Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
\frac {(3x^2-4) \left( \sqrt [3]{2x^2+1}+1
\right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
\right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 ... | $\lim_\limits{x\rightarrow\infty}
{\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}}$
$\lim_\limits{x\rightarrow\infty}
{\frac { \left( 3\,{x}^{2}-4 \right) \left( (2x^2+1)^{1/3}+1
\right) ^{2}}{ \left( 2\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve thi... | $\frac{c}{a+b} +\frac{a}{b+c}+\frac{b}{c+a}=\frac{11-a-b}{a+b}+\frac{11-b-c}{b+c}+\frac{11-a-c}{c+a}
=(11\cdot \frac{13}{17})-3=\frac{92}{17}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How does $\cos(x)\cdot\cos\left(\frac{3}{2}x\right)$ become $\frac{1}{2}\left(\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{5}{2}x\right)\right)$? How can you rewrite $\cos(x)\cdot\cos\left(\frac{3}{2}x\right)$ to $\frac{1}{2}\left(\cos\left(\frac{1}{2}x\right) + \cos\left(\frac{5}{2}x\right)\right)$?
What rules have... | Hint. One may use
$$\begin{align}
\cos a \cos b &=\frac12\left((\cos (a-b)+\cos (a+b)\right).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the set of values of x for which $\frac{x+1}{2x-3}<\frac{1}{x-3}$ Here's what I've done:
$\frac{x+1}{2x-3}<\frac{1}{x-3}$
$x+1<\frac{2x-3}{x-3}$
$(x+1)(x-3)<2x-3$
$x^2-2x-3<2x-3$
$x^2-4x<0$
$x(x-4)<0$
$0<x<4$
However this clearly fails because when $x$ is $2$, for example, the inequality fails.
Where have I gone w... | Beware of cross-multiplying when solving inequalities! You might be multiplying by a negative number – which would invalidate the inequality.
One way to proceed is to bring everything to one side and simplify:
$\quad\quad\quad\dfrac{x+1}{2x+3}\ <\ \dfrac1{x-3}$
$\implies\ 0\ <\ \dfrac1{x-3}-\dfrac{x+1}{2x+3}$
$\implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Differential equation system IVP appears to be wrong $Y(0) = (2,5)$ Let $A = \Bigg ( \begin{array}{cc} 1 & -1 \\ 1 & 3 \end{array} \Bigg )$
Need to solve the IVP: $(x(0),y(0)) = (2,5)$
Solving for the eigenvalues get the characteristic polynomial:
$\lambda^2 - 4 \lambda +4 \implies \lambda_1=\lambda_2 = 2$
for the eige... | I will give you a solution that uses matrix exponential of the jordan normal form. Generally, the solution of $y(t)=Ay(t), y(0)=(2,5)^{\top}=y_0$ is given by $y(t)=e^{At}y_0$, so we must focus on $e^{At}$. You already remarked that the characteristic equation has one root $\lambda=2$, but the corresponding eigenspace o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $ $$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx $$
Attempt:
I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.
| HINT:
Note that we can write
$$\begin{align}
\frac{1+x\cos(x)}{x(1-x^2e^{2\sin(x)})}&=(1+x\cos(x))\left(\frac{1}{x(1-x^2e^{2\sin(x)})}\right)\\\\
&=(1+x\cos(x))\left(\frac{1}{x}+\frac{xe^{2\sin(x)}}{1-x^2e^{2\sin(x)}}\right)\\\\
&=\frac1x+\cos(x)+\frac{(x+x^2\cos(x))e^{2\sin(x)}}{1-x^2e^{2\sin(x)}} \tag 1
\end{align}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Rational ODE $y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$ I really don't know how to start to solve this ODE:
$$y'=\frac{3x^2-2xy+2}{6y^2-x^2+3}$$
I know that somehow I have to isolate $y$ but how?
| Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving by induction that $n! < (\frac{n+1}{2})^n$ As an analysis homework I have to prove by induction that
$n! < (\frac{n+1}{2})^n : (2 \le n \in\mathbb{N})$
For $n = 2$ this is trivial, but for $n+1$ no matter how I transform the equation I can't seem to get $(\frac{n+2}{2})^{n+1}$ on the right-hand side. I'm sure t... | Without induction
and completely elementary:
$n!
= \prod_{k=1}^n k
$
so
$\begin{array}\\
n!^2
&= (\prod_{k=1}^n k)^2\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n k)\\
&= (\prod_{k=1}^n k)(\prod_{k=1}^n (n+1-k))\\
&= \prod_{k=1}^n k(n+1-k)\\
&= \prod_{k=1}^n (k(n+1)-k^2)\\
&= \prod_{k=1}^n \left(\dfrac{(n+1)^2}{4}-\dfrac{(n+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
write down the expression for $\sin (15°)$ using the double angle formula. show that $\sin 15^\circ=\frac {\sqrt3 -1}{2\sqrt2}$ using $\cos2A=1-2\sin^2A$
However I got $\sin 15^\circ= \sqrt{\frac {2-\sqrt 3}{4}}$ instead.
| $cos(2A)$ = $1 - 2Sin^2(A)$
$\Rightarrow$ $cos(30)$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ = $1 - 2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3}{2}$ - 1 = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{2}$ = $-2Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt3 - 2}{4}$ = $-Sin^2(15)$
$\Rightarrow$ $\frac{\sqrt{2 - \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$12^7+8^8$ divided by $13$ I Need to find what the remainder is when $12^7+8^8$ is divided by $13$
I have a solution, but don't know if it is right.
$12=-1\mod13$
$12^7=-1\mod13$
$8=8\mod13$
$8^2=6\mod13$
$8^4=10\mod13$
$8^8=9\mod13$
Then I did $-1\mod13+9\mod13=8\mod13$, so the remainder is $8$.
If someone could tell... | You are quite incorrect in your calculations. $8^2 \equiv -1 \pmod {13}$ in your second line.
However, there is an easier way you can proceed. $$12^{7}+8^8 \equiv (-1)^7+2^{24} \equiv -1+(2^{12})^2 \equiv -1+1 \pmod {13}$$ since $2^{12} \equiv 1 \pmod{13}$ from Fermat's Little Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $n < 2^n$ for all $n \geq 0$ using induction. Please verify for me?
Base case: $n = 0 $
$0 < 2^0$
$0 < 1$. This is true.
Inductive step: Suppose $n \geq 0$. Assume $P(k)$ is true if $k = n$. We must deduce that $P$ holds for $k+1$.
$n < 2^n$
$n - 2^n < 0$
$(n+1) - 2^{n+1}$
$n+1 - (2^n \times 2)$
$n+1 - (2^n + 2^n... | There is an easier way. If $n<2^n$ for some $n$, then $$n+1<2^n+1 \leq 2^n+2^n=2^{n+1}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear Differential Equation but with piece wise function ( Cant Solve ) Find a continuous solution satisfying the DE
$$y' + y = f(x),$$
where
\begin{align}
f(x) &= \begin{cases}
1, & 0 \leq x \leq 1 \\
-1, & x>1\text{.}\end{cases}\\
y(0)&=1\end{align}
I am new to the forum so sorry guys for the format mistakes but ... | First equation
(Update Fixed, thanks to RodrigodeAzevedo)
Consider $0 \leq x \leq 1$, then:
$$\begin{cases}
y' + y = 1 \\
y(0) = 1
\end{cases} \Rightarrow y(x) = 1 \Rightarrow y(1) = 1.$$
Now, solve the differential equation for $x > 1$:
$$\begin{cases}
y' + y = -1 \\
y(1) = 1
\end{cases} \Rightarrow y(t) = 2e^{-x+1} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What am I doing wrong in calculating the following limit? $$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ Dividing numerator and denominator by $x \neq0$
$$\frac{1+2/-2}{\sqrt{(6/4)+(1/-2)}-2/4}=\frac{0}{1/2}=0$$ but the limit is $4$ according to Wolfram Alpha?
| One may set $u=x+2$, then, as $x \to -2$, we have $u \to 0$, giving
$$
\frac{x+2}{\sqrt{6+x}-2}=\frac{u}{\sqrt{u+4}-2}\times\frac{\sqrt{u+4}+2}{\sqrt{u+4}+2}=\sqrt{u+4}+2 \to 4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to prove that A and B are similar Let be $$A=\begin{pmatrix}
\frac{-3}{2} & 2 & \frac{-1}{2} \\
\frac{-1}{2} & 0 & \frac{1}{2} \\
\frac{1}{2} & -2 & \frac{3}{2}
\end{pmatrix},
B=\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$$
Prove that A and B are similar.
I know if we can find a matrix $P$ so ... | The easiest way I know would be to show that $A^2$ and $B^2$ are nonzero, but $A^3=B^3=0$. There is only one $3$-by-$3$ nilpotent matrix of nilpotency index $3$, up to conjugation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding the matrix for a linear transformation on a vector space when the basis changes Let B={$u_1,u_2,u_3$} as basis of Vector Space V, and Let T: V→V be the linear operator defined by,
$$
[T]_B=\begin{bmatrix}
-3 & 4 & 7 \\
1 & 0 & -2 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
Find $[T]... | Here is a trick.
Let $B' = \begin{pmatrix} 1&1&1\\0&1&1\\0&0&1\end{pmatrix}$ This is the basis of B' in terms of the basis of B.
To change the basis of T.
$[T]_B = [B'^{-1}TB']_{B'}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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build absolute value equations know solution We have absolute value equations with unknown coefficients:
$$|x + a| = b$$
and we know the solutions:
$$x = 11 \text{ and } x = 5$$
We need to find $a$ and $b$. From
$$11 + a = b \\
5 + a = -b$$
we get $a = -8$ and $b = 3$.
But we can try another way:
$$11 + a = -b \\
5 + a... | We want to solve $|x + a| = b$ where we know that the solutions for $x$ are 11 and 5. Since it's an absolute value, $b$ must be positive and therefore $b > -b$. We might try to solve
$$11 + a = -b \qquad 5 + a = b$$
But $11 + a > 5 + a$ and $-b$ cannot be greater than $b$. So we are left with
$$11 + a = b \qquad 5 + a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arc... | First of all, $$\arccos2x\sqrt{1-x^2}=\dfrac\pi2-\arcsin2x\sqrt{1-x^2}$$
Now using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $, $$2\arcsin x=\begin{cases}\arcsin2x\sqrt{1-x^2} &\mbox{if } |x|\le\dfrac1{\sqrt2} \\ \pi-\arcsin2x\sqrt{1-x^2} & \mbox{if } x>\dfrac1{\sqrt2}\\-\pi-\arcsin2x\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Volume enclosed by $(x^2+y^2+z^2)^2=x$ I need to calculate the volume of solid enclosed by the surface $(x^2+y^2+z^2)^2=x$, using only spherical coordinates.
My attempt: by changing coordinates to spherical: $x=r\sin\phi\cos\theta~,~y=r\sin\phi\sin\theta~,~z=r\cos\phi$ we obtain the Jacobian $J=r^2\sin\phi$. When $\phi... | It's a solid of revolution. Let $h=\sqrt{y^{2}+z^{2}}$, then $(x^{2}+h^{2})^{2}=x$.
$\therefore \; h^{2}=\sqrt{x}-x^{2} \:$ where $\, 0\leq x \leq 1$.
\begin{align*}
V &= \int_{0}^{1} \pi h^{2} dx \\
&= \pi \int_{0}^{1} \left( \sqrt{x}-x^{2} \right) dx \\
&= \pi \left[ \frac{2}{3} x^{\frac{3}{2}}-
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Using generating function to solve initial value problems I have a hard exam coming up and something I've struggled with since week 1 of semester is initial value problems. How would I go about solving:
(a) $u_{n} - 7u_{n-1} = 3 * 7^n : u_0 = 4 $
(b) $u_{n} - 4u_{n-1} + 4u_{n-1} = 3*2^n : u_0 = 6/u_1 =1 $
(c) $u_{n} -... | Part (a) and (c) follow the form:
$$f_{n} - a f_{n-1} = b \, c^{n}.$$
Since the generating function method is being asked then that is what will be demonstrated next.
\begin{align}
\sum_{n=0}^{\infty} \left( f_{n+1} - a f_{n} \right) t^{n} &= \frac{b}{1-c t} \\
\sum_{n=1}^{\infty} f_{n} t^{n-1} - a F(t) &= \\
\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyl... |
Where I am wrong
In the following part :
$$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$
This is not correct.
$$\begin{align}f(\theta)&=\sin^4\theta+\cos^4\theta+\frac{1}{\sin^4\theta}+\frac{1}{\cos^4\theta}+4\\&=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta+\frac{(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Given $\tan 3x=4$, find the value of $\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$ Given $\tan 3x=4$, find the value of $S=\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$
I expanded each of $\tan (120+x)$ and $\tan (60+x)$ getting as
$$S=\tan^2 x+\left(\frac{\tan 120+\tan x }{1-\tan 120 \tan x}\right)^2+\left(\frac{\tan 60+\tan x }{1... | As $a=\tan x,$
$b=\tan(60^\circ+x),\tan3(60^\circ+x)=\cdots=\tan3x,$
and $c=\tan(120^\circ+x),\tan3(120^\circ+x)=\cdots=\tan3x$
The roots of
$$4=\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$$
$$\iff\tan^3y-12\tan^2y-3\tan y+4=0$$ are $a,b,c$
We need $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $a_1,a_{100}, a_i=a_{i-1}a_{i+1}$, what's $a_1+a_2$? I've been given the following puzzle
Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.
Find $a_1+a_2$.
I... | Note that since $a_i=a_{i-1}a_{i+1}$, then, for $i\geqslant 3$, we also have $a_{i-1}=a_{i-2}a_i$. Plug this in the first formula, we have $a_i=a_{i-2}a_ia_{i+1}\Rightarrow a_{i-2}a_{i+1}=1\Rightarrow a_{i+3}=\frac{1}{a_i}.$
Then, the sequence is in the form of $a_1=a,a_2=ab,a_3=b,a_4=\frac{1}{a},a_5=\frac{1}{ab},a_6=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| $$2^{3^{n+1}}+1 = (2^{3^n})^3 + 1 = (2^{3^n}+1)((2^{3^n})^2 - 2^{3^n} + 1).$$ Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $a+b+c=0$ what is the value of $\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$ Let $s=\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$.
If we use inequality $\frac{x^2}{a}+\frac{y^2}{b} \ge \frac{(x+y)^2}{(a+b)}$ we get $s \ge 0$ as $a+b+c=0$.
Again $s \le \frac{a^2}{bc }+\frac... | Since for $a+b+c=0$ we have $a^3+b^3+c^3=3abc$, we obtain:
$$\sum_{cyc}\frac{a^2}{2a^2+bc}=\frac{\sum\limits_{cyc}a^2(2b^2+ac)(2c^2+ab)}{\prod\limits_{cyc}(2a^2+bc)}=\frac{\sum\limits_{cyc}(4a^2b^2c^2+4a^3b^3+a^4bc)}{\sum\limits_{cyc}(3a^2b^2c^2+4a^3b^3+2a^4bc)}=$$
$$=\frac{\sum\limits_{cyc}(4a^2b^2c^2+4a^3b^3+a^4bc)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the second derivative of$f(x)=x^2\sqrt{4-x}$
Find the second derivative of the function following:
$$f(x)=x^2\sqrt{4-x}$$
Here I go...
$$f(x)= x^2(4-x)^{1\over 2}$$
\begin{align*}
f'(x) &= 2x(4-x)^{1\over 2}+{1\over 2}x^2(4-x)^{-{1\over 2}}(-1)\\
&= 2x(4-x)^{1\over 2} - {1 \over 2}x^2(4-x)^{-{1\over 2}}\\
&... | Notice that
$$f'(x)=\frac12(4-x)^{-1/2}(16x-5x^2)$$
Then
\begin{align*}
f''(x)&=\frac14(4-x)^{-3/2}(16x-5x^2)+(4-x)^{-1/2}(8-5x)\\
&=\frac{4(4-x)(8-5x)+(16x-5x^2)}{4(4-x)^{3/2}}\\
&=\frac{15x^2-96x+128}{4(4-x)^{3/2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply... | The common factor can be taken to front as coefficient in the numerator as well as in the denominator.
$$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{1}{(x+1)} \cdot [\frac{x+3}{(x-1)} - \frac{x-2}{(x+1)}] $$
$$=\frac{1}{(x+1)} \cdot \frac{(x^2+4 x+ 3) - (x^2- 3 x + 2)} {(x^2-1)} $$
$$=\frac{1}{(x+1)} \frac{(7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$3$ digits numbers in which digits are repeated.
Total number of $3$ digit number which can be formed by using the digits $1,2,3,4,3,2,1$
$\bf{My\; Try::}$ Total no. of $3$ digits numbers in which exactly $2$ digits are identical, are
$112,113,114,221,223,224,331,332,334$
So Total numbers of $3$ digits numbers are $\... | Yes, your solution is correct. In the second case, you want to find the number of 3 digit numbers with distinct digits where the digits are taken from $\{1,2,3,4\}$. This number is just $P(4,3)=4 \times 3 \times 2 = 24$ because the first digit can be chosen in 4 ways, the second in 3 ways (since the second digit can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Equilibrium in a system of nonlinear differential equations I have two questions about a specific system of differential equations. First, if a complex number can be an equilibrium point. Second, and related with the first question, how can I verify that $(0,0)$ is the only solution that satisfy $X'=0$ and $Y'=0.$ The ... | We have two coupled polynomial ODEs
$$\dot x = xy^3 - xy^2 \qquad \qquad \dot y = -y^3 - 3x^4$$
To find the equilibria, solve the system of polynomial equations
$$0 = xy^3 - xy^2 \qquad \qquad 0 = -y^3 - 3x^4$$
Using SymPy,
>>> from sympy import roots, solve_poly_system
>>> x, y = symbols('x y')
>>> p1 = x * y**3 - x *... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
| Use: $$a^2+b^2+c^2\ge ab+bc+ca$$
Proof:$$2(a^2+b^2+c^2)\ge 2(ab+bc+ca) \Leftrightarrow$$
$$\Leftrightarrow(a-b)^2+(b-c)^2+(c-a)^2\ge0$$
Then
$$x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xy^2z+x^2yz+xyz^2=$$
$$=xyz(x+y+z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can someone show me how to simplify $\sin\left(\arccos\left(\frac{x}{x+1}\right)\right)$ I am needing help learning how to simplify the following equation:
$$\sin\left(\arccos\left(\frac{x}{x+1}\right)\right)$$
Also any steps on how to get the answer would be greatly appreciated!
| as Rahul said... let $\arccos(\frac{x}{x+1})=\theta$ so $\cos\theta=\frac{x}{x+1}$ but you want $\sin\theta$ which is $\sqrt{1-\cos^2\theta}$.
*
*Remember $\arccos t$ is between $0$ and $\pi$, so $\sin(\arccos t)$ is positive.
so the answer would be $$\sqrt{1-(\frac{x}{x+1})^2}=\sqrt{1-\frac{x^2}{(x+1)^2}}=\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to find the expected value of this game? How can I create a function that figures out the expected value for a value that can change.
Example:
I can flip a (fair) coin $n$ number of times.
The pot starts at $\$1$.
If I lose, $50$ cents is added to the pot.
If I win, I take what is in the pot and the pot is reset to... | Let $e_{n, p}$ be the expected value over $n$ coin flips if the initial value of the pot is $p$.
$e_{1,p}$ is easy to calculate; $$e_{1,p} = \frac 12 \cdot p = \frac p2$$
Now given the independence of different flips, we can write in general
$$e_{n,p} = \frac 12(p + e_{n-1, 1}) + \frac 12 e_{n-1, p+1/2}$$
Using this w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx$ How can I evaluate the following integral?
$$I=\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx$$
I tried it with Wolfram Alpha, it gave me a numerical solution: $0.785398$.
Although I immediately know that it is equal to $\pi /4$, I fail to obtain the answer with pen... | There are 5 wonderful solutions already. I want to share with you one more alternative which is long but hopefully interesting.
Using $\sin 2x=2\sin x\cos x$ and multiplying both denominator and numerator by $1-\sin x \cos x$ rewrites$$
I=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos x+\sin x \cos ^2x}{1-\sin ^{2} x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Calculate the area of a sphere drilled by two cylinders.
Let $S$ be the sphere given by the equation $x^2+y^2 +z^2 =4$ cut with
$z \geq 0$. Now, we drill the semisphere that is left with two
vertical cylinders of radius $1$, whose axes are respectively on the
points $(0,1,0)$ and $(0,-1,0)$. Calculate the area ... |
Parametrization of the surface cut out by one cylinder
The image above shows the intersection of the hemisphere with the cylinder whose axis passes through $(0,1,0)$ and whose equation is
$$
x^2+(y-1)^2=1
$$
The surface cut out by the cylinder can be parametrized with Cartesian coordinates as
$$
\vec\Sigma=\left(x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$ Just a bit of background on the question:
When proving:
$$\cos\frac{\pi}{15}\cdot \cos\frac{2\pi}{15} \cdot \cos\frac{3\pi}{15}\cdot \cos\frac{4\pi}{15} \cdot \cos\frac{5\pi}{15} \cdot \cos\frac{6\pi}{15}\cdot \cos\frac... | Turn the double-angle formula for sine "inside out". Put in
$$\begin{align}\cos(A)& =\frac{\sin(2A)}{2·\sin(A)} \\[6pt]
\cos(2A)&=\frac{\sin(4A)}{2·\sin(2A)}
\end{align}$$
etc., and use telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \... | $$x^2-y^2-3(x-y)=u^2-v^2$$
$$(x-y)(x+y-3)=(u+v)(u-v)$$
Assuming $u^2-v^2\neq 0$, we have:
$$\dfrac{x-y}{u-v}=\dfrac{u+v}{x+y-3}=\dfrac{r}{s}$$ where $\gcd(r,s)=1$.
$$x-y=\dfrac{r(u-v)}{s}$$
$$x+y=\dfrac{s(u+v)}{r}+3$$
since $x,y$ are integers, and $u,v$ have the same parity, then there exist $p,q$ such that: $$u-v=ps $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Linear application Let $f:\mathbb{R}^3\to\mathbb{R}^3$ be a linear application and let $\{e_1,e_2,e_3\}$ the canonical basis of $\mathbb{R}^3$.
We know that $\operatorname{Im} f=\langle(1,1,3), (0,1,1)\rangle$ and that
$$2f(e_2)-f(e_3)=e_1-e_2+e_3$$
Which is the matrix associated to $f$ with respect the canonical basi... | Assume that
$$fe_1=\begin{pmatrix}1\\1\\3\end{pmatrix}\;\;,\;\;\;fe_2=\begin{pmatrix}0\\1\\1\end{pmatrix}\;\;,\;\;fe_3=\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}$$
Observe that we have already
$$\text{Im}\,f\subset\left\langle\;\begin{pmatrix}1\\1\\3\end{pmatrix}\;,\;\;\begin{pmatrix}0\\1\\1\end{pmatrix}\;\right\rangle$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of special tridiagonal determinant Let $A_{n}$ be the following tridiagonal determinant of order $n:$
\begin{vmatrix}
a_{0}+a_{1}& a_{1}& 0& 0& \cdots& 0& \quad0\\
a_{1}& a_{1}+a_{2}& a_{2}& 0& \cdots& 0& \quad0\\
0& a_{2}& a_{2}+a_{3}& a_{3}& \cdots& 0& \quad0\\
\vdots& \vdots& ... | Here is an observation that may save some labor later on. At first glance, the above recurrence relation only makes sense for $n>2$. But for $n=2$, the recurrence requires $A_2=(a_2+a_1)A_1-a_{1}^2 A_0$; since But $A_2=(a_0+a_1)(a_1+a_2)-a_1^2$ and $A_1=(a_1+a_0)$, the recurrence will be valid at $n=2$ if $A_0=1$. Simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A BBP-type series The BBP-type series
$$ \frac{\pi}{2} \, \left( \frac{\alpha^{2}}{5} \right)^{\frac{1}{4}} = \sum_{n=0}^{\infty} \left[ \frac{1}{10 n + 1} + \frac{\alpha}{10 n + 3} - \frac{\alpha}{10 n + 7} - \frac{1}{10 n + 9} \right],$$
with golden ratio $\alpha = \frac{1 + \sqrt{5}}{2}$ can be obtained by a parti... | Hint. One may recall the following series representation of the digamma function,
$$
\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{u+k}
\right)=\psi(u+1)+\gamma, \qquad u >-1,
$$ giving
$$
\sum_{k=0}^{\infty} \left( \frac{1}{k+a} - \frac{1}{k+b}
\right)=\psi(b)-\psi(a),\qquad a>0,\,b>0. \tag1
$$
From $(1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$ Let
$$A=
\begin{bmatrix}
-1 & 6 & -9 \\
-11 & 24 & -33 \\
-6 & 12 & -16 \\
\end{bmatrix}
$$
a) Without calculating $A^4$ prove that $A^4\in Span\{A,I\}.$
b) Write $A^n$ in a form of $a_nA+b_nI$
If matrix $A^4\in Span\{A... | The characteristic polynomial of $A$ is
$$ p_A(\lambda) = \det(\lambda I - A) = \lambda^3 - 7\lambda^2 + 16\lambda - 12 = (\lambda - 3)(\lambda - 2)^2. $$
Since part (b) asks you to show in particular that $A^2 \in \operatorname{span} \{ I, A \}$, the minimal polynomial of $A$ must be of degree at most two. Since the m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality problem, for positive $a,b,c$, if $abc=1$, then $\frac{1}{1+a+b^2}+\frac{1}{1+b+c^2}+\frac{1}{1+c+a^2}\leq1$ I need help or guidance in solving this inequality that I am battling for 3 days now. I have tried everything that comes to mind, but I am stuck. The inequality is as $$\sum_\textrm{cyc}\frac{1}{1+a+... | The following technique can deal with many of the above type of inequalities. For example, Problem 3 of IMO 2005 can be solved using the same idea (but a little bit easier). People participating in math Olympiads should keep it in mind.
For any $k$, applying Cauchy-Schwarz inequality we have
$$(1+a+b^2)(c^{2k} + a^{2k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Triple Integration problem, stuck. I have a doubt over this triple integral: I have to integrate
$$\iiint x\ \text{d}x$$
between
$$z = x^2 + y^2 ~~~~~~~ and ~~~~~~~ 4x + z = -2$$
But I feel like dumb because I cannot even start!
It's between $z = x^2 + y^2$ and $z = -4x - 2$.
If I equal them I get
$$x^2 + y^2 + 4x + ... | $$I=\int_{-2-\sqrt{2}}^{-2+\sqrt{2}}\int_{-\sqrt{2-(x+2)^2}}^{\sqrt{2-(x+2)^2}}\int_{-2-4x}^{x^2+y^2}x\,dzdydx$$
let $x=-2+r\cos \theta$ and $y=r\sin\theta$
| {
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"url": "https://math.stackexchange.com/questions/1827544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$ Show that $x^2 + y^2 + z^2 \ge 35$ if $x+3y+5z \ge 35.$
I have tried everything (proof by contradiction, etc.) but I can't seem to get it. The book didn't give any constraints whatsoever. Any hints would be appreciated. Thank you.
| I suggest a solution for who know Lagrange multiplier.
Let $f(x,y,z)=x^2+y^2+z^2$, and $g(x,y,z)=x+3y+5z=a\ge 35$. Then
\begin{align}
\nabla f(x,y,z)&=(2x,2y,2z)\\
\nabla g(x,y,z)&=(1,3,5)\ne \mathbf{0}
\end{align}
Let $\lambda$ be a real number such that $\nabla f = \lambda \nabla g$, then
$$
2x=\lambda,\quad 2y=3\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Does this equation have no solutions? The question is this :
The source from where I got this question was devoid of any answers to it, so I came here, this is how I proceeded :
LHS :
$((((({(x)^x})^{2x})^{3x})^{....x^2})^2 = (((((x)^{2x^2})^{3x})^{....x^2})^2 =...........= (x^{(x^x)x!})^2 = x^{2(x^x)x!} $
RHS :
$\sqr... | The equation can be written as:
$x^{2x^{x} x!}=\sqrt{x^{\frac{ 1}{x! x^{x}} }}$.
By squared up the two members, we get:
$x^{4x^{x} x!}=x^{\frac{ 1}{x! x^{x}}}$.
Since this is the equality between two powers, for them to be equal they must have the same basis and the same exponent; therefore, since the basics are the sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| More generally, you are looking to prove: $x^x > (x-2)^{x+1}$. This can be done by taking log both sides, and its easier. Consider $f(x) = x\ln x - (x+1)\ln (x-2)$ on $(11, \infty)$, and taking log we have: $f'(x) = \ln x + 1 - \ln(x-2) - \dfrac{x+1}{x-2}$. We have $f''(x) = \dfrac{1}{x} - \dfrac{1}{x-2}+ \dfrac{3}{(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| I would multiply by the conjugate of denominator $$a=\frac{3 \sqrt{3}-4}{7-2 \sqrt{3}}=\frac{1}{37} \left(13 \sqrt{3}-10\right)$$ $$b=\frac{3 \sqrt{3}-8}{1-2 \sqrt{3}}=\frac{1}{11} \left(13 \sqrt{3}-10\right)$$ this makes $a<b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| First you need $x \ne 0$ and $y \ne 0$ in order for your fraction $\dfrac{x^2+y^2}{2xy}$ to make sense.
Then as $\sin \in [-1,1]$ you need $x^2+y^2 \leq 2xy$ which means:
$x^2 -2xy+y^2 \leq 0$ ie $(x-y)^2 \leq 0$
Therefore you have solutions on $\theta$ only if $x = y \ne 0$, and those solutions are $\theta = \dfrac{\p... | {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt... | You have got yourself into a tangle by being too clever too soon, and squaring. It is better to simplify first. With careful simplification you don't need calculus or anything advanced at all.
To pick up a theme of Kenny Lau, put $u=x+\sqrt{x^2-10x+9}$. Then your inequality becomes $u\ge\sqrt{u}$, which is equivalent (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to show that this function $f(x,y,) <0$? I would like to show that the function $f(x,y) = -5 x^4 + 4 y^2 - 5 y^4 + x^2 (4 - 6 y^2)$ is less than $0$ for $1 < x^2 + y^2 <2$ (at least). Upon rearranging, I obtain
$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2 + x^2y^2)$
But then I don't know how to proceed to get $f(x,y)<0?
Th... | Let us write this expression under the form:
$$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2) + 4 x^2y^2$$
Let us convert this expression into polar coordinates. One gets:
$$F(r,\theta) = -5r^4 + 4r^2 + r^4 \sin ^2(2 \theta)$$
(using relationship $\sin (2 \theta) = 2 \sin \theta \cos \theta$).
Therefore:
$$F(r,\theta) = r^2 (r^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part. I just cannot seem to get the algebra to work in my favor.
Here is the original:
... | This should help:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{(i+3)(i+4)}&= \sum_{i=1}^k\frac{1}{(i+3)(i+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k}{4(k+4)}+\frac{1}{(k+4)(k+5)}\\[1em]
&= \frac{k(k+5)+4}{4(k+4)(k+5)}\\[1em]
&= \frac{(k+1)(k+4)}{4(k+4)(k+5)}\\[1em]
&= \frac{k+1}{4(k+5)}.
\end{align}
See if you can spot where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is $(2,5)$ the only solution? Find all pairs $(m,n)\in{\mathbb{N^2}}$ such that
$$(m^2-1)^3-n^2=2$$
Is $(2,5)$ the only solution?
| Maybe this helps
$(m^2-1)^3-n^2 =2\iff (m^2-1)^3-27=n^2-25$
$\iff (m^2-4)( (m^4-2m^2+1)+3(m^2-1)+9)=(n-5)(n+5)$
$\iff(m+2)(m-2)(m^4+m^2+7)=(n-5)(n+5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\... | Expanding
\begin{equation*}
\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1
\end{equation*}
we get
\begin{equation*}
(t-1^2)(t-3^2)(t-5^2)(t-7^2) - (t-3^2)(t-5^2)(t-7^2)x^2 - \text{ similar terms } = 0
\end{equation*}
This biquadratic in $t$ has four roots $2^2, 4^2, 6^2, 8^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846993",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
... | Your work, while a reasonable approach, does not lead to the answer. A slightly simpler approach does:
Let $p \in \Bbb{Z} = \lfloor x \rfloor$ and let $x= p +q$ with $0\leq q < 1$.
Then the left hand side is
$$L = 63p + \lfloor q\rfloor + \lfloor 2q\rfloor+ \lfloor 4q\rfloor+ \lfloor 8q\rfloor+ \lfloor 16q\rfloor+ \lfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.
| Using repeated differences and Newton's interpolation formula we get
$$
n^5-5n^3+4n = 120 \binom{n}{3} + 240 \binom{n}{4} + 120 \binom{n}{5}
$$
Although this identity suffices for answering the question, it also implies the simpler identity below:
$$
n ^5-5n^3+4n = 120 \binom{n+2}{5}
$$
which gives a crystal clear an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $... | If you insist on algebraic techniques then note $\ 3(3^{35}-5)\, =\, (3^{36}-1)-14$
and $\,\ 7 = \color{#c00}{3^2\!-\!3\!+\!1}\mid 3^6\!-\!1\mid 3^{36}-1\ $ so $\,7\,$ divides both summands above, so $\,7\mid 3^{35}\!-\!5$
where $\ \color{#c00}{x^2-x+1}\mid x^6-1\ $ since it divides the $\,\rm\color{#c00}{difference\ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Polynomial ,divides and Induction Proof?
$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n } \quad x-y \;\text{ divides }\; x^n-y^n \text{ prove by induction... | For the inductive step, suppose $x-y$ divides $x^n-y^n$ for some $n$. You have to prove that, under this hypothesis, $x-y$ divides $x^{n+1}-y^{n+1}$.
Hint:
Rewrite it as $$x^{n+1}-y^{n+1}=x(x^n-y^n)+xy^n - y^{n+1}$$
and make partial factorisations.
Some details:
By the induction hypothesis, $x^n-y^n=(x-y)q(x,y)$, so
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$ what is the value of $P\times Q\times R$ Problem
The roots of $ax^2+bx+c$ are 6 and $P$. The roots of $cx^2+bx+a$ are $Q$ and $R$
And we are asked to find $P\times Q\times R$ by using the identities:
$P(x)=Q(x)\times D(x)+R(x)$
where $P(x)... | Considering roots of $ax^2+bx+c$ as $\frac{-b\pm \delta}{2a}$ the roots of $cx^2+bx+a$ will be $\frac{-b\pm \delta}{2c}$. So, the product of all the four roots are $$6PQR=\frac{(-b+ \delta)(-b-\delta)}{2a}\times\frac{(-b+ \delta)(-b-\delta)}{2c}=\frac{4ac}{4ac}=1$$ Thus, we have $$PQR=\frac16$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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On the solution of the equation $z+ \alpha \left| z-1\right| + 2i = 0$ Question:-
Find the range of real number $\alpha$ for which the equation $z+ \alpha \left| z-1\right| + 2i = 0$; $z=x+iy$ has a solution. Also find the solution.
Attempt at a solution:-
On substituting $z=x+iy$ in the equation $z+ \alpha \left| z-1... | From $(1)$, $\alpha=0$ works. If $\alpha\not=0$, then
$$\sqrt{(x-1)^2+4}=\frac{x}{-\alpha}\gt 0\tag3$$
From $(2)$, for $-\frac{\sqrt 5}{2}\le \alpha\le\frac{\sqrt 5}{2}$ with $\alpha^2\not=1$,
$$x=\frac{\alpha^2\pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$$
Case 1 : For $\alpha=0$, $x=0$.
Case 2 : For $\alpha^2=1$, you'v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Finding the roots $x^4-4x^3-x^2-8x+4=0$ (contest math) So the problem is :
$x^4-4x^3-x^2-8x+4=0$, find all solutions
A tip that I have gotten, is to divide both sides by $x^2$. I've tried so, but I do not manage to see any further. Do anyone know how this tip could help me?
(Yes, I'm aware that the polynomial above can... | Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(x^2-5x+2)(x^2+x+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| If we "bunch up" $8$ and $9$ to get $89$, the rest of the numbers must add up to $55$. Bunching $6$ and $7, 5$ and $6$, and $4$ and $5$ will make the sum too high. If we bunch $3$ and $4$ to get $34$, so we have $34+5+6+7+89=141$ which needs $3$ more.
So $$1+2+34+5+6+7+89=144$$.
That's the only possible $89$ bunch u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\... | Since I am almost blind, I have a lot of problems reading the image.
Consider $$I=\int \frac{\sqrt{64 x^2-256}}{x}\,dx$$ What you apparently did is $x=2\sec(t)$, $dx=2 \tan (t) \sec (t)$ which make $$I=\int \tan (t) \sqrt{256 \sec ^2(t)-256}\,dt=16\int \tan (t) \sqrt{\tan ^2(t)}\,dt=16\int \tan^2 (t) \,dt$$ $$I=16\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$
How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$
using the following method :
note that :
$(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^... | One may write, as $n \to \infty$,
$$
\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\cdot\sqrt{1+\frac1n}\: \right)^{-1}\\
&= \frac{(-1)^n}{n}\left(1+\frac{(-1)^n}{\sqrt{n}}\left(1+\frac1{2n}+\mathcal{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and... | You van do it also by purely algebraic means:
$$\begin{align}
\frac{1}{\sqrt{n+1}+\sqrt{n}}&=\frac{1}{2\sqrt{n}}+\frac{\sqrt{n}-\sqrt{n+1}}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})}\\
&=\frac{1}{2\sqrt{n}}-\frac{1}{2\sqrt{n}(\sqrt{n+1}+\sqrt{n})^2}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{p... | If I rewrite your augmented matrix as follows:
$\begin{bmatrix}
1&0&0&0&\\
0&1&1&0&\\
0&0&0&1&\\
\end{bmatrix}$
$\begin{bmatrix}
a_1\\
a_2\\
a_3\\
a_4\\
\end{bmatrix}$
$= \begin{bmatrix}
4\\
-2\\
1\\
\end{bmatrix}$
Where $a_1, \cdots a_4$ are the coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| In general we have $\frac{\partial}{\partial x} \frac{f(x)}{g(x)}= \frac{\frac{\partial f(x)}{\partial x} \times g(x) - \frac{\partial g(x)}{\partial x} \times f(x)}{g^2(x)}$. So in the case of your question, set $f(x)= x$ and $g(x) = (4x - 3)^{\frac{1}{2}}$. And we have:
$\frac{\partial}{\partial x} (x) = 1$ and
$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Remainder when $5^{5555}$ is divided by $10000$. Find the remainder when $5^{5555}$ is divided by $10000$.
A step by step guide with explanation for a beginner student in modular arithmetic is needed.
| Start by factoring $10000 = 10^4 = 2^4 \times 5^4$.
So we want to know the results mod $2^4$ and mod $5^4$.
The second is easiest: $5555 > 4$ so $5^{5555} \equiv 0 \mod 5^4$.
Now mod $2^4 = 16$: $\gcd(2^4, 5) = 1$ and $\phi(16) = 8$, so $5^8 \equiv 1 \mod 2^4$. $5555 \equiv 3 \mod 8$, so $5^{5555} \equiv 5^3 \equiv 13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to decompose $x^3-1$ I need to decompose $x^3-1$, I know the Binomial theorem, and finding roots of a polynomial, how should I approach this?
| We can just use binomial theorem:
Observe
$$(x-1)^3=x^3-3x^2+3x-1$$
So $$\begin{align}x^3-1&=(x-1)^3+3x^2-3x\\
&=(x-1)^3+3x(x-1)\\
&=(x-1)((x-1)^2+3x)\\
&=(x-1)(x^2+x+1)\end{align}$$
Hope this helps.
P.S.
If factoring $x^2+x+1$ is needed, recall we have a formula for quadratic polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equ... | we get the curvature $$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$$
now $y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
$y''=\frac{-a^2y'^2-b^2}{a^2y}$
putting up the values in formula
we get
$$\kappa =\frac{\frac{a^2y'^2+b^2}{a^2y}}{\bigg[1+\frac{b^4}{a^4}{\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Derivative of $\tan^{-1}(f(x))$ What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated
| Easy way :
Put $$x=4\cos 2\theta$$
Then $$y=\tan ^{-1} \left(\frac{\sqrt{4+4 \cos 2\theta}+\sqrt{4-4 \cos 2\theta}}{\sqrt{4+4 \cos 2\theta}-\sqrt{4-4 \cos 2\theta}} \right)$$
$$y=\tan ^{-1} \left(\frac{2\sqrt{1+ \cos 2\theta}+2\sqrt{1- \cos 2\theta}}{2\sqrt{1+ \cos 2\theta}-2\sqrt{1- \cos 2\theta}} \right)$$
$$y=\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Diagonalizability of a given matrix I must find out under which conditions the matrix
$$A=
\left[\begin{array}{ccc|cc}& & & c_0 &\\
& & &c_1&\ddots\\
& & &c_2 &\ddots& c_0\\
& 0 & & \vdots & \ddots &c_1\\
& & & c_d & \ddots &c_2\\
& & & & \ddots & \vdots\\
& & & & & c_d\\
& & & && \end{array}\right]
$$
is ... | Note that your matrix $A$ is a block upper triangular matrix of the form
$$ A = \begin{pmatrix} 0_{d \times d} & B \\ 0 & D_{(d + 1) \times (d + 1)} \end{pmatrix} $$
where $D$ is also a block upper triangular matrix with $c_d$ on the diagonal. Hence, the characteristic polynomial of $A$ is
$$ p_A(x) = x^d(x - c_d)^{d+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| $$\begin{array}{rl} & \begin{bmatrix} -1\\ 1\end{bmatrix} \begin{bmatrix} -1\\ 1\end{bmatrix}^T - \begin{bmatrix} -5\\ 1\end{bmatrix} \begin{bmatrix} -5\\ 1\end{bmatrix}^T = \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix} - \begin{bmatrix} 25 & -5\\ -5 & 1\end{bmatrix}\\\\ &= \begin{bmatrix} -24 & 4\\ 4 & 0\end{bmatrix} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightar... | Hint: Note that
$$\left(2x^2-x\right)^2<4\left(x^4-x^3+x^2-x+1\right)\leq\left(2x^2-x+2\right)^2$$
for all $x\in\mathbb{R}$. The right-hand side becomes an equality if and only if $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit
Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.
My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ pro... | It seems me correct,the alternative short proof could be:
For any $\epsilon >0$ ,and by choosing $\quad \delta =min\left( 1,\epsilon \right) $ and note that if $\left| x-1 \right| <\delta $ the we will get
$$
\left| \frac { x^{ 2 }+3 }{ x+1 } -2 \right| =\left| \frac { { x }^{ 2 }-2x+1 }{ x+1 } \right| =\left| \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $x^6+x^4+x^3+x^2+1=0$? There is a hint in the question, use the factorization of $x^5+x+1$.
| Divide by $x^3$ and remember that
$$
x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
$$
so the equation becomes
$$
\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
+\left(x+\frac{1}{x}\right)+1=0
$$
Set $t=x+1/x$ and you get
$$
t^3-2t+1=0
$$
Can you go on? Note that $t=1$ is a solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove $ \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0 $ Prove that for $ 0<x< \pi $, $$ \quad S_n(x) = \sin x + \frac{ \sin3x }{3} + ... + \frac{ \sin((2n-1)x) }{2n-1} >0 \quad \forall n = 1,2,... $$
Having trouble with this problem. This is an olympiad-style question, so an answer that doesn't us... | If we want to avoid integration, we can obtain the result via summation by parts. Let
$$F_m(x) := \sum_{k = 0}^{m-1} \sin \bigl((2k+1)x\bigr).\tag{1}$$
Using the addition theorem for the cosine, we obtain
$$\sin(rx) = \frac{2\sin x \sin(rx)}{2\sin x} = \frac{\cos\bigl((r-1)x\bigr) - \cos \bigl((r+1)x\bigr)}{2\sin x}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
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Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$
Let $x,y,z>0$ and $x+y+z=1$. Prove that
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$
My work so far:
I use Titu's Lemma:
$$\frac{x^3}y+\frac{y^3}z+\... | Suppose $x \geq y \geq z$. Using the rearrangements inequality we have
$$ \frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x} \geq x^3 \cdot \frac{1}{x}+...+z^3 \cdot \frac{1}{z} =x^2+y^2+z^2 $$
since $x^3,y^3,z^3$ and $1/x,1/y,1/z$ have opposite ordering. The same thing applies to
$$ \frac{x^3}{z}+\frac{z^3}{y}+\frac{y^3}{x} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{... | $\frac{n}{n+2}-\frac{n}{2n+1}=n\int\limits_{n+2}^{2n+1}\frac{1}{x^2}dx<n(\frac{1}{n+1}^2+\frac{1}{n+2}^2+\dots + \frac{1}{2n}^2)< n\int\limits_{n+1}^{2n}\frac{1}{x^2}dx=\frac{n}{n+1}-\frac{n}{2n}$
Clearly the expressions on both sides go to $\frac{1}{2}$ when $n\to \infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| 2^n is never odd, so n^2 has to be odd to give a prime, n is odd
2^n with n odd is 4^k 2, 4^k = 4 mod 6, so
2^n = 2 mod 6 here
n is 1, 3 or 5 mod 6
n^2 is 1 or 3 mod 6
if n^2 = 1 mod 6, then 2^n + n^2 = 3 mod 6, which would mean that when divided by 6, then 3 would remain, meaning that 2^n + n^2 divides by 3 and is no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How can you prove Ceva's Theorem using vectors? Given the following layout, where $ \overrightarrow{CY} = \mathbf{a} $ and $ \overrightarrow{CX} = \mathbf{b} $:
Prove that $$ \frac{\overrightarrow{BX}}{\overrightarrow{XC}} \times \frac{\overrightarrow{CY}}{\overrightarrow{Y... | Our proof is in three parts. Define $ \overrightarrow{CB} = \alpha\overrightarrow{CX}$, $ \overrightarrow{CA} = \beta\overrightarrow{CY}$ and $ \overrightarrow{AB} = \gamma\overrightarrow{AZ}$. Also, let $ \overrightarrow{AX} = \lambda\overrightarrow{AP}$, $ \overrightarrow{BY} = \mu\overrightarrow{BP}$ and $ \overrigh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The indefinite integral $ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$ $$ \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}$$
$$ \int \frac{\sin \left(x +\alpha\right)}{\cos^3 x}{\sqrt\frac{ \cos x +\sin x }{\cos x - \sin x}}$$
$$ \int \frac{\s... | Let $$I = \int \frac{\sin (x +\alpha)}{\cos^3 x}{\sqrt\frac{\csc x + \sec x}{\csc x - \sec x}}dx$$
$$I = \cos \alpha\int \frac{\sin x+\cos x\cdot \tan \alpha}{\cos x}\cdot \sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
So $$I = \cos \alpha \int (\tan \alpha+\tan x)\sqrt{\frac{1+\tan x}{1-\tan x}}\cdot \sec^2 xdx$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$Then $|z_{1}+z_{2}+z_{3}|$
If $z_{1},z_{2},z_{3}$ are three complex number such that $|z_{1}| = |z_{2}| = |z_{3}| = 1$
and $\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.$The... | (You made a mistake when taking conjugates: You have $\bar k=\ldots\ $.)
Denote by $s_1$, $s_2$, $s_3$ the elementary symmetric functions of the $z_i$. It is easy to see that we may assume $s_3=1$. Then the $z_i$ are the solutions of the third degree equation
$$z^3-s_1z^2+s_2z-1=0\ .\tag{1}$$
The solutions of the equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Convergence/Divergence $\int_2^\infty \frac{4x^3+3x^2-x}{5x^5-2x^4+x^2-2}\ln x \, dx$ $$\int_2^\infty \frac{4x^3+3x^2-x}{5x^5-2x^4+x^2-2}\ln x\, dx$$
How should I approach this? can I look at $$\int_2^\infty \frac{\ln x}{x^2}\,dx \text{?}$$
| The integrand function is bounded between $\frac{4\log x}{5 x^2}$ and $\frac{84\log x}{65 x^2}$ on the interval $(2,+\infty)$, hence the integral is converging since
$$ \int_{2}^{+\infty}\frac{\log x}{x^2}\,dx = \frac{1+\log 2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate this integral, $\int_0^{\infty}\frac{\ln^2x}{1+x^2}dx$ $\displaystyle\int_0^{\infty}\dfrac{\ln^2x}{1+x^2}dx$
$x=\arctan\alpha$
$dx=\dfrac{1}{1+\alpha^2}d\alpha$
and try a lot items, but didnt arrived anywhere.
| Let us evaluate it via the Mellin transform
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:1608132}
\tag{2}
\end{equation}
where
\begin{equation}
f(x) = \frac{1}{1+x^{2}}
\label{eq:1608133}
\tag{3}
\end{equation}
Applying the Mellin transform, yields
\begin{equation... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Calculating the infinite sum of one over the odd numbers squared: $\sum_{i \ge 0} \frac{1}{(2i+1)^2}$ Can someone tell me how to calculate the following infinite sum?
$$
(1/1^2)+(1/3^2)+(1/5^2)+(1/7^2)+(1/9^2)+(1/11^2)+ \cdots
$$
Don't give me the answer. Can you tell me if this is a geometric series?
| To test if it's geometric, compare the ratio of adjacent terms. For example, $\frac{1/3^2}{1/1^2}$ and $\frac{1/5^2}{1/3^2}$. Are these two quantities equal?
Hint:
$$
\left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots\right) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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If $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then how are $x$ and $y$ related?
If $$\cos x + \cos y - \cos(x+y) = \frac{3}{2}$$ then
*
*(a) $x + y = 0$
*(b) $x = 2 y$
*(c) $x = y$
*(d) $2 x = y$
It is problem of trigonometry, and I have the solution of the problem. However, after seeing the so... | Let $z_1 = -\cos x + i\sin x$, $z_2 = -\cos y + i\sin y$. Then,
\begin{align*}
|1+z_1+z_2|^2 \geq 0 &\Rightarrow 3 -2\cos x - 2\cos y + 2\cos(x+y) \geq 0\\
&\Rightarrow \cos x + \cos y - \cos(x+y) \leq \frac{3}{2}
\end{align*}
Equality holds if and only if $1 + z_1 + z_2 = 0$ and hence $1+\bar{z_1}+\bar{z_2} =0$. Addin... | {
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"timestamp": "2023-03-29T00:00:00",
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solution of absolute value equation Question: If $x + |x| + y = 7$ and $x + |y| - y = 6$, then $x + y =?$
My solution:
I considered each cases of $x$ and $y$ i.e $x$ positive and $y$ positive, $x$ positive and $y$ negative, $x$ negative and $y$ positive, and finally $x$ negative and $y$ negative. After solving I found ... | Substitute $y=7-x-|x|$, giving an equation in a single unknown, $$x+|7-x-|x||-(7-x-|x|)=6.$$
Then for $x\le0$, $x=6$, which is impossible and for $x\ge0$, $x+|7-2x|-(7-2x)=6$.
You still have to distinguish $2x\le7$, giving $x=6$, which is impossible, and for $2x\ge7$, $x-14+4x=6$ or $$\color{green}{x=4,y=-1}.$$
Not mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| Let $$S = 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Also we know, $$(1-b)(1+b+b^2+b^3 \cdots +b^n-1) = 1-b^n$$
We now start feeling that having $(1-b)$ as a factor in each term in the $RHS$ will create a "nice" series.
So multiplying the $LHS$ & $RHS$ by $(1-b)$, we get, $$S(1-b) = (1-b)+a(1-b^2)+a^2(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability that n+1 toss is a head too? A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What ... | The probability of seeing $n$ heads in $n$ tosses of a fair coin is $2^{-n}$. Thus the total probability of seeing $n$ consecutive heads is $$\frac 12\times 2^{-n}+\frac 12 \times 1$$ Therefore, given that you have in fact observed $n$ consecutive heads the new estimate for the probability that you have the fair coin ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| Using the function $x^{1/4}$ is not a purely algebraic proof. Here is one. Explicitely, we'll prove $2^n>n^4$ for all $n>16$.
For that, we'll prove by induction that if $n\ge 16$ and $2^n\ge n^4$, then $2^{n+1}>(n+1)^4$.
For $n=16$, we have an equality: $2^{16}=16^4$.
Now suppose that, for some $n\ge 16$, we have $2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Integral ${\large\int}_0^1\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}$ Mathematica claims that
$${\large\int}_0^1\!\!\frac{dx}{(1+x^{\sqrt2})^{\sqrt2}}=\frac{\sqrt\pi}{2^{\sqrt2}\sqrt2}\cdot\frac{\Gamma\left(\frac1{\sqrt2}\right)}{\Gamma\left(\frac12+\frac1{\sqrt2}\right)},\tag{$\diamond$}$$
and it also confirms numerically.
Ho... | Using Euler's integral representation of the Gaussian hypergeometric function, and assuming $a>0$, we get
$$ \begin{align} \int_{0}^{1} \frac{dx}{(1+x^{a})^{a}} &= \frac{1}{a} \int_{0}^{1} \frac{u^{1/a-1}}{(1+u)^{a}} \, du \\ &= \frac{1}{a} \int_{0}^{1} u^{1/a-1} (1-u)^{(1+1/a)-1/a-1}(1+u)^{-a} \, du \\ &= \frac{1}{a} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow... | Suppose $1+2^{2^n}+2^{2^{n+1}}=7k$ and set, for simplicity, $a=2^{2^n}$. Then $2^{2^{n+1}}=a^2$ and $2^{2^{n+2}}=a^4$. Then
\begin{align}
&1+2^{2^n}+2^{2^{n+1}}=1+a+a^2 \\[4px]
&1+2^{2^{n+1}}+2^{2^{n+2}}=1+a^2+a^4
\end{align}
Then
$$
(1+a^2+a^4)-(1+a+a^2)=a^4-a=a(a-1)(a^2+a+1)=7ka(a-1)
$$
so
$$
1+a^2+a^4=7k+7ka(a-1)
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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What's the best bound for $\sum_{i=0}^{[\lg{n}]}{\binom{n}{2^{i}}}$ I want to find best lower bound and upper bound for:
$$\sum_{i=0}^{[\lg{n}]}{\binom{n}{2^{i}}}$$
| As @JackD'Aurizio states, the sum is dominated by the choices of $i$ for which $2^i$ is closest to $\frac{n}{2}$. Let $a = \left\lfloor \lg \frac{n}{2} \right\rfloor = \lfloor \lg n \rfloor - 1$. This gives the largest $i$ for which $2^i \le \frac{n}{2}$. The terms we should focus on are then $i = a$ and $i = a + 1$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simple determinant calculation
\begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix}
It is fairly easy I just want to reassure the steps are correct:
\begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix} Taking out $-2^{n}$ when $n$ is the number of rows/columns in this case $n=1$ (... | \begin{vmatrix}
1 & -2 & 2 \\
1 & -2 & -2\\
2 & -2 & 1
\end{vmatrix}
$$R'_1=R_1-R_2 \space R'_2=R_2-R_3 $$
$$\begin{vmatrix}
0 & 0 & 4 \\
-1 & 0 & -3\\
2 & -2 & 1
\end{vmatrix}
=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find ordered positive pairs of integers (x, y) that $x^2y^2-2(x+y)$ is a perfect square. Source: Q1 level Hard test 3 of Violympic National Round 2015-16 (Test 3/c/Q1)
(ended in May 2016).
I have looked for similar questions here on Math.SE, but NOTHING was close to mine.
The competition actually asked for the number ... | Let us write $x^2y^2 -2(x+y) = (xy-a)^2$. Then, $a(2xy-a) = 2(x+y)$. This means that $a$ must be even, because $2(x+y)$ is even, and $a$ and $2xy-a$ have the same parity.
To give you an intuition about our idea, let us think about the following question: What is the difference of squares of consecutive numbers, or con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
I determined that since $2a^2-1 \equiv 1,7 \pmod{8}$ we must have $b^2+2 \equiv 3 \pmod{8}$ in order for the fraction to be an integer. I didn't see how to find a contradiction from he... | $b^2+2 \equiv 2,3,6 \pmod{8}$ because $b^2+2$ satisfying $b^2+2 \equiv 3 \pmod{8}$ gets an inverse under module condition. Otherwise, $b^2+2$ will be an even which contradict to the parity of $2a^2-1$
Notice that $3 ^{-1} \equiv 3 \pmod 8$.
And $3 \times 1 \equiv 3^{-1} \pmod 8$, $7 \times 3^{-1} \equiv 5 \pmod 8$ none... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove this integral $\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$ Turns out this integral has a very nice closed form:
$$\int_0^\infty \frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\Gamma(1/4)^2 }{6 \sqrt{\pi}} \frac{a^3-b^3}{a^4-b^4}$$
I found it ... | Another way to do it, may be.
Assuming $a>0$ and $t>0$, we have
$$I_a=\int_0^t \sqrt{x^4+a^4}\, dx=a^2 t \, _2F_1\left(-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{t^4}{a^4}\right)$$ Expanding as Taylor series for infinitely large values of $t$, $$I_a=\frac{t^3}{3}+\frac{\sqrt{\frac{\pi }{2}} a^3 \Gamma
\left(\frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
Is this a new method for finding powers? Playing with a pencil and paper notebook I noticed the following:
$x=1$
$x^3=1$
$x=2$
$x^3=8$
$x=3$
$x^3=27$
$x=4$
$x^3=64$
$64-27 = 37$
$27-8 = 19$
$8-1 = 7$
$19-7=12$
$37-19=18$
$18-12=6$
I noticed a pattern for first 1..10 (in the above example I just compute the firs... | What you have discovered is a finite difference calculation. For any function $f$, in this case the third-power function
$$
f(n) = n^3
$$
we can define the forward difference, or forward discrete derivative:
$$
\Delta f(n) = f(n+1) - f(n) = 3n^2 + 3n + 1
$$
Likewise,
\begin{align*}
\Delta \Delta f(n) = \Delta^2 f(n) &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 1
} |
Inequality solving How must I proceed to solve inequalities of the following form:
$$-1≤r-1/r≤1$$
This is the final step for another problem that I'm attempting to solve and thus far, this is where I've gotten:
By taking the L.C.M and multiplying both sides by $r$ (given to take strictly positive (or zero) values), I ... | Let's consider two cases, $r > 0$ and $r < 0$.
Case 1
Suppose $r > 0$. Multiplying both sides of the inequality by $r$,
$$-r \leq r^2-1 \leq r.$$
Rearranging the left-side inequality,
$$r^2+r-1 \geq 0.$$
Using the quadratic equation to find the roots of $r^2+r-1$,
$$r = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
| We have $$81=3^4=(x+y)^4\\=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\=x^4+12x^2+54+12y^2+y^4\\=x^4+y^4+12(x^2+y^2)$$so once you solve for $x^2+y^2$, it should be straight-forward to solve for $x^4+y^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| If we have $y = \arcsin(2^{x + 1} 3^x / (1 + 36^x))$, then $$\sin y = \frac{2^{x + 1} 3^x}{1 + 36^x}$$ The expression on the right is still a daunting product/quotient, but it can be reduced to sums/differences using logarithms, giving $$\ln \sin y = (x + 1) \ln 2 + x \ln 3 - \ln(1 + 36^x)$$ Now it is routine to take t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.