Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | Hint:
Find the Jordan matrix, by which the exponential can always be found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
prove that $(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$
For $a, b, c$ positive reals prove that $$(ab+bc+ca)\bigg(\frac{a}{b(a^2+2b^2)}+\frac{b}{c(b^2+2c^2)}+\frac{c}{a(c^2+2a^2)}\bigg) \ge 3$$
I used the Cauchy-Swartz inequality in the LHS so I was left to prove tha... | Let $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$.
Hence, by Holder $(ab+ac+bc)\sum\limits_{cyc}\frac{c}{a(c^2+2a^2)}=(x+y+z)\sum\limits_{cyc}\frac{x^2}{y(x^2+2z^2)}=$
$=(x+y+z)\sum\limits_{cyc}\frac{x^3}{xy(x^2+2z^2)}\geq\frac{(x+y+z)^4}{3\sum\limits_{cyc}(x^3y+2x^2yz)}$.
Thus, it remains to prove that $(x+y+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I Show that, $\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{c\sqrt2}\right)$ Where $c^2=a^2+b^2$ is Pythagoras theorem. Sides a,b and c are of a right angle triangle.
Show that,
$$\arcsin\left(\frac{b}{c}\right)-\arcsin\left(\frac{a}{c}\right)=2\arcsin\left(\frac{b-a}{... | *
*$\arcsin \dfrac bc -\arcsin \dfrac ac =2\arcsin \dfrac{b-a}{c \sqrt 2}$
*$a,b,c \gt 0$
*$a^2 + b^2 = c^2$
Let's rewrite this as
$\dfrac 12 \left(\arcsin \dfrac bc -\arcsin \dfrac ac \right)
=\arcsin \dfrac{b-a}{c \sqrt 2}$
Since you said that sides $a, b, $ and $c$ are sides of a right triangle, then we know th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divisible by $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}$
Prove that
$$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$$
is divisible by
$$(a+b+c)^{3}-a^{3}-b^{3}-c^{3},$$
where $a,,b,c -$ integers, such that $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}\not =0$
My work so far:
$(a+b+c)^{3}-a... | Hint $1:$
Let $f(a)=(a+b+c)^{333}- a^{333} -b^{333} -c^{333}.$
$f(-b)=(-b+b+c)^{333} -(-b)^{333} -b^{333} -c^{333}=0$
So, by Factor theorem we get $(a+b)$ as a factor of the given expression.
Similarly, $(b+c)$ and $(c+a)$ are factors of the given expression.
Hint $2:$ Now, to prove the divisiblity of the expression b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x... | Without using L'Hospital,
$$\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}x$$
$$=-\lim_{x\to0}x^n\cdot\lim_{x\to0}\frac{\ln(1-x^{n+1})}{-x^{n+1}}+\lim_{x\to0}\dfrac{\ln(1-x)}{-x}$$
Use $\lim_{y\to0}\dfrac{\ln(1+u)}u=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding eigenvectors of a 3x3 matrix Struggling with this eigenvector problems. I've been using this SE article (Finding Eigenvectors of a 3x3 Matrix (7.12-15)) as a guide and it has been a very useful, but I'm stuck on my last case where $\lambda=4$.
Q: Find the eigenvalues $\lambda_1 < \lambda_2 < \lambda_3$ and cor... | The equations corresponding to that row-reduced form at the end are
$$
x - y/2 = 0 \\
y + 2z = 0
$$
Since $z$ is a free variable, you can pick $z = 1$ and back-substitute to get
$y = -2$, and then use this in the first equation to get $x = 1$. Of course, any nonzero multiple of this vector is also an eigenvector for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Largest root as exponent goes to $+\infty$ Let $a\geq 1$ and consider
$$
x^{a+2}-x^{a+1}-1.
$$
I am interested to see what is the largest root of this polynomial as $a\to +\infty$.
In order to find a root, we surely have to have
$$
x^{a+2}-x^{a+1}=x^{a+1}(x-1)=1.
$$
Hence, I guess we have to look for which $x$ we have... | I will show,
by elementary means that
the root is beteween
$1+\dfrac1{\sqrt{a+1}}$
and
$1+\dfrac1{a+1}$.
Since the root of
$x^{a+2}-x^{a+1}-1
$
is close to $1$,
let
$x = 1+y$.
Then we want
$1
=(1+y)^{a+2}-(1+y)^{a+1}
=(1+y)^{a+1}(1+y-1)
=y(1+y)^{a+1}
$.
Since
$(1+y)^{a+1}
\gt 1+y(a+1)
$,
$1
\gt y(1+y(a+1))
\gt y^2(a+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$
Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$
Attempt: If $m>n$ where $m, n$ are natural numbers,
$|a_m-a_n|=|(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2m})... | Note that
$$|a_m - a_n| = \sum_{k=n+1}^m \frac{1}{2k} > \frac{m-n}{2m} > \frac{1}{2}(1 - n/m).$$
For any $n \in \mathbb{N}$ we can choose $m > 2n$ and find
$$|a_m - a_n| > \frac1{4}.$$
Thus, it is not the case that for any $\epsilon > 0$ there exists $N$ such that $|a_m-a_n| < \epsilon$ for all $m > n > N$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$
$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{... | Hint;
Assume
$$ \theta = (n\pi)/7$$
$$\implies 4\theta = n\pi -3\theta$$
Take the sine on both the sides
An apply the required manipulations
You should get a cubic in $\sin^2 \theta$
Since we have defined out $\theta$, we are quite aware what the roots are going to be
So the numerical part in the end divided by the coe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP... | If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have
$$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$
from which:
$$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$
and
$$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Estimation the integration of $\frac{x}{\sin (x)} $ How to prove that integral of $\frac{x}{\sin (x)} $ between $0$ to $\frac{\pi}{2}$ lies within the interval $\frac{\pi^2}{4}$ and $\frac{\pi}{2}$ ?
| $$f'(x)=\frac{\sin x-x\cos x}{\sin^2x}=\frac{\cos x(\tan x-x)}{\sin^2x}$$
The function is increasing in the given domain since $\tan x>x$ for $x\in\left(0,\frac{\pi}{2}\right)$.
When $x\to 0$, $\frac{x}{\sin x}=1$ and at $x=\frac{\pi}{2}$, $\frac{x}{\sin x}=\frac{\pi}{2}$.
The minimum area is the area of rectangle with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that:
If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.
I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases}
a^4+a^3-1=0\\
b^4+b^3-1=0
\end{cases}$
whic... | The companion matrix of $p(z) = z^4 + z^3 - 1$ is
$$ A = \pmatrix{0 & 0 & 0 & 1\cr
1 & 0 & 0 & 0\cr
0 & 1 & 0 & 0\cr
0 & 0 & 1 & -1\cr}$$
This is a matrix whose eigenvalues are the roots of $p(z)$. The Kronecker product $A \otimes A$ is a $16 \times 16$ matrix; all produ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| Note that for $n\ge 2$ we have
$$\frac{n(n+1)}{n!}= \frac{1}{(n-2)!} + \frac{2}{(n-1)!}. $$
This lets us rewrite the sum as
$$
\frac{2}{0!} + \left(\frac{1}{0!} + \frac{2}{1!} \right)
+ \left(\frac{1}{1!} + \frac{2}{2!} \right)
+ \left(\frac{1}{2!} + \frac{2}{3!} \right)
+ \left(\frac{1}{3!} + \frac{2}{4!} \right)
+ \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Is there an easy way of showing $8\cos^3(12^o)-6\cos(12^o)=\phi$? Is there an easy way of showing (1)
(1)
$$8\cos^3(12^o)-6\cos(12^o)=\phi$$
with out substituting into the equation?
$8\left(\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]\right)^3-6\cdot\frac{1}{8}\left[\sqrt{6(5+\sqrt5)}+\sqrt5-1\right]=\phi$
Trig... | Recall that $$\cos 3x=4\cos^3 x- 3\cos x$$
$$\implies 2(4\cos^3 12^{\circ}-3\cos 12^{\circ})=2\cos36^\circ$$
Using $$\cos 36^\circ=\frac{1+\sqrt{5}}4$$
We get the that
$$8\cos^3 12^{\circ}-6\cos 12^{\circ}=\frac{1+\sqrt{5}}2=\phi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6$ for $a^2 + b^2 + c^2 + d^2 = 1$. For $a, b, c, d \in \Bbb R$ such that $a^2 + b^2 + c^2 + d^2 = 1$, show that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6.$$
The answer uses the mysterious identity $... | Note that $(a+b)^4+(a-b)^4=2a^4+12a^2b^2+2b^2$. Hence
$\begin{align}(a + b)^4 &+ (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4\\
&+ (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4\\
&=6(a^4+b^4+c^4+d^4)+12(a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2)\\
&= 6(a^2 + b^2 + c^2 + d^2)^2.\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ Examine convergence of $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx$ for $a, b > 0$. There are 2 problems. $|\sin(x)|^b = 0$ for $x = k \pi$ and $x^a = 0$ for $x = 0$. We can write $\int_0^{\infty} \frac{1}{x^a \cdot |\sin(x)| ^b}dx = \i... | Note that $\lim_{x \to 0} x^{-1} \sin x = 1$ and on $[0,1]$ we have $\sin 1 \leqslant x^{-1} \sin x \leqslant 1.$
Hence, with $a,b > 0$,
$$(\sin 1)^b x^{a+b} \leqslant x^a |\sin x|^b = x^{a+b}|x^{-1} \sin x |^b \leqslant x^{a+b},$$
and
$$\frac{1}{x^{a+b}} \leqslant \frac{1}{x^a |\sin x|^b} \leqslant \frac{(\sin 1)^{-b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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How to solve the equation $(2x + 1)(2x + 3) = 143$ without using the Quadratic Formula? I have been a bit stuck on this question.
The product of two consecutive odd numbers is $143$. Find the next numbers.
I have made this into:
$$
(2x+1)(2x+3)=143.
$$
I got $x_1 = -7$ and $x_2 = -5$ for this, which doesn't seem right.... | \begin{align*}
(2x + 1)(2x + 3) & = 143\\
2x(2x + 3) + 1(2x + 3) & = 143 && \text{expand}\\
4x^2 + 6x + 2x + 3 & = 143 && \text{apply the distributive law}\\
4x^2 + 8x + 3 & = 143 && \text{combine like terms}\\
4x^2 + 8x - 140 & = 0 && \text{subtract $140$ from each side of the equation}\\
x^2 + 2x - 35 & = 0 && \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 ... | A sphere with equal volume to a cube of side $a$ must have radius $r$:
$$\frac{4\pi r^3}{3} = a^3$$
So
$$r = \sqrt[3]{\frac{3}{4 \pi}}a$$
Now just take
$$\frac{4\pi r^2}{6a^2} = \frac{4\pi\sqrt[3]{\frac{9}{16 \pi^2}}a^2}{6a^2} = \frac{2}{3}\pi\sqrt[3]{\frac{9}{16 \pi^2}} = \sqrt[3]{\frac{9\cdot 8 \pi^3}{27 \cdot 16 \pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the prob... | Hint
$$\frac{a^2}{a^4+a^2+1}=\frac{a}{
(a^2+a+1)}\frac{a}{(a^2-a+1)}=\frac{1}{(\frac{a^2+1}{a}+1)}\frac{1}{(\frac{a^2+1}{a}-1)}=\frac{1}{((\frac{a^2+1}{a})^2-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Simplifying sum of powers of conjugate pairs The result of summing a conjugate pair of numbers each raised to the power $n$:
$$
(a + bi)^n + (a - bi)^n
$$
Produces a real number where $a + bi$ is a complex number.
Given the result is real, is there a simplified way to express the above expression in terms of $a$ and $b... | By De Moivre's Formula we have, for $a+bi=r(\cos \theta+i\sin \theta)$
$$(a+bi)^n+(a-bi)^n=\\r^n(\cos n\theta+i\sin n\theta)+r^n(\cos (-n\theta)+i\sin (-n\theta))=\\2r^n\cos n\theta$$
Since $\cos n\theta=\cos (-n\theta)$ and $\sin n\theta=-\sin(-n\theta)$
Now we use the fact that $$r=\sqrt{a^2+b^2}$$ and
$$L(a,b)=\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Limit of $\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}$ when $n\to\infty$ I have to show the convergence of the series
$$\lim\limits_{n \to \infty}a_n=\sum \limits_{k=1}^{n} \frac{2^kn+2n^2+k}{2^{k+1}n^2+2^kk}.$$ I am quite sure that the limit is 1.5. I wanted to show this by estimation for another series... | break it up.
$a_n=\sum \limits_{k=1}^{n} \frac{2^kn}{2^{k+1}n^2+2^kk} + \sum\limits_{k=1}^{n} \frac{2n^2}{2^{k+1}n^2+2^kk}+\sum\limits_{k=1}^{n} \frac{k}{2^{k+1}n^2+2^kk}.$
Since every element of each of those series is greater than zero,
the series for $a_n$ converges iff all three of the above converge, and diverges... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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All real values of $k$ in rational function
All real values of $k$ for which the range of function $\displaystyle f(x) = \frac{x-1}{k-x^2+1}$ does not contain the interval $\displaystyle \left[-1,-\frac{1}{3}\right].$
$\bf{}My\; Try::$ Let $\displaystyle y = \frac{x-1}{k-x^2+1}\Rightarrow ky-x^2y+y=x-1\Rightarrow x^2... | As you did, considering the discriminant, we get
$$(4k+4)y^2+4y+1\ge 0\tag1$$
Now, let us separate it into cases.
Case 1 : $k\lt -1$
$$(1)\iff (-4k-4)y^2-4y-1\le 0\iff \frac{1-\sqrt{-k}}{2(-k-1)}\le y\le \frac{1+\sqrt{-k}}{2(-k-1)}$$
So, we need
$$\frac{1+\sqrt{-k}}{2(-k-1)}\lt -1\quad\text{or}\quad \frac{1-\sqrt{-k}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that the last digit of $4n^5-5n^2+n$ is $0$
Prove that the last digit of $4n^5-5n^2+n$ is $0$ for all natural $n$
My attempt:
$$4n^5-5n^2+n\overset{?}\equiv 0\pmod {10}$$
using Fermat's little theorem
$$4n\cdot\pmod 5 -5n\pmod 2+n\overset ?\equiv 0 \pmod{10}$$
I am suck here
| You want to prove that $4n^5-5n^2+n\equiv 0\pmod {10}$.
So prove that $4n^5-5n^2+n\equiv 0\pmod {2}$, which is equivalent to $-n^2+n\equiv 0$, which is always true because $n^2\equiv n\pmod {2}$.
Then prove that $4n^5-5n^2+n\equiv 0\pmod {5}$, which is equivalent to $-n^5+n\equiv 0$, which is always true because $n^5\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite integration of an algebraic expression
Evaluate $$\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}$$
I think none of the properties of definite integral will be useful here so I think I will have to integrate. But I am unable to do so. Some hints on how to integrate. Thanks.
| Mathematica gives the indefinite integral:
$\int {1-x \over 1+x}{1 \over \sqrt{x+x^2+x^3}} dx =$
$-\frac{2 \sqrt[6]{-1} x^{3/2} \left(i \sqrt{1-\frac{(-1)^{2/3}}{x}}
\sqrt{\frac{x+\sqrt[3]{-1}}{x}} F\left(i \sinh
^{-1}\left(\frac{(-1)^{5/6}}{\sqrt{x}}\right)|(-1)^{2/3}\right)+\frac{2
\left(1+\sqrt[3]{-1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x.
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha g... | Let $z = \frac{1}{2}(n - x^2)$. Then the equation becomes $\Big(\frac{z}{x}\Big)^2 = z$ or better $z\Big(\frac{z}{x^2} - 1\Big) = 0$. Solutions are $z = 0$ and $z = x^2$. If $z = 0$, then $x = \pm \sqrt{n}$. If $z = x^2$, then $x = \pm \sqrt{n/3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you solve $x^2 - 4 \equiv 0 \mod 21$ There is an example in my textbook of how you solve:
$$ x^2 -4\equiv 0 \mod 21 \Leftrightarrow x^2-4\equiv 0 \mod 3 \times 7$$
and then 2 congruences can be formed out of this equation if:
$$x^2-4\equiv0 \mod 3 \\ x^2-4 \equiv 0 \mod 7$$
and from these 2 congruences result 2... | You need that $x^2\equiv 4\pmod{3}$ and $x^2\equiv 4\pmod{7}$. For instance, your $x_2=1$ doesn't satisfy $x_2^2\equiv 4\pmod{21}$.
You used or instead of and.
| {
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Number of solutions to $x+y+z = n$
If $\beta(n)$ is the number of triples $(x, y, z)$ such that $x + y + z = n$ and $0 \le z \le y \le x$, find $\beta(n)$.
Attempt:
I think there are many cases to look at to find $\beta(n)$. We know that the number of solutions without restriction is $\binom{n+2}{2}$, but we need one... | Let $a=z$, $b=y-z$, $c=x-y$. We are looking for the number of solutions of $c+2b+3a=n$ with $a,b,c\geq 0$, so the answer is given by:
$$[x^n](1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)$$
i.e. by:
$$ [x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)} \tag{1}$$
that can be recovered through partial fraction decompos... | {
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Work done in a conservative vector field If my vector field is:
$F=(1-\frac{x}{x^2+y^2})i-(\frac{y}{x^2+y^2})j$
How would I go about finding the work done between A(3,2) and B(4, -3)?
I have proved that the vector field is conservative:
$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=\frac{2xy}{(x^2+y^2)^2... | I'll elaborate the idea of @RobBland .
Let us find $f$, so that $\nabla f=\mathbf{F}$. We will do that as follows:
$$
f(x,y)=\int P(x,y)dx=\int\left(1-\frac{x}{x^2+y^2}\right)dx=x-\frac{1}{2}\ln(x^2+y^2)+c(y).\tag 1
$$
We assume $y$ to be constant while performing integration $(1)$. $f(x,y)$ satisfies $\frac{\partial f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does the Pearson correlation coefficient change under rotations I was reading on wikipedia about the pearson correlation coefficient. Assuming the data has zero mean it can be written as
$$
\rho = \frac{ \sum x_i y_i } {\sqrt{\sum x_i^2 \sum y_i^2}}
$$
The caption below this image says:
[...] Note that the ... | I suppose that $\sum_i x_i = \sum_i = y_i = 0$.
Moreover, $P_x = \sum_i x_i^2$, $P_y = \sum_i y_i^2$ and $C_{xy} = \sum_i x_i y_i$.
Then, the sample Pearson coefficient $\rho$ based on data $x_i$ and $y_i$ produced by random variables $X$ and $Y$ is:
$$\rho = \frac{C_{xy}}{\sqrt{P_x P_y}}.$$
Notice that:
$$P_{x'} = \su... | {
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Graphing $\frac{(x-3)}{(x^2-3x)}$ and $1/x$ For graphing the first equation, in the solved example given in the textbook, we proceeded as follows:
$\frac{(x-3)}{x^2-3x} \implies \frac{(x-3)}{x(x-3)} \implies \frac{1}{x}$ for $x \neq 3$.
Now the graph of this equation is exactly the same as that of $\frac{1}{x}$ except ... | $f(x)=\frac{x-3}{x^2-3x}=\frac{(x-3)}{x(x-3)}$, only when $x \neq 3$, as division by $0$ is not possible.
*
*If $x \neq 3$, then we can divide both sides by $(x-3)$ to get $f(x)=\frac{1}{x}$. For example, for $x=2$, we have $f(2)=\frac{(2-3)}{2(2-3)}=\frac{1}{2}$(obtained by cancelling $(2-3) \neq 0$ )
*If $x =3$, ... | {
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Orthogonal diagonalisation of a $4\times 4$ matrix Can somebody help me to orthogonally diagonalise the matrix $\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}$?
| The assiocated quadratic form is : $$q(a,b,c,d)=ad+bc+cb+da=2ad+2bc=\frac{1}{2}(a+d)^2-\frac{1}{2}(a-d)^2+\frac{1}{2}(b+c)^2-\frac{1}{2}(b-c)^2$$
So you can deduce that the eigen values are $1$ and $-1$, and by coming back to a matrix notation, you have :
$$\begin{bmatrix}0 & 0 & 0 & 1\\0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 ... | {
"language": "en",
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I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x ... | $4^{y+3x} = 64$
$\log_4 ({4^{y+3x}}) = \log_4 (64)$
$(y+3x)\log_4 {4} = 3$
$y + 3x = 3$
$\log_x (x+12) - 3\log_x (4) = -1$
$\log_x (\frac{x+12}{4^3}) = -1$
$\frac{x+12}{64} = x^{-1}$
$x^2 + 12x - 64 = 0$
$(x+16)(x-4) = 0$
$x = 4$, as $x > 1$ for $\log_x (x+12) - 3\log_x (4) = -1$
(Why must the base of a logarithm be a... | {
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"url": "https://math.stackexchange.com/questions/1807717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How did we derive this general term for the series? We have this series of numbers:
$1, 3, 6, 10, 15$
The general term can be described wit: $\frac{r(r + 1)}{2}$
Apparently the following series:
$1, 4, 10, 20, 35$
Can be described with $\frac{r(r + 1)(r + 2)}{6}$ based on the first series.
But I am not clear how this i... | $S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 1$
$S = 1 + 4 +10 + 20 + 35 \cdots + a_{n-1} + a_n \cdots - 2$
Subtracting 1 from 2
$0 = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1}) - a_n$
$a_n = 1 + ((4 - 1) + (10 - 4) + (20 - 10) + (35 - 20)+ \cdots + a_{n} - a_{n-1})$
$a_n =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Where did $\sqrt{x^2/x^2}$ come from in $\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} = \lim_{x \to -\infty}\frac{-1-1/x}{\sqrt{x^2/x^2}} = -1$? I'm reading a calculus book and I saw the following limit solution.
$$
\lim_{x \to -\infty}\frac{x+1}{\sqrt{x^2}} =
\lim_{x \to -\infty}
\left(\frac{x+1}{\sqrt{x^2}} \cdot \fr... | I guess it's an exercise meant to show how things might go wrong without being careful doing algebraic manipulations.
Since they're evaluating the limit at $-\infty$, it's not restrictive to assume $x<0$. Divide numerator and denominator by $-x$:
$$
\frac{x+1}{\sqrt{x^2}}=
\frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811234",
"timestamp": "2023-03-29T00:00:00",
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Prove $\forall n \geq 10, 2^n > n^3$ Prove $\forall n \geq 10, 2^n > n^3$
base case: $n = 10$
$2^{10} = 1024$
$10^3 = 1000$
$1024 > 1024$.
So $P(k)$ holds for $k = n$. We seek to show $P(k+1)$ holds:
We know $2^k > k^3$.
$2^k+3k^2+3k+1>k^3+3k^2+3k+1 = (k+1)^3$
$\iff 2^k+3k^2+3k+1>(k+1)^3$
Let us compare $k^3$ and $3k^... | I see your logic: you prove the chain
$$
2^{k+1}=2^k+2^k>k^3+k^3>k^3+3k^2+3k+1=(k+1)^3
$$
where the first inequality is due to the induction hypothesis and the second is because of the inequality $k^3>3k^2+3k+1$ ($k\geq 10$) for which you supply a separate proof. I think this is a good approach although you can write i... | {
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Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$
Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.
I've tried to use the cyclotomic polynomial as:
$$X^5-1=... | A different route to the factorization over the reals (obviously the end result is same as in Egreg's post, but I give the factors explicitly).
Let $p(x)$ be your polynomial. By a direct calculation we see that
$$
(x^2+\frac x2+1)^2=x^4+x^3+\frac94x^2+x+1=p(x)+\frac54 x^2.
$$
This calculation is aided by palindromic sy... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a matrix $P$ such that $P^T H P$ is a diagonal matrix with non integer eigenvalues I have this problem in my textbook:
Find a matrix $P$ such that $P^T H P$ is a diagonal matrix, where $H := \begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}$
My attempt:
$$\det(A-\lambda I) = 6 - 5\lambda + \lambda ^2 -1$$
$$= 5 - 5 \lam... | Just do the normal procedure of finding eigenvectors for the complex numbers.
$$\left[\begin{array}{cc|c}2-\frac{5-\sqrt5}{2}&1&0\\1&3-\frac{5-\sqrt5}{2}&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}4-5+\sqrt5&2&0\\2&6-5+\sqrt5&0\end{array}\right]\Rightarrow\left[\begin{array}{cc|c}-1+\sqrt5&2&0\\2&1+\sqrt5&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816501",
"timestamp": "2023-03-29T00:00:00",
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Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$
$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roo... | Note that $$c=t((t-2)^2+1),\ t\in \mathbb{R}$$, so according to the calculations, $c\in [a,b]$ where $$a=t((t-2)^2+1)|_{t=5/3}=\frac{50}{27},\ b=t((t-2)^2+1)|_{t=1}=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2 + 4 + 3\sin(ax+b)-2x=0$ has at least one real solution, ...
If the equation
$$x^2 + 4 + 3\sin(ax+b)-2x=0$$
has at least one real solution, where $a$ and $b$ belong to $(0,2\pi)$, then what can a possible value of $(a+b)$ be?
Can anyone say what exactly should be done in the question?
Value of $\sin (ax +... | $x^2 + 4 + 3sin(ax+b)-2x=0$ can be written as $(x-1)^2+3= -3sin(ax+b)$ now you can observe that left hand side will always be greater than or equal to 3, while right hand side will always be less than or equal to 3.
hence for consistent system both sides will be equal to 3 .that implies $x=1$ further $sin(ax+b)=-1$ put... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometry- Triangles Let $A,B,C$ by the angles of a triangle.
Then how to prove that
$$a^3cos(B-C) + b^3cos(C-A) +c^3cos(A-B) = 3abc$$
I divided both sides by $abc$ and then tried to open the cosine function but nothing worked. I also took the cube of sine in the triple angle identity but that too just made the fun... | Let $2R=k$
$a=k\sin A, b=k\sin B,c=k\sin C$ (sine rule)
$$ a^3 \cos (B-C)= a^2\cdot a\cos (B-C)=$$
$$= a^2\cdot k\sin A\cos (B-C)= ka^2\sin (180-(B+C))\cos (B-C)=$$
$$= ka^2\sin (B+C)\cos (B-C)=\frac{ka^2}{2} (\sin 2B+\sin 2C) = $$
$$=\frac{ka^2}{2} (2\sin B\cos B+2\sin C\cos C) = ka^2(\sin B\cos B+\sin C\cos C) = $$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$ $$\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\frac{5!}{9!}+\frac{6!}{10!}+\cdots$$
This goes up to infinity. Trying finite cases may help.
My Attempt:It seems that it is going to be $\frac{1}{18}$. My calc... | Write $$\begin{align} {\frac{n!}{(n+4)!}}&=\frac{1}{(n+1)(n+2)(n+3)(n+4)}=\\&=\frac{A}{(n+1)(n+2)(n+3)}-\frac{B}{(n+2)(n+3)(n+4)}\end{align}$$
which gives $(n+4)A-B(n+1)=1 \implies A=B=1/3$
Then we get a telescoping series:
$$\begin{align}\sum\limits_{n=0}^{\infty }{\frac{n!}{(n+4)!}}=\frac{1}{3}\sum\limits_{n=0}^{\inf... | {
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Prove inequality for $a, b , c, d$ for $a,b, c, d\ge 0$ How does one prove the following inequality:
$$ \frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}+\frac{d}{d^2+1}\le\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{d^2+1}+\frac{d}{a^2+1}$$
without much computation? Is there a trick?
| Yes, there is.
You can consider the numbers $a,b,c,d$ on one side and the numbers $\frac1{a^2+1}$, $\frac1{b^2+1}$, $\frac1{c^2+1}$, $\frac1{d^2+1}$ on the other, and apply the Rearrangement Inequality.
| {
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Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x... | Hint:
$1+2q+3q^2+\ldots+nq^{n-1}= (q+q^2+\ldots+q^{n})'$
| {
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In $\triangle ABC$ with $A = \frac{\pi}{4}$, what is the range of $\tan B\tan C$?
In a $\triangle ABC\;,$ If $\displaystyle A=\frac{\pi}{4}\;,$ and $\tan B\cdot \tan C = p\;,$ Then range of $p$
$\bf{My\; Try::}$ For a $\triangle ABC\;, A+B+C=\pi.$ So we get $\displaystyle A+B=\frac{3\pi}{4}$
So $$\tan(A+B)=-1\Rightar... | Like Triangular sides, if $\tan B=q,\tan C=\dfrac{q+1}{q-1}$
So, we have $p=\dfrac{q(q+1)}{q-1}$
$$\iff q^2+q(1-p)+p=0$$
As $q$ is real, the discriminant $$(1-p)^2-4p$$ must be $\ge0$
$$\implies p^2-6p+1\ge0$$
Now if $(x-a)(x-b)\ge0$ with $a\le b,$ we need $x\ge b$ or $x\le a$
| {
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Suppose that $a$ and $b$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
Suppose that $a$ and $b \in \mathbb{Z}^+$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
I have reduced the above formulation to these two cases. Assuming $b = a + k$. Proving that any of the below two implies that $a=b$ will be enough.
$$a^2b|(a+b)^3 - 3a... | Write $d=\gcd(a,b)$ and $a=dA$, $b=dB$, with $\gcd(A,B)=1$.
Then we can cancel $d^3$ on both sides of $a^2b \mid a^3+b^3$ and get $A^2B \mid A^3+B^3$.
This implies that $A^2 \mid B^3$ and $B \mid A^3$, and so $A=B=1$.
Indeed, $A^2 \mid B^3$ implies that every prime that divides $A$ divides $B$. Since $\gcd(A,B)=1$, we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove $\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$ $a,b,c \geqslant 0,$$ a+b+c=3$, and $(a+b)(b+c)(c+a) \neq 0$ , prove
$$\left(\frac{a+1}{a+b}\right)^a+\left(\frac{b+1}{b+c}\right)^b+\left(\frac{c+1}{c+a}\right)^c \geqslant 3$$
I try Bernouli's inequality... | Yuri Negometyanov gave
$$\left(\frac{a+1}{a+b}\right)^a = \left(1 + \frac{b-1}{a+1}\right)^{-a} \ge 1 - a\cdot \frac{b-1}{a+1}
= 2 - b + \frac{b-1}{a+1}. \tag{1}$$
This follows from the Bernoulli inequality $(1+x)^r \ge 1 + rx$ for $x > -1$ and $r \le 0$.
Using (1), it suffices to prove that
$$\frac{b-1}{1+a}+ \frac{c-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 1,
"answer_id": 0
} |
How to obtain this factorization of $x^4+4$? $x^4 + 4 = (x^2 + 2x +2)(x^2 - 2x +2)$
I am curious how would one obtain this factorization?
Clearly, once the factorization is known it is routine to verify it, however the hard part is how to find the factorization in the first place?
Thanks!
I observed that $(A+B)(A-B)=A... | You can try as follows (based on the $(a+b)^2=a^2+2ab+b^2)$):
\begin{align*}
x^4+4&=(x^2)^2+2^2\\
&=\big(x^2+2\big)^2-4x^2\\
&=\big(x^2+2\big)^2-(2x)^2\\
&=(x^2+2+2x)(x^2+2-2x)\\
&=(x^2+2x+2)(x^2-2x+2).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How to find range of $\frac{\sqrt{1+2x^2}}{1+x^2}$? How to find range of $$\frac{\sqrt{1+2x^2}}{1+x^2}$$ ?
I tried put it equal to $y$ and squaring but I'm getting $4$th degree equation.
| Let $\sqrt{1+2x^2}=u\ge1,\implies1+x^2=\dfrac{1+u^2}2$
$$\dfrac{\sqrt{1+2x^2}}{1+x^2}=\dfrac{2u}{1+u^2}=\dfrac2{u+\dfrac1u}$$
Now $u+\dfrac1u\ge2\sqrt{u\cdot\dfrac1u}=2$
Alternatively, let $\sqrt{1+2x^2}=\tan v$
Clearly, $\tan v\ge1+2\cdot0=1$ WLOG we can choose $\dfrac\pi4\le v<\dfrac\pi2\iff\dfrac\pi2\le2v<\pi$
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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An erroneous application of the Counting Theorem to a regular hexagon? I'm trying to count the unique orbits of a regular hexagon such that each vertex is either Black or White and each edge is either Red, Gree, or Blue. The group I've chosen to act on the hexagon is the dihedral group $D_7$, $$\{e,r,r^2,r^3,r^4,r^5,r^... | O negligence! Fit for a fool to fall by.
All credit to @ladisch for obviously making the obvious.
a) I should have used $D_6$ instead of $D_7$. So chop off the $r^6$ and $r^6s$ from the group and we get the dihedral group of order 12.
b) The conjugacy classes containing rotations only are adjusted acrodingly and $\{s,r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Algebraic Manipulation Given that
$ a^2 - b^2 = 60 $
and
$a - b = 6 $
$a + b = 10$
Find value of $a\cdot b$
I tried
$(a-b)^2 = 6^2 \longrightarrow a^2 - 2ab + b^2 = 36$
| We can arrange the equations as a system of equations as follows
$$\begin{cases} a-b=6\\ a+b=10 \end{cases}$$
Now we can add the two equations
$$(a-b)+(a+b)=6+10$$
$$ 2a=16 \rightarrow a=8 $$
$$ a-b=6,\ 8-b=6\rightarrow b=2 $$
So the solutions are $a=8$ and $b=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve in integers the equation $\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$
Solve in integers the equation
$$\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$$
My attempt:
I used http://www.wolframalpha.com/:
$x=-2; y=\{3,4,5,6,7,..\}$ or $x=-1, y=\{-10,-9,...\}$.
1) Let $\lfloor\frac{x^2-y^3}{x+y^... | If $x\ge0$ it is easy to show that the equation cannot hold. If $x=y=0$ that is obvious. So assume $x\ge0,y\ne0$. Then $x+y^2>0$, and we have $(x-y)(x+y^2)-(x^2-y^3)=xy(y-1)\ge0$, so dividing by $x+y^2$ we have $x-y\ge\frac{x^2-y^3}{x+y^2}$. The LHS is an integer, so we have $1+x-y>\lfloor\frac{x^2-y^3}{x+y^2}\rfloor$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?
| The inequation is equivalent to
$$x-\frac1{x+2}=\frac{x^2+2x-1}{x+2}=\frac{(x-(-1-\sqrt2))(x-(-1+\sqrt2))}{x-(-2)}<0.$$
For the expression to be negative, you need an odd number of negative factors.
$$\begin{array}{l|ccccccc}
&&-1-\sqrt2&&-2&&-1+\sqrt2\\
\hline x-(-1-\sqrt2)&\color{green}-&0&+&+&\color{green}+&+&+\\
x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solve $4 \times2^x+3^x=5^x$ without any sort of calculator Is there a way i can solve the following equation only by using high school mathematics?
$$4 \times2^x+3^x=5^x$$
I tried writing $5$ as $2+3$ but didn't get any result.
After that i tried to divide by $5^x$ and see how the function goes, but, unfortunately didn... | Assuming $x$ to be natural,
$$
5^x-3^x=(5-3)(5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^x\\
\implies (5^{x-1}+3\cdot5^{x-2}+3^2\cdot5^{x-3}+\cdots+3^{x-1})=4\cdot2^{x-1}
$$
Each term in LHS looks like $3^{n}5^{x-1-n}>2^n2^{x-1-n}=2^{x-1}$.
There are $x$ number of terms in LHS. If there are $4$ or mor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression
$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$
My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\... | $f'(\theta)=3(sin\theta-cos\theta)^2 + 2sin^2\theta = 6 (\sin\theta -\cos\theta )(\cos\theta + \sin\theta) + 4 \sin \theta \cos\theta$
$
=6 (\sin^2 \theta - \cos^2 \theta) + 4 \sin \theta \cos\theta =-6\cos(2\theta) +2\sin(2\theta)\ge 0 \implies \frac{2}{6} \tan (2\theta)\ge 1
$
$
\implies \tan(2\theta)\ge 3 \implies 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2 ... | Dividing everything by $3$, including the modulus, we get the equivalent congruence
$$2x+5y\equiv 3\pmod{6}.$$
It is convenient to rewrite this as $2x-y\equiv 3\pmod{6}$, or equivalently $y\equiv 2x-3\equiv 2x+3\pmod{6}$.
Now we have a parametric solution: $x\equiv t\pmod{6}$, $y\equiv 2t+3\pmod{6}$. To write it out a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\le... | In restricting $0 \le x < 2\pi$ you are were inadvertently also restricting $0 \le 2x < 2\pi$ which is not a stated restriction.
So $x = \pi$ is a solution as $\sin 2\pi$ does = $\sin \pi$ after all.
(Likewise $\sin 5\pi/3 = \sin 10\pi/3 = \sin 4\pi/3$).
So your restrictions are actually $0 \le 2x < 4\pi$ and for tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Prove that $\lim \limits_{x \to 5}\left(4x^2-7\right)=93$ So I first need to determine the limit and then prove it:
$\lim \limits_{x \to 5}\left(4x^2-7\right)$
So $L=93$
And thus $\left|f(x)-L\right|=\epsilon$ and $\left|x-c\right|=\delta$
Plugging in the values...
$\left|\left(4x^2-7\right)-93\right| \lt \epsilon$,... | the next step.
Find $\delta$ such that when $|x-5|<\delta, |4(x + 5)(x-5)| < \epsilon$
let $\delta = \min (1, \frac {\epsilon}{44})$
Why 44?
Suppose $x-5 = \delta, |4(x + 5)(x-5)| = \delta \le 1 \implies |4(\delta + 10) \delta| = 4 \delta^2 + 40 \delta$
If $\delta \le 1$ then $|4(x + 5)(x-5)| < 44 \delta$
$\delta = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to perform the following integration $\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$ $$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$$
How to simplify the expression given? I tried using formulas for $\cos 2x$ and $\cos 3x$. I also tried the using $\cos x + \c... | If you write $$\cos w = \frac{1}{2}\left(e^{iw}+e^{-iw}\right),$$ you find the numerator is $$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^5 = 16\cos^5(x)$$
The denominator is, likewise:
$$\frac{1}{2}\left(e^{ix}+e^{-ix}\right)^6=32\cos^6(x)$$
So the integrand is $\frac{1}{2\cos x}$.
This all comes because $1,5,10,10,5,1$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
| A very long way... (Let $I$ equal the integral)
$$I=\frac{1}{\sqrt{2}}\int\frac{x^2-1}{x^3\sqrt{(x^2-1/2)^2+1/4}}dx$$
Let $(x^2-1/2)=u/2$ meaning $x=\sqrt{1/2(u+1)}$ and $dx=\frac{1}{4\sqrt{1/2(u+1)}}$ Hence
$$I=\frac{1}{\sqrt{2}}\int\frac{u-1}{(u+1)^2\sqrt{u^2+1}}du$$
Then let $u=\tan(v)$ and $du=\sec^2(v)$ giving
$$I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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surface area of $\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 =\frac{1}{z^2}, 1I want to calculate the surface area of the surface that bounds the solid $$K=\left\{(x,y,z)\in R^3\,\mid\, x^2+y^2 \leq\frac{1}{z^2}, 1<z<3\right\}$$
I'm stuck with the differential surface area that I shall consider so that I can solve $S=\iin... | Here is how I would do it:
The surface can be parametrized as follows
\begin{cases}
x=x\\
y=y\quad\quad\quad\quad\quad\quad\quad(x,y)\in D=\{(x,y)\;|\;\frac{1}{9}\le x^2+y^2\le 1\}\\
z=\frac{1}{\sqrt{x^2+y^2}}\\
\end{cases}
You can plot this surface and its domain $D$ with WolframAlpha:
Now, the surface area is given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is conjugation of a positive semi-definite Hermitian matrix equal to conjugation by some rotation? Let $s \in GL_2(\Bbb R)$ be a symmetric positive definite matrix (is this roughly the stretching part of the polar decomposition of some other matrix $x=sk$?). Conjugate $s$ by the reflection $$\gamma=\begin{pmatrix} 1 & ... | Notation: For square matrices $s$ and $t$ of the same order, let $t$-conjugation of $s$ be denoted by $s^t = tst^{-1}$. Then $(s^{t_1})^{t_2} = s^{t_2 t_1}$ [note the reversal of order — this could be remedied by defining conjugation differently, but this answer will be consistent with the definition used in the questi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Convexity of $\log f(e^s)$ where $f$ is a polynomial Let $f(t)$ be a monic real polynomial such that $f(t) > 0$ for all $t \ge 0$. Suppose that $\log f(e^x)$ is strictly convex on $\mathbb{R}$, i.e.
$f(s^2) \cdot f(t^2) > f(st)^2$
for all positive real numbers $s \neq t$.
Can one show that
$$\frac{d^2}{dx^2}(\log f(e^... | This is false. Take $f(x) = x^4+bx^3+cx^2+bx+1$ where $-0.5<b<0.5$ and $c$ is the unique root of
$$216 b^2 + 108 b^4 - 324 b^2 c - 54 b^2 c^2 + 128 c^3 + b^2 c^3=0 \quad (\ast)$$
between $-1$ and $0$. The specific example I used to check this is $b=0.1$, $c\approx -0.2880429933200467$.
Then $\tfrac{d^2}{(dz)^2} \log f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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From $ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $ to $ \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}} $? Good evening to everyone. I have an expression that I don't know how to arrive at. $$ \frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1} $$ ... | The expression is not quite right. It is not true that
$$
\frac{\left(n\cdot \:n!+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+1\right)!+1}
= \frac{n+\frac{1}{n!}}{n+1+\frac{1}{n!}}.
$$
For example, if you plug in $n = 1$ you get
$\frac{4}{5}$ and $\frac{2}{3}$, not equal.
What is true is that
$$
\frac{\left(n\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\frac{G\ln 2}{2}$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( ... | \begin{align}
I&:=\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x)\,dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)\,dx\\
&=\int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\cos\left(x+\frac {\pi}4\right)\right)\ln(\cos x)\,dx - \int_0^{\tfrac{\pi}{4}}\ln\left(\sqrt{2}\,\sin\left(x+\frac {\pi}4\right)\right)\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
What is $e^{A}$ where A is an anti-diagonal matrix I am trying to get a closed form for the matrix produced by the following operation: $$e^A$$ where $A$ is an anti diagonal matrix, say, of size $2\times 2$:
$$A=\begin{pmatrix}
0 &b \\
c &0
\end{pmatrix}$$
Using Mathematica MatrixExp I got
$$ \left(
\begin{array}{cc}... | Recall that: $$e^A=\sum_{k=0}^{+\infty}\frac{A^k}{k!}.$$
Moreover, notice that in your case, one has: $$A^2=\begin{pmatrix}bc&0\\0&bc\end{pmatrix}.$$
Therefore, you can compute $e^A$ summing on even and odd integers. Indeed, one has: $$A^{2k}=\begin{pmatrix}b^kc^k&0\\0&b^kc^k\end{pmatrix},A^{2k+1}=\begin{pmatrix}0&b^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Is $77!$ divisible by $77^7$?
Can $77!$ be divided by $77^7$?
Attempt:
Yes, because $77=11\times 7$ and $77^7=11^7\times 7^7$ so all I need is that the prime factorization of $77!$ contains $\color{green}{11^7}\times\color{blue} {7^7}$ and it does.
$$77!=77\times...\times66\times...\times55\times...\times44\times...\... | Your solution is right on!
You might be careful how you present it to your professor. I think it is fine, but here is other words saying the same thing. (I usually try to avoid too many $\dots$ in number theory proofs.)
$$
11 = 1\cdot 11 \\
22 = 2\cdot 11 \\
33 = 3\cdot 11 \\
44 = 4\cdot 11 \\
55 = 5\cdot 11 \\
66 = 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off an... | Here is an easy way of going about the inductive step:
\begin{align}
\sum_{i=1}^{k+1}i(i+1)&= \sum_{i=1}^ki(i+1)+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2)\\[1em]
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3}\\[1em]
&= \frac{(k+1)(k+2)(k+3)}{3}.
\end{align}
See if you can determine how each line was obtained (where t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find range of $f(x)=3^x+5^x-8^x$ Find range of $f(x)=3^x+5^x-8^x$.
My attempt:
On observation one sees that $f(1)=0$.
On taking $g(x)=\left(\frac{3}{8}\right)^x+\left(\frac{5}{8}\right)^x-1$ and then observing that
$g'(x)=\left(\frac{3}{8}\right)^x \ln\left(\frac{3}{8}\right)+\left(\frac{5}{8}\right)^x \ln\left(\frac{5... | Considering the function $$f(x)=3^x+5^x-8^x$$ by inspection we have $$f(-\frac 12)=-\frac{1}{2 \sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\approx 0.67101$$ $$f(0)=1$$ $$f(\frac 12)=-2 \sqrt{2}+\sqrt{3}+\sqrt{5}\approx 1.13969$$ $$f(\frac 34)=-4 \sqrt[4]{2}+3^{3/4}+5^{3/4}\approx 0.86638$$ So, as already answered by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the equation of this hyperbola? What is the equation of the hyperbola that satisfies these conditions:
Asymptotes $y=2x$ and $y=-2x$, centre $(0,0)$, and the point $(1,1)$ lies on the curve.
This isn't a homework question; I study maths for the fun of it. The question is taken from the book "Delta Mathematics" ... | The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $y = \pm \frac{bx}{a}$.
Since $(1, 1)$ lies on your curve then $\frac{1}{a^2} - \frac{1}{b^2} = 1 \iff b^2 -a^2 = a^2b^2$.
But you know that $\frac{b}{a} = 2\iff b = 2a$. Plug this into the above equation to get ($a\neq 0$) $$3a^2 = 4a^4 \Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using row operations to compute the following 3x3 determinant
Use row operations to compute the following determinant $\begin{bmatrix}3&3&-3\\3&4&-4\\2&-3&-5\end{bmatrix}$
I know how to easily compute the determinant using $i - j + k$ method... The problem is I have put the matrix in LTF (Lower Triangular Form) and t... | The final matrix should be
$$
\begin{bmatrix}
3 & 3 & -3 \\
0 & -1 & 1 \\
0 & 0 & 24
\end{bmatrix}
$$
However, you have multiplied the determinant by $-1$ with the first operation and by $-3$ with the second one, so you get
$$
\frac{3\cdot(-1)\cdot24}{(-1)\cdot(-3)}=-24
$$
I use a different method, reducing the pivots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Embeddings of pure cubic field in complex field I know that the complex embeddings (purely real included) for quadratic field $\mathbb{Q}[\sqrt{m}]$ where $m$ is square free integer, are
*
*$a+b\sqrt{m} \mapsto a+b\sqrt{m}$
*$a+b\sqrt{m} \mapsto a-b\sqrt{m}$
So, norm of $a+b\sqrt{m}$ is $a^2-mb^2$.
Motivated by th... | The conjugates of $\sqrt[3]{n}$ are $\omega\sqrt[3]{n}$ and $\omega^2\sqrt[3]{n}$, where $\omega$ is a primitive cube root of unity. Therefore the norm of $a+\sqrt[3]{n}$ is
$$ (a+\sqrt[3]{n})(a+\omega\sqrt[3]{n})(a+\omega^2\sqrt[3]{n})=a^3+n$$
using the fact that $1+\omega+\omega^2=0$.
Edit: the three complex embeddin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Show that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0$
Show that for any real number $x$:
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$
$\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$
So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$
and $$4x^4+4x^2+1>4x^4+4x^2\Rightarrow (2x^2+... | If we consider $a=(1+\sin x),\, b=\cos(x),\, c=\frac{1}{2}$ we have that
$$ b^2-4ac = \cos^2(x)-2-2\sin(x)=-4\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)^4 \leq 0$$
hence $ax^2+bx+c$ is never negative, since $a\geq 0$ and $\Delta=b^2-4ac\leq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Locus of intersection point of perpendicular tangents Here is the question which I am referring to:
A tangent is drawn to the circle $(x-a)^2+y^2=b^2$ and a perpendicular tangent to the circle $(x+a)^2+y^2=c^2$, find locus of their point of intersection.
What I did:
First I supposed the intersection of the perpendicu... | There're two pairs of tangents for the two circles are perpendicular. Let the contact points on the circle be
$\begin{pmatrix} a+b\cos t \\ b\sin t \end{pmatrix}$,
$\begin{pmatrix} -a-c\sin t \\ c\cos t \end{pmatrix}$,
$\begin{pmatrix} a-b\cos t \\ -b\sin t \end{pmatrix}$ and
$\begin{pmatrix} -a+c\sin t \\ -c\cos t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?
| Indeed, the expansion of $(1+x)^{1/x}$ about $x=0$ is $e - \frac{ex}{2} + \frac{11e}{24} x^2 + O(x^3)$, so the limit is then $\dfrac{11e}{24}$.
--
To get this expansion, start with
$$(1+x)^{1/x} = \exp \left({1 - x/2 + x^2 / 3 - \dots}\right)$$
which you already found and put this into the power series of $\exp$ to get... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried... | Since $$\sin { x } =2\sin { \frac { x }{ 2 } } \cos { \frac { x }{ 2 } } \\ \cos { x } =\cos ^{ 2 }{ \frac { x }{ 2 } } -\sin ^{ 2 }{ \frac { x }{ 2 } } $$
so
$$\cos \left( x \right) +\sin \left( x \right) =\frac { 7 }{ 5 } \\ 5\left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P
If $a>b>0$ and $a^3 +b^3 +27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $P,Q(P<Q)$. Find the value of $4Q-aP$?
How will I begin with the solution just a hint would be enough.
| $$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$
therefore
\begin{cases}
a+b-9=0\\
\qquad\operatorname{or}\\
a=b=-9
\end{cases}
since $a>b>0$ thus
$$a+b-9=0$$
Set $f(x)=ax^2+bx-9$. We have $f(1)=a+b-9=0$, thus
$Q=1$ and $P=\frac{-9}{a}$
finally
$$4Q-aP=4+9=13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove by induction that $a^{4n+1}-a$ is divisible by 30 for any a and $n\ge1$ It is valid for n=1, and if I assume that $a^{4n+1}-a=30k$ for some n and continue from there with $a^{4n+5}-a=30k=>a^4a^{4n+1}-a$ then I try to write this in the form of $a^4(a^{4n+1}-a)-X$ so I could use my assumption but I can't find any $... | Don't factor entirely. Just factor enough to realize
$a^{4n+1} - a = a(a^{4n} - 1)=a(a - 1)(a^{4n-1} + a^{4n-2} + .... + a + 1)$.
Assume for $n = k$ that
$a^{4k + 1}-a= a(a^{4n} - 1)=a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)= 30M$
then $a^{4(k+1) + 1} - a = a(a-1)(a^{4k + 3} + ...)=$
$a(a-1)(a^{4k + 3} + a^{4k + 2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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find $f\in L^2([0,\pi])$ such that its $L^2$ distance to $\sin(x)$ and $\cos(x)$ are both bounded by specific constants I want to find all $f\in L^2([0,\pi])$ such that
$$
\begin{align}
\int_0^\pi\lvert f(x)-\sin(x)\rvert^2\,dx &\le \frac{4\pi}{9}\\
\int_0^\pi\lvert f(x)-\cos(x)\rvert^2\,dx &\le \frac{\pi}{9}\\
\end{al... | First, note that
\begin{align}
\|\sin-\cos\|^2&=\|\sin(x)\|^2+2\langle\sin,\cos\rangle+\|\cos\|^2 \\
& =\|\sin\|^2+\|\cos\|^2 \\
& = \|1\|^2=\pi.
\end{align}
You want
$$
\|f-\sin\|+\|f-\cos\| \le \frac{2}{3}\sqrt{\pi}+\frac{1}{3}\sqrt{\pi}=\sqrt{\pi}
$$
If you have the above, then
$$
\sqrt{\pi}=\|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Trigonometry Olympiad problem: Evaluate $1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$
Find the value of
$$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$
My attempt
I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$... | Use $\sin(\theta)=\sin(180^\circ-\theta)$. Let the summation as $S$. Then
\begin{align}
2S&=\sum_{k=1}^{90}(k\sin(2k^\circ)+(90-k)\sin((180-2k)^\circ)) \\
&=90\sum_{k=1}^{90}\sin(2k^{\circ})
\end{align}
Now use $-2\sin(2k^\circ)\sin(1^\circ)=\cos((2k+1)^\circ)-\cos((2k-1)^\circ)$, then
\begin{align}
S=45\sum_{k=1}^{90}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Show $\ln\left(\frac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\frac{a-b}{n}+\mathcal{O}\left(\frac{1}{n^2} \right) $
I would like to prove the following:
$$\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\dfrac{a-b}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) $$
My attempt
i tried thi... | One way could be $$\ln { \left( \frac { n+{ \left( -1 \right) }^{ n }\sqrt { n } +a }{ n+{ \left( -1 \right) }^{ n }\sqrt { n } +b } \right) } =\ln { \left( n+{ \left( -1 \right) }^{ n }\sqrt { n } +a \right) -\ln { \left( n+{ \left( -1 \right) }^{ n }\sqrt { n } +b \right) } } =\\ =\ln { \left( n\left( 1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$
Find $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$
My attempt:
$$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{3}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n-1}{3}}\right)^{2n+3}$$
Now ... |
$$\lim _{ n\to \infty } \left( \frac { n+2 }{ n-1 } \right) ^{ 2n+3 }=\lim _{ n\to \infty } \left( 1+\frac { 3 }{ n-1 } \right) ^{ 2n+3 }=\\ =\lim _{ n\to \infty }{ \left[ \left( 1+\frac { 1 }{ \frac { n-1 }{ 3 } } \right) ^{ \frac { n-1 }{ 3 } } \right] } ^{ \frac { 3 }{ n-1 } \left( 2n+3 \right) }=\lim _{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2)$ I have to solve the following problem but I don't know how to :
Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't ... | Assume $a\leq b\leq c$
Notice $\frac{a+b}{a+b-c}\in(1,2)\implies \frac{a+b-c}{a+b}\in(\frac{1}{2},1)\implies \frac{c}{a+b}\in(0,\frac{1}{2})$, but $\frac{c}{a+b}\geq\frac{c}{2c}=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the determinant of this $n \times n$ matrix in a clever way? Is there any clever and short way to find out the determinant of the following matrix?
\begin{bmatrix}
b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\
a_1 & 0 & 0 & \cdots & 0 & b_1 \\
0 & a_2 & 0 & \cdots & 0 & b_2\\
\vd... | $b_1 \det\begin{pmatrix}
0 & 0 & \cdots & 0 & b_1 \\
a_2 & 0 & \cdots & 0 & b_2\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
\end{pmatrix}
- a_1 \det\begin{pmatrix}
b_2 & b_3 & \cdots & b_{n-1} & 0 \\
a_2 & 0 & \cdots & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need th... | It obviously holds for $n=0$; now assume it holds for a general $n$. That is, we have $$(1-x^2)\frac{\mathrm{d}^{n+2}y}{\mathrm{d} x^{n+2}} - (2n+3)x\frac{\mathrm{d}^{n+1}y}{\mathrm{d}x^{n+1}} - (n+1)^2\frac{\mathrm{d}^ny}{\mathrm{d}x^n} = 0$$
Let us differentiate (all differentiation in this answer is with respect to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving an expression is an integer
Prove that $$A^2 \cdot \dfrac{\sum_{m=1}^{n}(a^{3m}-b^{3m})}{\sum_{m=1}^n(a^m-b^m)}-3A^2$$ is an integer if $a=\frac{k+\sqrt{k^2-4}}{2}, b=\frac{k-\sqrt{k^2-4}}{2}$, where $k>2$ is a positive integer, and $A = \dfrac{1}{\sqrt{k^2-4}}$ for all positive integers $n$.
I thought about... | Partial answer:
Observe that $a,b$ are the roots of the quadratic equation $x^2 - kx + 1 = 0$, and hence $ab = 1$.
This implies that
$$ \frac{\sum (a^{3m}- b^{3m})}{\sum (a^m - b^m)} - 3 = \frac{\sum (a^m - b^m)^3}{\sum (a^m - b^m)} $$
Next, we also have the recurrence relation
$$ k(a^m - b^m) - (a^{m-1} - b^{m-1}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also ... | Hint: $$\int\frac{1+x^{2}}{1+x^{4}}dx=\frac{1}{2}\int\left(\frac{1}{x^{2}+\sqrt{2}x+1}+\frac{1}{x^{2}-\sqrt{2}x+1}\right)dx$$ $$=\frac{1}{2}\int\left(\frac{1}{\left(x+\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}}\right)^{2}+\frac{1}{2}}\right)dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Solve the second order equation $$\frac{d^2u}{dt^2} = \begin{bmatrix} -5 & -1 \\ -1 & -5 \end{bmatrix}u
$$
with
$$
u(0)=\begin{bmatrix}1 \\ 0 \end{bmatrix}
$$
and
$$
u'(0) = \begin{bmatrix}0 \\ 0 \end{bmatrix}
$$
Okay I tried to create a $u$ vector but couldn't. And believe I can't get $\frac{du}{dt}$ by finding the ei... | An equivalent method without (explicitly) using eigenvalues, as the matrix is symmetric:
Writing $u=(u_1,u_2)^T$ the d.e. is equivalent to:
\begin{align}
\frac{d^2 u_1}{d t^2}&=-5u_1-u_2\\
\frac{d^2 u_2}{d t^2}&=-u_1-5u_2\\
\end{align}
Consider instead $u_1+u_2$, and $u_1-u_2$,
\begin{align}
\frac{d^2 }{d t^2}(u_1+u_2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
| WLOG, $a=b=1$. Then factoring the denominator, we decompose in simple fractions
$$\frac y{y^3+1}=A\frac1{y+1}+B\frac{2y-1}{y^2-y+1}+C\frac1{y^2-y+1}=\frac{(A+2B)y^2+(-A+B+C)y+(A-B+C)}{y^3+1}.$$
Identifying,
$$\frac y{y^3+1}=\frac13\frac1{y+1}-\frac16\frac{2y-1}{y^2-y+1}+\frac12\frac1{y^2-y+1}.$$
The first two terms hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Constructing the 11-gon by splitting an angle in five In "Angle Trisection, the Heptagon, and the Triskaidecagon", published in the American Mathematical Monthly in March 1988, Andrew Gleason discusses what regular polygons can be constructed with compass, straightedge and angle trisector. At the end of that article he... | The question essentially asks about transforming solvable equations from one form to another.
I. Cubic
Using just a linear transformation, the general cubic $P(x)=0$ can be transformed to the form,
$$y^3+3ay+b = 0\tag1$$
with solution,
$$y_k = 2\sqrt{-a}\;\cos\left(-\tfrac{2\pi\,k}{3}+\tfrac{1}{3}\,\arccos\big(\tfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Permutations and Combinations, Picking discs A bag contains 6 red, 5 blue and 4 green discs. If 3 discs are chosen at random, find the probability that there is at least one blue disc given that at least one red disc is chosen.
I thought that on the denominator is 6*14C2 as you pick one red and from the rest you pick 2... | A red disc is selected unless only blue and green discs are selected. Therefore, the number of selections in which at least one red disc is selected is
$$\binom{6 + 5 + 4}{3} - \binom{5 + 4}{3} = \binom{15}{3} - \binom{9}{3}$$
Selections that include both a blue disc and a red disc either contain one disc of each col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve the limit $\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$? Hi I got an examination at the school which was so arduous that I'm stumped.
This problem is the toughest for me :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\righ... | We have that
$$\begin{align*}
n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{\left(n+1\right)}{n}\right)
&=n^2\left(\exp\left(n\ln\left(1+\frac{1}{n\left(n+2\right)}\right)\right)-1-\frac{1}{n}\right)\\
&=n^2\left(\exp\left(n\left(\frac{1}{n\left(n+2\right)}+o\left(\frac{1}{n^3}\right)\right)\right)-1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \su... | You can make your approach work if you do the following:
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^{\color{red}{n-1}}+n^{\color{red}{n-1}}+\cdots+n^{\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\lef... | This step is wrong: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )$
It should be: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k} \right)=15(\sum_{k=0}^\infty\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Basic Joint probability distribution calculation example? (X, Y) has the Joint probability distribution:
\begin{array}{|c|c|c|}
\hline
X / Y & 1 & 2 \\ \hline
1& \frac{1}{4}& \frac{1}{3} \\ \hline
2& \frac{1}{6}& a \\ \hline
\end{array}
So $P(X=1\mid Y=1)=\frac{3}{5}$ and $P(X=2\mid Y=2)=\frac{3}{7}$.
My challeng... | $$
P(X=1|Y=1)=\frac{P(X=1, Y=1)}{P(Y=1)}= \frac{1/4}{1/6+1/4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is my solution of $e^{\sin x } - e^{- \sin x} - 4 = 0$ wrong? $$e^{\sin x } - e^{- \sin x} - 4 = 0$$
Substitute $e^{\sin x} = y$:
$$y - \frac{1}{y} - 4 = 0 \implies y^2 - 4y - 1 = 0$$
Solve for $y$:
$$y = 2 \pm \sqrt{5}$$
$e^{\sin x}$ can't be negative:
$$\therefore y = 2 + \sqrt{5} \implies e^{\sin x} = 2 + \sqr... | Let $e^{\sin x} = 2 + \sqrt{5}$. Then, $\sin{x}=\ln{(2+\sqrt{5})}>\ln{3}>1$, which isn't possible (since $\sin{x}\in[-1,1]$ for every $x$). Therefore, there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$
Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$
My work: $d|n^3+5n,n^3-5n^2+6n\implies\ d|5n^2-n,5n^2+25\implies d|n+25 $
Now I can't go further!!
| Let $\,f(n) = n^2+5.\,$ Then $\,d\mid \color{#c00}{f(n),n\!+\!25}\,\Rightarrow\, d\mid \overbrace{{f(n)\ \rm mod}\ (n\!+\!25) = f(-25)}^{\textstyle\!\! \color{#c00}{f(n)} - (\color{#c00}{n\!+\!25})q(n) = r(n)} = 630\, $ where we used the Remainder theorem $\ f(n)\equiv f(a)\pmod{n-a}$
Remark $\ $ It is simpler to work... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $a^{2013} + b^{2013} + c^{2013}$ Problem Statement
Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$
Suppose that the following conditions hold:
*
*$f(1)=0$
*the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$
$$\text{Find the value of} \: ... | The line
$$b = pq + pr + qr = p^2 + q^2 + r^2$$
is wrong.
$$b = pq + pr + qr =-( p^2 + q^2 + r^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Optimizing continued fraction of square root From this question, I learned that the square root of a number $n$ can be written as a continued fraction of the form:
$$\sqrt n=a+\frac{n-a^2}{a+\sqrt n}$$
where $a$ can have any value. By jumping to conclusions and testing, I believe that the optimal value for $a$ for a ra... | \begin{align*}
\frac{p_{n}}{q_{n}} &= \sqrt{n}+e_{n} \\
&= a+\frac{n-a^2}{a+\frac{p_{n-1}}{q_{n-1}}} \\
&= \frac{a\left(a+\frac{p_{n-1}}{q_{n-1}} \right)+n-a^2}
{a+\frac{p_{n-1}}{q_{n-1}}} \\
&= \frac{a \frac{p_{n-1}}{q_{n-1}}+n}{\frac{p_{n-1}}{q_{n-1}}+a} \\
&= \frac{a (\sqrt{n}+e_{n-1})+n}{\sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving a non-homogeneous linear recurrence relation So i have this non-homogeneous linear recurrence relation to solve:
$$a_{n}=2a_{n-1}-a_{n-2}+2^n+2$$
$a_{1}=7$ and $a_{2}=19$
I know that the non-homogeneous part is $2^n$ and i know how to solve homogeneous linear recurrence relations, but when i get a non-homogeneo... | Define:
$g_n := a_{n}-a_{n-1}$
Then $a_{n}=2a_{n-1}-a_{n-2}+2^n+2 \implies a_{n}-a_{n-1}=a_{n-1}-a_{n-2}+2^n+2 \\\ \implies g_n = g_{n-1} + 2^n +2 \implies g_n- g_{n-1} = 2^n +2$
$$g_n- g_{n-1} = 2^n +2 $$
$$g_{n-1}- g_{n-2} = 2^{n-1} +2 $$
$$.$$
$$.$$
$$.$$
$$g_3- g_{2} = 2^3 +2 $$
Since $g_2 = 19-7=12$, then $g_n = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
hypergeometric 2F2 function close form for specific case I am trying to find the closed form for the following
\begin{equation}
{_2F_2}(2,4;3,4+a;a)
\end{equation}
any help would be greatly appreciated. thank you
I find the closed form for
\begin{equation}
{_2F_2}(1,3;2,3+a;a)=1+\frac{a}{2}
\end{equation}
But don't suc... | $_2F_2(2,4;3,4+a;a)$
$=\sum\limits_{n=0}^\infty\dfrac{(2)_n(4)_na^n}{(3)_n(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(n+3)(2)_na^n}{3(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(n+2)(2)_na^n}{3(a+4)_nn!}+\sum\limits_{n=0}^\infty\dfrac{(2)_na^n}{3(a+4)_nn!}$
$=\sum\limits_{n=0}^\infty\dfrac{(2)_{n+1}a^n}{3(a+4)_nn!}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.