Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Coefficients of a polynomial function from $\Bbb{N}$ to $\Bbb{N}$ Let $f$ be a polynomial from $\Bbb{N}$ to $\Bbb{N}$.
So let $f(x) = a_0+a_1x+a_2x^2+..a_rx^r$
Then for every natural number $k$, $f(k) \in \Bbb{N}$.
But how does it imply that all the coefficients of $f(x)$ i.e. $a_0, a_1,..a_r$ will belong to $\Bbb{Q}$?... | It results that
$\begin{cases}
a_0+a_1+\ldots+a_i+\ldots+a_r=f(1)\\
a_0+2a_1+\ldots+2^ia_i+\ldots+2^ra_r=f(2)\\
a_0+3a_1+\ldots+3^ia_i+\ldots+3^ra_r=f(3)\\
......................................\\
a_0+ra_1+\ldots+r^ia_i+\ldots+r^ra_r=f(r)\\
a_0+(r+1)a_1+\ldots+(r+1)^ia_i+\ldots+(r+1)^ra_r=f(r+1)
\end{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$ Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\t... | For
$$M=\left(
\begin{array}{ccc}
a & b & c \\
c & a & b \\
b & c & a
\end{array}
\right)\implies |M|=a^3+b^3+c^3-3 a b c$$ So, replacing
$$\Delta=\tan ^3(x)+\tan ^3(x+h)+\tan ^3(x+2 h)-3 \tan (x) \tan (x+h) \tan (x+2 h)$$ Now, composing Taylor series around $h=0$ gives
$$\Delta=9 \left(\tan ^5(x)+2 \tan ^3(x)+\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
toom-cook algorithm matrix G For this toom-cook algorithm at https://arxiv.org/pdf/1803.10986v1.pdf#page=6 , how do I get the value 4/2 in the matrix G ?
| We have
$$G = \begin{bmatrix} x_0^0 N^0 & x_0^1 N^0 & x_0^2 N^0 \\ x_1^0 N^1 & x_1^1 N^1 & x_1^2 N^1 \\ x_2^0 N^2 & x_2^1 N^2 & x_2^2 N^2 \\ x_3^0 N^3 & x_3^1 N^3 & x_3^2 N^3 \end{bmatrix} = \begin{bmatrix} (1)( -\dfrac{1}{6}) & (1)(-\dfrac{1}{6}) & (1)(-\dfrac{1}{6})\\ (1)(\dfrac{1}{2}) & (2)(\dfrac{1}{2}))& (4)(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \lef... | We have
$$
\frac{bc-a^2}{c-b}+\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a}\geq \frac{bc-a^2}{c-b}-\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a}
$$
with equality only if $b=1$ and $c=\frac{1}{a}$. Now insert $b$ and $c$ to obtain
$$
\frac{\frac{1}{a}-a^2}{\frac{1}{a}-1}+\frac{\frac{1}{a^2}-a}{1-a}=\frac{1-a^3}{1-a}+\frac{1-a^3}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Substitution goes wrong while evaluating $\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx$ Im aware this integral has been evaluated here before, i started with
$$\int _0^{\infty}\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx=2\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx+2G$$
Solving the integral on the left is very e... | Letting $x=\tan \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta \\
&=-2 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta \\
&=-2\left(-\frac{\pi}{4} \ln 2+\frac{1}{2} G\right) \\
&=\frac{\pi}{2} \ln 2-G
\end{aligned}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate :- $\frac{(2020^2 - 20100)(20100^2 - 100^2)(2000^2 + 20100)}{10(2010^6 - 10^6)}$
Evaluate :- $\frac{(2020^2 - 20100)(20100^2 - 100^2)(2000^2 + 20100)}{10(2010^6 - 10^6)}$
What I Tried :- I couldn't think of any ways to factorise this expression . The denominator can be written as $10(2016^3 - 10^3)(2016^3 + ... | Let $2010=x$ and $10=y$.
Thus, for our expression we obtain: $$\frac{((x+y)^2-xy)(x^2y^2-y^4)((x-y)^2+xy)}{y(x^6-y^6)}=$$
$$=\frac{(x^2+xy+y^2)y^2(x^2-y^2)(x^2-xy+y^2)}{y(x^6-y^6)}=$$
$$=\frac{y(x^2-y^2)(x^4+x^2y^2+y^4)}{x^6-y^6}=y=10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Minimizing $x^2+y^2+z^2$ subject to $xy -z + 1 = 0$ via Lagrange multipliers $$\begin{array}{ll} \text{minimize} & f(x,y,z) := x^2 + y^2 + z^2\\ \text{subject to} & g(x,y,z) := xy - z + 1 = 0\end{array}$$
I tried the Lagrange multipliers method and the system resulted from has no solution. So I posted it to see if the... | This can be solved in at least two methods.
First, let's solve without Lagrange, using convenient changes of variables. Let $u=x+y, v=xy$. This results in $u^2=x^2+y^2+2xy=x^2+y^2+2v$.
We now need to minimize $u^2-2v+z^2$ under the constraint $v-z+1=0$. We can rearrange this constraint to be of the form $z=1+v$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could... | Assume $a>0$. Since the denominator is always positive, maximum value of $f$ must occur when $x>0$.
Hence for all positive $x$'s,
$$\frac{ax-1}{x^4-x^2+1} \le 1 \implies x^3-x+\frac{2}{x} \ge a$$
Rewrite it as
$$2+\left(2+x+\frac{2}{x}\right)(1-x)^2 \ge a$$
The equality holds for $x=1$ and $a=2$.
Similar argumentation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
What is the minimum value of $x+y$?
Suppose $x,y$ are positive real numbers that satisfy $$xy(x+2y)=2$$ What is the minimum value of $x+y$?
My Thoughts
I’ve attempted using arithmetic-geometric mean inequality and got:
$\frac{x+y+x+2y}{3} \geq \sqrt[3]{2}$
Therefore $2(x+y)+y \geq 3\sqrt[3]{2}$, then I got trapped.
... | Let $k$ be a minimal value of $x+y$.
Thus, $$x+y\geq k$$ or
$$\frac{2(x+y)^3}{xy(x+2y)}\geq k^3$$ or for $x=ty$
$$\frac{2(t+1)^3}{t^2+2t}\geq k^3$$ and since $$\min_{t>0}\frac{2(t+1)^3}{t^2+2t}=3\sqrt3,$$ which occurs for $t=\sqrt3-1,$ we obtain that $k=\sqrt3$.
Can you get now a full solution?
By the way, we can get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Roots of the Cubic equation The Cubic formula:
$ax^3+bx^2+cx+d=0$
With under the following conditions:
*
*$a \neq 0 $
*$a,b,c,d \in \Bbb{R} $
We can derive the following formula as a the root of $x$:
$u= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}+\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\... | Starting with the "depressed" cubic:
$y^3 + py + q = 0$
If you don't have a diminished cubic you can substitute $x = y - \frac {b}{3a}$ which will eliminate the $bx^2$ term from the original cubic.
Next we do a similar substitution:
$y = z-\frac {p}{3z}$
$(z - \frac {p}{3z})^3 + p(z-\frac {p}{3z}) + q = 0\\
z^3 - pz + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using ... | $$\left(\frac13+\frac1{3^2}+\frac1{3^3}+\cdots \right)\left(1+3+\frac53+\frac7{3^2}+\cdots\right)$$
\begin{aligned}=\left(\frac12\right)\left(1+3+ \frac33+\frac3{3^2}+...\\
+\frac23+\frac2{3^2}+\cdots\\+\frac2{3^2}+\cdots\\+\cdots\right)\\
=\frac12\left(4+\frac32\\+1\\+\frac13\\+\cdots\right)\\
=\frac12\left(4+\frac32... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we... | Also, note that the terms of the second sum decay exponentially: Using $\ln(1-x)=-x+O(x^2)$ for $x\to0$, $$\frac{n^{n^2}}{(n+1)^{n^2}}=\left(1-\frac1{n+1}\right)^{n^2}=\exp\left(n^2\ln\left(1-\frac1{n+1}\right)\right)=O(1)\exp\left(-n\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the meaning of "due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x=\frac1r$ is also a zero" I was studying this answer about factoring $x^4+x^3+x^2+x+1$:
https://socratic.org/questions/how-do-you-factor-x-4-x-3-x-2-x-1
The author says: "A cleaner algebraic approach is to no... | The list of coefficients of$$x^4+x^3+x^2+x+1$$is $(1,1,1,1,1)$, which is symmetric (if you reverse it, you will get the same list). In other words, it is a list of the type $(a,b,c,b,a)$. And if $r(\ne0)$ is a root of$$ax^4+bx^3+cx^2+bx+a,\tag1$$then$$ar^4+br^3+cr^2+br+a=0,$$and therefore$$a+\frac br+\frac c{r^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end... | Hint:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} & bc^{-1} & ca^{-1}\\
1 & 1 & 1\\
ba^{-1}& cb^{-1} & ac^{-1}
\end{pmatrix}
\end{equation*}
\begin{equation*}
= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} -ca^{-1} & bc^{-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove al... | The source of the problem is: https://math.stackexchange.com/questions/2825783/problems-regarding-inequality. Below are solutions for both parts, some were written by other users in the old thread above.
Part (a): Solution by Truth.
Let $a = \dfrac{2kx}{y+z}$, $b = \dfrac{2ky}{z+x}$, and $c = \dfrac{2kz}{x+y}$, where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))$ for an identity of Srinivasa Ramanujan Ramanujan is supposed to have given more than five thousand elegant results, a good number of them are yet to be proved or disproved.
Yesterday in the comment section of
Prov... | You can give Feynman's trick a shot.
\begin{align*}
I&=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I\left(a\right)&=\int _0^1\frac{\operatorname{arctanh} \left(a\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I'\left(a\right)&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| From showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
$$2\arctan x=\begin{cases}\arctan\dfrac{2x}{1-x^2}\text{ if } x^2\le1\\\pi+\arctan\dfrac{2x}{1-x^2}\text{ if } x^2>1\end{cases}$$
Again, from Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
$$\sin^{-1}(\sin2y)=\begin{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find the smallest possible value of an equation, where $a+b+c=3$ We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation:
$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$
I solved it in the following fashion:
$$
\begin{align}
A&=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1... | Your solution is wrong because after your first step we need to prove that:
$$6-(a^2+b^2+c^2)\geq3,$$ which is wrong for $a=2$.
We'll prove that $3$ is a minimal value.
Indeed, we need to prove that$$\frac{2(ab+ac+bc)}{abc}-(a^2+b^2+c^2)\geq3$$ or
$$\frac{2(ab+ac+bc)(a+b+c)}{3abc}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}\geq3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Median of triangle and tangents Let $AD$ be an altitude of any triangle $ABC$. Consider a circle with $AD$ as its diameter, cutting $AB$ and $AC$ at $P$ and $Q$ respectively. Let the tangents at $P$ and $Q$ meet at $X$. Prove that $AX$ bisects $BC$.
I have attempted the proof, using angles in the alternate segment, and... |
Pure coordinate proof:
$D:=(0,0),\,A:=(0,a),\,B:=(b,0),\,C:=(c,0),$
$O:=\left(0,\frac{a}{2}\right),\,P:=pB+(1-p)A,\,
Q:=qC+(1-q)A,\,X:=(x,y)$
$$\begin{cases}
OP^2=\frac{a^2}{4}\\
OQ^2=\frac{a^2}{4}\\
\overrightarrow{OP}\cdot\overrightarrow{XP}=0\\
\overrightarrow{OQ}\cdot\overrightarrow{XQ}=0\\
\end{cases}$$
$$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How many even three-digit numbers have distinct digits and have no digit $5$? How many even three-digit numbers have distinct digits and have no digit $5$?
The answer my teacher gave was $252$, but I don't see how she got that. I thought it would be $6\times 8 \times 5=240$ because the $3\text{rd}$ digit must be even $... | So, one is asking for the number of injections $f:\{1,2,3\} \to \{0,1,2,3,4,6,7,8,9\}$ where $f(1) \neq 0$ and $f(3) \in \{0,2,4,6,8\}$.
Ignoring the last two conditions, one gets $_{9}P_{3}=504$ possible numbers (possibly odd and/or having a leading zero).
If one requires the first digit to be nonzero, then one would ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating a limit without usage of Riemann sum
Evaluation of $$\lim_{n \rightarrow \infty}\bigg[\frac{1}{n}+\frac{1}{n+2}+\frac{1}{n+4}+\cdots \cdots +\frac{1}{3n}\bigg]$$
My work: Using Riemann sum
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{1}{n+2r}=\lim_{n\rightarrow\infty}\sum^{n}_{r=0}\frac{1}{1+2\frac{r}{n... | For even $n=2m$ we have
\begin{align}
\sum_{r=0}^{2m}\frac{1}{2m+2r}
&=\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+r}=\\
&=\frac{1}{2}\left(\sum_{r=1}^{3m}\frac{1}{r}-\sum_{r=1}^{m-1}\frac{1}{r}\right)=\frac{1}{2}(H_{3m}-H_{m-1}),
\end{align}
where
$$
H_n=\sum_{r=1}^n\frac{1}{r}
$$
are the Harmonic numbers.
Given the kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$ Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$
Can somebody check my solution? Thanks!!
Let $u=11xe^{2x}$
then $\frac{du}{dx} = \frac{d}{dx}(11xe^{2x})$
$=11(e^{2x}+2xe^{2x})$
and so $du=11e^{2x}(2x+1)dx$
Now, let $dv=\frac{1}{(2x+1)^2}dx$
Then $\int dv = \int \frac{1}{(2x+1)^... | When you have a short time to compute antiderivatives looking like
$$I=\int \frac {e^{k x}\,P_n(x)} {[Q_m(x)]^r} \,dx$$ think that a possible solution could be of the form
$$\frac {e^{k x}\,R_p(x)} {[Q_m(x)]^{r-1}}$$ where $P,Q,R$ are polynomials of different degrees.
Differentiate both sides to get
$$\frac {e^{k x}} {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$\cos\left(\frac{\pi}{5}\right)$ using De Moivre's Theorem This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:
Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).
[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see fro... | Start with something simpler, an expression for $e^{i\frac{\pi}5}$:
$$\left(\cos\frac{\pi}5+i\sin\frac{\pi}5\right)^5=-1$$
Write explicitly the terms of the expansion. We know that we can ignore therms in even powers of $i$. Using $z=\cos\frac{\pi}5$ and $\sin^2\frac{\pi}5=1-z^2$ one gets:
$$z^5-10z^3(1-z^2)+5z(1-z^2)^... | {
"language": "en",
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"question_score": "3",
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On the Convergence of Series Will Jagy has provided the answer for the following question, the series is convergent:
I come across with the following question about the convergence of series:
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{\log k}-\dfrac{1}{\log(k+1)}\right),
\end{align*}
where $\epsilon\in(0... | For $0 < t < 1,$
$$ t - \frac{t^2}{2} < \log (1+t) < t $$
$$ \frac{1}{k} - \frac{1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \frac{k}{2k^2} < \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a pr... | We have that
$$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x+3^x)(2^{2x}-6^x+3^{2x})}{(6^x)(2^x+3^x)}=\frac{2^{2x}-6^x+3^{2x}}{6^x}=\frac76$$
$$\iff\left(\frac23\right)^x+\left(\frac32\right)^x=\frac{13}6$$
$$\iff y+\frac1y=\frac{13}6 \iff y^2-\frac{13}6y+1=0$$
which leads to $y=\frac23,\frac32$ and $x=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equ... | Modulo $88$ one has $$123^{456} = 35^{456} = (35^2)^{228} = (-7)^{228} = ((-7)^6)^{38} = (-7)^{38} = ((-7)^6)^6 \times 49 = (-7)^6 \times 49 = -7 \times 49 = -343 = 9 \quad [88]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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Evaluate integral $\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$ Evaluate the integral
$$\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$$
Edit:
Here is Quanto's brilliant method of doing this problem with the details filled out:
SOLUTION:
First, lets complete the square $-9x^2+6x+15$:
$-9x^2+6x+15$
$=-9(x^2-\fr... | Completing the square in the denominator actually yields
$$\int \frac{x^2}{\left(-\left(3x-1\right)^2+\color{red}{16}\right)^{\frac{3}{2}}}dx$$
Substituting $x = \frac{4\sin(u)+1}{3}$ yields $$\frac{4}{27}\int \frac{\left(4\sin(u)+1\right)^2}{\left( -16\sin^2(u) + 16 \right)^{3/2}} \cos(u) du$$
Or since $1-\sin^2(x) = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I c... | $$
\frac{1}{\sin^4x+\cos^4x}=
\frac{1}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}=
\frac{1}{1-\frac{1}{2}(2\sin x\cos x)^2}=\\
\frac{2}{2-\sin^22x}=
\frac{2}{2-\frac{1}{2}(1-\cos4x)}=
\frac{4}{3+\cos4x}=
$$
If we now set
$$
t=\tan2x\qquad\implies\qquad dx=\frac{1}{2}\cdot\frac{1}{1+t^2}dt
$$
we have
$$
\int\frac{1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ under given conditions Consider $a,b,c \in \mathbb{R}-\{0\}$. A real valued function is defined as $$f(x,y,z)=2^y3^za^xb^yc^z+2^z3^xa^yb^zc^x+2^x3^ya^zb^xc^y$$ where $x,y,z \in \mathbb{Z}$ and $f(1,0,0)=4$ and $f(2,0,0)=6$. Find the maximum ... | $a+2b+3c=4$ ...(i)
$a^2+4b^2+9c^2=6$ ...(ii)
Given we have to find the maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ which has equal of $a^2, b^2, c^2$, we should maximize $a^2$ in (ii).
The minimum value of $a^2, b^2, c^2$ can be $0$ each. So individually, max value of
$a^2 = 6, b^2 = ... | {
"language": "en",
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"source": "stackexchange",
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Proving that inequality holds under condition.
Let $a$ and $b$ be positive numbers. Prove that inequality $$\frac{ax+by}{2} \leqslant \sqrt{\frac{ax^2+by^2}{2}}$$
holds for all real $x$ and $y$ only and only if $a+b \leqslant2$
Problem needs to be done using "basic" algebraic methods.
I tried expanding this into form... | Let $a+b\leq2$.
Thus, by C-S $$\sqrt{\frac{ax^2+by^2}{2}}=\sqrt{\frac{(a+b)(ax^2+by^2)}{2(a+b)}}\geq\frac{|ax+by|}{\sqrt{2(a+b)}}\geq\frac{|ax+by|}{2}\geq\frac{ax+by}{2}.$$
Let $a$ and $b$ are positives and $$\sqrt{\frac{ax^2+by^2}{2}}\geq\frac{ax+by}{2}$$ is true for any reals $x$ and $y$.
Thus, for $x=y=1$ we obtain... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Solve $ay''+by+c=0$ using separation of variables How could be
$$ay''+by+c=0$$
(for constants $a,b,c$) solved using separation of variables?
I tried using the method on an easier equation, namely $ay''+by'+c=0$:
$$\begin{align}a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c&=0,\, u=\frac{dy}{dx}\\a\frac{du}{dx}+bu+c&=0\\ \frac{1}{... | $$ay''+by+c=0$$
Substitute $p=\dfrac {dy}{dx}=y'$ then you have:
$$y''=\dfrac {dy'}{dx}=\dfrac {dp}{dx}=\dfrac {dp}{dy}\dfrac {dy}{dx}=p'p$$
Where $p'=\dfrac {dp}{dy}$.
The differential equation becomes:
$$app'+by+c=0$$
This is now a first order DE that is separable:
$$a\int p\;dp =-\int (c+by) \; dy$$
$$ap^2=-2cy-by^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n... | $$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$ Thus the given sum is $$S = \frac{1}{2} \sum_{n=1}^\infty \frac{n(n+1)}{n!} = \frac{1}{2} \sum_{n=1}^\infty \frac{n+1}{(n-1)!} = \frac{1}{2} \left(\sum_{n=1}^\infty \frac{n-1}{(n-1)!} + 2\sum_{n=0}^\infty \frac{1}{n!}\right).$$ The second term in parentheses is simply $2e$; the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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How do i form the inequality $ -5\le 3\cos x - 4\sin x\le5$? We know that
$-1\le \cos x\le1$ so we if we multiply 3 so we get $-3\le 3\cos x\le 3$ and we also know that $-1\le \sin x\le 1$ and and again we multiply 4 we get $-4\le 4\sin x\le 4$ and we can add these two inequality we get $-7\le 3\cos x - 4\sin x\le 7$ b... | Note that\begin{align}3\cos x-4\sin x&=5\left(\frac35\cos x-\frac45\sin x\right)\\&=5\left(\cos\alpha\cos x-\sin\alpha\sin x\right),\end{align}where $\alpha\in\Bbb R$ is such that $\cos\alpha=\frac35$ and that $\sin\alpha=\frac45$. Therefore$$3\cos x-4\sin x=5\cos(\alpha+x)\in[-5,5].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$ For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$
My proof is using SOS$:$
$${c}^{2}{a}^{2} {b}^{2}\Big( \sum a\B... | For $a,\,b,\,c$ are real numbers. We have
$$(a+b+c)^2 =(|a+b+c|)^2 \leqslant (|a|+|b|+|c|)^2,$$
$$\frac{a+b+c}{abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \leqslant \left |\frac{1}{ab} \right |+\left |\frac{1}{bc} \right |+\left |\frac{1}{ca} \right |=\frac{|a|+|b|+|c|}{|a||b||c|}.$$
So, we need to prove
$$4\left(\d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$ Let $a$ and $b$ be positive integers such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$$ Find $a + b.$
We let $\sqrt[3]a, \sqrt[3]b$ be $x,y,$ respectively. We expand LHS to get $$x... | Let $$\alpha = \sqrt[3]{a} + \sqrt[3]{b} = 1 + \sqrt{49 + 20 \sqrt[3]{6}}$$
We can compute the minimal polynomial of the RHS:
(1 + sqrt(49 + 20 * (6)^(1/3))).minpoly()
x^3 - 72*x - 336
and then applying Cardano's formula we get $\alpha = 2\cdot (6^{1/3} + 6^{2/3})$.
We can transform this into $a,b$ form with $2^3 = 8$... | {
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"url": "https://math.stackexchange.com/questions/3814269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that st... | For $n=1$ the last term on the lefthand side is $\frac1{1\cdot0}$, which is undefined. The induction has to start at $n=2$, and as you say, the statement is false for $n=2$. The fact that the induction step works (after you correct the sign error in your answer, which I suspect is a typo) means that the formula $\frac3... | {
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"source": "stackexchange",
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Collatz Conjecture: Is there a straightforward argument showing that there are no nontrivial 2 "step" repeats (where each "step" is an odd number) Let:
*
*$C(x) = \dfrac{3x+1}{2^w}$ where $w$ is the highest power of $2$ that divides $3x+1$
*$C_n(x) = C_1(C_2(\dots(C_n(x)\dots)) = \dfrac{3^n x + 3^{n-1} + \sum\limits... | You got
$$x = \frac{3 + 2^{w_1}}{2^{w_1 + w_2} - 9} \tag{1}\label{eq1A}$$
For $x$ to be a positive integer means the denominator must divide evenly into the numerator which means the denominator must also be less than or equal to the numerator.
For somewhat easier algebra, let
$$y = 2^{w_1} \tag{2}\label{eq2A}$$
so \eq... | {
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"url": "https://math.stackexchange.com/questions/3816569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2... | try :$$x^5+x+1=(ax^3+bx^2+cx+d)(Ax^2+Bx+C)$$and compare coefficients.!
Again intelligent guessing works the best
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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How to find the $n$-th derivative of the function $\log(1-x(1-x))$ I know how to find the $n$ derivative of $\log (1-x)$ which is $-\frac{(n-1)!}{(1-x)^n}$, but how to find the $n$-th derivative of the function $\log(1-x(1-x))$?
| By purely symmetry conditions we get that $x(1-x)$ is a parabola with a vertex at $x=\frac{1}{2}$ which means we can rewrite it as
$$x(1-x) = \frac{1}{4}-\left(x-\frac{1}{2}\right)^2$$
Then the log becomes
$$\log\left(\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\right)$$
Finding the $n$th derivative of this function at $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\sin80^\circ\sin20^\circ\sin x = \sin 10^\circ\sin60^\circ\sin(30^\circ+x)$
Solve:
$$\sin80^\circ\sin20^\circ\sin x = \sin 10^\circ\sin60^\circ\sin(30^\circ+x)$$
I tried cancelling $\sin10^\circ$
$$\sin80^\circ\cdot 2\cos10^\circ\cdot\sin x=\sin60^\circ\sin(30+x)$$
$$\frac{2\sin^280^\circ}{\sin60^\circ} = \f... | We have $$\frac{\sin80^{\circ}\sin20^{\circ}}{\sin10^{\circ}\sin60^{\circ}}=\frac{1}{2}\cot{x}+\frac{\sqrt3}{2}$$ or
$$\frac{4\sin80^{\circ}\cos10^{\circ}}{\frac{\sqrt3}{2}}-\sqrt3=\cot{x}$$ or
$$\cot{x}=\frac{8\cos^210^{\circ}-3}{\sqrt3}$$ or
$$\cot{x}=\frac{1+4\cos20^{\circ}}{\sqrt3}.$$
Now, $$\frac{1+4\cos20^{\circ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$... | $p=\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 + \cdots$
So $p+1=1+\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 +\cdots$
Now notice that $$(1+x)^n=1+nx+\frac{n(n-1)}... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ where $f$ is a real quadratic function? Let's consider the following function $f(x)=ax^2+bx+c$, where $a, b$ and $ c $ are all real constants. Is there any way to calculate the value of $$\sum_{n=1}^{\infty} \frac{1}{f(n)}\text{?}$$ I can find no sequences but that $\sum_{n=... | As @Mourad commented, rewrite
$$a n^2+b n +c=a (n-r)(n-s)$$ where $r$ and $s$ are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition
$$\frac 1{a n^2+b n +c}=\frac 1{a(r-s)} \left(\frac 1{n-r}-\frac 1{n-s} \right)$$ So, for the partial sum
$$S_p=\sum_{n=1... | {
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"url": "https://math.stackexchange.com/questions/3825222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ re... | The first test of factors using the rational root theorem yields
$$(x - 1) (x^6 + x^5 + 6 x^4 + 6 x^3 + 7 x^2 + 4 x + 7) = 0$$
From this we can see that, for one root, $X=1$.
From the co-factor, we can see that the function never crosses the $x$-axis so there are no more real roots.
Answer: $D)1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x... | Ok so by @player3236's hint , he said that taking $(x = y)$ will create a problem as then it would imply that $\frac{1}{a} + \frac{1}{a+x} = 0$ which is giving $\frac{1}{b} = 0$ and stopping me to move on.
For now, I am taking say $x = 1$ , $y = 2$ .
This gives :- $a = -\frac{2}{3}$ in the $3$rd equation .
Now in the $... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(... | We need
$$\left\{\begin{aligned} & x+1 > 0 \\& x+1 \ne \sqrt{x+1} \end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x > -1 \\& x \ne 0\end{aligned}\right. \quad (1)$$
Setting $t = \sqrt{x+1} > 0,$ then $x=t^2-1,$ now
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} = \frac{(t+1)^2}{t^2},$$
and
$$\frac{{x}^{2}+3x+18}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In which base $b$ is $(374)_b$ a perfect square? If you convert this number to base 10, we can obtain the expression $$3b^2+7b+4 = (b+1)(3b+4).$$
Since $\gcd(b+1,3b+4) = 1$, we further conclude that both $b+1$ and $3b+4$ are perfect squares.
So the problem is equivalent to finding $b$ that satisfies the Diophantine equ... | You're off to a good start; if $b+1$ and $3b+4$ are both perfect squares, then
$$3b+4=x^2\qquad\text{ and }\qquad b+1=y^2,$$
for some integers $x$ and $y$, and hence
$$x^2-3y^2=1.$$
This is a Pell equation, and its solutions are well known. I suggest to start from the Wikipedia page to understand how to find all integr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding an epsilon-delta proof of a sequence I'd like to write an epsilon-delta proof that
$$
\lim_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0
$$
so what I've done is,
$\lvert \frac{2x-3}{x^2-2x-5} \rvert < \frac{2x}{x^2-2x-5}$ if $x>3/2$.
$$
\frac{2x}{x^2-2x-5} < \frac{2x}{x^2-7x} < \epsilon
$$
$$
x > \frac{2}{\epsilon... | You are essentially correct, though you are going to want $x > \sqrt{6}+1$ to avoid division by $0$ and to ensure $\frac{2x-3}{x^2-2x-5}>0$ Later you will want $x >7$ for the same reason. and since $\frac2\epsilon+7>7>\sqrt{6}+1> \frac32 >1$ you can slightly simplify your $N(\epsilon)$ expression. While your exploratio... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding original function from composition of function If $f(f(x)) = x^2 + 2$, then find $f(11)$? Given that if $a>b$ then $f(a)>f(b)$
I got this question from a study group of which I am part of.
There the question was described as Let $x,f(x),a,b$ be positive integers and if $a>b$ then $f(a)>f(b)$ and $f(f(x)) = x^2 ... | So the function has to be defined for positive integers. Then, we must have $f(x)>x$, because $f(x)<x$ would imply $f(f(x)<f(x)<x$, but $x^2+2>x$, and $f(x)=x$ is equally impossible. Replacing $x$ by $f(x)$ shows that $f(f(f(x))=f(x)^2+2=f(x^2+2)$. Since $x<f(x)<f(f(x))=x^2+2$, we must have $1<f(1)<1^2+2=3$, i.e. $f(1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Dependent variable substitution of a differential equation. I am attempting to answer a question from a textbook. The question is as follows:
"Use the substitution $y = x^2$ to turn the differential equation $x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$ into a second order differential equation with c... | In general, you are incorrect that $\left(\frac{dx}{dt}\right)^2 = \frac{d}{dt}\left(x^2\right)$ since $\frac{d}{dt}\left(x^2\right) = 2x\left(\frac{dx}{dt}\right)$, so they're equal only in the special case where $2x = \frac{dx}{dt}$. Instead, multiply the original equation on both sides by $2$ to get
$$2x\frac{d^2x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral with the indicator function using spherical coordinates The integral of interest is:
$$I = \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} \boldsymbol{1}\left(\frac{1}{2}\frac{(x_2^2 - x_1^2) + (y_2^2 - y_1^2) + (z_2^2 -z_1^2)}{x_2-x_1} \in [0,1]\right) \nonumber \\
\times \boldsymbol{1}\left(2\mathrm{arcsin}\left(\fra... | You already made it to spherical coordinates. Using $r$ instead of $\rho$ and introducing polar coordinates for $t_3=\rho \cos \phi$ and $t_5=\rho \sin \phi$, the integral becomes $$I = \frac{1}{4} \int_0^{2\pi}\mathrm{d}\varphi\int_0^\pi\mathrm{d}\theta \int_0^\infty \mathrm{d}r\, r^2\sin \theta \int_0^{2\pi} \mathrm{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\... | It is not bad if you write
$$(x+1)^2+1=(x-a)(x-b)$$
$$\frac 1{(x+1)^2+1}=\frac 1{a-b}\left(\frac 1{x-a}-\frac 1{x-b} \right)$$ making that you face two integrals
$$I_c=\int \frac {\log^2(x)}{x-c}\,dx=\log ^2(x) \log (x-c)-2\int \frac{ \log (x) \log (x-c)}{x}\,dx$$
$$I_c=-2 \text{Li}_3\left(\frac{x}{c}\right)+2 \log (x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Given $|b-c|\leq a\leq b+c$ show that $\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$ I am trying to show the following:
Given that $|b-c| \leq a \leq b+c$ show that:
$$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$
So far I have done the following:
$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$
$1-1/(1+a) \leq 2 - 1/(1+b)... | I suppose $b,c\ge0$.
$$
\frac{b}{1 + b} + \frac{c}{1 + c} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow \frac{b + c + 2bc}{1 + b + c +bc} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow b + c + 2bc + ab +ca + 2abc\ge a + ab + ca + abc
$$
$$
\Rightarrow b + c + 2bc + abc \ge a
$$
Add
$$
b + c \ge a \quad \textrm{and} \quad bc(a +2)\ge0
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides ... | Please always try and draw a sketch. It helps in ensuring you have the right bounds and also understand the easy way to find the area.
Now in the sketch, you can see that there are two circles -
i) Centered at $(0,0)$ with radius $1$
ii) Centered at $(3/2, 0)$ with radius $3/2$
As they are both symmetric to X-axis, th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | You start by writing
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
but that's not a "sentence". It's just noun without a verb. What about $\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$? what are you saying about it? Are you claiming it is equal to something? To what?
Then you go from th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Collatz Conjecture: Does this argument show if a non-trivial cycle exists, the sum of powers of $2$ must be minimal power of $2 > 3^n$? What is wrong with this argument?
I am sure that I am misunderstanding something or there is a mistake in this argument. This argument is taken from the answer given to one of my ques... | You wrote in your last line
$x_{\text{min}} < \dfrac{1}{3(2^{\frac{1}{n}} - 1)} < \dfrac{1}{3}$
However, note for $n \gt 1$ that $2^{1/n} \lt 2 \implies 2^{1/n}
- 1 \lt 1$, so $3(2^{1/n} - 1) \lt 3$ and, thus, $\frac{1}{3\left(2^{1/n} - 1\right)} \gt \frac{1}{3}$. For example, $n = 10$ gives
$$\frac{1}{3\left(2^{0.1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $n$ is odd, then $n/2 + 1/2$ is always even?
If $n$ is odd, prove that $n/2 + 1/2$ is even.
Context: I'm a Statistician and the term $n/2 + 1/2$ showed up in the index of a summation when deriving the pdf of some Order Statistic:
$$
\sum_{j = (n+1)/2}^{n}...
$$
I realized that $n/2 + 1/2$ is always even if $n$ is ... | If $n$ is odd then $n$ is of the form $2k - 1$.
The number $\frac n2 +\frac 12$ is simply $\frac {n+1}2 = \frac {2k-1 +1}2 = \frac {2k}2 = k$. The expression $\frac n2 + \frac 12$ is simply the the index of which odd number $n$ is.
If $n = 1,3,5,7,9,......, 2k -1 ,.....$ then $\frac n2 + \frac 12 = 1,2,3,4,5,.....,k,.... | {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate combination with 2 variables? Combination formula with 1 variable is simple $nCr = n! / (r!(n-r)!)$
So to find a number of combinations for rolling a 6 ONCE out of 6 rolls is
$$\dfrac {6! }{ (1!(6-1)!)} = 6$$
But how do you find number of combinations for rolling a 6 ONCE AND rolling a 4 TWICE out of 6... | Assuming the die is fair, the probability of obtaining a 6, two fours, and three numbers other than 4 or 6 in that order is
$$\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^3$$
However, we have to multiply that number by the number of orders in which one 6, two 4s, and three numbers other th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\sqrt a + \sqrt b + \sqrt c$ only in terms of $p$ .
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$
$$ax^2 = by^2 = cz^2$$
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a + \sqrt b + \sqrt c$ in terms of $p$ .
What I Tried :- How do you use the information here? I assumed that :-
$$ax^2 = by^2... | proceeding from your method,$$a^2x^4=b^2y^4=c^2z^4=k^2$$
thus $a^2x^3+b^2y^3+c^2z^3=k^2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=p^5$
$k=p^3$
can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-... | Let $a_n=3^{2n +2}-8n -9$. Then the power series $f (z )=\sum_{k =0}^{\infty }a_k z^k$ can be written as
$$f(z)=\frac{b_0+b_1 z +b_2 z^2}{(1-c_0 z)(1-c_1 z)(1-c_2z)}$$
for some nonnegative integers $b_i$, $c_j$, and furthermore the common divisors of the $b_i$ are divisors of the $a_n$. Can you take it from there?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{\tan8°}{1-3\tan^2 8°}+\frac{3\tan24°}{1-3\tan^2 24°} + \frac{9\tan72°}{1-3\tan^2 72°} + \frac{27\tan216°}{1-3\tan^2 216°}=x\tan108°+y\tan8°$ $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8° .$$
Find the valu... | $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8°$$
Now, $$\frac {\tan 8°}{1-3\tan^2 8°}=\frac{1}{8}\left(\frac {8 \tan 8°}{(1-3\tan^2 8°)}+\tan 8°-\tan8° \right) \\
=\frac{1}{8}\left(-\tan8° \right)+\frac{1}{8}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Definite integrals of fractions of trig functions $$
\int _{\pi/4}^{\pi/3}\sin(x)\frac{\sin(x)+\cos^2(x)}{\sqrt{2}\sin(x)+\cos(x)}
\,\mathrm{d}x
$$
How do you solve for this integral?
I was considering using Weierstrass substitution or manipulating the integrand with respect to $\sec(x)$ or $\csc(x)$ but couldn't yiel... | Using Weierstrass substitution, I suppose that you did arrive at
$$-4\int \frac{t \left(t^4+2 t^3-2 t^2+2 t+1\right) } {\left(t^2+1\right)^3 \left(t^2-2 \sqrt{2} t-1\right) }\,dt=-4\int \frac{t \left(t^4+2 t^3-2 t^2+2 t+1\right) } {\left(t^2+1\right)^3 \left(t+\sqrt 3-\sqrt 2\right)\left(t-\sqrt 3-\sqrt 2\right) }\,dt$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve parameter function
Solve $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$ over parameter $m$
I dunno what to process but I want to show some of my work at least. I try to solve for $m$ firstly so I get
\begin{equation}
m_1=\frac{-2^{\sin{x}}+\sqrt{4-3×4^{\sin{x}}}}{2}\\m_2=\frac{-2^{\sin{x}}-\sqrt{4-3×4^{\sin{x}}}}{2}
\end... | $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$
Say, $y = 2^{\sin x}$ and so we have $y^2 + my + (m^2-1) = 0$ ...(i)
For equation to have any real solution, its discrimant has to be $\ge 0$ i.e.
$m^2 - 4(m^2-1) \ge 0 \implies - \frac{2}{\sqrt 3} \le m \le \frac{2}{\sqrt 3}$ ...(ii)
Now solving (i), $ (y + \frac{m}{2})^2 = 1 - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Euclidean algorithm for greatest integers x and y for common divisor (GCD) I have a problem with finding the gcd of two numbers: gcd(4620, 8190) = 210.
I did the following:
8190 / 4620 = 1 with remainder: 3570
4620 / 3570 = 1 with remainder: 1050
3570 / 1050 = 3 with remainder: 420
1050 / 420 = 2 with remainder: 210
42... | $$ \frac{ 8190 }{ 4620 } = 1 + \frac{ 3570 }{ 4620 } $$
$$ \frac{ 4620 }{ 3570 } = 1 + \frac{ 1050 }{ 3570 } $$
$$ \frac{ 3570 }{ 1050 } = 3 + \frac{ 420 }{ 1050 } $$
$$ \frac{ 1050 }{ 420 } = 2 + \frac{ 210 }{ 420 } $$
$$ \frac{ 420 }{ 210 } = 2 + \frac{ 0 }{ 210 } $$
Simple continued fraction tableau:
$$
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Summing the first $n$-terms of the series whose general term is $nx^{n-1}$ I suppose several of you know some fancy ways to establish the formula for the sum of the first $n$ terms of the geometric series $$1+x+x^{2}+x^{3}+ \ldots $$
Can you share below some of your fave ways to sum the first $n$ terms of the series
$$... | $1 + 2x + 3x^2 + \cdots + nx^{n-1}$
= $1 + x + x^2 + \cdots + x^{n-1} $
$+ x + x^2 + \cdots + x^{n-1} $
$+ x^2 + \cdots + x^{n-1} $
$+ \cdots + x^{n-1}$
$= \dfrac{x^n-1}{x-1}$
$+\dfrac{x^n-x}{x-1}$
$+\dfrac{x^n-x^2}{x-1}$
$+\cdots + \dfrac{x^{n}-x^{n-1}}{x-1}$
$=\dfrac{nx^n-(1+x+x^2+\cdots+x^{n-1})}{x-1}$
$=\dfrac{nx^n... | {
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"timestamp": "2023-03-29T00:00:00",
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If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x) (Sweden 1950) If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x)
I have been trying to do this question but I have not succeeded. I h... | You have $g(x) = \frac{P(x)}{x^2-4}$ and you want to find $P$.
First condition about $\lim_{x \to \infty}$ gives us the rank of $P$ together with coefficient. Indeed, if $P$ is arbitrary polynomial $a_nx^n + ... + a_1x + a_0$ where $a_n \neq 0$ then
$$ \lim_{x \to \infty} \frac{P(x)}{x^2-4} = \begin{cases} 0 & n \in \{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Advice on integrating $\int \frac{x^2 +n(n-1)}{(x \sin x +n \cos x)^2} dx$ Our integral is:
$$\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} dx$$
I considered that this can be turned into some sort of quotient differential:
$$ d \frac{u}{v} = \frac{ v du - u dv}{v^2}$$
Now, comparing this with our integral:
$$ v = x \... | $$I=\int \frac{x^2+n(n-1)}{(x\sin x+n \cos x)^2} dx$$
Multiply up and down by $x^{2n-2} \cos x$, then
$$I=\int \frac{(x^2+n(n-1))x^{2n-2} \cos x dx}{(x^n \sin x+n x^{n-1} \cos x)^2 \cos x}$$
Let $$(x^n \sin x+n x^{n-1} \cos x)=t \implies x^{n-2} \cos x(x^2+n(n-1)) dx=dt$$
$$\implies I=\int \frac{(x^2+n(n-1))x^{n-2} \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is ... | Also, $uvw$ helps here very well.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, after homogenization your last inequality it's $f(w^3)\geq0,$ where $f$ is a linear function,
which says that it's enough to prove it for the extreme value of $w^3$,
which by $uvw$ happens in the following cases.
*
*$w^3\ri... | {
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Generating function of Trinomial Coefficients
Let $f(n)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose k}{n-k\choose k}$. Show that $\sum_{n=0}^{\infty}f(n)x^n=\frac{1}{\sqrt{(1-3x)(1+x)}}$.
I managed to find the following recurrence relation for the trinomial coefficients in Wolfram: $(n+2)a_{n+2}=(2n+3)a_{n+1}+3(n... | Here is a direct computation of the generating function, using the binomial series:
\begin{align*}
\sum_{n=0}^\infty f(n)x^n\quad
&=\sum_{n=0}^\infty x^n\sum_{0\leqslant 2k\leqslant n}\frac{n! }{k!^2(n-2k)!}
=\sum_{k=0}^\infty\sum_{n=2k}^\infty\frac{n! \ x^n}{k!^2(n-2k)!}
\\\color{gray}{[\text{replace $n$ with $n+2k$}]... | {
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"source": "stackexchange",
"question_score": "5",
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Determine whether a recurrence relation converges and, if yes, find its limit. I have the sequence defined as $$ a_{n+1} = \dfrac{5a_n -6}{a_n -2} $$ for $n \geq 2$, $a_n \neq 2$ for all $n \geq 1$ and $a_n$ is a real number.
I want to determine whether the sequence converges and if it does, find it's limit.
There are ... | Let's experiment with some values of $n$.
\begin{align}
a_2 &= \frac{5a_1 - 6}{a_1 - 2} \\
a_3 &= \frac{19a_1 - 18}{3a_1 - 2} \\
a_4 &= \frac{77a_1 - 78}{13a_1 - 14} \\
a_5 &= \frac{307a_1 - 306}{51a_1 - 50}.
\end{align}
The coefficient of $a_1$ in the numerator generates the sequence $5, 19, 77, 307, \dots$ and the co... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Computing $\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz$ using Cauchy integral formula Let $\alpha(t) = re^{it}$ where $|a|<r<|b|$ and $t \in [0,2\pi]$. I'd like to compute
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz \ \ \ \ n, m \in \mathbb{N}.$$
It appears that the answer is
$$2\pi i (-1)^m{n + m -2 \choose n-1}\frac{1}{(b-... | Answering my own question, thanks to the comments.
Let $f(z) = \frac{1}{(z-b)^m}$, then by Cauchy integral formula
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = \int_{\alpha}\frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a).$$
Trying to find an expression of $f^{(n-1)}(z)$
\begin{align*}
f^{(1)} &= -m\frac{1}{(z... | {
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"source": "stackexchange",
"question_score": "2",
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On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part II (This question is an offshoot of this earlier post.)
Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of the aliquot divisors of $x$ b... | Too long to comment :
You used an inequality
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)\le \sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)$$
This inequality is true, but we can have
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\B... | {
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"source": "stackexchange",
"question_score": "2",
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find function with condition using Leibniz okay, so we have $f:(0, +\infty) \to (0, +\infty)$, with $f(1)=2$, such as $f'(\frac{1}{x})=\frac{1}{f(x)}$.
Now, I know we're supposed to use Leibniz's binomial theorem. And the answer to the problem is $$f(x)=(1+x)e^\frac{1-x}{2+2x}$$
Further than that, I've tried working wi... | Hint:
If
$$
f'\left(\frac{1}{x}\right)=\frac{1}{f(x)}
$$
then
$$
f'(x)=\frac{1}{f\left(\frac{1}{x}\right)}
$$
and
\begin{align}
\frac{d}{dx} \left(
f(x)f\left(\frac{1}{x}\right)\right)&=f'(x) f\left(\frac{1}{x}\right)-\frac{f(x)f'\left(\frac{1}{x}\right)}{x^2}\\
&=1-\frac{1}{x^2}\ .
\end{align}
Therefore, by integratin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find value of $\sec80^\circ-2\cos20^\circ-\dfrac{4}{\sec20^\circ}+\dfrac{1}{2\sin10^\circ}$ My attempt :
\begin{align*}
\dfrac{1}{\cos80^\circ}-2\cos20^\circ-4\cos20^\circ+\dfrac{1}{\sin10^\circ}\\
\left(\dfrac{1}{\sin10^\circ}+\dfrac{1}{2\sin10^\circ}\right)-6\cos20^\circ\\
\dfrac{3}{2\sin10^\circ}-6\cos20^\circ
\end{... | Use Werner Formulas, $2\sin B\cos A=\sin(A+B)-\sin(A-B)$
$$\dfrac{3(1-4\sin10^\circ\cos20^\circ)}{2\sin10^\circ}$$
$$=\dfrac{3(1-2(\sin(20+10)^\circ-\sin(20-10)^\circ))}{2\sin10^\circ}$$
$$=?$$
Generalization:
$$\dfrac1{2\sin x}-2\cos y=\dfrac{1-2(\sin(x+y)+\sin(x-y))}{2\sin x}$$
We need $\sin(x+y)=\dfrac12$
and $\sin... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How many real roots does $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$ have? A multiple choice questions asks me to find the number of real roots of $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$.
A brute force approach involving the factor theorem and algebraic long division gives $3$ real roots.
Is there a quicker way?
I w... | Using factorization it's very quickly here:
$$2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4=$$
$$=2x^5+2x^4-5x^4-5x^3+6x^3+6x^2-10x^2-10x+4x+4=$$
$$=(x+1)(2x^4-5x^3+6x^2-10x+4)=$$
$$=(x+1)(2x^4-4x^3-x^3+2x^2+4x^2-8x-2x+4)=$$
$$=(x+1)(x-2)(2x^3-x^2+4x-2)=(x+1)(x-2)(2x-1)(x^2+2),$$
which gives three real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Using differentiation to find the power series of a fairly tricky function!! (b) Use part (a) to find a power series for: $$f(x) = \frac{1}{(8+x)^3}$$
What is its radius of convergence?
From Using differentation to find a power series expression for a function We know that
$$\frac{1}{(8+x)^2}=\sum_{n=0}^\infty (-1)^{n... | Looks good to me.
Maybe it's "nicer" to write it as $$f(x)=\sum_{n=0}^{\infty}(-1)^n(n+1)(n+2)\frac{x^n}{2^{3n+10}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877312",
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"source": "stackexchange",
"question_score": "2",
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Define $f(x) = x^6 + x^5 + 3x^4 +x^3 + 3x^2 + x + 1$. Find the largest prime factor of $f(19) + 1$ (Homework) Define $f(x) = x^6 + x^5 + 3x^4 +x^3 + 3x^2 + x + 1$. Find the largest prime factor of $f(19) + 1$ This problem is from a homework set of my class at source: Alphastar.academy. I believe there a number of ways ... | Conveniently, $f(x)+1$ factors as
$$(x^2-x+1)(x^2+x+1)(x^2+x+2)$$
With $x=19$ this produces the three factors $343×381×382$, from which we work out that the largest prime factor is $191$ (of $382$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Initial value problem with separation of variables
I have this initial value problem. Also I am a beginner on the topic.
As far as I know, I should separating the variables and integrate and with using initial condition I should have get rid off from the constant(C).
But the problem is, the equation kinda looks scary ... | $$x^3y'-3x^2y=\dfrac {2x^7(4x^3-3y)}{3x^5+3x^3+2y}$$
Substitute $w=\dfrac y {x^3}$:
$$\left(\dfrac y {x^3}\right)'=\dfrac {2x(4-3y/x^3)}{3x^2+3+2y/x^3}$$
$$w'=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
Then take $v=x^2+1$
$$\dfrac {dw}{dx}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$\dfrac {dw}{dx^2+1}\dfrac {dx^2+1}{dx}=\dfrac {2x(4-3w)}{3x... | {
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"source": "stackexchange",
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If $x$ and $y$ are two linearly independent column $n$-vectors how can I find all the eigenvalues of $xx^{T}-yy^{T}$ If $x$ and $y$ are two linearly independent column $n$-vectors where $n\geq2$ .find all the eigenvalues of $xx^{T}-yy^{T}$
I know that because the matrix $xx^T-yy^T$ has rank $2$. So $n-2$ of the eigenva... | Using the identity
\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}
for $A \in F^{m \times n}$ and $B \in F^{n \times m}$,
we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \time... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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The process of using taylor series to evaluate limits. For instance we want evaluate this simple limit using taylor series :
$$L=\lim_{x\to 0}\frac{\sin x}{x^5}=\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}$$
In this case we only care about coefficient of $x^5$ to cancel ... | We can't ignore the other terms, we obtain indeed
$$\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}=\frac1{x^4}-\frac1{3!x^2}+\frac1{5!}+O(x^2) \to \infty$$
the limit is finite for
$$\lim_{x\to 0}\frac{\sin x-\left(x-\frac{x^3}{3!}\right)}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{5!}-\frac{x^... | {
"language": "en",
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"source": "stackexchange",
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Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians ... | I think, it's better to use $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$
$$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ because from the given easy to get $a^2$, $b^2$ and $c^2$.
Indeed, $$\frac{1}{3}\sqrt{2b^2+2c^2-a^2}=2\sqrt3,$$
$$\frac{1}{3}\sqrt{2a^2+2c^2-b^2}=2\sqrt2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$? I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?
| Using partial fraction decomposition (left as an exercise for the reader), notice that $$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$ Therefore, $$\int \frac{1}{x^8(x^2 + 1)} ~ dx = \int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2... | Expressions like $ \ a^2+ab+b^2 \ , \ a^2+ac+c^2 \ , \ b^2+bc+c^2 \ \ $ make me think of "differences of two cubes", so let's see what happens. If we call the progression $ \ a \ = \ b - d \ , \ b \ , \ c \ = \ b + d \ \ , $ then we may write
$$ \ a^2 \ + \ ab \ + \ b^2 \ \ = \ \ \frac{b^3 \ - \ a^3}{b - a} \ \ = \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892856",
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"source": "stackexchange",
"question_score": "1",
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Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$
Case $S(1)$ is true:
$$2((1)+2)\leq((1)+2)^{2}$$
$$6\leq9$$
Case $S(n)$ is true for all $n=1,2,...$
$$2(n+2)\leq(n+2)^{2}(i)$$
Case $S\left(n+1\right)$
$$2(n+3)\leq(n+3)^{2}(ii)$$
From (i)
$$2(n... | Note that$$2(n+2)-2(n+1)=2$$and that$$(n+3)^2-(n+2)^2=2n+5.$$So, if $2(n+1)\leqslant(n+2)^2$, you have\begin{align}2(n+2)&=2(n+1)+2\\&\leqslant(n+2)^2+2\\&\leqslant(n+2)^2+2n+5\\&=(n+3)^2.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894498",
"timestamp": "2023-03-29T00:00:00",
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Integrate $\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)$ This is question from Axler's Measure, Integration, & Real Analysis Problem 5.B.1, probability measures.
Let $\lambda$ denote Lebesgue measure on $[0,1]$. Show that
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\... | Note that $$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right), $$
Riemann and Lebesgue integrals coincide on finite intervals when the integrand is Riemann itegrable.
We can evaluate the inner integral as a Riem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894653",
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"source": "stackexchange",
"question_score": "2",
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Solving $\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$ without series expansion or L'Hopital's rule I want to find the following limit without using series expansion or L'Hopital's rule. I tried replacing $x$ with $2x$.
$$\begin{aligned}L_{1}&=\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3... | HINT
To simplify the derivation we can proceed starting to prove the followings
$$L_1: \frac{\cos x-1}{x^2}\to \frac12 \implies L_2: \frac{x\cos x-x+\frac12x^3}{x^5}\to \frac1{24}$$
$$L_3:\frac{x-\sin x}{x^3}\to \frac16 \implies L_4:\frac{3(x-\sin x)-\frac12 x^3}{x^5}\to -\frac1{40}$$
and then use that
$$\frac{2+\cos x... | {
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"source": "stackexchange",
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Find $\int_0^bg(x,u)\,du$ given $\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$ I encounter the following equation, which holds for any $x\in[0,b]$:
$$\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$$
I would like to solve for $G(x,b)=\int_0^b g(x,u)du$, subject to the condit... | We have \begin{align}\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)\,du&=\frac{x^2+b^2x+b^2-2b}{x-1}\tag1\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_x(x,u)\,du&=1-\frac{2b^2-2b+1}{(x-1)^2}\tag2\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_{xx}(x,u)\,du&=\frac{2(2b^2-2b+1)}{(x-1)^3}\end{align} so assume that $(x-1)^3g_{xx}(x,u)$ is constant. Then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Elementary Ways to Solve System of Exponential Equation Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation:
$$ \begin{cases}
3^{x+y}+2^{y-1}=23, \\
3^{2x-1}+2^{y+1}=43.
\end{cases} $$
I have tried to eliminate the exponent of 2 but it gets me
$$ 12 \cdot 3^{x + y} + ... | Hint:
$$(3^x)^2+(12\cdot3^y)3^x-405=0$$
The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$
For rational $3^x,$ we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say)
$$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$
$$\implies d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}... | $\begin{array}\\
\binom{n+1}{3}-\binom{n}{3}
&=\dfrac{(n+1)n(n-1)}{6}-\dfrac{n(n-1)(n-2)}{6}\\
&=\dfrac{n(n-1)((n+1)-(n-2))}{6}\\
&=\dfrac{3n(n-1)}{6}\\
&=\dfrac{n(n-1)}{2}\\
&=\binom{n}{2}\\
\end{array}
$
Therefore
$\sum_{i=1}^{n}\binom{i}{2}
=\sum_{i=1}^{n}(\binom{i+1}{3}-\binom{i}{3})
=\binom{n+1}{3}
$.
Note that
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902467",
"timestamp": "2023-03-29T00:00:00",
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Show $f(x,y)=\frac{x^3\sin(x-y)}{x^3-y^3}$ for $x\neq y$, $f(x,y)=0$ for $x=y$ is continuous at $(0,0)$. I know that once we bound $f$ for $x\neq y$ we are done because $f=0$ for $x=y$. I got that for $x\neq y$, $|f(x,y)|\le \left|\frac{x^3}{x^2+xy+y^2}\right|$ because $\left|\frac{\sin(a)}{a}\right|\leq 1$ for $a\neq ... | so following @Kelenner's hint, $x^2+xy+y^2 = \frac{x^2+y^2}{2} +\frac{(x+y)^2}{2} \geq \frac{x^2+y^2}{2}$, in particular it must be positive, and $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq |\frac{2x^3}{x^2}| =2|x|$, which tends to $0$ as $(x,y)\rightarrow(0,0)$.
We can also say that $|f(x,y... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{-\frac{1}{\sqrt 3}}^{\frac{1}{\sqrt 3}} \frac{x^4}{1-x^4} \cos^{-1} (\frac{2x}{1+x^2}) dx$ Note that the integral can(not) be simplified as
$$
2\int_0^{1/\sqrt 3} \frac{x^4}{1-x^4} \cos^{-1} \left(\frac{2x}{1+x^2}\right)dx
$$
Since $\cos^{-1}$ is not an even function. Let $x=\tan y$
$$
\implies
\int_0^{... | Hint:
$$\int_{-a}^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx$$
$$=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx+\int_0^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}dx=I_1+I_2$$
For the first integral set $x=-y,dx=-dy$
$$I_1=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx=-\int_a^0\dfrac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\frac{1}{3+\sin x+\cos x}$ I can easily do it with $\tan {\frac{x}{2}}$ substitution, but my problem requires $x\in(0,2\pi)$.
$\int \frac{1}{3+\sin x+\cos x}dx, x\in(0,2\pi)$
How do i solve it now ?
| Take $\sin x=2t/(1+t^2), \cos x=(1-t^2)/(1+t^2), tan(x/2)=t \implies 2 \sec^2(x/2) dx=dt$. Then
$$I=\int \frac{dx}{3+\sin x+ \cos x}=\frac{1}{4}\int \frac{dt}{t^2+t+2}=\frac{1}{4}\int \frac{dt}{(t+1/2)^2+7/4}==\frac{1}{2\sqrt{7}} \tan^{-1} \frac{2t+1}{\sqrt{7}}+C.$$
where $t=\tan(x/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{... | $$\tan \alpha =2 \implies \tan 2\alpha = \frac{2\cdot 2}{1-2^2}=-\frac{3}{4} \implies tan(\pi-2\alpha)=\frac{3}{4} \\\implies \cos (\pi-2\alpha)=\frac 35. \blacksquare$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$ If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$
So $x+y+z=0$
We need t... | Hint
Define
$$
x={a-b\over b-c}\quad,\quad y={b-c\over c-a}\quad,\quad z={c-a\over a-b}
$$
and try to minimize $x^4+y^4+z^4$ constrained to $xyz=1$.
Alternatively, you can also define
$$
x=\ln\left|{a-b\over b-c}\right|\quad,\quad y=\ln\left|{b-c\over c-a}\right|\quad,\quad z=\ln\left|{c-a\over a-b}\right|
$$
and minim... | {
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"url": "https://math.stackexchange.com/questions/3926587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proof verification of $6^n \geq n3^n$ n is a natural number. Prove $6^n \geq n3^n$ holds for every natural number.
$n = 1:$
$$6 \geq 3 $$
$n \rightarrow n + 1:$
$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$
Is my proof correct?
EDIT:
$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli in... | Suppose $6^n\ge n\cdot 3^n$ is true for $n$ and let's prove it for $(n+1)$.
$6^{n+1}=6\cdot 6^n\ge 6\left(n\cdot 3^n\right)=2\cdot 3\left(n\cdot 3^n\right)=2 n\cdot 3^{n+1}\ge (n+1)3^{n+1}$
proved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$ I have been asked in a exam to find the average of the number: $$n \sin n^\circ$$ for $n$=$2,4,6,\cdots,180$
I have tried a lot basically with sum product, or pairing the inputs...but at the end don't able to find any way to solve it, can someone h... | Since $\sin(180^\circ - \theta) = \sin(\theta)$, $\sin{90^\circ} = 1$, and $\sin{180^\circ} = 0$, we can write the sum as
$$
(2 \sin{2^\circ} + 178 \sin{2^\circ}) + (4 \sin{4^\circ} + 176 \sin{4^\circ}) + \ldots + (88 \sin{88^\circ} + 92 \sin{88^\circ}) + 90\text.
$$
To get the average, divide by the number of terms, $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the name of Fibonacci variation when $F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$, were $c$ is a constant, and $a >0, \ b >0, \ c>0$ I am trying to write $\log(n)$ algorithm for the above. I don't know if there is a specific name for the Fibonacci variation when:
$$F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$$
where... | Understanding that the function is null for negative $n$
$$
F(n) = aF(n - 1) + bF(n - 2) + c\quad \left| {\;F(n < 0) = 0} \right.
$$
so that its first values are
$$
\left\{ \matrix{
F(0) = c \hfill \cr
F(1) = \left( {a + 1} \right)c \hfill \cr
F(2) = \left( {a\left( {a + 1} \right) + b + 1} \right)c \hfill \cr ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3932764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Factoring $4x^{4} - x^{2}$ into $4x^{2}(x-1/2)(x+1/2)$ So i was trying to make an integral but i'm stuck with the easiest part.
How do you get this?
$$4x^{4} - x^{2} = 4x^{2}(x-1/2)(x+1/2)$$
I know how to take the roots $x = 0$, $x = \pm 1/2$,
but when putting it all together I get
$$x(x−1/2)(x+1/2)$$ and I can't rem... | I think you meant $4x^{4} - x^{2}$. In such case, you can factor $x^{2}$ out and apply the following identity:
\begin{align*}
a^{2} - b^{2} = (a+b)(a-b)
\end{align*}
At your case, you have $4x^{4} - x^{2} = x^{2}(4x^{2} - 1)$, where $a = 2x$ and $b = 1$. Consequently, we get that
\begin{align*}
4x^{4} - x^{2} = x^{2}(2... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
| Rearrange $2x^3 - 6x^2 + kx -1=0$ as $x(2x^2+k)=6x^2+1$, and square both sides to get the equation in $x^2$
$$x^6 +(k-9)x^4+(\frac{k^2}4-3)x^2-\frac14=0$$
Then
$$x_1^2+x_2^2+x_3^2= 6= -(k-9)$$
which yields $k=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$... | There is another solution.
Let $\sqrt[5]{16+\sqrt{x}}=1+t$, where $t>0$; therefore, $\sqrt[5]{16-\sqrt{x}}=1-t$. Hence
$$(1+t)^5+(1-t)^5=32,$$
or
$$10\,t^4+20\,t^2+2=32.$$
Immediately we have
$$t^4+2\,t^2-3=0,$$
and $t^2=1$. Since $t>0$, $t=1$ is the single solution. So we have
$$\sqrt[5]{16-\sqrt{x}}=0\Rightarrow \sqr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Evaluate $\sum\limits_{n=1}^{+\infty} \frac{\left( \frac{3-\sqrt{5}}{2} \right)^{n}}{n^{3}}$ Evaluate
$$\sum\limits_{n=1}^{+ \infty} \frac{ \left( \frac{3-\sqrt{5}}{2} \right)^{n} }{n^{3}}$$
We can use the Fourier series to calculate this sum, because it converges.
Also, we know that $\frac{3-\sqrt{5}}{2} = \frac{1}{\... | $$\sum_{n=1}^\infty \frac{a^n}{n^3}=\text{Li}_3(a)$$ Making $a=\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$, you just get a number
$$\text{Li}_3\left(\frac{1}{\varphi ^2}\right)=0.4026839629521090211599594481825111422197338\cdots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.