Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is $x=\tan\theta$ the only way for this integral? Evaluate $$\int\frac{dx}{(1+x^2)^4}$$ Now I did solve it, but I used the mentioned substitution and after a lot of converting into double angles, I did it. But, it doesn't look like a good approach. Is there a better way? Thanks.
| If you let $x=\tan\theta$ then
$$\begin{align}\int\frac{dx}{(x^2+1)^4}&=\int\cos^6\theta\,d\theta\\
&=\frac1{32}\int\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)d\theta\\
&=\frac1{32}\left(\frac16\sin6\theta+\frac32\sin4\theta+\frac{15}2\sin2\theta+10\theta\right)+C\\
&=\left(\frac16\left(32\sin^5\theta-32\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| Hint. One may apply
$$
\left( \frac1{f}\right)'=-\frac{f'}{f^2} \tag1
$$ with
$$
f(x)=x^2+3^3.
$$
Find $f'(x)$ then use $(1)$ and conclude with $\dfrac2{x^2+3^3}=2 \times \dfrac1{x^2+3^3}=2 \times \dfrac1{f}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Large Prime numbers Say we have $2^{690} + 345^4$ and we want to figure out whether this is a prime number.
I feel that we could break down the numbers into their respective prime factors (prime factorization) and use modular arithmetic to calculate whether these numbers are prime numbers or not.
$690 = 2\cdot 3 \cdo... | The factor lulu means is $17$. To show $$2^{690}+345^4\equiv 0\mod 17$$ you can reduce the exponents modulo $\phi(17)=16$ and the bases modulo $17$. This gives $2^2+5^4$. Since $5^2\equiv 8\mod 17$ and $5^4\equiv 8^2\equiv 13\equiv-4\mod 17$, we get $$2^2+5^4\equiv 4+(-4)=0\mod 17$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the integral $\int \frac{x^2+x}{(e^x+x+1)^2}dx$
Evaluate $$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$
I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
| One may observe that
$$
\begin{align}
\left(\frac{x+1}{e^x+x+1}\right)'&=\frac{1 \times(e^x+x+1)-(x+1)(e^x+1)}{(e^x+x+1)^2}
\\\\&=\frac{(e^x+x+1)-(x+1)(e^x+x+1)+x(x+1)}{(e^x+x+1)^2}
\\\\&=\frac{-x(e^x+x+1)}{(e^x+x+1)^2}+\frac{x^2+x}{(e^x+x+1)^2}
\\\\&=\frac{\left((e^x+x+1)'-(e^x+x+1)\right)(e^x+x+1)}{(e^x+x+1)^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the least positive integer $n$ for which $n!$ is divisible by $3^8$? What is the least positive integer $n$ for which $n!$ is divisible by $3^8$?
I am not sure how to tackle this problem without just putting in random low digits as $n$.
| An additional $3$ in the prime factorization can only occur, if the number is divisble by $3$. If it is divisible by $9$, $3^2$ occurs additionally, if it is divisible by $27$, $3^3$ occurs aditionally and so on.
$3!$ is divisble by $3$
$6!$ is divisble by $3^2$
$9!$ is divisible by $3^4$
$12!$ is divisible by $3^5$
$1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding and b in an equation with a floor function \begin{align}
a^2 + ba + c &= 0 & \text{{No real roots.}}\\
\lfloor a^2 \rfloor + ba + c &= 0 & \text{{At least one real roots.}}
\end{align}
Are there any values of b and c that will make the given number of roots correct?
The first thing I thought about doing... | \begin{align}
x^2 + bx + c &= 0 & \text{{No real roots.}}\\
\lfloor x^2 \rfloor + bx + c &= 0 & \text{{At least one real root.}}
\end{align}
Let's start by noting that the discriminant of $x^2 + bx + c$ is
$b^2 - 4c$. Hence $x^2 + bx + c$ has no real roots if and only if $b^2 - 4c < 0$, which is true if and on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.
| Let $c=\dfrac{\sqrt{2}}{3}$, $d=\dfrac{\sqrt{2}}{\sqrt{3}}$
With these notations, the solution is:
$A_1(0,0,1)$ (given), $A_2(2c,0,-1/3)$, $A_3(-c,d,-1/3)$ and $A_4(-c,-d,-1/3)$
The way I have done it:
1) As the center of mass is necessarily the origin, knowing that a regular tetrahedron has its center of mass at the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$ Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.
This is the last part of a problem that I am working on. So far, we hav... | Let $\sum_{n=1}^\infty \frac{1}{(2n-1)^3}=s$, then we have
\begin{eqnarray*}
\sum_{n=1}^\infty \frac{1}{n^3} &=& \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{1}{(2n-1)^3} \frac{1}{2^{3m}} \\
&=& \sum_{m=0}^\infty \frac{1}{8^m} \sum_{n=1}^\infty \frac{1}{(2n-1)^3} \\
&=& \frac{1}{1-\frac{1}{8}} s \\
&=& \frac{8}{7} s
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 ... | In general, if $\lim_{x \to \infty} f(x) = a$ and $\lim_{x \to \infty} g(x) = b$ then continuity of multiplication shows that $\lim_{x \to \infty} f(x)g(x) = ab$.
Write $${1 \over 2x-1} = {1 \over x(2-{1 \over x})} ={1 \over x} {1 \over 2 - {1 \over x}},$$ then since
$$\lim_{x \to \infty} {1 \over 2 - {1 \over x}} = ... | {
"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": 0
} |
If sum of n natural number is 20 then what is their max. product? If sum of n natural number is 20 then what is their max. product ?
| Let $s=20$ denote the given sum.
As almagest noted, $k(k+2)\lt(k+1)^2$ shows that the factors can differ by at most $1$. Thus there must be $j$ factors of $\left\lceil\frac sn\right\rceil$ and $n-j$ factors of $\left\lfloor\frac sn\right\rfloor$, with $j$ determined by
$$j\left\lceil\frac sn\right\rceil+(n-j)\left\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Presentations of rational numbers into product Prove or disprove:
For every rational number $q$ there exist rational numbers $q_1$, $q_2$, $q_3$ and $q_4$, such that:
*
*$q_1\times q_2\times q_3\times q_4 = q$
*$q_1+q_2+q_3+q_4 = 0$
My work so far:
I was able to solve it for $q_1$, $q_2$, $q_3$, $q_4$ and $q_5$:
... | It is a terrible solution (I obtained it by solving some diophantine equations):
$$q_1=\frac{(q-2^{10}\cdot3^{4})^2}{2^7\cdot3^4\cdot(q+2^9\cdot3^4)};$$
$$q_2=-4-\frac{q}{2^7\cdot3^4};$$
$$q_3 = \frac{2^3\cdot3\cdot(q+2^9\cdot3^4)}{q-2^{10}\cdot3^4};$$
$$q_4 = -\frac{2^{11}\cdot3^7\cdot q}{(q-2^{10}\cdot3^4)(q+2^{9}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Shortest distance between two lines in 3-dimensional space Can someone explain to me how to solve this question?
Find the shortest distance between the lines
$L_1 = \left\{t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} : t \in \mathbb{R}\right\}$ and $L_2 = \left\{s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + \begin{bmatrix} 1... | The difference vector is
$$
d = L_2(s) - L_1(t)
= s (1,2,3) + (1,0,0) - t (1,1,1)
= (s - t + 1, 2s - t, 3s - t)
$$
Then
$$
q = \lVert d \rVert^2 = d^2 = (s-t+1)^2 + (2s - t)^2 + (3s - t)^2
$$
Then the gradient is
$$
q_s =2(s-t+1)+ 4 (2s-t) + 6(3s-t) = 28s-12t + 2 \\
q_t = -2(s-t+1)-2(2s-t) -2(3s-t) = -12s+6t-2
$$
It ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$. Prerequisite
This problem is found in "Trigonometry" by I. M. Gelfand [in English].
It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as we... | Here's the strategy that I would use.
Note that
$$\sin 10^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ}{\sin 30^\circ}$$
And that
$$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 20^\circ\sin 30^\circ\sin 40^\circ \sin 50^\circ\sin 60^\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Simplifying a definite integral expression
If $a_1,a_2,a_3$ are the three values of a which satisfy the equation $$\int_{0}^ {\pi/2}(\sin x+a\cos x)^3dx-\frac{4a}{\pi-2}\int_{0}^{\pi/2}x\cos x dx=2$$ Find the value of $a_1+a_2+a_3$.
Now I did find the value of the second integral which comes out to be $\frac{\pi-2}{... | To find the value of the first integral,
using that
$$\sin^3\alpha=\frac{-\sin(3\alpha)+3\sin\alpha}{4}\quad\text{and}\quad\cos^3\alpha=\frac{3\cos\alpha+\cos(3\alpha)}{4}$$
gives
$$\begin{align}&\int_{0}^{\pi/2}(\sin x+a\cos x)^3dx\\&=\int_{0}^{\pi/2}(\sin^3x+3a\sin^2x\cos x+3a^2\sin x\cos^2x+a^3\cos^3x)dx\\&=\left[\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of two consecutive squares equal square $N^2 + (N+1)^2 = K^2$, find all solutions for $N<200$
I have done some factoring and also realized that $ K=[n\sqrt{2}]+1$ in eventual solutions, where $[x]$ denotes the greatest integer less than.
The problem is from Nordic math competition 1994.
| We want solutions to $k^2=n^2+(n+1)^2$. So $(k,n,n+1)$ is a Pythagorean triple. Moreover it is a primitive triple since $n,n+1$ are coprime. So we must be able to write it as $k=a^2+b^2,n=2ab,n+1=a^2-b^2$ for $n$ even or $k=a^2+b^2,n=a^2-b^2,n+1=2ab$ for $n$ odd.
If $n$ is odd, then $a^2-b^2+1=2ab$ and $a>b$, so we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of $p(x)\operatorname{csch}(x)$ I'd like to calculate the following integral
$$\int_{-\infty}^{+\infty}\frac{x^4 \left(\frac 1 {a^2+x^2} +\frac 1 {b^2+x^2}\right)}{\sinh^2(x\pi /c)} \, dx$$
where $a$, $b$ and $c$ are positive constants. Any suggestions? I probably have to use contour integrals, but I'm not sur... | Obviously two of the parameters are redundant, so let us just study:
$$\begin{eqnarray*} I(a) &=& \int_{0}^{+\infty}\frac{x^4}{\sinh^2(\pi x)}\frac{dx}{(x^2+a^2)}\\&=&-a^2\int_{0}^{+\infty}\frac{x^2\,dx}{(x^2+a^2)\sinh^2(\pi x)}+\int_{0}^{+\infty}\frac{x^2\,dx}{\sinh^2(\pi x)}\\&=&\frac{1}{6\pi}-a^2\int_{0}^{+\infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Where did I go wrong in finding maximum? The question in my book is given as:
If $x^2+y^2+z^2=1$ for $x,y,z$ belongs to all real numbers ($x,y,z$ are independent), then find the maximum of $x^3+y^3+z^3-3xyz$.
What I tried:
As all variables are independent I took the derivative of first equation and I got $x+y+z=0$.
... | I'm not sure why you are taking derivatives of $x^2+y^2+z^2=1$ and setting them to zeroes. $x^2+y^2+z^2=1$ is your constraint, and $x^3+y^3+z^3-3xyz$ is your function.
For this problem you can use Lagrange Multipliers.
Set $g(x)=x^2+y^2+z^2$ and $f(x)=x^3+y^3+z^3-3xyz$
$g_x=2x$ $f_x=3x^2-3yz$
$g_y=2y$ $f_y=3y^2-3xz$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the properties of the sum $\sum_{k=0}^n (-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}$ I have to show that $$\displaystyle\sum\limits_{k=0}^n (-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k} = \begin{cases} 1\ \text{if}\ n=0 \\ 0\ \text{if}\ n>0 \end{cases}$$
My try: I have tried to use snake oil method
$$\sum_{k=0}^n \binom{m+1}... | Note: Here we show that the snake oil method which was applied by OP does also work.
Let $S_m(n)$ denote the series
\begin{align*}
S_m(n)=\sum_{k=0}^n(-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}\qquad\qquad m,n\geq 0
\end{align*}
The following is valid for $m\geq 0$
\begin{align*}
S_m(n)=
\begin{cases}
1\qquad&\qquad n=0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$ Infinite product
$F_{n}:=[1,1,2,3,5,8,\cdots]$ and
$L_{n}:=[1,3,4,7,\cdots]$
for $n=1,2,3,\cdots$ respectively.
$\frac{1+\sqrt5}{2}=\phi$
Show that,
$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{... | Hint. One may use the following result,
$$
\prod_{n=1}^{\infty}\left(1+\frac{F_b}{F_{2^na+b}}\right)=\frac{1-(-1)^b\phi^{-2a-2b}}{1-\phi^{-2a}} \tag1
$$
which is proved by J. Sondow here (see proposition 1 on page 3), where $\phi=\dfrac{1+\sqrt5}{2}$ and where $F_{n}:=[1,1,2,3,5,8,\cdots]$ are the Fibonacci number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the spectrum of graphs for adjacency matrix Find the spectrum of graphs for adjacency matrix $(A)$ below :
$$ \left[
\begin{matrix}
0 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 0 \\
\end{matrix}
\right] ... | $(x-1)(x+1)^2(x-\frac{1-\sqrt{17}}{2})(x-\frac{1+\sqrt{17}}{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Computing the residue of a rational function The real integral I am trying to compute with residues/contour integration is
$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^3} \,dx$ For $a$ positive and by using the complex integral
$$\int_{C_R}\frac{z^2}{(z^2+a^2)^3} \,dz$$
For some arc bounding only the positive pole, $a ... | Computing the residue as @joriki suggested
\begin{align*}
\text{Res}(f,z_0)&=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[
(z-z_0)^nf(z)]\\
&\Rightarrow \text{Res}(f,ai)=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2}
[(z-ai)^{3}\frac{z^2}{(z^2+a^2)^3}]=
\frac{1}{2}\lim_{z\rightarrow ai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Can anyone help me with this finite sum? I have to calculate the sum
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1}$
We can re-write the sum as follows
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k-1+1}{(3^{k+1}-1)(3^k-1)}$
And then we obtain
$\displaystyle\sum_{k=1}^n \displaystyle\frac... | Telescopic!
$$\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \frac{1}{2}\left(\frac{1}{3^k-1}-\frac{1}{3^{k+1}-1}\right) $$
hence:
$$ \sum_{k=1}^{n}\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \color{red}{\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3^{n+1}-1}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convolution of Gaussian and error function I am trying to evaluate the following integral:
$$
\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}\Phi(x-t)dx
$$
where
$$
\Phi(y) = \frac{1}{2} + \frac{1}{2}erf\left(\frac{y}{\sqrt{2}}\right)
$$
I have tried integration by parts, a substitution, but nothing useful came out of this
| The key is to express as integral of a gaussian.
$$
I(t) = \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\Phi(x-t)~dx\\
= \int_{-\infty}^\infty \exp\left(-\frac{x^2}{2} \right)~\left[
1-\Phi(t-x)
\right]~dx\\
=\sqrt{2\pi}-\int_{-\infty}^\infty{
\exp\left(-\frac{x^2}{2} \right)~
\Phi(t-x)
~dx
}
$$
Plug in
$$
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How does one integrate $x^2 \frac{e^x}{(e^x+1)^2}$? How can I show this? $$ \int_{-\infty}^{\infty} x^2 \frac{e^x}{(e^x+1)^2} dx = \pi^2/3$$
I tried applying residuals, but the pole is of infinite(?) order.
| We have
$$\frac d{dx}\frac1{e^x+1}=\frac{-e^x}{(e^x+1)^2}$$
Also
$$\frac{e^x}{(e^x+1)^2}=\frac{e^{-x}}{(1+e^{-x})^2}$$
So
$$\begin{align}\int_{-\infty}^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx&=2\int_0^{\infty}x^2\frac{e^x}{(e^x+1)^2}dx=-2\int_0^{\infty}x^2\frac d{dx}\frac1{e^x+1}dx\\
&=\left.-2x^2\frac1{e^x+1}\right|_0^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric s... | Induction is fundamental to the structure of mathematics. The $``\cdots"$ in the expression $1+ 5+ 9 + \cdots +(4n-3)$ cannot be explained rigorously without using some form of induction. All of the picture proofs here have an implied $``\cdots"$ in them. Similarly, $\text{$\Sigma$-notation}$ and telescoping sums canno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 13
} |
limit $\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$ without Hospital Is it possible to find $$\lim_{x\to 0}\frac{\tan x-x}{x^2\tan x}$$ without l'Hospital's rule?
I have $\lim_{x\to 0}\frac{\tan x}{x}=1$ proved without H. but it doesn't help me with this complicated limit (however I'm sure I have to use it somehow).
I know... | $$L=\lim_{x\to 0}\frac{x-\tan x}{x^2 \tan x}=\lim_{x\to 0}\frac{\cos x-\frac{\sin x}{x}}{x^2}\cdot\frac{x}{\sin x}=\lim_{x\to 0}\frac{1-2\sin^2\frac{x}{2}-\cos\frac{x}{2}\frac{\sin(x/2)}{(x/2)}}{x^2} $$
gives $L=A+B$ where:
$$ A = -\frac{1}{2}+\lim_{x\to 0}\frac{1-\cos\frac{x}{2}}{x^2} = -\frac{1}{2}+\lim_{x\to 0}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$. I came across this problem and i could not solve it despite many tries.
Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^... | Using $a^2+b^2 =(a+b)^2-2ab$, we have
$3^{2008} +4^{2009}=(3^{1004})^2+(2^{2009})^2=(3^{1004}+2^{2009})^2 - 2 \cdot 2^{2009} \cdot 3^{1004}$
So we have $3^{2008} +4^{2009}=(3^{1004}+2^{2009})^2- 2^{2010} \cdot 3^{1004}$
Now, using $a^2-b^2=(a-b)(a+b)$, we have
$3^{2008} +4^{2009}=(3^{1004}+2^{2009}+2^{1005} \cdot 3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
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Solve first order partial diferential equation Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}... | $$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
FIRST METHOD :
The characteristics equations are :
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{du}{u^2}$$
First characteristic curve, from $\frac{dx}{x^2}=\frac{dy}{y^2} \quad\to\quad \frac{1}{x}-\frac{1}{y}=c_1$
Second characteri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.
| $$2^{3^m}+1=h\cdot 3^{m}\implies 2^{3^{m+1}}+1=(h\cdot 3^m-1)^3+1=\\=(h\cdot 3^m)((h\cdot 3^{m}-1)^2-(h\cdot 3^m-1)+1)=\\=(h\cdot 3^m)(h^2\cdot 3^{2m}-3h\cdot 3^m+3)$$ if $m\ge 1$, then $3$ divides $h^2\cdot3^{2m}-3h\cdot 3^m+3$
| {
"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": 1
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Show that $z = \ln (x^2+y^2) +2\tan^{-1}(y/x)$ satisfies the laplaces's equation.
I can get to the first partial derivative of $\partial z / \partial x$ of course, but after that I get this, which I don't know how to differentiate further... I am talking about c part of the question. Attaching my working...
| Differentiating once:
$$\frac{\partial z}{\partial x}\left( \ln(x^2+y^2)+2\tan^{-1}\left( \frac{y}{x}\right )\right )=\frac{2x}{x^2+y^2}+\frac{-2\frac{y}{x^2}}{\frac{x^2+y^2}{x^2}}=\frac{2x}{x^2+y^2}-\frac{2y}{x^2+y^2}=\frac{2x-2y}{x^2+y^2}$$
Differentiating again:
$$\frac{\partial z}{\partial x}\left ( \frac{\partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Systems of equations with unknown constant How do I solve this system? It says I must row reduce it to solve it (depending on parameter $a$).
$(8−a)x_1 + 2x_2 + 3x_3 + ax_4 = 2$
$x_1 + (9−a)x_2 + 4x_3 + ax_4 = 1$
$x_1 + 2x_2 + (10−a)x_3 + ax_4 = 2$
$x_1 + 2x_2 + 3x_3 + ax_4 = 2$
I have reduced it to the form
$$\left... | Permuting the columns,
$$\det \left[
\begin{array}{ccc|c}
1 & 2 & 3 & a\\
\hline
7-a & 0 & 0 & 0\\
0 & 7-a & 1 & 0\\
0 & 0 & 7-a & 0\\
\end{array}
\right] = \pm \det \left[
\begin{array}{c|ccc}
a & 1 & 2 & 3\\
\hline
0 & 7-a & 0 & 0\\
0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply... | $a=(x^2-1)=(x+1)(x-1)$
$b=(x^2+2x+1)=(x+1)^2=(x+1)(x+1)$
So, we have:
$$\frac{(x+3)}{a}-\frac{(x-2)}{b}$$
$$\frac{b(x+3)-a(x-2)}{ab}$$
$ab=(x+1)^3(x-1)$
$b(x+3)=(x+3)(x+1)(x+1)$
$a(x-2)=(x-2)(x+1)(x-1)$
Then,
$$\frac{(x+3)(x+1){(x+1)}-(x-2){(x+1)}(x-1)}{{(x+1)^3}(x-1)}$$
$$\frac{{(x+1)}[(x+3)(x+1)-(x-2)(x-1)]}{{(x+1)^{... | {
"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": 1
} |
Help with a limit using big O? $\lim_{x\to 0} \frac{4sin^{2}(\frac{x}{2})-x^{2}cos(\frac{x}{2})}{4x^{2}sin^{2}(\frac{x}{2})}$ is equal to $\frac{1}{24}$ apparently but I can't work it out.
My attempt:
$\lim_{x\to 0} \frac{4(\frac{x}{2}+O(x^{3}))^{2}-x^{2}(1-\frac{(\frac{x}{2})^{2}}{2}+O(x^{4}))}{4x^{2}(\frac{x}{2}+O(x^... | Using, almost as you did (but using one more term for the first expansion), $$\sin \left(\frac{x}{2}\right)=\frac{x}{2}-\frac{x^3}{48}+O\left(x^4\right)$$ $$\sin^2 \left(\frac{x}{2}\right)=\frac{x^2}{4}-\frac{x^4}{48}+O\left(x^5\right)$$ $$\cos \left(\frac{x}{2}\right)=1-\frac{x^2}{8}+O\left(x^4\right)$$ $$4\sin^2 \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let R be a Boolean ring with unity. Show that $0$ is the only nilpotent element and $1$ is the only unit. A ring is said to be Boolean if $x^2=x$ for all $x \in R$ .
I tried it this way: let $x \neq 0$ ,
$x^n=0$ and
$x^2=x$ so $x^{2n}=x^{n}=(-x)^n$ How do I show that $x=0$?
For unit part:
Let x be a unit so there exit... | For the first part, $x^2=x$ implies that $x^3=x^2x=x^2=x$ and similarly $x^n=x$ for all $n\geq 1$. Therefore if $x^n=0$ for some $n$ then $x=0$.
For the second part, since $x^2=x$ it follows that $x(1-x)=0$. Therefore if $xy=1$ then multiplying by $1-x$ we get $1-x=0$, so $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a ... | Notice that
$$ \left( \sqrt[3]{a+b}+\sqrt[3]{a-b} \right)^3= 2a + 3\cdot \sqrt[3]{a^2-b^2}\left(\sqrt[3]{a-b}+\sqrt[3]{a+b} \right)$$
Now with your question
Consider
$$x= \sqrt[3]{a+\sqrt {a^2+b^3}} +\sqrt[3]{a-\sqrt {a^2+b^3}} $$
Let $\sqrt{a^2+b^3} = c$
$$ x= \sqrt[3]{a+c} +\sqrt[3]{a-c} $$
$$ \Leftrightarrow x^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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What is the sum of the $81$ products in the $9 \times 9$ multiplication grid? What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.
Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?
| Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
} |
A variation of Ahmed's integral $\int_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $ Given that the closed form exist, evaluate the following Integral:
$$\int\limits_{0}^{1} \frac{(x^2+4)\sin^{-1}x}{x^4-12x^2+16} \, dx $$
| Since:
$$ \frac{z+4}{z^2-12 z+16} = \frac{\bar{\varphi}}{z-(2\bar\varphi)^2}+\frac{\varphi}{z-(2\varphi)^2}\tag{1}$$
the integral can be computed by exploiting the relations between the dilogarithm and the golden ratio, namely Landen's identity. A first step of partial fraction decomposition followed by integration by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the solution of the Cauchy problem for the pde Find the solution of the Cauchy problem for the pde
$x\dfrac{\partial z }{\partial x}+y\dfrac{\partial z}{\partial y}=z$
on $D=\{(x,y,z):x^2+y^2\neq 0,z>0\}$ with the initial condition $x^2+y^2=1,z=1$.
By Lagrange's Equations
$\dfrac{dx}{x}=\dfrac{dy}{y}=\dfrac{dz}{... | The set of characteristic equations is :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$$
With the condition $z=1$ on the curve $x^2+y^2=1$
So, it should be easy to satisfy this condition if this curve would be a characteristic curve. So, it is of interest to make appear the terme $x^2+y^2$ in a combination of the characteri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the value of k when $f(x) = x^3 -4x^2 + 6x + k$ If $ \ \ a ,b \ \ $and $\ \ c \ \ $ are roots of $ \ \ f(x) = x^3 -4x^2 + 6x + k \ \ $
and $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ \ \ $then find the value of $ \ k $
I have to tell you honestly that I cannot solve this ques... | We know: $a+b+c=4, ab+bc+ca=6, abc=-k$
Using the condition and $a^2+b^2+c^2=4$ you can get that:
$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}=1=\frac{1}{4-c^2}+\frac{1}{4-a^2}+\frac{1}{4-b^2}\Leftrightarrow -a^2b^2c^2+4(a^2b^2+b^2c^2+c^2a^2)-16(a^2+b^2+c^2)+64=a^2b^2+b^2c^2+c^2a^2-8(a^2+b^2+c^2)+48 \Leftright... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, whi... | First, you can simply $f(x)$ a little and write
$$ f(x) = \frac{x^2}{x-1} = \frac{x^2-1+1}{x-1} = (x + 1) + \frac{1}{x-1}. $$
Then we can calculate the derivative to obtain:
$$ f'(x) = 1 - \frac{1}{(x-1)^2} = \frac{x^2-2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}. $$
If the line passes through $(2,0)$, it has the form $y = m(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$? Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence of real numbers defined as follows: $x_1=1$ and $x_{n+1}=\frac{3+2x_n}{3+x_n} \forall n\in\mathbb{N}$ then can we find $\lim_{n\to \infty} x_n$ ?
| To see that the limit exist, consider
$\epsilon_{n+1}=\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_{n}} = \frac{\frac{3+2\frac{ (3+2 x_n)}{3+x_n}}{3+\frac{3+2 x_n}{3+x_n}}-\frac{3+2 x_n}{3+x_n}}{\frac{3+2 x_n}{3+x_n}-x_n}=\frac{3}{12+5x_n}$
Since $x_n>0$ from its definition, $0<\epsilon_{n+1} < 1 $, which mean it converges.
An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1... | I know you're asking if it converges, but unless I'm interpreting your answer incorrectly I think it's wrong. As others have pointed out before me you can see it converges via a limit comparison test to see $x\rightarrow\infty$. So here's how I get $\ln(4/3)$:
$$\int\dfrac{2}{x(x+1)(x-1)}dx$$
$$=2\int\dfrac{1}{x(x+1)(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\frac {\sin x}{1-\sin x}-\frac {\sin x}{1+\sin x}\equiv 2\tan^2 x$ I need assistance with proving the following identity:
$$\frac {\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x} \equiv 2\tan^2 x$$
What I have done so far is expanded them:
$$\frac {\sin x\;(1+\sin x)}{(1-\sin x)(1+\sin x)}-\frac {\sin x\;(1-\sin ... | You have
$$\frac{\sin x}{1-\sin x}-\frac{\sin x}{1+\sin x}=\frac {\sin x+\sin^2x}{1-\sin^2x} -\frac {\sin x-\sin^2x}{1-\sin^2x}$$
Then, use that $\frac{B}{A}-\frac CA=\frac{B-C}{A}$ to get $$\frac {\sin x+\sin^2x-(\sin x-\sin^2x)}{1-\sin^2x}=\frac{2\sin^2x}{1-\sin^2x}$$
Now use $1-\sin^2x=\cos^2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Convert $16x^2+56x-80=0$ to the form of $(\text{something})^2=D$ Sorry about this very basic question, I want to convert the
equation $16x^2+56x-80=0$ to the form $(\text{something})^2=D$, I know that the answer is $(4x+7)^2-129$, but how can I convert this without usnig calculator
| Here is an expansion which had helped me when I had come to know about completing the squares for the first time:
$ax^2+bx+c=0$
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + c = 0 $
$(\sqrt{a}x)^2 + (2\times \frac{b}{2\sqrt{a}}\times \sqrt{a}) x + (\frac{b}{2\sqrt{a}})^2 + c - (\frac{b}{2\sqrt{a}})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$
Could someone please explain to my why this is th... | $\lim_{x\to\infty} x(\sqrt{x^2+1}-x)=\lim_{x\to\infty} \sqrt{x^4+x^2}-x^2$
$=\lim_{x\to\infty}\frac{(\sqrt{x^4+x^2}-x^2)(\sqrt{x^4+x^2}+x^2)}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{x^4+x^2-x^4}{\sqrt{x^4+x^2}+x^2}$
$=\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| This is a variant on Servaes's answer. Note that $3^5=243=2\cdot11^2+1$. Using a binomial expansion and some extremely crude upper bounds, we find
$$\begin{align}
3\cdot9^{12}
&=3^{25}\\
&=(2\cdot11^2+1)^5\\
&=32\cdot11^{10}+80\cdot11^8+80\cdot11^6+40\cdot11^4+10\cdot11^2+1\\
&\lt32\cdot11^{10}+80\cdot11^8+11^8+11^8+... | {
"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": 1
} |
How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$? How can I develop $f(x) = (x^2+1)\cdot \arctan(x)$ into a Taylor series about $x_{0} = 0$ ?
I see that $f(x) = \frac{\arctan(x)}{\arctan'(x)}$ and I also know what the Taylor expansion of $\arctan(x)$ looks like about zero. But I... | The key here is that you want the same power of $x$ for all the terms of the sum.
\begin{align*}
(x^2+1)\arctan(x) & = (x^2+1) \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+3}}{2k+1}+\sum_{k=0}^{+\infty}(-1)^k\frac{x^{2k+1}}{2k+1}\\
& = \sum_{k=0}^{+\infty}(-1)^k\frac{x^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Bounding a series: $\frac{\pi}{2} < \sum_{n=0}^\infty \frac{1}{n^2 + 1} < \frac{3\pi}{2} $ I have the following statement -
$$\frac{\pi}{2} < \sum_{n=0}^\infty \dfrac{1}{n^2 + 1} < \frac{3\pi}{2} $$
So I tried to prove this statement using the integral test and successfully proved the lower bound. But when I tried to ... | Note that
$$ \sum_{n=0}^\infty \dfrac{1}{n^2 + 1}= \sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{n^2+1}dx>\sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{x^2+1}dx=\int_{0}^\infty\frac{1}{x^2+1}dx=\frac{\pi}{2}. $$
Also note
$$ \sum_{n=0}^\infty \dfrac{1}{n^2+1}=1+\sum_{n=1}^\infty \dfrac{1}{n^2 + 1}<1+\sum_{n=1}^\infty \dfrac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Antiderrivative of ${d^2 y \over dx^2} = 1-x^2$
At any point $(x,y)$ on a curve, ${d^2 y \over dx^2} = 1-x^2$, and an equation of the tangent line to the curve at the point $(1,1)$ is $y=2-x$. Find an equation of the curve.
This is what I've done
$${d^2 y \over dx^2} = 1-x^2
\\ \int dy' = \int (1-x^2)dx
\\y' = x- {x^... | Since the equation of the tangent line to the curve at $(1,1)$ is $y=-x+2$, we know that $dy/dx|_{(x,y)=(1,1)}=-1=1-\frac{1}{3}+C \Rightarrow C=-\frac{5}{3}$
Thus, $y=\frac{x^2}{2}-\frac{x^4}{12}-\frac{5}{3}x+D$ and we know that
$1=\frac{1}{2}-\frac{1}{12}-\frac{5}{3}+D\Rightarrow D=-\frac{9}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Convergence and Limit of a Recursive Sequence from the multiples of $a_n$ I'm having trouble with this recursive sequence problem. I'm supposed to find the limit, assuming that is it convergent, but I can't seem to get the answer.
$a_1 = 1, a_{n+1}= \frac {2a_n}{7+a_n} $
| We may just find the explicit form of $a_n$. By setting $a_n=\frac{p_n}{q_n}$ we have:
$$\begin{pmatrix} p_{n+1} \\ q_{n+1}\end{pmatrix}=\begin{pmatrix} 2 & 0 \\ 1 & 7\end{pmatrix}\begin{pmatrix} p_{n} \\ q_{n}\end{pmatrix}$$
where the characteristic polynomial of the involved matrix is $(x-2)(x-7)$.
By the Cayley-Hami... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that the determinant is $(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$
Prove that
$$
\begin{vmatrix}
1 & a^2 + bc & a^3 \\
1 & b^2 + ac & b^3 \\
1 & c^2 + ab & c^3 \\
\end{vmatrix}
=(a-b)(b-c)(c-a)(a^2 + b^2 + c^2 )$$
My work
I am getting - sign extra , please tell me why I ... | It is faster to exploit Gaussian elimination:
$$\begin{eqnarray*} \det\begin{pmatrix}1 & a^2+bc & a^3 \\ 1 &b^2 +ac & b^3 \\ 1 & c^2+ab & c^3\end{pmatrix}&=&\det\begin{pmatrix}1 & a^2+bc & a^3 \\ 0 &(b-a)(a+b-c) & (b-a)(b^2+ba+a^2) \\ 0 & (c-a)(a+c-b) & (c-a)(c^2+ca+a^2)\end{pmatrix}\\[0.2cm]&=&(b-a)(c-a)\det\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inequality with $\arctan$ I try to show that $x \cdot \arctan\left( \frac{1}{x^2} \right)$ is monotonically decreasing, but I can't solve this inequality with $\arctan$. Can somebody show me how to do this?
$$
x \in [1, \infty)
$$
$$
\arctan\left( \dfrac{1}{x^2} \right) \leq \dfrac{2x^2}{x^4+1}
$$
| Putting $x=1/y$, it is enough to show that, for $0 < y \leq 1$
\begin{equation*}
\arctan(y^2) \leq \frac{2y^2}{1+y^4}
\end{equation*}
If we let
\begin{equation*}
f(y) = \frac{2y^2}{1+y^4} - \arctan(y^2)
\end{equation*}
then
\begin{equation*}
f'(y) = \frac{2y(1-3y^4)}{(1+y^4)^2}
\end{equation*}
$f'(y)=0$ when $y = 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}$ What's the value of $$\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}?$$
I've tried the AM-GM inequality with no luck.
Also tried to right the inequality as:
$$\left... | Hint
$$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\frac{(2n+1)!}{{{2}^{n}}n!(n+1)!}$$
Note
$$\left( \frac{3}{2}\cdot \frac{5}{3}\cdot \cdots \frac{2n+1}{n+1} \right)=\left( \frac{2\times 4\times 6\times \cdots \times 2n}{{{2}^{n}}(1\times 2\times 3\times \cdots \times n)} \right)\left( \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x ... | Remember when you integrate both sides, there will be the constant of integration. You need to first find this constant of integration. That's where the $-1$ comes from. In actual fact, it is $\frac{-1}{n+1}$ but looks like $-1$ because it's been absorbed into the numerator.
You may like to try $x=0$ to find this $+C$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_ 1^2\frac{1}{x}dx$ with a Riemann Sum I'm trying to solve
$$ \int_ 1^2 \frac{1}{x} \ dx $$
using Riemann sums, however I'm having trouble solving it.
| $$
\begin{align}
\int_1^2\frac1x\,\mathrm{d}x
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac nk\cdot\frac1n\\
&=\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac2{2k}\right)\\
&=\lim_{n\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
As $ b$ varies range of $m(b)$ is? Let $f(x)=(1+b^2)x^2+2bx+1$ and let $m(b)$ be the minimum value of $f(x)$.As $b$ varies range of $m(b)$ is ?
I can find the minima by putting $f'(x)=0$ and hence find $m(b)$.But after that i'm not able to find range of $m(b)$.
| Finding minimum :
$$f(x)=(1+b^2)x^2+2bx+1$$
$$f(x)=(1+b^2)\left[x^2+\frac{2b}{1+b^2}x+\frac{1}{1+b^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1}{1+b^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1+b^2}{(1+b^2)^2}\right]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC) $? I found this identity in Wikipedia,
When $D$ is a $1 \times 1$ matrix, $B$ is a column vector, and $C$ is a row vector, then
$$\det{\begin{bmatrix} A & B \\ C & D\end{bmatrix}}=(D-1)\det(A) +\det(A-BC)= (D+1)\det(A) -\det(A+BC)$$
I wonde... | Observe that
\begin{equation*}
\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \begin{bmatrix}
A & B + 0\\
C & (D-1) + 1
\end{bmatrix}.
\end{equation*}
Therefore, using multilinearity of determinants
\begin{equation*}
\det\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
= \det \begin{bmatrix}
A & 0\\
C & D - 1
\end{bmatrix} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is... | Some asymptotics that should lead to a better bound with some additional work
From $p_n\sim n\ln n$ you can derive $\lfloor \sqrt{p_n} \rfloor\sim \sqrt{ n\ln n}$
Since $\sum_i \lfloor \sqrt{p_i} \rfloor$ diverges, the asymptotics above imply
$$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor \sim \sum_{i=1}^n\sqrt{ i\ln i} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation.
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
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)... | $$ax^{ 2 }+bx+c=0\\ 6p=\frac { c }{ a } \\ cx^{ 2 }+bx+a=0\\ \\ qr=\frac { a }{ c } \\ 6pqr=1\Rightarrow \quad pqr=\frac { 1 }{ 6 } $$
| {
"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": 1
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What is the minimum value for $(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$? The primary question was:
What is the minimum value for $(1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})$ if $a+b+c=1$ and $a,b,c\in\mathbb{R}^+$?
$\color{red}{\text{But sorry guys! I messed it up! my qu... | Since $a+b+c=1$, Re-write all $1$s to $a+b+c$, then
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\left(\frac{a+b+c}{a}-1\right)\left(\frac{a+b+c}{b}-1\right)\left(\frac{a+b+c}{c}-1\right)=\left(\frac{b+c}{a}\right)\left(\frac{a+c}{b}\right)\left(\frac{a+b}{c}\right)$$
Using AM-GM ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Proving $\frac{1}{\cos^2\frac{\pi}{7}}+ \frac {1}{\cos^2\frac {2\pi}{7}}+\frac {1}{\cos^2\frac {3\pi}{7}} = 24$ Someone gave me the following problem, and using a calculator I managed to find the answer to be $24$.
Calculate $$\frac {1}{\cos^2\frac{\pi}{7}}+ \frac{1}{\cos^2\frac{2\pi}{7}}+\frac {1}{\cos^2\frac{3\pi}{7... | Incomplete solution
The idea behind this solution is using Vieta's cubic formulas to rewrite the equality in a simpler, more manageable way. Letting $\cos^2\frac{\pi}{7}=\alpha$, $\cos^2\frac{2\pi}{7}=\beta$, and $\cos^2\frac{3\pi}{7}=\gamma$, we have this:
$$\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}=24$$
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Compute $E(X\mid X+Y)$ if $(X,Y)$ is centered normal with known covariance matrix
The random variable $(X,Y)$ has a two dimensional normal distribution with mean $(0,0)$ and covariance matrix $\begin{pmatrix}
4&2 \\
2&2
\end{pmatrix}$. Find $E(X\mid X+Y)$.
I am completely lost with this question.
| Let's pose $Z = X+Y$. The random variable $(X,Z)$ has a two dimensional normal distribution with mean $$(E(X),E(Z)) ) (E(X), E(X)+E(Y)) = (0,0)$$ and covariance matrix $$\pmatrix{E(X^2) & E(XZ)\\E(ZX) & E(Z^2)} = \pmatrix{E(X^2) & E(X^2)+E(XY) \\E(X^2) + E(YX) & E(X^2)+E(Y^2)+E(2XY)} = \\
\pmatrix{4 & 4+2 \\4 + 2 & 4 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Last digit on $3^{100}$ How to find the last digit on $3^{100}$?
Is there any proper method to solve such questions without calculator of course?
| $3^{100}=(3^4)^{25 }=81^{25}=(80 + 1)^{25}=a*80^{25}+b*80^{24}+.....+x*80^2+y*80 + 1$ where a,b,c,....,y are integer coeffients obtained by expanding $(80 +1)^{25} $. We can calculate these coefficients if we wanted to, but it's enough to know they are integers.
Notice, every term $k*80^j $ is divisible by 10 so
$3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 7
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Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $ $$ \int \frac{ x^7}{(x+1)}{dx} $$
$$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$
$$ \int { \left(x^6 - x^5 + x^4 - x^3 + x^2 -x +... | Same as your method : But try to use reduction formulas for higher degrees :
Let
$$I_n= \int \frac{x^n}{(x+a)}dx$$
$$I_n= \int \frac{(x+a)x^{n-1}-ax^{n-1}}{(x+a)}dx$$
$$I_n= \int x^{n-1}dx-a\int \frac{x^{n-1}}{(x+a)}dx$$
So $$I_n+aI_{n-1}=\frac{x^n}{n}$$
Now you have to find $I_7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove the Inequality $\frac{1}{1-x}-\frac{x(3-x)(2-x)(13x^4-50x^3+89x^2-84x+36)}{4(1-x)(2x(1-x))^2}<1$ Can anyone suggest any hints to prove the following inequality:
$$\frac{1}{1-x} - \frac{x(3-x)(2-x)(13x^4 - 50x^3 + 89x^2 - 84x + 36)}{4(1-x)(2x(1-x))^2} < 1,$$
for all $x \in (0,1)$?
| One may see that for the initial inequality to hold true it is sufficient to prove that
$$
1-\frac{(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)}{16 x(1-x)^2 }<0,\quad x \in (0,1), \tag1
$$ then setting
$$
\begin{align}
&f(x)=16 x(1-x)^2-(2-x) (3-x) \left(36-84 x+89 x^2-50 x^3+13 x^4\right)
\end{align}
$$ one g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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
| Assuming the right expression to be $$\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}$$
$$\tan^{-1}\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}=\dfrac\pi4+\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}$$
Let $\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}=y\ge0\implies0\le y<\dfrac\pi2$
$$\dfrac{4-x}{4+x}=\tan^2y,\dfr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $x \neq 0$ and $x = \sqrt{4xy - 4y^2}$, then how does expressing $x$ in terms of $y$ mean $x = 2y$? I have the equation $x=\sqrt{4xy - 4y^2}$, and I know that $x=2y$ when expressed in terms of $y$, but I'm not sure of the process to get there.
I know that
\begin{align}\sqrt{4xy - 4y^2} &= \sqrt{4y(x-y)}\\
&= 2 \sq... | $x =\sqrt{4xy + 4y^2}$
$x^2 = 4xy - 4y^2$
$x^2 - 4xy + 4y^2 = 0$
$x = \frac{4y \pm \sqrt{16y^2 - 16y^2}}{2}$
$x = \frac{4y \pm \sqrt{0}}{2}$
$x = 2y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
|
$${ \left( x-1 \right) }^{ 2 }-{ \left( x-5 \right) }^{ 2 }=\left( x-1-x+5 \right) \left( x-1+x-5 \right) =4\left( 2x-6 \right) =8\left( x-3 \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that this inequality doesn't hold Given $(a,b,c) \in \mathbb R^3_+$ show that atleast one of the real numbers $a(1-b)$, $ b(1-c)$ and $c(1-a)$ is less than or equal to 1\4.
I tried to show it by contradiction i.e Suppose that $$a(1-b) > \frac{1}{4} $$
$$b(1-c) > \frac{1}{4} $$$$c(1-a) > \frac{1}{4} $$
So as to en... | Suppose that for all $a,b,c$ $$a(1-b) > \frac{1}{4}\\b(1-c) > \frac{1}{4}\\c(1-a) > \frac{1}{4}$$ We then have $$a+b+c\gt \frac 34+(ab+ac+bc)=\frac 34+\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}$$ It follows that if $a=b=c$ we have $$12a\gt 15a^2+3\Rightarrow(15a-6)^2+45\lt 0$$ This is absurde.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $2^x+2^y+1=3^z$ in integers I have reasons to believe that there should be an elementary, relatively simple way to find all solutions of the equation in the title in positive integers $x>y$ and $z$. Any ideas?
| I think I was eventually able to recover (?) the elementary solution mentioned in the literature. I will show below that the only solutions in positive integers $x,y,z$ are $x=4,y=6,z=4$, $x=6,y=4,z=4$, and $x=y=z=2$.
Reducing $2^x+2^y+1=3^z$ modulo $3$, we conclude that both $x$ and $y$ are even; say, $x=2a$ and $y=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to solve $\int \frac{1}{x^2+4x+7} dx$? How to solve $\int \frac{1}{x^2+4x+7} dx$?
I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$
| $$\int \frac { 1 }{ (x+2)^{ 2 }+3 } dx=\int { \frac { dx }{ 3\left( 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } \right) } } =\frac { 1 }{ \sqrt { 3 } } \int { \frac { d\left( \frac { x+2 }{ \sqrt { 3 } } \right) }{ 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } } } =\\=\frac { 1 }{ \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Another way to solve $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ Let $| x^2 - 4x + 3 | + | x^2 - 6x + 8 | =\sqrt{4|x|^2 - 20x + 25}$ then number of integral values of $|x|$ is
My work
$=
\begin{cases}
2x^2 - 10x + 11= -2x + 5, & x\ge 1\\
-2x+5 = -2x+5 , & 1 \lt x \le 2
\\
-2x^2 + 10x - 11= -2x + ... | Note that $|x|^2 = x^2$ for all $x\in \mathbb R$. Using @juantheron's comment, notice that
$$|x^2-4x+3-(x^2-6x+8)| = |2x-5| = \sqrt{4x^2-20x+25}.$$
Finally, note that
$$(x^2-4x+3)(x^2-6x+8) = (x-1)(x-2)(x-3)(x-4).$$
So the set of solutions is $[1,2]\cup[3,4]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Getting the wrong result when solving a limit I correctly solved to following limit like this:
$$\lim_{x\to\infty} \left(\frac{1}{x^2}\right)^{\frac{2x}{x+1}}$$
$$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2x}{x(1+\frac{1}{x})}}$$
$$ = \lim_{x\to\infty} (\frac{1}{x^2})^{\frac{2}{(1+\frac{1}{x})}}$$
$$ = 0^2 = 0$$
Howe... | $$\frac{1-x^2}{x^2}$$ does not approach zero as $x$ approaches infinity, so you cannot simply write that $$\lim_{x\rightarrow\infty}\left(1+\frac{1-x^2}{x^2}\right)^{\frac{x^2}{1-x^2}}=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integration of root of quadratic over quadratic I need help to get a generalized solution for this type equations:
$$ \Gamma(v) = \int\frac{(1-v^2)^{\frac{1}{2}}}{a^{2}-b^{2}v^{2} }dv $$
NOTE: This is not a homework question. I am in high school and I was reading up some extra material and landed on this from another e... | Substituting $v=\sin\theta$ gives $\sqrt{1-v^2}=\cos\theta$, $dv=\cos\theta\,d\theta$ so we have
\begin{eqnarray}
\int\dfrac{\cos^2\theta}{a^2-b^2\sin^2\theta}\,d\theta&=\int\dfrac{1-\sin^2\theta}{a^2-b^2\sin^2\theta}\,d\theta\\
&=\dfrac{1}{b^2}\int\left[1-\dfrac{a^2-b^2}{2a}\left(\frac{1}{a-b\sin\theta}+\frac{1}{a+b\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of given limit when $f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\frac{1}{\tan 2^nx}$
Question Statement:-
If $\displaystyle f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\dfrac{1}{\tan 2^nx}$, then find the value of
$$\lim_{x\to 0} \bigg( (\cos x)^{g(x)}+\bigg(\frac{1}{\cos x} \bigg)^... | Basically, what you have to show (lab bhattacharjee gave you the hint while I was typing) is that $$f(x)=\cot (x)-\cot \left(2^n x\right)$$ which makes $$g(x)=\cot (x)$$ Now consider $$A=\cos(x)^{\cot(x)}\qquad , \qquad B=\bigg(\frac{1}{\cos (x)} \bigg)^{\frac{1}{\sin (x)}} $$ So $$\log(A)=\cot(x) \log(\cos(x))\qquad ,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Finding the coefficient of $x^r$ in an expansion. Suppose that the summation of the infinite series $$1+nx+\frac{n(n-1)}{2} x^2+\cdots+\frac{n(n-1)\cdots(n-r+1)}{r}x^r+\cdots$$ is equal to $(1+x)^n$ for $|x|<1$.
Show that the coefficient of $x^r$ in the expansion of $\frac{1+x+x^2}{(1-x)^2}$ is $3r $.
Hence show that... | In another approach, as inquired in the OP, we write the term of interest as
$$\begin{align}
\frac{1+x+x^2}{(1-x)^2}&=1-\frac{3}{1-x}+\frac{3}{(1-x)^2}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\frac{d}{dx}\sum_{n=0}^\infty x^n\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=1}^\infty nx^{n-1}\\\\
&=1-3\sum_{n=0}^\infty x^n+3\sum_{n=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a$ and $b$ are integers then $a^2 -4b-2$ does not equal zero. I was just wondering if the rough outline of this proof was logically rigorous.
If a and b are integers then $a^2 -4b-2$ does not equal zero.
Proof:
Suppose for the sake of contradiction that $a^2-4b-2=0$.
Then we have $a^2 = 2(2b+1)$. $a^2$ is even w... | I think the best way to think about this is to work modulo 4:
$0^2 \equiv 0$ mod 4
$1^2 \equiv 1$ mod 4
$2^2 \equiv 0$ mod 4
$3^2 \equiv 1$ mod 4
Clearly $4b + 2 \equiv 2$ mod 4. So indeed it is not possible that $a^2$ = 4b + 2. Similarly you can show $a^2$ - 4b - 3 is not equal to zero either.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving the fraction $\frac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is irreducible for $n \in \mathbb{N}$
Prove that the fraction $\dfrac{n^3 + 2n}{n^4 + 3n^2 + 1}$ is in lowest terms for every positive integer $n$.
I just don't know how to solve this. I tried polynomial division, expressing the gdc of the two terms as a linear c... | You can do it with linear combinations:
$$
(n^4+3n^2+1)-n(n^3+2n)=n^2+1 \tag{1}
$$
Then
$$
(n^3+2n)-n(n^2+1)=n \tag{2}
$$
and
$$
(n^2+1)-n\cdot n=1 \tag{3}
$$
Just work backwards if you need the linear combination, but this is already sufficient.
\begin{align}
1
&=\color{red}{(n^2+1)}-n\cdot\color{red}{n}
&&\text{by (3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Integrate $\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$
Integrate $$\int \frac{x\ln (x)}{(x^2-1)^{3/2}} dx$$.
My Try:
$$\int \frac{x\ln (x)}{x^3(1-1/x^2)^{3/2}} dx=\int \frac{\frac{1}{x^2}\ln (x)}{(1-\frac{1}{x^2})^{3/2}} dx.$$
Putting $x=\frac{1}{t}$,
I get
$$\int \frac{t^2\ln (\frac{1}{t})}{(1-t^2)^{3/2}(-t^2)} dt=\int ... | Check that
$$\int\frac1{t^{3/2}}dx=-2t^{-1/2}+C\implies \int\frac{f'(x)\,dx}{f(x)^{3/2}} =-2f(x)^{-1/2}+C$$
and now just observe that $\;2x=(x^2-1)'\;$ , so by parts:$${}$$
$$\begin{cases}&u=\log x,&u'=\cfrac1x\\{}\\&v'=\cfrac x{(x-1)^{3/2}},&v=-\cfrac1{\sqrt{x^2-1}}\end{cases}\;\;\implies$$$${}$$
$$\int\frac{x\log x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Real function with complex antiderivative $\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}$? Consider this indefinite integral (I'm interested in the interval $x>0$):
$$\int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}dx$$
By substitution:
$$u=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{x^2}}}, \qquad x=\fra... | I got the answer
$$\text{arctanh} = \frac12 \ln \frac{1+x}{1-x}$$
consider $\ln(x)$ where $x < 0 $
$$\ln(x) = \ln(|x| (-1))=$$
$$ = \ln(x) + \ln(-1)= \qquad \text{(properties of logs)} $$
$$= \ln(x) + \ln(e^{i \pi}) $$
$$= \ln(x) + i \pi$$
so in your integration, the log of a negative number generates a constant $i \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Interesting Logarithmic Integral: $\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $ Other than numerical approximation, how can we calculate the closed form of this integral?
$$\int_{0}^{1} \frac{\ln^2 x \ln^2(1+x)}{x} \;dx $$
| OK as I suggested in a comment here is an approach.Let us begin with the Taylor series of $\ln^2(1+x)$. It is known to be:
$$\ln^2 (1+x) = 2 \sum_{n=2}^{\infty} \frac{(-1)^n \mathcal{H}_{n-1}}{n} x^n$$
Then we have successively:
\begin{align*}
\int_{0}^{1} \frac{\ln^2 x \ln^2 (1+x)}{x} \, {\rm d}x &= 2 \int_{0}^{1} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does the limit exist for $\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}$? $$\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}\le \lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3}}=\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\s... | $$\displaystyle\lim_{x\to \infty} \frac{x^{\frac{3}{2}}\cdot \arctan(x)}{\sqrt{2x^3+4x^2+x}}=\lim_{x\to \infty} \frac{\arctan(x)}{\sqrt{2+4\frac{1}{x}+\frac{1}{x^2}}}=\frac{\pi}{\sqrt{2}\cdot2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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In ℕ⁺, can the sum of three squares equal the sum of two squares? Are there any examples where:
$a² + b² + c² = p² + q²\qquad {a, b, c, p, q ∈ ℕ⁺}\tag{1}$
If not, can $(1)$ be disproven?
| One need only to pick out 3 distinct naturals, with the only constraint that they aren't all even:
Suppose you pick $(p,q,r) \in \mathbb N^3, \ \ \ $ let $r \gt q \ \land \ r\gt p ,\ \ $ then there exists a natural number $N \not \equiv 2 \pmod 4 $, which can be converted to a difference of two squares in at least one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Counting multiple of $3$ in a triangular arrangement of the integers. I write
:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&17&\ldots \\
1&2&4&7&11&16&\ldots
\end{matrix}$$
Can we det... | The $n$ th column is the sequence of $n$ consecutive numbers starting at $1+\frac{n (n-1)}{2}$. Let's call $b$ this base value.
The sequence length is $n$. $n$ starts at 1 like the height of its column. $k$ is an integer.
Each 3 consecutive numbers sequence contains exactly one multiple of 3.
When starting with a base ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to turn the recurrence $a_n=2a_{n−1}−a_{n−2}$, with $a_0=−1$ and $a_1=1$ into a quadratic, to find the closed form? I am having trouble understanding how to turn the recurrence into a quadratic equation. What principles are involved, rules etc.?
| You can write the recurrence with matices:
$$
\begin{pmatrix}
a_{n-1} \\
a_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}
\begin{pmatrix}
a_{n-2} \\
a_{n-1} \\
\end{pmatrix}
$$
Then,
$$
\begin{pmatrix}
a_{n-1} \\
a_{n} \\
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
-1 & 2 \\
\end{pmatrix}^{n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 ... | You should consider one case:
If $y\ge0$ then $|y|=y$ , and
$$x+|y|-y=x=6$$
and
$$\underbrace{x+|x|}_{12}+y=7\implies y=-5$$
It is contradiction, hence $y<0.$ Similarly $x\le 0$ then $y=7$ whereas $x+\underbrace{|y|-y}_{0}=6$. So $x>0$ and $y<0$, we have
\begin{cases}
2x+y=7\\
x-2y=6
\end{cases}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum upto 20 terms $$\frac{1}{4}+\frac{1\times 2}{4\times 5}+\frac{1\times2\times 3}{4\times5\times6}+\frac{1\times2\times3\times4}{4\times 5\times 6\times 7}+\cdots(\operatorname{upto 20 terms})=?$$
I am trying like this...
$$\frac{1}{4}+\frac{1}{10}+\frac{1}{20}+\frac1{35}+\frac1{56}+\cdots $$
Here some term... | $$\begin{eqnarray*}\sum_{n=1}^{N}\frac{n!}{(n+3)!/6} &=& 6\sum_{n=1}^{N}\left(\frac{1}{(n+1)(n+2)(n+3)}\right)\\&=&3\sum_{n=1}^{N}\left(\frac{1}{(n+1)(n+2)}-\frac{1}{(n+2)(n+3)}\right)\\&=&\color{red}{\frac{1}{2}-\frac{3}{(N+2)(N+3)}}.\end{eqnarray*} $$
An equivalent approach is to use Euler's beta function. We have:
$... | {
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"timestamp": "2023-03-29T00:00:00",
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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!
| HINT:
$$\sum_{n=0}^\infty a^n\sum_{m=0}^n b^m=\sum_{n=0}^\infty a^n \frac{1-b^{n+1}}{1-b}$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $\binom{x+2}{5} = 126$ for $x$ How can we solve for $x$ in the equation $\binom{x+2}{5} = 126$? By expanding, we get $(x+2)(x+1)x(x-1)(x-2) = 126(5!)$, and by trial and error, we can see $x = 7$, but how can we solve for this without trial and error? Also, how do we know only one solution exists?
|
Use:
*
*$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$
*$$a!=\prod_{p=1}^{a}p=a\times(a-1)\times(a-2)\times\dots\times3\times2\times1$$
So, we get:
$$\binom{x+2}{5}=126\Longleftrightarrow\frac{(x+2)!}{5!((x+2)-5)!}=126\Longleftrightarrow$$
$$\frac{(x+2)!}{120(x-3)!}=126\Longleftrightarrow\frac{(x+2)!}{(x-3)!}=15120$$
Expan... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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... | Hint: Show that $2a^2-1$ is divisible only by primes $\equiv 1,7\pmod 8$, whereas $b^2+2$ must have a prime divisor not of that form. You already did the latter part.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube. Find three consecutive positive integers a,b and c such that a+b is a perfect square and b+c is a perfect cube.
Any ideas on how to begin?
| For all $x,y \in \mathbb{N}$, $x^3-y^2=2$ according to the given question, where $x=a+b$ and $y=b+c$. However, $x^3 \equiv 0,\pm1 \pmod4$ and $y^2 \equiv 0,1 \pmod4$.
So to satisfy the given equation, the following must hold: $$x^3-y^2\equiv 2 \pmod4$$
For this means that the required $x,y$ are such that $x^3 \equiv -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $x+y+z$ From Equations Including Radicals Let $(x,y,z)$ be the real solution to the system of equations
\begin{align*}
x+y &= \sqrt{4z-1} ,\\
y+z &= \sqrt{4x- 1} , \\
z+x &= \sqrt{4y-1} .
\end{align*}
Find $x+y+z.$
I could add all the equations up but that doesn't do any good. Thanks in advance!
| $$x+y=\sqrt{4z-1}\tag{1}$$
$$y+z=\sqrt{4x-1}\tag{2}$$
$$z+x=\sqrt{4y-1}\tag{3}$$
Adding $(1),(2)$ and $(3)$, we get
$$x+y+z=\sqrt{x-\frac14}+\sqrt{y-\frac14}+\sqrt{z-\frac14}$$
Let $a=\sqrt{x-\frac14}$,$b=\sqrt{y-\frac14}$ and $c=\sqrt{z-\frac14}$
Using these substitutions, we get
$$\left(a^2+\frac14\right)+\left(b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit of the difference between two exponential functions
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$
I did nothing by now.
| We can proceed as follows
\begin{align}
L &= \lim_{x \to \infty}(x + 3)^{1 + 1/x} - x^{1 + 1/(x + 3)}\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 1)\log(x + 3)}{x}\right) - \exp\left(\frac{(x + 4)\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\left\{\exp\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differentiating this inverse trigonometric function $$\sin^{-1}\left( \frac{2^{x+1}\cdot3^x}{1+36^{x}} \right)$$
Had this question for todays test but still cannot find out how to proceed.
| The function
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
is defined for every real $x$, because $|2x|<1+x^2$. We also have
$$
f'(x)=\frac{1}{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}
\cdot 2\frac{(1+x^2)-x\cdot 2x}{(1+x^2)^2}=
2\frac{1+x^2}{|1-x^2|}\frac{1-x^2}{(1+x^2)^2}
$$
so
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & \text{i... | {
"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": 0
} |
$\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$
If $ a, b, c \in R^{+}$ and $\frac{1}{a + 1}$ + $\frac{1}{b + 1}$ + $\frac{1}{c + 1}$ = 2, then $\sqrt{ab}$ + $\sqrt{bc}$ + $\sqrt{ac}$ $\leq$ $ \frac{3}{2}$.
I don't know how to start.
| The condition gives $\sum\limits_{cyc}\left(\frac{1}{1+a}-1\right)=-1$ or $\sum\limits_{cyc}\frac{a}{1+a}=1$.
Thus, by C-S $1=\sum\limits_{cyc}\frac{a}{1+a}\geq\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3+a+b+c}$, which gives what you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
homework - probability problem A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?
My Attempt:
The sample space for the above problem, $S = 6^3$.
Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5,... | You have to consider all the possible ways in which you can obtain $15$.
Those possibilities are:
$\{3, 6, 6 \}$, $\{6, 3, 6 \}$, $\{6, 6, 3 \}$
$\{5, 5, 5 \}$
$\{4, 5, 6 \}$, $\{4, 6, 5 \}$, $\{5, 4, 6 \}$, $\{5, 6, 4 \}$, $\{6, 4, 5 \}$, $\{, 5, 4 \}$
Total = $10$
There are only $2$ cases in which $4$ is the first el... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit with little-o It's a while I'm trying to verify this limit:
$$\lim_{x\rightarrow 0^+} \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0,$$
this is my attempt:
Since $x \rightarrow 0$ I can do the substitution:
$$\sqrt{1+x^2}=1+o(1)$$
$$\sin{x}=x(1+o(1))$$
and I obtain:
$$\frac{\log(x+1+o(1))-x... | You have
\begin{align}
\sqrt{1+x^2}&=1+\frac{x^2}2-\frac{x^4}8+o(x^6)\\ \ \\
\log(1+x)&=x-\frac{x^2}2+\frac{x^3}3+o(x^4)\\ \ \\
\sin x&=x-x^3/6+o(x^5)\\ \ \\
2^x&=1+x\log2+o(x^2)\\ \ \\
e^x&=1+x+o(x^2)
\end{align}
Then
\begin{align}
\frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}
&=\frac{\log(1+x+x^2/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(... | As $D=2\pi-(A+B+C),\cos D=\cos(A+B+C)$
$$\implies\cos A+\cos B+\cos C+\cos(A+B+C)$$
$$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+2\cos\dfrac{A+B+C-C}2\cos\dfrac{A+B+C+C}2$$ (using Prosthaphaeresis Formulas)
$$=2\cos\dfrac{A+B}2\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B+2cc}2\right)$$
Apply Prosthaphaeresis Formula on $$\cos\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the largest positive integer $n$ such that $2^n$ divides $3^{4096} −1$. I dont know how to start with this questions.It is an olympiad sum.
| HINT:
Observe that the highest power of $2$ that divides $3^2-1$ is $3$
Now $3^{2n}=(1+8)^n\equiv1\pmod8\implies3^{2n}+1\equiv2\pmod8\equiv2\pmod4$
So, the highest power of $2$ that divides $3^{2n}+1$ is $1$ for integer $n\ge0$
So, the highest power of $2$ that divides $3^{(2^2)}-1=(3^2-1)(3^2+1)$ will be $3+1$
So, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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.