Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| It is also true that $$ \rho \leq 1 $$
In the case that $\rho > 1,$ take $a=1, \; b = -1, \; c = 0.$ The polynomial becomes $2 - 2 \rho = 2(1 - \rho) < 0.$
In a similar style, we can do the original problem this way: if $\rho < -\frac{1}{2},$ take $a=1, \; b = 1, \; c = 1.$ The polynomial becomes $3 + 6 \rho < 3 - 3 = 0.$
Put still another way, the eigenvalues of
$$
\left(
\begin{array}{ccc}
1 & \rho & \rho \\
\rho & 1 & \rho \\
\rho & \rho & 1 \\
\end{array}
\right)
$$
are $$ 1-\rho, \; 1-\rho, \; 1 + 2 \rho $$
The quadratic form in the question is simply
$$
\left(
\begin{array}{ccc}
a & b & c \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & \rho & \rho \\
\rho & 1 & \rho \\
\rho & \rho & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
a \\
b \\
c \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
inequality in geometry
Given triangle $ABC$ has $BC=a$, $CA=b$, $AB=c$ and $M$ is a point of the triangle plane. Prove that:
\begin{align*}
\cos \dfrac{A}{2} \cdot MA+\cos \dfrac{B}{2} \cdot MB+\cos \dfrac{C}{2} \cdot MC \geq \dfrac{a+b+c}{2}.
\end{align*}
My attempts:
\begin{eqnarray*}
a \cdot \overrightarrow{IA}+b \cdot \overrightarrow{IB}+c \cdot \overrightarrow{IC}=\overrightarrow{0} \Rightarrow \dfrac{\cos \dfrac{A}{2}}{IA}\overrightarrow{IA}+\dfrac{\cos \dfrac{B}{2}}{IB}\overrightarrow{IB}+\dfrac{\cos \dfrac{C}{2}}{IC}\overrightarrow{IC}=\overrightarrow{0}??????
\end{eqnarray*}
with $I$ is center of inscribed circle of triangle $ABC$.
\begin{eqnarray*}
\cos \dfrac{A}{2} \cdot MA=\dfrac{\cos \dfrac{A}{2}}{IA} \cdot MA \cdot IA\geq \dfrac{\cos \dfrac{A}{2}}{IA} \cdot \overrightarrow{MA} \cdot \overrightarrow{IA}.
\end{eqnarray*}
The same as,
\begin{eqnarray*}
\cos \dfrac{B}{2} \cdot MB &\geq& \dfrac{\cos \dfrac{B}{2}}{IB} \cdot \overrightarrow{MB} \cdot \overrightarrow{IB}\\
\cos \dfrac{C}{2} \cdot MC &\geq& \dfrac{\cos \dfrac{C}{2}}{IC} \cdot \overrightarrow{MC} \cdot \overrightarrow{IC}.
\end{eqnarray*}
I am stuck here.
| Here is my proof, based on the main result that if $DEF$ is the pedal triangle of $M$ then we have: $$\cos\left(\frac{A}{2}\right)\cdot MA\ge \frac{AE+AF}{2}\qquad (1)$$
In short, we prove the above inequality by using the formula for $\sin\alpha -\sin\beta$ to show that $$\cos\left(\frac{A}{2}\right) \cdot MA = \dfrac{MF-ME}{2\sin\left(\dfrac{\widehat{MAF}-\widehat{MAE}}{2}\right)}\quad (2)$$
Then we use the following lemma: If $ABCD$ is a cyclic quadrilateral and $\widehat{B}=\widehat{D}=90^{\circ}$ then $$\dfrac{DA-DC}{BA+BC}\ge \sin\left(\dfrac{\widehat{ABD}-\widehat{CBD}}{2}\right) \qquad (3)$$
$(1)$ follows from $(2)$ and $(3)$, which gives us the original inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $s_{n} := 1 + 1/2 + \cdots + 1/n - \ln(n+1)$ and show that $s_{n} \leq 1 - 1/(n+1)$. This is in a Riemann integration section of a practice sheet from real analysis. I know that if $f(x) := \frac{1}{[x]} - \frac{1}{x+1}$, where $[*]$ is the floor function, then $\int_{1}^{n} f = s_{n}$. But I'm not sure where to go to get the quantity on the right-hand side of the inequality. I feel that I should compare $f \leq g$ for some $g$ where $\int_{1}^{n} g = 1- \frac{1}{n+1}$, but I am having trouble doing that!
| Induction does the trick
Assume it true for $n-1$, i.e.
$$1 + \ldots + \frac{1}{n-1} - \ln n < 1 - \frac{1}{n}$$
Then
$$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < \ln n + 1 - \frac{1}{n} + \frac{1}{n} - \ln (n+1) = 1 + \ln (\frac{n}{n+1}) = 1 + \ln(1 - \frac{1}{n+1})$$
Note that $$\ln (1-x) < -x$$ for $0 < x < 1$, hence
$$ 1 + \ln(1 - \frac{1}{n+1}) < 1 - \frac{1}{n+1}$$
We conclude
$$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < 1 - \frac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$$
So, maximum value $m$ is $\frac{7}{a+3b}+{\sqrt{3ab}}$ so $m=\frac{7}{2{\sqrt{3ab}}}+{\sqrt{3ab}}$ is $2{\sqrt{\frac{7}{2}}}$ so $2m^2=28$.
But it didn't satisfy equal condition so how do I get it?
| The mistake is when you apply AM-GM in the last step. Applying it you get $\ge$ inequality, which corresponds to the minimal value of l.h.s, not maximal.
Also $m \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab}$, not equal.
UPD: You did the following
$m = \frac{7}{a + 3b} + \sqrt{3ab} \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab} \ge 2\sqrt{\frac 72}$ and that's it. You get $2\sqrt{\frac 72}$ as maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$ Evaluate $$I=\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$$
My book gave the substitution for $$\int \frac{dx}{P\sqrt{Q}}$$ as
$\frac{Q}{P}=t^2$ when $P$ and $Q$ are quadratic expressions
So accordingly i used
$$\frac{x^2+x+1}{x^2-x+1}=t^2 \tag{1}$$ we get
$$\frac{(1-x^2) \, dx}{(x^2-x+1)^2}=t \,dt$$
Then
$$I=\int \frac{\sqrt{x^2-x+1}\:dt}{1-x^2}$$
By Componendo Dividendo in $(1)$ we get
$$\frac{x^2+1}{x}=\frac{t^2+1}{t^2-1}$$
But how to express integrand purely in terms of $t$?
| We have that $$\frac{x^2+1}{x}=\frac{t^2+1}{t^2-1}\tag{1}$$ so $$\frac{\sqrt{x^2-x+1}}{1-x^2}=\frac{\sqrt{x\left(\frac{t^2+1}{t^2-1}-1\right)}}{1-x\cdot\frac{t^2+1}{t^2-1}+1}=\frac{(t^2-1)\sqrt{\frac{2x}{t^2-1}}}{2(t^2-1)-x(t^2+1)}=\frac{\sqrt{2x(t^2-1)}}{2(t^2-1)-x(t^2+1)}$$ and you can further invoke $(1)$ to solve for $x$ to be substituted into the above expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Polynomial problem with two conditions I have to find $P(0)$ from the polynomial with minimum degree given that
$$(x-1)^3|(P(x)+1)$$
$$(x+1)^3|(P(x)-1)$$
Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively: $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$
$$P(-1)-1 =0; \ P'(-1)=0; \ P''(-1)=0$$
Furthermore since $P(1)=-1$ and $P(-1)=1$, also $P(1)=-P(-1)$
and $$P'(1)=P''(1)=P'(-1)=P''(-1)$$ I don't see how to use this stuff.
| Suppose $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that
\begin{align*}
(x+1)^3&{\,\mid\,}(P(x)-1)\\[4pt]
(x-1)^3&{\,\mid\,}(P(x)+1)\\[4pt]
\end{align*}
Equivalently, $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that
\begin{align*}
P(x)&=A(x)(x+1)^3+1\\[4pt]
P(x)&=B(x)(x-1)^3-1\\[4pt]
\end{align*}
for some polynomials $A,B\in\mathbb{Q}[x]$.
Necessarily, $A,B$ have the same degree.
Let $u=(x+1)^3$ and let $v=(x-1)^3$.
If $A,B\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $Au-Bv=-2$, then we have $Au+1=Bv-1$, hence $P=Au+1$ is a polynomial in $\mathbb{Q}[x]$ of least degree satisfying the specified requirements.
The equation $Au-Bv=-2$ is equivalent to $-\bigl({\large{\frac{A}{2}}}\bigr)u+\bigl({\large{\frac{B}{2}}}\bigr)v=1$, hence we have $A=-2a$ and $B=2b$, where $a,b\in\mathbb{Q}[x]$ are polynomials of least degree satisfying $au+bv=1$.
Since $u,v$ are relatively prime in $\mathbb{Q}[x]$, there are unique polynomials $a,b\in\mathbb{Q}[x]$ with
*
*$\deg(a) < \deg(v) = 3$$\\[4pt]$
*$\deg(b) < \deg(u) = 3$
such that $au+bv=1$, so these are the polynomials $a,b$ we want.
The polynomials $a,b$ can be found via the Extended Euclidean Algorithm.
Here are the steps . . .
\begin{align*}
(x+1)^3=(1)(x-1)^3+(6x^2+2)&\implies u-v=6x^2+2\\[4pt]
6(x-1)^3=(x-3)(6x^2+2)+16x&\implies (-x+3)u+(x+3)v=16x\\[4pt]
8(6x^2+2)=(3x)(16x)+16&\implies (3x^2-9x+8)u+(-3x^2-9x-8)v=16\\[4pt]
\end{align*}
from which we get
\begin{align*}
a&={\small{\frac{1}{16}}}(3x^2-9x+8)\\[4pt]
b&={\small{\frac{1}{16}}}(-3x^2-9x-8)\\[4pt]
\end{align*}
which yields
\begin{align*}
A&=-2a=-{\small{\frac{1}{8}}}(3x^2-9x+8)\\[4pt]
B&=2b={\small{\frac{1}{8}}}(-3x^2-9x-8)\\[4pt]
\end{align*}
hence
\begin{align*}
P&=Au+1\\[4pt]
&=\left(-{\small{\frac{1}{8}}}(3x^2-9x+8)\right)(x+1)^3+1\\[4pt]
&=-{\small{\frac{1}{8}}}(3x^5-10x^3+15x)\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansion... Also got no more clues.
| We can prove that
$$f^{(n)}(x) = - \frac {(2n - 3)!!\, (x - 2n)} {2^n (1 - x)^{(2n + 1)/2}}$$
for $n \ge 2$ by induction on $n$.
The base case is easy. For the inductive step,
$$\begin{align*}
\frac d {dx} f^{(n)} (x) & = - \frac {(2n - 3)!!\, 2^n (1 - x)^{(2n + 1)/2} + (2n - 3)!!\, (x - 2n) 2^n \frac {2n+1} 2 (1 - x)^{(2n - 1)/2}} {2^{2n} (1 - x)^{2n + 1}} \\
& = - \frac {(2n - 3)!!\, 2^{n-1} (1 - x)^{(2n - 1)/2} \, [2 (1 - x) + (x - 2n) (2n + 1)]} {2^{2n} (1 - x)^{2n + 1}} \\
& = - \frac {(2n - 3)!!\, (2n - 1)(x - 2n - 2)} {2^{n+1} (1 - x)^{(2n + 3)/2}} \\
& = - \frac {[2(n + 1) - 1]!!\, [x - 2(n + 1)]} {2^{n+1} (1 - x)^{[2(n + 1) + 1]/2}}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does there exist a right triangle with area 7 and perimeter 12? This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then
$$a^2 + b^2 = c^2,$$
$$a + b +c = 12,$$
$$ab = 14$$
It is easy (although a bit boring and long) to see that there are no real solutions to this system.
But I feel that there is a simple answer to this question - perhaps using the triangle inequality - but I just cannot see it.
| Given a general triangle with
inradius $r$, circumradius $R$
and semiperimeter $\rho=\tfrac12(a+b+c)$,
the lengths of its sides $a,b,c$ can be found as
three roots of the cubic equation
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R
&=0
\tag{1}\label{1}
.
\end{align}
For the right triangle
the equation \eqref{1}
can be simplified using a well-known condition
\begin{align}
R&=\tfrac12(\rho-r)
\tag{2}\label{2}
\end{align}
as follows:
\begin{align}
(x+r-\rho)(x^2-(r+\rho)x+2r\rho)
&=0
\tag{3}\label{3}
.
\end{align}
The first root, provided by the linear term in \eqref{3}
is the size of hypotenuse,
\begin{align}
c&=\rho-r
,
\end{align}
the other two sides must be
\begin{align}
a,b&=\tfrac12\left(r+\rho\pm\sqrt{r^2-6r\rho+\rho^2}\right)
.
\end{align}
Or, in terms of the area $S$ and perimeter $p$,
\begin{align}
c&=\tfrac12p-\tfrac{2S}{p}
,\\
a,b&=
\frac{4S+p^2\pm\sqrt{p^4-24Sp^2+16S^2}}{4p}
\tag{4}\label{4}
.
\end{align}
So, given $S=7$, $p=12$ we get
\begin{align}
c&=\tfrac{29}6
,\\
a,b&=\tfrac1{12}(43\pm\sqrt{167}\,i)
,
\end{align}
thus, the right triangle with declared properties is impossible.
Using \eqref{4} it is trivial to find out that
the right triangle with $S=7$, $p=14$
has $c=6$, $a,b=4\pm\sqrt2$.
And as a bonus, the right triangle with $p=13$
is also valid and has side lengths
\begin{align}
c&=\tfrac{141}{26}\approx 5.423
,\\
a&=\tfrac1{52}(197-\sqrt{953})\approx 3.195
,\\
b&=\tfrac1{52}(197+\sqrt{953})\approx 4.382
\end{align}
and is just slightly bigger than
the famous $3-4-5$ right triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 5
} |
Elliptic integral of a quartic I am trying to compute the following integral:
\begin{equation}
I(a) = \int_{-\infty}^{+\infty} \left[ 1-x^2 + \sqrt{(1-x^2)^2 + a} \right] dx
\end{equation}
where $a >0$.
I am pretty confident this integral is well defined, as it basically looks like a bell-shaped curve that behaves like $\frac{1}{x^2}$ as $x$ gets large.
More precisely, I am trying to get the first non-constant term of the Taylor expansion of $I(a)$ as $a \to 0$.
| With CAS help:
$$\int_{-\infty }^{\infty } \left(1-x^2+\sqrt{\left(1-x^2\right)^2+a}\right) \, dx=\\\frac{4}{3} \sqrt[4]{1+a} \left(2 E\left(\frac{1}{2}
\left(1+\frac{1}{\sqrt{1+a}}\right)\right)+\left(-1+\sqrt{1+a}\right) K\left(\frac{1}{2}
\left(1+\frac{1}{\sqrt{1+a}}\right)\right)\right)$$
for $a > 0 $.
where: $E$ and $K$ Elliptic Integral of the Second Kind and Complete Elliptic Integral of the First Kind
Mathematica code:
HoldForm[Integrate[1 - x^2 + Sqrt[(1 - x^2)^2 + a], {x, -Infinity, Infinity}] ==
4/3 (1 + a)^(1/4) (2 EllipticE[1/2 (1 + 1/Sqrt[1 + a])] +
(-1 + Sqrt[1 + a]) EllipticK[1/2 (1 + 1/Sqrt[1 + a])])] // TeXForm
Series[4/3 (1 + a)^(1/4) (2 EllipticE[
1/2 (1 + 1/Sqrt[1 + a])] + (-1 + Sqrt[1 + a]) EllipticK[
1/2 (1 + 1/Sqrt[1 + a])]), {a, 0, 2},
Assumptions -> a > 0] // Normal
Expanding with series at $a=0$
$I(a)=\frac{8}{3}+\frac{1}{2} a (1+4 \ln (2)+\log (4)-\ln (a))+\frac{1}{128} a^2 (17-24
\ln (2)-6 \ln (4)+6 \ln (a))+\text{...}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$dy/dx=\sqrt{x^2+y^2}$
$$\frac{dy}{dx}=\sqrt{x^2+y^2}$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(\frac{dy}{dx}-y)(\frac{dy}{dx}+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $\frac{1}{2}x^2=t$.
Second approach:
$y=x\sinh(u)$ and $x=e(t)$ yields $\frac{du}{dt} + \tanh(u)=e(t)$.
Sorry, I am not yet using the proper format.
| Similar to To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$:
Apply the Euler substitution:
Let $u=y+\sqrt{x^2+y^2}$ ,
Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$
$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=u-\left(\dfrac{u}{2}-\dfrac{x^2}{2u}\right)$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}=\dfrac{u}{2}+\dfrac{x^2}{2u}$
$\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}=\dfrac{u}{2}+\dfrac{x^2+2x}{2u}$
$(u^2+x^2)\dfrac{du}{dx}=u^3+(x^2+2x)u$
Let $v=u^2$ ,
Then $\dfrac{dv}{dx}=2u\dfrac{du}{dx}$
$\therefore\dfrac{u^2+x^2}{2u}\dfrac{dv}{dx}=u^3+(x^2+2x)u$
$(u^2+x^2)\dfrac{dv}{dx}=2u^4+(2x^2+4x)u^2$
$(v+x^2)\dfrac{dv}{dx}=2v^2+(2x^2+4x)v$
Let $w=v+x^2$ ,
Then $v=w-x^2$
$\dfrac{dv}{dx}=\dfrac{dw}{dx}-2x$
$\therefore w\left(\dfrac{dw}{dx}-2x\right)=2(w-x^2)^2+(2x^2+4x)(w-x^2)$
$w\dfrac{dw}{dx}-2xw=2w^2+(4x-2x^2)w-4x^3$
$w\dfrac{dw}{dx}=2w^2+(6x-2x^2)w-4x^3$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{dx}=-\dfrac{1}{z^2}\dfrac{dz}{dx}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{dx}=\dfrac{2}{z^2}+\dfrac{6x-2x^2}{z}-4x^3$
$\dfrac{dz}{dx}=4x^3z^3+(2x^2-6x)z^2-2z$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\left( \frac{x+2}{3-x}-1\right)=x\left(\frac{2x-1}{3-x}\right)=x\left(\frac{2x}{-x}\right)=-2x\rightarrow-\infty$$
The textbook reports that the limit is actually $5$. Why is my solution wrong?
| We have that for $x>3$
$$\log{\left|\frac{x+2}{3-x}\right|}=\log{\left(\frac{x+2}{x-3}\right)}=\log{\left(1+\frac{5}{x-3}\right)}$$
and therefore
$$x\log{\left|\frac{x+2}{3-x}\right|}=\frac{5x}{x-3}\frac{\log{\left(1+\frac{5}{x-3}\right)}}{\frac{5}{x-3}}\to 5\cdot 1=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to find the analytical representation of eigenvalues of the matrix $G$? I have the following matrix arising when I tried to discretize the Green functionοΌ now to show the convergence of my algorithm I need to find the eigenvalues of the matrix $G$ and show it has absolute value less than 1 for certain choices of $N$.
Note that the explicit formula for entry $(i,j)$ is $-i(N+1-j)$ when $i\le j$ and it is symmetric, so we can get the formulas for $i>j$ by interchanging $i$ and $j$ in the $i\le j$ case.
Any one has any ideas about how to find the analytical representation of eigenvalues of the matrix $G$, i,e, the eigenvalues represented by $N$? Thank you so much for any help!
$\begin{pmatrix}
- N & - N + 1 & -N+2 & -N+3 &\ldots & 1(-2) & 1(-1) \\
- N + 1 & 2( - N + 1) & 2(-N+2) & 2(-N+3) &\ddots & 2(-2) & 2(-1) \\
- N + 2 & 2( - N + 2) & 3(-N+2) & 3(-N+3) &\ddots & 3(-2) & 3(-1) \\
- N + 3 & 2( - N + 3) & 3(-N+3) & 4(-N+3) &\ddots & 4(-2) & 4(-1) \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
- 2 & 2(-2) & 3(-2) & 4(-2) &\ddots & ( - 1 + N)( - 2) & ( - 1 + N)( - 1) \\
- 1 & 2(-1) & 3(-1) & 4(-1) &\ldots & ( - 1 + N)( - 1) & N( - 1) \\
\end{pmatrix}$
| Eigenvalues of matrix $G$ are
$$\lambda_k(G)=-\tfrac{1}{4}(N+1)\sin^{-2}\left(\frac{k\pi}{2(N+1)}\right)\,,\quad k=1,\ldots,N\,.$$
To prove the statement, we start by writing elements of matrix $G$ as $g_{ij}=ij-(N+1)\min(i,j)$.
Next, we define matrix $L$ as a lower triangular matrix with all elements in the lower triangle equal to one, i.e. $l_{ij}=1$ for $i\geq j$ and $l_{ij}=0$ for $i<j$, and we define vector $\mathbf{1}$ as a vector of ones.
This enables us to write
$$G=(L\mathbf{1})(L\mathbf{1})^T-(N+1)LL^T = L(\mathbf{1}\mathbf{1}^T-(N+1)I)L^T\,.$$
Since $L^{-1}$ is bidiagonal matrix with $1$ on the diagonal and $-1$ on the subdiagonal, we hoped inverting $G$ would result in a matrix with more zero elements.
Using Sherman-Morrison formula to invert $\mathbf{1}\mathbf{1}^T-(N+1)I$, we obtain
$$G^{-1}=\frac{-1}{N+1}L^{-T}(I+\mathbf{1}\mathbf{1}^T)L^{-1} = \frac{-1}{N+1}\begin{bmatrix}\hphantom{-}2 & -1 & & 0\\-1 & \ddots & \ddots\\ & \ddots & \ddots & -1\\0 & & -1 & \hphantom{-}2\end{bmatrix}\,.$$
This matrix is a tridiagonal Toeplitz matrix, and there is analytical expression for eigenvalues of such matrices, see this link for example. In our case
$$\lambda_k(G^{-1}) = \frac{-2}{N+1}+\frac{2}{N+1}\cos\left(\frac{k\pi}{N+1}\right) = \frac{-4}{N+1}\sin^2\left(\frac{k\pi}{2(N+1)}\right)\,,\quad k=1,\ldots,N\,.$$
Now, our statement follows from eiganvalues of $G$ being reciprocals of eigenvalues of $G^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$
| Hint: It is $$\sqrt{4x^2+7x}=\sqrt{x^2\left(4+\frac{7}{x}\right)}=-x\sqrt{4+\frac{7}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $ x,y β\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y β\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3yΒ±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3Β±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3Β±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.
| Note that $\sqrt{25y^2+56}$ is an integer if and only if $(5y)^2+56=z^2$ for some integer $z$. This limits how big $5y$ can be. After all;
$$29^2=(28+1)^2=28^2+2\cdot28+1=28^2+57>28^2+56,$$
so surely $z^2<29^2$ and $(5y)^2<28^2$. This only leaves a few values of $y$ to try.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $\binom{n}{4}$ if $\binom{n}{6}=\binom{n}{9}$ I was trying to solve a question related to combinatorics as follows :
A group contains $n$ persons. If the number of ways of selecting $6$ persons is equal to the number of ways of selecting $9$ persons, then find the number of ways of selecting $4$ persons from the group.
I dont have a step by step solution for concluding that $n=15$.
I tried to solve for $n$ but I am stuck at $(n-9)(n-8)(n-7)=(1/504)$
| If $n<6$ then the number of ways to select $6$ people is the same as the number of ways to select $9$ people; in both cases this is $0$. For $n<6$ we can easily compute $\binom{n}{4}$:
$$\binom{5}{4}=5,\qquad\binom{4}{4}=1,$$
and $\binom{n}{4}=0$ when $n<4$.
If $n>6$, then by selecting $6$ people we leave $n-6$ people out. So the number of ways to select $6$ people is the same as the number of ways to select $n-6$ people. This suggests that $n-6=9$ and so $n=15$, and $\binom{15}{4}=1365$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Differentiate $\frac{x^3}{{(x-1)}^2}$
Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$
I start by finding the derivative of the denominator, since I have to use the chain rule.
Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:
$$\frac{d}{dx}(x-1)^{2}=2(x-1)$$
After having found the derivative of the denominator I find the derivative of the numerator, which is $3x^2$. With the two derivatives found I apply the quotient rule, which states that
$$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v'u-vu'}{v^2}$$
and substitute in the numbers
$$\frac{d}{dx}\frac{x^3}{(x-1)^2}=\frac{3x^2(x-1)^2-2x^3(x-1)}{(x-1)^4}$$
Can I simplify this any further?Is the derivation correct?
| You're mixing the product rule and the quotient rule. You can apply each of them, but not simultaneously.
*
*by the product rule: remember $\dfrac{\mathrm d}{\mathrm dx}\Bigl(\dfrac1{x^n}\Bigr)=-\dfrac n{x^{n+1}}$, so
\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)&=\dfrac{\mathrm d}{\mathrm dx}(x^3)\cdot\dfrac1{(1-x)^2}+x^3\dfrac{\mathrm d}{\mathrm dx}\dfrac1{(1-x)^2} \\
&= \frac{3x^2}{(1-x)^2}+\frac{2x^3}{(1-x)^3} =\frac{x^2\bigl(3(1-x)+2x\bigr)}{(1-x)^3}\\&
=\frac{x^2(3-x)}{(1-x)^3}.
\end{align}
*by the quotient rule:
$$\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)=\frac{3x^2(1-x)^2+x^3\cdot2(1-x)}{(1-x)^4}=\frac{x^2\color{red}{(\not1-\not x)}\bigl(3(1-x)+2x\bigr)}{( 1-x)^{\color{red}{\not4}\,3}}=\dots$$
*Since there are exponents, logarithmic differentiation may help make it shorter. It's simpler to use it with Lagrange's notations: set $f(x)=\dfrac{x^3}{(1-x)^2}$. Then
$$\frac{f'(x)}{f(x)}=\frac 3x+\frac2{1-x}=\frac{3-x}{x(1-x)}, \quad\text{so }\;f'(x)=\frac{f'(x)}{f(x)}\cdot f(x)=\dotsm $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the solution.
| Radius is exactly $\dfrac{72}{23}$, pretty neat. See the figure below, where dotted lines from the center of the circumcribing circle passes through the midpoints of the sides of the triangle.
Let : $B$ be the origin $(0,0)$, and the center of circle of unknown radius $r$ be $(x,y)$. Then we solve the following three equations to find $r$
$\dfrac{3}{2}+\sqrt{x^2+(\dfrac{3}{2}-y)^2}=r$
$2+\sqrt{(2-x)^2+y^2}=r$
$\dfrac{5}{2}+\sqrt{(2-x)^2+(\dfrac{3}{2}-y)^2}=r$
so that $(r, x, y)= \left(\dfrac{72}{23}, \dfrac{36}{23}, \dfrac{24}{23}\right)$
Note: The midpoints of the sides $(0,\dfrac{3}{2})$, $(2,0)$, and $(2,\dfrac{3}{2})$ are centers of circles of radii $\dfrac{3}{2}, 2, \dfrac{5}{2}$.
Given any two tangent circles, their centres and point of tangency are colinear.
Bonus: If right triangle : $\triangle ABC$ has rational sides, then $(r,x,y)$ are also all rationals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 1
} |
How to determine continuity at origin for given 2 variable function Consider the function,
$f(x,y)=\dfrac{x^3\cos\left(\frac1y\right)+y^3\cos\left(\frac1x\right)}{ x^2+y^2},\qquad\forall x,y\neq0$
And $f(x,y)= 0,\qquad\text{ if } xy=0$
Clearly this function is not continuous at either $x-$ or $y-$ axis
As, when approaching $x-$axis,
$f(x,y)=\dfrac{x^3\cos\left(\frac1y\right)}{ x^2} = x\cos\left(\dfrac1y\right) \neq 0$
But how do you tell at origin. Approaching O via path $y=mx$ does not prove that it is discontinuous.
| By inequalities and polar coordinates we have that
$$0\le \left|\frac{x^3\cos(1/y)+y^3\cos(1/x)}{ x^2+y^2}\right|\le \left|\frac{x^3\cos(1/y)}{ x^2+y^2}\right|+\left|\frac{y^3\cos(1/x)}{ x^2+y^2}\right|\le$$$$\le \left|\frac{x^3}{ x^2+y^2}\right|+\left|\frac{y^3}{ x^2+y^2}\right|=r\left(|\cos^3\theta|+|\sin^3\theta|\right)\to 0$$
For the limit when approching the $x$ axis at a point $x=a\neq 0$ we have that
$$\lim_{y\to 0} \frac{a^3\cos(1/y)+y^3\cos(1/a)}{ a^2+y^2}=\lim_{y\to 0} \frac{a^3\cos(1/y)}{ a^2+y^2}+\lim_{y\to 0} \frac{y^3\cos(1/a)}{ a^2+y^2}$$
and the second limit is equal to $0$ but the first one doesn't exist indeed for
*
*$y_n=\frac1{2\pi n}\to 0 \implies \frac{a^3\cos(2\pi n)}{ a^2+\frac1{4\pi^2 n^2}}\to a$
*$y_n=\frac1{(2n-1)\pi}\to 0 \implies \frac{a^3\cos((2n-1)\pi)}{ a^2+\frac1{(2n-1)^2\pi^2}}\to -a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding sum of a geometric series I am asked to find the summation of $1/3^n$ from $n=5$ to infinity.
I have done the calculation: $1/(1-r)$, for $r=1/3$, and received $1.5$. As this summation starts from $5$, I subtracted $3^0, 3^-1, 3^-2, 3^-3$ and $3^-4$ from $1.5$ and got $6.17e-3$. However, apparently this answer is wrong, and so is the answer 0.
I appreciate any help,
thank you!
| $$1,\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4} \ldots$$
You need to calculate $$S_{\infty}-S_5$$
$$=\frac{1}{1-\frac{1}{3}}-\frac{1(1-\frac{1}{3^5})}{1-\frac{1}{3}}$$
$$=\frac{\frac{1}{3^5}}{\frac{2}{3}}$$
$$=\frac{1}{2\cdot{3^4}}$$
$$=\frac{1}{162}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$
| Check again it should be
$$=\int \left(x^{-\frac{1}{2}}-4x^{-\frac{3}{2}}+10x^{-\frac{5}{2}}\right)dx$$
indeed $\frac{1}{x^2\sqrt x}=x^{-\frac{5}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.
| Hint:
In general for
$\displaystyle\dfrac{d\left(\dfrac{a\sin x+b\cos x}{c\sin x+d\cos x}\right)}{dx}$
write numerator$(a\sin x+b\cos x)$ as $A(c\sin x+d\cos x)+B\cdot\dfrac{d(c\sin x+d\cos x)}{dx}$
Compare the coefficients of $\sin x,\cos x$ to find $A,B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Sum the series $\frac{3}{1β
2β
4}+\frac{4}{2β
3β
5}+ \frac{5}{3β
4β
6}+...\text{(upto n terms)}$ $\frac{3}{1β
2β
4}+\frac{4}{2β
3β
5}+ \frac{5}{3β
4β
6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression.
Do I have to use something else? Or am I missing some easy manipulation?
| We have that
$$a_r=\frac{r + 2}{r (r + 1) (r + 3)} = \frac2{3 r}-\frac1{2 (r + 1)} - \frac1{6 (r + 3)}=$$$$= \frac1{2r}-\frac1{2 (r + 1)}+ \frac1{6r} - \frac1{6 (r + 3)}=$$
$$= \frac12\left(\frac1{r}-\frac1{r + 1}\right)+ \frac16\left(\frac1{r} - \frac1{r + 3}\right)$$
and therefore
$$\sum_1^n a_r=\frac12 \sum_1^n \left(\frac1{r}-\frac1{r + 1}\right)+\frac16\sum_1^n\left(\frac1{r} - \frac1{r + 3}\right)=$$
$$=\frac12\left(1-\frac1{n+1}\right)+\frac16\left(1+\frac12+\frac13-\frac1{n+1}-\frac1{n+2}-\frac1{n+3}\right)=$$
$$=\frac{29}{36}-\frac2{3(n+1)}-\frac1{6(n+2)}-\frac1{6(n+3)}=$$
$$=\frac{29 n^3+138 n^2+157 n}{36(n+1)(n+2(n+3)}$$
See also the related Sum the series $\frac{n}{1β
2β
3}+\frac{n-1}{2β
3β
4}+ \frac{n-2}{3β
4β
5}+...\text{(upto n terms)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$ QUESTION
Find Where The Curvature has an extremum ?
$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$$
MY APPROACH
$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}. . . . . (1)$$
$$\Rightarrow y^\frac{1}{2}=a^\frac{1}{2}-x^\frac{1}{2}$$
now differntiating both sides we will get:
$$\frac{1}{2}{y^\frac{-1}{2}}\frac{dy}{dx}=(-1)\frac{1}{2}x^\frac{-1}{2}$$
$$\Rightarrow \frac{dy}{dx}=-(\frac{y}{x})^\frac{1}{2}. . . . . .(2)$$
Now differentiating again with respect to x again:
$$\frac{d^2y}{dx^2}=-\bigg(\frac{1}{2}(\frac{y}{x})^\frac{-1}{2}.\frac{d}{dx}(\frac{y}{x})\bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{d}{dx}(\frac{1}{x})y-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{-y}{x^2}-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
now put the value of $\frac{dy}{dx}$ :
$$=\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{y}{x^2}-(\frac{y}{x})^\frac{1}{2}\frac{1}{x}\bigg)\Bigg)$$
After simplifying i got :
$$\frac{d^2y}{dx^2}=\frac{1}{2x}\bigg(\frac{y^\frac{1}{2}}{x^2}-1\bigg). . . . .(3)$$
but from simplification of equation (2) in terms of $a$ wil result :
$$\frac{dy}{dx}=1-(\frac{a}{x})^\frac{1}{2}$$
here i can easily simlify this to get $\frac{d^2y}{dx^2}$ i.e.
$$\Rightarrow\frac{d^2y}{dx^2}=\frac{\sqrt a}{2x\sqrt x}. . . . .(4)$$
I dont know why i am unable to reduce (3) to (4).May be there exists some calculation error,even thats not my question.
proceeding to find radius of curvature and curvature :
FROM FORMULA
$$\rho=\frac{\bigg(1+(\frac{dy}{dx})^2\bigg)^\frac{3}{2}}{\frac{d^2y}{dx^2}}$$
putting the value of (2) and (4):
$$\Rightarrow \rho=\frac{\bigg(1+\frac{y}{x}\bigg)^\frac{3}{2}2x\sqrt x}{\sqrt a}$$
$$\Rightarrow\rho=\frac{2(x+y)^\frac{3}{2}}{\sqrt a}$$
so curvature is
$$\frac{1}{\rho}=\kappa=\frac{\sqrt a}{2(2x+a-2\sqrt a\sqrt x)^3/2}$$
[notice that i have put y in terms of a nad x]
NOW BEGINS THE PROBLEM
for being extremum
$\frac{d\kappa}{dx} =0 $ and i have to check the sign of $\frac{d^2\kappa}{dx^2}$ :
as you can see in the image :
$$\frac{d\kappa}{dx}=\frac{3\sqrt a(2-\frac{\sqrt a}{\sqrt x})}{4(2x-2\sqrt x\sqrt a+a)^\frac{3}{2}}$$
Now letting this to zero we have :
$$2=\frac{\sqrt a}{\sqrt x}$$
thus i am getting $x=\frac{a}{4}$ as a critical point.
BUT the answer is given as $\frac{\sqrt 2}{a}$
you can even see by inspection $x=\frac{a}{4}$ is not a critical point.
please help and let me know where i have made mistake.
THIS IS MY HUMBLE REQUEST.
| the equation $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ is equivalent to $$y=x-2\sqrt{ax}+a,~~~0\leq x \leq a.$$
Hence, $$y'=1-\frac{a}{\sqrt{ax}},~~~y''=\frac{a}{2x\sqrt{ax}}.$$
Therefore, $$k=\frac{|y'|}{(1+y''^2)^{3/2}}=\frac{1}{2}\sqrt{\frac{a}{(2x-2\sqrt{ax}+a)^3}}.$$
Notice that $$2x-2\sqrt{ax}+a=2\left(\sqrt{x}-\frac{\sqrt{a}}{2}\right)^2+\frac{a}{2}\geq \frac{a}{2}$$ with the equality holding if and only if $\sqrt{x}=\dfrac{\sqrt{a}}{2}$, namely $x=\dfrac{a}{4}$. As a result, $k$ takes its maximum value $k=\dfrac{\sqrt{2}}{a}$ at $x=\dfrac{a}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum of Lagrange polynomials: $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $. Given $\{x_0,...,x_n\}$ I am asked to show that $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $.
I already showed that $\sum_{i=0}^{n}L_i(x)x_i^{j} = x^j$ for $j=1,...,n$ and that $\sum_{i=0}^{n}L_i(x)=1$ but I don't know how to use this to solve my problem.
If I use the same method by which I solve the other problems then I am basically looking for a polynomial P of degree less than n such that $P(x_i) = x_i^{n+1}$ and
$P(0) = (-1)^{n}x_0\cdot\cdots\cdot x_n$
| It can be proven with determinant form of Lagrange polynomial that interpolates $(x_0;y_0)$, $\dots$, $(x_n;y_n)$
$$
P(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & y_0 & y_1 & \cdots & y_n \\
x^n & x_0^n & x_1^n & \cdots & x_n^n \\
x^{n-1} & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
},
$$
presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation".
Obviously, setting all $y_i = x_i^{n+1}$ and $x = 0$ we'll get the expression you need, because
$$
P(x) = \sum_{i=0}^n\, y_i L_i(x) = \sum_{i=0}^n x_i^{n+1} L_i(x).
$$
So I just plug them in
$$
P(0) = (-1)
\frac{
\det
\begin{pmatrix}
0 & x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\
0 & x_0^n & x_1^n & \cdots & x_n^n \\
0 & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}
$$
Now consider Laplace expansion along the first column
$$
P(0) = (-1)^n
\frac{
\det
\begin{pmatrix}
x_0^{n+1} & x_1^{n+1} & \cdots & x_n^{n+1} \\
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
x_0 & x_1 & \cdots & x_n \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}
$$
and use properties of determinant
$$
P(0) = (-1)^n x_0 x_1 \dots x_n
\frac{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}.
$$
The latter finishes the proof
$$
P(0) = (-1)^n x_0 x_1 \dots x_n
$$
For more practical examples you may want to check "Workbook on mapping simplexes affinely", section "Lagrange interpolation".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to find such integers $y,b$ that $\left|-\frac{x_{0}}{n}-\frac{y}{b}\right|<\frac{1}{b\sqrt{n}}$ Given integers $x_{0},n$ with $x_{0}^{2}\equiv -1$ (mod $n$) then there are integers $y,b$ with $(y,b)=1,0<b\le\sqrt{n}$ and
$$\left|-\frac{x_{0}}{n}-\frac{y}{b}\right|<\frac{1}{b\sqrt{n}}$$
I tried to solve $|bx_{0}+ny|<\sqrt{n}$ for some integers $y,b$, but it seems do nothing.
Is there any hint that I can follow ?
Any comment and advise I will be appreciated.
|
Lemma. If $m\mid(x^2+1)$, then there exists $y,d$ such that
*
*$m=y^2+d^2$
*$xy\equiv d\pmod m$.
Proof.
Let $x^2+1=mn$.
If $m$ is prime, then $m=2$ or $m\equiv 1\pmod 4$, hence $m$ is the sum of two squares $m=y^2+d^2$.
Since $x^2\equiv -1\pmod{m}$, we have
$$(xy+d)(xy-d)=x^2y^2-d^2\equiv 0\pmod{m}$$
hence (wlog) $m\mid(xy-d)$.
Let $m_1m_2\mid(x^2+1)$ and assume $m_j=y_j^2+d_j^2$ and $xy_j\equiv d_j\pmod{m_j}$ for $j\in\{1,2\}$.
Then
$$m_1m_2=(y_1^2+d_1^2)(y_2^2+d_2^2)=(y_1y_2-d_1d_2)^2+(y_1d_2+d_1y_2)^2$$
and
$$0\equiv -(xy_1-d_1)(xy_2-d_2)\equiv (y_1d_2+y_2d_1)x+(y_1y_2-d_1d_2)\pmod{m_1m_2}$$
thus proving the lemma.
Proof of the main question.
The assertion is obvious for $n=1$, hence we assume $n>1$.
Let $x^2+1=mn$ and let $y,d$ as in the lemma.
Let $xy-d=bm$ and $a=bx-ny$.
Then
$$m(a^2+b^2-n)=n((mb-xy)^2-d^2)=0$$
proves $a^2+b^2=n$, hence $|a|,|b|\leq\sqrt n$.
If $b=0$, then $d=xy$ hence $m=y^2+d^2=y^2+x^2y^2=y^2(1+x^2)=y^2nm$ hence $y^2n=1$ from which $n=1$.
If $a=0$, then $bx=ny$, hence $x^2y-xd=x(xy-d)=xbm=nym=y(x^2+1)$ from which $y=-xd$ hence $m=y^2+d^2=x^2d^2+d^2=(x^2+1)d^2=mnd^2$ from which $nd^2=1$ that's $n=1$.
This proves $0<|a|,|b|<\sqrt n$ from which
$$\left|\frac xn-\frac yb\right|=\frac{|a|}{n|b|}<\frac 1{|b|\sqrt n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the smallest four digit number which is divisible by $15,25,40$ and $75$ I'm stuck on this question. My working:
\begin{align*}
15 & = 3 \cdot 5\\
25 & = 5^2\\
40 & = 2^3 \cdot 5\\
75 & = 3 \cdot 5^2
\end{align*}
LCM $= 600$
And I'm not sure what to do after this (if the above steps are right).
| The answer is $\boxed{1200}$. Following your method, we have
$$15 = 3 \cdot 5 \\ $$
$$25 = 5^2 \\$$
$$40 = 5^1 \cdot 2^3 \\$$
$$75 = 3 \cdot 25 $$
Thus, to find the LCM, we take the maximum exponents for each of the prime factors, and we obtain $600$. But, since we need a four-digit number, we can multiply by $2$ to obtain $1200$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$ is equicontinous on $[0,+\infty)$ Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$
*
*is equicontinous on $[0,+\infty)$.
*converges pointwise to $0$ on $[0,+\infty)$.
MY TRIAL
*\begin{align}\sqrt{x+4n^2\pi ^2}&=\sqrt{x+4n^2\pi ^2}\left[\dfrac{\sqrt{x+4n^2\pi ^2}+2n\pi}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=\left[\dfrac{x+4n^2\pi ^2+2n\pi\sqrt{x+4n^2\pi ^2}}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=\left[\dfrac{4n^2\pi ^2+2n\pi\sqrt{x+4n^2\pi ^2}+x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\right]\\&=2n\pi+\dfrac{x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\end{align}
As $n\to \infty,$
\begin{align}\dfrac{x}{\sqrt{x+4n^2\pi ^2}+2n\pi}\to 0\end{align}
Hence, $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\to 0$ as $n\to 0$
1.)
Let $x,y\in [0,+\infty)$ such that $|x-y|<\delta$. We show that \begin{align}\left|f_n(x)-f_n(y)\right|<\epsilon\end{align}
Now,
\begin{align}\left|f_n(x)-f_n(y)\right|=\left|\sin{\sqrt{x+4n^2\pi ^2}}-\sin{\sqrt{y+4n^2\pi ^2}}\right|\end{align}
I'm stuck here! Please, how should I go?
| By mean value theoremοΌ
$$\left|f_n(x)-f_n(y)\right|=\left|\sin{\sqrt{x+4n^2\pi ^2}}-\sin{\sqrt{y+4n^2\pi ^2}}\right|= |f_n'(\xi_n)||x-y|\leq|x-y|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$
Evaluate $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$ (where $ \lfloor \,\cdot\, \rfloor $ denotes greatest integer function).
Can anyone please give a detailed explanation of how to do this?
This is my first question on MathStack Exchange.
Thank You
| For a given $a\geq0$ there should $\exists k \in \mathbb{N}: \color{red}{k}\leq a^{n}<\color{red}{k}+1$ (i.e. $k=\left \lfloor a^n\right \rfloor$) then
$$\int\limits_0^a \lfloor x^n \rfloor dx=
\int\limits_0^1 \lfloor x^n \rfloor dx+
\int\limits_1^{\sqrt[n]{2}} \lfloor x^n \rfloor dx+
\int\limits_{\sqrt[n]{2}}^{\sqrt[n]{3}} \lfloor x^n \rfloor dx+...+
\int\limits_{\sqrt[n]{\color{red}{k}}}^{a} \lfloor x^n \rfloor dx\tag{1}$$
and we have
*
*for $x\in[0,1) \Rightarrow x^n\in[0,1) \Rightarrow \lfloor x^n \rfloor=0 \Rightarrow \int\limits_0^1 \lfloor x^n \rfloor dx=0$
*for $x\in[1,\sqrt[n]{2}) \Rightarrow x^n\in[1,2) \Rightarrow \lfloor x^n \rfloor=1 \Rightarrow \int\limits_1^{\sqrt[n]{2}} \lfloor x^n \rfloor dx=\sqrt[n]{2}-1$
*for $x\in[\sqrt[n]{2},\sqrt[n]{3}) \Rightarrow x^n\in[2,3) \Rightarrow \lfloor x^n \rfloor=2 \Rightarrow \int\limits_{\sqrt[n]{2}}^{\sqrt[n]{3}} \lfloor x^n \rfloor dx=2\left(\sqrt[n]{3}-\sqrt[n]{2}\right)$
*...
*for $x\in[\sqrt[n]{t},\sqrt[n]{t+1}) \Rightarrow x^n\in[t,t+1) \Rightarrow \lfloor x^n \rfloor=t \Rightarrow \int\limits_{\sqrt[n]{t}}^{\sqrt[n]{t+1}} \lfloor x^n \rfloor dx=t\left(\sqrt[n]{t+1}-\sqrt[n]{t}\right)$
*...
*for $x\in[\sqrt[n]{\color{red}{k}},a) \Rightarrow x^n\in[\color{red}{k},a^n) \Rightarrow \lfloor x^n \rfloor=\color{red}{k} \Rightarrow \int\limits_{\sqrt[n]{\color{red}{k}}}^{a} \lfloor x^n \rfloor dx=\color{red}{k}\left(a-\sqrt[n]{\color{red}{k}}\right)$
then $(1)$ becomes
$$\int\limits_0^a \lfloor x^n \rfloor dx=
\left(\sum\limits_{t=1}^{\color{red}{k}-1}t\left(\sqrt[n]{t+1}-\sqrt[n]{t}\right)\right) + \color{red}{k}\left(a-\sqrt[n]{\color{red}{k}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find this inverse matrix using Gauss-Jordan? I am trying to find the inverse matrix of
$$\begin{pmatrix} \ln\left(x\right) & -1\\ \:\:1 & \ln\left(x\right) \end{pmatrix}$$
using the Gauss-Jordan method. Using a different method I could already find that the inverse matrix is:
$$\frac{1}{\ln ^2\left(x\right)+1}\begin{pmatrix}\ln \left(x\right)&-\left(-1\right)\\ -1&\ln \left(x\right)\end{pmatrix}=\begin{pmatrix}\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}&\frac{1}{\ln ^2\left(x\right)+1}\\ -\frac{1}{\ln ^2\left(x\right)+1}&\frac{\ln \left(x\right)}{\ln ^2\left(x\right)+1}\end{pmatrix}$$
| Suppose that the matrix $A$ is given by
$$ \begin{bmatrix} \ln(x) & -1 \\ 1 & \ln(x) \end{bmatrix} \tag{1} $$
using Gaussian Elimination
$$ U=A, L=I \tag{2} $$
$$ \ell_{21} = \frac{u_{21}}{u_{11}} = \frac{1}{\ln(x)} \tag{3} $$
the point is to find the coefficient to zero the column
$$ u_{2,1:2} = u_{2,1:2} -\frac{1}{\ln(x)}u_{1,1:2} \tag{4}$$
$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} -\frac{1}{\ln(x)}\begin{bmatrix} \ln(x) & -1 \end{bmatrix} \tag{5}$$
which gives us
$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{6} $$
$$ u_{2,1:2} = \begin{bmatrix} 1 & \ln(x) \end{bmatrix} - \begin{bmatrix} 1 & -\frac{1}{\ln(x)}\end{bmatrix} \tag{7} $$
$$ u_{2,1:2} = \begin{bmatrix} 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{8} $$
updating the matrix
$$ U = \begin{bmatrix} \ln(x) & -1 \\ 0 & \frac{\ln^{2}(x)+1}{\ln(x)} \end{bmatrix} \tag{9} $$
$$ L = \begin{bmatrix} 1 & 0 \\ \frac{1}{\ln(x)} & 1 \end{bmatrix} \tag{10} $$
$$ A = LU \tag{11} $$
So you'd find $U^{-1}L^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\pi/4} y(\sin y+\cos y)^n dy.
\end{equation*}
Although some terms may be canceled, the summation of the others seems terrible.
Another method is integration by parts. Here I found the recurrence formula for $\sin^n x$:
Integrating $\int \sin^n{x} \ dx$
So the first step is to have
\begin{equation*}
\left.\left(-\frac{1}{n}x\cos x\sin^{n-1} x+\frac{n-1}{n}x\int \sin^{n-2} xdx\right)\right|_0^{\pi/2}-\int_0^{\pi/2}\left(-\frac{1}{n}\cos x\sin^{n-1} x+\frac{n-1}{n}\int_0^x \sin^{n-2} t dt\right)dx
\end{equation*}
the evaluations at $0$ and $\pi/2$ and the integral of $\cos x\sin^{n-1}x$ seem OK, but what about the second term? Any better method for this problem?
| $I_0=\frac{\pi^2}8$ and $I_1=1$ and for $n\ge2$,
$$
\begin{align}
I_n
&=\int_0^{\pi/2}x\sin^n(x)\,\mathrm{d}x\tag1\\
&=-\int_0^{\pi/2}x\sin^{n-1}(x)\,\mathrm{d}\cos(x)\tag2\\
&=(n-1)\int_0^{\pi/2}x\cos^2(x)\sin^{n-2}(x)\,\mathrm{d}x+\int_0^{\pi/2}\cos(x)\sin^{n-1}(x)\,\mathrm{d}x\tag3\\
&=(n-1)\int_0^{\pi/2}x\sin^{n-2}(x)\,\mathrm{d}x-(n-1)I_n+\frac1n\tag5\\
&=\frac{n-1}nI_{n-2}+\frac1{n^2}\tag5\\
\end{align}
$$
Explanation:
$(2)$: prepare to integrate by parts
$(3)$: integrate by parts
$(4)$: $\cos^2(x)=1-\sin^2(x)$ and the $\sin^2(x)$ part replicates $I_n$
$(5)$: add $(n-1)I_n$ to both sides and divide by $n$
Using $I_0$ and $(5)$, we can compute $I_{2n}$:
$$
\begin{align}
I_{2n}
&=\frac{(2n-1)!!}{(2n)!!}\left[\frac{\pi^2}8+\sum_{k=0}^{n-1}\frac{(2n-2k-2)!!}{(2n-2k-1)!!}\frac1{2n-2k}\right]\\
&=\frac{\binom{2n}{n}}{4^n}\left[\frac{\pi^2}8+\sum_{k=0}^{n-1}\frac{4^{n-k}}{\binom{2n-2k}{n-k}}\frac1{(2n-2k)^2}\right]
\end{align}
$$
Using $I_1$ and $(5)$, we can compute $I_{2n+1}$:
$$
\begin{align}
I_{2n+1}
&=\frac{(2n)!!}{(2n+1)!!}\left[1+\sum_{k=0}^{n-1}\frac{(2n-2k-1)!!}{(2n-2k)!!}\frac1{2n-2k+1}\right]\\
&=\frac1{2n+1}\frac{4^n}{\binom{2n}{n}}\left[1+\sum_{k=0}^{n-1}\frac{\binom{2n-2k}{n-k}}{4^{n-k}}\frac1{2n-2k+1}\right]
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solution for Simultaneous Logarithmic Equations Following are the Equations:
\begin{align}
\log x +\frac {\log(xy^8)}{((\log x)^2+(\log y)^2)} &= 2 \\
\log y + \frac{\log(x^8/y)}{((\log x)^2 +(\log y)^2)} &= 0
\end{align}
I tried substituting $\log x$ and $\log y$ with $a$ and $b$ but it results in a cubic equation with two variables .
| These are
$\log x +\dfrac{\log(x/y^8)}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{\log((x^8)y)}{(\log x)^2 +(\log y)^2)}
=0
$.
Expanding the logs,
$\log x +\dfrac{\log x-8\log y}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{8\log x+\log y}{(\log x)^2 +(\log y)^2)}
=0
$.
Letting
$\log x = a, \log y = b$,
$a +\dfrac{a-8b}{a^2+b^2}
= 2
$
and
$b+\dfrac{8a+b}{a^2 +b^2)}
=0
$.
The 8 is a mysterious constant,
so replace it by $c$.
$a +\dfrac{a-cb}{a^2+b^2}
= 2
$
and
$b+\dfrac{ca+b}{a^2 +b^2}
=0
$.
Now, solve.
Looks like we will get a cubic,
like you wrote.
$2(a^2+b^2)
=a(a^2+b^2) +a-cb
$
and
$0=
b(a^2+b^2)+ca+b
$.
From the first,
$a^2+b^2
=\dfrac{cb-a}{a-2}
$.
Putting this in the second,
$\begin{array}\\
0
&=b\dfrac{cb-a}{a-2}+ca+b\\
&=\dfrac{b(cb-a)+(ca+b)(a-2)}{a-2}\\
&=\dfrac{cb^2-ab+ca^2-2ca+ab-2b}{a-2}\\
&=\dfrac{cb^2+ca^2-2ca-2b}{a-2}\\
\end{array}
$
so,
if $a \ne 2$,
$0
=cb^2+ca^2-2ca-2b
$.
If $a=2$,
$0
=b(a^2+b^2)+ca+b
=b(4+b^2)+2c+b
$
so
$0
=b^3+5b+2c
$.
For $c=8$,
this has a negative real
(about -1.8771)
and two complex roots.
This looks like a mess,
so I am probably
doing something wrong,
so I'll stop here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$
So I have forgotten how to do this. Do I:
- reduce $A$ to reduced row-echelon form (why do I do this? Is it because it's easy to solve once you have pivot columns and free variables)?
- rewrite the system of equations
Is this right:
\begin{align}
A &= \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & -5 & -2 & 7 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 2 & 1 & -3 \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix} \\
&= \begin{bmatrix} 1 & 0 & \frac{1}{5} & \frac{-29}{5} \\ 0 & 1 & \frac{2}{5} & \frac{7}{5} \end{bmatrix}
\end{align}
so: $v_4 = t, v_3 = s, v_2 = \frac{-2}{5}s - \frac{7}{5}t, v_1 = \frac{-1}{5}s + \frac{29}{5}t$
$$\begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} = \begin{bmatrix} \frac{-1}{5}s + \frac{29}{5}t \\ \frac{-2}{5}s - \frac{7}{5} t \\ s + 0t \\ 0 + t \end{bmatrix} = s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix} $$
Is this the set of all $v$?
EDIT
I messed up, the first $\frac{7}{5}$ should be a $\frac{-7}{5}$
| You found the solutions to the homogeneous system $Tv = 0$.
The solutions to the system $Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ are of the form $$\left(\text{one particular solution to }Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right) \quad + \quad\left(\text{ any solution of the homogeneous system }\right)$$
We can guess one solution as $v = \begin{bmatrix} \frac45 \\ \frac35 \\ 0 \\ 0\end{bmatrix}$ so all solutions are of the form
$$v = \begin{bmatrix} \frac45 \\ \frac35 \\ 0 \\ 0\end{bmatrix} + s \begin{bmatrix} \frac{-1}{5} \\ \frac{-2}{5} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} \frac{29}{5} \\ \frac{-7}{5} \\ 0 \\ 1 \end{bmatrix}$$
for some $s,t \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve, thanks in advance
| Let $x = 2+\cos \theta$ where $0\leq \theta \leq \pi$. Then the equation is equivalent to
$$
(\sin\theta + 4)^{2} + (\cos\theta-3)^{2} = 26 + 8\sin\theta - 6\cos\theta = 26 + 10\sin(\theta+\alpha)
$$
where $-\pi/2 < \alpha <0$ satisfies $\tan \alpha = -3/4$. This attains minimum at $\theta = \pi/2 - \alpha$, which corresponds to $x=7/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$
$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$
I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows
$$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after that I am not able to figure out what to do. If I use double angle formula then
$$\frac{\frac{\sin2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$$ again i am clueless
| Nosrati's idea is good. However his second identity is wrong. Using identities
$$\sin x+\cos x=\sqrt{2}\cos(\dfrac{\pi}{4}-x), \sin x\cos x=\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]$$
and substitution $x\to\dfrac{\pi}{4}-x$ gives
\begin{eqnarray}
I&=&\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx\\
&=&\int_{0}^{\frac{\pi}{4}}\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}dx\\
&=&\frac14\int_{0}^{\frac{\pi}{4}}\frac{\sin^2(2x)}{\sqrt2\cos(\frac{\pi}{4}-x)(1-\frac12[\cos^2(x-\dfrac{\pi}{4})-1])}dx\\
&=&\int_{0}^{\frac{\pi}{4}}\frac{\frac14\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]^2}{\sqrt{2}\cos(\dfrac{\pi}{4}-x)(1-\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big])}dx\\
&=&\frac1{2\sqrt2}\int_{0}^{\frac{\pi}{4}}\frac{(1-2\cos^2x)^2}{\cos x(3-2\cos^2x)}dx\\
&=&\frac1{6\sqrt2}\int_{0}^{\frac{\pi}{4}}\bigg(\frac{(1-2\cos^2x)^2}{\cos x}+\frac{2\cos x(1-2\cos^2x)^2}{3-2\cos^2x}\bigg)dx.
\end{eqnarray}
Noting that
\begin{eqnarray}
\int\frac{(1-2\cos^2x)^2}{\cos x}dx&=&\int(4 \cos ^3(x)-4 \cos (x)+\sec (x))dx\\
&=&\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)+C\\
\int\frac{2\cos x(1-2\cos^2x)^2}{(3-2\cos^2x)}dx&=&2\int\frac{(2\sin^2x-1)^2}{1+2\sin^2x}d\sin x\\
&=&2\int\bigg(-3+2\sin^2x+\frac{4}{1+2\sin^2x}\bigg)d\sin x\\
&=&2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)+C\bigg).
\end{eqnarray}
So
\begin{eqnarray}
I&=&\frac1{6\sqrt2}\bigg(\frac13\sin(3x)-\sin x+\ln(\sec x+\tan x)\\
&&+2\bigg(-3\sin x+\frac{2}{3}\sin^3x+2\sqrt2\arctan(\sqrt2\sin x)\bigg)\bigg)\bigg|_0^{\pi/4}\\
&=&\frac{1}{6\sqrt2}(-3\sqrt2+\pi\sqrt2+\ln(\sqrt2+1)).
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Consider a partition P Consider a partition $P= \{ [0,\frac{2}{7}]\}, \{ [\frac{2}{7}, \frac{1}{2}\}], \{ [\frac{1}{2}, \frac{2}{3}]\}, \{ [\frac{2}{3}, \frac{5}{7}]\}, \{ [\frac{5}{7},1]\} $ of $[0,1].$. Compute $\overline{I}_P(f)$ and $\underline{I}_P(f)$ for $f(x)=(x-\frac{1}{2})^2$
Now I know that:
A partition $P $ of $I$: $P=\{J_k: 0 \leq k \leq N\}$
$ l(J_k)=x_{k+1}-x_k$, where l is the length.
$\overline{I}_P(f)= \sum_{k=0}^N sup_{x \in J_k}f(x) \cdot l(J_k)$
$\underline{I}_P(f)= \sum_{k=0}^N inf_{x \in J_k}f(x) \cdot l(J_k)$
With this information, I'm having a hard time starting this solution. I can't find examples on computation of partitions, and I really want to understand how to do it. Can someone provide an example similar to this or help me step by step?
| First, let us find the $l(J_k)$'s.
\begin{align*}
l(J_1) &= 2/7-0=2/7 \\
l(J_2)&=1/2-2/7=3/14 \\
l(J_3) &= 2/3 - 1/2 = 1/6 \\
l(J_4)&=5/7-2/3=1/21 \\
l(J_5) &= 1 - 5/7 = 2/7 \\
\end{align*}
Then the suprememum's (Because this seems to be a computation problem, I will simply use Wolfram Alpha):
\begin{align*}
\sup_{x\in J_1} f(x) &= 1/4 \\
\sup_{x\in J_2} f(x) &= 9/196 \\
\sup_{x\in J_3} f(x) &= 1/36 \\
\sup_{x\in J_4} f(x) &= 9/196 \\
\sup_{x\in J_5} f(x) &= 1/4 \\
\end{align*}
And now the infimum's:
\begin{align*}
\inf_{x\in J_1} f(x) &= 9/196 \\
\inf_{x\in J_2} f(x) &= 0 \\
\inf_{x\in J_3} f(x) &= 0 \\
\inf_{x\in J_4} f(x) &= 1/36 \\
\inf_{x\in J_5} f(x) &= 9/196 \\
\end{align*}
Then, multiplying and summing, we get (put answers as decimals so its easy to check):
$$
\overline{I}_P (f) = \frac{1}{4}\frac{2}{7} + \frac{9}{196}\frac{3}{14} + \frac{1}{36}\frac{1}{6} + \frac{9}{196}\frac{1}{21} + \frac{1}{4}\frac{2}{7} = .1595...
$$
and
$$
\underline{I}_P (f) = \frac{9}{196}\frac{2}{7} + 0\cdot\frac{3}{14} + 0\cdot\frac{1}{6} + \frac{1}{36}\frac{1}{21} + \frac{9}{196}\frac{2}{7} = .0276...
$$
Disclaimer: I kind of did this quickly so it is very possible I made a mistake somewhere in my calculations. Please call me out if you think I did that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First derivative of $f(x)= \frac{2}{x+1} +3$ I'm struggling to find the first derivative of $f(x)= \frac{2}{x+1} +3$ using the limit definition of derivative. I keep coming up with $f'(x) = \frac{-5}
{(x+1)^2}$ but I should be getting $f'(x) = \frac{-2}{(x+1)^2}$
\begin{align}
\lim_{x\to 0} &= \frac{(\frac{2}{x+h+1}+3)-(\frac{2}{x+1}+3)}{h}\\
\lim_{x\to 0} &= \frac{\frac{2(x+1)+3(x+1)-(2(x+h+1)+3(x+h+1))}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{2x+2+3x+3-2x-2h-2-3x-3h-3}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= \frac{\frac{-5h}{(x+h+1)(x+1)}}{h}\\
\lim_{x\to 0} &= (\frac{-5h}{(x+h+1)(x+1)})*\frac{1}{h}\\
\lim_{x\to 0 }&= \frac{-5}{(x+1)^2}\\
\end{align}
| In case you want to go by this definition.
In some cases it is more straightforward than the one you have used. If you put $x-a=h$ in the definition below you get the one you have.
There is an equivalent definition which says that function is differentiable at a point $a$ if $$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$ exists and we denote this limit as $f'(a)$
$$\lim_{x\rightarrow a}\frac{\frac{2}{x+1}+3-\frac{2}{a+1}-3}{x-a}$$
$$\lim_{x\rightarrow a}\frac{\frac{2}{x+1}-\frac{2}{a+1}}{x-a}$$
$$\lim_{x\rightarrow a}\frac{\frac{-2(x-a)}{(a+1)(x+1)}}{x-a}$$
$$\lim_{x\rightarrow a}\frac{-2}{(a+1)(x+1)}$$
$$f'(a)=\frac{-2}{(a+1)(a+1)}=\frac{-2}{(a+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$ but cant figure how to apply those here since the centre is not at origin.
| $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$
$$ {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} \right)=0$$
$$ \frac{2}{9}(x- 3)- \frac{2}{4}(y- 2){dy\over dx}= 0 $$
$$ \frac{4(x- 3)}{9(y- 2)}= {dy\over dx} $$
$${dy\over dx}\biggr|_{(-2,{14\over3})}= -\frac{5}{6} $$
Thus the equation of the tangent is
$$(y-\frac{14}{3})=\frac{-5}{6}(x+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize denominator and then find factor of nominator in a way that these two cancel each other.
factorizing denominator gives:
$$ \lim_{x \rightarrow -2}\frac{3x^2+ax+a+3}{(x-1)(x+2)} $$
Now if it is possible to find solution to $3x^2+ax+a+3=0$ when $x=-2$
$$ 3(-2)^2+a(-2)+(-2)+3=0 $$
$$ 12-2a-2+3=0 $$
$$ 2a=13 \iff a = \frac{13}{2} $$
factorizing nominator gives:
$$ 3x^2+\frac{13x}{2}+\frac{19}{2}=0 $$
$$ x= \frac{-\frac{13}{2}\pm \sqrt{(\frac{13}{2})^2-4\cdot 3 \cdot (\frac{19}{2})}}{2\cdot 3} $$
Only problem is this equation is never zero with $x\in \mathbb{R}$ since there is negative value under square root. Now there is contradiction between $a=\frac{13}{2}$ and that equation $3x^2+ax+a+3=0$ when $a=\frac{13}{2}$ and $x=-2$
$$ 3(-2)^2+\frac{13}{2}\cdot (-2)+\frac{13}{2}+3 \neq 0 $$
There is clearly something wrong but i cannot see where this went wrong.
| When you substituted $x=-2$ in $3x^2+ax+a+3$ you incorrectly put $(-2)$ in place of some of the $a$-s. The remaining part appears to be the follow-up of said mistake of algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Difficulty Finding $A^k$ Let $A=
\begin{bmatrix}
1& -1 & 1\\
0 & 1 & 1 \\
0 & 0 & 1\\
\end{bmatrix}$. Compute $A^k$.
My attempt
I'm trying to compute $A^k$ using this approach as follows:
$$
A=I+N=
\begin{bmatrix}
1& 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}+
\begin{bmatrix}
0& -1 & 1\\
0 & 0 & 1 \\
0 & 0 & 0\\
\end{bmatrix}
$$
with
$$
N^2=
\begin{bmatrix}
0& 0 & -1\\
0 & 0 & 0 \\
0 & 0 & 0\\
\end{bmatrix}, \, \text{and} \, \,
N^3=
\begin{bmatrix}
0& 0 & 0\\
0 & 0 & 0 \\
0 & 0 & 0\\
\end{bmatrix}
$$
Then,
$$
A^2=(I+N)^2=I+2N+N^2, \\
A^3=(I+N)^3=I+3N+3N^2, \\
A^4=(I+N)^4=I+4N+6N^2, \\
A^5=(I+N)^5=I+5N+10N^2, \\
A^6=(I+N)^5=I+6N+15N^2,
$$
By induction, we can see $A^k=(I+N)^k=I+kN+f[k]N^2$. But, I couldn't figure out what $f[k]$ is. Any help?
| How about this. Take the exponential function $e^{tA}$, where $t$ is some parameter
$$e^{tA}=\sum_{k=0}^\infty\frac{t^kA^k}{k!}=e^{t(I+N)}= e^{tI}e^{tN}=e^t\left[I+tN+\frac{(tN)^2}{2}\right]$$
where we used the matrix identity $e^{A+B}=e^Ae^B$ that is valid when
matrices $A$ and $B$ commute.
Since the functions $t^k$ are linearly independent, we obtain
$$A^k=I+kN+\frac{k(k-1)}{2}N^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
The tangent at $P$ to $y = x^2 - x^3$ meets the curve again at $Q$. Show that locus of midpoint of $PQ$ is $y=1-9x+28x^2-28x^3$
The tangent at a variable point $P$ of the curve $y = x^2 - x^3$ meets it again at $Q$. Show that the locus of the middle point of $PQ$ is $$y = 1 - 9x + 28x^2 - 28x^3$$
My approach
$$y^\prime=2x-3x^2$$
Equation of tangent: $$2\ x_1 -3\ x_1^2=\frac{y-\ x_1^2+\ x_2^3}{x-\ x_1}$$
From the equation of tangent:
$$(2\ x_1 -3\ x_1^2)(x-\ x_1)=y-\ x_1^2+\ x_2^3$$
Substituting the value of $y$ in the conic section we can find the value of $x$ but still it is getting complicated.
| Given $f(x) = x^2-x^3$ the tangent line at $x_0$ is
$$
y = f(x_0) + f'(x_0)(x-x_0)\to y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2
$$
Solving now for the intersection points we have
$$
y = x^2-x^3\\
y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2
$$
with the feasible solution $x = 1-2x_0$ so the sought locus is given by
$$
(x, y) = \frac 12((x_0, f(x_0))+(1-2x_0,f(1-2x_0))) = \frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0)
$$
Attached a plot showing $f(x)$ in blue, $\frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0)$ in dashed red and the tangent line.
NOTE
Eliminating $x_0$ in the equations
$$
(x,y) = \frac 12(1-x_0, 7(x_0^3-x_0^2)+2x_0)
$$
we obtain
$$
y = 1-9x+28x^2-28 x^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)$$
I have tried to factor out the $(1+\frac{1}{x})$ use the distributive property, and failed multiple times. Every online calculator I tried gives me different results.
| My first thought was to factor out the $\frac 6{x^2}$ but yours was to factor out the $1 + \frac 1x$. So we'll do it your way:
Factor out the $1 + \frac 1x$ and we get
$(1 + \frac 1x) (\frac {-6}{x^2} + \frac 6{x^3}(1 + \frac 1x))$.
Now factor out the $\frac 6{x^2}$ to get:
$\frac 6{x^2}(1 + \frac 1x) (-1 + \frac 1x(1 + \frac 1x))$
$\frac 6{x^2}(1 + \frac 1x)(-1 + \frac 1x + \frac 1{x^2})$
now, I can struggle with trying to factor $(-1+\frac 1x + \frac 1{x^2})$[1] but I must point out that if you claimed factorization and my demonstrated factorization were both true we'd have $-1 + \frac 1x + \frac 1{x^2} = \frac 1x(1+\frac 2x)$. WHich it doesn't.
Are you sure the question wasn't to solve for the $x$ where it does factor to that? In which case we'd solve for $-1 + \frac 1x + \frac 1{x^2} = \frac 1x(1+\frac 2x)$ or $-1 = \frac 1{x^2}$ which is $x =\pm i$ if we are allowed to use complex numbers... So... I guess that wasn't the question.
[1] Long story short; by the rational roots theorem we can't.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
determining the common difference of an arithmetic sequence and common ratio of a geometric sequence with related terms Coming from a finance guy, I understand how AP and GP work. However, I came upon a problem that combines the two and was stuck. Here it goes.
Given first term of AP and GP$=4$, and common ratio of GP is $8$ less than common difference of AP. The ratio of $3$rd term of AP to $3$rd term of GP is $7:16$. What is the common difference and common ratio?
I tried it this far:
given common ratio$=r$
difference$=d$
so $r=d-8$
for the AP; 3rd term$=4+2d$
and GP$=4(d-8)(d-8)$
so $(4+2d) : 4(d-8)(d-8) =7:16$
and I got this far..
some headway please..
| Here, assume that the AP is :
$$a, a+d, a+2d, ...$$
and, the GP is :
$$ a, ar, ar^2 ,...$$
Then from the given data -
$a = 4$, $ r = d - 8$ ( or $d = r + 8$) and $\frac {a + 2d}{2r^2} = \frac{7}{16}$.
The third equation above simplifies as,
$\frac{4+2d}{4r^2} = \frac{2+d}{2r^2} = {7\over 16}$
$\implies \frac{r+10}{2r^2} = frac{7}{16}$ ( by the second equation)
$\implies 7r^2 - 4r - 40 = 0$
or, $r = \frac{4 \pm \sqrt{16 + 4.7.40}}{2.7} = \frac{4\pm \sqrt{1136}}{14}$ (using the quadratic formula)
Thus, $r = 2.693$ or $-2.121$ (approximately)
Using the second equation above, $d = 10.693$. or $5.8782$ (approx.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$
Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$
| Hint:
If trigonometric substitution is not mandatory, for (where $a$ is an arbitrary constant )
$$\dfrac{\sqrt{4x^2+a}}{x^{2n-1}}=4^{n-1}\dfrac{\sqrt{4x^2+a}}{(4x^2)^n}\cdot4x$$
set $\sqrt{4x^2+a}=y,4x^2+a=y^2,y\ dy=4x\ dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Simplify $(3\log x) - (2\log x)$ How to simplify $(3\log x) - (2\log x)$? Would this become $(\log x )^ {\frac{3}{2}}$ or would this be just $3\log x-2\log x =\log x$? If so how to get $\log x$?
I was given this question: solve for $x$ if $\log x + \log x^2 +...+ \log x^n =n(n+1)$. But, the answer to my main question will also be enough. Thank you for trying!
| $$3\log x-2\log x$$
Here, we use two identities.
$$\log a^b = b\cdot\log a$$
$$\log_a b-\log_a c = \log_a \big(\frac{b}{c}\big)$$
Use the first identity to reach the following expression.
$$\implies \log x^3-\log x^2$$
Use the second identity to simplify.
$$\implies \log \big(\frac{x^3}{x^2}\big) \implies \boxed{\log x}$$
So no, you canβt get $(\log x)^{\frac{3}{2}}$...
For the next question, we have the following equation.
$$\log x+\log x^2+\log x^3+...+\log x^n = n\cdot(n+1)$$
Simplify the left hand side using the first identity.
$$\log x+2\cdot\log x+3\cdot\log x+...+n\cdot\log x = n\cdot(n+1)$$
$$(1+2+3+...+n)\cdot\log x = n\cdot(n+1)$$
$$\sum_{k = 1}^n k = n\cdot\frac{n+1}{2}$$
$$\biggr(\frac{n\cdot(1+n)}{2}\biggr)\cdot \log x = n\cdot(n+1)$$
$$\log x = \frac{n\cdot(n+1)}{\frac{n\cdot(1+n)}{2}}$$
$$\log x = \frac{n\cdot(n+1)\cdot 2}{n\cdot(1+n)}$$
$$\log x = 2$$
Use the definition of logarithms.
$$\log_b x = y \longleftrightarrow b^y = x$$
So, we get:
$$\boxed{x = 10^{2} = 100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $2^q+q^2$ is divisible by 3 where $q$ is a prime and $q\geq5$. I'm looking to prove that $2^q+q^2$ is divisible by $3$ where $q$ is a prime such that $q\geq5$.
I know that primes greater than five will be congruent to either $1\ (\text{mod}\ 3)$ or $2\ (\text{mod}\ 3)$, which means that the $q^2$-term will always be congruent to $1\ (\text{mod}\ 3)$ which simplifies the problem to finding the congruence of $2^q+1\ (\text{mod}\ 3)$. However, I'm unable to go any further as I'm unable to show that the congruence of $2^q\ (\text{mod}\ 3)$ is always equal to $2$.
| $2^{1} \equiv 2 \pmod{3}$;
$2^{2} \equiv 1 \pmod{3}$;
$2^{3} \equiv 2 \pmod{3}$;
$2^{4} \equiv 1 \pmod{3}$;
$2^{5} \equiv 2 \pmod{3}$;
$2^{6} \equiv 1 \pmod{3}$;
$...$
Note that when the exponent is odd, the remainder equals to $2$ and when the exponent is even, the remainder is $1$.
This pattern keeps repeating because multiplying number with a remainder of $1\pmod{3}$ by $2$ equals to $2\pmod{3}$, and multiplying again by $2$ gives us the remainder of $4$, which is $1\pmod{3}$.
Because $q$ is odd, $2^{q} \equiv 2 \pmod{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Antiderivative of $x\sqrt{1+x^2}$ I am attempting this problem given to me, but the answer key does not explain the answer. The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.
My attempt:
$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$
$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$
$f'(x)$ must equal $\sqrt{1+x^2}$, therefore $f(x)=\frac{2}{3}(1+x^2)^{\frac{3}{2}}$
But then I saw that if $g(x)=\frac{1}{3}x^2$, then $g'(x)=\frac{2}{3}x$, and $f(x)=(1+x^2)^{\frac{3}{2}}$, then $\frac{d}{dx}f(g(x))'=g'(x)f'(g(x))=(2/3)x(3/2)\sqrt{1+x^2}=x\sqrt{1+x^2}$.
But I saw this by chance; what would be a more general and effective way to do this? Should I check $this$, and then write $that$, and put together $those$, etc.? Or is the way I just 'saw' the way the numbers should fall together how one would normally approach this?
| $u$-substitution is more standard, but I can follow your method to see how it works out.
$$\begin{align*}
\frac d{dx}f(g(x)) &= g'(x) \cdot f'(g(x))\\
&= x\sqrt{1+x^2}
\end{align*}$$
Try $g'(x) = x$, then
$$g(x) = \frac12x^2+C_1$$
Try $C_1 = \frac12$ to get $g(x) = \frac12(1+x^2)$. Then
$$\begin{align*}
f'(x) &= \sqrt{2x}\\
&= \sqrt2 \cdot \sqrt x\\
f(x) &= \sqrt2\cdot\frac{2}3 x^{3/2}+C\\
f(g(x)) &= \sqrt2\cdot\frac23\left[\frac12(1+x^2)\right]^{\frac32}+C\\
&= \frac13(1+x^2)^{3/2}+C
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Calculate variance for a probability mass function In a family of ten people, the probability mass function of the number of people
who contracted the flu is given by
$$P_X(x) = K(2x + 9); x = 0,1,..., 10$$
Calculate the variance of the number of people with flu in the family.
My attempt:
The first thing I needed to find was "K". Thus $$\sum_{x=0}^{10}{P_X(x)=1}$$
$$\sum_{x=0}^{10}{K(2x+9)=1}$$
$$\frac{2k.10(10+1)}{2} + 90K = 1$$
$$K=\frac{1}{200}$$
And then use $V(X)=E(X^2)-{E(X)}^2=47.575-(6.325)^2=7.5693$
$$E(X)=\sum_{x=0}^{10}{\frac{x(2x+9)}{200}}=6.325$$
$$E(X^2)=\sum_{x=0}^{10}{\frac{x^2(2x+9)}{200}}=47.575$$
For y second attempt, I have $K=\frac{1}{209}$. What am I missing? Am I on the right path?
| Note that$$\sum_{x=0}^{10}9=9(11)=99$$
$$10K(11)+99K=1$$
$$110K+99K=1$$
$$209K=1$$
$$K=\frac1{209}$$
\begin{align}
E(X^2) &= K \sum_{x=0}^{10} x^2(2x+9)\\
&=K\left( 2\left(\frac{(10)(11)}{2}\right)^2+9\cdot\frac{(10)(11)(21)}{6}\right) \\
&= 10(11)K\left(\frac{10(11)}{2} + \frac{3(21)}{2} \right)\\
&=\frac{10(11)}{2}K\left(110+63 \right)\\
&=9515K
\end{align}
\begin{align}
E(X) &= K \sum_{x=0}^{10} x(2x+9)\\
&=K\left( 2\cdot\frac{(10)(11)(21)}{6}+9\left(\frac{(10)(11)}{2}\right)\right) \\
&= 10(11)K\left(7 + \frac{9}{2} \right)\\
&=\frac{10(11)}{2}K\left(23 \right)\\
&=1265K
\end{align}
$$Var(X)=E(X^2)-(E(X))^2=9515K-(1265K)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
2nd order differentiel equation I was wondering how I could solve
$$
f''=\left(1+x^4\right)f
$$
for some $f$.
I must admit that when it is 2nd order depending on $x$ I've more struggle to succeed, but here i've no clue ! Any hint please ?
| Hint:
Let $f=e^{ax^3}u$ ,
Then $f'=e^{ax^3}u'+3ax^2e^{ax^3}u$
$f''=e^{ax^3}u''+3ax^2e^{ax^3}u'+3ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u$
$\therefore e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=(1+x^4)e^{ax^3}u$
$u''+6ax^2u'+(9a^2x^4+6ax)u=(x^4+1)u$
$u''+6ax^2u'+((9a^2-1)x^4+6ax-1)u=0$
Choose $9a^2-1=0$ , i.e. $n=\dfrac{1}{3}$ , the ODE becomes
$u''+2x^2u'+(2x-1)u=0$
Let $t=bx$ ,
Then $b^2\dfrac{d^2u}{dt^2}+\dfrac{2t^2}{b}\dfrac{du}{dt}+\left(\dfrac{2t}{b}-1\right)u=0$
$\dfrac{d^2u}{dt^2}+\dfrac{2t^2}{b^3}\dfrac{du}{dt}+\left(\dfrac{2t}{b^3}-\dfrac{1}{b^2}\right)u=0$
Choose $b^3=2$ , i.e. $b=\sqrt[3]2$ , the ODE becomes
$\dfrac{d^2u}{dt^2}+t^2\dfrac{du}{dt}+\left(t-\dfrac{1}{\sqrt[3]4}\right)u=0$
Which relates to Heun's Triconfluent Equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
| Let's continue from your last step.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+2\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6}+k^2$$
By induction hypothesis, first three terms yileds an integer and $k^2$ is an integer.
Thus finally we have showed that $$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}$$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
probability of selection without replacement An urn contains 20 black marbles and 20 white marbles. Three marbles are chosen without replacement. What is the probability that the first marble is white given that the third marble was black?
| You could look at the various probabilities for the eight possibilities for the first three marbles,
but a quicker way is to use symmetry (each marble can be in any position) and say this is the same as the probability that the second marble is white given that the first marble was black, and that is $$\frac{20}{39}$$
coronermclarson came up with a different answer. I believe the long-winded answer is to look at the probabilities of the possible patterns for the first three marbles:
*
*$BBB$: $\frac{20}{40}\times \frac{19}{39} \times \frac{18}{38} = \frac{9}{78}$
*$BWB$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$
*$WBB$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$
*$WWB$: $\frac{20}{40}\times \frac{19}{39} \times \frac{20}{38} = \frac{10}{78}$
*$WWW$: $\frac{20}{40}\times \frac{19}{39} \times \frac{18}{38} = \frac{9}{78}$
*$WBW$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$
*$BWW$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$
*$BBW$: $\frac{20}{40}\times \frac{19}{39} \times \frac{20}{38} = \frac{10}{78}$
which add up to $1$, as they should
We are only interested in the first four of these which have the third black, making the probability that the first marble is white given that the third marble was black$$\dfrac{\frac{10}{78}+\frac{10}{78}}{\frac{9}{78}+\frac{10}{78}+\frac{10}{78}+\frac{10}{78}} = \dfrac{10+10}{9+10+10+10} = \dfrac{20}{39}$$ as before
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convergence for series failed using "Ratio Test" $$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$
Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$
which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$
the result for latter is q=1. seemingly this is unclear if divergent or convergent.
The answer for this is convergent....
| Hint:
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{1\cdot3\cdot5\cdots\cdot(2n-1)\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=2n+1$$
and not
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{2n+1}{2n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Maximum value on a circle I need to find the maximum value of a function on a circle: Let $C$ denote the circle of radius $6$ centered at the origin in the $xy$-plane. Find the maximum value of $x^2y$ on $C$. Where do I even start with this?
| What about polars? If we allow the circle $C(R)$ to be of arbitrary positive radius $R$, then for $(x, y) \in C(R)$,
$x = R\cos \theta, \tag 1$
$y = R \sin \theta, \tag 2$
so that
$x^2y = R^3 \cos^2 \theta \sin \theta; \tag 3$
this will take on a maximum value whenever $\cos^2 \theta \sin \theta$ does; we have
$\dfrac{d(\cos^2 \theta \sin \theta)}{d\theta} = -2\cos \theta \sin^2 \theta + \cos^3 \theta; \tag 4$
we set this to $0$ and obtain
$2\cos \theta \sin^2 \theta = \cos^3 \theta; \tag 5$
now,
$\cos \theta = 0 \Longrightarrow \theta = \pm \dfrac{\pi}{2}, \tag 6$
and also
$\cos^2 \theta \sin \theta = 0; \tag 7$
next we bear these facts in mind whilst we investigate
$\cos \theta \ne 0 \Longrightarrow 2\sin^2 \theta = \cos^2 \theta \Longrightarrow \tan^2 \theta = \dfrac{1}{2} \Longrightarrow \tan \theta = \pm \dfrac{1}{\sqrt 2}; \tag 8$
it is easy to see that the point in the first quadrant on the unit circle with $\tan \theta = 1/\sqrt 2$ is
$(\cos \theta, \sin \theta) = (x, y) =\left ( \dfrac{\sqrt 2}{\sqrt 3}, \dfrac{1}{\sqrt 3} \right ); \tag 9$
then it is also easy to see that the three other points on $C(1)$ such that $\tan \theta = \pm 1 / \sqrt 2$ are
$(\cos \theta, \sin \theta) = (x, y) = \left ( -\dfrac{\sqrt 2}{\sqrt 3}, \dfrac{1}{\sqrt 3} \right ), \; \left ( \dfrac{\sqrt 2}{\sqrt 3}, -\dfrac{1}{\sqrt 3} \right ), \; \left ( -\dfrac{\sqrt 2}{\sqrt 3}, -\dfrac{1}{\sqrt 3} \right ); \tag{10}$
to these four critical points $(\pm \sqrt 2 / \sqrt 3, \pm 1 \sqrt 3)$ of $\cos^2 \theta \sin \theta$ on $C(1)$, we must add the two found above where $\theta = \pi / 2, - \pi / 2 \equiv 3 \pi / 2$ and our function $\cos^2 \theta \sin \theta = 0$; since $\cos^2 \theta \ge 0$ everywhere, we look to the factor $\sin \theta$ as potentially determinative of the types of the critical points we have found--whether maxima or minima. For example, the points $(\pm \sqrt 2 / \sqrt 3, 1 / \sqrt 3)$ are both clearly local maxima since there and there only is $x^2y$ is positive; similarly the points $(\pm \sqrt 2 / \sqrt 3, -1 / \sqrt 3)$ are local minima since there $x^2y = \cos^2 \theta \sin \theta < 0$; finally, the point where $\theta = \pi / 2$, $\cos \theta = \cos^2 \theta \sin \theta = 0$ is seen to be a local minimum, since $\cos^2 \theta \sin \theta > 0$ slightly to either side of it; that is, for $0 < \vert \theta - \pi / 2 \vert < \epsilon$ for $\epsilon > 0$ sufficiently small; and a similar argument shows $\theta = -\pi / 2 \equiv 3\pi / 2$ is a local maximum.
Thus we have found all the critical points of $x^2y = \cos^2 \theta \sin \theta$ on $C(1)$, of these three are maxima as given above.
On the circle $C(R)$ the function $x^2y$ becomes $R^3 \cos^2 \theta \sin \theta$; the values of $x$ and $y$ scale with $R$; maxima and minima occur for the same $\theta$ values, however.
I leave it to the reader to supply the numerical details for the case $R= 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int\frac{x^3+1}{x^4+x^3+x^2+x}\,dx$ How do I integrate $\displaystyle\int\dfrac{x^3+1}{x^4+x^3+x^2+x}\,\mathrm dx$?
I tried by splitting the equation in two parts like $\dfrac{x^3}{x^4+x^3+x^2+x}$ and $\dfrac1{x^4+x^3+x^2+x}$ and then cancelling out the $x$ terms from the first part and then trying to integrate further.
But this did nothing much to it, and also the other part became more difficult to solve.
How do I solve this integration?
| We first realise that both our numerator and denominator are factorable.
We can factor the numerator and denominator as:
$$\frac{(x+1)(x^2-x+1)}{(x)(x^3+x^2+x+1)}.$$
We can now multiply both the numerator and denominator by $x-1$,
we will now get:
$$\frac{(x^2-1)(x^2-x+1)}{(x)(x^4-1)}.$$
We can also factor the denominator
$$\frac{(x^2-1)(x^2-x+1)}{(x)(x^2-1)(x^2+1)}.$$
The $x^2-1$ cancels out, leaving us with
$$\frac{x^2-x+1}{(x)(x^2+1)}.$$
We can now complete the square in the numerator,
$$\frac{(x-1)^2+x}{(x)(x^2+1)}.$$
We can now split the expression apart and evaluate each part separately
$$\frac{(x-1)^2}{(x)(x^2+1)}+\frac{x}{(x)(x^2+1)}.$$
In the second part of our expression $x$ cancels out, leaving us with
$$+\frac{1}{(x^2+1)}.$$
For the first part of our expression, we can expand the $(x-1)^2$ in the numerator
$$\frac{(x-1)^2}{(x)(x^2+1)}=\frac{x^2-2x+1}{(x)(x^2+1)}$$
We now split the sum up
$$\frac{x^2+1}{(x)(x^2+1)}-\frac{2x}{(x)(x^2+1)}$$
The first part of our expression is equal to $\dfrac{1}{x}$, the second part of our expression is just $-\dfrac{2}{x^2+1}.$
Therefore, our total expression evaluated together is
$$\frac{1}{x}-\frac{2}{x^2+1}+\frac{1}{(x^2+1)}=\frac{1}{x}-\frac{1}{(x^2+1)}.$$
These two expressions have elementary integrals.
\begin{align*}
\int \frac{1}{x}-\frac{1}{(x^2+1)}\,\mathrm dx&=
\int \frac{1}{x}\,\mathrm dx-\int \frac{1}{(x^2+1)}\,\mathrm dx\\
&=\boxed{\ln |x| - \arctan (x) + C.}
\end{align*}
Thats it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
show that $\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$ , $\forall n \mathbb \in{N}$
Show that $$\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$$ for all $n \in \Bbb{N}$
I tried with induction method but I can not find any results.
| $$\frac{1}{3n+1}+\frac{1}{3n+2}+\cdots+\frac{1}{5n+1} <\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+2}+\cdots+\frac{1}{3n+2}$$ $$=\frac{1}{3n+1}+\frac{2n}{3n+2}<\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$.
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It is now necessary to prove that $x$-coordinates of extrema of $f(x)$ fall between $a, b$ and $c$. Or otherwise stated,
$a<x_1<b<x_2<c$, where $x_1$ and $x_2$ are $x$-coordinates of extrema.
The problem is, however, in the fact that this should be proven without the usage of Rolle's Theorem. I can prove this fact for quadratic, I get $x=(a+b)/2$, which suggests that extremum is halfway between the roots. With cubic, however, it is much more difficult. Any ideas?
| I'm taking capital $A=1$ and then $a < b < c$ as the roots. we get the cubic
$$ x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3 \; , $$
where
$$ \sigma_1 = a + b + c, $$
$$ \sigma_2 = bc + ca + ab, $$
$$ \sigma_3 = abc. $$
The first derivative is $3 x^2 - 2 \sigma_1 x + \sigma_2,$ with roots
$$ \frac{\sigma_1 \pm \sqrt{\sigma_1^2 - 3 \sigma_2}}{3} \; .$$
Worth emphasizing that, with $a,b,c$ distinct, we get
$$ \sigma_1^2 - 3 \sigma_2 = a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right) $$
being strictly positive, so the two roots of $3 x^2 - 2 \sigma_1 x + \sigma_2$ are real and distinct.
The claim that $c > \frac{\sigma_1 + \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $c > \frac{\sigma_1}{3}$ along with
$$ \left( c - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(c-b) > 0 $$
The claim that $a < \frac{\sigma_1 - \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $a < \frac{\sigma_1}{3}$ along with
$$ \left( a - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(b-a) > 0 $$
The final claim is that $b$ lies between the critical points, in that the distance between $b$ and $\sigma_1/3$ is smaller than $\sqrt{\sigma_1^2 - 3 \sigma_2}/3.$ Indeed,
$$ \left( b - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-b)(a-b) < 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving that $(a,b)\cong [1,2)$. I aim to prove the proposition indicated in the title.
Is the following argument correct?
We now show that the map $h:(a,b)\to [1,2)$ defined as follows is a bijection.
\begin{align*}h(x) = \begin{cases}\frac{x-a}{b-a}+1&\text{ if }x\neq \frac{a+b}{2}\\1&\text{ if }x = \frac{a+b}{2}\\\end{cases}\end{align*}
Let $x\in(a,b)$. Assume for the purpose of contradiction that $h(x) = 1$ and $x\neq\frac{a+b}{2}$. Now $x>a$ and $b>a$, implying $\frac{x-a}{b-a}>0$, so $h(x) = \frac{x-a}{b-a}+1>1$. In summary $h(x)=1$ if and only if $x=\frac{a+b}{2}$.
Now let $x_1,x_2\in(a,b)$ and assume $h(x_1) = h(x_2)$ now if $h(x_1) = h(x_2)$ then $x_1 = x_2 = \frac{a+b}{2}$ and if $h(x_1) = h(x_2)\neq 1$, then both $x_1$ and $x_2$ are not equal to $\frac{a+b}{2}$, thus $h(x_1) = \frac{x_1-a}{b-a}+1 = \frac{x_2-a}{b-a}+1= h(x_2)$, implying $x_1 = x_2$.
Let $y\in[1,2)$. In the event $y = 1$, then $h(\frac{a+b}{2}) = 1$ and if $y\neq 1$, then given the injectivity of $h$, we may have $h(x) \neq 1$ only if $x\neq\frac{a+b}{2}$. Let $x = y(b-a)+a$, here $x\neq \frac{a+b}{2}$ as arguing to the contrary would imply that $y = \frac{1}{(b-a)}(\frac{a+b}{2}-a) = \frac{1}{2}$, which is impossible considering $y\in[1,2)$. Then given the definition of $h$ it is then apparent that $h(x) = y$. Thus the map in question is surjective.
$\blacksquare$
| First, note that your function is not surjective, since there is no value in $(a,b)$ that maps to $\frac32$. We know this because $\frac{x-a}{b-a}+1$ is continuous and strictly increasing on the set $(a,b)$. As such, we know that there is only one value on the set that can map to $\frac32$, which is $x=\frac{a+b}2$. However, $\frac{a+b}2$ maps to $1$.
Instead, we note that since the composition of bijective maps is bijective, we can split the problem into finding bijective maps of the following kind:
$$f: (a,b)\to(0,1)$$$$g: (0,1)\to[0,1)$$$$h:[0,1)\to[1,2)$$
Here are the mappings we will use $$f(x)=\frac{x-a}{b-a}$$$$g(x)=\begin{cases}\frac{n-1}n&\exists n\in\mathbb N,\;x=\frac n{n+1}\\x&\textrm{otherwise}\end{cases}$$$$h(x)=x+1$$
It's obvious that the composition of these functions gives you your desired function. Can you prove these functions are bijective?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ According to De Moivre's theorem, we get:
$$z^n= \cos{\left(n\arctan{\frac{4}{3}}\right)} + i\sin{\left(n\arctan{\frac{4}{3}}\right)}$$
Now, if for $n\in \mathbb{N}: z^n=1$, then the imaginary part of the expression above must be zero, therefore:
$$\sin{\left(n\arctan{\frac{4}{3}}\right)}=0 \iff n\arctan{\frac{4}{3}} = k\pi, \ \ \ k \in \mathbb{N}$$
And we get that for $z$ to be a root of unity for some natural number $n$, $n$ must be in the form:
$$n = \frac{k\pi}{\arctan{\frac{4}{3}}}, \ \ \ k \in \mathbb{N}$$
On the other hand, for $z^n=1$ it must be that:
$$\cos{\left(n\arctan{\frac{4}{3}}\right)} = 1 \iff n\arctan{\frac{4}{3}} = 2l\pi, \ \ \ l \in \mathbb{N}$$
and thus
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
By comparing those two forms of $n$, it must be the case that $k=2l$ and for $n$ to satisfy $z^n = 1$. What follows is that $n$ should be in the form
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
But, at the same time, $n$ must be a natural number. Should I prove now that such $n$ cannot even be a rational number, let alone a natural one? Or how should I approach finishing this proof?
| Hint: Look at the real and imaginary parts of $(3+4i)^n$, mod 5.
If $(3+4i)^n$ never has real and imaginary parts a multiple of 5, then $\left(\dfrac{3+4i}5\right)\!^{\Large n}$ never has real and imaginary parts an integer. (For $n\ge2$, anyway.) In particular, it never equals $1$, and $\dfrac{3+4i}5$ is not a root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
What can I say of x given these conditions? I know that $x^2-x$ is integer and also $x^n-x$ is integer for some $ n > 2$. Can I say that $x$ is integer? How can I show it? ($x \in \mathbb{R}$)
| Suppose that $x^2 - x = k \in \mathbf{Z}$. In order for $x$ to be real, either $k = 0$ and $x \in \{0,1\}$ is an integer, or $k > 0$. For any $k$ and $n$, long division shows that
$$x^n - x = F_{n,k}(x)(x^2 - x - k) + A_{n,k} x + B_{n,k} = A_{n,k} x + B_{n,k}.$$
In particular, for the LHS to be an integer, either $x$ must be an integer, or $A_{n,k}$ must equal zero. (This uses the fact that $x^2 - x = k$ implies that either $x$ is an integer or $x$ is irrational by Gauss' Lemma.)
I claim that $A_{n,k} > 0$ for $n > 2$ and $k > 0$, and so, if $x$ is real, the LHS is only an integer when $x$ is an integer.
Lemma: If $k > 0$, then $A_{n,k} > 0$ and $B_{n,k} > 0$ for all $n \ge 3$.
Proof: By induction on $n$. For $n = 3$ and any $k > 0$, we have
$$x^3 - x = (x + 1)(x^2 - x - k) + k x + k.$$
So $A_{n,k} = k > 0$, and $B_{n,k} = k > 0$.
Assume the result is true for $n$. Note that
$$x^{n+1} - x = x(x^n - x) + x^2 - x = x(x^n - x) + (x^2 - x - k) + k.$$
Hence
$$\begin{aligned}
x^{n+1} - x = & \ x F_{n,k}(x)(x^2 - x - k) + A_{n,k} x^2 + B_{n,k} x + (x^2 - x - k) + k \\
= & \ (x F_{n,k}(x) + 1)(x^2 - x - k) + A_{n,k} x^2 + B_{n,k} x + k \\
= & \ (x F_{n,k}(x) + 1)(x^2 - x - k) + A_{n,k} (x^2 - x - k) + A_{n,k}(x + k) + B_{n,k}x + k \\
= & \ (x F_{n,k}(x) + 1 + A_{n,k})(x^2 - x - k) + (A_{n,k} + B_{n,k})x + A_{n,k} \cdot k + k \end{aligned}$$
And thus (by induction on $n$, and using that $k > 0$)
$$A_{n+1,k} = A_{n,k} + B_{n,k} > 0,$$
$$B_{n+1,k} = A_{n,k} \cdot k + k > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How can I simplify the derivative further to match the correct answer? I've been stuck on trying to simplify my expression in order to match the correct answer but I can't seem to get the correct solution. Guidance towards the proper steps would be greatly appreciated!
The original question:
Differentiate and simplify completely
$$g(x)=\sqrt[3]{\frac{x^3-1}{x^3+1}}$$
My solution:
$$g^\prime(x) = \frac{2x^2}{\sqrt[3]{\left(x^3-1\right)^2}\sqrt[3]{\left(x^3+1\right)^4}}$$
I need help simplifying this to become this solution:
$$g^\prime(x) = \frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{(x+1)^2\left(x^2-x+1\right)^2(x-1)\left(x^2+x+1\right)} $$
Original image of solution.
| Just multiply by $1$ (to free the denominator from irrationality):
$$g^\prime(x) = \frac{2x^2}{\sqrt[3]{\left(x^3-1\right)^2}\sqrt[3]{\left(x^3+1\right)^4}}\cdot \frac{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}=\\
\frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\color{red}{(x^3+1)^2}\color{blue}{(x^3-1)}}=\\
\frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\color{red}{(x+1)^2\left(x^2-x+1\right)^2}\color{blue}{(x-1)\left(x^2+x+1\right)}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What can be said about the roots of $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$?
Let a, b and c be real numbers. Then the fourth degree polynomial in
$x$, $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$
(a) Has four complex (non-real) roots
(b) Has either four real roots or four complex roots
(c)Has two real roots and two complex roots
(d) Has four real roots
This question is from a book called 'Test of Mathematics at the $10+2$ Level' published by the Indian Statistical Institute
We can see that the expression factorizes to $(ax^2+bx+c)(a+bx+cx^2)$. If $\alpha,\beta$ are the roots of the first factor then $1/\alpha , 1/\beta$ are the roots of the second expression. And we know that complex roots occur in conjugation. So, we can easily understand that option (b) is a correct.
But today, I want to solve this problem in some different way in which it does not require us to factorize the expression in a hit and trial way as I did. Even if we have to factorize, we must use a proper way to find the factors. Please help.
| Hint: It is $$ac\left(x^2+\frac{1}{x^2}\right)+b(a+c)\left(x+\frac{1}{x}\right)+a^2+b^2+c^2=0$$
and now substitute $$t=x+\frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $F(x)=x^4-bx^3-11x^2+4(b+1)x+a$.Find $a,b$ Let $F(x)=x^4-bx^3-11x^2+4(b+1)x+a$. Find $a,b$
Itβs given that $F(x)$ is a complete square of a quadratic polynomial and $(x+2) $ is a factor of $F(x)$
My attempt :
I can write $F(x) = (x+2)^2(Ax+B)^2$
Then putting $x=-2$, I got $a=36$
And by putting $x=0$, I got $4B^2=36$
Then $B=+3$ or $B=-3$.
Then clearly $A=1$.
Now I have $F(x)=(x+2)^2(x\pm 3)^2$
Now when x=-3 I get b= 2/5
And when x=3 I get b= 2.
So $F(x)=(x+2)^2(x- 3)^2$.
How I can I properly select whether $B=-3$ or $B =3$
| You proved $F(x) = (x+2)^2(x\pm 3)^2$, and if you multiply out the terms, the coefficient of $x^2$ in $F$ will be
$$
\left[x^2\right]F(x) = 4 \pm 24 + 9 = 13 \pm 24
$$
but you know this must be $-11$, so the correct sign is $-$ and you end up with $$F(x) = (x+2)^2(x-3)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $(x \mod 7) - (x \mod 8) = 5$? I'm trying to solve $(x \mod 7) - (x \mod 8) = 5$ but no idea where to start. Help appreciated!
| I'm going to assume you are incorrectly assuming $\mod n$ refers to the remainder function and want to solve $(x \% 7) -(x\% 8) = 5$ where $a \% b$ is the unique remainder when $a$ is divided by $b$.
$x \% 7 = a$ means $x = 7k + a$ for some integer $k$ and integer $0 \le a < 7$. So $a = x - 7k$.
And $x \% 8 = b$ means $x = 8j + b$ for some integer $j$ and integer $0 \le b < 8$. So $b = x - 8j$
So $a - b = 8j - 7k = 5$. By Bezout Lemma solutions exist. Let's find them.
$8*1 - 7*1 = 1$ so $8*5 - 7*5 =5$ is such a solution and $x = 5*7 + a = 35+a$ and $x = 5*8 + b = 40+b$ which can work with $a = 5$ or $6$ and $b = 0$ or $1$.
So $x = 40$ or $41$ will be solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
Moreover,
$\theta=\tan^{-1}(1)=\frac{\pi}{4}$
Then, the polar form of $z$ is
$z=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Let $w\in\mathbb{C}$ such that $w^2=z$
Then
$w_k=\sqrt{\sqrt{2}}(\cos(\frac{\frac{\pi}{4}+2k\pi}{2})+i\sin(\frac{\frac{\pi}{4}+2k\pi}{2})$ for $k=0,1$.
Then the square roots, are:
$w_0=\sqrt[4]2(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8})=1,84+0,76i.$
$w_1=\sqrt[4]2(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8})=-1,84-0,76i.$
But, here i'm stuck because the $w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}\not = 1,84+0,76i$
| Just to expand on Dr. Sonnhard Graubner's answer, square-rooting the sum of the squares of these equations gives $A^2+B^2=\sqrt{2}$. (The right-hand side won't be $-\sqrt{2}$, for obvious reasons.) So $A^2=\frac{\sqrt{2}+1}{2},\,B^2=\frac{\sqrt{2}-1}{2}$. Although square-rooting these equations introduces $\pm$ signs, they're not independent; $AB>0$ so $A,\,B$ have the same sign. This gives the two desired results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute the following limit if they exist: $\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$ Compute the following limit if they exist:
$\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$
Attempt:
Given $\epsilon > 0$. Choose $\delta = \text{Not sure yet}$. Then $||(x, y, z) - (0,0,0)|| = \sqrt{x^2 + y^2 + z^2}$, and so $||(x, y, z) - (0,0,0)|| < \delta$
$\left| \frac{2x^2y c oz(z)}{x^2 + y^2} \right| \leq \frac{2x^2|y|}{x^2+y^2} \leq \frac{2 (x^2 + y^2)(\sqrt{x^2+y^2})}{x^2 + y^2} = 2 \sqrt{x^2 + y^2}$
Okay, I am stuck here. As you can see I need $\sqrt{x^2 + y^2 + z^2}$ in order for $\delta = \frac{\epsilon}{2}$.
Also what does it compute to?
| To compute the limit we have that
$$\left|\frac{2x^2 y \cos z}{x^2 + y^2}\right|\le \frac{2x^2 |y| }{x^2 + y^2}=2r\cdot(\cos^2 \theta|\sin \theta|) \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$
$A$ and $B$ are positive acute angles satisfying $3\cos^2A+2\cos^2B=4$ and $\dfrac{3\sin A}{\sin B}=\dfrac{2\cos B}{\cos A}$, then find the value of $A+2B$ ?
My Attempt
$\cos2B=2\cos^2B-1=3-3\cos^2A$ and $\sin2B=2\sin B\cos B=3\sin A\cos A$
\begin{align}
\sin(A+2B)&=\sin A\cos2B+\cos A\sin2B\\
&=\sin A.[3-3\cos^2A]+\cos A.[3\sin A\cos A]=3\sin A\\
\sin(A+2B)=3\sin A\implies \color{red}{?}
\end{align}
How do I prove that $A+2B=\dfrac{\pi}{2}$ from $\sin(A+2B)=3\sin A$ ?
What I know
\begin{align}
\cos(A+2B)&=\cos A\cos2B-\sin A\sin2B\\
&=\cos A.[3-3\cos^2A]+\sin A.[3\sin A\cos A]\\
&=3\cos A-3\cos^3A+3\sin^2A\cos A\\
&=3\cos A-3\cos A.[\cos^2A+\sin^2A]=0\implies \boxed{A+2B=\dfrac{\pi}{2}}
\end{align}
| Note that:
$$\begin{cases}\cos2B=2\cos^2B-1=3-3\cos^2A\\ \sin2B=2\sin B\cos B=3\sin A\cos A\end{cases} \Rightarrow \\
\cos^22B+\sin^22B=9-18\cos^2A+9\cos^4A+9\sin^2A\cos^2A \Rightarrow \\
1=9-18\cos^2A+9\cos^4A+9(1-\cos^2A)\cos^2A \Rightarrow \\
1=9(1-\cos^2A) \Rightarrow \\
\sin^2A=\frac19 \Rightarrow \\
\sin A=\frac13 \ \ (\text{because} \ 0<A<\frac{\pi}{2}).$$
You can plug this into your relation to get what you want:
$\sin(A+2B)=3\sin A=3\cdot \frac13=1 \Rightarrow A+2B=\frac{\pi}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ convergent?
$\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ : convergent?
My attempt
$$
(\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} = (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{n}}) + \cdots + (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n}})
$$
But I stuck here.
| Hint:
Note that for $1\leq j \leq \frac{n}{2}$ we have
$$ \frac{1}{\sqrt{j}} - \frac{1}{\sqrt{n}} \geq \frac{\sqrt{2}}{\sqrt{n}}- \frac{1}{\sqrt{n}} \geq \frac{1}{10} \cdot \frac{1}{\sqrt{n}} $$
Use this to find a lower bound of your sequence which divergeces.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to simplfy this indices question Grade 10 :$\frac{3^n-4 \cdot 3^{n-2}}{3^{n+1}-3^{n-1}}$ I am a student in grade 10 and i was given this question about indices
$$\frac{3^n-4 \cdot 3^{n-2}}{3^{n+1}-3^{n-1}}$$
All I am capable to do is expanding it out, i.e. from $3^{n-2}$ to $3 \cdot 3^{-2}$. I am unable proceed to the next step, please show step by step solution! Thank you!
| Welcome. You can write:
$3^n-4\times 3^{n-2}=3^n-3^{n-2}-3\times 3^{n-2}$
$3^{n+1}-3^{n-1}=3(3^n-3^{n-2})$
Then the fraction(F) reduces as follows:
$F=\frac{1}{3}-\frac{1}{3^{2-n}}\times(\frac{1}{3^{n+1}-3^{n-1}})=\frac{1}{3}-\frac{1}{8}=\frac {5}{24}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integration-by-part Here is the given question
$$\int{\frac{x^3}{\sqrt{1+x^2}}dx}$$
I solved using integration by part as follow:
$$
\int\frac{x}{\sqrt{1+x^2}}x^2\,dx = x^2\int\frac{x}{\sqrt{1+x^2}}\,dx - \int\biggl(\frac{dx^2}{dx}\int\frac{x}{\sqrt{1+x^2}}dx\biggr)dx\tag{i}\label{i}
$$
Solving for $\int \frac{x}{\sqrt{1+x^2}}$ and taking $1+x^2 = t \rightarrow 2xdx = dt \rightarrow xdx = \frac{dt}{2}$.
\begin{align}
&\int \frac{xdx}{\sqrt{1+x^2}} = \int\frac{dt}{2\sqrt{t}}\\
&\int\frac{dt}{2\sqrt{t}} = \frac{1}{2}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \rightarrow \sqrt{t} = \sqrt{1+x^2}
\end{align}
Pluggin in equation \eqref{i}
$$\int{\frac{x}{\sqrt{1+x^2}}x^2dx} = x^2\sqrt{1+x^2} - 2\int{x\sqrt{1+x^2}dx}\tag{ii}\label{ii}$$
Now solving for $\int{x\sqrt{1+x^2}dx}$.
Assuming $1+x^2 = t \rightarrow xdx = \frac{dt}{2}$
$$\frac{1}{2} \int\sqrt{t}dt \rightarrow \frac{\sqrt[3]{t}}{3} \text{ or }
\int{x\sqrt{1+x^2}dx} = \frac{\sqrt[3]{1+x^2}}{3}$$
Pluggin in equation
$$\int\frac{x}{\sqrt{1+x^2}}x^2dx = x^2\sqrt{1+x^2} - 2 \frac{\sqrt[3]{1+x^2}}{3} + C$$
But given answer is:
$$\frac{1}{3}(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$$
What is wrong with my solution?
| \begin{align}\displaystyle\int\dfrac{x^3}{\sqrt{1+x^2}}\,\mathrm dx&=\dfrac12\displaystyle\int\dfrac{x^2\mathrm d(x^2)}{\sqrt{1+x^2}}\\&=\dfrac12\displaystyle\int\dfrac{(1+x^2)-1}{\sqrt{1+x^2}}\mathrm d(x^2)\\&=\dfrac12\displaystyle\int\sqrt{1+x^2}\mathrm d(x^2)-\dfrac12\displaystyle\int\dfrac{\mathrm d(x^2)}{\sqrt{1+x^2}}\\&=\dfrac12\displaystyle\int\sqrt{1+x^2}\mathrm d(1+x^2)-\dfrac12\displaystyle\int\dfrac{\mathrm d(1+x^2)}{\sqrt{1+x^2}}\\&=\dfrac13(1+x^2)^{\frac32}-\sqrt{1+x^2}+C\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
| We have that
$$\prod_{k=1}^\infty \left(1+\frac{1}{k^3}\right)<3\iff \sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 3$$
and since $\forall x>0\, \log(1+x)<x$
$$\sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)=\log 2+\sum_{k=2}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 2+\sum_{k=2}^\infty \frac{1}{k^3}<\log 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
Die is rolled until 1 appears. What is the probability of rolling it odd number of times? Problem: Die is rolled until 1 appears. What is the probability of rolling it odd number of times?
So, so far I have this:
$\frac{1}{6}$ - this is a probability of rolling "1" on first try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6}$ - three tries (two times something else than "1" and "1" on the 3rd try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot
\frac{1}{6}$ - five tries
and so on...
By the looks of it, it seems I could come up with following formula
$$ p = \left( \frac{5}{6}\right)^{n-1} \cdot \frac{1}{6}$$
where p would be a probability for nth try.
However, I cannot think of any way to get probability of all odd number of tries and not sure if I am even on right track here.
| $p(1)=\frac 1 6 $ $P(not 1)=\frac 5 6$
1 on $1^{st}$ roll: $$\frac 1 6=\frac{5^0}{6^1}$$
1 on $3^{rd}$ roll: $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 1 6 = \frac {5^2} {6^3} $$
1 on $5^{th}$ roll: $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6\cdot \frac 1 6=\frac {5^4}{6^5}$$
keep going upto $\infty$: $$\frac{5^0}{6^1}+\frac{5^2}{6^3}+\frac{5^4}{6^5}+...\infty$$
use infinite GP summation: $$\frac 1 6 \cdot \frac {1}{ (1- \frac {5^2}{6^2})} $$
$$=\frac 6 {11} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proof by induction in inequalities $$\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$$ For all $n\ge 2$
Now this really a tough one for me.
The base case holds at $n = 2$
Then i replaced it with $p$ and then $p+1$
I got an inequality where i deduced that $p+1 = p$ and then plus a 1. But its not burging
| Assume that the inequality be equal for $n=p$ as you did
$$
\sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} \le {5 \over 4} - {1 \over {2p^{\,2} }}
$$
Then it ia also valid for $n=p+1$
$$
\sum\limits_{k = 1}^{p + 1} {{1 \over {k^{\,3} }}} = \sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} + {1 \over {\left( {p + 1} \right)^{\,3} }}
\le \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }}
$$
because in fact, we can write in sequence
$$
\eqalign{
& \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }} \cr
& - {1 \over {2p^{\,2} }} + {1 \over {\left( {p + 1} \right)^{\,3} }} \le - {1 \over {2\left( {p + 1} \right)^{\,2} }} \cr
& - {{\left( {p + 1} \right)^{\,2} } \over {p^{\,2} }} + {2 \over {\left( {p + 1} \right)}} \le - 1 \cr
& - \left( {p + 1} \right)^{\,3} + 2p^{\,2} \le - p^{\,2} \left( {p + 1} \right) \cr
& p^{\,2} \left( {p + 1} \right) \le \left( {p + 1} \right)^{\,3} - 2p^{\,2} \cr
& p^{\,3} + 3p^{\,2} \le \left( {p + 1} \right)^{\,3} = p^{\,3} + 3p^{\,2} + 3p + 1 \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I inserted some $n$'s to get an idea on how the sequence is going to look like.
I got:
\begin{align}
a_0&=3\\
a_1&=1\\
a_2&=\frac{7}{9}\approx 0.\overline{7}\\
a_3&=\frac{3}{4}=0.75
\end{align}
Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$
Therefore, I show that
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I am having problems when trying to prove the inequalities above:
\begin{align}
& a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\
& = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\
& = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\
& = \cdots \text{ not sure what steps I could do now}
\end{align}
Boundedness:
The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$:
\begin{align}
a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\
&=3
\end{align}
The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally.
Convergence
Assumption (s.a) $\lim_{ n \to \infty} a_n =1$
Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:
\begin{align}
\mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\
&= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\
&= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\
&= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\
&= \left|\frac{2-2n}{(n+1)^2}\right|\\
&= \cdots \text{(how to go on?)}
\end{align}
| An alternative for monotonicity:
Consider the corresponding (continuous) function $f(x) = \dfrac{x^2 + 3}{(x+1)^2}$. You can then check (using calc 1 methods) that $f'(x) > 0$ if $x \geq 3$. Hence $a_n$ is increasing for $n \geq 3$. You have already checked that $a_1 \leq a_2 \leq a_3$, so the sequence is monotone increasing.
Alternatively, to show boundedness, first show that $a_n$ converges to $1$, and then use the fact that every convergent sequence is bounded (you should think about why this last fact is true).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_1=F_2=1$, we suppose that $n \geq 3$.
$$\begin{align}
F_{n+1}&=F_n+F_{n-1} \\
F_{n+1}^2&=(F_n+F_{n-1})^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2
\end{align}$$
$$\begin{align}
F_{n-1} &=F_{n-2}+F_{n-3} \\
F_{n-1}^2&=(F_{n-2}+F_{n-3})^2=F_{n-2}^2+2F_{n-2}F_{n-3}+F_{n-3}^2
\end{align}$$
so that
$$
F_{n+1}^2-F_{n-1}^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2-F_{n-2}^2-2 F_{n-2} F_{n-3}-F_{n-3}^2
$$
How can we deduce that the latter is divisible by $4$?
Or do we show it somehow else, for example by induction?
| Note that$$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=F_n(F_n+2F_{n-1}).$$And if $2\mid F_n$, $4\mid F_n(F_n+2F_{n-1})$.
On the other hand\begin{align}F_{n+1}^3-F_{n-1}^3&=(F_{n+1}-F_{n-1})(F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2)\\&=F_n\bigl((F_n+F_{n-1})^2+F_n+F_{n-1}+F_{n-1}^2+F_{n-1}^2\bigr)\\&=F_n(F_n^2+3F_nF_{n-1}+3F_{n-1}^2).\end{align}And if $3\mid F_n$, $9\mid F_n(F_n^2+3F_nF_{n-1}+3F_{n-1}^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+b^2)\cos^2x-2ac\cos x+(c^2-b^2)=0\\
\implies\cos^2x-\frac{2ac}{a^2+b^2}\cos x+\frac{c^2-b^2}{a^2+b^2}=0
$$
$$
a\cos\alpha+b\sin\alpha=c\implies a\cos^2\alpha\cos\beta+b\sin\alpha\cos\alpha\cos\beta=c\cos\alpha\cos\beta\\
a\cos\beta+b\sin\beta=c\implies a\sin\alpha\sin\beta\cos\beta+b\sin\alpha\sin^2\beta=c\sin\alpha\sin\beta\\
c\cos(\alpha+\beta)=a\cos\beta+a\sin\alpha\cos\beta.(\sin\beta-\sin\alpha)+b\sin\alpha+b\sin\alpha\cos\beta(\cos\alpha-\cos\beta)\\
$$
I think its getting complicated to solve now. What is the simplest way to solve this kind of problems?
| I would like to present you a geometric explanation of what is happening.
If $\xi = \cos x$ and $\eta=\sin x$, then you rewrite your equation as:
$$
a\xi +b\eta=c,\qquad \xi^2+\eta^2=1.
$$
So you are trying to find intersection points of a line and a circle.
Simple geometry tells us that bisector $AF$ is perpendicular to line $DE$. If $\angle CAD = \alpha$ and $\angle CAE = \beta$, then $\angle CAF=(\alpha+\beta)/2$. On the other hand, coordinates of $F$ are
$$\xi_F = \pm \frac a{\sqrt{a^2+b^2}},\qquad \eta_F=\pm \frac b{\sqrt{a^2+b^2}}$$
because vector $(a,b)$ is normal to line $a\xi +b\eta=c$, and then we scale it so $F$ lies on a unit circle. We also have $\pm$ here, because bisector intersects circle in two points, and which of them is $F$ depends on the value of $c$.
$$\cos(\alpha+\beta)=\cos\left(2\frac{\alpha+\beta}2\right) = \cos^2\frac{\alpha+\beta}2 - \sin^2\frac{\alpha+\beta}2 = \xi_F^2 -\eta_F^2=\frac{a^2-b^2}{a^2+b^2}.$$
Notably, our answer doesn't depend on the choice of the signs, as $(\pm a)^2 = a^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determine the image of $S$ under $f(z)=\frac{z^2+2}{z^2+1}.$
Let $S$ be the region $\{z:0<|z|<\sqrt{2}, \ 0 < \text{arg}(z) <
\pi/4\}$. Determine the image of $S$ under the transformation
$$f(z)=\frac{z^2+2}{z^2+1}.$$
I'm facing some difficulties on problems of this nature that includes MΓΆbius transformations. First I can see that under $z\mapsto z^2$ doubles the argument and squares the modulus. So now we have that our $S$ is transformed to
$$S_1=\{z:0<|z|<2, \ 0 < \text{arg}(z) <
\pi/2\}$$
We can now let $M(z)=\frac{z+2}{z+1}.$ But here we need to parametrize the boundary of $S_1$. Thus we have that
$$\gamma_1=2it, \quad t\in[0,1]$$
$$\gamma_2=2t, \quad t\in[0,1]$$
$$\gamma_3=2e^{it}, \quad t\in[0,\pi/2]$$
So,
$$M(\gamma_1)=\frac{2it+2}{2it+1}=\frac{4t^2+2}{4t^2+1}-i\frac{2t}{4t^2+1},$$
$$M(\gamma_2)=\frac{2t+2}{2t+1}$$
$$M(\gamma_3)=\frac{2e^{it}+2}{2e^{it}+1}$$
substituting $t=0$ and $t=1$ in $M(\gamma_1)$ we get the two points $2$ and $\frac{6}{5}-i\frac{2}{5}$. This means that $\gamma_1$ is mapped to a circle or line going through those points.
Doing the same for $M(\gamma_2)$ we get that the two points are $2$ and $\frac{4}{3}$.
However, plotting these points dont really make sense. Also, I'm not sure how to handle $M(\gamma_3)$, it gets very messy. This leads me to assume that the method I'm using is very inefficient.
Any way around this?
| I think this might help:
$$f(z)=\frac{2+z^2}{1+z^2}=1+\frac{1}{1+z^2}$$
If $z=x+iy$:
$$\frac{1}{1+z^2}=\frac{1}{(x^2-y^2+1)+2ixy}=\frac{(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$
So:
$$f(z)=\frac{(x^2-y^2+1)^2+4x^2y^2+(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Algebraic proof to show product of three numbers is zero from a system Let $a, b, c$ be real numbers satisfying: $$(a+b)(b+c)(c+a)=abc$$
$$(a^3+b^3)(b^3+c^3)(a^3+c^3)=a^3b^3c^3$$
Prove that $abc=0$
My work: I tried to prove it by contradiction, assuming $abc\neq0$ so $a,b,c\neq0.$
Then I factored second equation to be $(a+b)(b+c)(a+c)(a^2-ab+b^2)(b^2-bc+c^2)(a^2-ac+c^2)=(abc)^3$. Then when you divide the second equation by the first equation, you get $(a^2-ab+b^2)(b^2-bc+c^2)(a^2-ac+c^2)=a^2b^2c^2$. At this point I think there maybe a AM-GM inequality I can use, but I'm stuck. Any suggestions or alternate ways to prove this?
| Look at the first equation as a quadratic equation in $a$. It has discriminant $(b^2+bc+c^2)^2-4bc(b+c)^2=(b^2-bc+c^2)-8b^2c^2$. So we have
$$
\begin{align*}
a^4b^4c^4&=(a^2b^2c^2)^2\\
&=\prod_{cyc}(b^2-bc+c^2)^2\\
&\geq\prod_{cyc}8b^2c^2\\
&=512a^4b^4c^4
\end{align*}
$$
which only works if $abc=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How is $2(\ln \tan x + c)$ simplified to $A \tan^2 x$ where $A =2c$ I'm trying to follow this reasoning:
$1/2 \ln(4 +y^2) = \ln(\tan x) + C$
$\ln(4 +y^2) = 2\ln(\tan x) + \ln A$ ( constant 2C = A)
$4 + y^2 = A \tan^2 x$
| $\frac{1}{2}ln(4+ y^2)= ln(tan(x))+ C$
Multiply on both sides by 2:
$ln(4+ y^2)= 2ln(tan(x))+ 2C$
Taking $A= e^{2C}$ (NOT A= 2C) we have $2C= ln(A)$ so
$ln(4+ y^2)= ln(tan^2(x))+ ln(A)= ln(Atan^2(x))$
Now take the exponential of both sides:
$e^{ln(4+ y^2)}= e^{ln(Atan^2(x))}$
$4+ y^2= Atan^2(x)$
I have used:
1) $2ln(x)= ln(x^2)$
2) $ln(a)+ ln(b)= ln(ab)$
3) $e^{ln(x)}= x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation .
Help appreciated.
I've made some research in MSE and found several questions which could be helpful:
1) $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=-\frac{17\pi^4}{1440}$
2) $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{\pi^6}{12960}$
3) $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}= \frac{1}{3}\zeta(6)$
|
The double summation is equal to
$$\frac{11\zeta(4)}{8}=\frac{11\pi^4}{720}.$$
Note that
$$\frac{m^2+n^2}{m n\left(m^2-n^2\right)^2}=
\frac{1}{2 mn(m+n)^2}+
\frac{1}{2 mn(m-n)^2}.$$
Now consider the Tornheim double sums:
$$T(a,b,c)= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}
\frac{1}{m^an^b(m+n)^c}.$$
Then
$$ \sum_{m,n=1\, m\neq n}^{\infty} \frac{1}{mn(m+n)^2}=T(1,1,2)-\sum_{m=1}^{\infty}\frac{1}{4m^4}=T(1,1,2)-\frac{\zeta(4)}{4},$$
$$\begin{align} \sum_{m=1}^{\infty} \sum_{n=1}^{m-1}
\frac{1}{mn(m-n)^2}
&= \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty}
\frac{1}{mn(m-n)^2}\\
&= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}
\frac{1}{nk^2(n+k)}=T(1,2,1)\end{align}$$
and again
$$ \sum_{m=1}^{\infty} \sum_{n=m+1}^{\infty}
\frac{1}{mn(m-n)^2}=T(1,2,1).$$
Hence
$$\begin{align}\sum_{m,n=1\, m\neq n}^{\infty}{\frac{m^2+n^2}{mn(m^2-n^2)^2}}
&=\frac{T(1,1,2)-\zeta(4)/4}{2} +T(1,2,1)\\
&=\frac{11\zeta(4)}{8}
\end{align}$$
where we used
$$T(1,1,2)=\zeta(4)/2\quad,\quad T(1,2,1)=5\zeta(4)/2$$
(see page 31 in The evaluation of Tornheim double sums).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $ $$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+1}= $$
$$\lim_{n \to\infty} \sum_{k=1}^{n} \arctan(\sqrt{k^2+2}+\sqrt2)-\arctan(\sqrt{k^2+2}-\sqrt2) $$
But the elements do not reduce :(
| Note that
$$\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\arctan\left(\frac{\left(\frac{k+1}{\sqrt{2}}\right)-\left(\frac{k-1}{\sqrt{2}}\right)}{\left(\frac{k+1}{\sqrt{2}}\right)\left(\frac{k-1}{\sqrt{2}}\right)+1}\right)=\arctan\left(\frac{k+1}{\sqrt{2}}\right)-\arctan\left(\frac{k-1}{\sqrt{2}}\right)\,.$$
Therefore,
$$\sum_{k=1}^n\,\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\arctan\left(\frac{n+1}{\sqrt{2}}\right)+\arctan\left(\frac{n}{\sqrt{2}}\right)-\arctan\left(\frac{1}{\sqrt{2}}\right)-\arctan\left(\frac{0}{\sqrt{2}}\right)\,.$$
Ergo,
$$\sum_{k=1}^\infty\,\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\pi-\arctan\left(\frac{1}{\sqrt{2}}\right)\approx 2.52611\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimize the expression This was a problem given in a no calculator math contest (HMMT):
What is the minimum value of $(xy)^2+(x+7)^2+(2y+7)^2$ where $x$ and $y$ are reals. My friends and I discussed about using calculus, but we reasoned that there must be a faster trick (calc would take too long). Any ideas?
| Hint:
\begin{align}
f(x,y) &= (xy)^2+(x+7)^2+(2y+7)^2 \\
&= (xy-2)^2+(x+2y+7)^2+45 \\
\end{align}
Details for the thoughts
By $(a+b+c)^2=a^2+b^2+c^2+2(bc+ca+ab)$
Hence,
\begin{align}
(x+7)^2+(2y+7)^2 &= x^2+14x+49+4y^2+28y+49 \\
&= (x^2+4y^2+49+14x+28y \color{red}{+4xy})+49 \color{red}{-4xy} \\
&= (x+2y+7)^2+49-4xy \\
\end{align}
Also,
$$(xy)^2-4xy+4=(xy-2)^2$$
Therefore,
\begin{align}
f(x,y) &= (xy-2)^2+(x+2y+y)^2+49-4 \\
&= (xy-2)^2+(x+2y+7)^2+45
\end{align}
Note that $(xy-2)^2 \ge 0 \, $ and $\, (x+2y+7)^2 \ge 0$, so $f(x,y)$ attains minimum when
$$
\left \{
\begin{align}
xy-2 &= 0 \\
x+2y+7 &= 0
\end{align}
\right.$$
which gives real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The $1997$ IIT JEE problem
Let $S$ be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$
This problem appeared in IIT JEE $1997$ (re-exam).I really do not know know how to approach this problem properly.I have managed to prove $a^2 + b^2 + c^2 +d^2 \leq 4$
but not the initial inequality.
| The smallest $a^2+b^2+c^2+d^2$ is when the quadrilateral has its 4 vertices on the midpoint of each side of square S. $a=b=c=d=0.5$, and the sum of the squares is $2$. The largest is simply achieved when the vertices of the quadrilateral lies on the vertices of square S. ie, both squares are the same. The sum of the squares of sides is $4$ then.
Now why exactly is this so? Let the sides of the square be $w, x, y, z$, and let the subscript 1 and 2 denote a section of the side (e.g. $w_1+w_2=w$). Then $a^2+b^2+c^2+d^2=w_2^2+x_1^2+x_2^2+y_1^2+y_2^2+z_1^2+z_2^2+w_1^2$. The restriction here lies in that $w+x+y+z$ is fixed, the perimeter of S. We know that $2\times a^2<(a-1)^2+(a+1)^2$, for any $a>=1$. Hence, since we are trying to achieve the smallest possible result, we would see each side of the square divided exactly in half, such that $w_1=w_2=...=0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Lograthmic equation $ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}$ solutions $$ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}.$$
I made the graph using wolfram alpha it is giving answer as
6. But how to solve it algebraically?
base of logarithm is 2.
Tried using taking Lcm but then two different log terms
are getting formed.
| $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2} = \frac{5}{12}$$
Rewrite the logs using $$\log_a b^c = c\log_a b$$
and factor. Then, simplify both sides.
$$\frac{1}{2}\cdot\frac{1}{\log_2\vert x-2\vert}+\frac{1}{2}\cdot\frac{1}{\log_2\vert x+2\vert} = \frac{5}{12}$$
$$\frac{1}{2}\bigg(\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert}\bigg) = \frac{5}{12}$$
$$\frac{1}{\log_2\vert x-2\vert}+\frac{1}{\log_2\vert x+2\vert} = \frac{5}{6}$$
$$\frac{\log_2\vert x-2\vert+\log_2\vert x+2\vert}{\log_2\vert x-2\vert\cdot\log_2\vert x+2\vert} = \frac{5}{6}$$
Set $\color{blue}{a = \log_2\vert x-2\vert}$ and $\color{purple}{b = \log_2\vert x+2\vert}$.
$$\frac{\color{blue}{a}+\color{purple}{b}}{\color{blue}{a}\color{purple}{b}} = \frac{5}{6}$$
Can you take it on from here? (Hint: Solve for possible values of $a$ and $b$. Then, plug in $\color{blue}{\log_2\vert x-2\vert = a}$ and $\color{purple}{\log_2\vert x+2\vert = b}$ and check for any extraneous solutions, in case there are any.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving characteristic equation for closed form solution from a given recurrence equation Provided a recurrence relationship, I converted the equation assuming $T(n) = r(n)$, and found the roots to the quadratic equation.
root 1: $r_1 = 2+\sqrt{3}$
root 2: $r_2 = 2-\sqrt{3}$
Resulting in the equation with the form $T(n) = c_1(r_1)^n + c_2(r_2)^n$
$$T(1): 1 = c_1(2+\sqrt{3})^1 + c_2(2-\sqrt{3})^1$$
$$T(3): 3 = c_1(2+\sqrt{3})^3 + c_2(2-\sqrt{3})^3$$
How do I go about solving for $c_1$ and $c_2$ in this particular instance. I'm used to an initial condition being 0, eliminating one of the $c$'s, then plugging the remaining into the second equation. But I am stuck here.
My end goal is to result with a closed form solution to a recurrence relationship from $T(n)$.
| For brevity, take
\begin{align}
\alpha &= 2+\sqrt{3} \\
\beta &= 2-\sqrt{3} \\
A &= c_1 \\
B &= c_2
\end{align}
Now,
\begin{align}
\alpha+\beta &= 4 \\
\alpha-\beta &= 2\sqrt{3} \\
\alpha \beta &= 1 \\
A\alpha+B\beta &= 1 \tag{1} \\
A\alpha^3+B\beta^3 &= 3 \tag{2}
\end{align}
$(2)-\beta^2 \times (1)$,
\begin{align}
A\alpha (\alpha^2-\beta^2) &= 3-\beta^2 \\
A\alpha (\alpha+\beta)(\alpha-\beta) &= 3-(4-4\sqrt{3}+3) \\
8A\alpha\sqrt{3} &= 4(\sqrt{3}-1) \\
A &= \frac{\sqrt{3}-1}{2\sqrt{3}(2+\sqrt{3})} \\
&= \frac{(\sqrt{3}-1)(2-\sqrt{3})}{2\sqrt{3}} \\
&= \frac{9-5\sqrt{3}}{6}
\end{align}
$\alpha^2 \times (1)-(2)$,
\begin{align}
B\beta (\alpha^2-\beta^2) &= \alpha^2-1 \\
B\beta (\alpha+\beta)(\alpha-\beta) &= (4+4\sqrt{3}+3)-3 \\
8B\beta\sqrt{3} &= 4(\sqrt{3}+1) \\
B &= \frac{9+5\sqrt{3}}{6}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ I need some help calculating this limit:
$$\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$$
I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Hopital. Thanks a lot.
| EDIT
I guess you need a basic method that preceeds the l'Hospital's rule.
Set $x=a^2,\; a>0.$ The limit rewrites
$$ \begin{aligned}\sqrt{a^2+a} - \sqrt{a^2-a}=&\;\left(\sqrt{a^2+a} - \sqrt{a^2-a}\right)\frac{\sqrt{a^2+a}+\sqrt{a^2-a}}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\
=&\;\frac{2a}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\=&\;
\frac{2}{\sqrt{1+a^{-1}}+\sqrt{1-a^{-1}}}\to 1\; {\text {as}}\; a \to \infty\end{aligned}$$
My first answer
$$ \begin{aligned}\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}}=&\;
\sqrt{\sqrt{x}(\sqrt{x}+1)} - \sqrt{\sqrt{x}(\sqrt{x}-1)}\\=&\;
\sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\\
=&\; \sqrt[4]{x}\left( \sqrt{\sqrt{x}+1}-\sqrt{\sqrt{x}-1}\right)\cdot\frac{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\;
\sqrt[4]{x}\cdot\frac{2}{\sqrt{\sqrt{x}+1}+\sqrt{\sqrt{x}-1}}\\=&\;
\frac{2}{\sqrt{1+x^{-1/4}}+\sqrt{1-x^{-1/4}}}\end{aligned}$$ from where the limit is $1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)\left(\frac{x}{5}\right)} \: dx = \frac{\pi}{2}$$
To begin with, I made a simple rearrangement
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{x^3} \: dx = \frac{\pi}{30}$$
From here I used the Sine/Cosine Identities
$$ \int_{0}^{\infty} \frac{\frac{1}{4}\left(-\sin\left(\frac{7}{15}x\right)+ \sin\left(\frac{13}{15}x\right) + \sin\left(\frac{17}{15}x\right) -\sin\left(\frac{23}{15}x\right) \right)}{x^3} \: dx = \frac{\pi}{30}$$
Which when expanded becomes
$$ -\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx +
\int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx -
\int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx
= \frac{2\pi}{15}$$
Using the property
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^3}\:dx = a^2 \int_{0}^{\infty}\frac{\sin(x)}{x^3}\:dx$$
We can reduce our expression to
$$\left[ -\left(\frac{7}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(\frac{17}{15}\right)^2 - \left(\frac{23}{15}\right)^2\right] \int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
Which simplifies to
$$ -\frac{120}{15^2}\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
And from which we arrive at
$$\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = -\frac{\pi}{4}$$
Is this correct? I'm not sure but when I plug into Wolframalpha it keeps timing out...
| As the other answers have pointed out, the integral does indeed diverge. But if want to assign a finite value to it, there are a couple ways to see that in fact $-\pi/4$ is the "right" value.
One is to take the integral not quite down to zero, but instead to $\epsilon$. If we do, then expand in a series in $\epsilon$, we get
$$\int_\epsilon^\infty\frac{\sin x}{x^3}\,dx=\epsilon^{-1}-\frac{\pi}{4}+O(\epsilon).$$
The leading term is the divergent $\epsilon^{-1}$, but if we ignore that then the next term is $-\pi/4$.
Another way to get the same value is to first extend the integral to $-\infty$. Since the integrand is even, we would expect
$$\int_0^\infty\frac{\sin x}{x^3}\,dx=\frac12\int_{-\infty}^\infty\frac{\sin x}{x^3}\,dx.$$
Of course, the right-hand integral diverges, as well. But the only problem point is when $x=0$. If we imagine $x$ as being in the complex plane, travelling from $-\infty$ to $\infty$, then we can just "go around" $0$ by curving $x$ slightly out into the complex plane, for example:
If we do, then we end up with the same answer of $-\pi/4$. I'm not sure what this sort of regularization is called, but it's frequently used in quantum field theory where divergent integrals abound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$ How to prove
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$
where $G$ is the Catalan's constant.
Attempt
For the first sum,
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\left\{\sum_{k=1}^{\infty}\frac{i^k}{(k+1)^2}H_{k}\right\},$$
which can be evaluated by using the formula in this post:
$$\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x),$$
but we cannot apply the similar approach to the second sum
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_k.$$
Then, I tried to write the sum as
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}\int_0^1\frac{2x^{2k}+x^k-3}{x-1}~\mathrm dx$$
and it become more complicated.
Edit:
Are we able to evaluate the sum directly (avoid calculating integrals and polylogs as much as possible)? The integral given by @Jack D'Aurizio is a bit complicated (see this post).
| this summation also has some relationship with integral identity, which asked almost the same time, see this, rewrite as
$$\frac3{2} \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x}$$
given some famous series
$$\arctan x = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}$$
$$\frac1{2} \arctan x \ln (1+x^2) = \sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n+1}}$$
on the left
$$\frac3{2} \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \mathrm{d}x} = 3\int_{0}^{1} {\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} x^{2n}} \>\mathrm{d}x} = 3\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{(2n+1)^2}}$$
on the right
$$\begin{aligned}
\int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x}
& = \int_{0}^{1} {\sum_{n=0}^{\infty} {\frac{(-1)^n}{2n+1}} x^{2n}\ln(1+x) \>\mathrm{d}x}\\
& = 2\ln2\sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}} - \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \int_{0}^{1} {\frac{1+x^{2n+1}}{1+x}}\mathrm{d}x}\\
& = 2G\ln2 - \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \int_{0}^{1} {\sum_{k=0}^{2n} {(-x)^k}}\mathrm{d}x}\\
& = 2G\ln2 + \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2} \sum_{k=1}^{2n+1} {\frac{(-1)^k}{k}}}\\
& = 2G\ln2 + \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}-H_{2n+1})}
\end{aligned}$$
unscramble this identity
$$\sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}+3H_{2n}-H_{2n+1})} = -2G\ln2$$
or
$$\begin{aligned}
\sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{n}+2H_{2n})}
& = \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^2}(H_{2n+1}-H_{2n})} - 2G\ln2\\
& = \sum_{n=0}^{\infty} {\frac{(-1)^n}{(2n+1)^3}} - 2G\ln2 = \frac{\pi^3}{32} - 2G\ln2
\end{aligned}$$
where you may need some particular values of Dirichlet Beta Function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Complex root of equation 1 let $a,b$ & $c$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $\frac a{a-2},\frac b{b-2}$ and $\frac c{c-2} $
hence or otherwise find value of $(a-2)(b-2)(c-2)$
| HINT
We have that $(x-a)(x-b)(x-c)=x^3-3x^2+1$ and then
*
*$a+b+c=3$
*$ab+bc+ca=0$
*$abc=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{HTHTHTHTH}$
Then we'll be left with 1 tail which can be placed at 6 places.
So number of cases = $6$
2) $4$ heads, $6$ tails
$\mathrm{HTHTHTH}$
Left with $3$ tails.
So all $3$ can be placed together at $5$ places.
Or $3 = 2+1$, in which case we have $5 \times 4 = 20$ cases.
Total number for #2 $= 20 + 5 = 25$
3) 3 heads, 7 tails.
$\mathrm{HTHTH}$
Left with $5$ tails.
All 5 can be placed together, so in that case we have $4$ cases.
Or $5 = 2+3= 1+4$
For $5 = 2+3 = 1+4$, we have $4\times 3$ cases each.
Therefore total number of cases for #3 $= 12+12+ 4 = 28$
4) 2 heads 8 tails.
$\mathrm{HTH}$
$7$ left
$7 = 7+ 0$ , $3$ cases.
$7 = 6+1= 5+2= 3+4$ , $6$ cases each
So number of cases for #4 = $6\times 3 + 3 = 21$
5) 1 head 9 tail, 10 cases
6) 0 heads, 1 case.
So total number of cases = $6+25+28+21+10 +1 = 91$
I tried thrice and got $91$ cases only.
So answer should be $P(E) = \dfrac{91}{2^{10}}$
But that's not the right answer.
Please tell me my mistake.
| You are missing the cases where the tails can be split into more than 2 groups.
i.e. 3 tails can be split into 1+1+1, yielding another 10 cases = $ {5 \times 4 \times 3\over 3 \times 2 \times 1} $ etc...
The final answer should be $144\over2^{10}$. There are 28 more cases for 3 heads and 15 more cases for 2 heads.
Key Flex's answer is a more straightforward way to calculate the answer. To think about it more intuitively, instead of breaking tails into differently sized groups to place between the heads, break the heads into individual groups of 1 to slot between the tails.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is $\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$? What's the result of:
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$$
Is it
$$\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|x|}=0$$
or
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}=\frac{1}{|{x}|}-\frac{1}{x}\frac{x^2}{|x|^2}=\frac{1}{|{x}|}-\frac{1}{x}=\left\{\begin{matrix}\frac{2}{x},x<0\\ 0,x>0 \end{matrix}\right.$$
| Your first conclusion is right since $$|x|^3=|x|^2\cdot |x|=x^2\cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3\ne x^3= |x|^2\cdot x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Show that these non-linear recursions produce integers only The recurrence is of third order:
Start with
\begin{align*}a_0(x)&=1\\ a_1(x)&=1\\ a_2(x)&=x \end{align*} and then
\begin{align*}a_{n+3}(x)&=\frac{a_{n+2}^2(x)-a_{n+1}^2(x)}{a_{n}(x)}.\end{align*}
The calculation and factorization of $a_n(x)$ for $3\le n \le 6$ gives the following results:
\begin{align*} a_3(x)&= (x+1)(x-1)\\
a_4(x)&=(x^2 + x -1)(x^2-x-1)\\
a_5(x)&=x(x^2 -2)(x^4-4x^2+2)\\
a_6(x)&=(x^6 +x^5-5x^4-4x^3+6x^2+3x-1)(x^6 -x^5-5x^4+4x^3+6x^2-3x-1)\\
\end{align*}
Questions:
*
*Any reference for this ?
*Show that $a_n(x)$ is an integer coefficient polynomial. It would be of
degree $F_{n+1} -1$, where $F_n$ is a Fibonacci number.
*Is it true that $a_n(x)$ divides $a_{2n+1}(x)$?
*For $k \in \mathbb N, k>1$, is there a general close form for $a_n(k)$?
For $k=2$, we obviously have $a_n(2)=F_{n+1}$.
For $k=3$, we have $a_n(3)=F_{2F_{n+1}}$.
What about $a_n(4)$?
| Nice recursion!! However, let us change the name and indexing of your sequence. Notice that if $\,U_n(x)$ is the Chebyshev polynomial of the second kind, then $\, U_n(x/2)$ is a polynomial in $x$ with integer coefficients. Define $\, P_n(x) := U_{F_n - 1}(x/2).\,$ Your sequence $\, a_n(x) = P_{n+1}(x).\,$ Please note the different indexing.
Check the following equations hold true for all integer $\,n$
$$ P_1(x) = P_2(x) = 1, \,\, P_3(x) = x,\,\,
P_n(x)P_{n+3}(x) = P_{n+2}^2(x) - P_{n+1}^2(x).$$
Both $\,U_{n-1}(\cos\theta) = \sin(n\theta)/\sin(\theta)\,$ and $\,F_n = U_{n-1}(i/2)/i^{n-1}\,$ are divisibility sequences,
hence $\,P_n(x)\,$ divides $\,P_{kn}(x)\,$ for all integer $\,n,k\,$ where $\,P_n(x)\ne 0.\,$ We also have
$\,P_n(x) = -(-1)^n P_{-n}(x)\,$ for all integer $\,n.\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
approximation of trigonometric functions in form $a + bx + cx^2 +...$ Is there any way to express the trigonometric functions as infinitely long polinomials? If so, how? If not why?
Obviously doing $x(x-\pi)(x-2\pi)(x-3\pi)...$ does not work as it doesn't match for values in between the zeroes.
| You already received good answers to the question.
In this area, we can do other things which are amazing. If you look here, being myself fascinated by the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which wasproposed, more than $1400$ years ago by Mahabhaskariya of Bhaskara I (a seventh-century Indian mathematician), I wrote
$$\sin(x)=\sum_{k=1}^n a_k [(\pi-x)x]^k$$ which respects the value of $0$ at $x=0$ and $x=\pi$. The first coefficients are
$$\left(
\begin{array}{cc}
k & a_k \\
1 & \frac{1}{\pi } \\
2 & \frac{1}{\pi ^3} \\
3 & \frac{12-\pi ^2}{6 \pi ^5} \\
4 & \frac{10-\pi ^2}{2 \pi ^7} \\
5 & \frac{1680-180 \pi ^2+\pi ^4}{120 \pi ^9} \\
6 & \frac{1008-112 \pi ^2+\pi ^4}{24 \pi ^{11}} \\
7 & \frac{665280-75600 \pi ^2+840 \pi ^4-\pi ^6}{5040 \pi ^{13}} \\
8 & \frac{308880-35640 \pi ^2+450 \pi ^4-\pi ^6}{720 \pi ^{15}}
\end{array}
\right)$$ The maximum error is at $x=\frac \pi 2$; for example, using $n=8$, the error is just $4.99 \times 10^{-13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating $\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{m-1} |mi-nj|$ I'd like to calculate $f(n,m)=\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{m-1} |mi-nj|$ for all $1 \leq n \leq N,\ 1 \leq m \leq M$. Straightforward brute force method runs in $O(N^2M^2)$ which is too slow. How to calculate all values in $O(NM)$?
|
We show the following is valid for positive integers $n,m$:
\begin{align*}
\sum_{i=1}^{n-1}\sum_{j=1}^{m-1}|mi-nj|=\frac{1}{6}\left(2m^2n^2-3m^2n+m^2-3mn^2+3mn+n^2-\left(\gcd(m,n)\right)^2\right)
\end{align*}
In the following we denote with $d=\gcd(m,n)$.
We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^{n-1}}&\color{blue}{\sum_{j=1}^{m-1}|mi-nj|}\tag{2}\\
&=2\sum_{i=1}^{n-1}\sum_{j=1}^{\lfloor mi/n\rfloor}(mi-nj)\tag{3}\\
&=2m\sum_{i=1}^{n-1}i\sum_{j=1}^{\lfloor mi/n\rfloor}1-2n\sum_{i=1}^{n-1}\sum_{j=1}^{\lfloor mi/n\rfloor}j\\
&=2m\sum_{i=1}^{n-1}i\left\lfloor\frac{m}{n}i\right\rfloor
-n\sum_{i=1}^{n-1}\left\lfloor\frac{m}{n}i\right\rfloor\left(\left\lfloor\frac{m}{n}i\right\rfloor+1\right)\tag{4}\\
&=\sum_{i=1}^{n-1}\left(2mi-n\left\lfloor\frac{m}{n}i\right\rfloor-n\right)\left\lfloor\frac{m}{n}i\right\rfloor\tag{5}\\
&=\sum_{i=1}^{n-1}\left(2mi-n\left(\frac{m}{n}i-\left\{\frac{m}{n}i\right\}\right)-n\right)\left(\frac{m}{n}i-\left\{\frac{m}{n}i\right\}\right)\tag{6}\\
&=n\sum_{i=1}^{n-1}\left(\frac{m^2}{n^2}i^2-\left\{\frac{m}{n}i\right\}^2-\frac{m}{n}i+\left\{\frac{m}{n}i\right\}\right)\\
&=\frac{m^2}{n}\sum_{i=1}^{n-1}i^2-n\sum_{i=1}^{n-1}\left\{\frac{m}{n}i\right\}^2-m\sum_{i=1}^{n-1}i+n\sum_{i=1}^{n-1}\left\{\frac{m}{n}i\right\}\\
&=\frac{m^2}{n}\frac{1}{6}(n-1)n(2n-1)-nd\sum_{i=0}^{n/d-1}\left(\frac{d}{n}i\right)^2\\
&\qquad -m\frac{1}{2}(n-1)n+nd\sum_{i=0}^{n/d-1}\left(\frac{d}{n}i\right)\tag{7}\\
&=\frac{1}{6}m^2(n-1)(2n-1)-\frac{d^3}{n}\frac{1}{6}\left(\frac{n}{d}-1\right)\frac{n}{d}\left(\frac{2n}{d}-1\right)\\
&\qquad-\frac{1}{2}mn(n-1)+d^2\frac{1}{2}\left(\frac{n}{d}-1\right)\frac{n}{d}\tag{8}\\
&\,\,\color{blue}{=\frac{1}{6}\left(2m^2n^2-3m^2n+m^2-3mn^2+3mn+n^2-d^2\right)}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (3) we use that positive and negative parts in (2) correspond to each other.
*In (4) we expand the inner sums.
*In (5) we rearrange the terms and factor out $\left\lfloor\frac{m}{n}i\right\rfloor$.
*In (6) we rewrite the expression using the fractional part $\{x\}=x-\lfloor x\rfloor$ of $x$.
*In (7) we expand the sums with linear and quadratic terms and we apply the identity
\begin{align*}
\sum_{i=1}^{n}f\left(\left\{\frac{m}{n}i\right\}\right)=d\sum_{i=0}^{n/d-1}f\left(\frac{d}{n}i\right)
\end{align*}
where $d=\gcd(m,n)$.
*In (8) we expand the sums and simplify the expression in the final step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the upper Riemann integral of following function? $f(x)$ is defined on $[a,b]$ as -
$= 0$ if $x β [a, b] β© Q$
$= x$ otherwise
Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.
| Since the irrationals are dense, we have $\sup_{x \in [x_{j-1},x_j]}f(x) = x_j$.
Hence, for any partition, $a = x_0 < x_1 < \ldots < x_n = b$, the upper Darboux sum is
$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) = \frac{1}{2} \sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \\ = \frac{1}{2} \sum_{j=1}^n(x_j^2 - x_{j-1}^2) + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2 = \frac{b^2-a^2}{2} + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2, $$
Since the last term on the RHS is nonnegative we have, for the upper integral,
$$\overline{\int_a^b} f(x) \, dx = \inf_{P}U(P,f) \geqslant \frac{b^2 - a^2}{2}$$
To prove we actually have $\inf_{P}U(P,f) = \frac{b^2 - a^2}{2}$, we show that for any $\epsilon > 0$ there exists a partition such that
$$U(P,f) < \frac{b^2 - a^2}{2} + \epsilon$$
For a uniform partition where $x_j - x_{j-1} = \frac{b-a}{n}$ for $j=1,2, \ldots, n$ we have
$$\frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2 = \frac{1}{2} n\frac{(b-a)^2}{n^2} = \frac{(b-a)^2}{2n}$$
Now choose $n > \frac{(b-a)^2}{2\epsilon}$ and we have
$$\begin{align}U(P,f) &=\frac{b^2 - a^2}{2} + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2\\ &= \frac{b^2 - a^2}{2}+ \frac{(b-a)^2}{2n} \\&< \frac{b^2 - a^2}{2} + \epsilon\end{align}$$
Therefore,
$$\overline{\int_a^b} f(x) \, dx = \inf_{P}U(P,f) = \frac{b^2 - a^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line y=x+2 it is the minimum sum of distance from the above given two points.
But from here I am not able to get the value of x and hence $\alpha$. Any suggestions?
| By letting $x=t+1$ we get an even function
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}.$$
Now we show that the minimum value is attained at $t=0$: we have to verify
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}\geq \sqrt{20}$$
or, after squaring,
$$4t^2+10+2\sqrt{4t^4-16t^2+25}\geq 20$$
that is
$$\sqrt{(5-2t^2)^2+4t^2}\geq 5-2t^2$$
which trivially holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x^3-1)(x^2-x+1)\\
&=(x-1)(x^2+x+1)(x^2-x+1).
\end{align}
If we set $Q'(x)$ equal to zero, then I get a real critical point of $x=1$. I also get complex critical points from $(x^2 -x +1)$. Since I'm asked if $Q(x)$ has any real roots, should I still plug in the complex critical points into $Q(x)$?
| egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $\ln(x+1)$:
$$\ln(x+1)=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-\ln(x+1)+O(x^7)+1.$$
As $x\to-1$ (here, the series expansion works), $\ln(x+1)$ grows much quicker than $x^7$, so no zeroes there. You can check from the original expression of $Q$ that there are no zeros as $x\to\infty$ as well. Of course this is just heuristics, but you can rigorously check this by finding the critical points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why?
| To avoid confusion with sign, in these cases I suggest to take $y=-x\to \infty$ then
$$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x =\lim_{y\to \infty}\sqrt{y^2-5y+3}-y $$
from here we can clearly see that the expression is an indeterminate form $\infty-\infty$ then we can proceed by
$$\sqrt{y^2-5y+3}-y=\left(\sqrt{y^2-5y+3}-y\right)\cdot \frac{\sqrt{y^2-5y+3}+y}{\sqrt{y^2-5y+3}+y}$$
Note that your first cancellation is not correct because you have considered $\sqrt{x^2+5x+3}=-x$ which is not true whereas in the second case we have cancelled out $x^2-x^2$ which is a correct step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\tag {ii}$$
For $(\mathrm i)$, I initially tried simplifying what was in the 2 brackets but ended up getting 1 + 1.
I then tried just multiplying out the brackets and got as far as $$1+ \sec^2x + \frac{2}{\cos^2x \sin^2x}$$
| $$(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)$$
$$=(2\sec^2x-1)(2\csc^2x-1)$$
$$=4\sec^2x\csc^2x-2(\sec^2x+\csc^2x)+1$$
Now use $\sec^2x+\csc^2x=\cdots=\sec^2x\csc^2x$
The second one has been solved by Taussig
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Put a matrix $A$ in Jordan Normal Form and find a $P$ such that $P^{-1}AP=J$ I have a linear algebra exam tomorrow and this is a frequent question.
$A=
\begin{pmatrix} 4 & 0 & 1 & 0 \\ 2 & 2 & 3 & 0 \\ -1 & 0 & 2 & 0 \\ 4 & 0 & 1 & 2
\end{pmatrix}$
$C_T(x)=(x-2)^2(x-3)^2$ so eigenvalues are $2$ and $3$.
$a_2=2$ so the sum of the sizes of Jordan blocks for eigenvalue $2$ equals $2=1+1$
$a_3=2$ so the sum of the sizes of Jordan blocks for eigenvalue $3$ equals $2=1+1$
$\\$
$g_2=dimE_2=dim \ ker \ (A-2I)$
$A-2I = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 2 & 0 & 3 & 0 \\ -1 & 0 & 0 & 0 \\ 4 & 0 & 1 & 0 \end{pmatrix}$
$ker(A-2I) = \left\{ \begin{pmatrix} 0 \\ y \\ 0 \\ t \end{pmatrix} |\ y,t \in \mathbb{R} \right\} = Span \left\{ \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix} \right\}$
Therefore $g_2=2$. So there are 2 Jordan blocks, and they must both be of size $1$.
I put this in Jordan Normal form and got:
$\\$ $\\$
$J = \begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2
\end{pmatrix}$
How do I find an invertible matrix $P$ such that $P^{-1}AP=J$?
I found the Jordan Normal Form using geometric and algebraic multiplicities along with the minimum polynomial, if that helps at all!
Thanks!
| The left two columns are just a basis of 2 eigenvectors. For 3, we take the far right vector as some $w$ such that $(A-3I)^2 w = 0 $ but $(A-3I) w \neq 0. $ Then the third column is $v = (A - 3I)w.$
$$
P =
\left(
\begin{array}{rrrr}
0 & 0 & 1&1 \\
1&0&-1&3 \\
0& 0 & -1& 0 \\
0&1&3&1
\end{array}
\right)
$$
determinant is $-1$ and
$$
P^{-1} =
\left(
\begin{array}{rrrr}
-3 & 1 & -4&0 \\
-1&0&2&1 \\
0& 0 & -1& 0 \\
1&0&1&0
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$ Given that $0\lt x\lt 2$ and $0\lt y\lt 2$ then find the minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$$
My try:
On factorisation we need minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {(y-2)^2+x^2} +\sqrt {(x-2)^2+(y-2)^2}$$
On seeing it for first time, the only thing that popped up was using the Minkowski inequality but I am not getting proper sequences for its application. I tried as much as I could to use this inequality but failed.
I tried substituting $x=2\cos \alpha$ and $y=2\cos \beta$ (where $\alpha, \beta \in \left(0,\frac {\pi}{2}\right)$ )on seeing the constraints on $x$ and $y$ but continuing it was very cumbersome so dropped the method.
Any help would be greatly appreciated.
P.S : It would be very great if someone hints at how I could use the Minkowski inequality efficiently. Thanks!!!
| You can still use Minkowski as:
$$\sqrt{(x+y)^2+(y-x)^2}+\sqrt{(2-y)^2+x^2}+\sqrt{(2-x)^2+(2-y)^2}\geq\sqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=\sqrt{20}.$$
One can check that the equality is attained when: $$(x,y) = \left(\frac 25,\frac 65\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - x_{1} \quad (*)$
Consider:
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 0, x_{3} \geq 4, x_{4} \geq 0 \quad (**)$
$x_{2} + x_{3} + x_{4} = 40 - x_{1}$ where $x_{2} \geq 5, x_{3} \geq 4, x_{4} \geq 6 \quad (***)$
Let $f$ is the function that compute the number of non-negative solutions of an equation.
$\implies f(*) = f(**) - f(***)$
Thus, the number of non-negative solutions of (*) is $\sum_{x_{1} = 2}^{8}( {40 - x_{1} + 3 - 1 \choose 3 - 1} - {25-x_{1}+3 - 1 \choose 3 - 1}) = 3045$
I found that the right answer is 210 by trying some programming script. But I don't know what was wrong with my solution. Please help me. Thank you!
| $x_1 + x_2 + x_4 \le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 \ge 40 -17 \ge 4$ so we can ignore the restriction on $x_3$.
$2 \le x_1 \le 6$ so there are $7$ values that $x_1$ can be, $x\le 4$ so there are $5$ values it can be. $x_4 \le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ there is only one option dependant on the other three options.
So there $7*5*6 = 210$ options.
Your solution dosn't take $x_3 \ge 4$ into account (which you can be seting it up so that the some is $36$ and not $40$-- I haven't done the math to figure it out but that will lower you answer significantly. Also by subtracting you are removing the cases with both $x_2 \ge 5$ and $x_4 \ge 6$ but not removing the cases where one or the other is.)
I think to fix your problem using inclusion exclusion you'd want
$\sum_{x_1=2}^8({{40 - 4 -x_1 + 3-1}\choose {3-1}} - {{40 - 4-5 -x_1 + 3-1}\choose {3-1}}-{{40 - 4-6 -x_1 + 3-1}\choose {3-1}}+{{40 - 4 -5-6-x_1 + 3-1}\choose {3-1}})=$
And I'm too lazy to finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.