Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding a function with certain properties I ran into a problem, and I'm not sure how to continue.
Problem: Let $f$ be a function such that $\sqrt {x - \sqrt { x + f(x) } } = f(x)$, for $x > 1$. In that domain, $f(x)$ has the form $\dfrac{a+\sqrt{cx+d}}{b}$, where $a$, $b$, $c$, $d$ are integers and $a$, $b$ are relati... | Putting $y=f(x)$, your equation becomes:
$y^4 - 2xy^2 + x^2 - x - y = 0$
This is hard to solve for $y$, but we have the suggestion that $y$ has the form $y=\frac{a+\sqrt{cx+d}}{b}$. This can be substituted into our polynomial and then simplified:
$\left(\frac{a+\sqrt{cx+d}}{b}\right)^4 - 2x\left(\frac{a+\sqrt{cx+d}}{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving that $A+A^2+A^3$ has eigenvalue $\lambda+\lambda^2+\lambda^3$ where $\lambda$ is an eigenvalue of $A$
I know that $$P= \begin{pmatrix} 4 & 5 & 1 \\ -9 & -6 & 0 \\ 8 & 4 & 2 \end{pmatrix}$$
I also know that
$$D= \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$$
Where $a,b,c$ are eigenvalues... | We know that a matrix $M^n = PD^nP^{-1}$. Then $A = PDP^{-1}$ so
$$
A + A^2 + A^3 = PDP^{-1} + PD^2P^{-1} + PD^3P^{-1} = P(D+D^2+D^3)P^{-1}
$$
You have already found $P$ so once you have $D$ you just add.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Why does $\frac{\frac yx - \frac xy}{\frac1x - \frac1y} = x + y $ and $\frac{\frac xy - \frac yx}{\frac1x - \frac1y} = -(x + y)$? As the titles states: Why does $\frac{\frac yx - \frac xy}{\frac1x - \frac1y} = x + y $ and $\frac{\frac xy - \frac yx}{\frac1x - \frac1y} = -(x + y)$?
I am only able to find $\frac{y^2 - x^... | Notice that both numerators are differences of squares:
$$
\frac{y^2 - x^2}{y - x} = \frac{(y - x)(y + x)}{y - x} = y + x
$$
and likewise:
$$
\frac{x^2 - y^2}{y - x} = \frac{(x - y)(x + y)}{y - x} = \frac{-(y - x)(x + y)}{y - x} = -(x + y)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the Sum of the Series: $1/(x+1) + 2/(x^2 + 1) + 4/(x^4 +1) +\cdots$ $n$ terms Find the sum of $n$ terms the following series:
$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$
$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$
But after that I'm not sure as to how to proceed
| note: having just seen Andre's lovely solution this is very much a hammer-and-chisel job! however there was some devil in the detail, and i did learn something from the exercise!
define:
$$
D_m=\prod_{k=1}^m (1+x^{2^{m-1}}) = \sum_{k=0}^{2^m-1}x^k
$$
and
$$
N_m = \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} \tag{1}
$$
and now def... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding the integral $\int_{0}^{\infty }\frac{1}{(4x-3)(4x-1)}\,dx$. Which method that will be effective for solving this integral?
| The integral does not converge because of the singularities.
To compute the integral between $a$ and $\infty$ for $a>\frac 34$, you can use the following method:
\begin{align}
\int_{a}^{\infty }\frac{1}{(4x-3)(4x-1)} dx
&= \lim_{N\to\infty} \int_{a}^N \frac{1}{(4x-3)(4x-1)} dx
\end{align}
Now use the fact that $$
\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/995535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
$ab$ divides $3^a+1$ and $3^b+1$ Find all positive integers $a,b$ such that $ab$ divides $3^a+1$ and $3^b+1$.
It is clear that $3$ cannot divide either $a$ or $b$, because $3$ doesn't divide $3^a+1$ or $3^b+1$.
$(a,b)=(1,1),(2,1),(1,2)$ all work. If $b=1$, the condition reduces to $a$ divides $3^a+1$ and $4$, so $a=1,... | Let us start with a lemma.
Lemma For integer $m>2$, $gcd(m^{a}-1,m^{b}-1)=m^{gcd(a,b)}-1$
Proof Suppose wlg that $a>b$. Then
\begin{align*}
gcd(m^{a}-1,m^{b}-1)= \enspace&gcd(m^{a}-1,m^{a-b}(m^{b}-1))\\
=\enspace&gcd(m^{a}-1,m^{a-b}-1)\\
=\enspace&gcd(m^{b}-1,m^{a-b}-1)
\end{align*}
since $gcd(x,y)=gcd(x,x-y)$. By rei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int_{1}^{\infty} \frac{x 3^x}{(3^x-1)^2}$ How do we prove that $$\int_{1}^{\infty} \frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x=\dfrac{3\log{3}-2\log{2}}{2\log^2{3}}$$
I have tried some substitutions such as $3^x=t$, but it didn't work out. The denominator is already factored. I do not know what to do next. Plea... | Integrating by parts ($u=x$, $dv=\frac{3^x}{(3^x-1)^2}dx$), $$\int x\frac{3^x}{(3^x-1)^2}dx=x\int\frac{3^x}{(3^x-1)^2}dx-\int\left[\frac{dx}{dx}\int\frac{3^x}{(3^x-1)^2}dx\right]dx$$
Now for $\displaystyle\int\dfrac{3^x}{(3^x-1)^2}dx,$ set $3^x-1=u$
Finally $\displaystyle\int\dfrac{dx}{3^x-1},$ set $3^x-1=v$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$ How to solve:
$$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$
What I have done:
I) $x < 0$:
Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$
II) $x > 0$:
$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$
$$\frac{|x+4|}{|ax+2|} > \frac1x$$
b... | You have already identified and made use of the points where sides of the inequality are undefined. So all that is needed is to solve for equality and then make determinations about the subintervals between the points of equality and points where values are undefined.
Equality happens when $x=\left|\frac{ax+2}{x+4}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + ... | Hint: Try re-doing it as two separate series, $\sum(1/9)^n$ and $\sum(7/9)^n.$ The first has first term $1/9$ and common ratio $1/9$ while the second has first term $7/9$ with common ratio $7/9.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Check: For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of transpositions For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of transpositions:
*
*$\begin{pmatrix} 1 & 3 & 5 & 4 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix... | I'm assuming you're using the right-sided convention, in which case they are right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
a two-variable cyclic power inequality $x^y+y^x>1$ intractable by standard calculus techniques If $x$ is in the open interval (0,1) and so is $y$, prove that $$x^y+y^x>1$$
A direct two-variable application of maxima and minima seems difficult.
| Let's suppose(if it's not true, change the place of $x,y$ in the following arguments) $$\dfrac{x^{y-1}}{y^{x-1}} \geq 1$$
Then we have $ (x^y + y^x)^\frac{1}{x} = (y^x (1+\dfrac{x^y}{y^x}))^\frac{1}{x} = y (1+\dfrac{x^y}{y^x})^\frac{1}{x} $
Then since $\frac{1}{x} > 1$, we have $$(x^y + y^x)^\frac{1}{x} = y (1+\dfrac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Polynomial function question If $f(x)$ is equal to $\frac{1}{x^3 + 3x^2 + x}$, find the smallest value of $n$ for which $f(1) + f(2) + ... F(n) = \frac{503}{2014}$. I tried noting that first initial values of f sum to $\frac{1}{5}$, and $503$ divided by $2014$ is approx $\frac{1}{4}$, so we're aiming for values after $... | Note that
$$0 < f(x) = \frac{1}{x^3 + 3x^2 + x} \leq \frac{1}{x} + \frac{1}{3x} + \frac{1}{x} = \frac{7}{3x},$$
since $x \in \mathbb{N}$.
Now,
$$f(1) = \frac{1}{5}$$
$$f(2) = \frac{1}{22}$$
$$f(3) = \frac{1}{57}$$
Consequently:
$$f(1) + f(2) = \frac{27}{110} = 0.2\overline{45}$$
$$f(1) + f(2) + f(3) = \frac{1649}{6270}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A thinking problem of limit from my teacher. Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}... | Since $n$ is supposed to be large, then the argument of the sine is small. So, start with $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ Now, replace $x$ by $\frac{\pi }{\sqrt{k+n^2}}$ and then $$\frac{1 }{\sqrt{k+n^2}}=\frac{1}{n}-\frac{k}{2 n^3}+\frac{3 k^2}{8
n^5}+O\left(\left(\frac{1}{n}\right)^6\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find an equation of the tangent line to the curve
Find an equation of the tangent line to the curve $2·(x^2+y^2)^2=25·(x^2−y^2)$ at the point
$(3,1)$.
Any hints?
| The more general way to solve this type of problem is with implicit differentiation:
We have $2(x^2+y^2)^2=25(x^2-y^2)$. Differentiating both sides w.r.t. $x$, we get: $$4(x^2+y^2)(2x+2yy')=50x-50yy'$$ $$\implies (8(x^2+y^2)y+50y)y'=50x-4(x^2+y^2)\cdot 2x$$ $$\implies y'=\frac {50x-8x^3-8xy^2}{8x^2y+8y^3+50y}$$ $$\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find minima of the multivariable function $\frac{4}{x^2+y^2+1}+2 xy$ $$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the domain: }1/5\leq x^2+y^2\leq 4$$
I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the l... | You know that $1/5\leq x^2+y^2\leq 4$, so:
$$\frac{4}{x^2+y^2+1} \geq \frac{4}{4+1}=\frac{4}{5}$$
In addiction by AM-GM:
$$2|xy| \leq x^2+y^2 \leq 4$$
$$|xy| \leq 2$$
And:
$$2xy \geq -4$$
So let's try to find such a $x,y$ that:
$$xy=-2,x^2+y^2=4$$
But for $x=-\sqrt{2}$ and $y=\sqrt{2}$ both equalities are satisfy, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The limit of a difference between a cube root and square root I've been trying to evaluate the limit of $$\lim_{x \to \infty}[(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}]$$ I've tried using the property $$x^6-y^6 =(x-y)(x^5+x^4y+\cdots+xy^4+y^5)$$
where $x = (x^3+x^2+1)^{1/3}$ and $y=(x^2+x)^{1/2}$. But I can't seem to 'clean up' ... | Using your property, you have
\begin{align}
(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}&=\frac{(x^3+x^2+1)^2-(x^2+x)^3}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-x^5-2x^4+x^3+2x^2+1}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-1-2/x+1/x^2+2/x^3+1/x^5}{(1+1/x+1/x^3)^{5/3}+\cdots+(1+1/x)^{5/2}}\\ \ \\
&\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove complex equation Prove the following $$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$$
for any integer n greater than 0. DO NOT USE ....
I believe that I can use mathematical induction. Any help would be greatly appreciated.
| You want to show that
$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$
Multiply by
$z-1$ and it becomes
$\frac{1}{z^n}
= 1 - (z-1)\sum_{k=1}^n\frac{1}{z^k}
$
or
$1-\frac{1}{z^n}
= (z-1)\sum_{k=1}^n\frac{1}{z^k}
$.
But
$\begin{array}\\
(z-1)\sum_{k=1}^n\frac{1}{z^k}
&=z\sum_{k=1}^n\frac{1}{z^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.
|AB| = a, P is midpoint of segment AB
| I believe we can use Descartes Circle Theorem. We can identify 4 circles:
*
*the circle $O_A$ with center at $A$
*circle with diameter $\overline{AP}$
*circle with diameter $\overline{PB}$
*unknown circle
Unfortunately, these 4 circles are not mutually tangent since the circle with diameter $\overline{AP}$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Non-linear system with all trajectories converging on the line $x=0$, rather than $(2,0)$? I have the following nonlinear system:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
Which I set up to $F=\dot{y}$
Giving the jacobian of transformation being:
$$\dot{F}= \begin{pmat... | It would be better to attack the problem directly as Amzoti has said:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
We want the critical points of this system, which we can find by taking:
$$\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
$(0,\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simple Trigonometric Problem Here is a Trig problem, and I am missing some understanding of some basic algebraic rule:
$if \sin\theta= \frac{m^2 +2mn}{m^2+2mn+2n^2}$
then prove that $\tan\theta = \frac {m^2 +2mn}{2mn+2n^2},$ In the picture below can anyone explain me how did they achieve the step from 1 to 2 highlighte... | $$\begin{align}
\sqrt{1-\frac{(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}&=\sqrt{\frac{(m^2+2mn+2n^2)^2-(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&=\sqrt{\frac{(2n(m+n))^2}{(m^2+2mn+2n^2)^2}}\tag{1}\\
&=\frac{2n(m+n)}{m^2+2mn+2n^2}\\
\end{align}$$
$\text{Explanation } 1$ $$(a+b)^2-a^2 = 2ab+b^2=b(2a+b)$$
Where $a=m^2+2mn$ , $b=2n^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding lowest possible value please help me with this problem:
Find the lowest possible value of
$$
x+y^3
$$
where both x and y are positive and x*y=1.
I know how to solve this one using my method, but I was suggested to use geometric and arithmetic mean. I have no idea how to solve this using them.
| Since $xy = 1$, observe that $x + y^3 = y^{-1} + y^3 = y^3 + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} \geq 4\sqrt[4]{\frac{1}{3^3}}$.
The equality holds if and only if $y^3 = \frac{y^{-1}}{3}$, that means for $y = \frac{1}{\sqrt[4]{3}} > 0$, therefore your lowest possible value of $x + y^3$ is $4\sqrt[4]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two eq... | My original answer did not use the nice symmetry. Kelenner gave a better answer. Based on that, I have made yet another answer, using essentially the same method as Kelenner did, but with a different definition of the same function which is (for me) easier to understand. When I first wrote this answer, I did not realiz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\displaystyle\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x} \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$
Evaluate the following limit: $$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt$$
Any advice on how to tackle this problem ?
| Substituting $t = xu$, we get
$$\int_0^x dt \frac{t^3 \log{(1-t)}}{t^4+4} = x^4 \int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} $$
so the limit is,
$$\lim_{x\to 0} \frac1{x}\int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} = \lim_{x\to 0} \frac1{x}\int_0^1 du \frac{-x u^4}{4} = -\frac1{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit $\lim_{x\to 0^-}| \left( 1+x^{3} \right)^{1/2}-1-x^{5} |/(\sin x-x)$ I am studying for my first calculus exam (well, it's half an exam), and of course we have to solve limits, without using L'Hospital rule, and using asymptotic analysis.
I can't solve this one $$\lim_{x\rightarrow 0^{-}}\frac{\left| \lef... | Look at
$f(x)
=\frac{ ( 1+x^{3} )^{\frac{1}{2}}-1-x^{5} }{\sin x-x}
$.
By "asymptotic analysis",
as $x \to 0$,
$(1+x^3)^{\frac12}
\approx 1+\frac12 x^3 + \frac{\frac12 (-\frac12)}{2}(x^3)^2
= 1+\frac12 x^3 - \frac18 x^6
$
and
$\sin(x)
\approx x-\frac{x^3}{6}+\frac{x^5}{120}
$
so
$\sin(x)-x
\approx -\frac{x^3}{6}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does $\log_a b = \log_\sqrt a \sqrt b$ can be a basic logarithm law? Does the following equation is true for all $ a,b\in{\mathbb R}$?
$$\log_a b = \log_\sqrt a \sqrt b$$
I have tried to proove this, and I didnt find any contradiction. Is it true?
EDIT
Thanks guys for the answers. I've just found a different proof wit... | It's true for all positive $b$, $a \neq 1$ as
$$\log_\sqrt{a} \sqrt{b} = \frac{ \ln \sqrt{b}}{\ln \sqrt{a}} = \frac{\frac{1}{2} \ln b}{\frac{1}{2} \ln a} = \frac{\ln b}{\ln a} = \log_a b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Maximizing sin(a-b) given a trig relation Suppose $a$, $b$ are acute angle measures such that $\tan a = 5\tan b$. Find the maximum value of $\sin(a-b)$.
$\sin(a-b)=4\sin b \cos a$, but I don't know what to do from here.
| Since $a,b\in(0,\frac\pi2)$ and $\tan$ is positive and strictly increasing there, the assumption $\tan a=5\tan b$ implies $a>b$. Thus $a-b\in(0,\frac\pi2)$ also, so $\sin(a-b)>0$, and we might as well maximize $\sin^2(a-b)$ instead. And since
$$ \sin^2(a-b) = 1 - \frac1{1+\tan^2(a-b)} $$
we might as well maximize $\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How prove this diophantine equation $x^2+y^2+z^3=n$ always have integer solution
show that: For any postive ineteger $n$,then the equation
$$n=x^2+y^2+z^3$$
always have integer solution
My idea: such as $n=1$,then we have
$$1=0^2+0^2+1^3$$
$$2=0^2+1^2+1^3$$
$$3=1^2+1^2+1^3$$
$$4=2^2+0^2+0^3$$
$$5=1^2+2^2+0^3$$
$... | My friend put this as an MAA Monthly problem, years ago. The comparison is that there are infinitely many numbers that have no expression as $x^2 + y^2 + z^9.$ This simple result defeated an existing conjecture; we sent it early to Robert C. Vaughan, so it made it into the second edition of his book The Hardy-Littlewoo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$ I realize that is a basic math problem, but I am still having problems with it.
The expression $$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$$ equals one of the following:
*
*$2\sqrt{3}-1$
*$\dfrac{6}{11}$
*$4\sqrt{3}$
*$12$
How do I simpl... | Hint:
First Rationalize the denominator of first term by multiplying numerator and denominator with denominator's conjugate and then use
$$(a+b)(a-b)=a^2-b^2$$
$$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}=\frac{\sqrt{3}}{2\sqrt{3}+1}\cdot\frac{2\sqrt{3}-1}{2\sqrt{3}-1} + \frac{\sqrt{3}}{11}=\frac{6-\sqrt3}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Quadratics with roots as integers; possible values of a Suppose $a$, $b$ are real numbers such that $a+b=12$ and both roots of the equation $x^2+ax+b=0$ are integers.
Determine all possible values of $a$.
I don't know how to go about doing this without long, messy casework. I tries $(x-s)(x-r)=x^2+ax+b$ and got $-r-... | Since $a + b = 12$
$$-(r+s) + rs = 12$$
Using Simon's favorite factoring trick:
$$(r-1)(s-1) - 1 = 12 \iff (r-1)(s-1) = 13$$
Only two possible solutions arise:
$r -1 = 1$ and $s -1 = 13$, therefore $r = 2$ and $s = 14$
$r -1 = -1$ and $s -1 = -13$, therefore $r = 0$ and $s = -12$
Therefore, $a = -16$ and $b = 28$
or
$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.
| Apply the substitution $x=sec(u)$, then $dx=\frac{\tan \left(u\right)}{\cos \left(u\right)}du$.
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{\tan \left(u\right)}{\cos \left(u\right)}du$$
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{1}{\cos^2 \left(u\right)}du$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 7
} |
Can all real polynomials be factored into quadratic and linear factors? So I understand how to do integration on rational functions with a linear and a quadratic denominator, and I understand how to do a partial fraction decomposition, but I was wondering what happens if the polynomial is higher degree and can't be fac... | For example, take $P(x) = x^n-1$. The factorization into complex linear factors is
$$(x^n-1) = \prod_{k=0}^{n-1}( x - (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) )$$
From the pairs of non-real conjugate roots $\{ (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) ,(\cos \frac{2 (n-k) \pi}{n} + i \sin \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 6,
"answer_id": 5
} |
Recursive Sequence solving for $f(200)$ Let $f$ be defined recursively by: $f(0)=5$ and $f(n+1)=3f(n)-2$. Find $f(200)$
I'm really confused how to go about solving this. Can someone help? Thank you!
| Expanding on
OC-Sansoo's answer:
Since
$f(n+1)=3f(n)-2$,
subtracting one from each side,
$f(n+1)-1
=3f(n)-3
=3(f(n)-1)
$.
Letting
$g(n)
=f(n)-1
$,
$g(n+1)
=3g(n)
$.
By induction,
$g(n)
=3^n g(0)
$
or
$f(n)-1
=3^n(f(0)-1)
=4\cdot 3^n
$
or
$f(n)
=4\cdot3^n+1
$.
In general,
if
$f(n+1)
=af(n)+b
$,
we want to find $c$
such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Application of Mergesort We have $8$ players and we want to sort them in $24$ hours.
There is one stadium. Each game lasts one hour.
In how many hours can we sort them??
I thought that we could it as followed:
$$\boxed{P1} \ \boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5} \ \boxed{P6} \ \boxed{P7} \ \boxed{P8} \\ ... | The idea of mergesort is to exploit the fact, that merging two pre-sorted lists is "cheap": if the lenght of list1 is $l_1$ and that of list2 is $l_2$ and $L=l_1+l_2$ then you need at most $L-1 $comparisions to make a sorted list. The minimum (in a simple algorithm) is just the length of the shorter list.
This idea c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ Prove by induction that
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}
$$
Let $n=r$, so that
$$
S_r=2-\frac{r+2}{2^r}
$$
Therefore
$$\begin{align}
S_{r+1}=S_r+\frac{r+1}{2^{r+1}}&... | Your mistake seems to be in your second-to-last line.
You clearly wanted to get a common denominator of $2^{r+1}$, but $2^r\cdot2^r=2^{2r}$, not $2^{r+1}$. You should have multiplied numerator and denominator by $2$, not $2^r$. Also, be careful with your signs.
You should get $$S_{r+1}=2+\frac{-2(r+2)}{2^{r+1}}+\frac{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\int_{-1}^{1} 4x \sqrt{1-x^{2}}\, dx = -3\pi$ What is of interest is the assertion
$$\int_{-1}^{1} 4x \sqrt{1-x^{2}}\,dx = -3\pi.$$
Since
$$\pi = 2\int_{-1}^{1}\sqrt{1-x^{2}}\,dx,$$
i.e. the area of a unit circle,
it suffices to prove that
$$\int_{-1}^{1}x\sqrt{1-x^{2}}\,dx = -\frac{3}{2}\int_{-1}^{1}\sqrt... | $x=\sin \theta \implies dx= \cos \theta \ d\theta$
$\displaystyle\int x \sqrt{1-x^2}\ dx=\displaystyle\int \sin \theta \cos^2\theta\ d\theta$
$\cos^2 \theta=z \implies -2\sin \theta\cos \theta\ d\theta=dz$
$\therefore \displaystyle\int \sin \theta \cos^2\theta\ d\theta=-\dfrac{1}{2}\displaystyle\int \sqrt{z}\ dz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Differentiation with the quotient rule I have the question:
Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$
Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$
I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?
... | The error is in
$$\frac1{\sqrt{2x+1}} \cdot \frac1{(2x+1)^2} = \frac1{(2x+1)^{2+\frac12}} = \frac1{(2x+1)^{\frac52}}$$
and
$$\frac1{\sqrt{2x+1}^2} \ne \frac1{(2x+1)^2}$$
so there is an error in the quotient rule.
(the exponent is positive in the denominator, not negative)
Tracing the errors through the equation we obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
simple trigonometric functions how would you solve this
$$\cos^2 x - 2\sin x \cos x - \sin^2 x = 0$$
I tried to simplify it but I got $\cos 2x - \sin 2x$. I can't simplify that further.
| Based on what you have done,
\begin{align*}
\cos^2x - 2\sin x\cos x - \sin^2x & = 0\\
\cos^2x - \sin^2x - 2\sin x\cos x & = 0\\
\cos(2x) - \sin(2x) & = 0\\
\cos(2x) & = \sin(2x)\\
1 & = \tan(2x)
\end{align*}
Observe that if you want values of $x$ in the interval $[0, 2\pi)$ that satisfy the equation, then $0 \leq 2x < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and t... | The integral in your last line can be simplified to
$$
\frac{16}3\int_0^2 (4-x^2)^{3/2}\,dx
$$
To do this, first use the substitution $x=2\sin\theta$, obtaining
$$
\frac{16}3\int_0^{\pi/2}(4-4\sin^2\theta)^{3/2}\cos\theta\,d\theta=
8\cdot\frac{16}3\int_0^{\pi/2}\cos^4\theta\,d\theta
$$
To complete the integral, use th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the series with unordered powers of $3$
Consider the following series: $$\sum_{n=1}^{\infty} a_n = 1/3+1+1/3^3+1/3^2+1/3^5+1/3^4+1/3^7+1/3^6 +\dots$$
Determine if it converges, and find the sum.
Here is what I got:
a) $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = (\frac{1}{3})^3 =1/27$
b) $\lim_{n\to\i... | Since $a_n=\dfrac{1}{3^n}$ when n is odd and $a_n=\dfrac{1}{3^{n-2}}$ when n is even,
A)$\;\;$ $\dfrac{a_{n+1}}{a_n}=\begin{cases}\frac{1}{27},&\mbox{if n is odd}\\\; 3,&\mbox{if n is even}\end{cases}.$ Therefore $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}$ does not exist.
B) $\;\;$$\displaystyle\left(a_n\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then,... | Knowing the answer, it is not hard to construct the right CS inequality to solve your problem:
$$52\cdot (13^2+12^2+18^2) \ge (13a+12b+18c)^2 = (7a+6\cdot 26)^2 \implies a \le \frac{26}7$$
It is easy to check equality is attained when $a = \frac{26}7, b = \frac{24}7, c = \frac{36}7$.
If you don't know the answer to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
compute sums of x,y given a condition
Problem: given that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=p$, then compute $x+y$
try: i tryed to solve by this way
$$\begin{align}
\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)&=p\\
1&=p\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)\\
p&=p^2\lef... | It is not possible. We have that $f(x):\mathbb{R}\to\mathbb{R}^+$ given by:
$$f(x)=x+\sqrt{x^2+1}$$
is bijective. So, assume that $p=f(2)\cdot f(3)=6+5 \sqrt{2}+3 \sqrt{5}+2 \sqrt{10}$.
Obviously $x=2,y=3$ is a solution of $f(x)\cdot f(y)=p$, giving $x+y=5$.
But another solution exists, given by:
$$x=y=f^{-1}\left(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
... | You can argue by contradiction, using the relation
$$\text{rank}(AB) \le \text{min}(\text{rang}(A),\text{rang}(B))$$
with $A= \begin{pmatrix}1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}$ and $B:=X$ a $4\times 3$ matrix with $\text{rank}(X)\le 3$.
Since then you would have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Identities for hypergeometric functions ${}_2F_1$ with z=1/2 Is there a closed form (or approximation) for a hypergeometric function of form:
$_2F_1(1,b+c;c;\frac{1}{2}) \quad \text{where} \; b,c \in \mathbb{N}$ ?
I researched all identities in http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/0... | Case $1$: $c\neq1$
Then $_2F_1\left(1,b+c;c;\dfrac{1}{2}\right)$
$=~_2F_1\left(b+c,1;c;\dfrac{1}{2}\right)$
$=\dfrac{1}{B(1,c-1)}\int_0^1\dfrac{(1-x)^{c-2}}{\left(1-\dfrac{x}{2}\right)^{b+c}}dx$
$=\dfrac{2^{b+c}(c-1)!}{0!(c-2)!}\int_0^1\dfrac{(1-x)^{c-2}}{(2-x)^{b+c}}dx$
$=2^{b+c}(c-1)\int_0^1\dfrac{(2-x-1)^{c-2}}{(2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to solve $\sin\theta +\sin3\theta =0$
Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$
Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta }{ 2 } \cos\frac { \theta -3\theta }{ 2 }\\
0&=2\sin\frac { 4\theta }{ 2 } \cos\frac { -2\theta }{ 2 } \\
0&=2\sin2\thet... | As far as I see, you didn't make any mistake in your calculation. You did, however, overcomplicate things.
For one, if $2\sin(\alpha)\cos(\beta) = 0$, you can get rid of the $2$ and just write $\sin(\alpha)\cos(\beta) = 0$.
The main mistake you made, however, was when you got to $2\sin(2\theta)\cos\theta$. From there o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!... | jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$\sin(x+x^3)=\sum_{n=0}^\infty \frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Product of a Continuous and Discrete Distribution
Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that
\begin{align*}
P(Y=y) = \begin{cases}
\frac{1}{2} & y = -1\\
\frac{1}{2} & y = 1\\
0 & otherwise
\end{cases}
\end{align*}
If $ Z = XY$, show that $Z \sim \mathcal{N}(0,1)$.
He... | Your mistake: you can't divide by $Y$ without knowing its sign (if it's negative, the inequality would reverse).
\begin{align*}
P(XY<z)
&= P(XY<z \mid Y=1) P(Y=1) + P(XY<z \mid Y=-1) P(Y=-1)\\
&=\frac{1}{2} P(X<z) + \frac{1}{2} P(-X<z)\\
&= \frac{1}{2} P(X<z) + \frac{1}{2} P(X>-z)\\
&= P(X<z).
\end{align*}
The last ste... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}... | Without AM-GM, you could expand the expression which, after simplifications, write $$f(x)=(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 =\csc ^2(x)+\sec ^2(x)+5$$ The derivative write $$f'(x)=2 \tan (x) \sec ^2(x)-2 \cot (x) \csc ^2(x)=-8 \sin (4 x) \csc ^4(2 x)$$ and cancels for $x=\frac{\pi}{4}, \frac{3\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Problem integrating when attempting a solution with the Poincaré Lemma
d) This is part I am having troubles with.
I get that
$$
\hat{\mathbb{X}}_t = \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \Phi_t^{-1}
= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) (x/t,y/t,z)
= (x/t,y/t,0)
$$
so,
$$\begin{ali... | Let $$I(x,y,z) = \int_0^1 \frac{2z^2 - t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}t\, dt.$$
Using the $u$-substitution $u = t^2x^2 + t^2y^2 + z^2$, we compute
\begin{align}I(x,y,z) &= \frac{1}{2(x^2+y^2)}\int_{z^2}^{r^2} \frac{2z^2 - (u - z^2)}{u^{5/2}}\, du\\
&= \frac{1}{2(x^2 + y^2)} \int_{z^2}^{r^2} \left(\frac{3z^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^7-5x^4-x^3+4x+1=0$ for $x$
Solve for $x$
$$x^7-5x^4-x^3+4x+1=0$$
This equation has been bugging me since the past few days. I have found, using the Rational Root Theorem that $x=1$ is a root of this equation. However, after dividing, I cannot solve the six degree equation thus generated. I have also trie... | $x^4(x^3 + 1) + (x^3 + 1) = 0$
$\implies(x^3 + 1)(x^4 + 1) = 0$
$\implies(x + 1)(x^2 - x + 1)(x^4 + 2x^2 + 1 - 2x^2) = 0$
$\implies(x + 1)(x^2 - x + 1)((x^2 + 1)^2 - 2x^2) = 0$
$\implies(x + 1)(x^2 - x + 1)(x^2 + 1 + (\sqrt 2)(x))(x^2 + 1 - (\sqrt 2)(x)) = 0$
$ x^2 - x + 1 = 0$
$\implies x = \frac{1 \pm \sqrt{1 - 4\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Finding the value of $y=b^2(3a^2+4ab+2b^2)$ if $a^2(2a^2+4ab+3b^2)=3$ and $a$ and $b$ are distinct zeros of $x^3-2x+c$
If $a$ and $b$ are distinct zeroes of the polynomial $x^3-2x+c$ and
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+4ab+2b^2)=y$$
Evaluate $y$
I tried for many hours but couldn't solve this question. Thi... | Since $a$ and $b$ are the solutions of $x^3-2x+c=0$ we have,
$a^3-2a+c=0$
&
$b^3-2b+c=0$
Subtracting the above two equations, we get,
$a^3-b^3-2(a-b)=0$
$\implies (a-b)(a^2+ab+b^2-2)=0$
$\implies a^2+ab+b^2=2$ (Since $a$ and $b$ are distinct)
Now,
$$a^2+ab+b^2=2$$
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric substitution and Integration of $\frac{1}{x^2\sqrt{x^2+1}} $ Regarding the integral
$$
\int \frac{dx}{x^2\sqrt{x^2 + 1}}
$$
I'm not sure what to do about the extra $x^2$ in the denominator. What can I do about it?
| $x=\cos \theta \implies dx=-\sin \theta\ d\theta$
$\therefore\displaystyle\int\dfrac{dx}{x^2\sqrt{x^2+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{\cos^2 \theta+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{2\cos^2 \left(\dfrac{\theta}{2}\right)}}\\=-\sqrt{2}\display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Showing convergence of recursive sequence $A_{n+1}=\frac 1 {1+A_n}$
Given : $\forall n\in\Bbb N,\quad A_{n+1} = \frac 1 {1+A_n}$ and $A_1 = 0$
Show the sequence converges and find its limit.
Briefly what I did was to create two sub-sequences with an index difference of 2. Doing that I got two recursive formulas that ... | The recurrence is of the form
$$x_{n+1} = \phi(x_n)$$
where $\phi(x) = \frac{1}{1+x}$. The map $\phi$ has two fixed points $x_{1,2}$, roots of the equation $x^2 + x -1=0$, that is $\frac{-1\pm \sqrt{5}}{2}$. Now, we can write $\phi(x)$ in the form
$$\frac{\phi(x) -x_1}{\phi(x) - x_2} = k \cdot \frac{x-x_1}{x-x_2}$$
$k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
An exercise from Knuth's book - Proving a formula by induction I would like to find a formula for this sum:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}
$$
The answer given (Knuth's book, The Art of Computer Programming Volume 1, Third Edition, Section 1.2.1... | You are very close.
Take two consecutive sums and subtract them (actually you will be adding as the signs alternate):
$$\frac{n+1}{4(n+1)^2+1}+\frac{n+2}{4(n+2)^2+1}=\frac{8n^3+36n^2+54n+27}{16n^4+96n^3+216n^2+216n+85}.$$
The numerator is $(2n+3)^3$, and the denominator is $(2n+3)^4+4$.
To make this derivation more sy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that the following polynomials are irreducible or not.
I want to show that:
1) $X^4+1$ is irreducible
The roots are the elements of
$$\left\{\frac{\pm 1+i}{\sqrt 2},\frac{\pm 1-i}{\sqrt 2}\right\}$$
therefore it's not a product of a polynomial of degree 1 and a polynomial of degree 3. Is there an easier way to... | $1)$ : $$f(x)=x^4+1$$
$$f(x+1)=x^4+4x^3+6x^2+4x+2$$
and the criterion of Eisenstein shows that $f(x+1)$ is irreducible. Therefore
$f(x)$ is also irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the points of intersection of a circle and a line In a test (of math in arabic language) we we're asked to find the points of intersection of a circle and a line. Their equation is given.
In the test I solved system of equations made of their equation and in the process I explain my line of thought using the wo... | The posted solutions are correct, below is a possible derivation.
Exercise 1
In this exercise you need to find the center and radius of this circle which depends on $m$.
$$x^2+y^2+(m+2)x-2my+m^2-36=0$$
Notice that $(y-m)^2=y^2+m^2-2my$ and rewrite as
$$x^2+(y-m)^2+(m+2)x-36=0$$
Add $0=1-1+m+\frac{m^2}4-m-\frac{m^2}4$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area of a circumcenter triangle equals area of medial triangle Let $X$, $Y$, $Z$ be the midpoints of sides $BC$, $AC$, $AB$ respectively in triangle $ABC$. Let $O_{A}$, $O_{B}$, and $O_{C}$ be the circumcenters of triangles $AZX$, $BXY$, and $CYZ$ respectively. Prove that the area of triangle $O_{A} O_{B} O_{C}$ equals... | As requested, here is a solution via coordinates. However, as mentioned above, I'd be interested in seeing a solution via pure geometry instead.
We assign coordinates $A(0,0)$, $B(b,0)$, $C(c,d)$ to the vertices of triangle $ABC$. This gives $X = ((b+c)/2, d/2)$, $Y = (c/2, d/2)$, $Z = (b/2, 0)$. $O_A$ has coordinates ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{... | let me try a change of variable
$$x = {1 + t^2 \over 1 - t^2},\ dx = {-4t \ dt \over (1-t^2)^2},\ x^2+1 = {2(1+t^4) \over (1-t^2)^2}, \ x^2-1 = {4t^2 \over (1-t^2)^2}$$
$$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \int {t^2 - 1 \over t^4 + 1} \ dt
= \int {1 \over t^2 + 1} dt - 2 \int {1 \over t^4 + 1} dt = \tan^{-1}t -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Maximising the Area of a Cyclic Quadrilateral In cyclic quadrilateral $ABCD$, $AB = AD$. If $AC = 6$ and $AB/BD = 3/5$, find the maximum possible value of $[ABCD]$.
(Source: SMT 2014)
If we let $AB=AD = 3x$ and $BD=5x$, from Ptolemy, we have $BC+CD=10$. From the triangle inequality, we can also get $BC+CD > BD \Rightar... | I've been exposed to a solution to this problem.
Since $AB=AD$, $\angle ABD = \angle ADB$. Furthermore, from the proprerties of a cyclic quadrilateral, $\angle ACB = \angle ADB$ and $\angle ACD = \angle ABD$. Therefore, $\angle ACB = \angle ADB = \angle ABD = \angle ACD$. Most importantly, $\angle ACB = \angle ACD$. No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
primes of the form $4k+3$ and sums of squares It is well-known that if $p$ is a prime of the form $4k+3$ and $p|x^2+y^2$ then $p|x$ and $p|y$. I forget what is the name of this result, and where can I find a proof (please provide a link).
| EDIT: note that your binary quadratic form $x^2 + y^2$ has discriminant $\Delta = -4.$
Somewhere you need to find a proof of the fact that, for prime $q \equiv 3 \pmod 4,$ we always have
$$ (-1|q) = -1. $$
For instance,
Niven and Zuckerman (and Montgomery) page 132, Theorem 3.1 part (5), says
$$ (-1|p) = (-1)^{(p-1)/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x... | After thinking a bit about some of what @Daniel Hast typed, I realized that the reason why, say, $\sqrt{(4x-1)^2} = 4x-1$ was that more generally, given that $\sqrt{x} = r$ such that $r^2 = x$, squaring $r$ in $\sqrt{x} = r$ produced $(\sqrt{x})^2 = r^2 = x$ and subsequently $(\sqrt{x})^2 = x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Integrate $I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$ I have a problem with this integral. It seems that solution has to be simple, but I couldn't find out.
$$I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$$
I tried using integration by parts and differentiating with regard to a, but neither helped.
| Observe that your integral is convergent in the usual sense for $|a|<1$. You may then write $t=\tan \dfrac{x}{2}$, giving $ \sin x=\dfrac{2t}{1+t^2}$.
Hence
$$
\begin{align}
I(a) &= \int_0^{\pi/2} \frac{dx}{1-a\sin x}\\\\
&= 2\int_0^{1} \frac{1}{1-a\dfrac{2t}{1+t^2}}\frac{dt}{1+t^2}\\\\
&= 2\int_0^{1} \frac{1}{(t-a)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate residue of $\cot z/z^4$ at $z=0$? How to evaluate residue of $\cot z/z^4$ at $z=0$?
As we know :
$$f(x)=f(0)+f'(0)x+f''(0)x^2/2+...$$
but $\cot(0)\to\infty$ or is undefined? I know that:
$$\tan x=x+x^3/3+2x^5/15+...$$
|
$$\frac{\cos z}{z^4\sin z}$$
$$\newcommand{\f}[2]{\frac{#1}{#2}}
\newcommand{\r}[1]{\frac{1}{#1}}
\newcommand{\array}[2]{\begin{array}{#1}#2\end{array}}
\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\s}[1]{\left[#1\right]}
\newcommand{\d}[0]{\ldots}
\newcommand{\ul}[1]{\underline{#1}}
\array{r|lll}{
&&\r{z^5}&-\b{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| Let
$$a=\lim_{x\to\infty}x\left(\sqrt{x^2+1}-x\right)$$
Now rationalize it:
$$a=\lim_{x\to\infty}\dfrac{x\left(x^2+1-x^2\right)}{\sqrt{x^2+1}+x}$$
Then simplify it:
$$a=\lim_{x\to\infty}\dfrac{x}{\sqrt{x^2+1}+x}$$
$$a=\lim_{x\to0}\dfrac{\dfrac1x}{\sqrt{\left(\dfrac1x\right)^2+1}+\dfrac1x}$$
$$a=\lim_{x\to0}\dfrac{1}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Solve $2(x+y)+xy=x^2+y^2$ where $x,y \in \mathbb{Z}$ Solve the equation: $$2(x+y)+xy=x^2+y^2$$
How should I go about solving this? Any guidance appreciated.
Thanks!
| $$8(x+y)=4(x^2-xy+y^2)\\8(x+y)=3(x-y)^2+(x+y)^2\\8z=3w^2+z^2\\16=3w^2+(z-4)^2$$
where $w=x-y$ and $z=x+y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Quadratic congruences in which modulus is divisible by constant term There are two similar congruences:
$$
x^2-6x\equiv16\pmod{512}\\
y^2-y\equiv16\pmod{512}
$$
It is easy to see that the $\gcd$ of all three parts in both of them is $16$, $x$ and $x-6$ are even and one of $y$ and $y-1$ is odd. After also noticing $x\no... | The first one is easy. You need $(x-8)(x+2)$ divisible by $512$. Both $x+8$ and $x-2$ are even, but only one of them can have higher powers of $2$ as factors. Thus, $x\equiv 8\pmod {256}$ or $x\equiv -2\pmod{256}$.
The second requires more care.
First assuming $y\equiv 0\pmod{16}$ you have $y=16y_0$ and $$16y_0^2=y_0+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Limit of function at infinity: $ \lim_{n\to \infty} \frac{a^n-b^n}{a^n+b^n} $ I'm learning the basics of Calculus. I got stuck in a very (seemingly) simple problem of evaluating a limit. The problem given is:
$$
\lim_{n\to \infty} \frac{3^n-2^n}{3^n+2^n}
$$
The answer given is $1$ (I have no idea how, or why).
I think ... | The other answers have already shown the standard trick. Here's how we might have proceeded if we hadn't noticed it applies.
One significant point we should understand is that as $n \to \infty$, $3^n$ dwarfs $2^n$ in size; the difference between $3^n$ and $3^n \pm 2^n$ should be negligible. So we estimate
$$ \frac{3^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugat... | Set $h=\dfrac1x$
$$4x^2+ax=\frac{4+ah}{h^2}\implies\sqrt{4x^2+ax}=\frac{\sqrt{4+ah}}{\sqrt{h^2}}$$
Now as $h\to0^+,h>0\implies\sqrt{h^2}=|h|=h$
$$\implies\lim_{x \to \infty} (\sqrt{4x^2+5x} - \sqrt{4x^2+x})=\lim_{h\to0^+}\frac{\sqrt{4+5h}-\sqrt{4+h}}h$$
$$=\lim_{h\to0^+}\frac{4+5h-(4+h)}{h(\sqrt{4+5h}+\sqrt{4+h})}=\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
find $\sum_{i=1}^\infty \frac{1}{n3^n}$ How to find $$\sum_{i=1}^\infty \frac{1}{n3^n}$$
Don't know how to start, any hints
A rigorous proof is also welcome
| Consider
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}.
\end{align*}
This series converges for all $|x|<1$. Differentiation wrt $x$ yields
\begin{align*}
\sum_{n=1}^\infty x^{n-1}=\sum_{n=0}^\infty x^n=\frac{1}{1-x},
\end{align*}
and integrating again gives
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}=\int\frac{1}{1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there... | More generally:
Given a quartic equation with the form
$x^4+bx^3+cx^2+(br)x+r^2=0$
you guarantee factors with the form
$(x^2+px+r)(x^2+qx+r)=0$
For then the cubic and linear terms are both matched by rendering
$p+q=b$
and we have only to match the quadratic terms which oeadsto
$pq=c-2r^2$
From these equations $p$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
"answer_id": 3
} |
Find Minimum Value of: $P=x^2+y^2+2xy+2-\frac{1}{xy}$ Given: $x,y>0$ and $x^2y-x+xy^2-y-3xy=0$
Find Minimum Value of:
$P=x^2+y^2+2xy+2-\frac{1}{xy}$
I found $\min P =\frac{71}{4}$ at $x=y=2$ but I cant prove that
Could some one help me ?
| It is sure that, without using Lagrange multipliers, the problem is quite hard.
But let us try using brute force. The constraint $x^2y-x+xy^2-y-3xy=0$ gives a quadratic equation in $y$; so we can express $y$ as a function of $x$ $$y_{\pm}=\frac{-x^2+3 x+1\pm\sqrt{x^4-6 x^3+11 x^2+6 x+1}}{2 x}$$ Replace $y_+$ in $P$ (wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How prove this inequality $(y-z)^2>8(y+z)$
For three distinct postive integer $x,y,z$ such
$$(x+y)(x+z)=(y+z)^2$$
show that
$$(y-z)^2>8(y+z)$$
My some idea: since
$$x^2+(y+z)x+yz-(y+z)^2=0$$
so
$$\Delta_{x}=(y+z)^2-4yz+4(y+z)^2=5(y+z)^2-4yz=m^2$$then I can't
| WLOG $y<z$, so let $z=y+a$, where $a>0$. Then the discriminant is
\begin{align*}5(y+z)^2-4yz&=5(2y+a)^2-4y(y+a)=5(4y^2+4ya+a^2)-4y^2-4ya\\
&=4(4y^2+4ya+a^2)+a^2=4(2y+a)^2+a^2=(4y+2a)^2+a^2\end{align*}
So if it's a square, then it's either the next square
$$(4y+2a+1)^2=(4y+2a)^2+2(4y+2a)+1$$
and
$$a^2=4(2y+a)+1\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Deriving inverse matrix formula If matrix $A$ is given with dimensions $2 \times2 $ then,
A is invertible if, and only, if $ad - bc \neq 0$:
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
How can this be derived?
I just need hints, I am not ... | Notice that
$$\begin{align}\begin{bmatrix}a & b\ |\ 1 & 0 \\c & d \ |\ 0 & 1\end{bmatrix} &\Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & \frac{ad - bc}{a} \ &|\ -\frac{c}{a} & 1\end{bmatrix} \Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & 1 \ &|\ -\frac{c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\\&\Leftrig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Efficient Way of finding possible sums in equation I am looking for an efficient way to find the following numbers numbers $a, b$
Let $x$ be 63 (as example)
$$\sum_{k = a}^b k = \sum_{l = 1}^b l - \sum_{j=1}^{a-1} j = \frac{b*(b+1)}{2}-\frac{a*(a-1)}{2} = x = 63$$
in this case for example
$$63 = 8+9+10+11+12+13 = \sum... | Given positive integer $x$, you want positive integers $a,b$ such that
$$ 2x = b(b+1) - a(a-1) $$
If $c = 2a-1$ and $d=2b+1$, this says
$$ (d-c)(d+c) = d^2 - c^2 = 8x $$
(with $c$ and $d$ both odd).
So look at factorizations of $8x$. If $8x = uv$ where $u$ and $v$ are positive integers, $u < v$,
we want $d-c = u$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.
Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$,
then $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$ y^2+ e^{2x}-2(e^x)y= 1$$
So if ... | $$y=\log(x+\sqrt{x^2+1})\implies x+\sqrt{x^2+1}=e^y\implies x^2+1=e^{2y}-2xe^y+x^2\implies$$
$$2e^yx=e^{2y}-1\implies x=\frac{e^{2y}-1}{2e^y}=\frac{e^y-e^{-y}}2=\sinh y (=\text{hyperbolic sine})$$
Also:
$$f'(x)=\left(1+\frac x{\sqrt{x^2+1}}\right)\frac1{x+\sqrt{x^2+1}}=\frac1{\sqrt{x^2+1}}$$
Thus, by the theorem of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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integration using partial fraction with repeating denominator So i need to integrate this $1/[(x^3)(x-1)]$, that means it could be decomposed into:
$$A/x + B/x^2 + C/x^3 + D(x-1)$$
Furthermore the resulting equation would then be:
*
*$1 = A(x^2)(x-1) + Bx(x-1) + C(x-1) + Dx^3$
*if i let $x=0$ then $C=-1$
*if i let... | Here's an easy way to do it:
$$\frac{1}{x^3(x-1)}=\frac{Ax^2+Bx+C}{x^3}+\frac{D}{x-1}$$
I chose to start at $x^2$ for the first one since the degree of the denominator is $3$, and on the other $x^0$ because the denominator has degree $1$ (so go all the way down starting with $x^{\text{denominator degree}-1}$).
Combine ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Permutation in discrete math Is the permutation
$$\begin{pmatrix} 1& 2 &3 &4 &5 &6&7 \\ 7 & 4 & 2 & 1 & 3 & 6 & 5 \end{pmatrix}$$
even or odd?
The product of disjoint cycles is $$\begin{pmatrix}1& 7&5&3&2&4\end{pmatrix}\begin{pmatrix}6\end{pmatrix}$$
and the transposition are $$\begin{pmatrix}1 & 7 \end{pmatrix}\begin... | Another way to check the parity of a permutation is to see how many pairs are out of order (this is the number of inversions). There are $11$:
$$
\{(7,4),(7,2),(7,1),(7,3),(7,6),(7,5),(4,2),(4,1),(4,3),(2,1),(6,5)\}
$$
Thus, this is an odd permutation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be? If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be?
I have seen such problems many times but no specific method to solve them.
I hope I find the general rule to solve them here.
| since
$$2^{x+2y}=2^x\cdot 4^y=9\cdot 27=3^5$$
and note
$$2^{2y-4x}=4^y/(2^x)^4=27/9^4=3^{-5}$$
so
$$2^{x+2y}\cdot 2^{2y-4x}=2^{(x+2y)+(2y-4x)}=3^5\cdot 3^{-5}=1=2^{0}$$
so $$\Longrightarrow x+2y+2y-4x=0$$
$$\Longrightarrow x+2y=-(2y-4x)$$
so
$$\Longrightarrow \dfrac{x+2y}{2y-4x}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{... | To evaluate the second series note:
$$ e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$
$$ \sum_{n=1}^\infty\frac{1}{(n+1)!}=\frac{1}{2!}+\frac{1}{3!}+\dots$$
Thus the second series is just $e-2$. If you combine the two series you get $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 8
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If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem :
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$
Solution :
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightar... | Let $f$ be an integer polynomial with $f(x)f(1/x) = f(x) + f(1/x)$. There are four solutions, $f = 0$, $f = 2$, and $f = 1 \pm x^n$.
Clearly $f = 0$ is a solution.
As in the question, we have:
$$ (f(x) - 1)(f(1/x) - 1) = 1 $$
Let $n$ be the degree of $f \neq 0$. Multiplying both sides of the above by $x^n$ gives:
$$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| Knowing that $(x+2)$ is a factor, you know that the solution will be of the form
$$x^3+4x^2+x-6=(x+2)(ax^2+bx+c).$$
Execute the product
$$(x+2)(ax^2+bx+c)=ax^3+(2a+b)x^2+(2b+c)x+(2c).$$
Identify the coefficients
$$a=1,2a+b=4,2b+c=1,2c=-6.$$
Solve and get
$$a=1,b=2,c=-3,$$
so that
$$x^3+4x^2+x-6=(x+2)(x^2+2x-3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Limit of $\left(\frac{x^2+x+1}{2x+1}\right)^{1/(x^2-1)}$ when $x\to1$ $$\displaystyle \lim_{x\to 1}\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$
I know that the two-sided limit of $\frac{1}{x^2-1}$ does not exist. I don't know what to do with $\frac{x^2+x+1}{2x+1}$ to get something else than $1^\infty$
| Let $y=x-1$. Then we have $$\require{cancel}\lim_{y\to0}\left(\frac{y^2+3y+3}{y^2-(y+1)^2+4y+4}\right)^{\frac{1}{y(y+2)}}=\lim_{y\to0}\left(\frac{y^2+3y+3}{2y+3}\right)^{1/(2y)}=\\ \lim_{y\to0}\left(\frac{\cancel{y}(y+3+3/y)}{\cancel{y}(2+3/y)}\right)^{1/(2y)},$$ which, letting $z=1/y$, yields $$\lim_{z\to\infty}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where does $r = 1 + 2\cos(\theta)$ have tangents? Where does:
$$r = 1 + 2\cos(\theta)$$
Have horizontal and vertical tangent lines?
$x = r\cos(\theta) = \cos(\theta) + 2\cos^2(\theta)$
$y = r\sin(\theta) = \sin(\theta) + 2\sin(\theta)\cos(\theta)$
$$dx/d\theta = -\sin(\theta) - 4\sin(\theta)\cos(\theta)$$
$$dy/d\theta ... | Note that $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1$ so
$$\frac{dy}{d\theta} = 0 \Leftrightarrow \cos\theta + 4\cos^2 \theta - 2 = 0$$
Now compute the roots of $4x^2+x-2$ to get equations for $\cos \theta$ and apply $\arccos$ to find $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do I need to use partial fractions to find $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$? I need to simplify $\frac{2n+1}{n^2(n+1^2)}$ as part of an exam question. The solution states $$\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$$
In the solution it does not state how this simplification w... | Note that $$\frac{2n+1}{n^2(n+1)^2} = \frac{(n^2 + 2n + 1)-n^2}{n^2(n+1)^2}$$
$$=\frac{(n+1)^2-n^2}{n^2(n+1)^2}$$
$$=\frac{(n+1)^2}{n^2(n+1)^2}-\frac{n^2}{n^2(n+1)^2}$$
$$=\frac1{n^2}-\frac1{(n+1)^2}$$
as desired. This approach is motivated by noticing that the original numerator is "almost" a recognizable perfect squa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Odd Primes Problem Proof Given the odd prime numbers,
Prove that if $x$ and $y$ are adjacent odd primes in this list, then $x + y$ has $3$ prime factors. The factors need not be distinct.
Here is an example I have provided:
$3 + 5 = 8 = 2 \cdot 2 \cdot 2$. Therefore, $8$ has $3$ repeated factors of $2$.
| Suppose $x<y$ are adjacent odd primes. Then $x+y = 2a$, $a\in\mathbb{Z}$.
$x<y \Rightarrow 2x<x+y<2y \Rightarrow 2x<2a<2z \Rightarrow x<a<y$
How $x$ and $y$ are adjacent primes, then $a$ is not prime. Suppose $a=b\cdot c$, with $b, c>1$.
So $x+y = 2\cdot b\cdot c$. Therefore $x+y$ has at least 3 prime factors
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it i... | $$\frac1n\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{1+x^2}\left(=\frac\pi4\;.\;\;Why?\right)$$
Just note that the above sum is
$$\sum_{k=1}^n f\left(\frac kn\right)(x_{k+1}-x_k)\,,\,\,\text{with}\;\;f(x)=\frac1{1+x^2}\;$$
and the partition of the unit interval
$$\left\{0<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How prove this inequality $\frac{e^x}{x+1}>\frac{\cos{x}}{\sin{x}+\sqrt{2}}$ Today,when I use wolf found this following inequality
let $x>-1$, show that
$$\dfrac{e^x}{x+1}>\dfrac{\cos{x}}{\sin{x}+\sqrt{2}}$$
I found this
I want $$\Longleftrightarrow (\sin{x}+\sqrt{2})e^x-(x+1)\cos{x}>0$$
let
$$f(x)=(\sin{x}+\sqrt{... | We know $\sqrt{2}+\sin x > 0$ since $|\sin x| \le 1$ and $\sqrt{2} > 1$.
$\displaystyle \frac{e^{x}}{x+1} \ge 1 \ge \frac{\cos x}{\sqrt{2}+\sin x}$, the later inequality is true since,
$1 \ge \sin (\frac{\pi}{4} - x) = \dfrac{\cos x - \sin x}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
\frac{x}{y} + \frac{y}{x} = 1
\end{align}
Here is what I did:
\begin{align}
&\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\... | we get by the quotient and the chaine rule
$$\frac{y-xy'}{y^2}+\frac{y'x-y}{x^2}=0$$ multiplying by $x^2y^2$ we obtain
$$x^2y-x^3y'+y'y^2x-y^3=0$$
solving for $y'$ we get
$$y'=\frac{x^2y-y^3}{x^3-xy^2}$$ if $x^3-y^2x\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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find a function satisfying the recurrence find a function satisfying the recurrence
$$F (n) = 2F (\sqrt{n}) + 1$$
replace $n$ by $2^m$
Thus getting the answer as $$F(n)=\frac{1}{2}c \log(n) + \log(n) - 1$$
Is this correct
| Looks good:
\begin{align}
2 F(\sqrt{n}) + 1
&=
2\left(
\left(\frac{c}{2} + 1 \right) \log(\sqrt{n}) - 1
\right) + 1 \\
&=
2\left(
\left(\frac{c}{2} + 1 \right) \frac{1}{2} \log(n) - 1
\right) + 1 \\
&=
\left(\frac{c}{2} + 1 \right) \log(n) - 1 \\
&= F(n)
\end{align}
You could replace the constant factor $c/2 + 1$ by a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}... | $\dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2n(n+1)} - \dfrac{1}{2(n+1)(n+2)}$
therefore,
$$\sum_{n = 1}^{\infty} \dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2*1*2} = \dfrac{1}{4}$$
we used the telescoping series formula $\sum_{n=1}^\infty (f(n) - f(n+1)) = f(1) - \lim_{n \to \infty} f(n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn... | Copied from my answer to this near-duplicate:
Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1} \stackrel{n\to\infty}\longrightarrow 1$$
This even allows you to obtain the v... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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recurrence relation with variable coefficients How to solve recurrence relation $y_n= y_{n-1}+ (n-1).y_{n-2}$ where $n$ is a variable ?
$y_n$ is a $n$th term, $y_{n-1}$ is $(n-1)$th term and $y_{n-2}$ is $(n-2)$th term.
| Hint
Let $y_{0}=a,y_{1}=b$,and let
$$f(x)=\sum_{n=0}^{\infty}y_{n}x^n$$
then we have
$$y_{n}x^n=xy_{n-1}x^{n-1}+x^2(n-2)y_{n-2}x^{n-2}+y_{n-2}x^n$$
$$\Longrightarrow \sum_{n=2}^{\infty}y_{n}x^n=x\sum_{n=2}^{\infty}y_{n-1}x^{n-1}+x^2\sum_{n=2}^{\infty}(n-2)y_{n-2}x^{n-2}+x^2\sum_{n=2}^{\infty}y_{n-2}x^{n-2}$$
$$\Longrig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequalites of triangle side with $abc = 1$ Let $a,b,c$ be the sides of a triangle with $abc=1$.
Prove that $$ \frac{\sqrt{b+c−a}}{a} + \frac{\sqrt{c+a-b}}{b} + \frac{\sqrt{a+b−c}}{c} \ge a+b+c $$
| By Holder $\left(\sum\limits_{cyc}\frac{\sqrt{b+c-a}}{a}\right)^2\sum\limits_{cyc}a^2(b+c-a)^2\geq(a+b+c)^3$.
Hence, it remains to prove that $abc(a+b+c)\geq\sum\limits_{cyc}a^2(b+c-a)^2$, which is
$\sum\limits_{cyc}(a-b)^2(a+c-b)(b+c-a)\geq0$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Triple Integral of $1/\sqrt{2 + x^2 + y^2 + z^2}$ over unit sphere I'm studying triple integrals (physics major), and I'm having trouble solving this little beast:
$$
\iiint_V \frac{1}{\sqrt{2 + x^2 +y^2 +z^2}} \,dx\,dy\,dz$$
where V is $$x^2+y^2+z^2=1$$
Of course we use spherical coordinates:
$$I = \iiint_V \frac{r^2... | Write
$$\frac{r^2}{\sqrt{a^2 + r^2}} = \sqrt{a^2 + r^2} - \frac{a^2}{\sqrt{a^2 + r^2}}$$
These two terms are now standard integrals, equal respectively to $\displaystyle \frac{1}{2} \left( r \sqrt{a^2 + r^2} + a^2 \ln(\sqrt{a^2 + r^2} + r) \right)$ and $-a^2 \cdot \ln(\sqrt{a^2 + r^2} + r)$
Hence
$$\int_0^1 \frac{r^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that {$r_n$} satisfies $r_n = 7r_{n-1} - 10r_{n-2}, n \geq 2$ Problem: For the sequence $r$ defined by
$$r_n = 3 \cdot 2^n - 4 \cdot 5^n, \ \ \ n \geq 0$$
Prove that {$r_n$} satisfies
$$r_n = 7r_{n-1} - 10r_{n-2}, \ \ \ n \geq 2$$
Can this problem be explained and broken down and show the process? I'd like to ... | To show that $r_n = 3\cdot 2^n - 4\cdot 5^n$ satisfy the equation then start by writing the equation as
$$r_n - 7r_{n-1} + 10r_{n-2} = 0$$
Now calculate $r_n, r_{n-1}$ and $r_{n-2}$ from the formula. This gives us
$$r_n = \color{red}{3\cdot 2^n - 4\cdot 5^n}$$
$$r_{n-1} = 3\cdot 2^{n-1} - 4\cdot 5^{n-1} = \color{blu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$ How would I go about solving this following limit?
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
My attempts:
Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.
I don't see any clever sub... | Let's proceed in the following manner $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x}\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x}\right) - e}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x} - 1\right) - 1}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp z - 1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
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Triple Angle Condition Let $ABC$ be a triangle with integral side lengths such that $\angle A=3\angle B$. Find the minimum value of its perimeter.
Essentially we want sinb, sin3b, sin4b to have rational ratios (manipulate the sine law). But then we don't actually need sinb to be rational, so that's pretty mysterious. A... | Let $x = \sin\angle B$. Using the identity $\sin 3\theta = -4\sin^3 \theta + 3\sin\theta$, $\sin\angle A = -4x^3+3x$.
Then $\sin\angle C = \sin (\pi - \angle B - \angle A) = \sin 4\angle B = 4\sin\angle B\cos\angle B(1-2\sin^2 \angle B) =4x\sqrt{1-x^2}(1-2x^2) $.
By the law of sines, $\frac{a}{b} = \frac{\sin\angle A}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ an... | As has already been explained at least twice, the best functions follow this pattern:
a continuous function $f$ with $f(0)=f(1)=0$
that approximates $y = h(x) = 1-\frac\pi8+\sqrt{x-x^2}$ for $0<x<1.$
I propose a family of functions for $n$ a positive integer,
$$f_n(x) = \sqrt{x-x^2} + \left(1-\frac\pi8\right)g_n(x),$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
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"answer_id": 5
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Find all primes $p$ such that $p^2-p+1$ is a perfect cube Find all primes $p$ such that $p^2-p+1$ is a perfect cube.
I found out that p is of the form $18n+1$ and $p=19$ is a solution but I am not getting anything further.
$p^2-p-(m^3-1)=0$
$1+4(m^3-1)=k^2$
$4m^3-3=k^2$
every square is $0,1,4 or 7(\mod9)$ and every cub... | you can work in $Z[1/2+\sqrt{-3}/2]$.we have unique factorization in this ring,and units are $1,-1,1/2+\sqrt{-3}/2,1/2-\sqrt{-3}/2$,and 2 is prime,yo can easily check this facts by the norm
$N(a+b\sqrt{-3})=a^2+3b^2$
we have
$4m^3=k^2+3=(k-\sqrt{-3})(k+\sqrt{-3})$,we can see $gcd((k-\sqrt{-3}),(k+\sqrt{-3})|2\sqrt{-3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$x^2+3$ has two zeros over ${\Bbb F}_p$ provided that $x^2+x+1\in{\Bbb F}_p[x]$ has two? The following is an exercise in abstract algebra:
If $p=1\pmod{3}$, then $x^2+x+1\in\Bbb{F}_p[x]$ has two zeros. Prove in this case that $-3$ is a quadratic residue mod $p$.
Showing that $(x^2+x+1)\mid (x^p-x)$, one can get the... | Complete the square. We have $x^2+x+1\equiv 0\pmod{p}$ iff $4x^2+4x+4\equiv 0\pmod{p}$ iff $(2x+1)^2\equiv -3\pmod{p}$.
Remark: Let $F$ be any field that has characteristic $\ne 2$. Then we can make precisely the same argument: We have $x^2+x+1=0$ if and only if $4x^2+4x+4=0$ if and only if $(2x+1)^2=-3$. So if $x^2+x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$ I need integer solutions of $x^2 + y^2 = z^2 + w^2$ parametrized. Can it be done? Thanks.
| First let us solve equation
$$
xy-zt=0.
$$
Let $(x,y,z,t)$ be any solution with $x\neq 0$, and let $u=\text{gcd}(x,z)>0$. Then $x=uv$ and $z=uw$ with $v,w$ coprime and $v\neq 0$. Then $xy-zt=0$ implies that $uvy=uwt$, or $vy=wt$. Because $v$ and $w$ are coprime, this implies that $t$ is divisible by $v$, so we can writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.