Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}\neq 0$ According to solution, $\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\frac{1}{2}$. Why it is so, when power of polynomial in denominator is greater than in numerator?
| If you change: $x+1=t$, then:
$$\lim_{x\to -1}\frac{x^3-2x-1}{x^4+2x+1}=\lim_{t\to 0}\frac{(t-1)^3-2(t-1)-1}{(t-1)^4+2(t-1)+1}=\lim_{t\to 0}\frac{t^3-3t^2+t}{t^4-4t^3+6t^2-2t}=\\
\lim_{t\to 0}\frac{t^2-3t+1}{t^3-4t^2+6t-2}=-\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral over set given by $x^2+y^2 \le 1, \frac {1}{\sqrt{3}} \le \frac{y}{x} \le \sqrt 3, z \le 1$ I want to compute the integral
$$\int_V x^2yz \, dx \, dy \, dz$$
over the set
$$V = \left\{ (x,y,z) \in \mathbb{R}^3_{>0} \mid x^2+y^2 \le 1, \frac {1}{\sqrt{3}} \le \frac{y}{x} \le \sqrt 3, z \le 1 \right\}$$
I thought the way the set is given motivates the substitution $u=x^2+y^2$, $v=\frac{y}{x}$ and $w=z$. Because then $V$ becomes a cuboid
$$V' = \left\{(u,v,w) \in \Bbb R^3_{\gt 0} \mid u \le 1, \frac{1}{\sqrt{3}} \le v \le \sqrt{3}, w \le 1 \right\} = [0,1] \times [\frac{1}{\sqrt{3}}, \sqrt{3}] \times [0,1]$$
which we can easily integrate over. However this doesn't work out as the integral in the end doesn't become easy. Can somebody tell me what has gone wrong? My computations are below.
To use the substitution rule we define the diffeomorphism
$$\Psi \left( \begin{bmatrix} x\\y\\z \end{bmatrix} \right) = \begin{bmatrix} u\\v\\w \end{bmatrix} = \begin{bmatrix} x^2+y^2\\ \frac{y}{x} \\z \end{bmatrix}$$
with inverse
$$\Phi \left( \begin{bmatrix} u\\v\\w \end{bmatrix} \right) = \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} \sqrt{\frac{u}{1+v^2}} \\ v\sqrt{\frac{u}{1+v^2}} \\ w \end{bmatrix}$$
We need to compute the correction factor with
$$\begin{aligned} D_{(u,v,w)}\Phi &= \begin{bmatrix} \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{1}{1+v^2} & \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{-2vu}{(1+v^2)^2} & 0 \\ \frac{v}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{1}{1+v^2} & \sqrt{\frac{u}{1+v^2}} + v \left( \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{-2vu}{(1+v^2)^2} \right) & 0 \\ 0 & 0 & 1 \end{bmatrix} \\&= \begin{bmatrix} \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} & \frac{-v\sqrt{u}}{(1+v^2)^{3/2}} & 0 \\ \frac{v}{2} \frac{1}{\sqrt{u(1+v^2)}} & \frac{\sqrt{u}}{(1+v^2)^{3/2}} & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$$
and
$$\begin{aligned} \det(D_{(u,v)}\Phi) &= \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{\sqrt{u}}{(1+v^2)^{3/2}} - \frac{v}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{-v\sqrt{u}}{(1+v^2)^{3/2}} \\&= \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{\sqrt{u}(1+v^2)}{(1+v^2)^{3/2}} \\&= \frac{1}{2} \frac{1}{1+v^2}\end{aligned}$$
Thereby with substitution
$$\begin{aligned} \int_V f \, dx \, dy \, dz &= \int_{V'} (f \circ \Phi)(u,v) \, \lvert \det(D_{(u,v)}\Phi) \rvert \, du \, dv \, dw \\&= \int_{V'} \frac{u}{1+v^2} v\sqrt{\frac{u}{1+v^2}}w \, \frac{1}{2} \frac{1}{1+v^2} \, du \, dv \, dw \end{aligned}$$
| As suggested if you make a sketch of the domain it seems a good idea use cylindrical coordinates
that is
$$\int_{\pi/6}^{\pi/3} d\theta \int_{0}^{1} dr \int_{0}^{1} r^4\cos^2\theta\sin\theta z \,dz$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit problem with summation: $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \dots + \frac{n}{n^2 +n}$ $\lim_{n\to\infty} \frac{1}{n^2 +n} + \frac{2}{n^2 +n} + \frac{3}{n^2 +n} + \frac{4}{n^2 +n} + \dots + \frac{n}{n^2 +n}$
question is when we take limit we can seperate things right ? So we can write $\lim_{n\to\infty} \frac{1}{n^2 + n}$ + $\lim_{n\to\infty} \frac{2}{n^2 + n}$ + .... $\lim_{n\to\infty} \frac{n}{n^2 + n}$
if we take limits one by one we get zeroes.
we get sum = 0
but if we do sum first than take limit
$\lim_{n\to\infty} \frac{\frac{n.(n+1)}{2}}{n^2 + n}$ with simplification we get 1/2
so did my first question wrong ? can't we take limits first than do the sum ?
| Two other approaches:
1.$$\sum_{k=1}^n \frac{k}{n^2+n}
= \frac{\sum_{k=1}^n k}{n^2+n}
= \frac{n(n+1)/2}{n^2+n}
=\frac{1}{2}$$
2.
Let $\Delta x = \frac{1}{n}$ and $x_k=k\Delta x$, then we rewrite as a right Riemann sum:
$$\sum_{k=1}^n \frac{k}{n^2+n}
= \underbrace{\frac{1}{1+1/n}}_{\to\, 1} \cdot\underbrace{\sum_{k=1}^n x_k \Delta x}_{\to\,\int_0^1x\;dx \,=\,\frac12}
\to \frac12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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General term of a sequence. So i have the following sequence:
${1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, ...}$
Where the number $i$ appears $i + 1$ times.
I would like to know the $n$-th term of this sequence. I tried to analise certain patterns within the sequence, but wasn´t able to conclude anything so far.
| Given $\color{red}1, 1, \color{red}2, 2, 2, \color{red}3, 3, 3, 3, \color{red}4, 4, 4, 4, 4, \color{red}5, ...$, first note that: $a_{T(k)}=k$, where $T(k)$ is a triangular number. Indeed:
$$a_{T(1)}=a_1=\color{red}1; a_{T(2)}=a_3=\color{red}2; a_{T(3)}=a_6=\color{red}3; a_{T(4)}=a_{10}=\color{red}4; a_{T(5)}=a_{15}=\color{red}5; \ ...$$
Hence:
$$T(k)\le n\le T(k+1)-1, a_n=k \iff \\
\frac{k(k+1)}{2}\le n\le \frac{(k+1)(k+2)}{2}-1 \iff \\
k^2+k-2n\le 0\le k^2+3k-2n \iff \\
\left\lceil \frac{\sqrt{8n+9}-3}{2}\right\rceil\le k\le \left\lfloor \frac{\sqrt{8n+1}-1}{2}\right\rfloor$$
Hence:
$$a_n=\left\lceil \frac{\sqrt{8n+9}-3}{2}\right\rceil \ \text{or} \ \left\lfloor \frac{\sqrt{8n+1}-1}{2}\right\rfloor.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, Hence:
$A=-\frac{1}{4}$ and $B=\frac{1}{4}$
But is this correct?
| Let $$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{A(s^2+2s+2)+B(s^2-2s+2)}{(s^2+2s+2)(s^2-2s+2)}$$
The solutions to $$s^2+2s+2=0\space\text{are}\space s=-1\pm i$$
Using $s=-1+i$, we find $-1+i=B(4-4i)$, yielding $B=-\frac 14$.
Moreover, the solutions to$$s^2-2s+2=0\space\text{are}\space s=1\pm i$$
Using $s=1+i$, we get $1+i=A(4+4i)$, yielding $A=\frac 14$. Hence we can say that:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}=\frac{1}{4(s^2-2s+2)}-\frac{1}{4(s^2+2s+2)}$$
Both graphed here
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Curve of center of third circle Given two non-intersecting circles (i.e. their centers are separated by a distance larger than the sum of their radii), it is always possible to place a third circle on top of (or below) these circles such that the third circle is tangent to the first two circles. In fact there are infinitely many such third circles.
My question is: what is the curve described by the centers of such third circles?
If the parameters of the first two circles are $x_1, y_1, r_1$ and $x_2, y_2, r_2$ it seems to me that the center of the third circle $(x,y)$ should satisfy the following
$$ \sqrt {(x-x_1)^2+(y-y_1)^2} - r_1 = \sqrt {(x-x_2)^2+(y-y_2)^2} - r_2$$
But the solution to this equation eludes me.
| HINT
Wlog we can assume
*
*$x_1=y_1=y_2=0$
*$r_1=1$
*$x_2=a$
*$r_2=r$
and consider
$$\sqrt {x^2+y^2} - 1 = \sqrt {(x-a)^2+(y)^2} - r$$
$$\sqrt {x^2+y^2} = \sqrt {(x-a)^2+y^2} +1- r$$
$${x^2+y^2} = (x-a)^2+y^2 +(1- r)^2+2(1-r) \sqrt {(x-a)^2+y^2}$$
$$x^2+y^2-x^2+2xa-a^2-y^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$
$$2xa-a^2 -(1- r)^2= 2(1-r) \sqrt {(x-a)^2+y^2}$$
$$[2xa-(a^2 +(1- r)^2)]^2= 4(1-r)^2 [(x-a)^2+y^2]$$
$$4x^2a^2-4xa(a^2 +(1- r)^2)+(a^2 +(1- r)^2)^2=4(1-r)^2x^2-8a(1-r)^2x+4a^2(1-r)^2+4(1-r)^2y^2$$
$$[4a^2-4(1-r)^2]x^2+[8a(1-r)^2-4a(a^2 +(1- r)^2]x-4(1-r)^2y^2=4a^2(1-r)^2-(a^2 +(1- r)^2)^2$$
Note that for $r=1$ we obtain
$$4a^2x^2-4a^3x=-a^4\implies 4x^2-4ax+a^2=0 \implies (2x-a)^2=0 \implies x=\frac a 2$$
For $r=2$ and $a =4$ we have
$$60x^2-240x-4y^2-225=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $2^{x-z} + 2^{y-z} = 1$ Suppose we have an equation $2^{x-z} + 2^{y-z} = 1$,where $x, y, z$ are integers from $0$ to $9$. $x, y, z$ can have same values. I guess $x$ and $y$ should have $9$ possible values like $x=y=0$ and $z=1$ and so on . Or should it have more values?
| Without loss of generality let $x \leq y$ ad write $1+2^{y-x}=2^{z-x}$. We can show that $x=y$ is necessary. Assume that opposite false, i.e. $y>x$, then left side is odd integer, therefore right side must be odd integer as well, in other words $z-x=0$. But that would mean $1+2^{y-x}=1$, or $2^{y-x}=0$, impossible. Therefore $x=y$ and so $2=2^{z-x}$, i.e. $z-x=1$. Thus we have found all solutions $x=y=z-1$ (subjected to the set constrains).
Another way to look this (perhaps more intuitive way) is to look at binary representations of $2^x+2^y=2^z$. Since all three numbers $2^x,2^y,2^z$ have just one $1$ in its binary representation, the sum on the left must make one $1$. The only way for this is if $x=y$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Centroid of a non-polygonal concave shape determined by an equilateral triangle an its incircle I have a diagram here of an equilateral triangle ABC, centre O, where circle centre O has tangents which are all three sides of the triangle ABC. M is the midpoint of AB, and F is the intersection of arc DE and line MC. I know each coordinate of the triangle, and the triangle has edges of length 1.
I need to calculate the centroid of the shape CDE, where the DE vertex is the arc passing through point F. I understand the centroid will be along the line CM, because the shape is symmetrical. I have no idea how to find the exact point though. One thought is that it's the midpoint of line FC, and another thought is
that it's the midpoint of the perpendicular bisector of line DE through C.
Are any of these presumptions right? Or is there no way of working out the centroid of it without actually having it in real life and using a plumbline?
| We can do this using calculus.
Coordinatize by placing the center of the circle, of radius $r$, at the origin. Generalizing slightly, I'll take $\angle COE = \theta$ instead of specifically $\pi/3$; thus,
$$C = r(0,\sec\theta) \qquad D = r(-\sin\theta,\cos\theta) \qquad E = r(\sin\theta,\cos\theta)$$
As OP notes, the centroid of region $CDFE$ lies on $\overline{CM}$, so its $x$-coordinate is $0$. One sees that the $y$-coordinate of that centroid must match that of the half-region $CFE$, which is bounded by $\overleftrightarrow{CE}$ ($f(x) = - x \tan\theta + r \sec\theta$) and the circle ($g(x) = \sqrt{r^2-x^2}$).
By the formula for the centroid of a bounded region,
$$\begin{align}
\bar{y} \cdot (\text{area}\;CFE) &= \frac12\int_{0}^{r\sin\theta}f(x)^2 - g(x)^2 \;dx \tag{1a}\\[4pt]
&= \frac12\int_{0}^{r\sin\theta}( x^2\tan^2\theta - 2 r x\tan\theta\sec\theta + r^2\sec^2\theta) - (r^2-x^2) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2\sin^2\theta - 2 r x\sin\theta + r^2 - r^2\cos^2\theta + x^2\cos^2\theta) \;dx \tag{1b}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} x^2 - 2 r x\sin\theta + r^2\sin^2\theta \;dx \tag{1c}\\[4pt]
&= \frac1{2\cos^2\theta}\int_{0}^{r\sin\theta} \left( x - r\sin\theta\right)^2 \;dx \tag{1d}\\[4pt]
&= \left.\frac1{6\cos^2\theta} \left( x - r\sin\theta \right)^3\;\right|_{0}^{r\sin\theta} \tag{1e}\\[4pt]
&= \frac{r^3\sin^3\theta}{6\cos^2\theta} \tag{1f}
\end{align}$$
(Note: We could get from $(1a)$ to $(1d)$ fairly immediately by observing that $f(x)^2-g(x)^2$ gives the "power", with respect to the circle, of a variable point along $\overline{CE}$. But I digress ...) Then, since
$$\begin{align}
\text{area}\;CFE &= \text{area of }\; \triangle COE - \text{area of sector}\;FOE \tag{2a}\\[4pt]
&= \frac12 \cdot r \cdot r\tan\theta - \frac12 r^2 \cdot \theta \tag{2b}\\[4pt]
&= \frac12 r^2 (\tan\theta - \theta) \tag{2c}
\end{align}$$
we have
$$\bar{y} = \frac{r\sin^3\theta}{3\cos\theta(\sin\theta-\theta \cos\theta)} \qquad\stackrel{\theta=\pi/3}{\to}\qquad
\frac{3r\sqrt{3}}{2(3\sqrt{3} -\pi)} = r\cdot 1.26454\ldots
\tag{$\star$}$$
Alternatively, we can use geometric decomposition.
Writing $\bar{p}$ for the $y$-coordinate of the centroid of $\triangle DCE$ and $\bar{q}$ for the $y$-coordinate of the centroid of sector $DFE$, we have
$$\bar{y} \cdot(\text{area} \;CDFE) = \bar{p}\cdot (\text{area}\; \triangle DCE) - \bar{q}\cdot (\text{area}\; DFE) \tag{3}$$
We "know" that a triangle's centroid is $1/3$ of the way up along a median, and its area is $1/2$-base-times-height, so
$$\begin{align}
\bar{p} \cdot (\text{area}\;DCE) &= \left( r\cos\theta + \frac13 r ( \sec\theta - \cos\theta ) \right) \cdot \frac12 \cdot 2r\sin\theta \cdot r(\sec\theta - \cos\theta) \tag{4a}\\
&= \frac{r \sin^3\theta}{3 \cos^2\theta} \left( 1 + 2 \cos^2\theta\right) \tag{4b}
\end{align}$$
Consulting a convenient list of centroids, we find
$$\bar{q}\cdot(\text{area}\;DFE) = \frac{4 r \sin^3 \theta}{3(2\theta - \sin 2\theta)}\cdot \frac{r^2}{2}(2\theta-\sin 2\theta) = \frac23 r^3 \sin^3 \theta \tag{5}$$
So, the right-hand side of $(3)$ is $(4b)-(5)$, which reduces to twice the value of $(1f)$. Since the area of $CDFE$ is likewise twice the value in $(2c)$, the "twice"s cancel, and $(3)$ yields the result shown in $(\star)$. $\square$
| {
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"url": "https://math.stackexchange.com/questions/2807143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $(1^2 + 2^2 + \cdots + n^2)^n > n^n (n!)^2$ I need to prove that $a_{n} = (1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2}=b_{n}$ for any natural number n.
Proof:
Note that by the arithmetic-geometric mean inequality,$\frac{1^{2} +2^{2} + ...+n^{2}}{n} > (1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2}$. Then $1^{2} +2^{2} + ...+n^{2} > n(1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{n/2} = n(n!)^{n}$. This implies that $a_{n} =(1^{2} +2^{2} + ...+n^{2})^{n} >n^{n}(n!)^{2n} \geq n^{n}(n!)^{n}= b_{n}$. Then we conclude that $a_{n} > b_{n}$ for all natural numbers $n$.
Is there anything wrong with my proof?
| It is correct but by AM-GM we have
$$\frac{1^{2} +2^{2} + ...+n^{2}}{n} \ge (1^{2}\cdot 2^{2}\cdot...\cdot n^{2})^{1/n} $$
from which we conclude
$$\implies (1^{2} +2^{2} + ...+n^{2})^n\ge n^n(n!)^2$$
and equality holds only for the case $n=1$.
| {
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"url": "https://math.stackexchange.com/questions/2808213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx$
Integrate $\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx$
$\newcommand{\intd}[1]{\,\mathrm{d}#1}$
My Attempt
Put $t=\tan x\implies\intd t=\sec^2x\intd x$
\begin{align*}
\int_0^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}dx&=\int_0^{\pi/2}\frac{\tan x\cdot\sec^2x}{(1+m^2\tan^2 x)\sec^2x}\intd x\\
&=\int_0^{\infty}\frac{t}{(1+m^2t^2)(1+t^2)}\intd t
\end{align*}
Put $t^2=y\implies 2t\intd t=\intd y$
\begin{align*}
\frac{1}{2}\int_0^\infty\frac{dy}{(1+m^2y)(1+y)}&=\frac{1}{2}\int_0^\infty\bigg[\frac{1}{1-m^2}\cdot\frac{1}{1+y}+\frac{m^2}{m^2-1}\cdot\frac{1}{1+m^2y}\bigg]dy\\
&=\frac{1}{2(1-m^2)}\int_0^\infty\frac{dy}{1+y}-\frac{m^2}{2(1-m^2)}\int_0^\infty\frac{dy}{1+m^2y}\\
&=\bigg[\frac{1}{2(1-m^2)}\log|1+y|-\frac{1}{2(1-m^2)}\log|1+m^2y|\bigg]_0^\infty\\
&=\bigg[\frac{1}{2(1-m^2)}\log|\frac{1+y}{1+m^2y}|\bigg]_0^\infty=\color{red}{\frac{1}{2(1-m^2)}\bigg[\frac{\infty}{\infty}-0\bigg]}\end{align*}
I think I am getting stuck here as the substitutions does not seem to give the solution ?
Doubt
$$
\lim_{y\to 0}\log|\frac{1+y}{1+m^2y}|=\log 1=0\\
\lim_{y\to \infty}\log|\frac{1+y}{1+m^2y}|=\lim_{y\to \infty}\log|\frac{\frac{1}{y}+1}{\frac{1}{y}+m^2}|=\log\frac{1}{m^2}
$$
So what really we are doing with definite integrals ?. Does this mean that we are actually computing the upper and lower limits and taking the difference to find the definite integral ?
Note: Expanding in terms of $\sin x$ and $\cos x$ gives the solution, no doubt about limit going to infinity, not looking for that.
| I think that you could have done it faster avoiding the second change of variable since, as you wrote,
$$
I=\int\frac{\tan (x)}{1+m^2\tan^2 (x)}\,dx=\int\frac{t}{(1+m^2t^2)(1+t^2)}\,dt
$$ Now, partial fraction decomposition gives
$$\frac{t}{(1+m^2t^2)(1+t^2)}=\frac{m^2 t}{\left(m^2-1\right) \left(1+m^2 t^2\right)}-\frac{t}{\left(m^2-1\right)
\left(1+t^2\right)}$$ making
$$I=\int\frac{t}{(1+m^2t^2)(1+t^2)}\,dt=\frac{\log \left(1+m^2 t^2\right)-\log \left(1+t^2\right)}{2(m^2-1)}$$ $$I=\frac 1{2(m^2-1)}\log\left(\frac{1+m^2t^2}{1+t^2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to determine the fourth coefficient in the power series expansion of $(1 + \ln(1-x))^{-1}$?
$\dfrac{1}{1+\ln(1-x)}=\sum\limits_{n=0}^{+\infty}a_n x^n$, then $a_4=$?
My approach:
\begin{align*}\dfrac{1}{1+\ln(1-x)}&=\sum\limits_{n=0}^{+\infty}(-1)^n(\ln(1-x))^n\\
&=\sum\limits_{n=0}^{+\infty}(-1)^n\left(\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\end{align*}
Then $$a_4=(-1)^1\cdot\dfrac{1}{4}+(-1)^2\cdot\left(\dfrac{1}{2}\cdot \dfrac{1}{2}+\binom{2}{1}\cdot 1\cdot\dfrac{1}{3}\right)+ (-1)^3 \cdot\binom{3}{1}\cdot 1\cdot1\cdot\dfrac{1}{2}+ (-1)^4\cdot 1\cdot1\cdot 1\cdot 1=\dfrac{1}{6}$$
But the answer is $\dfrac{11}{3}$, ignoring $(-1)^n$. What’s the problem?
| You left out a negative sign in the logarithmic series:
$$\begin{align*}
\dfrac{1}{1+\ln(1-x)}
&=\sum\limits_{n=0}^{+\infty}(-1)^n(\ln(1-x))^n\\
&=\sum\limits_{n=0}^{+\infty}(-1)^n
\left(-\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\\
&=\sum\limits_{n=0}^{+\infty}
\left(\sum_{m=1}^{+\infty}\dfrac{x^m}{m}\right)^n\end{align*}$$
and expanding this carefully gives the coefficient of $x^4$ as $\frac{11}3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2809574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Simplifying $\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$ How to simplify the expression and get a closed form?
$$\frac{\sin\alpha\cos^2\beta}{\sin\beta}+\frac{\sin\beta\cos^2\alpha}{\sin\alpha}$$
where $\alpha := \frac12(m+1)\theta$ and $\beta:=\frac12(m-1)\theta$; and where $m$ is an integer.
I wanna get rid of the denominators or make it a constant without $m$, but don't know how.
| Let $2a = (m+1) \theta$ and $2b = (m-1) \theta$ to obtain
$$f = \frac{\sin(a) \, \cos^{2}(b)}{\sin(b)} + \frac{\sin(b) \, \cos^{2}(a)}{\sin(a)}.$$
Now, consider the following:
\begin{align}
f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right)
\end{align}
Using $2 \sin^{2} x = 1 - \cos(2 x)$ and $2 \cos^{2} x = 1 + \cos(2 x)$, then
\begin{align}
f &= \frac{1}{\sin(a) \, \sin(b)} \, \left(\sin^{2}(a) \, \cos^{2}(b) + \sin^{2}(b) \, \cos^{2}(a) \right) \\
&= \frac{1}{4 \sin(a) \sin(b)} \, ( (1-\cos(2a))(1+\cos(2b)) + (1-\cos(2b))(1+\cos(2a)) ) \\
&= \frac{1 - \cos(2a) \cos(2b)}{2 \, \sin(a) \sin(b)} \\
&= \frac{2 - \cos2(a-b) - \cos2(a+b)}{4 \, \sin(a) \sin(b)} \\
&= \frac{\sin^{2}(a-b) + \sin^{2}(a+b)}{2 \, \sin(a) \sin(b)}
\end{align}
This can be seen in the form
$$f = \frac{\sin^{2}(m \theta) + \sin^{2}\theta}{\cos\theta - \cos(m \theta)}.$$
It becomes evident from $\sin(a) \, \sin(b)$ that there are problems (asymptotes) at
$$\theta \in \left\{ \frac{2 n \pi}{m \pm 1} \right\} \hspace{5mm} n \geq 0. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proving the following inequality without using AM-GM inequality.
Let $x,y,z \in \Bbb R^+$ such that $x+y+z=3$. Prove the inequality
$\sqrt x+\sqrt y+\sqrt z \ge xy+yz+zx$
I tried to prove that $\sqrt x+\sqrt y+\sqrt z-(xy+yz+zx)\ge 0$
I squared the equality, put the value of $xy+yz+zx$ (in terms of $x^2+y^2+z^2$).
Now I tried to prove that the above expression but failed.
Thanks for hints or solutions.
| Your idea works because for $\sqrt{x}=a$, $\sqrt{y}=b$ and $\sqrt{z}=c$ we obtain:
$$\sum_{cyc}(\sqrt{x}-xy)=\sum_{cyc}\left(\sqrt{x}-\frac{3-x^2}{2}\right)=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3)=$$
$$=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3-3(x-1))=\frac{1}{2}\sum_{cyc}(x^2+2\sqrt{x}-3x)=$$
$$=\frac{1}{2}\sum_{cyc}(a^4-3a^2+2a)=\frac{1}{2}\sum_{cyc}a(a+2)(a-1)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series and computation Let $F\colon \mathbb{C}\to\mathbb{C}$ and let $f\colon \mathbb{R}\to\mathbb{R}$ be the functions defined by:
$$ F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}$$
$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}$$
Show that
$$ F(z) =
\begin{cases}
\frac{e^z-1}{z} & \text{for $z\in\mathbb{C}\setminus \{0\}$} \\
1 & \text{if $z=0$}
\end{cases}$$
and show that $xf'(x) + f(x) = e^x$
It is trivial that for $F(z) = 0\ \text{if $z=0$}$
However, I am not sure how to show that for all $z\in\mathbb{C}$ $F(z) = \frac{e^z-1}{z}$.
In order to show that $xf'(x) + f(x) = e^x$ I know that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$ and $xf'(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n-1}\cdot x}{(n+1)!}=\sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$
$x'f(x) + f(x) = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} = \sum_{n=0}^{\infty}n\cdot \frac{x^{n}}{(n+1)!}$
Where do I go from here to get $e^x$?
| $$F(z) = \sum_{n=0}^{\infty} \frac{z^n}{(n+1)!}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{z^{n+1}}{(n+1)!}=\frac{1}{z}\sum_{\color{blue}{n=1}}^{\infty} \frac{z^{n}}{n!}=\frac{1}{z}(\color{blue}{e^z-1})$$
edit: For the second part,
$$xf'(x) + f(x)=x\sum_{n=0}^{\infty} \frac{nx^{n-1}}{(n+1)!}+\sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}=\sum_{n=0}^{\infty} \frac{(n+1)x^{n}}{(n+1)!}=e^x$$ by cancelling $(n+1)$ from the fraction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof an Inequality of a sequence $\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt \frac{5}{3}$ I have tried to amplify it like this
$$\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}} $$
which obviously too big. Any proper ways to amplify it ?
| Maybe you are overdoing it. Consider leaving a first few terms untouched and then "amplify" the rest like you did.
$$S < \dfrac{1}{2-1}+\dfrac{1}{2^2-1}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}=...?$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $ \int_{0}^{2\pi} \frac{1}{12+5 \cos(t)}dt$ without residues So I have to evaluate the next integral, without using the residue theorem, but writing the next integral as an integral of a complex function over the unit circle (see substitution below):
$ \int_{0}^{2\pi} \frac{1}{12+5 \cos(t)}dt$
I know what the answer is going to be: $\frac{2\pi}{\sqrt{119}}$, but I get the following:
substituting $z=e^{it}$, $dt = \frac{dz}{iz}$ and working out the denominator:
$$-2i \oint_{|z|=1} \frac{1}{5z^2+24z+5} \,dz$$
Partial fractions:
$$\frac{-10i}{2\sqrt{119}} \oint_{|z|=1} \frac{-1}{5z+\sqrt{119}+12} -\frac{1}{-5z+\sqrt{119}-12}\,dz$$
Now by the Cauchy Riemann Theorem, I get:
$\frac{-10i}{2\sqrt{119}} (\frac{-2\pi i}{5} - \frac{-2\pi i}{5}) = 0$
since here (https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula) $f(z)=1$ so the integral equals $\frac{1}{5}*f(a)*2\pi i = \frac{2\pi i}{5}$.
I thought maybe I confused some - signs, but I ran over my solutions multiple times. Perhaps someone could help me out :)
| We first replace the integral operator with $2\int_0^\pi dt$, based on the $t\mapsto 2\pi-t$ symmetry. On this integration range, $x:=\tan\frac{t}{2}$ is strictly increasing. Since $dt=\frac{2dx}{1+x^2}$ and $\cos t=\frac{1-x^2}{1+x^2}$, the integral is $4\int_0^\infty\frac{dx}{17+7x^2}=\frac{4}{7}\frac{\pi}{2}\sqrt{\frac{7}{17}}=\frac{2\pi}{\sqrt{119}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can you prove $1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$ by induction? Can you provide the steps and corresponding explanations to prove the following predicate by induction?
$$P(n) := 1^3 + 2^3+\cdots+(n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$
I've done some work on it myself by attempting to show that $\frac{k^4}{4} < \frac{(k + 1)^4}{4}$ for the RHS, but I don't understand exactly what I am doing.
Thank you.
Notice: This is not a homework question. I'm attempting to self-study Calculus over the Summer.
| Let's try a generalization.
$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1}
$
Base case.
For $n=1$
this is
$0 < \frac1{m} < 1
$
which is true.
Induction step.
Suppose true for $n$.
$P(n) := \sum_{k=1}^{n-1} k^{m-1} < \frac{n^{m}}{m} < \sum_{k=1}^{n} k^{m-1}
$
Want to show that
$P(n) \implies P(n+1)$.
First, the left inequality of $P(n+1)$,
which is
$\sum_{k=1}^{n} k^{m-1} < \frac{(n+1)^{m}}{m}
$.
$\begin{array}\\
\sum_{k=1}^{n} k^{m-1}
&=\sum_{k=1}^{n-1} k^{m-1}+n^{m-1}\\
&<\frac{n^m}{m}+n^{m-1}\\
&=\frac{n^{m}+mn^{m-1}}{m}\\
\end{array}
$
so we are done if
$n^m+mn^{m-1}
\lt (n+1)^m
$
or,
dividing by $n^m$,
$1+m/n
\lt (1+1/n)^m
$
and this follows from
Bernoulli's inequality.
Next, the right inequality of $P(n+1)$,
which is
$\sum_{k=1}^{n+1} k^{m-1} > \frac{(n+1)^{m}}{m}
$.
$\begin{array}\\
\sum_{k=1}^{n+1} k^{m-1}
&=\sum_{k=1}^{n} k^{m-1}+(n+1)^{m-1}\\
&>\frac{n^m}{m}+(n+1)^{m-1}\\
\end{array}
$
so we are done if
$\frac{n^m}{m}+(n+1)^{m-1}
\ge \frac{(n+1)^m}{m}
$
or
$n^m
\ge (n+1)^m-m(n+1)^{m-1}
=(n+1)^{m-1}(n+1-m)
$.
or
$1
\ge (1+1/n)^{m-1}(1-(m-1)/n)
$
or
$(1+1/n)^{m-1}
\le \frac1{1-(m-1)/n}
$.
This requires its own proof.
which we will do
by induction on $m$.
For $m=1$
this is
$1
\le 1$
which is true.
Suppose it is true for $m$
where $m < n-1$.
Then
$(1+1/n)^{m}
=(1+1/n)^{m-1}(1+1/n)
\le \frac1{1-(m-1)/n}(1+1/n)
$.
We want
$\frac1{1-(m-1)/n}(1+1/n)
\le \frac1{1-m/n}
$
or
$(1-m/n)(1+1/n)
\le 1-(m-1)/n
$
or
$1-(m-1)/n
\ge 1-(m-1)/n+m/n^2
$
which is true.
Therefore,
if $m < n-1$,
or $n > m+1$,
the right side is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conditional probability question involving balls w/o replacement A box contains 12 balls numbered 1 through 12. If 5 balls are selected one at a time from the box, without replacement, what is the probability that the largest number selected will be 9?
I want to just say $$\frac{9\times8\times7\times6\times5}{\binom{12}{5}}$$ but that is wrong and I don't know why.
| There are $8$ choices (i.e.: 1,2,3,4,5,6,7,8) which are less than $9$, $1$ choice which equals $9$ and $3$ choices (i.e.: 10,11,12) which exceed $9$. So, appplying the hypergeometric distribution formula, we readily get:
$$
\frac{\binom{8}{4}\binom{1}{1}\binom{3}{0}}{\binom{12}{5}}=\frac{\binom{8}{4}}{\binom{12}{5}}=\frac{\frac{5\cdot 6\cdot 7\cdot 8}{2\cdot 3\cdot 4}}{\frac{8\cdot 9\cdot 10\cdot 11\cdot 12}{2\cdot 3\cdot 4\cdot 5}}=\frac{5\cdot 6\cdot 7\cdot 8}{8\cdot 9\cdot 2\cdot 11\cdot 12}=\frac{5\cdot 7}{3\cdot 11\cdot 12}=\frac{35}{396}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ahmed's Integral Variation I was just wondering if anyone knows how to approach either (or both) of the following integrals. I'm not sure how much the bounds change between them, but regardless the integrands are the same. It's not exactly Ahmed's, but it seems pretty close. The Two Integrals
$ \int_1^\infty \dfrac{\tan^{-1}(\sqrt{2a^2+1})}{(1+3a^2)\sqrt{2a^2+1}}da \hspace{1cm} \int_0^1 \dfrac{\tan^{-1}(\sqrt{2a^2+1})}{(1+3a^2)\sqrt{2a^2+1}}da$
Feel free to give them a shot and thanks in advance. I came across both of these while trying to evaluate Coxeter's integral using tangent-half-angle substitution and then differentiation under the integral sign.
| $$\int_{0}^{1}\frac{\arctan\sqrt{2a^2+1}}{(1+3a^2)\sqrt{2a^2+1}}\,da=\int_{1}^{3}\frac{\arctan\sqrt{b}}{\sqrt{2}(3b-1)\sqrt{b(b-1)}}\,db=\sqrt{2}\int_{1}^{\sqrt{3}}\frac{\arctan(z)\,dz}{(3z^2-1)\sqrt{z^2-1}} $$
by Feynman's trick and the substitution $x\mapsto\frac{1}{z}$ equals
$$\sqrt{2}\int_{0}^{1}\int_{1/\sqrt{3}}^{1}\frac{z^2}{(3-z^2)(z^2+a^2)\sqrt{1-z^2}}\,dz\,da $$
or
$$\sqrt{2}\int_{0}^{1}\int_{\arctan(1/\sqrt{2})}^{\pi/2}\frac{\sin^2\theta}{(3-\sin^2\theta)(a^2+\sin^2\theta)}\,d\theta\,da $$
or, by partial fraction decomposition,
$$\sqrt{2}\int_{0}^{1}\frac{\pi}{\sqrt{6}}\cdot\frac{1}{3+a^2}\,da-\sqrt{2}\int_{0}^{1}\frac{\arctan\sqrt{\frac{2}{1+1/a^2}}}{(3+a^2)\sqrt{1+1/a^2}}\,da $$
which equals
$$ \frac{\pi^2}{18}-\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{\arctan\sqrt{\frac{2a}{a+1}}}{(3+a)\sqrt{a+1}}\,da = \frac{\pi^2}{18}-\frac{1}{2}\int_{0}^{1}\frac{\arctan\sqrt{z}}{(3-z)\sqrt{2-z}}\,dz $$
or
$$\frac{\pi^2}{18}-\int_{1}^{\sqrt{2}}\frac{\arctan\sqrt{2-z^2}}{1+z^2}\,dz\stackrel{\text{IBP}}{=}\frac{\pi^2}{18}+\frac{\pi^2}{16}-\frac{1}{2}\int_{1}^{2}\frac{\arctan\sqrt{z}}{(3-z)\sqrt{2-z}}\,dz.$$
By keep playing with the functional relations for the arctangent function, Feynman's trick and integration by parts the following conjectural identity should follow:
$$\int_{0}^{1}\frac{\arctan\sqrt{2a^2+1}}{(1+3a^2)\sqrt{2a^2+1}}\,da=\frac{13\pi^2}{288}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the value of $\cot(16)\cot(44)+\cot(44)\cot(76)-\cot(76)\cot(16)$ Find the value of $$S=\cot(16)\cot(44)+\cot(44)\cot(76)-\cot(76)\cot(16)$$
Note:All angles are in degrees
My method:
I used the identity
$$\tan(x)\tan(60+x)\tan(60-x)=\tan(3x)$$
So choosing $x=16$ we get
$$\tan76 \tan44 \tan 16=\tan48$$
hence
$$S=\frac{\tan(76)+\tan(16)-\tan(44)}{\tan(48)}$$
$$S=\frac{\frac{\sin(76)}{\cos(76)}-\frac{\sin(44)}{\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{\sin(32)}{\cos(76)\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{2\sin(32)}{2\cos(76)\cos(44)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
$$S=\frac{\frac{2\sin(32)}{\frac{-1}{2}+\cos(32)}+\frac{\sin(16)}{\cos(16)}}{\tan(48)}$$
can i need a clue from here?
| The idea is to exploit the identity $$\cot(\alpha+\beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot\beta}.$$ With $\alpha = 16^\circ$, $\beta = 44^\circ$, we get for instance $$\cot 16^\circ \cot 44^\circ = (\cot 16^\circ + \cot 44^\circ)\cot 60^\circ + 1.$$ Similarly, $$\cot 44^\circ \cot 76^\circ = (\cot 44^\circ + \cot 76^\circ) \cot 120^\circ + 1,$$ and $$\cot (-16^\circ) \cot 76^\circ = (\cot (-16^\circ) + \cot 76^\circ) \cot 60^\circ + 1.$$ Note $\cot (-\theta) = -\cot \theta$. All that is left to compute the sum of these three expressions, observing that $\cot 60^\circ = - \cot 120^\circ$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiation of 5^x I differentiated $5^x$ and got $x5^{x-1}.$
But the answer is $5^x\log_e 5.$
Why so?
| For a generalized answer to this question, you can use the following which works in cases of $5^x$ and $x^5$:
$$
\frac{\mathrm{d}}{\mathrm{d}x}\left(f(x)^{g(x)}\right) = f(x)^{g(x)-1}(g(x)f'(x)+f(x)\log(f(x))g'(x)
$$
So
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^5 &= x^{5-1}(5 \times 1 + x \log(x) \times 0) \\
&= x^4(5+0) \\
&= 5x^4
\end{align}
$$
And, likewise:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x} 5^x &= 5^{x-1}(x \times 0 + 5 \log(5) \times 1) \\
&= 5^{x-1}(0 + 5 \log(5)) \\
&= 5^{x} \log(5)
\end{align}
$$
Note $5 \times 5^{x-1} \equiv 5^x$
| {
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How are these Chernoff's bounds derived? In Esser, Kübler, May (2017), Chernoff's bounds are used to check that the Hamming weight of something could be part of a Binomial distribution.
The relevant parts are:
I looked up Wikipedia's description of the tail bounds for Binomial distributions, but Wikipedia notes that those work for $k \le np$, in this case $c \le \tau m$. But obviously, the authors define $c > \tau m$.
I want to work out what is happening here to both improve my understanding of the paper and allow me to tweak the variables and formulae to my end. So how are the authors applying Chernoff's bounds here?
| $\def\e{\mathrm{e}}\def\deq{\stackrel{\mathrm{d}}{=}}\def\peq{\mathrel{\phantom{=}}}$The best Chernoff's bound for binomial distribution will be derived first.
Suppose $X \sim B(m, p)$, then there exists i.i.d. $X_1, \cdots, X_m$ such that $X_k \sim B(1, p)$ and $X \deq \sum\limits_{k = 1}^m X_k$. For any $pm < c < m$, $a > 0$,\begin{align*}
P(X \geqslant c) &= P(\e^{aX} \geqslant \e^{ac}) \leqslant \e^{-ac} E(\e^{aX}) = \e^{-ac} (E(\e^{aX_1}))^m\\
&= \exp(-ac + m \ln(p\e^a + 1 - p)).
\end{align*}
Define $f(a) = -ac + m \ln(p\e^a + 1 - p)$, then $f'(a) = -c + \dfrac{mp\e^a}{p\e^a + 1 - p}$. Note that $f'(a)$ is increasing, $f'(0) = mp - c < 0$, and$$
f'(a) = 0 \Longleftrightarrow a = a_0 := \ln\left( \frac{1 - p}{p} · \frac{c}{m - c} \right),
$$
thus the logarithm of the best Chernoff's bound is $f(a_0)$. After rearraging terms,$$
f(a_0) = c \ln\frac{pm}{c} + (m - c) \ln\frac{(1 - p)m}{m - c}.
$$
Now, for trying to derive the given bounds, making substitution $x = \dfrac{c}{pm} - 1$, $y = \dfrac{1}{p} - 1$, then $0 < x < y$ and$$
c \ln\frac{pm}{c} + (m - c) \ln\frac{(1 - p)m}{m - c} = pm\left(-(1 + x) \ln(1 + x) + (y - x) \ln\frac{y}{y - x}\right).
$$
Define$$
g(x, y) = -(1 + x) \ln(1 + x) + (y - x) \ln\frac{y}{y - x}. \quad \forall 0 < x < y
$$
For the first bound, it suffices to prove $g(x, y) \leqslant -\dfrac{1}{3} \min(x, x^2)$ for $0 < x < y$. Note that\begin{align*}
g(x, y) &= -(1 + x) \ln(1 + x) + (y - x) \ln\left( 1 + \frac{x}{y - x} \right)\\
&\leqslant -(1 + x) \ln(1 + x) + (y - x) · \frac{x}{y - x}\\
&= x -(1 + x) \ln(1 + x).
\end{align*}
Define$$
G_1(x) = \frac{1}{x} (x -(1 + x) \ln(1 + x)),\quad G_2(x) = \frac{1}{x^2} (x -(1 + x) \ln(1 + x)).
$$
Because $G_1'(x) = -\dfrac{1}{x^2} (x - \ln(1 + x)) \leqslant 0$, then for $x > 1$,$$
G_1(x) \leqslant G_1(1) = 1 - 2\ln 2\\
\Longrightarrow g(x, y) \leqslant x -(1 + x) \ln(1 + x) \leqslant (1 - 2\ln 2)x.
$$
Next, $G_2'(x) = \dfrac{x + 2}{x^2} \left( \ln(1 + x) - \dfrac{2x}{x + 2} \right)$. Define $h(x) = \ln(1 + x) - \dfrac{2x}{x + 2}$, then$$
h'(x) = \frac{x^2}{(x + 1)(x + 2)^2} \geqslant 0 \Longrightarrow h(x) \geqslant h(0) = 0,
$$
which implies $G_2$ is increasing. Thus for $0 < x \leqslant 1$,$$
G_2(x) \leqslant G_2(1) = 1 - 2\ln 2\\
\Longrightarrow g(x, y) \leqslant x - (1 + x)\ln(1 + x) \leqslant (1 - 2\ln 2)x^2.
$$
Therefore, $g(x, y) \leqslant (1 - 2\ln 2) \min(x, x^2)$. Note that $1 - 2\ln 2 ≈ -0.38629 < -\dfrac{1}{3}$, thus for $p = τ$,$$
c \ln\frac{τm}{c} + (m - c) \ln\frac{(1 - τ)m}{m - c} \leqslant -\frac{1}{3} \min\left( \frac{c}{τm} - 1, \left( \frac{c}{τm} - 1 \right)^2 \right) · τm.
$$
For the second bound, i.e. $p = \dfrac{1}{2}$, it suffices to prove $g(x, 1) \leqslant -\dfrac{x^2}{2}$ for $0 < x < 1$. Define $G_3(x) = g(x, 1) + \dfrac{x^2}{2}$, then$$
G_3'(x) = \ln(1 - x) - \ln(1 + x) + x,\quad G_3''(x) = -\frac{x}{1 - x} - \frac{1}{1 + x} \leqslant 0,
$$
which implies $G_3'$ is decreasing. Thus $G_3'(x) \leqslant G_3'(0) = 0$, which implies $G_3$ is decreasing and $G_3(x) \leqslant G_3(0) = 0$. Therefore,$$
c\ln\frac{m}{2c} + (m - c)\ln\frac{m}{2(m - c)} \leqslant -\frac{1}{2} \left( \frac{2c}{m} - 1 \right)^2 · \frac{m}{2}.
$$
| {
"language": "en",
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} |
Proving trigonometric problem from given trigonometric equations If $p$ & $q$ are the solutions of $$a \cos x + b \sin x = c$$
Then how do I prove that, $$\cos (p + q) = \dfrac{a^2-b^2}{a^2 + b^2} $$
I tried all the adjustments I could think of, like dividing by $ \cos x $ and extracting $a$ and $b$ from the 2 equations. Also tried adding/subtracting and all the basics I know to no avail.
Any help is appreciated, thank you :)
| We have
$$c-a\cos x=b\sin x$$ and by squaring and rewriting,
$$c^2-2ac\cos x+a^2\cos^2x=b^2(1-\cos^2x),$$
$$(a^2+b^2)\cos^2x-2ac\cos x+c^2-b^2=0.$$
Using the Vieta's formulas, the product of the roots is
$$\cos p\cos q=\frac{c^2-b^2}{a^2+b^2}.$$
Repeating the same reasoning symmetrically,
$$\sin p\sin q=\frac{c^2-a^2}{a^2+b^2}.$$
Now by subtraction,
$$\cos(p+q)=\frac{a^2-b^2}{a^2+b^2}.$$
| {
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} |
Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?
Let $f(x)=\frac{1}{x^3+3x^2+3x+5}$, then what is $f^{(99)}(-1)$?
By letting $g(x)=x^3+3x^2+3x+5$, it is easy to see that $g'(-1) = g''(-1) = 0$, and so $f'(-1) = f''(-1) = f'''(-1) = 0$.
However, I have no idea how to compute $f^{(n)}(-1)$ for any positive integer $n$, and in particular for $n=99$.
Any idea would be appreciated.
| Let $t=x+1$. Then:
$$ f(x)= {1\over (x+1)^3+4} = {1\over 4}{1\over 1+{t^3\over 4}}=$$
$$ = {1\over 4}\Big(1-{t^3\over 4}+ \big({t^3\over 4}\big)^2-\big({t^3\over 4}\big)^3+...\Big)$$
So $$f^{(99)}(x) = -{1\over 4}{99!\over 4^{33}} + t(...)\implies f^{(99)}(-1) = -{99!\over 4^{34}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
| As James Arathoon mentioned in the comments, by the substitution $t= \sqrt[]{\frac {1-x^2}{x^2}}$ the integral is equal to:
\begin{align}
I:=\int^1_0 \frac{\log(x+\sqrt[]{1-x^2} )}{x}\,dx=\int^\infty_0 \frac{t\log\left( \frac{t+1}{\sqrt[]{t^2+1}}\right)}{t^2+1}\,dt
\end{align}
One can rewrite it a bit:
\begin{align}
I=\frac 1 2 \int^\infty_0 \frac{t\log\left( \frac{(t+1)^2}{t^2+1}\right)}{t^2+1}\,dt = \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2t}{t^2+1}\right)}{t^2+1}\,dt
\end{align}
Now define the following function $F:[0, 1]\to\mathbb R$ as follows:
\begin{align}
F(a) := \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2at}{t^2+1}\right)}{t^2+1}\,dt
\end{align}
Using Feynman's Trick one gets:
\begin{align}
F'(a) = \int^\infty_0 \frac{t^2}{(t^2+1)(t^2+2at+1)}\,dt
\end{align}
This integral is not very hard to compute, for instance one can do it by partial fraction decomposition or contour integration to get:
\begin{align}
F'(a) =\frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{2\ \sqrt[]{1-a^2}}
\end{align}
We know that:
\begin{align}
I = F(1) = \int^1_0 F'(a)\,da = \frac{1}{2}\int^1_0 \frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{\sqrt[]{1-a^2}}\,da
\end{align}
This looks a bit scary, but hey it is very innocent after setting $a=\cos(x)$, because then one gets:
\begin{align}
I = \frac{1}{2}\int^0_{\pi/2} \frac{\arctan\left(\tan(x)\right)}{\sin(x)}(-\sin(x))\,dx = \frac{1}{2}\int^{\pi/2}_0 x\,dx = \frac{\pi^2}{16}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
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} |
Calculating a flux integral
Let $$F=(xe^{xy}-2xz+2xy\cos^2 z, y^2\sin^2 z-y e^{xy}+y, x^2+y^2+z^2)$$ and $V$ be the solid in space bounded by $z=9-x^2-y^2$ and $z=0$. I am trying to compute the flux integral $\iint_{\partial V}F\cdot n \ dS$, $n$ being the outward unit normal.
Setting $r(x,y)=(x,y,9-x^2-y^2)$, I found that $r_x\times r_v=(2x,2y,1)$ and $$\iint_{\partial V}F\cdot n \ dS=\iint_D F\cdot r_x\times r_y \ dA$$ where $$F\cdot r_x\times r_y=2e^{xy}(x^2-y^2)-36x^2+4x^4+4x^2y^2+4x^2y\cos^2(9-x^2-y^2)+2y^3\sin^2(9-x^2-y^2)+2y^2;$$ after the substitution $x=r\cos t, \ y=r\sin t$ I get $$f(r,t)=r(F\cdot r_x\times r)=2r^3e^{r^2\sin t \cos t}(\cos^2 t-\sin^2 t)-36r^3\cos^2 4r^5\cos^2 t + \\4r^4\cos^2 t\sin t \cos^2{(9-r^2)}+2r^4\sin^3 t \sin^2(9-r^2)+2r^3\sin^2 t $$ and I need to compute $$\int_0^{2\pi}\int_0^3 f(r,t)drdt$$
I wonder whether I can further simplify $f(r,t)$? The current expression looks to cumbersome and it seems like a hassle to compute the integral if no simplifications can be made.
| If you are calculating the flux directly, then remember to normalize the normal to the surface. That is
$$\mathbf{n}=\frac{\mathbf{r}_x {\times} \mathbf{r}_y}{\Vert {\mathbf{r}_x {\times} \mathbf{r}_y} \Vert}$$
HINT: A simpler approach is to use the Divergence theorem:
$$\iint_R{\mathbf{F}.\mathbf{n}}dS = \iiint_V{\nabla \cdot\mathbf{F}dV}$$
| {
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"question_score": "2",
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$n,m$ are natural positive integers. F$(n,1)$=F$(1,n)=1$ F$(n,m+1)$+F$(n+1,m)=$F$(n+1,m+1)$ Write F($2,n$) and F($3,n$) as a function of $n$ How do I solve this problem? I do not know where to begin nor does anyone close to me. I am lost and I need help. Thank you so much for whoever helps.
| Using the rules: $F(n,1)=F(n,1)=1; F(n+1,m+1)=F(n,m+1)+F(n+1,m)$:
$$\begin{align}F(1,1)&=1;\\
F(1,2)&=F(2,1)=1;\\
F(2,2)&=F(1,2)+F(2,1)=2;\\
F(3,2)&=F(2,2)+F(3,1)=2+1=3;\\
F(2,3)&=F(1,3)+F(2,2)=1+2=3;\\
\vdots\end{align}$$
We get the Pascal triangle for $F(m,n)$:
$$\begin{array}{c|c|c|c|c|c|c}
m/n & 1 & 2 & 3 &4 &5&6&\cdots \\
\hline
1 & 1 & 1 & 1 &1&1&1 \\
\hline
2 & 1 & 2 & 3&4&5&6 \\
\hline
3 & 1 & 3 & 6&10&15&21\\
\hline
4 & 1 & 4 & 10&20&35&56\\
\hline
5 & 1 & 5 & 15&35&70&126\\
\hline
6 & 1 & 6 & 21&56&126&252\\
\hline
\vdots \\
\end{array}$$
Hence:
$$\begin{align}F(m,n)&= {m+n-2\choose n-1}=\frac{(m+n-2)!}{(n-1)!(m-1)!}.\\
F(2,n)&={n\choose n-1}=\frac{n!}{1!(n-1)!}=n.\\
F(3,n)&={n+1\choose n-1}=\frac{(n+1)!}{2!(n-1)!}=\frac{n(n+1)}{2}.\end{align}$$
Note: $F(m,n)=F(n,m)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the sum $\sum _{p=0}^{p=D}(-1)^{\frac{(p-1)(p-2)}{2}}\binom{D}{p}$ Calculate the sum $\sum _{p=0}^{p=D}(-1)^{\frac{(p-1)(p-2)}{2}}\binom{D}{p}$
I would like to calculate this sum. So far , I tried to write the factor $(-1)^{\dots}$ as the real part of a complex number but I can't still calculate it.
| For an integer $k$, $\binom{k-1}{2}=\frac{(k-1)(k-2)}{2}$ is even iff $k\equiv1\pmod{4}$ or $k\equiv2\pmod{4}$. Hence,
$$S_D:=\sum_{p=0}^D\,(-1)^{\binom{p-1}{2}}\,\binom{D}{p}=-1+\binom{D}{1}+\binom{D}{2}-\binom{D}{3}-\ldots\,.$$
We separate $S_D$ into even and odd partial sums
$$E_D:=-\sum_{p=0}^D\,(-1)^{p}\,\binom{D}{2p}=-1+\binom{D}{2}-\binom{D}{4}+\binom{D}{6}-\ldots$$
and
$$O_D:=\sum_{p=0}^D\,(-1)^{p}\,\binom{D}{2p+1}=\binom{D}{1}-\binom{D}{3}+\binom{D}{5}-\binom{D}{6}+\ldots\,.$$
It is clear via binomial expansion that $$E_D=-\left(\frac{(1+\text{i})^D+(1-\text{i})^D}{2}\right)=-\text{Re}\left((1+\text{i})^D\right)=-2^{\frac{D}{2}}\,\cos\left(\frac{\pi D}{4}\right)\,,$$ where $\text{i}:=\sqrt{-1}$.
Now, we note that
$$\begin{align}
S_D&=-1+\left(\binom{D}{1}+\binom{D}{2}\right)-\left(\binom{D}{3}+\binom{D}{4}\right)+\left(\binom{D}{5}+\binom{D}{6}\right)-\ldots
\\
&=-1+\binom{D+1}{2}-\binom{D+1}{4}+\binom{D+1}{6}-\ldots=E_{D+1}\,.
\end{align}$$
In other words, $S_D=E_{D+1}=-\text{Re}\left((1+\text{i})^{D+1}\right)=-2^{\frac{D+1}{2}}\,\cos\left(\frac{\pi (D+1)}{4}\right)$. That is,
$$S_D=\left\{
\begin{array}{ll}
(-1)^{\frac{D-4}{4}}\,2^{\frac{D}{2}}\,,&\text{if }D\equiv 0\pmod{4}\,,\\
0\,,&\text{if }D\equiv 1\pmod{4}\,,\\
(-1)^{\frac{D-2}{4}}\,2^{\frac{D}{2}}\,,&\text{if }D\equiv 2\pmod{4}\,,\\
(-1)^{\frac{D-3}{4}}\,2^{\frac{D+1}{2}}\,,&\text{if }D\equiv 3\pmod{4}\,.
\end{array}
\right.$$
On the other hand, it is also not difficult to see that $$O_D=\frac{(1+\text{i})^D-(1-\text{i})^D}{2\text{i}}\,.$$ That means $$\begin{align}S_D&=E_D+O_D=\frac{(-1-\text{i})(1+\text{i})^D+(-1+\text{i})(1-\text{i})^D}{2}\\&=-\left(\frac{(1+\text{i})^{D+1}+(1-\text{i})^{D+1}}{2}\right)=-\text{Re}\left((1+\text{i})^{D+1}\right)\,,\end{align}$$ which is the same as before.
| {
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"url": "https://math.stackexchange.com/questions/2836929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Minimum of the quartic $(x^2-1)^2+y^2$ using KKT conditions
Consider the following optimization problem.
$$\begin{array}{ll} \text{minimize} & (x^2-1)^2+y^2\\ \text{subject to} & x^2 - 4 \le 0\\ & x + y \le 0\end{array}$$
Using KKT conditions, find the optimal solution.
Solution: If one draw the region and the objective function then we clearly see that $\overline x=(\frac{1}{2},-\frac{1}{2})$ is the optimal solution.
And the rest it is just calculations and verifications of KKT conditions. So we can verify algebraically that $\overline x$ is the optimal solution.
Question:
Suppose we are not lucky enought to draw the region and objective function, so we are not able to find the solution geometrically.
What to do in this cases?
If you check KKT theorem (Confusion about definition of KKT conditions) , it does not explicitly says how to find the point that will be the optimal solution, it just states that if a point $x$ is fesible and f pseudoconvex at $x$ and there exists scalars such that ... then $x$ is optimal.
| Introducing some slack variables to cope with the inequalities we have the lagrangian
$$
L(x,y,\lambda,\epsilon) = (x^2-1)^2+y^2+\lambda_1(x+y+\epsilon_1^2)+\lambda_2(x^2-4+\epsilon_2^2)
$$
The stationary points are calculated by solving
$$
\nabla L = \left\{
\begin{array}{rcl}
4 x^3+2 \lambda_2 x+\lambda_1=4 x \\
\lambda_1+2 y=0 \\
\lambda_1 \epsilon_1=0 \\
\lambda_2 \epsilon_2=0 \\
\epsilon_1^2+x+y=0 \\
\epsilon_2^2+x^2=4 \\
\end{array}
\right.
$$
giving
$$
\left[
\begin{array}{c|ccc}
\text{label} & x & y & f(x,y)\\ \hline
A & -2 & 2 & 13 \\
B & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{3}{4} \\
C & 0 & 0 & 1 \\
D & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & \frac{3}{4} \\
E & 2 & -2 & 13 \\
F & -2 & 0 & 9 \\
G & -1 & 0 & 0 \\
\end{array}
\right]
$$
The minimum point is the interior point $G$
Attached a plot showing the stationary points found using the Lagrange multipliers technique.
NOTE
According to the KKT conditions the feasible points to consider are the labeled as $A,B,F, G$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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An inequality involving n-th real root of real numbers Given any $1\leq K\leq a$ and $1\leq K\leq b^{1/n}, $ $n$ is any positive integer, I was trying to prove
$$\frac{1}{1+a}+\frac{1}{1+b^{1/n}}-\frac{1}{1+(ab)^{1/(n+1)}}\leq \frac{1}{1+K},$$ where $x^{1/n}$ represents nth real root of $x.$
Will the induction principle help?
| The RHS of the inequality is decreasing with $K$, so for inequality to hold in general it is enough that it holds for the minimum value of the RHS, which is attained for the maximum $K=\sqrt[n]{b}\,$:
$$\require{cancel}
\begin{align}
\frac{1}{1+a}+\cancel{\frac{1}{1+\sqrt[n]{b}}}-\frac{1}{1+\sqrt[n+1]{ab}}\leq \cancel{\frac{1}{1+\sqrt[n]{b}}} \;\;&\iff\;\; \frac{1}{1+a} \le \frac{1}{1+\sqrt[n+1]{ab}} \\
&\iff\;\; \cancel{1}+\sqrt[n+1]{ab} \le \cancel{1} + a \\
&\iff\;\; \bcancel{a}b \le a^{n+\bcancel{1}} \\
&\iff\;\; \sqrt[n]{b} \le a
\end{align}
$$
Therefore the posted inequality does not hold in general, but the following reformulation does:
Given $1\leq K\leq \sqrt[n]{b} \le a\,$, $n \in \mathbb{N}\,$: $\;\;\displaystyle\frac{1}{1+a}+\frac{1}{1+\sqrt[n]{b}}-\frac{1}{1+\sqrt[n+1]{ab}}\leq \frac{1}{1+K}\,$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Determining $x + y + z$ $x$, $y$ and $z$ are different prime numbers;
$$x(z-y) = 18$$
$$y(z-x) = 40$$
Determine $x+y+z$.
I conceive of a way to solve this problem, which is to analyze the cases. For instance, Let's recall $x = 2$, then $z-y$ should be $9$. It will also give us $y = 5$. However, it doesn't satisfy the condition due to $z = 14$, which is not a prime number. How would you solve this question?
Regards!
| Given that the are different integers and prime the $x(z-y) = 18$ means $x = 2$ or $x=3$. $z > y$ and $z -y = 9$ or $z-y= 6$.
$y(z-x) =40$ mean $y = 2$ or $5$. $z> x$ and $z-x = 20$ or $z-x =8$.
There are only so many cases to check.
1: $x =2; y=5$; So $z-5=18;z-2 = 8$ so $z = 23=10$. This is impossible.
2: $x=3; y = 2$; So $z-2 = 6; z-3= 20$ so $z = 8 = 23$. This is impossible.
3: $x=3; y = 5$; So $x-6 =6; z-3=8$ so $z = 11$. This is possible.
As those are the only three options $x = 3; y = 5; z = 11$.
| {
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"source": "stackexchange",
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Find a real matrix $B$ such that $B^3 = A$
Given $$A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}$$ find a real, invertible matrix $B$ such that $B^3 = A$
I think I am doing something wrong here, so let me describe my attempt:
1) So I started off with diagonalizing the matrix $A$ with finding the eigenvalues $\lambda_1 = -8$ and $\lambda_2 = 1$ and the corresponding eigenvectors $ \vec v_1 = \begin{bmatrix}1 & 1\\0 & 0\end{bmatrix} = x + y = 0 \Rightarrow -x = y \Rightarrow \begin{bmatrix}1\\-1\end{bmatrix}$ and $ \vec v_2 = \begin{bmatrix}1 & -\frac{1}{2}\\0 & 0\end{bmatrix} = x - \frac{1}{2}y = 0 \Rightarrow 2x = y \Rightarrow \begin{bmatrix}1\\2\end{bmatrix}$
2) With that being done I proceeded with computing $D = \begin{bmatrix}-8 & 0\\0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 1\\-1 & 2\end{bmatrix}$ and check everything with $D = PAP^{-1}$
3) Now I thought I will simple find a diagonal matrix $M = PBP^{-1}$ and $M^3 = D$ and the easiest solution I came up with was $M = \begin{bmatrix}\sqrt[3]{-8} & 0\\0& 1\end{bmatrix}$ so basically $M = D^{\frac{1}{3}}.$ So that $B = PMP^{-1}$. But now come the tricky part, if I compute $B$ it results in a complex matrix not a real. //It is real!
Have I perhaps overlooked something here or miscalculated the solution for $B$?
Edit:
As Cameron pointed out my calculator and I totally failed as it was in complex mode and computed one of the non-real cube roots instead of -2. So $M = \begin{bmatrix}-2 & 0\\0 & 1\end{bmatrix}$ and consequentially $B = \begin{bmatrix}-1 & 1\\2 & 0\end{bmatrix}$
| Note: My goal here was to work the problem using triangular matrices.
Let
$
A = \begin{bmatrix}-5 & 3\\6 & -2\end{bmatrix}.
$
The eigenvalues for the matrix $A$ are $1$ and $-8$.
Set
$
\quad R = \begin{bmatrix}-2 & 1 \\ 0 & 1 \end{bmatrix} \quad \text{and} \quad S = \begin{bmatrix}-8 & 3 \\ 0 & 1 \end{bmatrix}
$
Observe that
$\tag 1 R^3 = S$
Set
$
\quad T = \begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix}
$
Observe that
$\tag 2 T^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$
and that
$\tag 3 T\,A\,T^{-1} = S$
Set
$\quad B = T^{-1} \, R \, T$
Using algebra it is easy to see that $B^3 = A$.
Calculating
$
\quad B = \begin{bmatrix}-1 & 1 \\ 2 & 0 \end{bmatrix}
$
| {
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Evaluating $\int_0^1\frac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$
$$\int_0^1\dfrac{3x^4+ 4x^3 + 3x^2}{(4x^3 + 3x^2 + 2x+ 1)^2}\, dx$$
Attempt:
If we write: $f(x)= x^4 + x^3+ x^2$, we get:
$$I = \displaystyle\int_0^1 \dfrac{3f(x)+x^3}{(f'(x)+1)^2}\, dx$$
I have no idea how to proceed. Integration by parts/ substitution can't help.
I don't need the full solution. Just a guiding hint would suffice.
| The hint of @achillehui to substitute $x\to y=\frac{1}{x}$ is valuable and should be an answer by its own.
We obtain
\begin{align*}
\color{blue}{\int_0^1\frac{3x^4+4x^3+3x^2}{\left(4x^3+3x^2+2x+1\right)^2}\,dx}
&=\int_1^\infty\frac{3y^{-4}+4y^{-3}+3y^{-2}}{\left(4y^{-3}+3y^{-2}+2y^{-1}+1\right)^2}y^{-2}\,dy\tag{1}\\ %fixed denominator here
&=\int_1^\infty\frac{3y^2+4y+3}{\left(y^3+2y^2+3y+4\right)^2}\,dy\tag{2}\\
&=-\left.\frac{1}{y^3+2y^2+3y+4}\right|_1^\infty\tag{3}\\
&\,\,\color{blue}{=\frac{1}{10}}
\end{align*}
Comment:
*
*In (1) we substitute $y=\frac{1}{x},\quad dy=-\frac{1}{x^2}dx$.
*In (2) we expand with $y^6$.
*In (3) we integrate by noting that $\frac{d}{dy}\left(y^3+2y^2+3y+4\right)=3y^2+4y+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
$A^2+B^2 +AB=36. B^2+C^2+BC=49. C^2+A^2+AC=64.$ Find $(A+B+C)^2$
$$A^2+B^2 +AB=36.\\ B^2+C^2+BC=49.\\ C^2+A^2+AC=64.$$ Find $(A+B+C)^2$.
I have tried it by using geometry I.e constructing a triangle and marking a point inside it which is making 120 ° and then using cosine rule
But have difficulty in solving further
Please use geometry
| If $A$, $B$, and $C$ are assumed to be positive real numbers, then consider a triangle $PQR$ with $p:=QR=7$, $q:=RP=8$, and $r:=PQ=6$. Then, the (internal) angles of this triangle are all less than $\frac{2\pi}{3}$. If $X$ is the Fermat point of the triangle $PQR$, then $XP=A$, $XQ=B$, and $XR=C$. Let $R'$ be a point on the opposite side of $QR$ with respect to $P$ such that $QRR'$ is an equilateral triangle. Then, $A+B+C=PR'$. This shows that
$$(A+B+C)^2=\left(PR'\right)^2=(PQ)^2+\left(QR'\right)^2-2\,(PQ)\,\left(QR'\right)\,\cos\left(\angle PQR'\right)\,,$$
where $PQ=6$, $QR'=QR=7$, and $\angle PQR'=\angle PQR+\frac{\pi}{3}$. You are left to determine the values of $\cos(\angle PQR)$ and $\sin(\angle PQR)$, as
$$\cos\left(\angle PQR'\right)=\cos\left(\angle PQR+\frac{\pi}{3}\right)=\frac{1}{2}\,\cos(\angle PQR)-\frac{\sqrt{3}}{2}\,\sin(\angle PQR)\,.$$
The answer is $(A+B+C)^2=\frac{149+63\sqrt{5}}{2}$.
| {
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} |
Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics. Evaluate $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle.
Similar question that appears just before this question in the text: Combinatorics problem involving series summation
| Using generating functions, which are widely used in combinatorics:
$$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$
which is the same as
$$a_n=a_{n-1}+\frac{n}{(n+1)!}$$
with generating function
$$f(x)=\sum\limits_{n}\color{red}{a_n}x^n
=a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\
a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\
a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\
a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\
a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\
a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$
or
$$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$
since $a_0=0$
$$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\
\frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$
as a result
$$a_n=1-\frac{1}{(n+1)!}, n\geq1$$
| {
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"question_score": "5",
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"answer_id": 2
} |
Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| As has been noted, you occasionally get powers of $2$ wrong; for example, $2\sin^2 x\cos^2 x\ne 2\sin^2 2x$, and even if it were $1-2\sin^2 x$ is $\cos 2x$, not $\cos\frac{x}{2}$. Let's talk about a better approach. Odd powers of $s:=\sin x,\,c:=\cos x$ have period $2\pi$, like $\tan\frac{x}{2}$; but even powers have period $\pi$, like $\tan x$. So let's use a slight variant on the usual Weierstrass substitution, viz. $t=\tan x$ so $dx=\frac{dt}{1+t^2},\,c^2=\frac{1}{1+t^2},\,s^2=\frac{t^2}{1+t^2}$. The integrand actually has period $\pi/2$ (can you see why?), so $$\int_0^{2\pi}\frac{dx}{s^4+c^4}=4\int_0^\infty\frac{(1+t^2)dt}{(1+t^2)^2-2t^2}=2\sum_\pm\int_0^\infty\frac{dt}{1+t^2\pm t\sqrt{2}}.$$I'll leave the rest to you.
| {
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"source": "stackexchange",
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"answer_id": 5
} |
Solving Quadratics Using Continued Fractions/Nonsimple to Simple Continued Fractions Let's say we want to find the continued fraction that solves the equation $x^2 - 2x - 1 = 0$.
Solution:
$$
x = 2 + \frac1x = 2 + \dfrac1{2 + \dfrac1x} = 2 + \dfrac1{2 + \dfrac1{2 + \dfrac1x}} = [2;\overline2]
$$
However, what happens if the quadratic is a bit more complicated, say, $2x^2 - 5x + 1 = 0$. If we use non-simple continued fractions to solve this, we get
$$
x = \frac52 -\dfrac{\frac12}{\frac52-\dfrac{\frac12}{\frac52-\ddots}} \stackrel{how?}= [ 2; \overline{3, 1, 1} ]
$$
My question is: How to convert non-simple continued fractions to simple continued fractions and in general how to solve quadratics using simple continued fractions?
| As far as i know it's possible to represent all quadratic equations by Simple Continued Fractions (SCF). The accepted answer is just fine but the common method to convert quadratics into SCF is slightly different and more intuative. For a second degree polynomial, we need a rational expression of $x$ as a quotitent so that we can recurse to obtain the SCF. Given the quadratic;
$
cx^2+(d-a)x-b = 0 \implies x=\frac{ax+b}{cx+d}
$
Our $2x^2-5x+1=0$ equation becomes a quotient of two linear equations $x=\lfloor\frac{6x-1}{2x+1}\rfloor$. So for what integer $x$ value, the floor of the fraction gives $x$?
As you may notice from the quadratic to quotitent conversion formula given above, we are free to chose $a$ and $d$ by keeping their difference the same.
As for starters, for the genaral case of $x=\frac{ax+b}{cx+d}$ equation, in order to obtain the coefficient $a_i$ I do like $\lfloor\frac{a}{c} - (\frac{d}{c} - \frac{b}{a})\rfloor$. However this is not very reliable. To be on the safe side we aim to make $\lfloor\frac{a}{c}\rfloor\cong\lfloor\frac{b}{d}\rfloor$. Like in this case you may guess it sometimes. Try $x=\lfloor\frac{4x-1}{2x-1}\rfloor$ which would suffice at $x=2$. One other point to remember is if $a,b,c,d>0$ then your $x$ value is for sure somewhere between $\frac{a}{c}$ and $\frac{b}{d}$ so playing with $a$ and $d$ by keeping the distance the same, if we can make these fractions close enough to have the same integer value then we are done. As a last resort you can do like $m=a-d$ and solve $\frac{a}{c}=\frac{b}{a-m}\implies a^2-ma-bc=0$ for $a$ by $\frac{m+\sqrt{m^2+4bc}}{2}$ as both $b,c$ and $m$ are known constants. Then simply$\lfloor\frac{a}{c}\rfloor$ is your first coefficient ($a_0$) to start with.
So with all this information at hand we can easily generate a regular (simple) continued fraction for second degree polynomials. Lets proceed with the solution of the question.
Step 1
For $2x^2-5x+1=0$
$x=\frac{6x-1}{2x+1}$
$a_0 =\lfloor\frac{6}{2} - (\frac{1}{2} - \frac{-1}{6})\rfloor = \lfloor\frac{7}{3}\rfloor = 2$
$x=2+\frac{1}{x_1}$.
This means we can now substitude $2+\frac{1}{x_1}$ at the place of $x$ in the above formula and move on.
Step 2
$2+\frac{1}{x_1}=\frac{6(2+\frac{1}{x_1})-1}{2(2+\frac{1}{x_1})+1}= \frac{11x_1+6}{5x_1+2}\implies\frac{1}{x_1}=\frac{11x_1+6}{5x_1+2}-2=\frac{x_1+2}{5x_1+2}\implies x_1=\frac{5x_1+2}{x_1+2};a_1=3$
Step 3
$3+\frac{1}{x_2}=\frac{5(3+\frac{1}{x_2})+2}{(3+\frac{1}{x_2})+2}= \frac{17x_2+5}{5x_2+1}\implies\frac{1}{x_2}=\frac{17x_2+5}{5x_2+1}-3=\frac{2x_2+2}{5x_2+1}\implies x_2=\frac{5x_2+1}{2x_2+2};a_1=1$
Step 4
One interesting part of this step is, calculating $a_3$ would result in an integer rather than a rational to floor. I think that marks the end of the period. But still..
$x_3=\frac{4x_3+2}{2x_3+3};a_1=1$
Step 5
Here we reach to what we calculated at Step 2 hence this repeats from here on.
$x_4=\frac{5x_4+2}{x_4+2};a_1=3$
Now that we have the coefficients at hand, we can say;
$2x^2-5x+1=0 \implies x=[2;\overline{3,1,1}]$
| {
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"source": "stackexchange",
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} |
Uniform Convergent and term by term differentiation Show that $f(x)= \sum_{n=1}^{\infty} \frac{1}{1+x^2+n^4x^2}$ is uniformly convergent for all real values of $x$.
Examine whether $f'(0)$ can be found by term-by-term differentiation.
| $\sum_{n=1}^{\infty} \frac{1}{1+x^2(n^4+1)}$ converges pointwise for all $x \ne 0$.
The convergence is not uniform because for $x_m = \frac1{\sqrt{m^4+1}}$ we have
$$ \sup_{x\in\mathbb{R}}\left|\sum_{n=m}^{\infty} \frac{1}{1+x^2(n^4+1)} \right| \ge \frac{1}{1+x_m^2(m^4+1)} = \frac12$$
which doesn't converge to $0$ as $m\to\infty$.
Your function $f$ is not defined at $0$ so $f'(0)$ doesn't make sense. However, term-wise differentiation at $x = 0$ gives
$$-\sum_{n=1}^\infty \frac{2x(n^4+1)}{(1+x^2(n^4+1))^2}\Bigg|_{x = 0} = \sum_{n=1}^\infty 0 = 0$$
| {
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"source": "stackexchange",
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} |
Cyclic Algebra. Suppose that $a,b,c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$ and $a^3 + b^3 + c^3 = 1$.
Find all possible value(s) of $a+b+c$.
My solution :
$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)$
$\Rightarrow a+b+c=1+c(1-c^2)+a(1-a^2)+b(1-b^2)$
$\Rightarrow a+b+c=1+c-c^3+a-a^3+b-b^3$
$\Rightarrow a+b+c=1+c+a+b-1$
$\Rightarrow a+b+c=c+a+b$
This led me to the same expression $a+b+c$.
Is there a smarter way to solve this problem?
| The numbers $a,b,c$ are the roots of the cubic
$$
(x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3
$$
where $e_1=a+b+c$, $e_2=ab+ac+bc$, $e_3=abc$.
Newton's identities tells us that
$$
e_{1}=p_{1},
\quad
2e_{2}=e_{1}p_{1}-p_{2},
\quad
3e_{3}=e_{2}p_{1}-e_{1}p_{2}+p_{3}
$$
for $p_1=a+b+c$, $p_2=a^2 + b^2 + c^2$, $p_3=a^3 + b^3 + c^3$.
Newton's identities with $p_1=s$, $p_2=1$, $p_3=1$ then give
$$
e_2 = \frac12(s^2-1),
\quad
e_3 = \frac16 (s - 1)^2 (s + 2)
$$
The discrimant of the cubic is then
$$
\Delta = \frac16(-s^6 + 9 s^4 - 8 s^3 - 21 s^2 + 36 s - 15)
= -\frac16 (s - 1)^2 (s^4 + 2 s^3 - 6 s^2 - 6 s + 15)
$$
The three roots of the cubic are real iff $\Delta \ge 0$ and this happens iff $s=1$ because $s^4 + 2 s^3 - 6 s^2 - 6 s + 15 \ge 0$ for all $s$.
(done with lots of help from WA)
| {
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"source": "stackexchange",
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Minima of $f(x)$
The miniumum value of $\left(1+ \dfrac{1}{\sin^n \alpha}\right)\left(1+ \dfrac{1}{\cos^n\alpha}\right)$ is?
Attempt:
I expanded the brackets and then differentiated and set the derivative equal to zero but it gets really complicated with that because of too many trig functions.
What is the efficient way to solve it?
The answer given is:
$(1+2^{\frac n2})^2$
| Cauchy Schwarz Inequality
$\displaystyle (a^2+b^2)(c^2+d^2)\geq (ac+bd)^2$
and equality hold when $\displaystyle \frac{a}{c} = \frac{b}{d}.$
Here I am assuming $\displaystyle \alpha \in \bigg(0,\frac{\pi}{2}\bigg).$
$$\Bigg[1+\frac{1}{\left(\sin^{\frac{n}{2}}\alpha\right)^2}\Bigg]\cdot \Bigg[1+\frac{1}{\left(\cos^{\frac{n}{2}}\alpha\right)^2}\Bigg]\geq \bigg[1+\frac{1}{\sin^{\frac{n}{2}}\alpha\cdot \cos^{\frac{n}{2}}\alpha}\bigg]^2$$
$$ = \bigg[1+\frac{2^{\frac{n}{2}}}{\left(\sin 2\alpha\right)^{\frac{n}{2}}}\bigg]^2 \geq \bigg(1+2^{\frac{n}{2}}\bigg)^2.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumption is correct then $x^2+1=(x-i)(x+i)$
Now for $x=\pm i$ using synthetic division it leaves remainder 0
Am I at the right direction, please tell
| $$
x^{20} + x^{15} + x^{10} + x^5 = x^5(x^5+1)(x^{10}+1)=x^5 \left((1 + x)(1 - x + x^2 - x^3 + x^4)\right) \left((1 + x^2) (1 - x^2 + x^4 - x^6 +
x^8)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Determine the line $R$
Determine the line $R$ that goes throught the point $p$, parallel to the plane $\alpha$ and dissects the line $L$
Given $p=\begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix}, \alpha \equiv \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ -1\end{bmatrix} + \rho\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix} + \tau\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}$
and $L \equiv \begin{cases} 5x_1 -x_2 -x_3 +2 = 0 \\ 12x_1 -3x_2 -2x_3 +4=0 \end{cases}$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $L\equiv \begin{bmatrix}2 \\ 4 \\ 0 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 1\\ 3 \end{bmatrix}$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$ but then I have to check parallelity with the plane $\alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $\alpha$ which it is.
So I think $R\equiv\begin{bmatrix}2 \\ 3\\ -1 \end{bmatrix}+ \omicron\begin{bmatrix}0 \\ -3 \\ 1 \end{bmatrix}$
Is this correct?
P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.
| Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).
Given $\alpha \equiv \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ -1\end{bmatrix} + \rho\color{blue}{\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}} + \tau\color{purple}{\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}}$
Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $\alpha$:
$$\vec n_\alpha = \color{blue}{\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}} \times \color{purple}{\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}}=\color{red}{\begin{bmatrix}2 \\ 4 \\ -5 \end{bmatrix}}$$
and $L \equiv \begin{cases} 5x_1 -x_2 -x_3 +2 = 0 \\ 12x_1 -3x_2 -2x_3 +4=0 \end{cases}$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $L\equiv \begin{bmatrix}2 \\ 4 \\ 0 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 1\\ 3 \end{bmatrix}$.
Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$L\equiv \begin{bmatrix}0 \\ 0 \\ 2 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix} \tag{$*$}$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$\underbrace{\begin{bmatrix}0 \\ 0 \\ 2 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}}_{\mbox{point on $L$}}-\underbrace{\begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix}}_{p}=\color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}}$$
Recall that we want this to be orthogonal to $\vec n_\alpha$, so:
$$\color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}} \cdot \color{red}{\begin{bmatrix}2 \\ 4 \\ -5 \end{bmatrix}} = 0 \iff \sigma = \ldots$$
This $\sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $\sigma$, you have:
$$R\equiv \begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix} + \mu \color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}}$$
Addition after comments. I wouldn't focus on the first coordinate but on the other two:
*
*$R_2 \to R_2-3R_1$
*$R_1 \to R_1+R_2$
$$\begin{bmatrix} 5 & -1 & -1 & -2 \\ 12 & -3 & -2 & -4 \end{bmatrix}
\sim \begin{bmatrix} 5 & -1 & -1 & -2 \\ -3 & 0 & 1 & 2 \end{bmatrix}
\sim \begin{bmatrix} 2 & -1 & 0 & 0 \\ -3 & 0 & 1 & 2 \end{bmatrix}$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
What is the least value of $\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$? What is the least value of this expression?
$$\tan^2 \theta + \cot^2 \theta + \sin^2 \theta + \cos^2 \theta + \sec^2 \theta+ \textrm{cosec}^2 \theta$$
Will putting $\theta=45^{\circ}$ give right answer?
| Let $s=\sin \theta$ and $c=\cos \theta$, then
\begin{align}
E &= \tan^2 \theta+\cot^2 \theta+\sin^2 \theta+
\cos^2 \theta+\sec^2 \theta+\csc^2 \theta \\
&= \frac{s^2}{c^2}+\frac{c^2}{s^2}+1+\frac{1}{c^2}+\frac{1}{s^2} \\
&= \frac{s^4+c^4+s^2c^2+s^2+c^2}{s^2c^2} \\
&= \frac{(s^2+c^2)^2-s^2c^2+1}{s^2c^2} \\
&= \frac{2-s^2c^2}{s^2c^2} \\
&= \frac{2}{\sin^2 \theta \cos^2 \theta}-1 \\
&= \frac{8}{\sin^2 2\theta}-1 \\
& \ge 7
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2856632",
"timestamp": "2023-03-29T00:00:00",
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Quick question about bound on certain function while computing limit I'm given the limit: $$\lim_{n\to\infty}\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}$$
The solution says:
Because exponentials grow faster than polynomials, there is some $n_0$ such that $\forall n\in\Bbb{N},n\geq n_0:n^2\leq n^3\leq n^4 \leq 2^n \leq 3^n \leq 4^n$ so that $$\forall n\in\Bbb{N},n\geq n_0:4\leq\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}\leq\sqrt[n]{6\cdot4^n}=4\sqrt[n]{6}$$ and by squeeze lemma it follows that the limit is $4$. I'm quite confused, where did we get the bound that the function is always greater than $4$ i.e. the first inequality.
| As is written in the comments we can see this through some straightforward replacements:
$$4$$
$$=\sqrt[n]{0+0+0+0+0+4^n}$$
$$\leq\sqrt[n]{n^2+n^3+n^4+2^n+3^n+4^n}$$
$$\leq\sqrt[n]{4^n+4^n+4^n+4^n+4^n+4^n}$$
$$=\sqrt[n]{6\cdot4^n}$$
$$=4\sqrt[n]{6}$$
Where for the last inequality we need $n>3$, since $n^4\leq 4^n$, fails for some small values of $n$, for $n=3$.
| {
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$[1+(\frac{1+i}{2})][1+(\frac{1+i}{2})^2][1+(\frac{1+i}{2})^{2^2}]...[1+(\frac{1+i}{2})^{2^n}]=(1-\frac{1}{2^{2^n}})(1+i)$,where $n\ge 2$
Show that $$\!\!\!\left[1+\left(\frac{1+i}{2}\right)\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^2\right]\!\!\!\left[1+\left(\frac{1+i}{2}\right)^{2^2}\right]\cdots\left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]\!\!\!=\left(1-\frac{1}{2^{2^n}}\right)(1+i)$$ for $n\ge 2$.
I took $\frac{1+i}{2}=\frac{1}{\sqrt2}e^{i\frac{\pi}{4}}$ and tried solving but i could not reach the RHS.Please help.
| Hint: Use the general case
$$(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})=\dfrac{(1-x^{2^{n+1}})}{1-x}$$
| {
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Solution to Putnam 2007 A-1
2007 A-1: Find the values of $\alpha$ for which the curves $y=\alpha x^2 + \alpha x + \frac{1}{24}$ and $x=\alpha y^2 + \alpha y + \frac{1}{24}$ are tangent to each other.
My solution: Notice that the second equation is obtained from the first by swapping $x$ and $y$. This happens when you reflect the graph of the first equation across the $x=y$ line. Hence, the two curves will intersect only once, and therefore be tangent to each other at that point of intersection, if each curve individually also intersects the $y=x$ line only once. Hence $x=\alpha x^2 + \alpha x + \frac{1}{24}$, on solving which we get $\alpha=\frac{43}{12},-\frac{17}{12}$.
This is not the answer or solution given on Kedlaya's putnam page. Where am I going wrong?
| Let $(x,y)=(a,b)$ be the point of tangency between the two curves. Thus,
$$b=\alpha a^2+\alpha a+\frac{1}{24}\text{ and }a=\alpha b^2+\alpha b+\frac{1}{24}\,.\tag{*}$$
We claim that $a=b$. To show this, suppose that $a\neq b$. By symmetry, $(x,y)=(b,a)$ is also a point of tangency of the two parabolas. By Bézout's Theorem, there are at most $4=2\cdot2$ common points of the two parabolas (which are algebraic curves of degree $2$), where points of tangency are counted with multiplicities greater than $1$. Thus, we have two distinct common points $(x,y)=(a,b)$ and $(x,y)=(b,a)$, each with multiplicity at least $2$. Consequently, there are no other common points. However, since $a\neq b$ and a parabolic curve is continuous, both parabolas must intersect the line $y=x$. By symmetry, this intersection point (indeed, if one intersection along the line $y=x$ existed, then there had to be at least $2$) is another common point of the parabolas, which is absurd. Therefore, $a=b$ must hold.
The slopes at $(x,y)=(a,b)$ of the two parabolas must be equal. For the graph $y=\alpha x^2+\alpha x+\dfrac{1}{24}$, the slope is $2\alpha a+\alpha$. For the other graph $x=\alpha y^2+\alpha y+\dfrac{1}{24}$, the slope is $\dfrac{1}{2\alpha b+\alpha}$ (whence $\alpha\neq 0$). That is,
$$\alpha^2\,(2a+1)(2b+1)=1\,.\tag{#}$$
We obtain $(2a+1)^2=(2a+1)(2b+1)=\dfrac{1}{\alpha^2}$. Thus,
$$2a+1=\pm\frac{1}{\alpha}\,,\text{ or }a=b=-\frac{1}{2}\pm\frac{1}{2\alpha}\,.$$
Now, plug this result into (*), we have
$$-\frac{1}{2}+\frac{s}{2\alpha} =\alpha \,\left(-\frac{1}{2}+\frac{s}{2\alpha}\right)^2+\alpha\,\left(-\frac{1}{2}+\frac{s}{2\alpha}\right)+\frac{1}{24}\,,$$
where $s=\pm1$. Thus,
$$\alpha^2-\frac{13}{6}\alpha+2s-1=0\,.$$
If $s=+1$, then
$$\alpha^2-\frac{13}{6}\alpha+1=\left(\alpha-\frac{3}{2}\right)\left(\alpha-\frac{2}{3}\right)\,,\text{ so }\alpha\in\left\{\dfrac{2}{3},\dfrac{3}{2}\right\}\,.$$
If $s=-1$, then
$$\alpha^2-\frac{13}{6}\alpha-3=0\,,$$
that is, $$\alpha=\frac{13\pm\sqrt{601}}{12}\,.$$
We have found four values of $\alpha$: $\dfrac{3}{2}$, $\dfrac{2}{3}$, $\dfrac{13-\sqrt{601}}{12}$, and $\dfrac{13+\sqrt{601}}{12}$. Clearly, these values work.
| {
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Double summation with a variable stopping point I am interested in calculating the following double summation:
$\sum_{n=2}^ \infty \sum_{k =0}^{n-2}\frac{1}{4}^k \frac{1}{2}^{n-k-2}$
I don't really know where to start, so I was hoping someone could point me to some resource where I could learn the terminology/methodology associated with solving such problems.
The summation arose when trying to describe the probability of never reaching a certain state in a finite markov chain. I really was just inquiring about the algebra which is why I didn't provide the context previously.
| This problem involves sums of geometric sequences for which there is a well-known formula:
$$\sum_{n=0}^{m-1} r^n = \frac{1-r^m}{1-r}.$$
Applying this to your summation and using a little algebra you get:
$$\begin{equation} \begin{aligned}
\sum_{n=2}^\infty \sum_{k=0}^{n-2} \Big( \frac{1}{4} \Big)^k \Big( \frac{1}{2} \Big)^{n-k-2}
&= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \sum_{k=0}^{n-2} \Big( \frac{1}{4} \Big)^k \Big( \frac{1}{2} \Big)^{-k} \\[6pt]
&= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \sum_{k=0}^{n-2} \Big( \frac{1}{2} \Big)^k \\[6pt]
&= \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} \cdot \frac{1-(\tfrac{1}{2})^{n-1}}{1-\tfrac{1}{2}} \\[6pt]
&= 2 \Bigg[ \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{n-2} - \sum_{n=2}^\infty \Big( \frac{1}{2} \Big)^{2n-3} \Bigg] \\[6pt]
&= 2 \Bigg[ \sum_{n=0}^\infty \Big( \frac{1}{2} \Big)^{n} - \frac{1}{2} \sum_{n=0}^\infty \Big( \frac{1}{4} \Big)^{n} \Bigg] \\[6pt]
&= 2 \Bigg[ \frac{1}{1-\tfrac{1}{2}} - \frac{1}{2} \cdot \frac{1}{1-\tfrac{1}{4}} \Bigg] \\[6pt]
&= 2 \Bigg[ 2 - \frac{1}{2} \cdot \frac{4}{3} \Bigg] \\[6pt]
&= 2 \Bigg[ 2 - \frac{2}{3} \Bigg] \\[6pt]
&= 2 \cdot \frac{4}{3} \\[6pt]
&= \frac{8}{3} = 2.6\bar{6}. \\[6pt]
\end{aligned} \end{equation}$$
| {
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Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?
| As an alternative, using your idea for decomposition, by Rearrangement inequality with
*
*$(a_1,a_2)=(\frac13,\frac{1}{(x^2+3)})$
*$(b_1,b_2)=\left(3(x^2+3),1\right)$
we have that
$$a_1b_1+a_2b_2=\frac13\cdot 3(x^2+3)+\frac{1}{(x^2+3)}\cdot 1= x^2+3+\frac{1}{(x^2+3)}\ge a_1b_2+a_2b_1=$$
$$=\frac13\cdot 1 +\frac{1}{(x^2+3)}\cdot 3(x^2+3)=\frac13+3=\frac{10}3$$
with equality for
$$a_1=a_2 \iff \frac13=\frac{1}{(x^2+3)}\iff x=0$$
therefore
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1\ge \frac{10}3-1=\frac 73$$
with the minimum attained at $x=0$.
| {
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For positive real numbers $x,y,z$, Prove that
For positive real numbers $x,y,z$, Prove that,
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq
\bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$
I have no idea how to begin this? I am getting a hint for A.M. - G.M. equality but numbers don't seem to fit.
Thanks for you time.
| HINT:
Try using $x$ times $x$, $y$ times $y$ , $z$ times $z$.
EDIT:
Seemingly people downvote hints so updating the full solution.
Using A.M. - G.M. inequality,
$$\frac{x+x+..(x \;\Bbb{times})+y+y+..(y \;\Bbb{times})+z+z+..(z \;\Bbb{times})}{x+y+z}\geq (x^xy^yz^z)^{\frac{1}{x+y+Z}}$$
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z $$
Using G.M. - H.M. inequality,
$$(x^xy^yz^z)^{\frac{1}{x+y+Z}} \geq \frac{x+y+z}{\frac{1}{x}+\frac{1}{x}+..(x \; \Bbb{times})+\frac{1}{y}+\frac{1}{y}+..(y \; \Bbb{times})+\frac{1}{z}+\frac{1}{z}+..(z \; \Bbb{times})}$$
$$ x^xy^yz^z \geq
\bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$
$$\bigg(\frac{x^2+y^2+z^2}{x+y+z}\bigg)^{(x+y+z)}\geq x^xy^yz^z \geq
\bigg(\frac{x+y+z}{3}\bigg)^{(x+y+z)}$$
| {
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Each member of a population dies with probability $\frac12$ each day, what is the probability that there will be exactly $1$ person alive? Suppose that there are $n$ people alive in a population. Due to a deadly disease, each person dies with probability $\frac12$ each day (and there are no births). What is the probability that there will be exactly one person alive at some time?
Thoughts:
Let $p_k$ be the probability that the population reaches exactly $1$ person given that there are currently $k$ people alive. Then $p_0 = 0$ and $p_1 = 1$.
The probability of going from $k$ people alive to $k - j$ being alive (where $0 \leq j \leq k$) is the probability that $j$ die:
$$
\binom{k}{j} \left(\frac{1}{2}\right)^j \left(\frac{1}{2}\right)^{k - j} = \binom{k}{j} \left(\frac{1}{2}\right)^k
$$
And using conditional probability we have the recursion:
$$
p_k = \frac{1}{2^k} \binom{k}{0} p_k + \frac{1}{2^k} \binom{k}{1} p_{k - 1} + \cdots + \frac{1}{2^k} \binom{k}{k - 1} p_1 + \frac{1}{2^k} \binom{k}{k} p_0,
$$
or
$$
(2^k - 1)p_k = \binom{k}{1} p_{k - 1} + \cdots + \binom{k}{k - 1} p_1.
$$
Is it possible to solve a recursion like this? Is there a better way to solve the puzzle?
| Partial solution. First find the probability that one specific person is the unique last survivor, then multiply by $n$.
Omegaman is the last to die on the $k+1$st day with probability
$$
p_k = \frac{1}{2^{k+1}} \left(1-\frac{1}{2^k}\right)^{n-1}
$$
so then the desired probability is
\begin{align}
q & = n\sum_{k=1}^\infty p_k \\
& = n\sum_{k=1}^\infty \frac{1}{2^{k+1}}
\left(1-\frac{1}{2^k}\right)^{n-1} \\
& = \frac{n}{2}
\sum_{k=1}^\infty \frac{1}{2^k}
\left(1-\frac{1}{2^k}\right)^{n-1}
% & = \frac{n}{2}
% \sum_{k=1}^\infty \frac{1}{2^k}
% \sum_{j=0}^{n-1} \binom{n-1}{j}\left(-\frac{1}{2^k}\right)^j \\
% & = \frac{n}{2}
% \sum_{j=0}^{n-1} \binom{n-1}{j} (-1)^j
% \sum_{k=1}^\infty \left(\frac{1}{2^k}\right)^{j+1} \\
% & = \frac{n}{2}
% \sum_{j=0}^{n-1} \binom{n-1}{j} (-1)^j
% \sum_{k=1}^\infty \left(\frac{1}{2^{j+1}}\right)^k \\
% & = \frac{n}{2}
% \sum_{j=0}^{n-1} \binom{n-1}{j} (-1)^j
% \frac{\frac{1}{2^{j+1}}}{1-\frac{1}{2^{j+1}}} \\
% & = \frac{n}{2}
% \sum_{j=0}^{n-1} \binom{n-1}{j} (-1)^j \frac{1}{2^{j+1}-1}
\end{align}
Still working out if there's a closed-form expression for this. I will point out that we can obtain
$$
q = \frac{n}{2} \sum_{j=0}^{n-1} \binom{n-1}{j} (-1)^j \frac{1}{2^{j+1}-1}
$$
if that counts as closed.
| {
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How to check for local extrema or saddle point given an semidefinite matrix I've computed the Hessian of a given function $f(a,b,c) = y-a\sin(bx-c)$ and got the following result:
$\begin{pmatrix}
0 & -x\cdot\cos(bx - c) & \cos(bx - c) \\
-x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\
\cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c)
\end{pmatrix}$
This matrix is positive semi-definite and thus one can not state for a given point $P=(a_i,b_i,c_i)$ if it is a local min, max or a saddle point. Is there any other way to explicitly determine if we have a loc. min, max or saddle point?
| With reference to the given matrix, we have that
$$\det(a\sin(bx - c))=a\sin(bx - c)$$
$$\begin{vmatrix}
0 & -x\cos(bx - c) \\
-x\cos(bx - c) & ax^2\sin(bx - c) \\
\end{vmatrix}=-x^2\cos^2(bx - c)$$
$$\begin{vmatrix}
0 & -x\cdot\cos(bx - c) & \cos(bx - c) \\
-x\cdot\cos(bx - c) & ax^2\cdot\sin(bx - c) & -ax\sin(bx - c) \\
\cos(bx - c) & -ax\cdot\sin(bx - c) & a\cdot\sin(bx - c)
\end{vmatrix}=$$
$$=x\cdot\cos(bx - c)(-ax\cos(bx - c)\sin(bx - c)+ax\cos(bx - c)\sin(bx - c))+\cos(bx - c)(ax^2\cos(bx - c)\sin(bx - c)-ax^2\cos(bx - c)\sin(bx - c))=0$$
therefore for $x^2\cos^2(bx - c)\neq 0$
*
*$a\sin(bx - c)\le 0$ the matrix is negative semidefinite
*$a\sin(bx - c)>0$ the matrix is indefinite
and for $x^2\cos^2(bx - c)= 0$
*
*$a\sin(bx - c)< 0$ the matrix is negative semidefinite
*$a\sin(bx - c)>0$ the matrix is positive semidefinite
Note that the condition for critical points implies
*
*$f_a=-\sin(bx - c)=0\implies \sin(bx - c)=0$
*$f_b=-ax\sin(bx - c)=0\implies \sin(bx - c)=0 \lor x=0$
*$f_c=a\sin(bx - c)=0\implies \sin(bx - c)=0$
therefore
*
*$\cos(bx - c)=\pm 1$
and the expression for the Hessian simplifies.
| {
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Evaluating $\int\sqrt{\frac{2-x}{x-3}} dx$
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
Need help in spotting my mistake:
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
$x-2 = t^2$
$\implies dx = 2t dt$
$$2 \int \sqrt{\dfrac{1}{1-t^2}} t^2dt$$
$t= \sin \theta $
$dt = \cos\theta d\theta$
$$\int (1- \cos 2\theta )d\theta $$
$=\theta - \dfrac{\sin 2\theta}{2}$
$$= \arcsin(t)- t\sqrt{1-t^2}$$
$= \arcsin (\sqrt{x-2})- \sqrt{(x-2)(3-x)}+C$
But answer given is: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$
| Your answer is absolutely fine, as you can check by differentiating.
If the given answer were correct as well, the equations
\begin{align}
C &= \frac{\pi}{2} \, , \\
\frac{\pi}{2} + C &= - \frac{\pi}{2} \, ,
\end{align}
which we obtain by evaluating both answers at $x=2$ and $x=3$, would have to be satisfied simultaneously for some $C \in \mathbb{R}$ . Clearly, this is impossible, so the given answer must be wrong.
| {
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Find the least value of the expression $x^2+2xy+2y^2+4y+7$ Find the least value of the expression
$x^2+2xy+2y^2+4y+7$
I am not able to solve this equation though i am able to differentiate it.
| The following one may be a more general solution without any trick. At first, let's recall a simple fact involving the quadric expression, which states that
$$ax^2+bx+c \geq \frac{4ac-b^2}{4a}~~~$$
holds for $a>0$ and with the equality
holding iff $x=-\dfrac{b}{2a}.$
Hence, we have
\begin{align*}
x^2+2y\cdot x+2y^2+4y+7 &\geq \frac{4 \cdot 1 \cdot (2y^2+4y+7)-(2y)^2}{4\cdot 1}\\&=y^2+4y+7\\&\geq \frac{4\cdot 1 \cdot 7-4^2}{4 \cdot 1}\\&=3,
\end{align*}
with the equality holding iff $x=-\dfrac{2y}{2 \cdot 1}$ and $y=-\dfrac{4}{2\cdot 1}$, namely, $x=2, y=-2.$
| {
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minimal time to reach given state. number of time steps unclear. Consider the discrete-time state-space realization
$$x(t+1)=Ax(t)+Bu(t), \qquad y(t)=Cx(t)$$
with
$$A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 2 & 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}, \quad C= \begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$$
and let $x(t,x_0,u) : \mathbb N \to \mathbb R^3$ denote the state $x(t)$ of the state-space realization $(A,B,C)$ at time $t \in \mathbb N$ corresponding to initial state $x(0)=:x_0$ and input function $u : \mathbb N \to \mathbb R$ such that $x(T,0,u)=x_f$ with $x_f=\begin{bmatrix} 1 & 1& 0 \end{bmatrix}^T$.
My approach:
The following sequence must be filled in until the state $x_i$ matches $x_f$:
$x_1 = Ax_0+Bu_0$
$x_2 = Ax_1+Bu_1=A^2x_0+ABu_0+Bu_1$
$x_3 = Ax_2+Bu_2=A^3x_0+A^2Bu_0+ABu_1+Bu_2$
$u_{1}, u_2$ and $u_3$ get determined using
$\begin{bmatrix}u_2 \\ u_1 \\ u_0 \end{bmatrix} = \begin{bmatrix} B & AB & A^2B \end{bmatrix}^{-1}(x_f-A^3x_0) = \begin{bmatrix} -1&1&-\frac{1}{2} \\ 1 & 0 & -\frac{1}{2}\\ 0& 0& \frac{1}{2}\end{bmatrix}\begin{bmatrix}1 \\ 1\\ 0 \end{bmatrix}=\begin{bmatrix}0\\1\\0 \end{bmatrix}$
And thus we get:
$x_1 = A \cdot 0 + B \cdot 0 = 0$
$x_2 = A^2 \cdot 0 + A \cdot B \cdot 0 +\begin{bmatrix} 0\\1\\0 \end{bmatrix} \cdot 1 = \begin{bmatrix} 0\\1\\0 \end{bmatrix}$
$x_3 = A^3 \cdot 0 + A^2 \cdot B \cdot 0 + \begin{bmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0\\1\\0 \end{bmatrix} \cdot 1 + \begin{bmatrix} 0\\1\\0 \end{bmatrix} \cdot 0 = \begin{bmatrix}1\\1\\0 \end{bmatrix}=x_f$
It is unclear to me how many time steps this took?
$x_3$ reached the correct state, so $3$ steps or only $2$ because from $x_1$ to $x_2$ is $1$ step and from $x_2$ to $x_3$ is another. But $u_2$ did not get used in the final step because it is $0$ so $2$ steps or $1$?
| The smallest number of time steps is the smallest $n$ such that $x_f - A^n\,x_0$ lies in the span of $\begin{bmatrix}B & A\,B & \cdots A^{n-1} B\end{bmatrix}$. Namely this implies that there exits a vector $u$ (containing the inputs) such that
$$
x_f - A^n\,x_0 = \begin{bmatrix}B & A\,B & \cdots A^{n-1} B\end{bmatrix} v.
$$
For your system, assuming $x_0 = 0$, the $n$ which satisfies this is two. If the top element of $v$ is zero just means the not actuated system dynamics after the first input will bring the state to $x_f$ after one more time step. But without that additional (not actuated) time step the system will not end up in $x_f$, so the number of time steps is still $n$.
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Prove that $\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$ Prove that if $a,b,c$ are the lengths of the edges of a given triangle, then the following inequality holds:
$\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$.
I have tried so far using the Ravi substitution, but the expression got even uglier. I have also tried making the substitutions: $x = a + b$, $y = b + c$, $z = a + c$, and after using $3(a^2 + b^2 + c^2) \geq (a + b + c)^2$, this resulted in showing that: $x + y + z \geq \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$, where we also know (from triangle inequality) that: $z \geq \frac{x+y}{3}$ and the other 2 analogues, but at this point I got stuck.
| Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, we need to prove that
$$8(u^2-uv+v^2)x^3+6(3u^3-2u^2v+uv^2+3v^3)x^2+$$
$$+(11u^4+2u^3v-15u^2v^2+26uv^3+11v^4)x+2(u^5+u^4v-3u^3v^2+u^2v^2+5uv^4+v^5)\geq0,$$
which is obvious.
Also, the $uvw$'s technique helps, but it's very complicated here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that, for all $m\geq 2$, there exists $C_m>0$ such that, for all $k\geq 1$, $\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$ Define $\phi:\mathbb{N}\to \mathbb{N}$ by $\phi(k)=k^m$, where $m\geq 2$ is some fixed number. I want to investigate, if there exists $C_m>0$ such that $$\frac{\phi(k+2)-\phi(k+1)}{\phi(k+1)-\phi(k)}=\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$$ for all $k\geq 1$. Since
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=\frac{\left ( \frac{1+2/k}{1+1/k} \right )^m-1}{1-\left ( \frac{1}{1+1/k} \right )^m},
$$
we define $f:(0,1]\to\mathbb{R}$ by
$$
f(x)=\frac{\left ( \frac{1+2x}{1+x} \right )^m-1}{1-\left ( \frac{1}{1+x} \right )^m}
$$
I suspect that $f(x)$ is increasing, and so $f(1/x)$ must be decreasing (through some calculator graphs), which implies that
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq \frac{(1+2)^m-(1+1)^m}{(1+1)^m-1^m}=\frac{3^m-2^m}{2^m-1}=:C_m
$$
for all $k\geq 1$. The question is, how do I show the monotonicity of $f(x)$ or $f(1/x)$ in a simplest way? Determining $f'$ and then checking if it's greater than $0$ on $(0,1]$ takes a lot time.
| Using $$a^m-b^m=(a-b)\left(a^{m-1}+a^{m-2}b+a^{m-3}b^2+...+ab^{m-2}+b^{m-1}\right)$$
For a fixed $m$ we have:
$$\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=
\frac{(k+2)^{m-1}+(k+2)^{m-2}(k+1)+...+(k+2)(k+1)^{m-2}+(k+1)^{m-1}}{(k+1)^{m-1}+(k+1)^{m-2}k+...+(k+1)k^{m-2}+k^{m-1}}=\\
\left(\frac{k+2}{k+1}\right)^{m-1}\frac{1+\frac{k+1}{k+2}+...+\left(\frac{k+1}{k+2}\right)^{m-1}}{1+\frac{k}{k+1}+...+\left(\frac{k}{k+1}\right)^{m-1}} \rightarrow 1, k\rightarrow\infty$$
and any sequence with a finite limit is bounded.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof verification of an exercise involving a functional equation Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function and $a \in \mathbb{R}$ such that
$$f(m+n) = f(m) + f(n) + a$$
$$f(2) = 10, f(20) = 118$$
Find $a$ and $f$.
I found this exercise at the beginning of a Real Analysis textbook. I've never solved a functional equation before, but here's my solution (attempt):
i) Using induction it's easy to verify that for $m, n \in \mathbb{N}$ we have $f(m \cdot n) = m (f(n) + a)$, since
$$f((m+1)n) = f(mn + n) = f(mn) + f(n) + a$$
ii) Then $118 = f(20) = f(10 \cdot 2) = 10 (f(2) + a) = 10 (10 + a) \Rightarrow a = \frac{9}{5}$
iii) Then $f(0) = f(0 + 0) = f(0) + f(0) + a \Rightarrow f(0) = -a = -\frac{9}{5}$.
We also get $10 = f(2) = f(1+1) = f(1) + f(1) + \frac{9}{5} = 2 f(1) + \frac{9}{5} \Rightarrow f(1) = \frac{41}{10}$
iv) Then finally we can define $f$ recursively by $f(0) = -\frac{9}{5}$ and $$f(n+1) = f(n) + f(1) + \frac{9}{5} = f(n) + \frac{41}{10} + \frac{9}{5} = f(n) + \frac{59}{10}$$
EDIT
Then thanks to the user lulu, the real pattern should be $f(m \cdot n) = m f(n) + (m-1) a$ instead. Using this in ii) then gives $a = 2$. Then we get in iii) that $f(0) = -2$ and $f(1) = 4$, so $f$ is defined by $f(n+1) = f(n) + 6$. And now it's pretty obivous that $f(n) = 6n - 2$ solves the equation.
| We have
$$f(m+n) = f(m) + f(n) + a$$
Hence
$$f(2n)=2f(n)+a$$
Thus
$$f(0)=2f(0)+a$$
Then
$$f(0)=-a$$
Now
\begin{align}
f(20)
&= 2f(10) + a \\
&= 2(f(8)+f(2)+a) +a \\
&= 2(2f(4)+a) +2f(2) +3a \\
&= 4(2f(2)+a) +2f(2) +5a \\
&= 10f(2) + 9a \\
118 &= 100 + 9a \\
a & =2
\end{align}
So
$$f(0)=-2, \ f(1)=\frac{10-a}{2}=4$$
And
$$f(2n)=2f(n)+2=2(f(n)+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$? How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$?
My try:
All possible $4$-digit numbers $= 7 \cdot 7 \cdot 7 \cdot 6$
$4$-digits numbers NOT containing $3, 5 = 5 \cdot 5 \cdot 5 \cdot 4$
Answer= $7 \cdot 7 \cdot 7 \cdot 6-5 \cdot 5 \cdot 5 \cdot 4 =1558$
Is that OK ? If not, please explain my mistake.
Thank you.
| Since you only care about 4 states: contains neither, contains 3 not 5, contains 5 not 3, contains both; you can use a transition matrix:
$$\begin{bmatrix} 4 & 1 & 1 & 0 \end{bmatrix}
\begin{bmatrix}
5 & 1 & 1 & 0 \\
0 & 6 & 0 & 1 \\
0 & 0 & 6 & 1 \\
0 & 0 & 0 & 7 \end{bmatrix}^3
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$
This way, if you wanted to change it to "how many 55 digit numbers..." all you have to do is change the $3$ in the exponent to a $54$.
Interesting that if you diagonalize the matrix you the same expression as N.G.Taussig's answer: $6 \times 7^3 - 2 \times 5 \times 6^3 + 4 \times 5^3$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
| After you render $x^2=y^4+y^2-41$, complete the square on the right side. Multiplying by $4$ to eliminate fractions then gives
$(2x)^2-(2y^2+1)^2=-165$
Then from the difference of squares factorization the following are both divisors of $165$, with product $-165$:
$2x+2y^2+1$
$2x-2y^2-1$
Their sum is then $4x$ which must be the sum of two divisors of $165$, one positive and one negative, with the proper product; thus
$4x=\pm(165-1)=\pm 164, x=\pm 41$
Or
$4x=\pm(55-3)=\pm 52, x=\pm 13$
Or
$4x=\pm(33-5)=\pm 28, x=\pm 7$
Or
$4x=\pm(15-11)=\pm 4, x=\pm 1$
And we try all these to find integer roots for $y$, finding that only $x=\pm 7$ makes it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Algebraic trick to map $|z|<2$ Suppose that we want to find the image of the region $|z|<1$ under the mapping $w=\frac z{z+1}$. Since $z=\frac{-w}{w-1}$ (and assuming $w= u+iv$) we should have $\left|\frac w{w-1}\right|=\left|\frac{u(u-1)+v^2-iv}{(u-1)^2+v^2}\right|<1$ or equivalently
$$u^2(u-1)^2+v^4+2v^2u(u-1)+v^2<(u-1)^4+v^4+2v^2(u-1)^2.$$
Now we can use the following algebraic trick to write
$$(u-1+1)^2(u-1)^2+2v^2(u-1+1)(u-1)+v^2<(u-1)^4+2v^2(u-1)^2$$
which implies that
$$(u-1)^2+2(u-1)(u-1)^2+2v^2(u-1)+v^2<0$$
or
$$((u-1)^2+v^2))(2u-1)<0.$$
Thus the image would be $2u-1<0$.
Now I wonder if there is some similar algebraic trick for $|z|<2$. In this case we have
$$u^2(u-1)^2+v^4+2v^2u(u-1)+v^2<4(u-1)^4+4v^4+8v^2(u-1)^2$$
I've tried similar method but couldn't arrive to something useful. Could anyone help me in this case? Thanks!
| For a direct algebraic proof, start the same way as in the first case with $\,z = \dfrac{w}{1-w}\,$, then:
$$
\begin{align}
|z| \lt 2 \quad&\iff\quad |w|^2 \lt 4 \cdot|1-w|^2 \\
&\iff\quad w \bar w \lt 4 \cdot(1-w)(1-\bar w) \\
&\iff\quad 3 w \bar w - 4 w - 4 \bar w \color{red}{+\frac{16}{3}-\frac{16}{3}} + 4 \gt 0 \\
&\iff\quad 3\left(w-\frac{4}{3}\right)\left(\bar w-\frac{4}{3}\right) \gt \frac{4}{3} \\
&\iff\quad \left|w-\frac{4}{3}\right|^2 \gt \frac{4}{9} \\[5px]
&\iff\quad \left|w-\frac{4}{3}\right| \gt \frac{2}{3}
\end{align}
$$
Therefore the image is the exterior of the circle of radius $\,2/3\,$ centered at $\,4/3\,$.
(Geometrically, the locus of points in the plane with constant ratio $\,\left|\frac{w}{w-1}\right| = 2\,$ between the distances to the two fixed points $\,0, 1\,$ is a circle of Apollonius having as diameter the points that divide the segment between $\,0\,$ and $\,1\,$ in ratio $\,2:1\,$ internally and respectively externally. In this case the endpoints of that diameter are $\,2/3$ and $\,2\,$, so the circle is centered at $\,(2+2/3)/2=4/3\,$ and has a radius $\,(2-2/3)/2=2/3\,$, which is of course the same as found above.)
| {
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"source": "stackexchange",
"question_score": "3",
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Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| Or you can write $x=y+z$ for some $z\in Z$ (that is $z=x-y$). So we get a quadratic equation on $y$:
$$ 3y^2z+y(3z^2-2)+z^3-1=0$$
which has a discriminant a perfect square: $$d^2 = (3z^2-2)^2-12z(z^3-1)=-3z^4-12z^2+12z+4$$
So we have $$3z^4+12z^2\leq 12z+4$$
and this can not be true for a lot of integers $z$ (in fact only for $0$ and $1$)...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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"answer_id": 0
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Number theory - Find the number satisfying the condition. Find a 3 digit number which equals to 4 times the product of its digits.
My approach:
I considered the Number to be $\overline{ABC}$ then wrote the relation $$100A+10B+C=4A\times B\times C$$ But i don't know how to proceed further. Please help.
| Long solution without programming.
$100A+10B+C=4\times A\times B\times C (1)$
Divide both parts of (1) by 4
$25A+\frac{5}{2}B+\frac{C}{4}=A\times B\times C (2)$
From (2)
Based on Daniel Fischer suggestion
$C=2$ (2) goes to $25A+\frac{5}{2}B+\frac{2}{4}=2\times A\times B =>
\frac{50A+5B+1}{4}=A\times B$ and $50A+5B+1$ can not be divided by 4
$C=6$ (2) goes to $25A+\frac{5}{2}B+\frac{3}{2}=6\times A\times B =>
\frac{50A+5B+3}{12}=A\times B$ and $50A+5B+3$ can not be divided by 2
$C$ must be 4 or 8 and $B$ must be even.
If $c=8$ (2) can be rewritten as $25A+\frac{5}{2}B+2=8 \times A\times B (3')$
or
$\frac{25}{8}A+\frac{5}{16}B+\frac{1}{4}=A\times B (3'')$
(3'') can not be solved for for integers A and B because of sum plus $\frac{1}{4}$ and B must be divisible by 16 (wrong).
Try $C=4$
$25A+\frac{5}{2}B+1=4 \times A\times B (4')$ or
$\frac{25}{4}A+\frac{5}{8}B+\frac{1}{4}=A\times B (4'')$
Try (4'') with $B=2,4,6,8,0$
$B=0$ (4'') impossible
$B=2$ (4'') goes to $\frac{25}{4}A+\frac{5}{4}+\frac{1}{4}=2\times A (4''')$ or
$25A+6=8A => 17A= -6$- impossible
$B=4$ (4'') goes to $\frac{25}{4}A+\frac{10}{4}+\frac{1}{4}=4\times A (4''')$ or
$25A+11=16A => 9A= -11$- impossible
$B=6$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}6+\frac{1}{4}=6\times A (4''')$ or
$25A+16=24A => A= -16$- impossible
$B=8$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}8+\frac{1}{4}=8\times A (4''')$ or
$25A+21=32A => 7A= 21 => A=3$ - possible
384
Question, comments, edits?
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a$, $b$, $c$, $d$ be the roots of $x^4 + x + 1 = 0$. Find $a^4 + b^4 + c^4 + d^4$.
Let $p(x) = x^4 + x + 1 = 0$, and let $a$, $b$, $c$, $d$ be its roots.
Find $a^4 + b^4 + c^4 + d^4$.
I have no idea how to start solving this problem.
| As an alternative, let consider
$$(x-a)(x-b)(x-c)(x-d)=$$
$$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$
then
*
*$S_1=a+b+c+d=0$
*$S_2=ab+ac+ad+bc+bd+cd=0$
*$S_3=abc+abd+acd+bcd=-1$
*$S_4=abcd=1$
and by Newton's sums we have that
*
*$P_1=a+b+c+d=S_1=0$
*$P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2=0$
*$P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3=-3$
*$P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4=-4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values for $a,b,c,d$
Given $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$ for any
real number $x$ and $y$ , find the value of $a,b,c,d$
| $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$
R.H.S only has $x^3$ and $y^2$, therefore all the terms in L.H.S must cancel out except for $x^3$ and $y^2$.
So let's pair the terms that may cancel out.
$$x^3+y^2+(axy^2+7xy^2)+(3xy+dxy)+4x^2y-bx^cy$$
From here you can conclude that,
$a=-7,d=-3$
Also for $4x^2y$ to cancel $c=2$ and $b=4.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it. Is $\sum_{n=1}^\infty \frac{n^2+1}{n!2^n}$ convergent? If yes, evaluate it.
Using the ratio test, I could show that the series is convergent. What is an easy way to evaluate it? I was thinking about estimating it from below and above by series with equal value but coulnd't find any familar series that I could use for this.
| First of all, you can separate the sum in 2 convergent series :
$$ \sum_{n=1}^\infty \frac{1}{2^n n!} + \sum_{n=1}^\infty \frac{n^2}{2^n n!} $$
The first term is equal to $e^{1/2}-1$ (see exponential as a sum) and we can write the second as :
$$ \sum_{n=1}^\infty \frac{n^2}{2^n n!} = \sum_{n=1}^\infty \frac{n}{2^n (n-1)!} = \frac{1}{2} \sum_{n=1}^\infty \frac{n-1 + 1}{2^{n-1} (n-1)!}$$
As previously, we can separate and apply the same method :
$$ \sum_{n=1}^\infty \frac{n^2}{2^n n!} = \frac{1}{2}(e^{1/2} + \frac{1}{2} e^{1/2})$$
All together, we have :
$$ \sum_{n=1}^\infty \frac{n^2 + 1}{2^n n!} = e^{1/2} - 1 + \frac{1}{2}(e^{1/2} + \frac{1}{2} e^{1/2}) = \frac{7}{4}e^{1/2} - 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all functions if $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\in\mathbb{R}$. Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$$f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$$
for all $x,y\in\mathbb{R}$.
If we put $x=0$ and mark $a=f(0)$ we get $$2a=af(y)+ya$$
so for $a\ne 0$ we get $f(y)=2-y$ for all $y$. Now say $a=0$, so $f(0)=0$. Leting $y=0$ we get $\boxed{f(x^2) = xf(x)}$. If we put this in starting equation we get $$xf(x) +f(xy)=f(x)f(y)+yf(x)+xf(x+y)\;\;\;\;(*)$$
From boxed equation we see that $f$ is odd:
Edit
$$-xf(-x) = f((-x)^2)= f(x^2)=xf(x) \implies f(-x)=-f(x)$$
Leting $y=-x$ in $(*)$ we get: $$xf(x)-f(x^2) =-f(x)^2-xf(x)\implies \underline{xf(x)= -f(x)^2}$$
Edit 28. 06. 2022
Let $G$ be a set of all real $x$ such that $f(x)=-x$. Clearly $0\in G$.
Suppose exists $x,y\in G$ so that $x \neq 0$. Here we can assume both $x,y$ are nonnegative, since $f(-x)=-f(x) = -(-x)$. Then we have $$-x^2+f(xy) =xf(x+y)$$ Then we have 4 possibilities:
*
*$f(xy)=0$ and $f(x+y)=0$ so $-x^2=0$ which is not true.
*$f(xy)=0$ and $f(x+y)=-x-y$ so $-x^2=-x^2-xy$ which is true only if $y=0$.
*$f(xy)=-xy$ and $f(x+y)=0$ so $-x(x+y)=0$ which is not true since $x,y$ are nonegative and $x\ne0$.
*$f(xy)=-xy$ and $f(x+y)=-x-y$ so $-x^2-xy=x(-x-y)$ which is true.
So if $x,y\in G$ then also $x+y\in G$ wich means $G$ is aditive subgroup of $(\mathbb{R},+)$ if$|G|\geq 2$.
What to do now? Is this $G$ usefull at all?
| $x f(x) = f(x)^2$ says either $f(x) = 0$ or $f(x) = x$. Now if $f(x)=x$ and $f(y)=y$, $f(x^2) = x f(x) = x^2$, and your equation says
$$ x^2 + f(xy) = 2 x y + x f(x+y) $$
But this does not work with any combination of $f(x+y)=0$ or $f(x+y)=x+y$ and $f(xy)=0$ or $f(x+y)=xy$ unless $x=0$ or $y=0$ or $x=y$
or $x=2y$. Ultimately the conclusion should be that $f(x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
I can't solve $x-2<\frac{1}{x-1}$ We have to solve this inequation.
$x-2<\frac{1}{x-1}$.
So we try $(x-1)(x-2)<1$ what's wrong ?
Please, give me a hint.
| $$x-2<\frac { 1 }{ x-1 } \\ x-2-\frac { 1 }{ x-1 } <0\\ \frac { { x }^{ 2 }-3x+1 }{ x-1 } <0\\ \frac { \left( x-\frac { 3-\sqrt { 5 } }{ 2 } \right) \left( x-\frac { 3+\sqrt { 5 } }{ 2 } \right) \left( x-1 \right) }{ { \left( x-1 \right) }^{ 2 } } <0\\ \left( x-\frac { 3-\sqrt { 5 } }{ 2 } \right) \left( x-\frac { 3+\sqrt { 5 } }{ 2 } \right) \left( x-1 \right) <0\\ x\in \left( -\infty ,\frac { 3-\sqrt { 5 } }{ 2 } \right) \cup \left( 1,\frac { 3+\sqrt { 5 } }{ 2 } \right) \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Summing a series having a geometric component
If $$\alpha=\frac {5}{2!3}+\frac {5\cdot 7}{3!3^2}+\frac {5\cdot 7\cdot 9}{4!3^3}+\dots $$ Find the value of $\alpha^2+4\alpha $.
The possible options are:
*
*21
*23
*25
*27
I think $$\alpha=\sum \frac {(2n+4)!}{2^{n+2}(n+1)!(n+2)!3^{n+1}}$$ but cannot think of what else to do.
| So we have that
$$\alpha=\sum_{n=0} \frac{(5+2n)!!}{3^{n+2}(2+n)!}$$
Using the fact that $(2n)!=2^n(2n-1)!!(n!)$, we have that
$$\alpha=\sum_{n=0} \frac{(2n+6)!}{2^{n+3}3^{n+2}(n+2)!(n+3)!}$$
$$\alpha=\frac{1}{2}\sum_{n=0} \binom{2n+6}{n+3}\left(\frac{1}{6}\right)^{n+2}(n+3)$$
Using the generating function of central binomial coefficients, we know
$$\sum_{n=0}^\infty \binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$$
Define the function
$$f(x)=\sum_{n=0}^{\infty}\binom{2n+6}{n+3}x^{n+3}=\frac{1}{\sqrt{1-4x}}-1-2x-6x^2$$
We know
$$f'(x)=\sum_{n=0}^\infty \binom{2n+6}{n+3}x^{n+2}(n+3)=\frac{2}{\left(\sqrt{1-4x}\right)^3}-2-12x$$
Hence,
$$\alpha=\frac{1}{2}f'\left(\frac{1}{6}\right)=3\sqrt{3}-2$$
We can then deduce that $\alpha$ is a root of $(x+2)^2=27$, or $x^2+4x-23=0$. Hence, $\alpha^2+4\alpha=\boxed{23}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Geometry problem (find QS) Could someone help with this one? Been trying it for days, no success until now.
Consider the picture above, in which $ABCD$ is a square, $\angle DCQ = 45^\circ$, $\overline{AS} = 12$ and $\overline{DS} = 6$. Calculate the length of $\overline{QS}$.
(a) $6+ 3\sqrt{6}$
(b) $2+3\sqrt{6}$
(c) $6+\sqrt{6}$
(d) $3+6\sqrt{6}$
(e) $3+3\sqrt{6}$
| Hint.
Given $L$ the square side and $\theta$ the square rotation
$$
L\cos(\theta)+L_0\cos(\theta+45^{\circ})=12\\
\left(L\sin(\theta)\right)^2+\left(L_0\cos(\theta+45^{\circ})\right)^2 = 6^2
$$
solving this you have the values for $L$ and $\theta$
so
$$
QS = L\sin(\theta)+L_0\sin(\theta+45^{\circ})
$$
NOTE
Here $L_0 = \frac{L}{\sqrt 2}$
To solve this we have
$$
\frac{3}{2} L \cos (\theta )-\frac{1}{2} L \sin (\theta )=12\\
\frac{1}{2} L^2 \sin ^2(\theta )-\frac{1}{2} L^2 \cos ^2(\theta )-\frac{1}{2} L^2 \sin (\theta ) \cos (\theta )+\frac{3 L^2}{4}=36
$$
or
$$
\frac{3 c}{2}-\frac{s}{2}=\frac{12}{L}\\
\frac{s^2}{2}-\frac{1}{2} c^2-\frac{1}{2} s c+\frac{3}{4}=\frac{36}{L^2}
$$
solving for $s, c$ we have
$$
\left(
\begin{array}{cc}
c & s \\
\frac{12}{L}-\frac{\sqrt{\frac{3}{10}} \sqrt{-L^2 \left(L^2-144\right)}}{L^2} & -\frac{3 \left(\sqrt{30} \sqrt{-L^2 \left(L^2-144\right)}-40
L\right)}{10 L^2} \\
\frac{\sqrt{\frac{3}{10}} \sqrt{-L^2 \left(L^2-144\right)}}{L^2}+\frac{12}{L} & \frac{3 \left(40 L+\sqrt{30} \sqrt{-L^2
\left(L^2-144\right)}\right)}{10 L^2} \\
\end{array}
\right)
$$
now using $c^2+s^2=1$ we have
$$
\left\{
\begin{array}{rcl}
-\frac{48 \sqrt{\frac{6}{5}} \sqrt{-L^2 \left(L^2-144\right)}}{L^3}+\frac{720}{L^2}-4&=&0 \\
\frac{48 \left(75 L+\sqrt{30} \sqrt{-L^2 \left(L^2-144\right)}\right)}{5 L^3}-4&=&0 \\
\end{array}
\right.
$$
giving
$$
L = \left\{
\begin{array}{c}
\sqrt{\frac{1}{5} \left(13-2 \sqrt{6}\right)} \\
\sqrt{\frac{468}{5}+\frac{72 \sqrt{6}}{5}} \\
\end{array}
\right.
$$
finally substititing the values for $c,s,L$ in
$$
QS = \frac{c L}{2}+\frac{3s L}{2}
$$
we obtain
$$
QS = 6+3\sqrt 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\frac{3}{\sqrt{-5x^2-4\sqrt{5}x+2}}$
Integrate $\frac{3}{\sqrt{-5x^2-4\sqrt{5}x+2}}$
For this question I first factored out the 3, seeing as its a constant
$$3\int\frac{1}{\sqrt{-5x^2-4\sqrt{5}x+2}}dx$$
I then noticed that this integral has a similar form to the integral of $\arcsin(x)$
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin(\frac{x}{a})+C, \vert x \vert \lt a$$
The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4\sqrt{5}x+2$$ to $$a^2-x^2$$
Any help or ideas would be highly appreciated!
| Hint: Write
$$-5x^2-4\sqrt{5}x+2=6-(\sqrt{5}x+2)^2$$
and let substitution $\sqrt{5}x+2=\sqrt{6}\sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given that $z=\cos\theta+i\sin\theta$, show that $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
Given that $z=\cos\theta+i\sin\theta$, show that
$Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$
For this question I had to show that the real part of $\frac{z-1}{z+1}=0$
To find that I first substituted $z$ with $\cos\theta+i\sin\theta$ to get
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}$$
I then multiplied by the conjugate of the denominator
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\cdot\frac{\cos\theta-i\sin\theta+1}{\cos\theta-i\sin\theta+1}$$
Which, when expanded, gives me
$$\frac{\cos^2\theta+\sin^2\theta+2i\sin\theta-1}{\cos^2\theta+\sin^2\theta+2i\cos\theta+1}$$
After that,
$$\frac{2i\sin\theta}{2\cos\theta+2}=\frac{i\sin\theta}{\cos\theta+1}$$
How do I proceed?
Edit: Just realized (after being told by a commenter) that $\frac{i\sin\theta}{\cos\theta+1}$ is imaginary. I won't delete the question. Ínstead I'll leave it here, as a testament to my stupidity, for everyone to see. May God have mercy on my soul during my exam.
| Yet another way: $|z| = 1 \iff z\bar z=1 \iff \bar z = \dfrac{1}{z}\,$, then using that $\,|\operatorname{Re}(z)| \le |z| \le 1\,$:
$$
\left(\frac{z-1}{z+1}\right)^2=\frac{z^2-2z+1}{z^2+2z+1}=\frac{z-2+\dfrac{1}{z}}{z+2+\dfrac{1}{z}} = \frac{z+\bar z - 2}{z+\bar z +2} = \frac{2\operatorname{Re}(z)-2}{2\operatorname{Re}(z)+2} = \frac{\operatorname{Re}(z)-1}{\operatorname{Re}(z)+1} \;\le\; 0
$$
Since the square of $\,\dfrac{z-1}{z+1}\,$ is a non-positive real, $\,\dfrac{z-1}{z+1}\,$ is either $\,0\,$, or purely imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ Prove that $f(x)=(1+x)^\frac{1}{x}+(1+\frac{1}{x})^x \leq 4$ for all $x>0.$
We have $f(x)=f(\frac{1}{x}), f'(x)=-\frac{1}{x^2}f'(\frac{1}{x}),$ so we only need to prove $f'(x)＞0$ for $0 < x < 1.$
| From a friend who doesn't want to use math.se...
Write the inequality as
$$
\frac{1}{1+x} \left(1+x\right)^{1+\frac{1}{x}} + \frac{x}{1+x} \left(1+\frac{1}{x}\right)^{1+x} \leq 4.$$
Let $f(x) =(1+x)^{1+\frac{1}{x}}$. If $f$ is concave, then we are done since this means that
$$4=f(1)=f((1- \alpha)x + \alpha y)) \geq (1- \alpha) f(x) + \alpha f(y) = \frac{1}{1+x} \left(1+x\right)^{1+\frac{1}{x}} + \frac{x}{1+x} \left(1+\frac{1}{x}\right)^{1+x}$$ with $\alpha = \frac{x}{1+x}, 1-\alpha = \frac{1}{1+x},$ and $y = \frac{1}{x}$. So, we will show that $f$ is concave.
Note that if $f(x) = e^{g(x)}$ and $g''(x) + g'(x)^2 \leq 0$ for all $x$, then $f$ is concave.
So, consider $g(x) = (1+\frac{1}{x}) \log (1+x).$ Then we have that
$$g''(x) + g'(x)^2 \leq 0 \leftrightarrow \log(1+x) \leq \frac{x}{\sqrt{1+x}}$$
Note that $\lim_{x \to 0^+} \log(1+x) = \lim_{x \to 0^+} \frac{x}{\sqrt{1+x}}=0.$
By AM-GM,
$$\frac{1}{1+x} = \frac{d}{dx} \log(1+x) \leq \frac{x+2}{2(1+x)^{\frac{3}{2}}} = \frac{d}{dx} \frac{x}{\sqrt{1+x}} $$
So, $$\log(1+x) = \int_0^x \log (1+x) ' dx \leq \int_0^x \left(\frac{x}{\sqrt{1+x}}\right)' dx = \frac{x}{\sqrt{1+x}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 1
} |
Numbers in Different Bases (reversed representation after conversions) The problem is to find a 3-digit natural number in base ten, where its depiction in base seven is the reverse of its depiction in base nine. 248 is one answer, but then there remains the fact that a 3-digit natural number in base ten can become a 4-digit natural number in base seven or nine. It would be really appreciated if someone could explain how 248 is the only possible answer and that it doesn't work when the 3-digit natural number in base ten becomes a four digit natural number in base seven or nine.
Thanks :)
| Suppose the number is
$n = 729a+81b+9c+d = 343d+49c+7b+a$ with $1 \leq a \leq 6,$
$0 \leq b \leq 6,$ $0 \leq c \leq 6,$ and $1 \leq d \leq 6,$
that is, a four-digit number in both bases nine and seven.
Since this must be a three-digit number in base ten,
$729a < n < 1000,$ which implies that $a < 2.$ Therefore $a = 1$
and $n > 729.$
Similarly, $343d < n < 1000$ implies $d < 3.$
But $49c + 7b+a < 343,$ hence $343d + 343 > n > 729$ implies $d > 1.$
Therefore $d = 2.$
That is, $n = 729 + 81b + 9c + 2 = 686 + 49c + 7b + 1,$
from which $40c = 44 + 74b,$
that is, $20c = 22 + 37b.$
First proof
From $c \leq 6$ it follows that $37b = 20c - 22 \leq 98$
and therefore $b \leq 2.$
But then $20c = 22 + 37b \leq 96$ implies $c \leq 4,$
which implies $37b = 20c - 22 \leq 58,$ so $b \leq 1.$
But then $20c = 22 + 37b \leq 59$ implies $c \leq 2,$
which implies $37b = 20c - 22 \leq 18,$ so $b = 0.$
But then $20c = 20, $ which has no integer solution.
Therefore no such number $n$ exists.
Second proof
By trial and error or any other means, find that $20(27) = 22 + 37(14).$
Therefore the solutions of the Diophantine equation
$20c = 22 + 37b$ are $b = 20k+14,$ $c = 37k+27$ where $k$ is an integer.
No such solution satisfies $0 \leq b \leq 6,$ so no such number $n$ exists.
Third proof
From $20c = 22 + 37b$ we conclude $b$ is even.
Let $b = 2m,$ then $20c = 22 + 74m,$ so $10c = 11 + 37m,$
and $0 \equiv 1 - 3m \pmod{10}.$
This has a unique solution, $m \equiv 7 \pmod{10}.$
But $0 \leq b \leq 6$ implies $0 \leq m \leq 3,$ a contradiction.
Therefore no such number $n$ exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Laurent series expansion of $\sqrt{(z-1)(z-2)}, |z|>2$
Find Laurent series expansion of $\sqrt{(z-1)(z-2)}$, $|z|>2$.
Using the The Binomial Theorem: $\pm z(z-1)^{1/2}(z-2)^{1/2} = \pm z[\sum_{k=0}^\infty\binom{1/2}{k}(-1)^{k}z^k][\sum_{k=0}^\infty\binom{1/2}{k}(-2)^{k}z^k]$.
I have a problem how multiply these two sums.
The answer in the book: $c_n$ = $\binom{1/2}{k}$
+ $2\binom{1/2}{k-1}\binom{1/2}{1}$ + $2^2\binom{1/2}{k-2}\binom{1/2}{2}$ + … + $2^n\binom{1/2}{k}$ and expansion is $\pm [c_0z-c_1+\frac{c_2}{z}-\frac{c_3}{z^2}+...]$
Could someone, please, help to get the right answer?
I think Cauchy product of the series can help to find coeffitients.
|
The idea was correct:
$z(z-1)^{1/2}(z-2)^{1/2} =\pm z\sqrt{(1-\frac{1}{z})}\sqrt{(1-\frac{2}{z})}$,
then using the The Binomial Theorem:
$\sqrt{(1-\frac{1}{z})} = [\sum_{k=0}^\infty\binom{1/2}{k}(-1)^{k}z^{-k}] = \sum_{k=0}^\infty(-1)^{k}\binom{1/2}{k}z^{-k}$ and
$\sqrt{(1-\frac{2}{z})} = [\sum_{k=0}^\infty\binom{1/2}{k}(-2)^{k}z^{-k}] = \sum_{k=0}^\infty(-1)^{k}2^k\binom{1/2}{k}z^{-k}$
Now we have : $\pm z[\sum_{k=0}^\infty(-1)^{k}\binom{1/2}{k}z^{-k}][\sum_{k=0}^\infty(-1)^{k}2^k\binom{1/2}{k}z^{-k}]$
Using Cauchy product: $[\sum_{k=0}^\infty a_kx^n * \sum_{k=0}^\infty b_kx^n] = \sum_{k=0}^\infty c_kx^n$,
and $c_0 = a_0b_0, c_1 = a_1b_0 + a_0b_1 $ etc
$c_n = \sum_{k=0}^n a_{n-k}b_k $.
Then the coeff. $c_n = \sum_{k=0}^n(-1)^{n-k}\binom{1/2}{n-k}(-1)^k2^k\binom{1/2}{k} = (-1)^n\sum_{k=0}^n2^{k}\binom{1/2}{n-k}\binom{1/2}{k}$.
The Laurent expansion: $\pm[c_0z - c_1 + \frac{c_2}{z^2} - ... ]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$ I want to find the following limit:
$$\lim_{x\rightarrow -\infty}{e^{\frac {1}{2-x}}\cdot\frac{x^2+2x-1}{x-2}}-x$$
This is what I do. I change the variable $t=-x$ and I have the following limit:
$$\lim_{t\rightarrow +\infty}{e^{\frac {1}{2+t}}\cdot{\frac{t^2-2t-1}{-t-2}}+t}=\lim_{t\rightarrow+\infty}{h(x)}$$
We have $e^{\frac{1}{2+t}}\rightarrow1$ for $t\rightarrow+\infty$
Therefore I think (this is the passage I'm less sure about)
$$h(x)\sim \frac{t^2-2t-1}{-t-2}+t=\frac{t^2-2t-1+t(-t-2)}{-t-2}=\frac{t^2-2t-1-t^2-2t}{-t-2}=\frac{-4t-1}{-t-2}\sim{\frac {-4t}{-t}}\rightarrow4$$
The solution should actually be $3$. Any hints on what I'm doing wrong?
| Although $\lim\limits_{x\to-\infty}e^{\frac1{2-x}}=1$, we cannot simply replace it by $1$ in the limit because it is actually $1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)$
For $x$ near $\pm\infty$,
$$
\begin{align}
e^{\frac1{2-x}}\frac{x^2+2x-1}{x-2}-x
&=\overbrace{\left(1+\frac1{2-x}+O\!\left(\frac1{(2-x)^2}\right)\right)}^{e^{\frac1{2-x}}}\overbrace{\left(x+4-\frac7{2-x}\right)\vphantom{\left(\frac1{()^2}\right)}}^{\frac{x^2+2x-1}{x-2}}-x\\
&=\frac x{2-x}+4+O\!\left(\frac1{2-x}\right)
\end{align}
$$
Now let $x\to-\infty$.
Simply replacing $e^{\frac1{2-x}}$ by $1$ misses the $\frac x{2-x}$ in the formula above and leaves us only with $4$ in the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to integrate $ \int \frac{\sin {3x} + \cos{3x}}{ \sin^3 x + \cos^3 x } dx\ $ I want to know how to integrate this function I have tried many things many substitutions but none works.
I even try to expand the numerator by $\sin3x$ and $\cos3x$ properties and tried to convert higher powers to multiple angles
| Use
\begin{equation}\cos\left(3x\right)=\cos^3\left(x\right)-3\cos\left(x\right)\sin^2\left(x\right)
\end{equation}
and
\begin{equation}
\sin\left(3x\right)=3\cos^2\left(x\right)\sin\left(x\right)-\sin^3\left(x\right)
\end{equation}
accompanied with
\begin{align}
\sin\left(x\right)&=\dfrac{\tan\left(x\right)}{\sec\left(x\right)} \\
\cos\left(x\right)&=\dfrac{1}{\sec\left(x\right)} \\
\sec^2\left(x\right)&=\tan^2\left(x\right)+1
\end{align}
You will get
\begin{equation}
{\int}{{\sec^2\left(x\right)}}{{\left(-\dfrac{\tan^3\left(x\right)+3\tan^2\left(x\right)-3\tan\left(x\right)-1}{\tan^5\left(x\right)+\tan^3\left(x\right)+\tan^2\left(x\right)+1}\right)}}\,\mathrm{d}x
\end{equation}
Use the change of variable $u = \tan (x)$ then $dx = \frac{1}{\sec^2 (x)} du$, you will get
\begin{equation}
{\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u
\end{equation}
But
\begin{equation}
{\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u
={\int}\dfrac{u^3+3u^2-3u-1}{\left(u+1\right)\left(u^2+1\right)\left(u^2-u+1\right)}\,\mathrm{d}u
\end{equation}
Now perform partial fraction decomposition, we get
\begin{equation}
{\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u
={\int}\left(\dfrac{10u-5}{3\left(u^2-u+1\right)}-\dfrac{4u}{u^2+1}+\dfrac{2}{3\left(u+1\right)}\right)\mathrm{d}u
\end{equation}
That is
\begin{equation}
{\int}\dfrac{u^3+3u^2-3u-1}{u^5+u^3+u^2+1}\,\mathrm{d}u
=\dfrac{5}{3}\underbrace{{}{\int}\dfrac{2u-1}{u^2-u+1}\,\mathrm{d}u}_{\ln \vert u^2-u+1 \vert}-
\underbrace{{\int}\dfrac{u}{u^2+1}\,\mathrm{d}u}_{\dfrac{\ln\vert u^2+1\vert}{2}}+
{{\dfrac{2}{3}}} \underbrace{{\int}\dfrac{1}{u+1}\,\mathrm{d}u}_{\ln\vert u+1\vert}
\end{equation}
Finally plug back $u = \tan x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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How to find $\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}$ without L'Hopital's Rule.
How would you find $\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}$ without L'Hopital's Rule?
The way the problem is set up, it makes me think I would try and use the fact that
$\displaystyle\lim_{x \to 0}\frac{1-\cos(x)}{x}=0$ or $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}=1$
So one idea I did was to multiply the top and bottom by $2x$ like so:
$\displaystyle\lim_{x \to 0}\frac{1-\cos(2x)}{\sin^2{(3x)}}\cdot\frac{2x}{2x}$.
Then I would let $\theta=2x$:
$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\sin^2{(\frac{3}{2}\theta)}}\cdot\frac{\theta}{\theta}$
which would let me break it up:
$\displaystyle\lim_{\theta \to 0}\frac{1-\cos(\theta)}{\theta}\cdot \frac{\theta}{\sin^2(\frac{3}{2}\theta)}$
So I was able to extract a trigonometric limit that is zero or at least see it. The second part needed more work. My immediate suspicion was maybe the whole limit will go to zero, but when checking with L'Hopital's Rule, I get $\frac{2}{9}$..... :/
| Notice that
$$
\frac{1 - \cos t}{t^2} = \frac{1 - \cos t}{t^2} \cdot \frac{1 + \cos t}{1 + \cos t} = \frac{1 - \cos^2 t}{t^2(1 + \cos t)} = \frac{\sin^2 t}{t^2} \cdot \frac{1}{1 + \cos t}.
$$
Hence
$$
\lim_{t \to 0}\frac{1 - \cos t}{t^2} = \frac{1}{2}.
$$
Returning back to your problem:
$$
\frac{1 - \cos(2x)}{\sin^2 (3x)} = \frac{1 - \cos(2x)}{x^2} \cdot \frac{x^2}{\sin^2(3x)} = 4 \cdot \frac{1 - \cos(2x)}{(2x)^2} \cdot \frac{1}{9} \cdot \frac{(3x)^2}{\sin^2(3x)}.
$$
Now take the limit...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
what is the difference between selection and division of identical and distinct objects and which these will be used in this problem? I have seen these formulas in my textbook:
(i)The number of ways of selecting one or more items from ${n}$ distinct items is
${2^n - 1}$.
(ii)The number of ways of selecting zero or more items from a group of ${n}$ distinct items
is ${n+1}$.
(iii)The number of ways of dividing ${n}$ identical items among ${r}$ persons,each one of whom receives at least one item is ${^{n-1}C_{r-1}}$.
(iv)The number of dividing ${n}$ identical items among ${r}$ persons,each of them who can receive 0,1,2, or more items is ${^{n+r-1}C_{r-1}}$.
So what is the main difference between difference between division and selection of identical and distinct objects.
I get confused when I see problems involving some kind of selection of distinct or identical objects.
For example:-
Q.1 In a shop there are 5 types of ice creams available.A child buys six ice-creams.How many number of different ways are there for child to buy six ice-creams?
In the above problem I thought that there unlimited ice-creams of 5 types.
If a child wants to choose from them it must involve selection of icecreams.
But in my textbook I see that
The number of different ways the child can buy 6 ice-creams is same as the number of dividing ${6}$ identical items among ${5}$ persons,each of them who can receive 0,1,2, or more items is ${^{6+5-1}C_{5-1}}$=${^{10}C_{4}}$.
I am confused how can division be involved here when it there are unlimited number of objects that are to be divided.
| Imagine there are $3$ distinct items: $A,B,C$.
(i) The number of ways of selecting one or more items from $3$ distinct items is:
$${3\choose 1}+{3\choose 2}+{3\choose 3}=2^3-1=7 \ \ \ (\text{note:} \sum_{k=0}^n {n\choose k}=2^n).$$
Indeed, the outcomes are:
$$A,B,C,AB,AC,BC,ABC.$$
Now imagine there are $3$ identical items: $A,A,A$.
(iv) The number of dividing $3$ identical items among $2$ persons, each of them who can receive 0,1,2, or more items is:
$${3+2-1\choose 2-1}={4\choose 1}=4 \ \ \ (\text{explained below})$$
Indeed, the outcomes are:
$$\{AAA,0\}, \{AA,A\}, \{A,AA\}, \{0,AAA\}.$$
Explanation: let $x_1,x_2$ be the number of identical items the two persons receive, respectively. Then, the equation is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
1-method: Stars and bars. Consider the number of stars as number of items to be given to a person and a bar to switch from one person to another person. For example, several examples:
$$\begin{align}**|* &\ \ (\text{the first person gets $2$, the second gets $1$})\\
|*** &\ \ (\text{the first person gets $0$, the second gets $3$})\\
*|** &\ \ (\text{the first person gets $1$, the second gets $2$})\\
***| &\ \ (\text{the first person gets $3$, the second gets $0$})
\end{align}$$
Note that it is basically a combination of $4$ symbols (items) taken $1$ (bar) or $3$ (stars) at a time:
$${4\choose 1}={4\choose 3}=4.$$
2-method: Generating functions. The equation to be solved is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
Let the equation
$$(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3)=x^{3}$$
represent the number of items (indicated on the exponents) the two persons (indicated by the brackets) can receive. For example:
$$x^0\cdot x^3=x^3 \ \ (\text{the first person gets $0$, the second gets $3$})\\
x^2\cdot x^1=x^3 \ \ (\text{the first person gets $2$, the second gets $1$})\\
x^3\cdot x^0=x^3 \ \ (\text{the first person gets $3$, the second gets $0$})\\
x^1\cdot x^2=x^3 \ \ (\text{the first person gets $1$, the second gets $2$})\\$$
Now if we consider the sum of $x^3$ we get $4x^3$. So, the problem reduces to finding the coefficient of $x^3$ in the expansion of the left-hand side:
$$\begin{align}[x^3](1+x+x^2+x^3)^2&=[x^3]\left(\frac{1-x^4}{1-x}\right)^2= \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)\\
&=[x^3](1-x^4)^2(1-x)^{-2}= \ \quad \quad \quad \quad \quad \quad (2)\\
&=[x^3]\sum_{k=0}^2 {2\choose k}(-x^4)^k\cdot \sum_{k=0}^{\infty} {-2\choose k}(-x)^k= \ \ (3)\\
&=[x^3]{2\choose 0}(-x^4)^0\cdot {-2\choose 3}(-x)^3= \quad \quad \quad \quad (4)\\
&={2+3-1\choose 3}= \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (5)\\
&={4\choose 3}.\end{align}$$
where:
(1) inside brackets: the sum of four terms of GeomProg.
(2) simple algebra.
(3) negative binomial series.
(4) considering only the terms, whose exponents add up to $3$.
(5) negative binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.
| As an alternative, we can also try to solve the problem by a geometrical intepretation, indeed the constraints represent a circle which is the intersection between
*
*the plane $x+y+z=10$
*the sphere $x^2+y^2+z^2=36$
therefore the center of the circle is located at
$$OC=\left(\frac{10}{3},\frac{10}{3},\frac{10}{3}\right)$$
and the radius of the circle is
$$R^2=36-|OC|^2 \implies R^2=\frac83 \implies R=\frac{2\sqrt 6}{3}$$
To parametrize the circle let consider at first the circle in $x-y$ plane and centered in the origin
$$x^2+y^2=\frac83 \iff \left(\frac{2\sqrt 6}{3}\cos \alpha, \frac{2\sqrt 6}{3} \sin \alpha,0\right)$$
and since the rotation matrix around $u=\left(\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2},0\right)$ of an angle $\theta=\arccos \left(\frac{\sqrt 3}{3}\right)$ is given by
$$M = \begin{bmatrix}
\frac12+\frac{\sqrt 3}{6} & -\frac12 +\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\
-\frac12 +\frac{\sqrt 3}{6} & \frac12+\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\
-\frac{\sqrt 3}{3} & -\frac{\sqrt 3}{3} & \frac{\sqrt 3}{3}
\end{bmatrix}$$
the intersection circle can be parametrized as follow
*
*$x(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 3}{6}\right)\cos \alpha +\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 3}{6}\right)\sin \alpha
=\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(1+\frac{\sqrt 3}{3}\right)\cos \alpha +\left(-1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$
*$y(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 6}{6} \right)\cos \alpha +\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 6}{6}\right)\sin \alpha
=\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(-1+\frac{\sqrt 3}{3}\right)\cos \alpha +\left(1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$
*$z(\alpha)=\frac{10}{3}-\frac{2\sqrt 2}{3}\left(\cos \alpha +\sin \alpha \right)=\frac{2\sqrt 2}{3}\left(\frac{5\sqrt 2}{2}-\cos \alpha -\sin \alpha \right)$
and finally we obtain
$$xyz(\alpha)=\frac{8}{27}\left(\sqrt 2 \cos (3\alpha )-\sqrt 2 \sin (3\alpha)+110\right) \quad \alpha \in [0,2\pi)$$
that is
$$xyz(\alpha)=\frac{8}{27}\left[ 2 \sin \left(\frac{\pi}4-3\alpha\right)+110\right] \quad \alpha \in [0,2\pi)$$
and therefore
$$32\le xyz \le \frac{896}{27}$$
form the plot of we can see that the maximum and the minimum values are reached each one in threee different points of the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating the integral $\int_0^{\infty}\frac{x^3}{x^2+a^2}\,\mathrm{d}x$
If $\displaystyle\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x=\large\displaystyle\dfrac1{ka^6}$, then find the value of $\displaystyle\dfrac{k}{8}$.
I tried a lot but finally stuck at an intermediate form :
$$\begin{align}
&\int_0^{\infty}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x, \text{with}\, {x^2=t},{2x~\mathrm{d}x=\mathrm{d}t}\\
&=\frac12\int_0^{\infty}\dfrac{(x^2)(2x)}{x^2+a^2}\,\mathrm{d}x=\frac12\int_0^{\infty}\dfrac{t}{t+a^2}\,\mathrm{d}t=\frac12\int_0^{\infty}\dfrac{t+a^2-a^2}{t+a^2}\,\mathrm{d}t\\
&=\frac12\left[\int_0^{\infty}\mathrm{d}t-\int_0^{\infty}\dfrac{a^2}{t+a^2}\,\mathrm{d}t\right]=\frac12\left[t|_0^{\infty}-a^2\ln(a^2+t)|_0^{\infty}\right]
\end{align}$$
| $$I={\displaystyle\int}\dfrac{x^3}{x^2+a^2}\,\mathrm{d}x$$
Substitute $u=x^2+a^2$ thus $\mathrm{d}x=\dfrac{1}{2x}\,\mathrm{d}u$
$$I=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{2}}}{\displaystyle\int}\dfrac{u-a^2}{u}\,\mathrm{d}u$$
$$I={\dfrac{1}{2}\displaystyle\int}\left(1-\dfrac{a^2}{u}\right)\mathrm{d}u$$
$$I=\dfrac{1}{2}{\displaystyle\int}1\,\mathrm{d}u-\dfrac{1}{2}\class{steps-node}{\cssId{steps-node-2}{a^2}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
$$I=\dfrac{u}{2}-\dfrac{a^2\ln\left(u\right)}{2}+c$$
$$I=\left(\dfrac{x^2+a^2}{2}-\dfrac{a^2\ln\left(x^2+a^2\right)}{2}\right)\biggr|_{x=0}^{\infty}$$
The integral is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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About the proof that every real number in the unit interval is the limit of a sequence of dyadic numbers
Given $x \in (0,1)$, show there exists a sequence $(x_n) \subset \{0,1\}$ such that $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$.
After running into difficult in trying to solve this problem, I found this MSE post, which, to my chagrin, gave me more difficulties. I'm trying to follow John Ma's reasoning. Here's how I understand it:
If $x < \frac{1}{2}$, the choose $x_1 =0$ and therefore $|x-\frac{x_1}{2}| = |x| = x < \frac{1}{2}$, and in fact $x-\frac{x_1}{2} \ge 0$, so in this case we can find $x_1$ for which $0 \le x - \frac{x_1}{2} < \frac{1}{2}$. If, however, $x \ge \frac{1}{2}$, then choose $x_2 =1$, and $|x-\frac{x_1}{2}| < \frac{1}{2}$ holds if and only if $- \frac{1}{2} < x - \frac{1}{2} < \frac{1}{2}$ if and only if $0 < x < 1$. Since $0 < x < 1$ is in fact true, so must $|x-\frac{x_1}{2}| < \frac{1}{2}$. Also, we have $x - \frac{x_1}{2} \ge 0$ since we assumed $x \ge \frac{1}{2}$. Hence, in either case we found $x_1 \in \{0,1\}$ for which $0 \le x-\frac{x_1}{2} < \frac{1}{2}$ holds. Now assume that $x_1,...x_n \in \{0,1\}$ have been chosen such that $0 \le x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^n}$. From this we want to show that it is possible to choose $x_{n+1} \in \{0,1\}$. If $x - \sum_{k=1}^n \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$, then choose $x_{n+1} = 0$ and we obtain $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} < \frac{1}{2^{n+1}}$. If, however, $x - \sum_{k=1}^n \frac{x_k}{2^k} \ge \frac{1}{2^{n+1}}$, choose $x_{n+1} = 1$. Then $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k} = (x - \sum_{k=1}^n \frac{x_k}{2^k}) - \frac{1}{2^{n+1}}$
But here's the problem: why is $x - \sum_{k=1}^{n+1} \frac{x_k}{2^k}$ nonnegative? If that can't be shown, then I cannot appeal to Squeeze lemma to show $x = \sum_{n=1}^\infty \frac{x_n}{2^n}$. I could really use some help. Thanks in advanced!
| The following argument offers an alternative proof; the contrast between different proofs can be instructive.
For any $k$, define
$\tag 1 R_k = \{\sum_{n=1}^k \frac{x_n}{2^n} \; | \; x_n \in \{0,1\} \land \sum_{n=1}^k \frac{x_n}{2^n} \le x\}$.
This is a finite set with at most $2^k$ elements.
Proposition 1: For every $k$,
$\tag 2 x - \text{max}(R_k) \le 2^{-k} $
Proof
To get a contradiction, assume that $x_n$ defines the maximum number in $R_k$ and
$\tag 3 x - \sum_{n=1}^k \frac{x_n}{2^n} \gt 2^{-k} $
It can't be true that $x_n = 1$ for all $n$, since then $x \gt 1$. But then since
$\tag 4 \sum_{n=1}^k \frac{x_n}{2^n} + 2^{-k} \le x $
the $2^{-k}$ number can be 'absorbed' into the sigma expression form, so that a another number can be found in $R_k$ greater than the number defined with the $x_n$. $\quad \blacksquare$
It is now easy, using $\text{(2)}$, to show that the increasing sequence $\text{max}(R_k)$ converges to $x$.
Note: Although induction wasn't explicitly used, the $\sum$ notation (an 'induction machine') is carrying the proof construction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find $a,b,c \in \Bbb N$ such that $n^3+a^3=b^3+c^3$ Is it true that for every $n \in \Bbb N$, there exist $a,b,c \in \Bbb N$ satisfing
$$
n^3+a^3=b^3+c^3,
$$
where $\operatorname{gcd}(a,b,c)=1$ and $b,c\ne n$?
I checked all positive integers less than 1000, it seems true but I don't know how to prove it.
We have $n^3+(3 n^3 + 3 n^2 + 2 n)^3=(3 n^3 + 3 n^2 + 2 n + 1)^3 - (3 n^2 + 2 n + 1)^3$, but there is a negative number on the right-hand side.
| A quick check shows that we have
$$
n^3 + (3 n^3 - 3 n^2 + 2 n)^3 = (3 n^3 - 3 n^2 + 2 n -1)^3 + (3 n^2 -
2 n + 1)^3
$$
Edit 1: If I did not make any mistakes this is the only cubic parameterization. There are no quadratic ones.
For $n\in\mathbb N \implies n\geq 1$, writing $a,b,c$ as
$$
\begin{align}
a &= 3 n^3 - 3 n^2 + 2 n = n (2 n^2 - 1) + (n - 1)^3 + 1\\
b &= 3 n^3 - 3 n^2 + 2 n -1 = n (2 n^2 - 1) + (n - 1)^3\\
c &= 3 n^2 - 2 n + 1 = 2n^2 + (n-1)^2
\end{align}
$$
we see that $a,b,c \in \mathbb N$ and clearly $c\neq n$. Finally, since
$$
a - b = 1,
$$
we must always have
$$
\gcd(a,b,c) = 1
$$
Therefore the statement is true.
This is found via a parametric search of
$$
\begin{align}
a &= a_3 n^3 + a_2 n^2 + a_1 n + a_0\\
b &= b_3 n^3 + b_2 n^2 + b_1 n + b_0\\
c &= c_3 n^3 + c_2 n^2 + c_1 n + c_0
\end{align}
$$
In particular the coefficients of $n^9$ and $n^0$ satisfies
$$
\begin{align}
a_3^3 &= b_3^3 + c_3^3\\
a_0^3 &= b_0^3 + c_0^3
\end{align}
$$
so by Fermat's Last Theorem we can assume one of $\{a_3,b_3,c_3\}$ and one of $\{a_0,b_0,c_0\}$ to be $0$ to speed things up a bit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Let a, b and c be positive real numbers satisfying Let a, b and c be positive real numbers satisfying
$\frac{1}{a+2019}$ + $\frac{1}{b+2019}$ +$\frac{1}{c+2019}$ = $\frac{1}{2019}$
Show that abc ≥$4038^3$.
My first impression is to use arithematic mean ≥ geometric mean.
| Hint: Your equation can be written as $$4076361(a+b+c)+16460345718-abc=0$$
Substituting $$x=2019$$ you will get $$2x^3+x^2(a+b+c)-abc=0$$
So we have $$abc=2x^3+x^2(a+b+c)$$ and
$$\frac{abc-2x^3}{x^2}=a+b+c\geq 3\sqrt[3]{abc}$$ and from here we get
$$(abc-2x^3)^3\geq 27x^6abc$$
Expanding
$$(abc)^3-6x^3(abc)^2-15abcx^6-8x^9\geq 0$$
with $$abc=t$$ we get the function
$$f(t)=t^3-6t^2x^3-15tx^6-8x^9$$
So now we must compute the positive zero of $$f(t)$$
and we get $$t=abc\geq 65841382872=4038^3$$ and $$f(t)\geq 0$$ is fulfilled.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If I subtract 1 from the n,n entry of a Pascal Matrix, why does the determinant become zero? In problem 19.2, how did the author reach this statement: "Since the n, n entry multiplies its cofactor positively, the overall determinant drops by 1 to become 0."
I was able to solve it a different way, but I feel that my approach was less efficient. I'd like to understand the solution's approach.
Here is what I did:
$det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&19\end{bmatrix}) = det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&20\end{bmatrix}) - det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix})$.
and
$ det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix}) = (-1)^3 * det(\begin{bmatrix}0&0&0&1\\1&1&1&1\\1&2&3&4\\1&3&6&10\end{bmatrix}) = (-1)^3 * (-1) * det(\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix})$.
therefore
$det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\1&4&10&19\end{bmatrix}) = 1 - 1 = 0 $.
It feels like the solution skipped doing this step, I don't get it.
| Note: I don't have enough reputation to comment so I will try to make my point with your equations.
Your approach seems to be more efficient than the answers here to be honest. As I understand you only need to show the determinant below is equal to 1.
$
det(\begin{bmatrix}1&1&1&1\\1&2&3&4\\1&3&6&10\\0&0&0&1\end{bmatrix})=1
$
Since all pascal matrices has a determinant of 1 this includes:
$
det(\begin{bmatrix}1&1&1\\1&2&3\\1&3&6\end{bmatrix})=1
$
Thus the first equation can be shown easily. I think this is as intuitive and efficient as it gets.
| {
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"url": "https://math.stackexchange.com/questions/2918309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial equation, cannot solve for $x$
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$\begin{align}
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\
a &= 3 \\[0.2ex]
b &= -6 \\
c &= -6
\end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$\begin{align}
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{(-6)^2-4(3)(-6)}}{2a} \\[0.8ex]
x&=\frac{-6\pm\sqrt{36-72}}{2a}
\end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + \sqrt{3}$ and $x = 1 - \sqrt{3}\,$ per my textbook's given solutions?
| so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-(4)(3)(-6)}}{6}=1\pm\frac{\sqrt{108}}{6}=1\pm\sqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
| {
"language": "en",
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"question_score": "2",
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Find the line through $P = (2,4,7) $ that intersects and is perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $
Find the line through $P = (2,4,7) $ that intersects and is
perpendicular to the line : $x = -1 + t, y = -2 + 3t, z = 4 $
I set up a system of equations and obtained :
$-1 + t = x_0 +2 \\ -2 + 3t = y_0 + 4 \\ 4 = z_0 + 7$
Solving for each and putting them I obtain the vector $ v_1 = <t -3, 3t-6, -3>$ I then set the direction vector as $ v_2 =<1,3,4>$
Since $v_1 \cdot v_2 = 0 $ means they are orthogonal I obtain :
$<t -3, 3t-6, -3> \cdot <1,3,4> = 0 $
Multiplying out and solving for t I get $t = 10/21$
Would I now just substitute my given value for t back into $t_1$ ?
| 1. The plane $P_1$ which includes the line
$$\ell_0: x = -1 + t, y = -2 + 3t, z = 4 $$
and the point $p=(2,4,7)$ is
$$\boxed{3x-y-z=-5}$$
2. The plane $P_2$ perpendicular to the line $\ell_0$ psses through point $p=(2,4,7)$ is
$$\boxed{x+3y=14}$$
3. Intersection of planes $P_1$ and $P_2$ is desired line that is
$$\boxed{\dfrac{x-14}{-3}=y=\dfrac{-z+47}{10}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality involving positive real numbers Let $x_1,x_2,\cdots x_n,x_{n+1}$ be any real numbers greater than or equal to $1$.
Then for $n\ge 2,$ I was trying to verify the validity of the inequality
$$\frac{n-1}{n}\sum_{k=1}^n\frac{1}{1+x_k}+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n}\ge \frac{n}{n+1}\sum_{k=1}^{n+1}\frac{1}{1+x_k}+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}.$$ May I kindly seek your suggestions?
Same result can be tried when $x_1,x_2,\cdots x_n,x_{n+1}$ are non-negative real numbers less than or equal to $1$.
| The inequality is equal to
$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n} \ge \frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)+\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}$$
We know that $\frac{n-1}{n}>\frac{n}{n+1}$ hence
$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)\ge\frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)$$
Since $1>\frac{n}{n+1}$
$$\frac{1}{1+x_{n+1}}\ge\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)$$
Since $a\le b \implies \frac{1}{a} \ge \frac{1}{b}$ hence
$$\frac{1}{1+x_{n+1}}\ge\frac{1}{1+x_1.x_2.\cdots.x_n.x_{n+1}}$$
Adding the three equations above gives us the inequality we seek. $\ _\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\sin^8\theta+\cos^8\theta=\frac{17}{32}$, find the value of $\theta$ using de Moivre's theorem. If
$$\sin^8\theta+\cos^8\theta=\frac{17}{32}$$
find the value of $\theta$ using de Moivre's theorem. I tried a lot exapanding using Binomial theorem and taking real part and equating it given value.
| I try to give a solution with the given constraint.
Using de Moivre's formula we will show
$$
\begin{aligned}
\cos^4 t +\sin^4t &=\frac 1{4}\Big[\ \cos 4t+3\ \Big]\ ,
\\
\cos^8 t +\sin^8t &=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
\end{aligned}
$$
There is no prize of beauty we can win for the proof, so let us use $c:=\cos t$, $s=\sin t$ as shortcuts for an easy typing. The strategy is to write $\cos 4t$ and $\cos 8t$ in terms of an even polynomial in $c,s$, with no mixed monomials.
Let us get explicit relations:
$$
\begin{aligned}
\cos 4t
&=\text{Real part of }(\cos 4t +i\sin 4t)\\
&=\text{Real part of }(\cos t +i\sin t)^4\quad\text{(de Moivre)}\\
&=c^4-6c^2s^2+s^4\\
&=c^4-3(\underbrace{(c^2+s^2)^2}_{=1}-c^4-s^4)+s^4\\
&=4(c^4+s^4)-3\ ,\\
&\qquad\text{showing the first formula.}\\[2mm]
2c^2s^2
&=(c^2+s^2)^2-c^4-s^4=1-c^4-s^4\ ,\\
4c^4s^4
&=(1-c^4-s^4)^2\\
&=1+c^8+s^8-2c^4-2s^4+2c^4s^4\ ,\qquad\text{ so}\\
2c^4s^4
&=1+c^8+s^8-2c^4-2s^4\ ,\qquad\text{ so}\\[2mm]
\cos 8t
&=\text{Real part of }(\cos 8t +i\sin 8t)\\
&=\text{Real part of }(\cos 4t +i\sin 4t)^2\qquad\text{(de Moivre)}\\
&=\cos^24t-\sin^24t\\
&=2\cos^24t-1\qquad\text{(or going straightforward here...)}\\
&=2(4c^4+4s^4-3)^2
\\
&=\dots
\ .
\end{aligned}
$$
Some further lines are showing the claimed formula for $\cos 8t$. (The term in $c^4s^4$ can be split as above to have only monomials in $c$, or in $s$, or constant ones.)
From the given relation we have to solve the equation in $t$ (instead of $\theta$, which is too hard to type):
$$
\frac {17}{32}
=\frac 1{64}\Big[\ \cos 8t+28\cos 4t +35\ \Big]\ .
$$
Possibly using de Moivre,
we write again
$\cos 8t = 2\cos^2 4t-1$ and obtain an equation of second degree in $u:=\cos 4t\in[-1,1]$,
$$
\frac {17}{32}
=\frac 1{64}\Big[\ 2u^2+28u +34\ \Big]\ .
$$
Now $17/32$ cancels on both sides, we get $2(u^2+17u)=0$, equivalently (for $|u|\le 1$) $u=0$, i.e. $4t$ is an odd multiple of $\pi/2$, i.e. $t$ is an odd multiple of $\pi/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Simplify $\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$ to $\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$ I'm trying to bring this expression:
$$\frac{5}{6}\log\left(\frac{5}{4}\right) - \frac{1}{6}\log(2)$$
To this one:
$$\log\left(\frac{5}{4}\right) - \frac{1}{6}\log\left(\frac{5}{2}\right)$$
Where $\log$ is the natural algorithm.
I know the two expressions are equal (checked with wolfram) but I really can't find the correct passages... Could you please help me?
| Alt. hint: by brute force, using only $\,\log \frac{a}{b} = \log a - \log b\,$ and $\,\log a^n = n \log a\,$:
$$\small
\frac{5}{6}\log\frac{5}{4} - \frac{1}{6}\log 2 = \frac{1}{6}\left(5 \log 5 - 5 \log 4-\log 2\right) = \frac{1}{6}\left(5 \log 5 - 10 \log 2-\log 2\right) = \frac{1}{6}\left(5 \log 5 - 11 \log 2\right)
$$
Now do the same for $\,\log \frac{5}{4} - \frac{1}{6}\log \frac{5}{2}\,$ and compare.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Probability of drawing objects: Combinations vs Permutations
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is
Attempt 1
$$
P(\text{exactly 1 red ball})=\frac{^5C_1 \cdot ^3C_2}{^8C_3}=\frac{5 \cdot 3}{\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}}=\frac{5 \cdot 3 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}=\frac{15}{56}
$$
Attempt 2
\begin{align*}
P(\text{exactly 1 red ball}) & = P(RBB)+P(BRB)+P(BBR)\\
& =\frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 5 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 2 \cdot 5}{8 \cdot 7 \cdot 6}\\
& =3 \cdot \frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}\\
& =\frac{15}{56}
\end{align*}
Though I get same result in both methods, I think first method uses combinations(order does not matter) and second uses permutation(order matters), how can I differentiate the two ?
| In Attempt 1 the underlying assumption is that you have a random mechanism selecting a $3$-element subset from a given $8$-element set, whereby all $3$-element subsets are equiprobable. You then count the number of favorable $3$-element subsets.
In Attempt 2 the underlying assumption is that you have a random mechanism selecting an element from a given finite set, whereby all elements are equiprobable, and you apply this mechanism three times. Finally you compute the probability of the desired outcome.
The reason that you obtain the same result in both cases is the following: Both mechanisms select (in one, resp., in three steps) a $3$-element subset of the given $8$-element set, uniformly over all ${8\choose3}$ such subsets.The first mechanism does so by its specification, and the second "by symmetry".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a geometric series for $\frac{1}{(1 + x)^2}$ I'm asked to find a geometric series for $f(x) = \frac{1}{(1 + x)^2}$, I integrate it first and I get $F(x) = \int{\frac{1}{(1 + x)^2}dx} = \frac{1}{-2(1 - (-x))}$
Since $\frac1{1-x} = \sum_{n=0}^\infty x^n$, $F(x) = \frac{-1}2\sum_{n=0}^\infty (-x)^n = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2}x^n$
If I derive that, $f(x) = \sum_{n=0}^\infty\frac{(-1)^{n+1} n x^{n-1}}{2} = \sum_{n=1}^\infty\frac{(-1)^n (n + 1) x^n}{2}$
But my textbook says that $f(x) = \sum_{n=0}^\infty (-1)^n (n + 1) x^n$
Obviously those 2 things are pretty different (notice the index). What did I get wrong?
| HINT
The geometric series is given by
\begin{align*}
\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \ldots = \sum_{n=0}^{\infty}x^{n}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation: $\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$ How would I go about solving the equation below:
$\arcsin(2x^2 −1) + 2\arcsin x = -\pi/2$
After appyling sin to both sides I end up with:
$(2x^2-1)(\sqrt{1-x^2}^2 - x^2) + 2x \sqrt{1 -(2x^2-1)^2}\sqrt{1-x^2}= -1$
| A method using derivatives: put $f(x) := \arcsin (2x^2-1)+2\arcsin x$. Observe that $f$ is defined on $[-1,1]$, it is continuous and $f(0) = -\pi/2$. Let us compute its derivative:
\begin{align}
f'(x) & = \frac{4x}{\sqrt{1-(2x^2-1)^2}}+\frac{2}{\sqrt{1-x^2}} \\
& = \frac{2x}{\lvert x \rvert \sqrt{1-x^2}}+\frac{2}{\sqrt{1-x^2}}=
\begin{cases}
0, \text{ if } -1 < x < 0 \\
\frac{4}{\sqrt{1-x^2}} \text{ if } 0 < x < 1.
\end{cases}
\end{align}
So $f$ is constantly $-\pi/2$ on $[-1,0]$. For $x > 0$ its derivative is positive, so $f$ increases strictly and you do not get other solutions to your equation.
A remark on your method: if you decide to apply $\sin$ to both sides be aware that you are actually solving more than one equation, because $-1 = \sin \left(-\pi/2 + k\pi\right)$, with $k \in \mathbb{Z}$. So when you get the final solutions of $\sin(\arcsin(2x^2-1)+2\arcsin x)=-1$ you should check which ones are also solutions of, for example, $\arcsin (2x^2-1)+2\arcsin x = 3\pi/2$. You see, $x=1$ satisfies the latter equation, but not the former.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find real $x$ that satisfies $\frac{1}{1+x}>\frac{x}{x-1}$ $\dfrac{1}{1+x}>\dfrac{x}{x-1}$
Multiply both sides with $(x+1)(x-1)$.
$x-1>x(x+1)$
Subtract $(x-1)$ from both sides.
$0>x^2+1$
This seems to have no real answers, although I have been told there should be. What am I doing wrong?
| Writing your inequality in the form
$$\frac{1}{x+1}-\frac{x}{x-1}>0$$ this is
$$-\frac{x^2+1}{x^2-1}>0$$ or $$\frac{x^2+1}{x^2-1}<0$$
Can you finish?
The only solution we get for $$x^2-1<0$$ since $$x^2+1>0$$ holds for all real numbers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of discrete non-uniform random variable and a Bernoulli is non-uniform Let $X$ be a non-uniform discrete random variable over $\{0,1,\dots,n\}$, and $Y$ be a Bernoulli random variable, not necessarily independent of $X$.
Is this information sufficient to conclude that $X + Y$ is non-uniform over $\{0,1,\dots,n+1\}$? If not, what would be a counter-example?
Edit: And what if $Y$ is independent of $X$?
| I'll find all solutions to the independent case. Let $P(X=i) = p_i$ and $P(Y=1) = r$. Then consider the probability generating functions of $X$ and $Y$. These are $f_X(z) = p_0 + p_1 z + p_2 z^2 + \cdots + p_n z^n$ and $f_Y(z) = (1-r) + rz$. Then the pgf of $Z = X + Y$ is
$$ f_Z(z) = f_X(z) f_Y(z) = (p_0 + p_1 z + \cdots + p_n z^n) ((1-r) + rz). $$
For $X + Y$ to be uniform this must be $(1 + z + \cdots + z^{n+1}) / (n+2)$. So we want to know: are there nonnegative real $p_0, p_1, \cdots, p_n, r$ such that, for all $z$,
$$ (p_0 + p_1 z + \cdots + p_n z^n) ((1-r) + rz) = {1 + z + \cdots + z^{n+1}\over {n+2}}. $$
Now $1 + z + \cdots + z^{n+1} = (1-z^{n+2})/(1-z)$, so it suffices to find $p_0, p_1, \ldots, p_n, r$ such that
$$ (p_0 + p_1 z + \cdots + p_n z^n) ((1-r) + rz) (1-z) = {1 - z^{n+2} \over (n+2)}$$
To be equal as polynomials, the two sides must have the same zeroes. The zeroes of the right-hand side are the $(n+2)$ roots of unity.
If $n$ is odd, then only one of these is real. But the left-hand side has at least two real roots, coming from the factors $((1-r) + rz)$ and $1-z$, so we have no solution.
If $n$ is even then the right-hand side has two real roots $z = \pm 1$. The right-hand side has the real root $z = 1$ coming from the factor $1-z$. If $r = 1/2$, it will also have the root $z = -1$ coming from the factor $(1-r) + rz = (1/2 + z/2)$. So any solution has $r = 1/2$.
Then we get
$$ (p_0 + p_1 z + \cdots + p_n z^n) = {1-z^{n+2} \over (n+2) (1-z)(1+z)/2} $$
or after some rearrangement
$$ (p_0 + p_1 z + \cdots + p_n z^n) = {2 \over n+2} {1-z^{n+2} \over 1-z^2}. $$
Doing the division on the right-hand side we get
$$ p_0 + p_1 z + \cdots + p_n z^n = {2 \over n+2} (1 + z^2 + z^4 + \cdots + z^n) $$
and so we conclude that $X$ must be uniform on $\{ 0, 2, 4, \ldots, n \}$.
So the only solutions when $X$ and $Y$ are independent have $X$ uniform on $\{ 0, 2, 4, \ldots, n \}$ for even $n$, and $Y$ uniform on $ \{0, 1\}$. If $n = 4$ this is the example that kimchi lover gave.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\sum_{n=0}^\infty \frac1{(4n^2 - 1)^2}$ How do I find the value of the following infinite series?
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} $$
My attempt at a solution:
$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} = \sum_{n=0}^\infty \frac{1}{((2n-1)(2n+1))^2} = \sum_{n=0}^\infty \left(\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)\right)^2 = \frac{1}{4}\sum_{n=0}^\infty \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)^2 $$
I then tried to compute the partial sums of this series, but with no luck. Does anyone else know how to do it?
| Since
$$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} $$
by Parseval's theorem we have
$$ \pi=\int_{-\pi}^{\pi}\left|\sin x\right|^2\,dx =2\pi\cdot\frac{4}{\pi^2}+\frac{16}{\pi}\sum_{n\geq 1}\frac{1}{(4n^2-1)^2}$$
hence by rearranging:
$$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \color{red}{\frac{\pi^2}{16}-\frac{1}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Find $a_1$ so that $ a_{n+1}=\frac{1}{4-3a_n}\ ,n\ge1 $ is convergent Let $ \left\{ a_n \right\} $ be a recursive sequence such that $$a_{n+1}=\frac{1}{4-3a_n}\quad,n\ge1 $$
Determine for which $a_1$ the sequence converges and in case of convergence find its limit.
The problem is from the book 'Problems in Mathematical Analysis I by W.J.Kaczor'.
| Suppose that this sequence does converge to A. Then we must have $A= \frac{1}{4- 3A}$. Then $A(4- 3A)= 4A- 3A^2= 1$. $3A^2- 4A+ 1= (3A- 1)(A- 1)= 0$. A is either 1 or 1/3.
If $a_1> 1$ the sequence clearly converges to 1. If $a_1\le 1/3$ if clearly converges to $\frac{1}{3}$. It's a little harder to show, but still true, that if $1/3< a_1< 1$ then the sequence converges to 1/3: if $\frac{1}{3}< a< 1$ then $1< 3a< 3$ so that 0< 3a-1< 2. But for $\frac{1}{3}< a< 1$, $a- 1< 0$. That is, 3a-1 is positive while a- 1 is negative so that $(3a- 1)(a- 1)= 3a^2- 4a+ 1< 0$. Then $3a^2- 4a= a(3a- 4)> 1$ and $a> \frac{1}{3a- 4}$. That is, for $\frac{1}{3}< a_1< 1$ the sequence is decreasing to $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$ is true.
which means $\dfrac{k^3 + 2k}{3}$ is divisible by $3$ and $\dfrac{k^3 + 2k}{3}=p$ for some integer $p$.
Now we need to show that $P(k+1)$ is true.
$(k+1)^3+2(k+1)$ and we will show that this divisible by $3$.
Proof:
$(k+1)^3+2(k1)=k^3+3k^2+3k+1+2k+2$
$.$
$.$
$.$
$.$
$=3(p+k^2+k+1)$
As $p+k^2+k+1$ is an integer we have that $(k+1)^3+2(k+!)$ is divisible by $3$.
Is my above attempt correct?
Did I show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer?
| Yes, you have proved the statement.
Note: Natural numbers are positive integers, so the base case is actually $1$. You made a few typing errors. Please define what $p$ is. (A simple statement like where $p=P(x)$ works)
| {
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Evaluate expression Let $a$ and $b$ be positive real numbers such that $$ a^4 + 3ab + b^4 = \frac{1}{ab}$$ Evaluate $ {(\frac{a}{b})}^{\frac{1}{3}} + {(\frac{b}{a})}^{\frac{1}{3}} - {(2+\frac{a}{b})}^{\frac{1}{2}}$. I tried in five different ways but I can't to solve. I suppose that the expression it is not constant...
| Let us take first $a=b=1/\sqrt 2$. Then the equation $F(a,b)=1$ is satisfied for
$$
F(a,b) = ab(a^4+3ab+b^4)\ ,
$$
and for this special choice the expression we need to evaluate, using the function
$$
G(a,b) =
\left(\frac ab\right)^{\frac 13}
+
\left(\frac ba\right)^{\frac 13}
-
\left(2+\frac ab\right)^{\frac 12}\ ,
$$
is $G(1/\sqrt 2, 1/\sqrt 2) = 1+1-\sqrt 3$.
The implicit equation $F(a,b)=1$ has many other solutions with $a\ne b$,
if $F(a,b)=1$, then $F(b,a)=1$, too,
and we obtain obviously different values $G(a,b)\ne G(b,a)$.
The problem suggests however that $G(a,b)$ is constant on $F(a,b)=1$, this is not the case.
Note:
Here is a parametrization that works to have $F=1$, and the values for $G$ can be obtained explicitly:
$$
F\left(\
\sqrt{\frac{t^5}{t^4 + 1}}
\ ,\
\sqrt{\frac 1{(t^4 + 1) t}}
\ \right)
$$
$$
=
{\left(\frac{t^{10}}{{\left(t^{4} + 1\right)}^{2}} + 3 \, \sqrt{\frac{t^{5}}{t^{4} + 1}} \sqrt{\frac{1}{{\left(t^{4} + 1\right)} t}} + \frac{1}{{\left(t^{4} + 1\right)}^{2} t^{2}}\right)} \sqrt{\frac{t^{5}}{t^{4} + 1}} \sqrt{\frac{1}{{\left(t^{4} + 1\right)} t}}
$$
$$
=1\ .
$$
| {
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Show that $\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+...}}}}=\sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}$
Assuming that $a>b^2$ show that
$$\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+...}}}}=\sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}$$ (corrected)
This problem listed in a contest-math preparation book with the tag Russian-IMO-Longlist 1999.
It is not in the problem statement, but I'm assuming that $a,b \in \Bbb R$ and $a,b>0$. I tried two routes, without much progress:
(a) developed $x=\sqrt{a-b\sqrt{a+bx}}$, as the solution $x$ must be a fixed point of $f(x)=\sqrt{a-b\sqrt{a+bx}}$, but the development does not appear promising (or the required algebra is beyond my capabilities).
(b) evaluated $f(x)=\sqrt{a-b\sqrt{a+bx}}$ on $x=\sqrt{a-\frac{3b^2}{2}}-\frac{b}{4}$ with the hope of getting $f(x)=\sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}$. Perhaps a naive approach.
Not great results from both approaches.
Hints and solutions are welcomed. Sorry if this is a duplicate.
| Simplifying $x = \sqrt{a-b\sqrt{a+bx}}$ gives
\begin{equation*}
\frac{a^2}{b^3}-\frac{2 a x^2}{b^3}-\frac{a}{b}+\frac{x^4}{b^3}-x
= \frac{1}{b^3}(x^2+bx+b^2-a)(x^2-bx-a) = 0,
\end{equation*}
so that
\begin{equation*}
x = -\frac{b}{2}\pm\sqrt{a-\frac{3b^2}{4}}\text{ or }
x = \frac{b}{2}\pm\sqrt{a + \frac{b^2}{4}}.
\end{equation*}
The two possibilities with minus signs are negative (since $a>b^2$), which we reject since clearly $x>0$. Further, since $x < \sqrt{a}$, we also reject $x = \frac{b}{2} + \sqrt{a+\frac{b^2}{4}}$, leaving only
$$x = \sqrt{a-\frac{3b^2}{4}}-\frac{b}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Double summation with improper integral So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$
Here is my approach:
First evaluate the integral:
$$ \frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$
This can be done using integration by parts and we get:
$$ \frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$
We can do this $ b $ times until we get:
$$ \frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$
and hence we end up with:
$$ \frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$
Now we can apply the sum of GP to infinity formula and we get:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}} = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$
This is a telescoping series and we end up with $$ \frac{1}{a-1} = \frac{1}{2-1} = 1.$$
Do you guys have any other ways of solving this problem? Please do share it here.
| Also note:
$$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\int_0^\infty\frac{x^b}{e^{ax}b!}dx=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{b!}\int_0^\infty x^be^{-ax}dx$$
now let:
$$u=ax$$
$$dx=\frac{du}{a}$$
so:
$$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{a*b!}\int_0^\infty \left(\frac{u}{a}\right)^be^{-u}du=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{a^{-(b+1)}}{b!}\int_0^\infty u^be^{-u}du$$
and we know that:
$$(n-1)!=\Gamma(n)=\int_0^\infty e^{-t}t^{n-1}dt$$
so our summation now simplifies to:
$$S=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{a^{-(b+1)}}{b!}b!=\sum_{a=2}^\infty\sum_{b=1}^\infty\frac{1}{a^{b+1}}=\sum_{a=2}^\infty\sum_{c=2}^\infty\frac{1}{a^{c}}=\sum_{a=2}^\infty\left(\sum_{c=1}^\infty\frac{1}{a^c}-\frac{1}{a}\right)=\sum_{a=2}^\infty\sum_{c=1}^\infty\frac{1}{a^c}-\sum_{a=2}^\infty\frac{1}{a}$$
I know this is correct up to the second summation on the final line but after this I am not sure.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 2
} |
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