Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A limit involving the Hurwitz zeta function I want to show that
$$ \lim_{s \to 1} \left( \zeta(s,a) - \frac{1}{s-1} \right) = - \psi(a)$$
where $\zeta(s,a)$ is the Hurwitz zeta function and $\psi(a)$ is the digamma function.
The only approach I can think of is to use the integral representation
$$\zeta(s,a) = 2 \int_{0... | Everything is correct except in the beginning where you take a limit of the integral representation. You conclude that $\lim_{s\rightarrow 1}\frac{a^{1-s}-1}{s-1}=1$, whereas by l'Hopital's rule, the correct answer is $-\ln(a)$ which should cancel the other log that you get later.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find AB where A= matrix and B=matrix $A=\left[\begin{array}{ccc} 2&1&0\\0&3&-1
\end{array}\right]$
$B=\left[\begin{array}{cc}a&1\\1&b\\b&a\end{array}\right]$
Matrices
Find $AB$
| $$AB=\left[\begin{array}{ccc} 2&1&0\\0&3&-1
\end{array}\right]\left[\begin{array}{cc}a&1\\1&b\\b&a\end{array}\right]$$
If you know basic matrix multiplication, you should know
$$\begin{align}
AB =& \left[\begin{array}{cc}
A\left[\begin{array}{c}a\\1\\b\end{array}\right]&
A\left[\begin{array}{c}1\\b\\a\end{array}\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/503028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots ... | Hint: $6 \mid (n-1)(2n-1)$ implies that $n$ is odd and cannot be divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$
To solve it I used induction but it is leading me nowhere my attempt was as follows:
Lets assume the inequality i... | $$f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}=H_{2n}-H_n $$
is an increasing function, since
$$ f(n+1)-f(n) = \frac{1}{(2n+1)(2n+2)}>0. $$
It follows that $\forall n>1,\; f(n)\geq f(2)=\frac{7}{12}>\frac{13}{24}$.
Additionally, by Riemann sums
$$ \lim_{n\to +\infty}f(n)=\lim_{n\to +\infty}\sum_{k=1}^{n}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Laurent series of a trigonometric function Find the Laurent Serie(and residue) aroud $z_0=0$ of the function
$f(z) = \frac{1}{1-\cos z}$.
Progress:
It looks very trivial but it seems to get complicated so I'll only try with 3-4 terms:
We can use that $\cos z = 1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0... | Since we're interested only around $\,z=0\,$ we can try the following using what you've already done:
$$\frac1{1-\cos z}=\frac1{\frac{z^2}2\left(1-\frac{z^2}{12}+\mathcal O(z^4)\right)}=\frac2{z^2}\left(1+\frac{z^2}{12}+\frac{z^4}{12^2}+\ldots\right)=$$
$$\frac2{z^2}+\frac16+\frac{2z^2}{12^2}+\ldots$$
We used, of cours... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral of $ \frac{\sqrt{x+1}}{x} $. Im looking for a way to calculate:
$$ \int{\frac{\sqrt{x+1}}{x}} dx $$
I substituted $ u = \sqrt{x+1} $ and then $ dx = 2 * \sqrt{x+1} du $ so I have:
$$ \int{\frac{u}{x} * 2 \sqrt{x+1} du } $$
$ x = u^2-1 $
and
$$ 2 \int{\frac{u^2}{u^2-1} } du = 2 \int{\frac{u^2}{(u-1)(u+1)} }... | What you did was correct. To finish it off, notice that
$$\dfrac{u^2}{u^2-1} = \dfrac{(u^2-1)+1}{u^2-1} = 1 + \dfrac{1}{u^2-1}$$
The integral can then be computed using partial fractions:
$$\dfrac{1}{u^2-1} = \dfrac{1}{(u-1)(u+1)} = \dfrac{1}{2} \left( \dfrac{1}{u-1} - \dfrac{1}{u+1} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be o... | Consider cases:
*
*The last digit is $0$. Then you have six choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $6 \cdot 5 \cdot 4 \cdot 1 = 120$ four digit even numbers that end in $0$.
*The last digit is $2$, $4$, or $6$. Since the first digit cannot be z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/511261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| Suppose $gcd(a,b)=1$ then you have $ax+by=1$, cubing this, we get $(ax+by)^3=1$
i.e., $a^3x^3+b^3y^3+3a^2x^2by+3axb^2y^2=1$
i.e., $a^2(ax^3+3x^2by)+b^2(by^3+3axy^2)=1$
does this imply $gcd(a^2,b^2)=1$???
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 2
} |
A basic doubt on a infinite series problem I see in Rudin the following statement is claimed for the following convergent series:
$$1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$
If $s$ is the sum of this series then $$s \lt 1 -\frac{1}{2} + \frac{1}{3}$$. How is that possible to tell ... | $$S_{2k+1} = 1-\frac{1}{2} + \frac{1}{3} - \left(\frac{1}{4} - \frac{1}{5}\right) - \ldots -\left(\frac{1}{2k} - \frac{1}{2k+1}\right) < 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right)$$ and $S_{2k+1}\to s$ imply
$$s \leqslant 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right) < 1-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/513336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Distributing amounts of different objects to distinct people I have the following exercise and I am wondering if the subsequent solution of mine is correct.
Alan, Peter, Steve, and Donald have a total of 11 chocolate bars that
they are going to divide among themselves. They have 9 Snickers (S), 1
Mars (M) and 1 Bo... | You seem to get something like $3824$, which looks far too many: my answer is about half that.
So looking at possible issues with your answer:
*
*In case 3 you seem to have changed a 6 into a 1, so case 3 does not give the same answer as case 2; I am unclear about whether this is deliberate or not
*I think you ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The Fejer kernel has this $\sin$ closed form. Let $D_N$ be the $N$th Dirichlet kernel, $D_N = \sum_{k = -N}^N w^k$, where $w = e^{ix}$. Define the Fejer kernel to be $F_N = \frac{1}{N}\sum_{k = 0}^{N-1}D_k$. Then $$F_N = \frac{1}{N}\frac{\sin^2(N x/2)}{\sin^2(x/2)}$$.
So far I have $D_k = \frac{w^{k+1} - w^{-k}}{w-1... | We can prove the equality for the Fejér kernel in the following way. Using the formula for the geometric progression and the fact that $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ for each $\theta\in\mathbb R$,
\begin{align*}
D_k(x)
&=e^{-ikx}\sum_{s=0}^{2k}e^{isx}\\
&=e^{-ikx}\frac{1-e^{ix(2k+1)}}{1-e^{ix}}\\
&=\frac{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Although this is not induction, I still find it quite amusing that Leibniz (inventing the differential calculus almost simultaneously with Newton) solved these kind of problems with some sort of difference method with great similarity to his differential calculus:
First we define the differences of a sequence: For inst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$ Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$.
You can use L'Hospitale, or Maclaurin, etc
| Hint: Use $\tan x = x+\dfrac{x^3}{3}+o(x^3)$ and $\sin x = x-\dfrac{x^3}{6}+o(x^3)$ to obtain $\tan(\tan x) = x+\dfrac{2x^3}{3}+o(x^3)$ and $\sin (\sin x) = x-\dfrac{x^3}{3}+o(x^3)$.
Hence $$\dfrac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x} = \dfrac{x^3+o(x^3)}{\frac{x^3}{2}+o(x^3)} \to 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x... | The number which leaves (5, 4, 3, 2) mod (6, 5, 4, 3) is one less than the one that leaves a residue of (0, 0, 0, 0), mod (6, 5, 4, 3). So one finds $n=\operatorname{lcm}(6,5,4,3)-1$ to get 59, which is the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
$x,y,z$ positive real numbers , $x+y+z=3$ $\implies x^4y^4z^4(x^3+y^3+z^3)≤3$ If $x,y,z$ are positive real numbers with $x+y+z=3$ then how to prove (without using calculus) that $\space$ $x^4y^4z^4(x^3+y^3+z^3)≤3$ ?
| Well, I tried and failed to find an answer made of sugar and spice. Here is a frogs and snails answer.
As karafka noted, the AM-GM inequality easily implies that $xyz\leq 1$. So let $\epsilon\geq 0$ be such that $xyz=1-\epsilon$ .
Let us order the variables so that $x\leq y\leq z$. We must have $z\geq 1$ and $x\leq 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/520129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Justify $\gcd$ of $f(x) = x^3 - 6x^2 + x + 4$ and $g(x) = x^5 - 6x +1$ Let $f(x) = x^3 - 6x^2 + x + 4$ and $g(x) = x^5 - 6x +1$. Using Euclidean algorithm I find $\gcd[f(x), g(x)] = 1$. How could I JUSTIFY that $h(x) = 1$ is the ACTUAL $\gcd$ of $f(x)$ and $g(x)$?
Thanks.
| Actually Euclidean Algorithm is a really good proof.
One other way is to factorize one of the polynomials into linear polynomials. To do the find the zeroes of that polynomials. I'll use the polynomial $f(x)$ so the zeroes are at $x = 1$, $x=\frac 12 (5 \pm \sqrt{41})$.
So we can factorize it as:
$$x^3 - 6x^2 + x + 4 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of a sequence $\frac{1}{1+n^3}$ How can I prove by integral test that the sequence $1+ \dfrac{1}{1+2^3} + \dfrac{1}{1+3^3} + \dots + \dfrac{1}{1+n^3}$ is convergent?
Thank you. Is there a way that I can integrate $\dfrac{1}{1+n^3}$ ?
| I will guess that the $n$-th term of the sequence is
$$\frac{1}{1+1^3}+\frac{1}{1+2^3}+\frac{1}{1+3^3}+\cdots+\frac{1}{1+n^3}.$$
There is indeed a closed form for $\int \frac{1}{1+t^3}\,dt$. However, the details are somewhat unpleasant. It is simpler to note that the $n$-th term is less than $\frac {1}{1^3}+\frac{1}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Con... | Presented below is a self-contained evaluation. With
$\int_0^1 \frac{\ln t}{1+t^2}dt =-G$
\begin{align*}
I & = \int_0^1 \frac{\ln x \ln (1-x^4 )}{1+x^2}dx \\
& = \int_0^1 \ln (1-x^4 ) d\left(\int_1^x \frac{\ln t}{1+t^2}dt \right) \overset{IBP}=\int_0^1 \frac{ 4x^3}{1-x^4} \underset{t=xs }{\left(\int_0^x \frac{\ln t}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 3
} |
If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightar... | I am new here so don't know how to type math equation but I am providing the solution.
2005+x=y^2 ..(1)
2005+y=x^2 ..(2)
After subtracting 2nd equation from first
x-y=y^2-x^2
x-y=(x-y)(x+y)
(x-y)+(x-y)(x+y)=0 (we change the sign x-y to y-x here)
(x-y)(x+y+1)=o
because as mentioned x and y are different integers s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
√2+√3 is irrational I’m trying to prove that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ is irrational.
I have already proved that $\sqrt{2}+\sqrt{3}$ is irrational. Should I use a similar approach as below or is there a different way?
Proof: $\sqrt{2}+\sqrt{3}$ is irrational
Assume $\sqrt{2}+\sqrt{3}$ is rational so $\sqrt{... | You have a shorter proof: if $\sqrt{2}+\sqrt{3}=\frac{p}{q}$, where $p\in\mathbb Z$ and $q\in\mathbb N$, $q\neq 0$, then $5+\sqrt{6}=\frac{p^2}{q^2}$. So, $\sqrt{6}=\frac{p^2-5q^2}{q^2}$ is rational, which is known to be false.
For $\sqrt{n}$, with $n\in\mathbb N$, you have the following alternative:
1) $\sqrt{n}$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/525172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$ Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$.
This is what I've managed:
$x_... | You can use that
$$\cos (2x) = \cos^2 x - \sin^2 x,$$
and hence
$$1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}.$$
Then write
$$\sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Area of a circle inscribed in a rhombus? Let's say we have a rhombus with diagonals $a$ and $b$, which contains an inscribed circle. How can we find the area of that circle in terms of $a$ and $b$?
|
Obviously, the radius of inscribed circle is also a height $h=OH$ of the right triangle $\triangle AOB$. To find it, use equations for triangle's area
$$
S_{\triangle AOB} = \frac 12 \frac a2 \frac b2 = \frac {ab}8 = \frac 12 ch
$$
where $c = AB$ is a hypotenuse. So $r = h = \frac {ab}{4c} = \frac {ab}{4\sqrt{\frac {a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Partial Fractions and power of a factor with $x^2$ I just started working with partial fractions and hit a wall with splitting this one:
$$ \frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2} $$
I get here:
$$ \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}$$
Then on to:
$$ (Ax + B)(x + 2) + (Cx ... | You don't have any equations there.
Presumably, you start with
$$\frac{3x^2 + 2x + 1}{(x + 2)(x^2 + x + 1)^2}
= \frac{Ax + B}{(x^2 + x + 1)^2} + \frac{Cx + D}{x^2 + x + 1} + \frac{E}{x + 2}
$$
then multiply both sides
by $ (x + 2)(x^2 + x + 1)^2$
to get
$$3x^2 + 2x + 1
=(Ax + B)(x + 2) + (Cx + D)(x^2 + x + 1)(x + 2) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality $\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}} \le n$ I was proving a statement and have finished all the other part, the remaining boiling down to proving the inequality below (if I proved the other part correctly):
$$\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}} \le n$$
where $1 \le i \le k \le j \... | Proof: Define $s=\sum^{k}_{i=1}\sum^{n}_{j=k}{{1}\over{j - i + 1}}$. We prove the required inequality by factoring out common and distinct summand $\frac{1}{d+1}$, where $d=j-i$. The region of summation $\{1,2,...,k\}\times\{k,k+1,...,n\}$ is partitioned into 3 parts, with 2 isosceles right triangles sandwiching 1 para... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Use L'Hôpital's rule to evaluate this limit $$\lim_{x\to0} \left(\frac{a^x + b^x}{2}\right)^{1/x} = (ab)^{1/2}$$
I tried taking logarithm to both sides, but got stuck
I got to
e^ (lim (a^x loga + b^x logb)/(a^x + b^x)
| Note that $ \ln \left( \left( \dfrac {a^x+b^x}{2} \right)^{1/x} \right) = \dfrac {1}{x} \cdot \ln \left( \dfrac {a^x+b^x}{2} \right) $.
Now, consider $ \lim \left( \dfrac {\ln \left( \dfrac {a^x+b^x}{2} \right)}{x} \right) $.
The derivative of the numerator is $ \dfrac {a^x \ln a + b^x \ln b}{a^x+b^x} $.
The derivati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/529832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$. Let $f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$.
Then Max. of $f(x,y)$.
$\underline{\bf{My\;Try}}::$ We can convert into Complex no. form ...
Let... | let $A(0,1),B(3,4),C(0,0),D(1,0),P(x,y)$, we want $PA+PB-PC-PD$ have max ,but $PA-PC\le AC$, the "=" will hold when $P$ is on $AC$, $PB-PD \le BD$,the "=" will hold when $P$ is $BD$, so MAX is $AC+BD=1+\sqrt{(3-1)^2+(4-0)^2}=1+2\sqrt{5}$,when $P(0,-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How ... | It seems like the way to attack this that doesn't require guessing is this: Start with $N (N-1) (N-2) (N-3) = N^4 + 6 N^3 + 11 N^2 + 6 N = M^2 - 1$ Since the product is "near" $N^4$, $M$ has to be "near" $N^2$. And it's very likely that $M$ is a polynomial in $N$. So set $M = N^2 + aN + b$. Then $M^2-1 = N^4 + 2aN... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/532737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
Express $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've been trying with no success expressing this functions.
a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\s... | Hint:
You my want to take a look at the half angle formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^... | $$\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}+\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}=1$$
$$\implies\dfrac{\cos x\cos{\dfrac y2}}{\cos\left(x-\dfrac y2\right)}=1-\dfrac{\cos y\cos{\dfrac x2}}{\cos\left(y-\dfrac{x}{2}\right)}$$
$$\implies\frac1{1+\tan x\tan\frac y2}=\frac{\sin y\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How do I prove that $(1+\frac{1}{2})^{n} \ge 1 + \frac{n}{2}$ for every $n \ge 1$? How do I prove that $(1+\frac{1}{2})^{n} \ge 1 + \frac{n}{2}$ for every $n \ge 1$
My base case is $n=1$
Inductive step is $n=k$
Assume $n=k+1$
$(\frac{3}{2})^{k} \times \frac{3}{2} \ge (1 + \frac{k+1}{2})$
I'm not sure how to proceed.
| Let $P(n):(1+x)^n\ge 1+nx$
Clearly $P(n)$ is true for $n=1$
Let $P(n)$ is true for $n=m\implies (1+x)^m\ge 1+mx$
$\implies (1+x)^{m+1}\ge (1+mx)(1+x)=1+(m+1)x+mx^2\ge 1+(m+1)x$ if $m\ge0$ and $x$ is real
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/535948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Use mathematical induction to prove that for any $k \in\mathbb N , \lim (1+k/n)^n = e^k$. Use mathematical induction to prove that for any $k \in \mathbb N, \lim (1+k/n)^n = e^k$.
I already used monotone Convergence Theorem to prove $k=1$ case. Do I just need to go through the same process to show $k$? If not, could yo... | First a useful bound: Suppose $\theta_n \ge 0$ is such that $\lim_n n \theta_n = 0$, then $\lim_n (1+ \theta_n)^n = 1$. To see this, note $(1+\theta_n)^n = \sum_{k=0}^n \binom{n}{k} \theta^k \le \sum_{k=0}^n n^k \theta^k \le \frac{1}{1-n \theta_n}$.
Suppose the result is true for $k$ (I am taking $k=1$ as already prov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/536443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\binom{12}0+\binom{12}2+\ldots+\binom{12}{12}$ using binomial theorem
Solve the sum:
$$
{12 \choose 0}+
{12 \choose 2}+
{12 \choose 4}+
{12 \choose 6}+
{12 \choose 8}+
{12 \choose 10}+
{12 \choose 12}
$$
using the binomial theorem.
I know the binomial theorem:
$$ \left(a+b\right)^2 = \sum_{k=0}^{n} {n \c... | With $a = 1, b = -1$ you get that
$$
\sum_{i= 0}^{12}(-1)^i\binom{12}{i} = (1-1)^{12} = 0
$$
and $a = 1, b = 1$ gives you
$$
\sum_{i = 0}^{12}\binom{12}{i} = (1 + 1)^{12} = 2^{12}
$$
Add these two together and see what terms are left.
Edit
Expanding, another way to write the $b = -1$ equation over is
$$
0 = \binom{12}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/536888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| Write the integrand as a sum of fractions and use the polylogarithm function $\mathrm{Li}_2:$
$$f(x):=\int\frac{\ln(x)}{x^2-1}dx=\int\frac{1}{2}\left(\frac{\ln(x)}{x-1}- \frac{\ln(x)}{x+1}\right) dx \\
=\frac{1}{2} \int \frac{\ln(x) dx}{x-1}- \frac{1}{2} \int \frac{\ln(x)dx }{x+1}
=-\frac{1}{2}\mathrm{Li}_2(1-x) -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 8,
"answer_id": 3
} |
How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| We'll using the following notation:
$$s_1 = x + y + z$$
$$s_2 = x^2 + y^2 + z^2$$
$$s_3 = x^3 + y^3 + z^3$$
And we'll express $x,y,z$ as root of the cubic equation $f(x) = ax^3 + bx^2 + cx + d$
From the Vieta's formulas we know that $d = -xyz$
From the condition we have $d = - s_2$ and $s_1 = s_3$
Now use Newton's iden... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/538252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of theta. Let $X$ have the density function $f(x) = \dfrac{3x^2}{\theta^3}$ for $0 < x < \theta$, and $f(x) = 0$ otherwise. If $P\{X > 1\} = 7/8$, find the value of $\theta$.
I don't know
| We know that $\Pr(X\le 1)=\dfrac{1}{8}$. But
$$\Pr(X\le 1)=\int_0^1 \frac{3x^2}{\theta^3}\,dx,$$
which is $\dfrac{1}{\theta^3}$. Thus we need to solve the equation $\dfrac{1}{\theta^3}=\dfrac{1}{8}$.
Remark: Alternately, we could have used the equation
$$\int_1^\theta \frac{2x^2}{\theta^3}\,=\frac{7}{8}.$$
The integra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\int_0^{\frac{\pi}{2}}x\cot(x)dx$ and $ \lim_{m \rightarrow \infty}\log\left( e^{2m}\left(\frac{(2m-1)!!}{(2m+1)^m}\right)^2\right)$. I'm trying to evaluate the integral, but in doing so have stumbled upon the limit, which I don't know whether it exists, and if so how to resolve it (and whether I've derived the relati... | IMHO, it will be simpler if one integrate the integral by parts. Using answers from the question Evaluate: $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$, we have
$$\int_0^{\frac{\pi}{2}} x \cot x dx = \int_0^{\frac{\pi}{2}} x (\log \sin x)' dx
= \Big[x \log \sin x\Big]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction that $3^{4n + 2} + 1$ is divisible by $10$ Prove by induction: $3^{(4n+2)} + 1$ is divisible by $10$.
My basic step: $3^{(4n+2)} + 1$, where $n = 1$ gives me $3^6 + 1 = 730$, which is divisible by $10$. However, then I have to do the induction hypothesis and I am kind of stuck because I do not have a... | Suppose $3^{4k+2} +1$ is divided by 10, then we need to show $3^{4(k+1)+2} +1$ is also divided by 10. Note that:
\begin{align*}
\ 3^{4(k+1)+2} +1 &= 3^{4k+2+4}+(81-80)
\\ &= 81\cdot3^{4k+2} + 81 - 80
\\ &= 81\cdot(3^{4k+2}+1) - 80
\\ &= 81\cdot10m -80\ldots\ldots (\text{where}~3^{4k+2}+1 = 10m ~~\text{for some integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/543717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the limit of $( 1 + a + a^2 + \ldots+ a^n)/(1 + b + b^2 + \ldots+ b^n)$ I have some problems finding the limit of $$\frac{ 1 + a + a^2 +\cdots + a^n}{1 + b + b^2 + \cdots + b^n}.$$
$0\le a,b \le +∞$
Here is what I got :
Forcefuly factorize $a^n$ and $b^n$ :
$$ \frac{a^n ( 1 + \frac1{a} + \frac1{a^2} + ...... | $\displaystyle\lim_{n\rightarrow\infty}\frac{1+a+\cdots+a^n}{1+b+\cdots+b^n}=\displaystyle\lim_{n\rightarrow\infty}\frac{1-b}{1-a}.\frac{1-a^{n+1}}{1-b^{n+1}}=\frac{1-b}{1-a}$ if $1<a\leq b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution verification: $\lim\limits_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} = -1$ I am trying to find the following limit
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2}$$
and I did the following steps:
\begin{align}
\require{cancel}
&\lim_{x\rightarrow \infty} \frac{\sqrt{x}+x^2}{2x-x^2} \\
&\lim_... | Looks good! I would be a bit more precise about "the top portion goes to $0$," but your reasoning is just fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/546922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
By definition, how is a prime number represented? Even numbers can be easily represented as $2n$. Odd numbers as $2n+1$. An exactly divisible operation can be defined as $n = dq$.
But, is there an specific way of representing a prime number, obtained by a proof of some sort?
| From Jones, J., Sato, D., Wada, H. and Wiens, D. (1976). Diophantine representation of the set of prime numbers. American Mathematical Monthly, 83, 449-464.
The set of prime numbers is identical with the set of positive values taken on by the polynomial
$(k+2)(1-(wz+h+j-q)^2-((gk+2g+k+1)\cdot(h+j)+h-z)^2-(2n+p+q+z-e)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/547359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $?
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $
| We know that $(x+z)y=48$. We are looking for $(x+z)+y$.
$$
x+z=1;y=48\implies (x+z)+y=49\\
x+z=2;y=24\implies (x+z)+y=26\\
x+z=3;y=16\implies (x+z)+y=19\\
x+z=4;y=12\implies (x+z)+y=16\\
x+z=6;y=8\implies (x+z)+y=14\\
$$
we could switch the values, and get the other choices, but the sums and products remain the same.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/548410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Problems while solving the cubics Some time ago, I got a question of the form $\sqrt{a+(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} }$, which after cubing I realized that $(p-q)^\frac{1}{3}+(p+q)^\frac{1}{3} = -a $. That set my instincts, and I figured out that that all cubic of the form $x^3 = a + bx$ must have a solution of t... | Better late than never. This is strongly related to this answer of mine, when I proved a general form for all solutions of $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n$$ for rational numbers. Here, we should work backwards: we have $$N=\left(\frac{9\sqrt{6} - 19}{6\sqrt{6}}\right)^{1/3} + \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A problem on indefinite integration $$\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx$$
I tried some substitutions, but none succeeded in simplifying the expression. Please help.
| Once we know the answer,
$$\displaystyle\int\frac{x^4-2}{x^2\sqrt{x^4+x^2+2}}dx =\int\frac{x^4-2}{x^3\sqrt{\frac{x^4+x^2+2}{x^2}}}dx=\int\frac{x-\frac2{x^3}}{\sqrt{x^2+1+\frac2{x^2}}}dx$$
Put $\displaystyle x^2+1+\frac2{x^2}=u^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/554395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$
$H_n$ denotes the harmonic numbers.
| using the following identity :
$$\displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$
multiply both sides by $\frac{\ln^2x}{x}$ then integrate both sides w.r.t $x$ from $0$ to $1$, we have:
\begin{align*}
S&=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\int_0^1x^{n-1}\ln^2x\ dx=\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/555266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 4,
"answer_id": 2
} |
How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\d... | Starting with $x+y+z =3$
From the relation between $AM - GM$:
$\frac{x+yz}2 \ge \sqrt{xyz}$
$\Rightarrow \frac 2{x+yz} \le \frac 1{\sqrt{xyz}}$
$\Rightarrow \frac{2x}{x+yz} \le \frac x{\sqrt{xyz}}$
Similarly for others it can be shown that
$ \frac {2y}{y+xz} \le \frac y{\sqrt{xyz}}$
$\frac {2z}{z+xy} \le \frac z{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Probability Problem on Divisibility of Sum by 3 From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem. I would welcome a good solution.
| Note that our set has $34$ members equivalent to $1$ modulo $3$ (that is, with remainder $1$ upon division by $3$), $33$ equivalent to $2$, and $33$ equivalent to $0$.
Suppose we draw three numbers at random to create our subset. In order to get a sum divisible by $3$, we could draw any of the following sets of remaind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
if $f(\frac{x+y}{2})=\frac{1}{2}[f(x)+f(y)], f(0)=0,f(1)=1$ then$f(\frac{1}{22})=?$ let function $f:[0,1]\to [0,1]$,and such $f(0)=0,f(1)=1$,
and foy any $0\le x\le y\le 1$,then we have
$$f\left(\dfrac{x+y}{2}\right)=\dfrac{1}{2}[f(x)+f(y)]$$
Question 1
Find the value $f(\dfrac{1}{22})$
Qusetion 2: Find the
$$f(\dfrac... | As long as you consider $f$ at rational places, no continuity assumption is needed. Since 1. is a special case of 2., let's solve that:
Let $n\in\mathbb N$ and $x_i=\frac in$, $0\le i\le n$.
Then for $0<i<n$ we have $x_i=\frac{x_{i-1}+x_{i+1}}2$ and hence $f(x_i)=\frac{f(x_{i-1})+f(x_{i+1})}2$, i.e. $$\tag1f(x_{i+1})=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Elementary proof that $\pi < \sqrt{5} + 1$ I wanted to show that
$$ \frac{\pi}{4\phi} < \frac{1}{2} $$
Where $\phi$ is the golden ratio. I have confirmed the results numerically, and by
simple algebra the inequality simplifies down to
$$
\pi < \sqrt{5} + 1
$$
This is a weaker relation than what was shown here. Prove t... | Start with an unit circle inscribed inside a $2 \times 2$ square. One can chop off 4 right angled isosceles triangle whose shorter side has length $2 - \sqrt{2}$ from the four corners. This will turn the square into a octagon with the unit circle inscribed inside it. By comparing the area of the circle and the octagon,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
} |
Which step is wrong in this proof
Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original eq... | if $x^2+x+1=0$ then $(x-1)(x^2+x+1)=0$ thus $x^3-1=0$, thus: $x^3=1$.
if ... then .... is not a logical equivalence but only a logical implication.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x +... | Other solution:
If you know that the set of diffrential equation: $$(E) \quad y''+y=0$$ solutions is a two dimensinal space, then since the three applications : $x \mapsto \sin(x+a)$ where $a \in\{\alpha,\beta,\gamma \}$ are solutions of $(E)$, there are linearly dependantes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
$a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$ $a;b;c\in \mathbb{R}^+$. Prove : $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a+b+c}{\sqrt{a^2+b^2+c^2}} \geq 3+\sqrt{3}$
Thanks :)
I have proved that :
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3... | use this known lema:
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{9(a^2+b^2+c^2)}{(a+b+c)^2}$$
proof:By applying the know inequality
$$(x+y+z)^3\ge\dfrac{27}{4}(x^2y+y^2z+z^2x+xyz)$$
for
$x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$,we get
$$\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)^3\ge\dfrac{27}{4}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/562510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Concave Convex $e^{-x}-e^{-2x}$ My main goal is to see here if my discusison on concave/convex is good enough.
Problem
Find the extreme points (max and/or min) for $$f(x)=e^{-x}-e^{-2x}.$$ Also discuss concavity/convexity.
Attempt
Extreme points
Solve $f'(x) = 0 -e^{-x} + 2e^{-2x} = 0$
Multiply by $-e^{2x}$ and ge... | !
$f′′(\ln(4))=0$
$f′′$ is negative to the left of $\ln(4)$
and positive to the right
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/563726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the Unknown Coordinate when given Distances from two points I have this problem that I don't know how to do.
Locate the two points that are $5$ units from $(5,-3)$ and square root of $41$ units from$(-2,6)$
| You need the intersection of two circles.
Let's say $A=(5,-3)$ and $B=(-2,6)$, $r_1=5$ and $r_2=\sqrt{41}$.
Now
$$c_1:(x-5)^2+(y+3)^2=5^2 \quad and \quad c_2:(x+2)^2+(y-6)^2=\sqrt{41}^2$$
Now you have to find you radical axis of $c_1,c_2$:
\begin{eqnarray*}
c_1-c_2&:x^2-10x+y^2+6y+9=0 \\
&-[x^2+4x+y^2-12y-1=0]\\
&\Rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Fast way to integrate $\frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$ in unit square I am looking for a fast way to integrate $$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx \,dy$$ using standard techniques ( no complex analysis and no functional analysis).
I am aware that wolframalpha spits out a solution, but this one is ... | Start with the inside integral:
$$
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx
$$
Let $x=y\tan\theta$, so that $dx = y\sec^2\theta\,d\theta$ and $x^2+y^2 = y^2\tan^2\theta+y^2 = y^2\sec^2\theta$. As $x$ goes from $0$ to $1$, $\theta$ goes from $0$ to $\arctan(1/y)$, so the integral becomes
\begin{align}
& \phantom{={}}\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/567412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \... | HINT:
Use $\log a+\log b+\log c=\log(abc) $
and then put $a=\tan A$ etc. to find
$\displaystyle \sum\tan A=\prod\tan A\implies A+B+C=n\pi$ where $n$ is any integer (See here)
and $\displaystyle\frac{2a}{1-a^2}=\frac{2\tan A}{1-\tan^2A}=\tan2A$
$\displaystyle\implies \sum\tan2A=\prod\tan2A$ as $2A+2B+2C=2n\pi$
Now a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
| I prefer @math110's solution, but here's a brute force method using complex exponentials,
with
$$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \qquad \sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$
We define
$$a := e^{ix} \qquad b := e^{iy} \qquad c := e^{iz}$$
so that
$$p = \frac{\cos x + \cos y + \cos z}{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$ How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?
What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$
for values $x>1$. $C = 3/2, k = 1$
Is this right?
| We have
$$
f(x)= \frac{x^4 + x^2 +1 }{x^3 +1} \leq C x,
$$
for $x \geq \sqrt{\frac{C}{3(C-1)}}$ and $C>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question on Convergence of Improper integrals Question is to check which of the following improper integrals are convergent?
$$\int _1^{\infty} \frac{dx}{\sqrt{x^2+2x+2}}$$
$$\int _0^5 \frac{dx}{x^2-5x+6}$$
$$\int _0^5\frac{dx}{\sqrt[3]{7x+2x^4}}$$
I was having a stupid mindset that :
"$\textbf{Improper integrals shoul... | Hint: For the issue concerning convergence it sufficient and necessary to understand only the two following types of integrals for $\alpha \in \mathbb{R}$:
$$ \int_{0}^1 x^\alpha d x , \qquad \int_{1}^\infty x^\alpha d x.$$
E.g.
$$ \int_{1}^\infty ( (x+1)^2 + 1)^{-1/2} d x = \int_{0}^\infty ( x^2 + 1)^{-1/2} d x \geq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/577096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to speedup evaluation of hypergeometric ${}_3 F_2(1)$? I need to make a table of ${}_3 F_2\left(\frac{a}2+\frac14, \frac{a}2+\frac34, \frac{a+b}2;\; a+1,\frac{a+b}2+1;\;1\right)$ for integer $a, b,$ $0\le a\le N_1$, $0\le b\le N_2$, with precision of 50 decimal places in mantissa.
Currently I'm trying to use Wolfra... | Hint:
$_3F_2\left(\dfrac{a}{2}+\dfrac{1}{4},\dfrac{a}{2}+\dfrac{3}{4},\dfrac{a+b}{2};a+1,\dfrac{a+b}{2}+1;1\right)$
$=\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{a}{2}+\dfrac{1}{4}\right)^{(n)}\left(\dfrac{a}{2}+\dfrac{3}{4}\right)^{(n)}\left(\dfrac{a+b}{2}\right)^{(n)}}{(a+1)^{(n)}\left(\dfrac{a+b}{2}+1\right)^{(n)}n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuc... | Using Aron D'souza's idea further we can get:
$$L^2-\frac{1}{2}< \sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}$$
$$\sqrt[3]{\frac{1}{3}+\sqrt[3]{\frac{1}{3}+\dots}}=\frac{1}{2} \left(1+\sqrt{\frac{7}{3}} \right)$$
$$L<\sqrt{1+\frac{1}{2} \sqrt{\frac{7}{3}}}=1.328067$$
To find the next bound we will need to solve:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 4,
"answer_id": 2
} |
Recursive sequence of functions Recursive sequence of functions: $f_{n+1}= \sqrt{x+f_n}$, $f_1(x)= \sqrt{x}$.
this sequence is monotonic, but what is bounding it? thanks
| So you have defined $$f_n(x) = \underbrace {\sqrt{x + \sqrt {x + \dots \sqrt x}}}_{\text {n square roots}}$$ and you say that it is "monotonic", suggesting some comparison like
$$f_{n+1} \ge f_n$$
which, correct me if I'm wrong, I'm going to assume means $\forall x\, f_{n+1}(x) > f_n(x)$.
A bound (..but not a least upp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$
Thanks in advance for any help.
| Heres a nice combinatorial proof: Lets say you have $n+1$ kids, and want to form a committee of three. Order the kids $a_1, a_2, \cdots, a_{n+1}$. There are $\dbinom{n+1}{3}$ ways to form the committee. On the other hand, if $a_1$ is the first person on the committee, we need to choose two more, in $\dbinom{n}{2}$ ways... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 3
} |
Fixed sum of combinations When you have combinations where numbers are $0,1,2,\dots,m$, meaning we have $n=m+1$ and $k$, is there a way to see how $k$ of them sum up to a given number?
For the sake of simplicity I have the numbers $0,1,2...,7$ (so $n=8$), and $k=3$. I need to find how much of these combinations with ... | Write $m$ as
$$\underbrace{1 + 1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m 1s}}$$
This isn't the only sum, however. Other possible sums are $$3 + 3 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-6 1s}}$$ or $$2 + 1 + \underbrace{1 + 1 + 1 + 1 + \cdots + 1 + 1}_{\text{m-5 1s}} + 2$$ Notice, however, if we decompos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/585712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
evaluation of $\int\frac{x^5}{x^5+x+1}dx$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx$
$\bf{My\; Try::}$ $\displaystyle \int\frac{x^5}{x^5+x+1}dx = \int\frac{\left(x^5+x+1\right)-(x+1)}{x^5+x+1}dx = x-\int\frac{x+1}{x^5+x+1}dx$
Now Let $\displaystyle I = \int\frac{x+1}{x^5+x+1}dx = \int \frac{x+1}{(x^2+x+1)\cdot (x^3-x^2+... | Note that\begin{align}
&\frac{x^5}{x^5+x+1}
=1-\frac{2x+3}{7(x^2+x+1)}+
\frac{2x^2-x-4}{7(x^3-x^2+1)}\\
=& \ 1-\frac{2x+1}{7(x^2+x+1)}-\frac{2}{7(x^2+x+1)} +
\frac{2(3x-2x)}{21(x^3-x^2+1)} +
\frac{x-12}{21(x^3-x^2+1)}
\end{align}
Then
$$\int \frac{x^5}{x^5+x+1} dx
=x -\frac1{21}\ln\frac{(x^2+x+1)^3}{(x^3-x^2+1)^2}
-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I prove that $\lim_{n\to+\infty}\frac{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}{\sqrt{n}}=?$
let sequence $\{a_{n}\}$ such $a_{1}=1$,and
$$a_{n+1}a_{n}=n,n\ge 1$$
show that
$$2\sqrt{n}-1\le\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}<\dfrac{5}{2}\sqrt{n}-1$$
(2): I consider we... | observe that for $a_{n}$ to satisfy the inequality $a_{n}+\dfrac{n}{a_{n}}<\dfrac{5}{2}\sqrt{n}$
we need $\dfrac{\sqrt{n}}{2} <a_{n}<2\sqrt{n}$.
this can be proved by induction: first observe that it is true for $a_{1}$ and $a_{2}$ then suppose it is true for any n ($a_{n}$) then prove it is true for n+1 ($a_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum value of the expression. Find the minimum value of the expression. $x,y,z \in R$
$\sqrt{x^2+1}+ \sqrt {4+(y-z)^2} + \sqrt{1+ (z-x)^2} + \sqrt{9+(10-y)^2}$
| $S = d(E, A) + d(A,B) + d(B, C) + d(C,D) \\ \therefore S = \sqrt{10^2+7^2}=\boxed{\sqrt{149}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let n be a positive integer. Prove that: Let n be a positive integer. Prove that:
$\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$
| Case $1$: Let $n=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m$$
We now have $\sqrt{4n+2} = \sqrt{4m^2+2} \in [2m,2m+1]$. Hence, $\lfloor \sqrt{4n+2} \rfloor = 2m$
Case $2$: Let $n+1=m^2$, where $m \in \mathbb{Z}^+$. We then have $$\lfloor \sqrt{n} + \sqrt{n+1}\rfloor = 2m-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/587734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Need Help With Strong Induction Let $(X_n)$ be a sequence given by the following recursion formula:
$$X_1 = 3, X_2 = 7,\text{ and }X_{n+1} = 5X_n - 6X_{n-1}$$
Prove that for all $n\in\Bbb N$, $X_n = 2^n + 3^{n-1}$.
Attempt:
For $n = 1$, we have $2^1 + 3^0 = 3 = a_1$ TRUE
For $n = 2$, we have $2^2 + 3^1 = 7 = a_2$ TRUE... | Yes, go on. Use $2^k=2\cdot 2^{k-1}$ and associate the terms:
$$X_{k+1}=5\cdot(2^k+3^{k-1})\ -\ 6\cdot(2^{k-1}+3^{k-2}) = 2^{k-1}\cdot(10-6)+3^{k-2}\cdot(15-6)=\\
=4\cdot 2^{k-1}+9\cdot 3^{k-2}=2^{k+1}+3^k\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/588945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$ Simplify $\sqrt{3+2\sqrt{2}}-\sqrt{4-2\sqrt{3}}$
To do it I have see it that we have basically $\sqrt{2}$ and $\sqrt{3}$ that is we can write it as,
$$\sqrt{\sqrt{3}\sqrt{3}+\sqrt{2}\sqrt{2}\sqrt{2}}-\sqrt{\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{3}}$$
But ... | $$\sqrt{3+2\sqrt2}=\sqrt{3+\sqrt8}=\sqrt\frac{3+\sqrt{9-8}}{2}+\sqrt\frac{3-\sqrt{9-8}}{2}=\sqrt2+1$$
$$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt12}=\sqrt\frac{4+\sqrt{16-12}}{2}-\sqrt\frac{4-\sqrt{16-12}}{2}=\sqrt3-1$$
Now we have:
$$\sqrt{3+2\sqrt2}-\sqrt{4-2\sqrt3}=\sqrt2+1-(\sqrt3-1)=\sqrt2+2-\sqrt3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/591482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Is it always possible to find a $t>0$, such that $\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|dxIs it always possible to find a $t>0$, such that
$$\int_{0}^{t}|\sum_{k=1}^{n}\cos kx|\,dx<C~~~?$$
where $C$ is independent of $n$.
Here is my idea:
We know that
\begin{align}
\sum_{k=1}^{n}\cos kx&=\frac{\sin(n+\frac{1}{2})x}{2\sin... | $$\vert \cos kx \vert \leq 1 \implies \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq n \implies \int_0^t \bigg \vert \sum_{k = 1}^n \cos kx \bigg \vert \leq nt \implies nt < C \implies t < \frac{C}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can ... | Notice
$$(x^2 + y^2)^2 = 64 \implies x^4 + y^4 + 2(xy)^2 = 64$$
and
$$ (x + y )^2 = 4 \implies x^2 + y^2 + 2xy = 4 \implies 2xy = 4 - 8 = -4 \implies xy = -2
$$
$$ \therefore x^4 + y^4 = 64 - 2(xy)^2 = 64 - 2(-2)^2 = 56 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequalities In Algebra So the problems ask to find Find all values of $x$ for which
$$\frac{x}{x-4}<\frac{x-5}{x+1}.$$
So the solution requires moving both fractions to one side, finding a common denominator, combining, then factoring. Then you have a set of possible solutions for $x$ and you go on from there.
(If a... | $$\frac x{x-4}<\frac{x-5}{x+1}\iff \frac x{x-4}-\frac{x-5}{x+1}<0\iff 10\frac{x-2}{(x-4)(x+1)}<0$$
$$\text{ Multiplying the numerator & the denominator by } \frac{(x-4)^2(x+1)^2}{10} \text{ which is}>0$$
$$\frac x{x-4}<\frac{x-5}{x+1}\iff (x-4)(x+1)(x-2)<0$$
so we need odd number of factors $<0$
If all of the three a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Joint distribution of RVs involving rolls of die We roll a die until we get $4$ fives. Let $X$ be the number of rolls needed for the first $5$ and let $Y$ be the number of rolls needed to get the fourth five. What is the joint probability mass function of $X,Y.$
My attempt: $P(X=x \cap Y=y)=$ ${y-1}\choose{3}$$\left({\... | You need to do it in this way . Let $x$ be the number of tries required to get the first $5$ then $p_X(x) = (\frac{5}{6})^{x-1}\frac{1}{6}$ After this you need $3$ more $5$ . You can treat this as a completely new event , because it doesn't depend on what happened earlier .
Now let $y$ be the total number of trials ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}... | The convergence of $$\sum_{n=1}^N (-1)^n \dfrac{n^2}{\sqrt{n^5+1}}$$ can be concluded based on alternating test. The general result is termed as generalized alternating test or Dirichlet test and is based on Abel partial summation. We will prove the generalized statement, though this is a bit of a overkill for this pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/596053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Power series of trigonometric functions Problem statement: Determine those $x$, for which the power series is convergent and determine the sum.
$$f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$$.
Progress:
I have difficulty to handle the $2n$ inside the sum. I can clearly see, since $\sin x = \sum_{n=1}... | This is similar to Andre's observation but more explicit.
We have $f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$,
thus we can kill the $2n$ term by integrating under summation, giving us
$$
\begin{align*}
\int f(x) dx &= C + \frac{x^2}{2} + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n}}{(2n-1)!}
\\
&= C ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove using mathematical induction pt 2 Assumed that i asked a question like 30 min ago thinking i got the hang of this, seems not.
So $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac12n(6n^2-3n-1) \text{ for all } n\in\mathbb N$$
This time it seems way harder with the squares.
so i did the steps and got stuck on the 3rd step(Again)... | $$1^2+4^2+7^2+\dots+(3n-2)^2=\frac{1}{2}n(6n^2-3n-1), \forall n\in N$$
$$1^2+4^2+7^2+\dots+(3n-2)^2+(3n+1)^2=\frac{1}{2}n(6n^2-3n-1))+(3n+1)^2 $$
$$=\frac{1}{2}(6n^3-3n^2-n+2(3n+1)^2)=\frac{1}{2}(6n^3+15n^2+11n+2))= $$
$$\frac{1}{2}(6n^3+12n^2+6n+3n^2+5n+2)=$$
$$\frac{1}{2}(6n(n^2+2n+1)+3n^2+3n+2n+2)=$$
$$\frac{1}{2}(6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$... | $$T=\sqrt{2\sqrt{2\sqrt{2}}}...\\
\frac{T^2}{2}=T\\$$
T is a nonzero real number so:$$
T=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/601045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
How to find the limit of this equation? I have a Calc final coming up soon and am currently reviewing. I am currently stumped on finding the limit of as $x \rightarrow 1$, of $(1-\sqrt{2x^2 -1}) / (x-1)$.
Don't you have to multiply by the conjugate? Any help is seriously appreciated. I know the answer is $-2$, but I... | Here's a way to compute the limit without using derivatives but by multiplying by the conjugate:
\begin{align*}
\lim_{x\to1} \frac{1 - \sqrt{2x^2-1}}{x-1} \cdot \frac{1 + \sqrt{2x^2-1}}{1 + \sqrt{2x^2-1}}
&= \lim_{x\to1} \frac{1 - (2x^2-1)}{(x-1)(1 + \sqrt{2x^2-1})} \\
&= \lim_{x\to1} \frac{-2x^2 + 2}{(x-1)(1 + \sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$ How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x ... | The calculation will be easier if you showed zeros:
Subtract the first row from the second and third row and develop relative the first column we find
$$\det A=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{vmatrix}=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
0& \cos x(1-\cos x) & -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/602731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$ Find
$$\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$$
I got
$$\frac{2^{2n+1}(2n^2+n+1)-1}{(2n+1)(2n+2)}$$
but the answer is
$$\frac{2^{2n+1}(2n^2-n+1)-2}{(2n+1)(2n+2)}$$
Thanks for the help...
| $$\frac r{r+2}\binom{2n}r=\left(1-\frac2{r+2}\right)\binom{2n}r=\binom{2n}r-2\cdot\underbrace{\frac1{r+2}\binom{2n}r}_1$$
$$\text{Now, }\underbrace{\frac1{r+2}\binom{2n}r}_1=(r+1)\cdot\frac{(2n)!}{(2n-r)!\cdot(r+2)(r+1) \cdot r!}$$
$$=\frac{r+1}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{\{2n+2-(r+2)\}!\cdot(r+2)!} =\underbrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/603925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$ Evaluate the integral $$\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$$
| The answer given by Gordon is very good for computing the integral.
But it does not provide much why it works. The following figure might help with that
This is the integral for $\alpha=\sqrt{2}$. It looks as if the area under the function
is exactly half of the dashed rectangle... A good guess is therefore
$$
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/605673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 5,
"answer_id": 3
} |
Prove that :$\frac{1}{a+b} +\frac{1}{b+c} +\frac{1}{c+a}\ge \frac{4}{a^2+7} +\frac{4}{b^2+7} +\frac{4}{c^2+7}$ Let $a,b,c>0$ and satisfying $a^2+b^2+c^2=3$.
Prove that :$\dfrac{1}{a+b} +\dfrac{1}{b+c} +\dfrac{1}{c+a}\ge \dfrac{4}{a^2+7} +\dfrac{4}{b^2+7} +\dfrac{4}{c^2+7}$
| Use AM-GM we have :
$\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{2}{\sqrt{(a+b)(b+c)}}\ge\frac{4}{a+2b+c}\ge\frac{4}{\frac{a^2+1}{2}+b^2+1+\frac{c^2+1}{2}}=\frac{8}{(a^2+b^2+c^2)+4+b^2}=\frac{8}{b^2+7}$
Similar, the inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
| Alternatively $\rm\ \dfrac{2}{a-5} = \color{#c00}4+\color{#0a0}{3\sqrt2}\, <\, \color{#c00}5+\color{#0a0}{3\sqrt3} = \dfrac{2}{b-5}\ $ since $\rm\, \color{#c00}{4<5},\,\ \color{#0a0}{3\sqrt2 < 3\sqrt 3} $
Flipping gives $\rm\ a-5\, >\, b-5 \ \Rightarrow\ a > b\ \ $ QED
Remark $\ $ This is not pulled out of a hat. Rath... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Mod calculation
$$10x + 20 \equiv 11 \pmod{23}$$
So I know the answer is $6 \mod{23}$. I understand that we subtract $20$ from each side. But how do we do extended gcd on this?
Can you explain the steps involved?
| $$10x+20 \equiv 11 \pmod{23} \implies 10x \equiv -9 \pmod{23} 10x \equiv -9+23 \pmod{23}$$
$$10x \equiv -9+23 \pmod{23} \implies 10x \equiv 14 \pmod{23} \implies 5x \equiv 7 \pmod{23}$$
$$5x \equiv 7 \pmod{23} \implies 5x \equiv 7+23 \pmod{23} \implies 5x \equiv 30 \pmod{23}$$
Hence,
$$x \equiv 6 \pmod{23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
prove statement. cyclic sum complex numbers If $(a+b)^3 = (b+c)^3 = (c+a)³\,\,\,\,a,b,c\,\,\in\mathbb{C},\,\,a\neq b\neq c$, show that $a^3 = b^3 = c^3$.
Some hints would be great.
| As $\displaystyle a\ne c, a+b\ne b+c\implies \frac{a+c}{b+c}\ne1$
So, from $\displaystyle(a+b)^3 = (b+c)^3 = (c+a)^3\implies a+b=\omega(b+c)=\omega^2(c+a)$ where $\omega$ is one of the two complex cube root of unity
From $\displaystyle a+b=\omega(b+c),a+b(1-\omega)-c \omega=0\ \ \ \ (1)$
From $\displaystyle\omega(b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ a... | $$\sin2x= \frac{1}{2}$$
$$2x=n\pi+(-1)^n\pi/6$$
Therefore
$$x=\frac{n \pi}{2} + (-1)^n \frac{ \pi}{12} \text{ , } n=Z$$
$f(x)=\sin2x$
$f'(x)=2\cos2x$
where $m=f'(x)$ and c can be find by putting x and $y=\frac{1}{2}$
$y_{1}=\sqrt{3}x-\frac{\sqrt{3}\pi-6+12k\pi\sqrt{3}}{12} $ and $y_{2}=-\sqrt{3}x-\frac{5\sqrt{3}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Cantor set in base 3. I'm trying to prove the cantor set $C$ is equivalent to the set of all numbers with ternary expansion of $2$'s and $0$'s. That is:
Let $A_0=[0,1]$. $A_n$ is defined to equal $A_{n-1}$ with it's middle third removed. let $C=\bigcap_{n\in\mathbb{N}} A_n$.
Prove $C =\{x=0.a_1a_2a_3...| a_n\in\{0,2\}... | Let
$F_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$F_2 = [0,\frac{1}{3^2}] \cup [\frac{2}{3^2},\frac{3}{3^2}] \cup [\frac{6}{3^2},\frac{7}{3^2}] \cup [\frac{8}{3^2},1]$
and so on.
The Cantor set is then $C=\bigcap^{\infty}_{k=1} F_k$.
Each $x \in C$ can be written in the form $ x = \frac{a_0}{3} + \frac{a_1}{3^2} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\righ... | Now take out the common factor $\displaystyle\cos\frac{B-C}2$
and apply Prosthaphaeresis Formulas on $\displaystyle\sin\frac{2A-B-C}2+\sin\frac{B-C}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/611324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value. Let $x$ be in the set of real numbers $\mathbb{R}$ and let $f(x)=|2x-1|-3|2x+4|+7$ be a function, write $f(x)$ without the absolute value.
I thought of it this way: $$f(x)=\begin{ca... | The absolute value has a discontinuous slope change at zero. Hence, f(x) has discontinuous slope changes at -2 from $|2x+4|$ and at 1/2 from $|2x-1|$. So there are three cases: $-\infty < x \le -2$, $-2 < x \le 1/2 $ and $1/2 < x < \infty$.
Now for $x$ sufficiently negative $|2x+4| = -2x-4$ and we know that $|2x+4|$ ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$ Prove that for any odd natural number $n$, the number $1^{2007}+2^{2007}+\cdots+n^{2007}$ is not divisible by $n+2$.
| By taking modulo $n + 2$, and partition the terms as groups of two
each,
$$\begin{align*}
1^{2007}+2^{2007}+3^{2007}+\cdots n^{2007} \\
\equiv1+\big\lfloor 2^{2007}+(n)^{2007}\big\rfloor+\cdots +\cdots \big\lfloor \ \Big(\frac{n+1}{2}\Big)^{2007}+\Big(\frac{n+3}{2}\Big)^{2007}\big\rfloor\\ \equiv1+\big\lfloor 2^{2007}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/617428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont... | HINT:
$$\frac{4x+13}{2x^2+7x+5}=0\implies 4x+13=0$$
$$\frac{4x+13}{2x^2+7x+5}<0 \iff (4x+13)(2x^2+7x+5)<0$$
$$\iff(4x+13)(2x+5)(x+1)<0\iff\left(x+\frac{13}4\right)\left(x+\frac52\right)(x+1)<0$$
So, we need odd number of factor(s) to be $<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Solving inequation with two absoulte values I need to solve the following inequation:
$$
|x| \cdot |x-1|-1>-x\\
$$
I cant get the correct result.
I tried to solve it like this:
$$
|x| \cdot |x-1|-1>-x
$$
I know that I can write $|x \cdot y|=|x| \cdot |y|$, so:
$$
\begin{eqnarray}
|x(x-1)|-1&>&-x\\
|x^2-x|-1&>&-x \qquad... | You should distinguish three cases:
*
*when $x<0$, $|x|=-x$ and $|x-1|=1-x$;
*when $0\le x\le 1$, $|x|=x$ and $|x-1|=1-x$;
*when $x>1$, $|x|=x$ and $|x-1|=x-1$.
Thus you have to solve
\begin{align}
&\begin{cases}
(-x)(1-x)-1>-x\\
x<0
\end{cases}\\
&\begin{cases}
x(1-x)-1>-x\\
0\le x\le 1
\end{cases}\\
&\begin{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve a 2nd order non-homogeneous linear recurrence? I have a problem in solving this equation :
$x_{n+2} + 3\ x_{n+1} + 2\ x_{n} = 5 \times 3^n $
given that $x_{0} = 0$ and $x_{1} = 1$.
I solved the homogeneous associated equation and got $v_{n} = c_{1} \times (-1)^{n} + c_{2} \times (-2)^{n}$ (where $c_{1}$ an... | Use generating functions directly. Define $G(z) = \sum_{n \ge 0} x_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognize a few sums:
$$
\frac{G(z) - x_0 - x_1 z}{z^2} + 3 \frac{G(z) - x_0}{z} + 2 G(z)
= 5 \frac{1}{1 - 3 z}
$$
As partial fractions:
$$
A(z) = \frac{1}{4} \cdot \frac{1}{1 - 3 z} - \frac{1}{4} \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving quadratic system If $a,b,c\in \mathbb{R}$ satisfy the system
$a^2+ab+b^2=9$;
$b^2+bc+c^2=16$;.
$c^2+ac+a^2=25$.
Find $ab+ac+bc$
| If you subtract the first equation from the second, you get $c^2-a^2 + bc - ab = 7,$ so $(c-a)(c+a +b) = 7.$ You get similar equations (all with an $a+b+c$ factor) when you subtract the second from the third, etc, which gives you a linear system in $a, b, c$ (you know $a+b+c \neq 0,$ from the first sentence).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the pos... | As $\displaystyle10\equiv1\pmod3, 10^n\equiv1\pmod3$ for integer $n\ge0$
As $\displaystyle a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod{m\cdot c }$ where $a,b,m,c$ are integers
$\displaystyle10\cdot10^n\equiv10\pmod{3\cdot10}\equiv10\pmod6$ as $6|30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Determine the smallest positive value of x(in degrees) for which: $\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$ Determine the smallest positive value of x(in degrees) for which:
$\tan(x+100^{\circ}) = \tan(x+50^{\circ})\tan (x)\tan(x-50^{\circ})$
I tried to apply the formula of $\tan(A+B) = \frac... | Given $\displaystyle \tan(x+100^0) = \tan(x+50^0)\cdot \tan (x)\cdot \tan(x-50^0)$
$\displaystyle \Rightarrow \frac{\tan(x+100^0)}{\tan(x-50^0)}\Rightarrow =\tan(x+50^0)\cdot \tan(x^0)$
$\displaystyle \Rightarrow \frac{\sin(x+100^0)\cdot\cos(x-50^0)}{\cos(x+100^0)\cdot \sin (x-50^0)}=\frac{\sin(x+50^0)\cdot\sin(x)}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/621213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
probability using permutations what is the probability that if we pick n natural numbers then their product's unit digit contains $2,4,6,8$.and i have no idea how to start ....so please mention all the steps...
i thought that last digit may be $0-9$ and fav cases are $4$..so prob should be $4/10=2/5$..but that is wro... | We assume that our $n$ numbers are independently chosen, and that each has final digit equally likely to be $0,1,2,3,\dots,9$.
The product ends in $2$, $4$, $6$, or $8$ precisely if (i) the digits $0$ or $5$ are not chosen and (ii) the digits are not all odd.
The probability $0$ or $5$ are never chosen is $\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$ let $x,y,z\ge 0$,and such
$$xy+yz+xz=xyz+2$$
show that
$$x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$$
my try: let $x+y+z=p,xy+yz+xz=q, xyz=r$
then
$$q=r+2$$
show that
$$p^2-2q+r(p-2)\ge 4$$
then I can't,Thank you
| WLOG, $z=$Min{$x,y,z$},$z=\dfrac{2-xy}{x+y-xy}, t^2=xy,p=x+y, \implies p^2 \ge 4t^2,p\ge2t$
$z\ge 0 \implies (2-t^2)(p-t^2)\ge 0 \implies p\ge 2, t^2 \le2 $.
(if $p<2, t^2\le \dfrac{1}{2} \implies z>1$, but $z\le \dfrac{x+y}{2} < 1, $ contradict.)
(if $t^2>p\ge2 ,z=\dfrac{t^2-2}{t^2-p}<\dfrac{p}{2} \iff t^2 >\dfrac{p^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
how to integrate : $\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$ Need to compute:
$$\int_{x^2}^{x^3}\frac{dt}{\sqrt{1+t^4}}$$
I tried to use partial fraction but got a messy algebra.
thanks.
| For any $x \in [0,\infty]$, let $I(x)$ and $J(x)$ be the integrals
$$
I(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}
\quad\text{ and }\quad
J(x) = \int_{x^2}^{x^3} \frac{dt}{\sqrt{1+t^4}} = I(x^3) - I(x^2)
$$
Notice under the substitution $t = \frac{1}{u}$,
$$\frac{dt}{\sqrt{1+t^4}} = - \frac{du}{\sqrt{1+u^4}}$$
We find $I(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find this $DF=?$ In diamond $ABCD$,such $\angle B=\dfrac{\pi}{3}$,and the point $E$ in on $BC$.such
$BE=3CE$,and the point $F$ is on $DE$,such $\angle AFC=\dfrac{2\pi}{3}$
Find $$DF=?$$
My try: since
$$\angle B+\angle AFC=\pi$$
so
$A,B,C,F$ is cyclic
and follow I can't
| Let's draw another equilateral triangle $BCG.$
Let $O$ be the center of that triangle.
Let's show that points $D, E$ and $O$ are collinear $\Longleftrightarrow$ points $D, F$ and $O$ are collinear.
$\overrightarrow{DE}=\overrightarrow{DC}+\overrightarrow{CE}=\overrightarrow{DC}+\dfrac{1}{4}\overrightarrow{CB},$
$\overr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/626785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"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.