Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating $\cos\frac\pi4$ from the half-angle formula gives $\sqrt{\frac12}$ instead of $\frac{\sqrt{2}}2$. What went wrong? I am using the formula $$\cos\left(\frac x2\right)=\sqrt{\frac{1+\cos(x)}2}$$ to find $\cos\left(\frac{pi}{4}\right)$ but it does not give me the correct result.
$$
\cos\left(\frac{\pi}{4}\right)
= \sqrt{\frac{1+\cos\left(\frac{\pi}{2}\right)}2}
= \sqrt{\frac{1+0}2}
= \sqrt{\frac12}
$$
This contradicts $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $. What did I do wrong?
| Nothing: $\dfrac{\sqrt2}2=\dfrac{\sqrt2}{\sqrt4}=\sqrt{\dfrac24}=\sqrt{\dfrac12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1} u + C$.
Sympy fails on the definite integral and gives me this weird RootSum expression for the indefinite integral:
$$\operatorname{RootSum} {\left(t^{4} \left(4096 \zeta^{8} - 8192 \zeta^{6} + 4096 \zeta^{4}\right) + t^{2} \left(256 \zeta^{6} - 384 \zeta^{4} + 128 \zeta^{2}\right) + 1, \left( t \mapsto t \log{\left (- 512 t^{3} \zeta^{6} + 768 t^{3} \zeta^{4} - 256 t^{3} \zeta^{2} - 32 t \zeta^{4} + 32 t \zeta^{2} - 4 t + u \right )} \right)\right)}$$
Wolfram Alpha gives me the following for the indefinite integral :
$$\begin{align}
& \frac{\frac{1}{a_1}\tan^{-1} \frac{u}{a_1} - \frac{1}{a_2}\tan^{-1} \frac{u}{a_2}}{4\zeta\sqrt{\zeta^2-1}} + C \\
\\
a_1 &= \sqrt{2\zeta^2-2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b-c}\\
a_2 &= \sqrt{2\zeta^2+2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b+c}\\
\end{align}$$
(with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$) but I'm a bit lost how it got there, and then I'm not exactly sure what to do if $\zeta \le 1$ (is the formula still valid?!)
edit: OK, partial fraction expansion is sloooowwwwly coming back to me. It looks like $a_1a_2 = 1$ and $a_1{}^2 + a_2{}^2 = 4\zeta^2-2$, so I guess they used the expansion
$$
\frac{1}{u^{4} + \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$
| Let $I$ denote the integral. Then using the identity
$$ u^4 + (4\zeta^2-2)u^2 + 1 = u^2 \left( ( u - u^{-1} )^2 + 4\zeta^2 \right), $$
we may write
$$ I = 2 \int_{0}^{\infty} \frac{u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{1} $$
Now applying the substitution $u \mapsto u^{-1}$,
$$ I = 2 \int_{0}^{\infty} \frac{1}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{2} $$
Averaging $\text{(1)}$ and $\text{(2)}$, we get
\begin{align*}
I
&= \int_{0}^{\infty} \frac{1 + u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \frac{1}{t^2 + 4\zeta^2} \, \mathrm{d}t \tag{$t=u-u^{-1}$} \\
&= \frac{\pi}{2|\zeta|}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
I need help with this probability question I am working on a probability question, and I'm getting stuck at what to do next. Here's the question:
You and a friend are playing a game with a fair coin. You go first, and you each take turns flipping the coin, and record what the coin lands on. A player wins the game when they get two consecutive heads. What is the probability that you win?
Here's what I tried:
If you get H and the friend gets T, then the probability you get H is $\frac18$. Similarly, the friend could get H and you would win if you get H, so that's another $\frac18$. If you get T and your friend gets H, then if you get H, your friend must get T for you to win, and you could get H on your third flip, giving $\frac{1}{32}$. If you get T and your friend gets T, then you could get H, and so on. As you can see, the cases get out of control, and I don't know how to actually use these cases, since there's an infinite amount. Can someone help?
| We consider the last $2$ results of tossing the coin: yours $a$ and your friend's $b$ (let heads be $1$ and tails $0$), suppose now is your turn again.
Then we consider $6$ states of the game: $A(0,0),\,B(0,1),\,C(1,0),\,D(1,1)$, $E$ -- the player that have just tossed the coin wins, $F$ -- the state the game continiues after winning/losing (for easy summation after).
Let $\mathbf{x}_n=(p(A),p(B),p(C),p(D),p(E),p(F))^T$ after tossing the coin $n$ times.
It's clear, that $\mathbf{x}_2=\left(\frac14,\frac14,\frac14,\frac14,0,0\right)$ and $\mathbf{x}_{n+1}=A\mathbf{x}_n$ where
$$A=\begin{pmatrix}
\frac12&0&\frac12&0&0&0\\
\frac12&0&0&0&0&0\\
0&\frac12&0&\frac12&0&0\\
0&\frac12&0&0&0&0\\
0&0&\frac12&\frac12&0&0\\
0&0&0&0&1&1
\end{pmatrix}$$
It's left only to compute $A=SJS^{-1}$ and perform summation $M=S\left(
\sum\limits_{n=0}^{\infty}J^{2n+1}
\right)S^{-1}$ so the answer will be $(0,0,0,0,1,0)M\mathbf{x}_2$ (as was done in these answers: 1, 2, 3). WA gives
$$S=\begin{pmatrix}
0 & -\frac{i}{2} & \frac{i}{2} & 0 & \frac{1}{4} (1 + \sqrt{5}) & \frac{1}{4} (1 - \sqrt{5})\\
0 & \frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{4} (-3 - \sqrt{5}) & \frac{1}{4} (-3 + \sqrt{5})\\
0 & -\frac{1}{2} + \frac{i}{2} & -\frac{1}{2} - \frac{i}{2} & 0 & \frac{1}{4} (-3 - \sqrt{5}) & \frac{1}{4} (-3 + \sqrt{5})\\
0 & \frac{i}{2} & -\frac{i}{2} & 0 & \frac{1}{2} (2 + \sqrt{5}) & 1 - \frac{\sqrt{5}}{2}\\
-1 & -1 - \frac{i}{2} & -1 + \frac{i}{2} & 0 & \frac{1}{4} (-3 - \sqrt{5}) & \frac{1}{4} (-3 + \sqrt{5})\\
1 & 1 & 1 & 1 & 1 & 1\end{pmatrix}\\
J=\operatorname{diag}\left(
0,-\frac{i}{2},\frac{i}{2},1,\frac{1}{4} (1 - \sqrt{5}),\frac{1}{4} (1 + \sqrt{5})
\right)$$
And the answer is $\frac{14}{25}$ (computations here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving the existence of infinitely many numbers with a given property in a non-increasing sequence.. $\textbf{Question:}$
Let $a_1, a_2, a_3,\cdots$ be a nonincreasing sequence of positive real numbers such that
$a_n \ge a_{2n} + a_{2n+1}$ for all $n \ge 1$.
Show that there exist infinitely many positive integers $m$ such that
$2m \cdot a_m > (4m − 3) \cdot a_{2m−1}.$
I haven't made any good progress in this problem.I noticed that,if $a_n=a_{n-1}$ infinitely many times,then it holds true.Any kind of hint or full solution both are appreciated.
| I think this works do tell me if you found an error
Notice that $2^{n+1} - 2^{n} = 2^{n}$ we will use this. Now let on contrary the above condition is true for only finitely many positive integers, thus there exist $ K $ such that for all $m \geq K$ we have
$$2m\cdot a_m \leq (4m-3)\cdot a_{2m-1}$$ Now consider the index of the sequence $(a)_N$ from
$(2^{r} \cdot m + 1)_{r \in N}$ for sufficiently large $m \geq K$. Now consider
$$2 \cdot (2^{r} \cdot m + 1) \cdot a_{2^r \cdot m + 1} \leq (4 \cdot (2^r \cdot m + 1) -3) \cdot a_{2^{r+1} \cdot m + 1}$$
which gives
$$2 \cdot (2^{r} \cdot m + 1) \cdot a_{2^r \cdot m + 1} \leq (2^{r+2} \cdot m + 1) \cdot a_{2^{r+1} \cdot m +1}$$
hence
$$ \frac{a_{2^{r+1} \cdot m + 1}}{a_{2^r \cdot m + 1}} \geq \frac{2 \cdot (2^r \cdot m + 1)}{(2^{r+2} \cdot m +1)}$$
now consider the inequalities for $ r \in \{0, 1, 2, 3, \cdots , n\}$
we can multiply them to get
$$\frac{a_{2^{n+1} \cdot m + 1}}{a_{m+1}} \geq 2^{n+1} \cdot (m+1) \cdot (2m+1) \cdot \left( \frac{1}{(2^{n+1} \cdot m + 1) \cdot (2^{n+2} \cdot m + 1) } \right)$$
hence
$$ a_{2^{n+1} \cdot m + 1} \geq a_{m+1} \cdot (m+1) \cdot (2m+1) \cdot \left( \frac{1}{(2^{n+1} \cdot m + 1)} - \frac{1}{(2^{n+2} \cdot m + 1)} \right) \cdot \frac{1}{m}$$
Let $$ p_{n + 1} \coloneqq \left( \frac{1}{(2^{n+1} \cdot m + 1)} - \frac{1}{(2^{n+2} \cdot m + 1)} \right) \cdot \frac{(m+1) \cdot (2m+1)}{m} $$
hence we have
$$ a_{2^{n} \cdot m + 1} \geq a_{m+1} \cdot p_{n}$$
now also from the first condition we have
$$a_{n} \geq a_{2n} + a_{2n+1}$$ putting $n = 2^{r-1} \cdot m$ we get
$$a_{2^{r-1} \cdot m} \geq a_{2^r \cdot m} + a_{2^r \cdot m + 1}$$
hence
$$a_{2^{r-1} \cdot m} - a_{2^r \cdot m } \geq a_{2^r \cdot m + 1}$$
summing both side from $r = 1 $ to $ r = n $ we get
$$\sum_{r = 1}^{n} a_{2^r \cdot m + 1} \leq a_{m} - a_{2^n \cdot m}$$
$$\Rightarrow a_m \geq \sum_{r = 1}^{n} a_{2^r \cdot m + 1} + a_{2^ n \cdot m} $$
$$\Rightarrow a_m \geq \sum_{r = 1}^{n} a_{2^r \cdot m + 1} + a_{2^ {n+1} \cdot m} $$
because $(a_n)$ is non increasing hence we have
$$a_{m} \geq \sum_{r = 1}^{n+1} a_{2^r \cdot m + 1} $$
$$a_{m} \geq \sum_{r = 1}^{n+1} a_{2^r \cdot m + 1} \geq \sum_{r = 1}^{n+1} (a_{m + 1} \cdot p_{r}) $$
giving
$$a_m \geq \frac{a_{m+1} \cdot (m+1) \cdot (2m+1)}{m} \cdot \left(\frac{1}{2m+1} - \frac{1}{2^{n+2}\cdot m + 1} \right)$$\
now as $n \rightarrow \infty$ we get
$$a_{m} \cdot m \geq a_{m+1 \cdot (m+1)}$$
$$\Rightarrow a_{m} \cdot m \geq a_{m+1 \cdot (m+1)} \geq a_{m+2} \cdot (m+2) \cdots \geq a_{2m-1} \cdot (2m-1)$$
Notice this happens only for sufficiently large $m \geq K$
$$2 \cdot a_{m} \cdot m \geq a_{2m-1} \cdot (4m-2) > a_{2m-1} \cdot (4m-3) $$
contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\sum_{j=n}^{n+i}\frac{\binom{n+i}j}{2^{n+i}}=0$ I found the following question online: How can I prove that
$$\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\sum_{j=n}^{n+i}\frac{\binom{n+i}j}{2^{n+i}}=0$$
?
One notices that the inner sum is equal to the probability $\mathsf P\left(\mathrm B\left(n+i;\frac12\right)\geq n\right)$, where $\mathrm B$ denotes the binomial distribution. Using Hoeffding's inequality, one gets $\mathsf P\left(\mathrm B\left(n+i;\frac12\right)\geq n\right)\le\exp\left(-\frac{(n-i)^2}{2(n+i)}\right)$, i.e.
$$\tag1\label1\frac1n\sum_{i=0}^{n-1}\sum_{j=n}^{n+i}\frac{\binom{n+i}j}{2^{n+i}}\le\frac1n\sum_{i=0}^{n-1} \exp\left(-\frac{(n-i)^2}{2(n+i)}\right).$$
Based on numerical experiments, the right-hand side converges to $0$. If you apply $\exp(-x)\le\frac{1}{1+x}$, you get $$\tag2\label2\frac1n\sum_{i=0}^{n-1} \exp\left(-\frac{(n-i)^2}{2(n+i)}\right)\le\frac1n\sum_{i=0}^{n-1} \frac{1}{1+\frac{(n-i)^2}{2(n+i)}},$$
and the right-hand side still seems to converge to $0$. However, it is 2am so I lack the stamina to find a proof for this. I am asking for a sketch of proof that either the right-hand side in \eqref{1}, or even better, the right-hand side in \eqref{2} converges to $0$.
Note: Here, I answered a similar question.
| We have
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1} \frac{1}{1 + \frac{(n-i)^2}{2(n+i)}}
&= \frac{1}{n}\sum_{i=0}^{n-\lfloor \sqrt{n} \rfloor} \frac{1}{1 + \frac{(n-i)^2}{2(n+i)}}
+ \frac{1}{n}\sum_{i=n+1-\lfloor \sqrt{n} \rfloor}^{n-1} \frac{1}{1 + \frac{(n-i)^2}{2(n+i)}}\\
&\le \frac{1}{n}\sum_{i=0}^{n-\lfloor \sqrt{n} \rfloor} \frac{1}{0 + \frac{(n-i)^2}{2(n+n)}}
+ \frac{1}{n}\sum_{i=n+1-\lfloor \sqrt{n} \rfloor}^{n-1} \frac{1}{1 + 0}\\
&= 4 \sum_{i=0}^{n-\lfloor \sqrt{n} \rfloor}\frac{1}{(n-i)^2} + \frac{\lfloor \sqrt{n} \rfloor - 1}{n}\\
&= 4 \sum_{m=\lfloor \sqrt{n} \rfloor}^n \frac{1}{m^2} + \frac{\lfloor \sqrt{n} \rfloor - 1}{n}.
\end{align}
From $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}$,
we know that $\lim_{n\to \infty} 4 \sum_{m=\lfloor \sqrt{n} \rfloor}^n \frac{1}{m^2} = 0$.
Also, clearly, $\lim_{n\to \infty} \frac{\lfloor \sqrt{n} \rfloor - 1}{n} = 0$.
The desired result follows. (Q. E. D.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove there are finitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y^2}$ Prove there are finitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y^2}$ where $d$ is a non-square natural number. I know there're infinitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y}$ according to Dirichlet's approximation theorem. Any hints?
| I will show that
there are only a finite number of solutions
if
$|x-y\sqrt{d}|
\lt \dfrac1{y\,f(y)}
$
where
$f(y) \to \infty$,
$f(1) > 0$,
and $f'(y) > 0
$.
If
$f(y) = y$
(this question)
then
$y
\lt \sqrt{d}+\sqrt{d+1}
$.
If
$f(y) = y^c$
with $c > 0$
then
$y
\le (4\sqrt{d})^{1/c}
$.
If
$f(y) = \ln(y)$
then
$y
\le e^{4\sqrt{d}}
$.
If $d$ is not a square then
$|x^2-dy^2|
\ge 1$
so
$1
\le |x^2-dy^2|
=|(x-y\sqrt{d})(x+y\sqrt{d})|
$
so
$|x-y\sqrt{d}|
\ge|\dfrac1{x+y\sqrt{d}}|
$.
If
$|x-y\sqrt{d}|
\lt \dfrac1{y^2}
$
then
$-\dfrac1{y^2}
\lt x-y\sqrt{d}
\lt \dfrac1{y^2}
$
and
$\dfrac1{y^2}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y^2
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y^2})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$y^2-2y\sqrt{d}
\lt 1
$
so that
$(y-\sqrt{d})^2
=y^2-2y\sqrt{d}+d
\lt d+1
$
so
$y
\lt \sqrt{d}+\sqrt{d+1}
$.
More generally,
if
$|x-y\sqrt{d}|
\lt \dfrac1{y^{1+c}}
$
where $c > 0$
then
$-\dfrac1{y^{1+c}}
\lt x-y\sqrt{d}
\lt \dfrac1{y^{1+c}}
$
and
$\dfrac1{y^{1+c}}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y^{1+c}
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y^{1+c}})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$1
\gt y^{1+c}-2y\sqrt{d}
= y^{1+c}(1-\dfrac{2\sqrt{d}}{y^c})
$
so that,
if
$y^c
\gt 4\sqrt{d}
$
then
$1
\gt \dfrac{y^{1+c}}{2}
$
which is false.
Therefore
$y \le (4\sqrt{d})^{1/c}
$.
Even more,
if
$|x-y\sqrt{d}|
\lt \dfrac1{y\ln(y)}
$
then
$-\dfrac1{y\ln(y)}
\lt x-y\sqrt{d}
\lt \dfrac1{y\ln(y)}
$
and
$\dfrac1{y\ln(y)}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y\ln(y)
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y\ln(y)})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$1
\gt y\ln(y)-2y\sqrt{d}
= y\ln(y)(1-\dfrac{2\sqrt{d}}{\ln(y)})
$
so that,
if
$\ln(y)
\gt 4\sqrt{d}
$
or
$y
\gt e^{4\sqrt{d}}
$
then
$1
\gt \dfrac{y\ln(y)}{2}
$
which is false.
Therefore
$y \le e^{4\sqrt{d}}
$.
This works for
$|x-y\sqrt{d}|
\lt \dfrac1{y\,f(y)}
$
where
$f(y) \to \infty$,
$f^{(-1)}(y) \to \infty$,
$f(1) > 0$,
and $f'(y) > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of infinite series by considering Maclaurin Series for $e^x$ I want to find:
$ \sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!}$
We have:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^n(n!)}$
which we could integrate on both sides to get:
$ \int e^\frac{x}{3} dx = \sum_{n = 0}^{ \infty } \frac{ x^{n+1} }{3^n({n+1}!)}$
Now, I'm just missing the numerator $1505n+1506$, which I cannot for the life of me see how to proceed... Any hints would be appreciated!
$\textbf{Problem resolved using hint provided below:}$
$\sum_{n = 0}^{ \infty } \frac{1505n + 1506}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505(n+1) + 1}{3^n(n+1)!} = \sum_{n = 0}^{ \infty } \frac{1505}{3^nn!} + \sum_{n = 0}^{ \infty } \frac{1}{3^n(n+1)!}$
The first term can be computed using:
$e^\frac{x}{3} = \sum_{n = 0}^{ \infty } \frac{x^n}{3^nn!}$
, where we set $x = 1$ to get: $1505e^\frac{1}{3}$
The second term can be computed by taking the definte integral of $e^\frac{x}{3}$:
$ \intop\nolimits_{0}^{1} e^\frac{x}{3} dx = \sum_{n = 0}^{ \infty } \frac{ 1^{n+1} }{3^n({n+1}!)} - \frac{ 0^{n+1} }{3^n({n+1}!)} = \sum_{n = 0}^{ \infty } \frac{1}{3^n{(n+1)}!} = 3e^\frac{1}{3}-3$
Sum = $1508e^\frac{1}{3}-3$
| $$S=\sum_{n=0}^{\infty} \frac{an+b}{3^n (n+1)!}= \sum_{n=0}^{\infty} \frac{(a(n+1)+b-a}{3^n (n+1)!}=a\sum_{n=0}^{n}\frac{3^{-n }}{n!}+(b-a)\sum_{n=0}^{\infty} \frac{3^{-n}}{(n+1)!}$$
$$\implies S=ae^{-3}+3(b-a)\sum_{n=0}^{\infty}\frac{3^{-(n+1)}}{(n+1)!}=ae^{-3}+3(b-a)[e^{-3}-1].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show the $\arcsin$ identity: $ \arcsin(1 - 2x) + 2\arcsin(\sqrt{x}) = \pi / 2$ Can somebody find an elementary proof of the following identity:
$$
\arcsin ( 1 - 2x) + 2 \arcsin(\sqrt{x}) = \frac\pi2
$$
I noticed it while solving the following integral:
$$ I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} = -2 \sqrt{1 - x}\sqrt{x} + \int \frac{\sqrt{1 - x}}{\sqrt{x}}
$$
where the first equality follows after applying an integration by parts with $f = \sqrt{x}$ and $\mathrm{d} g = 1/\sqrt{1 - x}$. For the sake of simplicity we omit $\mathrm{d}x$ in each integral. Adding the integral to itself:
\begin{align} 2I = \int \frac{\sqrt{x}}{\sqrt{1 - x}} &= -2 \sqrt{1 - x}\sqrt{x} + \int \left(\frac{\sqrt{1 - x}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{1- x}}\right) \\&= -2 \sqrt{1-x}\sqrt{x} + \int\frac{1}{\sqrt{x}\sqrt{1-x}}\end{align}
The last integral on the RHS evaluates to $\arcsin(1 - 2x)$, so $$I = - \sqrt{1-x}\sqrt{x} + \frac{1}{2}\arcsin(1 - 2x) + C.$$
On the other hand, the integral can also be evaluated by applying a $u$-sub with $u = \sqrt{x}$. We find that:
$$
I = 2 \int\frac{u^2}{\sqrt{1 - u^2}} =2 \left(-u \sqrt{1 - u^2} + \int\sqrt{1 - u^2}\right) = - u\sqrt{1 - u^2} + \arcsin u + C_2
$$
So then it follows that $I$ is also equal to $-\sqrt{x}\sqrt{1-x} + \arcsin{\sqrt{x}} + C_2$. Equate the results of the two methods and plug in a random point to find $C - C_2$ and the subsequent identity.
| Let $f(x)=\arcsin ( 1 - 2x) + 2 \arcsin\sqrt x$ where $x\in[0,1]$ so that \begin{align}f'(x)&=\frac1{\sqrt{1-(1-2x)^2}}\cdot(-2)+2\cdot\frac1{\sqrt{1-\sqrt x^2}}\cdot\frac1{2\sqrt x}\\&=-\frac2{\sqrt{4x-4x^2}}+\frac1{\sqrt{x(1-x)}}\\&=-\frac1{\sqrt{x(1-x)}}+\frac1{\sqrt{x(1-x)}}=0\end{align} for all $x\in(0,1)$. Hence in this interval, we have that $f$ is constant. Since $f(0)=f(1/2)=f(1)=\pi/2$, we conclude that $$\arcsin ( 1 - 2x) + 2 \arcsin\sqrt x=\frac\pi2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Asymptote of $x^2 + y^2 -3xy = 1$ How can I find the asymptote of the hyperbola $x^2 + y^2 -3xy = 1$?
I tried to convert the equation to $y=±\sqrt{1+3xy−x^2}$ and mark $x=\infty$ and I got $y=\pm\sqrt{3xy−x^3}$ but I don't know how to continue.
| We obtain $$y=\frac{3x+\sqrt{(9x)^2-4(x^2-1)}}{2}$$ or
$$y=\frac{3x-\sqrt{5x^2+4}}{2}.$$
Now, we'll work with $y=\frac{3x+\sqrt{5x^2+4}}{2}.$
For $y=\frac{3x-\sqrt{5x^2+4}}{2}$ we'll get a same result.
$$\lim_{x\rightarrow+\infty}\frac{y}{x}=\frac{1}{2}\lim_{x\rightarrow+\infty}\left(3+\sqrt{5+\frac{4}{x^2}}\right)=\frac{3+\sqrt5}{2}$$ and
$$\lim_{x\rightarrow+\infty}\left(y-\frac{3+\sqrt5}{2}x\right)=\lim_{x\rightarrow+\infty}\frac{\sqrt{5x^2+4}-\sqrt5x}{2}=\lim_{x\rightarrow+\infty}\frac{2}{\sqrt{5x^2+4}+\sqrt5x}=0.$$
Id est, $y=\frac{3+\sqrt5}{2}x$ is an asymptote.
Now, $$\lim_{x\rightarrow-\infty}\frac{y}{x}=\frac{1}{2}\lim_{x\rightarrow-\infty}\left(3-\sqrt{5+\frac{4}{x^2}}\right)=\frac{3-\sqrt5}{2}$$ and
$$\lim_{x\rightarrow-\infty}\left(y-\frac{3-\sqrt5}{2}x\right)=\lim_{x\rightarrow-\infty}\frac{\sqrt{5x^2+4}+\sqrt5x}{2}=\lim_{x\rightarrow-\infty}\frac{2}{\sqrt{5x^2+4}-\sqrt5x}=0.$$
Id est, $y=\frac{3-\sqrt5}{2}x$ is an asymptote.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding $\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$ with Fubini's theorem I have to solve the following double integral
$$\iint_{A}\frac{dx\,dy}{(1+x^2)(1+x^2 y^2)}$$
with $A= \left[0,+\infty\right[ \times [0,1].$ So far I've tried to solve it integrating w.r.t. $y$ first.
$$\iint_0^1\frac{dy\,dx}{(1+x^2)(1+x^2 y^2)} = \int_0^\infty\frac{1}{1+x^2}\int_0^1\frac{dy}{1+x^2 y^2} \, dx. $$
I've solved the internal integral by substitution, remembering that $\int\frac{du}{1+u^2}=\arctan u$
Substitution:
$$x^2 y^2= u^2 \to y=\frac{1}{x}u \to dy=\frac{1}{x}du.$$
$$y=0 \to u=0, \qquad y=1 →u=x$$
So:
\begin{align}
& \int_0^∞\frac{1}{1+x^2} \left( \int_0^x\frac{1}{x}\frac{1}{1+u^2}\,du \right) \, dx \\[8pt]
= {} & \int_0^\infty\frac{1}{x}\frac{1}{1+x^2}[\arctan u] \, dx \\[8pt]
= {} & \int_0^\infty\frac{\arctan x}{x(1+x^2)} \, dx.
\end{align}
Now I have to solve this last integral with the Fubini's theorem but I don't know how to do it.
| Following Michael Hardy's suggestion, I will use partial fractions. We will have
$$\frac{1}{(1+x^2)(1+x^2y^2)}=\frac{1}{1-y^2}\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)$$
So
$$
\begin{eqnarray}
\iint_{A=[0,\infty[\times[0,1]}\frac{dx\,dy}{(1+x^2)(1+x^2y^2)}&=&\int_0^1\frac{1}{1-y^2}\left[\int_0^\infty\left(\frac{1}{1-x^2}-\frac{y^2}{1+x^2y^2}\right)dx\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\arctan x|_0^\infty-y\arctan (xy)|_0^\infty\right]dy \\
&=&\int_0^1\frac{1}{1-y^2}\left[\frac{\pi}{2}-y\frac{\pi}{2}\right]dy \\
&=&\frac{\pi}{2}\int_0^1\frac{dy}{1+y}=\left.\frac{\pi}{2}\ln|1+y|\right|_0^1 \\
&=&\frac{\pi}{2}\ln 2
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$
| If we call our sum $S$, then we see that
\begin{align*}
S&=\sum_{n=1}^{\infty}\left(\frac{1}{4n+1}-\frac{1}{4n}\right)\\
&=-\frac{1}{4}\left(\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+\frac{1}{4}}\right)-\gamma+\gamma\right)
\end{align*}
where $\gamma$ is the Euler-Mascheroni constant. The reason that I have done this reformulation of the problem is that there is a well known function called the digamma function defined by
$$\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$$
where $\Gamma(s)$ is the gamma function. It is well known that
$$\psi(s+1)=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+s}\right)-\gamma$$
and so we get now that
$$S=-\frac{1}{4}\left(\psi(5/4)+\gamma\right)$$
Gauss's formula relates values of the digamma function at rational numbers to a finite number of elementary functions, namely that
$$\psi\left(\frac{r}{m}\right)=-\gamma-\ln(2m)-\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right)+2\sum_{n=1}^{\lfloor (m-1)/2\rfloor}\cos\left(\frac{2\pi nr}{m}\right)\ln\left(\sin\left(\frac{\pi n}{m}\right)\right)$$
This formula is only valid for $r<m$, and so by using the reccurence relation
$$\psi(s+1)=\psi(s)+\frac{1}{s}$$
we get that
\begin{align*}
\psi\left(\frac{5}{4}\right)&=\psi\left(\frac{1}{4}\right)+4\\
&=-\gamma-\ln\left(2\left(4\right)\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\sum_{n=1}^{\lfloor(3)/2\rfloor}\cos\left(\frac{2\pi n}{4}\right)\ln\left(\sin\left(\frac{\pi n}{4}\right)\right)+4\\
&=-\gamma-\ln\left(8\right)-\frac{\pi}{2}\cot\left(\frac{\pi}{4}\right)+2\cos\left(\frac{\pi}{2}\right)\ln\left(\sin\left(\frac{\pi}{4}\right)\right)+4
\end{align*}
substituting for our basic trig values and using properties of natural logs this simplifies to
$$\psi\left(\frac{5}{4}\right)=-\gamma-3\ln\left(2\right)-\frac{\pi}{2}+4$$
Plugging this back into our main formula for $S$ in terms of the digamma function yields that
$$S=\frac{1}{8}\left(6\ln\left(2\right)+\pi-8\right)$$
which completes our calculation. In general, this method can be used to solve any sum in form
$$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+c}\right)$$
where $c$ is rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Prove $\cos(A + B) \cos(A − B) = \cos^2A − \sin^2B$ I have a question regarding trigonometric identities. The question I am currently struggling to understand is:
Prove that $$\cos(A + B) \cos(A − B) = \cos^2A − \sin^2B$$
When approaching this problem I know that there is
$$\cos2A = \cos^2A - \sin^2A$$
but how would I apply this here, or is it completely wrong way of approaching it, or should I try
$$\cos(A+B)=\cos A \cos B - \sin A \sin B$$
Many Thanks.
| \begin{align*}
\cos(x+y)\cdot\cos(x-y)&=
(\cos x\cos y-\sin x\sin y)(\cos x\cos y+\sin x\sin y)\\
&=\cos^2 x\cos^2 y-\sin^2 x\sin^2 y\\
&=\cos^2 x(1-\sin^2 y)-(1-\cos^2 x)\sin^2 y\\
&=\cos^2 x-\cos^2 x\sin^2 y-\sin^2 y+\cos^2 x\sin^2y\\
&=\cos^2 x-\sin^2y
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
How to find integer solutions of the equation $x(x+9)=y(y+6)$ where x,y are integers? I found that the equation becomes $$(2x+9)^2-(2y+6)^2=45$$. And $45 = 9^2-6^2 = 7^2-{2^2}$. From here I found these solutions $(x,y)=(0,0), \ (0,-6), \ (-9,0),\ (-9,-6),\ (-1,-2),\ (-1, 4), \ (-8,-2), \ (-8,-4)$.
Does this equation has finite numbers of integer solutions and what would be better approach to solve such problems?
| The equation in the title (which is different from the one in your text) can be written as
$$(x^2-y^2)+6(x-y)=0 \implies (x-y)(x+y+6)=0.$$
Now either $x=y$ or $x+y+6=0$.
So it has solutions as $\{(a,a) \, | \, a \in \Bbb{Z}\} \cup \{(a,-6-a)\, |\, a \in \Bbb{Z}\}$.
After you modified your original problem
The equation can be written as
$$(2x+2y+15)(2x-2y+3)=45.$$
Now factor $45=ab$ and solve the system
\begin{align*}
2x+2y+15&=a\\
2x-2y+3&=b
\end{align*}
To get
$$x=\frac{a+b-18}{4} \quad \text{ and } \quad y=\frac{a-b-12}{4}.$$
So test the factors $a,b$ of $45$ such that these quantities are integers.
\begin{align*}
ab&=45\\
a+b & \equiv 2 \pmod{4}\\
a-b & \equiv 0 \pmod{4}\\
\end{align*}
Hopefully you can take it from here and see for example $a=9,b=5$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $2x^3-9x^2-10x+13$ for $x=3+\sqrt{5}$. Is there an efficient way to tell that this reduces to $1$? I was helping my sibling with a math problem from a past year paper for a competitive exam. It requires us to evaluate this cubic expression for a given value of $x$ which has an $a+b$ form where $b$ is a square root, as shown:
I essentially plugged in $x=a+b$ in the expression and expanded each term and finally got the answer as $1$. But this does not seem like the fastest way to do this, especially because there is only 1 minute given to solve each multiple choice question.
Is there a better way to reduce the original expression that gives the answer as $1$ or suggests that the square root term is going to evaluate to $0$?
Thanks!
| Let $x=3+\sqrt{5}$.$\;$Then
\begin{align*}
&
x=3+\sqrt{5}\\[4pt]
\implies\;&
x-3=\sqrt{5}\\[4pt]
\implies\;&
(x-3)^2=5\\[4pt]
\implies\;&
x^2-6x+9=5\\[4pt]
\implies\;&
x^2-6x+4=0\\[4pt]
\end{align*}
Dividing $2t^3-9t^2-10t+13$ by $t^2-6t+4$ by polynomial long division, we get
$$2t^3-9t^2-10t+13=(2t+3)(t^2-6t+4)+1$$
hence
$$2x^3-9x^2-10x+13=(2x+3)(x^2-6x+4)+1=(2x+3)(0)+1=0+1=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| You can pull out a $\tan x$ factor and
$$\frac{\tan x-\sin x}{x^3}=\frac{\tan x}x\frac{1-\cos x}{x^2}=\frac{\tan x}x\frac{2\sin^2\dfrac x2}{x^2}.$$
This is enough to conclude
$$\to\frac12.$$
Direct L'Hospital is manageable
$$\frac{\tan x-\sin x}{x^3}\to\frac{\tan^2x+1-\cos x}{3x^2}\to\frac{2\tan x(\tan^2x+1)+\sin x}{6x}\to\frac{2+1}6,$$
but easier after pulling $\tan x$,
$$\frac{\tan x}x\frac{1-\cos x}{x^2}\to\frac{\tan x}x\frac{\sin x}{2x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
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How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$
*
*if i use $x=\sin^2\theta$
$$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$
*if i use $x=\tan^2\theta$
$$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$
Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.
Thanks
| You can split the integral into two parts
$$\int \frac{(\sqrt{1+x}+\sqrt{1-x})}{2x} \, dx=\frac{1}{2}\left[\int \frac{\sqrt{1+x}}{x}\,dx+\int \frac{\sqrt{1-x}}{x}\,dx\right].$$
Solve these separately as follows:
\begin{align*}
\int \frac{\sqrt{1+x}}{x}\,dx & =\int \frac{t^2}{(t^2-1)}\,dt && (\text{ let } 1+x=t^2) \\
& =\int \frac{t^2-1+1}{(t^2-1)}\,dt\\
& =\int 1 \, dt+\int \frac{1}{(t^2-1)}\,dt\\
& =t+\frac{1}{2}\left[\int \frac{1}{(t-1)}\,dt-\int \frac{1}{(t+1)}\,dt\right]\\
&=t+\ln\frac{|t-1|}{|t+1|}+c\\
&=\sqrt{1+x}+\ln\frac{|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|}+c\\
\end{align*}
Observe that the second part is pretty much the same. If you use $x=-u$, then
$$\int \frac{\sqrt{1-x}}{x}\, dx=\int \frac{\sqrt{1+u}}{u}\, du.$$
So you can write the answer without any further computation.
$$\int \frac{\sqrt{1-x}}{x}\, dx=\sqrt{1\color{red}{-x}}+\ln\frac{|\sqrt{1\color{red}{-x}}-1|}{|\sqrt{1\color{red}{-x}}+1|}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_2=(b+c-a)(c+a-b)=c^2-(a-b)^2$, leading to (after manipulation):
$4a^2b^2 \geq (c^2-(a-b)^2)((a+b)^2-c^2)=(a+b+c)(a+b-c)(b+c-a)(c+a-b)$
$4a^2b^2\leq(a^2+b^2)^2$, clearly, so initial inequality proven.
However, I just realised AM-GM only holds for non-negative reals, and my $a_1$ and $a_2$ are both non-negative only if $b-a\leq c\leq b+a$.
Is there another way to prove this using AM-GM, or an extension of the same idea that covers the cases where $c\gt b+a \gt b-a$ or $b+a \gt b-a \gt c$? Also, why does the proof so cleanly yield the result?
| So your proof works fine when $a+b\geq c, b+c\geq a, c+a\geq b$. Note that, at most one term among $a+b-c,b+c-a,c+a-b$ can be negative otherwise, one among $a,b,c$ will become negative. Thus, when exactly one term among $a+b-c,b+c-a,c+a-b$ is negative, RHS of inequality becomes negative. However, LHS is always non-negative and so, the inequality follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic? Let $a_{n} = \sqrt{n^{2}+n} - n$, for $n\in\textbf{N}$. Show that $a_{n}$ converges as $n\to\infty$. What is the limit? Is the sequence $(a_{n})_{n=1}^{\infty}$ monotonic?
MY ATTEMPT
The answer to the first question is $a_{n}\to 1/2$ as $n\to\infty$. Indeed,
\begin{align*}
\lim_{n\to\infty}\sqrt{n^{2}+n} - n = \lim_{n\to\infty}\frac{n}{\sqrt{n^{2} + n} + n} = \lim_{n\to\infty}\frac{1}{\sqrt{1 + 1/n} + 1} = \frac{1}{2}
\end{align*}
To test for monotonicity, we can try to study the behavior of the quotient:
\begin{align*}
\frac{a_{n+1}}{a_{n}} & = \frac{\sqrt{(n+1)^{2} + n + 1} - n - 1}{\sqrt{n^{2} + n} - n}\\\\
& = \frac{(n+1)^{2} + n + 1 - (n+1)^{2}}{n^{2} + n - n^{2}}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}\\\\
& = \frac{n+1}{n}\times\frac{\sqrt{n^{2} + n} + n}{\sqrt{(n+1)^{2} + n + 1} + n + 1}
\end{align*}
But then I get stuck, because the first factor is greater than one and the second is smaller than one.
Can someone please finish my attempt or provide an alternative approach?
| Since the sequence$$\left(\sqrt{1+\frac1n}\right)_{n\in\Bbb N}$$is decreasing and since you proved that$$\sqrt{n^2+n}-n=\frac1{\sqrt{1+\frac1n}+1},$$your sequence is increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this combinatorially $\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+\cdots+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$? $n,m,k$ are natural numbers.
$$\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+\cdots+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$$
I need to prove this combinatorially but I can't think of a story, how can I approach this? I thought about starting from the left side
| ${n \choose r} + {n \choose r+1} = {n+1 \choose r+1}$
${n \choose k} = {n+1 \choose k+1} - {n \choose k+1}$ .....(1)
${n+1 \choose k} = {n+2 \choose k+1} - {n+2 \choose k+1}$.....(2)
.
.
.
.
.
${n+m \choose k} = {n+m+1 \choose k+1} - {n+m\choose k}$.....(m)
(1) + (2).............(m) implies
$\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+...+\binom{n+m}{k} = \binom{n+m+1}{k+1}-\binom{n}{k+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is it possible to prove that $P(x)$ is unique?
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then here are my questions:
i) What is the general form of the $P(x)$ ?
ii) What is the minimal degree of polynomial for $P(x)$ ?
iii) Is the polynomial $P(x)$ unique? Can we prove this uniqueness?
I have an example for $P(x)$.
$$P(x)=
6x^2-4x-10$$
Can we prove that, this polynomial is unique?
| My work:
$$\begin{cases} P(x)=k_1(x)(x+1) \\ P(x)=k_2(x)(x-2)+6 \end{cases} \Longrightarrow (x+1)k_1(x)=k_2(x)(x-2)+6 \Longrightarrow (x+1)k_1(x)=k_2(x)(x+1)-3(k_2(x)-2) \Longrightarrow (k_2(x)-2) \mod (x+1)=0 \Longrightarrow k_2(x)=g(x)(x+1)+2 \\ \\ k_1(x)= \dfrac{k_2(x)(x-2)+6}{x+1} =\dfrac{(g(x)(x+1)+2) \times (x+1)+2)(x-2)+6}{x+1}=\dfrac{x(x+1)g(x)-2g(x)(x+1)+2x+2}{x+1}=\dfrac{(x+1)(xg(x)-2g(x)+2)}{x+1}=xg(x)-2g(x)+2=g(x)(x-2)+2 \\ P(x)=(x+1)(xg(x)-2g(x)+2)=g(x)x^2-x(g(x)-2)-2g(x)+2$$
$$\color{red}{\boxed{{P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2}}}$$
here, $g(x)$ is an any polynomial.
$$ \color{blue}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=0 \end{cases} \Longrightarrow P(x)=2x+2}$$
Random example:
$$\color{green}{\begin{cases} P(x)=g(x)x^2-x(g(x)-2)-2g(x)+2 \\ g(x)=x^2-x+1 \end{cases} \Longrightarrow P(x)=x^4-2x^3+3x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?
| Here is a solution using polynomials. Note that $$\begin{align}S&:=\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}
\\&=(a^2+b^2+c^2)\left(\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(a-b)}\right)
\\&\phantom{abc}-2\left(\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-c)(c-a)}+\frac{c^2}{(c-a)(c-b)}\right)\,.\end{align}$$
Consider
$$p(x):=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$$
and
$$q(x):=a^2\frac{(x-b)(x-c)}{(a-b)(a-c)}+b^2\frac{(x-c)(x-a)}{(b-c)(b-a)}+c^2\frac{(x-a)(x-b)}{(c-a)(c-b)}\,.$$
Observe that $p(x)$ and $q(x)$ are polynomials of degree at most $2$ such that $p(t)=1$ and $q(t)=t^2$ for $t\in\{a,b,c\}$. Therefore, $p(x)=1$ and $q(x)=x^2$ identically. If $[x^k]f(x)$ denotes the coefficient of $x^k$ in a polynomial $f(x)$, then we have
$$\begin{align}S&=(a^2+b^2+c^2)\Big([x^2]p(x)\Big)-2\Big([x^2]q(x)\Big)\\&=(a^2+b^2+c^2)\cdot 0-2\cdot 1=-2\,.\end{align}$$
Alternatively, note that
$$S=-\frac{P(a,b,c)}{Q(a,b,c)}\,,$$
where
$$P(a,b,c):=(b^2+c^2-a^2)(b-c)+(c^2+a^2-b^2)(c-a)+(a^2+b^2-c^2)(a-b)$$
and
$$Q(a,b,c):=(b-c)(c-a)(a-b)\,.$$
Note that when two variables are equal, $P(a,b,c)=0$. Thus, $P(a,b,c)$ is divisible by $Q(a,b,c)$. This shows that $$P(a,b,c)=k\,Q(a,b,c)$$ for some constant $k$. Since $P(-1,0,+1)=4$ and $Q(-1,0,+1)=2$, we conclude that $k=2$, whence $S=-k=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$ is an integer.
Prove that $$\frac{(3 a+3 b) !(2 a) !(3 b) !(2 b) !}{(2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}}$$ is an integer for all pairs of positive integers $a, b$
(American Mathematical Monthly)
My work -
$
v_{p}((3 a+3 b) !(2 a) !(3 b) !(2 b) !)=\sum_{k \geq 1}\left(\left\lfloor\frac{3 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 a}{p^{k}}\right\rfloor+\left\lfloor\frac{3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{2 b}{p^{k}}\right\rfloor\right)
$
and
$
\begin{array}{l}
v_{p}\left((2 a+3 b) !(a+2 b) !(a+b) ! a !(b !)^{2}\right) \\
\quad \quad=\sum_{k \geq 1}\left(\left\lfloor\frac{2 a+3 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+2 b}{p^{k}}\right\rfloor+\left\lfloor\frac{a+b}{p^{k}}\right\rfloor+\left\lfloor\frac{a}{p^{k}}\right\rfloor+2\left\lfloor\frac{b}{p^{k}}\right\rfloor\right)
\end{array}
$
now
With the substitution $x=\frac{a}{p^{k}}, y=\frac{b}{p^{k}},$ we have to prove that for any nonnegative real numbers $x, y$ we have
$\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor$
I tried putting $\{x\}+\lfloor x\rfloor=x$ and $\{y\}+\lfloor y\rfloor=y$ and i get things in terms of fractional parts but i am not able to prove after that ....
thankyou
| By your work we need to prove that:
$$\lfloor 3 x+3 y\rfloor+\lfloor 2 x\rfloor+\lfloor 3 y\rfloor+\lfloor 2 y\rfloor \geq\lfloor 2 x+3 y\rfloor+\lfloor x+2 y\rfloor+\lfloor x+y\rfloor+\lfloor x\rfloor+2\lfloor y\rfloor,$$
where $\{x,y\}\subset[0,1).$
Let $x\in\left[\frac{i}{6},\frac{i+1}{6}\right)$, where $i$ is an integer, $0\leq i\leq 5$ and $y\in\left[\frac{j}{6},\frac{j+1}{6}\right)$, where $j$ is an integer, $0\leq j\leq 5$.
Thus, after considering these $36$ cases (maybe a bit of less) we obtain that our inequality is true.
For example, let $x\in\left[\frac{1}{3},\frac{1}{2}\right)$ and $y\in\left[\frac{1}{2},\frac{2}{3}\right).$
Thus, $$2.5\leq3x+3y<3.5,$$
$$\frac{2}{3}\leq2x<1,$$
$$\frac{3}{2}\leq3y<2,$$
$$1\leq2y<\frac{4}{3},$$
$$\frac{13}{6}\leq2x+3y<3,$$
$$\frac{4}{3}\leq x+2y<\frac{11}{6},$$
$$\frac{5}{6}\leq x+y<\frac{7}{6}$$ and
$$[x]=[y]=0.$$
Id est, it's enough to prove that $$[3x+3y]+1+1\geq2+1+[x+y],$$ which is true because $$[3x+3y]\geq2$$ and $$[x+y]\leq1.$$
The rest is the similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3744935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Number of positive integral solutions for $\frac {xy}{x+y} = 2^33^45^6$
I didnt know how to approach so I first tried with small numbers $\frac {xy}{x+y} = k$ (k=2,3, etc.) but all I could decode was min x = k+1, max x = k(k+1), and x=y=2k is the middle solution.
| $\frac {xy}{x+y} = k$
${xy}=kx + ky$
$x(y-k)= ky +(k^2-k^2)$
$(x-k)(y-k) = k^2$
The number of solutions for this equation will be the same as the number of solutions of the equation $xy=k^2$
Given, $k=2^33^45^6$ so $k^2=2^63^85^{12}$
Number of ways of expressing it as a product of 2 numbers = (6+1)(8+1)(12+1) = 819
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\sin^{-1}(\cos 2)-\cos^{-1}(\sin 2) +\tan^{-1}(\cot 4) -\cot^{-1}(\tan 4)+\sec^{-1}(\csc 6)-\csc^{-1}(\sec 6)$ The given expression simplifies to
$$\sin^{-1}(\sin 2)-\cos^{-1}(\cos 2)+\tan^{-1}(\tan 4)-\cot^{-1}(\cot 4)+\sec^{-1}(\sec 6)-\csc^{-1} (\csc 6)$$
$$=(\pi-2)-2+(4-\pi)-(2\pi-4)+(2\pi-6)-(2\pi-6)$$
$$=-2\pi+4$$
But the given answer is $5\pi-16$. I rechecked all the the principle branches, and they all seem to be right. Where did use the wrong value?
| I don't get your simplification. How did $\sin^{-1}(\cos 2) - \cos^{-1}(\sin 2)$ become $\sin^{-1}(\sin 2) - \cos^{-1}(\cos 2)$? I think the simplification part is off.
We must identify which quadrant each of these lie in. For example, $2$ lies in the second quadrant, because $\pi > 2 > \frac \pi 2$. $4$ lies in the third quadrant, and $6$ lies in the fourth quadrant.
With this in mind, $\sin^{-1}(\cos 2) = \frac{\pi}{2} - 2$. Similarly, $\cos^{-1}(\sin 2) = 2 - \frac \pi 2$, because the inverse cosine function is restricted to $(-\frac{\pi}{2},\frac{\pi}{2}]$.
*
*$\tan^{-1}(\cot 4) = \frac{3 \pi}{2} - 4$.
*$\cot^{-1} (\tan 4) = \frac{3 \pi }{2} - 4$.
*$\sec^{-1}(\csc 6) = \frac{5 \pi}{2} - 6$.
*$\csc^{-1}(\sec 6) = 6 - \frac{3 \pi}{2}$.
(You can convince yourself of these and finish the problem).
With the simplified expression :
*
*$\sin^{-1}(\sin 2) = \pi- 2$
*$\cos^{-1}(\cos 2) = 2$
*$\tan^{-1}(\tan 4) = 4-\pi$.
*$\color{green}{\cot^{-1}(\cot 4) = 4-\pi}$. Explanation : the principal region for $\cot^{-1}$ is $[-\frac \pi 2, \frac \pi 2]$, and $\cot 4$ is positive because $4$ is in the third quadrant. Thus, $\cot^{-1}(\cot 4)$ lies in the first quadrant, so the angle is positive, hence $4-\pi$ (and not $\pi -4$, which is a negative angle).
*$\sec^{-1}(\sec 6) = 2\pi - 6$.
*$\color{green}{\csc^{-1}(\csc 6) = 6-2\pi}$. Explanation : the principal region of $\csc^{-1}$ is $[-\frac {\pi}{2} , \frac{\pi}{2}]$ , and $6$ belongs in the fourth quadrant, so is already in the prescribed region, as long as we subtract $2\pi$, which gives $6 - 2\pi$ (and since the cosecants of both these angles are negative, this makes sense).
Now evaluating gives the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Evaluate $\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$ I am supposed to evaluate
$$\int \:\frac{3x^5+13x^4+32x^3+8x^2-40x-75}{x^2\left(x^2+3x+5\right)^2}\:dx$$
I started using partial fractions
$$3x^5+13x^4+32x^3+8x^2-40x-75=x\left(x^2+3x+5^2\right)^2A+\left(x^2+3x+5^2\right)^2B+x\left(x^2+3x+5^2\right)\left(Cx+D\right)+x^2\left(Ex+F\right)$$
I managed to get to
$$\:\int \:\left(\frac{2}{x}-\frac{3}{x^2}+\frac{x+1}{x^2+3x+5}+\frac{4x}{\left(x^2+3x+5\right)^2}\right)\:dx$$
Am i on the right track? is there an easier way to simplify the original integral?
| An alternative method to treat multiple roots is the following (see Application to symbolic integration):
$$
\frac{3 x^5 + 13 x^4 + 32 x^3 + 8 x^2 - 40 x - 75}{x^2(x^2+3x+5)^2}=\frac{a}{x}+\frac{bx+c}{x^2+3x+5}+\frac{d}{dx}\left[\frac{e}{x}+\frac{fx+g}{x^2+3x+5}\right]
$$
from which we get the linear system
$$
\left\{
\begin{align}
&a+b=3,\\
&6 a+3 b+c-e-f=13,\\
&19 a+5 b+3 c-6 e-2 g=32,\\
&30 a+5 c-19 e+5 f-3 g=8,\\
&25 a-30 e=-40,\\
-&25 e=-75
\end{align}
\right.
$$
Finally, we obtain
$$
\int\left(\frac{2}{x}+\frac{x-\frac{1}{11}}{x^2+3 x+5}\right)dx+\frac{3}{x}+\frac{-\frac{12}{11}x-\frac{40}{11}}{x^2+3 x+5}
$$
The subsequent integration is pretty easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Showing that $a$, $b$, $c$, $d$ are in geometric progression iff $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$
If the real numbers $a$, $b$, $c$, $d$ are in geometric progression, show that
$$
\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}
$$
Prove that the converse also holds.
The simplest way I could think of is making/assuming a general G.P
Let $r$ be the common ratio and $a$ be the first term, then
$$b=a r, c=a r^{2}, d=a r^{3}$$
after multiplying a lot of terms, many times, (skipping significant steps here so it is more readable)
$\begin{aligned} \mathrm{LHS} &=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}\right)\left(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right) \\ &=a^{2}\left(1+r^{2}+r^{4}\right) \cdot a^{2} r^{2}\left(1+r^{2}+r^{4}\right) \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} \\ &=\left(a^{2} r+a^{2} r^{3}+a^{2} r^{5}\right)^{2} \\ &=\left(a \cdot a r+a r \cdot a r^{2}+a r^{2} \cdot a r^{3}\right)^{2} \\ &=(a b+b c+c d)^{2}=R H S \end{aligned}$
Once after expanding each term, and getting no where, I realised that I had already gotten the answer, all I had to do was take the terms inside the and then I had the RHS. But it all was so tedious and took multiple attempts. Though, this method guarantees that converse holds,
Can it be done more elegantly?
Edit: I am in highschool (and just a little more interested in math) So I don't have knowledge about the much spoken Cauchy's identity.
| Q. If $a, b, c, d$ are in G.P., prove that
$\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} .$
Sol. Let $r$ be the common ratio of the G.P. $a, b, c, d$.
Then $b=a r ; c=a r^{2}$ and $d=a r^{3}$.
$\begin{aligned} \therefore \quad \text { L.H.S. } &=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right) \\ &=\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}\right)\left(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right) \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} \\ \text { And, } R \cdot H . S_{.} &=(a b+b c+c d)^{2} \\ &=\left(a^{2} r+a^{2} r^{3}+a^{2} r^{5}\right)^{2} \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} . \end{aligned}$
Hence, $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving a non-homogeneous inequality with $x,y,z>0$ For $x,y,z>0.$ Prove: $$\frac{1}{2}+\frac{1}{2}{r}^{2}+\frac{1}{3}\,{p}^{2}+\frac{2}{3}\,{q}^{2}-\frac{1}{6} Q-\frac{3}{2} r-\frac{2}{3}q-\frac{1}{6}pq-\frac{5}{3} \,pr\geqslant 0$$
where $$\Big[p=x+y+z,q=xy+zx+yz,r=xyz,Q= \left( x-y \right) \left( y-z \right)
\left( z-x \right)\Big ]$$
My SOS proof$:$ $$\text{LHS}=\frac{1}{12}\,\sum \left( 3\,{z}^{2}+1 \right) \left( x-y \right) ^{2}+\frac{1}{6} \sum\,y
\left( y+z \right) \left( x-1 \right) ^{2}+\frac{1}{2}\, \left( xyz-1
\right) ^{2} \geqslant 0$$
By the way$,$ there is an C-S proof :D Who can find$?$
| Since by AM-GM $$1+x^2y^2z^2\geq2xyz$$ and $$\sum_{cyc}x^2y^2\geq\sum_{cyc}x^2yz,$$ it's enough to prove that:
$$\sum_{cyc}x^2y^2-\sum_{cyc}(x^2y+xyz)+\sum_{cyc}x^2\geq0,$$ which is true by AM-GM twice:
$$\sum_{cyc}(x^2y^2+x^2)\geq2\sum_{cyc}x^2y\geq\sum_{cyc}(x^2y+xyz).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the abundancy index of divisors of odd perfect numbers and a possible upper bound for the special/Euler prime (Note: This post is an offshoot of this earlier question.)
The topic of odd perfect numbers likely needs no introduction.
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Euler proved that an odd perfect number $n$, if one exists, must have the form
$$n = p^k m^2$$
where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Here is my question:
Does $p \leq P$ follow from $$I(p^k)+I(m^2) \leq 3 - \bigg(\frac{p-1}{p(p+1)}\bigg),$$ if $p^k m^2$ is an odd perfect number with special prime $p$, where we set
$$0 < \varepsilon = \frac{p-1}{p(p+1)}$$
and $P$ is some finite constant?
MY ATTEMPT
Notice that the inequality
$$I(p^k)+I(m^2) \leq 3 - \bigg(\frac{p-1}{p(p+1)}\bigg)$$
holds in general, since
$$\bigg[I(p^k) - \frac{2p}{p+1}\bigg]\bigg[I(m^2) - \frac{2p}{p+1}\bigg] \geq 0$$
follows from
$$I(p^k) < I(m^2) = \frac{2}{I(p^k)} \leq \frac{2}{I(p)} = \frac{2p}{p+1}.$$
Furthermore, note that we obtain the upper bound
$$\varepsilon = \frac{p-1}{p(p+1)} = \bigg(1 - \frac{1}{p}\bigg)\bigg(\frac{1}{p+1}\bigg) < \frac{1}{p + 1} \leq \frac{1}{6},$$
since $p$ is prime with $p \equiv 1 \pmod 4$ implies that $p \geq 5$.
We now compute for $p$ in terms of $\varepsilon$:
$$\varepsilon p^2 + p(\varepsilon - 1) + 1 = 0$$
$$p = \frac{(1-\varepsilon) \pm \sqrt{(1-\varepsilon)^2 - 4\varepsilon}}{2\varepsilon}$$
This gives
$$p = \frac{(1-\varepsilon) \pm \sqrt{{\varepsilon}^2 - 6\varepsilon + 1}}{2\varepsilon}.$$
Quoting verbatim from [Remark 11, page 5 of The Abundancy Index of Divisors of Odd Perfect Numbers by Dris (JIS, 2012)]:
Remark 11. As remarked by Joshua Zelinsky in 2005: "Any improvement on the upper bound of $3$ would have (similar) implications for all arbitrarily large primes and thus would be a very major result." (e.g. $L(p) < 2.99$ implies $p \leq 97$.) In this sense, the inequality
$$\frac{57}{20} < I(p^k) + I(m^2) < 3$$
is best-possible.
I tried using WolframAlpha to get the global maxima of the two functions
$$p(\varepsilon_1) = \frac{(1-\varepsilon_1) + \sqrt{{\varepsilon_1}^2 - 6{\varepsilon_1} + 1}}{2\varepsilon_1}$$
and
$$p(\varepsilon_2) = \frac{(1-\varepsilon_2) - \sqrt{{\varepsilon_2}^2 - 6{\varepsilon_2} + 1}}{2\varepsilon_2},$$
but the outputs were not helpful.
Alas, this is where I get stuck.
| This answer proves the following three claims :
Claim 1 : There is no $x$ such that $\dfrac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16$
Claim 2 : $\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16\iff 0\lt x\le \dfrac{2}{15}$
Claim 3 : $\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$
Note that $$x^2 - 6x + 1\ge 0\quad\text{and}\quad 0\lt x\le\frac 16\iff 0\lt x\le \frac 16$$
Claim 1 : There is no $x$ such that $\dfrac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16$
Proof :
Suppose that there is such an $x$. Then, we have
$$\begin{align}&\frac{1-x - \sqrt{x^2 - 6x + 1}}{2x}\ge 5\quad\text{and}\quad 0\lt x\le \dfrac 16
\\\\&\implies \sqrt{x^2 - 6x + 1}\le 1-11x\quad\text{and}\quad 0\lt x\le \dfrac 16
\\\\&\implies\sqrt{x^2 - 6x + 1}\le 1-11x\quad\text{and}\quad 0\lt x\le \dfrac 16\quad\text{and}\quad 0\le 1-11x
\\\\&\implies x^2 - 6x + 1\le (1-11x)^2\quad\text{and}\quad 0\lt x\le \dfrac 1{11}
\\\\&\implies x\ge\frac{2}{15}\quad\text{and}\quad 0\lt x\le \dfrac 1{11}\end{align}$$
which is impossible.
Claim 2 : $\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\ $ and $\ 0\lt x\le \dfrac 16\iff 0\lt x\le \dfrac{2}{15}$
Proof :
$$\begin{align}&\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}\ge 5\quad\text{and}\quad 0\lt x\le \dfrac 16
\\\\&\iff \sqrt{x^2 - 6x + 1}\ge 11x-1\quad\text{and}\quad 0\lt x\le \dfrac 16
\\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\bigg(x^2-6x+1\ge (11x-1)^2\quad\text{and}\quad \frac{1}{11}\lt x\le\frac 16\bigg)
\\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\bigg(0\le x\le\frac{2}{15}\quad\text{and}\quad \frac{1}{11}\lt x\le\frac 16\bigg)
\\\\&\iff 0\lt x\le\frac{1}{11}\quad\text{or}\quad\frac{1}{11}\lt x\le\frac{2}{15}
\\\\&\iff 0\lt x\le\frac{2}{15}\end{align}$$
Claim 3 : $\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$
Proof : Since $\displaystyle\lim_{x\to 0^+}(1-x + \sqrt{x^2 - 6x + 1})=2$, we get
$$\displaystyle\lim_{x\to 0^+}\dfrac{1-x + \sqrt{x^2 - 6x + 1}}{2x}=+\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7? Find the remainder $1690^{2608} + 2608^{1690}$ when divided by 7?
My approach:-
$1690 \equiv 3(\bmod 7)$
$1690^{2} \equiv 2(\bmod 7)$
$1690^{3} \equiv-1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*1690}{7}$..so on]
$\left(1690^{3}\right)^{869} \cdot 1690 \equiv(-1)^{869}1690 \quad(\mathrm{mod} 7)$
$1690^{2608} \equiv -1690 \quad(\mathrm{mod} 7)$....(1)
again for $2608$
$2608 \equiv 4(\bmod 7)$
$2608^{2} \equiv 2(\bmod 7)$
$2608^{3} \equiv1 \quad(\mathrm{mod} 7)$[ quite easy to determine , $\frac{2*2608}{7}$..so on]
$\left(2608^{3}\right)^{563} \cdot 2608 \equiv(1)^{563}2608 \quad(\mathrm{mod} 7)$
$2608^{1690} \equiv 2608 \quad(\mathrm{mod} 7)$...(2)
Now applying property
adding (1) + (2),
$1690^{2608} + 2608^{1690}=918 \quad(\mathrm{mod} 7)$
$\boxed{1690^{2608} + 2608^{1690} \equiv 1 \quad(\mathrm{mod} 7)}$
Is my approach best? or Anyother approach is there comparatively better than it
| A shorter way would be to use Fermat's little theorem, plus the fact that $1690 = 7k+3$ and $2608 = 7k-3$
$$1690^6 \equiv 1 \text{ (mod 7)} \implies 1690^{2608} \equiv 1690^4 \text{ (mod 7)} \equiv 3^4 \text{(mod 7)} \equiv 4 \text{(mod 7)}$$
Similarily,
$$2608^6 \equiv 1 \text{ (mod 7)} \implies 2608^{1690} \equiv 2608^4 \text{ (mod 7)} \equiv 3^4 \text{(mod 7)} \equiv 4 \text{ (mod 7)}$$
Adding the remainders, $4 + 4 \text{ (mod 7)} \equiv 1 \text{ (mod 7)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the probability to form a triangle with the three pieces of the stick?
On a stick $1$ meter long is casually marked a point $X \sim U[0,1]$. Let $X=x$, is also marked a second point $Y\sim U[x,1]$.
1) Find the density of $(X,Y)$ showing the domain.
$$\rightarrow \quad f_{XY}(x,y)=\frac{1}{1-x}\mathbb{I}_{[0,1]}(x)\mathbb{I}_{[x<y<1]}(y)$$
2) Say if $X$ and $Y$ are independent or not, and compute $\operatorname{Cov}(X,Y)$.
$$\rightarrow f_Y(y)=-\log(1-y)\mathbb{I}_{[0,1]}(y)\Rightarrow f_X(x)f_Y(y)\neq f_{XY}(x,y)\\
\Rightarrow X\text{ and }Y\text{ are not independent}$$
$$\rightarrow \operatorname{Cov}(X,Y)=-\frac{1}{6}$$
3) Now we assume to break the stick in the points $X$ and $Y$, and to form a triangle with the pieces that we have. Remembering that in a triangle the sum of the lengths of two sides must be greater than the length of the third side, what is the probability to form a triangle with the three pieces of the stick?
I'm stuck on point 3). How would you fix it?
Thanks in advance for any help.
| Let $S$ be the region in the $xy$-plane defined by the constraints
$$
\left\lbrace
\begin{align*}
0 < x < 1\\[4pt]
x\le y < 1\\[4pt]
\end{align*}
\right.
$$
Then the joint density function of the random variables $X,Y$ is given by
$$
f(x,y)=
\begin{cases}
{\Large{\frac{1}{1-x}}}&\text{if}\;\,(x,y)\in S\\[4pt]
0&\text{otherwise}\\
\end{cases}
$$
Then we get
\begin{align*}
E[X]&=\int_0^1\int_x^1 x\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{1}{2}\\[4pt]
E[Y]&=\int_0^1\int_x^1 y\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{3}{4}\\[4pt]
E[XY]&=\int_0^1\int_x^1 xy\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{5}{12}\\[4pt]
\end{align*}
hence $X,Y$ are not independent since
$$
E[X]{\,\cdot\,}E[Y]
=
\frac{1}{2}{\,\cdot\,}\frac{3}{4}
=
\frac{3}{8}
\ne
\frac{5}{12}
=
E[XY]
$$
For the covariance, we get
$$
\text{Cov}(X,Y)
=
\int_0^1\int_x^1
\left(
\Bigl(x-\frac{1}{2}\Bigr)
\Bigl(y-\frac{3}{4}\Bigr)
\right)
\!
\Bigl(\frac{1}{1-x}\Bigr)
\;dy\;dx
=
\frac{1}{24}
$$
The potential triangle has side lengths $a,b,c$ where
$$
\left\lbrace
\begin{align*}
a&=x\\[4pt]
b&=y-x\\[4pt]
c&=1-y\\[4pt]
\end{align*}
\right.
$$
hence noting that $a+b+c=1$, the triangle inequalities are satisfied if and only if
$0 < a,b,c < {\large{\frac{1}{2}}}$.
Computing $P\bigl(a \ge {\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(a\ge\frac{1}{2}\Bigr)
=
\int_{\large{\frac{1}{2}}}^1\int_x^1 \frac{1}{1-x}\;dy\;dx=\frac{1}{2}
$$
Computing $P\bigl(b\ge{\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(b\ge\frac{1}{2}\Bigr)
=
\int_0^{\large{\frac{1}{2}}}\int_{{\large{x+{\large{\frac{1}{2}}}}}}^1 \frac{1}{1-x}\;dy\;dx
=
\frac{1}{2}-\frac{1}{2}\ln(2)
$$
Computing $P\bigl(c\ge{\large{\frac{1}{2}}}\bigr)$, we get
$$
P\Bigl(c\ge\frac{1}{2}\Bigr)
=
\int_0^{{\large{\frac{1}{2}}}}
\int
_
{\large{x}}
^
{{\large{\frac{1}{2}}}}
\frac{1}{1-x}\;dy\;dx
=
\frac{1}{2}-\frac{1}{2}\ln(2)
$$
In the region $S$, we have $0 < a,b,c < 1$, hence since $a+b+c=1$, at most one of $a,b,c$ can be at least ${\large{\frac{1}{2}}}$.
It follows that the probability that $a,b,c$ qualify as the side lengths of triangle is given by
\begin{align*}
&
1
-
\left(
P\Bigl(a\ge\frac{1}{2}\Bigr)
+
P\Bigl(b\ge\frac{1}{2}\Bigr)
+
P\Bigl(c\ge\frac{1}{2}\Bigr)
\right)
\\[4pt]
=&
1
-
\left(
\left(
\frac{1}{2}
\right)
+
\left(
\frac{1}{2}-\frac{1}{2}\ln(2)
\right)
+
\left(
\frac{1}{2}-\frac{1}{2}\ln(2)
\right)
\right)
\\[4pt]
=&
-\frac{1}{2}
+
\ln(2)
\approx .193
\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate: $\int \frac{x}{\left(x^2-4x-13\right)^2}dx$.
Integrate:
$$\int \frac{x}{\left(x^2-4x-13\right)^2}dx$$
Here's my attempt:
I first completed the squares for the denominator:
$$\left(x^2-4x-13\right)^2=(x-2)^2-17 \implies \int \frac{x}{\left(\left(x-2\right)^2-17\right)^2}dx$$
I then used $u$-subsituition:
$$u=x-2 \implies \int \frac{u+2}{\left(u^2-17\right)^2}du = \int \frac{u}{\left(u^2-17\right)^2}du+\int \frac{2}{\left(u^2-17\right)^2}du$$
The first part of the new integral is quite simple:
$$\int \frac{u}{\left(u^2-17\right)^2}du=\frac{-1}{2(u^2-17)}$$
Then I did the second part:
$$\int \frac{2}{\left(u^2-17\right)^2}du = -\frac{1}{2\left(u^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|u+\sqrt{17}\right|-\frac{1}{68\left(u+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|u-\sqrt{17}\right|-\frac{1}{68\left(u-\sqrt{17}\right)}\right) = -\frac{1}{2\left(\left(x-2\right)^2-17\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) = -\frac{1}{2\left(x^2-4x-13\right)}+2\left(\frac{1}{68\sqrt{17}}\ln \left|x-2+\sqrt{17}\right|-\frac{1}{68\left(x-2+\sqrt{17}\right)}-\frac{1}{68\sqrt{17}}\ln \left|x-2-\sqrt{17}\right|-\frac{1}{68\left(x-2-\sqrt{17}\right)}\right) + C, C \in \mathbb{R}$$
Is this working out correct? I'm not really sure how WolframAlpha works, so I didn't check it on there.
| Making the problem more general
$$I=\int \frac x {(x^2+ax+b)^2}\,dx \qquad \text{with} \qquad a^2-4b \neq 0$$ Let $r$ and $s$ be the roots of the quadratic (whatever they could be - real or complex) to make
$$ \frac x {(x^2+ax+b)^2}= \frac x {(x-r)^2 \, (x-s)^2}$$ Using partial fraction decomposition
$$ \frac x {(x-r)^2 \, (x-s)^2}=\frac{r+s}{(r-s)^3}\left(\frac 1{x-s}-\frac 1{x-r} \right)+\frac 1{(r-s)^2 }\left(\frac{r}{(x-r)^2}+\frac{s}{(x-s)^2}\right)$$ The first part is simple. For the second piece, you have two integrals
$$J_k=\int \frac k {(x-k)^2}\,dx=\int \frac {dy}{(y-1)^2}=-\frac 1{y-1}=-\frac{k}{x-k}$$ Just combine all the pieces and, at the end, replace $r$ and $s$ by their values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Sum of Continued Fractions Let $x$ be a positive integer.
Consider the following sum (maybe there is a better notation with continued fractions, but I am not aware of it):
$\frac{1}{x} + \frac{1}{x- \frac{1}{x}} + \frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}}} +\frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} - \frac{1}{x- \frac{1}{x} - \frac{1}{x-\frac{1}{x}}}} + \dots $
My question is: How many summands do I need such that the sum is at least $x$? $x^2$ is a trivial upper bound here, because every summand is $\geq \frac{1}{x}$. Is there a way to get a better bound here?
| The number of steps needed, $s(x)$, is the sequence A140949 in OEIS. Not sure about the background. Also there is not much information there. Perhaps the auther Gil Broussard can tell you more (assuming that he is not you).
However it does look like there is some pattern going on here.
The value of $s(x)$ is very close to $x^2 / 2$. In fact, the sequence $(2s(x) - x^2)_{x \geq 0}$ looks like this:
$$0, 1, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7$$
This looks quite interesting and is probably a logarithmic growth. Thus $x^2 / 2$ is a fairly good approximation of $s(x)$. Not sure about the theory behind.
Also not sure what happens for non-integer real numbers $x$.
I will update this answer if I am motivated enough to look deeper into the problem. That is, if nobody gives a better reference.
It would also help if you could tell us your motivation, e.g. where does this problem come from.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $2a^3+27c=9ab$
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $$2a^3+27c=9ab$$
So far, I let the roots of $x^3+ax^2+bx+c=0$ be $r_1, r_2,$ and $r_3$. $r_1=r_2-d$ and $r_3=r_2+d$ because they form an arithmetic sequence with $d$ being the difference. the sum of the roots is $-a$. So, $r_2=-a/3$. We can let the product of the roots be $-c$. So, $(r_2-d)(r_2)(r_2+d)=-c$. Plugging in $r_2=-a/3$ we get $(-a/3-d)(-a/3)(-a/3+d)$. How do I continue with this method?
EDIT: I used hamam_abdallah's hint to get $\frac{-a^3}{27} + \frac{ad^2}{3} = -c$ what do i do after applying vieta's formulas?
| Hint
Finally note that
$$(-a/3-d)(-a/3+d) +(-a/3)(-a/3+d)+(-a/3-d)(-a/3)=b$$
gives
$$(-a/3-d)(-a/3+d)= b-(-a/3)(-a/3+d)-(-a/3-d)(-a/3)=b+ (a/3)(-a/3+d-a/3-d)$$
Thus
$$(-a/3-d)(-a/3+d)=b- (2a^2/9)$$
Therefore
$$-c=(-a/3-d)(-a/3)(-a/3+d)=(-a/3)\left( b- (2a^2/9) \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Let $p$ be a prime number for which also $p^2+2$ is a prime. Show that then $p^3+2$ is also a prime.
Let $p$ be a prime number for which also $p^2+2$ is a prime. Show that then $p^3+2$ is also a prime.
Computing few first primes I got:
$p=2$: $2^2+2=6$ (not satisfying the condition)
$p=3$: $3^2+2=11$ (satisfying the condition)
$p=5$: $5^2+2=27$ (not satisfying the condition)
$p=7$: $7^2+2=51$ (not satisfying the condition)
so I would have a reason to believe that only the case $p=3$ will satisfy this condition, but how would I go about showing this rigorously?
| If $3 \nmid p$, then $p \equiv \pm 1\pmod{3}$ and so $3 \mid (p^2+2)$, with $p^2+2>3$. So $p^2+2$ is prime can only hold for $p=3$. We note that both $3^2+2$ and $3^3+2$ are primes. $\blacksquare$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding various things given Cos(a-b)+Cos(b-c)+Cos(c-a) = $\frac{-3}{2}$ I have given the task to find $\sin^6(a)+\sin^6(b)+\sin^6(c)$. It is also given $a , b$ and $c$ are real .
Further given down is only my attempt , its your choice you want to read it not . I just want a solution which could be completing my solution or give a better and simpler method . Any help like another attempt , links , hints etc are appreciated. Solution can include use of complex numbers or trigonometry. It would be appreciated if you don't use tools higher than high school level .
First thing I noticed is he above equation can be written as $[\sin(a)+\sin(b)+\sin(c)]^2\ +\ [\cos(a)+\cos(b)+\cos(c)]^2\ =\ 0$
Which implies $\cos(a)+\cos(b)+\cos(c)=0$ .....(1) and also $\sin(a)+\sin(b)+\sin(c)=0$ ....(2) .
$NA =\cos(Na)+ i\sin(Na)$
$NB =\cos(Nb)+ i\sin(Nb)$
$NC =\cos(Nc)+ i\sin(Nc)$
Where $N$ is an arbitrary natural number and $i$ is the square root of $-1$.
From (1) and (2)
$A+B+C=0 .. (3)$
Which implies $\frac{1}{AB} + \frac{1}{BC} + \frac{1}{BC}=0$
Taking conjugate both sides
$AB+BC+CA=0$ ....(4)
So $A²+B²+C²=0$ .....(5) also as $(A+B+C)^2=0$
We can also by repeating same procedure $A^4+B^4+C^4=0$ ....(6)
Now $\cos²(a)+\cos²(b)+\cos²(c)$ = $\frac{1}{2}[3-\cos(2a)-\cos(2b)-\cos(2c)]$ = $\frac{3}{2}$
As by (5) we know $\cos(2a)+\cos(2b)+\cos(2c)=0$
Same way $\sin^2(a)+\sin^2(b)+\sin^2(c)= \frac{3}{2}$
Also $\cos^2(2a)+\cos^2(2b)+\cos^2(2c)=\frac{3}{2}$ as $\cos(4a) + \cos(4b) + \cos(4c) = 0 $ by (6)
As $[\cos(a) + \cos(b) + \cos(c)]^2 = 0 $, so $\cos(a)\cos(b)+ \cos(b)\cos(c)+\cos(c)\cos(a)= \frac{-3}{4} $
Same way $\cos(2a)\cos(2b)+\cos(2b)\cos(2c) +\cos(2c)\cos(2a)= \frac{-3}{4} $
$[\sin^2(a)-\sin^2(b)]^2+ [\sin^2(b)-\sin^2(c)]^2+ [\sin^2(c)-\sin^2(a)]^2$
= $\frac{1}{4} \left([\cos(2a)-\cos(2b)]^2+ [\cos(2b)-\cos(2c)]^2 +[\cos(2c)-\cos(2a)]^2\right)$
= $\frac{9}{8}$
$l³+m³+n³=(l+m+n)(l²+m²+n²-lm-mn-nm) + 3lmn $
$l=\sin²(a),\:m=\sin²(b)$ and $n=\sin²(c)$
I have already found $l+m+n$ and $l²+m²+n²-lm-mn-nm$ above. So I am stuck with finding $lmn$.
Please help me find $lmn$ or provide another easier solution as you can see this method is really lenghty and tedious . Please also make out any mistakes I have made .
| From your notes:
$$\cos a+\cos b+\cos c=0 \Rightarrow \cos^2a+\cos^2b+2\cos a\cos b=\cos^2c\\
\sin a+\sin b+\sin c=0\Rightarrow \sin^2a+\sin^2b+2\sin a\sin b=\sin^2c\\
$$
Adding the two:
$$\cos(a-b)=-\frac12.$$
Similarly:
$$\cos(b-c)=-\frac12\\
\cos(c-a)=-\frac12$$
One solution is: $a=180^\circ,b=60^\circ,c=-60^\circ$.
Hence, the sum of sixth powers of sines is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Cubes as the sum of odd integers It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!
| The proposed identity says that
$n^3
=\sum_{k=0}^{n-1}(n(n-1)+1+2k)
$.
(Figuring out how to write this
is the hard part.)
The right side is
$\sum_{k=0}^{n-1}(n(n-1)+1+2k)
=n(n(n-1)+1)+2\sum_{k=0}^{n-1}k
=n^3-n^2+n+n(n-1)
=n^3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find coordinates of a point Q on the graph $\sin (x) + \cos (y) = 0.5$ given that the gradient of its tangent is perpendicular to point P. Note:
Point $P$ is on the $y$-axis and above the $x$-axis
$\frac{-\pi}{6}\le x \le\frac{7\pi}{6}$
$\frac{-2\pi}{3}\le y\le\frac{2\pi}{3}$
What I have done so far:
Solving for $P$:
$$x = 0
\\ \sin (0) + \cos (y) = 0.5
\\ 0 + \cos (y) = 0.5
\\ y= \pm\frac{\pi}{3} $$
For $P$, $y \gt 0$
$\therefore y = \frac{\pi}{3}$
Solving for $\frac{dy}{dx}$:
$$\sin(x) + \cos(y) = 0.5
\\ \cos(x) - \sin(y)\frac{dy}{dx} = 0$$
$\therefore \frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}$
Derivative at $P$:
$$\frac{dy}{dx} = \frac{\cos(x)}{\sin (y)}
= \frac{\cos(0)}{\sin(\frac{\pi}{3})}
= \frac{2}{\sqrt3}$$
As for the gradient of the tangent line at $Q$ is perpendicular to that at $P$:
$\frac{dy}{dx} = \frac{-\sqrt3}{2}$
How do I solve for the coordinates of $Q$ after this?
| Good work. Note the point $Q$ lies on the curve as well as the tangent line. So:
$$\begin{cases}\frac{\cos x}{\sin y}=-\frac{\sqrt{3}}{2}\\ \sin x+\cos y=0.5 \end{cases} \Rightarrow$$
From the first:
$$\cos^2x=\frac34(1-\cos^2y)\Rightarrow \cos y=\pm\sqrt{1-\frac43\cos^2x}$$
Now sub it to the second:
$$1-\frac43\cos^2x=\frac14-\sin x+\sin^2x\Rightarrow \\
4\sin^2x+12\sin x-7=0 \Rightarrow \sin x=\frac12.$$
Referring to the given constraints, the final answer is:
$$x=\pi-\frac{\pi}{6},y=\frac{\pi}{2}$$
| {
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Definite integral of $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$ using the substitution of $x=2\tanθ$. Can someone help with the integral $\int_{-2}^{2} \frac{5}{(x^2+4)^2}\,dx$?
I'm supposed to find the definite integral for this using the substitution $x=2\tanθ$.
This is what I've done so far:
$$\longrightarrow \frac{dx}{dθ}=2\sec^2θ$$
$$\longrightarrow x=2 \rightarrow θ=\frac{\pi}{4}$$
$$\longrightarrow x=-2 \rightarrow θ=-\frac{\pi}{4}$$
$$\therefore \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2θ}{(4\tan^2θ+4)^2}\,dθ$$
Using; $$t=\tan^2θ+1,$$
$$=\int_{2}^{2} \frac{10}{32t^2\sqrt{t-1}}\,dt$$
After this, I don't know how to finish. Does anyone know how to finish?
*Note: The first substitution is the one the exercise is telling me to use, the other is one I used myself.
| Instead of applying another substitution, just simplify the expression after your first substitunion.
$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2{\theta}}{{\left(4\tan^2{\theta}+4\right)}^2} \; d \theta $$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{10\sec^2{\theta}}{16{\left(\tan^2{\theta}+1\right)}^2} \; d \theta$$
$$=\frac{5}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sec^2{\theta}}{{\sec^4{\theta}}} \; d \theta$$
$$=\frac{5}{8}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2{\theta} \; d \theta$$
Using the angle reduction formula for $\cos^2{\theta}$:
$$=\frac{5}{16} \left(\theta+\frac{1}{2}\sin{(2\theta)}\right) \bigg\rvert_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$$
$$=\frac{5\left(\pi+2\right)}{32}$$
| {
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Ideals $(X^2+1)$ and $(X^2+1, 7)$ of polynomial ring $\mathbb{Z}[X]$ How can I show that generated ideals $(X^2+1)$ and $(X^2+1, 7)$ of polynomial ring $\mathbb{Z}[X]$ are a prime ideal and a maximal ideal, respectively?
| Hint
Are you able to check that
$$\Bbb{Z}[x]/\langle x^2+1, 7\rangle =\{ax+b\, | \, a,b \in \Bbb{Z}_7, \, x^2+1 \equiv 0\}?$$
This is a finite set with $7^2=49$ elements. If you want to check this is a field, you need to check the properties that define a field. One such property is to see if this has any zero divisor (finite integral domain is a field).
\begin{align*}
(ax+b)(cx+d) \equiv 0 \\
acx^2+(ad+bc)x+bd \equiv 0 \\
(ad+bc)x+(bd-ac) \equiv 0 && (\because x^2 \equiv -1)
\end{align*}
This implies
\begin{align*}
ad+bc & \equiv 0 \pmod{7}\\
bd-ac& \equiv 0 \pmod{7}.
\end{align*}
Squaring and adding gives us
\begin{align*}
(ad+bc)^2+(bd-ac)^2 & \equiv 0 \pmod{7}\\
(ad)^2+(bc)^2+(bd)^2+(ac)^2 & \equiv 0 \pmod{7}\\
(a^2+b^2)(c^2 + d^2)& \equiv 0 \pmod{7}
\end{align*}
See if you can check that at least one of $ax+b$ or $cx+d$ has to be the zero element?
Similar approach can be helpful for the first part if you can show that
$$\Bbb{Z}[x]/\langle x^2+1\rangle =\{ax+b\, | \, a,b \in \Bbb{Z}, \, x^2+1 \equiv 0\}$$
| {
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-n\pi i-\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}+\frac{1}{\cos\left(n\pi +\frac{\pi}{4}\right)+i\sin\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\sqrt{2}}\frac{1}{\frac{1}{\sqrt{2}}(\cos n\pi -\sin n\pi )+\frac{i}{\sqrt{2}}(\cos n\pi +\sin n\pi)}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\cos n\pi +i\cos n\pi}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}\end{align}$$
How should I proceed?
| HINT: from your last line, consider the two different cases, according with $n$ odd and even.
| {
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What is the probability for $x$ to be positive only? If $x,y,z\in \mathbb R$ and $x+y+z=5,\, xy+yz+zx=3$, what is the probability that $x>0$ ? $$(a)\quad\frac3{16}\qquad (b)\quad\frac5{16}\qquad (c)\quad\frac{13}{16}\qquad (d)\quad \frac{15}{16}$$
I've tried forming a cubic equation and then trying to analyze its roots.
How to approach these type of questions?
| Given that $x+y+z=5$ and $xy+yz+zx=3$, we find that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=19$.
Now we find the range of $x$. For this, we write $y+z=5-x$ and $y^2+z^2=19-x^2$.
Now by a basic application of A.M-G.M Inequality, or Titu's Lemma, we can write $y^2+z^2 \geq \dfrac{(y+z)^2}{2} \implies 19-x^2 \geq \frac{(5-x)^2}{2} \iff (x+1)\left(x-\frac{13}{3}\right)\leq 0$.
Therefore $x \in \left[-1,\frac{13}{3}\right]$, thus the probability that $x>0$ is $\dfrac{\frac{13}{3}}{1+\frac{13}{3}}=\boxed{\frac{13}{16}}$.
| {
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Challenging Problem: $\int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big)$ $$ \int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big) $$ where $a,b,c,d,e$ are positive integers and $c$ and $e$ are square free numbers. Find $a+b+c+d+e$
My Attempt
$$ \int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =
\int_{-\infty} ^{\infty} \frac{16x \sin{3x} }{(2x-\sqrt{2}i-\sqrt{2})(2x-\sqrt{2}i+\sqrt{2})(2x+\sqrt{2}i-\sqrt{2})(2x+\sqrt{2}i+\sqrt{2})}dx =$$
by partial fractions
$$=\int -\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}+\sqrt{-2}\big) } +\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}-\sqrt{-2}\big) }+\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}+\sqrt{-2}\big) }-\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx=\frac{i}{2}\int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx +\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x+\sqrt{2}-\sqrt{-2}\big) }dx+\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}+\sqrt{-2}\big) }dx-\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx$$
Solving
$$\int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx$$ Substitute $u=2x+\sqrt{2}+\sqrt{-2}\longrightarrow \frac{du}{dx}=2$
$$\Rightarrow \int \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }=\frac{1}{2} \int \frac {\sin(\frac{3u}{2}-\frac{3i}{\sqrt{2}}-\frac{3}{\sqrt{2}})}{ u }du$$
applying the addition formula
$$=\int\frac {\cos \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \sin \big(\frac{3u}{2}\big) -\sin \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \cos \big(\frac{3u}{2}\big) }{ u }du$$ I understand the integral will not have an antidervative with sin(x) and cos(x) over x but, how do you proceed from here?
| Since the $\Im z>0$ roots of $z^{4}+1$ are $\frac{\pm1+i}{\sqrt{2}}$ and $\lim_{z\to a}\frac{z-a}{f\left(z\right)-f\left(a\right)}=\frac{1}{f^{\prime}\left(a\right)}$,$$\begin{align}\int_{\mathbb{R}}\frac{x\sin3x}{x^{4}+1}dx&=\Im\int_{\mathbb{R}}\frac{xe^{3ix}}{x^{4}+1}dx\\&=\Im\left(2\pi i\sum_{\pm}\lim_{z\to\frac{\pm1+i}{\sqrt{2}}}\frac{\left(z-\frac{\pm1+i}{\sqrt{2}}\right)ze^{3iz}}{z^{4}+1}\right)\\&=\Re\left(\frac{\pi}{2}\sum_{\pm}\lim_{z\to\frac{\pm1+i}{\sqrt{2}}}\frac{e^{3iz}}{z^{2}}\right)\\&=\frac{\pi}{2}e^{-3/\sqrt{2}}\Re\sum_{\pm}e^{\pm\left(\frac{3}{\sqrt{2}}-\frac{\pi}{2}\right)i}\\&=\pi e^{-3/\sqrt{2}}\sin\frac{3}{\sqrt{2}}.\end{align}$$Although Wolfram Alpha doesn't get it in such an elegant form, it does agree with the above value of $\approx0.320952$. Compared with $\pi^ae^{-b/\sqrt{c}}\sin\frac{d}{\sqrt{E}}$ (I've capitalised the last unknown for disambiguation) gives$$a=1,\,b=3,\,c=2,\,d=3,\,E=2\implies a+b+c+d+E=11.$$
| {
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Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$ There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$ is divisible by $x^3$ our polynomial is in the form of $ax^5+bx^4+cx^3.$ However, I'm not very sure how our second condition comes into use. Any help?
| A degree $5$ polynomial divisible by $(x+1)^3$ is of the form
$$(x+1)^3(ax^2+bx+c)$$
Expand, subtract $2$, and see when this is a multiple of $x^3$.
| {
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Identify the function represented by $\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)}$ So first I wrote it out in the terms, and I got
$\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)} = \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...$
I know the power series for $\displaystyle ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ which is similar to the derivative of the power series from above, as
$\displaystyle \frac{d}{dx}(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$
My question is, where do I go from here? How would I make it so the series is similar ln(1+x)? Or am I even going down the right route when it comes to solving this problem? Any help would be appreciated.
| You're very close. Since $$\log (1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots,$$ we have $$- \log (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots.$$ Then integration term by term gives $$- \int \log(1-x) \, dx = \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{12} + \frac{x^5}{20} + \cdots.$$ So all you need to do is figure out how to perform the integration. Do be careful, since the RHS should be zero when $x = 0$, so an antiderivative on the LHS should also be zero when $x = 0$.
| {
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Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!
| The given expression can be written as $\ (x-1)^2+(y+1)^2=2^2$
$\ x-1=2cos\theta$, $\ y+1=2sin\theta$
$\ x^2+y^2+4\sqrt2=(2cos\theta +1)^2+(2sin\theta-1)^2+4\sqrt2$
$\ =6+4\sqrt2+4(cos\theta-sin\theta)$
Therefore maximum value$\ =6+8\sqrt2$
| {
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Homographic function: alternative proofs to obtain $ad-bc$ Considering the function,
$$y=\frac{ax+b}{cx+d}\tag1$$
If $c = 0 \wedge d\neq 0$, the function represents a straight line of equation
$$y=\frac ad x+ \frac bd$$
If $c ≠ 0$ and $ad = bc$ the function represents a horizontal straight line. In fact, if
$$ad = bc \tag 2$$
we will have
$$ad/c = bc/c \iff ad/c = b$$
The coordinates of the point $P_0(-d/c,a/c)$ represent the asymptotes of hyperbola $(1)$. The importance of $(2)$ is due to the reason that if $ad-bc \neq 0$, using the traslation $\tau$,
$$\tau: \begin{cases}
X=x+\dfrac dc & \\
Y=y-\dfrac ac
\end{cases}
$$
I will obtain an equilater hyperbola. In fact
$$Y+\frac{a}{c}=\frac{a\Big(X-\frac{d}{c}\Big)+b}{c\Big(X-\frac{d}{c}\Big)+d}$$
$$Y=\frac{aX-\frac{ad}{c}+b}{cX-d+d}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b}{cX}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b-aX}{cX}$$
Hence:
$$Y=\frac{-\frac{ad}{c}+b}{cX}\Rightarrow XY=-\frac{ad}{c^2}+\frac{b}{c}\Rightarrow XY=k$$
with $$k=\frac{bc-ad}{c^2}$$
$$XY=k \tag 3$$
Starting from $(1)$ how can I create the condition quickly (step by step) $$\boxed{\color{orange}{ad-bc}} \quad ?$$
different from my proof?
| If $c\neq 0$, then $$y=\frac ac+\left(\frac{ax+b}{cx +d}-\frac ac\right)=\frac ac+\frac{bc-ad}{c(cx +d)}.$$
If $d\neq 0$, then $$y=\frac bd+\left(\frac{ax+b}{cx+d}-\frac bd\right)=\frac bd+\frac{(ad-bc)x}{d(cx+d)}.$$
| {
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Inequality with a High Degree Constraint This question-
Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$
The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes it difficult.
On trying C-S to manage-
$$ \left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3}\right)\left(x^5 + y^5 + z^5\right) \ge 9 \Rightarrow
\left(x^2y+y^2z+z^2x\right)^2\geq9 \Rightarrow x^2y+y^2z+z^2x\geq3 $$Still gives a third degree inequality and not a useful fifth degree.
How can I do it and solve the problem?
| Using the AM-GM inequality, we have
$$\frac{30 x^4}{y^3} +7x^{10}+y^{10}+16x^5y^5 \geqslant 54\sqrt[54]{\left(\frac{x^4}{y^3}\right)^{30} \cdot (x^{10})^7 \cdot y^{10} \cdot (x^5y^5)^{16}} = 54x^5.$$
Similar
$$\frac{30 y^4}{z^3} +7y^{10}+z^{10}+16y^5z^5 \geqslant 54y^5,$$
and
$$\frac{30 z^4}{x^3} +7z^{10}+x^{10}+16z^5x^5 \geqslant 54z^5.$$
Therefore
$$30\left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \right) +8(x^5+y^5+z^5)^2 \geqslant 54(x^5+y^5+z^5).$$
So
$${x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \geqslant .3$$
| {
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Let $\frac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$ Let $\dfrac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$
My attempt:
\begin{align*}
6\tan A(1-\tan^2A)&=\tan^4A+2\tan^2A+1\\
12\tan^2A&=\tan2A\tan^4A+2\tan2A\tan^2A+\tan2A\\
0&=\tan2A(\tan^4A)+(2\tan2A-12)\tan^2A+\tan2A\\
\because\tan2A&\in\mathbb{R}\\
\therefore \tan2A&\leqslant3
\end{align*}
From $\dfrac{1}{2}<\cos2A<1$ gives $0\leqslant\tan2A<\sqrt{3}$
Alfter using 2 inequality, I still can't find the exact value of $\tan2A$
| Note that the given eqn can be reduced to :
$\begin{align} & 6\tan A (1- \tan ^2 A) =(1+\tan^2A)^2 \\ \implies & 6\sin A \cos 2A \sec^3A=\sec^4A \\ \implies & 6\sin A\cos 2A\cos A=1\\ & \implies 3\sin 2A \cos 2A=1\\ & \implies \sin4A=2/3\\ & \implies \frac{2t}{1+t^2}=2/3 \\\end {align} $, where $t=\tan2A$. Solve for $t$. Can you take it from here?
| {
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Calculate list of all pairs $(x,y)$ The Number Theory question I am trying to solve is :-
Write $650$ as a product of irreducible elements in $\mathbb Z[i]$, then list all of the pairs $(x,y) \in$ $Z^2$ with $0 \le x \le y$ such that $x^2 + y^2 = 650$.
The way I approached this question is I computed the prime factorization of $650$ which is $2 \cdot 5^{2} \cdot 13$. Then I converted it into irreducible elements in $\mathbb Z[i]$ that came out to be $(1+i)(1-i)(2+i)^{2}(2-i)^{2}(3+2i)(3-2i)$. I am not able to get the second part of the question in which I need to list all the pairs $(x,y)$. I was able to compute the number of solution using Fermat's Sum of Two Squares which is $24$ but I am not able to list the pairs.
Can someone please help me with that part?
| You need to multiply the values $(1+i),(1-i),(2+i),(2+i),(2-i),(2-i),(3+2i),(3-2i)$ in ways that make complimenary pairs. where $(x+yi)(x-yi)=x^2 + y^2 = 650$
Note that $(a+ bi)(c+di) = (ac-bd) + (bc+da)i$ while $(a-bi)(a-di)= (ac -bd)-(bc+da)i$
So the way to make complimentary pairs are $x+yi = (1+ i)(2\pm i)(2\pm i)(3\pm 2i)$ and $x-yi = (1- i)(2\mp i)(2\mp i)(3\mp 2i)$. In all of these $x^2 + y^2 = 650$.
So there should be $8$ such pairs of $(x,y)$.
$(1+i)(2+i)(2+i)(3+2i)= (1+i)(3+4i)(3+2i)= (-1+7i)(3+2i)=-17+19i$ so $17^2 + 19^2=650$ and $(x,y) =(17,19)$.
$(1+i)(2+i)(2+i)(3-2i) = (-1+7i)(3-2i) = 11 +23i$ so $11^2 + 23^2 = 650$ and $(x,y)=(11,23)$.
And so on.
$(1+i)(2+i)(2-i)(3-2i)=(1+i)5(3-2i)=5(5+i)=25 + 5i$ so $25^2 + 5^ = 650$ and $(x,y)=(5,25)$.
But the is duplicated with $(1+i)(2-i)(2+1)(3-2i)$.
$(1+i)(2+i)(2-i)(3+2i)=5(1+i)(3+2i) = 5(1+5i)$ which is a permutation.
$(1+i)(2-i)^2(3\pm 2i)$ are the final $2$. $(3-4i)(1+i)(3\pm 2i)=(7-i)(3\pm 2i)= \{23+11i;19-17i\}$.
Hmmm... not entirely sure how I could have avoided the duplicates....
But we have three such pairs $(5,25), (11,23), (17,19)$
| {
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Finding $a$ such that $ \sqrt{\frac32x^2-xy+\frac32y^2}=x\cos a+y\sin a$ has at least one solution other than $(0,0)$ Find all values of the parameter $ a $ from the interval $ [0, 2 \pi) $, for which the equation
$$ \sqrt{\dfrac{3}{2}x^2 - xy + \dfrac{3}{2}y^2} = x \cos a + y \sin a $$
has at least one solution $ (x, y) $ other than $ (0,0) $.
| We know that $x\cos a+y\sin a\leq \sqrt{x^2+y^2}$ and so, $$ \sqrt{\dfrac{3}{2}x^2 - xy + \dfrac{3}{2}y^2} \leq\sqrt{x^2+y^2}\implies (x-y)^2\leq0\implies x=y$$
Now its trivial from here imho.
| {
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Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$ I was trying this question with factorization and other similar methods,
Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\geq a\sqrt{2a^2+bc}+b\sqrt{2b^2+ca}+c\sqrt{2c^2+ab}
\end{array}$$
This is one of Hoojoo-Lee's Inequality. This seems very intuitive at first as if we square each term, $$ 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2} \geq 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2bc} \Rightarrow \sum_{cyc}{a^2b^2} \geq \sum_{cyc}{a^2bc} $$ which is quite clear. I noticed it but can not exploit it. May be taking the square on each side could help? But I couldn't find a solution.
Please help!
| I think the key tools here are Cauchy-Schwarz and AM-GM. I just bound one term on the LHS. The other two are similar:
$$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=\sum_{cyc}\sqrt{(a^4+\frac{a^2b^2}{2})+(b^4+\frac{a^2b^2}{2})}\ge$$$$\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{b^4+\frac{a^2b^2}{2}}\right)$$$$=\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{a^4+\frac{a^2c^2}{2}}\right)\ge\sqrt{2}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}\sqrt{a^4+\frac{a^2c^2}{2}}\right)^{1/4}$$$$\ge\sqrt{2}\sum_{cyc}\sqrt{a^4+\frac{a^2bc}{2}}=\sum_{cyc}\sqrt{2a^4+a^2bc}$$
I use Cauchy-Schwarz at the first and third inequality, and AM-GM at the second.
| {
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We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .
What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
| If you happen to know eigenvalues/eigenvectors, your issue can be written :
$$\underbrace{\begin{pmatrix}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{pmatrix}}_A\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=r\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}$$
meaning that $r$ must be an eigenvalue of matrix $A$.
The eigenvalues of $A$ are $3$ and $-1$ ($-1$ being a
triple eigenvalue, but the text allows to take it only once) associated to resp. eigenvectors:
$$\begin{pmatrix}a\\b\\c\\d\\\end{pmatrix}=\begin{pmatrix}1\\1\\1\\1\\\end{pmatrix},\begin{pmatrix}1\\1\\1\\-3\\\end{pmatrix}$$
fulfilling the condition that neither $a$, nor $b,c,d$ are zero (being in the denominator of the initial equation)
Therefore the answer is $3+(-1)=2$.
| {
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How can I study the convergence of the improper integral $\int_{2}^{\infty} \frac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$? I need to study the convergence of the following improper integral:
$$\int_{2}^{\infty} \dfrac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$$
I did the following:
$$ -\dfrac{\pi}{2} < \arctan(x+1) < \dfrac{\pi}{2} \\
\implies -\dfrac{\pi}{2} + x < \arctan(x+1) +x < \dfrac{\pi}{2} +x \\
\implies \dfrac{-\dfrac{\pi}{2} + x}{2^x+3^x} < \dfrac{\arctan(x+1) +x}{2^x+3^x} < \dfrac{\dfrac{\pi}{2} + x}{2^x+3^x} \\
$$
I planned to integrate the inequality and then using the comparison criterion to proof its convergence. However, the idea did not work for me.
| You have a polynomial function over an exponential one. This dies fast enough at $\infty$ to integrate there.
| {
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Convergence of $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ The problem is:
prove $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely for all $z\neq 0$, but does not converge uniformly near $z=0$.
Proof:
for all $z\neq 0$
$$\left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|
= \left|
2^n\left(
-\frac{(\frac{1}{3^nz})^3}{3!}+\frac{(\frac{1}{3^nz})^5}{5!}\dots
\right)
\right|, \ \ \ \ (1)$$
$\exists~ N$ such that when n>N, $\frac{1/z}{3^n}<1$, so that (1) is less than
$$\left|
2^n\left(
\frac{|\frac{1}{3^nz}|^3}{3!}+\frac{|\frac{1}{3^nz}|^5}{5!}\dots
\right)
\right|
<\left|
\frac{2^n}{3^nz}\left(
\frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|
<\frac{2^n}{3^n}\left|
\frac{1}{z}\left(
\frac{|\frac{1}{3^nz}|^2}{1-|\frac{1}{3^nz}|^2}
\right)
\right|,
$$
$\forall~ \epsilon, \exists~ N_1>-\log(\epsilon^{1/2} z^{3/2})$, such that when $n>N_2=\max\{N, N_1\}$, $|\frac{1}{z}||\frac{1}{3^nz}|^2<\epsilon$, and so (1) is less $\frac{2^n}{3^n}\epsilon$.
Therefore, we have $N_2(\epsilon)$ satisfying that $\forall~ p,$
$$\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}\right|
<\sum_{n=N_2}^{N_2+p} \left|2^n\sin\frac{1}{3^nz}-2^n\frac{1}{3^nz}\right|\ \ \ \ (2)\\
<\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}\epsilon
\leq 2\epsilon,
$$
and so $\sum_{n=1}^\infty 2^n\sin\frac{1}{3^nz}$ converges absolutely. (A step seems to be missing. One should, instead of $\sum_{n=N_2}^{N_2+p}\frac{2^n}{3^n}$, use something like 2/z (plus a constant), which is the limit of the former.)$\blacksquare$
A possibly trivial question is whether it is proper to prove this way: given n sufficiently large, $u_n<f(n)\epsilon$ (different from that $u_n<\epsilon$, or that $u_n/f(n)<\epsilon$ and so $u_n<f(n)\epsilon$).
There are other questions that I may post somewhere else.
The following is a proof that the series doesn't converge uniformly. It's unnecessarily for answering my questions, but I put it here for completeness of proof.
Proof:
However large $N_2$, we can find $n_0$>$N_2$ and $z=\frac{2}{\pi 3^{n_0}}$ such that $2^{n_0}\sin\frac{1}{3^{n_0}z}=2^{n_0}$, and so (say the limit function is $f(z)=\frac{2}{z}+C$, where $C$ is a constant; I suddenly realize C seems also to be a function of z, that could cause some issues),
$$\left|\sum_{n=1}^{\infty} 2^n\sin\frac{1}{3^nz}-f(z)\right|
>\left|\sum_{n=N_2}^{N_2+p} 2^n\sin\frac{1}{3^nz}-\sum_{n=N_2}^{N_2+p} 2^n\frac{1}{3^nz}-C\right|>|2^{n_0}-\frac{2}{z}-C|>\epsilon.$$
| One easily proves the result even uniformly for $z \in K$ compact that doesn't contain $0$ since for all such there is $A=A_K >0, |z| \ge A, z \in K$; then pick $n_K, (8/9)^n \le A$ or equivalently $(2/3)^n \le A(3/4)^n, n \ge n_K$ and use that for $|w| \le 1/2, |\sin w| \le C|w|$ which can be easily seen since $|\sin w /w|$ is bounded and continuos there;
then for $n \ge n_K, z \in K, |3^nz| \ge A3^n \ge 2$ since $A(9/8)^n \ge 1$ by our choice of $n_K$, hence $|\sin (1/(3^nz)| \le C/|3^nz|$ or $2^n|\sin (1/(3^nz)| \le C(3/4)^n$ again by our choice of $n_K$
So one has uniformly in $z \in K$ that $\sum_{n=1}^\infty |2^n\sin\frac{1}{3^nz}| \le f_K(z)+ \sum_{n \ge n_K}C(3/4)^n$ where $f_K$ is the finite sum of the absolute values of the terms from $1$ ro $n_K-1$ so it is a fixed (for $K$ fixed) continuos function on $K$, so in particularly the OP series converges absolutely and uniformly on $K$
For part $2$ and $z=1/3^N$ the $N$ term of the series is $2^N/\sin 1$ so is unbounded with $N$ which shows that the partial series cannot be uniformly bounded for the sequence $z_N \to 0$, so in particular, they cannot converge uniformly
| {
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An uncertainty for $a^3+b^3+c^3-3abc$ I have a doubt regarding $a^3+b^3+c^3-3abc$.
For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see-
$$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$
AM-GM inequality ensures that equality holds if and only if all variables are equal.
So, equality holds if and only if $a=b=c$, which is trivial.
But we see by factoring that equality holds also if $a+b+c=0$. As the inequality is nothing more than a mere AM-GM, so equality should hold where the inequality ensures us it holds. But it also holds for $a+b+c=0$.
How is it possible and if it is, how can I find such equality cases?
| The Arithmetic-Geometric Mean inequality works for only non-negative real numbers. So, for non-negative real numbers equality holds if and only if $a=b=c.$
| {
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$\sum_{r=0}^{19} (r+1)^4 \binom{20}{r+1}=\lambda. \, 2^{16}$
If $$\sum_{r=0}^{19} (r+1)^4 \binom{20}{r+1}=\lambda. \, 2^{16}$$ Find $\lambda$.
I wrote it as $20\sum_{r=0}^{19} (r+1)^3\binom{19}{r+1}$. Then expanding cube will do the work, but that will be really tedious. Any short way?
| Hint:
Set $r+1=n$
$$\sum_{n=1}^{20}n^3\binom{19}n$$
Now write $n^3=n(n-1)(n-2)+An(n-1)+Bn$ so that $$n^3\binom{19}n=n(n-1)(n-2)\binom{19}n+An(n-1)\binom{19}n+Bn\binom{19}n$$
$n=1\implies B=1$
$n=2\implies2^3=2A+2B\iff B=?$
| {
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Prove two series are equal
Prove $I=J$, where:
$$I=\left\{\frac{4}{\pi^6}\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=1}^{\infty}\frac{1}{n^2m^2\sqrt{n^2+m^2}}\left(\pi\frac{e^{\pi\sqrt{n^2+m^2}}+e^{-\pi\sqrt{n^2+m^2}}}{e^{\pi\sqrt{n^2+m^2}}-e^{-\pi\sqrt{n^2+m^2}}}-\frac{1}{\sqrt{n^2+m^2}}\right)\right\}^{-1},$$
and
$$J=12\pi^2\left(\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\right)^{-1}.$$
My atempt: We have
\begin{align*}
\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+1)^2}+\frac{1}{(3n+2)^2}\right\}&=\displaystyle\sum_{n=1}^{\infty}\left\{\frac{1}{(3n-2)^2}+\frac{1}{(3n-1)^2}+\frac{1}{(3n)^2}\right\}-\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n)^2}\\
&=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}-\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n)^2}\\
&=\frac{8}{9}\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{4\pi^2}{27}\\
\end{align*}
and:
\begin{align*}
\displaystyle\sum_{n=0}^{\infty}\frac{1}{(3n+2)^2}&=4\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+4)^2}\\
&=4\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(6n+1)^2}+\frac{1}{(6n+4)^2}\right\}-4\displaystyle\sum_{n=1}^{\infty}\frac{1}{(6n+1)^2}\\
&=4\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+1)^2}-\frac{1}{(6n+1)^2}\right\}\\
\end{align*}
So :
\begin{align*}
\frac{4\pi^2}{27}&=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(3n+1)^2}+4\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{(3n+1)^2}-\frac{1}{(6n+1)^2}\right)\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\\
&\implies \frac{1}{12\pi^2}\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}=\frac{1}{81}\\
&\implies 12\pi^2\left(\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\right)^{-1}=81\\
\end{align*}
Finally $ J=81$, but how to prove $I=81$?
Any help for how to calculate $I$? Thanks in advance
| The first step is to recognize the fraction involving $e^{\pi\sqrt{n^2+m^2}}$ as $\coth (\pi\sqrt{n^2+m^2})$. Then we can use the Mittag-Leffler expansion of $\coth$,
$$
\pi\coth(\pi z) = \frac{1}{z} + 2 \sum_{n=1}^\infty \frac{z}{z^2+n^2}\ .
$$
Applying this to $z = \sqrt{m^2+n^2}$, the stuff inside the parantheses in $I$ can be rewritten as
$$
\frac{1}{\sqrt{m^2+n^2}} + 2\sum_{p=1}^\infty \frac{\sqrt{m^2+n^2}}{m^2+n^2+p^2} - \frac{1}{\sqrt{m^2+n^2}}\ .
$$
After canceling out the $1/\sqrt{m^2+n^2}$, note that the $\sqrt{m^2+n^2}$ factor outside of the parentheses also gets canceled.
Thus
$$
\frac{\pi^6}{4}I^{-1} = \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{2}{m^2n^2(m^2+n^2+p^2)}\ .
$$
To evaluate this sum, note that if we permute the dummy variables $m\mapsto n$, $n\mapsto p$, $p\mapsto m$, the summand can be rewritten as
$$
\frac{2}{n^2p^2(m^2+n^2+p^2)}\ .
$$
Permutting again turns the summand into
$$
\frac{2}{p^2m^2(m^2+n^2+p^2)}\ .
$$
Since relabeling the dummy variables clearly doesn't change the value of the series, we could add all three representations together and get
$$
\begin{align}
3\left(\frac{\pi^6}{8I}\right) &= \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \left(\frac{1}{m^2n^2(m^2+n^2+p^2)} +\right.\\
& \quad\left. \frac{1}{n^2p^2(m^2+n^2+p^2)} + \frac{1}{p^2m^2(m^2+n^2+p^2)}\right) \\
&= \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{p^2+m^2+n^2}{m^2n^2p^2(m^2+n^2+p^2)}\\
& = \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{1}{m^2n^2p^2} \\
& = \left(\sum_{m=1}^\infty \frac{1}{m^2}\right)^3 \\
& = \left(\frac{\pi^2}{6}\right)^3
\end{align}
$$
From this we can easily solve for $I = 81$.
| {
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Why If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ If $x^2=-\frac{1}{3}$ then $x=\pm\frac{i\sqrt{3}}{3}$ according to my text book.
I understand that $x=\pm\sqrt{-\frac{1}{3}}$ but do not fully grasp how to express this as $\pm\frac{i\sqrt{3}}{3}$
I understand that $i$ is $\sqrt{-1}$
As far as I can get with my understanding starting from:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{\sqrt{-3}}$$
$$x=\pm\frac{i}{\sqrt{3}i}$$
$$x=\pm\sqrt{3}$$
Where did I go wrong and how can I arrive at $x=\pm\frac{i\sqrt{3}}{3}$
[Edit]
From the comments I now know that $-\frac{1}{3}$ = $\frac{-1}{3}$
Taking that back into my working I still arrive at a different answer:
$$x=\pm\sqrt{-\frac{1}{3}}$$
$$x=\pm\sqrt{\frac{-1}{3}}$$
$$x=\pm\frac{\sqrt{-1}}{3}$$
$$x=\pm\frac{i}{3}$$
How can I get to $x=\pm\frac{i\sqrt{3}}{3}$
| What you did wrong was the fact that
$$-\frac13=\frac{-1}{3}=\frac{1}{-3}$$
and NOT
$$-\frac13=\frac{-1}{-3}$$
After the editing the problem is:
$$\sqrt{-\frac13}=\sqrt{\frac{-1}{3}}=\frac{\sqrt{-1}}{\sqrt{3}}=\frac{i}{\sqrt{3}}=\frac{i}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}}=\frac{i\sqrt{3}}{3}$$
| {
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| $x(x-8)=17$ is a true (but not useful) statement.
But $ab = m$ most certainly does !NOT! mean $a =m$ or $b=m$.
Consider $3\cdot 5 = 15$ where neither $3=15$ nor $5 = 15$. Or consider $\frac 2{5}\cdot {85}2 = 17$ where neither $\frac 12$ nor $\frac {85}2$ equal $17$.
In fact $x(x-8)=17$ CAN'T mean $x =17$ unless $x-8=1$. But $17-8 \ne 9$. And $x-8$ CAN'T equal $17$ unless $x =1$ and $1-8 \ne 17$.
So your solution is definitely wrong.
We could true $x = \frac {17}{x-8}$ and $x-8 = \frac {17}x$ neither of which are equal to $1$ or $17$ but .... that doesn't get us anywhere. (Also this assumes $x, x-8$ are neither $0$... which we can assume because if either were $0$ their product would have to be $0$ and not $17$.)
.....
We can't to $ab=m$ means $a=m$ or $b=m$ because that's just garbage.
But we can do $ab = 0$ means $a=0$ or $b=0$. Why does this work for $0$ and not for $m$?
Because $0\cdot anything = 0$ always happens while $m \cdot x = m$ doesn't always happen. And because $ab=m$ where $a\ne m; b\ne m$ can happen $ab = 0$ where $a\ne 0; b\ne 0$ can NOT happen.
....
So we could solve this by factoring $x^2 -8x + 17$ to $(x+a)(x+b)$ and having $(x+a)(x+b) = 0$. so $x+a =0$ or $x+b = 0$.
The only trouble is .... we don't know how to factor $x^2 - 8x + 17$......
But we can complete the square. If we have $x(x-8) = 17$ we can say anything about $x$ or $x-8$. But if instead we had $(x+a)(x+a) = M$ where $x+a$ and $x+a$ are the same things and not different things. the we can conclude if $(x+a)^2 = M$ then $x+a = \pm \sqrt M$.
And we can find an $x^2 -8x + something\ that\ isn't\ 17$ is a perfect square so we can do
$x^2 - 8x + 17 =0$
$x^2 - 8x= -17$
$x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = -17 ++something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square$
So what is $something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square$?
If $x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = (x+ \ who \ knows)^2$ then
$x^2 - 8x +something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = x^2 + 2\ who \ knows\ x + \ who\ know^2$.
So $-8x = 2\ who\ knows\ x$ so $\ who\ knows = \frac {-8}2 = -4$.
And $something\ that\ isn't\ 17\ but\ which \makes\ this \side\ a\ perfect\ square = \ who\ know^2= (-4)^2 = 16$.
.....
SO.....
$x^2 -8x + 17 = 0$
$x^2 - 8x = -17$
$x^2 - 8x + 16 = -17 +16$
$(x-4)^2 = -1$.
Now that means $x-4 = \pm \sqrt{-1}$.
AND $\sqrt{-1} = i$! THAT IS WHERE THE $i$ CAME FROM!
And so $x - 4 =\pm i$
And $x = 4 \pm i$. So $x = 4+i$ or $x = 4-i$.
| {
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How to find out whether these are the same? In the wikipedia article for Cubic equation, the root can be obtained by:
$-\frac{1}{3a}(b+C+\frac{\Delta_0}{C})$
Where $\Delta_0=b^2-3ac$ and $C=\sqrt[3]{\frac{\Delta_1\pm\sqrt{\Delta_1^2-4\Delta_0^3}}{2}}$. Also, $\Delta_1=2b^3-9abc+27a^2d$.
In another website, there's another root solution:
$$\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})+\sqrt{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}+\sqrt[3]{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})-\sqrt{(-\frac{b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}-\frac{b}{3a}$$
I have put the latter in Wolphram|Alpha to evaluate it. the $\Delta_1$ can be seen in it; but I have no idea how to find out it and the previous solution are the same.
| Define:
$$\begin{align*} x_N &= -\dfrac{b}{3a} \quad \text{(average of all 3 roots, x-value of inflection point)} \\
\\
\delta^2 &= \dfrac{b^2-3ac}{9a^2} \quad \mathrm{(x \; distance^2 \; from \;} x_N \; \text{to the 2 turning points)}\\
\\
y_N &= f(x_N) = \dfrac{2b^3}{27a^2}-\dfrac{bc}{3a} +d \quad \text{(y-value of inflection point)}\\
\\
h &= 2a\delta^3 \quad \mathrm{(y \; distance \; from \;} y_N \; \text{to the 2 turning points)} \\
\end{align*}$$
(See figure 1 in this paper by Nickalls: http://www.nickalls.org/dick/papers/maths/cubic1993.pdf)
The second expression you presented can then be written as
$$x_N + \sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) } + \sqrt[3]{\dfrac{1}{2a}\left(-y_N - \sqrt{y_N^2 - h^2}\right) } $$
or, for $h \ne 0$,
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} + \sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }} \right) $$
In the first expression you presented, we have
$$\begin{align*} \Delta_0 & = 9a^2 \delta^2 \\
\\
\Delta_1 &= 27a^2 y_N \\
\\
C &= -3a \sqrt[3]{\dfrac{1}{2a}\left(-y_N \mp \sqrt{y_N^2 - h^2}\right) }\\
\end{align*}$$
so that expression becomes
$$ x_N + \sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) } + \dfrac{\delta^2}{\sqrt[3]{\dfrac{1}{2a}\left(-y_N + \sqrt{y_N^2 - h^2}\right) }}$$
or, for $h \ne 0$,
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} + \dfrac{1}{\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} }\right) $$
which after multiplying the numerator and denominator of that last term in the parentheses by $$\sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }}$$
becomes
$$x_N + \delta\left(\sqrt[3]{\dfrac{-y_N}{h} + \sqrt{\dfrac{y_N^2}{h^2} - 1 }} + \sqrt[3]{\dfrac{-y_N}{h} - \sqrt{\dfrac{y_N^2}{h^2} - 1 }} \right) $$
So yes indeed, those two expressions for the roots of the cubic you found are equivalent.
Now I encourage you to throw away all of that classical solution for the roots of a cubic, and instead learn Nickalls' approach as presented by Nickalls and built upon by Holmes:
http://www.nickalls.org/dick/papers/maths/cubic1993.pdf
https://users.math.msu.edu/users/newhous7/math_235/lectures/cubic_gc_holmes.pdf
| {
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Is there an efficient way of showing $\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx = 2$? Hi I recently came across the following integral:
$$
\int_{-1}^{1} \ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx
$$
When an integral calculator finds the antiderivative of the equation (if it even can) it comes out as this crazy formula.
Anyways the definite integral happens to be equal to 2 and I was wondering if there might be an elegant way of showing that this is the case.
Thank you.
| $$I=\int_{-1}^1\ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx=2\int_{0}^1\ln\left(\frac{2(1+\sqrt{1-x^2})}{1+x^2}\right)dx$$
$$=2\int_0^1 \ln(2)dx+2\int_0^1\ln(1+\sqrt{1-x^2})dx-2\int_0^1\ln(1+x^2)dx$$
$$=2I_1+2I_2-2I_3$$
$$I_1=\boxed{\ln(2)}$$
$$I_2\overset{IBP}{=}x\ln(1+\sqrt{1-x^2}))|_0^1+\int_0^1\frac{x^2}{\sqrt{1-x^2}(1+\sqrt{1-x^2})}dx$$
$$\overset{x=\sin\theta}{=}\int_0^{\pi/2}\frac{\sin^2\theta}{1+\cos\theta}d\theta=\int_0^{\pi/2}\frac{1-\cos^2\theta}{1+\cos\theta}d\theta$$
$$=\int_0^{\pi/2}(1-\cos\theta)d\theta=\boxed{\frac{\pi}{2}-1}$$
$$I_3\overset{IBP}{=}x\ln(1+x^2)|_0^1-\int_0^1\frac{2x^2}{1+x^2}dx$$
$$=\boxed{\ln(2)-\left(2-\frac{\pi}{2}\right)}$$
Combining the boxed results gives $2$.
| {
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Integration with surds, problem taking out $x^2$ this my first post so I apologise for any conventional structure features I might be violating.
I'm trying to evaluate the indefinite integral:
$$\int\sqrt{x^2-2x^4}dx$$
I recognise that this question wants me to simplify the expression $\sqrt{x^2-2x^4}$ into $x\sqrt{1-2x^2}$
but what confused me was the fact that the absolute value of $x$ here was not taken out instead to obtain the solution:
$$-\frac{1}{6}(1-2x^2)\sqrt{1-2x^2}+C$$
I thought the simplification would have been:
$$|x|\sqrt{1-2x^2}$$
For which the indefinite integral would take on two separate values, one for each branch that changes the sign of $x$. At first, I suspected the natural domain of the square root implicitly stated that $x\geq0$ but at a closer inspection, I calculated the domain restriction as follows:
$$-\frac{\sqrt{2}}{2}\leq x \leq\frac{\sqrt{2}}{2}$$
Could somebody so kindly advise me where I have made a mistake? Any help will be greatly appreciated!
| As you noted, for $0\leq x\leq1/\sqrt{2},$ the factor $x^2$ can be taken out as $x,$ and we obtain
$$
-\frac{1}{6}\sqrt{(1-2x^2)^{3}}+C^+.
$$
For $-1/\sqrt{2}\leq x\leq0$ the factor $x^2$ can be taken out as $-x,$ and we obtain
$$
+\frac{1}{6}\sqrt{(1-2x^2)^{3}}+C^-.
$$
Note that, if we erroneously take the first one for the whole interval, we end with the absurd conclusion
$$
\int_{-1/\sqrt{2}}^{+1/\sqrt{2}}\sqrt{x^2-2x^4}dx=-\frac{1}{6}\sqrt{(1-2x^2)^{3}}\Biggr|_{-1/\sqrt{2}}^{+1/\sqrt{2}}=0
$$
If we would like to use the result to calculate a definite integral over an interval, we need a primitive that is a continuous function over that interval.
To this end, the two constants can be chosen independently and it is convenient to choose them so to have a unique function continuous in $[-1/\sqrt{2},+1/\sqrt{2}].$ If we choose
$$
C^+=+1/6,\\
C^-=-1/6
$$
we have
$$
\int\sqrt{x^2-2x^4}dx=F(x)=\begin{cases}
-\frac{1}{6}\sqrt{(1-2x^2)^{3}}+\frac{1}{6} & 0\leq x\leq 1/\sqrt{2} \\
+\frac{1}{6}\sqrt{(1-2x^2)^{3}}-\frac{1}{6} & -1/\sqrt{2} \leq x \leq 0
\end{cases}
$$
with $F$ continuous also in $x=0,$ so that, for example
$$
\int_{-1/\sqrt{2}}^{1/\sqrt{2}}\sqrt{x^2-2x^4}dx=\left(0+\frac{1}{6}\right)-\left(0-\frac{1}{6}\right)=\frac{1}{3}>0.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Probability of selecting a poker hand I am trying to solve a probability problem about five-card poker hand. I have access to the answer which is different from what I had come up with. The question is: What is the probability that a five-card poker hand has exactly two cards of same value, but no other cards duplicated?
My answer to this question was as follows:
$\binom{13}{1} \binom{4}{2} \binom{48}{1}\binom{44}{1} \binom{40}{1}$.
Which means:
*
*First select a card number then select its two suits ie. $\binom{13}{1} \binom{4}{2}$. These will be the two cards of same value.
*Select three other cards which are not duplicate as: $\binom{48}{1}\binom{44}{1} \binom{40}{1}$.
The correct answer doesn't match my answer. This answer is provided in book AOPS and is as: $\binom{13}{1} \binom{4}{2}\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$.
So question is, what am I doing wrong? Thanks
| You and the book count differently how to select the three remaining cards. Your answer is:
$$ \binom{48}{1}\binom{44}{1} \binom{40}{1} = 48 \cdot 44 \cdot 40 = 4^3 \cdot 12 \cdot 11 \cdot 10$$
The book answer is:
$$\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} = 4^3 \cdot \frac{12\cdot 11\cdot 10}{3!}$$
They differ by a $3!$ factor, which is precisely the number of permutations of three distinct objects. This suggests that you are considering the order of the three remaining cards.
| {
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"source": "stackexchange",
"question_score": "4",
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How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition? I have a function defined as follows:
$$
f(x)= \frac{\sqrt{x+1}}{2-x}
$$
I tried to calculate the derivative using the limit definition using four methods, but I was unsuccessful in any. Could someone help me calculate it and explain the method?
$$
1) \lim_{h\to 0} =\frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h
$$
$$
2)\lim_{z\to x} =\frac{\frac{\sqrt{z+1}}{2-z}-\frac{\sqrt{x+1}}{2-x}}{z-x}
$$
$$
3)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u=\sqrt{x+1}
$$
$$
\lim_{h\to 0} =\frac{\frac{u+h}{3-(u+h)^2}-\frac{u}{3-u^2}}h
$$
$$
4)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u={x+1};
$$
$$
\lim_{h\to 0} =\frac{\frac{\sqrt{u+h}}{3-(u+h)}-\frac {\sqrt{u}}{3-u}}h
$$
| We have that
$$\frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h=\frac{\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}\right)\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)}{h\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)}=$$
$$=\frac{\frac{(x+h)+1}{(2-(x+h))^2}-\frac{x+1}{(2-x)^2}}{h\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)}=\frac{\frac{((x+h)+1)(2-x)^2-(x+1)(2-(x+h))^2}{(2-(x+h))^2(2-x)^2}}{h\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)}=\ldots$$
and since (the cancellation in red is the crucial step)
$$((x+h)+1)(2-x)^2-(x+1)(2-(x+h))^2=$$
$$=\color{red}{(x+1)(2-x)^2}+h(2-x)^2\color{red}{-(x+1)(2-x)^2}+2h(x+1)(2-x)-h^2(x+1)=$$
$$=h(2-x)^2+2h(x+1)(2-x)-h^2(x+1)$$
we obtain
$$\ldots=\frac{\frac{h(2-x)^2+4h(x+1)(2-x)+h^2(x+1)}{(2-(x+h))^2(2-x)^2}}{h\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)}=\frac{(2-x)^2+2(x+1)(2-x)-h(x+1)}{\left(\frac{\sqrt{(x+h)+1}}{2-(x+h)}+\frac{\sqrt{x+1}}{2-x}\right)\left((2-(x+h))^2(2-x)^2\right)}\to$$
$$\to \frac{(2-x)^2+2(x+1)(2-x)}{\left(\frac{\sqrt{x+1}}{2-x}+\frac{\sqrt{x+1}}{2-x}\right)(2-x)^4}=\frac{(2-x)+2(x+1)}{2\sqrt{x+1}(x-2)^2}=\frac{x+4}{2\sqrt{x+1}(x-2)^2}$$
| {
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Find the ratio of the area of the original triangle to the area of the equilateral triangle at Gergonne point If $G$ is the Gergonne point of $\triangle PQR$ and $A, D, E$ are the contact points of the incircle such that $\triangle PGE$ is equilateral. What is the ratio of area $\dfrac{\triangle PQR}{\triangle PGE}$?
My attempt:
Let's put this in $X$-$Y$ coordinates with $C$ as origin and line $PQ$ being parallel to the $X$-axis and $EC$ being the $Y$-axis ($C$ is the center of the incircle and radius $r$).
$\angle APQ = \angle PER = \alpha = 60^0$; Say, $PG = EG = PE = 1$ unit
Coordinates of points: $E(0,-r), P(-1,-r), C(0,0)$.
The equation of the line $PR$: $y + r = m(x+1)$
As line $PR$ is tangent to the circle, the perpendicular from center $C$ will be equal to the radius.
$r = \dfrac{r - m}{\sqrt{m^2+1}}$ or, $m = \dfrac{2r}{1-r^2}$
So, the equation of the line $PR$: $y + r = \dfrac{2r}{1-r^2} (x+1)$ ...(1)
Equation of line $ER$: $y + r = -x \tan\alpha$ ...(2)
Equating (1) and (2), we get the intersection point $R (x_0,y_0)$
[$\dfrac{-2r}{2r+(1-r^2)\tan \alpha}, \dfrac{2r\tan \alpha}{2r+(1-r^2)\tan \alpha} - r$], where $\tan \alpha = \tan 60^0 = \sqrt3$
Equation of line $QR$:
$y - y_0 = m_0 (x - x_0)$
Again equating perpendicular distance from the center and the radius
$r = \dfrac{m_0x_0 - y_0}{\sqrt{m_0^2+1}}$
or, $(x_0^2-r^2)m_0^2 - 2x_0y_0m_0 + (y_0^2 - r^2) = 0$
As there are two lines passing through $(x_0,y_0)$ that are tangent to the circle ($PR$ and $QR$), the quadratic equation will give us two values of $m_0$, one of them being $m_0 = \dfrac{2r}{1-r^2}$ as we have established before.
This is how far I reached. I tried to replace values of ($x_0, y_0$) in the quadratic equation and we already know one of the roots but it is becoming very tedious to find the slope of line $QR$.
Once I know the slope, I can find the point $A$ and as we know the equation of line $PA$, I should be able to find the value of $r$ as well as coordinates of points $P, Q, R$. That will lead to us finding the area of $\triangle PQR$ and the ratio to $\triangle PGE$ (area of $\triangle PGE = \dfrac {\sqrt3}{4}$ units).
Any suggestions on how to simplify the solution? Are there any known properties of gergonne point that I am missing which makes the solution simpler?
| $\begin{array}{} λ(C,r) & ω(P,1) & C=(0,0) \\ P=(-1,-r) & E=(0,-r) \\ \text{ΔPGE is equilateral} & PG=EG=EP=1 & G=(\frac{-1}{2},\frac{\sqrt{3}}{2}-r) \\ \end{array}$
$\begin{array}{} λ∩ω=D & PG∩λ=A & DG∩PE=Q \\ AQ⟂AC & m_{AQ}·m_{AC}=-1 & PD∩EG=R \\ x^2+y^2=r^2 & (x+1)^2+(y+r)^2=1 & x+r·y+r^2=0 \\ \end{array}$
$\begin{array}{} y=\frac{-(x+r^2)}{r} & D=\left( \frac{-2r^2}{r^2+1},\frac{r(1-r^2)}{r^2+1} \right) \end{array}$
$\begin{array}{} PG∩λ=A & \left| \begin{array}{} x & y & 1 \\ -1 & -r & 1 \\ \frac{-1}{2} & \frac{\sqrt{3}}{2}-r & 1 \\ \end{array} \right|=0 & y=x·\sqrt{3}+\sqrt{3}-r & x^2+y^2=r^2 \end{array}$
$\begin{array}{} x_{A}=\frac{1}{4}\left( \sqrt{3r^2+2\sqrt{3}r-3}+\sqrt{3}r-3 \right) & y_{A}=x_{A}·\sqrt{3+\sqrt{3}-r} \end{array}$
$\begin{array}{} x_{Q}=\frac{2r}{\sqrt{3}-r} & y_{Q}=-r \end{array}$
$\begin{array}{} m_{AQ}·m_{AC}=-1 & ⇒ & -\sqrt{3r^2+2r\sqrt{3}-3}+3r^2-9+(\sqrt{3r^2+2r\sqrt{3}-3})\sqrt{3}r)=0 & ⇒ \\ r=\frac{7\sqrt{3}}{9}≈1.3471506 \\ \end{array}$
$ratio=\frac{ΔPQR}{ΔPGE}=\frac{\frac{1}{2}\left| \begin{array}{} -1 & \frac{7\sqrt{3}}{9} & 1 \\ 7 & \frac{-7\sqrt{3}}{9} & 1 \\ \frac{-21}{10} & \frac{119\sqrt{3}}{90} & 1 \\ \end{array} \right| }{\frac{\sqrt{3}}{4}}=\frac{168}{5}=33.6$
| {
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Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)^2 + (ab-c)^2 + abc \geq 0 $$
This proves the inequality for $abc\geq 0$.
I want to prove this Inequality for $abc\lt0$. But I couldn't find a solution for $abc\lt0$.
Any extensions for $abc\lt0$ are thankfully accepted.
| Another way.
It's enough to prove our inequality for non-negatives $a$, $b$ and $c$.
Now, since $$3(a^2-a+1)^3-a^6-a^3-1=(a-1)^4(2a^2-a+2)\geq0,$$ by Holder we obtain:
$$\prod_{cyc}(a^2-a+1)\geq\prod_{cyc}\sqrt[3]{\frac{a^6+a^3+1}{3}}\geq\frac{1}{3}(a^2b^2c^2+abc+1).$$
Now, let $a\leq0$, $b\geq0$ and $c\geq0.$
Thus, after replacing $a$ on $-a$ we need to prove that:
$$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2-abc+1,$$ which follows from the previous inequality:
$$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq$$
$$\geq3\sum_{cyc}(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2+abc+1\geq a^2b^2c^2-abc+1.$$
| {
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Show there are 1977 non-similar triangles such that$\frac{\sin X+\sin Y+\sin Z}{\cos X+\cos Y+\cos Z}=\frac{12}7$and$\sin X\sin Y\sin Z=\frac{12}{25}$
Show that there are 1977 non-similar triangles whose angles $X$, $Y$, $Z$ satisfy the conditions
$$\begin{align}
\frac{\sin{X}+ \sin{Y}+ \sin{Z}}{\cos{X}+\cos{Y}+ \cos{Z}}=\frac{12}{7} \tag1\\[4pt]
\sin{X}\sin{Y} \sin{Z}=\frac{12}{25} \tag2
\end{align}$$
My attempt: $$\begin{align} \sin{X}+ \sin{Y}+ \sin{Z}=4\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}} \\ \cos{X}+ \cos{Y}+ \cos{Z}=1+4\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}\\ X+Y+Z=\pi \\ \sin{X} \sin{Y}\sin{Z}=8\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}}\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}=\frac{12}{25}\end{align}$$ Here after what to do to find the required results.
How to show such points which satisfy these conditions?
Please, help. Thanks in advance.
| Using standard notation,
given $\triangle ABC$ with angles $\alpha,\beta,\gamma$,
side lengths $a,b,c$, semiperimeter $\rho$,
radius $r$ of the inscribed circle,
and radius $R$ of the circumscribed circle,
\begin{align}
\frac{\sin\alpha+ \sin\beta+ \sin\gamma}
{\cos\alpha+ \cos\beta+ \cos\gamma}
&=\frac{12}{7}
\tag{1}\label{1}
,\\
\sin\alpha\sin\beta\sin\gamma
&=\frac{12}{25}
\tag{2}\label{2}
.
\end{align}
Using known identities,
\begin{align}
\sin\alpha+ \sin\beta+ \sin\gamma
&=
\frac\rho R=u
\tag{3}\label{3}
,\\
\cos\alpha+ \cos\beta+ \cos\gamma
&=
\frac rR+1=v+1
\tag{4}\label{4}
,\\
\sin\alpha\sin\beta\sin\gamma
&=
\frac{\rho r}{2R^2}=\tfrac12\,uv
\tag{5}\label{5}
,
\end{align}
we ca rewrite \eqref{1}-\eqref{2}
in terms of parameters $u=\rho/R,\,v=r/R$ as
\begin{align}
\frac u{v+1}&=\frac{12}{7}
\tag{6}\label{6}
,\\
\tfrac12\,uv
&=\frac{12}{25}
\tag{7}\label{7}
.
\end{align}
The system \eqref{6}-\eqref{7}
has just two solutions,
\begin{align}
u &= -\frac{24}{35},\quad v = -\frac75
\tag{8}\label{8}
,\\
u &= \frac{12}5,\quad v =\frac25
\tag{9}\label{9}
,
\end{align}
and obviously, only positive is valid,
so, there is only one type of triangle with given properties.
Solution of the cubic equation
\begin{align}
x^3-2u\,x^2+(u^2+v^2+4v)\,x-4uv
&=0
\tag{10}\label{10}
,\\
x^3-\frac{24}5\,x^2+\frac{188}{25}\,x-\frac{96}{25}
&=0
\tag{11}\label{11}
\end{align}
gives a unique triplet of side lengths of triangle
with $R=1$,
which satisfies \eqref{1} and \eqref{2}
\begin{align}
a&=\frac65,\quad b=\frac85,\quad c=2
\tag{12}\label{12}
.
\end{align}
As we can see, this triangle is similar to the famous $3-4-5$
right-angled triangle.
Indeed, we have
\begin{align}
\sin\alpha&=\frac35,\quad\sin\beta=\frac45,\quad\sin\gamma=1
\tag{13}\label{13}
,\\
\cos\alpha&=\frac45,\quad\cos\beta=\frac35,\quad\cos\gamma=0
\tag{14}\label{14}
,
\end{align}
\begin{align}
\sin\alpha+\sin\beta+\sin\gamma
&=
\frac{12}5
\tag{15}\label{15}
,\\
\cos\alpha+ \cos\beta+ \cos\gamma
&=
\frac75
\tag{16}\label{16}
,\\
\frac{\sin\alpha+ \sin\beta+ \sin\gamma}
{\cos\alpha+ \cos\beta+ \cos\gamma}
&=\frac{12}{7}
\tag{17}\label{17}
,\\
\sin\alpha\sin\beta\sin\gamma
&=
\frac{12}{25}
\tag{18}\label{18}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804078",
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"source": "stackexchange",
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} |
Distributing $5$ different balls to $4$ different persons We have to find the number of ways of distributing $5$ different balls to $4$ different persons.
Clearly, the answer is $4^5$ as each ball can be given to any of the $4$ persons. However, I wanted to calculate it using a different method.
I assumed that $a$ balls are given to first person, $b$ to second , $c$ to third and $d$ to fourth person. So we have that $a+b+c+d=5$ where $0 \leq a,b,c,d \leq 5$.
But counting the solutions to the above equation assumes that balls are identical. So I tried to find the number of distributions each permutation of $(a,b,c,d)$ produces. That will be $\displaystyle \binom{5}{a}\cdot \binom{5-a}{b}\cdot \binom{5-a-b}{c} $ which equals $\dfrac{5!}{a!b!c!d!}$.
So, now we need to sum this value over all $a,b,c,d$ satisfying $a+b+c+d=5$. Now there are $\displaystyle \binom{8}{3}=56$ solutions to the equation. So there will be $56$ terms in that summation. So how do we do that?
| Consider the partitions $5$ into $4$ parts
\begin{eqnarray*}
(5,0,0,0),(4,1,0,0),(3,2,0,0),(3,1,1,0),(2,2,1,0),(2,1,1,1).
\end{eqnarray*}
These have symmetry factors $4,12,12,12,12,4$ respectively (which adds upto $56$ as you state)
Now the balls can be distributed in each case and multiply in the symmetry factors ...
\begin{eqnarray*}
4 \times \frac{5!}{5!0!0!0!} + 12 \times \dfrac{5!}{4!1!0!0!} + 12 \times \dfrac{5!}{3!2!0!0!} + 12 \times \dfrac{5!}{3!1!1!0!} + 12 \times \dfrac{5!}{2!2!1!0!} + 4 \times \dfrac{5!}{2!1!1!1!} =1024= 4^5.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| How about
$$\frac{\sqrt x-\sqrt2}{\sqrt{x-2}}
=\frac{x-2}{\sqrt{x-2}(\sqrt x+\sqrt2)}
=\frac{\sqrt{x-2}}{\sqrt x+\sqrt2}\to0
$$
as $x\to2^+$. You've successfully tackled the rest.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Bound $\sum_{k=1}^{n/2 } \frac{1}{ k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k} ) \right)}$ I am looking to find an upper bound on
\begin{align}
\sum_{k=1}^{n/2 } \frac{1}{ k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k} ) \right)}.
\end{align}
I am interested in the bound that is tight when $n$ is large.
I was thinking that for $k \in[1,n/2]$
\begin{align}
\log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k} ) \ge \log 4-\log(1+8 \log \log4 )=c,
\end{align}
where the minimum is attained at $k=n/2$ (I think this is true).
Then,
\begin{align}
\sum_{k=1}^{n/2 } \frac{1}{ k \left( \log \frac{2n}{k}-\log(1+\frac{4}{k} \log \log\frac{2n}{k} ) \right)} \le \frac{1}{c} \sum_{k=1}^{n/2 } \frac{1}{ k} \le \frac{1}{c} (\log(n/2)+1).
\end{align}
Can one do better?
| Start by bounding
\begin{align}
\sum_{k=1}^{n/2} \frac{1}{ k \log \frac{2n}{k+4\log \log\frac{2n}{k}}} \leq \sum_{k=1}^{n/2} \frac{1}{ k \log \frac{2n}{k+4\log \log2n}} \, .
\end{align}
Now for simplicity set $c=c(n)=4\log \log 2n$ and rearrange to $$\sum_{k=1}^{n/2} \frac{1}{k+c} \frac{1+\frac{c}{k}}{\log 2n - \log(k+c)} \, .$$
The second term is bounded by $$C\sum_{k=1}^{n/2} \frac{c}{k(k+c)} \leq C\sum_{k=1}^\infty \frac{c}{k(k+c)} \leq Cc\int_1^\infty\frac{{\rm d}k}{k(k+c)}=C\log(c+1)={\cal O}(\log \log \log n)$$ for some constant $C>0$.
The first term becomes $$\sum_{k=1}^{n/2} \frac{1}{k+c} \frac{1}{\log 2n - \log(k+c)} \leq \int_0^{n/2} \frac{{\rm d}k}{k+c}\frac{1}{\log 2n-\log(k+c)} \\
= \log \log \frac{2n}{c} - \log \log \frac{4n}{n+2c} = \log \log \frac{n}{2} + {\cal O}(\log \log\log \log n) \, .$$
$\log \log \frac{n/2}{\log n/2}$ is a good approximation.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ For $a,b,c\geq 0$, no two of which are $0$, prove that:
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\dfrac{a^2+b^2+c^2}{ab+bc+ca}}$$
This inequality actually came up as an accident when I tried to combine 2 known results, and after many testings on computer it still remains true, but there's still no original proof yet. Hope everyone enjoy and have some good ideas for it.
Here's that 2 known results:
$$\dfrac{a^2+b^2+c^2}{ab+bc+ca}\geq \prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)$$
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)}$$
The second one can be proved by direct Karamata's inequality, but it may also inspire some ideas for the original one too.
See the following links:
https://artofproblemsolving.com/community/u410204h2218857p16854913
https://artofproblemsolving.com/community/c6h487722p5781880
https://artofproblemsolving.com/community/u414514h2240506p17302184
| As Michael Rozenberg it's just a comment .Due to homogeneity we can assume that $a=1$ and $0<b,c\leq 1$ we have :
$$\sqrt{\dfrac{1}{b+c}+\dfrac{b}{c+1}}+\sqrt{\dfrac{b}{c+1}+\dfrac{c}{1+b}}+\sqrt{\dfrac{c}{1+b}+\dfrac{1}{b+c}}\geq 2+\sqrt{\dfrac{b^2+c^2+1}{b+bc+c}}\quad (1)$$
We can also assume that $b+c=k=\operatorname{constant}$ and try the substitution :
$$x=\dfrac{b}{c+1}$$
$$y=\dfrac{c}{1+b}$$
$$z=\dfrac{1}{(1+b)(c+1)}$$
$(1)$ becomes :
$$\sqrt{\dfrac{1}{k}+x}+\sqrt{x+y}+\sqrt{y+\dfrac{1}{k}}\geq 2+\sqrt{(\frac{x}{z}+\frac{y}{z}-k+1)\dfrac{z}{1-z}}$$
With the constraint $z(k+1)+xy=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
$2$-girl problem with bias of $P(G) = 0.75$ and $P(B) = 0.25$ A family has 2 children. The probability of a child being a girl is 0.75. We pick one of them at random and find out that she is a girl. What is the probability that all their children are girls?
My solution : It is given that $\text{P(child is a girl)}$ = $\text{P(G)} = 0.75$
$\text{P(B)} = 0.25$
There are four possibilities in this family : $\{ \text{BB, GG, BG, GB} \}$
P(BB) = $0.25 \times 0.25$
P(GG) = $0.75 \times 0.75$
P(GB) = $0.75 \times 0.25$
P(BG) = $0.25 \times 0.75$
Now question mentions that we pick one of them at random and find out that she is a girl. Hence we can reduce the sample space by removing the case P(BB).
Sample space is now $\{ \text{ GG, BG, GB} \}$
$\text{P(GG | one of the child is a girl)}$ = $\Large \frac{0.75^2}{0.75^2 + 2 \times (0.75 \times 0.25) }$
$\text{P(GG | one of the child is a girl)} = 0.60$
The given answer in my university quiz is 0.67.
Explanation:
The sample space is Ω = {BB, BG, GB, GG}
Let G_r be the event that a randomly chosen child is a girl
From the data given,
$P(G) = 0.75$ and $P(B) = 0.25$
So, $P(GG)=9/16, P(GB)=P(BG)=3/16, P(BB) = 1/16$
Now,
$P(G_r|BB) = 0$
$P(G_r|BG) = P(G_r|GB) = 0.75$
$P(G_r|GG) = 1$
We would like to find $P(GG|G_r)$
$P(GG|G_r) = $$\large \frac{P(G_r|GG)P(GG)}{P(G_r)}$
$\Large = \frac{1.\frac{9}{16}}{P(G_r|BB)P(BB) + P(G_r|BG)P(BG) + P(G_r|GB)P(GB) + P(G_r|GG)P(GG)}$
$\Large = \frac{\frac{9}{16}}{0.\frac{1}{16} + 0.75.\frac{3}{16} + 0.75.\frac{3}{16} + 1.\frac{9}{16}}$
Solving, we will get $P(GG|G_r) = \frac{2}{3} = 0.67$
| Suppose a family has two children and we pick a child at random. Our sample space is:
*
*GG, pick first child
*GG, pick second child
*GB, pick first child
*GB, pick second child
*BG, pick first child
*BG, pick second child
*BB, pick first child
*BB, pick second child
If we assume that we had an equal chance of picking the older and younger children, and with the given probabilities for each child to be a gender, we have the probabilities for each of the eight of:
*
*$\frac{3}{4}\frac{3}{4}\frac{1}{2}=\frac{9}{32}$
*$\frac{3}{4}\frac{3}{4}\frac{1}{2}=\frac{9}{32}$
*$\frac{3}{4}\frac{1}{4}\frac{1}{2}=\frac{3}{32}$
*$\frac{3}{4}\frac{1}{4}\frac{1}{2}=\frac{3}{32}$
*$\frac{1}{4}\frac{3}{4}\frac{1}{2}=\frac{3}{32}$
*$\frac{1}{4}\frac{3}{4}\frac{1}{2}=\frac{3}{32}$
*$\frac{1}{4}\frac{1}{4}\frac{1}{2}=\frac{1}{32}$
*$\frac{1}{4}\frac{1}{4}\frac{1}{2}=\frac{1}{32}$
Now, we are told that when we picked a random child, we picked a girl. This rules out cases 4,5,7, and 8
We want:
$$\begin{align}P(GG|chose\,girl)&=\frac{P(1)+P(2)}{P(1)+P(2)+P(3)+P(6)}\\&=\frac{\frac{9}{32}+\frac{9}{32}}{\frac{9}{32}+\frac{9}{32}+\frac{3}{32}+\frac{3}{32}}\\&=\frac{9+9}{9+9+3+3}\\&=\frac{18}{24}\\&=\frac{3}{4}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge a+b+c +\frac{3n}{a^2+b^2+c^2}$$
or
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge ab+bc+ca +\frac{3n}{a^2+b^2+c^2}$$
i dont know what to do next. Any ideas preferably using am-gm.
source Samin Riasat Basics in Olympiad ineq.
| We can use also, the following much more weaker estimation:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc},$$ which is just $$\sum_{cyc}(a^2c-b^2a)^2\geq0.$$
Thus, it's enough to prove that:
$$\frac{(a^2+b^2+c^2)(a+b+c)}{ab+ac+bc}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc)$ again.
Thus, we need to prove that:
$$k+\frac{9}{k(k+2)^2}\geq\frac{6}{k+2},$$ which is true by AM-GM:
$$k+\frac{9}{k(k+2)^2}\geq2\sqrt{k\cdot\frac{9}{k(k+2)^2}}=\frac{6}{k+2}.$$
| {
"language": "en",
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"question_score": "2",
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Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ actually exist? I have seen questions of this type being solved as follows :
$\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this expression be $y$. So
$$\sqrt{x+y} = y \implies x+y = y^2 \implies y^2-y-x=0$$
Using the quadratic formula, we obtain the value of $y$ as :
$$\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-x)}}{2(1)} = \dfrac{1\pm\sqrt{1+4x}}{2} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$$
This has been used to solve $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ in my Mathematics textbook.
Now, this method would work perfectly, assuming that a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exists. If a value for this expression does not exist, this would be similar to Numberphile's popular $\displaystyle\sum_{n=1}^\infty n = \dfrac{-1}{12}$ which is doubtlessly wrong.
So, does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exist? Why/why not?
Thanks!
| Kitchen's Calculus of One Variable gives a rigorous solution although is not exactly the same question (Exercise 8, Section 3-3).
To show that $\sqrt{x}$, $\sqrt{x+\sqrt{x}}, \cdots $ converges to $\frac{1+\sqrt{1+4x}}{2}$, let the nth term be $a_n$. Define $h_n=\frac{1+\sqrt{1+4x}}{2}-a_n$. You can prove by induction that
$$0<h_n\le \frac{\frac{1+\sqrt{1+4x}}{2}-\sqrt{x}}{(\frac{1+\sqrt{1+4x}}{2}+\sqrt{x})^{n-1}}.$$
Now it is easy to see that $h_n\rightarrow 0$ by pinching since the denominator above is greater than 1 when $x>0$ .
| {
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"question_score": "5",
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} |
Separable equation with $u$-substitution I'm trying to solve this separable equation.
Answer: x^3-y^3=c (Apparantly my calculus book only have answer for every second assignment)
$$\frac{dy}{dx}=\frac{3y-1}{x}\\
\frac{dy}{dx}=\frac{3y-1}{x} \Rightarrow \frac{1}{3y-1}dy=\frac{1}{x}dx\\
u-substition\ where\ u=3y-1\ and\ \frac{du}{3}=dy\\
\int\frac{1}{3}\frac{1}{u}=\int\frac{1}{x}dx\Rightarrow \frac{1}{3}\int\frac{1}{u}=\int\frac{1}{x}dx \\
\frac{1}{3}ln(u)+c=ln(x)+c\Rightarrow \frac{1}{3}ln(3y-1)+c=ln(x)+c\Rightarrow \sqrt[3]{3y-1}=x+c\Rightarrow x^3-3y=c$$
I know I did something wrong, but I can't see what. Can you tell me where I broke the rule of maths?
EDIT:
Is this a valid answer?
$$\frac{1}{3}ln(3y-1)+c=ln(x)+c\Rightarrow ln(3y-1)+c=3ln(x)+3c\Rightarrow\\
3y-1=e^{3ln(x)}e^{3c} \Rightarrow 3y-1=x^3c\Rightarrow 3y=x^3c+1\Rightarrow y=\frac{x^3c}{3}+\frac{1}{3}$$
| You didn't break any rules. However, you've added the integration constant $+c$ on both sides after integration, which cancels out..
From $\ln(3y-1)^{\frac{1}{3}}=\ln(x)+c$ we obtain $(3y-1)^{\frac{1}{3}}=xe^{c}$ and thus $3y-1=x^{3}e^{3c},$ which gives the solution $y=c_{1}x^{3}+\frac{1}{3}$
where $c_{1}=e^{3c}=constant.$
Then it's always good to check that the solution satisfies the given equation. In this case we have $\frac{dy}{dx}=3c_{1}x^{2}$ and $\frac{3y-1}{x}=\frac{3(c_{1}x^{3}+\frac{1}{3})-1}{x}=\frac{3c_{1}x^{3}}{x}=3c_{1}x^{2}$ as required.
| {
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"source": "stackexchange",
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Sum of squares of consecutive triangular numbers Find a general formula for:
$t^2_1 + t^2_1 ... t^2_n$.
I tried evaluating:
$1/4[1(2)^2+2^2(3)^2 ... (n)(n+1)]$ then I used sum of cubes to evaluate this.
$(1^3) +(1^3 +2^3) ... (1^3 + 2^3 ... + n^3)$
$n(1^3) + (n-1)(2^3) ... + n^3(1)$
I also tried grouping terms like:
$n(1)(1^2+n^2) +(n-1)(2)((n-1)^2+2^2)) ...$
but I didn't see anything else
| The most transparent and general way to deal with these questions is with generating functions. The humble geometric series identity says:
$$1+x+\cdots+x^{n-1} = \frac{1-x^n}{1-x}$$
Note that $\frac{d^2}{dx^2} x^i = i(i-1) x^{i-2}$. We can get a sum of triangular number coefficients by differentiating the above twice, dividing by $2$, and setting $x=1$. You want squares of triangular numbers though. For that, just notice
$$\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2}\right) x^i = x^2\frac{d^2}{dx^2} x^2 i(i-1)x^{i-2} = i^2 (i-1)^2 x^i. $$
Thus
$$\begin{align*}
\sum_{i=1}^n \left(\frac{i(i+1)}{2}\right)^2
&= \sum_{i=0}^{n+1} \left(\frac{i(i-1)}{2}\right)^2 \\
&= \sum_{i=0}^{n+1} \left.\left(\frac{i(i-1)}{2}\right)^2 x^i\right|_{x=1} \\
&= \frac{1}{4} \sum_{i=0}^{n+1} \left.\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} x^i\right)\right|_{x=1} \\
&= \frac{1}{4} \left.\left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} \sum_{i=0}^{n+1} x^i\right)\right|_{x=1} \\
&= \frac{1}{4} \lim_{x \to 1} \left(x^2 \frac{d^2}{dx^2} x^2 \frac{d^2}{dx^2} \frac{1-x^{n+2}}{1-x}\right) \\
&= \cdots \\
&= \frac{n (1 + n) (2 + n) (1 + 3 n (2 + n))}{60}
\end{align*}$$
Here $\cdots$ hides the mess. A few remarks are in order:
*
*No special identities are required--no random Faulhaber formula, sum of squares, sum of cubes, etc. You just need to churn through routine calculus.
*The step right before the messy calculation tells you all you really to know: it's very unlikely the final answer will have a nice form, since that messy nest of quotient rule applications won't.
*If you really desperately do want to know the explicit final answer, a computer can tell it to you with no margin for error or additional effort: in Mathematica,
*If you decide you want variations on this, e.g. cubes of triangular numbers rather than squares, it's often trivial to modify the argument and have a computer tell you the new answer.
| {
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Let $a$ be an integer. Prove that the following equation cannot have more than one integer $x^4+7x^3+(a+2)x^2-11x+a=0$ Let $a$ is a interger. Prove that the following equation cannot have more than one integer $$x^4+7x^3+(a+2)x^2-11x+a=0$$
$$x^4+7x^3+(a+2)x^2-11x+a=0$$
Or $$-\frac{x^4+7x^3+2x^2-11x}{x^2+1}=x^2+7x+1-\frac{18x+1}{x^2+1}=a$$
When $a\in \mathbb{Z}$ that means $$(x^2+1)\mid (x^4+7x^3+2x^2-11x)$$
Or $$(x^2+1)\mid (18x+1)\rightarrow 18x+1\ge x^2+1$$
Or $$0\le x\le 18$$
Now i tried all value and got only $x=0$ and $x=18$ are roots that satified.
$$x=0\rightarrow a=0 \text{ or } x=18\rightarrow a=-450 $$
Now i tried to solve this equation egain with $a=0$ and $a=-450$ and each value of $a$ get only one integer root of $x$ and it is done.
I don't know if my solution is correct.Help me check it and give me some solution,ty..
| Using Vieta's formulas helps.
Suppose that the equation has an odd root $x$. Then, LHS of the equation$$\underbrace{x^4+7x^3+2x^2-11x}_{\text{odd}}+a(\underbrace{x^2+1}_{\text{even}})$$
is odd, which contradicts that $x$ is a root.
So, we see that if the equation has an integer root, then it is even.
Now, suppose that the equation has four roots $2b,2c,p+qi,r-qi$ where $b,c\in\mathbb Z$ and $p,q,r\in\mathbb R$.
Then, by Vieta's formulas, we get
$$2b+2c+p+r=-7\tag1$$
$$4bc+(2b+2c)(p+r)+(p+qi)(r-qi)=a+2\tag2$$
$$2(b+c)(p+qi)(r-qi)+4bc(p+r)=11\tag3$$
It follows from $(1)$ that $p+r$ is an integer.
Also, it follows from $(2)$ that $(p+qi)(r-qi)$ is an integer.
Then, LHS of $(3)$ is even, and RHS of $(3)$ is odd, which is a contradiction.
| {
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Finding lim sup I am unable to find a way to solve this. I have tried using definition of lim sup but got no where.
Consider the series: $\sum^\infty_{n=1} a_n = \frac{1}{2}+\frac{1}{3}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\cdots$
Then $\limsup_{n\to\infty} \frac{a_{n+1}}{a_n}=$
Edit: This is the work I have done yesterday night. My apologies for not showing my work.
I have looked at the sequence $\frac{a_{n+1}}{a_{n}}$. I observed that it looks like $\frac{2}{3}, \frac{3}{2^2}, \frac{2^2}{3^2},\frac{3^2}{2^3},\cdots$.
So, the general terms look like $(\frac{2}{3})^n$ for $n$ odd and $\frac{1}{2}(\frac{3}{2})^n$ for $n$ even.
I looked at the supremums of the tail sequences and then decided to find their infimum. The sequence as I found has $\infty$ as the supremum of each tail sequence since the sequence for $n$ even is monotonically increasing. Thus I inferred that the limsup should be $\infty$. However, the answer was given to be $\frac{1}{2}$.
This sequence I found is of course a mistake as the exponents I have taken are wrong.
The sequence actually looks like $\frac{a_{n+1}}{a_n}=\cases{(\frac{2}{3})^{\frac{n+1}{2}}& n is odd\\ \frac{1}{2}(\frac{3}{2})^\frac{n}{2}& n is even}$
But still, the sequence of supremum of the tails consists only of $\infty$ and hence the infimum of the supremums is $\infty$. What am I missing?
| Your argument is perfectly fine and the "bandit" in this one is the author of the problem for making it unclear on how $a_{n}$ is to be expressed. Something I have found true for some authors is that; when in doubt a Mathematics textbook is always written in "three's".
If instead of treating $a_{n}$ to be a sequence of alternating reciprocal powers of $2$ and $3$, we take $a_{n}$ to be a sequence of equal powers, as in $a_{n}=\frac{1}{2^{n}}+\frac{1}{3^{n}}$, then we get an alternate (and in hindsight a true) description of the series in question. By the following sequence of calculations we reach $\lim \sup \frac{a_{n+1}}{a_{n}}=\frac{1}{2}$:
First by rewriting $a_{n}$ on a common denominator we get
$a_{n}=\frac{3^{n}+2^{n}}{6^{n}}$ and so
$$\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}+3^{n+1}}{6(2^{n}+3^{n})}.$$
Dividing numerator and denominator by $3^{n+1}$ we arrive at
$$\frac{a_{n+1}}{a_{n}}=\frac{(\frac{2}{3})^{n+1}+1}{2((\frac{2}{3})^{n}+1)}.$$
Now since
$$(\frac{2}{3})^{n},(\frac{2}{3})^{n+1}\to 0 \ \text{as} \ n \to \infty$$
we get that
$$\lim \frac{a_{n+1}}{a_{n}}=\lim \frac{(\frac{2}{3})^{n+1}+1}{2((\frac{2}{3})^{n}+1)}=\frac{1}{2}.$$
Because $\lim \sup x_{n} = \lim x_{n}$ for a sequence $x_{n}$ which converges, we can infer that
$$\lim_{n\to\infty} \sup \frac{a_{n+1}}{a_{n}}=\frac{1}{2}$$.
| {
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Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.
I have tried the question like this.
\begin{align*}
\frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x+ 48y &= 7xy\\
48x &= 7xy-48y\\
48x &= y\cdot(7x-48)\\
y &= \frac{48\cdot x}{7x-48}\\
\end{align*}
As $y$ is a positive number the denominator have to be $<0$, so $x \ge 7$. Now If I put $x =7$
I get $y = \frac{48\cdot 7}{49-48} = 336$. But
lcm(7,336) is 336
and $\frac{7+336}{336} \neq \frac{7}{12}$. Where am I making mistake? the hcf is not 4 for (7,336)
second method I Tried was,
$x = 4a, y =4b.$
now ,
\begin{align*}
\frac{4a + 4b}{4\cdot a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{1}{a}+\frac{1}{b} &= \frac{7}{12}\\
\frac{1}{a} &= \frac{7}{12} - \frac{1}{b}\\
\frac{1}{a} &= \frac{7b-12}{12b}\\
or\\
a &= \frac{12b}{7b-12}
\end{align*}
so $b\ge 2$ to get an integer, now for $b=2, a= 12. \Rightarrow x=48, y=4$, also if I choose b=2, the hcf is not 4, so if I put $b=3, a = 4 \Rightarrow (12, 16)$ , here the hcf and lcm are,
$4$ and $48$ respectively.and $\frac{12+16}{48} = \frac{7}{12}$.
What is the mistake I am making. ?
| For your first method since hcf(x,y)=4 then $4|x$ and $4|y$ so clearly $x=7$ is not a possibility since $4\not|7.$ So you can only choose $x$ of the form $4k$ for some $k\in\mathbb N.$
In fact the possible integer solutions are given by $(x,y)=(-48,6),(6,-48),(7,336),(7,48)$ and $(12,16).$ Then checking, the correct one is $(12,16).$
For your second method you have $x=4a$ and $y=4b$ then $$\frac{a+b}{ab}=\frac{7}{12}=\frac{7}{3\cdot 4}=\frac{3+4}{3\cdot 4}$$
so clearly you can have $(a,b)=(3,4),(4,3)$ (and the other solutions are $(2,12)$ and $(12,2)$ which are not valid since they have a common divisor of $2$).
So setting $(a,b)=(3,4)$ we have $(x,y)=(12,16)$ and setting $(a,b)=(4,3)$ you have $(x,y)=(16,12).$
| {
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Prove that limit $\lim \limits_{(x,y,z) \to (0,0,0)} \frac{xyz\tan(2x+y-z)}{(2x+y)^2-z^2}$ doesn't exist Show that the Limit doesn't exist
$\lim \limits_{(x,y,z) \to (0,0,0)} \frac{xyz\tan(2x+y-z)}{(2x+y)^2-z^2}$
I try to replace all of x, y, and z with what they are actually approaching. Please help me to prove that the Limit doesn't exist
| As suggested we have
$$\frac{\tan(2x+y-z)}{2x+y-z}\cdot\frac{xyz(2x+y-z)}{(2x+y)^2-z^2}$$
with $\frac{\tan(2x+y-z)}{2x+y-z}\to 1$ and
$$\frac{xyz(2x+y-z)}{(2x+y)^2-z^2}=\frac{xyz(2x+y-z)}{(2x+y-z)(2x+y+z)}=\frac{xyz}{2x+y+z}$$
which tends to zero for $(x,y,z)=(0,t,t)$ with $t \to 0$ but for $(x,y,z)=(t,-t,-t+t^3)$ we obtain
$$\frac{xyz}{2x+y+z}=\frac{t^3-t^5}{t^3}=1-t^2 \to 1$$
therefore the limit doesn't exist.
| {
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Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$ I am trying to compute the Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$. When I compute the Cesàro means, I get the following sequence
$$\left(1, 1, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \frac{6}{9}, \frac{7}{10}, \frac{8}{11}, \frac{8}{12}, \cdots\right)$$
Where does this sequence converge to? Is it $\frac{2}{3}$? I fail to see the pattern of this sequence. If I can just rewrite the sequence into a more general form, then I might be able to compute the limit.
| The pattern is ,,, (2k/(3k), (2k+1)/(3k+1), (2k+1)/(3k+2))...$k=2,3,...$.
So the limit is $2/3$ as you guessed.
| {
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Proving that $f(x) = 6\ln(x^{11}-4) -2$ is one-to-one Please verify my proof, and if there are any mistakes please explain.
Prove that that this function is one-to-one: $f(x) = 6\ln(x^{11}-4) -2$.
Suppose $f(x_1) = f(x_2)$
$\implies 6\ln(x_1^{11}-4) -2 = 6\ln(x_2^{11}-4) -2$
Then we have, $\ln(x_1^{11}-4) - \ln(x_2^{11}-4) = 0$
Using quotient rule for logarithms, $ \ln \frac{x_1^{11}-4}{x_2^{11}-4} = 0\implies e^0 = \frac{x_1^{11}-4}{x_2^{11}-4}$
Thus, $x_1^{11}-4 = x_2^{11}-4 \implies x_1 = x_2$.
So the function $f(x) = 6ln(x^{11}-4) -2$ is one-to-one
| Note that $\log a=\log b\implies a=b $. This reduces two steps.
| {
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Square root inequality $\sqrt {x-z} \geq \sqrt x -\sqrt{z} $ Does the following inequality hold?
$$\sqrt {x-z} \geq \sqrt x -\sqrt{z} \ , $$
for all $x \geq z \geq 0$.
My justification
\begin{equation}
z \leq x \Rightarrow \\ \sqrt z \leq \sqrt {x} \Rightarrow \\ 2\sqrt z \sqrt z \leq 2\sqrt z\sqrt {x} \Rightarrow \\ 2 z \leq 2\sqrt z\sqrt {x} \Rightarrow \\ z - 2\sqrt z\sqrt {x} + x \leq x - z \Rightarrow \\ (\sqrt x -\sqrt z )^2 \leq x - z \Rightarrow \\ \sqrt x -\sqrt z \leq \sqrt {x - z}
\end{equation}
| \begin{equation}
\qquad\sqrt {x-z} \ge \sqrt x -\sqrt{z}\\
\implies (\sqrt {x-z})^2 \geq (\sqrt x -\sqrt{z})^2\\
\implies x-z\ge x-2\sqrt{xz}+z\\
\implies x-z - x-z\ge-2\sqrt{xz}\\
\implies -2z\ge -2\sqrt{xz}\\
\implies -z\ge -\sqrt{xz}\\
\text{ subtracting both sides from both sides reverses the relationship}\\
\implies \sqrt{xz}\ge z\\
\implies xz\ge z^2\\
\implies x\ge z\quad \land\quad x-z\ge 0\\
\implies x\ge\ z\ge 0
\end{equation}
| {
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Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
| This would be my "napkin" heuristic:
Since $\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right)$ is the absolute value of the tail of a convergent series, it tends to zero. Therefore,
$$\begin{align*}\limsup_{n\to\infty} n&\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty}\, (n+1)\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ &+\lim_{n\to\infty}\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty} n\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right)\end{align*}$$
From here, we find that if the limit in question exists, it must equal $\frac{1}{2}.$
| {
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Prove that 5 lines are concurrent, and find the expression for the position vector of the point they all go through. Pentagon $ABCDE$ is inscribed in a circle centered at the origin. Define the lines \begin{align*}
\ell_{ABC} &= \text{Line through the centroid of $\triangle ABC$ perpendicular to $\overline{DE}$},\\
\ell_{BCD} &= \text{Line through the centroid of $\triangle BCD$ perpendicular to $\overline{AE}$}, \\
\ell_{CDE} &= \text{Line through the centroid of $\triangle CDE$ perpendicular to $\overline{AB}$}, \\
\ell_{DEA} &= \text{Line through the centroid of $\triangle DEA$ perpendicular to $\overline{BC}$}, \\
\ell_{EAB} &= \text{Line through the centroid of $\triangle EAB$ perpendicular to $\overline{CD}$}. \\
\end{align*}
These are lines going through the centroid of a triangle formed by three consecutive vertices, perpendicular to the line segment formed by the other two vertices. Here's $\ell_{ABC}$ in the picture:
Prove that $\ell_{ABC}, \ell_{BCD}, \ell_{CDE},\ell_{DEA}$ and $\ell_{EAB}$ are concurrent, and find the expression for the position vector of the point they all go through.
I truly have no idea how to approach this problem. Please help!
| WLOG, say the center of the circle ($O$) is at the origin. Vertices of the pentagon $ABCDE$ are represented by position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ and $\overline{e}$.
Centroid of $\triangle ABC, \, \overline {g} = \frac{\overline{a} + \overline{b} + \overline{c}}{3}$
Line $DE = \overline{d} - \overline{e}$
As points $A, B, C, D, E$ are concyclic with center at $O$
$|\overline{a}|^2 = |\overline{b}|^2 = |\overline{c}|^2 = |\overline{d}|^2 = |\overline{e}|^2$ ...(i)
If a point $P$ with position vector $\overline{p} \,$ is on the perpendicular line from the centroid of $\triangle ABC$ to the line $DE$,
$(\overline{p}-\overline{g}) \cdot (\overline{d} - \overline{e}) = 0$
Based on (i) one of the ways for the dot product to be zero is
$(\overline{p}-\overline{g}) = n_1 (\overline{d}+\overline{e}) \,$ (you can easily show why $\overline{p} = \overline{g}$ will not give you the concurrent point by symmetry)
$\overline{p}-\overline{g} = \overline{p}-\frac{\overline{a} + \overline{b} + \overline{c}}{3} = n_1 (\overline{d}+\overline{e})$ ...(ii)
Similarly,
$\overline{p}-\frac{\overline{b} + \overline{c} + \overline{d}}{3} = n_2 (\overline{e}+\overline{a})$ ...(iii)
From (ii)-(iii), you get one solution when $n_1 = n_2 = \frac{1}{3}$ and
$\overline {p} = \frac{\overline{a} + \overline{b} + \overline{c} + \overline{d} + \overline{e}}{3}$
Now we need to prove this point is the point of concurrency for other $3$ lines too. So we take the lines from centroids of $\triangle CDE, \triangle DEA, \triangle EAB$ through point $\overline {p}$ and show each of them is perpendicular to the line segment made by other two vertices.
$(\overline{p}- \frac{\overline{c} + \overline{d} + \overline{e}}{3}) \cdot (\overline{a} - \overline{b}) = 0$
$(\overline{p}- \frac{\overline{d} + \overline{e} + \overline{a}}{3}) \cdot (\overline{b} - \overline{c}) = 0$
$(\overline{p}- \frac{\overline{e} + \overline{a} + \overline{b}}{3}) \cdot (\overline{c} - \overline{d}) = 0$
which is easy to show given (i).
| {
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Given $a_{n} = 3a_{n-1}-2a_{n-2}$, $a_0=0$, $a_1 = 2$. Show $a_n = 2(2^n-1)$ by induction This is my problem:
The recurrence relation is \begin{aligned} a_{n} = 3a_{n-1}-2a_{n-2} \end{aligned}
This is given that $a_{0}=0, a_{1}=2$.
From the above information I calculated
\begin{aligned} a_{2} &=3(a_{2 - 1}) - 2(a_{2 - 2}) = 3(a_{1}) - 2(a_{0}) = 3(2) - 2(0) = 6 \\
a_{3} &=3(a_{3 - 1}) - 2(a_{3 - 2}) = 3(a_{2}) - 2(a_{1}) = 3(6) - 2(2) = 14 \\
a_{4} &=3(a_{4 - 1}) - 2(a_{4 - 2}) = 3(a_{3}) - 2(a_{2}) = 3(14) - 2(6) = 30 \end{aligned}
I have been trying to solve this recurrence relation for quite some time and have come up with the solution
\begin{aligned} a_{n} = 2 * (2^{n} - 1) \end{aligned}
I am having trouble proving this solution by induction.
My Attempt:
Base Cases:
$$a_0 = 0 = 2(2^0-1)\\a_1 = 2 = 2(2^1-1)$$
Inductive Hypothesis:
Assume that $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$
Inductive Step:
$$a_{k+1} = 3a_{(k+1)-1}-2a_{(k+1)-2} = 3a_{k}-2a_{k-1}$$
By the inductive hypothesis $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$,
$$a_{k+1}= 3 ( 2 (2^k-1) - 2 (2(2^{k-1}-1)).$$
From here it is just simplifying but I cannot get it to simplify to the correct expression.
Any ideas?
| Assume it holds for $n=k$ (so it also holds for $n<k$), then for the induction step
\begin{align}
a_{k+1}&=3a_{k}-2a_{k-1}\\
&=3(2(2^{k}-1))-2(2(2^{k-1}-1))\\
&=3(2^{k+1}-2)-2(2^{k}-2)\\
&=3\cdot2^{k+1}-6-2^{k+1}+4\\
&=2\cdot2^{k+1}-2\\
&=2(2^{k+1}-1)
\end{align}
as required.
| {
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Evaulate $ \lim_{n\to\infty} \frac{\sum_{k=1}^n\left[x^k\right]}{ x^n }$ I tried using Squeeze theorem :
$$x\le[x]\lt x+1$$
$$x^2\le[x^2]\lt x^2+1 $$
$$\vdots$$
$$x^n\le[x^n]\lt x^n+1$$
Therefore :
$$\frac{x+x^2+\cdots+x^n}{x^n}\le\frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} }\lt \frac{x+x^2+\cdots+x^n}{x^n}+\frac{n}{x^n}$$
$${ \underbrace{\frac{x \times (1-x^n)}{x^n(1-x)}}_{L_1} } \le\frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} }\lt \underbrace{\frac{x \times (1-x^n)}{x^n(1-x)} + \frac{n}{x^n}}_{L_2} $$
$L_1=\cfrac{x- x^{n+1}}{x^n-x^{n+1}} \times \cfrac{x^{-n}}{x^{-n}}= \cfrac{x^{1-n}-x}{1-x}$ and $L_2=L_1 + \cfrac{n}{x^n}$
For ${ \left| x \right| } \gt 1$ , ${ \lim_{n\to\infty} L_1 }={ \lim_{n\to\infty} L_2 }=\frac{x}{x-1}$
Hence we conclude that: ${ \lim_{n\to\infty} \frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} } }= \frac{x}{x-1}$ for ${ \left| x \right| } \gt 1$
And it is obvious ${ \lim_{n\to\infty} \frac{[x]+[x^2]+\cdots+[x^n]}{ x^{n} } }= n$ for $x=1$ .
But I don't know how to evaluate the limit for $-1\lt x \lt 0$ and $0 \lt x \lt 1$ and $x=-1$
| Hint:
*
*For $0<x<1$, $[x^n]=0$ for all $n\geq1$
*For $-1<x<0$, $[x^n]=-1$ if $n$ is odd and $0$ otherwise.
*For $x=1$ the expression is $n$
*For $x=-1$ the expression is $0$ if $n$ is even and $1$ of $n$ is odd.
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"answer_id": 0
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Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Is $f(x) = \frac{2x}{1+2x^2}$ uniformly continuous on $\mathbb{R}$? Please give your explanation.
I've tried this and got stuck
\begin{align}|f(x) - f(u)|
&= \Big|\frac{2x}{1+2x^2} - \frac{2u}{1+2u^2}\Big| \\
&=\Big|\frac{(-4xu+2)(x-u)}{(1+2x^2)(1+2u^2)}\Big| \\
& = \Big|\frac{-4xu+2}{(1+2x^2)(1+2u^2)}\Big| |x-u| \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{(1+2x^2)(1+2u^2)} \Big) |x-u| \\
& \leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + \frac{2}{1+2u^2} \Big) |x-u| \\
&\leq \Big(\frac{|-4xu|}{(1+2x^2)(1+2u^2)} + 2\Big) |x-u|.\end{align}
So I confused to make $$\frac{|-4xu|}{(1+2x^2)(1+2u^2)}$$ to be less than some positive number. What should I do next?
| Finish what you started using
$$\frac{4|xu|}{(1+2x^2)(1+2u^2)}= \frac{4|xu|}{1+2x^2+2u^2 + 4x^2u^2} \leqslant \frac{2|x||u|}{x^2 + u^2} \leqslant 1$$
The last inequality follows from $(|x| - |u|)^2 \geqslant 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Being $z=x+yi$ how can I factorize the polynomial $z^4+1$ as a product of real quadratic polynomials? Being $z=x+yi$ how can I factorize the polynomial $z^4+1$ as a product of real quadratic polynomials?
I don't really understand what I am asked to do. How can I start with it?
| So you understood the hint
\begin{eqnarray*}
(z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=0}z^2+1.
\end{eqnarray*}
To do the next one in your comment ... Factorise $z^6+1=(z^2+1)(z^4-z^2+1)$
\begin{eqnarray*}
(z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=-1}z^2+1.
\end{eqnarray*}
So
\begin{eqnarray*}
z^6+1=(z^2+1)(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to derive the equation $I-\hat x \hat x^T= (\hat x^T E)^T(\hat x^T E)$ I encountered the above question in the paper. I have no idea how to obtain the right entry.
Any help would be greatly appreciated!
$$I-\hat x \hat x^T= (\hat x^T E)^T(\hat x^T E)$$
where $\hat x = \frac{x}{\vert\vert x \vert\vert}, x\in \mathbb R^2$, and $E=\left[ {\begin{array}{*{20}{c}}
0&1\\
{ - 1}&0
\end{array}} \right]$
| You now that $|\hat x|=1$, so we can write $$\hat x=\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}$$
Then $$\begin{align}I-xx^T&=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}-\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}\begin{pmatrix}\cos\theta &\sin\theta\end{pmatrix}\\&=\begin{pmatrix}1-\cos^2\theta&-\sin\theta\cos\theta\\-\sin\theta\cos\theta&1-\sin^2\theta\end{pmatrix}\\&=\begin{pmatrix}\sin^2\theta&-\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{pmatrix}\end{align}$$
For the other side of the equation
$$x^TE=\begin{pmatrix}\cos\theta &\sin\theta\end{pmatrix}\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}=\begin{pmatrix}-\sin\theta &\cos\theta\end{pmatrix}$$
Transpose it, do the last multiplication, and compare to the left hand side
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?
To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.
Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.
A family of solutions is then given by $x=m$, $y=m^{2}z - m$ where $z$ takes on any positive integer.
How do I go about proving the striking observation: There are no positive integer solutions $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?
| Similar to finishing
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers?
where I had an acceptable bound but needed help from Gerry Myerson to improve to the sharp bound.
We have $$ (xz+1)(yz+1) = a z^3 + 1 $$
This becomes
$$ a z^3 - xyz^2 - (x+y)z=0$$
or
$$ a z^2 - xyz - (x+y) = 0 $$
We get
$$ z = \frac{ xy + \sqrt{ x^2 y^2 + 4a(x+y) } }{2a} $$
Let me also record
$$\color{fuchsia}{ z(az-xy) = x+y }$$
which follows directly from $ a z^2 - xyz - (x+y) = 0 $
Note also the simple
$$\color{fuchsia}{ z \leq x+y }$$
It is necessary to have square discriminant to get a rational value for $z,$ take
$$ w^2 = x^2 y^2 + 4a(x+y) $$
We have $$ w > xy $$ and
$$ w \equiv xy \pmod 2. $$
Therefore we can define an integer $t,$ when it all works, with
$$ w = xy+2t $$
Now $$ z = \frac{xy+w}{2a} = \frac{xy+xy+2t}{2a} = \frac{2xy+2t}{2a} = \frac{xy+t}{a} $$
$$ z = \frac{xy+t}{a} $$
There are always three flavors for any $a$
$$ t=a-1 \; , \; y = 1 \; , \; x = a^2 - 3a +1 \; , \; z = a-2 $$
$$ t=1 \; , \; y = 2a-1 \; , \; x = 2a +1 \; , \; z = 4a $$
$$ t=1 \; , \; y = a+1 \; , \; x = a^2 +a -1 \; , \; z = a^2+2a $$
From
$$ x^2 y^2 +4a(x+y) = (xy+2t)^2 $$
we get $$ t xy - ax -ay + t^2 = 0, $$
$$ t^2 xy - tax -tay + t^3 = 0, $$
$$ \color{red}{(tx-a)(ty-a) = a^2 - t^3} $$
IF $a > 1$ and $t = a + \delta$ with $\delta \geq 0,$ we find
$$ ((a+\delta)x-a)((a+\delta)y-a) = a^2 - (a+\delta)^3 < 0 $$
since $a>1.$
However, the left hand side is non-negative, which is a contradiction.
$$ \color{red}{ t \leq a-1} $$
I will fill in the (lengthy) details in a bit.
I always have $x \geq y \geq 1$
IF $$ \color{blue}{ a^{2/3} < t \leq a-1} $$
we get
$$ (tx-a) (a-ty) = t^3 -a^2 > 0 $$
so $a-ty >0,$ $ty - a < 0,$
$$ ty < a $$
$$ y < \frac{a}{t} < a^{1/3} $$
$$ a - ty \geq 1 $$
$$ tx-a \leq t^3 - a^2 $$
$$ tx \leq t^3 - (a^2 - a)$$
$$ x \leq t^2 - \frac{a^2 - a}{t} $$
DETAIL: As $t$ increases, $t^2$ increases, while $\frac{1}{t}$ decreases. Then $\frac{-1}{t}$ increases. We have $a \geq 2$ so that $a^2 - a > 0,$ so that
$\frac{-(a^2-a)}{t}$ increases. Together, $t^2 - \frac{a^2 - a}{t}$ increases and takes its maximum value at $t=a-1,$ that being $ a^2 - 3a + 1.$
Thus
$$ \color{magenta}{x \leq a^2 - 3a + 1}$$
$$xy + t < a^{7/3} -3a^{4/3} + a + a^{1/3} -1 $$
$$ z < a^{4/3} -3a^{1/3} + 1 + a^{-2/3} -\frac{1}{a} $$
$$ \color{red}{ z < a^{4/3} } $$ when $a^{2/3} < t \leq a-1$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
I always have $x \geq y \geq 1$
IF $$ \color{blue}{1 \leq t < a^{2/3} }$$
$$ (tx-a) (ty-a) = a^2 - t^3 > 0$$
$$ (tx-a) \leq a^2 -t^3$$
$$ tx \leq a^2 + a - t^3 < a^2 + a$$
$$ x \leq \frac{a^2 + a}{t} $$
Meanwhile
$$ t^2 xy - ta(x+y)= -t^3 < 0 $$
$$txy < ta(x+y) \leq 2ax $$
$$ ty < 2a
$$ y < \frac{2a}{t} $$
Together
$$ xy < \frac{2 a^3 + 2a^2}{t^2} $$
$$ xy + t < \frac{2 a^3 + 2a^2}{t^2} + a^{2/3} $$
$$ z < \frac{2 a^2 + 2a}{t^2} + \frac{1}{a^{1/3}} $$
IF $t \geq 2$
then $z < \frac{a^2 + a}{2}$
IF $t=1$ we have $$ (x-a)(y-a) = a^2 - 1 > 0 $$
If $x>a$ then $y>a.$ Then $y-a \geq 1$ and $x-a \leq a^2 - 1$
When $t=1$ we have $x \leq a^2 + a - 1.$
In general, when we have real $p \geq 1, q \geq 1,$ and $pq=c,$
the maximum of $p+q$ occurs when $p=1$ and $q=c$
so that $p+q \leq 1+c$
With $ (x-a)(y-a) = a^2 - 1 $ we get $x-a+y-a \leq a^2.$ Thus
$$ x+y \leq a^2 + 2a$$
With $t=1,$ we know $z = x+y.$ With $t=1$
$$ \color{red}{z \leq a^2 + 2a } $$
DETAIL $$\color{fuchsia}{ z(az-xy) = x+y }$$
and $$ z = \frac{xy+t}{a} $$
so that when $t=1,$ we get $az = xy+1$ or $az-xy = 1,$ so that $z(az-xy) = x+y$ tells us $z=x+y,$ when $t=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
| When you factorize the polynomial you'll get the following products:
$$(x^2 - 2 x + 4) (x^2 + 2 x + 4) (x^4 - 4 x^2 + 16) = 0 $$
And therefore you get the following roots by computing quadratic and quartic equations:
$-1-i\sqrt{3}$
$-1+i\sqrt{3}$
$1-i\sqrt{3}$
$1+i\sqrt{3}$
$-\sqrt{2-2i\sqrt{3}}$
$\sqrt{2-2i\sqrt{3}}$
$-\sqrt{2+2i\sqrt{3}}$
$\sqrt{2+2i\sqrt{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Show that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$ I cam across a question in The Book of Proof that states:
Show that $$\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$$
Which I found the answer to be:
Assume n ≥ 3. Then:
$$\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2} = \binom{n - 2}{3} + \binom{n-2}{2}+ \binom{n - 1}{2} = \binom{2}{2} + \binom{3}{2}+...+\binom{n-1}{2}$$
I have just learned about the binomial theorem, and I am not sure how we got to that answer.
I understand that $\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2}$ because that is the sum of the two previous rows in the Pascal triangle. But I can't find a simple explanation for the steps that follow or for why we assumed $n \geq 3$
| Maybe you know the possible ways to $x_1+x_2+x_3+...+x_k=n$ when $x_i \in \mathbb{Z^+}$ is $\left(\begin{array}{c}n+k-1\\ k-1\end{array}\right)$ now suppose we need to find possible way to $$x_1+x_2+x_3 \leq n$$ first we take apart the states
$$x_1+x_2+x_3 =0 \to \left(\begin{array}{c}0+3-1\\ 3-1\end{array}\right)\\
x_1+x_2+x_3 =1 \to \left(\begin{array}{c}1+3-1\\ 3-1\end{array}\right)\\
x_1+x_2+x_3 =2\to \left(\begin{array}{c}2+3-1\\ 3-1\end{array}\right)\\
...\\
x_1+x_2+x_3 =n\to \left(\begin{array}{c}n+3-1\\ 3-1\end{array}\right)$$ sum of them is $$ \left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}2\\ 2\end{array}\right)+\cdots+\left(\begin{array}{c}n+2\\ 2\end{array}\right)$$ second way to solve is to add a variable, to turn inequality to make equation like below
$$x_1+x_2+x_3 \leq n \mapsto x_1+x_2+x_3 +\bf{x_4}= n$$ it summarize all above ways to one states $$x_=0 \to x_1+x_2+x_3=n\\x_4=1 \to x_1+x_2+x_3=n-1\\\vdots$$ so $$x_1+x_2+x_3+x_4=n \to \left(\begin{array}{c}n+4-1\\ 4-1\end{array}\right)=\left(\begin{array}{c}n+2\\ 3\end{array}\right)$$ so finally $$
\left(\begin{array}{c}2\\ 2\end{array}\right)+\left(\begin{array}{c}3\\ 2\end{array}\right)+\left(\begin{array}{c}4\\ 2\end{array}\right)+...+\left(\begin{array}{c}n+2\\ 2\end{array}\right)=\left(\begin{array}{c}n+3\\ 3\end{array}\right)$$ hope it helps you
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt$ I am preparing for a master's degree entrance exam. One of the questions from the past exam asks to show the following formula. Could someone provide some hint?
$$\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt = \frac{\pi}{6} (e^{-x} - e^{-2x})$$ for $x \geq 0.$
Thank you!
| Let $f(t) = \frac{1}{(1+t^2)(4+t^2)}.$ Its Fourier transform is
$$
\hat{f}(x)
= \int_{-\infty}^{\infty} \frac{e^{-itx}}{(1+t^2)(4+t^2)}dt
= \int_{-\infty}^{\infty} \frac{1}{3}\left( \frac{1}{1+t^2} - \frac{1}{4+t^2} \right) e^{-itx} dt \\
= \frac{1}{12} \int_{-\infty}^{\infty} \left(2 \frac{2\cdot 1}{1+t^2} - \frac{2\cdot 2}{4+t^2} \right) e^{-itx} dt
= \frac{1}{12} \left( 2 \cdot 2\pi\,e^{-|x|} - 2\pi\, e^{-2|x|} \right) \\
= \frac{\pi}{6} \left( 2e^{-|x|} - e^{-2|x|} \right)
,
$$
where rule 208 and rule 105 from the third column of these tables have been used.
Taking the derivative we get
$$
\hat{f}'(x)
= \int_{-\infty}^{\infty} \frac{(-it)\,e^{-itx}}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{6} \left( -2e^{-|x|}\operatorname{sign}(x) + 2 e^{-2|x|}\operatorname{sign}(x) \right) \\
= -\frac{\pi}{3} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
$$
so
$$
\int_{-\infty}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= -\operatorname{Re} \int_{-\infty}^{\infty} \frac{(-i)t\,e^{-itx}}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{3} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
$$
and
$$
\int_{0}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{1}{2} \int_{-\infty}^{\infty} \frac{t\,\sin(tx)}{(1+t^2)(4+t^2)}dt
= \frac{\pi}{6} \left( e^{-|x|} - e^{-2|x|} \right) \operatorname{sign}(x)
.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
For every $x\in[\frac{3}{2}, 5]$ prove that: $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2<71.25$ For every $x\in[\frac{3}{2}, 5]$ prove that: $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2<71.25$
I proved this by saying that $(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1})^2=(\sqrt{2}*\sqrt{x-\frac{3}{2}}+\sqrt{\frac{3}{2}}*\sqrt{10-2x}+2\sqrt{x+1})^2$
$\le [(\sqrt{2})^2+(\sqrt{\frac{3}{2}})^2+2^2][(\sqrt{x-\frac{3}{2}})^2+(\sqrt{10-2x})^2+(\sqrt{x+1})^2$ (B.C.S)
$=71.25$
It took me too long till I used $\sqrt{2}, \sqrt{\frac{3}{2}}, 2$ as the numbers for the second bracket used in BCS. Could you please explain to me why I should have immediately intuitively thought of using them and not any other set of numbers?
| We'll choose positives $\alpha$, $\beta$ and $\gamma$ such that after using C-S we'll get:$$\left(\sqrt{2x-3}+\sqrt{15-3x}+2\sqrt{x+1}\right)^2=$$
$$=\left(\sqrt{\alpha}\sqrt{\frac{2x-3}{\alpha}}+\sqrt{\beta}\sqrt{\frac{15-3x}{\beta}}+\sqrt{\gamma}\sqrt{\frac{4x+4}{\gamma}}\right)^2\leq$$
$$\leq(\alpha+\beta+\gamma)\left(\frac{2x-3}{\alpha}+\frac{15-3x}{\beta}+\frac{4x+4}{\gamma}\right)=71.25,$$
For which we need: $$\frac{2}{\alpha}-\frac{3}{\beta}+\frac{4}{\gamma}=0$$ and
$$(\alpha+\beta+\gamma)\left(\frac{-3}{\alpha}+\frac{15}{\beta}+\frac{4}{\gamma}\right)=\frac{285}{4}.$$
The first gives $$\beta=\frac{3\alpha\gamma}{2(2\alpha+\gamma)}$$ and we obtain:
$$\left(\alpha+\frac{3\alpha\gamma}{2(2\alpha+\gamma)}+\gamma\right)\left(\frac{-3}{\alpha}+\frac{15}{\frac{3\alpha\gamma}{2(2\alpha+\gamma)}}+\frac{4}{\gamma}\right)=\frac{285}{4}$$ or
$$(2\alpha-\gamma)(96\alpha^2+7\alpha\gamma-28\gamma^2)=0$$ and we can choose $$\gamma=2\alpha,$$ which gives $$\beta=\frac{3}{4}\alpha$$ and for $\alpha=2$ we obtain your solution.
Easy to see that $71.25$ is not a maximal value.
Indeed, the equality should be occur for
$$(\sqrt{\alpha},\sqrt{\beta},\sqrt{\gamma})||\left(\sqrt{\frac{2x-3}{\alpha}},\sqrt{\frac{15-3x}{\beta}},\sqrt{\frac{4x+4}{\gamma}}\right),$$
which gives $$\frac{\sqrt{2x-3}}{\alpha}=\frac{\sqrt{15-3x}}{\beta}=\frac{\sqrt{4x+4}}{\gamma}$$ or
$$\frac{\sqrt{2x-3}}{\alpha}=\frac{\sqrt{15-3x}}{\frac{3}{4}\alpha}=\frac{\sqrt{4x+4}}{2\alpha}.$$
From the first and the third fractions we obtain $x=4$, which not so plays with the second fraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Solving a repeated eigenvalue ODE I am trying to solve the system:
$$
\bar{x}'
=\left(\begin{array}{rr}-8 & 4 \\ 0 & -8\end{array}\right)\bar{x}+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right),\space \bar{x}(0)=\left(\begin{array}{rr}1 \\ 3 \end{array}\right)
$$
So far I am confident that with the repeated roots $-8$, we have the general solution:
$$
\bar{x}(t)=C_1\left(\begin{array}{rr}1\\ 0\end{array}\right)e^{-8t}+C_2\left(\begin{array}{rr}1\\ 0\end{array}\right)te^{-8t}+C_2\left(\begin{array}{rr}0\\ \frac{1}{4}\end{array}\right)e^{-8t}$$
Then I pick the particular solution:
$\bar{x}_p=\bar{a}e^{-8t}+\bar{b}te^{-8t}+\bar{c}t^2e^{-8t}$
$\bar{x}_p'=-8\bar{a}e^{-8t}+\bar{b}(e^{-8t}-8te^{-8t})+\bar{c}(2te^{-8t}-8t^2e^{-8t})$
Then equating: $\bar{x}_p'=A\bar{x}_p+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right)$
$e^{-8t}$: $-8\bar{a}+\bar{b}=A\bar{a}+\left(\begin{array}{rr}3 \\ 1 \end{array}\right)$
$te^{-8t}$: $-8\bar{b}+2\bar{c}=A\bar{b}$
$t^2e^{-8t}$: $-8\bar{c}=A\bar{c}$
Solving $-8\bar{c}=A\bar{c}$:
$-8c_1+4c_2=8c_1$, so $c_2=0$
$-8c_2=-8c_2$, so $c_1=\alpha$
$\bar{c}=\left(\begin{array}{rr}\alpha \\ 0 \end{array}\right)$
Solving $-8\bar{b}+2\bar{c}=A\bar{b}$:
$-8b_1+4b_2=-8b_1+2\alpha$
$-8b_2=-8b_2+0$
$\bar{b}=\left(\begin{array}{rr}\beta \\ \frac{alpha}{2} \end{array}\right)$
But I am no where close to finding $\bar{a}$, $\bar{b}$ and $\bar{c}$ to finish off my solution as I have too many free variables. Can anyone see where I am going wrong as I am slightly confused? Or offer a better slicker method as mine seems quite clunky...
| We are given
$$x' = Ax + g =
\left(\begin{array}{rr}-8 & 4 \\ 0 & -8\end{array}\right)x+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right),\space x(0)=\left(\begin{array}{rr}1 \\ 3 \end{array}\right)$$
We find eigenvalues / eigenvectors and have a complementary solution
$$ x_c(t)=e^{-8t}\left(c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 0 \\ \dfrac{1}{4} \end{pmatrix} \right)\right)$$
Because of the eigenvalues and the non-homogenous terms, we choose
$$x_p(t) = e^{-8t}(\vec a + \vec b t + \vec ct^2)$$
This gives
$$x_p'(t) = e^{-8t}((-8 \vec a + \vec b) + (-8\vec b + 2 \vec c) t + (-8\vec c)t^2)$$
We have
$$x_p'(t) = A x_p(t) + \vec g \implies \vec g = x_p'(t) - A x_p(t)$$
Writing this out (the exponential terms divide out)
$$\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} b_1-4a_2 \\ b_2 \end{pmatrix}+ \begin{pmatrix} 2 c_1 - 4 b_2 \\ -2c_2 \end{pmatrix}t + \begin{pmatrix} -4c_2 \\ 0 \end{pmatrix}t^2$$
From this, we get
$$a_1 = a_2 = c_2 = 0, b_1 = 3, b_2 = 1, c_1 = 2$$
We can now write
$x(t) = x_c(t) + x_p(t) =e^{-8t}\left(c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 0 \\ \dfrac{1}{4} \end{pmatrix} \right)\right) + e^{-8t}\begin{pmatrix} 2t^2+ 3t \\ t \end{pmatrix}$
Using this $x(t)$ with the IC, we get
$$x(t) = e^{-8t}\begin{pmatrix} 2t^2+15t+1 \\ t+3 \end{pmatrix}$$
Note: We used Undetermined Coefficients to solve this system and that is only one approach, there are many other methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
| We have that
$$x^3-4x^2-4x+16=x(x^2-4x+4)-8x+16=x(x-2)^2-8(x-2)=$$
$$=(x-2)(x^2-2x-8)=(x-2)(x+2)(x-4)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $ I have been trying to derive
$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$
with $G$ being the the Catalan constant.
I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.
| $$I=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx$$
$$=\underbrace{\int_{0}^{1}\frac{log(1-x)log(x)}{1+x^2}dx}_{I_1}+\underbrace{\int_{0}^{1}\frac{log(1-x)log(1+x)}{1+x^2}dx}_{I_2}$$
Let's start evaluating $I_2$ using Leibniz Rule:
$$I_2={\int_0^1}\frac{log(1-x)log(1+x)}{1+x^2}dx=-{\int_0^1}{\int_0^1}{\int_0^1}\frac{x^2}{(1+x^2)(1-yx)(1+xz)}dydzdx$$
$$={\int_0^1}{\int_0^1}{\int_0^1}dxdydz\\\left(\frac{(1+yz)+x(y-z)}{(1+y^2)(1+z^2)(1+x^2)}-\frac{y}{(1+y^2)(y+z)(1-yz)}-\frac{z}{(1+z^2)(y+z)(1+zx)}\right)$$
$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\underbrace{\frac{log(1-y)}{(1+y^2)(y+z)}}_{y\rightarrow z\\ z\rightarrow y}+\frac{log(1+z)}{(1+z^2)(y+z)}\right)$$
$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\frac{log\left(\frac{1-z}{1+z}\right)}{(1+z^2)(y+z)}\right)$$
$$=\int_0^1\left(\frac{\frac{\pi^2}{16}+\frac{log^2(2)}{2}}{1+z^2}+\underbrace{\frac{log\left(\frac{1-z}{1+z}\right)log\left(\frac{1+z}{z}\right)}{1+z^2}}_{z\rightarrow \frac{1-x}{1+x}}\right)dz$$
$$=\frac{\pi}{4}\left(\frac{\pi^2}{16}+\frac{log^2(2)}{4}\right)+log(2)\int_0^1\frac{log(x)}{1+x^2}dx-\underbrace{\int_{0}^{1}\frac{log(x)log(1-x)}{1+x^2}dx}_{I_1}$$
$$=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)-I_1$$
Hence:
$$I=I_1+I_2=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx
=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)\ \blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$, and so $$\int\frac{2x+2}{x^2+2x+3}\,dx=\int\frac{du}{u}=\ln{|x^2+2x+3|}.$$
Next consider the other integral. Put $t\sqrt{2}=x+1$. Then $dx=\sqrt 2\,dt$. Now
\begin{align*}
\int\frac{x+2}{x^2+2x+3}\,dx&=\int\frac{(x+1)+1}{(x+1)^2+2}\,dx\\
&=\int\frac{t\sqrt 2+1}{2t^2+2}\sqrt 2\,dt\\
&=\frac{1}{2}\int\frac{2t+\sqrt 2}{t^2+1}\,dt\\
&=\frac{1}{2}\left(\int\frac{2t}{t^2+1}\,dt+\sqrt 2\int\frac{1}{t^2+1}\,dt\right)\\
&=\frac{1}{2}\left(\ln{|t^2+1|}+\sqrt 2\arctan t\right)
\end{align*}
and hence this is equal to
$$\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+1}{2}+1\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)=\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)$$
therefore
\begin{align*}
\int\frac{x}{x^2+2x+3}\,dx&=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx\\
&=\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C.
\end{align*}
apparently, the correct answer is $\frac{(\ln|x^2+2x+3|)}{2}-\frac{\sqrt{2}\arctan{\frac{(x+1)}{\sqrt{2}}}}{2}+C.$ what went wrong?
| It seems that you made no mistake. Actually, your answer and the “correct” one are one and the same, since$$\frac12\log|x^2+2x+3|-\frac12\log\left|\frac{x^2+2x+3}2\right|$$is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the meaning of the expression $\sqrt[3]{19 - a} + \sqrt[3]{10 - a}$, if $\sqrt[3]{19 - a} - \sqrt[3]{10 - a} = 1$.
Find the meaning of the expression $\sqrt[3]{19 - a} + \sqrt[3]{10 - a}$, if $\sqrt[3]{19 - a} - \sqrt[3]{10 - a} = 1$.
What i tried:
$$\sqrt[3]{19 - a} = m $$
$$\sqrt[3]{10 - a} = n $$
$$ m - n = 1 $$
$$ m = 1 + n $$
$$ (1 + n) + n = 1 + 2n =$$
$$ ... $$
| Also, $$m^3-n^3=\left(\sqrt[3]{19-a}\right)^3-\left(\sqrt[3]{10-a}\right)^3=19-a-(10-a)=9,$$ which gives $$m^2+mn+n^2=9$$ or
$$1+3mn=9$$ or
$$mn=\frac{8}{3},$$ which gives $$m^2+2mn+n^2=9+\frac{8}{3}$$ or
$$(m+n)^2=\frac{35}{3},$$ which gives $$m+n=\sqrt{\frac{35}{3}}$$ or $$m+n=-\sqrt{\frac{35}{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.