Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find major & minor axes of the ellipse $10x^2+14xy+10y^2-7=0$?
What are major & minor axes of the ellipse: $10x^2+14xy+10y^2-7=0$ ?
My trial:
from given equation: $10x^2+14xy+10y^2-7=0$
$$10x^2+14xy+10y^2=7$$
$$\frac{x^2}{7/10}+\frac{xy}{7/14}+\frac{y^2}{7/10}=1$$
I know the standard form of ellipse: $\frac{... | The ellipse we have is in the form $S(x, y) = ax^2 + by^2 + 2hxy + c$. To remove the $xy$ term we rotate the ellipse by the angle $\tan (2\theta) = \dfrac{2h}{a - b}$. Which is same as rotating coordinate system by $-\theta$.
So if $(X', Y')$ is coordinates in new coordinate system then we can use
$${x \choose y} = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculate the triple integral: $\iiint_W(x^2+y^2)z\ dx\,dy\,dz$ I am trying to solve the following triple integral:
$$
\iiint_W(x^2+y^2)z\,dx\,dy\,dz \\
W=\{(x,y,z) \in \mathbb{R}: x^2+y^2+z^2 \le 9; x^2+y^2\le 1; x \ge 0; y \ge 0; z \ge 0\}
$$
From which I know that there are two surfaces:
$$
x^2+y^2+z^3 = 9 \rightarr... | The integral is getting calculated in 1st octant therefore:
$$\iiint_W(x^2+y^2)z\,dx\,dy\,dz=\iiint_W R^5\sin^{3}\theta\cos\theta dR d\theta d \phi$$also we know that for $0<\theta<\theta_0$ where $\theta_0=\sin^{-1}\frac{1}{3}$ we have $0<R<3$ and for $\frac{\pi}{2}>\theta>\theta_0$ we have $0<R<\frac{1}{\sin\theta}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is this a correct derivation of completing the square? $x^2 + bx$
$=x^2 + bx + c - c$
$=(x + k)^2 - c$
$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$
This implies:
$2k = b$, so $k = b/2$, and:
$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$
So to complete the square we are making the transformation:
$x^2 + bx \implies (x + b/2)... | Okay. You are trying to derive.
As you want to derive $x^2 + bx = (x + k)^2 + c$ I'd start with that line:
"we want $x^2 + bx = (x+k)^2 +c$ and we need to find the $c$ and $k$ that will make it so.
You started instead with
$x^2 + bx = x^2 + bx + c - c = (x+k)^2 -c$ with, I guess (????) then assumption that $(x+k)^2 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can I calculate the remainder of $3^{2012}$ modulo 17? So far this is what I can do:
Using Fermat's Little Theorem I know that $3^{16}\equiv 1 \pmod {17} $
Also: $3^{2012} = (3^{16})^{125}*3^{12} \pmod{17}$
So I am left with $3^{12}\pmod{17}$.
Again I'm going to use fermat's theorem so: $ 3^{12} = \frac{3^{16}}{3^... | Just do it.
$3^4 = 81 \equiv -4$.
$3^{12} \equiv (3^4)^3 = (-4)^3 \equiv -81 \equiv 4 \mod 17$.
For insight:
You know $3^{16}\equiv 1 \mod 17$ so $3^{8}\equiv \pm 4$ so $3^4 \equiv \pm 1, \pm \sqrt{-1}$. So $-1 \equiv 16$ one of the $\sqrt {-1} \equiv 4\mod 17$. (the other is $13$). This should tell you to try t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Matrix Row Reduction - What am I doing wrong? In this homework problem, I'm asked to find the solution to the following augmented matrix that represents a system of linear equations. Extracting the solution from the matrix isn't a problem, so I'll just show you guys how I got to my reduced echelon form.
$\left[\begin{... | As you replace row $2$ with sum of row $1$ and row $2$, you have forgotten about the first entry which is non-zero.
You should have added row $2$ to row $1$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2... | Here is an alternative approach that uses a triple integral.
We begin by noting that
$$\frac{1}{n} = \int_0^1 x^{n - 1} \, dx, \quad \frac{1}{2n + 1} = \int_0^1 y^{2n} \, dy, \quad \frac{1}{n + 1} = \int_0^1 z^n \, dz.$$
The sum can therefore be written as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{2n(2n + 1)(2n + 2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$... | The polar of $(3,-5)$ w.r.t. the ellipse is $$\frac{3x}{25}+\frac{-5y}{16}-1=0.$$ Intersecting the polar and the ellipse you get the tangent points of the two tangents through $(3,-5).$
They are $(\frac{375\sqrt{41}+1200}{769},\frac{144\sqrt{41}-2000}{769})$ and $(\frac{-375\sqrt{41}+1200}{769},\frac{-144\sqrt{41}-2000... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Region(area) enclosed between line and parabola The problem is to find the area of the region enclosed by the given parabola and line:
$y=-x^2+3x-1$
$y=x-2$
I tried solving the definite integral, but the roots (points of intersection of the parabola and line) on the graph were too weird and I ended up with the wrong so... | This is a method where you do not need to find the actual roots from the graph, from which the problem of "weird points of intersection" will be completely eradicated. Trust me.
From $-x^2+3x-1=x-2$, $x^2-2x-1=0$.
Letting $a$ and $b$ (where $a<b$) be the solutions of $x^2-2x-1=0$, $a+b=-2$ and $ab=1$, due to the root-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sq... | To simplify the derivation you can use the binomial series and note that
$$\sqrt{n+\sqrt{n}}= \sqrt{n}\left( \sqrt{1+\frac1{\sqrt{n}}} \right)=\sqrt{n}\left( 1+\frac1{2\sqrt{n}}+o\left(\frac1{\sqrt{n}}\right) \right)=\sqrt{n}+\frac12+o(1)$$
thus
$$\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)=\sqrt{n}+\frac12+o(1)-\sqrt{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Why is major axis of this ellipse is along what appears to be minor axis? Given any ellipse in standard form, like
$$x^2/5+y^2/10=1$$
is always elongated towards $y$ as its denominator is larger or $b>a$.
But consider this ellipse
$$\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1$$
This is elongated to... | The two cases are similar
*
*for $x^2/5+y^2/10=1$ when $x=0$ we have the max elongation towards $y$
(that is $x=0$)
*for $\frac{\left(x+2y\right)^2}{5}+\frac{\left(2x-y\right)^2}{20}=1
$ when $x+2y=0$ we have the max elongation towards $2x-y$
(that is $x+2y=0$)
Indeed note that
$$x+2y=0\implies \frac{\left(0\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute the 100th power of a given matrix So, I was asked this question in an interview.
Given a matrix
$$M = \begin{bmatrix} 0 & 1& 1 \\ 1& 0& 1 \\ 1&1& 0\end{bmatrix},$$
find $M^{100}$.
How does one approach this question using pen and paper only?
| Here's another trick. Denote
$$
\mathbf 1 = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1& 1 & 1\end{bmatrix}, $$
and let $I=\mathrm{diag}(1,1,1)$ be the identity matrix. Then by the binomial formula
$$
M^{100}=(\mathbf 1 - I)^{100}=\sum_{k=0}^{100} \binom{100}{k}(-1)^k \mathbf 1^{100-k}, $$
so we are led to consider... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding all solutions to a trigonometric equation I was solving a simple trigonometric equation for my brother, going this path:
$16\sin^2(2x)\cos^2(2x) = 3$
${[4\sin(2x)\cos(2x)]}^2 = 3$
Applied the formula for $\sin(2a) = 2\sin(a)\cos(a)$ "backwards"
$[2\sin(4x)]^2 = 3$
$\sin(4x) = \frac{\sqrt{3}}{2}$
Then the solut... | Due to taking the square root, $$\sin(4x)=\pm\frac{\sqrt3}2\implies 4x=\frac\pi3, \frac{2\pi}3, \frac{4\pi}3, \frac{5\pi}3, \frac{7\pi}3, \frac{8\pi}3, \frac{10\pi}3, \frac{11\pi}3$$ so $$\boxed{x=\frac\pi{12}, \frac\pi6, \frac\pi3, \frac{5\pi}{12} \frac{7\pi}{12},\frac{2\pi}3, \frac{5\pi}6, \frac{11\pi}{12}}$$ are the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove the slope of the asymptote of a hyperbola Consider a hyperbola that opens horizontally (left to right):
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
How do we prove that the slopes of the asymptotes are $\pm \frac{b}{a}$?
| Trying to solve it myself:
Rearranging for $y$ I get:
$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
$$a^2(y-k)^2 = b^2(x-h)^2 - a^2b^2$$
$$y = k \pm \sqrt{\frac{b^2(x-h)^2 - a^2b^2}{a^2}}$$
$$y = f(x) = k \pm \frac{b \sqrt{-a^2 + h^2 + x^2 - 2 h x}}{a}$$
So then the slope is
$$\lim_{x\to\infty}\frac{f(x)}{x}$$
or
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalues of symmetric $\mathbb{R}^{p\times p}$ matrix I want to prove that
$$A^{(p)}
=
\begin{pmatrix}
a & 1... | The last diagonal matrix you are considering will be $(p-1)\times(p-1)$, because it results from applying Laplace expansion twice on a $(p+1)\times (p+1)$ matrix. Looking at the exponent of $(\lambda-a+1)$, you do not seem to have taken it into account.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by pa... | hint
$$\frac {1}{(x^2+a^2)^n}=\frac {1}{a^2}\frac {a^2+x^2-\frac {2x}{2}x}{(a^2+x^2)^n} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the ... | Hint: The sum is equal to
$$S_n=1^3+2^3+3^3+...+n^3=\dfrac{n^2(n+1)^2}{4}$$
The $n^\text{th}$ term in your representation is given as
$$S_{n(n+1)/2}-S_{n(n-1)/2}.$$
The indices are just the partial sums for
$$1+2+...+n = \dfrac{n(n+1)}{2}$$
$$1+2+...+(n-1)=\dfrac{n(n-1)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Inequality : $ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} +\sqrt{\frac{c}{a+1}} \le \frac{3}{2} $ $a+b+c =1$ , $a, b, c>0 $
Prove
$$ \sqrt{\frac{a}{b+1}} + \sqrt{\frac{b}{c+1}} + \sqrt{\frac{c}{a+1}} \le \frac{3}{2}$$
When I meet the inequality with square root symbols, I don't have any idea. It's my tragedy. I kno... | In eliminating square roots, a well chosen CS is your friend, (and for higher roots, often Holder). Note by Cauchy-Schwarz inequality,
$$\sum_{cyc} (a+1) \cdot\sum_{cyc} \frac{a}{(a+1)(b+1)}\geqslant \left(\sqrt{\frac{a}{b+1}}+\sqrt{\frac{b}{c+1}}+\sqrt{\frac{c}{a+1}} \right)^2$$
Hence it is enough to show that
$$\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1... | Put differently, if $\alpha+\beta+\gamma=\frac \pi2$, then
$$ \tag1\sin^2\alpha+\sin^2\beta+\sin^2\gamma+2\sin\alpha\sin\beta\sin\gamma=1.$$
Let's investigate how the left hand side aries if we keep $\gamma $ fixed, that is, we consider
$$\tag2 \sin^2(\alpha+t)+\sin^2(\beta-t)+\sin^2\gamma+2\sin(\alpha+t)\sin(\beta-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\int \frac{dx}{x+x^2}$ without using partial fractions? I can easily rewrite it as $\frac{1}{x} - \frac{1}{x+1}$, but is it possible to solve this without partial fractions?
| Let $\displaystyle y=1+\frac{1}{x}$. Then $\displaystyle \frac{dy}{dx}=-\frac{1}{x^2}$.
\begin{align*}
\int\frac{dx}{x+x^2}&=-\int\frac{\frac{-1}{x^2}dx}{1+\frac{1}{x}}\\
&=-\ln\left(1+\frac{1}{x}\right)+C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which one is correct $\sum_{n=1}^{k} 2^{n}× n^2$ or $\sum_{n=1}^{k} (2^{n}× n^2)$? I have $2$ questions about notation.
Question 1.
Suppose, numbers are as follows.
$\left\{3\right\}\longrightarrow7$
$\left\{3,6\right\}\longrightarrow9$
$\left\{3,6,7\right\}\longrightarrow11$
$\left\{3,6,7,5\right\}\longrightarrow15$
... |
A word of warning to question 2: The subscript $n$ in
\begin{align*}
\sum_{n=1}^k 2^n+n^2
\end{align*}
does not tell anything about the scope of the sigma operator $\Sigma$.
*
*The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. It is given by the expression that follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
FInd the limit without the l’Hospital’s rule FInd the limit without the l’Hospital’s rule:
$$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x}-\sqrt{x+2}}$$
I have tried multiplying the equation with:
$\frac{\sqrt{2x}+\sqrt{x+2}}{\sqrt{2x}+\sqrt{x+2}}$ and got stuck with
$\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{2x-... | You've made some errors in arithmetic. The solution involves doing both of the things you tried:
$$\begin{align*}
\frac{3-\sqrt{x^2+5}}{\sqrt{2x} - \sqrt{x+2}} &= \frac{(3^2 - (x^2+5))( \sqrt{2x} + \sqrt{x+2})}{(2x - (x+2))(3 + \sqrt{x^2+5})} \\ &= \frac{(4-x^2)(\sqrt{2x} + \sqrt{x+2})}{(x-2)(3 + \sqrt{x^2+5})} \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A proof in combinatorics (prove by any method) I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following by using any method?
$$\left( \begin{array}{c} n \\ 1\ \end{array} \right) + 3 \left( \begin{array}{c} n \\ 3\ \end{array} \right) +5 \left( \begin{arra... | Putting the right-hand side to the left we have
\begin{align*}
\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\cdots=
\begin{cases}
1\qquad n=1\\
0\qquad n>1
\end{cases}
\end{align*}
or in more compact notation using Iverson brackets:
\begin{align*}
\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}=[[n=1]]
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0... | The condition $f(-x)=-f(x)$, i.e.,
$$|-x+a|-|-x+2|+b|-x+5|=-(|x+a|-|x+2|+b|x+5|)$$
is equivalent to
$$|x+a|+|-x+a|+b|x+5|+b|-x+5|=|x+2|+|-x+2|$$
which can be rewritten as
$$(|x+|a||+|x-|a||)+b(|5+x|+|5-x|)=|x+2|+|x-2|$$
Now think of each side of this equation as the formula for a continuous function whose derivative fa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Mathematical proof that $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ (Table proof provided) My question is how to express my solution to $m^4+8$ is not a cube of an integer if $13$ does not divide $m$ as a mathematical proof.
I understand that to prove by contradiction, the result is to be a cube of ... | Your method is perfectly fine for this. Just for fun, I'm giving an alternative method in which you can make do with only a table of squares mod 13.
n 0 1 2 3 4 5 6 7 8 9 10 11 12
n^2 0 1 4 9 3 12 10 10 12 3 9 4 1
$$m^4+8 \equiv n^3 \mod 13\\
(m^4+8)^4 \equiv n^{12} \mod 13$$
From Fermat's littl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $ { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $ Prove that one root of the Equation
$${ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x $$
is
$$x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) $$
My progress:
After simplyfying the given Equation I got
$$ x^8-16x^6+88x^4-192x^2+x+140=0$$
Then I t... | I created this equation by the following way.
$$3^n\equiv1,3,9,8,5,15,7,2,6,...\!\pmod{19}$$
Now, let
$$x_1=2\left(\cos\frac{2\cdot1\pi}{19}+\cos\frac{2\cdot8\pi}{19}+\cos\frac{2\cdot7\pi}{19}\right),$$
$$x_2=2\left(\cos\frac{2\cdot3\pi}{19}+\cos\frac{2\cdot5\pi}{19}+\cos\frac{2\cdot2\pi}{19}\right)$$ and
$$x_3=2\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem,... | Replace $x$ with $3+e$
If $x = 3 + e$ then $z(3-e) = 9; z=\frac 9{3-e}$ and $(3+e)(6-y)=9$ while $y(6- \frac 9{3-e}) = 9$ or $y(2-\frac 3{3-e}) = 3$.
So $y = \frac 3{2-\frac 3{3-e}} = 6-\frac 9{3+e}$
Let solve for $e$
$ \frac 3{2-\frac 3{3-e}}= \frac {3(3-e)}{2(3-e) - 3}=3\frac {3-e}{3-2e}=6-\frac9{3+e}=3(2 - \frac 3{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$a,b,c$ are digits. İf $a+2b+3c=40$ then what is the largest value for $a+b+c$? I am finding it hard to understand the method and the logic behind this.
I read that we have to give the one with the smallest coefficient the biggest value ,why is this . ?
| Assume you have found an optimal triple $(a,b,c)$ with $a,b,c\in\{0,\ldots,9\}$.
Note that $a+2b+3c=40$ implies $3c=40-a-2b\ge 40-9-18=13$ and so $c>4$.
Also, $2b=40-a-3c\ge 40-9-27=4$ and so $b\ge 2$.
If $a<8$ in our optimal solution, then (as $b\ge 2$ and so $b-1$ is still a digit), then $(a+2,b-1,c)$ is a valid tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $n$ be a 6-digit number, perfect square and perfect cube. If $n-6$ is not even or a multiple of 3, find $n$ Let $n$ be a 6-digit number, perfect square and perfect cube. If $n-6$ is not even or a multiple of 3, find $n$.
My try
Playing with the first ten perfect squares and cubes I ended with:
The last digit of $n ... | Note that the required number is both a square and a cube, so it must be a sixth power. Already $10^6=1000000$ has seven digits and $5^6=15625$ has only five digits, so that leaves us with $6^6,7^6,8^6,9^6$ to test.
Furthermore, we are given that $n-6$ is not even and not a multiple of 3, which implies that $n$ itself ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Joukowski mapping a circle onto an ellipse problem The Joukowski mapping is defined by
$$\displaystyle w = J(z) = \frac{1}{2} \left(z + \frac{1}{z}\right)$$
where $ z = x + yi $
Show that
$J$ maps the circle $|z| = r (r > 0, r \neq 1)$ onto the ellipse
$$ \frac{u^2}{\left[\frac{1}{2}\left(r + \frac{1}{r}\right)\right]... | HINT...You are almost there: you have $$u+iv=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac{1}{r(\cos\theta+i\sin\theta)}\right)$$
$$=\frac 12\left(r(\cos\theta+i\sin\theta)+\frac 1r(\cos\theta-i\sin\theta)\right)$$
So now you can get $\cos\theta$ and $\sin\theta$ in terms of $u$ and $v$ and use the Pythagorean identity... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equa... | simplifying the given equation we get
$$-5\cos(x)^2-5\cos(x)+4=0$$
with $$\cos(x)=t$$ we get the quadratic equation
$$-5t^2-5t+4=0$$
solving this we get
$$t_{1,2}=-\frac{1}{2}\pm\frac{\sqrt{105}}{10}$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Partial fraction decomposition of $\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$ So I'm working on this problem:
$$\frac{5z^4 + 3z^2 + 1}{2z^2 + 3z + 1}$$ where $z = x + yi$
It might be that my brain is just blanking right now but, how would one find the partial fraction decomposition of this expression? I tried long divisio... | First of all, you should divide $5z^4+3z^2+1$ by $2z^2+3z+1$, getting$$5z^4+3z^2+1=\left(\frac52z^2-\frac{15}4z+\frac{47}8\right)\left(2z^2+3z+1\right)-\frac{111}8z-\frac{39}8.$$So$$\frac{5z^4+3z^2+1}{2z^2+3z+1}=\frac52z^2-\frac{15}4z+\frac{47}8-\frac{\frac{111}8z+\frac{39}8}{(2z+1)(z+1)}.$$Now, find numbers $a$ and $b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove polynomial identity $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots+1).$ I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below:
$(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$
| This one cries out for a simple proof by induction:
If $k = 1$ we evidently have
$x^1 - 1 = (x^1 - 1)(1), \tag 1$
and if $k = 2$:
$x^2 - 1 = (x -1)(x + 1) = (x - 1)\left ( \displaystyle \sum_0^1 x^i \right ); \tag 2$
if we now suppose that the formula holds for some positive $m \in \Bbb Z$,
$x^m - 1 = (x - 1) \left ( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2... |
Prove that $n^2-2n+7$ is even then $n+1$ is even.
If $n^2-2n+7$ is even, then $n^2-2n$ is odd.
For all even numbers, this does not work since $(even)^2-2(even)$ always results in an even number.
Therefore, the number $n$ must be an odd number. Since $n$ is odd, $n+1$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Different ways of solving inequation - different solutions? After 'solving' the following exercise, I entered the solution in a website that gave me the graph and I noticed that It wasn't the same as the solution it gave me for the 'original' inequation.
The inequation is:
$5-\frac{4-5x}{2}<3\left(2x-1\right)$
And I so... | With inequalities you need to be careful because when you multiply for negative value inequality reverses. For this reason the first attempt gives a wrong result and the second was correct.
Note that you can also use first method but you need to distiguish the cases, notably
*
*for $2x-1>0$
$$5-\frac{4-5x}{2}<3\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_... | This solution is similar to an inductive proof with respect to $m$.
Let the sum of interest be
$$s_{n,m}=\frac{1}{\binom{n}{m}} \sum _{k=1}^m \frac{1}{k} \binom{n-k}{n-m}$$
For $m=1$ we find
$$s_{n,1}=\frac{1}{n}$$
The difference of $s$ with respect to $m$ turns out to be surprisingly simple:
$$s_{n,{m+1}}-s_{n,m}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ $A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$
here after row reduction
$\begin{bmatrix}1&0&2&1\\0&1&-1&2\\0&0&1&2\\0&0&0&0\end{bmatrix}$
clearly determinant is zero
but how can I find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$... | HINT
Take a vector in $\ker(A^2)$ in the general form $v=sv_1+tv_2$ then solve for $Av\neq 0 $ to find coefficients $s$ and $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to construct a triangle given two sides and their bisector? Suppose I have two triangle sides $AB$ and $AC$, and the length of the angle bisector of $A$. How can I construct (straightedge and compass) the triangle? (This question is from one of the earlier Moscow Math Olympiads.)
|
Let
the length
of the angle bisector of $A$
$|AD|=d$
and the side lengths
$|BC|=a,|AC|=b,|AB|=c$,
$|BD|=m$.
Then according to
Stewart's Theorem
for $\triangle ABC$,
\begin{align}
c^2(a-m)+b^2m
-a(d^2+(a-m)m)
&=0
\tag{1}\label{1}
.
\end{align}
By the law of sines,
\begin{align}
\triangle ABD:\quad
\frac{m}{\sin\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Adding Absolute value to a complex number: $ z+| z|=2+8i$ I would like to know my error in this problem.
Find the complex number such that:
$$ z+|z|=2+8i$$
So far, I have:
$$
\begin{split}
a+bi+\sqrt{a^2+b^2} &= 2 + 8i\\
a^2-b^2+a^2+b^2&=4-64\\
2a^2 -b^2 + b^2&=-60\\
a^2&=-30
\end{split}
$$
But I should end up with
$$a... |
$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$
$$a^2-b^2+a^2+b^2=4-64$$
This is where you went wrong, it looks like you tried to square both sides and take the real part but you made the wrong assumption that all the cross-multiplication terms would be imaginary.
$$a+bi+\sqrt{a^2+b^2} = 2 + 8i$$
$$a^2-b^2+(a^2+b^2) + 2abi + 2a\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
A difficult improper integral
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$
I got $2$ as a result. Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance
| As an alternative expression, we have
$$ \int_{1}^{\infty} \frac{dx}{x^2(e^x + 1)}
= C + \int_{0}^{1} \left( \frac{1}{e^x + 1} \right)'' \log x \, dx $$
where $C$ is given by
\begin{align*}
C
&= \frac{1}{4} + \frac{1}{1+e} + \frac{\gamma}{4} + \frac{\log 2}{3} - 3\log A \\
&\approx 0.14803096668067398473\cdots.
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplification algebraic of a cube root. I am trying to simplify this:
$$\frac{1000}{\pi \cdot (\frac{500}{\pi})^{\frac{2}{3}}}$$
and I think it becomes:
$$2 \cdot \sqrt[3]{\frac{500}{\pi}}$$
I basically thought we cube root the $\frac{500}{\pi}$ and then multiply the denominator by $\frac{500}{\pi}$ which could cancel... | I would do it like this:
$$\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}\cdot\dfrac{\sqrt[3]\pi}{\sqrt[3]{\pi}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{250000\pi}}{\pi}}=\dfrac{1000}{\sqrt[3]{250000\pi}}=\dfrac{1000}{50\sqrt[3]{2\pi}}=\dfrac{20}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac... | Denote the required sum value by $S$. Multiplying by $i$ we get
$$iS = \sum_{k=1}^{10}\bigg(\cos \frac{2\pi k}{11} + i\sin \frac{2\pi k}{11}\bigg)$$
Now in the summation introduce one extra term corrsponding to $k=11$ whose value will be 1. We get LHS to be $iS+1$. The RHS now is a 11-term summation which is actual... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the smallest distance between point and ellipsoid
Find the smallest distance between the points on the ellipsoid $x^2+2y^2+z^2=16$ and the point $(0,0,1)$.
Answer:
Let $(x,y,z)$ be the closest point of the ellipsoid. The distance is
$$\sqrt{x^2+y^2+(z-1)^2}$$
Let $f(x,y,z)=x^2+y^2+(z-1)^2$. Construct the Lagrang... | Alternative way:
We have found that $x=0$, and we need $|z| \le 4$,
\begin{align}x^2+y^2+(z-1)^2&=\frac{16-z^2}2+(z-1)^2 \\
&=\frac{16+z^2-4z+2}{2}\\
&=\frac{(z-2)^2-4+16+2}{2}\\
&=\frac{(z-2)^2}{2}+7\end{align}
Note that $|2| \le 4$, and hence the smallest distance is $\sqrt{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can... | We all know that the quadrangle of maximal area inscribed in a disc is a square, and that this square covers ${2\over\pi}$ of the area of the disc. Your ellipse has semiaxes $2$ and $3$, hence area $6\pi$. The largest rectangle inscribed in this ellipse therefore has area
$$\leq{2\over\pi}\cdot 6\pi=12\ ,$$
and this va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
... | Using the Extended Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&4&2\\\hline
1&0&1&-4&9\\
0&1&-5&21&-47\\
47&9&2&1&0\\
\end{array}
$$
shows that $21\cdot9-4\cdot47=1$ so we want $63\cdot9-12\cdot47=3$ which can be reduced, by subtracting $47$ from $63$ and $9$ from $12$, to
$$
16\cdot9-3\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find $\lim_\limits{(x,y)\to (0,0)}\frac{x^\alpha y^4}{x^2+y^4}$ where $\alpha > 0$ How can we find the following limit? $$\lim_{(x,y)\to(0,0)}\frac{x^\alpha y^4}{x^2+y^4}\qquad \alpha>0$$
By using the polar coordinate, we get
$$\lim_{r\to 0}r^{\alpha+2}\frac{\cos^\alpha\theta \sin^4\theta}{\cos^2\theta+r^2\sin^4{\theta... | From the AM-GM inequality we have $$\left|\frac{x^ay^b}{x^2+y^4}\right|\le \frac12 |x|^{a-1}|y|^{b-2}$$
where we assume that either $a$ is such that $x^a\in \mathbb{R}$ for $x$ in a neighborhood of $0$ or that the limit is taken as $(x,y)\to (0^+,0)$.
And equality holds when $x=\pm y^2$. Hence, we find
$$\lim_{(x,y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdo... | You should simplify a 2, a 3, and a 5. This is equivalent to find the number of positive divisors of
$$
\frac{2079000}{30} = 69300 = 2^2 \cdot 3^2\cdot 5^2\cdot 7\cdot 11
$$
which is
$
3^3 \cdot 2^2.
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
The logarithmic inequality $$ \log_{3x+7}{(9+12x+4x^2)}+ \log_{2x+3}{(6x^2+23x+21)} \ge 4$$
The logarithmic inequality is defined for: $ x \in (-3/2, -1) \cup(-1, \infty)$.
First, I supposed that my solutions are in the interval $(-3/2, -1)$. Following this interval for which the logarithmic function $t$ are decreasing... | Hint:
$$9+12x+4x^2=(2x+3)^2$$
$$ 6x^2+23x+21=(2x+3)(3x+7)$$
$$4\le\log_{3x+7}(2x+3)^2+\log_{2x+3}(2x+3)(3x+7)=2\log_{3x+7}(2x+3)+1+\log_{2x+3}(3x+7)$$
As $\log(y)$ is real for $y>0,$
we need $2x+3, 3x+7>0\implies x>$max$(-3/2,-7/3)=-3/2$
Now if $\log_{3x+7}(2x+3)=a,$
$$4\le2a+1+\dfrac1a$$
$$\iff0\le2a+\dfrac1a-3=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How I can solve $s(n)=n+s(n-1)$ by iteration method? $$
s(n)=
\begin{cases}
0, \text{if $n=0$}\\
n+s(n-1), \text{if $n>0$}\\
\end{cases}
$$
Using the relation
\begin{align}
s(n) &= n+s(n-1) \\
&= n+n-1+s(n-2) \\
&= n+n-1+n-2+s(n-3) \\
&= \dots \\
&= n+n-1+n-2+\dots+1 \\
&= \dfrac{n^2+n}2 \\
&= \theta(n^2)
\end{align}
I... | $s(n)=n+s(n-1)$
$s(n-1)=n-1 + s(n-2)$
$s(n-2)=n-2 + s(n-3)$
$s(n-3)=n-3 + s(n-4)$
.
.
.
$s[n-(n-1)]=n-(n-1) + s(0)$
⇒ $s(n)=n\times n -\frac{(n-1)(n-1+1)}{2}=n^2-\frac{(n)(n-1)}{2}=\frac{n^2+n}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
is $\ c= \big( \overline{x+yi} \big)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $? is $\ c= \bigg( \overline{x+yi} \bigg)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $?
EDIT
$c$ is complex, but not real.
| You have $c=a+ib$, then $c^{\frac 12}=[r(\cos x+i \sin x)]^{\frac 12}$, where $r=\sqrt{a^2+b^2}$ and $x=arctan(\frac{b}{a})$
Now by Moivre's formula
$c^{\frac 12}=
r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin
\left( \dfrac {x+2\pi k}{2}\right )\right ],k=0,1$.
Finally when calculating the conju... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Seating $m$ people in $m^2$ chairs at a round table I am having difficulty with the following counting problem.
Suppose there are $m^2$ indistinguishable chairs at a round table. There are $m$ people to be seated. Each arrangement is equally likely. Let $X_m$ denote the number of pairs of people sitting adjacent to o... | First, it makes no difference whether the chairs are distinguishable or not because each configuration in the scenario of indistinguishable chairs corresponds to exactly $m^2$ configurations in the distinguishable scenario (Note that there are $m^2$ cyclic permutations from $\{1, \cdots, m^2\}$ to itself). Thus in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Significance of the matrix $A^2$? Say you have destinations/places, $A,B,C,D$ and $E$ that are points that you can travel to and from. The only ways to travel between this is by the following schematic:
$A \rightarrow B \\ A \rightarrow E \\ B \rightarrow C \\ B \rightarrow D \\ C \rightarrow A \\ D \rightarrow A \\ D ... | The square of an adjacency matrix like $A_c$ can be interpreted as the number of paths of length $2$. For example, the $2$ in the $2$nd row and $1$st column shows that there are two paths of length exactly $2$ from $B$ to $A$ ($B \to C \to A$ and $B \to D \to A$). Correspondingly $A_c^2 + A_c$ gives you all the paths o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing a function grows like a certain power of $n$ Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?
| Convert the polynomial in the representation with Gamma function
$$
\eqalign{
& b(n) = \left( \matrix{
n \cr
n/2 + 1/2 \cr} \right) = {{\Gamma \left( {n + 1} \right)} \over {\Gamma \left( {n/2 + 3/2} \right)\Gamma \left( {n/2 + 1/2} \right)}} = \cr
& = {{\Gamma \left( {n + 1} \right)} \over {\left( {n/2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Derive Newton QUADRATE (high precision) of the form $\int_a^b f(x)dx ≈ w0 f(a)+w1 f(b)+w2 f′(a)+w3 f′(b)$ Derive a formula of the form
$$\int_a^b f(x)dx ≈ w0 f(a)+w1 f(b)+w2 f′(a)+w3 f′(b)$$
that is exact for polynomials of the highest degree possible.
For a simpler version like $$\int_0^1 f(x) dx ≈ w1f(0) + w2f(x1)$$... | You know that the formula can't be exact for $4^{\text{th}}$ degree polynomials because
$$\int_a^b(x-a)^2(x-b)^2dx\ne0$$
Because the integrand is positive in $(a,b)$, but your quadrature formula predicts it is $0$. Thus all you need is a basis for $\mathcal{P}_3$, the vector space of polynomials of degree at most $3$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$
If $\alpha,\beta,\gamma$ be a variable and $k$ be a constant such that $a\tan\alpha+b\tan\beta+c\tan\gamma=k$.Then find minimum value of $\tan^2\alpha+\tan^2\beta+\tan^2\gamma$ is
Try: Using Cauchy Schwarz Inequality:
$(a^2+b^2+c^2)(\tan^2\alpha+\tan^2\beta+\ta... | A bit of Geometry.
$x=\tan \alpha$; $ y= \tan \beta$; $z=\tan \gamma$.
Plane: $ax+by +cz=k.$
$d^2 := x^2+y^2+z^2$ is the squared distance from the origin to a point $(x,y,z)$.
Need to find the (perpendicular= minimal) distance of the plane from the origin.
$\vec n = \dfrac{1}{a^2+b^2+c^2}(a,b,c)$ is the unit normal of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Catalan's constant I am studying the integral $$\int_1^\infty\!\ \frac{\ln(x)}{1+x^{2}}$$ and ran across this representation of :
$$ \frac{1}{1+x^{2}}=\frac{1}{\frac{1}{x^{2}}(1-\frac{1}{x^{2}}+\frac{1}{x^{4}}-\frac{1}{x^{6}}+\frac{1}{x^{8}}...)}$$
where the denominator in the original integral has been rewritten to t... | The correct representation is:
$$
\frac{1}{1+x^2}=\frac{1}{x^2}\frac{1}{1+x^{-2}}=\frac{1}{x^2}\sum_{k=0}^\infty\left(-\frac{1}{x^2}\right)^k,
$$
which you incorrectly put in the denominator of RHS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying calculations for hitting probabilities using communicating classes When given a rather elaborate transition matrix for 6 state Markov chain, for example
$$\begin{bmatrix} 0&0&\frac{1}{2}&0&0&\frac{1}{2} \\ \frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&0\\ \frac{1}{3}&0&\frac{1}{3}&0&0&\frac{1... | Partition the state space $S=\{1,2,3,4,5,6\}$ into $S=A\cup B\cup C$ with $A=\{1,3,6\}$, $B=\{2,4\}$, $C=\{5\}$. Then $\mathbb P(X_{n+1}\in A\mid X_n\in A) = 1$ and $\mathbb P(X_{n+1}\in C\mid X_n\in C)=1$, so $A$ and $C$ are closed communicating classes, and since $\mathbb P(X_{n+1}\in A\mid X_n\in B)>0$ and $\mathbb ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Stuck on this square root conjugate problem I have to multiply the following:
$((x+h)\sqrt{x+h} - x\sqrt{x}) * ((x+h)\sqrt{x+h} + x\sqrt{x})$
I tried to use $(a - b)(a + b) = a^2 + b^2$ so:
$((x+h)\sqrt{x+h} - x\sqrt{x})((x+h)\sqrt{x+h} + x\sqrt{x}) = (x^2 + h^2)(x + h) - x^2(x)$
and then:
$x^3 + x^2h + h^2x + h^3 - x^... | Your error is in the first line.
You should have (take $a = (x+h)\sqrt{x+h}$ and $b = x\sqrt{x}$ and apply $(a - b)(a + b) = a^2 - b^2$):
\begin{align}
\big((x+h)\sqrt{x+h} - x\sqrt{x}\big)\big((x+h)\sqrt{x+h} + x\sqrt{x}\big) &= \color{blue}{(x + h)^2}(x + h) - x^2(x) \\
&= (x+h)^3 - x^3 \tag{1}
\end{align}
and NOT:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of sum formula, no induction $$\sum_{k=1}^n k=\frac{n(n+1)}2$$
So I was trying to prove this sum formula without induction. I got some tips from my textbook and got this.
Let $S=1+2+\cdots+n-1+n$ be the sum of integers and $S=n+(n+1)+\cdots+2+1$ written backwards. If I add these $2$ equations I get $2S=(1+n)+(1+n... | Method 1: (requires you to consider whether $n$ is odd or even.)
$S = 1 + 2 + ...... + n$.
Join up the first to term to the last term and second to second to last and so on.
$S = \underbrace{1 + \underbrace{2 + \underbrace{3 +....+(n-2)} + (n-1)} + n}$.
$= (n+1) + (n+1) + .....$.
If $n$ is even then:
$S = \underbrace{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Is this sequence of functions uniformly convergent on [0, 2] ?? Define a sequence of functions $f_n : [0,2] \to \Bbb R$ as:
$$f_n(x) = \frac {1-x} {1+x^n}$$
Is this sequence of functions uniformly convergent on $[0,2]$?
| Let $f_n(x)=\frac{1-x}{1+x^n}$ and $f(x)=\begin{cases}1-x&,0\le x\le 1\\\\0&,1\le x\le 2\end{cases}$
Clearly we have
$$\lim_{n\to \infty}f_n(x)=f(x)$$
Furthermore, we see that
$$|f_n(x)-f(x)|=\begin{cases}\frac{(1-x)x^n}{1+x^n}&,0\le x\le 1\\\\\frac{x-1}{1+x^n}&,1\le x\le 2\end{cases}$$
Next, we have the following es... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculus, how to find the value of delta in $\epsilon$-$\delta$ limit proof
Problem: Show that $\lim_{x\to 4}x^3=64$.
My steps:
$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
$\sqrt[3]{64-\epsilon}<x<\sqrt[3]{64+\epsilon}$
$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt... | Note that in you proof is ok from here
$$\sqrt[3]{64-\epsilon}-4<x-4<\sqrt[3]{64+\epsilon}-4$$
you should set
$$\delta=min\{4-\sqrt[3]{64-\epsilon},\sqrt[3]{64+\epsilon}-4\}=\sqrt[3]{64+\epsilon}-4$$
and what you find is the optimal value for $\delta$.
Note that by binomial expansion
$$\sqrt[3]{64+\epsilon}-4=4\sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am... | You know that $2^{36}\equiv 1$, since $2$ is a primitive root. Hence you may conclude that $(2^{12})^3\equiv 1$, and also $(2^{24})^3\equiv (2^{36})^2\equiv 1^2$. These correspond to $26$ and $10$ respectively. Of course $2^{36}=1$.
Now, suppose that $(2^a)^3\equiv 1$. Then $2^{3a}\equiv 1$, so $36|3a$, or $12|a$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
If $\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.
I tr... | $\def\peq{\mathrel{\phantom{=}}{}}$Denote $γ = α + β$, then$$
\frac{\cos α}{\cos β} + \frac{\sin α}{\sin β} = -1 \Longrightarrow \sin γ = \sin(α + β) = -\sin β\cos β.
$$
Thus,$$
\sin(γ - β) = \sin γ\cos β - \cos γ\sin β = -\sin β\cos^2 β - \cos γ\sin β,\\
\cos(γ - β) = \cos γ\cos β + \sin γ\sin β = \cos γ \cos β - \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find limit of $\frac{2x+7}{\sqrt{x^2+2x-1}}$ (check my steps please..)
Compute the limit $$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}$$
Here are my steps:
$$\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}=\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}\cdot\frac{-1/x}{-1/x}=\frac{\displaystyle \lim_{x\to-\infty}... | You are right. Note that\begin{align}\lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}&=\sqrt{\lim_{x\to-\infty}\left(1+\frac2x-\frac1{x^2}\right)}\\&=\sqrt{1}\\&=1.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\... | $$d\ln c=0 \implies \ln(c)=K$$
Another Hint
$$x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$$
$$(x + xy^2)dx + (y + yx^2)dy=0$$
It's an exact differential...
$$\frac {\partial P}{\partial y}=\frac {\partial Q}{\partial x} \implies 2xy=2xy$$
$$
\begin{cases}
f(x,y)=\int x+xy^2dx \\
f(x,y)=\int y+yx^2dy
\end{cases}
$$
Therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Find $\tan a \cdot\tan b$ given equation involving $\sin x$ and $\cos x$
Given:
$$(9\sin a + 44\cos a) \cdot (9\sin b + 44\cos b) = 2017$$
Find $\tan a \cdot \tan b$.
I factored the given equation, but I don't know how to proceed. Note that $9^2 + 44^2 = 2017$, that should be helpful in some way.
| $$
(9\sin a + 44\cos a) (9\sin b + 44\cos b) = 2017
$$
Since $9^2+44^2 = 2017,$ you have
$$
\left(\frac 9 {\sqrt{2017}} \right)^2 + \left( \frac{44}{\sqrt{2017}} \right)^2 = 1
$$
and so for some value of $\theta$ you have
$$
\cos\theta = \frac 9 {\sqrt{2017}}, \quad \text{and} \quad \sin\theta = \frac{44}{\sqrt{2017}}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Twin primes of the form $n^2+1$ and $N^2+3$? Assume that there are infinity many primes of the form $n^2+1$ and there are infinity many primes of the form $N^2+3$ , Then could we show that there are infinity primes of the form $n^2+1$ and $N^2+3$ (twin primes ) ?
Edit: I have edit the question just to show that $n... | A (trivial) insight...It's a lot easier to show which primes $n^2+1$ can't produce. Consider: $n^2+1 = n^2-1+2 = (n-1)(n+1)+2$; if $n^2+1 = p$, $p$ prime, then $(n-1)(n+1)+2 = p$, or $(n-1)(n+1) = p - 2$. Clearly $p$ can never be the larger of any two twin prime pairs because $p-2$ has factors $(n-1)$ and $(n+1)$. Fur... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Radius of a circle inscribed in a right triangle is given, the hypotenuse is given, find the legs? The radius of the circle inscribed in the right triangle is 4, the hypotenuse is 20cm. How do I find the legs? Any help would be appreciated!
| For the right triangle $ABC$
with
the hypotenuse $|AB|=c=20$ and
the radius of inscribed circle $r=4$,
the radius of circumscribed circle is known to be
found as
\begin{align}
R&=\tfrac{c}2=10
,
\end{align}
and the semiperimeter
\begin{align}
\rho&=\tfrac12(a+b+c)
=r+c
,
\end{align}
Considering the side lengths o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)^2=x+(x+1)f(x+2)$, what is $f(1)$? Suppose $f$: $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}$ and $f(x)^2 = x + (x+1)f(x+2)$, what is $f(1)$? Or more in general, what is $f(x)$?
The motivation behind this problem is that I want to find what the number of this nested radical $\sqrt{1+2\sqrt{3+4\sqrt{5+6\sqrt{7+8.... | The functional equation for $f(x)$ actually implies functional
relations also for all its derivatives
$$
\left\{ \matrix{
f(x)^{\,2} = x + \left( {x + 1} \right)f(x + 2) \hfill \cr
2f(x)f'(x) = 1 + f(x + 2) + \left( {x + 1} \right)f'(x + 2) \hfill \cr
2f'(x)^{\,2} + 2f(x)f''(x) = 2f'(x + 2) + \left( {x + 1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Proof by induction of series summation I understand the core principles of how to prove by induction and how series summations work. However I am struggling to rearrange the equation during the final (induction step).
Prove by induction for all positive integers n,
$$\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2$$
After bot... | $$\frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^2+(k+1)(k+1)^2=(k+1)^2\left(\frac14k^2+k+1\right)=(k+1)^2\left(\frac14k^2+\frac44(k+1)\right)=(k+1)^2\left(\frac14(k^2+4(k+1))\right)=\frac14(k+1)^2\left(k^2+4(k+1))\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove the relation is an equivalence relation. Problem
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \vert(a^2 + b) $. Prove that $R$ is an equivalence relation.
This is what I have so far.
Claim:
Define the relation $R$ on the set of natural numbers as $(a,b) \in R
> \iff 2 \mid(a^2... | If $2\mid a^2+b$ and $2\mid b^2+c$, then $2\mid a^2+b+b^2+c$. But $b+b^2=b(b+1)$, which is the product of two consecutive natural numbers, and therefore it's an even number. So, since $2\mid a^2+b+b^2+c$ and since $2\mid b^2+b$, $2\mid a^2+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Minimum of $\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}$ Let $0\leq x,y,z<1$ and $x^2+y^2+z^2=1$. What is the minimum value of
$$\frac{x}{\sqrt{1-x^2}}+\frac{y}{\sqrt{1-y^2}}+\frac{z}{\sqrt{1-z^2}}?$$
From the condition, the point $(x,y,z)$ lies on a sphere with radius $1$. At the equality poin... | You may convert it to become a function of two variables, the find the minimum value of that function. To do this, use
$$
z = \sqrt{1-(x^{2}+y^{2})}
$$
And substitute this in
$$ \frac{x}{\sqrt{1-x^{2}}} + \frac{y}{\sqrt{1-y^{2}}} +
\frac{z}{\sqrt{1-z^{2}}}$$
so the function will be
$$
f(x,y) = \frac{x}{\sqrt{1-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\l... | Let $$\text{P}=27\, \left( a+b \right) ^{2} \left( b+c \right) ^{2} \left( c+a
\right) ^{2}-64\,abc \left( a+b+c \right) ^{3}$$
We have$:$ $$\text{P} =\sum\limits_{cyc}c \left( 44\,a{b}^{2}+19\,abc+49\,{c}^{2}a+5\,b{c}^{2}+27\,{c}^{3}
\right) \left( a-b \right) ^{2} \geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
interval in which $h(x)$ is increasing If $h(x)=2g(2x^3-3x^2)+g(6x^2-4x^3-3)\forall x\in\mathbb{R}$ and $g''(x)>0\forall x\in\mathbb{R}$. Then find values of $x$ for which $h(x)$ is increasing
solution i try
$h(x)=2g'(2x^3-3x^2)(6x^2-6x)-g'(6x^2-4x^3-3)(12x^2-12x)$
From $g''(x)>0$ implies $g'(x)$ is increasing function... | Jacky.
$$ h(x) = 2g(2x^3−3x^2)+g(6x^2−4x^3−3) $$
Now let $g(2x^3−3x^2) = g(a)$ and $g(6x^2−4x^3−3)=g(b).$ So the 1st deriv of $h(x)$ is :
$$h'(x) = 12 \left[ g'(a)(x^2 - x) + g'(b)(x-x^2) \right] $$
$$ = 12 (x^2-x)\left[ g'(a) - g'(b) \right] $$
Now $h'(x)$ will be positive when
$$ x^2-x > 0 \:\: \text{and} \:\: g'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\fra... | HINT :
There is no parameter $a$ in the formula to be proved. So, first transform the initial equation into an equation where $a$ will be immediately eliminated by differentiation:
$$\frac{\sqrt{1-x^2}+\sqrt{1-y^2}}{x-y}=a$$
Differentiate and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$... | If $8R^2=a^2+b^2+c^2$ then putting $$\begin{cases}R=\dfrac{r^2+s^2}{2}\\a=r^2-s^2\\b=2rs\\c=r^2+s^2\end{cases}$$ it is verified the identity
$$8(\dfrac{r^2+s^2}{2})^2=(r^2-s)^2+(2rs)^2+(r^2+s^2)^2\iff2(r^2+s^2)^2=2(r^2+s^2)^2$$
This show that $a,b,c$ satisfy the well known parametrics of the Pythagorean triples (when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
solving for positive integer solutions $x^2 = yz,$
$x + y + z = 61$
positive integral solutions.
On wolframalpha, i could find the solution, but not how to solve.
I tried by taking a single equation
$xz + x^2 +z^2 - 61z = 0$
Tried solving for both x & z separately but did not find a way.
Please help.
EDIT: This is a ... | I wanted to see that a more algebraic approach looked like
$$\begin{align}
&\left[x^2=yz \quad | \quad z=61-x-y\right]\implies x^2=61y-xy-y^2 \\
&\qquad x^2+xy+\frac{y^2}{4}-\frac{y^2}{4}=61y-y^2 \\
&\qquad \left(x+\frac{y}{2}\right)^2=61y-\frac{3}{4}y^2 \\
&\qquad \left(2x+y\right)^2=244y-3y^2 \\
&\qquad -\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding Pointwise Limit Of These Functions Sequences of functions $f_n, g_n : [0, \infty) \to \Bbb R$, defined as
$f_n(x) = \dfrac{x}{1+x^n}$ and
$g_n(x) =
\begin{cases}
1
& \text{if } x > \frac{1}{n} \\
nx & \text{$0\leq x\leq1/n$}
\end{cases}
$
Find the Pointwise Limit.
Solution Attempt
Pointwise... | Your result for $g_n$ is correct.
The first comment suggests that $f_n$ is not correct. OP has the correct pointwise limit for $x \ge 1$. To fix the solution, first observe that $f_n(0) = 0$. When $0 < x < 1$,
$$\frac{x}{1+x^n} = \frac{1}{\dfrac1x + x^{n-1}} \xrightarrow[n\to\infty]{} x,$$
so the pointwise limit of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where’s the error in my factorial proof? On one of our tests, the extra credit was to find which number you would take out from the set $\{1!,2!,3!,...(N-1)!,N!\}$ such that the product of the set is a perfect square, for even $N$ My answer was as follows:
Assume $N$ is even. First note that $(n!)=(n-1)!\cdot n$.
Appl... | On this line:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2))$$
I believe you missed the first $2$. It should be:
$$E^2 O^2 2 \cdot 4 \cdot 6 \cdots N = E^2 O^2 \cdot 2 \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot (2 \cdot 4) \cdots (2 \cdot (N/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion I want to solve this problem using Taylor expansions.
I tried
\begin{align*}
1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;,
\end{align*}
where $R_{2n}(x)$ is the remainder function.
For $n=2... | For $x>0$,
$$\cos x > 1 -\frac{x^2}2$$
$$\begin{aligned}1-(\cos x) &< 1-\left(1 -\frac{x^2}2\right)\\
1-\cos x&< \frac{x^2}2 +\frac{x^3}6\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Two disjoint random events You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.
How do I create a table of pr... | If X is the number of ones seen in two rolls.
Each roll has a $\frac 13$ chance of seeing a $1$
$P(X = 2) = (\frac {1}{3})^2\\
P(X = 0) = (1-\frac {1}{3})^2\\
P(X = 1)= 2(\frac 13)(\frac 23)$
If Y is the number of $4's$ seen in two rolls.
Each roll has a $\frac 16$ chance of seeing a $4$
$P(Y = 2) = (\frac {1}{6})^2\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Conditions on coefficients of positive semidefinite matrix with certain symmetries I have a real matrix, with certain symmetries, defined as
$
A = \left( {\begin{array}{*{20}{c}}
1-x&a&b&c\\
a&x&d&b\\
b&d&x&a\\
c&b&a&1-x
\end{array}} \right),
$
with $x,a,b,c,d \in \mathbb{R},{\rm{~ }}0 \le x \le 1$.
I want to obtain co... | Essentially the same approach:
the real symmetric matrix $A$ is positive if and only if the polynomial in $\lambda$
$$\det(\lambda I +A)$$ has all coefficients $\ge 0$. The nice thing in this case is that this polynomial of degree $4$ factors into two polynomials of degree $2$. The equivalent condition is that both the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
prove this inequality by $abc=1$ Let $a,b,c>0$ and $abc=1$,show that
$$(a^{10}+b^{10}+c^{10})^2\ge 3(a^{14}+b^{14}+c^{14})$$
since
$$LHS=\sum \left(a^{20}+\dfrac{2}{a^{10}}\right)$$
it is prove
$$\sum_{cyc}\left(a^{20}+\dfrac{2}{a^{10}}-3a^{14}\right)\ge 0$$ not easy,and I found $x^{20}+2x^{-10}-3x^{14}$http://www.wolf... | We start with two identities that we can call Lamé-type identites:
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)=P$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))=Q$$
Furthermore we have :
$$(a+b)(b+c)(c+a)=\frac{(a+b+c)^3-a^3-b^3-c^3}{3}$$
Or
With the ident... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How do I determine this integral? $\int_{0}^{+\infty}\sin^2(1/x)\frac{dx}{(4+x^2)^2}$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}$$
$${\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1-\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1\over 2(4a^2+x^2)^2}-{\cos\left({2a\over x}\rig... | What I suggest is to write
$$
I\left(a\right)=\int_{0}^{+\infty}\frac{\sin^2\left(\displaystyle \frac{a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x \text{ and }J\left(a\right)=\int_{0}^{+\infty}\frac{\cos^2\left(\displaystyle\frac{a}{x}\right)}{\left(4a^2+x^2\right)^2}\text{d}x
$$
Then you have
$$
I\left(a\right)+J\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
... | you are given:
$$ 1^2 + 2^2 + ... + (k-1)^2 < \frac{k^3}{3} $$
by induction hypothesis. Now, what he is doing is
$$ 1^2 + 2^2 + ... + (k-1)^2 {\color{red} + } \color{red}{k^2} < \frac{k^3}{3} {\color{red} + } \color{red}{k^2} $$
Now, you want to prove that
$$ 1^2 + 2^2 + ... + (k-1)^2 + k^2 < \frac{(k+1)^3}{3} $$
But,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it. Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.
I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M... | Your proof is fine, to simplify note that
$$\frac {2n^2+10n+5}{n^2}=2+\frac{10}{n}+\frac5{n^2}$$
then it suffices to prove that $\frac1n \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{... | Let $x=a\cos t$
Comparing with $\cos3t$ formula $$\dfrac{a^3}4=\dfrac{3a}3, a=?$$
$$a\cos3t=-1$$
$$3t:360^\circ n\pm120^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Fourier Series of Absolute Value I think I understand how to evaluate Fourier series, but I would love to have someone check this to make sure I am doing this correctly.
Find the Fourier series of the function $f(x) = 11 + \lvert 6x \rvert$.
First, we find $a_0$.
\begin{align}
a_0 &= \bigg\langle f,\frac{1}{\sqrt{2}... | Guide:
Note that in the $a_0$, don't forget the braces.
$$a_0 = \frac1{\pi}\int_{-\pi}^\pi \frac1{\sqrt2} \color{blue}(11+ |6x|\color{blue}) \, dx$$
Also, note that $11+|6x|$ is an even function, hence $b_k=0$.
For $a_k$,
$$a_k = \frac{2}{\pi} \int_0^\pi \cos (kx) (11+6x)\, dx,$$
I think integration by parts should wor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $ \log_2(9x+2)=\log_3(16x+3). $
Solve the equation
$$
\log_2(9x+2)=\log_3(16x+3).
$$
It is easy to see that $x=0$ is a solution but how to prove that there are no more solutions?
My idea was prove that the function $f(x)=\log_2(9x+2)-\log_3(16x+3)$ is monotone for $x \geq 0$. We have
$$
f'(x)... | YES, we can solve this equation without using derivative.
here I solved this equation by a method without using derivative.
$\log_2(9x+2)=\log_3(16x+3)=y $(say)
$\log_2(9x+2)=y \implies2^y=9x+2$
multiply this equation by 16$\implies2^{y+4}=144x+32$.........(1)
second part of given equation is $\log_3(16x+3)=y\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
... | $\newenvironment{gmatrix}{\left\lgroup\begin{matrix}}{\end{matrix}\right\rgroup}$Here it is a general way of finding Jordan normal forms. It applies to your question because all eigenvalues of $A$ are real.
Step 1: Computing the invariant factors of $λI - A$ to get its elementary divisors.
The monic greatest common div... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How many possibilities are there for at least $k$ consecutive heads to show up out of $n$ tosses? Consider $n$ coin tosses. In how many ways can we have at least $k$ consecutive heads? Call this number $f(n,k)$. Is there a general expression for it? Or at least tight upper and lower bounds?
For example take $n=5,k=2$. ... |
We consider the binary alphabet $V=\{H,T\}$. We are looking for the number $g(n,k)$ of strings of length $n$ having runs of $H$ at most length $k-1$. The wanted number is $$f(n,k)=2^n-g(n,k)$$
Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that a + b + c = 0 => a x b = b x c = c x a If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \in D^3$ it is proved that $ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \Rightarrow \overrightarrow{a} \t... | It looks from the comments that you did not understand the details
that are missing from your proof, so here is a start.
$\overrightarrow{a} \times ( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} ) = \overrightarrow0$, hence
$\overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{a} \times \ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diagonalization of a Matrix. Having Trouble.. This is for Continuous-time Markov chains but I'm having trouble with the linear algebra.
Formal Definitions:
If Q is diagonalizable, then so is $e^{tQ}$, and the transition function can be expressed in terms of the eigenvalues and eigenvectors of $Q$. Write $Q=SDS^{-1}$, w... | There's a typo. Reverse engineering the given answer we have:
$$
\begin{align}
Q&=PDP^{-1}\\
&=\begin{pmatrix}
-1 & 0 & 1 \\
-3 & 1 & 1 \\
3 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-4 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
-1/4 & 0 & 1/4 \\
-3/2 & 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Real roots of an equation A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that... | Let $f(x)=(x^3+1)/2$. Then $f$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ with inverse $g(x)=(2x-1)^{1/3}$. As you already remarked, the original equation can be written as
$$((x^3+1)=2(2x-1)^{1/3}.$$
This equation is equivalent to $f(x)=g(x)$ or $f(f(x))=x.$
Now note that if $f(x)=x$, that is $x^3-2x+1=0$ then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
| In your first reformulation, you should get $\sqrt[15]{2^{4 \cdot 3} \cdot 16}$ and in the last part note that $\sqrt[a]{c^b} = c^{\frac{b}{a}}$ and not $c^{\frac{a}{b}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
$ Q(x_{1}) = x_{2}+x_{3} $ , $ Q(x_{2}) = x_{1}+x_{3} $ , $ Q(x_{3}) = x_{1}+x_{2} $ , $ Q(x_{1}x_{2}x_{3}) = 6 $ Let $ P(X) = X^3+7X^2+3X+1 $, with the roots $ x_{1},x_{2},x_{3} \in \mathbb{C} $
Let $ Q $ be a third grade polynomial with the following properties :
$ Q(x_{1}) = x_{2}+x_{3} $
$ Q(x_{2}) = x_{1}+x_{3}... | We have $\sum_i x_i=-7,\,\prod_i x_i = -1$ so $Q(x_i)=-7-x_i,\,Q(-1)=6$. Define $R(x):=Q(x)+x+7$ so $R(x_i)=0,\,R(-1)=12$. Since $P(-1)=4$, $R=3P$ and $Q(x)=3P(x)-x-7=3x^3+21x^2+8x-4$. Thus $\sum_i y_i = -7$ and $\sum_i P(y_i)=\sum_i \frac{y_i+7}{3}=7+\frac{1}{3}\sum_i y_i=\frac{14}{3}$, since each root of $Q$ satisfie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2757403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probabili... | Let $T$ be the number of tosses it takes to get $2$ balls in $1$ of the bins. If you are still tossing balls after $i$ tosses it is because none of the bins has more than $1$ ball in it, so there is $1$ ball in $i$ bins
$$ \Pr(T=i+1|T>i) = \frac{i}{n} .$$
Next, using Bayes' rule
$$ \Pr(T=i+1|T>i) = \frac{\Pr(T=i+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Why does this relation of binomials hold? Why does $\sum_{m=n}^N m\binom {m-1}{n-1}=\sum_{m=n}^N n \binom mn=n \binom{N+1}{n+1}$ is true? Is there some special formula for it?
| The first equality is due to the equality in @OlivierOloa's answer. The second one is a bit trick to obtain.
*
*Change $\binom nn$ to $\binom {n+1}{n+1}$.
*Group the first two term.
*Apply the equality $\binom nk + \binom n{k+1} = \binom {n+1}{k+1}$ to condense the leftmost two terms into one.
*Repeat (3) until ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Infinite positive integer solutions of the equation: $x^2+x+1=(y^2+y+1)(n^2+n+1)$ Could anybody help me some ideas on the below problem:
Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that
$$x^2+x+1=(y^2+y+1)(n^2+n+1).$$
Thanks in advance.
| To solve the Diophantine equation.
$$x^2+x+1=(z^2+z+1)(y^2+y+1)$$
It is necessary to use the solutions of the Pell equation. $p^2-(z^2+z+1)s^2=\pm1$
Then the solutions can be written as follows.
$$x=\mp((z+1)p^2+(z^2+z+1)(zs-p)s)$$
$$y=\mp((2z+1)p-(z^2+z+1)s)s$$
For positive you need to take decisions at $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Which $z\in\mathbb{C}$ satisfy the equation $|z-i|=\sqrt{2}|\bar{z}+1|$? Let $S$ be the set of all complex numbers $z$ satisfying the rule $$|z-i|=\sqrt{2}|\bar{z}+1|$$
Show that $S$ contains points on a circle.
My attempt,
By substituting $z = x + yi$, and squaring both sides. But I can't get the circle form.
| Let $z=x+iy$. Note that $\overline{z}=x-iy$.
$$\begin{align}
|z-i|&=\sqrt{2}|\overline{z}+1|\\
|z-i|^2&=2|\overline{z}+1|^2\\
|x+i(y-1)|^2&=2|(x+1)-iy|^2\\
x^2+(y-1)^2&=2(x+1)^2+y^2)\\
x^2+4x+2+y^2+2y-1&=0\\
(x+2)^2+(y+1)^2&=2^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
In a triangle ABC if $B-C = \frac\pi 4$ and .......... What will be the value of $\frac {a+b+c}{a}$
In a triangle ABC if $B-C = \frac\pi 4$ . Consider the following determinant.
|-2 cosC cosB |
| |
|cosC -1 cosA | = P
| |
|cosB cosA -1 |
If $ P +... | ►The function $f(x)$ where $x=C$ solution of your problem can be find out without using your determinant. The variable $x$ should be such that $0\lt x\lt\dfrac{3\pi}{8}(=135^{\circ})$, as you can verify by drawing triangles appropriate to the problem, and the function is increasing from $2$ to $\infty$. You have
$$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.