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Why can we use inspection for solving equation with multiple unknowns? In our algebra class, our teacher often does the following:
$a + b\sqrt{2} = 5 + 3\sqrt{2} \implies \;\text{(by inspection)}\; a=5, b = 3
$
I asked her why we can make this statement. She was unable to provide a satisfactory answer. So I tried proving it myself.
$a + b\sqrt{2} = x + y\sqrt{2}$. We are required to prove that $a = x$, and $b = y$. Manipulating the equation, we get $\sqrt{2}(b - y) = x - a$, or $\sqrt{2} = \frac{x-a}{b-y}$. Expanding this, we get $\sqrt{2} = \frac{x}{b-y} + \frac{a}{b-y}$. I tried various other transformations, but nothing seemed to yield a result.
| We want to show that the only rational solutions $a,b,x,y$ of
$$a + b\sqrt{2} = x + y\sqrt{2}$$
are given by $a = x$ and $b = y$.
(Note that if you allow for real values of $a,b,x,y$, then for example $a = x + \sqrt{2}$, $b = y - 1$ would be a solution, too.)
If $y = b$, then obviously $a = x$.
Now assume $y \neq b$. Then
$$a + b\sqrt{2} = x + y\sqrt{2}\\ \implies a-x = (y-b)\sqrt{2}\\ \implies (a-x)^2 = 2(y-b)^2\\ \overset{y - b \neq 0}{\implies} 2 = \left(\frac{a-x}{y-b}\right)^{\!2}.$$
So $2$ is the square of a rational number. This contradicts the fact that $\sqrt{2}$ is irrational.
By the above argument, we have shown that the only rational solution of $a + b\sqrt{2} = 0$ is $a = b = 0$.
In terms of linear algebra, this property is formulated as "$1$ and $\sqrt{2}$ are $\mathbb Q$-linearly independent".
From this it follows that any representation $a + b\sqrt{2}$ with $a,b\in\mathbb Q$ uniquely determines $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3}$
| Hint:
The expansion of $(x+h)^\frac{4}{3}$ is
$$h\sqrt[3]{x+h} + x\sqrt[3]{x+h}$$
So then you will have
$$\lim_{x \rightarrow \infty}\frac{h\sqrt[3]{x+h} + x\sqrt[3]{x+h} - x^{\frac{4}{3}}}{h}$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $ \int_{25\pi/4}^{53\pi/4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx $ How to evaluate the integral
$$\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?$$
| For any integrable function $f:\mathbb{R}\to\mathbb{R}$ with period $T$ and any $a,b\in\mathbb{R}$ we have
$$
\int_{b}^{b+T} f(x)dx=\int_{a}^{a+T}f(x)dx\tag{1}
$$
And as the consequence for $k\in\mathbb{N}$
$$
k\int_{b}^{b+T} f(x)dx=\int_{a}^{a+kT}f(x)dx\tag{2}
$$
Also for any integrable $f$ we have
$$
\int\limits_{-c}^c f(x)dx=\int\limits_{0}^{c}(f(x)+f(-x))dx\tag{3}
$$
Now we turn to evaluation
$$
\begin{align}
I_1
&=\int_{25\pi/4}^{29\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&=\int_{\pi/4+3\cdot 2\pi}^{5\pi/4+3\cdot 2\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&\overset{(1)}{=}\int_{\pi/4}^{5\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&=\int\limits_{\pi/4}^{3\pi/4}\left(\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}+\frac{1}{(1+2^{\cos (x+\pi/2)})(1+2^{\sin (x+\pi/2)})}\right)dx\\
&=\int\limits_{\pi/4}^{3\pi/4}\frac{dx}{1+2^{\cos x}}\\
&=\int\limits_{\pi/4}^{\pi/2}\left(\frac{1}{1+2^{\cos x}}+\frac{1}{1+2^{\cos (\pi-x)}}\right)dx\\
&=\int\limits_{\pi/4}^{\pi/2}dx\\
&=\frac{\pi}{4}\\
\end{align}
$$
$$
\begin{align}
I_2
&=\int_{29\pi/4}^{53\pi/4}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&=\int_{29\pi/4}^{29\pi/4+3\cdot 2\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&\overset{(2)}{=}3\int_{-\pi}^{\pi}\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}dx\\
&\overset{(3)}{=}3\int\limits_{0}^{\pi} \left(\frac{1}{(1+2^{\cos x})(1+2^{\sin x})}+\frac{1}{(1+2^{\cos (-x)})(1+2^{\sin (-x)})}\right)dx\\
&=3\int\limits_{0}^{\pi}\frac{dx}{1+2^{\cos x}}\\
&=3\int\limits_{-\pi/2}^{\pi/2}\frac{dx}{1+2^{\cos (x+\pi/2)}}\\
&\overset{(3)}{=}3\int\limits_{0}^{\pi/2}\left(\frac{1}{1+2^{\cos (x+\pi/2)}}+\frac{1}{1+2^{\cos (-x+\pi/2)}}\right)dx\\
&=3\int\limits_{0}^{\pi/2}=\frac{3\pi}{2}\\
\end{align}
$$
Finally,
$$
I=I_1+I_2=\frac{7\pi}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Splitting field of $f=x^6+x^5+1$, orders of elements, and minimal polynomials.
Let $F$ be the splitting field over $\Bbb{Z}_2$ of $f=x^6+x^5+1$, an irreducible polynomial over $\Bbb{Z}_2$.
We know that:
$\bullet$ The Galois group of $F$ over $\Bbb{Z}_2$ is cyclic and is generated by the Frobenius autormorphism.
$\bullet$ The lattice of the subgroups of $Gal(F/\Bbb{Z}_2)$ is isomorphic to that of $\Bbb{Z}_6$. Let's denote the one of degree 2 over $\Bbb{Z}_2$ as $\Bbb{Z}_2(\omega)$ and the one of degree 3 over $\Bbb{Z}_2$ as $\Bbb{Z}_(\beta)$.
$\bullet$ There are four intermediate fields with orders $|F|=64$, $|\Bbb{Z}_2(\beta)|=8$, $|\Bbb{Z}_2(\omega)|=4$, and $|\Bbb{Z}_2|=2$ as additive groups.
a) For each intermediate field $E$ (including $F$), what are the possible multiplicative orders of the units $\gamma$ for which $E=\Bbb{Z}_2(\gamma)$?
b) What is the minimal polynomial of an element of order $9$?
a) Since fields are also additive subgroups, the orders of the multiplicative groups is just Euler's totient function. So $|F^{\times}|=\phi(63)=(7-1)(3-1)(3)=36$, $|\Bbb{Z}_2(\beta)|=2^2=4$, $|\Bbb{Z}_2(\omega)|=2$, and $|\Bbb{Z}_2|=2$.
Orders of the units must divide the orders of the multiplicative groups, so for the possibilities, we have (let $F=\Bbb{Z}(\alpha)$)
$|\alpha|=2,3,6,4,9$
$|\beta|=2,4$
$|\omega|=2$
$1$ is not an option, because otherwise we would have $\Bbb{Z}_2(\gamma)=\Bbb{Z}_2$.
b) We know that the minimal polynomial of that element (call it $\epsilon$), must divide $x^9-1$.
We also know that $x^9-1=\Psi_1(x)\Psi_3(x)\Psi_9(x)$, where $\Psi_k(x)$ is the $k$th cyclotomic polynomial.
Now we need to find $\Psi_9(x)$ in order to use this.
$x^9-1=(x-1)(x^2+x+1)\Psi_9(x)$
$\implies \Psi_9(x) = \frac{x^9-1}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{(x-1)(x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1)}{(x-1)(x^2+x+1)}$
$\implies \Psi_9(x) = \frac{x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1}{x^2+x+1}$.
However, when I tried to do long division, I kept getting a remainder of -2...and I'm not sure why. Doesn't $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$ for sure need to be divisible by $x^2+x+1$? Because we know for sure that $x^9-1$ equals $\Psi_1(x)\Psi_3(x)\Psi_9(x)$, right?
Thank you in advance
| $(x^9-1)/(x-1)$ is not $x^8+x^7-x^6+x^5-x^4+x^3-x^2+x-1$, but $x^8+x^7+\cdots +x+1$.
Indeed,
$$
x^9-1=(x^6+x^3+1)(x^2+x+1)(x-1),\quad \frac{x^9-1}{x-1}=(x^6+x^3+1)(x^2+x+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
solve the differential equation ${dy \over dx} + qy^4+y=0$ Solve the differential equation ${dy \over dx}+qy^4+y=0$
the initial condition is $y(x=1)=-\frac 12$ with considering $q \ll 1$.
don't consider the powers of $q$ which are higher than $2$.
You will use Expansion.(expand the function $y=f(x)$ with the powers of the $q$)
| In the ODE $\displaystyle \quad\frac{dy}{dx} + qy^4 + y = 0\quad$ treat $x$ as a function of $y$, we have:
$$
x(y) = \int dx = - \int \frac{dy}{qy^4 + y} = -\frac13\int \frac{d(y^3)}{y^3(qy^3+1)}
= -\frac13 \int \left(\frac{1}{y^3} - \frac{q}{qy^3+1}\right)d(y^3)\\
= -\frac13 \left(\log(y^3)-\log(qy^3+1)\right) + K
= -\frac13 \log\left(\frac{-y^3}{qy^3+1}\right) + K'
$$
where $K$ and $K'$ are integration constants. Since $y(x) = -\frac12$ at $x = 1$, we get $$K' = 1-\frac13 \log(8 - q) \quad\implies\quad
x = 1 -\frac13 \log\left(\frac{(8-q)(-y^3)}{1+qy^3}\right)
$$
This is equivalent to
$$\begin{align} \frac{-y^3}{1+qy^3} = \frac{1}{8-q}e^{3(1-x)}
\iff & -y^3 = \frac{\frac{1}{8-q}e^{3(1-x)}}{1 + \frac{q}{8-q}e^{3(1-x)}}
=\frac{1}{(8-q)e^{3(x-1)} + q }\\
\iff & \quad y\, = -\left((8-q)e^{3(x-1)} + q\right)^{-1/3}\\
\iff & \quad y\, = -\frac12 e^{1-x} \left(1- \frac{q}{8}( 1 - e^{3(1-x)})\right)^{-1/3}
\end{align}$$
Expand to first order of $q$, we have:
$$y = -\frac12 e^{1-x} \left( 1 + \frac{q}{24} ( 1 - e^{3(1-x)}) + O(q^2)\right)\tag{*1}$$
If one only want/need the first order approximation of $y(x)$ in $q$,
there is an alternate way. One can rewrite the ODE into an intego differential
equation and simplify it along the way using the given initial conditions:
$$\begin{align}
& \frac{dy(x)}{dx} + y(x) + qy(x)^4 = 0\\
\iff &\frac{dy(x)}{dx} + y(x) = - qy(x)^4\\
\iff & \frac{d}{dx} ( e^{x-1} y(x) ) = -q e^{x-1} y(x)^4\\
\iff & e^{x-1}y(x) = -\frac12 - q \int_{1}^{x} e^{t-1} y(t)^4 dt\\
\iff & y(x) = -\frac12 e^{1-x} - q\int_{1}^{x} e^{t-x} y(t)^4 dt\tag{*2}
\end{align}$$
The R.H.S of $(*2)$ tell us up to $O(q)$, $y(x)$ is just $-\frac12 e^{1-x}$.
One can substitute this back into the integral of R.H.S and get the next order of approximation in $q$:
$$\begin{align}y(x) = & -\frac12 e^{1-x} - q \int_{1}^{x} e^{t-x} \left( -\frac12 e^{1-t} \right)^4 dt + O(q^2)\\
= & -\frac12 e^{1-x} - \frac{q e^{4-x}}{16}\int_{1}^{x} e^{-3t} dt + O(q^2)\\
= & -\frac12 e^{1-x} - \frac{q e^{4-x}}{48}(e^{-3} - e^{-3x}) + O(q^2)\\
= & -\frac12 e^{1-x}\left ( 1 + \frac{q}{24}(1 - e^{3(1-x)})\right) + O(q^2)\\
\end{align}$$
This is the same result $(*1)$ we obtained before by solving the full initial value problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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finiding a ratio in Trapezoid
Given $ABCD$ a right angle trapezoid, $\measuredangle A=90^{0}$, $AB\parallel DC$.
$O$ is the point of intersection of the diagonals.
$E$ is a point on $BC$ such that: $AB=BE$, $DC=CE$.
$AD=12$, $BC=13$
Need to find the ratio $AO:OC$.
Tried to compute the area and divided it the partial area by the triangles but didn't help.
Any ideas?
Thanks!
| The given $ABCD$ trapazoid with $\angle BAD = 90^\circ$ is depicted in the following figure with given data:
It is given that $AD=12, \ BC=13$, and $E$ is situated on $BC$ in such a way that $AB=BE$ and $DC=CE$. If you draw a perpendicular line to $CD$ from the point $B$, which meets $CD$ at point $F$, then $AB = DF$. Moreover, the $\triangle BCF$ is a Pythagorian triangle where $BC = 13$ and $BF = 12$, then $FC^2 = 13^2 - 12^2 = 5^2$. Therefore, $FC = 5$.
If $AB= a$, then $DF = a$. Thus, since $CE = DC$, then $CE = DF + FC = a + 5$. Similarly, since $BE = AB$, $BE = a$. Therefore, $BC = a + (a + 5) = 13$ and hence, $a= \frac{13-5}{2}= 4$.
As shown in the diagram, using geometric knowledge in parrelel lines, you can show $\triangle ABO$ and $\triangle CDO$ are similar. Hence,
$$\frac{BO}{DO} = \frac{AO}{CO} = \frac{AB}{CD} = \frac{4}{4 + 5} = \frac{4}{9}$$
Consider the right triangle, $\triangle ABD$. Thus, $BD^2 = AB^2 + AD^2= 4^2 + 12^2= 16 \times 10$. Therefore, $BD = 4\sqrt{10}$.
Also, consider the right triangle, $\triangle ACD$. Thus, $AC^2 = DC^2 + AD^2= 9^2 + 12^2= 15^2$. Therefore, $AC = 15$.
Therefore, you can also calculate the distance of $AO, OC, BO,$ and $OD$ easily now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
| I provide one more solution, where we don't use sines and cosines.
First, some preparation.
We all know that
$$
\tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}.
$$
Then
$$
\cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot (y) - \cot (x)},
$$
or
$$
\cot (x) \cot (y) = \cot(x-y) (\cot(y) -\cot(x)) - 1
$$
Especially,
$$
\cot (2x) = \frac {\cot^2 (x) -1} {2 \cot (x)},
$$
which yields
$$
\cot^2(x) - 1 = 2\cot(x)\cot(2x).
$$
Also,
$$
\csc^2(x) = \frac {\sin^2(x) + \cos^2(x)} {\sin ^2(x)} = 1 + \cot^2(x).
$$
Now let's start. Let $x = \pi /7$.
$$
P := \csc^2(x) + \csc^2(2x) + \csc^2(4x) = 3 + \cot^2(x) + \cot^2(2x) + \cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(\cot(x) \cot(2x) + \cot(2x) \cot (4x) + \cot (4x) \cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $\cot(\pi \pm y) = \mp\cot (y)$, we have
$$
Q = \cot(x)\cot(2x) - \cot(2x) \cot (3x) - \cot(3x) \cot (x).
$$
Now,
\begin{align*}
Q &= 1+ \cot(x)(\cot(x) - \cot(2x)) - \cot(x) (\cot(2x) -\cot(3x)) - \cot(2x)(\cot(x) - \cot (3x))\\
&= 1+ \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + \cot(2x) \cot(3x)\\
&= \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + (\cot(2x)-\cot(3x))\cot(x) \\
&= \cot(x)(\cot(x) -2\cot(2x))\\
&= \cot^2(x) - 2\cot(2x) \cot(x)\\
&=1.
\end{align*}
Therefore $P = 6+2Q=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that: $\sqrt{x^2+21}+\sqrt{2y^2+14}+\sqrt{z^2+91}\ge 19$ Let $x, y, z$ be real number such that $xy+yz+zx=11$. Prove the inequality:
$$\sqrt{x^2+21}+\sqrt{2y^2+14}+\sqrt{z^2+91}\ge 19$$
I think that inequality can be solved by Minkowski. Equality holds if only is $(x;y;z)=(2;1;3)$...But I couldn't continue...
| we can prove this inequality $x,y,z$ are positive numbers.
By cauchy-Schwarz inequality have
$$\sqrt{\dfrac{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}{n}}\ge\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}$$
where $a_{i}>0,i=1,2\cdots,n$\
then
$$\sqrt{x^2+21}=5\sqrt{\dfrac{\dfrac{x^2+1}{5}+1+1+1+1}{5}}\ge 5\dfrac{\sqrt{\dfrac{x^2+1}{5}}}{5}=\sqrt{\dfrac{x^2+1}{5}}+4$$
$$\sqrt{2y^2+14}\ge \sqrt{2}\cdot 2\sqrt{2}\sqrt{\dfrac{2\dfrac{y^2+1}{2}+6}{8}}\ge 4\cdot\dfrac{2\sqrt{\dfrac{y^2+1}{2}}+6}{8}$$
$$\sqrt{z^2+91}=10\sqrt{\dfrac{\dfrac{z^2+1}{10}+9}{10}}\ge 10\cdot\dfrac{\sqrt{\dfrac{z^2+1}{10}}+9}{10}$$
then
$$LHS\ge \sqrt{\dfrac{x^2+1}{5}}+\sqrt{\dfrac{y^2+1}{2}}+\sqrt{\dfrac{z^2+1}{10}}+(4+3+9)\ge 3\sqrt[3]{\dfrac{(x^2+1)(y^2+1)(z^2+1)}{100}}+16$$
since
$$(x^2+1)(y^2+1)(z^2+1)=(x^2+1)[(y+z)^2+(yz-1)^2]\ge [x(y+z)+(yz-1)]^2=(xy+yz+xz-1)^2\ge 100$$
by this way: we only prove this $x,y,z$ are positive numbers, becase
$$x\longrightarrow |x|,y\longrightarrow |y|,z\longrightarrow |z|$$
then we have
$$|xy|+|yz|+|zx|\ge xy+yz+xz\ge 11$$
and I very like this inequality, who creat? Thank you
| {
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"timestamp": "2023-03-29T00:00:00",
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Cyclic polynomial proof 2. So here's my second question for the day,
Q. Manipulate the given equality
$$a^2b^2(bc-a^2)+b^2c^2(ca-b^2)+c^2a^2(ab-c^2)$$$$ = a^2b^2(b^2-ac)+b^2c^2(c^2-ab)+c^2a^2(a^2-bc)$$
to obtain the following:
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)=abc(a+b)(b+c)(c+a)$$
| Rearrange the LHS as
$$a^2b^2(bc)+b^2c^2(ca)+c^2a^2(ab) - (a^2b^2(a^2)+b^2c^2(b^2)+c^2a^2(c^2)) = A - B$$
Rearrange the RHS as
$$a^2b^2(b^2)+b^2c^2(c^2)+c^2a^2(a^2) - (a^2b^2(ac)+b^2c^2(ab)+c^2a^2(bc)) = C - D$$
So now we have
$$A - B = C - D$$
Exchange
$$A + D = B + C$$
First we have the following
$$(x+y)(y+z)(x+z) = x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz$$
We have
$$A+D = a^2b^2(bc)+b^2c^2(ca)+c^2a^2(ab) + a^2b^2(ac)+b^2c^2(ab)+c^2a^2(bc)$$
$$= (a^2 b^2+c^2a^2)(bc)+(a^2b^2+b^2c^2)(ca)+(c^2a^2+b^2c^2)(ab)$$
$$= abca(b^2+c^2) + abcb(a^2+c^2) + abcc ( a^2+b^2)$$
$$= abc(ab^2+ac^2+a^2b+bc^2+a^2c+b^2c)$$
$$= abc(ab^2+a^2b+a^2c+ac^2+b^2c+bc^2)$$
$$= abc((a+b)(b+c)(a+c)-2abc)$$
$$= abc (a+b)(b+c)(a+c) - 2(abc)^2$$
On the other hand
$$B + C = a^2b^2(a^2)+b^2c^2(b^2)+c^2a^2(c^2) +
a^2b^2(b^2)+b^2c^2(c^2)+c^2a^2(a^2)$$
$$= a^2b^2(a^2+b^2)+b^2c^2(b^2+c^2)+c^2a^2(a^2+c^2)$$
$$= a^4b^2 + a^2 b^4 + b^4c^2 + b^2c^4 + a^4 c^2 + a^2c^4$$
$$= (a^2+b^2)(b^2+c^2)(a^2+c^2) - 2 a^2 b^2 c^2$$
By cancelling the $2a^2b^2c^2$ term from each side you have the required identity.
| {
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Evaluating $\int_0^\infty \frac{\cos(ax)-e^{-ax}}{x \left(x^4+b^4 \right)}dx$ How can we evaluate
$$\int_0^\infty \frac{\cos(ax)-e^{-ax}}{x \left(x^4+b^4\right)}dx \quad a,b>0$$
using Complex Analysis? This problem was given in a Complex Analysis book which I was reading. The answer given in it is
$$\frac{\pi}{4b^4}e^{-ab/\sqrt{2}}\sin \left(\frac{ab}{\sqrt{2}} \right)$$
Which function and contour should we consider?
| The following is an evaluation that does not use contour integration.
Let $ \displaystyle I(a) = \int_{0}^{\infty} \frac{\cos (ax) - e^{-ax}}{x(x^{4}+b^{4})} \ dx$.
Then differentiating under the integral sign, $$ \begin{align} I^{(4)}(a)+b^{4}I(a) &= \int_{0}^{\infty} \frac{\cos(ax)-e^{-ax}}{x} \ dx \\ &= \text{Ci}(ax) - \text{Ei}(-ax) \Bigg|^{\infty}_{0} \\ &= \lim_{x \to \infty} \left( \frac{\sin ax}{ax} + \mathcal{O}(x^{-2})\right) -\lim_{x \to 0^{+}} \Big(ax + \mathcal{O}(x^{2}) \Big) \\ &= 0 \end{align}$$ where $\text{Ci}(x)$ is the cosine integral and $\text{Ei}(x)$ is the exponential integral.
The characteristic equation of the above linear homogeneous differential equation is $r^{4}+b^{4}=0$, which has roots $ r= \frac{b}{\sqrt{2}} ( \pm 1 \pm i )$.
The general solution of the differential equation is therefore $$ \begin{align} I(a) &= C_{1}e^{ab/ \sqrt{2}} \cos \left(\frac{ab}{\sqrt{2}} \right) + C_{2}e^{ab/ \sqrt{2}} \sin \left(\frac{ab}{\sqrt{2}} \right)+C_{3}e^{-ab/ \sqrt{2}} \cos \left(\frac{ab}{\sqrt{2}} \right)\\ &+C_{4}e^{-ab/ \sqrt{2}} \sin \left(\frac{ab}{\sqrt{2}} \right). \end{align}$$
But since $I(a)$ remains finite as $a \to \infty$, $C_{1}$ and $C_{2}$ must be zero.
To find the constants $C_{3}$ and $C_{4}$ we can use the initial conditions $I(0)=0$ and $$ \begin{align} I'(0) = \int_{0}^{\infty} \frac{1}{x^{4}+b^{4}} \ dx &= \frac{1}{b^{4}} \int_{0}^{\infty} \frac{1}{(\frac{x}{b})^{4}+1} \ dx \\ &= \frac{1}{b^{3}} \int_{0}^{\infty} \frac{1}{u^{4}+1} \ du \\ &= \frac{1}{b^{3}}\frac{\pi}{4} \csc \left( \frac{\pi}{4}\right) \\ &= \frac{\pi \sqrt{2}}{4b^{3}}. \end{align} $$
To satisfy the first initial condition, $C_{3}$ must be zero.
And using the second initial condition we have $$ \frac{\pi \sqrt{2}}{4b^{3}} = C_{4}\frac{b}{\sqrt{2}},$$
which implies $$C_{4} = \frac{\pi}{4b^{4}}.$$
Therefore,
$$ I(a) = \int_{0}^{\infty} \frac{\cos (ax) - e^{-ax}}{x(x^{4}+b^{4})} \ dx = \frac{\pi}{{\color{red}{2}}b^{4}} e^{-ab/ \sqrt{2}} \sin \left(\frac{ab}{\sqrt{2}} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
Diophantine equation involving prime numbers : $p^3 - q^5 = (p+q)^2$ Find all pairs of prime nummbers $p,q$ such that $p^3 - q^5 = (p+q)^2$.
It's obvious that $p>q$ and $q=2$ doesn't work, then both $p,q$ are odd. Assuming $p = q + 2k$ we conclude, by the equation, that $k|q^3 - q - 4$ because $\gcd(k,q)=1$ (else $p$ is not prime) and $k=1$ has no solution.
I also tried to use some modules, but I couldn't.
| $$p^3-q^5=(p+q)^2=p^2+2pq+q^2$$
So, $p^3-q^5\ge 0$, so $p^3\ge q^5$, so $p>q$.
$$p^2(p-1)=p^3-p^2=q^5+2pq+q^2=q(q^4+2p+q)$$
Hence, $p^2$ divides $q^4+2p+q$ (as it can't divide q) and so $q$ divides (p-1).
So $p^2\le q^4+2p+q$, so $p\le q^2+1$. We can then verify easily that $q=2$ has no solution for $p$.
But $p$ is prime, and $q^2+1$ and $q^2$ are not so :
$$q<p\le q^2-1$$
So let $p=aq+b$ with $1\le a< q$ and $1\le b<q$.
We have $$q^5=p^3-(p+q)^2=(b^3-b^2)+q.(\dots)$$
Hence $q$ divides $b^2(b-1)$, so $b=1$
Hence $p=aq+1$ with $1\le a< q$
We have $$q^5=p^3-(p+q)^2=q.(3a-2(a+1))+q^2.(\dots)$$
Hence, $q$ divides $a-2$, so $a=2$.
So $p=2q+1$
$$q^5=q^2.(8q+12-9)$$
$$0=q^3-8q-3=(q-3).(q^2+3q+1)$$
The only positive integer solution is $q=3$ and so $p=7$.
| {
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} |
Asymptotic expansion of a function $\frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx$ How to find the asymptotic expansion of the following function for large values of $z$.
$$f_{3/2}(z) = \frac{4}{\sqrt \pi} \int_0^\infty \frac{x^2}{1 + z^{-1} e^{x^2}}dx $$
I have to get something like (in the book)
$$ f_{3/2}(z) = \frac{4}{3 \sqrt \pi} \left[ (\ln z)^{3/2} + \frac{\pi^2}{8} (\ln z)^{-1/2} + \dots \right ] $$
the problem comes from physics in determining the chemical potential for Fermi gas.
| For simplicity let's write $\ln z = \lambda$. The integral we're trying to approximate is
$$
I(\lambda) = \int_0^\infty \frac{x^2}{1+\exp(x^2-\lambda)}\,dx
$$
as $\lambda \to \infty$.
Begin by making the change of variables $x^2-\lambda = y$ to get
$$
\begin{align}
I(\lambda) &= \frac{1}{2} \int_{-\lambda}^\infty \frac{\sqrt{y+\lambda}}{1+e^y}\,dy \\
&= \frac{\sqrt{\lambda}}{2} \left( \int_{-\lambda}^0 \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy + \int_0^\infty \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy \right). \tag{1}
\end{align}
$$
The asymptotics for the second integral can be found by simply expanding the square root in a power series
$$
\sqrt{1+y/\lambda} = 1 + \frac{y}{2\lambda} - \frac{y^2}{8\lambda^2} + \cdots
$$
and integrating term-by-term to get
$$
\int_0^\infty \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy = \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right). \tag{2}
$$
Getting the asymptotics for the first integral is a little more tricky. Making the change of variables $y = -\lambda u$ yields
$$
\int_{-\lambda}^0 \frac{\sqrt{1+y/\lambda}}{1+e^y}\,dy = \lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du. \tag{3}
$$
By invoking the dominated convergence theorem we may expand the denominator as a geometric series in $e^{-\lambda u}$ and interchange the order of integration and summation to get
$$
\begin{align}
\lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du &= \lambda \sum_{n=0}^{\infty} (-1)^n \int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du \\
&= \frac{2\lambda}{3} + \lambda \sum_{n=1}^{\infty} (-1)^n \int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du.
\end{align}
$$
By Watson's lemma we have
$$
\int_0^1 \sqrt{1-u} \ e^{-n\lambda u}\,du = \frac{1}{n\lambda} - \frac{1}{2n^2\lambda^2} + O\left(\frac{1}{n^3 \lambda^3}\right), \tag{4}
$$
and upon substituting this into the above sum we find that
$$
\begin{align}
\lambda \int_0^1 \frac{\sqrt{1-u}}{1+e^{-\lambda u}}\,du &= \frac{2\lambda}{3} + \lambda \sum_{n=1}^{\infty} (-1)^n \left[
\frac{1}{n\lambda} - \frac{1}{2n^2\lambda^2} + O\left(\frac{1}{n^3 \lambda^3}\right)\right] \\
&= \frac{2\lambda}{3} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n} - \frac{1}{2\lambda} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} + O\left(\frac{1}{\lambda^2}\right) \\
&= \frac{2\lambda}{3} - \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right). \tag{5}
\end{align}
$$
Recalling from $(3)$ that this is equal to the first integral in $(1)$, we substitute this and $(2)$ into $(1)$ to get
$$
\begin{align}
I(\lambda) &= \frac{\sqrt{\lambda}}{2} \left[ \frac{2\lambda}{3} - \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right) + \ln 2 + \frac{\pi^2}{24\lambda} + O\left(\frac{1}{\lambda^2}\right) \right] \\
&= \frac{1}{3} \lambda^{3/2} + \frac{\pi^2}{24} \lambda^{-1/2} + O\left(\lambda^{-3/2}\right).
\end{align}
$$
This concludes the answer since
$$
f_{3/2}(z) = \frac{4}{\sqrt{\pi}} \,I(\ln z).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
| \begin{align}
&\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
{1 \over 9}\sum_{k = 0}^{\infty}{1 \over \left(k + 1/3\right)\left(k + 2/3\right)}
=
{1 \over 9}\,{\Psi\left(1/3\right) - \Psi\left(2/3\right) \over 1/3 - 2/3}
\\[3mm]&=
{{1 \over 3}\left\lbrack \Psi\left(2 \over 3\right) - \Psi\left(1 \over 3\right)\right\rbrack}
=
{1 \over 3}\,\pi\ {\rm cotan}\left(\pi \over 3\right)
=
{1 \over 3}\,\pi\ {\sqrt{3} \over 3}
\\[5mm]&
\end{align}
$\Psi\left(z\right)$ is the Digamma function. In the last step I use
$\Psi\left(z\right) - \Psi\left(1 - z\right) = -\pi\,{\rm cotan}\left(\pi\,z\right)$ with
$z = 1/3$.
$$
\begin{array}{|c|}\hline\\
{\large\quad\sum_{k = 1}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} - {1 \over 2}
=
{\sqrt{3} \over 9}\,\pi - {1 \over 2}\quad}
\\
\\
\hline
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476354",
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"source": "stackexchange",
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Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried:
$$\begin{align*}
2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\
a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\
a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\
2a_n - a_{n-1} & = \frac{1}{2} \\
a_n = \dfrac{2a_{n-1} + 1}{4}
\end{align*}$$
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
| You could use generating functions. Put $$f(z) = \sum_{n\ge 1} a_n z^n.$$
Summing your recurrence for $n\ge 2$ and multiplying by $z^n,$ we get
$$ f(z) - \frac{1}{4} z = \frac{1}{4} \sum_{n\ge 2} n z^n
- \frac{1}{2} z \sum_{n\ge 2} z^{n-1} \sum_{k=1}^{n-1} a_k.$$
Simplify to obtain
$$ f(z) - \frac{1}{4} z = \frac{1}{4} \frac{z^2(2-z)}{(1-z)^2}
- \frac{1}{2} z \frac{1}{1-z} f(z).$$
Now solve for $f(z).$ This yields
$$ f(z) = \frac{1}{2} \frac{z}{(1-z)(2-z)}
= \frac{1}{2} \frac{1}{1-z} - \frac{1}{2} \frac{1}{1-z/2}.$$
Finally read off the coefficients, which is now easy, to get
$$a_n = \frac{1}{2} \left(1 - \frac{1}{2^n}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| Hint: Write $$x+\sin \frac1x = x\left(1+\frac1x \sin \frac1x\right)$$ and use
$$\sin \frac1x=\frac1x - \frac{1}{3!x^3} + \frac{1}{5!x^5}+ \ldots, \quad x \to \infty.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
Now how to proceed further in this problem... please suggest thanks....
| Hint:
$$2^{x+2}\cdot3^{\frac{3x}{x-1}}=3^2\iff 2^{x+2}=3^{2-\frac{3x}{x-1}}=3^{-\frac{x+2}{x-1}}\iff \begin{cases}x+2=0\\{}\\x-1=-\log_23\end{cases} \;\;(\text{why?)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? $(x+3)^2 \ge 0$
Am I missing the point in anyway?
| We are given $\frac{1}{x^2 + 6 x + 10}$ which we can easily rearrange to $\frac{1}{(x+3)^2 +1}$.
If we consider the maximum value of $(x+3)^2 + 1$ then clearly for large positive or negative numbers this can become arbitrarily large and will approach $+\infty$ for large positive or negative $x$. We are dividing 1 by a potentially large positive number so the result will get close to but will always be more than zero.
We have thus shown:
$$0 \lt \frac{1}{x^2 + 6 x + 10}$$
Now consider the minimum value of $(x+3)^2 + 1$, Since $(x+3)^2$ can not be negative for real $x$ the minimum is 1 at $x = -3$. And 1 divided by 1 is 1 so we have proved the other side of this inequality
$$0 \lt \frac{1}{x^2 + 6 x + 10} \le 1$$
And the resultant function will look like this:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Where's the mistake in this proof of Euler's reflection formula? $$\frac{\sin(\pi x)}{\pi}=x\prod_{r=1}^\infty\left(1-\frac{x^2}{r^2}\right)$$
$$\Gamma(x)=\frac{1}{x}e^{-\gamma x}\prod_{r=1}^\infty\left(\frac{r}{x+r}\right)e^{\frac{x}{r}}$$
$$\Gamma(x) \Gamma(1-x)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$
Assume $\Gamma(x) \Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, then, for $x \notin \mathbb{Z}$:
$$\frac{1}{x}\prod_{r=1}^\infty\left(\frac{r^2}{r^2-x^2}\right)=\frac{1}{x(1-x)}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{r^2}{(x+r)(1-x+r)}\right)e^{\frac{1}{r}}$$
$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$
$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}\prod_{r=2}^\infty\left(\frac{1}{r-x}\right)$$
$$1=e^{-\gamma}\prod_{r=1}^\infty e^{\frac{1}{r}}$$
Thus $-\gamma+\sum_{r=1}^\infty \frac{1}{r}=0$, which is ridiculous. Where did I go wrong?
| In the line
$$\prod_{r=1}^\infty\left(\frac{1}{r-x}\right)=\frac{1}{1-x}e^{-\gamma}\prod_{r=1}^\infty\left(\frac{1}{1-x+r}\right)e^{\frac{1}{r}}$$
you have divergent products. If one wants to give meaning to it, the only somewhat sensible meaning is $0 = 0$. Of course dividing that equation by $0$ does not produce anything meaningful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Alternating binomial sum with intervals of two Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum?
The first few values are $1, 0, -2, -4, -4$.
| Note that
$$(1+x)^n = \dbinom{n}0 + x \dbinom{n}1 + x^2 \dbinom{n}2 + x^3 \dbinom{n}3 + \cdots + x^n \dbinom{n}n$$
Hence,
$$(1+i)^n = \dbinom{n}0 + i \dbinom{n}1 + i^2 \dbinom{n}2 + i^3 \dbinom{n}3 + \cdots + i^n \dbinom{n}n$$
where $i^2=-1$.
Hence,
$$\text{Real part of }(1+i)^n = 1 - \dbinom{n}2 + \dbinom{n}4 - \dbinom{n}6 \pm \cdots$$
Also note that
$$(1+i)^n = 2^{n/2} e^{i n \pi/4} = 2^{n/2} \left(\cos(n \pi/4) + i \sin(n \pi/4)\right)$$
Hence,
$$\text{Real part of }(1+i)^n = 2^{n/2} \cos(n \pi/4)$$
This gives us
$$\boxed{\color{blue}{1 - \dbinom{n}2 + \dbinom{n}4 - \dbinom{n}6 \pm \cdots = 2^{n/2} \cos(n \pi/4)}}$$
| {
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If $a,b,c > 0$ and $a+b+c = 1$, then $(\tfrac{1}{a} −1)(\tfrac{1}{b} −1)(\tfrac{1}{c} −1) \geq 8$
If $a, b, c$ are positive real numbers and $a+b+c = 1$, prove that
$$\left(\frac{1}{a} −1\right)\left(\frac{1}{b} −1\right)\left(\frac{1}{c} −1\right) \geq 8.$$
Thank you.
| This following is my new methods:
let $$x=1-\dfrac{1}{a},y=1-\dfrac{1}{b},z=1-\dfrac{1}{c}$$
then
$$a+b+c=1\Longrightarrow \dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=1$$
$$\dfrac{x}{x+1}=
\dfrac{1}{1+y}+\dfrac{1}{1+y}\ge2\sqrt{\dfrac{1}{(1+y)(1+z)}}\cdots (1)$$
$$\dfrac{y}{1+y}\ge2\sqrt{\dfrac{1}{(1+x)(1+z)}}\cdots (2)$$
$$\dfrac{z}{1+z}\ge 2\sqrt{\dfrac{1}{(1+x)(1+y)}}\cdots (3)$$
$(1)\times (2)\times (3)$
$$xyz\ge 8$$
I very like this methods, this methods can solve some hard inequaity,such as
USA USAMO 1998
Problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343867&sid=f8230676af97b2afacd78d0ccae6ad68#p343867
| {
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Proving $ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $ for $0Let $a,b,c$ be positive real numbers between $0$ and $1$ ,i.e., they lie in the closed interval $[0,1]$. Prove that :
$$ \frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leq 1 $$
| Let $f(a,b,c)=\sum\limits_{cyc}\frac{a}{b+c+1}+\prod\limits_{cyc}(1-a)$.
Hence, $\frac{\partial^2f}{\partial a^2}=\frac{2b}{(a+c+1)^3}+\frac{2c}{(a+b+1)^3}\geq0$,
which says that $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $$\max_{\{a,b,c\}\subset[0,1]}f=\max_{\{a,b,c\}\subset\{0,1\}}f=f(1,1,1)=1.$$
Done!
| {
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find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$ Let $a,b,$ be real numbers such that $a\ge 4b$ .
Find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$.
This problem is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=553680
| The minimum is either on the interior of the feasible set, or on the boundary.
The gradient of the objective is
$$\left[2(a-b)+2(a-1), -2(a-b)+2(b-1)\right] = \left[4a-2b-2, 4b-2a-2\right],$$
and so the unconstrained minimum is the infeasible point $(1,1)$. Therefore the minimum is on the boundary. Substituting $a=4b$ gives
$$\min_b\ (3b)^2 + (4b-1)^2 + (b-1)^2 = \min_b\ 26b^2-10b+2.$$
The gradient is $52b-10$, so $b=\frac{5}{26}$ and the minimum is $\frac{27}{26}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving $\gcd( m,n)$=1 If $a$ and $b$ are co prime and $n$ is a prime, show that:
$\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a+b$ is a multiple of $n$
Also enlighten me why $n$ has to be prime so that $\frac{a^n+b^n}{a+b}$ and $a+b$ have no common factor unless $a$+$b$ is a multiple of $n$?
| If you like modular arithmetic, note that $$\tag 1 a\equiv -b\mod a+b$$
Now, since $n$ is odd $$\frac{{{a^n} + {{b }^n}}}{{a + b}}=\frac{{{a^n} - {{\left( { - b} \right)}^n}}}{{a - \left( { - b} \right)}} = \sum\limits_{k = 0}^{n - 1} {{a^k}{{\left( { - b} \right)}^{n - k-1}}} $$ and using $(1)$ $$\frac{{{a^n} + {{b }^n}}}{{a + b}} = \sum\limits_{k = 0}^{n - 1} {{a^k}{a^{n - k - 1}}} = \sum\limits_{k = 0}^{n - 1} {{a^{n - 1}}} = n{a^{n - 1}}$$
Now since $(a,b)=1$ we have that $$(a,a+b)=(a,a+b\mod a)=(a,b)=1\implies (a^{n-1},a+b)=1$$ so it all reduces to $$(n,a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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what to do next recurrence relation when solving exponential function? find gernal solution of :$a_n = 5a_{n– 1} – 6a_{n –2} + 7^n$
Homogeneous solution:
$$a_n -5a_{n– 1} + 6a_{n –2} = 7^n$$
put $a_n=b^n$:
$$b^n -5b^{n– 1} + 6b^{n –2} =0
\\b^{n-2} (b^2-5b^{} + 6b) =0
\\b^2-5b^{} + 6b =0
\\(b-2)(b-3)=0\\
b=2,3$$
$$a^h_{(n)} = C_1 3^n+ C_2 2^n$$
Particular solution:
Since RHS is exponent so $a^p_{(n)} = da^n$
put $a^p_{(n)}$ in $a_n -5a_{n– 1} + 6a_{n –2} = 7^n$
$$da^n -5da^{n– 1} + 6da^{n –2} =7^n$$
| Use generating functions... Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence so there aren't subtractions in indices:
$$
a_{n + 2} = 5 a_{n + 1} - 6 a_n + 49 \cdot 7^n
$$
Multiplply by $z^n$, sum over valid indices ($n \ge 0$), and recognize the resulting sums:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 5 \frac{A(z) - a_0}{z} - 6 A(z) + 49 \frac{1}{1 - 7 z}
$$
Solve for $A(z)$, write as partial fractions:
$$
A(z)
= \frac{49}{20} \cdot \frac{1}{1 - 7 z}
+ \frac{4 a_1 - 8 a_0 - 49}{4} \cdot \frac{1}{1 - 3 z}
- \frac{5 a_1 - 15 a_0 - 49}{5} \cdot \frac{1}{1 - 2 z}
$$
Everything in sight is just a geometric series:
$$
a_n
= \frac{49}{20} \cdot 7^n
+ \frac{4 a_1 - 8 a_0 - 49}{4} \cdot 3^n
- \frac{5 a_1 - 15 a_0 - 49}{5} \cdot 2^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $\int_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\,\mathrm dx$ $$\large\int \limits_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\mathrm dx$$
TL;DR: Is there an elegant way of integrating this? I've reduced it to a series, detailed below, but the closed form eludes me, and the only solution I've seen uses a rabbit-out-of-the-hat approach.
For the series,
$$\begin{align} \int \limits_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\,\mathrm dx&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int \limits_{0}^{\frac{\pi}{2}} \sin^{2k}(x)\ln(\sin(x))\,\mathrm dx\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int_{0}^{\frac{\pi}{2}} \sin^{2k}(x)\ln(\sin(x))\,\mathrm dx\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int_{0}^{1} u^{2k}\ln(u)\frac{\mathrm du}{1-u^2}\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k} \left(\int_{0}^{1} u^{2k}\ln(u)\,\mathrm du+ \sum \limits_{j=1}^\infty \frac{(2j-1)!!}{(2j)!!} \int_{0}^{1} u^{2(k+j)}\ln(u)\,\mathrm du \right)\\
&=\sum \limits_{k=1}^\infty \frac{1}{k} \left( \frac{1}{(2k+1)^2}+\sum \limits_{j=0}^\infty \frac{(2j-1)!!}{(2j)!!} \frac{1}{(2(k+j)+1)^2} \right)\end{align}$$
, which converges! Both summations are sort-of-justified as $|\sin(x)|, |u|\le1$,with equality only reached at one of the limits of integration, not in between.
| Use the Fourier series
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k \\
\ln(\cos(x)) = -\ln(2) - \sum_{k=1}^\infty (-1)^k \frac{\cos(2kx)}k$$
and expand the integrand to
$$\begin{align*}
\ln(\sin(x)) \ln(\cos(x)) &= \ln^2(2) + \ln(2) \sum_{k=1}^\infty \frac{1+(-1)^k}2 \frac{\cos(2kx)}k \\[1ex]
& \qquad+ \left(\sum_{k=1}^\infty \frac{\cos(2kx)}k\right) \left(\sum_{\ell=1}^\infty (-1)^\ell \frac{\cos(2\ell x)}\ell\right) \\[2ex]
&= \ln^2(2) + \ln(2) \sum_{k=1}^\infty \frac{1+(-1)^k}2 \frac{\cos(2kx)}x \\[1ex]
& \qquad + \sum_{k=1}^\infty \frac{(-1)^k \cos^2(2kx)}{k^2} + 2 \sum_{k\neq\ell} \frac{(-1)^\ell \cos(2kx) \cos(2\ell x)}{k\ell}
\end{align*}$$
Now integrate. We have for all non-negative integers $k,\ell$
$$\int_0^{\pi/2} \cos(2kx)\cos(2\ell x) \, dx = \begin{cases}\dfrac\pi2 & \text{if }k=\ell=0 \\ \dfrac\pi4 & \text{if }k=\ell>0 \\ 0 & \text{if }k\neq\ell\end{cases}$$
so upon swapping integrals with sums, we find
$$\begin{align*}
\int_0^{\pi/2} \ln(\sin(x))\ln(\cos(x)) \, dx &= \frac\pi2 \ln^2(2) + (0\times\ln(2)) + \frac\pi4 \sum_{k=1}^\infty \frac{(-1)^k}{k^2} + (2\times0) \\[1ex]
&= \boxed{\frac\pi2 \ln^2(2) - \frac{\pi^3}{48}}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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} |
Solving $y^2(x^2+1) +x^2(y^2+16) =448$ $y^2(x^2+1) +x^2(y^2+16) =448$
The task is to find all solutions in integers $(x,y)$. This is the fourth question of rmo 1st stage.The solution here is not complete. I have tried to solve unable to.
enter link description here
| First, this equation is symmetric under $x\to-x$ and $y\to-y$, independently. Therefore we can restrict attention to $x,y>0$ and such a solution will represent four actual solutions (we easily see that $x=0$ or $y=0$ are not possible integer solutions).
Solve with the quadratic formula to find
$$y=4\frac{\sqrt{28-x^2}}{\sqrt{1+2x^2}}.$$
The allowed integer values of $x$ are $1$, $2$, $3$, $4$, $5$. Check these by hand: $x=1$ and $x=3$ give integer values of $y$, so there are eight solutions total:
$$(\pm 1,\pm 12)$$
$$(\pm 3,\pm 4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a non-recursive formula for a recursively defined sequence So I have a recursive definition for a sequence, which goes as follows:
$$s_0 = 1$$
$$s_1 = 2$$
$$s_n = 2s_{n-1} - s_{n-2} + 1$$
and I have to prove the following proposition: The $n$th term of the sequence defined above is $s_n = \frac{n(n+1)}{2} + 1$.
To prove this, I started it off by induction. The base case is for $n = 0$ and its true since the non-recursive and recursive results match. I assumed the following hypothesis to be true for some $n = k$: that the $n$th term of the sequence is $s_k = \frac{k(k+1)}{2} + 1$. Then, in the induction step, we need to show that $s_{k + 1} = \frac{(k+1)(k+2)}{2} + 1$.
Using the recursive definition of the sequence, we have that $s_{k+1} = 2s_k - s_{k-1} + 1$, so we can use the hypothesis to replace $s_k$ and $s_{k-1}$ by their non-recursive formulas:
$$ s_{k+1} = 2 (\frac{k(k + 1)}{2} + 1) - (\frac{(k-1)k}{2} + 1) + 1$$
After simplifying i get
$$ s_{k+1} = \frac{k(k+3)}{2} + 2 $$
which is clearly wrong. Can someone point out what I'm doing wrong and where I can go with this?
EDIT: The answer I have given is correct, except that we need to simplify further to get the form we want:
$$ \frac{k(k+3)}{2} + 2 = \frac{k^2 + 3k + 4}{2}$$
after expansion and common denominators, and then this is clearly equal to
$$ \frac{(k+1)(k+2)}{2} + 1 = \frac{k^2 + 3k + 2}{2} + 1 = \frac{k^2 + 3k + 4}{2} $$
| Your work is fine.
$${s_{k + 1}} = \frac{{k(k + 3)}}{2} + 2 = \frac{{k(k + 3) + 2}}{2} + 1 = \frac{{{k^2} + 3k + 2}}{2} + 1 = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2} + 1$$
As GitGud has pointed out, you should assume that the formulas for $s_k$ and $s_{k-1}$ hold true, since you need them both for your recurrence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:
Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$
But then?
| To do it inductively, we also need a base case: $0 \times 1 \times 2=0$ is divisible by $3$. Then, if $3$ divides $k(k+1)(k+2)=k^3+3k^2+2k$ then
\begin{align*}
(k+1)(k+2)(k+3) &= k^3+6k^2+11k+6 \\
&= (k^3+3k^2+2k)+3(k^2+3k+2).
\end{align*}
We know $k^3+3k^2+2k$ is divisible by $3$, by the inductive hypothesis, and $3(k^2+3k+2)$ is divisible by $3$ since $k^2+3k+2$ is an integer. Therefore, by induction, $k(k+1)(k+2)$ is divisible by $3$ for all $k \geq 0$.
Note, if we want to prove it for negative $k$ too, we'd need to perform "induction in reverse".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically.
Suggestions?
| Although another answer is hardly needed at this point, here is yet another way of thinking about this. Note that, for any real numbers $x$ and $y$, one has
$$ \begin{pmatrix}
1 & x \\ 0 & 1 \\
\end{pmatrix} \cdot
\begin{pmatrix}
1 & y \\ 0 & 1 \\
\end{pmatrix} =
\begin{pmatrix}
1 & x+y \\ 0 & 1 \\
\end{pmatrix}.$$
We have learned that:
*
*The set of matrices of the form $\begin{pmatrix}
1 & \text{stuff} \\ 0 & 1 \\
\end{pmatrix}$ is closed under multiplication.
*To multiply two such matrices, you just need to add the stuff in the top right corner.
As a corollary, we conclude that:
$$
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}^n
=
\underbrace{
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}
\cdots
\begin{pmatrix}
1 & 1 \\ 0 & 1 \\
\end{pmatrix}
}_{n \text{ times}}
=
\begin{pmatrix}
1 & \underbrace{1+ \ldots + 1}_{n \text{ times}} \\ 0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & n \\ 0 & 1 \\
\end{pmatrix}.
$$
Remark: If you like, you can think of the map
$$ x \mapsto \begin{pmatrix}
1 & x \\ 0 & 1 \\
\end{pmatrix}$$
as being a homomorphism from the real numbers under addition to the $2 \times 2$ invertible real matrices under multiplication.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 2
} |
Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive real roots $=3$
But are they distinct $?$
Answer :- number of distinct real roots $ =1$
| If $$f(x)=(x-a)^3+(x-b)^3+(x-c)^3$$ then $$f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2$$ Since $a,b,c$ are distinct real numbers $f'(x) > 0$ for all $x\in\mathbb{R}$ and therefore $f$ is strictly increasing and therefore it has only one real root.
EDIT: The last statement is true since $f$ is a polynomial function of degree $3$ ($a_0>0$) so $\lim_{x\to-\infty} f(x) = -\infty$, $\lim_{x\to\infty} f(x) = \infty$ and $f$ is continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What are the values for which the series converges? Determining the values of $a$ and $b$ so that the series $\sum a_n$ converges, where
$$a_n=\ln n-a\ln(n+1)+b\ln(n+2)$$
| \begin{align}
a_{n}
&=
\ln\left(n\right) - a\ln\left(n + 1\right) + b\ln\left(n + 2\right)
\\[3mm]&=
\ln\left(n\right)
-
a\left[
\ln\left(n\right) + {1 \over n} - {1 \over 2n^{2}} + \cdots
\right]
+
b
\left[
\ln\left(n\right) + {2 \over n} - {2 \over n^{2}} + \cdots
\right]
\\[3mm]&=
\left(1 - a + b\right)\ln\left(n\right)
+
\left(-a + 2b\right){1 \over n} + \left({1 \over 2}a - 2b\right){1 \over n^{2}} +
\cdots\,,
\qquad
n \gg 1
\\[5mm]&
\end{align}
$$
1 - a + b = 0\,,
\quad
- a + 2b = 0
\qquad\Longrightarrow\qquad
\color{#ff0000}{\large a = 2\,,\quad b = 1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Expressing as a sum of squares. Suppose a number can be expressed as a sum of two squares. If I know it's prime factorization, what is the easiest way to find two such numbers?
For e.g., let us consider the number $97^5 \cdot 641^3 \cdot 3^8$, how can I find $a$ and $b$ such that $a^2 + b^2 = 97^5 \cdot 641^3 \cdot 3^8$?
| First you must do it for its prime factors by trial, then you can use Brahmagputa's identity.
$$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2+2abcd-2abcd$$
$$=(ac+bd)^2+(ad-bc)^2$$
Notice that if $p$ is a prime number and there exist $a$ and $b$ such that $a^2+b^2=p$ then without loss of generality we can assume that $a>b$ therefore $$\displaystyle 2b^2<a^2+b^2=p \implies b^2 <\frac{p}{2} \implies b \leq \lfloor\sqrt{\frac{p}{2}}\rfloor $$ So you only need to check the first $\lfloor\sqrt{\frac{p}{2}}\rfloor$ numbers to find the smaller number $b$ in $a^2+b^2=p$, and once you've found it the other one has already been found from $a^2 = p-b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The discriminant of a cubic extension Let $K=\mathbb{Q}[\sqrt[3]{7}]$. I want to find the discriminant $d_K$ of the number field $K$.
I have computed $\operatorname{disc}(1,\sqrt[3]{7},\sqrt[3]{7^2})=-3^3\cdot 7^2$. I know that $d_K$ must divide this quantity, and may differ form it by a square, hence we have 3 possibilities: $d_K=-3\cdot 7^2$ or $-3^3$ or $-3^3\cdot 7^2$. Since $7$ ramifies in $K$, I can exclude the case $-3^3$. Now, how can I exclude also $-3\cdot 7^2$?
I need this because I want to prove that $\mathscr{O}_K=\mathbb{Z}[\sqrt[3]{7}]$. Is it possible to prove this without the discriminant?
| Yes, it is possible to prove it independent of computing the discriminant.
It suffices to check that $\mathbb{Z}[\sqrt[3]{7}]$ is normal (why?). To see this, we a priori need to check that $M\mathbb{Z}[\sqrt[3]{7}]_M$ is principal for all $M$ maximal in $\mathbb{Z}[\sqrt[3]{7}]$ (note $\mathbb{Z}[\sqrt[3]{7}]$ is already Noetherian and dimension $1$, so this is checking for Dedkind domain, which is equvalent). But, by standard theory related to Dedekind-Kummer it suffices to check that $M\mathbb{Z}[\sqrt[3]{7}]_M$ is principal only when $f'(\sqrt[3]{7})\in M$, where $f(x)$ is the min poly of $\sqrt[3]{7}$ (i.e. $x^3-7$).
But, evidently $f'(\sqrt[3]{7})=3\sqrt[3]{7}^2$. So, evidently if $M$ contains $f'(\sqrt[3]{7})$ it contains $\sqrt[3]{7}f'(\sqrt[3]{7})=3\cdot 7$. So, $M$ either lies above $3$ or $7$.
Using more standard Dedekind-Kummer theory we know that the maximals lying above $3$ are $(3,\sqrt[3]{7}+2)$ (since $x^3-7=(x+2)^3$ in $\mathbb{F}_3[x]$) and the maximals lying above $7$ are $(7,\sqrt[3]{7})$ (since $x^3-7=x^3$ in $\mathbb{F}_7[x]$).
The latter is already principal (why?), and so we need only check that the first is principal in its localization. Namely, we need to check that $(3,\sqrt[3]{7}+2)\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$ is principal. But, it's easy to check that
$$(\sqrt[3]{7}+2)^2=3\cdot(5+4\sqrt[3]{7}+2\sqrt[3]{7^2})\quad(\ast)$$
But, note that $5+4\sqrt[3]{7}+2\sqrt[3]{7^2}\notin (3,\sqrt[5]{7}+2)$ else evidently $5\in(3,\sqrt[3]{7}+2)$ which is false since $(3,\sqrt[3]{7}+2)\cap\mathbb{Z}=(3)$. Thus, we see that $5+4\sqrt[3]{7}+2\sqrt[3]{7^2}$ is invertible in $\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$ and thus $(\ast)$ obviously implies that
$$(3,\sqrt[3]{7}+2)\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}=\sqrt[3]{7}\mathbb{Z}[\sqrt[3]{7}]_{(3,\sqrt[3]{7}+2)}$$
Thus, by prior comment we see that $\mathbb{Z}[\sqrt[3]{7}]$ is integrally closed, and thus $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{7}]$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$ Find the coefficient of $x^{4}$ in $(1+x)^{\frac{1}{3}}$
I try to use the general binomial coefficient to do this
so I will first find ${{\frac{1}{3}} \choose k}$
which is equal to ${\frac{(\frac{1}{3})(\frac{1}{3}-1) ... ({\frac{1}{3}-k+1})}{k!}}$
and I take out $(-1)^k$ and I will get ${\frac{(\frac{-1}{3})(\frac{-1}{3}+1) ... ({\frac{1}{-3}+k-1})}{k!}}$ which is equal $(-1)^k {{\frac{1}{-3}+k-1} \choose k!}$
back to $x^{4}$ in $(1+x)^{\frac{1}{3}}$
$(1+x)^{\frac{1}{3}} = {\sum_k{{\frac{1}{-3}+k-1}\choose{k!}}(-1)^kx^k}$
So when to find $[x^4]$ it is just equal to ${{\frac{1}{-3}+4-1} \choose {4!} } (-1)^4$ ?
| We have $(1+x)^\alpha = \sum_{k=0}^\infty \binom{\frac{1}{3}}{k} x^k$, for $|x|<1$.
Hence the coefficient of $x^4$ is $\binom{\frac{1}{3}}{4} =\frac{(\frac{1}{3}) (\frac{-2}{3}) (\frac{-5}{3}) (\frac{-8}{3}) }{4!} = - \frac{10}{243}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...
$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$
How can I get that?
| Welcome to StackExchange! You may use LaTeX-Formulas by using dollar signs \$ x^2 \$ $\longrightarrow{}$ $x^2$.
As a first step I used the substitution $u:=\sqrt{2-x^2}$ which implies $ \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2x}{2\sqrt{1-x^2}}=\frac{x}{u}$. So we have to solve $\int \sqrt{2-u}\cdot u\ \mathrm{d}u$. This can be done by using partial integration (http://en.wikipedia.org/wiki/Integration_by_parts, use $v'=\sqrt{2-u}$), which yields $-2/15 (2-u)^ {3/2} (4+3 u)+c$. This should be equivalent to your result, if you replace $u$ by $\sqrt{1-x^2}$.
| {
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction?
I think I'm missing some key point.
| Assume $x^2+xy+y^2<0$. Adding and subtracting $xy$ on the left-hand side gives $x^2+2xy+y^2-xy=(x+y)^2-xy<0$, and therefore $0\leq(x+y)^2<xy$. Conversely, $x^2+xy+y^2<0$ implies $xy<-(x^2+y^2)\leq0$. Combining these, we have $$xy<0<xy,$$ a clear contradiction.
| {
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What is wrong with this proof of: $2+2 = 5$ I have seen this image and surprised that we can prove $2 + 2 = 5$. can any one tell me what is wrong with this image.
Prove that, $2+2=5$.
We know that, $2+2=4$
$$\begin{align}\Rightarrow2+2&=4-\dfrac92+\dfrac92\\\,\\
&=\sqrt{\left(4-\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{16-2\cdot4\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{-20+\left(\dfrac92\right)^2}+\dfrac92\\
&=\sqrt{(5)^2-2\cdot5\cdot\dfrac92+\left(\dfrac92\right)^2}+\dfrac92\qquad\qquad\qquad\\
&=\sqrt{\left(5-\dfrac92\right)^2}+\dfrac92\\
&=5-\dfrac92+\dfrac92\\
&=5\\\,\\&\therefore\,2+2=5\text{ (Proved)}\\
\end{align}$$
| Hint: Given any real number $x,$ we have $$\sqrt{x^2}=|x|.$$
| {
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Divisibility of Fibonacci numbers This question is inspired by a Project Euler problem I was working on. Noticing something that did not make sense led me to the conclusion that for all primes $p$ ending in $1$ or $9$, the $(p-1)$st Fibonacci number is divisible by $p$. I haven't proven it, but it's the only conclusion that makes sense. I'd give more information on how I arrived there but I'm afraid it might spoil the problem.
I have tested this for the first few values using an online Fibonacci calculator
$$F_{10}=55=5\times11$$
$$F_{18}=2584=136\times19$$
$$F_{28}=317811=10959\times29$$
$$F_{30}=832040=26840\times31$$
Why is this true, assuming it's true at all?
| It's true because $5$ is a quadratic residue modulo an odd prime $p \neq 5$ if and only if $p \equiv \pm 1 \pmod{5}$. The relevance of $5$ can be seen from Binet's formula
$$F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi},$$
where
$$\varphi = \frac{1+\sqrt{5}}{2};\quad \psi = \frac{1-\sqrt{5}}{2} = 1 - \varphi = -\frac{1}{\varphi}.$$
Using $\varphi - \psi = \sqrt{5}$, we can expand
$$\begin{align}
2^{p-1}F_p &= \frac1{2\sqrt{5}} \left(\sum_{k=0}^p \binom{p}{k}\sqrt{5}^k - \sum_{k=0}^p \binom{p}{k}(-\sqrt{5})^k\right)\\
&= \sum_{k=0}^{\frac{p-1}{2}} \binom{p}{2k+1}5^k\\
&\equiv 5^{\frac{p-1}{2}} \pmod{p}.
\end{align}$$
Now $2^{p-1}\equiv 1 \pmod{p}$, and $5^{\frac{p-1}{2}} \equiv \left(\frac{5}{p}\right)\pmod{p}$, where $\left(\frac{a}{p}\right)$ is the Legendre symbol, that is $\left(\frac{a}{p}\right) = 1$ if $a$ is a quadratic residue modulo $p$, $= -1$ if $a$ is a quadratic non-residue modulo $p$.
A similar computation shows that
$$F_{p+1} \equiv \frac{1 + \left(\frac{5}{p}\right)}{2} \pmod{p}.$$
So if $\left(\frac5p\right) = -1$, that is, if $p \equiv \pm 2 \pmod 5$, then $F_{p+1}$ is a multiple of $p$, and if $\left(\frac5p\right) = 1$, that is, $p \equiv \pm 1 \pmod 5$, then $F_p \equiv F_{p+1} \equiv 1 \pmod{p}$, and hence
$$F_{p-1} = F_{p+1} - F_p \equiv 0 \pmod{p}.$$
| {
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An awful factorisation question. If $a+b+c = 0$ show that $$(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)$$
I have tried substituting the values but it gets too complicated.
Can anyone please help with the method? I have been trying for 30 minutes.
Thanks!
| Let $x=2a-b$, $y=2b-c$, and $z=2c-a$. Then $x+y+z=0$, so $LHS=x^3+y^3+z^3=x^3+y^3+(-x-y)^3=3xy(-x-y)=3xyz=RHS$.
| {
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prove that $\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$ How can I prove that if $x$ and $y$ are positive then
$$\lfloor x\rfloor\lfloor y\rfloor\le\lfloor xy\rfloor$$
| I am using the notation $\{x\}$ to mean the fractional part of $x$.
$$
x = \lfloor x \rfloor + \{x\}, \;\; y = \lfloor y \rfloor + \{y\} \\
xy = \lfloor x\rfloor \lfloor y \rfloor + \lfloor x\rfloor \{y\} + \{x\}\lfloor y\rfloor + \{x\} \{y\} \\
\lfloor xy\rfloor=\lfloor x\rfloor \lfloor y\rfloor + \lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\\
\text{but } x\ge0,\;\;y\ge0\\
\lfloor\lfloor x\rfloor\{y\}\rfloor + \lfloor\{x\}\lfloor y\rfloor\rfloor\ge0\\
\therefore \lfloor xy\rfloor\ge\lfloor x\rfloor\lfloor y\rfloor$$
| {
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Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| $ax^2+bx+c = a(x^2+\frac{bx}{a}+\frac{c}{a})=a((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2})$
Equate this to zero
NOTE: $a\neq0, $ if it is then the equation will not be quadratic.
you get $(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}$
which gives the desired $$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
| {
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How is this matrix called (two diagonals)? I need to write an algorithm for solving this matrix but I wanted to first make a search online and that's why I need its name.
| You can rearrange terms, to get the following system
$$
\left [ \begin{array}{ccccccc}
d_1 & a_{2n+1} \\
a_1 & d_{2n+1} \\
& & d_2 & a_{2n} \\
& & a_2 & d_{2n} \\
& & & & \ddots \\
& & & & & d_n & a_{n+2} \\
& & & & & a_n & d_{n+2} \\
& & & & & & & d_{n+1}
\end{array}\right ] \left [ \begin{array}{c}
x_1 \\ x_{2n+1} \\ x_2 \\ x_{2n} \\ \vdots \\ x_n \\ x_{n+2} \\ x_{n+1}
\end{array}\right ] = \left [ \begin{array}{c}
b_1 \\ b_{2n+1} \\ b_2 \\ b_{2n} \\ \vdots \\ b_n \\ b_{n+2} \\ b_{n+1}
\end{array}\right ]
$$
which is equivalent to solving $n$ $2\times 2$ linear systems and $1$ algebraic equation of $1^{st}$ order.
| {
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Prove: $3(a^4+b^4+c^4)+48\ge 8(a^2b+b^2c+c^2a)$ Let $a, b, c$ - real numbers. Prove that $3(a^4+b^4+c^4)+48\ge8(a^2b+b^2c+c^2a)$
| Consider the following inequality obtained by AM-GM inequality:
$$
2a^4+b^4+16=a^4+a^4+b^4+16\geq 4\sqrt[4]{16a^8b^4}=8ba^2
$$
Writing down similar inequalities for other pairs we get:
$$
2b^4+c^4+16\geq 8cb^2\\
2c^4+a^4+16\geq 8ac^2
$$
It is enough to sum up all these inequalities and we get the desired inequality.
| {
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clarification for the solution of $(z^3=|z|^2 +4)$!? $(z-2)(z^2+z+2)=0$
so the solutions are
$$z=2;$$$$z^2+z+2=0;$$
$$z=-(1+-\sqrt{-7})/2$$
I removed the absolute value for imposing a solution in the field of real, but only because I saw it on wolfram alpha, now I ask you! I can always impose a solution in the real numbers to remove the absolute value? and if yes I have to discard all imaginary values?
| Let us proceed in the conventional way:
Let $z=r(\cos\theta+i \sin\theta)$ where $r$ is real
$\implies |z|=r$ and $z^3=\{r(\cos\theta+i \sin\theta)\}^3=r^3\cos3\theta+ir^3\sin3\theta$ (using de Moivre's formula)
$\implies r^3\cos3\theta+ir^3\sin3\theta=r^2+4$
Clearly $r\ne0$
Equating the imaginary terms $r^3\sin3\theta=0\implies \sin3\theta=0\implies 3\theta=n\pi$ where $n$ is any integer
$\theta=\frac{n\pi}3$ where $n=0,1,2$ (A more general set of values of $n$ has been discussed here)
If $n=0,2,\cos3\theta=1$ and $r^3-r^2-4=0\implies r=2$ (as you have found)
$\implies z=2(\cos0+i \sin0)=2$ and $z=2(\cos\frac{2\pi}3+i\sin\frac{2\pi}3)=-1+\sqrt3i$
If $n=1,\cos3\theta=-1\implies z=-2(\cos\frac\pi3+i \sin\frac\pi3)=-1-\sqrt3i$
| {
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Spot my error in solving a linear system I almost always get the unit matrix if I try to get to an row reduced echelon form. I probably always make a mistake. Can you spot the error? What illegal operations could a beginner do while trying to solve a linear system?
\begin{array}{}
1 & 1 & 3 \\
-4 & -3 & -8 \\
-2 & -1 & -2 \\
1 & 2 & 7 \end{array}
2.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 1 & 4 \\
0 & -1 & -4 \end{array}
3.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 1 & 4 \\
0 & 0 & 0 \end{array}
4.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 0 & 2 \\
0 & 0 & 0 \end{array}
5.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 6 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
6.
\begin{array}{}
1 & 1 & 3 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
7.
\begin{array}{}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
8.
\begin{array}{}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \end{array}
| Your very first step on the very first row operation, unfortunately
| {
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What is the center of mass of the region bounded by $y=x^2$ and $y=4$?
English: That the centroid of the area bounded by $y=x^2$ and $y=4$? The answer is $(0,\frac{12}{5}$) I would like to understand the calculations!
$$A=\int_{-2}^{2}|4-x^2|dx=\frac{32}{3}$$
Em Português: Qual o centróide da região limitada por $y=x^2$ e $y=4$? A resposta é $(0,\frac{12}{5}$) Gostaria de entender os cálculos!
| Center of mass is the "average point". For example, if I just gave you 2 points, $p_1$ and $p_2$, and asked for the center of mass, the answer would be $\frac {p_1 + p_2} 2$.
I'm assuming that your object has uniform density; in other words, it's not made of steel on one side and cotton on the other.
Center of mass then becomes the "average point" of your object: add up the points, then divide by the area.
$\begin {align}\frac {\int_{\omega} {\begin{bmatrix} x \\ f(x) \end{bmatrix}}}
{\int_{\omega} {1}} \end{align}$
where $\omega$ is the area you are integrating over (note, this is assuming a finite surface and uniform density).
The numerator is the sum of all the points in your area:
$\begin {align}\int_{x=-2}^{x=2} \int^{y=4}_{y=x^2} {{\begin{bmatrix} x \\ y \end{bmatrix}\ dy}\ dx} \end{align}$
$x$ is a constant when you are integrating across $y$.
$\begin {align}\int_{x=-2}^{x=2} {\left. \begin{bmatrix} xy \\ \frac 1 2 y^2 \end{bmatrix}\right |_{y=x^2}^{y=4}\ dx} \end{align}$
$\begin {align}\int_{x=-2}^{x=2} { \begin{bmatrix} 4x - x^3 \\ 8 - \frac 1 2 x^4 \end{bmatrix}\ dx} \end{align}$
$\begin {align} \left. \begin{bmatrix} 2x^2 - \frac 1 4 x^4 \\ 8x - \frac 1 {10} x^5 \end{bmatrix} \right |_{x=-2}^{x=2}\end{align}$
$\begin {align} \begin{bmatrix} 0 \\ \frac {128} 5 \end{bmatrix} \end{align}$
The denominator is the area:
$\begin {align}\int_{x=-2}^{x=2} \int^{y=4}_{y=x^2} {{1\ dy}\ dx} \end{align}$
$\begin {align}\int_{x=-2}^{x=2} {4 - x^2 \ dx} \end{align}$
$\frac {32} 3 $
So numerator over denominator is:
$\begin{bmatrix} 0 \\ \frac {128} 5 \end{bmatrix} \cdot \frac 3 {32} = \begin{bmatrix} 0 \\ \frac{12} 5\end{bmatrix}$
| {
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sums of legendre symbol I am studying the sums of Legendre symbol. It's easy to prove that for $p\equiv 1 \pmod 4$,
$$\sum_{0<n<p/2}\left(\frac{n}{p}\right)=0.$$
by writing the sum over all residue classes and then using the fact that sum over $0$ to $p/2$ is same as $p/2$ to $p$ (by a change of variable). But the same idea is not working with the following two equalities:
(1) If $p\equiv 3 \pmod 8$, then
$$\sum_{0<n<p/4} \left(\frac{n}{p}\right) =0.$$
(2) If $p\equiv 7 \pmod 8$, then
$$\sum_{p/4<n<p/2} \left(\frac{n}{p}\right) =0.$$
Thanks in advance for any help.
| (1) If $p \equiv 3 \pmod{8}$ then $(\frac{-1}{p})=(\frac{2}{p})=-1$. Let $p=8k+3$. Observe that
\begin{align}
\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})}& =\sum_{1 \leq n \leq 4k+1}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k}{(\frac{2n}{p})}+\sum_{1 \leq n \leq 2k+1}{(\frac{2n-1}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{p+1-2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{2k+1 \leq n \leq 4k+1}{(\frac{2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+(\frac{-1}{p})(\frac{2}{p})\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})} \\
& =-\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}+\sum_{2k+1 \leq n \leq 4k+1}{(\frac{n}{p})}
\end{align}
Thus
\begin{equation}
\sum_{0<n<\frac{p}{4}}{(\frac{n}{p})}=\sum_{1 \leq n \leq 2k}{(\frac{n}{p})}=0
\end{equation}
(2) If $p \equiv 7 \pmod{8}$ then $(\frac{-1}{p})=-1, (\frac{2}{p})=1$. Let $p=8k+7$. Observe that
\begin{align}
\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}& =\sum_{1 \leq n \leq 4k+3}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k+1}{(\frac{2n}{p})}+\sum_{1 \leq n \leq 2k+2}{(\frac{2n-1}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{1 \leq n \leq 2k+2}{(\frac{p+1-2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})\sum_{2k+2 \leq n \leq 4k+3}{(\frac{2n}{p})} \\
& =(\frac{2}{p})\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}+(\frac{-1}{p})(\frac{2}{p})\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})} \\
& =\sum_{1 \leq n \leq 2k+1}{(\frac{n}{p})}-\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}
\end{align}
Thus
\begin{equation}
\sum_{\frac{p}{4}<n<\frac{p}{2}}{(\frac{n}{p})}=\sum_{2k+2 \leq n \leq 4k+3}{(\frac{n}{p})}=0
\end{equation}
| {
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How to solve step by step $\frac{2n+1}{n+2} < 8$ Given something like $\frac{2n+1}{n+2} < 8$ how do you solve that step by step to get all the possible intervals?
| Observe that $\displaystyle \lim_{n+2\to0^-}\frac{2n+1}{n+2}\to+\infty$ and $\displaystyle \lim_{n+2\to0^+}\frac{2n+1}{n+2}\to-\infty$
If $n+2\ne0,(n+2)^2>0$ for real $n$
$$\frac{2n+1}{n+2}<8\iff \frac{2n+1}{n+2}-8<0$$
Now, $$\frac{2n+1}{n+2}-8=-\frac{6n+15}{n+2}=-3 \frac{(2n+5)}{n+2}$$
$$\implies -3 \frac{(2n+5)}{n+2}<0\iff\frac{(2n+5)}{n+2}>0$$
Method $1:$
If $2n+5>0\iff n>-\frac52, n+2>0\iff n>-2\implies n>-2$
or if $2n+5<0\iff n<-\frac52, n+2<0\iff n<-2\implies n<-\frac52$
Method $2:$
Multiplying either sides by $(n+2)^2>0$ assuming $n+2\ne0$
$$ (2n+5)(n+2)>0 \iff\left(n+\frac52\right)(n+2)>0$$
We know, if $(x-a)(x-b)>0$ where $a,b$ we shall have $x<a$ or $x>b$
| {
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from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$ How can you go from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$? I mean how to prove that $1-\sin x = 2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$?
| Use $\displaystyle \sin x=\cos\left(\frac\pi2-x\right)$ and $\displaystyle\cos2y=1-2\sin^2y$
| {
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Solve the following Diffrential Equation $(3y-7x+7)dx-(3x-7y-3)dy=0$ I want to solve the following equation:
$$(3y-7x+7)dx-(3x-7y-3)dy=0$$
I will need two new variables? or I can solve it with 1, for example set expression as $z$?
What you are suggesting?thanks.
| Answer taken from
http://in.answers.yahoo.com/question/index?qid=20080830232129AAdNlmN
$$(3x−7y−3)dy=(3y−7x+7)dx \\
\frac{dy}{dx}=\frac{3y−7x+7}{3x−7y−3} $$
Now, let $x=x’+h$ and $y=y’+k$
Therefore
$$\frac{dy’}{dx’} =\frac{3y’−7x’+3k−7h+7}{3x’−7y’+3h−7k−3} $$
Putting $h=1$ and $k=0$, we get
$$\frac{dy’}{dx’}=\frac{3y’−7x’}{3x’−7y’} $$
Now, let $y’=vx’$ so that $$\frac{dy’}{dx’}=v+x’\frac{dv}{dx’}$$
We have
$$ v+x’ \frac {dv}{dx’} = \frac {3vx’−7x’}{3x’−7vx’} $$
$$x’ \frac {dv}{dx’} = \frac {3v−7}{3−7v} − v $$
$$x’ \frac {dv}{dx’} = \frac {3v−7−3v+7v²}{3−7v} $$
$$x’ \frac {dv}{dx’} = \frac {7v²−7}{3−7v} $$
$$ \frac{dx’}{x’} = \frac {3−7v}{7v²−7} dv $$
Integrating both sides,
$$\int\frac{dx’}{x’}=\int\frac{3 dv}{7v²−7}−\int\frac{7v dv}{7v²−7} \\
\int\frac{dx’}{x’}=\frac{3}{7} \int\frac{dv}{v²−1}−\int\frac{v dv}{v²−1} \\
ln|x’|+c=\frac{3}{7} \frac{1}{2} ln\frac{|v−1|}{|v+1|}
−\frac{1}{2} \int\frac{v dv}{v²−1} \\
ln|x’|+c=\frac{3}{14} ln\frac{|v−1|}{|v+1|}
−\frac{1}{2} ln|v²−1| \\
ln|x’|+c=\frac{3}{14} ln\frac{|y’−x’|}{|y’+x’|}
−\frac{1}{2} ln|\frac{(y’)^2−(x’)^2}{(x’)^2}| $$
(since $v=\frac{y’}{x’}$)
$$ ln |x’|+c=\frac{3}{14} ln|y’−x’|−\frac{3}{14} ln|y’+x’|−
\frac{1}{2} ln|y’+x’|−\frac{1}{2} ln|y’−x’|
+\frac{1}{2} ln|(x’)^2| \\
ln |x’|+c=−\frac{2}{7} ln|y’−x’|−\frac{5}{7} ln|y’+x’|+\frac{1}{2} ln|(x’)^2| \\
2ln|y’−x’|+5ln|y’+x’|=C \\
2ln|y−x+1|+5ln|y+x−1|=C $$
(since $y’=y$ and $x’=(x−1)$
| {
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Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\sum_{cyc}{\sqrt[4]{a^2b}})^2}{a+b+c}$$ but I don't obtain anything.
| Since $\left(1/\sqrt{x}\right)^{\prime\prime} = \frac{3}{4 x^{5/2}} > 0$ for $x > 0$, we have that $f(x)=\frac{1}{\sqrt{x}}$ is convex.
Then, by Jensen's inequality we have:
$$ \tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2} $$
Here $9=(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \geq 3\times (ab+bc+ca)$ $(*)$, so we actually get $3\geq ab+bc+ca$ and therefore:
$$ \tfrac{1}{a+b+c}\times \left( \frac{a}{\sqrt{b}} + \frac{b}{\sqrt{c}} + \frac{c}{\sqrt{a}} \right) \geq \left( \frac{ab+bc+ca}{a+b+c} \right)^{-1/2} \geq \left( \frac{3}{a+b+c} \right)^{-1/2} = 1 $$
Since we have $a+b+c=3$, we are done.
$(*)$ This follows from $2\times \{a^2+b^2+c^2-(ab+bc+ca)\} = (a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0$
| {
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Let p, q, r be distinct primes greater than 3, and let n = pqr. Show that if $x \in \mathbb{Z}$ satisfies $x^{2} \equiv 9\mod{n}$
then $x \equiv ±3 \mod{p}$, $x \equiv ±3 \mod{q}$ and $x \equiv ±3 \mod {r}$.
I'm not sure what to do. Any help is appreciated.
Thanks.
| Let $x \in \mathbb{Z}$, and $n=pqr$, such that $x^{2} \equiv 9 \mod {n} $, ie, $\exists k\in \mathbb{Z}$ such that
\begin{eqnarray}
x^{2}-9 &=& nk\\
(x-3)(x+3)&=&nk
\end{eqnarray}
Now
*
*P.D. $x\equiv 3\mod{p}$
\begin{eqnarray}
(x-3)(x+3)&=&nk\\
(x-3)(x+3)&=&pqrk\\
(x-3)(x+3)&=&p(qrk)\\
x&\equiv&\pm3\mod{p}
\end{eqnarray}
*P.D. $x\equiv 3\mod{q}$
\begin{eqnarray}
(x-3)(x+3)&=&nk\\
(x-3)(x+3)&=&pqrk\\
(x-3)(x+3)&=&q(prk)\\
x&\equiv&\pm3\mod{q}
\end{eqnarray}
*P.D. $x\equiv 3\mod{r}$
\begin{eqnarray}
(x-3)(x+3)&=&nk\\
(x-3)(x+3)&=&pqrk\\
(x-3)(x+3)&=&r(pqk)\\
x&\equiv&\pm3\mod{r}
\end{eqnarray}
| {
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Integration by parts, error on Primitive of $\sqrt{1-x^{2}}$ Problem: Integrate $\sqrt{1-x^{2}}$
Attempt:
| The usual method uses the substitution $x=\sin\theta$ followed by the identity $\cos^2\theta=\frac{1}{2}(\cos 2\theta+1)$. However, since the OP used integration by parts, let's do it that way. It is perhaps a little harder, but introduces a useful technique.
Let our integral be $I$, and let $g'(x)=1$ and $f(x)=\sqrt{1-x^2}$. Then we get
$$I=x\sqrt{1-x^2}+\int \frac{x^2}{\sqrt{1-x^2}}\,dx.$$
Rewrite $\frac{x^2}{\sqrt{1-x^2}}$ as $\frac{1-(1-x^2)}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-x^2}}-\sqrt{1-x^2}$. We get
$$I=x\sqrt{1-x^2} +\int \frac{1}{\sqrt{1-x^2}}\,dx-I.$$
But $\frac{1}{\sqrt{1-x^2}}$ has $\arcsin x$ as an antiderivative. Bring the second $I$ to the other side, divide by $2$, and don't forget about the $+C$.
| {
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Probability of getting Required Sum
What is the probability of when four dice rolled together once,and getting a sum of Thirteen
If we do by just calculating all possible values of sum, then it will take more time; so we can solve the above problem as Multinomial Coefficents of sum, i.e.:
$$
x_1+x_2+x_3+x_4 = 13\;\text{ where },1\leq x_i \leq 6\;\;\forall\;1\leq i \leq 4.
$$
I got stuck that form the above equation how can we obtain $N_{\text{ways to get $\sum=13$}}$ or is there any easy method to find this kind of problems?
| I assume that you already know how to calculate the probability for the sum of two dices. One gets the following table.
$$
\begin{array}{r|c}
k & p(x_1+x_2=k) \\
\hline{} \\
2 & 1/36 \\
3 & 2/36 \\
4 & 3/36 \\
5 & 4/36 \\
6 & 5/36 \\
7 & 6/36 \\
8 & 5/36 \\
9 & 4/36 \\
10 & 3/36 \\
11 & 2/36 \\
12 & 1/36 \\
\end{array}
$$
Now we can calculate
$$
p(x_1+x_2+x_3+x_4=13)=$$
$$\begin{array}{}
p( ((x_1+x_2=2) \land (x_3+x_4=11) ) &\lor \\
((x_1+x_2=3) \land (x_3+x_4=10)) &\lor \\
((x_1+x_2=4) \land (x_3+x_4=9)) & \lor \\
\cdots &\lor \\
((x_1+x_2=11) \land (x_3+x_4=2))) &=
\end{array}
$$
$$
\begin{array}{}
p(x_1+x2=2)\cdot p(x_3+x_4=11) &+ \\
p(x_1+x2=3)\cdot p(x_3+x_4=10) &+ \\
p(x_1+x+2=5)\cdot p(x_3+x_4=9)& + \\
\cdots &+ \\
p(x_1+x2=11)\cdot p(x_3+x_4=2)&=
\end{array}$$
$$\frac{1}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{4}{36} + \frac{4}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{6}{36} + \\
\frac{6}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{4}{36} +
\frac{4}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{1}{36}=$$
$$\frac{(2+6+12+20+30 ) \cdot 2}{36 \cdot 36}=\frac{140}{36 \cdot 36 }=\frac{35}{324}$$
| {
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Proof with mathematical induction that $ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $ Proof with mathematical induction.
I have the following induction problem:
$ (\frac{n}{n+1})^2 + (\frac{n+1}{n+2})^2 + ... + (\frac{2n - 1}{2n})^2 \le n - 0.7 $
This property applies to all $n \ge 1$. I've shown that it is true for $n = 1$ and $n = k$, however I can not show for $n = k+1$.
This is how I start for $n = k+1$:
$ (\frac{k+1}{k+2})^2 + (\frac{k+2}{k+3})^2 + ... + (\frac{2k - 1}{2k})^2 + (\frac{2k}{2k + 1})^2 + (\frac{2k + 1}{2k + 2})^2 \le k + 1 - 0.7 $
Any ideas?
| You are almost there. At the end, you wrote the difference between the two sums as a single complicated fraction. Maybe you have miscalculated something there, maybe not, anyway, it is not easy to see through. It becomes much easier if we proceed in steps, pairing like things with like things. The difference is
$$\begin{align}
\left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{n}{n+1}\right)^2
&= \left(\frac{2n}{2n+1}\right)^2 + \left( \frac{2n+1}{2n+2}\right)^2 - \left(\frac{2n}{2n+2}\right)^2\\
&= \left(\frac{2n}{2n+1}\right)^2 + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - \frac{(2n+1)^2 - (2n)^2}{(2n+1)^2} + \frac{(2n+1)^2-(2n)^2}{(2n+2)^2}\\
&= 1 - (4n+1)\left(\frac{1}{(2n+1)^2}-\frac{1}{(2n+2)^2}\right)\\
&< 1.
\end{align}$$
| {
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Does there exist a prime number within the interval? Conjecture
$\forall p_{n}\in \mathbb{P} : n\geq3, \: \exists p_{m}\in \mathbb{P} : 3p_{n} - 4 \geq p_{m} > \sqrt{2(p^2_{n+1} - 1)} $
How would you go about proving/disproving this?
| This is true indeed. First consider $n$ s.t. $p_n \geq 29$. It has been proven that for $m \geq 3$, there is a prime in the interval $(m, \frac{4(m+2)}{3})$, see here, Corollary 2.2.
Since $p_n \geq 3$, there is a prime in $(p_n, \frac{4(p_n+2)}{3})$, so $p_{n+1}<\frac{4(p_n+2)}{3}$. Now $$\sqrt{2(p_{n+1}^2-1)}<\sqrt{2}p_{n+1}<\frac{4\sqrt{2}(p_n+2)}{3}<\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil$$
Since $\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil>\frac{8\sqrt{2}}{3} \geq 3$, there is a prime in $\left(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil, \frac{4(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil+2)}{3}\right)$.
Thus $\exists p_m$ s.t.
$$\sqrt{2(p_{n+1}^2-1)}<\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil<p_m<\frac{4(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil+2)}{3}<\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3}$$
Now since $p_n \geq 29$, we have
$$(3p_n-4)-(\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3})=(3-\frac{16 \sqrt{2}}{9})p_n-(8+\frac{32\sqrt{2}}{9})>0$$
Thus
$$\sqrt{2(p_{n+1}^2-1)}<p_m<\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3}<3p_n-4$$
It remains to check $5 \leq p_n<29$.
Checking small cases:
$p_n=5$: The interval is $(\sqrt{96}, 11]$ so $p_m=11$ works.
$p_n=7$: The interval is $(\sqrt{240}, 17]$ so $p_m=17$ works.
$p_n=11$: The interval is $(\sqrt{336}, 29]$ so $p_m=29$ works.
$p_n=13$: The interval is $(\sqrt{576}, 35]$ so $p_m=31$ works.
$p_n=17$: The interval is $(\sqrt{720}, 47]$ so $p_m=47$ works.
$p_n=19$: The interval is $(\sqrt{1056}, 53]$ so $p_m=53$ works.
$p_n=23$: The interval is $(\sqrt{1680}, 65]$ so $p_m=61$ works.
| {
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Geometric mean proof with two tangents There are two tangent lines on $f(x) = \sqrt{x}$ each with the $x$-value $a$ and $b$ respectively.
I need to prove that $c$, the $x$ value of the point at which the two lines intersect each other, is equal to $\sqrt{ab}$, the geometric mean of $a$ and $b$.
I have been trying many different ways of doing this question and I keep getting stuck.
| The tangent line at $(a,\sqrt{a})$ has equation $y=\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}$.
The tangent line at $(b,\sqrt{b})$ has equation $y=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$.
Set $\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$ and solve for $x$.
| {
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Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ I am trying to prove that
$$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx = -\frac{\pi^3}{48}-\frac{\pi}{8}\log^2 2 +G\log 2$$
where $G$ is the Catalan's Constant. Numerically, it's value is $-0.199739$.
Using the substitution $x=\tan \theta$, it can be written as
$$
\begin{align*}
I &= \int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos 2\theta) d\theta-2\int_0^{\frac{\pi}{4}}\theta \tan \theta \log(\cos \theta)d\theta \end{align*}
$$
Can anyone suggest a good approach to evaluate it?
| Expanding the inverse tangent in logarithms, writing $\frac{x}{1+x^2}=\Re\frac1{x-i}$, and expanding $\log(1-x^2)=\log(1-x)+\log(1+x)$, each of the resulting four indefinite integrals has a closed form. Each term is amenable to automatic integration, (an example), which means that after taking limits, slogging through simplifications and special values, such as those found here, the closed form can be computed.
For example, for the term above,
$$ \int_0^1\frac{\log(1-ix)\log(1-x)}{x-i}\,dx =
-\frac{K\pi }{4}-\frac{17 i \pi ^3}{384}-\frac{1}{2} i K \log2+\frac{13}{192} \pi ^2 \log2+\frac{3}{32} i \pi (\log2)^2-\frac{(\log2)^3}{48}+3 \,\text{Li}_3({\textstyle\frac{1+i}{2}})-\frac{45 \zeta(3)}{32}.
$$
Now, the integrand of the integral in question is the real part of the sum
$$ \frac i2 \frac{\log(1-ix)\log(1-x)}{x-i} - \frac i2\frac{\log(1+i x)\log(1-x)}{x-i}+\frac i2\frac{\log(1-ix)\log(1+x)}{x-i}-\frac i2\frac{\log(1+ix)\log(1+x)}{x-i},
$$
where each term has a closed form for its integral, as above, in terms of $\pi$, $K$, $\log 2$ and $\text{Li}_3$.
After sufficient simplification, the integral of that sum is
$$\begin{aligned} &\int_0^1 \frac{\arctan x\log(1-x^2)}{x-i}\,dx = \\
&-\frac{1}{4} i K\pi -\frac{\pi ^3}{48}+\frac{1}{32} i \pi ^2 \log2-\frac{1}{8} \pi (\log2)^2+K \log2+\frac{7}{32} i \zeta(3), \end{aligned}$$
of which the real part gives the answer
$$ -\frac{\pi ^3}{48}-\frac{1}{8} \pi (\log2)^2+ K \log2$$
| {
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How to find the equation of a line, tangent to a circle, that passes through a given external point I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.
Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How would I go about finding one of the equations of the lines tangent to the circle?
I haven't started calculus, so I ask for advice fitting for someone starting a topic like this.
| We can use the formula for the distance from a point to a line. The distance between a line $ax + by + c = 0$ and a point $(x_0, y_0)$ is
$$\left| \frac{a x_0 + b y_0 + c}{\sqrt{a^2+b^2}} \right|.$$
The line of tangency that passes the point $(4, -3)$ has equation $y+3 = m(x-4)$ for some slope $m$. So we just need to find $m$. Note that this equation can be written as
$$mx - y -4m -3 = 0.$$
The distance between the center of the circle $(-2, -7)$ and the tangent line $mx - y -4m -3 = 0$ can be found from the formula above. This distance also equals to the radius of the circle, which is $5$. Hence
$$\left| \frac{m(-2) + (-1)(-7) -4m -3}{\sqrt{m^2 + 1}} \right| = 5.$$
Rearranging we have
$$|-6m + 4| = 5 \sqrt{m^2+1}. $$
Squaring and rearranging, we get the quadratic equation
$$11m^2 - 48m - 9 = 0$$
with solutions
$$m = \frac{24\pm15\sqrt{3}}{11}.$$
Therefore, the required equations for the tangent lines are
$$y+3 = \frac{24+15\sqrt{3}}{11}\cdot(x-4)$$
and
$$y+3 = \frac{24-15\sqrt{3}}{11}\cdot(x-4).$$
| {
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"source": "stackexchange",
"question_score": "14",
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Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
| Consider the polynomial
$$(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2+b\lambda-c\tag{*1}$$
We know
$$\begin{cases}a = x + y +z\\ b = xy + yz + xz \\ c = x y z\end{cases}$$
Substitute $x, y, z$ for $\lambda$ in $(*1)$ and sum, we get
$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$
This is equivalent to
$$\begin{align}
x^3+y^3+z^3 - 3xyz
= & x^3+y^3+z^3 - 3c\\
= & a(x^2+y^2+z^2) - b(x+y+z)\\
= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)
\end{align}$$
| {
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"question_score": "20",
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True or false: $a^2+b^2+c^2 +2abc+1\geq 2(ab+bc+ca)$ Is this inequality true?
$a^2+b^2+c^2 +2abc+1\ge2(ab+bc+ca)$, where $a,b,c\gt0$.
Can you find a counterexample for this or not?
| The following substitution transforms the constrained inequality into an unconstrained one:
$$a = t^2 \gt 0$$
$$b = u^2 \gt 0$$
$$c = v^2 \gt 0$$
The original inequality becomes:
$$F(t, u, v) = t^4+u^4+v^4 +2t^2u^2v^2+1 - 2*(t^2u^2 + t^2v^2 + u^2v^2) \ge 0$$
The new expression is fully symmetric in $t, u $ and $v$.
Therefore, the minimum must be symmetric as well, which means $t=u=v$.
$$F(t,t,t) = -3t^4 + 2t^6 + 1$$
$$\frac{\delta F}{\delta t} = -12t^33 + 12 t^5 = 12t^3(t^2-1)$$
For $t=1$ we get an extremum $F(1,1,1)=0$ which confirms our original inequality.
The following steps should convince us that the solution is in fact symmetric:
To be at the minimum of the left-hand-side, the partial derivatives have to be zero.
$$\frac{\delta F}{\delta t} = 4t^3 + 4tu^2v^2 - 4tu^2 - 4tv^2 = 0$$
$$\frac{\delta F}{\delta u} = 4u^3 + 4t^2uv^2 - 4t^2u - 4uv^2 = 0$$
$$\frac{\delta F}{\delta v} = 4v^3 + 4t^2u^2v - 4t^2v - 4u^2v = 0$$
this leads to:
$$t(t^2 - v^2 + u^2(v^2 - 1)) = 0$$
$$u(u^2 - v^2 + t^2(v^2 - 1)) = 0$$
$$v(v^2 - u^2 + t^2(u^2 - 1)) = 0$$
All variables have to be greater than zero.
$$t^2 - v^2 + u^2(v^2 - 1) = 0$$
$$u^2 - v^2 + t^2(v^2 - 1) = 0$$
$$v^2 - u^2 + t^2(u^2 - 1) = 0$$
The first two of these equations combine to:
$$ (t^2 - u^2)v^2 = 0$$
The last two equations give us:
$$ t^2(u^2+v^2-2) = 0$$
The first and the third equation yield:
$$t^2 - u^2 +u^2(v^2-1) + t^2(u^2-1)$$
Assuming $t=u$ we get:
$$t^2(t^2+v^2-2) = 0$$
or
$$v^2 = 2 - t^2$$
Inserting this we get
$$- t^4 + 3t^2 -2 = 0$$
This can be factored into
$$(t-1)(t+1)(t^2-2)=0$$
For $t^2-2=0$ we get $a=2$, $b=2$ and $c=0$ which violates the constraints.
Therefore, $a=b=c=1$ is the global minmum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/544218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How to efficiently compute $17^{23} (\mod 31)$ by hand? I could use that $17^{2} \equiv 10 (\mod 31)$ and express $17^{23}$ as $17^{16}.17^{4}.17^{3} = (((17^2)^2)^2)^2.(17^2)^2.17^2.17$ and take advantage of the fact that I can more easily work with powers of ten ($17^2 \equiv 10 (\mod 31), (17^2)^2 \equiv 100 (\mod 31) \equiv 7 (\mod 31)$, etc.). While this makes the computation easier, I'm thinking there might be a better way to do this.
| Modulo $31$ we have:
$$17^1 \equiv 17$$
$$17^2 \equiv 10 \text{ (because $289$ mod $31$ is $10$)}$$
$$17^4 \equiv (10)^2 \equiv 7 \text{ (because $100$ mod $31$ is $7$)}$$
$$17^8 \equiv (7)^2 \equiv 18 \text{ (Similar)}$$
$$17^{16} \equiv (18)^2 \equiv 14 $$
And $17^{23} \equiv 17^{16} \cdot 17^4 \cdot 17^2 \cdot 17^1$.
So $17^{23} \equiv 14 \cdot 7 \cdot 10 \cdot 17$. We note that $7\cdot 10 = 70 \equiv 8 \mod 31$, so $17^{23} \equiv 14 \cdot 8 \cdot 17$, and $8\cdot 17 = 136 \equiv 12 \mod 31$, so $17^{23} \equiv 14 \cdot 12 \mod 31$, and $14 \cdot 12 = 168 \equiv 13 \mod 31$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Proving $\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12$ How can I prove that
$$\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}$$
| $$\begin{aligned}\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)
&=\lim _{x\to 0}\left(\frac{\ln \left(x+1\right)-\ln \left(x+\sqrt{x^2+1}\right)}{\ln \left(x+1\right)\ln \left(x+\sqrt{x^2+1}\right)}\right)
\\&=\lim _{x\to 0}\left(\frac{x-\frac{1}{2}x^2+o\left(x^2\right)-\left(x+o\left(x\right)\right)}{\left(x-\frac{1}{2}x^2+o\left(x^2\right)\right)\left(x+o\left(x\right)\right)}\right)
\\&=\lim _{x\to 0}\left(\frac{-\frac{x^2}{2}+o\left(x\right)}{x^2-\frac{x^3}{2}+o\left(x\right)}\right)
\\&=\color{red}{-\frac{1}{2}}\end{aligned}$$
Solved with Taylor expansions
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find a matrix $X$ given $X^4$ Find the matrix $X$ such that
$$X^4=\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
This problem I can't work,and I think let the matrix the eigenvalue is $\lambda$,then
$\lambda^4$ is $$\begin{bmatrix}
3&0&0\\
0&3&1\\
0&0&0
\end{bmatrix}$$
eigenvalue?Thank you for your help.
| You can look for a solution of the same form as $X$: calculating $$\begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix} = a \cdot \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}$$ shows that $$\begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}^4 = a^3 \begin{pmatrix} a & 0 & 0 \\ 0 & a & b \\ 0 & 0 & 0\end{pmatrix}.$$ In particular, you can take $a = 3^{1/4}$ and $b = 3^{-3/4}$ for one possibility of $X$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| let $x=z+u,y=z+v,\to u\ge v \ge0$
$x^3y^2+y^3z^2+z^3x^2- xyz(x^2+y^2+z^2)=(u^2-uv+v^2)z^3+3u^2vz^2+uv(u^2-v^2)z+3u^2v^2z+u^3v^2 \ge0$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.
But what about the ordinary generating function for $ \displaystyle \binom{3n}{n}$?
According to Wolfram Alpha, $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$
Any suggestions on how to prove this?
EDIT:
Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$
I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$
where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.
Evaluating that contour integral would appear to be quite difficult.
| Standard conversion to hypergeometric series and use of the duplication
and triplication formulas for the $\Gamma$-function yields
\begin{equation}
\sum_{n\ge 0}\binom{3n}{n}z^n
=
\sum_{n\ge 0}\frac{\Gamma(3n+1)}{\Gamma(2n+1)}\frac{z^n}{n!}
=
\sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(n+1)}
{\Gamma(n+1/2)\Gamma(n+1)}\frac{z^n}{n!}
\frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}}
\end{equation}
\begin{equation}
=
\frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)}
{\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{z^n}{n!}
\frac{(2\pi)^{1/2} 3^{3n+1/2}}{2\pi2^{2n+1/2}}
\end{equation}
\begin{equation}
=
\sqrt{3/(4\pi)}
\frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{\Gamma(n+1/3)\Gamma(n+2/3)\Gamma(1/2)}
{\Gamma(n+1/2)\Gamma(1/3)\Gamma(2/3)}\frac{(2^{-2} 3^3z)^n}{n!}
\end{equation}
\begin{equation}
=
\sqrt{3/(4\pi)}
\frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)} \sum_{n\ge 0}\frac{(1/3)_n (2/3)_n}
{(1/2)_n}\frac{(2^{-2} 3^3z)^n}{n!}
\end{equation}
\begin{equation}
=
\sqrt{3/(4\pi)}
\frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/2)}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z)
\end{equation}
\begin{equation}
=
\sqrt{3}
\frac{\Gamma(1/3)\Gamma(2/3)}{2\pi}{} _2F_1(1/3, 2/3; 1/2; 2^{-2}3^3z)
\end{equation}
Furthermore by equation 15.1.18 of Abramowitz/Stegun
this Gaussian Hypergeometric Function can be reduced by
\begin{equation}
_2F_1(a,1-a;1/2;\sin^2z)=\frac{\cos[(2a-1)z]}{\cos z}
\end{equation}
with parameter $a=1/3$. Furthermore $\Gamma(1/3)\Gamma(2/3) = 2\pi/\sqrt{ 3}$
according to OEIS sequence A073006.
| {
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"source": "stackexchange",
"question_score": "21",
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Integral $\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}\mathrm dx$$
| I just want to give two alternate forms of
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac34-\Im\operatorname{Li}_2\frac{3i}4.$$
First by using $(12)$ from here we get
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\frac{3}{2}-6\,G+4\,\Im\operatorname{Li}_2\frac{i}{2}+2\,\Im\operatorname{Li}_2\frac{i}{3},$$
where $G$ is the Catalan's constant.
The second one comes from the relationship between inverse tangent integral and Clausen's integral.
$$\int_0^\infty\frac{\operatorname{arccot}\left(\sqrt{x}-2\,\sqrt{x+1}\right)}{x+1}dx=\pi\ln\left(\frac{3}{4}\right)+\arctan\left(\frac{3}{4}\right)\ln\left(\frac{4}{3}\right)-\frac{1}{2}\Im\left(\operatorname{Li}_2\left(\alpha^2\right)+\operatorname{Li}_2\left(-\frac{1}{\alpha^2}\right)\right),$$
where $\alpha=\frac{1+2i}{2+i}=\frac{4}{5}+\frac{3}{5}i$. Note that $\arctan\frac{3}{4}=\arcsin\frac{3}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 4,
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Residue/Contour integration problem Supposedly,
$\displaystyle\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right)$, $a>0$.
Using Residues/Contour integrals, I have
$\displaystyle\int_{-\infty}^\infty \frac{e^{iax}}{x^4+1}dx=\int_{-\infty}^\infty \frac{\cos ax}{x^4+1}dx+i\int_{-\infty}^\infty \frac{\sin ax}{x^4+1}dx=\frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left(\cos\frac{a}{\sqrt{2}}+i\sin\frac{a}{\sqrt{2}}\right)$
I don't see how to draw the final conclusion.
| Here is an answer that uses a quarter-circle instead of a semicircle. Start by observing that
$$\int_{-\infty}^\infty \frac{\cos(ax)}{x^4+1} dx
= 2\int_0^\infty \frac{\cos(ax)}{x^4+1} dx
= 2 \times\Re\left(\int_0^\infty \frac{e^{iax}}{x^4+1} dx\right).$$
Now use a contour consisting of three segments to evaluate the integral, a line segment $\Gamma_1$ along the real axis from the origin to $R$, a quarter circle $\Gamma_2$ to $iR$ on the imaginary axis and another line segment $\Gamma_3$ along the imaginary axis back to the origin.
Clearly along $\Gamma_1$ we get the desired integral in the limit. Along $\Gamma_2$ we can parameterize with $z=Re^{it},$ applying the M-L inequality to get
$$\left|\int_{\Gamma_2} \frac{e^{iaz}}{z^4+1} dz\right|\le
\frac{\pi}{2} R\times
\max_{0\le t\le \pi/2}
\left|\frac{e^{ia(R\cos(t)+iR\sin(t))}}{R^4 e^{4it} +1} \right|
\\ \le \frac{\pi}{2} R\times
\max_{0\le t\le \pi/2} \frac{e^{-aR\sin(t)}}{R^4-1}
\le \frac{\pi}{2} R\times \frac{1}{R^4-1}
\to 0 \quad\text{as}\quad R\to \infty.$$
This integral vanishes in the limit and will not be contributing. Along $\Gamma_3$ we can parameterize with $z=it$ where $0\le t\le R,$ getting
$$\left|\int_{\Gamma_3} \frac{e^{iaz}}{z^4+1} dz\right|=
\left|-\int_0^R \frac{e^{-at}}{t^4+1} i\;dt\right|\le
\int_0^R \left|\frac{e^{-at}}{t^4+1}\right|dt
<\int_0^R e^{-at}dt \le C,$$
so this integral converges. This is all we need to know since it is purely imaginary and hence does not contribute to the value we are trying to compute.
Applying the Cauchy Residue Theorem to this contour, there is only one pole inside it, the one at $z=e^{\pi i/4} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2},$ and we conclude that the desired value is
$$2 \times 2\pi \times
-\Im\left(\mathrm{Res}\left(\frac{e^{iaz}}{z^4+1};z=e^{i\pi/4}\right)\right).$$
Now the residue is
$$\left.\frac{e^{iaz}}{4z^3}\right|_{z=e^{i\pi/4}} =
- \frac{1}{4} e^{i\pi/4}
\exp\left(ia \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)\right)
\\= - \frac{1}{4}
\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)
\exp\left(ia\frac{\sqrt{2}}{2} - a\frac{\sqrt{2}}{2}\right)
\\= - \frac{1}{4} \exp\left( - a\frac{\sqrt{2}}{2}\right)
\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)
\left(\cos\frac{a}{\sqrt{2}}+i\sin\frac{a}{\sqrt{2}}\right).$$
Extracting the imaginary part from this residue we get for our end result the value
$$2 \times 2\pi \times \frac{1}{4} \exp\left( - a\frac{\sqrt{2}}{2}\right)
\frac{\sqrt{2}}{2}\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right) =
\frac{\pi}{\sqrt{2}} \exp\left( - a\frac{\sqrt{2}}{2}\right)
\left(\cos\frac{a}{\sqrt{2}}+\sin\frac{a}{\sqrt{2}}\right).$$
| {
"language": "en",
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} |
Imaginary numbers and polynomials question I have a task which I do not understand:
Consider $w = \frac{z}{z^2+1}$ where $z = x + iy$, $y \not= 0$ and $z^2 + 1 \not= 0$.
Given that Im $w = 0$, show that $| z | = 1$.
Partial solution (thanks to @ABC and @aranya):
If I substitute $z$ with $x + iy$ then we have $w = \frac{x + iy}{x^2+2xyi-y^2+1}$ or written slightly different $w= \frac{x + iy}{x^2-y^2+1+2xyi}$ or $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}$.
In this format we can multiply with complex conjugate $w= \frac{x + iy}{(x^2-y^2+1)+(2xyi)}\cdot\frac{(x^2-y^2+1)-(2xyi)}{(x^2-y^2+1)-(2xyi)}$.
Then we get $w=\frac{x^3+xy^2-x^2yi-y^3i+x+iy}{(x^2+^2+1)^2-(2xyi)^2} = \frac{(x^3+xy^2+x)+(-x^2yi-y^3i+iy)}{(x^2+^2+1)^2+(2xy)^2} = \frac{(x^3+xy^2+x)+i(-x^2y-y^3+y)}{(x^2+^2+1)^2+(2xy)^2}$.
And as stated above, imaginary part of $w = 0$ meaning $\frac{-x^2y-y^3+y}{(x^2+^2+1)^2+(2xy)^2} = 0$.
As denominator can't be zero, means nominator is zero $-x^2y-y^3+y=0$. Or $y\cdot(-x^2-y^2+1)=0$. As $y\not=0$ implies $-x^2-y^2+1=0$ or $x^2+y^2=1$.
What now?
| $$w=\frac z{z^2+1}=\frac{|z|^2\overline z+z}{|z^2+1|^2}$$
and thus
$$\text{Im}\,w=0\iff \text{Im}\,z\cdot\left(|z|^2-1\right)=0\stackrel{\text{since Im}\,z=y\neq 0}\iff |z|^2=1\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite continued fraction expansion How can we find the first six partial quotients of the infinite continued fraction expansion of $\sqrt[3]2$?
I know how to do this by expanding when we have a square root function... but I"m not sure what to do with a cubic root.
| In response to André Nicolas's post, here's one way to use exact rational arithmetic. Let $x_0 = 2^{1/3}$.
First note that for rationals $a, b, c$, we can check the sign of $a x_0^2 + b x_0 + c$. The nontrivial case is where $a \ne 0$ and $b^2 - 4 a c > 0$. Let
$r_1 < r_2$ be the roots of $a x^2 + b x + c$, which in this case are real
and distinct. Then $a x_0^2 + b x_0 + c$ has the same sign as $a$ iff either
$x_0 < r_1$ or $x_0 > r_2$, i.e. $2 < r_1^3$ or $2 > r_2^3$. Express these in terms of square roots, isolate the square roots: you get inequalities of the form
$\sqrt{b^2-4ac} < R$ or $\sqrt{b^2-4ac} > R$, where $R$ is rational. If $R > 0$, square both sides and you have a rational expression to check.
Now let the continued fraction be $x_0 = [a_0; a_1, a_2, \ldots]$ and the
$k$'th remainder be $x_k = [a_k; a_{k+1}, a_{k+2}, \ldots]$. We have
$a_k = \lfloor x_k \rfloor$ (which, by the previous paragraph, can be computed
using rational arithmetic),
and $x_{k} = a_k + 1/x_{k+1}$ so
$x_{k+1} = 1/(x_k - a_k)$. Now if $x_k = c_k x_0^2 + d_k x_0 + e_k$,
$$x_{k+1} = \dfrac{1}{c_k x_0^2 + d_k x_0 + e_k - a_k} = c_{k+1} x_0^2 + d_{k+1} x_0 + e_{k+1}$$
where $c_{k+1},d_{k+1},e_{k+1}$ are rational expressions in $c_k,d_k,e_k-a_k$.
In this case
$$\eqalign{a_0 &= 1 \cr
x_1 &= 2^{2/3} + 2^{1/3} + 1\cr
a_1 &= 3\cr
x_2 &= \dfrac{3}{10} 2^{2/3} + \dfrac{2}{5} 2^{1/3} + \dfrac{1}{5}\cr
a_2 &= 1\cr
x_3 &= \dfrac{4}{3} 2^{2/3} + \dfrac{5}{3} 2^{1/3} + \dfrac{4}{3} \cr
a_3 &= 5\cr}$$
etc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/563326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
What is the summation of the following expression? What's the summation of the following expression;
$$\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}$$
The solution is said to $$2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)$$
But I'm getting $$\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right).$$
How is this possible?
$$\sum_{k=1}^{n+3}\left(2 \times\frac{1}{4}\right)^{k}\left(\frac{1}{4}\right)^n\left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k\left(\frac{1}{4}\right)^k \left(\frac{1}{4}\right)^{-k}\rightarrow \left(\frac{1}{4}\right)^n\left(2^{n+3}-1\right)$$
| Note that
$$
\left(\frac{1}{2}\right)^{k}\left(\frac{1}{4}\right)^{n-k}=\left(\frac{1}{4}\right)^n 2^k
$$
So the sum in question simplifies to
$$
\left(\frac{1}{4}\right)^n\left(\sum_{k=1}^{n+3} 2^k\right)
$$
Now by the geometric series formula
$$
\sum_{k=1}^{n+3} 2^k=2\sum_{k=0}^{n+2} 2^k=2(2^{n+3}-1)
$$
So the total sum is
$$
2\left(\frac{1}{4}\right)^n(2^{n+3}-1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/563499",
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"source": "stackexchange",
"question_score": "1",
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A square matrix A is invertible if and only if det A ≠ 0. Use the theorem above to find all values of k for which A is invertible $$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}?$$
I did a sample question before this one:
$$\begin{pmatrix} k & k & 0 \\ k^2 & 16 & k^2 \\ 0 & k & k \end{pmatrix}?$$
And was able to get k ≠ -2√2, 0, 2√2. Not sure exactly how, can anyone help me guide my way through these two questions?
| Do the same thing you did before. First, take the determinant of
$$\begin{pmatrix} k & k & 0 \\ k^2 & 25 & k^2 \\ 0 & k & k \end{pmatrix}$$
to obtain the equation $25k^2-2k^4$. So if this equation is zero, then the matrix is not invertible. Then
$$
\begin{align}
25k^2-2k^4&=0 \\
k^2(25-2k^2)&=0 \\
k^2(5+\sqrt{2}k)(5-\sqrt{2}k) &=0
\end{align}
$$
So $k=0,\frac{5}{\sqrt{2}}$ or $-\frac{5}{\sqrt{2}}$. Therefore, if $k$ is any real number except $0,\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}$ then the matrix is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566121",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Infinite sums - order of terms say we had an infinite sum $ \displaystyle \sum_{n=0}^\infty a_n $ where $ a_n = \dfrac{1}{2} +1 + \dfrac{1}{8} + \dfrac{1}{4} + \dfrac{1}{32} + ... $ this is obviously a rearrangment of $\displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} $ but does the order of the sum matter? I know by the general theorem that as $\displaystyle \sum_{n=0}^\infty \dfrac{1}{2^n} $ converges absolutely then any rearrangement also converges to the same sum - but how would I go to prove that $a_n$ converges WITHOUT using that theorem? e.g. using the root test
also, why is it for some sums - the order matters, while others it does not
| If a series involves only positive (or at least non-negative) terms, then any rearrangement of the series has the same limit in the sum. One way to prove this is to note that if $a_n \ge 0$ for all $n$, then
$$\sum\limits_{n = 0}^{\infty} a_n = \sup_{A \text{ finite}} \sum_{n \in A} a_n$$
To prove that this series converges directly, however, note that we can just compute
\begin{align*}
1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{4} + \frac{1}{32} + \frac{1}{16} + \dots &= \frac{2}{2} + \frac{1}{2} + \frac{1}{8} + \frac{2}{8} + \frac{2}{32} + \frac{1}{32} + \dots \\
&= \frac{3}{2} + \frac{3}{8} + \frac{3}{32} + \dots \\
&= \frac{3}{2} \left(1 + \frac{1}{4} + \frac{1}{16} + \dots \right) \\
&= \frac{3}{2} \frac{1}{1 - \frac{1}{4}} \\
&= 2
\end{align*}
where we've used a geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/568646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$ Regarding this problem, I conjectured that
$$ I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx = 4 \pi \operatorname{arccot} \sqrt{ \frac{2r + 2\sqrt{r^{2} - s^{2}}}{s^{2}} - 1}. $$
Though we may try the same technique as in the previous problem, now I'm curious if this generality leads us to a different (and possible a more elegant) proof.
Indeed, I observed that $I(r, 0) = 0$ and
$$\frac{\partial I}{\partial s}(r, s) = \int_{0}^{\infty} \left\{ \frac{2\sqrt{y}}{(r-s)y^{2} + 2(2-r)y + (r+s)}+\frac{2\sqrt{y}}{(r+s)y^{2}+ 2(2-r)y + (r-s)} \right\} \,\mathrm dy, $$
which can be evaluated using standard contour integration technique. But simplifying the residue and integrating them seems still daunting.
EDIT. By applying a series of change of variables, I noticed that the problem is equivalent to prove that
$$ \tilde{I}(\alpha, s) := \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{ 1 + 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}}{ 1 - 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}} \right) \, \mathrm dx = 4\pi \alpha $$
for $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ and $s > 1$. (This is equivalent to the condition that the expression inside the logarithm is positive for all $x \in \Bbb{R}$.)
Another simple observation. once you prove that $\tilde{I}(\alpha, s)$ does not depend on the variable $s$ for $s > 1$, then by suitable limiting process it follows that
$$ \tilde{I}(\alpha, s) = \int_{-\infty}^{\infty} \log \left( \frac{ 1 + 2x \sin\alpha + x^{2}}{ 1 - 2x \sin\alpha + x^{2}} \right) \, \frac{\mathrm dx}{x}, $$
which (I guess) can be calculated by hand. The following graph may also help us understand the behavior of this integral.
| So, following the procedure I outlined here, I get for the transformed integral:
$$I(r,s) = \int_0^{\infty} dv \frac{4 s \left(v^2-1\right) \left(v^4-(4 r-6) v^2+1\right)}{v^8+4 \left(2 r-s^2-1\right) v^6 +2 \left(8 r^2-8 r-4 s^2+3\right) v^4 +4 \left(2 r-s^2-1\right) v^2 +1} \log{v} $$
Note that this reduces to the integral in the original problem when $r=3$ and $s=2$. Then we see that the roots of the denominator satisfy the same symmetries as before, so we need only find one root of the form $\rho e^{i \theta}$ where
$$\rho = \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}} + \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}-1}$$
and
$$\theta = \arctan{\sqrt{\frac{2 \left (r+\sqrt{r^2-s^2}\right )}{s^2}-1}}$$
Using the same methodology I derived, I am able to confirm your conjecture.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/568807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 2,
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} |
Find all real solutions for $16^{x^2 + y} + 16^{x + y^2} = 1$ Find all $x, y \in \mathbb{R}$ such that:
$$16^{x^2 + y} + 16^{x + y^2} = 1$$
The first obvious approach was to take the log base $16$ of both sides:
$$\log_{16}(16^{x^2 + y} + 16^{x + y^2}) = 0$$
manipulating did not give any useful result. The next thing I tried was getting some bounds on $x$ and $y$:
If $x, y \geq 0$,
$$16^{x^2 + y} + 16^{x + y^2} \geq 2$$
So, $x, y \le 0$. Trying to obtain a lower bound was not fruitful.
Also, in general, I have a lot of difficulty solving such problems which require all solutions to a certain equation.
Whatever I do is almost always contrary to what the actual solution is and the solution itself involves some bizarre counter-intuitive manipulations or methods. Some tips on how to approach such problems will be helpful for me. Thanks.
| Using $$\bf{A.M\geq G.M}$$,
$\displaystyle \frac{16^{x^2+y}+16^{x+y^2}}{2}\geq \left(16^{x^2+y}\cdot 16^{x+y^2}\right)^{\frac{1}{2}}$
$\displaystyle 16^{x^2+y}+16^{x+y^2}\geq 2\cdot \left\{2^{4\left(x^2+y+x+y^2\right)}\right\}^{\frac{1}{2}} = 2\cdot \left\{2^{(2x+1)^2+(2y+1)^2-2}\right\}^{\frac{1}{2}}\geq 1$
and equality hold when $16^{x^2+y} = 16^{x+y^2}\Rightarrow x^2+y = x+y^2\Rightarrow \displaystyle x=y = -\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Strange mistakes when calculate limits I have difficulties with calculating the following limits. W|A gives the correct answers for both of them:
$$
\lim_{x \to +\infty} \sqrt{x} \cdot \left(\sqrt{x+\sqrt x} + \sqrt{x - \sqrt x} - 2\sqrt x\right) = \lim_{x \to +\infty} \sqrt{x^2+x\sqrt x} + \sqrt{x^2-x\sqrt x} - 2x = \\
= \lim_{x \to +\infty} x\left(\sqrt{1+ \frac {\sqrt x}{x}} + \sqrt{1 - \frac{\sqrt x}{x}} - 2\right) = \left[t = \frac 1x\right] = \lim_{t \to 0+} \frac{\sqrt{1+\sqrt t} - 1 +\sqrt{1-\sqrt t} - 1}{t} =
\left[c = \sqrt t\right] = \lim_{c \to 0+} \frac{\sqrt{1+c} - 1 +\sqrt{1-c} - 1}{c^2} = \frac 12c + \left(-\frac 12 c\right) = 0
$$
It's because:
$$
\lim_{c \to 0+} \frac{\sqrt{1+c}-1}{c} = \left|k = \sqrt{1+c}\right| = \lim_{k \to 1+} \frac{k-1}{k^2-1} = \lim_{k \to 1+} \frac{1}{k+1} = \frac 12
$$
The second part is similar to it. But the answer is -$\frac 14$.
And the second limit is:
$$
\lim_{x \to 0} \left(\frac{e^x - \sin x}{\sqrt{1 - 2x} + \log(x+1)}\right)^\cfrac{1}{x^2} = \exp\left(\lim_{x \to 0} \frac{1}{x^2}\left(\frac{e^x-\sin x}{\sqrt{1-2x} + \log(x+1)}-1\right)\right) =
\exp\left(\lim_{x \to 0}\frac{e^x - \sin x - \sqrt{1-2x} - \log(x+1)}{x^2} \cdot \frac{1}{\sqrt{1-2x}+\log(x+1)}\right) = \exp\left(\lim_{x \to 0} \left(\frac{e^x-1}{x \cdot x} - \frac{\sin x}{x \cdot x} - \frac{\sqrt{1-2x}-1}{x \cdot x} - \frac{\log(x+1)}{x \cdot x}\right)\right) = \exp\left(\lim_{x \to 0} \left(\frac 1x - \frac 1x + \frac 1x - \frac 1x\right)\right) = e^0 = 1
$$
The answer is $e^\frac 32$. Where am I wrong?
| In the first line of the first one:
$\sqrt{x}\sqrt{x + \sqrt{x}} = \sqrt{x^2 + x\sqrt{x}}$, not $\sqrt{x^2 + x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determining if a vector is in the column space of a matrix Hi there I'm having some trouble with the following problem:
I have a $3\times3$ symmetric matrix
$$
A=\pmatrix{1+t&1&1\\ 1&1+t&1\\ 1&1&1+t}.
$$
I am trying to determine the values of $t$ for which the vector $b = (1,t,t^2)^\top$ (this is a column vector) is in the column space of $A$.
I think I'm fairly aware of how to go about it, forming the augmented matrix $[A|b]$ and basically using row ops to find a solution with which I could solve for the value(s) of $t$. But I've been trying this and have no luck. May I be missing something?
Thank you
| Gaussian elimination is not difficult in this case:
\begin{align}
\left[\begin{array}{ccc|c}
1+t & 1 & 1 & 1 \\
1 & 1+t & 1 & t \\
1 & 1 & 1+t & t^2
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
1 & 1+t & 1 & t \\
1+t & 1 & 1 & 1
\end{array}\right]
\\
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2(1+t)
\end{array}\right]
\end{align}
If $t\ne0$ we can go on:
\begin{align}
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & t & -t & t-t^2 \\
0 & -t & -2t-t^2 & 1-t^2-t^3
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & 1 & 1+t & t^2 \\
0 & 1 & -1 & 1-t \\
0 & 0 & -3t-t^2 & 1+t-2t^2-t^3
\end{array}\right]
\end{align}
If $t=-3$, the last row becomes
$$
\begin{array}{ccc|c}
0 & 0 & 0 & 7
\end{array}
$$
so the system has no solution.
If $t\ne-3$, the system has a solution.
If $t=0$, we get, from the place we stopped at,
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
$$
and the system has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Calculate the limit of the function Please help me to calculate the limit of the function. I do not know where to start.
Thank you.
| Let
$$
f(x)= \left[\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2-\frac{\arctan(x^2)}{1-x^2}-2x^3+1\right]
$$
and $$g(x)=-\frac{1}{\left(1-\operatorname{e}^{2x^2}\right)\sinh(3x^2)}.$$
We have to find the limit for $x\to0$ of the function $\varphi(x)=[f(x)]^{-\frac{1}{g(x)}}$
$$
\lim_{x\to 0}\varphi(x)=\lim_{x\to 0}[f(x)]^{g(x)}
$$
For $x\to 0$
for the first term in the bracket we have
$$
\begin{align}
\sin(x+x^2)&\sim x+x^2-\frac{(x+x^2)^3}{6}\sim x+x^2-\frac{x^3}{6}-\frac{x^4}{2}+o(x^4)\\
%
\frac{1}{1-x^2}&\sim 1+x^2+x^4+o(x^4)\\
%
\frac{\sin(x+x^2)}{1-x^2}&\sim \left(x+x^2-\frac{x^3}{6}-\frac{x^4}{2}\right)\left( 1+x^2+x^4\right)\sim x+x^2+\frac{5}{6}x^3+\frac{x^4}{2}+o(x^4)\\
\left(\frac{\sin(x+x^2)}{1-x^2}\right)^2&\sim \left(x+x^2+\frac{5}{6}x^3+\frac{x^4}{2}\right)\sim x+2x^2+\frac{8}{3}x^4+o(x^4)
\end{align}$$
For the second term
$$
\begin{align}
\arctan(x^2)&\sim x^2+x^4+o(x^4)\\
\frac{\arctan(x^2)}{1-x^2}&\sim \left(x^2+x^4\right)\left( 1+x^2+x^4\right)\sim x^2+2x^4+o(x^4)\\
\end{align}
$$
So the Taylor expansion for $f(x)$ at the order 4 is
$$
f(x)\sim x+2x^2+\frac{8}{3}x^4-(x^2+2x^4)-2x^3+1\sim 1+\frac{2}{3}x^4+o(x^4)
$$
Now let find the Taylor expansion for $g(x)$. Observing that
$$
\begin{align}
1-\operatorname{e}^{2x^2}&\sim 1-\left(1-2x^2-\frac{(2x)^2}{2}\right)=-2x^2+2x^4+o(x^4)\\
\sinh(3x^2)&\sim 3x^2+\frac{(3x^2)^3}{6}\sim 3x^2+o(x^4)
\end{align}
$$
we find
$$
g(x)\sim-\frac{1}{6x^4+o(x^4)}
$$
Putting all together we have
$$
\varphi(x)
\sim\left[1+\frac{2}{3}x^4\right]^{-1/(6x^4)}=\exp\left({-\frac{1}{6x^4}}\log\left(1+\frac{2}{3}x^4\right)\right)
$$
and observing that
$$
\log\left(1+\frac{2}{3}x^4\right)\sim \frac{2}{3}x^4+o(x^4)
$$
we finally find for $x\to 0$
$$
\varphi(x)\sim\exp\left(-\frac{1}{6x^4}\frac{2}{3}x^4\right)\to \operatorname{e}^{-1/9}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Triangle ratio of areas This is a photo that was originally posted on Google Plus. I would like to know how to solve for S. I started by splitting S into two parts S1 and S2 by drawing a line from A to M.
I also know that I should use the fact that since the triangles share the same base, the ratio of areas = ratio of heights.
What I have so far:
5/15 =
8/18 =
Not sure what to equate these two equations to.
| Let $x$ be the area of the triangle $AML$, and $y$ of $AMK$. Note that as $MBC$ and $LBC$ have the same bases, their heights must be in the ratio of $2/3$ and so must be the segments
$$\frac{MC}{LC}=\frac{2}{3}\Rightarrow\frac{LM}{LC}=\frac{1}{3}.$$
By a similar argument, we get
$$\frac{MB}{KB}=\frac{10}{18}=\frac{5}{9}\Rightarrow\frac{KM}{KB}=\frac{4}{9}.$$
Now thinking in the segment $AB$ as the base and about the triangles $AMB$ and $ABC$, we have
$$\frac{x+5}{23+x+y}=\frac{ML}{CL}=\frac{1}{3}.$$
Also, thinking of $AC$ as the base, we have
$$\frac{8+y}{23+x+y}=\frac{MK}{BK}=\frac{4}{9}.$$
Solving for $x$ and $y$, yields $x=10$ and $y=12$. So $S=x+y=22$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/573933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Double Integral $\int\limits_0^b\int\limits_0^x\sqrt{a^2-x^2-y^2}\,dy\,dx$ What is the best method for evaluating the following double integral?
$$
\int_{0}^{b}\int_{0}^{x}\,\sqrt{\,a^{2} - x^{2} - y^{2}\,}\,\,{\rm d}y\,{\rm d}x\,,
\qquad a > \sqrt{\,2\,}\,\,b
$$
Is there exist an easy method?
My try:
$$\int_0^b\int_0^x\sqrt{a^2-x^2-y^2}dy\,dx=\int_0^{\frac{\pi}{4}}\int_0^{b\sec(\theta)}r\sqrt{a^2-r^2}dr\,d\theta$$
$$=\int_0^{\frac{\pi}{4}}\frac{-1}{3}\left[(a^2-r^2)\sqrt{a^2-r^2}\right]_0^{b\sec(\theta)}d\theta$$
$$=\frac{1}{3}\int_0^{\frac{\pi}{4}}\left[a^3-(a^2-b^2\sec^2(\theta))\sqrt{a^2-b^2\sec^2(\theta)}\right]d\theta$$
but evaluating above integral is very difficult and antiderivative is very complexity! see here.
| In a sense "Polar coordinates does not help much since the integrand likes them while the domain does not".
It helps to know about partial integration, that is
$$\int f'(x)g(x)dx = f(t)g(t)- \int f(x)g'(x)dx +C $$
In addition one should know about the derivatives of $\arcsin$ and $\arctan$.
Now to the integral, first we simplify
\begin{eqnarray}\int_0^x\sqrt{a^2-x^2-y^2}dy &=& \int_0^x\sqrt{a^2-x^2}\sqrt{1-\left(\frac{y}{\sqrt{a^2-x^2}}\right)^2}dy \\ &=&\int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy\tag{1}\end{eqnarray}
Next we use partial integration
\begin{eqnarray}I=\int_0^\eta\sqrt{1-t^2}dt &=& \left[t\sqrt{1-t^2}\right]_0^\eta-\int_0^\eta t\cdot\left(\frac{-t}{\sqrt{1-t^2}}\right)dt\\&=& \eta\sqrt{1-\eta^2}+\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt
\end{eqnarray}
and then $$\int_0^\eta \frac{t^2}{\sqrt{1-t^2}}dt = \int_0^\eta \frac{-(1-t^2)}{\sqrt{1-t^2}}dt+\int_0^\eta \frac{1}{\sqrt{1-t^2}}dt=-I +\arcsin{\eta}$$
Thus
$$I=\frac12\left(\eta\sqrt{1-\eta^2} +\arcsin{\eta} \right)\tag{2}$$
which we use in (1) to get
\begin{eqnarray}
\int_0^{x/\sqrt{a^2-x^2}}(a^2-x^2)\sqrt{1-t^2}dy
&=& \frac{(a^2-x^2)}2\Big(\frac{x}{\sqrt{a^2-x^2}}\sqrt{1-\frac{x^2}{a^2-x^2}} \\
&&+\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \Big)\\
&=&\frac{x}{2}\sqrt{a^2-2x^2}
+\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}} \end{eqnarray}
It follows that
$$\int_0^b \int_0^x\sqrt{a^2-x^2-y^2}dy\,dx= \int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx+ \int_0^b
\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx$$
and since
$$\int_0^b\frac{x}{2}\sqrt{a^2-2x^2}dx= \frac{1}{12}(a^2-2b^2)^{3/2}$$
it suffice to calculate
\begin{eqnarray}\int_0^b
\frac{(a^2-x^2)}2\arcsin{\frac{x}{\sqrt{a^2-x^2}}}dx&=&\frac1{2a^2}\int_0^b
\frac{(1-(x/a)^2)}2\arcsin{\frac{(x/a)}{\sqrt{1-(x/a)^2}}}dx\\
&=&\frac1{2a}\int_0^{b/a}
(1-t^2)\arcsin{\frac{t}{\sqrt{1-t^2}}}dt\end{eqnarray}
which can be calculated in the same fashion as we did with $I$ above - I leave this fun part to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
Squaring both sides leads to a complete different result I came across with this equation:
$v^2 \cos (2 \alpha) \sqrt{v^2 \sin^2 \alpha+2gh}+v \sin \alpha [v^2 \cos (2 \alpha)-2gh]=0$
or
$2 g h \sin \alpha - v^2 \cos(2\alpha) \sin \alpha = v \cos(2 \alpha) \sqrt{v^2 \sin^2 \alpha + 2gh}$
Now this equation before squaring gives $$\alpha = \arccos \left(\sqrt{\frac{2 g h + v^2}{2 g h + 2v^2}} \right)$$
But if we square its sides:
$4 g^2 h^2 \sin^2 \alpha-4ghv^2\sin^2 \alpha \cos(2\alpha)+v^4 \cos^2(2\alpha) \sin^2 \alpha = v^2 [\cos^2(2\alpha)](v^2 \sin^2 \alpha + 2gh)$
The solution would be:
$$\alpha=\arccos \left(\frac{gh}{2v^2+2gh} \right)$$
Which is a whole different result. I'm wondering why squaring ruins the result? I do know that this operation may introduce extra solutions, but here we have a complete new one.
| We have $$v^2\cos2\alpha(\sqrt{v^2 \sin^2 \alpha+2gh}+v\sin\alpha)=2gh v\sin\alpha$$
Rationalizing the numerator & cancelling $v\ne0$
$$v\cos2\alpha\frac{2gh}{\sqrt{v^2 \sin^2 \alpha+2gh}-v\sin\alpha}=2gh \sin\alpha$$
Cancelling $2gh\ne0$ & using $\cos2\alpha=\cos^2\alpha-\sin^2\alpha$
$$v(\cos^2\alpha-\sin^2\alpha)=\sin\alpha(\sqrt{v^2 \sin^2 \alpha+2gh}-v\sin\alpha)$$
$$\implies v\cos^2\alpha=\sin\alpha\sqrt{v^2 \sin^2 \alpha+2gh}$$
Squaring we get $$ v^2(1-\sin^2\alpha)^2=\sin^2\alpha(v^2 \sin^2 \alpha+2gh)$$
Cancel $\displaystyle v^2\sin^4\alpha$ and use $\displaystyle \cos^2\alpha=1-\sin^2\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$ find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
| $$y^9+3y^6+3y^3-16y+9=0\iff(y-1)(y^2 + y - 1)(y^6 + 2y^4 + 2y^3 + 4y^2 + 2y +9) = 0$$
Three real-valued roots: $$y = 1, y = \dfrac{-1 \pm \sqrt 5}{2}$$
I noticed by inspection that $y = 1$ solves the equation. Then, using polynomial division, and manipulation with the resulting quotient, was able to find the quadratic factor. The sixth degree polynomial has no real roots.
Additional tip: Sometimes it helps to graph the equation to learn how many real roots there are: look for points intersecting the x-axis. Here's the graph of your 9th-degree polynomial:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Fixed points of group action Let $G=SL_2(Z)$ work on $\{z \in C | Im(z) \gt 0\}$ by $\begin{pmatrix}a&b \\ c&d\end{pmatrix}(z)=\frac{az+b}{cz+d}$.
First of all I want to find which elements of G leave $\zeta_3$ fixed.
Writing it out gives me $c\zeta_3^2+(d-a)\zeta_3-b=0$. If I now use $\zeta_3=e^{\frac{2\pi i}{3}}$ i find $a+c-d=0, -a+2b+c+d=0$ and together with $ad-bc=1$ (det=1) and the fact that a,b,c,d are from Z I can calculate the solutions with a computer. However there must be a much more elegant solution to this problem. Help would be much appreciated
Next I want to show that the action has no fixed points if |a+d|>2. Thus $cz^2+(d-a)z-b=0$ has no solutions (with Im(z)>0) when |a+d|>2
Solutions are $z=\frac{a-d}{2c}+\frac{\sqrt{(d-a)^2+4bc}}{2c}$, which has imaginary part $\sqrt{|(d-a)^2+4bc|}/2c$. But nowhere I can see the |a+d| coming in, so I'm stuck here?
Thanks in advance
| You don't need a computer. From $a+c-d=0$, you can substitute $d = a+c$ into $-a+2b+c+d = 0$ to get $2(b+c) = 0$. Thus you are at
$$\begin{pmatrix}a&b\\-b&a-b \end{pmatrix}.$$
Now the determinant gives you $a(a-b) + b^2 = 1$, or $(2a-b)^2 + 3b^2 = 4$, so you have
*
*$b = 0$, and $a = \pm 1$, which gives the identity, or
*$b = 1$, and $a \in \{0,1\}$, or
*$b = -1$, and $a\in \{0,-1\}$ which gives the same maps as the previous.
So, up to multiplication by $-1$:
$$\begin{pmatrix} 1&0\\0&1\end{pmatrix}: z\mapsto z;\quad \begin{pmatrix} 0 & -1\\1&1\end{pmatrix}: z \mapsto -\frac{1}{z+1};\quad \begin{pmatrix}-1&-1\\1&0 \end{pmatrix}: z \mapsto -\left(1+\frac1z\right).$$
We have the fixed points
$$\frac{a-d \pm \sqrt{(a-d)^2+4bc}}{2c} = \frac{a-d \pm\sqrt{a^2+d^2 -4(ad-bc)}}{2c} = \frac{a-d\pm\sqrt{a^2+d^2-4}}{2c},$$
and for $c \neq 0$, they are real if $a^2+d^2 \geqslant 4$, which is the case unless $\lvert a\rvert \leqslant 1$ and $\lvert d\rvert \leqslant 1$.
| {
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For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \leq b \leq c$ because of the symmetry.
$a \leq b \leq c$ implies that $0 \leq b-a \leq c-a$. Since both sides are positive and $y=x^2$ is an increasing function for positive numbers we conclude that $(b-a)^2 \leq (c-a)^2$ and since $(a-b)^2=(b-a)^2$ we obtain $(a-b)^2 \leq (c-a)^2$.
Therefore:
$\min\{(a-b)^2,(b-c)^2,(c-a)^2\}=$ $$\min\{(b-c)^2,\min\{(a-b)^2,(c-a)^2\}\}=
\min\{(a-b)^2,(b-c)^2\}.$$
So, we have to prove the following inequality instead:
$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
Now two cases can happen:
*
*$$(a-b)^2 \leq (b-c)^2 \implies |a-b| \leq |b-c|=c-b \implies b-c \leq a-b \leq c-b \implies 2b-c \leq a \implies b \leq \frac{a+c}{2}.$$
*$$(b-c)^2 \leq (a-b)^2 \implies |c-b| \leq |a-b|=b-a \implies a-b \leq c-b \leq b-a \implies c \leq 2b-a \implies b \geq \frac{a+c}{2}.$$
So, this all boils down to whether $b$ is greater than the arithmetic mean of $a$ and $b$ are not.
Now I'm stuck. If $a,b,c$ had been assumed to be positive real numbers it would've been a lot easier to go forward from this step. But since we have made no assumptions on the signs of $a,b$ and $c$ I have no idea what I should do next.
Maybe I shouldn't care about what $\displaystyle \min\{(a-b)^2,(b-c)^2\}$ is equal to and I should continue my argument by dealing with $(a-b)^2$ and $(b-c)^2$ instead. I doubt that using the formula $\displaystyle \min\{x,y\}=\frac{x+y - |x-y|}{2}$ would simplify this any further. Any ideas on how to go further are appreciated.
| If we assume that $a\leqslant b\leqslant c$, then it suffices to show$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}.$$
*
*If $c-b\geqslant b-a$, denote $c-b=y,b-a=x$, then $y\geqslant x\geqslant 0$, we just ned to show$$x^2\leqslant \frac{a^2+(a+x)^2+(a+x+y)^2}{2}\iff 3a^2+(4x+2y)a+(y^2+2xy)\geqslant 0.$$Notice that $$\Delta=(4x+2y)^2-4\times 3\times(y^2+2xy)=16x^2-8y-8y^2\leqslant 0,$$since$y\geqslant x\geqslant 0$,so $3a^2+(4x+2y)a+(y^2+2xy)\geqslant 0$ for all $a\in\mathbb{R}$.
*The same method can be applied to solve the condition that $c-b\leqslant b-a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying identities, trigonometry I am completely stuck. I cannot come up with a proof for this identity. I am ready to pull my hair out. $$\sin(x) \left(\tan(x)+ \frac 1{\tan(x)}\right) = \sec(x)$$
| Hint: Remember $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $1 + \tan^2(x) = \sec^2(x)$.
Edit to reflect full answer:
Because we know that both $\tan(x)$ and $\sec (x)$ can be written in terms of $\sin(x)$ and $\cos(x)$, we will rewrite them. Remembering:
$$\tan(x) = \frac{\sin(x)}{\cos(x)} \ \ \mathrm{and } \ \ 1 + \tan^2(x) = \sec^2(x)\ \ .$$
We have our original formula rewritten as :
\begin{align}
\sin(x) \left(\tan(x)+ \frac 1{\tan(x)}\right) &=\sin(x) \left(\tan(x)+ \tan^{-1}(x)\right) \\
&=\sin(x) \left(\frac{\sin(x)}{\cos(x)}+ \left(\frac{\sin(x)}{\cos(x)}\right)^{-1}\right)\\
&= \sin(x) \left(\frac{\sin(x)}{\cos(x)}+ \frac{\cos(x)}{\sin(x)}\right)\\
&=\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right)\ .
\end{align}
From here there are two techniques we can use to prove this identity.
Technique 1:
In this technique we try to put our identity into the form $1 + \tan^2(x) = \sec^2(x)$, then get to $\sec(x)$
We can see that the only thing necessary for the $\frac{\sin^2(x)}{\cos(x)}$ term to be equal to $\tan^2(x)$ it to divide by $\cos(x)$, and then for the $\cos(x)$ term to equal $1$ we also need to divide by $\cos(x)$. This means we need to multiply our whole expression by $\frac{1}{\cos(x)}$, but because we can not change our expression, we must multiply by one $1$ (because one is the identity for multiplication), therefore we must multiply by $\frac{\cos(x)}{\cos(x)}$.
Now we preform the multiplication:
$$\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{\cos(x)}{\cos(x)} = \left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{1}{\cos(x)} \cdot \cos(x) $$
We can distribute the $ \frac{1}{\cos(x)}$ to get:
\begin{align}
\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{1}{\cos(x)} \cdot \cos(x) &= \left(\frac{\sin^2(x)}{\cos^2(x)}+ \frac{\cos(x)}{\cos(x)}\right) \cdot \cos(x)\\
&= \left(\tan^2(x)+ 1\right) \cdot \cos(x) \ \ .
\end{align}
Now we can use the identity $1 + \tan^2(x) = \sec^2(x)$:
$$\left(\tan^2(x)+ 1\right) \cdot \cos(x) = \sec^2(x) \cdot \cos(x)$$
and because $ \sec(x) = \frac{1}{\cos(x)}$:
\begin{align}
\sec^2(x) \cdot \cos(x) &= \left(\frac{1}{\cos(x)} \right)^2 \cdot \cos(x)\\
&= \frac{1}{\cos^2(x)} \cdot \cos(x)\\
&= \frac{1}{\cos(x)}\\
&= \sec(x)
\end{align}
As desired.
Technique 2:
Here again we start with the expression $\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right)$. Now $\sec(x)$ is a single term, and our expression has two terms, so we will try to condense our expression down to one term by putting the fractions under common denominators:
\begin{align}
\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) &= \left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\cdot \frac{cos(x)}{\cos(x)} \right)\\
&= \left(\frac{\sin^2(x)}{\cos(x)}+ \frac{cos^2(x)}{\cos(x)} \right)\\
&= \left(\frac{\sin^2(x) + cos^2(x)}{\cos(x)} \right)\ .
\end{align}
We now can use the Pythagorean identity ($\sin^2(x) + \cos^2(x) = 1$):
\begin{align}
\left(\frac{\sin^2(x) + cos^2(x)}{\cos(x)} \right) &= \frac{1}{\cos(x)} \\
&= \sec(x)
\end{align}
as desired.
Remarks:
While both of the techniques are valid proofs, the second technique is (in my personal opinion) more desirable because it is both shorter and it is more intuitive. While both proofs use the Pythagorean identity (the first uses an indirect Pythagorean identity, attained by dividing the Pythagorean identity by $\cos^2(x)$), the second uses it without needing to do a large amount of manipulation to attain the applicable form. If I were presented with this problem on an exam I would most likely use the second technique (although the first came to mind first, as reflected in my hint), but like I said, either is valid.
| {
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An Urn with 60 marbles A urn contains 60 marbles 10 red 10 white 10 blue 10 green 10 brown and 10 orange. I draw from the urn 6 times. What is the probability I have exactly 5 varieties?
Here's what I've got,
There are $10^5$ ways I can draw 5 different marbles. So the number of ways I am unable to draw the 6th variety should be $9 \cdot 5$ ways? Thus $\dfrac{10^5 \cdot 45}{C(60,10)}$ is the probability of having exactly 5 varietys?
| Think of the marbles as distinct (write an ID number on each). We assume the drawing is done without replacement.
There are $\binom{60}{6}$ equally likely ways to choose $6$ marbles.
Now we count the "favourables." If there are $5$ types, we must have $2$ of one type, and $1$ each of $4$ other types.
There are $\binom{6}{1}$ ways to choose the colour we will have $2$ in. For each of these ways, there are $\binom{10}{2}$ ways of choosing the $2$ marbles. For each of these ways, there are $\binom{5}{4}$ ways to choose the other $4$ colours that will be represented. And for each such choice, there are $10^4$ ways to choose the actual marbles, for a probability of
$$\frac{\binom{6}{1}\binom{10}{2}\binom{5}{4}10^4}{\binom{60}{6}}.$$
| {
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Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| Solution from Thomas Mildorf notes
first we homogenize inequality with multipliying $(abc)^\frac{1}{3}$ and we get the following inequality $$a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{4}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{1}{3}c^\frac{4}{3}\leq a^2+b^2+c^2$$ then we can apply $A.O\geq H.O$ $$\frac{a^2+a^2+a^2+a^2+b^2+c^2}{6}\geq a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}$$ and we can get in the same manner other term and after adding them we are done
| {
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Proving a non-linear diophantine equation has no solutions How can I prove that
$x^2 + y^2 + z^2 + 3(x+ y + z) + 5 = 0$
Has no integer solutions?
| Assume that $(x,y,z) \in \mathbb{Z}^3$ satisfies $x^2+y^2+z^2 + 3(x+y+z) + 5 = 0$. For simplicity, let $f(X,Y,Z) = X^2+Y^2+Z^2 + 3(X+Y+Z) + 5$. Then $f(x,y,z) \equiv 0 \pmod{2}$. Notice that $a^2 \equiv a \pmod{2}$ implies $x^2+y^2+z^2\equiv 3(x+y+z)\pmod{2}$, and so $f(x,y,z) \equiv 5 \equiv 1 \not\equiv 0 \pmod{2}$. Contradiction.
| {
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Integration with substitution I want to integrate (2x+1)/((x^2 - 6x + 14)^3)
I'm guessing you use substitution but im unsure what to substitute, is it best to make u = x^2 or X^2-6x or even x^2 - 6x + 14
I find that it makes turning the top of the fraction into terms of u very difficult
| $$\begin{align} \int \frac{2x + 1} {(x^2 - 6x + 14)^3} & = \int \frac{2x - 6 + 6 + 1}{(x^2 - 6x + 14)^3}\,dx \\ \\ & = \int\dfrac{2x - 6}{(x^2 - 6x + 14)^3}\,dx + \int \frac{7}{(x^2 - 6x + 14)^3}\,dx \\ \\ &= \int\dfrac{\overbrace{2x - 6}^{du}}{(\underbrace{x^2 - 6x + 14}_{u})^3}\,dx + \int \frac{7}{(\underbrace{x^2 - 6x + 9}_{(x - 3)^2} + \underbrace{5}_{(\sqrt 5)^2})^3}\,dx \end{align}$$
The left-most integral: using $u = x^2 - 6x + 14 \implies du = 2x - 6, \,dx$
so we integrate $\displaystyle \int \frac {du}{u^3}$.
The right-most integral, we can use the substitution $x - 3 = \sqrt 5\tan \theta \implies dx = \sqrt 5 \sec^2 \theta\,d\theta$, to obtain the integral $$\displaystyle 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \tan^2 \theta + 5)^3}= 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 (\tan^2 \theta +1))^3} = 7\int \frac{\sqrt 5 \sec^2 \theta \,d\theta}{(5 \sec^2\theta)^3} = \cdots$$
| {
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Integrate the following integral using partial fractions I want to integrate
$$\frac{1}{(1-u^2)^2}. $$
I have used the difference of two squares to get
$$\frac{1}{2(1-u^2)} + \frac{1}{2(1+u^2)} $$
and then integrated it to get
$$\frac{1}{2ln|1-u^2|} + \frac{1}{2ln|1+u^2|}.$$
Just wondering if this is correct? thanks
| If you want to use partial fraction decomposition, then note that: $$\dfrac 1{(1-u^2)^2} = \dfrac 1{[(u-1)(u+1)]^2} = \dfrac 1{(u-1)^2(u+1)^2} $$
$$= \dfrac{A}{(u-1)} + \dfrac{B}{(u -1)^2} + \dfrac{C}{(u+1)} + \dfrac D{(u+1)^2}$$
Now try solving for the needed constant terms: $A,\, B, \,C,\, D$.
Spoiler I (to check your solutions to $A, B, C, D$)
$$A = -\dfrac 14, \;B = C = D = \dfrac 14.\quad$$
Spoiler II
$$\int \dfrac {du}{(1-u^2)^2} = \dfrac 14 \left(\int \dfrac{-du}{(u-1)} + \int \dfrac{du}{(u -1)^2} + \int \dfrac{du}{(u+1)} + \dfrac {du}{(u+1)^2}\right)\\ \\ = \dfrac 14 \left(-\ln|u - 1| - \dfrac{1}{u - 1} + \ln|u+1| - \dfrac{1}{u+1}\right) + C \\ \\ = \dfrac 14\left( \ln \left|\dfrac{u+1}{u-1}\right| - \dfrac{2u}{u^2 - 1}\right) + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble with factoring polonomial to the 3rd degree I am having trouble factoring this problem:
$\displaystyle{-x^{3} + 6x^{2} - 11x + 6}$
I know the answer but i can't figure out how it is done with this.
I have tried by grouping and is doesn't seem to work. Can someone show me how to do this.
| $$\begin{align}
& -x^3+6x^2-11x+6 \\
=& -x^3+1+6x^2-11x+5 \\
=& (1-x^3)+6x^2-6x-5x+5 \\
=& (1-x)(1+x+x^2)-6x(1-x)+5(1-x) \\
=& (1-x)(1+x+x^2-6x+5) \\
=& (1-x)(x^2-5x+6) \\
=& (1-x)(x^2-2x-3x+6) \\
=& (1-x)(x(x-2)-3(x-2)) \\
=& (1-x)(x-2)(x-3)
\end{align}$$
| {
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How find this invertible matrix $C=\left[\begin{smallmatrix} A&B\\ B^T&0 \end{smallmatrix}\right]$ let matrix $A_{n\times n}$,and $\det(A)>0$, and the matrix $B_{n\times m}$,and such $rank(B)=m$,and let
$$C=\begin{bmatrix}
A&B\\
B^T&0
\end{bmatrix}$$
Find this Invertible matrix
$C^{-1}$
my try: I found this matrix Invertible matrix $C$,it must find $B^TAB$ Invertible matrix.But I can't
Thank you for your help
| Let
$ C^{-1} = \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} $ with appropriate sizes (i.e. $X$ is $n \times n$, $Y$ is $n \times m$, $Z$ is $m \times n$ and $W$ is $m \times m$). Then,
$C C^{-1} = \begin{bmatrix} A & B \\ B^T & 0 \end{bmatrix} \begin{bmatrix} X & Y \\ Z & W \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 & I_m \end{bmatrix}$
Hence, we have the following equations:
$\begin{align}
AX + BZ &= I_n \\
AY + BW &= 0 \\
B^T X &= 0 \\
B^T Y &= I_m
\end{align}$
From the first and third equations:
$\begin{align}
X + A^{-1}BZ &= A^{-1}\\
B^T X + B^T A^{-1}BZ &= B^T A^{-1}\\
Z &= (B^T A^{-1}B)^{-1}B^T A^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ X = A^{-1} - A^{-1}B (B^T A^{-1}B)^{-1}B^T A^{-1} $
From the second and forth equations:
$\begin{align}
Y + A^{-1}BW &= 0\\
B^T Y + B^T A^{-1}BW &= 0\\
B^T A^{-1}BW &= -I_m\\
W &= -(B^T A^{-1}B)^{-1}
\end{align}$
Now, if we put this in the second equation we get
$ Y = A^{-1}B(B^T A^{-1}B)^{-1} $
which is consistent with the forth equation.
Note that $A$ and $B^T A^{-1}B$ must be invertible.
| {
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How to integrate the bump functions,i.e,$\int_a^{b}e^{-\frac{1}{x-a}+\frac{1}{x-b}}dx$,where $aSince $$\lim_{x\to{a}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=\lim_{x\to{b}}e^{-\frac{1}{x-a}+\frac{1}{x-b}}=0,$$ $e^{-\frac{1}{x-a}+\frac{1}{x-b}}$ is continuous on the interval $[a,b]$ (taking $0$ if $x=a$ or $b$).
So the integral $\int_a^{b}e^{-\frac{1}{x-a}+\frac{1}{x-b}}dx$ makes sense.But i do not know how to compute this integral?
In particularly, taking $a=0$ and $b=1$,we just need to compute $\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}dx$.
My thought:Considering another integration $$I(\epsilon)=\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}e^{-\epsilon{(\frac{1}{x^2}-\frac{1}{(x-1)^2}})}dx,$$
take the derivative of this with respect to $\epsilon$(assuming it converges uniformly),we get $$I'(\epsilon)=-\int_0^{1}e^{-\frac{1}{x}+\frac{1}{x-1}}e^{-\epsilon{(\frac{1}{x^2}-\frac{1}{(x-1)^2}})}d_{-\frac{1}{x}+\frac{1}{x-1}}.$$
I can not continue,and i do not know whether it work.Can you provide me some methods?
| $$\int_a^be^{-\frac{1}{x-a}+\frac{1}{x-b}}~dx$$
$$=\int_{a-\frac{a+b}{2}}^{b-\frac{a+b}{2}}e^{-\frac{1}{x+\frac{a+b}{2}-a}+\frac{1}{x+\frac{a+b}{2}-b}}~d\left(x+\dfrac{a+b}{2}\right)$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^{-\frac{1}{x+\frac{b-a}{2}}+\frac{1}{x-\frac{b-a}{2}}}~dx$$
$$=\int_{-\frac{b-a}{2}}^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_{-\frac{b-a}{2}}^0e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_\frac{b-a}{2}^0e^\frac{b-a}{(-x)^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx+\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=2\int_0^\frac{b-a}{2}e^\frac{b-a}{x^2-\frac{(b-a)^2}{4}}~dx$$
$$=2\int_0^\infty e^\frac{b-a}{\left(\frac{b-a}{2}\tanh x\right)^2-\frac{(b-a)^2}{4}}~d\left(\dfrac{b-a}{2}\tanh x\right)$$
$$=(b-a)\int_0^\infty e^{-\frac{b-a}{\frac{(b-a)^2}{4}\text{sech}^2x}}~d(\tanh x)$$
$$=(b-a)\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}~d(\tanh x)$$
$$=(b-a)\left[e^{-\frac{4\cosh^2x}{b-a}}\tanh x\right]_0^\infty-(b-a)\int_0^\infty\tanh x~d\left(e^{-\frac{4\cosh^2x}{b-a}}\right)$$
$$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh x\cosh x\tanh x~dx$$
$$=8\int_0^\infty e^{-\frac{4\cosh^2x}{b-a}}\sinh^2x~dx$$
$$=8\int_0^\infty e^{-\frac{2(\cosh2x+1)}{b-a}}\dfrac{\cosh2x-1}{2}dx$$
$$=4e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~dx$$
$$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh2x}{b-a}}(\cosh2x-1)~d(2x)$$
$$=2e^{-\frac{2}{b-a}}\int_0^\infty e^{-\frac{2\cosh x}{b-a}}(\cosh x-1)dx$$
$$=2e^{-\frac{2}{b-a}}\left(K_1\left(\dfrac{2}{b-a}\right)-K_0\left(\dfrac{2}{b-a}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/585768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Fibonacci sequence proof Prove the following:
$$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$
For $n \ge 2$
Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$
So, here are my first steps:
$$ \begin{align} & \frac 12(f_{3k+2}-1) + f_{3k+3} = \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} + f_{3k+1} \\
& = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) + f_{3k+1} \\
& = f_{3k+2}-1 + 2 \cdot f_{3k+1}
\end{align} $$
And the fun stops around here. I don't see how to get to the conclusion: $\frac 12(f_{3k+5}-1) \\ $. Any help from this point would be great.
| define
$$B_n = \sum_{k=1}^{3n} f_k$$
LEMMA $$B_n = f_{3n+2} - 1$$
PROOF by induction.
(A) for $n=1$ we have $f_1+f_2+f_3 = 1+1+2=4 = 5-1=f_5-1$
(B) suppose true for $n$, then:
$$B_{n+1} = B_n +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= f_{3n+2} - 1 +f_{3n+1} +f_{3n+2}+f_{3n+3} \\
= (f_{3n+1} +f_{3n+2}) +(f_{3n+2}+f_{3n+3})-1\\
= f_{3n+3} +f_{3n+4} -1\\
= f_{3n+5} -1\\
= f_{3(n+1)+2} -1 \\
$$
qed
now define $S_n = \sum_{k=1}^n f_{3k}$
clearly $B_n=2S_n$ since $f_1+f_2+f_3 = 2f_3$ etc.
and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do you calculate this sum $\sum_{n=1}^\infty nx^n$? I can not find the function from which I have to start to calculate this power series.
$$\sum_{n=1}^\infty nx^n$$
Any tips?.
Thanks.
| We can do it another way.
$S = x + 2x^2 + 3x^3 + \ldots $
It can be written as
$ \Rightarrow S = (x + x^2 + x^3 + \ldots)+(x^2 + x^3 + \ldots)+(x^3 + \ldots)+\ldots $
$\Rightarrow S = (x + x^2 + x^3 + ...)+x(x + x^2 + ...)+x^2(x + ...) + \ldots $
$\Rightarrow S = ( 1+x+x^2+ .. )\times( x+x^2+.. )$
$\Rightarrow S = \frac{1}{1-x}\times\frac{x}{1-x}$
$\Rightarrow S = \frac{x}{(1-x)^2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Change of Variables in differential equation Given the equation $zZ''(z) + Z'(z) + \alpha^2Z(z) = 0$ use the change of variables $x = \sqrt{\frac{z}{a}}$ where $a$ is a constant to map the problem to the differential equation $Z''(x) + \frac{1}{x}Z'(x) + \gamma^2 Z(x) = 0,$ where $\gamma = 2\alpha \sqrt{a}$
Attempt:
I think I am struggling to get the chain rule correct for this particular case. Here is my attempt: $$\frac{d}{dz} = \frac{d}{dx} \frac{dx}{dz} = \frac{1}{2\sqrt{za}}\frac{d}{dx} = \frac{1}{2xa} \frac{d}{dx}$$ This means by a similar argument $$\frac{d^2}{dz^2} = \frac{1}{4x^2a^2}\frac{d^2}{dx^2}$$
I am not really sure where to go from here.
Thanks.
| $$ \frac{dZ}{dz} = \frac{dZ}{dx}\frac{dx}{dz}=\frac{dZ}{dx}\frac{1}{2ax}$$
$$ \frac{d^{2}Z}{dz^{2}}=\frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}], $$
but $dz=2axdx$, so
$$ \frac{d}{dz}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{2ax}\frac{d}{dx}[\frac{dZ}{dx}\frac{1}{2ax}] = \frac{1}{4a^{2}x^{2}} \frac{d^{2}Z}{dx^{2}} -\frac{1}{4a^{2}x^{3}} \frac{dZ}{dx}. $$
Now
$$ z\frac{d^{2}Z}{dz^{2}} + \frac{dZ}{dz} + \alpha^{2}Z =0$$ becomes
$$ \frac{d^{2}Z}{dx^{2}} + \frac{1}{x}\frac{dZ}{dx} + 4a\alpha^{2}Z=0. $$
Cheers,
Paul Safier
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the nearest value of $y^2 - y$? I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question.
$\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} $, what is the
nearest value of $y^2 - y$?
$\textbf {(1) } 1 \qquad \textbf {(2) } \sqrt5 \qquad \textbf {(3) } 4 \qquad \textbf {(4) } 2\sqrt5$
I worked the problem and got $$y^2 - y = 5+ \sqrt{5-\sqrt{5+\sqrt{5-...}}} -\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} \\ = 5+ \sqrt{5-y} - y = 5-y + \sqrt{5-y}$$
What do I do next?
| Let's let
$$y=\sqrt{5+\sqrt{5-{\sqrt{5+{\sqrt5-\cdots}}}}}$$
and
$$z=\sqrt{5-\sqrt{5+{\sqrt{5-{\sqrt5+\cdots}}}}}$$
and assume that both limits exist. Note that
$$y^2=5+z\quad\text{and}\quad z^2=5-y$$
Hence
$$\begin{align}
y^2-y&=(5+z)-y\\
&=5-(y-z)\\
&=5-{y^2-z^2\over y+z}\\
&=5-{(5+z)-(5-y)\over y+z}\\
&=5-{z+y\over y+z}\\
&=4
\end{align}$$
Added 5/6/14: I thought I'd go ahead and add a proof that the two limits actually do exist. I'm not sure this is the simplest proof -- I'll be happy to see something slicker -- but here goes.
Define the sequence of $y_n$'s by $y_n=\sqrt{5+\sqrt{5-y_{n-1}}}$ with $y_1=\sqrt{5+\sqrt{5}}$, and let $z_n=\sqrt{5-y_n}$. It's easy to show (by induction) that $2\lt y_n\lt4$ for all $n$, so it suffices to show that the sequence of $y_n$'s has a limit.
I'll do so by showing that it's a Cauchy sequence, and I'll do that by showing that $\sum|y_n-y_{n-1}|$ converges.
We have $y_n^2-5=\sqrt{5-y_{n-1}}$, hence
$$y_n^2-y_{n-1}^2=\sqrt{5-y_{n-1}}-\sqrt{5-y_{n-2}}={y_{n-2}-y_{n-1}\over\sqrt{5-y_{n-1}}+\sqrt{5-y_{n-2}}}$$
hence
$$|y_n-y_{n-1}|={|y_{n-2}-y_{n-1}|\over|y_n+y_{n-1}|\left(\sqrt{5-y_{n-1}}+\sqrt{5-y_{n-2}}\right)}\lt{|y_{n-1}-y_{n-2}|\over|2+2|(1+1)}={1\over8}|y_{n-1}-y_{n-2}|$$
where we've used the inequalities $2\lt y_n\lt 4$ (for all $n$) in the denominator. It follows that $\sum|y_n-y_{n-1}|$ is bounded by a geometric series with ratio $1/8$, hence converges.
Added 6/29/16: For a nice example of what can go wrong if you assume a limit exists but don't bother to check if it actually does, see this problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator.
$$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$
| We need to find the value of $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{\dots}}}}}$.
Step 1: Let $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$
Step 2: Square both sides.
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y^2$$
Step 3: Recall that $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y$. So:
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=7y$$
Step 4: Rewrite the equation.
$$7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=y^2$$
$$7y=y^2$$
$$y^2-7y=0$$
Step 5: Solve for $y$.
$$y^2-7y=0$$
$$y(y-7)=0$$
$$y=0, \ 7$$
It is impossible that $y=0$. So, $y=7$.
$$\displaystyle \boxed{\therefore \sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\dots}}}}}=7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "69",
"answer_count": 7,
"answer_id": 5
} |
why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x tends to 2 but then i separate the sums and simplify but the calculator gives me another result here's what i did
$ \frac{\sqrt{x-2} +\sqrt{x-2}}{\sqrt{x^2 -4}} $
$\frac{\sqrt{x-2}}{\sqrt{x^2 -4}} + \frac{\sqrt{x-2}}{\sqrt{x^2 -4}} $
$ \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} + \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} $
simplifying
$ \frac{1}{\sqrt{x+2}} + \frac{1}{\sqrt{x+2}} $
this is when x tends to $ 2^+ $
$ \frac{2}{\sqrt{4}} = 1 $
but the calculator gets me $ \frac{\sqrt{2}}{2} $
where did i go wrong?
| Both of those expressions approach zero, so they have the same limit at 2. But they are far enough apart, when $x>2$, that dividing by $\sqrt{x^2-4}$ gives a different answer. The difference becomes 'something tiny' divided by 'something else tiny', which could be anything.
In this case, as B.S. says, Use the equation $(\sqrt{x}-\sqrt{2})=(x-2)/(\sqrt{x}+\sqrt{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Let $a;b;c>0$. Prove : $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$ Let $a;b;c>0$. Prove :
$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}+\frac{a^{2}}{a+b}$
P/s : Only use AM-GM and Cauchy-Schwarz
In the cases only use AM-GM and Cauchy-Schwarz; i don't have any thoughts about this problem !!
| We need to prove that $$\sum\limits_{cyc}\frac{a^2-b^2}{b+c}\geq0$$ or $$\sum\limits_{cyc}(a^2-b^2)(a+b)(a+c)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)(a^2+ab+ac+bc)\geq0$$ or
$$\sum\limits_{cyc}(a^2-b^2)a^2\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a^4+b^4}{2}-a^2b^2\right)\geq0,$$
which is true by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/589678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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