prove the following inequality:
$$\frac{a^2 + bc}{\s... | $\sum_{cyc}{\frac{a^2+bc}{\sqrt{2a^2(b+c)}}}\geq\sum_{cyc}{\frac{ab+ac}{\sqrt{2a^2(b+c)}}}$……①
WLOG $a\leq b\leq c$
$LHS-RHS=\sum_{cyc}{\frac{(a-b)(a-c)}{\sqrt{2a^2(b+c)}}}$
$=\frac{(c-a)(c-b)(b-a)}{\sqrt2}(\frac{1}{a(c-b)\sqrt{b+c}}-\frac{1}{b(c-a)\sqrt{a+c}}+\frac{1}{c(b-a)\sqrt{a+b}})$……②
$b(c-a)\geq c(b-a),\sqrt{a+... | {
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"url": "https://math.stackexchange.com/questions/4400380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ I've already shown irreducible. I know I need to get intermediate fields since they correspond to subgroups (which I can use for isomorphisms). But how do I get the intermediate fields here?
edit: I'm thinking perhaps t... | Let $K$ be the splitting field of $x^4 - 16x^2 + 4$
$x^2 - 16x + 4$ has roots $8 \pm \sqrt{60}$. Therefore, the roots of $x^4 - 16x^2 + 4$ are $\pm \sqrt{8 \pm \sqrt{60}}$. This is a degree 4 extension since we have $[K : \mathbb{Q}(\sqrt{60})] = 2$ and $[\mathbb{Q}(\sqrt{60}) : \mathbb{Q}] = 2$.
The Galois group is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ 3^7\cdot (a^9+b^9+c^9)+1\geq 12\cdot (a^3+b^3+c^3)$ Let $a,b,c>0$ s.t. $a+b+c=1$. Show that $$ 3^7\cdot (a^9+b^9+c^9)+1\geq 12\cdot (a^3+b^3+c^3)$$
I tried to apply the next formula:$$a^3+b^3+ c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$$
Also I made substitutions $ab+bc+ac=x$ and $abc=y$ but I am stuck.
| Hint: Show that $ 3^7 a^9 + a \geq 12 a^3$ using AM-GM.
Corollary: The result follows by summing up the cyclic inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating improper integral $\int_0^1 \frac{\log(x)}{x+\alpha}\; dx$ for small positive $\alpha$ Let $\alpha$ be a small positive real number.
How do I obtain
$$ I = \int_0^1 \frac{\log(x)}{x+\alpha}\; dx = -\frac{1}{2}(\log\alpha)^2 - \frac{\pi^2}{6} - \operatorname{Li}_2(-\alpha)$$? Maxima told me the result, but I ... | Assuming, for $u\leq 1$,
\begin{align}\text{Li}_2\left(u\right)&=-\int_0^1 \frac{u\ln x}{1-ux}dx\\
\text{Li}_2\left(1\right)&=\zeta(2)=\frac{\pi^2}{6}
\end{align}
Therefore, $0<\alpha<1$
\begin{align}\int_0^1 \frac{\log(x)}{x+\alpha}\; dx+\text{Li}_2\left(-\alpha\right)&=\underbrace{\frac{1}{\alpha}\int_0^1 \frac{\log(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all solutions to $a^3 + b^3 = p^4,$ where $p$ is prime, and $a,b$ are natural Initially, I tried to break the equation down into $(a+b)(a^2-ab+b^2)$, then break it down into two cases:
I: $(a+b) = p, (a^2-ab+b^2) = p^3$, which yields to a contradiction that $p^2 > p^3.$
II: $(a+b) = p^2, (a^2-ab+b^2) = p^2$, which... | You are on the right track when you render $a+b=p^2,a^2-ab+b^2=p^2$.
You should then plug in
$a^2-ab+b^2=\frac14(a^2+2
ab+b^2)+\frac34(a^2-2ab+b^2)$
$=\frac14(a+b)^2+\frac34(a-b)^2.$
Then conclude that with $a+b=p^2,a^2-ab+b^2=p^2$, you must have
$p^2\ge\frac14p^4.$
From this $p\le2$, thereby limiting your solution can... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a limit as product of cos $$\lim\limits_{n\to\infty}\cos\frac{1}{n\sqrt{n}}\cos\frac{2}{n\sqrt{n}}\cdots\cos\frac{n}{n\sqrt{n}}$$
It is not hard to prove that the limit exists, but is that possible to calculate the limit? Suggestions are welcome!
| The infinite product is a bit difficult to work with, so to make things easier we compute the log of the product instead. Our limit becomes
$$\lim_{n\to\infty}\sum^n_{i=1}\ln\cos\frac{i}{n^{\frac{3}{2}}}$$
We use a Taylor series to compute $\ln\cos x$ around $x=0$. We know $\cos x=1-\frac{x^2}{2}+O\left(x^4\right)$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to evaluate $\int _{0}^{\frac{\pi}{2}} \frac{\mathrm dx}{a+\cos x}$ using contour integration? I want to evaluate:
$\displaystyle \int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \tag*{}$
With my basic knowledge of Cauchy Residue theorem:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{\mathrm dx}{a + \cos x} &=... | I doubt you can use contour integration to handle this since you can't obtain a closed curve. Noting
$$ \cos x=\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}$$
one has, under $\tan(\frac x2)\to t$,
\begin{eqnarray}
&&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \\
&=&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx... | {
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"url": "https://math.stackexchange.com/questions/4404852",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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How can I solve the following 2nd Order DE? $$y''-\frac{1}{x}y'=2x\cdot cos(x)$$
For the homogeneous part I multiplied through with $x^2$ and got a second order Cauchy Euler equation with the general solution: $$y_h (x)=A x^2 +B$$
Then for the particular solution I tried using the method of undetermined coefficients bu... | This reduces the problem to a first order Diff Eq.
$u= y'\\
xu' - u = 2x^2\cos x$
Choose a candidate for the particular solution that could work.
$u_p = x\sin x$
"Generalize" this by adding terms that might come up in the derivative that we hope will cancel out. Give every term a coefficient.
$u_p = Ax\sin x + Bx\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n} = \sum_{r=1}^{n}\frac{2^r-1}{r}$ Good day,
I was solving this problem:
Show that $$\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n} = \sum_{r=1... | You need the two identities:
$\displaystyle {n-1\choose k} + {n-1\choose k-1} = {n\choose k}$ and $\displaystyle \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$
Let $\displaystyle S_n = \sum_{r=1}^{n} \frac{1}{r}{\binom{n}{r}}$ and note that (by def.) $\displaystyle {n-1\choose n} =0. $
$\begin{aligned} \sum_{r=1}^{n} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. A... | Edit: Much simpler answer: \begin{align}J_n(a)=\int_0^a\frac{\cos(n-2)x}{\cos^nx}\,dx&=\int_0^a\frac{\cos(n-1)x\cos x+\sin(n-1)x\sin x}{\cos^nx}\,dx\\&=\int_0^a\frac{\cos(n-1)x}{\cos^{n-1}x}-\frac{(-\sin x)\sin(n-1)x}{\cos^nx}\,dx\\&=\frac1{n-1}\int_0^a\frac d{dx}\frac{\sin(n-1)x}{\cos^{n-1}x}\,dx\\&=\frac{\sin(n-1)a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Number theory problem (my solution included). Let n be a natural number greater than 1. We call a natural number $a>2$ ,$n-$decomposable if $a^n-2^n$ is divisible by all numbers of the form $a^d + 2^d$, where $d|n$ and $1\le d<n$. Find all $n$ for which there is an $n$-decomposable number.
I got:
$n$ is decomposable $\... | The latter part of your proof doesn’t work - it only shows that $3$ is not $n$-decomposible (or by extension, odd numbers are not $n$-decomposible.) It doesn’t answer whether there is any $n$-decomposible number.
And, indeed, there are other $n.$
If $n=p$ is an odd prime, then the only condition is $a+2$ being a factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For $n \geq 2$, show that $\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$ Good day,
Can someone help me with giving hints for this problem:
Show that for $n \geq 2, n \in \mathbb{Z}$, $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$$
I tried $$\sum_{r = 1}^{n} r \sqrt{\binom{n}... | By Cauchy-Schwarz,
\begin{align*}
\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} &\le \left(\sum_{r = 1}^{n} r^2\right)^{1/2}\left(\sum_{r = 1}^{n} \binom{n}{r}\right)^{1/2} \\
&= \left((2^n-1)\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right)\right)^{1/2} \\
&< \left(2^{n-1}\cdot2\left(\frac{n^3}{3} + \frac{n^2}{2} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the limits while changing limit of an infinite sum into integral. I was solving the following question.
Find the following limit.
$$\lim_{n\to \infty}\dfrac1n \left(\dfrac{1}{1 + \sin\left(\dfrac{\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{2\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{3\pi}{2n}\righ... | Remark:$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n} f(\frac{k}{n})=\int_0^1 f(x)dx$$
$$= \lim_{n\to \infty}\dfrac1n \left(\sum_{k=1}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)
=\\ \lim_{n\to \infty}\dfrac1n \left(-1+\sum_{k=0}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)=\\
\lim_{n\to \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probability of drawing 4 aces when drawing 5 cards from a regular deck of cards. I cannot seem to wrap my head around how to calculate the probability of drawing four ace when drawing five cards from a deck of cards. My intuition tells me the math below but its wrong for some reason...
$
\frac{\frac{4}{52}\frac{3}{51}\... | The problem can be solved in two ways. We can either take the order of selection into account or not take it into account.
Taking the order of selection into account: There are $52 - 4 = 48$ ways to select the card which is not an ace. There are $5!$ ways to arrange the four aces and that card. There are $P(52, 5) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the number of rearrangements of AABBBCCDDD where there are no two consecutive As or Bs I tried using inclusion exclusion where set $A$ is all the rearrangements where two $A$s are consecutive and set $B$ where two $B$s are consecutive. However, I got $8400$ which is incorrect. I think it has to do with there being... | This answer is rather a supplement which could be used as crosscheck for manual calculations. We consider a $4$-ary alphabet built from letters $\mathcal{V}=\{A,B,C,D\}$. Words which do not have any consecutive equal letters are called Smirnov words. A generating function for Smirnov words is given as
\begin{align*}
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Inverse Derivative: Error/Intuition Problem is to find $[f^{-1}(4)]'$ given $f(x)=\frac{x^3+7}{2}$.
Way One: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}$. Therefore, $\sqrt[3]{2x-7}=y$ (i.e. our inverse function). So, $f^{-1}(x)=(2x-7)^{\frac{1}{3}}$. So, $\frac{d}{dx}f^{-1}(x)=\frac{2}{3}(2x-7)... | It was tricky for me too so I started doing these problems the following way. You know that $f(f^{-1}(x)) = x$ so differentiating both sides gives you $$f'(f^{-1}(x)) \cdot \frac{d}{dx} f^{-1}(x) = 1 \implies \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x)) }.$$ Now $f^{-1}(4) = 1$ so $$\frac{d}{dx} f^{-1}(4) = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Any hint on how to prove that the two given conditions may not be fulfilled simultaneously? I have these two conditions for $0<a<2\pi$ and $b>0$ and real.
$$ \sin \left(\frac{\pi a}{2 (\pi -a)}\right)=\frac{a }{2 \pi -a}\;\sin \left(\frac{\pi (a-2 \pi )}{2 (a-\pi )}\right)+\frac{4 b }{2 \pi -a}\;\sin \left(\frac{\pi ... | By the hint given by @Blue , considering $x:=\frac{\pi ^2}{2 (\pi -a)}$ my equations will be
$$ \sin \left( x-\frac\pi2\right)=\frac{a }{2 \pi -a}\;\sin \left(x+\frac\pi2\right)+\frac{4 b }{2 \pi -a}\;\sin \left(x\right) $$
$$ \cos \left(x-\frac\pi2\right)=\frac{a}{2 \pi -a} \;\cos \left(x+\frac\pi2\right)\qquad\qqua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Attempt to prove an inequality by induction I am trying to prove an inequality by induction which is as follows:
$$\frac{(2n)!}{2^{2n}\cdot (n!)^2} \le \frac{1}{\sqrt{3n + 1}}$$
Base Case, i.e, for n =1,
$$\frac{(2(1))!}{2^{2(1)}\cdot ((1)!)^2} \le \frac{1}{\sqrt{3(1) + 1}}$$
$$\implies \frac{2!}{2^{2}\cdot (1)^2} \le ... | I think you messed something up in the computation.
If you want to prove this using induction, what you need to show is that
\begin{equation*}
\frac{1}{\sqrt{3n+1}}\frac{(2n+1)(2n+2)}{4(n+1)^2}\leq \frac{1}{\sqrt{3n+4}}.
\end{equation*}
After simplifying this means
\begin{equation*}
\frac{(2n+1)^2}{(2n+2)^2}\leq \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Computing $\int_{-1}^{1} \frac{\ln (1+y)}{y} d y$ with simple method. Using the series
$$
\ln (1+y)=\sum_{n=0}^{\infty} \frac{(-1)^{n} y^{n+1}}{n+1} \text { for }|y|<1
$$
to convert the integral into $$
\begin{aligned}
\int_{-1}^{1} \frac{\ln (1+y)}{y} d y &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} \int_{-1}^{1} y^{n} ... | An elementary solution
\begin{align}
\int_{-1}^{1} \frac{\ln (1+y)}{y} &{d y}
\overset{y\to -y} = \frac12 \int_{-1}^{1} \frac{\ln \frac{1+y}{1-y}}{y} d y
\overset{y\to \frac{1+y}{1-y}}=\int_0^\infty
\frac{\ln y}{y^2-1}dy\\
=& \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy
=\frac\pi2 \int_0^1 \frac1{\sqrt{1-x^2}}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$ For $a\in\mathbb R$, I want to evaluate the integral
$$ I = \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx.$$
I tried to integration by parts by considering $\left( \tan^{-1}(x) \right)... | We proceed by FShrike's comment. We have
$$\frac{\partial I}{\partial a} = \int \frac{1}{x^2+1} \frac{1}{(a-x)^2+1} dx = \frac{2\pi}{a^2+4}.$$
Since $I(a=0)=0$, we obtain
$$I(a) = \int_0^a \frac{2\pi}{a^2+4}da = \pi \tan^{-1}\frac a2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our resu... | Since @Quanto provided a detailed and very good answer, I shall make the story short.
Working directly the problem of the difference between two consecutive terms
$$I_{2n+1}-I_{2n}=-\frac{3 }{32 } \frac{1 }{2 n+1}\Phi \left(-1,1,\frac{4n+3}{2}\right)$$ where appears the Lerch transcendent function.
We can also write it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
If a prime divides an integer of the form $n^2+1$ then ... This question is from my problem set in number theory. I tried it a week ago and again today but both times I couldn't solve it.
Problem: If a prime p divides an integer of the form $n^2+1$ , then $p\not\equiv 3 \pmod{4}$.
I thought of attempting this problem ... | You're right on assuming that $p\equiv 3\pmod{4}$ and then trying arrive at a contradiction. Here's how it may work out.
Suppose that $p\equiv 3\pmod{4}$ and $p$ divides $n^2+1.$ If $p$ divides $n^2+1,$ then we have;
$$n^2+1\equiv 0\pmod{p}$$
$$n^2\equiv -1\pmod{p}.$$
Now, since we have assumed that $p\equiv 3\pmod{4}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Expected value of the distance of sample average from the overall average Let $\mathbf{y}_1, \mathbf{y}_2, \dots, \mathbf{y}_N$ be a sequence of $N$ vectors in $\mathbb{R}^d$ and $\bar{\mathbf{y}}_N$ be the overall average, i.e.,
$$
\bar{\mathbf{y}}_N=\frac{1}{N}\sum_{i=1}^{N}\mathbf{y}_i.
$$
Also, let $\bar{\mathbf{y}... | Let $\mathbf{Y} = \begin{pmatrix} \mathbf{y}_1 & \cdots & \mathbf{y}_N \end{pmatrix} \in \mathbb{R}^{n \times N}$. Then notice that
\begin{align*}
\overline{\mathbf{y}} - \overline{\mathbf{y}}_N = \mathbf{Y}\left(n^{-1} \mathbf{1}_A - N^{-1}\mathbf{1}_N\right)
\end{align*}
where $\mathbf{1}_N, \mathbf{1}_A \in \mathbb{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The number of integer points on the curve $(7x-1)^2+(7y-1)^2=n$ The number of integral solutions to the equation
$$x^2+y^2=n$$
is defined to be $r_2(n)$ and if $n=2^ap_1^{a_1}\dots p_k^{a_k}q^2$ where $p_i\equiv 1\mod 4$ and $q$ is the product of primes which are $3\mod 4$, then
$$r_2(n)=4(a_1+1)\dots(a_k+1).$$
If we r... | Comment: We may use following equation as a particular case:
$$(a-1)^2+(a+1)^2=b^2+1\space\space\space\space\space\space(1)$$
This equation can have infinitely many solution; if a and b satisfy this equation then considering following identity we can have subsequent solutions:
$$(2b+3a-1)^2+(2b+3a+1)^2=(3b+4a)^2+1\spac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show that $\cot\left(\frac{\pi}{4}+\beta\right)+\frac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$ Show that $$\cot\left(\dfrac{\pi}{4}+\beta\right)+\dfrac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$$
I'm supposed to solve this problem only with sum and difference formulas (identities).
So the LHS is $$\dfrac{\cot\dfrac{\pi}{4}... | The alternative approach is to stay with your original approach and complete it.
$\displaystyle \frac{4\cot\beta}{1-\cot^2\beta}
= \frac{\frac{4}{\tan(\beta)}}{1 - \frac{1}{\tan^2(\beta)}}
= \frac{\frac{4}{\tan(\beta)}}{\frac{\tan^2(\beta) - 1}{\tan^2(\beta)}}
$
$\displaystyle = ~\frac{4\tan(\beta)}{\tan^2(\beta) - 1}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Difficulty evaluating $\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$ In evaluating
$$\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$$ I did not have a situation where the polynomial under the square root has lower degree. Neither trigonometric substitutions nor variable changes help, at least how I ... | The limits of integration and the form of the radical in the denominator strongly suggest $x = 2\sqrt{2}\sinh(u)$. Performing this substitution gives and expanding in partial fractions gives
$$
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\infty\frac{du}{1 + 2^{7/2}\sinh^3(u)}
=\frac{1}{6}\sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Question on Lagrange multipliers method
My understanding
Finding the minimum of the function using Lagrange multipliers method
\begin{align}
f(x,y,z) &= (y^2+z^2-x^2)^2+4x^2(x+y)^2 \\
x+y+z &= 1 \\
x &\leq y \\
x, y, z &\geq 0
\end{align}
My approach is to let
$$h(x,y,z)=f-\lambda g$$ where $~g=x+y+z-1=0$.
Geometrical... | By setting $\lambda=4\mu$ you get
$$
\left\{ \begin{array}{l}
- x\left( {y^2 + z^2 - x^2 } \right) + 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = \mu \\
y\left( {y^2 + z^2 - x^2 } \right) + 2x^2 \left( {x + y} \right) = \mu \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $|1 - (1+\frac{z-1}{n})^m| \leq 1$ where $|z|\leq 1$ and $mHow to prove
$$\Big|1 - \big(1+\frac{z-1}{n}\big)^m\Big| \leq 1$$
where $z\in\mathbb{C}$, $|z|\leq 1$ and $m,n\in\mathbb{N}$ and $m<n$.
I have run numerical experiments and believe the inequality is correct.
| Based on @onriv's nice answer and user3750444's comment therein:
(Without using Maximum Modulus Principle)
Let $w = 1 + \frac{z - 1}{n}$
and $r = |w|$. We have
\begin{align*}
\left|1 - w^m\right|
&= |1 - w|\cdot |1 + w + w^2 + \cdots + w^{m-1}|\\
&\le |1 - w|\cdot (1 + r + r^2 + \cdots + r^{m-1})\\
&\le |1 - w|\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdo... | Here is another approach
Define
$$
b_m=\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}\tag1
$$
By assumption, we have the convergence of
$$
\sum_{m=0}^\infty b_m=\sum_{k=2}^\infty\frac1{a_k}\tag2
$$
Next,
$$
\begin{align}
\sum_{k=2^m+1}^{2^{m+1}}a_kk^p\overbrace{\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}}^{b_m}
&\ge\left(\sum_{k=2^m+1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)... | With respect to your first question, I don't think induction would be a method that is very helpful here. The problem would be to relate different functions $f(n)$. This is more something that asks for generating functions.
\begin{eqnarray}
f(n)
& = & \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4453585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solution of $\theta$ when $\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$ I came across this trigonometry problem. If, $$\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$$
What is the value of $\theta$
I got the solution that $\theta$ will be $\frac{\pi}{3}$ by expanding the equation and turning into,
$$\sin^4(\theta)+\sqrt3\si... | $$\sqrt{3}\,t^4 + 8\,t^3 - \sqrt{3} =\left(t-\frac{1}{\sqrt{3}}\right) \left(\sqrt{3} t^3+9 t^2+3 \sqrt{3} t+3\right).$$
Using the hyperbolic method for the cubic gives for the real root
$$t=-\sqrt{3}-2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)$$ from which
$$\theta=2\pi -2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$? For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$?
My thoughts: We know that $(1-\frac{x}{n})^n\leq e^x$, so the numerator is $\leq e^{2x}$. So, we can play with $\frac... | Assume that $n \ge 2$.
Let
$$I_n := \int_0^n \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x.$$
Using $\ln(1 + u) \le u$ for all $u > -1$, we have, for all $0 \le x < n$,
\begin{align*}
\ln\left(1 - \frac{x}{n}\right)
&= \ln \left(1 - \frac{\sqrt n}{n}\right) + \ln\left(1 - \left(1 - \frac{\sqrt n}{n}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality cas... | We first prove the following lemma.
Lemma. Let $\lambda>0$. Then
$$\sup_{x>0}\left(\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+\frac{\lambda}x}}\right)=\begin{cases}\frac2{\sqrt{1+\sqrt\lambda}}&\text{if }\lambda<4\\\sqrt{\frac{\lambda}{\lambda-1}}&\text{if }\lambda\geq 4.\end{cases}$$
Proof. Fixing $\lambda$ constant, let the l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Showing that $x^3 + y = y^3 + x$ is an equivalence relation I am asked to prove that: $x^3 + y = y^3 + x$ is an equivalence relation. So far I have the following:
*
*Reflexive:
$m^3 +m = m^3 +m$
*Symmetric:
$m^3 + n = n^3 + m \rightarrow n^3 + m = m^3 + n$ Then:
$n^3 + m = m^3 +n$ From hypothesis
*Transitivity ... | Alternative approach:
$x \sim y \iff x^3 + y = y^3 + x \iff (x^3 - x) = (y^3 - y).$
Then, $\{ ~x \sim y ~~~~\text{and}~~~~ y \sim z ~\} \implies $
$(x^3 - x) = (y^3 - y) ~~~~\text{and}~~~~ (y^3 - y) = (z^3 - z).$
This implies that $(x^3 - x) = (z^3 - z) \iff x \sim z.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$... | I think OP approach is the simplest one because it can be done just in mind without paper writing. One of possible another ways
$$A=xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=x^2(y-z)+x(y^2-z^2)+yz(y-z)$$
$(y-z)$ is common factor of $(y-z)$ and $(y^2-z^2)=(y-z)(y+z)$, so we can factor it out:
$$A=(y-z)(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\a... | For this problem, I like your work, through the conclusion that
$$\tan(\alpha) = \frac{\sqrt{5} - 1}{2}. \tag1 $$
In my opinion, the simplest approach to complete the problem is to forgo any attempt at elegance, and simply use the following identities (one of which you have already referred to):
*
*$\displaystyle \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Remainder of $\frac{3x^{2019}+5x^{1019}-7x+4}{x^2-1}$ I don't understand how I should go about solving the following question:
Find the remainder when polynomial $f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.
I tried to use the factor theorem, but I never encountered a problem with a divisor which, in this ca... | Degree of the remainder less then degree of quotient then remainder is linear , $R(x)=ax+b$
$f(x)=3x^{2019}+5x^{1019}-7x+4=Q(x)(x-1)(x+1)+ax+b$
$f(1)=a+b=5$
$f(-1)=-a+b=3$
then $a=1,b=4$
$R(x)=x+4$
OR
To find remainder when
$f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.
By using remainder theorem
$x^2-1=0$
then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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