Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to integrate $\sqrt{x^2-1}$ How do you integrate the following integral? I have tried u-substitution, but it doesn't work out.
$$\int(x^2-1)^{0.5}dx$$
| $I=\int 1\cdot\sqrt{x^2-1}dx$
Put $x = \sec y$
then $dx = \sec y.\tan y dy$
$= \int \sqrt{\sec^2 y-1}. \sec y.\tan y dy$
$= \int \sqrt{\tan^2 y}. \sec y.\tan y dy$
$= \int \tan y. \sec y. \tan y dy$
$= \int \tan^2 y. \sec y dy$
$= \int (\sec^2 y - 1). \sec y dy$
$= \int \sec^3 y dy - \int \sec y dy$
$= \int \sec^3 y dy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is every parallelogram a rectangle ??
Let's say we have a parallelogram $\text{ABCD}$.
$\triangle \text{ADC}$ and $\triangle \text{BCD}$ are on the same base and between two parallel lines $\text{AB}$ and $\text{CD}$, So, $$ar\triangle \text{ADC}=ar\triangle \text{BCD}$$
Now the things those should be noticed are that... | As already hinted in comments and answers, you appear to have
applied Heron's formula
to the areas of the triangles $\triangle ADC$ and $\triangle BCD$.
Then observing that the only variables in Heron's formula are
the three sides of the triangle and its area,
and that three of these quantities in the formula for $\tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 9,
"answer_id": 1
} |
The function $f (n) = (1 + 1 / n) ^ {n+1}$ is decreasing I cannot prove that the function $$f (n) = \left(1 + \frac1n\right) ^ {n + 1},$$ defined for every positive integer $n$, is strictly decreasing in $n$. I already tried to prove by induction and also tried to prove by calculating the difference between $f (n + 1)$... | Take the fraction
$$
\frac{f(n)}{f(n-1)}=\left(\frac{n+1}{n}\right)^{n+1}\bigg/\left(\frac{n}{n-1}\right)^n=\left(\frac{(n-1)(n+1)}{n^2}\right)^n\frac{n+1}{n}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}.
$$
Using the induction, show that for $n\ge2$ and $x>-1$, $x\ne0$ (if $x=0$ then there would be a trivial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
$t^3$ is never equal to
Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to:
(A). $4t+3$
(B). $8t+5$
(C). $10t+3$
(D). $6t+5$
We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I do... | $t^2 = ax + b$
$t^3 = cx +d$
where $a$, $b$, $c$, $d$ are integers.
See 2 cases:
*
*$t^2=at+b$ has an integer root $\implies$ t is integer
*$t^2=at+b$ has no integer root $\implies$ roots are $p+\sqrt{q}$ and
$p−\sqrt{q}$, where $p$, $q$ are rational. You can see that if one
of these roots is a root of $x^3=cx+d$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$
Let $a, b, c, d > 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.
I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
| Using this inequality is probably the quickest way
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq \frac{(a+b+c+d)^2}{b+c+d+a}=a+b+c+d$$
Otherwise, using AM-GM we have:
$$\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{d}+d+\frac{d^2}{a}+a \geq 2\sqrt{\frac{a^2}{b}b}+2\sqrt{\frac{b^2}{c}c}+2\sqrt{\frac{c^2}{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Curious relation between $e$ and $\pi$ that produces almost integers I have seen in an article, without proof, that the following expression involving $e$ and $\pi$ is an almost integer very close to 1:
$ e^{-\frac{\pi}{9}} + e^{-4\frac{\pi}{9}} + e^{-9\frac{\pi}{9}} + e^{-16\frac{\pi}{9}} + e^{-25\frac{\pi}{9}} + e^{-... | Using the identity provided by Robert Israel (which can be derived using the Poisson summation formula), we have $$\sum_{k = -\infty}^{\infty}e^{-k^2\tfrac{\pi}{n}} = \sum_{k = -\infty}^{\infty}\sqrt{n}e^{-n\pi k^2}.$$
We can bound the second summation by: $$\sqrt{n} \le \displaystyle\sum_{k = -\infty}^{\infty}\sqrt{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$ The result below has been disproven.
Let $n \in \mathbb{N}$. Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$.
We know that $p \neq 2,5$. We need to have \begin{align*}5^n &\equiv 2 \pmod{p}\\2^n &\equiv 5 \pmod{p}.\end{align*} Thi... | $\gcd(5^{65} - 2, 2^{65} - 5) = 12681 = 3^2 \cdot 1409$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A ... | The equation
$$\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0$$
is equivalent to $p_A(-x)=0$, where $p_A$ is the characteristic polynomial of
$$A=\begin{pmatrix}
0 & a & b & c \\
a & 0 & b & c \\
m &n & 0 &p \\
m& n& p& 0
\end{pmatrix};$$hence the roots of your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
$a \cot A + b \cot B + c \cot C = 2(R+ r)$ With usual notations, prove that in a triangle $ABC$:
$$a \cot A + b \cot B + c \cot C = 2(R+ r)$$
I know $R = (abc)/(4\Delta)$ and $r = \Delta/s$.
I could not understand how to start it.
| \begin{align}
a \cot A + b \cot B + c \cot C &= 2R\Bigg[\sin A \frac{\cos A}{\sin A} + \sin B \frac{\cos B}{\sin B} + \sin C \frac{\cos C}{\sin C}\Bigg]\\
&= 2R\big[\cos A + \cos B +\cos C\big]\\
&= 2R\Bigg[2\cos \frac{A+B}{2}\cos \frac{A-B}{2} + 1-2\sin^2 \frac{C}{2}\Bigg]\\
&=2R\Bigg[1+ 2\sin \frac{A}{2}\sin \frac{B}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$ Let $a,b \in \mathbb{R}^*$ Show that :
$$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$
Let $a,b\in\mathbb{R}^*$
let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've :
$$
\begin{aligned}
3+\frac{a^... | We have: $1+\dfrac{a^2}{b^2} \ge \dfrac{2a}{b}, 1+\dfrac{b^2}{a^2} \ge \dfrac{2b}{a}$. Adding these we have the desire inequality. Thus the inequality is strict.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
finding value of indefinite integration of $\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$ finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$
$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$
$\displ... | HINT: $$y^{10}+1= \left( {y}^{2}+1 \right) \left( {y}^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1
\right)
$$
further we get
$$\frac{y^7-y^5+y^3-y}{y^{10}+1}=1/5\,{\frac {y \left( 4\,{y}^{6}-3\,{y}^{4}+2\,{y}^{2}-1 \right) }{{y}
^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1}}-4/5\,{\frac {y}{{y}^{2}+1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{... | Since $a,b,c$ are sides of a triangle we may write $(a,b,c) = (y+z,z+x,x+y)$
with $x,y,z>0$. The difference between the two sides is then
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2).
$$
By cyclic symmetry we may assume $x \geq z$ and $y \geq z$, and write
$x = (1 + \alpha) z$ and $y = (1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would... | Hint:
$$\frac{2}{\sqrt{x^2+4}-x}=\frac{2(\sqrt{x^2+4}+x)}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{... | Set $x=\frac{\sinh^2u}{\sqrt{5}}$ and $dx=\frac{2\sinh u\cosh u}{\sqrt{5}}$
So $$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx=\int \frac{6\sinh u\cosh u}{\sqrt{5}\cosh^6 u}du -\int\frac{2\sinh^3u \cosh u}{\sqrt{5}\cosh^6 u} \, du = \int \frac{6\sinh u}{\sqrt{5}\cosh^5 u} \, du -\int \frac{2(1+\cosh^2 u ) \sinh u}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Is there any double series that cannot exchange the sum? I want to find a sequence $(u_{n,p})_{(n,p)\in\mathbb{N}^2}$ that satisfied:
$$
\sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p} ~\text{is convergent}
$$
$$
\sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p} ~\text{is convergent too}
$$
but
$$
\sum_{p=0}^{\infty}\sum_{n=... | Let $u_{mn} = 1/(m^2 - n^2)$ if $m \neq n$ and $u_{mn} =0$ if $m = n.$
Note that
$$\frac{\pi^2}{12} = \sum_{n=1}^\infty \sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} \neq \sum_{m=1}^\infty \sum_{n=1, n \neq m}^\infty \frac{1}{m^2 - n^2} = - \frac{\pi^2}{12}.$$
By anti-symmetry, the double sum changes sign with an in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\... | using $$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$ and so on and plugging these equations in your equation and factorizing we get
$$-1/2\,{\frac { \left( c+a-b \right) \left( -c+a-b \right) \left( -2
\,c+a+b \right) }{bca}}
=0$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integral of a rational function I need a closed form for the following integral
$$\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$
where $a,b \in \mathbb{C}$ and $k \in \mathbb{N}$. That $t+b$ makes the integral complicated for me.
Any simple approaches for that ?
I tried wolfram |Alpha but it fails. A final solutio... | Use partial fractions to do the calculation. Let
$$ \frac{t+b}{(t^2+a)(k^2+(t+b)^2)}=\frac{At+B}{t^2+a}+\frac{Ct+D}{(t+b)^2+k^2}$$
and then one has
$$ (At+B)((t+b)^2+k^2)+(Ct+D)(t^2+a)=t+b.$$
It is not hard to obtain
\begin{eqnarray}
A&=&\frac{k^2-a-b^2}{a^2+2 a b^2-2 ak^2+b^4+2 b^2 k^2+k^4},\\
B&=&\frac{a b+b^3+b k^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$... | I'd do it differently: since $A$ and $B$ are not both zero, we can divide by $-\sqrt{A^2+B^2}$, getting
$$
\frac{-A}{\sqrt{A^2+B^2}}x+\frac{-B}{\sqrt{A^2+B^2}}-\frac{C}{\sqrt{A^2+B^2}}=0
$$
and we can choose $\alpha$ so that
$$
\begin{cases}
\cos\alpha=-\dfrac{A}{\sqrt{A^2+B^2}}\\[6px]
\sin\alpha=-\dfrac{B}{\sqrt{A^2+B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
If $a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$, then $a^2+b^2 \ge 2$ Prove that if
$$a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$$
then $$a^2+b^2 \ge 2.$$
| From the conditions we have $\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\geq2$.
Thus, it remains to prove that $a^2+b^2\geq\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}$, which is $a^2b^2\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \fra... | Let $f(x)=\sum_{k=1}^n\frac{x^k}{k}$. Then, we can write
$$f(x)=\sum_{k=1}^n \int_0^x t^{k-1}\,dk=\int_0^x \frac{1-t^n}{1-t}\,dt$$
so that $f(2)=\sum_{k=1}^n\frac{2^k}{k}=\int_0^2 \frac{1-t^n}{1-t}\,dt$.
Next, we choose $0<\delta<1$ and write
$$\begin{align}
\frac{n}{2^n}\int_0^2 \frac{1-t^n}{1-t}\,dt&=\frac{n}{2^n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$ Let $x,y$ be real numbers such that :
$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.
Prove that :
$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.
I tried taking $x=y$. It simplifies everything a l... | Let $x =\sinh(a), y=\sinh(b)$.
Then
$$(\sinh(a)+\cosh(b)) (\cosh(a) +\sinh(b))=1 \\
\sinh(a)\cosh(a)+\sinh(b)\cosh(b)+\cosh(a)\cosh(b)+\sinh(a)\sinh(b)=1 \\
\sinh(a+b)+\cosh(a+b)=1
$$
Now,
$$1=\cosh^2(a+b)-\sinh^2(a+b)=\left( \sinh(a+b)+\cosh(a+b)\right)\left( \cosh(a+b)-\sinh(a+b)\right) \\=\left( \cosh(a+b)-\sinh(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $ \sin \alpha = \frac 45 $ and $ \cos \beta = \frac{5}{13} $, prove that $ \cos \frac{\alpha-\beta}{2} = \frac{8}{\sqrt {65}} $ I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$
But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.
Is the question wrong ?
| HINT:
\begin{align}
\cos(\alpha/2-\beta/2)&=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}=\sqrt{(1+\cos\alpha\cos\beta+\sin\alpha\sin\beta)/2}\\
\end{align}
$$ \cos\alpha =\frac35, \cos\beta = \frac{5}{13},\, \sin\alpha=\frac45,\, \sin\beta =\frac{12}{13}, $$
plug in to get $ 8/\sqrt{65}$
For other quadrants calculate other th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$ How can I prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$, probably using basic number theory?
| Below is a somewhat not so elementary approach:
Notice $2016=36\times56$, and by Fermat's little theorem, for any $37\not\mid a$, we have $a^{36}\equiv1\pmod{37}$.
So $(10^{2014} + 10^{2015} + 2)^{2016} − 1\equiv-1\text{ or } 0\pmod{37}$.
But we know $10^{2016}\equiv1\pmod{37}$ so $10^{2015}\equiv10^{-1}\equiv2^{-1}\ti... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$.
My answer is $k<1$ or $k<\frac{1}{5}$, but the answer sheet says $k<\frac{1}{5}$ or $k>1$.
What have I done wrong? Please help.
What I'... | We have that the discriminant $$\Delta _{f (x)} =(1-3k)^2-4k^2 =(5k-1)(k-1) $$ We can analyse this into three cases:
Case $(1) $: If $k <\frac {1}{5} $, then surely $(5k-1) $ and $(k-1) $ are negative. So, $\Delta $ is positive. (Acceptable)
Case $(2) $: If $k\in (\frac {1}{5},1) $, then $(5k-1) $ is positive but $(k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.
| Let $a^2=x$.
Hence, $\frac{1+3x}{(1+x)^2}\leq\frac{9}{8}$ it's $(3x-1)^2\geq0$.
The equality occurs for $x=\frac{1}{3}$, which says that the answer is $\frac{9}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine if function is convex I want to determine if the function
$$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$
is convex on $\mathbb{R}^2$. Any ideas on how to prove convexity for two variable functions?
| $$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$
$$\nabla f=\begin{bmatrix} e^x+2x-2y+2 \\ e^y-2x+8y+3\end{bmatrix} $$
$$\nabla^2 f=\begin{bmatrix} e^x+2 & -2 \\ -2 & e^y+8\end{bmatrix} $$
Since
$$e^x+2>0$$
and $$(e^x+2)(e^y+8)-2^2=8e^x+2e^y+16-4=8e^x+2e^y+12>0$$
The function is convex.
Another way to see this is to observe t... | {
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"timestamp": "2023-03-29T00:00:00",
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If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$?
Prove that if $-1<a<1$, then :
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3 $
I do not know how to even approach this problem. Those fourth roots confuse me a lot.
Any help would be appreciated.
Also I would like to know how the left ... | It's wrong of course. Try $a=0$.
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\leq3$ is true.
Indeed, by P-M $(x+y+z)^4\leq27(x^4+y^4+z^4)$.
This inequality follows also from Holder and more:
$$(1+1+1)^3(x^4+y^4+z^4)\geq(x+y+z)^4$$
Hence, $$\left(\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\right)^4\leq27(1-a^2+... | {
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Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\b... | You have one mistake.
$\frac{(a^2+b^2)^2}{(ab)^2}$ this term should be
$\frac{(a^2+b^2)^2}{(2ab)^2}$ then its easily solve.
| {
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Find all prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$ Find prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$
$x^2-y^2-z^2$ is not factorable, so what should I do? My mind is blank here. Any help is appreciated.
| We have that $$a^2\equiv 1\lor 0\pmod{4}$$
We also have that $$2017\equiv 1\pmod{4}$$.Now try to choose $x^2,y^2,z^2$ so that you get $1\pmod{4}$.You can see that the only solution is if $y^2\equiv 0\pmod{4}$ and $z^2\equiv 0\pmod{4}$ but since $z,y$ are prime we have that $z=y=2$ so the only solution is
$x^2-8=2017,x... | {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt... | Write
$|z-(-1+i\sqrt{3})|=|z-(1-i\sqrt{3})|$
This is the locus of all $z$ which thier diatance from two points $A=-1+i\sqrt{3}$ and $B=1-i\sqrt{3}$ is equal. It halfs the fragment $AB$ which should be $y=\sqrt{3}x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral of a two variable function $e^{-(5x^2-6xy+5y^2)}$ How to compute $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(5x^2-6xy+5y^2)} \, dx \, dy \;?$$
I have seen the integral first time. I can't find any hint how to even proceed? Kindly HELP. Thank You!
| Try to use the fact that the integral of the density of the normal distribution is 1:
$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx=1.$$
Here $-(5x^2-6xy+5y^2)=-5(x-\frac{3}{5}y)^2-\frac{16}{5}y^2$, so our integral is
$$\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}\int_{-\infty}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$ How should I determine the following limit?
$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$
| Direct approach:
Let $f(x) := \cos\sqrt{x} - \cos\sqrt{x - 1}$.
By the mean value theorem, for two values $a, b \in \mathbb R$, there exists $t\in\mathbb [a, b]$ such that $\cos(b) - \cos(a) = \cos'(t)(b - a)$. Thus,$$
|f(x)|=|\cos\sqrt{x} - \cos\sqrt{x - 1}| \\
=|\cos'(t)(\sqrt{x} - \sqrt{x - 1})| \\
=|-\sin(t)|\cdot... | {
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"timestamp": "2023-03-29T00:00:00",
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series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$
We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$
This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\fra... | $$\dfrac2{2n(2n+1)(2n+2)}=\dfrac{2n+2-2n}{2n(2n+1)(2n+2)}=\dfrac1{2n(2n+1)}-\dfrac1{(2n+1)(2n+2)}$$
$$=\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)$$
$$\sum_{n=2}^\infty\left(\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)\right)$$
$$=\left(\dfrac12-\dfrac13+\dfrac14-\dfrac15+\c... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a Reduction Formula I'm trying to find a reduction formula for:
$${I}_n=\int \frac{x^{n}}{\sqrt{ax+b}}dx$$
This is where I have gotten to so far:
$${I}_n= \frac{2x^{n}\sqrt{ax+b}}{a}-\frac{2n}{a} \int \frac{x^{n-1}(ax+b)}{\sqrt{ax+b}}dx$$
I'd appreciate a little nudge in the right direction :)
| You're almost there. Simply note that for $I_n=\int \frac{x^n}{\sqrt{ax+b}}\,dx$, we have
$$\begin{align}
I_n&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\sqrt{ax+b}\,dx\\\\
&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\frac{ax+b}{\sqrt{ax+b}}\,dx\\\\
&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\left(a\int \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $z_1^2+z_2^2$ is real, $z_1(z_1^2-3z_2^2)=2$, and $z_2(3z_1^2-z_2^2)=11$, then find $(z_1^2+z_2^2)^2$
If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.
I have tried man... | $$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$
Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$
Now Add and Subtract these two equation, we get
$$(z_{1}+iz_{2})^3 = 2+11i$$
$$(z_{1}-iz_{2})^3 = 2-11i$$
So $$(z^2_{1}+z^2_{2})^3 = ... | {
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"source": "stackexchange",
"question_score": "4",
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Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6 $ is divisible by 8 for all $n \in N$.
Base: $ n = 0 $
$ 8 | 11 + 3 - 6 $ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.
So we get $ 1... | Setup the same as your current work:
$\dots$
$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$
$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$
$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115633",
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"source": "stackexchange",
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If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadrati... | To find the remainder of the long division by $x^2-2x+k$ you can keep replacing $x^2$ with $2x-k$ repeatedly, until getting the remainder of degree $1\,$:
$$
\begin{align}
x^4-6x^3+16x^2-25x+10 & = (2x-k)^2 - 6x(2x-k)+16(2x-k)-25x+ 10 \\
& = -8x^2 + (-4k +6k +32-25)x+k^2-16k+10 \\
& = -8(2x-k) + (2k+7)x+k^2-16k+10 \\... | {
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"source": "stackexchange",
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Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also
$$
(100 a+10b... | $(a+b+c)$ must be a square, so it can only be $1,4,9,16$ or $25$.
By your inequalities we have $a+b+c=1,4$ or $9$.
notice that $(100a+10b+c)^2\geq 100^2$.On the other hand $1^5$ and $4^5$ are too small.
We conclude that $a+b+c=9$.
So now we must have $(100a+10b+c)=\sqrt{9^5}=3^5=243$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does the discriminant tell us how many zeroes a quadratic equation has? The quadratic formula states that:
$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The part we're interested in is $b^2 - 4ac$ this is called the discriminant.
I know from school that we can use the discriminant to figure out how many zeroes a quad... | Well if $a \ne 0$
$ax^2 + bx + c = 0 \iff$
$x^2 + \frac bax = -\frac ca \iff$
$x^2 + \frac bax + \frac {b^2}{4a^2} = \frac {b^2}{4a^2}- \frac ca \iff $
$x^2 + 2*\frac {b}{2a}x +(\frac {b}{2a})^2 = \frac {b^2 - 4ac}{4a^2} \iff $
$(x + \frac b{2a})^2 = \frac {b^2 - 4ac}{4a^2} \iff $
$x + \frac b{2a} = \pm \sqrt{\frac {b^... | {
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"timestamp": "2023-03-29T00:00:00",
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Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to sh... | No tedious algebraic manipulation required!
Consider the following experiment: say that we choose a sequence $x_1,x_2,x_3$ from $\{1,\ldots,n\}$, without replacement. What is the probability that $x_1$ is the smallest of the three numbers?
On the one hand, this probability is $\frac{1}{3}$ by symmetry. Why? Given any... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\s... | If $\arctan x=y,-\dfrac\pi2\le y\le\dfrac\pi2, x=\tan y$
$\implies\dfrac{2x}{1+x^2}=\sin2y$
$\implies\arcsin\dfrac{2x}{1+x^2}=\begin{cases} 2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1 \\\pi-2\arctan x & \mbox{if } 2y>\dfrac\pi2\iff x>1 \\-\pi-2\arctan x & \mbox{if } 2y<-\dfrac\pi2\iff x<-1\end{ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Explain how these answers were computed from this graph and equation? this question was given on a previous exam:
Below is a graph of a function y = f(x) with a few key points.
Draw a graph of y = 3f(2x) + 4, with the corresponding points shown in the graph above.
And the answer key is given as the points:
(-1/2,10), ... | Suppose you have a function defined by some equation $y=f(x)$ as in this problem and suppose each $y$ in the equation is replaced by a linear expression $ay+b$ where $a$ and $b$ are constants, $a\ne0$. Suppose also that each $x$ in the equation is replaced by a linear expression $cx+d$ where $c\ne0$ and $d$ are constan... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
| You can proceed as follows:
$$a+2b+3c=15\tag1$$
From $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{3} \ge \sqrt[3]{6abc}$$
$$\implies \frac{15}{3} \ge \sqrt[3]{6abc}$$
$$\implies 5^3 \ge 6abc$$
Again from $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{2} \ge \frac{a+2b}{2} \ge \sqrt{2ab}$$
$$\frac{15}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz=1+2xyz$. Prove that:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$$
The equality occurs for $x=y=1$ and $z=0$.
I tried Lagrange Multipliers and more, but I don't see a proof.
| Short proof.
Clearly $xy+yz+zx \ge 1$
We have by AM-GM
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$
$$\ge x+y+z+ \frac{4xy}{x+y}+\frac{4yz}{y+z}+\frac{4zx}{z+x} $$
$$ \ge x+y+z + \frac{4(xy+yz+zx)}{x+y+z} \ge x+y+z+\frac{4}{x+y+z} \ge 4$$
The proof is complete.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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value of $2\lambda^2-1 $ in trigonometric expression if $\displaystyle \frac{\sin^3 \beta}{\sin(\beta - 3 \alpha)} = \frac{\cos^3 \beta}{\cos(\beta-3 \alpha)} = \lambda\;,$ then $2\lambda^2-1 = $
options
$(a)\; \lambda \sin \beta$
$(b)\; \lambda \cos \beta$
$(c)\; 2\lambda \cos \beta$
$(d)\; \lambda \tan \beta$
Attempt... | HINT:
Clearly we need to eliminate $\alpha$
Expand $\sin(\beta-3\alpha),\cos(\beta-3\alpha)$ to form two simultaneous equations in $\sin(3\alpha),\cos(3\alpha)$
Solve for $\sin(3\alpha),\cos(3\alpha)$
Use $\sin^2(3\alpha)+\cos^2(3\alpha)=1$ to eliminate $\alpha$
| {
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"url": "https://math.stackexchange.com/questions/2128686",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2... | You got the two $x$ points right, but you make a mistake in the next step. The correct version should be:
$$\int_1^2[\text{top function}-\text{bottom function}]\ dx$$
Here, the top function is $4x-x^2-1$ and the bottom function is $x+1$, hence
$$\begin{align}\text{Area}&=\int_1^2[(4x-x^2-1)-(x+1)]\ dx\\&=\int_1^2[4x-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Polynomial with no three distinct rational roots Prove that $x^3-2x^2-2x+a$, where $ a\in \mathbb{R} $, has no three distinct rational roots.
| Suppose
$$f(x) = x^3 - 2x^2 - 2x + a = (x - P)(x - Q)(x - R)$$
where $P,Q,R$ are rational numbers.
Our goal is to derive a contradiction.
If any of $P,Q,R$ is zero, then $a = 0$, hence $f(x) = x(x^2 - 2x - 2)$, contradiction, since $x^2 - 2x - 2$ has no rational roots.
Thus, assume $P,Q,R$ are nonzero.
By Vieta's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $x^2 \equiv -1 \pmod{365}$ solvable? I know that a similar question already exists, but I have a different question to ask.
We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.
My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \... | We have $365 = 5 \times 73$. The congruence becomes $x^2 = -1 \mod 5$ and $x^2 = -1 \mod 73$.
We have if $p = 1 \mod 4 \implies x^2 = -1 \mod p$ has exactly $2$ solutions.
Thus $x^2 = -1 \mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \mod 73$ has solutions $y_0,y_1$.
The original solutions satisfies either:
$x = x_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Limit of function which is product of $x^x $ primitives I'm trying to analytically derive the following limit: $$ \lim_{x\to \infty}\left(\frac{1}{8}\right)\left(e^{1-\frac{\sqrt{x}}{2}}\right)\left(\frac{x}{x-\frac{\sqrt{x}}{2}+1}\right)^{x-\frac{\sqrt{x}}{2}+1} $$
I've found it extremely difficult because, as you app... | $$\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) \left( \frac { x }{ x-\frac { \sqrt { x } }{ 2 } +1 } \right) ^{ x-\frac { \sqrt { x } }{ 2 } +1 }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding thi... | One developes an intuition that as $n^2$ becomes large the $+5$ and $-3$ become negligible and the whole thing becomes but how to prove it...
Well we can is eliminate the difference of square roots by multiplying by the compliment so we can probably do the same for cube roots.
I.e. $(\sqrt[3]a - \sqrt[3]b)(\sqrt[3]a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}dx$ $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}$$
Thought for a while but cannot seem to find how do i procced with this integral.
Another variant of this problem is $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin^{2}{x}\cos^{2}{x}}$$ where you just seper... | Notice that
\begin{eqnarray}
\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}&=&
\dfrac{\sin^3x}{\sin x\cos x}+
\dfrac{\cos^3x}{\sin x\cos x}\\
&=&\dfrac{\sin^2x}{\cos x}+\dfrac{\cos^2x}{\sin x}\\
&=&\dfrac{1-\cos^2x}{\cos x}+\dfrac{1-\sin^2x}{\sin x}\\
&=&\dfrac{1}{\cos x}-\dfrac{\cos^2x}{\cos x}+\dfrac{1}{\sin x}-\dfrac{\sin^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
To find value of $\tan \frac{\theta}{2} (1+\sec \theta)(1+\sec 2\theta)(1+\sec 4\theta)\dotsm(1+\sec 2^n\theta)$ I have to find value of $f_{n} \theta = \tan \frac{\theta}{2} (1+\sec \theta)(1+\sec 2\theta)(1+\sec 4\theta)\dotsm(1+\sec 2^n\theta)$.
I tried to use $1 + \cos\theta = 2 \cos^2 \frac{\theta}{2}$, there wer... | We have $$f_0 \theta = \tan \frac {\theta}{2}(1+ \sec \theta) = \frac {\sin \theta/2}{\cos \theta/2} \frac {2\cos^2 \theta/2}{\cos \theta} = \frac {2\sin \theta/2 \cos \theta/2}{\cos \theta} = \tan \theta $$ Similarly, $$f_1 \theta = \tan \theta \frac {2\cos^2 \theta}{\cos 2\theta} = \tan 2\theta $$ $$f_2 \theta = \tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Group of Rotations.
Attempt:
$$\frac{(2^6 + 6 \cdot 2^3 + 3 \cdot 2^4 + 8 \cdot 2^2 + 6 \cdot 2^3)}{24} = 10$$
| $10$ appears correct for distinct cubes allowing rotation mapping. There are:
- $2$ solid-colour cubes $(B^6,R^6)$,
- $2$ cubes with one face different $(B^5R, BR^5)$,
- $2$ (opposite-face and adjacent-face) options for each of the $4{+}2$ colourings $B^4R^2, B^2R^4$
- and $2$ similar options for the evenly-split colou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove that $(xy+2)^2+(x-1)^2+(y-1)^2 > 1$ for real $x,y$ How do I prove that $$(xy+2)^2+(x-1)^2+(y-1)^2 > 1$$ for all real $x,y$? I tried setting the partial derivative to $0$ but I couldn't get it to work.
| Using @quasi 's original comment, consider the function
$$
f(x,y)=(xy+2)^2+(x-1)^2+(y-1)^2.
$$
If we take partial derivatives of $f$, we get
\begin{align*}
\frac{\partial f}{\partial x}&=2(xy+2)y+2(x-1)\\
\frac{\partial f}{\partial y}&=2(xy+2)x+2(y-1).
\end{align*}
At minima, these are equal to zero, so we consider
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Help with the limit of a function in the $\frac{0}{0}$ case The limit is this one:
$$\lim_{x \rightarrow 0}\frac{(1+2x)^\frac{1}{x}-(1+x)^\frac{2}{x}}{x}$$
I have found that both $(1+2x)^\frac{1}{x}$ and $(1+x)^\frac{2}{x}$ tend to $e^2$, so the numerator tends to 0. I think that the book said that the result of this... | Note that we use here two well known limits $$\lim _{ \quad x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1$$
$$\lim _{ x\rightarrow 0 } \frac { (1+2x)^{ \frac { 1 }{ x } }-(1+x)^{ \frac { 2 }{ x } } }{ x } =\lim _{ x\rightarrow 0 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Maximize $P=a^2+b^2+c^2+ab+ac+bc$
For real numbers $a, b, c$ that satisfy $a + b + c = 6$ and $0\leq a,b,c \leq 4$,
maximize $P=a^2+b^2+c^2+ab+ac+bc$.
My try:
$$\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$$ $$(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^... | Let $f(x)=x^2$ and $a\geq b\geq c$.
Hence, $f$ is a convex function and $(4,2,0)\succ(a,b,c)$.
Thus, by Karamata
$$\sum_{cyc}(a^2+ab)=a^2+b^2+c^2+\frac{36-a^2-b^2-c^2}{2}=\frac{1}{2}(a^2+b^2+c^2)+18\leq$$
$$\leq\frac{1}{2}(4^2+2^2+0^2)+18=28.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute $5!25! \mod 31$ For an exercise, I was asked to compute $5!25! \mod 31$. I noticed that $5! = 120 \equiv -4 \equiv 27 \mod 31$. Therefore we have that
$$5!25! \equiv 27 \cdot 25! \mod 31.$$
Because of the congruence of Wilson, I also know that $30! \equiv -1 \mod 31$.
We have that $30! \equiv 30 \cdot 29 \cdo... | I think your answer is ok, but I would rewrite it as follows
$$
5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\equiv
(-26) \cdot (-27) \cdot (-28) \cdot (-29) \cdot (-30)
\pmod{31}.
$$
Thus
$$
5! \cdot 25! \equiv (-1)^{5} \, 30! \equiv (-1) \cdot (-1) = 1 \pmod{31},
$$
using as you did Wilson's Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Generating function of algebraic equations modulo prime powers Given an integer polynomial $Q(x_1,x_2,...,x_m)$, let $N(p,n)$ denote the number of solutions to the equation $$Q(x_1,...,x_m) \equiv 0$$ modulo $p^n.$
We can consider the ordinary generating functions $$F(p,x) = \sum_{n=0}^{\infty} N(p,n) x^n.$$ I find exp... | Yes, it is always rational. It is related to the Igusa zeta function by the formula $$F(p,x) = \frac{1 - x p^d \zeta_{Igusa}(d-s)}{1 - x p^d},$$ and $\zeta_{Igusa}(s)$ is known to be a rational function in $p^{-s}.$ This is shown in the article here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How find this seqence $a_{n}$ closed form if such Strange condition
Let $\{a_{n}\}$ such
$$a_{0}=a_{1}=a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=4,a_{6}=10,a_{7}=18$$
and such
$$\begin{vmatrix}
a_{n+8}&a_{n+7}&a_{n+6}\\
a_{n+5}&a_{n+4}&a_{n+3}\\
a_{n+2}&a_{n+1}&a_{n}
\end{vmatrix}=2\cdot (n+1)!$$
Find the $a_{n}$ colsed f... | The first calculated terms are : $1,1,1,2,3,4,10,18,28,80,162,280,880$
So the closed formula appear to be $\bbox[5px,border:2px solid]{\begin{cases}a_0=1\\a_1=1\\a_2=1\\a_n=(n-1)a_{n−3}\quad n\ge3\end{cases}}$
Let see what the determinant look like under the recurrence hypothesis until $a_{n+7}$.
$D=\begin{array}{l}
+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Partial Simplified Proof for the prime version of the Catalan Conjecture I have found an elementary approach which seems to work in proving that Catalan's conjecture is true for specific primes. I am attempting to generalize this approach to see how far it will go.
I am presenting it here with the hopes of finding a m... | I couldn't find any errors so your approach seems to be correct.
Note, though, that your solution is more general in that it also works for any $x$ being a positive power of $2$, plus for all odd $y \gt 1$, not just odd primes. For the first case, this is because your proof just requires the LHS to be a power of $2$. F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Derive the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$ So, I was trying to derive the Laplace Tranform of the Modified Bessel Function Io(at) using its summation formula. I arrived at the answer $$\frac{1}{s}+\frac{a^2}{2s^3}+\frac{3a^4}{8s^5}+\frac{5a^6}{16s^2}+\frac{35a^8}{128s^9}+\frac{63a^{10}}{256s^{11}}+\cdots$$... | Laurent series are quite "similar" to Taylor series.
Considering that you want the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$, consider that $$\frac{1}{\sqrt{s^2-a^2}}=\frac 1 s \frac{s}{\sqrt{s^2-a^2}}=\frac 1 s\frac 1{\sqrt{1-\frac{a^2}{s^2}}}$$ Now, by Taylor series or generalized binomial theorem $$\frac 1{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)... | The MacLaurin expansion of $\sin3x$ up to order $3$ is
$$\sin3x = 3x - \frac{9x^3}{2} + \mathcal O(x^5)$$
Plugging in the limit we obtain
$$\lim_{x \to 0} \frac{(3 + a)x + (b - \frac92)x^3}{x^3} =0$$
If the numerator is not zero then the limit is infinite, because the denominator goes to $0$ as order $3$. Therefore it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove $B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$ Prove the equality $$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$$ for $p > 0, q > 0$
I think the first step is
$$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx = \int \limits_0^1 \frac{x^{p-1}}{(1+x)^{... | Note that
$$\begin {align}
2\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx &=\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx+\int_1^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} }{(1+x)^{p+q}}dx+\int_0^\infty \frac{x^{q-1}}{(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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For $a,b,c<1$. Minimize $M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$ For $0<a,b,c<1$ and $ab+bc+ca=1$, minimize $$M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$$
My Try:
From $ab+bc+ca=1\Leftrightarrow \sqrt{3}\leq a+b... | We will prove that $$M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2(a^2+b^2+c^2)\geq -2+\sqrt{3} (*)$$
or equivalently
$$\left ( \frac{a^2}{b}-2a+b \right )+\left ( \frac{b^2}{c}-2b+c \right )+\left ( \frac{c^2}{a}-2c+a \right )+(a+b+c-\sqrt{3})-2(a^2+b^2+c^2-1)\geq 0.$$
We have
$$\sum_{cyc} \left (\frac{a^2}{b}-2a+b\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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The number of 3 digit numbers of the form xyz such that $x
The number of 3 digit numbers of the form $xyz$ such that $x<y$ and $z\leq y$ is N such that n is a 3 digit number of the form $abc$ then find $a+c-b$
My approach:
Case 1:
$x<y$
$z<y$
$x,y,z\neq 0$
Choosing 3 digits out of 9 numbers: $\binom 93$, greatest ... | Your case 1 needs supplementing with case 1a) $x=z$, which gives $\binom 92$ additional cases, with the larger choice $=y$ and the smaller $=x=z$ -- another $36$ cases.
Then $$\overset{\text{case 1}}{2\cdot \binom 93} + \overset{\text{case 1a}}{\binom 92} + \overset{\text{case 2}}{\binom 92} + \overset{\text{case 3}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the determinant and possible sizes of a matrix $A$, which satisfies the equation $A^2 - 2A + 4E = 0$ Let $A \in M_n(\mathbb{R})$ satisfy the equation:
$$A^2 - 2A + 4I = 0$$
(a) What is $\det A$? (3 points)
(b) What are possible sizes of a matrix $A$? (6 points)
My solution
We can derive 2 useful equations from the... | Using eigenvalues: Let $x$ be an eigenvector of $A$ associated with $\lambda$. Note that
$$
0 = 0x = (A^2 - 2A + 4I)x = (\lambda^2 - 2\lambda + 4)x
$$
so, each $\lambda$ must be a root of $x^2 - 2x + 4$, which is to say that each eigenvalue is either $1 + \sqrt{3}i$ or $1 - \sqrt{3} i$. Since $A$ is real, its eigenva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$. Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.
I have shown it is not possible for $a=4$, but not for any a greater than $4$.
| We see that
$$\frac{x+y}{xy}=\frac{2^a}{7^a}$$
and that leads to
$$7^a(x+y)=2^a(xy)$$
now let $\gcd(x,y)=d$, and write $x=dx_0$ and $y=dy_0$.
We get:
$$7^a(dx_0+dy_0)=2^a(dx_0dy_0)$$
or, rewritten:
$$7^a(x_0+y_0)=2^a(dx_0y_0)$$
note that since $\gcd(x_0,y_0)=1$, we know $\gcd(x_0y_0,x_0+y_0)=1$; thus, $x_0$ and $y_0$ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes... | We have,
$$\lim_{n \to \infty} \frac{\arctan(\frac{1}{n})}{\frac{1}{n}}=1$$
This follows easily by first the change of variables $\frac{1}{n}=h$ then by Taylor series.
We also have that $a_n=\arctan (\frac{1}{n}) \geq 0$ for all $n \geq 1$. Hence by the limit comparison test your series $\sum_n a_n$ diverges with comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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If coefficients of the Quadratic Equation are in AP find $\alpha+\beta +\alpha\beta$. $ax^2+bx+c=0$ is a quadratic equation with integer roots $\alpha,\beta$ and the positive coefficients are in AP.Find $\alpha+\beta +\alpha\beta$.
My attempt:
$\alpha+\beta +\alpha\beta=\frac {-b}{a}+\frac {c}{a}$ and we know $\frac{a+... | First of all we need to that if the roots are integer then $S=\alpha+\beta+\alpha\cdot \beta$ is also integer then
$$S=\frac{c-b}{a}=\frac{b-a}{a}=\frac{b}{a}-1\to a|b$$
and once $2b=a+c$ we get that $a|c$ so we can write the terms $(a,b,c)$ like $(a,a(1+r),a(1+2r))$ and then
$$S=r$$
We also have the discriminant:
$$\D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly ap... | By AM-GM $2x^3+1\geq3x^2$ and $3y^4+1\geq4y^3$.
Thus, $$2x^3+3y^4+2\geq3x^2+4y^3$$ or
$$x^3+y^3\leq2+3(x^3+y^4-x^2-y^3)\leq2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If $|a-b| \leq C$ and $|x-y|\leq D$, is there a way to bound $\left|\frac{a}{x} - \frac{b}{y}\right|$? If $|a-b| \leq C$ and $|x-y|\leq D$, is there a nice way to bound $\left|\frac{a}{x} - \frac{b}{y}\right|$?
I made an attempt using Triangle inequality but to no avail.
\begin{align*}\left|\frac{a}{x} - \frac{b}{y}\ri... | Think about this intuitively. Consider $|\frac{a}{x} - \frac{b}{y}|$. Let's just keep $a$ and $b$ fixed for now. Consider $|x - y| \leq D$, which basically means $-D \leq x-y \leq D$. If we choose $x$ and $y$ from $(-\frac{D}{2}, \frac{D}{2})$, then we will always have $x-y$ in $(-D,D)$ (why?).
Okay, but let's fix ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
trigonometry equation - $\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$ how can i find all possible solutions to this equation?
$\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$
I've tried writing it all as a sum in the form $\sin^m(x)\cos^n(x) + ... sin(x)$, with just $\sin(x)$ and $\... | Let $\sin x=a$, $\sin2x=b$,
and $\sin3x=c$
Expand out
$(a+b+c)^3$
Then subtract $a^3, b^3$ and $c^3$, which yields
$3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + 3 b^2 c + 3 b c^2=0$.
The long expression can be factored nicely into :
$3(a+b)(b+c)(a+c)$, which equal to $0$.
The factor $3$ makes no difference, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2179486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ I'm preparing a mathematical olympiad and our group is stuck in this problem. Here is all we've done:
First, let's rewrite the equation
$$p^2-p-1=q^3\Rightarrow p(p-1)=(q+1)(q^2-q+1)$$
It's obvious that $q$ must be less than $p$. Then, $p|(q^2-q+1)$... | You already have
$$q^2-q+1=pk\quad\text{and}\quad p-1=k(q+1)$$
where $k$ is a positive integer.
Eliminating $p$ gives
$$q^2-q+1=(kq+k+1)k,$$
i.e.
$$q^2+(-1-k^2)q-k^2-k+1=0$$
to have
$$q=\frac{k^2+1+\sqrt{k^4+6k^2+4k-3}}{2}$$
Note here that we get, for $k\gt 3$,
$$k^2+3\lt \sqrt{k^4+6k^2+4k-3}\lt k^2+4$$
from which we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
| Let $$I=\int_0^1\frac{x\ln(1+x)}{1+x^2}dx$$
and $$I(a)=\int_0^1\frac{x\ln(1+ax)}{1+x^2}dx.$$
Note that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^2}{(1+x^2)(1+ax)}dx$$
$$=-\frac{\pi}{4}\frac{1}{1+a^2}+\frac12\ln(2)\frac{a}{1+a^2}+\frac{\ln(1+a)}{a}-\frac{a\ln(1+a)}{1+a^2}.$$
$$\Longrightarrow I=-\frac{\pi}{4}\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
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Inclusion/Exclusion - Assigned Seating Arrangements Question. Four people sit down for dinner at a round table. They have assigned seating. What is the probability that 1, and only 1, person sits in their assigned seat.
My try:
We fix one of the members to a seat that is not theirs. Suppose the events $A_i, i=1,2,3$ a... | Assume that person 1 is in his seat. Then 2,3,4 should not be in their assigned seats and this can be done in 2 ways (derangement on three symbols). Thus the number of ways in which only one is in the assigned seat is $4 \times 2 = 8$. Thus the required probability is $\frac{8}{24} = \frac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Multinomial Verification Here's my question, which is similar to the one I asked a couple weeks ago:
Find the coefficient of $x^4$ in the expansion of $(1+x+x^2+x^3)^7$.
I was suggested to use the Binomial Theorem, but it would become very tedious and just decided to use the Multinomial Theorem instead. So I need to... | $$\begin{eqnarray*}[x^4](1+x+x^2+x^3)^7 &=& [x^4](1-x^4)^7\cdot\frac{1}{(1-x)^7}\\&=&[x^4]\left(\sum_{k=0}^{7}(-1)^k \binom{7}{k}x^{4k}\right)\cdot\left(1 + 7 x + 28 x^2 + 84 x^3 + 210 x^4+\ldots\right)\end{eqnarray*}$$
hence the answer is given by $\binom{7}{0}\cdot 210-\binom{7}{1}\cdot 1 = \color{red}{203}$ as predi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How Do You Find The Point Of Intersection Of $2$ Vectors? Two lines $\textbf{v} = \begin{pmatrix} 7 \\ -3 \\ 1 \end{pmatrix} + \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix} t$ and $\textbf{w} = \begin{pmatrix} 8 \\ -1 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ -4 \\ 0 \end{pmatrix} u$ intersect. What point do they interse... | We require the $x,y,z$ coordinates to be equal at the point of intersection, so we solve the following set of equations:
$(1)$ $$7-2t = 8+u$$
$(2)$ $$-3+5t = -1-4u$$
$(3)$ $$1+ t = -1$$
Thus from $(3)$ we get $t=-2$
Then from $(1)$ we get $7+4 = 8+u \Rightarrow u = 3$
Then we must check that this satisfies equation $(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the roots of quadratic polynomial and factor into linear factors I'm looking at Calculus by Adams, 7th Ed. page 43 Example 3b. It shows how to find the roots of $9x^2-6x+1$ by using the quadratic formula to arrive at the double root $\frac{1}{3}$. That I understand.
However, I don't understand how the factoring (... | The factor theorem says only that $x-c$ is a factor of a polynomial $p(x)$ if and only if $p(c)=0$. What this exactly means is that there is a polynomial $q(x)$ such that $p(x) = (x-c)q(x)$. The theorem doesn't say, but it's required that $q(x)$ has degree one less than $p(x)$.
Applied to a quadratic polynomial such as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculating Euler Number limit Please, so far I did
$$\lim_{x\to +\infty}\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}},$$ but I can write
$$\frac{x^2-x+1}{x+2}=1+\frac{x^2-2x-1}{x+2}=1+\frac{1}{\frac{x+2}{x^2-2x-1}}.$$
But
$$\lim_{x\to +\infty}\frac{x+2}{x^2-2x-1}=0,$$ so I can not use
$$e =\lim_{N\to \infty}(1+\fr... | Notice that for sufficiently large $x$,
\begin{align}
1 < \left(\frac{x^2-x+1}{x+2}\right)^\frac{1}{x-1}
& = \left(x-3+\frac{7}{x+2}\right)^\frac{1}{x-1} \\
& < (x-1)^\frac{1}{x-1}
\end{align}
which has a well-known limit as $x \to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Teacher's proof of sequence convergence that I don't understand. Prove $a_n = \frac{n^2-2}{n^2 +2n + 2}$ converges to 1.
Solution:
WTS $\exists L \in R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in N$, if $n > N$, then $|a_n - 1| < \epsilon$.
Let $\epsilon > 0$ be arbitrary
Choose N = $\max(\frac{6}{... | Here's a much easier solution:
We have
$$1 > \frac{n^2 - 2}{n^2 + 2n + 2} > \frac{n^2 - 4}{n^2 + 4n + 4} = \frac{(n - 2)(n + 2)}{(n + 2)^2} = \frac{n - 2}{n + 2},$$
thus
$$\left\vert \frac{n^2 - 2}{n^2 + 2n + 2} - 1 \right\vert < \left\vert \frac{n - 2}{n + 2} - 1\right\vert,$$
so it suffices to show that
$$\lim_{n\to\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$
I think if we let $x=\tan A;y=\tan B;z=\tan C$, then
$$P\Leftrightarrow \... | For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3\sqrt3}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{\sqrt{x^2+xy+xz+yz}}\leq\frac{3\sqrt3}{2\sqrt{xy+xz+yz}}$$ or
$$\sum_{cyc}\sqrt{x+y}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}.$$
But by C-S $$\left(\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$ What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$
I know about partial fractions:
$$\int\frac{x\cdot dx}{(x^3+1)^2} $$
$$= \int\frac{x\cdot dx}{((x+1)(x^2-x+1))^2} $$
$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3... | HINT:
First integrate by parts,
$$3I=\int\dfrac1x\cdot\dfrac{3x^2}{(1+x^3)^2}dx=-\dfrac1{x(1+x^3)}+\int\dfrac{dx}{x^2(1+x^3)}$$
Now use Partial fraction $$\dfrac1{x^2(1+x^3)}=\dfrac Ax+\dfrac B{x^2}+\dfrac C{x+1}+\dfrac{Dx+E}{x^2-x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a... | Since $19$ is prime, the first $18$ powers of $10$ (starting with $10^0$) will produce all of the remainders $1, 2, 3, \ldots, 18$ (although not in that order) when each power is divided by $19.$ Hence the sum of these $18$ powers of $10,$ namely $111111111111111111,$ satisfies
\begin{align}
111111111111111111 &\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
exponent problem
Can anyone answer this question? $$3^{n+2} + (3^{n+3} - 3^{n+1})= ?$$
I really want to know how to solve this one, our solution can't seem to agree with the problem's answer thank you!
| If the answer is $3/8$ probably the question is:
$$\frac{3^{n+2}}{3^{n+3}-3^{n+1}}=\frac{3^2\cdot3^n}{3^3\cdot3^{n}-3\cdot3^{n}}=\frac{3^2\cdot3^n}{3^n(3^3-3)}=\frac{3^2}{27-3}=\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the expression of a power series. So I have the infinite sum:
$$\sum_{n=1}^\infty (-1)^{n-1} nx^{2n}$$
I have found in a previous part that the convergence interval is $(-1,1)$.
I know that the sum should equal
$$\frac {x^2}{{(x^2 +1)}^2}$$
when $|x|\lt1$
But I'm not sure how I can show this from the sum.
| \begin{align}
\sum_{n=1}^\infty (-1)^{n-1} nx^{2n} = x^2 - {}2x^4 + 3x^6 - 4x^8 + 5x^{10} - \cdots
\end{align}
$$
\begin{array}{cccccccccc}
n=1 & & n=2 & & n=3 & & n=4 & & n=5 & & \cdots \\[10pt] \hline
x^2 & + & (-x^4) & + & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt]
& & (-x^4) & + & x^6 & + & (-x^8) & + & x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove this Generalizing AM-GM inequality
Let $n\ge 2$ and $a_{i} \ge 0,i=1,2,\cdots,n$, show that
$$(n-1)^{n-1}(a^n_{1}+a^n_{2}+\cdots+a^n_{n})+n^na_{1}a_{2}\cdots a_{n}\ge (a_{1}+a_{2}+\cdots+a_{n})^n$$
When $n=2$,
$$a^2_{2}+a^2_{2}+4a_{1}a_{2}=(a_{1}+a_{2})^2+2a_{1}a_{2}\ge (a_{1}+a_{2})^2$$
When $n=3$, it is
$$4... | Maybe this can help you, it seems very similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Asymptotics of $\Sigma x^n/n^p$ I am looking for asymptotics of
$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$
in terms of $N$, where $x>1$ and $p$ is a positive number.
So I am looking for a $O(f(N))$ type result. For example what is the approximate value of
$\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is p... | We find
\begin{align*}
\sum_{n=1}^N \frac{x^n}{n^p}
&= \sum_{n=1}^\infty \frac{x^n}{n^p}
- \sum_{n=N+1}^\infty \frac{x^n}{n^p} \\
&= \sum_{n=1}^\infty \frac{x^n}{n^p} - x^{N+1}\sum_{k=0}^\infty \frac{x^k}{(k+N+1)^p}
\qquad (\textrm{let } n=N+1+k) \\
&= \mathrm{Li}_p(x) - x^{N+1}\Phi(x,p,N+1)
\qquad \textrm{(p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $999\times \mathrm{abc}=\mathrm{def123}$ in decimal system, then find $\mathrm{a,b,c,d,e\ and\ f}.$ Consider the following multiplication in decimal system:
$$999\times \mathrm{abc}=\mathrm{def123}$$
then find the value of digits $\mathrm{a,b,c,d,e\ and\ f}.$
Here $\mathrm{abc}$ means not $(a\times b\times c$), $\ma... | Using the expansion of a natural number into powers of $10$ your equation becomes
\begin{equation*}
999\times 10^2a+999\times 10b+999c=10^5d+10^4e+10^3f+123.\tag{1}
\end{equation*}
This implies that $c=7$ because the final digit of the right-hand side of $(1)$ is $3 $, so should be the left-hand side's; and since we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Le... | Hint: prove the following first (e.g. by squaring), then add up all the combinations thereof:
$$
\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}} \le \sqrt{2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Given is a linear mapping and a basis. Determine the transformation matrix
Given is $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with
$f:(x,y,z) \rightarrow (x+2y+z, y+z, -x+3y+4z)$. Determine the
transformation matrix in terms of the basis $B= \left\{\begin{pmatrix}
1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\... | Perhaps a more straightforward way of computing the matrix is by using the relation
\begin{equation} [T]_\mathcal{B} = P^{-1} T P, \end{equation}
where $[T]_\mathcal{B}$ is the matrix that you computed,
$$P = (b_1\quad b_2\quad b_3) = \begin{pmatrix} 1 & 0 & -1 \\
0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}, $$
and $$T = \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
To find factor of a polynomial equation
One of the factors of $4x^2+y^2+14x-7y-4xy+12$ is equal to
*
*$2x-y+4$
*$2x-y-3$
*$2x+y-4$
*$2x-y+3$
Step $1$:
$4x^2+y^2-4xy$ can be simplified as $(2x-y)^2$
Step $2$:
$14x-7y$ can be simplified as $7(2x-y)$
and finally
$(2x-y) (2x-y+7) + 12$
I can able to factor to thi... | We can write the given equation as the product of two smaller polynomials:
\begin{align}
\rlap{4x^2+y^2+14x−7y−4xy+12} \\
&= (ax+by+c)(dx+ey+f) \\
&= adx^2+aexy+afx+bdxy+bey^2+bfy+cdx+cey+cf \\
&= \rlap{adx^2+bey^2+(af+cd)x+(bf+ce)y+(ae+bd)xy+cf}
\end{align}
We can equate coefficients to get the following:
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
limit of an indetermined sequence I'm trying to find out the limits of a sequence of the type $\frac{\infty}{\infty}$, but I got stuck and am starting to get frustrated. The sequence in question is: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}$$ I tried looking at its parts and fig... | I get the limit to be $-1$.
Note that $\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)} =
\frac{ n^2( \sqrt{1+{3 \over n} + {1 \over n^2}} - \sqrt{1+{3 \over n} - {1 \over n^2}})}{n\ln(\frac{1+{1 \over n}}{1 + {2 \over n}})}$.
We have $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = -1 $ and
$\lim_{x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Want less brutish proof: if $a+b+c=3abc$ then $\frac1a+\frac1b+\frac1c\geq 3$ In the course of showing for positive $a,b,c\,$ that $a+b+c=3abc$ implies $$\frac1{2a+b} + \frac1{2b+c} + \frac1{2c+a} \geq 1$$ I needed to show that (given that same constraint)
$$
\frac1a+\frac1b+\frac1c\geq 3$$
I have a proof which is ... | I will use that
$$a+b+c=3abc \Leftrightarrow \frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3$$
By AM-GM
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\frac{4}{ab}\\
\left(\frac{1}{a}+\frac{1}{c}\right)^2\ge\frac{4}{ac}\\
\left(\frac{1}{b}+\frac{1}{c}\right)^2\ge\frac{4}{bc}$$
so
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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proving $\cos (A+B)>0$, if given angles $A$ and $B$ If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$
Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right... | Let $\displaystyle \alpha=\sin^{-1}\left(\frac{6}{11}\right)=\frac{A}{3}$ and let $\displaystyle\beta=\cos^{-1}\left(\frac{4}{9}\right)=\frac{B}{3}$
Then since $\frac{1}{2}<\frac{6}{11}<\frac{\sqrt{2}}{2}$ we have $\frac{\pi}{6}<\alpha<\frac{\pi}{4}$.
And since $0<\frac{4}{9}<\frac{1}{2}$ we have $\frac{\pi}{3}<\beta<\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Exponential map of Beltrami-Klein model of hyperbolic geometry In the Betrami-Klein model of hyperbolic geometry, geodesics are represented as straight lines. Hence the exponential map of a tangent vector $\mathbf{v}$ at a point $\mathbf{p}$ is $\mathbf{p} + \lambda \mathbf{v}$, where $\lambda$ is a scalar that depends... | The hyperbolic plane doesn't have a center. However, the Klein model does. Because of the isotropy of the hyperbolic plane we can do our calculations from the center of the model. All the results will be the same. Let $p=(0,0)$ and let $\|v\|$ be arbitrary, and let $v$ be originated at $p$ (see the blue vector). $r=p+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Use substitution of $x =\tan\theta$ to show that Use the substitution $x = \tan\theta$ to show that $$\int\frac {1 - x^2} {(1+x^2)^2} dx = \int \cos2\theta\ d\theta $$
I'm a bit lost on how to handle this question, I have tried subtituting $d\theta/dx = 1 / \sec^2\theta$ but I still don't reach the answer.
| I have done with your substitution yet it is so boring
$$\int \frac { 1-x^{ 2 } }{ (1+x^{ 2 }) } dx=\int { \frac { 1-\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } \frac { 1 }{ \cos ^{ 2 }{ \theta } } d\theta =\int { \left( 1-\tan ^{ 2 }{ \theta } \right) d\theta =\int { \frac { \cos ^{ 2 }{ \theta -\sin ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the Rate of change of f (x^2) given rate of change of f (x) Find the average rate of change of $f(x^2)$ on the interval [1,4] given that the average rate of change of $f (x)$ equals 9
on interval [1,16]?
This question has two different "answers" according to two different teachers
The first give answer ... | Let $g(x) = f(x^2)$.
The average rate of change of a function $f$ on an interval $[a,b]$ is $\dfrac{f(b) - f(a)}{b-a}$.
The average rate of change of $f$ on the interval $[1,16]$ is $$\frac{f(16) - f(1)}{16 - 1} = \frac{f(16) - f(1)}{15}.$$
You are told this rate of change equals $9$.
The average rate of change of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Is every integer odd $p$ not divisible by $3$ a sum of difference of powers of $2$ and $3$ or vice versa? Does every odd integer $p$ not divisible by $3$ have the form
$2^a+3^b$, $2^a-3^b$, and or $3^b-2^a$ for integers $a$ and $b$?
Please show proofs if possible. Thanks for help.
| It can be verified by a finite if tedious computation that none of the equations
\begin{align}
2^a + 3^b \equiv 133 \pmod {683} \\
2^a - 3^b \equiv 133 \pmod {683} \\
3^b - 2^a \equiv 133 \pmod {683}
\end{align}
have a solution $(a,b)$, and therefore $133$ cannot be written as sum or difference of a power of $2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ i... | Revised proof:
We have $(a-b)(a^2 + b^2 + ab + a + b + 1) = (2b)^3$
From OP, we see that if $p$ divides $d$ then $p$ divides $2b \implies p \mid 2$ or $p \mid b$.
Assume $p$ doesn't divide $2$.
$p \mid b \implies p \mid a$ but then $p$ does not divide $a^2+b^2+ab +a+b+1$.
Therefore $p=1$ or $p=2$. If $p=1$ then $a-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 2
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Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1 = -\frac{3}{2}\sec \theta $$
$$2\sec... | No, the factoring is not correct - it should be $\left(2\sec\theta +4\right)\left(\sec\theta- \frac12\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
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