Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. ... | The coefficients of the polynomial is real, hence complex roots comes in conjugate pairs.
If $$\sum_{i=0}^n a_i z^i = 0$$
We can take conjugate and see that
$$\sum_{i=0}^n a_i \bar{z}^i = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Exponential diophantine equation $2^x+7^y=9^z$. The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions:
\begin{eqnarray*}
(x_1,y_1,z_1)&=&(3,0,1),\\
(x_2,y_2,z_2)&=&(1,1,1)\\
(x_3,y_3,z_3)&=&(5,2,2).\\
\end{e... | This is a PARTIAL answer. What remains to show is that there are not an infinite number of solutions for $x=1$. Assume $x\ge 3$. Then as in Conner's answer,
$$2^x = (3^z-7^v)(3^z+7^v).$$
This gives for some positive integer $a$ the equation
$$2^a(3^z-7^v)=(3^z+7^v).$$
Rearranging terms gives
$$(2^a-1)3^z = (2^a+1)7^v,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By con... | The equation $ \ \cos(5 \theta) \ = \ 0 \ $ has the set of angle solutions $ \ 5 \theta \ = \ \frac{\pi}{2} \ + \ k·\pi \ \Rightarrow \ \theta \ = \ \frac{(2k + 1) \ · \ \pi}{10} \ \ , \ $ arranged on the unit circle as presented in the graph below. Upon solving the polyomial equation in cosine which is equivalent to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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how to express the $n$th power of the cosine as a series of cosines? Which is the correct way for expressing the $n$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n=
\frac{1}{2^n}\left... | Here is a way to fix the factor of $2$ while still summing from $0$ to $n$
(setting aside the question of whether that is the best way to do the sum).
You have
$$
\cos^n(\theta) =
\frac{1}{2^n}\left(
z^{-n} + \binom n1 z^{-(n-2)} + \binom n2 z^{-(n-4)} + \cdots
+ \binom n2 z^{n-4} + \binom n1 z^{n-2} + z^n \right).
$$... | {
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"url": "https://math.stackexchange.com/questions/4300731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$ $$\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$$ where $a = 0$ and $b= \ln 2$
I just need a little hint to start. Please don't answer it completely. Just a hint.
Edit : Originally the question was to find the arc length of the curve $f(x) = m \sinh x$ on the interval $[0, \ln 2]$. Bu... | If you do not know yet about elliptic integrals, you do not have much choice.
Start using
$$\cosh^2(x)=\frac 1 2(1+\cosh(2x))$$ which makes
$$1+m^2\cosh^2(x)=\frac{1}{2} m^2 \cosh (2 x)+\frac{m^2}{2}+1=\frac{1}{2} m^2\Big[1+\frac{2}{m^2}+\cosh (2 x) \Big]$$
Now, use the series expansion of $\cosh(2x)$. This will give
$... | {
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"timestamp": "2023-03-29T00:00:00",
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About an inequality wich have a link with $\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2$ Hi it's a follow up of show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$:
Problem :
Let $a,x>0$ then (dis)prove :
$$\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\... | Conjectures :
Using the comment above and the fact that $r(x)$ increases along the positive real axis we have the first conjecture :
Let $x>0$ then we have :
$$x\cdot r\left(\left(\frac{4x^{2}}{\left(x+2\right)^{2}}\right)^{\frac{-1}{x}}\left(x-\sqrt{\frac{4x^{2}}{\left(x+2\right)^{2}}}\right)^{-1}\right)\geq 2$$
After... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$
Here the sol... | Rewrite $2x+3y+1\le a(x+2)+b(y+3)$ as $(2-a)x+(3-b)y\leq 2a+3b-1$. Solve for the intersection of the line $(2-a)x+(3-b)y=2a+3b-1$ with $x^2+y^2=1$ to get that the $x$-coordinates of the intersection are $$\frac{-2a^2-3ab+5a+6b-2\pm\sqrt{-(b-3)^2(3a^2+12ab+8b^2-12)}}{a^2-4a+b^2-6b+13}.$$ If this has two real solutions, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$g=\rho^2 (d x^2 + dy^2) \Rightarrow g = ds^2 +\tanh^2 s d\theta^2$ The question is from the 10th page of Topping's Lectures on Ricci flow, the calculation about Hamilton's cigar soliton.
Consider $R^2$ with metric
$$
g=\rho^2 (dx^2 + dy^2),~~~\rho^2= \frac{1}{1+x^2 + y^2}
$$
where $dx^2 = dx \otimes dx $. Topping st... | As you noted, in polar coordinates, the metric reads
$$
g = \frac{1}{1+r^2}\mathrm{d}r^2 + \frac{r^2}{1+r^2}\mathrm{d}\theta^2
$$
Heuristically, you are looking for $s$ such that $\mathrm{d}s^2 = \frac{1}{1+r^2}\mathrm{d}r^2$, which can be stated as ${\mathrm{d}s} = \frac{1}{\sqrt{1+r^2}}{\mathrm{d}r}$. So let's define... | {
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"url": "https://math.stackexchange.com/questions/4310531",
"timestamp": "2023-03-29T00:00:00",
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Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using ... | $9-3\sqrt[3]{3}+\sqrt[3]{9} = a^2 -ab+b^2$
$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}} \cdot \frac{3+\sqrt[3]{9}}{3+\sqrt[3]{9}}=\frac{12+4\sqrt[3]{9}}{30}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\p... | From what you have deduced, we can apply here the Faà di Bruno's formula
$$
\begin{align}
I_n(a) &= \frac{(-1)^{n-1}\pi}{(n-1)!} \frac{d^{n-1}}{da^{n-1}} \left(\frac{1}{\sqrt{a^2 - 1}}\right) \\
&= \frac{(-1)^{n-1}\pi}{(n-1)!} \frac{d^{n-1} \sqrt{b(a)}}{da^{n-1}} \\
&= \sum_{}\frac{(n-1)!}{\prod_{k=1}^{n-1}m_k! k!^{m_k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the easier way to find the circle given three points? Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, if $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ then there will be a circle passing through them. The general form of the circle is $$x^2 + y^2 + dx + ey ... | Move the circle so that it passes through the origin (subtract a point from all three). The equation loses a coefficient,
$$x^2+y^2+dx+ey=0$$
which is easier to solve. Then reverse-translate.
$$-\frac d2=\dfrac{\begin{vmatrix}x_1^2+y_1^2&y_1\\x_2^2+y_2^2&y_2\end{vmatrix}}{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Proving that equation has at least one solution in the interval (-1,1) We are given two numbers $a>0$ and $b>0$ and the equation given is
$$ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0 $$
Prove that this equation has at least one solution in the interval $(-1,1)$. Now I can let
$$ f(x) = \frac{a}{x^3 + 2x^2 - ... | The two cubics do not have a common root.That can be exploited as follows:
Let $g(x)=a(x^{3}+x-2)+b(x^{3}+2x^{2}-1)$. Then $g(1)=2b>0$ and $g(-1)=-4a<0$. Hence, there exists $x \in (-1,1)$ such that $g(x)=0$. At this point $x$ neither $x^{3}+x-2$ nor $x^{3}+2x^{2}-1$ can be $0$: If $x^{3}+x-2=0$ we get $g(x)=b(x^{3}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people ... | The number of ways is
$$\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2} =6\times35+4\times35+21 = 371$$
"At least two women" means either $2$, $3$ or $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Spivak Calculus, Ch. 5 Limits: Finding the smallest $b$ in $|x-3|Consider the function $f(x)=x^2$.
In the text of chapter 5 of Spivak's Calculus, he goes through the following argument to show that $f(x)$ approaches the limit $9$ near $x=3$.
We want that
$$|x^2-9|=|x-3||x+3|<\epsilon$$
Assume $|x-3|<1$.
$$\implies 2<x<... | Let $x = 3+h$ then the problem is to estimate $h$, now you need $$-\varepsilon< (3+h)^2 - 9<\varepsilon$$
which is same as saying $ 9-\varepsilon < (3+h)^2 < 9+\varepsilon$, Now we need to check cases if $\varepsilon < 9$ and othercases, using that you will be able to get the interval for $h$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve this nonhomegenous ode $y''+4y = \cos(2x)$ Solve the given nonhomogenous linear ODE by variation of parameters or undetermined coefficients $y'' + 4y = \cos(2x)$.
A general solution is $y_1 = \cos(2x), y_2 = \sin(2x)$, and the Wronksin determinant is equal to 2.
Plugging this into the equation for the method of v... | Null space solution
$m^2 +4 =0$
$\implies m=\pm 2i$
Hence, $ y_N = C_1 \sin 2x + C_2 \cos 2x$.
Since, $\cos2x$ already in null space solution, take particular integral as $$y_p =x(A \sin 2x +B \cos 2x) $$
Then, \begin{align} y'_p &{=A \sin 2x +B \cos 2x +2x(A \cos 2x -B \sin 2x )}\\ \end{align}
$y''_p =4A \cos 2x -4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Number of line of curvature meets at one point I am now consider the surface given by $$f=(x,y,x^3-3xy^2)$$
And I have been asked to prove that there are three line of curvature meet at origan point.
I can compute that
\begin{align*}
f_x&=(1,0,3x^2-3y^2)\\
f_y&=(0,1,-6xy)\\
f_x\times f_y&=(-3(x^2-y^2),6xy,1)\\
... | Principal directions at $(x,0)$ (for $x\ne 0$) are $(1,0)$ and $(0,1)$.
The computations are very difficult unless you happen to recognize that $x^3-3xy$ is the real part of $(x+iy)^3$. Thus, this surface has rotational symmetry about the origin, with rotations of $\pm 2\pi/3$. If one does the computation with a polar ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $A$, $B$ be square matrices of order $2$ such that $|I_2 + AB| = 0$. Prove that $|I_2 + BA| = 0$. In this question, I denote
$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$
$$B=\begin{bmatrix}e&f\\g&h\end{bmatrix}.$$
So, from $|I_2 + AB| = 0$, I'll have:
$$(ae+fc+1)(gb+hd+1) = (ga+hc)(eb+fd).$$
If $|I_2 + BA| = 0$ (whi... | It is just a corollary (taking $n = m = 2$) of the result below:
For $A \in F^{m \times n}, B \in F^{n \times m}$, it follows that
\begin{equation}
\det(I_{(m)} + AB) = \det(I_{(n)} + BA). \tag{$*$}
\end{equation}
To prove identity $(*)$, consider the block matrix:
\begin{align}
\Delta = \begin{pmatrix} I_{(m)} & A \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4333104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are ... | We have the following : \begin{aligned}\frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}&=\frac{\left(1+x-1\right)^{5}}{\left(1+x\right)^{18}}-\frac{\left(1+x-1\right)^{11}}{\left(1+x\right)^{18}}\\ &=\frac{\sum\limits_{k=0}^{5}{\left(-1\right)^{k}\binom{5}{k}\left(1+x\right)^{5-k}}}{\left(1+x\right)^{18}}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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minimum polynomial of $\sqrt[3]{3}+\sqrt[3]{5}$ over $\mathbb {Q}$ I have tried putting $x=\sqrt[3]{3}+\sqrt[3]{5}$ but i got stuck because after elevating to the third power I obtain again cube roots and I am stuck . How can I solve?
| \begin{align}
x &= \sqrt[3]{3} + \sqrt[3]{5} \\
x^3 &= 3 + 3\sqrt[3]{3\cdot 3 \cdot 5} + 3\sqrt[3]{3 \cdot 5 \cdot 5} + 5 \\
x^3 - 8 &= 3\sqrt[3]{3 \cdot 5}(\sqrt[3]{3} + \sqrt[3]{5}) \\
x^3 - 8 &= 3\sqrt[3]{15}x\\
(x^3-8)^3 &= 405 x^3 \\
x^9 - 24x^6 - 213 x^3 - 512 &= 0.
\end{align}
Therefore the minimum polynomial o... | {
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"timestamp": "2023-03-29T00:00:00",
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prove that $(p,mp]$ will always contain at least $y-1$ cubes
If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.
This works for small examples, such as for $n=7, m=4$ wh... | Lemma (without proof): Show that $ ( n, mn ] $ contains at least $y$ cubes iff
$$ \lfloor \sqrt[3]{mn} \rfloor - \lfloor \sqrt[3] {n} \rfloor \geq y.$$
Corollary: The problem can be restated to:
For $ p \geq n \geq 1$, and $ m \geq 1$, show that
$$ \lfloor \sqrt[3]{mp} \rfloor - \lfloor \sqrt[3] {p} \rfloor \geq \lf... | {
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Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\... | Using your parametrization we have:
$$\begin{align}\iint_{D_1}\sqrt{x^2-y^2}dxdy&=\int_0^1dx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{x^2-y^2}dy\\&=\int_0^1xdx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{1-\left(\frac{y}{x}\right)^2}dy \:\: \blacktriangle\end{align}$$
Now, suppose we know that (there's pleny of proofs of this fact on the... | {
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"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1... | Another approach, by solving a linear 2-term recurrence relation.
Let $ y_i = x_i - \frac{ 1-x_1}{ n-1}$.
Then, the equations become
$ny_i - y_{i+1} = 0$ for $ i = 1$ to $k-1$ and $y_1 + ny_k = 2 - \frac{n+1}{n-1} ( 1 - x_1)$,
which then become
$ y_{i+1} = n^{i} y_1 $ for $ i = 1 $ to $k-1$ and $y_1 (1+n^k) = 2 - \fr... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of this
Let $a,b,c,d$ be nonnegative integers, such that $a+b+c+d=4$. Find the minimum value of $$\sum_{cyc}\frac{a}{b^3+4}.$$
My textbook says that the answer is $2/3$ achieved at $(2,2,0,0)$, but my method show that it is $1/17$ here it is
Since $a+b+c+d=4$, we have $a,b,c,d\le 4$ so $b^3+4\l... | You have a mistake in replacement. Actually, the point $(a,b,c,d)=(4,0,0,0)$ leads to
$$
\sum_{cyc}\frac{a}{b^3+4}=\frac{4}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{4^3+4}=1.
$$
What you found, is the value of $\sum_{cyc}\frac{a}{a^3+4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
How to find the ratio of area in increasing order for the given figure?
Consider the attached image for the diagram. I'm interested in finding the area of the 4 parts so formed as the ratio of increasing order. Here line segment BD is diagonal for square and AE is so-called median for the square as it divides the line... | The sketch given below might help you.
$\underline{\text{Added at the request of @Utkarsh}}$
Let us denote the point of intersection of $AE$ and $BD$ as $V$. We draw a line segment perpendicular to the side $AB$ through $V$ to cut the side $CD$ at $U$. Another line segment is drawn through $V$ perpendicular to the sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $\cos^{10}x + \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$ If $\cos^{10}x+ \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$. I've tried solving using index manipulation but no way. I think we're to use $\sin^2 x+\cos^2 x=1$. But I don't know how. Need help please.
| Here is one solution. Let $A=\sin^2 x\cos^2 x$
$$\sin^4+\cos^4=\sin^2+\cos^2-2\sin^2 x\cos^2 x=1-2A$$
$$\sin^6+\cos^6=\sin^4+\cos^4-A(\sin^2 +\cos^2 x)=1-3A$$
$$\sin^8+\cos^8=\sin^6+\cos^6-A(\sin^4 +\cos^4 x)=1-4A+2A^2$$
$$\sin^{10}+\cos^{10}=\sin^8+\cos^8-A(\sin^6 +\cos^6 x)$$
$$=1-5A+5A^2$$
Thus we have $$5A^2-5A+1=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the area $ S_ {OGBH}$ in the figure below For reference:
In the figure, calculate area $ S_ {OGBH} $ if the triangle area $ABC=9\ \mathrm{m^2}$ and $AB=3\ \mathrm m$ and $AO = 2\ \mathrm m$. (Answer: $5\ \mathrm{m^2}$)
My progress:
Draw $BO.$
Let $AG = x$ and $HO=GO=R.$
$BG = BH.$
$S_{BGO} = S_{BOH}$
$\displays... | Say radius of the semicircle is $r$, Then $OG = OH = r$. Now drop a perp from $B$ to $AC$. Say the foot of perp is $E$.
$\triangle ABE \sim \triangle AOG$ with hypotenuse $3$ and $2$ respectively.
So, $S_{AOG} = \frac 49 S_{ABE}$ and $BE = \dfrac{3r}{2}$
Next, using similarity of $\triangle COH$ and $\triangle CBE$,
$S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)$ How to show that if $a_k,b_k,c_k\in \mathbb{R}^+$ for which $a_kc_k-b_k^2\geq 0$ for $k=1,2$, then
$$
(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)?
$$
I tried to check some well-known identities such as Young's inequality, but I failed to make it work.
| HINT
You can start with the RHS as follows
\begin{align*}
(a_{1} + a_{2})(c_{1} + c_{2}) & = a_{1}c_{1} + a_{1}c_{2} + a_{2}c_{1} + a_{2}c_{2}\\\\
& \geq b^{2}_{1} + a_{1}c_{2} + a_{2}c_{2} + b^{2}_{2}
\end{align*}
Now you can apply the AM-GM inequality in order to obtain
\begin{align*}
a_{1}c_{2} + a_{2}c_{1} & \geq 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I prac... | $\require{cancel}$I'd us a combination putting over a common denominator and factoring.
Using that $foo = \sqrt{foo}^2$ we can factor $\sqrt{n+1}$ out of $(n+2)\sqrt{n+1} + (n+1)\sqrt{(n+2)}= \color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]$
So
$1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $a$ and $b$ be positive integers such that $an + 1$ s a cube if and only if $ bn + 1$ is a cube. Prove that $a = b.$ Let $a$ and $b$ be positive integers such that $an + 1$ is a cube if and only if $bn + 1$ is a cube. Prove that $a = b.$
By choosing $p_n^3 \equiv 1 \mod b$ , we find that there are infinitely many n... | Here's a fairly short proof.
There are infinitely many positive integers $n$ for which $abn+1$ is a cube $x^3$ (any $x\equiv 1\pmod{ab}$ works). Then, $a^2n+1$ and $b^2n+1$ must both be cubes; say $a^2n+1=y^3$ and $b^2n+1=z^3$. We have
$$(y^3-1)(z^3-1)=(a^2n)(b^2n)=(abn)^2=(x^3-1)^2;$$
rearranging gives
$$y^3+z^3-2x^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Finding $\sin 2x$ from transforming $\sin^4 x+ \cos^4 x = \frac{7}{9}$ using trigonometric identities While studying for the first exams of the year, the following question found on one of Kognity's questionbank was extremely challenging for myself, a pre calculus student. The current topic is that of trigonometric equ... |
$\displaystyle \sin^4(x) + \cos^4(x) = \frac{7}{9}.$
$\sin(2x) = ~$ ?
$\underline{\text{Preliminary Results}}$
$$\cos(2x) = 2\cos^2(x) - 1 \implies 2\cos^2(x) = \cos(2x) + 1.\tag{R-1}$$
$$\cos^4(x) = \frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3]. \tag{R-2}$$
$$\sin^4(x) = \frac{1}{8} \times [\cos(4x) - 4\cos(2x) + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A sum that's possibly equal to the Euler-Mascheroni Constant $\sum_{n=1}^\infty \frac{\ln n!}{n^3}$ The following interesting sum seems to approach the Euler-Mascheroni constant $\gamma$.
$$\sum_{n=1}^\infty \frac{\ln n!}{n^3} \overset{?}{=} \gamma$$
I've looked at the different ways to express the Euler-Mascheroni con... | It is known that for $n\geq 1$,
$$
\log n! > \left( {n + \frac{1}{2}} \right)\log n - n + \log \sqrt {2\pi } + \frac{1}{{12n}} - \frac{1}{{360n^3 }}.
$$
Thus
\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{{\log n!}}{{n^3 }}} & > \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n^2 }}} + \frac{1}{2}\sum\limits_{n = 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
About integration of $\frac{1}{(1+a^2 x^2)(1+x^2)}$ in differentiation under integral sign Consider the function $F(a)=\int_0^{+\infty} \frac{\arctan(ax)}{x(1+x^2)} dx$. I assumed $a \ge 0$ because $F$ is odd, I applied the theorem of differentiation under the integral sign and arrive to $F'(a)=\int_0^{+\infty} \frac{1... | If $a=\pm1$ your fraction becomes $\frac 1{(1+x^2)^2}$ The partial fraction decomposition is modified in the case of a repeated root in the denominator. Over the reals this one is already decomposed. Alpha gives the integral of this as
$$\int \frac 1{(1+x^2)^2} dx= \frac 12 \left(\frac x{(x^2 + 1)} + \arctan(x)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because th... | Using Cauchy-Bunyakovsky-Schwarz,
\begin{align*}
\frac{ab^2}{a^2 + b^2} + \frac{bc^2}{b^2 + c^2} + \frac{ca^2}{c^2 + a^2}
\ge \frac{(b + c + a)^2}{\frac{a^2 + b^2}{a} + \frac{b^2 + c^2}{b} + \frac{c^2 + a^2}{c}}
= \frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}.
\end{align*}
It ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be th... | We can think of the seating assignment as a random permutation of $7$ items (the three males, the three females, and the empty seat). This random permutation puts a female in the first chair with probability $\frac37$. Conditional on having done that, there are $2$ females left, so one of them ends up in the last chair... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What is wrong with this projection onto two basis vectors? Given
\begin{align}
b_1 &= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\
b_2 &= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \\
v &= \begin{bmatrix} 3 \\ 2 \end{bmatrix}
\end{align}
it is clear that
\begin{align}
v &= b_1 + b_2 \\
&= \begin{bmatrix} b_1 & b_2 \end{bmatrix} ... | The flaw lies in the fact that equation $v = \left(v \cdot \frac{b_1}{||b_1||}\right)\frac{b_1}{||b_1||} + \left(v \cdot \frac{b_2}{||b_2||}\right)\frac{b_2}{||b_2||}$ is valid only when $b_1$ and $b_2$ are orthogonal. (It also requires $v$ to be in the span of $\{b_1,b_2\}$, although that is satisfied here.)
Side note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the maximum value of $x^2y^3$ subject to the condition that $3x+2y=1$ I am trying to use G.M.$\leq $ A.M.as follows
$(\frac{2x^2y^3}{4})^\frac{1}{5} \leq \frac{x+2x+y+\frac{y}{2} +\frac{y}{2}}{5}$
$\implies (\frac{x^2y^3}{2})^\frac{1}{5} \leq \frac{1}{5}$
$\implies$ $x^2y^3\leq \frac{2}{5^5}$.
Is it correct ?
| For positive $x$ and $y$, you may use AM-GM:
$$\frac{1}{5}=\frac{3x+2y}{5}=\frac{\frac{3x}{2}+\frac{3x}{2}+\frac{2y}{3}+\frac{2y}{3}+\frac{2y}{3}}{5}\ge \sqrt[5]{\frac{2}{3}x^2y^3}$$
From where one gets $x^2y^3\le\frac{3}{2\cdot 5^5}=\frac{3}{6250}$
The equality takes place when $\frac{3x}{2}=\frac{2y}{3}$, which coupl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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... | Remark (Theorem of transformation for one random variable): Let be $X$ a continuous r.v. with pdf given by $f_X$ and let be Y=g(X) with g a diffeomorphism, then the pdf $f_Y$ is given by:
$f_Y(y)=f_X(g^{-1}(y))\cdot|{{d\over{dy}}g^{-1}(y)}|$.
If $Y=g(X)={{1+X}\over{1-X}}$ then $X=g^{-1}(Y)={{1-Y}\over{1+Y}}$. We find... | {
"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": 3
} |
Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way... | I've approached the problem wrongly, as it turned out. Below is the concise solution I came up with and haven't seen elsewhere.
Let's start out from analyzing initial data:
*
*To get $1^3$, add $1$ odd number, starting from the $1$st odd number.
*To get $2^3$, add $2$ odd numbers, starting from the $2$nd odd number.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Is there an elegant method to find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for positive $x$?
Find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for $x\gt0$.
In some textbook, the problem is usually tackled by calculus.... | We are going to find the minimum value of}
$$ S(x)=\dfrac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$$
geometrically using Triangle Inequality.
$$\displaystyle \begin{aligned} S(x)&=\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}} +\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\&=\sqrt{x^{2}+\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve for theta , $0<\theta<90^\circ$
Solve $$\cos60=\frac{\cos(2θ)+\frac32\sin(2θ))}{\sqrt{4-\sin^2 (2θ)}},$$ where $0<θ<90^\circ $
Question: Is the following solution correct?
Solution attempt:
$$\sqrt{4-\sin^2 (2θ)}=2\cos(2θ)+3\sin(2θ)$$
$$4-\sin^2 (2θ)=(2\cos(2θ)+3\sin(2θ))^2$$
$$4-\sin^2 (2θ)=4\cos^2 (2θ)... | When you arrive to $$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$ you should conclude as follows:
Since $\theta \in (0,90^\circ)$, then $2\theta \in (0,180^\circ)$, so $\sin(2\theta)\ne 0$.
From the other equation, $\tan(2\theta) = -2$, so $2\theta = \arctan(-2) + 180^\circ k$ for some integer $k$. Then $\theta = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculating probability of values for multiple dice rolls where sum is equal or greater than 4 Assuming we have 6 sided dice and play game of rolling dice each turn until we get sum value of at least 4.
For example we know that game will always finish with value less or equal than 9 because on 1. turn max value we can ... | You know that the final sum will be in the range of $4$ through $9$. Determine all possible rolls that yield the final result. Add up the probabilities as each event will be disjoint.
Example: to get to 4, you have the following roll patterns:
$$4 \\ 3, 1 \\ 2, 2 \\ 2, 1, 1 \\ 1, 3\\ 1,2,1 \\1, 1, 2 \\ 1, 1, 1, 1$$
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral replacement I have the following integral
$φ=\frac{M}{\sqrt{2m}}\int_{ }^{ }\frac{dr}{r^{2}\sqrt{E-\frac{M^{2}}{2mr^{2}}-\frac{kr^{2}}{2}}}$
As a result, I want to get the following result: $r=\frac{p}{1-ε\cosφ}$ This satisfies the ellipse equation in polar coordinates
I know I have to get an integral that red... | $E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{1}{2}kr^2=\frac{1}{2}m\dot{r}^2+\frac{1}{2} \frac{ L^2}{mr^2} +\frac{1}{2}kr^2$
$k/m=\omega^2$
$\frac{2E}{m}=\dot{r}^2+\frac{L^2}{m^2r^2}+\omega^2r^2$
$\frac{2E}{m}r^2=r^2\dot{r}^2+\frac{L^2}{m^2}+\omega^2r^4$
$\frac{2E}{m}r^2-\omega^2r^4-\frac{L^2}{m^2}=r^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$ really fast? Context:
*
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.
*Taking the major and minor axes of an ellips... | I've found a quicker way:
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
Now, $(2)\times8-(1)$:
$$\frac{200}{p}+\frac{16}{q}-\frac{4}{p}-\frac{16}{q}=8-1$$
$$\frac{196}{p}=7$$
$$p=\frac{196}{7}$$
$$p=28$$
Now, inputting $p=28$ in $(1)$,
$$\frac{4}{28}+\frac{16}{q}=1$$
$$\frac{16}{q}=\frac{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Compute $\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$ with the residue theorem $$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$$
I calculated the above integral several times using the residue theorem. The answer I get is $\pi / 2$ but this answer is not in the options please help
*
*$\frac{\pi i}{2}$
*$\frac{-\pi i}{2}$... | You have:
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
After this you didn't apply the Residue Theorem correctly. Instead of
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)+\pi i\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
matrix rearrangement that preserves row/column sum I have the following symmetric matrix
$$
\begin{pmatrix}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9 \\
\end{pmatrix}
$$
whose rows and columns sum to 9, 45 and 27. I would like to transform this matrix such that the diagonal is zero, but the row and columns still sum to ... | As you want to obtain the column/row sum, the symmetry and positivity you can start by combinatorics. Fortunately you don't have many possibilities. Let's start with the first row/column. By having zero on the diagonal, the numbers $5$ and $3$ have to be replace by two which sum up to $9$
$$
\begin{pmatrix}
0 & 0 & 9 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
radius of circle tangent to the parabola y=x^2 In the following diagram, what's the amount of r=?
I supposed the center of the small circle to be $(0,y)$. So I have
its distance from $(x, x^2)$ is $(x-0)^2+(y-x^2)^2=1400^2$ this question must have just one answer so the discriminant of the equation must be zero. But t... | You have the lower circle of radius $1400$. Assuming its center is $(0, a)$ then its equation is
$ x^2 + (y - a)^2 = 1400^2 $
At the tangency points $y = x^2 $, therefore
$ x^2 + (x^2 - a)^2 = 1400^2 $
Expanding, this becomes,
$ x^4 + x^2 (1 - 2 a) + a^2 - 1400^2 = 0 $
Since we want only one intersection point (actuall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find area of $x^2+axy+y^2=1$, $|a|\leq1$ I was wondering how to find the area of
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
I have solved
$$\rho^{2}(\theta)=\frac{1}{1+\frac{a}{2}\sin 2\theta}$$
However, integrating this equation using trigonometric substitution is very cumbersome.
| Solving
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
for $y$, one has
$$ y_{1,2}=\frac12(-ax\pm\sqrt{a^2x^2-4(x^2-1}))=\frac12(-ax\pm\sqrt{4-(4-a^2)x^2})$$
where $4-(4-a^2)x^2\ge0$ or $|x|\le\frac{2}{\sqrt{4-a^2}}$. So the are is
\begin{eqnarray}
A&=&\int_{-\frac{2}{\sqrt{4-a^2}}}^{\frac{2}{\sqrt{4-a^2}}}|y_1-y_2|dx\\
&=&\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving a probability inequality Given three probability density functions $p$, $q$ and $h$ of Normal distributions. $p$ and $h$ are fixed and I want to solve for $q$, is it possible to identify a condition on $q$ apart from the trivial cases q=h or q=p such that:
$$\int_{z} \frac{p(z)}{q(z)} \: h(z) \: dz \leqslant 1$... | Another approach: WLOG assume $p \sim N(a, \sigma_1^2)$ and $h \sim N(b, \sigma_2^2)$ where $0<\sigma_1^2 \leq \sigma_2^2$. Define $q\sim N(r, \sigma_2^2)$ for some $r \in \mathbb{R}$ that we will use to achieve the result.
Then
\begin{align}
\frac{h(z)}{q(z)} &= \exp\left(-\frac{(z-b)^2}{2\sigma_2^2}\right)\exp\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Volume between paraboloid $x^2 +y^2 -4a(z+a)=0$ and sphere $x^2 + y^2 +z^2 =R^2$ I'm trying to obtein the volume via triple integral but think I'm setting the wrong radius. The solid in particular is bounded by the sphere $x^2 + y^2 +z^2 =R^2$ and above the parabolloid $x^2 +y^2 -4a(z+a)=0$ (consedering $R>a>0$). I'm s... | First we find the intersection of the sphere with the paraboloid.
Since $x^2 + y^2 = 4a(z + a)$ then $4a(z + a) + z^2 = R^2$ so $z^2 + 4az + 4a^2 -R^2 = 0$. Then $$z = \frac{-4a \pm \sqrt{16a^2 + 4R^2 - 16a^2}}{2} = -2a \pm R$$ so the intersection is at height $R-2a$, as you said. The maximum radius would occur at the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
calculate one determinant to show 【National College Entrance Exam, old China】
$ \omega =\frac{1 \pm \sqrt{3} i } {2}$
Please calculate and show:
$ \begin{vmatrix} 1 &\omega& \omega^2 & 1\\ \omega& \omega^2 & 1 &1\\ \omega^2& 1& 1 & \omega \\ 1& 1& \omega& \omega^2\\ \end{vmatri... | One approach is to note that this matrix (which I will refer to as $A$) is a circulant matrix. As is explained on the page, we can thereby deduce that the eigenvalues of $A$ are of the form
$$
\lambda_j = 1 + i^j + \omega^2 i^{2j} + \omega i^{3j}, \quad i = \sqrt{-1}, \quad j = 0,1,2,3.
$$
So, we find that the eigenval... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$.
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$
My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^... | A solution using elementary calculus.
First observe by changing variables $x\mapsto\frac{x}{2}$ that the question becomes finding the maximum value of $4|cx^2+bx+a|$ on $[-1,1]$. Let's set $f(x)=ax^2+bx+c$ and $g(x)=cx^2+bx+a$. Also note that by swapping between $f$ and $-f$ (and $g$ and $-g$ respectively), we can assu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Maclaurin series of $\ln \left( 1+\frac{\ln(1+x)}{1+x} \right)$ So the first thing I done was
$$\begin{align}\ln(1+x)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3+o(x^3)\\&=(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)\end{align}$$
I've never seen this done but I'm pretty sure I can do this.
Now I want to divide:
$$(1+... | Following @jjagmaths approach we show
\begin{align*}
\color{blue}{\ln\left(1+\frac{\ln(1+x)}{1+x}\right)}
&\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}\sum_{m=1}^{n}\frac{1}{m}
\sum_{{k_1+\cdots +k_m=n}\atop{k_1,\ldots,k_m\geq 1}}H_{k_1}\cdots H_{k_m}x^n}\tag{1}\\
&=x-2x^2+\frac{11}{3}x^3\color{blue}{-\frac{163}{24}}x^4+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What could be these polynomials? Trying to build an approximation, I encountered the following polynomials
$$\left(
\begin{array}{cc}
1 & 1 \\
2 & x-3 \\
3 & x^2-30 x+45 \\
4 & x^3-273 x^2+1575 x-1575 \\
5 & x^4-2460 x^3+43470 x^2-132300 x+99225 \\
6 & x^5-22143 x^4+1123650 x^3-8004150 x^2+16372125 x-9823275 \\
... | $\{-1575, -132300, -8004150, -431531100,...\}$:
$\dfrac{35}{1024}\bigg(5 - 3^{2 n + 8} + 5^{2 n + 7} - 7^{2 n + 6}\bigg)$
$\{1, -30, 1575, -132300, 16372125, -2809456650,...\}$:
$-(-1)^{n + 1} 2^{2 n} (n + 1) Pochhammer\bigg(\dfrac{3}{2}, n\bigg) Pochhammer\bigg(\dfrac{5}{2}, n\bigg)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Orthogonal projection of $D=\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix}$ on $W=\{A\in M_{2\times 2}(\mathbb{R})~|~Trace(A)=0\}$ in product space $V$. This was an assigned homework problem (Turned in 4/14):
$V=M_{2\times 2}(\mathbb{R})$ with the inner product $\langle A,B\rangle = Trace(B^t,A),~D=\begin{pmatrix}1 & 1\\ -... | In vector form
$D = [1, 1, -3, 1]^T $
And a basis for $W$ is
$\{ [1, 0, 0, -1]^T , [0, 1, 0, 0]^T , [0, 0, 1, 0]^T \} $
The orthogonal space is spanned by
$N = [1, 0, 0, 1]^T$ (which corresponds to the identity matrix)
Therefore,
$D - Proj(D) = c_1 N $
$ D \cdot N - 0 = c_1 (N \cdot N) $
Hence $c_1 = \dfrac{D \cdot N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help with a surface integral Let G(x, y, z) = $\ (1-x^{2}-y^{2})^{\frac{3}{2}}$
Evaluate the surface integral:
$\int\int_{S}^{}G(x, y,z)dS$
where S is the hemisphere:
z =$\ (1-x^{2}-y^{2})^{\frac{1}{2}}$
This is my work thus far:
$\int_{}^{}\int_{S}^{}G(x,y,z)dS=\int_{}^{}\int_{R}^{}(1-x^{2}-y^{2})^{\frac{3}{2}}\sqrt[]... | Since,
$$\boxed{\iint_{S}f(x,y,z)\,{\rm d}S=\iint_{D}f(x,y,g(x,y))\sqrt{z_{x}^{2}+z_{y}^{2}+1}\,{\rm d}A}$$
So, we have
\begin{align*}
\iint_{S}(1-x^{2}-y^{2})^{3/2}\, {\rm d}S&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{z_{x}^{2}+z_{y}^{2}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\left(-\frac{x}{\sqrt{1-x^{2}-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trouble with a triple integral on a region bounded by a sphere and two planes
I would like to compute the integral $\int_A zdzdydx,$ where $A$ is the region bounded by the sphere $x^2+y^2+z^2=R^2,$ plane $\frac{x}a+\frac{y}b=1$ and coordinate planes (which doesn't contain the origin on its boundary and is in the first... | Proceed in cylindrical coordinates as follows
\begin{align}
&\int_0^{\pi/2}\int_{r(\varphi)}^R\int_0^{\sqrt{R^2-r^2}}z \>rdzdrd\varphi\\
= & \> \frac12 \int_0^{\pi/2}\int_{\frac{ab}{a\sin\varphi+b\cos\varphi}}^R (R^2-r^2)r dr d\varphi\\
= & \> \frac12 \int_0^{\pi/2}\left( \frac14R^4–\frac12\frac{a^2b^2R^2}{(a\sin\varp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
minimizing the distance between values
Find all values of $a$ that minimize the expression $|a- x_1| + |a-x_2|+\cdots + |a-x_n|$, where $x_1\leq x_2\leq \cdots \leq x_n$.
I know the minimum is achieved when $a=x_{\lfloor n/2\rfloor + 1}$ if $n$ is odd and when $a\in [x_{\lfloor n/2\rfloor}, x_{\lfloor n/2\rfloor + 1... | This is a long-winded comment.
It is not an answer.
The approach that I would suggest is a minor variation of the approach that you described in your posting. Because (as you indicated) it is tedious, I will simply explore the individual cases of $n = 9$ and $n = 10$, for illustrative purposes.
Further, I will map o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding closed form expression for $n^{th}$ term of sequence with generating function $F(x)$? I have been asked the above question regarding the generating function
$$F(x) = \frac{x^2(1-x)}{(1-x)^3}$$
I have no idea what procedure this type of question follows. The solution gives that $F(x)$ can be written as
$$\frac{x... | Simplify and then use partial fraction decomposition to obtain
\begin{align}
\sum_{n=0}^\infty a_n x^n
&= \frac{x^2(1-x)}{(1-x)^3} \\
&= \frac{x^2}{(1-x)^2} \\
&= 1 - \frac{2}{1-x} + \frac{1}{(1-x)^2} \\
&= 1 - 2\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty (n+1) x^n \\
&= 1 + \sum_{n=0}^\infty (n-1) x^n,
\end{align}
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$, for positive values with $xyz=1$, without expansion?
Let, $x,y,z>0$ and $xyz=1$, then prove that
$$\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$$
I know that $$x^2+y^2+z^2\geq x+y+z$$ by Cauchy-Sc... | Here is a solution using only various mean inequalities:
First use AM-GM:
$$2\frac{x^2+y^2+z^2}{x+y+z}+3\sqrt[3]{\frac{xy+yz+zx}{x+y+z}}\ge5\sqrt[5]{\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}}$$
Then using $xyz=1$:
$$\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}=\frac{(x^2+y^2+z^2)^2}{(x+y+z)^4}(x+y+z)\left(\frac{1}{x}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$p$ ways to write $p$ as sum of primes I hope this question is valid as I'm just curious.
in a tweet from AlgebraFact I read the following:
"There are 17 ways to write 17 as a sum of primes": $17, 2+2+13, 3+3+11, 3+7+7, 5+5+7, 2+2+2+11, 2+3+5+7, 2+5+5+5, 2+2+3+3+7, 2+2+3+5+5, 3+3+3+3+5, 2+2+2+2+2+7, 2+2+2+3+3+5, 2+3+3+... | Here is an approach to prove that there are only finitely many numbers with this property. We only pay attention to expressions of the form $m \cdot 2 + n \cdot 3 +p\cdot 5 +7=N$. The point is that the number of expressions of this type grows faster than $N$, so there will eventually be far too many expressions of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Constructing a right triangle given $a+b-c$ and $\alpha$ The exercise is to construct a right triangle given $a+b-c$ and $\alpha$.
I know we then have $\beta=90^\circ -\alpha$. I tried to draw the right triangle $\triangle ABC$ and find where I can use $a+b-c$. By adding $a$ to $b$ to get $\overline{DA}, |DA|=a+b$, I ... | Alternative approach.
It is assumed that $c$ is the hypotenuse, that $a$ is the side opposite the angle $\alpha$, and that the value $(a+b-c)$ equals a given fixed value $T$.
Then, you have the following two equations in the two unknowns, $a,b$.
*
*$c^2 = (a + b - T)^2 = a^2 + b^2.$
*$\displaystyle \frac{a}{b} = \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $abc=1$ ,and $n$ is a natural number,prove $ \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2} $ If $a, b, c$ are distinct positive real numbers such that $abc=1$,and $n$ is a natural number,prove
$$
\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \g... | As can be found here and here , we have that
$$
P_n = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} =
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
$$
Now by AM-GM, it follows that
$$
P_n =
\displaystyle\sum_{\substack{n_1, n_2, n_3 \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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solve the recurrence relation $h_n - 0h_{n - 1} - 3h_{n - 2} + 2h_{n - 3}= 0$ $h_n - 0h_{n - 1} - 3h_{n - 2} + 2h_{n - 3}= 0$
$h_0=2, h_1=0, h_2=7$, and n≥3
is given
here is what I did
$x^3-3x+2=0$
$h_n=a1^n+b(-2)^n$
roots of the polynomial are 1 and -2 but there are 3 equations($h_0,h_1,h_2$) and 2 variables so we nee... | What you did is correct.
The characteristic equation is
$ x^3 - 3 x + 2 = 0 $
By inspection, one of the roots is $x = 1 $. By synthetic division,
$ (x^3 - 3 x + 2) = (x - 1) (x^2 + x - 2) = (x - 1)^2 (x + 2) $
So the roots are $ 1, 1, -2 $
Therefore, the solution to this homogeneous difference equation is
$ h_n = A + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $a_n=n(a_{n-1}+a_{n-2})$ where $a_0=1,a_1=2$ using generating functions I am trying to solve a recurrence relation,$a_n=n(a_{n-1}+a_{n-2})$ where $a_0=1,a_1=2$, using generating functions. So, I did: let $$A(x)=\sum_{n\geq 0}a_n\frac{x^{n}}{n!}$$
$$\sum_{n\geq 0}a_{n+2}\frac{x^{n+2}}{(n+2)!}=\sum_{n\geq 0}a_{n+1}... | A standard way to do this does end up with a differential equation.
Define
$$
A(x) := \sum_{n=0}^\infty a_n \frac{x^n}{n!}
$$
We need similar sums with $na_{n-1}$ and $na_{n-2}$.
Compute
$$
\sum_{n=1}^\infty n a_{n-1} \frac{x^n}{n!}
= \sum_{n=1}^\infty a_{n-1}\frac{x^n}{(n-1)!}
= \sum_{n=0}^\infty a_n\frac{x^{n+1}}{n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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What is the value of $a_1a_2\cdots a_{2019}$? Let $a_1=\frac 34$ and for any $n\geq2$ $4a_n=4a_{n-1}+\frac {2n+1}{1^3+2^3+\cdots n^3 }$. What is the value of $a_1a_2\cdots a_{2019}$?
I tried $1^3+2^3+\cdots +n^3=\frac {n^2(n+1)^2}{4}$ and
I wroted
$4(a_n-a_{n-1})=\frac {2n+1}{1^3+2^3+\cdots n^3 }$.
Then
$$4(a_2-a_1)=\f... | Observe $$a_n - a_{n-1} = \frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2 (n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}. \tag{1}$$ So $$a_m - a_1 = \sum_{n=2}^m a_n - a_{n-1} = \frac{1}{2^2} - \frac{1}{(m+1)^2},$$ hence $$a_m = 1 - \frac{1}{(m+1)^2} = \frac{m(m+2)}{(m+1)^2}. \tag{2}$$ Therefore, $$\prod_{m=1}^{201... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Proving $\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$ I would like to prove, that for $m \ge 2$ we have the following inequality:
$$\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$$
My thoughts
This inequality is e... | Gautschi's inequality states that
$$
x^{1 - s} < \frac{{\Gamma (x + 1)}}{{\Gamma (x + s)}}
$$
for $x>0$ and $0<s<1$. Let $m>1$. Taking $x=\frac{m-1}{2}$ and $s=\frac{1}{2}$ gives
$$
\sqrt {\frac{{m - 1}}{2}} < \frac{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}}{{\Gamma \left( {\frac{m}{2}} \right)}},
$$
i.e.,
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$? here's my question
If $y^2=ax^2+bx+c$ then prove that $y^3[d^2y/dx^2]$ is a constant .
I have solved this using the conventional method, taking square root, differentiating w.r.t to x and using chain and quotient rule
But can't think of some alter... | Using implicit derivation:
$$
(y^2)^\prime = 2yy^\prime = 2ax+b \Rightarrow y^\prime=\frac{2ax+b}{2y}
$$
$$
(y^2)^{\prime\prime} = 2(y^\prime)^2+2yy^{\prime\prime} = 2a \Rightarrow y^{\prime\prime}=\frac{a-(y^\prime)^2}{y}
$$
Thus
\begin{align}
y^3y^{\prime\prime}=y^2(a-(y^\prime)^2) &= y^2(a-\left(\frac{2ax+b}{2y}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\t... | Note that the hyperbolic relation $y^2-2xy+1=1$ is equivalent to $y=x\pm\sqrt{x^2-1}$. And one of those branches, $y=x-\sqrt{x^2-1}$ is equivalent to $y=\frac{1}{x+\sqrt{x^2-1}}$ (if you rationalize its denominator). And once you take the logarithm of $\frac{1}{x+\sqrt{x^2-1}}$, you have the negative of the logarithm o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$? How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$?
My Attempt:
$f'(x)=\frac{\frac x{\sqrt{1-x^2}}-\arcsin x}{x^2}$
Since $0\le x\le1\implies0\le\sqrt{1-x^2}\le1$
And $0\le\arcsin x\le1.57$
But not able to show that $f'(x)\ge0$
| You can also find a series expansion of $\arcsin(x)$ as follows:
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\dots$$
Put $x=\frac{t^2}{1-t^2}$, and we have
$$\frac{1}{\sqrt{1-t^2}}=1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\dots$$
Now integrating we have
$$\arcsin(x)=\int_0^x\frac{1}{\sqrt{1-t^2}}dt=x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Let $a$,$b$ be the roots of $x^2-6cx-7d=0$, and let $c$, $d$ be the roots of $x^2-6ax-7b=0$. Find $b+d$.
Let $a$,$b$ be the roots of the equation $x^{2}-6 c x-7 d=0$ and let $c$,$d$ be the roots of the equation $x^{2}-6ax-7 b=0$. $a,b,c,d$ are distinct. Find value of $b+d$.
Using the relation between roots and coeffi... | Alternative approach:
From the original equations you have that:
*
*$[E_1:] ~ a + b = 6c \implies b = 6c - a.$
*$[E_2:] ~ c + d = 6a \implies d = 6a - c.$
Using the two results above, you also have, from the original equations that:
*
*$[E_3:] ~ 6ac - a^2 = ab = -7d = -7(6a - c) = 7c - 42a.$
*$[E_4:] ~ 6ac - c^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $ I put this equation into Symbolab and it produced me a very complex result
Basically this is the result but I believe there is a simpler way to solve this question
$$
\frac{2}{3\sqrt{3}}\left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)+\sin \left(2\arctan... | Under $x+\frac12=\frac{\sqrt3}{2}\tan\theta$ or $\theta=\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$, one has
\begin{eqnarray}
&&\int \frac{1}{(x^2+x+1)^2}dx\\
&=& \int \frac{1}{((x+\frac12)^2+\frac34)^2}dy\\
&=& \frac{8\sqrt3}{9}\int\frac{1}{\sec^4\theta}\sec^2\theta d\theta\\
&=& \frac{8\sqrt3}{9}\int\cos^2\theta d\theta\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all odd primes $p$ such that $3p-8$ is equal to the cube of a positive integer I recently tackled this problem from one of my country's regional competitions. My solution is as follows:
Let $$3p - 8 = n^3$$ From this we know that $$p = \frac{n^3 + 8}{3}$$ It follows that $$p = \frac{(n+2)(n^2 - 2n + 4)}{3}$$
Now, ... | I'd say that in getting the factorization of $p$ that you did is a crux of the problem, but you still had more math to do. You did not exhaust all possibilities. However, you could note the following:
Claim 1. If $a$ and $b$ are integers that are at least $4$ each, then either $\frac{ab}{3}$ is nonintegral or composite... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\frac{3\pi}{2}-\alpha\right)$ Prove that $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\dfrac{3\pi}{2}-\alpha\right)$$
The RHS is equal to $\tan\alpha,$ so we are to show $$\dfrac{\sin4\... | In fact, using
$$ 1-\cos 2x=2\sin^2x, 1+\cos2x=2\cos^2x $$
one has
\begin{eqnarray}
&&\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{\sin4\alpha(1-\cos4\alpha)}{1-\cos^24\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{1-\cos4\alpha}{\sin4\alpha}\cdot\dfrac{\cos2\alpha}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to simplify the calculations of reflecting a ray about an ellipse
I wrote the script that made these images several days ago, the segments each depict a ray of light, as the light hits the boundary of the ellipse, it is reflected by the ellipse according to the laws of reflection, and the reflected ray of light... | I assume that the ellipse is given by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
We are given a starting point $\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}$ inside the ellipse and the ray direction vector $\begin{pmatrix} u \\ v \end{pmatrix}$.
*
*The point of the intersection is
$$\tag{1} \begin{pmatrix} x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$ without Hopital rule I want to evaluate this limit without applying Hopital rule,$$\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$$
After expanding the denominator I got,
$$\frac{1}{12}\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+... | use $(a^6 - b^6) = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$
Let $a = \sqrt [3]{x^2 - x - 1}, b = \sqrt{x^2 - 3x + 3}$
that is where we have $\frac {a-b}{x-2}$ we multiply top and bottom by $(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$ giving
$\frac {a^6-b^6}{(x-2)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | Holy unsimplified long expressions Batman.
$$
\begin{align}
a^2+b^2
&=(a+b)^2-2ab\\
&=45
\end{align}
$$
and
$$
\begin{align}
a^6+b^6
&= (a^2+b^2)^3-3(ab)^2(a^2+b^2)\\
&= (a^2+b^2)((a^2+b^2)^2-3(ab)^2)\\
&= 45\times(45^2-12)\\
&=90585\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 6
} |
Getting two different answers on differentiating $\cos^{-1}(\frac{3x+4\sqrt{1-x^2}}{5})$
Question given in my book asks to find $\frac{dy}{dx} $ from the following equation.$$y=\cos^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right)$$
My Attempt:
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45... |
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45 = \sin\alpha$.
Instead, just let $$\alpha=\arccos\frac35$$ so that $\frac45$ indeed equals $\sin\alpha\,$ (no need for any $\pm$ sign).
*
*Putting $x = \cos\beta\implies \sqrt{1-x^2} = \sin\beta$.
Instead, let $$\beta=\arccos x$$ so t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Showing that switching the order of variables gives same sum so as to simplify the required expression for exact sum If $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$ , then if we want to show that $S(x,y) = S(y,x)$ so as to get the simplification by adding both to get the exac... | Your double series has non-negative terms and $\sum_{x\geq 0}\sum_{y\geq 0}=\sum_{y\geq 0}\sum_{x\geq 0}$ is granted by Fubini's theorem. Very crudely
$$ \sum_{x=0}^{M}\sum_{y=0}^{N}\frac{x+y+xy}{5^x(5^x+5^y)}\stackrel{\text{AM-GM}}{\leq}\sum_{x=0}^{M}\sum_{y=0}^{N}\frac{(1+x)(1+y)}{2\cdot 5^x 5^{\frac{x+y}{2}}}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac... | To answer the first discrepancy with the textbook, note that
$$\left( \frac{1 - \cos(2x)}{2} \right)^2
= \frac{1- 2 \cos(2x) + \cos^2(2x)}{4}$$
Hence the middle term indeed has a factor of $-1/2$.
Textbooks are just as prone to errors as anything else, being made by humans; it happens. Not much one can do but contact t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Should the vertical "at bar" go before or after the function being differentiated? Suppose I want to calculate the derivative of a long function at a particular point $a$. Is it more common to write
$$
\frac{d}{dx} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x)\right) \Bigg|_{x=a}
$$
or
$$
\frac... | I agree with MathGeek. Use the first one
$$
\frac{d}{dx} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x)\right) \Bigg|_{x=a}
$$
(in mathematics).
However, I do not speak for engineering or physics. The second notation
$$
\frac{d}{dx}\Bigg|_{x=a} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $a\ge b \ge c \ge d \ge 1 \implies x^{4} -ax^{3}-bx^{2}-cx-d $ has no integer root. If $a,b,c,d$ are positive integers such that $a\ge b\ge c \ge d\ge 1$, prove that
$$ x^{4} -ax^{3}-bx^{2}-cx-d $$
has no integer root.
Attempt:
If there is an integer solution $x=x_{0}$, then
$$x_{0}^{4} = a x_{0}^{3}+ b x_{0}^{2... | The case $x=0$ is trivial.
For $x>0$, you have $x^4=(ax+b)x^2+(cx+d)$, with $ax+b\ge cx+d>1$. Let's say that $A=ax+b$ and $C=cx+d$. That means that $x^4=Ax^2+C$ and $C$ must be divisible by $x^2$ but $A\ge C\ge x^2$ so you can't find a solution.
For $x<0$, we have $A\le C\le0$ so $Ax^2+C$ will never be positive. So onc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the measure of the relationship $\frac{1}{r_1} - \frac{1}{r} $in the figure below In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:
$\triangle... | In a right triangle, $~b + c = 2r + a$, where $a$ is hypotenuse. If $s$ represents sub-perimeter, then by law of cotangent,
$ \displaystyle \frac{s_{\triangle ABC} - b}{r} = \frac{s_{\triangle ABD} - 2}{r_1}$
$ \displaystyle \frac{s_{\triangle ABC} - c}{r} = \frac{s_{\triangle ACD} - 2}{r_1}$
Adding both,
$~\displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\frac{x^2 + 4x}{2x - 1} < \frac{-4}{2x-1}$? I want to solve the following inequality $\frac{x^2 + 4x}{2x - 1} < \frac{-4}{2x-1}$ but I know that it is undefined when $x = 1/2,$ so I can not multiply both sides by $2x -1.$ Is there any suggestion of how to solve it? I also have an idea of transferring one ... | The basic principle is
$$\frac ab<0 \iff ab<0.$$
From $\frac{x^2+4x}{2x-1}<\frac{-4}{2x-1}$ we have
$$\frac{x^2+4x+4}{2x-1}<0,$$
which is equivalent to
$$(x^2+4x+4)(2x-1)<0.$$
However, $x^2+4x+4=(x+2)^2\geq 0$. Hence, $x\neq -2$ (otherwise $x^2+4x+4=0$) and $2x-1<0$. Therefore, the solution is $x<\frac12$ and $x\neq-2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(... | First, notice that both of the methods you followed are incorrect because the sum is larger than 3/2. Hence it can't be less than epsilon (for values small enough). I propose a different approach.$$$$
Notice that for all $n \geq 1$ :
$$|\sqrt{n^2+3n-3}-n-\frac{3}{2}| = -\sqrt{n^2+3n-3}+n+\frac{3}{2}$$
Now we're looking... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the range of $\frac {2x^2+x-3}{x^2+4x-5}$ I solved for the range as follows:
Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x.
$$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got
$$(4y-1)^2-4(y-2)(3-5y) \ge 0$$
Which boiled down to
$$(6y-5)^2 \ge 0$$
And this... | Your quadratic equation has $x=1$ as a solution, for any $y$. Indeed, for $y\neq2$ the quadratic formula gives
$$x=\frac{-(4y-1)\pm(6y-5)}{2(y-2)}$$
$$x=\frac{-10y+6}{2y-4}=\frac{-(5y-3)}{y-2}\qquad\text{or}\qquad x=\frac{2y-4}{2y-4}=1$$
and for $y=2$ your equation reduces to $7x-7=0$.
The original function was
$$y=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Product of matrix and vectors in orthogonal subspaces Let $p < n$, and let $A \in \mathbb{R}^{n \times n}$ be rank $p$. Let $V \in \mathbb{R}^{n \times n}$ be an orthogonal matrix with columns $V = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$.
Suppose that for any linear combination of the first $p$ columns of $V$... | $A=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}$
$\operatorname{Rank} A=p=2$
$V=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$
Then $\begin{align}A(c_1v_1+c_2v_2) &=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\\ &=\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\neq 0 [\text{ as $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For an arbitrary $n$, evaluate $\left(z^n+\frac{1}{z^{n}}\right)\left(z^n +\frac{1}{z^n}+1\right)$ Let me put you in context, I have the following problem:
Let $z$ be a complex number such that $$\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1.$$
For an arbitrary $n$, evaluate
$$\left(z^n +\frac{1}{z^n}\right... | Let $\left(z+\frac{1}{z}\right)=x$
Then $\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=(x)(x+1)=x^2+x$
$x^2+x = 1 \to x^2 + x -1 = 0$
By the fundemental theorem of algebra, we know this has exactly two solutions. By the quadratic formula $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ we know that those solutions are $\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)... | Let $F(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$. Then your polynomial is $f(x)=F'(x)$. Because $a,b,c,d,e$ are all distinct, by Rolle's theorem, between any two consecutive roots of $F(x)$ will be a root of $f(x)$. Thus, $f(x)$ will have 4 distinct real roots because $F(x)$ has 5 distinct real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as we... | Let's first try to find a recursive relation of $I_n$. In fact using integration by part,
\begin{align}
I_{n} &= \int_0^\infty \frac{1}{(x^2 + a^2)^n}\mathrm dx\\
&= \int_0^\infty \frac{x^2 + a^2}{(x^2 + a^2)^{n+1}}\mathrm dx\\
&= \frac{1}{2n} I_{n} + a^2 I_{n+1}.
\end{align}
So $$I_{n+1} = \frac{1}{a^2} \frac{2n-1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Solve $x^2 dy + (xy+y^2) dx = 0$ if $y=1$ when $x=1$. Using the given equation I arrived at the following expression:
$$\frac{dy}{dx} = - \left[\frac{y}{x} + \left(\frac{y}{x}\right)^2\right] \tag{1}$$
Since this is a homogenous equation, I assumed $\frac{y}{x}=v$ giving:
$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$
Substit... |
$$|\frac{y}{y+2x}| = \frac{C^2}{x^2} \tag{2}$$
In your Eq.$(2)$, if you remove the absolute value, you have
$$\frac{y}{y+2x} =\pm \frac{C^2}{x^2} $$
But your initial condition $x=1,y=1$ exculdes the "$-$" sign solution. So you only need to take the positive solution, hence
$$\frac{y}{y+2x} = \frac{C^2}{x^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !}... | Even the antiderivative exists $$I_n=\int \frac{d x}{\left(1+\sin ^{2} (x)\right)^{n}}$$
$$I_n=-\frac{\sqrt{-\sin ^2(x)} \sqrt{\cos ^2(x)} \csc (x) \sec (x) }{2 \sqrt{2} (n-1)\,\left(1+\sin^2(x)\right)^{n-1} }\times $$
$$F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{1}{2} \left(1+\sin
^2(x)\right),1+\sin ^2(x)\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle ... | $$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x =& \int_{0}^{\frac{\pi}{4}} \frac{d(\sin x-\cos x)}{5^{2}-4^{2}(\sin x-\cos x)^{2}} \\
=& \frac{1}{40}\left[\ln \left| \frac{4 (\sin x-\cos x)+5}{4(\sin x-\cos x)-5} \right|\right]_{0}^{\frac{\pi}{4}}\\
=& \frac{\ln 9}{40}
\end{aligned}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$
I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt ... | Observe that,
If $x=0$ or $y=0$ or $z=0$, you will get $x=y=z=0$. So, we can assume $x,y,z\neq 0$.
$$\begin{cases}\frac 1x+\frac 1y=\frac {\sqrt 2}{2}\\
\frac 1y+\frac 1z=\frac {\sqrt 3}{3}\\
\frac 1x+\frac 1z=\frac {\sqrt 5}{5}\end{cases}\implies \frac 1x+\frac 1y+\frac 1z=\frac {\frac {\sqrt 2}{2}+\frac {\sqrt 3}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\int_0^\infty\frac{\ln{(x^8+x^4+2x^2+1)}}{x^2+1}dx$ and $\int_0^\infty\frac{1}{x^2+1}\arctan{\left(\frac{x^2+1}{x^4}\right)}dx$ I derived some integrals a while ago
$$I(t)=\int_0^\infty\frac{\ln{(x^4+t(x^2+1))}}{x^2+1}dx$$
$$I(i)=\int_0^\infty\frac{\ln{(x^4+i(x^2+1)}}{x^2+1}dx=2\pi\ln{(1+\sqrt{i}+\sqrt{i+2\sqrt{i}})}$... | We could even compute the antiderivative
$$I=\int\frac{\log{(x^8+x^4+2x^2+1)}}{x^2+1}dx=\sum_{i=1}^4\int\frac{\log(x^2-r_i)}{x^2+1}dx$$
where
$$r_1=\frac{-i-\sqrt{-1-4 i}}{2}\qquad \qquad r_2=\frac{-i+\sqrt{-1-4 i}}{2} $$
$$r_3=\frac{i-\sqrt{-1-4 i}}{2} \qquad \qquad r_4=\frac{i+\sqrt{-1-4 i}}{2} $$
and $$I_a=\int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Writing an integral in terms of special functions I would like to find an expression for the following integrals, using special functions or otherwise; I suspect they can be written in terms of elliptic functions but I've played with it for a while and haven't arrived at anything useful. The integrals are:
$$
\mathcal{... | Since $\cos^2(x) = 1 - \sin^2(x)$ both $I_1$ and $I_2$ can be rewritten as some constant multiple of an antiderivative of the form:
\begin{align}
I(a,b) =\int \frac{x\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x
\end{align}
for some constants $a,b$. Now, from the definition of the Elliptic integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int^{\infty}_0\frac{\ln^2(x)}{\left(x^2+1\right)^2}\text{ d}x$ I want to integrate $$I=\int^{\infty}_0\frac{\ln^2(x)}{\left(x^2+1\right)^2}\text{ d}x=\frac{\pi^3}{16}$$ using contour integration. I set up the function $$f(z)=\frac{\ln^3(z)}{\left(z^2+1\right)^2}$$
and used a keyhole contour with a branch ... | The residue at $-i$ should be
$$\mathop{\mathrm{Res}}_{z = -i}\frac{\ln^3(z)}{\left(z^2+1\right)^2}=\lim_{z\to -i}\frac{\text{d}}{\text{d}z}\left[\frac{\ln^3(z)}{\left(z-i\right)^2}\right]=\lim_{z\to -i}\frac{\ln^{2}(z)\left(3z-3i-2z\ln(z)\right)}{z\left(z-i\right)^{3}}=\frac{27\pi^3}{32}+\frac{27\pi^2 i}{16}$$
where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | A polynomial approach.
Note that $a_1,a_2=\sqrt{3\pm\sqrt 5}$ are two of the roots of $x^4-6x^2+4=0.$ The other roots are $-a_1,-a_2.$
So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$
Now, $a_1a_2=\sqrt{4}=2.$ If $S=a_1+a_2,$ then this factorization becomes:
$$x^4-6x^2+4=(x^2-Sx+2)(x^2+Sx+2)=x^4+(4-S^2)x^2+4.$$
So $4-S^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
} |
Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$
If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$
I genuinely have no clue on how to proceed. Is it proved using repeated... | Remark: Once we know the equality case $x = 1, y = 2, z = 3$, we apply AM-GM.
Using AM-GM, we have
\begin{align*}
&30\cdot x + 12 \cdot (y/2)^2
+ 6\cdot (z/3)^3 + 18 \cdot \frac{1}{xy/2} + 6\cdot \frac{1}{yz/6} + 12 \cdot \frac{1}{zx/3}\\
\ge\,& 84\sqrt[84]{x^{30}\cdot (y/2)^{24}\cdot (z/3)^{18} \cdot \left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
numerical calculation of 3d integral $\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2} $ I have to calculate the integral
$$I=\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2} $$
where $r^2 = x^2 + y^2 + z^2 $.
My strategy is to split the integral domain into two parts: the unit sphere and the supplement domain. From the unit sphere, I get... | $$\iiint_{[-1,1]^3}\frac{d\mu}{x^2+y^2+z^2} = \iint_{[-1,1]^2}\frac{2}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$
equals
$$ 8\iint_{[0,1]^2}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$
or
$$ 16\iint_{0\leq y\leq x\leq 1}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x... | there are just two flavors of these; For integer $k$ we are looking for integer points on the arc of
$$ x^2 - kxy + y^2 + x + y = 1 -k $$
between the lines $ky = 1 + 2x$ and $kx = 1+2y$
For $k > 7$ the arc includes a point $(t,t)$ with $1 < t < 2$
Calculation will show that, for $x=2,$ we have $y < 1,$ and the (l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
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