Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove: $\overline{(\frac{z_1}{z_2})}=\frac{\overline{z_1}}{\overline{z_2}}$
Prove: $$\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}$$
$$\overline{\left(\frac{z_1}{z_2}\right)}=\overline{z_1\cdot z_2^{-1}}=\overline{z_1}\cdot \overline{z_2^{-1}}=\overline{z_1}\cdot \overline{z_2}^{-1}=\... | Let's do this one in two parts.
If $z$ and $w$ are complex numbers, then $\overline{zw}=\overline{z}\,\overline{w}$.
Suppose $z=a+bi$ and $w=c+di$ with $a,b,c,d$ real numbers. Then
$$
\overline{zw} = \overline{(a+bi)(c+di)} = \overline{ac-bd + i(ad+bc)}
= ac-bd-i(ad+bc)
= (a-bi)(c-di)
= \overline{z}\,\overline{w}.
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Looking for a more efficient way to solve the Straight Line Equation problem.
If the straight line through the point $P(3,4)$ makes an angle $\dfrac{\pi}{6}$ with the $x$ axis and meets the line $12x +5y + 10=0$ at Q, find the length of PQ.
My method:
Equation of the given line is;
$y- \sqrt3x+3\sqrt3-4=0$ // using ... | The given line contains the point $P=(3,4)$ and has slope $m=\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$ so its equation is
$$ -\sqrt{3}x+3y-12+3\sqrt{3}=0\tag{1} $$
The second line has equation
$$ 12x+5y+10=0\tag{2} $$
The two equations intersect at a point $Q$ and we wish to find the distance from $P$ to $Q$.
The coordina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$
Find the absolute maximum and minimum of function $f(x,y)=x^2+3y^2-y$ over the region $x^2+2y^2\le2$.
I know how to find maxima/minima if the constraint was the boundary of the circle (by Lagrange's multiplier method... | Rewrite constraint as: $\dfrac{x^2}{2} + y^2 \le 1\implies x =r \sqrt{2}\cos \theta, y = r\sin \theta, 0 \le r \le 1, 0 \le \theta \le 2\pi\implies f(x,y) = f(r,\theta) = 2r^2\cos^2 \theta+3r^2\sin^2 \theta- r\sin \theta\implies f_r = 0 = f_{\theta}\implies 4r\cos^2\theta+6r\sin^2\theta-\sin \theta = 0 = -4r^2\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Computing greatest common divisor and determining whether Diophantine equation has solutions Question:
(1) Compute gcd$(224, 12 + 2^{10})$
(2) Describe in words the set of all integers $ n$ for which the equation $ \ 4x + ny = 1$ has integer solutions.
(3)
Describe in words the set of all integers $ \ n$ for which th... | $224= 4\cdot 7\cdot 8$ and $12+2^{10} = 4(3+2^8)=4\cdot 7\cdot 37$.
So $gcd(224,12+2^{10}) =28$.
The rest is fine.
Ok, ok, we can factor $$2^8+3 = 2^8-2^2+4+3= 2^2(2^6-1)+7 = 4(2^3-1)(2^3+1)+7 =4\cdot 7\cdot 9+7 = 7(36+1) =7\cdot 37$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
| Hint :divide into cases.
$$\forall x\geq 0 \to R.H.S=L.H.S \\2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \to \\2^{x+1}-2^x=2^x-1+1\\2(2^x)-2^x=2^x \checkmark\\ x\in[0,\infty)$$
Then $$for \space -\leq x\leq0 \to 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 \implies x=-2\\
\to 2^{x + 1 } - 2^x = 1-2^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ oc... | Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
We encode the digits
\begin{align*}
2,3,4,5 \qquad\text{as}\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\si... | The given inequality is equivalent to $\cos^2\left(\frac{2\pi}{9}\right)>\frac{4}{7}$, or to $\cos\left(\frac{4\pi}{9}\right)>\frac{1}{7}$.
The minimal polynomial of $\alpha=\cos\left(\frac{4\pi}{9}\right)$ over $\mathbb{Q}$ can be easily derived from $\Phi_9(x)=x^6+x^3+1$, and it is given by $p(x)=1-6x+8x^3$. Since $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form $ z=\frac{1+it}{1-it}$ where $t\in \mathbb{R}$. If $|z|=1$, $z\neq-1$, show that $z$ may be expressed in the form
$$ z=\frac{1+it}{1-it},$$
where $t\in \mathbb{R}$.
I don't know, how to begin. I started with the given conditions. Given that $|z|=1 \text{... | Note that
$\dfrac{1 + it}{1 - it} = \dfrac{(1 + it)^2}{(1 + it)(1 - it)} = \dfrac{(1 + it)^2}{1 + t^2} = \dfrac{(1 - t^2) + 2it}{1 + t^2} = \dfrac{1 - t^2}{1 + t^2} + \dfrac{2it}{1 + t^2}; \tag 1$
since
$(1 - t^2)^2 + (2t)^2 = 1 - 2t^2 + t^4 + 4t^2 = 1 + 2t^2 + t^4 = (1 + t^2)^2, \tag 2$
there is a unique $\theta \in (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
integral involving floor function $\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $ i need to find the result of this integral
$$\int_0^1 \left \lfloor{ (a^{n}x) ^{ \frac{n}{2} }}\right \rfloor dx $$
with $$a \in \mathbb{N}$$
i tried to transform it to a finite sum and i found this:
$$ \frac{1}{a^{... | Let's put
$$
\eqalign{
& y = \left( {a^{\,n} x} \right)^{\,n/2} \cr
& x = {{y^{\,2/n} } \over {a^{\,n} }} \cr
& dx = {2 \over {na^{\,n} }}y^{\,2/n - 1} dy \cr
& u = y(1) = a^{\,n^{\,2} /2} \cr}
$$
then
$$
\eqalign{
& I(a,n) = \int_{x = 0}^1 {\left\lfloor {\left( {a^{\,n} x} \right)^{\,n/2} } \right\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Generating function in combinatorics: combining ordinary and exponential generating functions I want to make a generating function for a combinatorics problem with $2$ different simultaneous constraints. In the one variable cases, one of the constraints would use an ordinary generating function, the other an exponentia... | If we let $x$ be the enumerator for the sum of digits and $y$ be the enumerator for the number of digits then we can form the generating function for occurrences of $4$
$$f(x,y)=(x^4)\frac{y}{1!}+(x^4)^2\frac{y^2}{2!}$$
Powers of $y$ count the number of $4$s in the string and powers of $x$ are multiples of $4$.
Similar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Hard Integral $\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$ I am trying to calculate the integral $$\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}$$
My idea is to use and try to calculate and prove that
$\int\limits_{-1}^{1}{\frac{\arcta... | It seems the integral is designed to intentionally benefit from the following facts:
$$\log \frac{1+x^{2}}{2}=\log(\frac{1}{2}(x+i)(x-i))=\log(\frac{x+i}{\sqrt 2})+\log(\frac{x-i}{\sqrt 2})$$
and
$$\arctan (x)= \frac{1}{2}i\log(1-i x)-\frac{1}{2}i\log(1+i x)= \frac{1}{2i}(\log(\frac{x+i}{\sqrt 2})-\log(\frac{x-i}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2401903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$
For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$
I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes
$... | If you let $B=xy+yz+zx$, by AM-GM, $B \ge 3$ and
$$A \ge \frac{3B}{B+3}=\frac{3}{1+3/B} \ge \frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving the differential equation $y''=y'^2+1$ I would like help solving the next differential equation.
$$y''=y'^2+1$$
I tried subsitiuting $p=y'$ and got to $\frac{1}{2}\ln|p^2+1|=\frac{1}{2}\ln|y|+\ln|c|$
but I don't know how to continue and maybe made a mistake.
any ideas?
| $$y''=y'^{ 2 }+1\\ { y }'=p\left( y \right) ,y''=pp'\\ \\ p{ p }'={ p }^{ 2 }+1\\ \int { \frac { pdp }{ { p }^{ 2 }+1 } } =\int { dy } \\ \frac { 1 }{ 2 } \int { \frac { d\left( { p }^{ 2 }+1 \right) }{ { p }^{ 2 }+1 } } =y+{ C }_{ 1 }\\ \ln { \left( { p }^{ 2 }+1 \right) =2\left( y+{ C }_{ 1 } \right) } \\ { p }^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
I... | Here is an alternative proof that doesn't use induction - just for fun!
Let $a \in \mathbb{Z}$. Observe that $a^3 - a = a(a^2 - 1) = (a-1)a(a+1)$ is the product of three consecutive integers, hence divisible by three. Adding $3a$ to the initial expression does not alter divisibility by $3$.
In particular, $a^3 - a + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$, $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.
It is my equation. I subtracted the second equation from the first one and found $(a-... | Hint: let $\,u=\sqrt{a}\,$ and $\,v=\sqrt{b}\,$, then it's given that $\,u^3+v^3=183\,$ and $\,u^2v+uv^2=182\,$, and the problem asks for $\frac{9}{5}(u^2+v^2)$.
*
*$\;(u+v)^3=u^3 + v^3 + 3(u^2v+ uv^2) = 183 + 3 \cdot 182 = 729 \implies u+v = 9\,$
*$\;uv= (u^2v+uv^2)/(u+v) = 182 / 9 \,$
*$\;u^2+v^2 = (u+v)^2 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a so... | Let us consider the function $f(\theta)=\log\sin\theta$ on the interval $\theta\in\left(0,\frac{\pi}{2}\right)$. It is a concave function
since $f''(\theta)=-\frac{1}{\sin^2\theta}<0$, hence by Jensen's inequality
$$ f(a)+f(b)+f(c) \leq 3\,f\left(\frac{a+b+c}{3}\right) $$
holds for any $a,b,c\in\left(0,\frac{\pi}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2405360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Why am i getting an incorrect result (0/0)? The given pair of equations are:
$$
3x-2y=0\\
kx+5y=0
$$
Clearly,here, $a_1=3$, $a_2=k$, $b_1=-2$, $b_2=5$, $c_1=0$ and $c_2=0$.
Now, here, I have to find a value of $k$ for which the given system of equations has infinite solutions.
Hence, $\frac{a_1}{a_2} = \frac{b_1}{b_2} ... | First, if an equation has no unique solution,
$Δ = 0$,
$3×5 - (-2)k=0$
$2k=15$
Using Gaussian Elimination,
$ \left[\begin{array}{rr|r}3 & -2 & 0 \\k & 5 & 0 \\\end{array}\right]$~
$ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k & 5 & 0 \\\end{array}\right]$~$ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k + 7.5 & 0 & 0 \\\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determine all pairs $(a,b)$ of positive integers such that $ab^2+b+7$ divides $a^2b+a+b$
Determine all pairs $(a,b)$ of positive integers such that $ab^2 + b + 7$ divides $a^2b + a +b$.
I guess the trivial solution is $a=b=7$.
Also any $a$ and $b$ which are divisible by $7$ will seem to suit.
we can also write
$b(\fr... | $$%%ab^2 + b + 7 \ | \ a^2b + a + b$$
We know that:
$$
ab^2 + b + 7 \ | \ a(ab^2 + b + 7)
=
a^2b^2+ab+7a
;
$$
$$
ab^2 + b + 7 \ | \ b(a^2b + a + b)
=
a^2b^2+ab+b^2
;
$$
so we can conclude
$$
ab^2 + b + 7 \ | \
\Big[
a(ab^2 + b + 7) - b(a^2b + a + b) \Big]
\ \ \ \ \ \ \ \
\Longrightarrow
\\
ab^2 + b +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$. I am stuck with the following problem :
Find the value of $$\lim_{x \to 0}\left (\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$$
My try : Let $$p=\lim_{x \to 0} \left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}\implies \log p=\lim_{x \to 0}... | From the Taylor expansion of $\sin x$ around $0$, namely $\sin u = u -\frac{u^3}{6} + o(u^3)$; and that of $\ln(1+u)$, specifically $\ln(1+u)=u+o(u),$ we get
$$\begin{align}
\left( \frac{\sin x}{x}\right)^{1/x^2}
&= \left(1-\frac{x^2}{6} + o(x^2)\right)^{1/x^2}
= e^{\frac{1}{x^2}\ln\left(1-\frac{x^2}{6} + o(x^2)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A question regarding minimum value of... $a+b+c=2$
$a,b,c \in \mathbb{R}^+$
What is The min value of $1/a+4/b+9/c$?
I have tried to use geometric shapes and aritmetic harmonic geometric means but unfortunately ı am going nowhere?
What do you suggest?
| Lagrange multiplier
$$f(a,b,c,k)=\frac{1}{a}+\frac{4}{b}+\frac{9}{c}+k (a+b+c-2)$$
Solve the system $f'_x=0,f'_y=0,f'_z=0,f'_k=0$
That is
$$k-\frac{1}{a^2}=0,k-\frac{4}{b^2}=0,k-\frac{9}{c^2}=0,a+b+c-2=0$$
there are $3$ solutions but only one leads to the minimum and respects constraints
$a= \frac{1}{3},b=\frac{2}{3},c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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An olympic mathematics problem regarding Cauchy-Schwarz This question was asked in Turkish National Maths Olympiad in 2008.
For all $xy=1$ we have $((x+y)^2+4)\cdot ((x+y)^2-2) \ge A\cdot(x-y)^2$.
What is the maximum value $A$ can get?
My efforts regarding this problem;
$(x+y)^2-8 \ge A\cdot(x-y)^2$
Using the propert... | Let $x^2+y^2=m$.
Thus, $$(m+6)m\geq A(m-2)$$ or
$$m^2+(6-A)m+2A\geq0,$$
for which we need $$(6-A)^2-8A\leq0,$$
which gives $2\leq A\leq18$.
For $A=18$ we get $m=6$ or $x^2+y^2=6$, which says that the equality occurs.
Thus, $18$ is our answer.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Finding the Coefficient of $(x^3 + 2y^2)^n$ containing $x^{18} y^{12}$ I was helping somebody when I scratched my head because of this question. It goes like
this:
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
My work:
I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 =... |
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
You want the coefficient of the term $\binom{6+6}{6}(x^3)^{6}(2y^2)^6$, so ....$$C~=~ 2^6 \binom{6+6}{6} ~=~59\,136$$
Therefore: okay $\checkmark$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or ... | $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20(50)^n+18(12)^n\equiv-18(50)^n+18(12)^n\equiv-18(12)^n+18(12)^n\equiv0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
| Another way to solve $A \sin\theta + B \cos\theta = C$ is as follows.
Divide by $\sqrt{A^2 + B^2}$. Then, set $A / \sqrt{A^2 + B^2} = \sin\xi$
and $B / \sqrt{A^2 + B^2} = \cos\xi$.
This gives:
$$\sin\xi \sin\theta + \cos\xi \cos\theta = C / \sqrt{A^2 + B^2}.$$
So, the solution is given by
$$\cos(\theta - \xi) = C / \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \g... | I think our inequality is true for all reals $x$, $y$ and $z$ such that $(x-y)(x-z)(y-z)\neq0$.
Indeed, if $z=0$ then
$$\sum_{cyc}\frac{(2x-y)^2}{(x-y)^2}=\frac{(2x-y)^2}{(x-y)^2}+5\geq5.$$
Let $xyz\neq0$, $\frac{2x-y}{x-y}=a$, $\frac{2y-z}{y-z}=b$ and $\frac{2z-x}{z-x}=c$.
Thus, $(2-a)x=(1-a)y$, $(2-b)y=(1-b)z$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2419508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suita... | You have shown that $y$ is in the range of the function
$$y = \frac{x^2 + 3x + 1}{x + 1}$$
if $x$ is a real-valued root of the quadratic equation
$$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$
For $x$ to be a real-valued root of the quadratic equation, its discriminant must be nonnegative. The discriminant is
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose... | You have "$f(x) = f(y)$" implies that at least one of "$x=y$" or "$2(x^2 + x y+ y^2)+3 = 0$" is true. If you can show that the second is never possible for any $x,y \in \mathbb{R}$, you would be done.
Now $x^2 + xy + y^2$ has what minimum value? Can it ever be as small as $-3/2$?
EDIT
In your case 2, you go off the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly f... | Lagrange multipliers will give the solution
\begin{eqnarray*}
L=ab+bc+ca+ \lambda (a+2b+c-4)
\end{eqnarray*}
Differentiating gives
\begin{eqnarray*}
b+c+ \lambda =0 \\
c+a +2\lambda =0 \\
a+b+ \lambda =0 \\
\end{eqnarray*}
Add these equation together and substitute to get $ \lambda = \frac{b-4}{2}$ now substitute this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ende... | I made the subsitution $y=\frac{1}{x}$. The limit as $y \rightarrow 0$ falls straight out as $\frac{a-b}{2}$. You'll need to use binomial expansion of $\sqrt{1+p} = 1 + \frac{p}{2}+...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Show that $a_{n+1} =a_n + \frac{1}{3^n}$ defines a Cauchy sequence I just need help.
Let ${a_n}$ be the sequence defined by the relations $a_1 = 1$ and $a_{n+1} =a_n + \frac{1}{3^n}$ for $n = 1, 2,....$ Prove that ${a_n}$ is a Cauchy sequence using definition.
| It's easy to see that
$$a_n = \sum_{k=0}^{n}\frac{1}{3^k}$$
The RHS is the $n$'th partial sum of a convergent geometric series (with limit $3/2$), so $a_n$ is a convergent sequence, hence a Cauchy sequence.
Edit after the OP added "using definition":
Let $m > n$. Then
$$\begin{aligned}
a_m - a_n &= \sum_{k=n+1}^{m}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing $ \mathbb{Q}[\sqrt[4]{2}]$ is a field Here i want to prove
\begin{align}
\mathbb{Q}[\sqrt[4]{2}] = \{ a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8} \mid a,b,c,d\in\mathbb{Q}\}
\end{align}
this is a field. To do that i want to prove
\begin{align}
\frac{1}{a + b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}... | Idea: use that $\sqrt[4]{2} = \sqrt{ \sqrt{2}}$, so first rationalize to a denominator in $\mathbb{Q}(\sqrt{2})$, and then finish off.
$$\frac{1}{a+b \sqrt[4]{2} + c \sqrt[4]{4} + d \sqrt[4]{8}}= \frac{1}{a + c \sqrt{2} + (b + d \sqrt{2}) \sqrt[4]{2}}= \\
=\frac{a + c \sqrt{2} - (b + d \sqrt{2}) \sqrt[4]{2}}{(a + c \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0$, compute $ab+cd$
Consider that $\{a,b,c,d\}\subset \Bbb R$ and it is known that
$$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0.$$
Compute the value of $ab+cd$. (Edit: originally written with a typo as $ab+bd$),
This is from a list of training problems used within the prep... | Since
$$ac+bd=0\implies a=-\frac{bd}{c}; c\neq0$$
Then
$$1=a^2+b^2=b^2[\frac{d^2}{c^2}+1]=b^2\cdot\frac{c^2+d^2}{c^2}=\frac{b^2}{c^2}$$
$$\implies c=\pm b\implies d=\mp a$$
The solution is given by
$$(a,b,c,d)=(a,b,\pm b,\mp a)$$
So
$$ab+cd=ab+(\pm b)(\mp a)=ab-ab=0$$
(For $c=0$, the case is trivial.)
Another solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$ In the paper A Probabilistic Algorithm for k-SAT Based on Limited Local Search and Restart, by Uwe Schöning, I fail to understand an identity used in a proof:
$$\sum_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z... |
We consider the generating functions
\begin{align*}
A(z)=\frac{1}{\sqrt{1-4z}}\quad\text{and}\qquad B_2(z)=\frac{1-\sqrt{1-4z}}{2z}
\end{align*}
and derive for non-negative integer $r$ the coefficients of
\begin{align*}
\left(B_2(z)\right)^r\qquad\text{ and }\qquad A(z)\left(B_2(z)\right)^r
\end{align*}
We app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that the ellipsoid $E = \{ (x,y,z) \in R^3 \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \}$ is convex I know what an ellipsoid is, but I do not know how to approach how to prove that one is convex. Any advice would be greatly appreciated.
| Let $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ be two points in the ellipsoid. We have to prove that $(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2,\lambda z_1+(1-\lambda)z_2)$ also lies in the ellipsoid, where $0 \le \lambda \le 1$.
Let $$A = \frac{\left(\lambda x_1+(1-\lambda)x_2\right)^2}{a^2}+\frac{\left(\lambda ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't unde... | Note: $(n+1)^2 = n^2 + 2n +1$.
So $(2n-1)^2 - (2n)^2 = 4n^2 -4n +1 - 4n^2 = -4n+1$.
So $\sum_{k=1}^{n=2m} (-1)^{k-1}k^2 = \sum_{j=1}^m [(2j-1)^2 - (2j)^2]=\sum_{j=1}^m [-4j +1]= -4(\sum j) + m = -4*\frac {m(m+1)}2 + m = -2m(m+1) +m=m(-2(m+1) + 1)=m(-2m-1)=-m(2m+1)= -\frac{n(n+1)}{2}$
..... if $n$ is even.
If $n=2m+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
} |
Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gi... | $x=\arctan 2$ means $\tan x=2$ thus
$\sin x = \dfrac{\sqrt{\sin^2 x}}{\sqrt{1}}=\dfrac{\sqrt{\sin^2 x}}{\sqrt{\sin^2 x+\cos^2 x}}=\dfrac{\sqrt{\frac{\sin^2 x}{\cos^2 x}}}{\sqrt{\frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}}}=\dfrac{\tan x}{\sqrt{\tan^2 x+1}}$
$\sin x=\dfrac{2}{\sqrt 5}$ and $\cos x =\sqrt{1-\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 5
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$... | For $n=1$ the formula holds,
$$S_n=1^2+3^2+5^2+\cdots+(2n-1)^2$$
$$S_{n+1}=1^2+3^2+\cdots+n^2+(2n+1)^2$$
$$\bbox[5px,border:2px solid red]{S_{n+1}-S_n=(2n+1)^2}$$
Now if,
$$S_n=\frac{n(2n-1)(2n+1)}{3}$$$$$$
$$S_{n+1}=\frac{(n+1)(2n+1)(2n+3)}{3}$$$$$$
$$S_{n+1}-S_n=\frac{(n+1)(2n+1)(2n+3)}{3}-\frac{n(2n-1)(2n+1)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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$\{a,b,c\}\subset \Bbb R$, $a^2-ab=1,b^2-bc=1, c^2-ca=1$. Find $abc(a+b+c)$.
Consider that $\{a,b,c\}\subset \Bbb R$ and it is known that
$$a^2-ab=1,\ b^2-bc=1,\ c^2-ca=1\ \ \text{(cond.1)}$$
Find $abc(a+b+c)$.
This question is from OBM 2007 (Brazilian Math Olympiad). Sorry if it is a duplicate. I will present ... | My approach (not necessary simpler than yours):
Let $S=abc(a+b+c)$ then:
$$S=a^2bc+b^2ac+c^2ab=(1+ab)bc+(1+bc)ac+(1+ca)ab=ab+bc+ac+S$$
So: $ab+bc+ac=0$
Now add the three constraints
$$a^2+b^2+c^2-(ab+bc+ca)=3$$
So $a^2+b^2+c^2=3$
Then:
$$0=(ab+bc+ac)^2=a^2b^2+b^2c^2+c^2a^2+2S=a^2(1+bc)+b^2(1+ac)+c^2(1+ab)+2S=3+3S$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a m... | There is a trouble with this question.
There are non-real solutions for example $a\approx 3.12128,\space b,c\approx -0.56064\pm 1.47835i$.
On the other hand, in $\mathbb R$ as in $\mathbb C$, if $a=2$ then the system
$$-1+(y-2)^3+(z-2)^3=0\\y^2+z^2=2\\y+z=0$$ is incompatible and if $b$ or $c$ are equal to $2$ then th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Determine for what values of $x$ the given series converges The given series is
$\sum_{n=1}^∞
(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n}
)x^{n} $.
I tried it by using Cauchy Root Test as follows-
Let
$y=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{... | Since both $\frac{1}{1-x}=\sum_{n\geq 0}x^n$ and $-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$ have radius of convergence equal to one, the same holds for their (Cauchy) product
$$ \sum_{n\geq 1} H_n x^n = -\frac{\log(1-x)}{1-x}. $$
In particular the LHS is convergent for any $x\in(-1,1)$ and clearly is not convergent for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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determinant of the $7\times7$ matrix How to find determinant of the following $7\times7$ matrix
\begin{bmatrix}
a&1&0&0&0&0&1&\\
1&a&1&0&0&0&0&\\
0&1&a&1&0&0&0&\\
0&0&1&a&1&0&0&\\
0&0&0&1&a&1&0&\\
0&0&0&0&1&a&1&\\
1&0&0&0&0&1&a&
\end{bmatrix}
| Let $e_0, e_1, \ldots, e_6$ be the basis column vector with entries $1$ at $q^{th}$ row and zero otherwise. i.e
$$(e_q)_p = \begin{cases}1, & p = q\\0, & \text{ otherwise }\end{cases},\quad 0 \le p, q \le 6$$
Let $A$ be the matrix at hand. Let $L = (\ell_{pq})$ and $R = (r_{pq})$ be the $7 \times 7$ matrices given by
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Polynomials with integration problem
Are there polynomials $P,Q$ with real coefficients satisfying the equalities
$$\int_0^{\ln n}\frac{P(x)}{Q(x)} \, dx
= 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$
for each integer $n\ge 2$?
I don't know how I show the existence of such polynomials or nonexistence. I took some... | Suppose there existed such $P,Q$. It is easy to see that $\deg P\le \deg Q$; otherwise the integral will blow up too quickly. This implies that $\frac{P(1/z)}{Q(1/z)}$ is a rational function which does not blow up at $0$, and hence has a Taylor expansion $\frac{P(1/z)}{Q(1/z)} = a_0+a_1z+a_2z^2+\dots$ valid for all sma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the e... | $0\le |\dfrac{x^5+y^2}{x^4 +|y|}| \le$
$\dfrac{|x^5| +|y^2|}{|x^4+|y||} \le$
$\dfrac{|x^5|}{x^4} + \dfrac{y^2}{|y|} =$
$|x| + |y|.$
And now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding the product $\prod_{k=2}^n \left(1 - \frac{1}{k}\right)$ using Induction I'm trying to figure out how to prove by induction the following statement:
$$
\prod_{k=2}^n \left(1 - \frac{1}{k}\right) = \frac{1}{n}.
$$
| Inductive hypothesis:
Assume that
$$
\prod_{k = 1}^m \left( 1 - \frac{1}{k}\right) = \frac{1}{m}
$$
holds true for some $m$.
Then we compute
$$
\left(1 - \frac{1}{m+1} \right) \cdot \prod_{k = 1}^m \left( 1 - \frac{1}{k}\right) = \left(1 - \frac{1}{m+1} \right) \cdot \frac{1}{m} = \left(\frac{1}{m} - \frac{1}{(m + 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the area of a triangle using the sides lengths
The sides lengths of a triangle $a,$ $b$ and $c$ verify: $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=12b.$$ The task is to find the area of the triangle.
I'm trying to apply the heron's formula: $$\dfrac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}} {4}.$$
How do i get to heron... | I think in you equation ,must be $24b$ $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b\\
\sqrt{a-24} + b^2 -24b+144+|c-12\sqrt{3} |=0\\
\sqrt{a-24} + (b-12)^2+|c-12\sqrt{3} |=0
\\\underbrace{\sqrt{a-24}}_{\geq 0} +\underbrace{(b-12)^2}_{\geq 0} + \underbrace{|c-12\sqrt{3} |}_{\geq 0} =0$$ some of positive expressions are ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\... | $$\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)^3=$$
$$=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}} \sqrt[3]{18-5\sqrt{13}}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36+3\sqrt[3]{-1}\left(\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}\right)=$$
$$=36-3\left(\sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Quadratic formula for complex numbers Consider the equation
$$
az^2+bz+c=0,
$$
where $a,b$ and $c$ are complex numbers and $a\ne 0$. Applying usual operations on $\mathbb{C}$ we have the following:
$$
az^2+bz+c=0\iff z^2+\frac{b}{a}z = -\frac{c}{a}\iff z^2+\frac{b}{a}z + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} -\frac{4ac... | Let's call $\sigma$ the square root of $z$ which means that $\sigma$ such as $\sigma^2=z$.
Thus, if $\rho = b^2 - 4ac$ and $\sigma^2 = \rho$ then
\begin{align*}
& \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\\
\iff & \left(z + \frac{b}{2a}\right)^2 = \frac{\rho}{4a^2}\\
\iff & z + \frac{b}{2a} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Statistics Question involving Mode : Given the set of data $1,1,2,2,2,3,3,x,y$ .... I was studying for some quizzes when a wild question appears. It goes like this:
Given the set of data $1,1,2,2,2,3,3,x,y$, where $x$ and $y$ represent two different integers. If the mode is $2$, which of the following is true?
A. If $... | When you read something like:
If $x=a$ or $b$, then $y=c$
What this means is that for the values of $x$, $y$ must equal $c$ in order to satisfy the condition that the sequence has a mode of 2. And if at least one value of $x$ does not satisfy the condition for $y$, then the entire thing is false.
For example, $A$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
... | Let $\{ e_1,e_2,e_3 \}$ be the canonical (standard) basis of $\mathbb{R}^3$. Note that for any $(x,y,z) \in \mathbb{R}^3$ we can write $(x,y,z)=ze_1+ye_2+ze_3$. Hence $T(x,y,z)=T(xe_1+ye_2+ze_3)$, and using lineality $$T(x,y,z)=T(xe_1)+T(ye_2)+T(ze_3)=xT(e_1)+yT(e_2)+zT(e_3)$$ and by hypotesis $$T(x,y,z)=x(3,1,4)+y(-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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In how many ways can $5$ balls of different colours be placed in $3$ boxes of different sizes if no box remains empty? 5 balls of different colours are to be placed in 3 boxes of different sizes. Each box can hold all 5 balls. The number of ways in which we can place the balls in the boxes so that no box remains empty.... | There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to dist... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to find the closed form of $\int_{0}^{\infty}{(e^{-x}+x-1)^2\over x(e^{x\over n}-1)}\mathrm dx$ We partially guessed the closed form of $(1)$ to be
$$\int_{0}^{\infty}{(e^{-x}+x-1)^2\over x(e^{x\over n}-1)}\mathrm dx=f(n)+{\pi^2\over 6}n^2+\ln{{2n\choose n}}\tag1$$
where $n\ge1$
$f(1)=-2$
$f(2)=-6$
$f(3)=-11$
$f(4)... | Hint by user reuns:
$$\begin{align}
\int_0^{\infty } \frac{(\exp (-x)+x-1)^2}{x \left(\exp \left(\frac{x}{n}\right)-1\right)} \, dx & =\int_0^{\infty } \frac{(\exp
(-x)+x-1)^2 \sum _{k=1}^{\infty } \exp \left(-\frac{k x}{n}\right)}{x} \, dx \\
&=\sum _{k=1}^{\infty } \int_0^{\infty }
\frac{(\exp (-x)+x-1)^2 \exp ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because o... | Any rational function can always be integrated by decomposing it into partial fractions. In this case this decomposition is particularly simple.
\begin{eqnarray}
\frac{1}{x^4-\sqrt{3} x^2+1} &=& \frac{1}{x^2-\theta_+} \cdot \frac{1}{\theta_+-\theta_-} + \frac{1}{x^2-\theta_-} \cdot \frac{1}{\theta_--\theta_+}\\
\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Rationalize a fraction. Rationalize the denominator $\frac{1}{2^{\frac{1}{3}} + 3^{\frac{1}{3}} + 4^{\frac{1}{3}}}$. Is there a short solution for this task ? Thanks in advance.
| I found the minimal polynomial for $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}$
$x^9-18 x^7-27 x^6+108 x^5+162 x^4-459 x^3-972 x^2-81=0$
The I divided all terms by $81x$ and I got
$$\frac{1}{x}=\frac{x^8}{81}-\frac{2 x^6}{9}-\frac{x^5}{3}+\frac{4 x^4}{3}+2 x^3-\frac{17 x^2}{3}-12 x$$
Then I substituted $x$ with $\sqrt[3]{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the area of a triangle inscribed in the ellipse (Unicamp Mathematics Olympiad) Let $\zeta$ denote the Carteasian region given by $\zeta = \{(x,y): \frac{x^2}{4} + \frac{y^2}{9} = 1\}$. Let $A = (0,3), B = (x,y), C = (z,w)$ be points such that $A,B,C \in \zeta%$ and $\Delta ABC$ is equilateral. What is the area of ... | The points of the ellipse are parameterised by $(2 \cos \theta, 3 \sin \theta)$ & by symmetry it is clear that the points $B$ and $C$ will have coordinates $(2 \cos \theta, 3 \sin \theta)$ & $(-2 \cos \theta, 3 \sin \theta)$. So the requirement of equilateralness gives
\begin{eqnarray*}
(4 \cos \theta)^2= (4 \cos \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\... | By your idea
$$\lim_{x\rightarrow2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(6-x-4)}{(\sqrt{6-x}+2)(3-x-1)}$$
$$=\lim_{x\rightarrow2}\frac{(\sqrt{3-x}+1)(2-x)}{(\sqrt{6-x}+2)(2-x)}=\lim_{x\rightarrow2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{2}{4}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Problems involving LOCUS Okay so I'm really having a hard time finding the locus of curves or point of intersection of curves.
The problem : I am able to get the geometrical condition in most of the questions but then I have no idea what to do to get the locus.
The question :A variable straight line drawn through the p... | Let $M(a,b)$ be a point on our locus and $C\left(\frac{4}{3},\frac{4}{3}\right)$.
Thus, $$m_{MC}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}$$ and we have an equation of $AM$:
$$y-\frac{4}{3}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}\left(x-\frac{4}{3}\right),$$
which for $x=0$ gives
$$y=\frac{4}{3}-\frac{4}{3}\cdot\frac{b-\frac{4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with d... | It can be done by differentiation but it is much easier to use $x^2 \geq 0$.
Assuming $a,b,c $ are positive
\begin{eqnarray*}
a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \\
\end{eqnarray*}
Expand these out
\begin{eqnarray*}
ab(a+b)+bc(b+c)+ca(c+a) \geq 6abc \\
\end{eqnarray*}
Now divide by $abc$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Solve differential equation $yy'(yy'-2x)=x^2-2y^2$.
$$yy'(yy'-2x)=x^2-2y^2$$
I've tried to divide by $y^2$ and substitude $\frac{x}{y}=z$, but it led to:
$$y'^2-2zy'=z^2-2$$
$$(y'-z)^2=2(z^2-1)$$
$$|y'-z|=\sqrt{2}\sqrt{z^2-1}$$ hence
$y'=\sqrt{2}\sqrt{z^2-1}+z$ or $y'=-\sqrt{2}\sqrt{z^2-1}+z$ and I don't know any w... | With $y^2=u$ then $2yy'=u'$ so
\begin{align}
2yy'(2yy'-4x)&=4x^2-8y^2 \\
u'(u'-4x)+4x^2&=8x^2-8u \\
(u'-2x)^2&=8(x^2-u) \\
-d(x^2-u)&=2\sqrt{2}\sqrt{x^2-u^2}dx \\
\sqrt{x^2-u}&=C-\sqrt{2}x \\
x^2-y^2=&(C-\sqrt{2}x)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $E(X)$ and the PMF of Number of Flips to Get $2$ Consecutive Heads or Tails with a Coin Flips Consider flipping a fair coin until either two consecutive heads appear or two consecutive tails appear, and let $X$ denote the total number of flips required to obtain the $2nd$ consecutive heads or the $2nd$ consec... | The Probability Mass Function (PMF)
Let $ X $ be the random variable that counts the number of tosses until we get two successive tosses of the same type.
We're basically dealing with Binary Chains which each chain has the probability of $ \frac{1}{ {2}^{n} } $ where $ n $ is the length of the chain.
The probability $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Binomial Coefficient Question In the expression
$((. . . ((x − 2)^2 − 2)^2 − 2)^2 − · · · − 2)^2 − 2)^2$
,
there are $k$ pairs of parentheses, where k is a positive integer. In terms of $k$, find the coefficient of $x^2$
after expanding and collecting the terms
So far I have attempted to look for a recursion which woul... | Hint. Let $F_k(x)$ the the function at the $k$-th step, then
$$F_k(x)=a_kx^2+b_kx+4+o(x^2)$$
Then $F_1=x^2-4x+4$ implies that $a_1=1$ and $b_1=-4$.
Moreover
\begin{align*}a_{k+1}x^2+b_{k+1}x+4+o(x^2)&=F_{k+1}(x)=(F_k(x)-2)^2\\
&=(a_kx^2+b_kx+4+o(x^2)-2)^2\\
&=(4a_k+b_k^2)x^2+4b_kx+4+o(x^2).
\end{align*}
Therefore $b_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the probability that a team roster contains Ted and Vanessa A school Science Fair team will consist of 2 male students, two female students and one alternate. The alternate can be of either sex. There are $4$ males (including Ted) and $5$ females (including Vanessa) trying out for the team. A team roster is a l... | A team roster consists of three males and two females or two males and three females. Since there are four male and five female students available, the number of possible team rosters is
$$\binom{4}{3}\binom{5}{2} + \binom{4}{2}\binom{5}{3}$$
As you observed, if both Ted and Vanessa are selected, a team with three ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is ... | But it is not true, say $n=2$. :(
If $n$ is odd then $2^n+1 = (2+1)(2^{n-1}-...+1) =3k$, then is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to find $\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}$? How to find
$$
\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}?
$$
My try : $$(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=(x-1)$$
So we have :
$$\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1} \cdot\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\... | Setting $$t=\sqrt[6]{x}$$ then we have $$t^6=x$$ and we get
$$\frac{t^{12}+t^6+t^3-3}{t^2-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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A fencing area question. A farmer creates a rectangular pen by using one side of a barn as one side of the pen and using fencing for the other three sides. The farmer has $80~\text{ft}$ of fencing, and the side of the barn is $40~\text{ft}$ long. If $x$ represents the length of the fenced side of the pen that is parall... | You correctly found that
$$A(x) = x\left(40 - \frac{1}{2}x\right)$$
Since we require that the area enclosed by the fence is at least $600~\text{ft}^2$ (you omitted the exponent),
\begin{align*}
x\left(40 - \frac{1}{2}x\right) & \geq 600\\
40x - \frac{1}{2}x^2 & \geq 600\\
x^2 - 80x & \leq -1200\\
x^2 - 80x + 1600 & \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+... | You have
\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}
Case 1: Assume $xyz\ne 0.$ Then $x^2=y^2=z^2=\dfrac{1}{2\lambda}.$ So $$1=x^4+y^4+z^4=\dfrac{3}{4\lambda^2}$$ and you'll get $\lambda$ and thus $x,y,z.$
Case 2: Assume $z=0,xy\ne 0.$ Then $x^2=y^2=\dfrac{1}{2\lambda}.$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac... | $x-1$ doesn't divide $(x+1)^2$ exactly, so there must also be a remainder term of lower degree than $x-1$, i.e.
$$ \frac{(x+1)^2}{x-1} = ax+b + \frac{c}{x-1}. $$
Then you find
$$ x^2+2x+1 = ax^2+(b-a)x+(c-b), $$
giving three equations in three unknowns (you expect a unique solution anyway, so you should have the same n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Proving properties of rings None of these are complete and I am looking for some guidance.
*
*In any ring if $ab=-ba$ then $(a+b)^2=(a-b)^2=a^2+b^2$.
So assuming $ab=-ba$ to be true, then squaring both sides gives $(ab)^2=(-ba)^2$, i.e. $abab=-ba-ba$. Not all rings are commutative under multiplication so I d... | 1:
$(a + b)^2 = a^2 + ab + ba + b^2 = a^2 + ab - ab + b^2 = a^2 + b^2; \tag 1$
$(a - b)^2 = a^2 -ab - ba + b^2 = a^2 - ab + ab + b^2 = a^2 + b^2; \tag 2$
2:
$(a - b)(a + b) = a^2 - ab + ab - b^2 = a^2 - b^2 = 0, \tag 3$
which works since integral domains are commutative. Now if
$a \ne b, \tag 4$
then
$a - b \ne 0, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Combination of $n$ letter out of $3n$.
Show that the number of combinations of $n$ letters out of $3n$ letters of which $n$ are $a$s , $n$ are $b$s and the rest are unequal is $(n+2)2^{n-1}$.
My approach...
The number of ways of distributing $n$ distict object in $k$ distinct box is $\binom{n+k-1}{k-1}$ now when $k=2... | Hint: The required number is coefficient of $x^n$ in $$[(1+x+x^2+\cdots x^n)^2(1+x)^n].$$
Do some simplification and find that the coefficient of $x^n$ is $$2^n\binom {n+2-1}{1} - \binom n1 2^{n-1} \binom{n+1-1}{0} = (n+2)2^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2481407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Polynomial sum: $x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$ Find sum of the expression,
$$x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$$
where $x,y$ are real numbers and $n$ is a natural number.
| It is:
$$\frac{x^{n+1}-y^{n+1}}{x-y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
For which value of $k$ is the following matrix diagonalizable? \begin{pmatrix}7&k\\ \:0&7\end{pmatrix}
I could not figure out how to derive the eigenvalues and eigenvectors of the matrix above because of the letter $k$. How am I supposed to deal with a value in terms of $k$? And how would I be able to find out if the m... | When $k \neq 0:$
A column vector that is not sent to zero by
$$
\left(
\begin{array}{cc}
0 & k \\
0 & 0
\end{array}
\right)
$$
is
$$
\left(
\begin{array}{c}
0 \\
1
\end{array}
\right),
$$
and
$$
\left(
\begin{array}{cc}
0 & k \\
0 & 0
\end{array}
\right)
\left(
\begin{array}{c}
0 \\
1
\end{array}
\right) =
\left(
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\frac{6n^2+3n}{2n^2-5}$ converges to 3. Prove the following:
$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$
Here's my solution:
Consider the function $\frac{6n^2+3n}{2n^2-5}$.
Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$
The upper bound for th... | To get an upper bound, you want to replace the numerator by something big and the denominator by something small. So instead of $2n^2-5\lt4n^2$, you should use $2n^2-5\gt2n^2$ (or even just $n^2$). Aside from that, everything is basically OK.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could ... | Here is a larger counterexample : $a=6,b=2,n=4$, then $n+1$ is not a multiple of $\binom{an}{bn}$.
Another one is given by $a=5,b=2,n=5$.
Also, note that $\binom{an}{bn} = \binom{an}{(a-b)n}$, therefore replacing $b$ above with $a-b$ would also work.
I am still thinking about $b > 2$(other than taking $b \to a-b$ obvio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2492260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to find ord$[a]_m$ What are the steps to find minimum $k$ such that $\left([a]_m\right)^k=[1]_m$? In short, how to find ord$[a]_m?$
Specifically I need to find mod$\left( 173^{607},1147\right)$ but since $\varphi(1147)=1080>607$ I can't use Euler-Fermat's Th. on this one, and I don't feel like square-and-multiplyi... | Since $1147=31\cdot 37$, by Chinese Remainder theorem, you have a ring isomorphim
$$\psi:\mathbb Z/1147\mathbb Z\longrightarrow\mathbb Z/31\mathbb Z\times\mathbb Z/37\mathbb Z$$
Thus
$$\psi(173^{607}+1147\mathbb Z)=(173^{607}+31\mathbb Z,173^{607}+37\mathbb Z)$$
But
\begin{align*}
173^{607}
&\equiv 18^7\\
&=2^7\cdot 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this identity of summation of ramanujan? In the first book of ramanujan him find this identity:
Let k , n $\in\mathbb{N}$* and define $A_k=3^k\left(n+\frac{1}{2}\right)-\frac{1}{2}$. the if r is a positive interger
$$\sum_{k=n+1}^{A_r}\frac{1}{k}=r+2\sum_{k=0}^{r-1}\left(r-k\right)\sum_{j=A_{k-1}+1}^{A_k}\... | First we need the result from entry $2$ of the reference. Lets start with the parial fractions
\begin{eqnarray*}
\frac{1}{(3j)^3-3j} =\frac{1/2}{3j-1}-\frac{1}{3j}+\frac{1/2}{3j+1}.
\end{eqnarray*}
So
\begin{eqnarray*}
1+2 \sum_{j=1}^{n} \frac{1}{(3j)^3-3j} =1+ \sum_{j=1}^{n} \left( \frac{1}{3j-1}\color{red}{+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $ Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $
Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $
Answer:
$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \... | That looks fine!
To generalise a bit, if $a \equiv1(\mathrm{mod}n)$, you see that $a^k \equiv 1(\mathrm{mod}n)$, and so $a^{k^l}\equiv 1(\mathrm{mod}n)$ whatever $k$ and $l$ are :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
let $f(x) = \left\lbrace\begin{array}{ll} x^2+3x, &x \geq1\\ x^2-3x+6, & x<1 \end{array}\right.$ then find the limit… Let
$f(x) =
\begin{cases}
x^2+3x, &x \geq1\\[2ex]
x^2-3x+6, & x<1
\end{cases}$
Then find the
$$\lim_{h \to0} \frac{f(1)-f(1-h^2)}{h^2}=?$$
My Try :
$$f(1)=4$$
$$f(1-h^2)=(1-h^2)^2-3(1-h^2)+6=h^4-h^2+... | It should be observed that the limit in question is left hand derivative of $f$ at $1$ and hence it is equal to $2x-3$ evaluated at $x=1$. Thus the desired limit is $-1$. It should be also observed that the above works because $f$ is continuous at $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to differentiate $\sqrt[5]{1/x}$ Hey guys I could use some help with this derivative:
$$\sqrt[5]{1/x}$$
This is what I have so far:
$$=-\dfrac{1}{5}\left(\dfrac{1}{x}\right)^{-6/5}$$
Having trouble simplifying this to my given solution so I don't know if it is correct.
Given solution:
$$-\frac{1}{5x\sqrt[5]{x}}$$
| Here's my take on it with each step spelled out as clearly as possible:
$$
\left(\sqrt[5]{\frac{1}{x}}\right)'=
\left(x^{-\frac{1}{5}}\right)'=
-\frac{1}{5}x^{\left(-\frac{1}{5}-1\right)}=
-\frac{1}{5}x^{\left(-\frac{1}{5}-\frac{5}{5}\right)}=
-\frac{1}{5}x^{-\frac{6}{5}}=\\
-\frac{1}{5\sqrt[5]{x^6}}=
-\frac{1}{5\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Proof by Contradiction: let a,b $\in$Z . Prove that if 3 $\nmid$ a and 3 $\nmid$ b then $3 | a^2 - b^2$ Just wondering if this works for this question, book had a different answer and I couldn't find another answer for the question.
Assume, to the contrary, that 3 | a and 3 | b, then a = 3k, and b = 3x for x,k $\in$ Z,... | Proof by contradiction:
Suppose 3 does not divide $a^2-b^2$ the we must have:
$ 3 ł (a-b)(a+b)$
Suppose:
$a+b=3k_1 + r_1; r_1<3 ⇒ r_1 =1 , 2$
$a-b=3k_2 +r_2; r_2<3 ⇒ r_2=1, 2$
$2a=3(k_1 +k_2) + r_1 +r_2$
$2b=3(k_1-k_2) +r_1 -r_2$
1): $r_1 =1 $ and $r_2 =1$ ⇒ $2b=3(k_1-k_2) +1 - 1=3(k_1-k_2)$ ; cntradicts 3łb
2):... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2}... | Side comment: $\frac{\sqrt{m-1} + \sqrt{m+1}}2 \ge \sqrt{\sqrt{m^2 - 1}}>\sqrt m$ by $AM-GM$ theorem.
So $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} \approx 3\sqrt{x+1}$ but slightly larger.
$x \le \sqrt{x} + \sqrt{x+1} + \sqrt{x+2}< 3\sqrt{x+1} < x+1$ implies $x^2 < 9x + 9 < x^2 + 2x +1$ (the first inequality is definite, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such ... | (Not using the combinatorial identity but without induction)
You can start saying that $$(k+1)^{3}-k^3=3k^2+3k+1$$
then you take the sum in both sides to get:
$$\sum_{k=1}^{n}\bigg((k+1)^{3}-k^3\bigg)=\sum_{k=1}^{n}\big(3k^2+3k+1\big)$$
The LHS is a telescopic sum and RHS will give you the sum you are looking for:
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$) Prove if $ x^{11}\equiv 21$ (mod $59$) then $x^{9}\equiv 35$ (mod $59$).
I know by Fermatt's Little Theorem, $x^{58} \equiv1$ (mod 59), and I know that $x^{58} = (x^{11})^5(x^3)$, but I dont know where to go from here. What do I do next?
| As you figured out
$$(x^{11})^5 \equiv (21)^5 \equiv 3 \pmod{59} \Rightarrow (x^{11})^5x^3 \equiv 3x^3 \pmod{59} \Rightarrow \\
1 \equiv 3x^3 \pmod{59} \Rightarrow 1 \equiv 3^3x^9 \pmod{59}\Rightarrow \\
35 \equiv (35\cdot 27) \cdot x^9 \equiv x^9 \pmod{59}$$
since
$$35\cdot 27 \equiv 1 \pmod{59}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Determining the limit of a function containing $(x-1)^5$ without l'hopital`s rule. Find the limit of $$\begin{equation*}
\lim_{x \rightarrow 0}
\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}
\end{equation*}$$
Shall I use the binomial theorem?
Any hint will be appreciated!
| Ok, now that OP's conditions are better known, let's try again.
Numerator is
$$(x-1)^5+(1+5x)\equiv x^5-5x^4+10x^3-10x^2+5x-1+(1+5x)$$$$\equiv x^5-5x^4+10x^3-10x^2+10x$$
so the given expression is
$$\frac{x^5-5x^4+10x^3-10x^2+10x}{x^5+x^2}\equiv\frac{x^4-5x^3+10x^2-10x+10}{x^4+x}.$$
Now, numerator does not approach $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing: $\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$ I would like to compute the exact value of the integral below.
$$\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx$$
I have proved the convergence already. but failed to the residues theorem in other to get the exact value. It will be great if som... | In fact
\begin{eqnarray}
&&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\
&=&\int_0^1\sin(x^2)dx+\int_0^1\frac{\sin(\frac{1}{x^2})}{x^2}dx.
\end{eqnarray}
For the second integral, under $x\to\frac1x$, one has
\begin{eqnarray}
&&\int_0^1\frac{x^2\sin(x^2)+\sin(\frac{1}{x^2})}{x^2}dx\\
&=&\int_0^1\sin(x^2)dx+\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
| May be, you could make life a bit simpler using $x=y+a$ which makes
$$\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}=\frac{\sqrt{a+y}-\sqrt{a}+\sqrt{y}}{\sqrt{y (2 a+y)}}$$ Now, use the generalized binomial theorem or Taylor series around $y=0$ to get
$$\sqrt{a+y}=\sqrt{a}+\frac{y}{2 \sqrt{a}}+O\left(y^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}$ How do we show that $(1)$
$$(-1)^n\int_{0}^{\pi/4}(\tan{x})^{2n}\ln(\cot{x})\mathrm dx=G+\sum_{k=1}^{n}{(-1)^k\over (2k-1)^2}?\tag1$$
$n\ge0$ and G is the Catalan's constant
$u=\tan{x}$ then $\mathrm \cos^2{x... | This is a partial solution. I can take the integral as far as expressing it in terms of polygamma functions of order one (the trigamma function) but are currently unable to simplify it further into the form the OP gives in terms of Catalan's constant $G$ and a finite sum term.
Let
$$I_n = \int^{\pi/4}_0 \tan^{2n} x \ln... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$ Through Induction I tried to prove that:
$$\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$$
I first let $n=1$, so that on the left hand side and the right hand side you get 1.
Then I tried to prove that this also works when ... | i think it must be
$$\sum_{k=1}^{n+1}(-1)^{k-1}k^2=\sum_{k=1}^n(-1)^{k-1}k^2+(-1)^n(n+1)^2$$
and we get $$(-1)^{n-1} \frac {n(n+1)}{2}+(-1)^n(n+1)^2=(-1)^n\frac{(n+1)(n+2)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
By using Wallis’s product, prove $\zeta(2)=\frac{\pi^2}{6}$ $\sin x = x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\cdots$
This equal is used in both the solution of Basel’s problem and the proof of Wallis’s product.
So I wonder if I can prove one by using the other.
Help me solve this problem... | You get the $\,$ Wallis product $\,$, if you set $\,\displaystyle x=\frac{\pi}{2}\,$ .
Basel problem :
The coefficient for $\,z\,$ of the infinite product $\,(1+a_1 z)(1+a_2 z)...\,$ is $\,a_1+a_2+...\,$ .
Therefore with $z:=-x^2$ and using the series of sine we get $\,\displaystyle\frac{1}{3!}=\frac{1}{(1\cdot \pi)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substi... | May be somehow tricky but I love this kind ...
$$\quad{\frac{2x}{(x+2)^2} = \\\frac{2(x)}{(x+2)^2} = \\
\frac{2(x+2)-4}{(x+2)^2} = \\
\frac{2(x+2)}{(x+2)^2} +\frac{-4}{(x+2)^2} = \\
\frac{2}{(x+2)} +\frac{-4}{(x+2)^2} \\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 1
} |
Sum of series $\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}$ I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there i... | We have the series were $n<m$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1}- (\frac {1}{1-m} + \frac {1}{2-m} +\frac {1}{3-m}+\cdots+ \frac {1}{-1}))$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{1} + \frac {1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:
$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$
The given expression can be rewritten as:
$$a+b+c-ab-ac-bc = S_1-S_2 \tag{1}$$
Then from the given condition:
$$\begin... | We'll prove that $\frac{3}{4}$ is a minimal value.
Let $c=0$.
Hence, $(1-a)(1-b)=0$ and $\sum_{cyc}a(1-c)=a+b-ab=1>\frac{3}{4}.$
Now, let $abc\neq0$, $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z$.
Thus, $xyz=1$, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y}$, $c=\frac{z}{1+z}$ and we need to prove that
$$\sum_{cyc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating Laurent series expansion I have to calculate the Laurent series expansion of $$f(z) = \frac {2z−2}{(z+1)(z−2)}$$ in $1 < |z| < 2$ and $|z| > 3$.
For first annulus, I know I must manipulate the given expression to contain terms $1/z$ and $z/2$ so that some expansion is valid for $|\frac1{z}|<1$ and $|\fr... | The function
\begin{align*}
f(z)&=\frac{2z-2}{(z+1)(z+2)}\\
&= \frac{4}{3}\left( \frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\
\end{align*}
is to expand around the center $z=0$ in $1<|z|<2$ and $|z|>3$.
Since there are simple poles at $z=-1$ and $z=2$ we have to distinguish three regions of convergen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Probability that the minimum is $2$ and the maximum is $5$ after rolling a die $4$ times
Suppose that a fair $6$-sided die is rolled $4$ times. Letting $N$ be
the minimum number of spots observed and $X$ be the maximum number of
spots observed, give the value of $P(N= 2, X= 5)$
For this to happen, we need all $4$... | Your enumeration counts some rolls repeatedly. For instance a 2 2 4 5 roll is counted twice, on account of the two different choices for 2.
I prefer an inclusion-exclusion-style argument for this, which yields a probability of
$$
\frac{4^4-2*3^4+2^4}{6^4} = \frac{55}{648} \approx 0.0849
$$
Here the $4^4$ is all rolls... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is the polynomial $\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6$ irreducible over $\Bbb Q$? How do I approach proving irreducibility/reducibility in a polynomial with rational coefficients? Can I apply Eisenstein in some way?
Is the polynomial $\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6$ irreducible over $\Bbb... | Replace $x$ by $2y$ gives
$$\frac{1}{64}x^6 + 3x^4 - \frac{1}{4}x^3 - x + 6 = y^6+48y^4-2y^3-2y+6$$
Now apply Eisenstein criterion with $p=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to solve a quartic equation using trigonometric power identitiy? According to this post, we can solve a cubic equation
$$t^3+pt+q=0$$
by the trigonometric identity
$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$
So I've tried to solve the quartic equation
$$t^4+pt^2+qt+r=0$$
using the identity
$$8\cos^4\theta-4\cos2\t... | it`s possible to solve this system. Just rewrite
$$\cos 3\phi = \cos\phi(2\cos2\phi-1)$$
and
$$\sin3\phi \sin\phi=\sin^2(\phi)(2\cos 2\phi+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Trig integral evaluations with Complex analysis I am currently trying to evaluate this integral with residues but I'm not quite there. The questions asks me to show
$$\int_0^{2\pi} \frac{\sin^2 \theta}{5 + 3\cos\theta}\ d\theta = \frac{2\pi}{9} $$
So I let $z = e^{i\theta}$ and so $\frac {dz}{iz} = d\theta$
I let $\sin... | $\int_0^{2\pi} \frac{\sin^2 \theta}{5+3\cos\theta} \ d\theta$
$z = e^{i\theta}\\
dz = ie^{i\theta}\ d\theta\\
d\theta = \frac 1{i z} \ dz$
$\sin \theta = \frac {1}{2i}(z - z^{-1})\\
\sin^2 \theta = \frac {1}{4}(2-z^2 - z^{-2})\\
\cos\theta = \frac 1{2}(z + z^{-1})$
$\oint_{|z| = 1} \frac{2-z^2 - z^{-2}}{(4iz)(5 + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find angle between $2$ points in $3D$ space? I think I am not able to convey what I need so here is the edited question:
As you can see in the image, a point on the left is at $(-3,-3,0)$ and the point on the right is at $(3,3,0)$.
And when I draw the cylinder it spawns at $(0,0,0)$, the center point of the ... | Angle between $(a,b,c)$ and $(x,y,z)$ is given by,
$$\cos\theta =\frac{ax+by+cz}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}$$
The unit vector from $(a,b,c)$ to $(x,y,z)$ will be$$\frac{a-x}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}i+\frac{b-y}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2}}j+\frac{c-z}{\sqrt{a^2+b^2+c^2}\sqrt{x^2+y^2+z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Question related to the simplification of a nested radical $\sqrt[n]{\frac{\sqrt[n]{a}}{a^2}\dots}$
Given that $a\in \mathbb R^+\setminus\{0\}$ and $n\in \mathbb N$, $n>2$, find the denominator of the irreducible fraction that represents the exponent for $a$ resulting from the simplification of the following nested ra... | Your $S(n)$ is correct.
For $n=3$, it is easy to see that the asked denominator is $3^3$.
For $n\ge 4$, we have
$$\begin{align}&\frac{1-4n+2n^2+n^n(1+n-n^2)}{(1-n)^2}\\\\&=\frac{-n^n(n^2-2n+1+n-2)+2n^2-4n+2-1}{(n-1)^2}\\\\&=-n^n+2+\frac{-n^{n-1}(n^2-2n+1-1)-1}{(n-1)^2}\\\\&=-n^n+2-n^{n-1}+\frac{n^{n-1}-1}{(n-1)^2}\\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To evaluate $\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$ Evaluate $$\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$$
I tried taking log on both sides and then using L'Hospital rule but its giving complex results.Are there any simpler methods to approach this?
| From this answer, we can see that the Taylor coefficients of $\tan(x)$ expanded around $0$ will be positive, which implies that every truncation of the Taylor series acts as a lower bound for $\tan(x)$ on $[0, \pi/2)$. Therefore, $$\tan(x)\geq x+\frac{x^3}{3}$$ for $x\in [0, \pi/2)$. As $\tan(x)/x$ is an even function,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$
Find the real roots for $\displaystyle \sqrt[4]{386-x}+\sqrt[4]{x}=6.$
Question from a Math Olympiad (ES, 2005). Answer: $(3\pm \sqrt{2})^4$.
My attempt: I will make my attempt below, but I think the approach might be too complicated... are there... | I would simply let $x=u^4$ and hope that $386-u^4=(6-u)^4$ is easy to factor. Expanding and collecting first to
$$2u^4-24u^3+216u^2-864u+910=0$$
we can hope for a factorization
$$u^4-12u^3+108u^2-432u+455=(u^2-au+b)(u^2-cu+d)$$
from which it would be easy to identify the real roots. (Hoping for an integer root would l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$ Let $\tan{x}\tan{y}=-\frac{1}{2}$,find the minumin of the value $$f=(1+\sin^2{x})(1+\sin^2{y})$$
My ugly solution:
$$f=\dfrac{1+\sin^2{x}}{\sin^2{x}+\cos^2{x}}\cdot\dfrac{1+\sin^2{y}}{\sin^2{y}+\cos^2{y}}=\dfrac{2m^2+1}{m^2+1}\cdot\dfrac{2n^2+1}{n^2+1}$$where... | Let $\tan^2x=a$ and $\tan^2y=b$.
Thus, $ab=\frac{1}{4}$ and by AM-GM we obtain:
$$(1+\sin^2x)(1+\sin^2y)=(2-\cos^2x)(2-\cos^2y)=$$
$$=\left(2-\frac{1}{1+a}\right)\left(2-\frac{1}{1+b}\right)=\frac{(2a+1)(2b+1)}{(1+a)(1+b)}=$$
$$=\frac{2+2(a+b)}{\frac{5}{4}+a+b}=2-\frac{\frac{1}{2}}{\frac{5}{4}+a+b}\geq$$
$$\geq2-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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.