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Q regarding finding sum of first 2002 terms Q : A sequence of integers $a_{1}+a_{2}+\cdots+a_{n}$ satisfies $a_{n+2}=a_{n+1}$ $-a_{n}$ for $n \geq 1$. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. Find the sum of the first 2002 terms.
My questions regarding this problem are:
*
*What will be the 1st term ?
Since they say n ≥1. So , do we say 1st term of the sequence has n = 2.
$a_{2+2} = a_{2+1} - a_2$
*It will be great if u could please share different-different ways u can solve it. Also , a very important point is that : What are you thinking at every step while solving this Q. This really helps me a lot because then I can also know what is the way you’re trying to find the solution. Like if it’s a method you’re using , did you already know about it ? Or you’re thinking of ways to find the solution i.e how are u thinking to find ways to solve ?
3)What I have tried:
*
*Sum of 1000th - 1003th term:
1003 - ( -999) = 2002.
*
*What we need to find is the sum of next 999terms. I’m not able to solve further than that.
*Also , please share if you what are the other method that we can use ?
Thank you.
| M:1 = Telescoping
I can see we can’t do anything after this. So , I use telescoping method w/ the help of user dxiv.
$\begin{aligned} a_{n+2} &=a_{n+1}-a_{n} \\ \therefore \quad a_{n+3} &=a_{n+2}-a_{n+1} \\ &=a_{n+1}-a_{n}-a_{n+1} \\ &=-a_{n} \end{aligned}$ $$ \begin{array}{ll}a_{n+4} & =a_{n+3}-a_{n+2} \\ \therefore \quad a_{n+5} & =a_{n+4}-a_{n+3} \\ \therefore \quad a_{n+4} & +a_{n+5}=a_{n+4}-a_{n+2} \\ \therefore \quad a_{n+5} & =-a_{n+2} \\ \therefore \quad a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5} & \\ & =a_{n}+a_{n+1}+a_{n+2}-a_{n}+a_{n+3}-a_{n+2}-a_{n+2} \\ & = & a_{n+1}+a_{n+3}-a_{n+2}=0\end{array} $$
Let $S_{n}$ denotes the sum of first $n$ terms
$$
\begin{aligned}
S_{999} &=S_{6 \times 166+3}=S_{3} &(\because \text { every '6' consecutive }\\
S_{1003} &=S_{6 \times 167+1}=S_{1}^{\circ} &\text { terms has sum zero. }) \\
S_{2002} &=S_{6 \times 333+4}=S_{4} & \\
S_{2002} &=S_{4}=a_{1}+a_{2}+a_{3}+a_{4} \\
&=S_{3}-a_{1} \\
&=1003-(-999) \\
&=2002
\end{aligned} \quad\left(\because a_{4}=-a_{1} \text { since } a_{n+3}=-a_{n}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are there any other "involutive" (a la the orthocenter) points on the Euler line? Throughout, given a triangle $T$ let $G(T)$ and $H(T)$ be the centroid and orthocenter of $T$, respectively. For $r\in\mathbb{R}$ and $p\in\mathbb{R}^2$, let $h_{p:r}$ be the homothety with focus $p$ and factor $r$ (with the understanding that $h_{p:0}(q)=p$ for all $q\in\mathbb{R}^2$). So, for example, the Euler line of a triangle $T$ is $$\{h_{G(T): r}(H(T)): r\in\mathbb{R}\}.$$
I've asked before about "generalized triangle centers" satisfying the same involutive property as the orthocenter. This question is a much more local version of that one: roughly speaking, are there any such generalized triangle centers (besides the orthocenter itself) which always lie on the Euler line?
Precisely, say that an Eulerian triangle center function is a continuous function $f:D\rightarrow\mathbb{R}$ with the following properties:
*
*$D$ is a dense open connected subset of $(\mathbb{R}^2)^3$.
*Both $D$ and $f$ are $S_3$-invariant: for each permutation $\sigma\in S_3$ and $(p_1,p_2,p_3)\in D$ then $(p_{\sigma(1)},p_{\sigma(2)},p_{\sigma(3)})\in D$ and $f(p_1,p_2,p_3)=f(p_{\sigma(1)},p_{\sigma(2)},p_{\sigma(3)})$.
*Both $D$ and $f$ are similarity-invariant: if $\triangle pqr$ is similar to $\triangle p'q'r'$ and $(p,q,r)\in D$, then $(p',q',r')\in D$ and $f(p,q,r)=f(p',q',r')$.
If $f$ is an Eulerian triangle center function, let $$\hat{f}:D\rightarrow \mathbb{R}^2: (p,q,r)\mapsto h_{G(\triangle pqr): f(p,q,r)}(H(\triangle pqr))$$ be the corresponding appropriately-scaled point on the Euler line. So, for example, the constant map ${\bf 1}:(p,q,r)\mapsto 1$ recovers the orthocenter $\hat{\bf 1}: (p,q,r)\mapsto H(\triangle p,q,r)$ (modulo issues re: degenerate triangles).
My question is:
Is there a non-constantly-$1$ Eulerian triangle center $f:D\rightarrow\mathbb{R}$ such that, for comeager-many triples of points $(p,q,r)\in(\mathbb{R}^2)^3$, we have $$\hat{f}(p, q, \hat{f}(p,q,r))=r$$ (so basically, it's not the orthocenter but it satisfies the above-mentioned involutive property)?
If the answer is positive, my next question is how many there are - up to the obvious equivalence relation of "agree on a dense subset of the intersection of their domains." I tentatively suspect the answer is negative, however.
| A point $P$ on the Euler line has barycentric coordinates we can parameterize as
$$(a^2 + b^2 - c^2) (a^2 - b^2 + c^2)-p\,(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4)\;:\;\cdots\;:\;\cdots$$
(with the second and third coordinates derived cyclically from the first); here, $p$ is the dilation factor of the circumcenter with respect to the orthocenter. Straightforward symbol-crunching shows that (barring degeneracies in the triangle) the corresponding center of $\triangle PBC$ is $A$ if and only if $p=0$, which makes $P$ the orthocenter.
For the specific symbol-crunching, let the vertices of the triangle have Cartesian coordinates
$$A=(0,0) \qquad B = (c,0) \qquad C = (b\cos A, b\sin A)$$
Then $P$ is given by
$$\begin{align}
x &=\frac{-a^2 + b^2 + c^2 + (a^2-b^2)p}{
2 c} \\[10pt]
y &=
\frac{a b ((-a^2+b^2+c^2)(a^2-b^2+c^2) + p(a^4 - 2 a^2 b^2 + b^4 + a^2 c^2 + b^2 c^2 - 2 c^4)}{
2(-a+b+c)(a+b-c)(a-b+c)(a+b+c)r}
\end{align}$$
where $r$ is the circumradius. Then, the corresponding point of $\triangle PBC$ is more of a mess to calculate, since $P$ is already much more complicated than $A$, and since we have to replace $b^2\to|PC|^2$ and $c^2\to|PB|^2$ in the barycentric formulas; luckily, Mathematica doesn't see this as too much of a burden, and dutifully churns-out coordinates of the new point.
$$\begin{align}
x &= p\;\frac{
\left(\begin{array}{l}
\phantom{+} 2 (a^4 b^2 - 2 a^2 b^4 + b^6 + 2 a^2 b^2 c^2 - 2 a^2 c^4 -3 b^2 c^4 + 2 c^6) \\
-p(a^6 + 2 a^4 b^2 - 7 a^2 b^4 + 4 b^6 - 3 a^4 c^2 + 12 a^2 b^2 c^2 -
b^4 c^2 - 5 a^2 c^4 - 10 b^2 c^4 + 7 c^6) \\
+ p^2(a^6 - 3 a^2 b^4 + 2 b^6 - 2 a^4 c^2 + 6 a^2 b^2 c^2 - b^4 c^2 -
2 a^2 c^4 - 4 b^2 c^4 + 3 c^6)
\end{array}\right)}{2 c (-(a^2 + b^2 - c^2) (a^2 - b^2 + c^2) + p(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4))} \\[1em]
y &= p\cdot ab\;\frac{\begin{array}{l}
\phantom{\cdot}\left(
-a^4 + 2 a^2 b^2 - b^4 + 2 b^2 c^2 - c^4
+ p(a^2 - b^2 - a c + c^2) (a^2 - b^2 + a c + c^2) \right) \\
\cdot\left(
-2 b^2 (a^2 - b^2 + c^2)
+ p(a^4 + a^2 b^2 - 2 b^4 - 2 a^2 c^2 + b^2 c^2 + c^4) \right)
\end{array}}{
2 (-a + b + c) (a + b - c) (a - b + c) (a + b +
c) (\cdots)}
\end{align}$$
(Barring degeneracies in the triangle) These coordinates simultaneously vanish (so that this secondary point is the required $A$) when and only when $p$ itself vanishes; that is, whenn $P$ is the orthocenter. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I expand $(1+x)^{1/x}$ for small $x$? The binomial expansion
$$(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2}x^{2}+...$$
didn't work because of the $n$ term being undefined at $x=0$.
Taylor expansion doesn't work either since it too would depends on an undefined $1/x$ term.
How does one do it?
| Let's consider the function
$$
f(x)=\begin{cases}
\dfrac{\log(1+x)}{x} & x>-1, x\ne0 \\[6px]
1 & x=0
\end{cases}
$$
Then $f$ is everywhere differentiable and its Taylor expansion at 0 is
$$
f(x)=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\dotsb
$$
Now we have $(1+x)^{1/x}=e^{f(x)}$ (with continuous extension at $x=0$), so we can apply the series for $e^x$.
Say we want to find the Taylor expansion up to degree $3$, for simplicity, so we need
$$
\exp\Bigl(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+o(x^3)\Bigr)
$$
and we get
$$
e\Bigl(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{1}{2}\Bigl(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}\Bigr)^2+\frac{1}{6}\Bigl(-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}\Bigr)^3+o(x^3)\Bigr)
$$
and so
$$
e\Bigl(1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^2}{8}-\frac{x^3}{6}-\frac{x^3}{48}+o(x^3)\Bigr)
$$
and, eventually,
$$
(1+x)^{1/x}=e-\frac{ex}{2}+\frac{11ex^2}{24}-\frac{7ex^3}{16}+o(x^3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
2015 Cambridge Entrance Examination Q6
*
*Show : $\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{2}{1+\sin(x)}\\$. Then evaluate$\int\frac{dx}{1+\sin(x)}$
*Show : $\int_{0}^{\pi}x \space f(\sin(x)) \space dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x)) \space dx$. Then evaluate $\int_{0}^{\pi}\frac{x}{1+\sin(x)}$
*Evaluate: $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}dx$
The first trigonometric identity is quite easy to prove.
$$\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{1}{\cos^2\left(\frac{1}{2}\left((\frac{\pi}{2}-x\right)\right)}=\frac{1}{1/2\left(1+\cos\left(\frac{\pi}{2}-x\right)\right)}=\frac{2}{1+\sin(x)}$$
And the integral just involves substitution the integrand with the secant term, which alloys us to take advantage of the fact that the derivative of the tangent is secant squared.
The first second integral were tricky at first but allowing $y=\pi-x$ did the trick. However evaluting the integral in the third part...
$$\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin(x))^2}$$
... wasn't as simple as I thought. I'm really not sure where to begin or how to utelise the integral properties proved in the previous parts. Thanks
| The beauty of this problem is that it used an allied function $\sin(x)$ with phase $\pi$
If you could see the graph of $y = E(x)\times A(x)$ it can help you understand the complexity of the problems
Basically if $E(x)$ is an odd function then this can be solved:
*
*Solution
$\int_0^\pi x^nf(\sin x)dx$ can be solved if $n $ is an odd
Given: $\int_0^\pi x f(\sin x) dx = \frac \pi2\int_0^\pi f(\sin x)dx$
*
*To find $\int_0^\pi \frac {2x^3-3\pi x^2}{(1+\sin x)^2}dx = \int_0^\pi (2x^3-3\pi x^2) f(\sin x)$
Take
$$\begin{align*}
\int_0^\pi x^3 f(\sin x)dx
& = \int_0^\pi(\pi-x)^3f(\sin x)dx\\
& \text { by expansion.......and using the given property }\\
& \implies \int_0^\pi x^3f(\sin x)dx = \int_0^\pi \left(\frac {\pi^3}{2}-\frac{3\pi^3}4 + \frac {3\pi x^2}{2} \right)f(\sin x)dx
\end{align*}$$
Coming back to our original problem:
$$I = \int_0^\pi \frac {2\left(-\frac {\pi^3}{2} + \color{blue}{\frac {3\pi x^2}{2}}\right) - \color{blue}{3\pi x^2}}{(1+\sin x)^2}dx$$
Now, I believe you can solve $\int_0^\pi\text {constant }\frac {1}{(1+\sin x)^2}dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| To solve
$$
\left(\frac{z-3+2i}{z}\right)^4=1\tag1
$$
Using De Moivre, we want to find $\theta$ so that
$$
\begin{align}
1
&=(\cos(\theta)+i\sin(\theta))^4\tag{2a}\\
&=\cos(4\theta)+i\sin(4\theta)\tag{2b}
\end{align}
$$
which is $\theta\in\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}$
$$
\frac{z-3+2i}z=\cos\left(\tfrac{k\pi}2\right)+i\sin\left(\tfrac{k\pi}2\right)\tag3
$$
for $k\in\{1,2,3\}$ (we won't get a solution for $k=0$). That is,
$$
z=\frac{3-2i}{1-\cos\left(\frac{k\pi}2\right)-i\sin\left(\frac{k\pi}2\right)}\tag4
$$
We get the three solutions
$$
\left\{\frac{5+i}2,\frac{3-2i}2,\frac{1-5i}2\right\}\tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is not that brought me in a right way.
Do you have any idea how to solve this one?
| Let $ t = \tan \left(y\right)$ and $ u = \tan \left(x\right)$. One has $\sin \left(y\right) = \pm \frac{t}{\sqrt{1+{t}^{2}}}$ and $\cos \left(x\right) = \pm \frac{1}{\sqrt{1+{u}^{2}}}$. The second equation then implies
\begin{equation}2 {t}^{2} = \left(1+{u}^{2}\right) \left(1+{t}^{2}\right)\end{equation}
but according to the first equation, we have $ u = 1-t$, hence the quadric equation
\begin{equation}\renewcommand{\arraystretch}{1.5} \begin{array}{cc}&2 {t}^{2} = \left(1+{\left(1-t\right)}^{2}\right) \left(1+{t}^{2}\right)\\
\Longleftrightarrow &\left(t-1\right) \left({t}^{3}-{t}^{2}-2\right) = 0
\end{array}\end{equation}
This equation has two real solutions $ {t}_{0} = 1$ and $ {t}_{1} = \frac{1}{3} \left(d+1+\frac{1}{d}\right)$ with $ d = \sqrt[3]{3 \sqrt{87}+28}$.
The first solution gives
\begin{equation}x = k {\pi} + 2 n \pi, \quad y = \frac{{\pi}}{4}+ k {\pi}\end{equation}
The second solution gives
\begin{equation}y = \arctan \left({t}_{1}\right)+k {\pi} , \quad x = \arctan \left(1-{t}_{1}\right)+k {\pi}+2 n {\pi}\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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What is the solution to $a_n =a_{n-1}^2+a_{n-2}^2, a_0, a_1>1 $? I saw this on Quora
and have made
very little progress.
What is the solution
(exact or asymptotic) to
$a_n
=a_{n-1}^2+a_{n-2}^2,
a_0, a_1>1
$?
I would be satisfied
with a good asymptotic analysis.
Heck,
I would be happy with an
asymptotic form
of the log of the solution.
A lower bound is easy.
Assuming $a_1 > a_0 > 1$,
$a_n>a_{n-1}^2
\gt a_{n-2}^4
... > a_{n-k}^{2^k}
...
\gt a_1^{2^{n-1}}
\gt a_0^{2^{n}}
$.
A reasonable upper bound seems harder.
$a_{n-1} > a_{n-2}^2
$
so
$a_n
=a_{n-1}^2+a_{n-2}^2
\lt a_{n-1}^2+a_{n-1}
$
so
$a_n+\frac14
\lt a_{n-1}^2+a_{n-1}+\frac14
=(a_{n-1}+\frac12)^2
$
or
$a_n+\frac12
\lt (a_{n-1}+\frac12)^2+\frac14
$.
If
$b_n
= a_n+\frac12
$,
$\begin{array}\\
b_n
&\lt b_{n-1}^2+\frac14\\
&\lt (b_{n-2}^2+\frac14)^2+\frac14\\
&= b_{n-2}^4+\frac12 b_{n-2}^2+\frac1{16}+\frac14\\
\end{array}
$
Not sure where to go
from this.
Another possibility
is to notice that
once $a_n$ gets large,
for any $c > 0$
there is an
$n(c)$ such that
for $n \ge n(c)$,
$a_n > 1/c$ so
$\begin{array}\\
a_n
&=a_{n-1}^2+a_{n-2}^2\\
&\lt a_{n-1}^2+a_{n-1}\\
&= a_{n-1}^2(1+1/a_{n-1})\\
&\lt (1+c)a_{n-1}^2\\
&\lt (1+c)((1+c)a_{n-2}^2)^2\\
&=(1+c)^3a_{n-2}^4\\
&<(1+c)^3((1+c)a_{n-3}^2)^4\\
&=(1+c)^7a_{n-3}^8\\
& ...\\
&<(1+c)^{2^k-1}a_{n-k}^{2^k}\\
& ... \text{up to } n-k = n(c), k=n-n(c)\\
&<(1+c)^{2^{n-n(c)}-1}a_{n(c)}^{2^{n-n(c)}}\\
\end{array}
$
Don't see how to do better.
| For any $n \ge 1$, we have:
\begin{align*}
a_{n+1} &= a_n^2+a_{n-1}^2
\\
a_{n+1} &= a_n^2\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right)
\\
\log a_{n+1} &= 2\log a_n + \log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right)
\\
\dfrac{1}{2^{n+1}}\log a_{n+1} &= \dfrac{1}{2^n}\log a_n + \dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right)
\end{align*}
This is enough to show that $\dfrac{1}{2^n}\log a_n$ is non-decreasing, and thus, either converges to some limit or is unbounded.
Summing the previous equation from $n = 2$ to $n = N-1$ yields, $$\dfrac{1}{2^N}\log a_N = \dfrac{1}{4}\log a_2 + \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right).$$
It is easy to check that for all $n \ge 2$, we have $a_{n-1} > 1$, and thus, $a_n = a_{n-1}^2+a_{n-2}^2 > a_{n-1}^2 > a_{n-1}$. Hence, we can bound $$0 \le \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right) \le \sum_{n = 2}^{N-1}\dfrac{1}{2^n}\log 2 = \dfrac{1}{2}\log 2,$$ and thus, $$\dfrac{1}{4}\log a_2 \le \dfrac{1}{2^N}\log a_N \le \dfrac{1}{4}\log a_2 + \dfrac{1}{2}\log 2.$$
Since $\dfrac{1}{2^n}\log a_n$ is bounded (and previously shown to be non-decreasing), $\dfrac{1}{2^n}\log a_n$ converges to some number between $\dfrac{1}{4}\log a_2$ and $\dfrac{1}{4}\log a_2 + \dfrac{1}{2}\log 2$.
We can probably get tighter bounds if we are more careful about bounding the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How do I evaluate $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$ How do I evaluate the following integral when where $C$ is the square with vertices at $\pm2, \pm2+4i$,
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$$
Using Cauchy integral:
$\frac{z^{2}}{z^{2}+4}=\frac{z^2}{(z+2i)(z-2i)}=\frac12\frac{z^2}{z+2i}+\frac12\frac{z^2}{z-2i}$ then $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4} \implies \frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}$
$\frac{z^2}{z+2i}$ is analytic on and inside $C$, hence we can apply Cauchy theorem and for the second term we use Cauchy integral formula,
$$\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z+2i}+\frac{1}{2 \pi i} \oint_{C} \frac12\frac{z^2 d z}{z-2i}=0+\frac{1}{4\pi i}2\pi i \times f(2i)=-2$$
Using Residue Theorem:
$2i$ is the only isolated singularity in $C$.
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}=\frac{1}{2 \pi i} 2\pi i \times \text{Res}(f,2i)=i$$
I get correct answer for Residue Theorem but couldn't understand where I do wrong when using Cauchy integral.
It will be great help if someone clear me when to use which method to find the integral.
| I will perform the first computation in a slightly different manner:
$$
\begin{aligned}
\frac{1}{2 \pi i} \oint_{C} \frac{z^2 }{z^2+4}\; dz
&=
\frac{1}{2 \pi i} \oint_{C} \frac{(z^2 + 4) - 4}{z^2+4}\; dz
\\
&=
\frac{1}{2 \pi i} \oint_{C} dz
+
\frac{1}{2 \pi i} (-4)\oint_{C} \frac1{z^2+4}\; dz
\\
&=
0
+
\frac{1}{2 \pi i} (-4)\oint_{C} \frac 1{4i}\left(\frac1{z-2i} - \frac1{z+2i}\right)\; dz
\\
&=
\frac{1}{2 \pi i} \cdot i\oint_{C} \frac{dz}{z-2i}
-
\frac{1}{2 \pi i} \cdot i\oint_{C} \frac{dz}{z+2i}
\\
&= i-0\ .
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding bounds for a function I would like to show that
$$\frac{1}{\pi^2}<\int_{\pi/2}^\pi\frac{\sin x}{x^3}<\frac{3}{2\pi^2}.$$
We know that $\frac{\sin x}{x^3}\leq\frac{1}{x^3}$ and integrating gives $\int_{\pi/2}^\pi\sin x/x^3\leq3/(2\pi^2)$. I don't know how to make $\leq$ into $<$. On the other hand, $\frac{1}{\pi^3}\leq\frac{\sin x}{x^3}$ and integrating gives $1/\pi^3\leq\int_{\pi/2}^\pi\sin x/x^3$. But $1/\pi^3<1/\pi^2$ so this clearly is not what I want.
Any suggestions in how to fix these issues?
|
I don't know how to make $≤$ into $<$
Notice that $\frac{\sin(x)}{x^3}$ and $\frac{1}{x^3}$ are only equal at $\frac{\pi}{2}$. So if we split $\left[\frac{\pi}{2} , \pi \right]$ as $\left[\frac{\pi}{2} , \frac{3\pi}{4} \right] \cup \left[\frac{3\pi}{4} , \pi \right]$ for example, we can thus assert that
\begin{align}
\frac{\sin(x)}{x^3}\le \frac{1}{x^3} \text{ on }\left[\frac{\pi}{2} , \frac{3\pi}{4} \right]\\
\frac{\sin(x)}{x^3}\mathbin{\color{red}{<}} \frac{1}{x^3} \text{ on }\left[\frac{3\pi}{4}, \pi \right]
\end{align}
And hence
\begin{align*}
\int_{\frac{\pi}{2}}^\pi\frac{\sin x}{x^3} \, \mathrm{d}x &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{\sin x}{x^3} \, \mathrm{d}x + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x \\
& \le \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}\frac{1}{x^3} \, \mathrm{d}x + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x \\
& =\frac{10}{9 \pi^2} + \int_{\frac{3\pi}{4}}^{\pi}\frac{\sin x}{x^3} \, \mathrm{d}x\\
&\mathbin{\color{red}{<}}\frac{10}{9 \pi^2} + \int_{\frac{3\pi}{4}}^{\pi}\frac{1}{x^3} \, \mathrm{d}x\\
& = \frac{3}{2\pi^2}
\end{align*}
as desired.
On the other hand $\frac{1}{\pi^3}\leq\frac{\sin x}{x^3}$.
This is not the inequality you want. Notice that on $\left[\frac{\pi}{2}, \pi \right]$ we know $\sin(x)$ is concave, which means it can be bounded from below by the line passing through the endpoints. This line turns out to be
$$
y\ =\ -\frac{2}{\pi}\left(x-\pi\right)
$$
So we can thus say that
$$
\frac{-\frac{2}{\pi}\left(x-\pi\right)}{x^{3}} \le \frac{\sin(x)}{x^3} \text{ on }\left[\frac{\pi}{2} , \pi \right]
$$
And since $ \int_{\frac{\pi}{2}}^\pi\frac{-\frac{2}{\pi}\left(x-\pi\right)}{x^{3}} \, \mathrm{d}x =\frac{1}{\pi^2} $ this gives the desired result.
Lastly, if you're again worried about the $\le$ instead of $<$, then you can use the same trick of splitting the interval and then evaluating the integrals separately.
| {
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"url": "https://math.stackexchange.com/questions/4368515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $a,b,c,d,e$ be five numbers satisfying the following conditions...
Let $a,b,c,d,e$ be five numbers satisfying the following conditions: $$a+b+c+d+e =0$$ and $$abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=33$$ Find the value of $$\frac{a^3+b^3+c^3+d^3+e^3}{502}$$
My Approach:
$$(a+b+c+d+e)^3 = \sum_{a,b,c,d,e}{a^3} + 3\sum_{a,b,c,d,e}{a^2b} + 6\sum_{a,b,c,d,e}{abc} $$
Taking $\mod (a+b+c+d+e)$, $$(a+b) ≡ -(c+d+e)$$ $$ab(a+b) ≡ -ab(c+d+e)$$ $$\sum{a^2b} ≡ -ab(c+d+e) -bc(a+d+e) -cd(a+b+e)-... = -\sum_{a,b,c,d,e}{ab(c+d+e)} = -3\sum_{a,b,c,d,e}{abc}$$ Therefore, $\sum{a^2b} = p(a,b,c,d,e) . (a+b+c+d+e) - 3\sum_{a,b,c,d,e}{abc}$
Since,
$(a+b+c+d+e) = 0$ $$\sum{a^3} = (3×3 -6)\sum{abc} = 3×33 = \color{red}{99}$$
But the answer key shows: $$\frac{\sum{a^3}}{\color{blue}{502}} = 99$$
Where is my mistake?
| OP's result is correct. For verification, Newton's identity $\,p_3=e_1^3-3e_1e_2+3e_3\,$ for the sum of cubes gives the same result directly (where the sums are the symmetric sums over $a,b,c,d,e$):
$$
\sum a^3 = \left(\sum a\right)^3 - 3\,\left(\sum a\right)\left(\sum ab\right) + 3 \left(\sum abc\right) = 0 - 3 \cdot 0 \cdot \left(\sum ab\right) + 3 \cdot 33 = 99
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{(x,y)\to(0,0)}\dfrac{x^2+y^2}{x^4+y^4}$
Evaluate the limit: $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{x^2+y^2}{x^4+y^4}$
To solve this, I converted it to polar coordinate and got: $\displaystyle\lim _{r\to0}\left(\frac{1}{r^2(\sin^4\theta+cos^4\theta)}\right)=\infty$
But after putting this on WolframAlpha, it tells me that this limit does not exist.
Who is wrong here?
| We can also see that :
$$\dfrac{x^2 + y^2}{x^4 + y^4} \geq \dfrac{x^2 + y^2}{(x^2 + y^2)^2} = \dfrac{1}{x^2 + y^2} \underset{(x, y) \to (0, 0)}{\to} +\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.
Proof: $$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})
$$$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$
As $\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$
The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.
I.e. $$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$
$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$
and the RHS will reduce down to $\sqrt{3}$. Hence LHS=RHS.
Some things that I've noticed about this method of proof:
*
*It could be used to (incorrectly) prove that $$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$
So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?
*
*Instead of proving (*), wouldn't this method of proof actually prove that? $$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$
for some $k\in \mathbb{Z}$ which we must find. In this case being when $k=0$.
| You have$$3\arccos\left(\frac5{\sqrt{28}}\right)\in[0,3],$$since $\frac{\sqrt3}2<\frac5{\sqrt{28}}<1$, and therefore$$3>3\frac\pi6>3\arccos\left(\frac5{\sqrt{28}}\right)>0.$$You also have$$0\leqslant3\arctan\left(\frac{\sqrt3}2\right)<3\arctan\left(\sqrt3\right)=\pi,$$and therefore$$\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\in\left[0,\frac\pi3+1\right].$$What you did shows that$$\tan\left(\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\right)=\tan\left(\frac\pi3\right).$$But the only number in $\left[0,\frac\pi3+1\right]$ whose tangent is $\tan\left(\frac\pi3\right)$ is $\frac\pi3$. So, your proof is indeed correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given real numbers $a,b,c >0$ and $a+b+c=3$. Prove that $\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$ My first way is to put the inequality under the same degree, so I multipled both sides to $(a+b+c)$, however it leaded to a hard to solve result. Can anyone help me with this problem?
| For the current inequality of $ \sum \frac{a^2}{b} \geq 3 + \frac{4}{3} \max ( a-b)^2 $:
*
*Replace 3 with $ a+b+c$. The reason for this choice is the "well-known" inequality: $ \sum \frac{ a^2}{b} \geq \sum a $. We're then asking how much more leeway there is in the difference of these terms.
*We can write the difference as a as Sum of Squares (Figure out how to do this before looking at the hint.). Thus, we want to show that
$$ \sum \frac{ (a-b)^2 } { b} \geq \frac{4}{3} \max ( a-b)^2.$$
This is just Cauchy Schwarz:
*
*Numerator: $[ \sum | a - b | ] ^2= [2 ( \max (a,b,c) - \min (a, b, c) )]^2 = 4 \max((a-b)^2)$.
*Denominator: $a + b + c = 3$.
Equality holds in the Cauchy Schwarz iff:
*
*For $ a \geq b \geq c$: $\frac{ a-b}{b} = \frac{b-c}{c} = \frac{a-c}{a} , a+b+c = 3$ $\Rightarrow (a, b, c) = (1, 1, 1) , (\frac{3}{2} , \frac{ 3 \sqrt{5} - 3 } { 4} , \frac{ 9 - 3 \sqrt{5}}{ 4} ). $
*For $ a \geq c \geq b$: $\frac{ a-b}{b} = \frac{c-b}{c} = \frac{a-c}{a} , a+b+c =3 $, which leads to the same cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorizing a quartic expression to show that it is a perfect square.
Show that $\frac{a^4+b^4+(a+b)^4}{2}$ is a perfect square.
I tried this,
$$\frac{a^4+b^4+(a+b)^4}{2}$$
$$\frac{a^4+b^4+(a^2+b^2+2ab)^2}{2}$$
$$\frac{2a^4+2b^4+4a^2b^2+2(a^2b^2+2a^3b+2ab^3)}{2}$$
$$a^4+b^4+2a^2b^2+ab(ab+2a^2+2b^2)$$
$$(a^2+b^2)^2+ab(ab+2a^2+2b^2)$$
What can I do next?
| If $a^4+2a^3b+3a^2b^2+2ab^3+b^4$ is a square, it has to be of the form $(xa^2+yab+zb^2)^2$ for some coefficients $x,y,z$.
Then $a^4+2a^3b+3a^2b^2+2ab^3+b^4=x^2a^4+2xya^3b+(y^2+2xz)a^2b^2+2yzab^3+z^2b^4$.
Comparing coefficients we get $x^2=1, 2xy=2, y^2+2xz=3,2yz=2,z^2=1$.
Solving this system we get $x=y=z=1$ and $x=y=z=-1$, so $a^4+2a^3b+3a^2b^2+2ab^3+b^4=(a^2+ab+b^2)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$ $a,b,c$ are reals $ >0$ prove that $$\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$$
The inequality is homogeneous so assum WLOG $a+b+c=3$, the inequality is equivalent to $$\sum_{cyc}\frac{(3-2c)^2}{(3-c)^2+c^2} \ge 3/5$$
Set $$f(x)= \frac{(3-2x)^2}{(3-x)^2+x^2} $$
We wish to prove that for $0<x<3$ we have $f(x)\ge 1/5$... but unfortunately this is false, so what should I do?
| Hint: Show $f(x)+\frac{18}{25}(x-1)\geqslant \frac15$ for all $x\in (0,3)$.
| {
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Evaluate $I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$ I am trying to calculate this integral:$$I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$$.
I tried to find the antiderivative but it didn't exist.
So i changed variable by set $t=\frac{x-1}{x+1}$ due to factor numerator is $x(x-1)$ and it led to:$$I=2\int_{-1}^{0}\frac{t^2+t}{(1-t)^3\ln{\frac{t+1}{1-t}}}dx$$ and it seems more harder.
The result from Wolfram Alpha is ok, but i don't know how to evaluate this result. Need some hints or advices from everyone. Thank you.
| Another approach can be as below:
First, by this link https://en.wikipedia.org/wiki/Frullani_integral one can easily obtain: $$\int_0^1 \frac{x^{a-1}-x^{b-1}}{\ln x}dx=\int_0^\infty \frac{e^{-bt}-e^{-at}}{t}dt=\ln\frac{a}{b},$$
where $a,b\gt0.$
Second, we have a well-known relation that states: $$\frac{\pi}{2}=(\frac{2}{1}\times\frac{2}{3})(\frac{4}{3}\times\frac{4}{5})(\frac{6}{5}\times\frac{6}{7})(\frac{8}{7}\times\frac{8}{9})\times\cdots . $$
For example you can see this relation in this link https://en.wikipedia.org/wiki/Pi .
Third, notice that: $\frac{1}{1+x}=1-x+x^2-x^3+\cdots . $
Now, we get: $$I=\int_0^1\frac{(x^2-x)(1-x+x^2-x^3+\cdots)}{\ln x}dx=\sum_{k=0}^\infty (-1)^k\ln\frac{k+3}{k+2}.$$
The last sum is indeed:$$I=\ln\frac{3}{2}-\ln\frac{4}{3}+\ln\frac{5}{4}-\ln\frac{6}{5}+\cdots=\ln\frac{9}{8}+\ln\frac{25}{24}+\ln\frac{49}{48}+\cdots =\ln\frac{4}{\pi}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Which solution to $\int \frac{x^3}{(x^2+1)^2}dx$ is correct? I tried solving the following integral using integral by parts :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
but I got a different answer from Wolfram Calculator This is the answer that I got :
$$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$
I am wondering if I am wrong or the calculator if I am where did I do something wrong.
Here is my solution :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
$$\int x^2\frac{x}{(x^2+1)^2}dx$$
$\implies u =x^2 \qquad u'=2x$
$\displaystyle\implies v'=\frac{x}{(x^2+1)^2}\qquad v=\int\frac{x}{(x^2+1)^2}dx$
$$\int\frac{x}{(x^2+1)^2}dx$$
$\implies \zeta=x^2+1$
$\implies\displaystyle \frac{d\zeta}{2}=x$
$$\boxed{v=\frac{1}{2}\int\frac{d\zeta}{\zeta^2}=-0.5\zeta^{-1}=\frac{-1}{2(x^2+1)}}$$
$$\int x^2\frac{x}{(x^2+1)^2}dx=x^2\frac{-1}{2(x^2+1)}-\underbrace{(\int \frac{-2x}{2(x^2+1)}dx)}_{-\frac{1}{2}\ln(x^2+1)}$$
$$\boxed{\int \frac{x^3}{(x^2+1)^2}dx=\frac{-x^2}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C}$$
| BOTH are correct since the two answers differ by a constant.
$$
\begin{aligned}
\int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int x^{2} d\left(\frac{1}{x^{2}+1}\right) \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\
&=-\frac{x^{2}+1-1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\
&=-\frac{1}{2}+\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{1} \\
&=\frac{1}{2\left(x^{2}+1\right)}+\frac{1}{2} \ln \left(x^{2}+1\right)+C_{2},
\end{aligned}
$$
where $C_{2}= C_{1} -\frac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2+ac+bd \geq \sqrt{3}$.
Solution:
Make the inequality homogenous by multiplying both sides by $ad-bc=1$. [That seems sensible.] Take everything onto one side so we now want to show
$$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}.$$
[That's also a reasonable thing to do. The trouble is coming next...]
Now play around until you notice the left hand side can be written as
$$\frac{1}{4}(2a+c-\sqrt{3}d)^2 + \frac{1}{4}(2b+d+\sqrt{3}c)^2 \tag{2}.$$
[What??]
I played around for a fair while and didn't get to this. I have a suspicion that this question was created by reverse-engineering. What thought processes get you from (1) to (2), without knowing (2) beforehand? How can you get to the solution without pulling a rabbit out of a hat?
| Rather than $\sqrt 3 $ here is the matrix algorithm for coefficient $1.$ I will try $\sqrt 3$ in a few minutes
Positivity is shown in matrix $D.$ It is then matrix $Q$ that fills in the linear terms, as in: double your form (coefficient $1$) is
$$ 2 \left( a + \frac{c}{2} - \frac{d}{2} \right)^2 + 2 \left( b + \frac{c}{2} + \frac{d}{2} \right)^2 + \left(c \right)^2 + \left( d\right)^2 $$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\
- \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\
0 & 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
2 & 0 & 1 & - 1 \\
0 & 2 & 1 & 1 \\
1 & 1 & 2 & 0 \\
- 1 & 1 & 0 & 2 \\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Allowing the coefficient $\sqrt 3$ to be replaced by variable $x$ we get
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\frac{ 1 }{ 2 } & \frac{ x }{ 2 } & 1 & 0 \\
- \frac{ x }{ 2 } & \frac{ 1 }{ 2 } & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & \frac{3-x^2}{2} & 0 \\
0 & 0 & 0 & \frac{3-x^2}{2} \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & \frac{ 1 }{ 2 } & - \frac{ x }{ 2 } \\
0 & 1 & \frac{ x }{ 2 } & \frac{ 1 }{ 2 } \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrrr}
2 & 0 & 1 & - x \\
0 & 2 & x & 1 \\
1 & x & 2 & 0 \\
- x & 1 & 0 & 2 \\
\end{array}
\right)
$$
with resulting expansion
$$ 2 \left( a + \frac{c}{2} - \frac{dx}{2} \right)^2 + 2 \left( b + \frac{cx}{2} + \frac{d}{2} \right)^2 + \left( \frac{3-x^2}{2} \right) \left(c \right)^2 + \left( \frac{3-x^2}{2} \right) \left( d\right)^2 $$
Once we set $ x = \sqrt 3$ we get
$$ 2 \left( a + \frac{c}{2} - \frac{d\sqrt3}{2} \right)^2 + 2 \left( b + \frac{c\sqrt3}{2} + \frac{d}{2} \right)^2 $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
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"timestamp": "2023-03-29T00:00:00",
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An exercise on the implicit function theorem I am trying to learn the implicit function theorem and this is one exercise about it; I have solved it and would be grateful for any feedback on my solution, thanks.
Let $f\begin{pmatrix}x\\ y\\ z\end{pmatrix}=x y^2+\sin(xz)+e^z$ and $\textbf{a}=\begin{bmatrix}1\\ -1\\ 0 \end{bmatrix}$.
(a) Show that the equation $f=2$ defines $z$ as a $\mathcal{C}^1$ function $z=\phi\begin{pmatrix}x\\ y\end{pmatrix}$ near $\textbf{a}.$
(b) Find $\frac{\partial\phi}{\partial x}\begin{pmatrix}1\\-1\end{pmatrix}$ and $\frac{\partial\phi}{\partial y}\begin{pmatrix}1\\-1\end{pmatrix}.$
(c) Find the equation of the tangent plane of the surface $f^{-1}(\{2\})$ at $\mathbf{a}$ in two ways.
What I have done:
(a) Let $F=f-2$; $F$ is a $\mathcal{C}^1$ function, $F(\mathbf{a})=0,\ DF\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{bmatrix}y^2+z\cos(xz) &2xy & x\cos(xz)+e^z\end{bmatrix}$ and $DF(\mathbf{a})=DF\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}=\begin{bmatrix}1 & -2 & 2\end{bmatrix}$ so in particular $\frac{\partial F}{\partial z}(\mathbf{a})=2\neq 0$ thus there exists a neighborhood V of $\begin{pmatrix}1\\-1\end{pmatrix}$ and $W$ of $0$ and a $\mathcal{C}^1$ function $\phi:V\to W$ so that $z\in W\Leftrightarrow z=\phi\begin{pmatrix}x\\ y\end{pmatrix},\ \begin{pmatrix}x\\ y\end{pmatrix}\in V.$
(b) $$\frac{\partial\phi}{\partial x}=-\frac{\frac{\partial F}{\partial x}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}{\frac{\partial F}{\partial z}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}=-\frac{1}{2}$$ and $$\frac{\partial\phi}{\partial y}=-\frac{\frac{\partial F}{\partial y}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}{\frac{\partial F}{\partial z}\begin{pmatrix}1\\ -1\\ 0\end{pmatrix}}=-\frac{-2}{2}=1.$$
(c) Tangent plane: $$\begin{bmatrix}1 & -2 & 2\end{bmatrix} \begin{bmatrix}x-1\\ y+1\\ -z\end{bmatrix}=x-1-2(y+1)-2z=0\Leftrightarrow x-2y-2z=3$$
| (a) and (b) are fine. As regards (c), there is a minor error in your work: the tangent plane should be
$$\begin{bmatrix}1 & -2 & 2\end{bmatrix} \begin{bmatrix}x-1\\ y+1\\ z-0\end{bmatrix}=x-1-2(y+1)+2z=0\Leftrightarrow x-2y+2z=3.$$
Here it is another way to find the tangent plane. Since $\phi$ is differentiable at $\begin{pmatrix}1\\-1\end{pmatrix}$, then
$$\phi\begin{pmatrix}x\\ y\end{pmatrix}=\phi\begin{pmatrix}1\\-1\end{pmatrix}+
\nabla \phi\begin{pmatrix}1\\-1\end{pmatrix}\cdot \left( \begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}1\\-1\end{pmatrix}\right)+o\left(\sqrt{(x-1)^2+(y+1)^2}\right)$$
Hence the required tangent plane is
$$z= \phi\begin{pmatrix}1\\-1\end{pmatrix}+
\nabla \phi\begin{pmatrix}1\\-1\end{pmatrix}\cdot \left( \begin{pmatrix}x\\ y\end{pmatrix}-\begin{pmatrix}1\\-1\end{pmatrix}\right)=
0+
\begin{pmatrix}-\frac{1}{2}\\1\end{pmatrix}\cdot \begin{pmatrix}x-1\\ y+1\end{pmatrix}\\
=-\frac{1}{2}(x-1)+(y+1)$$
which is equivalent to the equation given above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability for Ben, Amos and Carl Three men Amos, Ben and Carl share an office at work with a single telephone. Calls call in at random with the proportions of $\dfrac{1}{2}$ for Amos, $\dfrac{1}{3}$ for Ben and $\dfrac{1}{6}$ for Carl. For any incoming, calls, the probabilities that it will be picked up by Amos, Ben, and Carl are $\dfrac{1}{2}$, $\dfrac{3}{10}$ and $\dfrac{1}{5}$ respectively. For calls arriving during working hours, find the probability that (i) a call is not picked up by the person being called, (ii)
a call is for Ben given that a call is not picked up by the person being called.
For (i), I tried to get P(proportion for Amos & not picked up) or P(proportion for Ben & not picked up) or P(proportion for Carl & not picked up)
$= \dfrac{1}{2} * \dfrac{1}{2} + \dfrac{1}{3} * \dfrac{1}{2} + \dfrac{1}{6} * \dfrac{5}{6} = 0.5546 $
But the answer for (i) is $\dfrac{37}{60}$, what went wrong?
How do I do for (ii)?
Your help is appreciated. Thanks
| You got one number wrong. Your idea is correct, we have
\begin{align*}
\def\P{\mathbf P}\P(\text{call picked up by wrong person})
&=\underbrace{ \frac 12}_{\text{call for A}}\cdot \underbrace{\frac 12}_{\text{call not picked by A}} +
\underbrace{ \frac 13}_{\text{call for B}}\cdot \underbrace{\frac 7{10}}_{\text{call not picked by B}}+
\underbrace{ \frac 16}_{\text{call for C}}\cdot \underbrace{\frac 45}_{\text{call not picked by C}}\\
&= \frac 14 + \frac 7{30} + \frac 4{30}\\
&= \frac{30 + 28 + 16}{120}\\
&= \frac{37}{60}
\end{align*}
For (ii), we have
\begin{align*}
\P(\text{call for B}\mid\text{wrong person}) &= \frac{\P(\text{call for B}\cap \text{wrong person})}{\P(\text{wrong person})}\\
&= \frac{\frac 12 \cdot \frac 12}{\frac{37}{60}}\\
&= \frac{15}{37}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating the spherical harmonic of θ=π/2 This is a very simple question, yet I'm not sure how to approach it. I want to calculate the spherical harmonic:
$$ Y_{l m}^{*}(\theta = \pi/2, \phi) $$
I know the general formula:
$$ Y_{l m}^{*}(\theta, \phi)=\sqrt{\frac{2 l+1}{4 \pi} \frac{(l-m) !}{(l+m) !}} P_{l}^{m}(\cos \theta) e^{-i m \phi} $$
But for $\pi/2$ I need to calculate the associate Legendre polynomial of 0 ($P_{l}^{m}(\cos \pi/2)$), which I'm not sure how.
The Rodrigues' formula is not clear to me for the case of $x=0$:
$$ P_{l}^{m}(x)=\frac{(-1)^{m}}{2^{l} l !}\left(1-x^{2}\right)^{m / 2} \frac{d^{l+m}}{d x^{l+m}}\left(x^{2}-1\right)^{l} $$
Any guidance on how to calculate it for that special case would be appreciated.
| For each choice of $l$ and $m$, you can find a closed form solution for $P_l^m(x)$ as a function of $x$. Typically, I would start there and only then evaluate it at a particular value of $x$ such as $x = 0$.
That said, if you prefer we can find a formula for $P_l^m(0)$ in terms of $l$ and $m$
Instead of starting from the Rodrigues' formula, we could just use this closed form [1] of the associated Legendre polynomials:
$$P_l^m(x)=(-1)^{m} \cdot 2^{l} \cdot (1-x^2)^{m/2} \cdot \sum_{k=m}^l \frac{k!}{(k-m)!}\cdot x^{k-m} \cdot \binom{l}{k} \binom{\frac{l+k-1}{2}}{l}$$
where $\binom \alpha k$ is the generalized binomial coefficient:
$$\binom \alpha k = \frac{1}{k!} \prod_{i=0}^{k-1} (\alpha-i)$$
In particular, for $x = 0$ and $k > m$ we have $x^{k-m} = 0$. Therefore, the only term in the sum of $k$ which may be non-zero is the $k = m$ term.
Now, if we naively plug in $x = 0$ and $k = m$, we'd get $x^{k-m} = 0^0$, which is indeterminate. But there are a few reasons I feel confident the value it takes here should be $1$. For one thing, I can see that this will give me the correct values for the first few associated Legendre polynomials, such as $P_0^0(x) = 1$. For another thing, I know that $P_l^m(x)$ for fixed $l$ and $m$ should be a continuous function in $x$, and clearly any small but non-zero $x$ to the $0$-th power is $1$.
But if we wanted to be rigorous, we could go back to the derivation of this closed form and see for ourselves that we could have instead produced a series over $k$ from $m + 1$ to $l$, and with a separate term in front (playing the same role as our $k = m$ term did previously). This would give us a closed-form solution like this:
$$P_l^m(x) = (-1)^{m} \cdot 2^{l} \cdot (1-x^2)^{m/2} \left[m! \binom{l}{m} \binom{\frac{l+m-1}{2}}{l} + \sum_{k=m+1}^l \frac{k!}{(k-m)!}\cdot x^{k-m} \cdot \binom{l}{k} \binom{\frac{l+k-1}{2}}{l} \right]$$
Now all terms of the sum are positive powers of $x$, so at $x = 0$ the whole summation goes away and we're left with this:
$$P_l^m(0) = (-1)^{m} 2^{l} m! \binom{l}{m} \binom{\frac{l+m-1}{2}}{l}$$
After replacing the binomial coefficients using the definition above and making some cancelations, we have:
$$P_l^m(0) = \frac{(-1)^{m} 2^{l}}{(l-m)!} \prod_{i=0}^{l-1} \left(\frac{l+m-1}{2} - i\right)$$
We can check a few values just to be sure. For example:
$$P_{3}^{1}(x)=-\tfrac{3}{2}(5x^{2}-1)(1-x^2)^{1/2} \\
P_{4}^{2}(x)=\frac{15}{2}(7x^2 - 1)(1-x^2)$$
so $P_{3}^{1}(0) = \frac{3}{2}$ and $P_{4}^{2}(0) = - \frac{15}{2}$
From the above formula, we have:
$$P_3^1(0) = \frac{(-1)^{1} 2^{3}}{(3-1)!} \prod_{i=0}^{3-1} \left(\frac{3+1-1}{2} - i\right) = \frac{-8}{2!} \prod_{i=0}^{2} \left(\frac{3}{2} - i\right) \\ = -4 \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{-1}{2} = \frac{3}{2}$$
and
$$P_4^2(0) = \frac{(-1)^{2} 2^{4}}{(4-2)!} \prod_{i=0}^{4-1} \left(\frac{4+2-1}{2} - i\right) = \frac{16}{2!} \prod_{i=0}^{3} \left(\frac{5}{2} - i\right) \\ = 8 \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \frac{-1}{2} = - \frac{15}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon? The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?
Solution:
Looking at the coefficients, we are immediately reminded of the binomial expansion of ${\left(x+1\right)}^{4}$.
Modifying this slightly, we can write the given equation as ${\left(z+i\right)}^{4}=1+i=2^{\frac{1}{2}}\cos \frac {\pi}{4} + 2^{\frac{1}{2}}i\sin \frac {\pi}{4}$
$\star$
We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: $z^{4}=2^{\frac{1}{2}}$
Because the roots of this equation are created by rotating $\frac{\pi}{2}$ radians successively about the origin, the quadrilateral is a square.
We know that half the diagonal length of the square is ${\left(2^{\frac{1}{2}}\right)}^{\frac{1}{4}}=2^{\frac{1}{8}}$
Therefore, the area of the square is $\frac{{\left( 2 \cdot 2^{\frac{1}{8}}\right)}^2}{2}=\frac{2^{\frac{9}{4}}}{2}=2^{\frac{5}{4}}$
After the $\star$ I become completely lost. "We can apply a translation of $-i$ and a rotation of $-\frac{\pi}{4}$ (both operations preserve area) to simplify the problem: $z^{4}=2^{\frac{1}{2}}$" Can you please show me exactly how this is achieved, perhaps visually? I am especially confused about what in the equation gets edited to obtain a $- \pi/ 4$ rotation.
| If we translate (preserves area) to the new variable $w=z+i$ we get $w^4=1+i$. Solutions are vertices of a square with side length $2^{5/8}$. Hence the area would be $2^{5/4}$.
| {
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Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}$, we have $xyz=1$ and using $abc=1$, the inequality can be written as:
$$\dfrac{x}{\sqrt{x+y}}+\dfrac{y}{\sqrt{y+z}}+\dfrac{z}{\sqrt{z+x}}\le \dfrac{xy+yz+zx}{\sqrt{2}}$$
I'm trying to use Cauchy-Schwarz:
$$LHS\le\sqrt{(x+y+z)(\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x})}$$ but now I have to prove $$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}\le\dfrac{3}{2}$$ because $ab+bc+ca\ge\sqrt{3(a+b+c)}$, but I can't prove it. Can anyone give me a hint? Not necessarily a complete solution.
By the way, I also relized a problem that seems quite similar to the above problem $\sqrt{\frac{2 x}{x+y}}+\sqrt{\frac{2 y}{y+z}}+\sqrt{\frac{2 z}{z+x}} \leq 3$ if $x,y,z>0$ (Vasile Cirtoaje) (and then we can use $3\le xy+yz+zx$ ?Hope it helps)
| Remark: As Calvin Lin pointed out, we can just deal with $a, b, c$, without the substitutions.
We have
\begin{align*}
&\sum_{\mathrm{cyc}}
\sqrt{\frac{ab}{bc^2 + 1}} \\
=\,& \sum_{\mathrm{cyc}}
\sqrt{\frac{ab ab}{(bc^2 + 1)ab}}\\
=\,& \sum_{\mathrm{cyc}}\frac{ab}{\sqrt{ab + bc}}\\
\le\,& \sqrt{(ab + bc + ca)\left(\frac{ab}{ab + bc} + \frac{bc}{bc + ca} + \frac{ca}{ca + ab}\right)} \tag{1}\\[5pt]
=\,&\sqrt{\frac{(ab + bc + ca)ab}{ab + bc} + \frac{(ab + bc + ca)bc}{bc + ca} + \frac{(ab + bc + ca)ca}{ca + ab}}\\[5pt]
=\,& \sqrt{ab + \frac{ca^2}{a + c} + bc + \frac{ab^2}{b + a} + ca + \frac{bc^2}{c + b}}\\[5pt]
\le\,& \sqrt{ab + \frac{\frac{(a + c)^2}{4}a}{a + c} + bc + \frac{\frac{(b + a)^2}{4}b}{b + a} + ca + \frac{\frac{(b + c)^2}{4}c}{c + b}}\tag{2}\\[5pt]
=\,&\sqrt{\frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)}
\end{align*}
where we have used the Cauchy-Bunyakovsky-Schwarz inequality in (1),
and $ca \le \frac{(c + a)^2}{4}$ etc. in (2).
It suffices to prove that
$$\frac{(a + b + c)^2}{2} \ge \frac{1}{4}(a^2 + b^2 + c^2) + \frac54(ab + bc + ca)$$
or
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
which is true.
We are done.
| {
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How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx$
Using the same method, I couldn't find my asked integral. Are there any ways to connect them? Any help would be appreciated.
| The similar problem was posted on AoPS the other day. There are nice answers posted. For the sake of completeness, I would like to add the solution, based on approach developed by Yaroslav Blagouchine; it is convenient for solving the problems with a specific symmetry by means of the integration along a rectangular contour in the complex plane.
Let
$$I=\int_0^\infty \frac{\ln{x}}{e^{x}+e^{-x}+1}dx=\int_0^\infty \frac{\ln{x}}{2\cosh x+1}dx$$
$$=\lim_{a\to0}\frac{1}{2}\int_0^\infty\frac{\ln(x^2+a^2)}{2\cosh x+1}dx=\frac{1}{4}\lim_{a\to0}\Re\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx$$
Let's consider
$$I(a)=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx=\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln(a-2\pi i t)}{\cosh 2\pi t+\frac{1}{2}}dt$$
$$=\frac{\pi}{2}\ln2\pi\int_{-\infty}^\infty\frac{dt}{\cosh 2\pi t+\frac{1}{2}}+\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t+\frac{1}{2}}dt=I_1+I_2(a)$$
Now, we move to the complex plane and consider the closed rectangular contour $-R\,\to R\,\to (R+i)\,\to (-R+i)\,\to -R;\,\,R\to \infty$; counter clockwise.
Let's also consider the following integral along this contour. The integrand has simple poles at the points $z=\frac{i}{3}$ and $\frac{2i}{3}$.
$$\frac{\pi}{2}\ln2\pi\oint\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}dz=I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi \,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}$$
$$I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi\frac{2\pi i}{\sqrt 3\pi i}\Big(e^{-\frac{\beta}{3}}-e^{-\frac{2\beta}{3}}\Big)$$
(We also have to add side integrals - along $R\to R+i$ and $-R+i\to-R$, but these integrals $\to 0$ at $R\to\infty$). Taking the limit $\beta\to 0$, we find
$$\boxed{\,\,I_1(0)=I_1=\frac{\pi}{3\sqrt 3}\ln2\pi\,\,}$$
To evaluate $I_2(a)$, we notice that
$$\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}=\frac{\ln\Gamma\big(\frac{a}{2\pi}-it+1\big)-\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
$$=\frac{\ln\Gamma\big(\frac{a}{2\pi}-i(t+i)\big)}{\cosh 2\pi (t+i)-\frac{1}{2}}\,-\,\frac{\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
(We used the fact that $\cosh (x+2\pi i)=\cosh x$). Adding two integrals (along $R\to R+i$ and $-R+i\to-R$ - these integrals $\to 0$ at $R\to\infty$), we can present $I_2(a)$ in the form of the integral along the same rectangular contour
$$I_2(a)=-\frac{\pi}{2}\oint\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}dz=-\frac{\pi}{2}\,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}=-\frac{\pi}{\sqrt 3}\Big(\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)-\ln\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)\Big)$$
$$\boxed{\,\,I_2(a)=\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)}{\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)}\,\,}$$
Coming back to our initial integral
$$I=I_1+\Re\,I_2(0)=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{2}{3}\big)}{\Gamma\big(\frac{1}{3}\big)}$$
Using the reflection formula for gamma-function $\Gamma\big(\frac{2}{3}\big)\Gamma\big(\frac{1}{3}\big)=\frac{\pi}{\sin\frac{\pi}{3}}=\frac{2\pi}{\sqrt3}$
$$\boxed{\,\,I=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{2\pi}{\sqrt3}-\frac{2\pi}{\sqrt 3}\ln\Gamma\Big(\frac{1}{3}\Big)=-0.126321...\,\,}$$
| {
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.
| Since you assume that $m > 0,$ this result of your calculations is good:
$$ AP = \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag1$$
Here's where you get in a bit of trouble:
$$ BP \stackrel?= \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag2$$
You want $AP - BP = 1,$ and I think the best interpretation of the problem statement interprets $AP - BP$ as the difference of two positive lengths (rather than a negative length subtracted from a positive length). Moreover, $AP$ must be the greater of the two lengths in order for the difference to be positive.
The problem with Equation $(2)$ is that if $c > 0$ then the expression on the right side of the equation is negative. A better equation is:
$$ BP = \left\lvert \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}\right\rvert.$$
A more useful correct equation is
$$ BP = \begin{cases}
\dfrac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c \geq 0, \\[1ex]
\dfrac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c < 0.
\end{cases} \tag3$$
The $c < 0$ case still looks shaky because of the (apparent) possibility that
$m^2 + 4c < 0,$ which would make the square root undefined, but what actually happens is that for very large negative $c$ the value of $m$ also will be large.
Equation $(1)$, on the other hand, is good because with $m > 0$ you are guaranteed that the expression on the right-hand side of the equation is positive,
and because the expression on the right-hand side is larger than either of the two expressions on the right-hand side of Equation $(3)$,
so you have chosen the correct expression for $AP$ in either case.
The two cases in Equation $(3)$ can (and I think should) be considered separately.
In the $c \geq 0$ case we have
\begin{align}
1 &= AP - BP \\
&= \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}
- \frac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} \\
&= m \sqrt{m^2 + 1}
\end{align}
and therefore
$$ m^4 + m^2 - 1 = 0, $$
for which the only solution (since $m^2$ must be positive) is
$$ m^2 = \frac12(\sqrt5 - 1). $$
In the $c < 0$ case, on the other hand, your further calculations are correct, and $m^2$ is the positive root $v$ of the quadratic equation
$$ v^2 + (4c + 1) v + 4c - 1 = 0, $$
that is,
\begin{align}
m^2 &= \frac{-(4c + 1) + \sqrt{(4c + 1)^2 - 4(4c - 1)}}{2} \\
&= \frac{-4c - 1 + \sqrt{(4c - 1)^2 + 4}}{2}.
\end{align}
You cannot eliminate $c$ from the solution in this case because the slope of the line actually does depend on how negative $c$ is.
With a $y$-intercept very far down the negative $y$ axis you need a steep slope in order to intersect the parabola.
My hunch is that you were supposed to solve the case $c \geq 0.$
This could have been stated explicitly, or it could have been implied by stating that $P$ is between $A$ and $B.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rewriting $\cos^4 x \sin^2 x $ with exponent no higher than $1$ I'm having some trouble finishing this one off.
Rewrite with exponent no higher than $1$:
$$\cos^4 x \sin^2 x$$
The answer is:
$$\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
So I started like this:
$$\cos^4 x \sin^2 x = \frac{1+\cos(2x)}{2}\frac{1+\cos(2x)}{2}\frac{1-\cos(2x)}{2}$$
$$= \frac{1}{8}\left(\{1+\cos(2x)\}\{1^2 - \cos^2(2x)\}\right)$$
$$\frac{1}{8}\left(\{1 + \cos(2x)\}\sin^2(2x)\right)$$
$$\frac{1}{16}\left(1 + \cos(2x)\{1-\cos(4x)\}\right)$$
Now this is where I start to get lost:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \cos(2x)\cos(4x) \right)$$
I really can't find a way from here - I try this, but not sure if this is the right path.
$$\require{cancel} \cancel{\frac{1}{16}\left(1 - \cos(4x) + \cos(2x)\{1 - \cos(4x)\} \right)}$$
Completing thanks to help below:
$$\frac{1}{16}\left(1 - \cos(4x) + \cos(2x) - \left(\frac{\cos(6x) + \cos(-2x)}{2}\right)\right)$$
$$\frac{1}{32}\left(2 - 2\cos(4x) + 2\cos(2x) - \cos(6x) - \cos(2x)\right)$$
$$=\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
| $\cos x = \frac {e^{ix} + e^{-ix}}{2}\\
\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\cos^4 x\sin^2 x = \frac {(e^{ix} + e^{-ix})^4(e^{ix} - e^{-ix})^2}{-64}\\
\frac {(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})(e^{2ix} - 2 + e^{-2ix})}{-64}\\
\frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix}+2e^{-4ix} + e^{-6ix}}{-64}\\
\frac {\cos 6x + 2\cos 4x - \cos 2x - 2}{-32}\\
\frac {-\cos 6x - 2\cos 4x + \cos 2x+2}{32}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand... like just adding up the fractions until I get to $\frac{7}{8}$ and got $n=7$. Surely there must be a better way. Help appreciated here! Thanks, I really appreciate it.
| There is a certain name for this type of sum...
First we can use partial fractions method:
$$
\frac{1}{k(k+1)} = \frac{A}k+\frac{B}{k+1} \quad \implies
$$
$$
1 = Ak+B(k-1) \quad \forall k
$$
So if we set $k=1$ we obtain that $A=1$.
If we set $k=0$ then $B = -1$ this means that
$$
\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}{k+1}
$$
Now the sum:
$$
\sum_{k=1}^s \frac{1}{k(k+1)} = \sum_{k=1}^s \frac{1}{k}-\frac{1}{k+1}
$$
But certain terms in this series cancel out...
For example if $s =3$ then
$$
\sum_{k=1}^3 \frac{1}{k}-\frac{1}{k+1} = \frac{1}{1}-\frac{1}{2} + \frac{1}{2}-\frac{1}{3}+ \frac{1}{3}-\frac{1}{4} = 1-\frac{1}{4}
$$
Do you see the pattern?
| {
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"timestamp": "2023-03-29T00:00:00",
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Height of an irregular tetrahedron with an equilateral base and lateral faces making angles $60^\circ$, $60^\circ$, $80^\circ$ with that base
An irregular tetrahedron has a base that is an equilateral triangle of side length $10$. The lateral faces make angles of $60^\circ, 60^\circ$ and $80^\circ$ with the base. Find the height of the tetrahedron.
So, one way I thought I could solve this problem is using coordinate geometry, specifically attaching a reference frame to the base, and writing the equations of the three planes that represent the three lateral faces, and then solving the linear system for the apex coordinates.
| Given a tetrahedron with base $ABC$ and apex $D$, let
*
*$E$ be the orthogonal projection of $D$ onto the plane holding $ABC$.
*$h = |DE|$ will be the height of tetrahedron.
*$\theta_A / \theta_B / \theta_C$ be the angle between faces $DBC$ / $DCA$ / $DAB$ and base $ABC$.
*$\ell_A / \ell_B / \ell_C$ be the distance of $E$ to edges $BC$ / $CA$ / $AB$.
As long as all $\theta_A, \theta_B, \theta_C < 90^\circ$, $E$ lies inside $ABC$.
Furthermore, we have
*
*$\ell_A = h \cot\theta_A$, $\ell_B = h \cot \theta_B$ and $\ell_C = h\cot\theta_C$
*$|BC|\ell_A + |CA|\ell_B + |BC|\ell_C = 2\verb/Area/(ABC)$
For the tetrahedron at hand, we have $|AB| = |BC| = |CA| = 10$ and $\verb/Area/(ABC
) = \frac{\sqrt{3}}{4}(10)^2$.
This leads to
$$\ell_A + \ell_B + \ell_C = 5\sqrt{3}$$
and as a result,
$$\begin{align}
h = \frac{\ell_A + \ell_B + \ell_C}{\cot\theta_A + \cot\theta_B + \cot\theta_C}
&= \frac{5\sqrt{3}}{2\cot(60^\circ) + \cot(80^\circ)} = \frac{15}{2 + \sqrt{3}\cot(80^\circ)}\\
&\sim 6.506442514261543\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Generating function for $a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$ Let $a_n$ be a sequence following the recurrence relation $$a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$$ with initial conditions $a_0 = a_1 = 1$.
We have to find the generating function for $a_n$ that does not contain an infinite series.
Let $f(x) = \sum_{k=0}^{n} a_k$. We know that $$\sum_{k=0}^{n-2} a_ka_{n-k-2}x^{n-2} = \left(\sum a_k x^k \right) \left(\sum a_{n-k-2} x^{n-k-2}\right)$$
After that, I am not able to proceed.
Can someone help?
Note that $a_2 = 1$ and $a_3 = 2$. Thus simplifying from the answer below:
$$x^2f^2(x) -f(x) + (1+x) = 0.$$
And solving for the quadratic function we have $$f(x) = \frac{1-\sqrt{1-4x^2(x+1)}}{2x^2}$$
| The expansion of $\sqrt{1+x}$ is $$\sum_{k=0}^{\infty}\binom{1/2}{k}x^k = 1 + \sum_{k=1}^\infty\frac{(1/2)(1/2 - 1) \cdots (1/2 - k + 1)}{k!}x^k = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots.$$ Then the expansion of $\sqrt{1 - 4x^2(x + 1)}$ is \begin{align*}1 + \frac{1}{2}(-4x^2(x+1)) - \frac{1}{8}(-4x^2(x+1))^2 + \frac{1}{16}(-4x^2(x+1))^3 - \frac{5}{128}(-4x^2(x+1))^4 + \cdots\end{align*} so $1 - \sqrt{1 - 4x^2(x + 1)}$ is \begin{align*}- \frac{1}{2}(-4x^2(x+1)) + \frac{1}{8}(-4x^2(x+1))^2 - \frac{1}{16}(-4x^2(x+1))^3 + \frac{5}{128}(-4x^2(x+1))^4 + \cdots\end{align*} and $\frac{1 - \sqrt{1 - 4x^2(x + 1)}}{2x^2}$ is \begin{align*}- \frac{1}{4x^2}(-4x^2(x+1)) + \frac{1}{16x^2}(-4x^2(x+1))^2 - \frac{1}{32x^2}(-4x^2(x+1))^3 + \frac{5}{256x^2}(-4x^2(x+1))^4 + \cdots\end{align*} which simplifies to $$1 + x + x^2(x+1)^2 + 2x^4(x+1)^3 + 5x^6(x+1)^4 +\cdots = 1 + x + x^2 + 2x^3 + 3x^4 + \cdots$$ and the coefficients are the terms you are looking for.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Has the sum of 4 cubes problem been proven? Today in class, my professor was lecturing on the sum of 3 cubes and sum of 4 cubes problems. Namely, can every number be written as the sum of 3 (or 4) cubes? He discussed their origins and showed a few examples, and showed how difficult they could be to find for certain numbers (such as 33 or 42 for the sum of 3 cubes). He said we would not cover their proofs in the course because they were "beyond the scope of the course." When I went to look them up, however, it seems as though they are open problems and have not been proved. I don't think my professor would get this wrong, so I'm a bit confused. I would appreciate any clarification. If they have been proven, where can I see the proofs?
| First:You can use this identity for finding numbers which are the sum of three cubes:
$$(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-z)(z-x)$$
For example:
$(3-5)^3+(5-7)^3+(7-3)^3=3(3-5)(5-7)(7-3)= 48$
Second : we solve this problem to find a number which it's cube is the sum of three cubes:
$$x^3+y^3+z^3=u^3$$
Let $u=-t$ we have:
$$x^3+y^3+z^3+t^3=0\space\space\space\space(1)$$
This equation has infinitely many solutions(positive or negative), as you will see they make a set of particular numbers which means not every cube can be written as the sum of three cubes.
Suppose $a, b, c, d , \alpha, \beta, \gamma, \delta $ are two groups of four numbers that satisfy equation (1) . Choose $k$ such that numbers $a+k\alpha, b+k\beta, c+k\gamma, d+k\delta$ also satisfy equation (1), or we can have:
$$(a+k\alpha)^3+(b+k\beta)^3+(c+k\gamma)^3+(d+k\delta)^3=0$$
we expand each term; considering groups (a, b, c , d ) and $(\alpha, \beta, \gamma, \delta)$ both satisfy the equation i.e.:
$a^3+b^3+c^3+d^3=0$
$\alpha^3+ \beta^3+ \gamma^3+ \delta^3=0$
We have:
$3a^2k\alpha+3ak^2\alpha^2+3b^2k\beta+3bk^2\beta^2+3c^2k\gamma+\3ck^2\gamma^2+3d^2k\delta+3dk^2\delta^2=0$
Or:
$3k[(a^2\alpha+b^2\beta +c^2\gamma+d^2\delta)+k(a\alpha^2+b\beta^2+c\gamma^2+d\delta^2)]=0$
This relation can be zero if one of it's factors is zero. Equating each factor to zero gives two values for k; one is $k=0$(which is not of our interest because it means we do not add anything to numbers a, b, c and d), second is:
$k=-\frac{a^2\alpha +b^2\beta+c^2\gamma+d^2\delta}{a\alpha^2+b\beta^2+c\gamma^2+d\delta^2}\space\space\space (2)$
If we have two groups of solutions we can find a new group of four numbers as the solution to equation (1).For this we k times of numbers of first group to numbers of second group , provided k is found by relation (2). To do this we need to have a group of solutions, say $(x, y, z, t)=(3, 4, 5, -6). To find second group let:
$\alpha, \beta, \gamma, \delta)=(r, -r, s, -s)$
clearly these numbers satisfy equation (1). Putting these values in (2) we get:
$$k=\frac{7r+11s}{7r^2-s^2}$$
So we have:
$a+k\alpha=\frac{28r^2+11rs-3s^2}{7r^2-s^2}$
$b+k\beta=\frac{21r^2-11rs-4s^2}{7r^2-s^2}$
$c+k\gamma=\frac{35r^2+7rs+6s^2}{7r^2-s^2}$
$d+k\delta=\frac{-42r^2-7rs-5s^2}{7r^2}$
In this way general form of solutions, considerin all numerators are equal, can be:
$x=28r^2+11rs-3s^2$
$y=21r^2-11rs-4s^2$
$z=35r^2+7rs+6s^2$
$t=-42r^2-7rs-5s^2$
For example take $r=s=1$ you get:
$(x, y, z, t)=(1, 6, 8, 9)$
$1^3+6^3+8^3=9^3$
| {
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"url": "https://math.stackexchange.com/questions/4420427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$, so $n^2-4 = k(2n+1)$, for some $k \in \Bbb Z$. But, I did not be able to find $n$ from here.
Any ideas? Thanks in advanced.
| Write $k=2n+1$ then $n=(k-1)/2$ so $$3n^2+4n+5= {3(k^2-2k+1) + 8(k-1)+20\over 4} ={3k^2+2k+15\over 4}$$ and thus $$4k\mid 3k^2+2k+15\implies k\mid 15$$
Now you have only few values of $k$...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate the definite integral $\int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$? I am struggling with this integral:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$
What I tried so far:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$
$\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{\cos^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
$\displaystyle =\int _{0}^{\frac{\pi }{2}}\frac{-\sin^{2} x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
$\displaystyle =\frac{1}{2}\int _{0}^{\frac{\pi }{2}}\frac{\cos 2x\ln(\tan x)}{1-\sin x\cos x}\mathrm{d} x$
The answer should come out to be $\dfrac{-7\pi^2}{72}$.
Any help will be appreciated.
| Since
$$
\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x=\int_0^{\infty} \frac{\ln t}{(1-t+t^2)(1+t^2)}dt,
$$
consider
$$
\begin{aligned}
\mathscr{I}(s)&=\int_0^{\infty} \frac{t^s}{(1-t+t^2)(1+t^2)}dt
\\&=\int_0^{\infty} \frac{t^{s-1}}{1-t+t^2}dt-\int_0^{\infty} \frac{t^{s-1}}{1+t^2}dt
\\&=\int_0^{\infty} \frac{t^{s-1}}{1+t^3}dt+\int_0^{\infty} \frac{t^{s}}{1+t^3}dt-\int_0^{\infty} \frac{t^{s-1}}{1+t^2}dt.
\end{aligned}
$$
With Beta function, we have
$$
\int_{0}^{\infty}\frac{t^{s-1}}{1+t^{a}}dt=\frac{\pi \csc(\frac{\pi s}{a})}{a}
$$
thus
$$
\mathscr{I}(s)=\frac{\pi \csc(\frac{\pi s}{3})}{3}+\frac{\pi \csc(\frac{1}{3} \pi (s+1))}{3}-\frac{\pi \csc(\frac{\pi s}{2})}{2}.
$$
In conclusion,
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}}\frac{\ln \tan x}{1-\tan x+\tan^2 x}dx&=\lim_{s\to 0}\frac{\partial }{\partial s}\mathscr{I}(s)
\\&=\lim_{s\to 0}\left(-\frac{\pi^{2} \csc(\frac{\pi s}{3}) \cot(\frac{\pi s}{3})}{9}-\frac{\pi^{2} \csc(\frac{1}{3} \pi s+\frac{1}{3} \pi) \cot(\frac{1}{3} \pi s+\frac{1}{3} \pi)}{9}+\frac{\pi^{2} \csc(\frac{\pi s}{2}) \cot(\frac{\pi s}{2})}{4}\right)
\\&=-\frac{7 \pi^{2}}{72}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A pattern of periodic continued fraction I am interested in the continued fractions which $1$s are consecutive appears.
For example, it is the following values.
$$
\sqrt{7} = [2;\overline{1,1,1,4}] \\
\sqrt{13} = [3;\overline{1,1,1,1,6}]
$$
In this article, let us denote n consecutive $1$s as $1_n$.
Applying this, the above numbers would be as follows.
$$
\sqrt{7} = [2;\overline{1_3,4}] \\
\sqrt{13} = [3;\overline{1_4,6}]
$$
While investigating these numbers, the following pattern was found experimentally.
$$ \sqrt{F(n)^2m^2-(F(n)^2-L(n))m+\frac{(F(n)-1)(F(n)-3)}{4}-\frac{F(n-3)-1}{2}}
=\left[F(n)m-\frac{F(n)-1}{2};\ \overline{1_{n-1},\ 2\left(F(n)m-\frac{F(n)-1}{2}\right)}\right] $$ ($m,n \in \mathbb{N},\ n\equiv\pm1\ (mod3),\ n > 3,\ $$F(n)$ is Fibonacci number, $L(n)$ is Lucas number)
I confirm that it works correctly when $n$ and $m$ are single digits.
If you find a proof or counterexample, please let me know.
(2022/04/13 edit)
A general expression was derived. I think the expression I found is that special case. The condition is that the inside of the square root is always an integer.
Here are some concrete examples.
$$\begin{array}{|c|c|}
\hline
n & pattern \\ \hline
4 & \sqrt{9m^2-2m} = [3m-1;\overline{1_3,2(3m-1)}] \\ \hline
5 & \sqrt{25m^2-14m+2} = [5m-2;\overline{1_4,2(5m-2)}] \\ \hline
7 & \sqrt{169m^2-140m+29} = [13m-6;\overline{1_6,2(13m-6)}] \\ \hline
8 & \sqrt{441m^2-394m+88} = [21m-10;\overline{1_7,2(21m-10)}] \\ \hline
\end{array}$$
| I think your claims are correct but quite needlessly complicated. The theorem at the end of this answer shows a result which is both simpler to write out and more general.
Let $(F_n)_{n\geq 0}$ be the standard Fibonacci sequence, defined by $F_0=0,F_1=1$ and $F_{n+2}=F_n+F_{n+1}$ for $n\geq 1$.
Let $f(x)=\frac{1}{1+x}$, and $f^n=f\circ f \circ \ldots \circ f$ ($n$ times). It is easy to check by induction that
$$
f^n(x)=\frac{\big(F_{n+1}-F_n\big)+\big(2F_n-F_{n+1}\big)x}{F_n+(F_{n+1}-F_n)x} \tag{1}
$$
Now let $a\geq 1$ be an integer. If we put $g(x)=f(\frac{1}{a+x})$,
$$
g(x)=\frac{\big(F_{n+1}-F_n\big)(x+a)+\big(2F_n-F_{n+1}\big)}{F_n(x+a)+F_{n+1}-F_n} \tag{2}
$$
The roots of $g(x)=x$ are therefore defined by the equation
$$
x^2+ax-\bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg)=0 \tag{3}
$$
This is quadratic whose roots are $-\frac{a}{2}\pm \sqrt{\Delta}$ where
$$
\Delta = \frac{a^2}{4} + \bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg) \tag{4}
$$
For $n\geq 3$, we have $F_{n+1}=\frac{3}{2}F_{n}+\frac{1}{2}F_{n-3}$ and hence $F_{n+1}\geq \frac{3}{2}F_{n}$. It follows from (4) that $\Delta \geq \frac{a^2}{4} + 4\bigg(\frac{3}{2}(a-1)+2-a\bigg) \gt \frac{a^2}{4}$, so that the largest root $\alpha$ of $g(x)=x$ is positive. Thus :
Theorem. For any $n\geq 3$ and $a\geq 2$, there is a unique positive number whose continued fraction is $[\overline{1_{n},\ a}]$. This number is
$$
\alpha = -\frac{a}{2} + \sqrt{\frac{a^2}{4} + \bigg(\frac{F_{n+1}}{F_n}(a-1)+2-a\bigg)} \tag{5}
$$
Update. When $a$ is of the form $a=F_n(2m+1)+1$, it is straightforward to compute that
$$
\begin{array}{lcl}
\Delta &=& \frac{4F_n^3m^2 + (8F_nF_{n+1} + (4F_{n}^3 - 4F_n^2))m + (4F_nF_{n+1} + (F_n^3 - 2F_n^2 + 5F_n)}{4F_n} \\
&=& F_n^2 m^2 + (2F_{n+1} + F_{n}^2 - 1)m + F_{n+1} + \frac{F_n^2-2F_n+5}{4}
\end{array}
$$
So that $\Delta$ is an integer iff $F_n^2-2F_n+5$ is divisible by $4$. This is easily seen to be the case when $n\not\equiv 2$ modulo $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$, where $x \in \mathbb R$ and $a \neq 0$. Throughout this derivation, I will use the definition: $\sqrt{x^2}=|x|$.
Here is my derivation, and I have placed a $\color{red}{\dagger}$ next to the part that I would like some clarificaiton about:
$ax^2+bx+c =0 \iff x^2+\frac{bx}{a}+\frac{c}{a}=0 \iff (x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})=0 \iff (x+\frac{b}{2a})^2+(\frac{4ac-b^2}{4a^2})=0 $
Bringing the right summand over to the right side of the equation:
$(x+\frac{b}{2a})^2=(\frac{b^2-4ac}{4a^2}) \iff \left| x+\frac{b}{2a}\right|=\sqrt{b^2-4ac}\cdot\sqrt{\frac{1}{4a^2}} \iff \left| x+\frac{b}{2a}\right|=\left|\frac{1}{2a} \right| \cdot \sqrt{b^2-4ac} \quad \quad \color{red}{\dagger}$
My confusion stems from how the final expression above is equivalent to the syntax "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" .
For $\color{red}{\dagger}$, we have 4 total cases:
*
*$ a \lt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \lt 0$ and $x+\frac{b}{a} \geq 0$
*$ a \gt 0$ and $x+\frac{b}{a} \lt 0$
*$ a \gt 0$ and $x+\frac{b}{a} \geq 0$
Case 1: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
Case 2: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 3: the conditions imply that $x=\frac{-b-\sqrt{b^2-4ac}}{2a}$
Case 4: the conditions imply that $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$
From the four scenarios, is the way we get to "$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$" simply by noting that for a fixed $a$, we have $x=\frac{-b+\sqrt{b^2-4ac}}{2a} \text{ or } x=\frac{-b-\sqrt{b^2-4ac}}{2a}$? ...where the '$\text{ or }$' here is denoting the logical or.
At this point, we define "$x=\pm \alpha$" as meaning $x = \alpha \text { or } x=-\alpha$...therefore meaning that $x=\beta\pm \alpha$ is equivalent to $x=\beta+\alpha \text{ or } x=\beta - \alpha$.
Is that the proper understanding?
| An appealing way of understanding why $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be written as $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ is to take a set perspective.
The statement $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be thought of as saying:
$x \in S$ where $S:=\left\{x\in \mathbb R:\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \right\}$
The statement $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ can be though of as saying:
$ x \in T$ where $T:=\left\{x\in \mathbb R:x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}\right\}$
The objective is to show that $x \in S \rightarrow x \in T$ and $x \in T \rightarrow x \in S$.
There is a lemma that follows directly from the definition of $|\cdot|$, which reads as:
$|x|=C \iff x=C \text{ OR } x= -C \quad (*_1)$
To prove the $\rightarrow$ direction, simply exhaust all possible values of $x \in \mathbb R$ by splitting it into the two cases of $x \geq 0$ and $x \lt 0$. The $\leftarrow$ direction is trivial.
Prove:$\quad x \in S \rightarrow x \in T$
By assumption, we have that $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} $. Taking the square root of both sides, and applying the definition of $\sqrt{\cdot}$, we have:
$$\left |x + \frac{b}{2a} \right|=\sqrt{\frac{b^2 - 4ac}{4a^2}}$$
Applying our lemma $(*_1)$, we have the logical statement:
$$x+\frac{b}{2a}=\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad\text { OR }\quad x+\frac{b}{2a}=-\sqrt{\frac{b^2 - 4ac}{4a^2}} \quad (*_2)$$
The symbol $\pm$ is defined to capture the meaning of $(*_2)$ and is equivalently written as:
$x+\frac{b}{2a}=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}$
Subtraction gives us: $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$, which means that $x \in T$.
Prove:$\quad x \in T \rightarrow x \in S$
By assumption, we have that $x=(\pm)\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$. This means that $x= \sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}\text { OR } x=-\sqrt{\frac{b^2 - 4ac}{4a^2}}-\frac{b}{2a}$.
In the first case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$.
In the second case, we have that $\left(x+\frac{b}{2a}\right)^2=\left(-\sqrt{\frac{b^2 - 4ac}{4a^2}} \right)^2=\frac{b^2 - 4ac}{4a^2}$, which means that $x \in S$. Therefore, in all cases, we have that $x \in T$.
In conclusion, we have: $x \in S \iff x \in T$, which means that the two statements:
(1) $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$
(2) $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$
are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration of a piecewise defined discontinuous function In a proof i posted recently on this site (link) i made the mistake of thinking that a function $f$, which is bounded on a closed interval $[a,b]$, would assume its infimum at a certain $x$ in the domain of $f$. I was presented with a counter example which, for my ability, seemed rather complicated. So i tried to find an easier counter example. I found the following function:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
Here, $ inf \{ f(x) : 0 \le x \le 1 \} = 0 $ but $f(x) \neq 0$ for all $x$.
I guess this is suitable as a counter example. Then i came up with the idea of trying to integrate this function. So came to the following proposition:
Proposition: Let $f$ be a function defined on $[a,b]$ as follows:
$$ f(x) = \begin{cases}
x & x > 0 \\
1 & x = 0
\end{cases} $$
This function is integrable with:
$$ \int_0^b = \frac{b^2}{2}$$
Proof: I use a partition $ P = \{t_0, ... , t_n \} $ of $[a,b]$ with
$$ t_i - t_{i-1} = \frac{b}{n} $$
$$ t_i = \frac{ib}{n} $$
$$ t_{i-1} = \frac{(i-1)b}{n} $$
$$ m_i = inf \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{(i-1)b}{n} $$
$$ m_1 = inf \{ f(x) : t_0 \le x \le t_1 \} = 0 $$
$$ M_i = sup \{ f(x) : t_{i-1} \le x \le t_i , i \neq 1 \} = \frac{ib}{n} $$
$$ M_1 = sup \{ f(x) : t_0 \le x \le t_1 \} = 1 $$
Then we have
$$ L(f, P) = \sum_{i=2}^n m_i \cdot \frac{b}{n} + m_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{(i-1)b}{n} \cdot \frac{b}{n} + 0 \cdot \frac{b}{n} = \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j = \frac{b^2}{2} \cdot \frac{n-1}{n}$$
and
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{ib}{n} \cdot \frac{b}{n} + 1 \cdot \frac{b}{n} = \Biggl[ \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j+1 \Biggl] + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
For the difference of the upper and lower sums this results in
$$ U(f, P) - L(f, P) = \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} = \frac{b^2}{2} \cdot \frac{n+1}{n} - \frac{b^2}{2} \cdot \frac{n-1}{n} + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{2}{n} + \frac{b}{n} = \frac{b^2+b}{n}$$
So in order get $ U(f, P) - L(f, P) < \epsilon$ we can choose $ n > \frac{b^2+b}{\epsilon} $. Thus $f$ is integrable and since
$$ \frac{b^2}{2} \cdot \frac{n-1}{n} \le \frac{b^2}{2} \le \frac{b^2}{2} \cdot \frac{n+1}{n} + \frac{b}{n}$$
and the integral is unique, if it exists, we have
$$ \int_0^b = \frac{b^2}{2}$$
as required. $ \blacksquare $
All these equations and manipulations are quite complex regarding my ability. So might anyone tell me if this is correct or point me towards my mistakes? Thanks in advance.
| In retrospection i think i found a mistake in my proof. I used $\sum_{j=1}^{n-1} j+1 = \frac{n(n+1)}{2}$ which is false, since $\sum_{j=1}^{n-1} j+1 = \frac{n^2 + n -2}{2}$.
With this discovery and the hint of @Andrea S. i found
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \frac{ib}{n} \cdot \frac{b}{n} + 1 \cdot \frac{b}{n} = \Biggl[ \frac{b^2}{n^2} \cdot \sum_{j=1}^{n-1} j+1 \Biggl] + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{(n-1)(n+2)}{n^2} + \frac{b}{n} = \frac{b^2}{2} \cdot \frac{(n-1)}{n} \cdot \frac{n+2}{n} + \frac{b}{n}$$
and
$$ L(f, P) = \frac{b^2}{2} \cdot \frac{n-1}{n}$$
Now, since $\lim_{n\to \infty} \frac{(n-1)}{n} = \lim_{n\to \infty} \frac{n+2}{n} = 1$ and $\lim_{n\to \infty} \frac{b}{n} = 0$ we have
$$\lim_{n\to \infty}L(f,P)=\lim_{n\to \infty}U(f,P_n)=\frac{b^2}{2}$$
as required. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I find the eigenvalues of this $2n \times 2n$ matrix? The matrix I am dealing with is of the form below.
$$\begin{pmatrix} 1-\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}}\\ -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& \cdots & -\frac{1}{\sqrt{2^n}} \\ -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}}& \cdots & -\frac{1}{\sqrt{2^n}}\\-\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}}\\ \vdots & \vdots& \vdots& \vdots & \ddots & \vdots\\ -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}} \end{pmatrix}$$
It's a $2n \times 2n$ matrix whose $(1,1)$-th entry is $1 - \frac{1}{\sqrt{2^n}}$ and all the others $- \frac{1}{\sqrt{2^n}}$. How can I solve this eigenvalue problem?
| Here's a similar approach to what KBS suggests. Begin with the observation that we can write $M = e_1e_1^T - \alpha 11^T$ for some $\alpha \in \Bbb R$. From there, we can write $M = AB$, with
$$
A = \pmatrix{e_1 & \alpha \mathbf 1}, \quad B = \pmatrix{e_1 & - \mathbf 1}^T.
$$
From the fact that $AB$ and $BA$ have the same non-zero eigenvalues, conclude that every eigenvalue of $M$ is either equal to $0$ or is an eigenvalue of
$$
BA = \pmatrix{e_1 & - \mathbf 1}^T\pmatrix{e_1 & \alpha \mathbf 1} = \pmatrix{1 & \alpha\\ -1 & -\alpha n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $(a^2 - b^2)^2$ $ \ge $ $4ab(a-$ $b)^2$ An inequality problem from Beckenbach and Bellman:
Show that $(a^2 - b^2)^2 \ge 4ab(a-b)^2$
The given answer is simply
Equivalent to $(a - b)^4 \ge 0$
I have tried two approaches, one which agrees with the given answer, and the other which does not.
Approach one. (Agrees with answer)
\begin{align}
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\
(a^2 - b^2)^2 - 4ab(a-b)^2 & \ge 0\\
((a+b)(a-b))^2 - 4ab(a-b)^2 & \ge 0\\
(a+b)^2(a-b)^2 - 4ab(a-b)^2 & \ge 0\\
(a-b)^2((a+b)^2 - 4ab) & \ge 0 \\
(a-b)^2 (a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 (a-b)^2 & \ge 0\\
(a - b)^4 & \ge 0
\end{align}
Approach Two
\begin{align}
(a^2 - b^2)^2 & \ge 4ab(a-b)^2\\
((a+b)(a-b))^2 & \ge 4ab(a-b)^2\\
(a+b)^2(a-b)^2 & \ge 4ab(a-b)^2\\
(a+b)^2 & \ge 4ab\\
(a^2 -2ab + b^2) &\ge 0 \\
(a-b)^2 & \ge 0
\end{align}
Could someone point out where the second approach is going wrong?
| Note that
$$
a^2 - b^2 = (a + b) (a - b)
$$
Thus,
$$
(a^2 - b^2)^2 = (a + b)^2 (a - b)^2
$$
Thus, the given inequality
$$
(a^2 - b^2)^2 \geq 4 a b (a - b)^2
$$
is equivalent to
$$
(a + b)^2 (a - b)^2 \geq 4 a b (a - b)^2
$$
or
$$
(a - b)^2 \left[ (a + b)^2 - 4 a b \right] \geq 0
$$
or
$$
(a - b)^2 \left[ a^2 + b^2 - 2 a b \right] \geq 0
$$
or
$$
(a - b)^2 (a - b)^2 \geq 0
$$
or
$$
(a - b)^4 \geq 0
$$
which is always true.
Hence, we proved the given inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4430076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving that $f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$ is strictly positive for odd integers $n\geq 5$
Let $n\geq 5$ be an odd integer. I'd like to prove that the function $$f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$$ is strictly positive on $\mathbb{R}$.
(I already figured out that the numerator is divisible by the denominator, so $f_n(x)$ is in fact a polynomial of degree $n-3$.
I tried to find the minimum by differentiation, but I have difficulty finding the location and the value of the minimum. Another thing I know is that $f_n(x)$ is palindromic, but it doesn't seem to give any inspirations. Does anyone have ideas?
Thanks in advance!
| First, we have
$$f_5(x) = 5(x^2 + x + 1) > 0, \, \forall x \in \mathbb{R}.$$
Second, for $k\ge 2$, let
\begin{align*}
g_k(x) &:= f_{2k+3}(x) - f_{2k+1}(x)\\[5pt]
&= \frac{(x + 1)^{2k + 3} - (x + 1)^{2k + 1} - x^{2k + 3} + x^{2k + 1}}{x(x + 1)}\\
&= \frac{(x + 1)^{2k}[(x + 1)^3 - (x + 1)] - x^{2k}(x^3 - x)}{x(x + 1)}\\
&= \frac{(x + 1)^{2k}x(x + 1)(x + 2) - x^{2k}x(x - 1)(x + 1)}{x(x + 1)}\\
&= (x + 1)^{2k}(x + 2) - x^{2k}(x - 1).
\end{align*}
We can prove that $g_k(x) \ge 0$ for all $x\in \mathbb{R}$.
If $-2 < x < 1$, clearly $g_k(x) \ge 0$.
If $x \le -2$, we have $g_k(x) = (1 - x)(-x)^{2k} - (-x - 2)(-x - 1)^{2k} \ge 0$ since $1 - x > - x - 2 \ge 0$
and $- x > - x - 1 \ge 1$.
If $x \ge 1$, we have $g_k(x) \ge 0$
since $x + 1 > x \ge 1$
and $x + 2 > x - 1 \ge 0$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
I have checked on google for $ \sum_{n=1}^\infty \frac{1}{n^2}$ and i do not suppose to know that is equal to $\frac{\pi^2}{6}$.
and I checked how The sum can be given explicitly of $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$
and the result was show that needs the Fourier series and we haven't learned that yet.
So, I have no idea how to prove that equal to $\frac{3}{4}S$ without know the exactly the sum
Thanks.
| Hint: Note that you get to this series by subtracting the even $n$ terms from the original.
Further, note that for any $n$,
$$
\frac{1}{(2n)^2}=\frac{1}{4}\cdot\frac{1}{n^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{1\le iLet $n \ge 2$ be a an integer and $x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$
Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}$$
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\sum_{1\le i<j\le 1}\frac{x_ix_j}{(1-x_i)(1-x_j)}=\frac{1}{2}\left(\sum_{k=1}^n\frac{x_k}{1-x_k}\right)^2-\frac{1}{2}\sum_{k=1}^n\frac{x_k^2}{(1-x_k)^2}$$
This formula was derived in @GCab's Answer.
First let $a_k=x_k/(1-x_k)$
so we want to prove
$$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le \frac{n(n-1)}{(2n-1)^2}$$
But since $$\frac{x_k}{1-x_k}<2x_k\implies \sum_{k=1}^na_k<1 \quad (1)$$
Hence $$\left(\sum_{k=1}^na_k\right)^2\le \sum_{k=1}^na_k$$
Meaning $$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le\sum_{k=1}^na_k(1-a_k)$$
Now consider the following function $$f(x)=\frac{x}{1-x}\left(1-\frac{x}{1-x}\right)$$
$f$ is concave on $(0,1)$ and by the tangent line trick we have $$f(x)\le f'(a)(x-a)+f(a)$$
set $a=1/2n$ to get $$a_k(1-a_k)\le\frac{4n^2\left(2n-3\right)}{\left(2n-1\right)^3}\left(x_k-\frac{1}{2n}\right)+ \frac{2(n-1)}{(2n-1)^2}$$
Now we sum to finish $$\sum_{k=1}^na_k(1-a_k)\le \frac{2n(n-1)}{(2n-1)^2}$$
Maybe by tweaking $(1)$ a little bit we can get rid of this factor of $2$
| Hint:
Indicating as $S_n , \, T_n$
$$
\begin{array}{l}
T_n = \sum\limits_{1 \le i < j \le n} {a_i a_j } \\
S_n = \sum\limits_{1 \le i,j \le n} {a_i a_j }
= \sum\limits_{1 \le i \le n} {\sum\limits_{1 \le j \le n} {a_i a_j } }
= \sum\limits_{1 \le i \le n} {a_i } \sum\limits_{1 \le j \le n} {a_j }
= \left( {\sum\limits_{1 \le i \le n} {a_i } } \right)^2 \\
\end{array}
$$
then you also have that
$$
\begin{array}{l}
S_n = \sum\limits_{1 \le i,j \le n} {a_i a_j }
= \sum\limits_{1 \le i < j \le n} {a_i a_j } + \sum\limits_{1 \le i = j \le n} {a_i a_j }
+ \sum\limits_{1 \le j < i \le n} {a_i a_j } = \\
= 2T_n + \sum\limits_{1 \le i \le n} {a_i ^2 } \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Geodesics on a pseudosphere I am trying to show the following:
Let
\begin{equation}
\gamma(t) = \begin{pmatrix}
\frac{1}{t} \\
\sqrt{1 - \frac{1}{t^2}} + \cosh^{-1}(t)
\end{pmatrix}
\end{equation}
for $t \geq 1$ be the unit-speed parametrization of the tractrix and
\begin{equation}
\sigma(u,v) = \begin{pmatrix}
\frac{1}{u} \cos v \\
\frac{1}{u} \sin v \\
\sqrt{1 - \frac{1}{u^2}} + \cosh^{-1}(u)
\end{pmatrix}
\end{equation}
the surface of revolution obtained by rotating the tractrix around the z-axis. Now I want to show that the geodesics on the pseudosphere (tractroid in this case) are represented in the $(u,v)$-coordinates by arcs of circles centered on the $v$-axis.
The first thing that I noted is that the curve is not unit-speed parametrized.
So I calculated the coefficients of the first fundamental form (for a surface patches)
\begin{align*}
E(u,v) &= \dot{f}^2(u) + \dot{g}^2(u) = \frac{3 + u^2}{u^2(u^2 - 1)}, \\
F(u,v) &= \langle \sigma_u(u,v), \sigma_v(u,v) \rangle = 0, \\
G(u,v) &= \frac{1}{u^2} = f^2(u)
\end{align*}
and now the geodesic equations are
\begin{align*}
\frac{d}{dt}\left( E \dot{u} + F \dot{v} \right) &= \frac{1}{2}\left( E_u \dot{u}^2 + 2 F_u \dot{u} \dot{v} + G_u \dot{v}^2 \right), \\
\frac{d}{dt}\left( F \dot{u} + G \dot{v} \right) &= \frac{1}{2}\left( E_v \dot{u}^2 + 2 F_v \dot{u} \dot{v} + G_v \dot{v}^2 \right)
\end{align*}
therefore
\begin{align*}
\frac{d}{dt} \left(\frac{3 + u^2}{u^2(u^2 - 1)} \dot{u} \right) &= -\frac{4 u \dot{u}^2 }{(u^2 - 1)^2} - \frac{\dot{v}^2 - 3 \dot{u}^2}{u^3}, \\
\frac{d}{dt}\left( \frac{1}{u^2}\dot{v} \right) &= 0
\end{align*}
solving the second gives
\begin{equation*}
\dot{v} = C u^2
\end{equation*}
which can also be obtained by clairauts theorem
where $\psi$ is the angle between $\dot{\gamma}$ and the meridians of the surface
\begin{equation*}
\dot{v} = \frac{\sin \psi}{f} = u \sin \psi
\end{equation*}
and $f \sin \psi$ being constant gives
\begin{equation*}
f^2 \dot{v} = \frac{1}{u^2} \dot{v} = f \sin \psi \stackrel{!}{=} \mbox{const}.
\end{equation*}
Plugging this $\dot{v}$ into the first geodesic equation leaves us with
\begin{equation*}
\frac{d}{dt} \left(\frac{3 + u^2}{u^2(u^2 - 1)} \dot{u} \right) = \frac{\dot{u}\left(3 - \frac{4 u^4}{(u^2 - 1)^2} \right)}{u^3} - C^2 u.
\end{equation*}
From this I can't figure out how the geodesics are arcs of circles centered on the $v$-axis. So I tried using the fact that
\begin{equation}
I(\dot{\gamma},\dot{\gamma}) \stackrel{!}{=} 1
\end{equation}
which gives
\begin{equation}
E \dot{u}^2 + G \dot{v}^2 = 1
\end{equation}
or
\begin{equation}
\dot{u}^2 \frac{3 + u^2}{u^2 - 1} + C^2 u^4 = u^2
\end{equation}
thus
\begin{equation}
\dot{u} = \pm \frac{\sqrt{u^2(u^2C^2 - 1)}}{\sqrt{ \frac{3 + u^2}{1 - u^2} }}.
\end{equation}
From this I would get an expression for $\frac{\dot{v}}{\dot{u}}$ hence by separation of variables $(v - v_0) = (\ldots)$ where the right hand side involves an elliptic integral.
I would expect that from $\frac{\dot{v}}{\dot{u}} = F(u)$ where $F$ is some function depending only on $u$, I can use separation of variables to get something like
\begin{equation}
(v-v_0)^2 + u^2 = \frac{1}{C}
\end{equation}
but from the geodesic equation and the ODE it is not possible to obtain such a form. If I would switch the sign from $\cosh^{-1}(t)$ to $-\cosh^{-1}(t)$ this would work.
I would be thankful if anybody can give me a hint.
| Comment
If $ (f, v) $ are polar coordinates it is in hyperbolic geodesic representation that have circles passing through origin and centered on the x-axis... or any radial line but not the lines on 2-D surface in 3-space.
| {
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Computing limits using Taylor expansions and $o$ notation on both sides of a fraction Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
I will use a concrete example to demonstrate the confusion I have, but the question is probably more generally applicable. By using Taylor expansions, compute $\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2}$.
\begin{equation}
\cos x^2 = 1 - \frac{x^4}{2} + o(x^5) \\
\sin x^2 = x^2 + o(x^4)
\end{equation}
$$
\lim_{x \to 0} \frac{1 - \cos x^2}{x^2 \sin x^2} = \lim_{x \to 0} \frac{1 - (1 - \frac{x^4}{2} + o(x^5))}{x^2 (x^2 + o(x^4))} = \lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)}
$$
What would be an easy way to solve that, without flattening the fraction by the application of case $5$ of the theorem $7.8$ above, while using the provided definition for $o$ notation (and without using a more advanced methods, like L'Hopital's rule)?
I saw somewhere $\lim_{x \to 0} \frac{\frac{1}{2} + \frac{o(x^5)}{x^4}}{1 + \frac{o(x^6)}{x^4}} = \frac{1}{2}$, without explanation, suggesting it should be a trivial matter, but I'm not sure why that would be the case.
Appendix in how I solved that, relying only on the definition and the $T7.8$.
One way to proceed could be to use case $3$ from Theorem $7.8$, in reverse (i.e. $o(x^6) = x^4 o(x^2)$)
$$
\lim_{x \to 0} \frac{\frac{x^4}{2} + o(x^5)}{x^4 + o(x^6)} = \lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)}
$$
Then I could apply case $5$ from the above theorem and applying case $4$ of the theorem (i.e. $o(o(x^2)) = o(x^2)$), to get $\frac{1}{1+o(x^2)} = 1 - o(x^2)$.
After applying case $2$ with $c = -1$, we get:
$$
\lim_{x \to 0} \frac{\frac{1}{2} + o(x)}{1 + o(x^2)} = \lim_{x \to 0} (\frac{1}{2} + o(x)) (1 + o(x^2)) = \frac{1}{2}
$$
| A half is the correct answer. For instance, something that is smaller than $ x^5 $ or $ x^6 $ is neglected in comparison to something that grows (or decreases) at $x^4$.
Remember that $o(x)$ is just a notation for something else, it may be a large polynomial approximation, so as it happens to be in this case.
| {
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Proof: $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \leq \cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}}{2}+\cos \frac{\mathrm{C}}{2}$ Let $\mathrm{I} \subseteq \mathbf{R}$ be an interval and
$\mathrm{f}: \mathrm{I} \rightarrow \mathbf{R}$.
We have this inequality for any $x, y, z \in I$,
$$
f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right) \geq f(x)+f(y)+f(z) .
$$
If A, B, C are the measures of the angles of a triangle $ \mathrm {ABC} $, expressed in radians, show that:
$$\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \leq \cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}}{2}+\cos \frac{\mathrm{C}}{2}$$
I have to prove the inequality by using f.
I don't know what "function" to take in order to prove inequality.
| Following up on my comment, let $\,f(t) = \sin(t)\,$ and $\,x = B+C\,$, $\,y = C+A\,$, $\,z = A+B\,$.
*
*$f(x) = \sin \left(B+C\right) = \sin \left(\pi - A\right) = \sin \left(A\right)$
*$f\left(\frac{y+z}{2}\right) = f\left(\frac{(C+A)+(A+B)}{2}\right) = f\left(\frac{(A+B+C) + A}{2}\right) = f\left(\frac{\pi}{2} + \frac{A}{2}\right) = \cos\left(\frac{A}{2}\right)$
Then: $\quad\quad\quad\quad f\left(\frac{x+y}{2}\right)+f\left(\frac{y+z}{2}\right)+f\left(\frac{z+x}{2}\right) \geq f(x)+f(y)+f(z) \quad\quad\quad\quad\quad\text{(1)} \\ \iff\quad \cos \frac{A}{2} + \cos \frac{B}{2} + \cos \frac{C}{2} \ge \sin A + \sin B + \sin C$
What remains to be proved is inequality $\,(1)\,$ itself. That follows because $\,\sin(t)\,$ is concave on $\,[0, \pi]\,$, so $\,f\left(\frac{x+y}{2}\right) \ge \frac{f(x)+f(y)}{2}\,$ by Jensen's inequality. Writing it also for $\,(y,z)\,$ and $\,(z,x)\,$ then adding the three inequalities gives $\,(1)\,$.
| {
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How can I prove that the perimeter is at most 60? Problem: Let $\Delta$ be a triangle in the plane. Let $P$ be the perimeter of the triangle and $A$ be the area. Let $a,b,c$ be the length of the sides and suppose they are positive integers. Suppose finally that $A=P$. How can I prove that $P \leq 60$?
My attempt: I tried using Erone's formula getting:
$$a+b+c = \sqrt{\frac P 2 \left(\frac{P}{2}-a\right)\left(\frac P 2-b\right)\left(\frac P 2 - c\right)}$$
from which I obtained:
$$4P=-P^3 + 4 (a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+abc)$$
and I noticed that $abc \leq \left(\frac{P}{2}\right)^2$ but I am not able to continue. Any help will be appreciated.
Remark: as suggested in the comments, defining $u=a+b-c$, $v=a+c-b$ and $w=b+c-a$ we get $16(u+v+w)=uvw$.
Noticing that $uvw$ should be even and that none of $u,v,w$ can be odd we can write $u=2k$, $v=2h$, $w=2n$ in such a way we deduce $4(k+h+n)=khn$.
| $$hnk=4(h+n+k), h,n,k\in\mathbb{N}$$
$$(hk-4)n=4(h+k)$$
WLOG $h\geq n \geq k > 0$. Then
$$(hk-4)k \leq 4(h+k) \Rightarrow hk^2-8k \leq 4h \Rightarrow hk^2-8h\leq 4h\Rightarrow hk^2\leq12h\Rightarrow k\leq 3$$
$$n=\frac{4(h+k)}{hk-4}$$
At $k=1$:
$$n=\frac{4h+4}{h-4}=4+\frac{20}{h-4}$$
$$h+n+k=h+5+\frac{20}{h-4}=9+(h-4)+\frac{20}{h-4}$$
$$(h-4)|20 \Rightarrow (h-4)+\frac{20}{h-4}\leq 21 \Rightarrow h+n+k\leq 30$$
At $k=2$:
$$n=\frac{4h+8}{2h-4}=2+\frac{8}{h-2}$$
$$h+n+k=h+4+\frac{8}{h-2}=6+(h-2)+\frac{8}{h-2}$$
$$(h-2)|8 \Rightarrow (h-2)+\frac{8}{h-2}\leq 9 \Rightarrow h+n+k\leq 15$$
At $k=3$:
$$n=\frac{4h+12}{3h-4}=1+\frac{h+16}{3h-4}$$
$$n\geq k=3 \Rightarrow \frac{h+16}{3h-4}\geq 2\Rightarrow h+16\geq 6h-8 \Rightarrow h\leq \frac{24}{5}<5$$
$$h\leq 4\Rightarrow h+n+k\leq 4+4+3=11$$
Then $h+n+k\leq 30$ in all possible cases.
Perimeter of triangle $$P=a+b+c=u+v+w=2(h+k+n)\leq 60$$
| {
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$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$? I stumble upon the expression
$$
\sum_{k=0}^\infty
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}
$$
and it seems to converge to
$$
\frac{3-\sqrt{5}}{2}
$$
Do they equate ? How to prove that ?
Not sure if it's helpful :
$\frac{1}{k+1}\binom{3k+1}{k} , \; k\ge0$ is the OEIS sequence $A006013$.
I only have fundamental knowledge in combinatorics so I could only check it numerically that the convergence seems to be true . However I wouldn't mind learning new theories .
| Some thoughts:
We use the integral representation
$$\binom{3k + 1}{k} = \frac{1}{2\pi}\int_{-\pi}^\pi (1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^{3k + 1}(2^{-1}\mathrm{e}^{\mathrm{i}t})^{-k}\mathrm{d} t. \tag{1}$$
(Note: Similar to https://functions.wolfram.com/GammaBetaErf/Binomial/07/02/)
We have
\begin{align*}
&\sum_{k=0}^\infty
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}\\[6pt]
=\,& \frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=0}^\infty \frac{1}{k + 1}2^{-(3k + 2)}(1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^{3k + 1}(2^{-1}\mathrm{e}^{\mathrm{i}t})^{-k}\, \mathrm{d} t\\[6pt]
=\,& \frac{1}{2\pi}\int_{-\pi}^\pi
\frac{1 + z}{4}\sum_{k=0}^\infty \frac{1}{k + 1} \left(\frac{(1 + z)^3}{8z}\right)^k\Big\vert_{z = 2^{-1}\mathrm{e}^{\mathrm{i}t}}\, \mathrm{d} t \\[6pt]
=\,& \frac{1}{2\pi}\int_{-\pi}^\pi
\frac{-2z}{(1 + z)^2}
\ln \left(1 - \frac{(1 + z)^3}{8z} \right)\Big\vert_{z = 2^{-1}\mathrm{e}^{\mathrm{i}t}}\,\mathrm{d} t \tag{2}\\[6pt]
=\,& \frac{3 - \sqrt 5}{2} \tag{3}
\end{align*}
where we have used
$$\sum_{k=0}^\infty \frac{1}{k + 1} a^k = - \frac{\ln(1 - a)}{a}, \quad 0 < |a| < 1$$
and
$\frac{1}{32} \le \frac{(1 - 1/2)^3}{4} \le |\frac{(1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^3}{8\cdot 2^{-1}\mathrm{e}^{\mathrm{i}t}}| \le \frac{(1 + 1/2)^3}{4} = 27/32 < 1$ in (2).
Proof of the integral representation (1):
Using Cauchy integral formula, we have
$$\Big[(1 + z)^{3k + 1}\Big]^{(k)}(0) = \frac{k!}{2\pi \mathrm{i}}
\oint\limits_{|z| = 1/2} \frac{(1 + z)^{3k + 1}}{z^{k + 1}} \mathrm{d} z$$
which results in (with the substitution $z = 2^{-1}\mathrm{e}^{\mathrm{i} t}$)
$$\binom{3k + 1}{k} = \frac{1}{2\pi \mathrm{i}}
\oint\limits_{|z| = 1/2} \frac{(1 + z)^{3k + 1}}{z^{k + 1}} \mathrm{d} z
=
\frac{1}{2\pi \mathrm{i}}
\int_{-\pi}^\pi \frac{(1 + 2^{-1}\mathrm{e}^{\mathrm{i} t})^{3k + 1}}{(2^{-1}\mathrm{e}^{\mathrm{i} t})^{k + 1}} 2^{-1}\mathrm{e}^{\mathrm{i} t} \mathrm{i}\, \mathrm{d} t.$$
The desired result follows.
| {
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Other approaches to evaluate $\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}$
$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=?$$
I evaluated the limit by using the Hopital rule,$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=4^x\lim_{h\to0}\frac{4^h+4^{-h}-2}{h^2}=4^x\lim_{h\to0}\frac{\ln(4)(4^h-4^{-h})}{2h}=4^x\lim_{h\to0}\frac{(\ln4)^2(4^h+4^{-h})}{2}=4^x\times4(\ln2)^2=4^{x+1}(\ln2)^2$$
I want to learn other ideas to solving this problem, so can you please evaluate the limit with other approaches?
| A possible way is using the Taylor expansion
$$2\cosh t = e^t+e^{-t} = 2+t^2 + o(t^2)$$
Hence,
\begin{eqnarray*}\frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}
& = & 4^x\frac{e^{h\ln 4}+e^{-h\ln 4}-2}{h^2}\\
& = & 4^x\cdot \frac{2+h^2\ln^2 4 + o(h^2)-2}{h^2} \\
& = & 4^x\cdot\ln^2 4 + o(1) \\
& \stackrel{h\to 0}{\longrightarrow} & 4^x\cdot\ln^2 4
\end{eqnarray*}
| {
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Possible positions of the knight after moving $n$ steps in Chessboard. Problem
There is a knight on an infinite chessboard. After moving one step, there are $8$ possible positions, and after moving two steps, there are $33$ possible positions. The possible position after moving n steps is $a_n$, find the formula for $a_n$.
I found this sequence is http://oeis.org/A118312
But I can't understand this Recurrence Relation
$$a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3}, \quad\quad n\geq3$$
Can someone give the intuition for this relationship?
| Mordechai Katzman demonstrates in section $3$ of his paper Counting monomials (pages $5$ - $8$) that
$$a_n = \begin{cases}
1 \quad \quad \quad \quad \quad \; \, n = 0 \\
8 \quad \quad \quad \quad \quad \; \, n = 1 \\
33 \quad \quad \quad \quad \quad n = 2 \\
1 + 4n + 7n^2 \quad \; \, n \ge 3
\end{cases}$$
We can now prove by induction that
$$a_n = 3a_{n-1} - 3a_{n-2} + a_{n-3} = 1 + 4n + 7n^2, \quad\quad n\geq3 \tag{1}$$
To test whether $(1)$ holds for $n \ge 3$, we need to define
$$a_0 = 1 + 4(0) + 7(0)^2 = 1$$
$$a_1 = 1 + 4(1) + 7(1)^2 = 12$$
$$a_2 = 1 + 4(2) + 7(2)^2 = 37$$
For the base cases, we have
$$a_3 = 3a_2 - 3a_1 + a_0 = 3\cdot37 - 3\cdot12 + 1 = 1 + 4(3) + 7(3)^2 = 76$$
$$a_4 = 3a_3 - 3a_2 + a_1 = 3\cdot76 - 3\cdot37 + 12 = 1 + 4(4) + 7(4)^2 = 129$$
$$a_5 = 3a_4 - 3a_3 + a_2 = 3\cdot129 -3\cdot76 + 37 = 1 + 4(5) + 7(5)^2 =196$$
Now, we must prove using $(1)$ that $$a_{n+1} = 3a_n - 3a_{n-1} + a_{n-2} = 1 + 4(n+1) + 7(n+1)^2 = 7n^2 + 18n + 12$$
Substituting for $a_n, a_{n-1}$ and $a_{n-2}$, we get
\begin{align}
a_{n+1} &= 3\left(1 + 4n + 7n^2\right) -3\left(1 + 4(n-1) + 7(n-1)^2\right) + \left(1 + 4(n-2) + 7(n-2)^2\right)\\
& = 7n^2 + 18n + 12
\end{align}
$\blacksquare$
| {
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How to find the probability of drawing a certain two cards from a pack Could I please ask for help on the last part of this question:
Two cards are drawn without replacement form a pack of playing cards. Calculate the probability:
a) That both cards are aces
b) that one (and only one) card is an ace
c) That the two cards are of different suits
d) Given that at least one ace is drawn, find the probability that the two cards are of different suits.
Here's my attempt (for parts a, b, and c I get the answer given in the book):
Let:
$A =$ Event that both cards are aces
$B =$ Event that one and only one card is an ace
$C =$ Event that the two cards are of different suits.
a) $P(A) = \frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$ (as must pick an ace AND another ace)
b) $P(B) = \frac{4}{52} \cdot \frac{48}{51} + \frac{48}{52} \cdot \frac{4}{51} = \frac{32}{221}$ (as can pick ace then not ace, or not ace than ace)
c) $P(C) = \frac{13}{52} \cdot \frac{39}{51} \cdot 4 = \frac{13}{17}$ (as can pick any given suit first, followed by not that same suit, and this can be done in four ways, one for each suit).
d) Let $D =$ Event that at least one ace is drawn.
$P(D) = P(A) + P(B)$ (because at least one ace is drawn only "if both cards are aces" or "one and only one card is an ace")
so $P(D) = \frac{1}{221} + \frac{32}{221} = \frac{33}{221}$
Now, I need to calculate $P(C \mid D) = \frac{P(C \cap D)}{P(D)}$
So if I can calculate $P(C \cap D)$ then I can divide this by $P(D)$ to get the answer.
I (wrongly it appears!) reasoned like so:
To end up with two cards where "at least one is an ace and both are of different suits" you can only have this by either having "the first card be an ace and the second a card of a different suit from that ace" OR having "the first card be of a certain suit and the second card an ace of another suit".
Let @ stand for any suit.
so
\begin{align*}
P(C \cap D) & = P(\text{ace of @}) \cdot P(\text{not @}) + P(\text{@}) \cdot P(\text{ace not of @})\\
& = \frac{4}{52} \cdot \frac{39}{51} \cdot 4 + \frac{13}{52} \cdot \frac{3}{51} \cdot 4\\
& = \frac{5}{17}
\end{align*}
Well this leads to $P(C \mid D) = \frac{5}{17} \cdot \frac{221}{33} = \frac{65}{33}$.
Answer given in book is $\frac{25}{33}$.
Thanks for any help.
|
Two cards are drawn without replacement from a standard deck of playing cards. Given that at least one ace is drawn, find the probability that the two cards are of different suits.
Method 1: We correct your attempt.
As you observed, the probability that at least one ace is drawn is
$$\Pr(\text{at least one ace is drawn}) = \frac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51} = \frac{33}{221}$$
If we draw two aces, they must be of different suits. This event can occur in $4 \cdot 3$ ways.
If we draw one of the four aces with the first draw, there are $51 - 3 - 12 = 36$ cards left in the deck which are not aces and are of a different suit than the ace. Hence, the number of ways of drawing an ace and then drawing a non-ace of a different suit is $4 \cdot 36$.
If we draw one of the $48$ non-aces with the first draw, there are $3$ aces left in the deck with a different suit than the non-ace we drew first. Hence, the number of ways of drawing a non-ace and then drawing an ace of a different suit is $48 \cdot 3$.
Hence, the probability that the two cards are of different suits and at least one ace is drawn is
$$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51} = \frac{25}{221}$$
Hence, the probability that the two cards are of different suits given that at least one ace is drawn is
\begin{align*}
\Pr(\text{of different suits} \mid & \text{at least one ace is drawn})\\
\qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\
\qquad & = \frac{\dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51}}{\dfrac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51}}\\
\qquad & = \frac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}\\
\qquad & = \frac{25}{33}
\end{align*}
Method 2: We work with combinations to avoid considering the order in which the cards are drawn, which simplifies the calculations.
The probability of drawing at least one ace when two cards are drawn is
$$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}$$
since we must either select two of the four aces or one of the four aces and one of the $48$ non-aces while selecting two of the $52$ cards in the deck.
The probability that the two cards are of different suits and at least one ace is drawn is
$$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}} = \frac{25}{221}$$
since either two aces are drawn or one ace and one of the $36$ non-aces that are of a different suit than the ace are drawn when two cards are drawn from the deck of $52$ cards.
The probability of drawing two cards of different suits given that at least one ace has been selected is
\begin{align*}
\Pr(\text{of different suits} & \mid \text{at least one ace is drawn})\\
\qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\
\qquad & = \frac{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}}}{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}}\\
\qquad & = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}\\
\qquad & = \frac{25}{33}
\end{align*}
Note: Observe that in both methods, after we cancel the common denominators, the numerator reduces to the number of cases in which at least one ace is drawn and cards are of different suits and the denominator reduces to the number of cases in which at least one ace is drawn. This is because the sample space for the conditional probability that the cards are of different suits given at least one ace is drawn is the set of cases in which at least one ace is drawn.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) .$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{d x}{x^{6}+1} \\
&=\frac{1}{2} \oint_{\gamma} \frac{d z}{z^{6}+1} \\
&=\frac{1}{2} \cdot 2 \pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^{6}+1} , z_k \right)
\end{aligned}
$$
with its simple poles at $ z_{k}=e^{\frac{(2 k+i) \pi}{6} i}$, where $k=0,1,2$.
Now lets evaluate the residues of $\frac{1}{z^{6}+1} $ at $ z_{k}$.
$$
\operatorname{Res}\left( \frac{1}{z^{6}+1} , z_{k}\right)=\frac{1}{6 z _k^{5}}=\frac{z_{k}}{6 z_{k}^{6}}=-\frac{z_{k}}{6}
$$
Putting the residues back yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{6}+1} d x &=\pi i\left(-\frac{1}{6}\left(z_{0}+z_{1}+z_{2}\right)\right) \\
&=-\frac{\pi i}{6}\left(e^{\frac{\pi}{6}}+e^{\frac{\pi}{2} i}+e^{\frac{5 \pi}{6} i}\right) \\
&=-\frac{\pi i}{6}\left(\frac{\sqrt{3}}{2}+\frac{1}{2} i+i-\frac{\sqrt{3}}{2}+\frac{1}{2} i\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
Similarly, we can deal with the integral in general
$$I_{2n}=\int_{0}^{\infty} \frac{1}{x^{2n}+1} d x$$
with the same contour $\gamma$ and $n $ simple poles whose residues are
$$
\operatorname{Res}\left( \frac{1}{z^{2n}+1} , z_{k}\right)=\frac{1}{2n z _k^{2n-1}}=\frac{z_{k}}{2n z_{k}^{2n}}=-\frac{z_{k}}{2n}
$$
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{x^{2 n}+1} d x&=\pi i \sum_{k=0}^{n-1} \operatorname{Res}\left(\frac{1}{z^{2 n}+1}, z_k\right) \\
&=\pi i \sum_{k=0}^{n-1}\left(-\frac{z _k}{2 n}\right)\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} z_{k}\\&=-\frac{\pi i}{2 n} \sum_{k=0}^{n-1} e^{\frac{2 k+1}{2 n} \pi i}\\ &=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{1-e^{\frac{\pi i}{n}(n)}}{1-e^{\frac{\pi i}{\pi}}}\\
&=-\frac{\pi i}{2 n} e^{\frac{\pi i}{2 n}} \cdot \frac{2}{1-e^{\frac{\pi}{n} i}}\\
&=-\frac{\pi i}{n} \frac{e^{\frac{\pi i}{2 n}}}{1-e^{\frac{\pi}{n} i}}\\&= -\frac{\pi i}{n} \cdot \frac{1}{e^{-\frac{\pi i}{2 n}}-e^{\frac{\pi i}{2 n}}}\\&= \frac{\pi}{2 n} \csc \frac{\pi}{2 n}
\end{aligned}
$$
However, I can’t use the same contour to deal with the odd one
$$\int_{0}^{\infty} \frac{1}{x^{2 n-1}+1} d x$$
as the simple pole $-1$ is on $\gamma_1$.
My Question: How can we evaluate the odd one?
| This can be calculated explicitly, by the fomula $B(t,1-t)＝π/\sin πt$, where $B$ is the Beta function.
| {
"language": "en",
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"question_score": "3",
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$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ I found this problem in a high school text book.
Let $ \displaystyle s = \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001}$. Find $s$.
How I tried:
Observe that $T_n = \frac{n^2}{(2n-1)(2n+1)}$. Here, $T_n$ is the $n$th term of the sequence. So, we need to find the value of $\sum_{n=1}^{500}\frac{n^2}{(2n -1)(2n+1)}$. We can find its value with Telescope Cancellation Method, but it requires breaking the expression we got into simpler terms.
How to simplify $T_n = \frac{n^2}{(2n-1)(2n+1)}$ ?
| $$T_n = \frac{n^2}{4n^2-1} = \frac{1}{4}\left( \frac{4n^2-1+1}{4n^2-1} \right) = \frac{1}{4}\left( 1 + \frac{1}{4n^2-1} \right)=\frac{1}{4}\left( 1 + \frac{1}{(2n-1)(2n+1)} \right). $$
Decomposing with partial fractions, we have:
$$ \frac{1}{(2n-1)(2n+1)} = \frac{\frac{1}{2}}{2n-1} - \frac{\frac{1}{2}}{2n+1} = \frac{1}{2}\left( \frac{1}{2n-1} - \frac{1}{2n+1} \right). $$
And we see that most of the terms in $\displaystyle\sum_{n=1}^{500} T_n$ will cancel, apart from a few at either end.
| {
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Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfied.
Alternatively, if $\ell_1\leq\ell_2\leq\ell_3$ are the values of $a,b$ and $c$ ordered in ascending order, then the triangle exists iff $\ell_1+\ell_2\geq\ell_3$.
Interestingly, the three triangle inequalities can be recast into a single quartic polynomial inequality. Let $0,x,y\in\mathbb R^2$ be the three vertices of the triangle, with $\|x\|=a,\,\|y\|=b$ and $\|x-y\|=c$. Then $c^2=\|x-y\|^2=\|x\|^2-2\langle x,y\rangle+\|y\|^2=a^2+b^2-2\langle x,y\rangle$. Therefore $x^Ty=\langle x,y\rangle=\frac{1}{2}(a^2+b^2-c^2)$ and
$$\pmatrix{x^T\\ y^T}\pmatrix{x&y}=\frac{1}{2}\pmatrix{2a^2&a^2+b^2-c^2\\ a^2+b^2-c^2&2b^2}.\tag{1}$$
The RHS of $(1)$ must be positive semidefinite because the LHS is a Gram matrix. Conversely, if the RHS is indeed PSD, it can be expressed as a Gram matrix. Hence we obtain $x$ and $y$ and the triangle exists.
As $2a^2$ and $2b^2$ are already nonnegative, the RHS of $(1)$ is positive semidefinite if and only if $(2a^2)(2b^2)-(a^2+b^2-c^2)^2\geq0$, by Sylvester's criterion. That is, the triangle exists if and only if
$$-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\geq0.\tag{2}$$
This polynomial inequality can be derived by more elementary means. See circle-circle intersection on Wolfram MathWorld. The geometric explanation for the necessity of $(2)$ is given by Heron's formula, which states that the square root of the LHS is four times the area of the triangle.
Since both $(0)$ and $(2)$ are necessary and sufficient conditions for the existence of the required triangle, the two sets of conditions must be equivalent to each other. Here are my questions. Is there any simple way to see why $(0)$ and $(2)$ are equivalent? Can we derive one from the other by some basic algebraic/arithmetic manipulations?
| The quartic polynomial in (2) may be shown by algebra to equal the product
$(a+b+c)(-a+b+c)(a-b+c)(a+b-c),$
so for nonnegative $a,b,c$ the triangle inequality implies that the quartic expression is nonnegative.
But the quartic as defined above is also nonnegative if two or all four of the factors above are negative. So we have to exclude these alternatives to prove equivalence.
Assume that $a+b-c$ and $a-b+c$ are negative. Simply adding these up gives $2a<0$, contradicting the hypothesis that $a$ is nonnegative. Similar contradictions occur if we try other pairs of factors being negative:
$(-a+b+c<0, a-b+c<0) \to (c<0)$
$(-a+b+c<0, a+b-c<0) \to (b<0)$
Thus by indirect proof the equivalence of (0) and (2) is established.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
This is related to the following problem:
Problem 2: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}.$$
Problem 2 is in "Problems From the Book", 2008, Ch. 2, which was proposed by Vasile Cartoaje.
See: Prove that $\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)} \le \frac{n(n-1)}{2(2n-1)^2}$
Background:
I proposed Problem 1 when I tried to find my 2nd proof for Problem 2.
It is not difficult to prove that
$$\frac{1}{(2n-1)^4} + \frac{16n^2(n-1)^2}{(2n-1)^4}\cdot \frac{x_ix_j}{1-x_i-x_j}
\ge \frac{x_ix_j}{(1-x_i)(1-x_j)}.$$
(Hint: Use $\frac{x_ix_j}{(1-x_i)(1-x_j)}= 1 - \frac{1}{1 + x_ix_j/(1-x_i-x_j)}$
and $\frac{1}{1+u} \ge \frac{1}{1+v} - \frac{1}{(1+v)^2}(u-v)$
for $u = x_ix_j/(1-x_i-x_j)$ and $v=\frac{1}{4n(n-1)}$. Or simply
$\mathrm{LHS} - \mathrm{RHS} = \frac{(4x_ix_jn^2 - 4x_ix_j n + x_i + x_j - 1)^2}{(2n-1)^4(1-x_i-x_j)(1-x_i)(1-x_j)}\ge 0$.)
To prove Problem 2, it suffices to prove that
$$\frac{1}{(2n-1)^4}\cdot \frac{n(n-1)}{2} + \frac{16n^2(n-1)^2}{(2n-1)^4}\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac{n(n-1)}{2(2n-1)^2} $$
or
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18.$$
For $n=2, 3, 4$, the inequality is true.
For $n=5, 6$, numerical evidence supports the statement.
Any comments and solutions are welcome and appreciated.
| Write $p_i = 2x_i$ and note that $\sum_i p_i = 1$. Then
\begin{align*}
1 + \sum_i \frac{p_i^2}{1 - p_i}
&= \sum_i \frac{p_i}{1 - p_i} \\
&= \sum_{i,j} \frac{1}{2} \left( \frac{1}{1 - p_i} + \frac{1}{1 - p_j} \right) p_i p_j \\
&\geq \sum_{i,j} \left( \frac{2}{2-p_i-p_j} \right) p_i p_j. \tag{by AM–HM}
\end{align*}
Rearranging this inequality, we get
$$ 1 \geq \sum_{i \neq j} \frac{2p_i p_j}{2 - p_i - p_j} = 8 \sum_{i < j} \frac{x_i x_j}{1 - x_i - x_j},$$
completing the proof.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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} |
The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
x^2 && \hbox{} \\[-3pt]
x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
\underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
2 +i.x^2
\end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
| Maybe all has been said already.
Polynomial with real coefficients:
1)Complex roots occur in pairs, one is the complex conjugate of the other.
(Complex root theorem)
$x_1=1+i,$ $x_2=1-i;$
2)By inspection the real root is $x_3=-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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Solving equation involving roots and powers . I'm trying to solve this equation :
$\sqrt{3}\sqrt{237x^2 + \frac{224}{x^2}}x^7 + \frac{35293}{222}x^8 + \frac{2}{999}\sqrt{3}{(\sqrt{237x^2 + \frac{224}{x^2}})}^3x^5 - \frac{44968}{111}x^4 + \frac{12544}{333}=0$.
It looks pretty complicated to me, but the computer gives me exact solutions. For example one real solution is $\frac{2}{\sqrt{3}}$, and another : $-\frac{2}{3^{\frac{3}{4}}\sqrt[4]{7}}$ and so on.
Does anyone know how to obtain the exact solutions to the above equation? Thanks in advance!
EDIT:
This came from fooling around with functions that describe surfaces spanned by the roots of polynomials $x^4+c_2x^2+c_3x+c_4$ in 3 dimensions. So here $r_1+r_2+r_3+r_4=0$. Newton's identity $r_1^4+r_2^4+r_3^4+r_4^4 = 4$ (the 4 here is randomly chosen) can thus be represented as a 2D surface in 3D. With some other transformations this gives :
$-\frac{1}{3}\sqrt{6}\sqrt{3}x_1x_2^2x_3 + \frac{1}{9}\sqrt{6}\sqrt{3}x_1x_3^3 + \frac{7}{12}x_1^4 + \frac{1}{2}x_1^2x_2^2 + \frac{1}{2}x_2^4 + \frac{1}{2}x_1^2x_3^2 + x_2^2x_3^2 + \frac{1}{2}x_3^4 = 4$ .
This surface looks somewhat like an octahedron. To find the extrema on the surface I used Lagrange multipliers leading to the equation in question. So I guess it's not very surprising that this has an exact solution. But I was surprised that the computer could find them so easily..
|
how to obtain the exact solutions to the above equation?
Not something you'd want to do by hand, but the equation can be reduced to a quartic in $\,x^4\,$, which just "happens" to factor nicely.
Let $\,y = \sqrt{237x^2 + \dfrac{224}{x^2}}\,$ then, after eliminating the denominators, the original equation can be written as the following system (with the restriction $\,y \ge 0\,$ for the real solutions):
$$
\begin{cases}
\begin{align}
p(x,y) &= 317637\,x^8 + 1998\sqrt{3}\,x^7 y + 4\sqrt{3}\,x^5 y^3 - 809424\,x^4 + 75264 &= 0
\\ q(x,y) &= 237\,x^4 - x^2y^2 + 224 &= 0
\end{align}
\end{cases}
$$
Eliminating $\,y\,$ between the two equations using resultants gives the following, courtesy WA:
$$
\begin{align}
0 = \text{res}_y(p,q) &= -7203 x^6 (13150431 x^{16} - 72718560 x^{12} + 97023744 x^8 \\ &\quad\quad\quad\quad\quad - 16990208 x^4 + 786432)
\\ &= -7203 x^6 (3 x^2 - 4) (3 x^2 + 4) (9 x^4 - 32) (189 x^4 - 16) (859 x^4 - 96)
\end{align}
$$
[ EDIT ] $\;$ The same result can be derived directly by eliminating the radicals from the original equation (just move the two radicals to one side, then square the equation and collect the terms), but the calculations are laborious, which is what I meant by "wouldn't want to do it by hand".
| {
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Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular points $v = \pm 1, \pm i$ all lie inside the contour $G$. Thus, from this theorem (*), we have
\begin{align*}
\int_{G}{\dfrac{1}{v^4-1} \text{d}v} &= \dfrac{1}{4} \left( \int_{G}{\dfrac{1}{v-1}\text{d}v} - \int_{G}{\dfrac{1}{v+1}\text{d}v} + \int_{G}{\dfrac{i}{v-i}\text{d}v} - \int_{G}{\dfrac{i}{v+i}\text{d}v} \right) \\
&= \dfrac{1}{4}\left( 2\pi i - 2\pi i + i\left( 2\pi i \right) - i \left( 2\pi i \right) \right) \\
&= \dfrac{1}{4} \left( 0 \right) \\
&= 0
\end{align*}
(*) Theorem: Let $C$ be a simple closed contour with a positive orientation such that $v_0$ lies interior to $C$, then $\displaystyle{\int_{C} {\dfrac{dv}{(v-v_0)^n}} = 2\pi i}$ for $n =1$ and $0$ when $n \neq 1$ is an integer.
Is that proof correct? If so, could you also point out if there are still theorems I have to mention to make it more accurate?
I'm trying to solve (perhaps overthink) this with the other approach:
We see that it is analytic except at $\pm 1$ and $ \pm i$.
Also, we can apply deformation of the contour $G$ by forming a leaf-like contour and forming the respective circles $C_1, C_2, C_3,$ and $C_4$. As shown here:
The integration can then be evaluated as
$$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} $$
And,
$$\int_{C_n}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4} \left( \int_{C_n}{\dfrac{1}{v-1}\text{d}v} - \int_{C_n}{\dfrac{1}{v+1}\text{d}v} + \int_{C_n}{\dfrac{i}{v-i}\text{d}v} - \int_{C_n}{\dfrac{i}{v+i}\text{d}v} \right) $$
Note that when $v_n$ lies exterior to $C_n$, then by Cauchy-Goursat theorem, $\displaystyle{\int_{C_n}{\dfrac{dv}{v-v_n}} = 0}$.
Thus, for $n = 1,$,
$$\int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0-0 + i(2\pi i)- 0) = \dfrac{- \pi }{2} $$
for $ n = 2,$
$$\int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}( 2\pi i - 0 + 0-0) = \dfrac{ \pi i}{2}$$
for $ n = 3,$
$$\int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 - 0 + 0 - i (2\pi i) ) = \dfrac{\pi }{2}$$
for $ n = 4,$
$$\int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 -(2\pi i) + 0-0 ) = \dfrac{ - \pi i }{2}$$
Therefore, $$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{- \pi }{2} + \dfrac{ \pi i}{2} + \dfrac{\pi }{2} + \dfrac{ - \pi i }{2} = 0$$
Did I just overcomplicate it? Is my first proof already enough? If any of these proofs are correct, could you also point out if there are still theorems I have to mention for them to make it more accurate?
| Since OP's first solution works just fine, I will provide yet another solution:
We "inflate" the contour $G$ so it becomes a CCW-oriented circle of radius $r > 1$.
Then
$$ \int_{G} \frac{\mathrm{d}z}{z^4 - 1}
= \int_{|z| = r} \frac{\mathrm{d}z}{z^4 - 1}
\stackrel{(w=1/z)}{=} \int_{|w|=\frac{1}{r}} \frac{-\mathrm{d}w/w^2}{w^{-4} - 1}
= - \int_{|w|=\frac{1}{r}} \frac{w^2}{1 - w^4} \, \mathrm{d}w $$
In the last integral, $\frac{w^2}{1-w^4}$ has no poles inside the circle $|w| = \frac{1}{r}$ since $\frac{1}{r} < 1$. Therefore the integral evaluates as $0$ by the Cauchy's integral theorem.
Remark. This is an example of the "residue at infinity".
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof $ \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos^{2}x} - 4 = \infty $ I want to prove that
$$ \lim_{x \to \frac{\pi}{2}} \left(\frac{1}{\cos^{2}x} - 4\right) = \infty ,$$
is my proof correct?
Proof:
Given $ M \ge 1$, choose $ \delta = \arccos\left(\sqrt{\frac{1}{M}}\right) - \frac{\pi}{2}$.
Suppose $ 0 \lt \left|x - \frac{\pi}{2}\right| \lt \delta $.
Therefore:
$ x - \frac{\pi}{2} \lt \arccos(\sqrt{\frac{1}{M}}) - \frac{\pi}{2} $
$ x \lt \arccos(\sqrt{\frac{1}{M}}) $
$ \cos\left(x\right) \lt \sqrt{\frac{1}{M}}$
$ \cos^{2}\left(x\right) \lt \frac{1}{M}$
$ \frac{1}{\cos^{2}\left(x\right)} \gt M $
$ \frac{1}{\cos^{2}\left(x\right)} - 4 \gt M $
| Given $M > 0$ we solve
$\vert\frac{1}{cos^2x} - 4\vert \geqslant M$ with $x \in [0, 2\pi]$
$$\frac{1}{cos^2x} \geqslant M + 4 \quad\vee\quad \frac{1}{cos^2x} \leqslant 4 - M$$
WLOG we assume $M > 4$ $$cos^2x \leqslant \frac{1}{M + 4} \quad\wedge\quad x \neq \frac{\pi}{2}$$
$$-\frac{1}{\sqrt[]{M + 4}} \leqslant cosx \leqslant \frac{1}{\sqrt[]{M + 4}}$$
Set $k = \arccos\left(\frac{1}{\sqrt[]{M + 4}}\right)$ and find $k \leqslant x < \frac{\pi}{2} \quad\vee\quad \frac{\pi}{2} < x \leqslant \pi - k$
So $A = [k, \frac{\pi}{2}) \quad\cup\quad (\frac{\pi}{2}, \pi - k]$ with $0 < k < \frac{\pi}{2}$
A is a neighborhood of $\frac{\pi}{2}$ and $0 < \delta \leqslant \frac{\pi}{2} - k$
| {
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Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$
Assuming $a,b,c>0$, show that
$$\Big(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Big)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq \Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big).$$
I know from CS that $ab+bc+ca \leq a^2+b^2+c^2$ and $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$ so the exponent is less than 3 and the second term in parenthesis greater than $1$, but I can't manage to convert this information, might work on the right hand side but seems like I'm missing a classical inequality since I'm a very beginner in this.
I noticed this inequality is symmetrical and homogeneous, maybe assuming $a+b+c=1$ could be useful...
| I use $abc$'s method .
Let $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$ then the problem is :
$$\frac{3u3v^2}{w^3}-1\geq \left(1+\frac{9u^2-6v^2}{3v^2}\right)^{\frac{9u^2}{9u^2-6v^2}}$$
A bit of algebra and the inequality is linear in $w$ and as it's homogeneous we can assume $a=b=1$ so we need to show :
$$\Big(1+\frac{2+c^2}{1+2c}\Big)^{\frac{(2+c)^2}{2+c^2}} \leq 2\Big(1+\frac{1}{c}\Big)\Big(1+c\Big)$$
We use logarithm the inequality now is :
$$\frac{(2+c)^2}{2+c^2}\ln\Big(1+\frac{2+c^2}{1+2c}\Big)\leq \ln(2)+\ln(1+c)+\ln\Big(1+\frac{1}{c}\Big)$$
We introduce the function :
$$f\left(x\right)=\frac{(2+x)^{2}}{2+x^{2}}\ln\left(1+\frac{2+x^{2}}{1+2x}\right)-\left(\ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)\right)$$
Now I haven't an idea to conclude except that we have for $x\in(0,10]$:
$$\ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)\geq\left(\left(\ln\left(2\right)+\frac{\ln\left(2\right)}{x}+\ln\left(2\right)\ln\left(xe\right)\right)\left(\frac{x}{x+1}\right)+\frac{1}{x+1}\cdot3\ln\left(2\right)\right)\geq \frac{(2+x)^{2}}{2+x^{2}}\ln\left(1+\frac{2+x^{2}}{1+2x}\right)$$
Wich is easier I think .
Edit :
the inequality is :
$$\left(1+\frac{2\left(1+2x\right)}{2+x^{2}}\right)\ln\left(1+\frac{\left(2+x^{2}\right)}{1+2x}\right)\leq \ln\left(2\right)+\ln\left(1+\frac{1}{x}\right)+\ln\left(x+1\right)$$
Setting $a=\frac{\left(1+2x\right)}{2+x^{2}},b=\frac{1}{x},c=1+x$ :
$$\left(1+2a\right)\ln\left(1+a^{-1}\right)\leq \ln\left(1+b\right)+\ln\left(2\right)+\ln\left(c\right)$$
With the constraint $$a=\frac{\left(c+b^{-1}\right)}{2+b^{-2}}$$
Wich is easier with derivatives .
| {
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A curve that intersects $2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ with multiplicity $3$ or more let $C=2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ be a curve in $\mathbb{C^2}$. Find a conic that intersects $C$ in $(2,2)$ with intersection multiplicity $\geq 3$.
Since $C_x(2,2) \neq 0$ and $C_y(2,2) \neq 0$ it is a nonsingular point for $C$. We have defined the multiplicity of intersection of two curves in the projective plane so I imagine I need to find a conic $D=a_{00}x^2+2a_{01}xy+2a_{02}xz+a_{11}y^2+2a_{12}yz+a_{22}z^2$ such that eliminating $z$ from $D$ and $C^h$, where $C^h$ is the homogenization of $C$, the factor $(x-y)^3$ divides such polynomial.
I computed such conditions but they are quite unusable in this form so I opted for a trial-and-error approach, but all of the conics I tried have i.m. $2$, like the circle $(x-\frac{7}{4})^2+(y-3)^2=\frac{17}{16}$.
Any idea or hint? Thank you.
| If it is not required the conic should be smooth, you could take the union of the tangent to $C$ at $(2,2)$ and any other line that passes through $(2,2)$.
| {
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
~
EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.
| Sometimes while finding maxima for positive integers you will have to implement some other techniques along with the traditional ones.In your problem you get $xy=\frac{-5(x-10)^2+500}{3}$.Note that $500\equiv 2$(mod 3) and $-5\equiv 1$(mod 3).So we must have $(x-10)^2\equiv 1$(mod 3).So x-10 can't be a multiple of 3.So for minimum value x=11 and thus xy=165
| {
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How to prove this summation of floor function I am supposed to show that
$$
\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots = n
$$
For all postive integers $n \geq 1 $. (Note that this summation is not infinite, since $0 < \frac{n+2^k}{2^{k+1}} < 1$ for large $k$.
I can't seem to find a good way to tackle this exercise.
| So the first idea is to consider the base 2 expansion of $n$ (this is natural because you have the power of 2 everywhere). We then write
$$
n=\sum_{k=\geq 0} a_k 2^k, \quad a_k\in\{0,1\}
$$
Then we can actually compute each of the terms in your sum:
$$
\lfloor \frac{n+2^s}{2^{s+1}}\rfloor=\lfloor \sum_k (a_k +\delta_{k,s})2^{k-(s+1)}\rfloor=\sum_{k\geq s+1} a_k 2^{k-(s+1)}+\lfloor \frac{a_s+1}{2}+\sum_{k< s} a_k 2^{k-(s+1)}\rfloor
$$
The last sum inside the floor function can be bounded by:
$$
\sum_{k< s} a_k 2^{k-(s+1)}\leq \frac{1}{2^{s+1}}\sum_{k<s} 2^k=\frac{1}{2^{s+1}}\frac{2^{s}-1}{2-1}=\frac{1}{2}-\frac{1}{2^{s+1}}<\frac{1}{2}
$$
Then we can just notice that if $a_s=0$ the whole term inside the floor function gives $1/2+\sum \dots<1/2+1/2$ hence the floor will be $0$. Similarly if $a_s=1$ one gets that the floor term is $1$ hence we simply have.
$$
\lfloor \frac{n+2^s}{2^{s+1}}\rfloor=\sum_{k\geq s+1} a_k 2^{k-(s+1)}+a_s
$$
Now the results follows from taking the sum over s, reversing the order of summation and using the geometric series (as I did above). I will leave you fill the details.
| {
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Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then the problem is pretty straightforward.
However, I don't understand why the inequality of the hint is true? A simple example shows that the inequality doesn't hold! There are no further information given.
What am I missing? Is $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$ simply nonsense? Or did the professor forgot to mention some more assumptions?
| I think I got it right this time!!
$e-\sum_{0}^{n}\frac{1}{k!}<\frac{1}{n!n}$ is equivalent to $ e-e+\sum_{n+1}^{\infty }\frac{1}{k!}<\frac{1}{n!n}$ equivalent to
$\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+2)}+.....<\frac{1}{n!n}$ and hence to
$\frac{1}{n!}({\frac{1}{(n+1)}}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+....)<\frac{1}{n!n}$
But $\frac{1}{(n+1)(n+2)}<\frac{1}{(n+1)^{2}}, \,\,\frac{1}{(n+1)(n+2)(n+3)}<\frac{1}{(n+1)^{3}} $ etc so it suffices to show setting $a=\frac{1}{n+1}$
that $a+a^{2}+a^{3}+....\leq \frac{1}{n}\Leftrightarrow \frac{1}{1-a}-1\leq \frac{1}{n}$. But this is
$\frac{n+1}{n}-1=\frac{1}{n}$. QED
| {
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Minimum value of $ab+bc+ca$ Given that $a,b,c \in \mathbb{R^+}$ and $(a+b)(b+c)(c+a)=1$
Find the Minimum value of $ab+bc+ca$
My try: Letting $x=a+b, y=b+c, z=c+a$ we get
$$a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2},c=\frac{y+z-x}{2}$$
$$xyz=1$$
Now the problem is to minimize $$ab+bc+ca=\frac{xy+yz+zx}{2}-\frac{x^2+y^2+z^2}{4}$$
Now by $A.M-G.M$ we have
$$\frac{xy+yz+zx}{2}\geq \frac{3}{2}\times \sqrt[3]{x^2y^2z^2}=\frac{3}{2}$$
But I am stuck for $\frac{-(x^2+y^2+z^2)}{4}$
| With this separation you can't find a minimum ; since $x^2+y^2+z^2$ is not bounded. Just take the sequence $(x_n,y_n,z_n)=(n,n,1/n^2)$, $n\in\mathbb{N}^\ast$, so that $x_ny_nz_n=1$, then $x_n^2+y_n^2+z_n^2=2n^2+1/n^4$ is not upper bounded. You may maximize the 'whole' function $x^2+y^2+z^2-2(xy+yz+zx)$ by Lagrange multiplier (for example).
Lagrange multiplier
| {
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}D\sec^2\theta d\theta
= \frac{1}{D^3}\int_{-\infty}^\infty \cos^2 \theta d\theta$$
$$=\frac{1}{D^3}[\frac{\theta}{2} + \frac{\sin {2\theta}}{4} + C]_{-\infty}^\infty$$
Put $\theta = \arctan{\frac{x}{D}}$;
$$=\frac{1}{D^3}[\frac{\arctan{\frac{x}{D}}}{2} + \frac{\sin {(2\arctan{\frac{x}{D})}}}{4} + C]_{-\infty}^\infty$$
We get the integral as $\frac{\pi}{2D^3}$. Since $\lim_{x \to +\infty} \arctan(x) = \frac{\pi}{2}$ and $\lim_{x \to -\infty} \arctan(x) = \frac{-\pi}{2}$
I feel something isn't right here. Can anyone point the mistake please. Thank you very much.
| Do you know about complex analysis? Integrals like this became trivial:
We consider the contour integral
$$\oint_C \frac{1}{(z^2+D^2)^2}dz=\int_{\Gamma_R}\frac{1}{(z^2+D^2)^2}dz +\int_{[-R,R]}\frac{1}{(z^2+D^2)^2}dz $$
where $C$ is a semicircle in the upper half plane. By the residue theorem, we need only consider the pole at $z=Di$. As the radius of the semicircle goes to infinity, we find that
$$\lim_{R\to\infty}\int_{\Gamma_R}\frac{1}{(z^2+D^2)^2}dz=0$$
so
$$\lim_{R\to\infty}\int_{[-R,R]}\frac{1}{(z^2+D^2)^2}dz=\int_{-\infty}^{\infty}\frac{1}{(z^2+D^2)^2}dz=\oint_C \frac{1}{(z^2+D^2)^2}dz=2\pi i Res(f,Di)=2\pi i\lim_{z\to Di}\frac{d}{dz}((z-Di)^2f(z))=\frac{\pi}{2D^3}$$
| {
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Complex Analysis to solve this integral? $\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$ Complex Analysis time! I need some help in figuring out how to proceed to calculate this integral:
$$\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$$
I tried to apply what I have been studying in complex analysis, that is stepping into the complex plane. So
$$\sin(x) \to z - 1/z$$
$$\cos(x) \to z + 1/z$$
Obtaining
$$\int_{|z| = 1} \frac{\ln(z^2-1) - \ln(z)}{\sqrt{z^4 + 3z^2 + 1}} \frac{\text{d}z}{i}$$
I found out the poles,
$$z_k = \pm \sqrt{\frac{-3 \pm \sqrt{5}}{2}}$$
But now I am confused: how to deal with the logarithms? Also, what when I have both imaginary and real poles?
I am a rookie in complex analysis so please be patient...
| @Hans-André-Marie-Stamm, I hope you don't mind that I was unable to solve this problem using Complex Analysis, but here's a method that relies on the Beta Function and some algebric work.
$$\begin{align}I&=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx;\ \cos(x)\rightarrow y\\&=\frac{1}{2}\int_{0}^{1}\frac{\log\left(1-y^2\right)}{\sqrt{1-y^4}}dy =\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{I_1}+\underbrace{\frac{1}{4}\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{I_2}\end{align}$$
$$\begin{align}I_1=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(1-y^4\right)}{\sqrt{1-y^4}}dy}_{y=z^{1/4}}=\frac{1}{16}\int_{0}^{1}z^{1/4-1}\frac{\log\left(1-z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{16}\lim_{t \rightarrow 1/2}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{4},t\right)=\frac{1}{16}\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi^{(0)}\left(\frac{1}{2}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
$$\begin{align}I_2=&\frac{1}{4}\underbrace{\int_{0}^{1}\frac{\log\left(\frac{1-y^2}{1+y^2}\right)}{\sqrt{1-y^4}}dy}_{y=\sqrt{\frac{1-\sqrt{z}}{1+\sqrt{z}}}}=\frac{1}{32}\int_{0}^{1}z^{1/4-1}\frac{\log\left(z\right)}{\sqrt{1-z}}dz\\=&\frac{1}{32}\lim_{t \rightarrow 1/4}\frac{d}{dt}\mathfrak{B}\left(\frac{1}{2},t\right)=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
Gathering both results:
$$\begin{align}I&=\frac{1}{32}\mathfrak{B}\left(\frac{1}{2},\frac{1}{4}\right)\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\\&=\frac{1}{32}\frac{\Gamma\left(1/4\right)\Gamma\left(1/2\right)}{\Gamma\left(3/4\right)}\left[\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right]\end{align}$$
This result can be simplified if one applies Gamma's Reflection Formula, and Digamma's Reflection and Multiplication Formulas, obtaining:
$$I=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx=\frac{\log(2)-\pi}{16\sqrt{2\pi}}\Gamma^2\left(\frac{1}{4}\right)$$
| {
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Root in $(1,2]$ of Equation $x^n-x-n=0$ Consider the equation $x^n-x-n=0$ with $n\in\mathbb{N},n\geq2.$
$a)$ Show that this equation has exactly one solution $u_n\in(1,2].$
$b)$ Show that the sequence $\left\{u_n\right\}$ is decreasing.
$c)$ Determine $L=\lim_{n\rightarrow\infty}u_n.$
Here is what I've done so far on this problem.
$a)$
Let $f_n(x)=x^n-x-n,$ then clearly $f_n$ is continuous and $f_n(1)=-n<0.$
Let $g(n)=f_n(2)=2^n-2-n,$ then $g'(n)=2^n\cdot\ln2-1>0$ as $n\geq2,$ hence
$$f_n(2)=g(n)\geq g(2)=2^2-2-2=0.$$
Furthermore, $\frac{d}{dx}f_n(x)=nx^{n-1}-1\geq1-1=0$ as $n\geq2$ and $x\geq1.$
Therefore, $f_n$ is non$-$decreasing on $[1,2]$ and $f_n(1)<0\leq f_n(2),$ and the Intermediate Value Theorem implies the unique root $u_n$ as desired.
$c)$ Note that $u_n^n-u_n-n=0\Leftrightarrow u_n=\sqrt[n]{u_n+n}$ and for $a>0$ fixed,
$$\lim_{n\rightarrow\infty}\ln(\sqrt[n]{a+n})=\lim_{n\rightarrow\infty}\frac{\ln(a+n)}{n}=\lim_{n\rightarrow\infty}\frac{1}{a+n}=0$$
hence
$$1=\lim_{n\rightarrow\infty}\sqrt[n]{1+n}\leq\lim_{n\rightarrow\infty}\sqrt[n]{u_n+n}\leq\lim_{n\rightarrow\infty}\sqrt[n]{2+n}=1$$
so $L=1.$
I'm currently stuck with part $b),$ so any hints/ideas/comments are appericated. Thank you!
| First, I shall prove a stronger claim.
Claim: $u_n \geq 1 + \frac{2}{n}$ for $n \geq 2$.
Proof: Applying IVT on $\left(1 + \frac{2}{n}, 2\right)$, we simply have to evaluate $f_n\left(1 + \frac{2}{n}\right)$. Now,
$$
f_n\left(1 + \frac{2}{n}\right) = \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{2}{n}\right) - n
$$
For $n \geq 7$, notice that $f_n\left(1 + \frac{2}{n}\right) \leq e^2 - (n + 1)$, so the inequality holds. For $2\leq n\leq 6$, simply substitute the values in.
Claim: $\frac{\sqrt{n^2 + 4} - (n - 2)}{2} \leq 1 + \frac{2}{n}$.
Proof: The inequality is equivalent to
$$\begin{align*}
n\sqrt{n^2 + 4} - n(n - 2) &\leq 2n + 4 \\
n\sqrt{n^2 + 4} &\leq n^2 + 4 \\
n^2(n^2 + 4) &\leq (n^2 + 4)^2 \\
4(n^2 + 4) \geq 0
\end{align*}$$
which holds for all $n > 0$.
Now, Apply IVT on $(1, u_{n - 1})$. In particular, let $u = u_{n - 1}$. Then,
$$\begin{align*}
u^{n - 1} - u - (n - 1) = 0 &\implies u^{n - 1} = u + (n - 1) \\\\
u^n - u - n &= u(u + (n - 1)) - u - n \\
&= u^2 + (n - 2)u - n
\end{align*}$$
Now, note that this quadratic has positive root
$$
u = \frac{-(n - 2) + \sqrt{(n - 2)^2 + 4n}}{2} = \frac{-(n - 2) + \sqrt{n^2 + 4}}{2} \leq 1 + \frac{2}{n}
$$
Since $u \geq 1 + \frac{2}{n}$, we have that $f(u) = u^n - u - n > 0$. Clearly, $f(1) = -n < 0$, so applying IVT gives part (b).
| {
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$ 20! = (2^{2} \cdot 5)(19)(2 \cdot 3^{2})(17)(2^{4})(3 \cdot 5)(2 \cdot 7)(13)(2^{2} \cdot 3)(11)(2 \cdot 5)(3^{2})(2^{3})(7)(2 \cdot 3) (5) (2^{2}) (3) (2) $$
$$ = 2^{18} (5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $$
so it is left to compare $2^{42}$ and $(5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $.
We write the prime factors nicely as:
$$ 3^{8}5^{4}7^{2}(11) (13)(17)(19) $$
Notice
$(3)(11) > 2^{5}$,
$(13)(5)>2^{6}$,
$(19)(7) >2^{7}$,
$17 > 2^{4}$, so we now focus on
$$3^{7}5^{3}7 = 2187(5^{3})7 > 2048(5^{3})7 = 2^{11}875 >2^{11}512 = 2^{20} $$
So we have that the prime factors is larger than $2^{42}$.
| Here is a way to go for the solution using only elementary computations.
(And building partial products that get bigger than and closer to powers of two. I wanted to use first very close approximations like $3\cdot 18\cdot 19=1026>1024$, but there is no need to be so economical at the beginning, and very generous at the end...)
$$
\begin{aligned}
3\cdot 12 &= 36 \\
&> 32 =2 ^5\ ,\\
5\cdot 13 &= 65 \\
&> 64 =2 ^6\ ,\\
6\cdot 7\cdot 9\cdot 11 &= 6\cdot 11\cdot 7\cdot 9
=66\cdot 63=(64+2)(64-1)=64^2 +64-2
\\
&> 64^2 = (2^8)^2=2^{12}\ ,
\\
15\cdot 17\cdot 14\cdot 19
&=
(256-1)(16-2)(16+3)=
(256-1)(256+16-6)
\\
&>(256-1)(256+2)=256^2 + 256-2\\
&>256^2=(2^8)^2=2^{16}\ ,
\\
10\cdot 18\cdot 20 &= 3600
\\
&>2048=2^{11}\ ,
\\[3mm]
&\text{Putting all together:}
\\[3mm]
20!
&=2\cdot 4\cdot 8\cdot 16
\cdot(3\cdot 12)
\cdot(5\cdot 13)
\cdot(6\cdot 7\cdot 9\cdot 11)
\\
&\qquad\qquad\qquad
\cdot(14\cdot 15\cdot 17\cdot 19)
\cdot(10\cdot 18\cdot 20)
\\
&>2^{\displaystyle1+2+3+4+5+6+12+16+11}
\\
&= 2^{\displaystyle60}\ .
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I solve $\int \frac{20}{(x-1)(x^2+9)}dx$ I've been trying to solve the following integral:
$\int \frac{20}{(x-1)(x^2+9)}dx$
Sadly I'm kinda new to resolving fractional integrals and I'm not sure which method(s) I should use to solve it.
I've tried using partial fractions but I'm doing something incorrectly or maybe this method isn't the best suited for this case.
I've tried using partial fractions. Here is what I've got so far
| We should resolve the integrand as below: $$\frac{20}{(x-1)\left(x^{2}+9\right)} \equiv \frac{A}{x-1}+\frac{B x+C}{x^{2}+9}$$
Then $$20 \equiv A\left(x^{2}+9\right)+(B x+C)(x-1)$$
Putting $x=1$ yields
$$
\begin{aligned}
20=A(10) & \Rightarrow A=2 \\
(B x+C)(x-1) &=20-2\left(x^{2}+9\right) \\
&=2-2 x^{2} \\
&=-2(x+1)(x-1) \\
\therefore \quad B x+C &=-2(x+1) \\
\therefore \frac{20}{(x-1)\left(x^{2}+9\right)} &=\frac{2}{x-1}-\frac{2(x+1)}{x^{2}+9} \\
\int \frac{20}{(x-1)\left(x^{2}+9\right)} d x &=2 \int \frac{d x}{x-1}-\int \frac{2 x d x}{x^{2}+9}-2 \int \frac{d x}{x^{2}+9}
\end{aligned}
$$
Wish it helpful for you to continue!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I can then just apply the definition of the area of a surface
$$A(S) = \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy$$
I consider the parametrization $\Phi (x,y) = (x, y, \sqrt{x^2 + y^2 - 1}) \ $. Then $$\begin{cases} \Phi_x = (1,0,\displaystyle \frac{-x}{\sqrt{x^2 + y^2 - 1}}) \\ \Phi_y = (0,1,\displaystyle \frac{-y}{\sqrt{x^2 + y^2 - 1}})\end{cases} \Longrightarrow \Phi_x \times \Phi_y = (\frac{x}{\sqrt{x^2 + y^2 - 1}},\frac{y}{\sqrt{x^2 + y^2 - 1}},1)$$
Then
$$|| \Phi_x \times \Phi_y||= \displaystyle \sqrt{\frac{x^2}{x^2 + y^2 - 1} + \frac{y^2}{x^2 + y^2 - 1} + 1} = \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1} $$
As we work in a symettric surface, we'll consider $z \in [0, \sqrt 3]$ and simply multiply the result by two. Then, the parametrization goes from $D$ to $\mathbb R^3$, $\Phi : D \subset \mathbb R^2 \rightarrow \mathbb R^3$, being $D$ the following domain
$$D = \lbrace (x,y) \in \mathbb R^2 : 1 \leq x^2 + y^2 \leq 4 \rbrace$$
Thus we get
$$A(S) = 2 \cdot \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy = 2 \cdot \iint_D \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1}\ dx \ dy $$
Using polar coordinates, $\begin{cases} x = r \cdot \cos \theta \\ y = r \cdot \sin \theta \end{cases} : r \in [1,2] \ \& \ \theta \in [0, 2\pi]$ we get the following integral
$$A(S) = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \displaystyle \sqrt{\frac{r^2 \cos^2 \theta + r^2 \sin^2 \theta}{r^2 \cos^2 \theta + r^2 \sin^2 \theta - 1} + 1} \ dr \ d\theta = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr \ d\theta$$
$$ = 4 \pi \cdot \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr $$
The problem is that I reach the integral above that I don´t know how to tackle. I think I may have done something wrong along the process since this is a question extracted from an university exam where no computers nor calculators were avilable. Any help?
| The surface is
$ x^2 + y^2 - z^2 = 1 $
Its standard parameterization is
$ P = (x, y, z) = ( \sec t \cos s , \sec t \sin s , \tan t ) $
So the surface area is
$ \text{A} = \displaystyle \int_{t = - \frac{\pi}{3} }^{ \frac{\pi}{3} } \int_{s = 0}^{2 \pi} \| P_t \times P_s \| \ d s \ d t $
And we have
$ P_t = (\sec t \tan t \cos s , \sec t \tan t \sin s , \sec^2 t ) $
$ P_s = (- \sec t \sin s , \sec t \cos s , 0 ) $
So that
$ P_t \times P_s = ( - sec^3 t \cos s , - \sec^3 t \sin s , \sec^2 t \tan t) $
And
$ \| P_t \times P_s \| = | \sec^2 t | \sqrt{ \sec^2 t + \tan^2 t } = \sec^2 t \sqrt{ 2 \tan^2 t + 1 } $
Therefore, the surface area is (using the substituting $u = \tan t $
$ \text{Area} = 2 \pi \displaystyle \int_{u = -\sqrt{3}}^{\sqrt{3}} \sqrt{ 2u^2 + 1} \ du $
Using the trigonometric substitution $ \sqrt{2} u = \tan \theta $ , then the integral becomes,
$ \displaystyle \int \dfrac{1}{\sqrt{2}} \sec^3 \theta d \theta $
From the tables,
$ \displaystyle \int \sec^3 \theta \ d \theta = \dfrac{1}{2} \bigg( \sec \theta \tan \theta + \ln \bigg| \sec \theta + \tan \theta \bigg| \bigg) $
Evaluating this between $\theta_1 = \tan^{-1}( -\sqrt{6} )$ and $\theta_2 = \tan^{-1}( \sqrt{6} )$ gives
$ \displaystyle \int_{u = -\sqrt{3}}^{\sqrt{3}} \sqrt{ 2u^2 + 1} \ du = \dfrac{1}{2\sqrt{2}} \bigg( 2 \sqrt{42} + \ln\bigg( \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7} - \sqrt{6} } \bigg) \bigg) $
Therefore, the area is
$ \text{Area} = \dfrac{\pi}{\sqrt{2}}\bigg( 2 \sqrt{42} + \ln\bigg( \dfrac{\sqrt{7}+\sqrt{6}}{\sqrt{7} - \sqrt{6} } \bigg) \bigg) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge4$
Let $a,b,c>0$. Prove that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{ab+bc+ca}{a^2+b^2+c^2}\ge4$$
I know $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge 3$ but $\dfrac{ab+bc+ca}{a^2+b^2+c^2}\le1$. And then I try $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge \dfrac{(a+b+c)^2}{ab+bc+ca}$ but still stuck. Any help please, thank you.
| Alternative approach with hint :
We have the inequality for $a,b,c>0$:
$$\frac{\left(ab+bc+ca\right)}{a^{2}+b^{2}+c^{2}}+\left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)\geq \left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)+\frac{9}{\left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)^{2}}$$
Remains to show the inequality for $x\ge 3$ :
$$x+9/x^2\geq 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Let $x = (0, 1)$ and $y = (−2, a)$ be two vectors in $\Bbb R^2$, where $a$ is a real number.
Problem:Let $x = (0, 1)$ and $y = (−2, a)$ be two vectors in $\Bbb R^2$, where $a$ is a real number.
Attempt: Please be nice.
(a) Compute the quantity $\frac{x · y}{||x||||y||}$, in terms of $a$.
$$
x·y=0(-2) + 1(a)=a
$$
$$
||x||=\sqrt{0^2+1^2}=1
$$
$$
||y||=\sqrt{-2^2+a^2}=\sqrt{4+a^2}
$$
$$
\cos θ=\frac{x · y}{||x|| ||y||}=\frac{a}{\sqrt{4+a^2}}
$$
$$
θ=\cos^{-1}\frac{a}{\sqrt{4+a^2}}
$$
(b) Determine all values of $a$ for which the angle between $x$ and $y$ is $\frac{π}{3}$(radians).
Using a reference angle, where I have <-2,a> with an angle of $\frac{1}{2}$ (because $\frac{π}{3}$ is 60 degrees and in unit cirlce that's $\frac{1}{2}$)
$$
tan(\frac{1}{2})=\frac{a}{-2}
$$
$$
(-2)tan(\frac{1}{2})=\frac{a}{-2}(-2)
$$
$$
(-2)tan(\frac{1}{2})=a
$$
$$
=-0.0174
$$
$$
0=a
$$
You guys have been giving me hints. But I just can't figure it out. a component is still missing. I don't know anymore.
$$
cos(\frac{1}{2})=\frac{a}{\sqrt{4+a^2}}
$$
$$
(\sqrt{4+a^2})cos(\frac{1}{2})=\frac{a}{\sqrt{4+a^2}}(\sqrt{4+a^2})
$$
$$
(\sqrt{4+a^2})cos(\frac{1}{2})=a
$$
EDITED
| The angle and the dot product are related by $\cos\theta=(x\cdot y)/(\lVert x\rVert\lVert y\rVert)$. We're given $\theta=\pi/3$, and we know $\cos(\pi/3)=1/2$. And you've calculated $(x\cdot y)/(\lVert x\rVert\lVert y\rVert)=a/\sqrt{4+a^2}$. So we have this equation:
$$\frac12=\cos(\pi/3)=\cos\theta=\frac{x\cdot y}{\lVert x\rVert\lVert y\rVert}=\frac{a}{\sqrt{4+a^2}}$$
$$\frac12=\frac{a}{\sqrt{4+a^2}}$$
Squaring:
$$\left(\frac12\right)^2=\frac{a^2}{\sqrt{4+a^2\,}^2}$$
$$\frac14=\frac{a^2}{4+a^2}$$
Multiplying by $4$:
$$1=\frac{4a^2}{4+a^2}$$
Multiplying by $(4+a^2)$:
$$4+a^2=4a^2$$
Subtracting $a^2$:
$$4=3a^2$$
Dividing by $3$:
$$\frac43=a^2$$
$$a^2=\frac43=\left(\frac{2}{\sqrt3}\right)^2$$
It is a general fact that, if $a=b$ or $a=-b$, then $a^2=b^2$; and conversely, if $a^2=b^2$, then either $a=b$ or $a=-b$. I think you can find a proof of this elsewhere. Thus:
$$a=\pm\frac{2}{\sqrt3}$$
So we have two possible answers, but we need to check that they actually work:
$$\frac12\overset?=\frac{\pm2/\sqrt3}{\sqrt{4+(\pm2/\sqrt3)^2}}$$
$$=\frac{\pm2/\sqrt3}{\sqrt{4+(4/3)}}$$
$$=\frac{\pm2/\sqrt3}{\sqrt{16/3}}$$
$$=\frac{\pm2/\sqrt3}{4/\sqrt{3}}$$
$$=\pm\frac12$$
Thus only the upper sign works: $a=+2/\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I show that $x$ and $y $ must have the same length where $ x+y $ and $x-y $ are non-zero vectors and perpendicular?
Problem: Let $x$ and $y$ be non-zero vectors in $\mathbb{R}^n$.
(a) Suppose that $\|x+y\|=\|x−y\|$. Show that $x$ and $y$ must be perpendicular.
(b) Suppose that $x+y$ and $x−y$ are non-zero and perpendicular. Show that $x$
and $y$ must have the same length.
Attempt:
(a)\begin{align*}\|x+y\|^2 & =\|x-y\|^2 \\
(x+y)\cdot (x+y) & =(x-y)\cdot (x-y) \\
\|x\|^2+\|y\|^2+2x\cdot y & =\|x\|^2+\|y\|^2-2x\cdot y \\
2x\cdot y & =-2x\cdot y \\
x\cdot y & =0.
\end{align*}(b)\begin{align*}(x+y)\cdot (x-y) & =0 \\
\|x\|^2-x\cdot y+x\cdot y-\|y\|^2 & =0 \\
\|x\|^2-\|y\|^2 & =0.
\end{align*}Are these attempts correct?
EDITED
| $a)$ Using polarization identity
$4\langle x, y\rangle =\|x+y\|^2-\|x-y\|^2=0$
$\langle x, y\rangle =x\cdot y=0$
$b)$ Again using polarization identity
$\begin{align}4\langle x+y, x-y\rangle &=\|(x+y)+(x-y)\|^2-\|(x+y)-(x-y)\|^2\\&=4(\|x\|^2 -\|y\|^2)\end{align}$
Given $\langle x+y, x-y\rangle=0$ implies $\|x\|=\|y\|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Is the $(3,4,5)$ triangle the only rectangular triangle with this property? While solving a loosely related exercise, by luck I found out that the $(3,4,5)$ triangle has the following property:
The product of the lengths ($\sqrt{2}$ and $\sqrt{5}$) of the two shorter line segments from a corner to the center of the inscribed circle equals the length ($\sqrt{10}$) of the longest one.
Somewhat satisfied with this, for me own new found, result I now wonder if any other rectangular $(a,b,c)$ triangles have this particular property. $(a,b,c)$ does not need to be a Pythagorean triple (but it would be extra nice).
It tried some straightforward algebraic equations but failed to find answer ... Maybe finding non rectangular such triangles is easier, but ideally I ask for rectangular ones.
update
Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$?
is already linked to this one but I take the freedom to explicitly mention it here in post for following reasons
*
*the question asked there is about generalizing answer given here
*the answers to both questions always left some exercises for reader
*myself I am not able (I continue to try) to do these exercises
Maybe someone can fully write out the missing gaps.
| In a Pythagorean right triangle $\triangle ABC$, we know that $a^2 + b^2 = c^2$ where $a, b, c$ are positive integers. We also know that $|\triangle ABC| = rs$, where $r$ is the inradius and $s = (a+b+c)/2$ is the semiperimeter. Thus we have $$\begin{align}
r &= \frac{ab}{a+b+c} \\
&= \frac{ab}{a+b+\sqrt{a^2+b^2}} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{(a+b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{2ab} \\
&= \frac{1}{2}(a+b-c) \\
&= s-c.
\end{align}$$ Denoting $I$ as the incenter, the respective distances from the incenter to the vertices are $$IA = \sqrt{r^2 + (s-a)^2}, \\ IB = \sqrt{r^2 + (s-b)^2}, \\ IC = \sqrt{r^2 + (s-c)^2} = r \sqrt{2}.$$
Then assuming $a < b < c$, we require $IB \cdot IC = IA$, or $$\begin{align}
0 = IB^2 \cdot IC^2 - IA^2 = \left(r^2 + (s-b)^2\right)(2r^2) - \left(r^2 + (s-a)^2\right).
\end{align}$$ I leave it as an exercise to show that this condition is nontrivially satisfied if and only if $b = (a^2-1)/2$, hence $a$ must be an odd positive integer for $b$ to be an integer. Then $c$ will automatically be an integer since $$c^2 = a^2 + b^2 = \left(\frac{a^2+1}{2}\right)^2.$$ Therefore, the solution set is parametrized by the triple
$$(a,b,c) = \bigl(2r+1, 2r(r+1), 2r(r+1)+1\bigr), \quad r \in \mathbb Z^+,$$ where $r$ is the inradius of such a triangle. In particular, this leads to the triples
$$(3,4,5), \\ (5,12,13), \\ (7,24,25), \\ (9,40,41), \\ \ldots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find angle $\alpha$ in equation containing weighted sum of $\sin(\alpha)$ and $\cos(\alpha)$. The context of this question comes from calculating the Fidelity drop of quantum systems after applying a rotation with some arbitrary angle $\alpha$ on the state vector of the system, see equation below:
$$\begin{eqnarray}
\Delta F = &sin(\alpha)&(\frac{(a_{1} + a_{4})(b_{2} + b_{3}) - (a_{2} + a_{3})(b_{1} + b_{4})}{2}) \nonumber \\
&-&cos(\alpha)(\frac{(b_{2} + b_{3})^{2} + (a_{2} + a_{3})^{2} - F}{2}) \nonumber \\
&+&(\frac{(b_{2} + b_{3})^{2} + (a_{2} + a_{3})^{2} - F}{2}) \nonumber
\end{eqnarray}$$
$\Delta F$, $F$, $a_i$ and $b_j$, $1\leq i,j\leq 4$ are known.
Could someone explain to me how I could get the angle $\alpha$, if that is even possible?
I will be very grateful for your help!
I have another similar problem, but with a weighted sum of $(cos(\alpha))^{2}$, $(sin(\alpha))^{2}$, and $cos(\alpha)sin(\alpha)$ in which I could also get some help.
| Let us have the general (and prettier) equation $$B=m\sin\alpha -n\cos \alpha+K.$$
Transpose $K$ and divide the whole equation by $\sqrt{m^2+n^2}$:$$\frac{B-K}{\sqrt{m^2+n^2}}=\frac{m}{\sqrt{m^2+n^2} }\sin\alpha-\frac{n}{\sqrt{m^2+n^2} }\cos\alpha$$
Now, $\Bigg |\dfrac {m}{\sqrt{m^2+n^2}}\Bigg |<1$ and so there must exist some angle A such that $\cos A= \dfrac {m}{\sqrt{m^2+n^2}}$ and also $\sin A= \dfrac {n}{\sqrt{m^2+n^2}}$ and $\tan A=\dfrac nm$. We have now,
$$\frac{B-K}{\sqrt{m^2+n^2}}=\sin\alpha\cos A-\cos\alpha\sin A=\sin (\alpha-A).$$ Thus, $$\alpha=A+\arcsin\left(\frac{B-K}{\sqrt{m^2+n^2}}\right)$$$$=\arctan\frac nm+ \arcsin\left(\frac{B-K}{\sqrt{m^2+n^2}}\right).$$
EDIT: Comparing the general equation to the OP’s equation, we have $K=n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?
| I wouldn't say this method helps you in solving it in 30 seconds, but I think it can help you in simplifying the calculations so that the integral can be computed faster
First consider by usage of the Binomial Theorem
$$(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(1)^k(-x)^{5-k}=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{5-k}$$
from which you can get by multiplying by $x^4$
$$x^4(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{9-k}$$
applying the integral to the sum
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{x^{10-k}}{10-k} \bigg)_{0}^{1}$$
which will give
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{1}{10-k} \bigg)$$
Expanding the sum gives
$$-\binom{5}{0}\bigg(\frac{1}{10}\bigg)+\binom{5}{1}\bigg(\frac{1}{9}\bigg)-\binom{5}{2}\bigg(\frac{1}{8} \bigg)+\binom{5}{3}\bigg(\frac{1}{7}\bigg)-\binom{5}{4}\bigg(\frac{1}{6} \bigg)+\binom{5}{5}\bigg(\frac{1}{5} \bigg)$$
simplifying will give
$$\frac{-1}{10} +\frac{5}{9} -\frac{10}{8} +\frac{10}{7} -\frac{5}{6} +\frac{1}{5}=\frac{1}{1260}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Which interval is correct here$?$ The equation $$2\textrm{sin}^2\theta x^2-3\textrm{sin}\theta x+1=0$$ where $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$ has one root lying in the interval
$(0,1)$
$(1,2)$
$(2,3)$
$(-1,0)$
I know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in general odd number of roots of the equation $f(x)=0$ lie between $a$ and $b$. But I am not able to use this piece of information here, maybe because of two variables. I also tried to assume $\textrm{sin}\theta x$ as $y$ but nothing good came out.
Any help is greatly appreciated.
EDIT Answer given is $(1,2)$
| Use the quadratic formula to solve for $x$:
$$x_\pm = \frac{3\sin\theta\pm\sqrt{9\sin^2\theta-8\sin^2\theta}}{4\sin^2\theta} = \begin{cases}\frac{1}{\sin\theta} & \text{ or } \\ \frac{1}{2\sin\theta} & \end{cases}$$
because $\sin\theta\in\left( \frac{1}{\sqrt 2},1\right)$ for $\theta\in\left( \frac{\pi}{4},\frac{\pi}{2}\right)$. Thus, the two roots of $x$ lie in the intervals
$$x_+ = \frac{1}{\sin\theta}\in \left( 1,\sqrt 2\right) \subseteq (1,2) \\
x_- = \frac{1}{2\sin\theta}\in \left( \frac{1}{2},\frac{1}{\sqrt 2}\right) \subseteq (0,1)$$
The claim that $x$ only has one root is wrong.
| {
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"url": "https://math.stackexchange.com/questions/4495411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Inverse in Modular Exponent Properties I have a question about modular exponentiation that I would be very grateful to get some help with.
Assuming we have the values $x, a, r$ and the inverse of $a$ as $-a$ all under $mod \:N$. I know that the following property holds:
$(({x^a} \:mod \: N)^r \: mod \: N)^{-a} \: mod \: N == x^{a \cdot r \cdot -a} \: mod \: N$
What I am trying to understand is if $(({x^a} \:mod \: N)^r \: mod \: N)^{-a} \: mod \: N == x^r \: mod \: N$
I thought this property would hold since $a\cdot-a \: mod \: N= 1$ but I may be missing something here since I get a different result when testing this. I think it may be because $a \cdot -a$ is only equal to $1$ when under the modular operation, which it never is in the exponent, but also I could be way off. Any help would be appreciated!
| You are confusing your self.
Short answer we do NOT have $a^{m} \equiv a^{m\pmod N} \pmod N$. Mod equivalences don't work on exponents as remainders are not preserved over exponents. And simple example we try will fail. Take for instance $2\pmod 5$. then $2^2 \equiv 4\pmod 5$ and $2^3\equiv 8\equiv 3\pmod 5$ and $2^4\equiv 16\equiv 1 \pmod 5$ and $2^5\equiv 32 \equiv 2 \pmod 5$ and $2^6 \equiv 64 \equiv 4\pmod 5$. We do NOT have $32=2^5\equiv 2^{5\mod 5}\equiv 2^0\equiv 1 \pmod 5$ nor do we have $64\equiv 2^6\equiv 2^{6\mod 5} \equiv 2^1 \equiv 2 \pmod 5$.
We just don't.
But we do have something close.
Fermat's Little thereom: If $p$ is prime and $a$ is not a multiple of $p$ then $a^{p-1}\equiv 1 \pmod p$ and so;
$a^{k\pmod {p-1}} \equiv a \pmod p$ and so if we have $k \equiv -a \pmod {p-1}$ then we do have $a^a \cdot a^r \cdot a^k \equiv a^a\cdot a^r\cdot a^{-a}\equiv a^{a+r-a\pmod {p-1}}\equiv a^r \pmod p$.
For example. Take $\mod 13$. Not that $4 \equiv -8 \pmod{12}$. So we ought to have $a^4 \cdot a^4\cdot a^8\equiv a^4 \pmod {13}$. And we do. For example if $a =3$ and $r = 5$ we have $a^4=3^4=81\equiv 3\pmod {13}$. And $3^8=6561\equiv 9\pmod{13}$. So $3^4\cdot 3^r \cdot e^8 \equiv 81\cdot 3^r \cdot 6561 \equiv 3\cdot 3^r \cdot 9\pmod 13 \equiv 27\cdot 3^r\equiv 1\cdot 3^r\pmod {13}$
We also have Eulers th: that if $\gcd(n,b) =1$ then $b^{\phi(n)} = 1$ and if then we would have $b^{a\pmod{\phi(n)}}\cdot b^r \cdot b^{-a\pmod{\phi(n)}} \equiv b^r\pmod{\phi(n)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the measure of angle $x$ in the triangle below?
Any point $D$, which is interior to the triangle $\triangle ABC$, determines vertex angles
having the following measures:
$m(BAD) = x,$
$m(ABD) = 2x,$
$m(BCD) = 3x,$
$m(ACD)= 4x,$
$m(DBC) = 5x.\\$
Find the measure of $x$.
(Answer:$10^\circ$)
My progress:
$\dfrac{AB}{AD}=\dfrac{\sin(180−3x)}{\sin(2x)}\\
\dfrac{AC}{AD}=\dfrac{\sin(11x)}{\sin(4x)}
AB=AC \implies
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}$
but it is a complex equation to solve. Is there another way?
| $\angle(CDB)=5x<\pi\implies x<\dfrac{\pi}{5} $
$\therefore 0<x<\dfrac{\pi}{5}\\
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}\\
\dfrac{\sin(3x)}{ {\sin(2x)}}=\dfrac{\sin(11x)}{2{\sin(2x)}\cos(2x)}\\
2\sin(3x)\cos(2x)=\sin(11x)\\
\sin(5x)+\sin(x)=\sin(11x)\\
\sin(5x)=\sin(11x)-\sin(x)\\
{\sin(5x)}=2\cos(6x){\sin(5x)}\\
\cos(6x)=\dfrac{1}{2}\\
6x=\dfrac{\pi}{3}\\
x=\dfrac{\pi}{18}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing the value of $(x+y)^4$ if $x^4+y^4=5$ and $x^2+xy+y^2=10$
Let $x^4+y^4=5$ and $x^2+xy+y^2=10.$ Find $(x+y)^4.$
First, I tried expanding $(x+y)^4$ using the binomial theorem to get $5+4x^3y+6x^2y^2+4xy^3,$ so simplifying I got $5+4xy(x^2+y^2)+6(xy)^2.$ Then I rearranged the given equation to get $x^2+y^2=10-xy,$ so the expansion becomes $5+4xy(10-xy)+6(xy)^2.$ I further simplified to get $2x^2y^2+40xy+5,$ but I'm not sure how to continue off here. Factoring this expression doesn't seem to help. May I have some help? Thanks in advance.
| We have $95 = (x^2+xy+y^2)^2 - (x^4+y^4) = 2x^3 y + 2xy^3 + 3x^2 y^2,$ which you may recognize. Any further and I'd just give away the entire solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality $3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}$ I tried to group the summands so that I could decompose them into multipliers, but nothing worked...
$$3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}\Leftrightarrow 3^{x^2+6x+9}+\frac{1}{9}\leq 3^{x^2-2}+3^{6x+9}$$
$$3^{x^2+6x+7}+1\leq 3^{x^2-4}+3^{6x+7}$$
How to solve it further?
| Hint: You can rewrite the inequality as follows: $(3^{6x+11}-1)(3^{x^2}-1) \le 0$. Can you take it from here?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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} |
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
| What about this : We have for example $$x^{10}+x^{12}=x^{10}(x^2+1)\equiv 0\mod x(x^2+1)$$
This way we can also cancel $11-13,14-16,15-17,18-20$. It remains $x^{19}$ for which you can use $x^6\equiv x^2$ giving $x^{18}\equiv x^6$ hence $x^{18}\equiv x^2$ hence $x^{19}\equiv x^3\equiv -x$ $\mod (x^3+x)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a horizontal line that divides a modulus function's area into 2 equal halves
How to find the horizontal line $y=k$ that divides the area of $y=2-\lvert x-2\rvert $ above the $x$-axis into two equal halves using Calculus?
I know of methods other than Calculus to solve this, but my question requires to solve it using Calculus (Definite Integration).
I've tried to solve it using calculus after watching youtube video solutions of the same type of question where the function of interest were parabolas or other curves. Still couldn't get to the right solution. I'm assuming I've made mistakes in taking the limits of integration as this an absolute value function and not a parabola.
How to solve this question using calculus? If there are ways to solve this by integrating the function with respect to $y$ (instead of $x$) which I have seen people doing while solving questions regarding parabola functions, then do let me know!
Note: I'm a high school senior who has been sitting with this problem since morning & I seriously cannot seem to get through and get over it. Helping me solve this question will broaden my knowledge of Calculus & will clear my misconceptions.
I've attached an image to the post as well.
The answer is $y=2-\sqrt{2}$
Thanks
| The area above the $x$ axis is
$$
A =\int_0^4 2 - \lvert x-2 \rvert \, \mathrm{d}x = \int_0^2 2 -(2-x)\, \mathrm{d}x + \int_2^{4}2 -(x-2) \, \mathrm{d}x = 4 \tag{1}
$$
Now, if we draw a horizontal line $\color{green}{y=k}$ on top of the function $y= 2 - \lvert x-2 \rvert$ it splits the latter area in $2$ halves:
In the above diagram the line $\color{green}{y=k}$ is drawn in green, and the $2$ halves are drawn in $\color{orange}{\text{orange}}$ and $\color{blue}{\text{blue}}$ respectively. The question asked is
Find the horizontal line $\color{green}{y=k}$ that divides the area of $y= 2 - \lvert x-2 \rvert$ above the $x$-axis into two equal halves.
Which in the diagram above translates to finding a $k$ such that
$$
\text{Area of }\color{orange}{\text{orange}} \text{ half}= \text{Area of }\color{blue}{\text{blue}} \text{ half} = \frac{A}{2} \overset{(1)}{=} \frac{4}{2} = 2 \tag{2}
$$
So now the question becomes writing the $\color{orange}{\text{orange}}$ and
$\color{blue}{\text{blue}}$ areas using calculus, then setting each of those equal to $2$, and then (hopefully) solving for $k$ from these equations using calculus techniques.
*
*We first calculate some key points relevant to calculating the $\color{orange}{\text{orange}}$ and $\color{blue}{\text{blue}}$ areas. Namely, what are the coordinates of the intersections of $y=k$ with $y = 2 - \lvert x-2 \rvert$?
We achieve these coordinates by setting $y=k$ equal to $y = 2 - \lvert x-2 \rvert$ and solving for $k$ $$k = 2 - \lvert x-2 \rvert \implies \begin{cases} k = 2- (2-x) &\implies x = \color{purple}{k}\\ k = 2 -(x-2) & \implies x =\color{purple}{4-k} \end{cases}$$
*Once you have the coordinates, how can you write an integral for the area of the $\color{blue}{\text{blue}}$ half?
We first notice that we can split the $\color{blue}{\text{blue}}$ region in $3$ parts: $A,B$ and $C$: Each of these regions is an area between the $x$ axis and a function, so we can calculate each of the areas $A$, $B$ and $C$ as integrals as follows $$\color{blue}{\text{Blue}}\text{ area}=\underbrace{\int_0^k \color{red}{2-\lvert x-2\rvert}\, \mathrm{d}x}_{A} + \underbrace{\int_k^{4-k} \color{green}{k} \, \mathrm{d}x}_{B} +\underbrace{\int_{4-k}^4 \color{red}{2-\lvert x-2\rvert}\, \mathrm{d}x}_{C} \tag{3}$$
*Since equation $(2)$ tells us that $\color{blue}{\text{Blue}}\text{ area} = 2$, how can we combine this with the previous step to get an equation for $k$?
Since $\color{blue}{\text{Blue}}\text{ area} = 2$, combining equations $(2)$ and $(3)$ we get \begin{align}&2 = \int_0^k 2-\lvert x-2\rvert\, \mathrm{d}x + \int_{k}^{4-k} k \, \mathrm{d}x + \int_{4-k}^4 2-\lvert x-2\rvert\, \mathrm{d}x \\ \implies & 2 = \int_{0}^{k} x \, \mathrm{d}x + k((4-k) -k) + \int_{4-k}^{4} 4-x\, \mathrm{d}x \\ \overset{\color{purple}{u = 4-x}}{\implies} & 2 = \frac{k^2}{2} + 4k - 2k^2 + \int_{0}^{k} \color{purple}{u}\, \mathrm{d}\color{purple}{u} \\ \implies & 2 = - \frac{3}{2} k^2 + 4k + \frac{k^2}{2} \\ \implies & 2 = 4k -k^2\end{align}
*How can we solve the previous equation to obtain a value for $k$?
Since the previous equation is a quadratic equation in $k$, it can be solved using the quadratic formula. We thus get \begin{align} &2 = 4k -k^2 \\ \implies & k^2 - 4k +2 =0 \\ \implies & k = \frac{4 \pm \sqrt{16 - 4(1)(2)}}{2} = \frac{4 \pm \sqrt{2 \cdot 2^2}}{2} = \frac{4 \pm 2\sqrt{2 }}{2} = 2 \pm \sqrt{2} \end{align} Now, for the line $y=k$ to split the function $y= 2 - \lvert x - 2\rvert$ into the $\color{orange}{\text{orange}}$ and $\color{blue}{\text{blue}}$ halves from the first diagram, $k$ must be between $0$ and $2$ since $2$ is the highest $y$ value of $y= 2 - \lvert x - 2\rvert$. Out of the two possibilities of sign choice in $2 \pm \sqrt{2}$ only $2 - \sqrt{2}$ is between $0$ and $2$, so we can conclude that the value of $k$ we want is $$ \boxed{y = k = 2 - \sqrt{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum^n_{k=0} k^2 {n\choose k} = (n+n^2)2^{n-2}$ We can start with the expression of the binomial expansion, $\sum^n_{k=0} {n\choose k} x^ky^{n-k}= (x+y)^n$. Setting $x=y=1$ gives $\sum^n_{k=0} {n\choose k} = 2^n$
Differentiating both sides with respect to $x$ gives $\sum^n_{k=0} k{n\choose k} x^{k-1}y^{n-k} = n(x+y)^{n-1}$, and substituting $x=y=1$ gives $\sum^n_{k=0} k{n\choose k} = n2^{n-1}$
Differentiating again gives $\sum^n_{k=0} k^2{n\choose k} x^{k-2}y^{n-k} = (n^2-n)(x+y)^{n-2}$, and substituting $x=y=1$ gives $$\sum^n_{k=0} k^2{n\choose k} = (n^2-n)2^{n-2}$$
However, the identity is $$\sum^n_{k=0} k^2{n\choose k} = (n^2+n)2^{n-2}$$
I don't seem to have lost a sign somewhere, where is my mistake?
| You have written the second derivative of $x^{k}$ as $k^{2}x^{k-2}$. It is $k(k-1)x^{k-2}$.
If you add $n2^{n-1}$ to $(n^{2}-n)2^{n-2}$ you get $(n^{2}+n)2^{n-2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}
Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}
Is it correct? How to compute $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
| An approach using the Lagrange inversion theorem: let $f(z)$ be analytic around $z=z_0$ with $f'(z_0)\neq0$; then $w=f(z)$ has an inverse $z=g(w)$ analytic around $w=f(z_0)$ with $$g^{(n)}\big(f(z_0)\big)=\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z-z_0}{f(z)-f(z_0)}\right)^n.\qquad(n>0)$$
We apply this to $f(z)=\dfrac{\sqrt{z}-1}{\sqrt{z}+1}$ and $z_0=1$, thus $g(w)=\left(\dfrac{1+w}{1-w}\right)^2$ and $$\lim_{z\to1}\frac{d^n}{dz^n}(1+\sqrt{z})^{2n+2}=g^{(n+1)}(0)=\color{blue}{4(n+1)!(n+1)}$$ since $g(w)=1+\displaystyle\frac{4w}{(1-w)^2}=1+4w\frac{d}{dw}\frac1{1-w}=1+4\sum_{n=1}^\infty nw^n$ for $|w|<1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ and
$\frac{1}{x}=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$
How can I find $x^{2022}+\frac{1}{x^{2022}}$?
| The solutions you got are (primitive) eighth roots of unity, $\zeta_8=e^{2\pi i/8},\dfrac 1{\zeta_8}=\bar {\zeta _8}$.
Now $$\zeta _8^{2022}=(\zeta _8^8)^{252}\cdot \zeta _8^6=1\cdot \zeta _8^6=\zeta_8^6$$. And $\dfrac 1{\zeta_8^{2022}}=\zeta_8^{-6}$.
So $$\zeta_8^6+\zeta_8^{-6}=-i+i=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c^2}=\frac{s^4}{c^2}=s^2t^2.$$
And denominator is,
$$\frac{2t}{1-t^2}-2t=\frac{2t^3}{1-t^2}.$$
So $$\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}=s^2t^2\times\frac{1-t^2}{2t^3}=\sin^2x\times \frac1{\tan 2x}=\frac{1-\cos 2x}{2\tan 2x}.$$
I'm looking for alternative approaches to simplify the expression.
| $\begin{align}\tan^2 x-\sin^2 x&=\sin^2 x(\sec^2 x-1) \\&=\sin^2 x\tan^2x\space \tag1\end{align}$
$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\tan 2x-2\tan x=\tan 2x\tan^2 x\space \tag2$$
From (1) and (2),
$$\begin{align}\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}&=\frac{\sin^2 x\tan^2x}{\tan 2x\tan^2 x}\\&=\frac{\sin^2x}{\tan 2x}\end{align}$$
which is equal to the given answer. (put $\sin^2x=\frac{1}{2}(1-\cos 2x)$)
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove if $\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}=\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}=\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}$, then $x=y=z$. Let $x$, $y$, $z$ be real numbers satisfying $$
\begin{align}
&\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}\\
=&\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}\\
=&\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}.
\end{align}$$
Prove that $x=y=z$.
I tried assuming $x>y>z$, $x>y=z$,$x<y<z$, etc., but none of the directions work. Please help me solve this problem.
| The solution of tehtmi is wonderful, and I have a similar approach.
For each parameter $t \in \{x, y, z\}$ and each $1 \leq i \leq 3$, let $t_i = \sqrt{t + i}$. For example $x_2 = \sqrt{x + 2}$. So we have:
\begin{align*}
&x_1 + y_2 + z_3\\
=\ &y_1 + z_2 + x_3 \label{1}\tag{$*$}\\
=\ &z_1 + x_2 + y_3
\end{align*}
Suppose $x = \min\{x, y, z\}$. Note that the function $f(t) = \sqrt{t + m} - \sqrt{t + n}$ for all $m > n$ is strictly decreasing. Thus
\begin{alignat*}{2}
y_2 - y_1 &\leq x_2 - x_1 &&\implies x_1 + y_2 \leq y_1 + x_2\\
z_3 - z_2 &\leq x_3 - x_2 &&\implies z_3 + x_2 \leq z_2 + x_3\\
&\ &&\stackrel{+}{\implies} x_1 + y_2 + z_3 \leq y_1 + z_2 + x_3
\end{alignat*}
But by \eqref{1} the equal case has occurred, and the equal case occurs only for $x = y = z$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "23",
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Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1 & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;
\lambda_2 = 4
$$
Now if I try to calculate the eigenvectors
For $\lambda_1$ I get:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}0 \\ -x_1+3 x_2 \end{pmatrix}
$$
and for $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}
$$
I see (and can compute using a symbolic calculation program) that there are two eigenvectors:
$$
e1 = \begin{pmatrix}0 \\ 1\end{pmatrix}
$$
$$
e2 = \begin{pmatrix}3 \\ 1\end{pmatrix}
$$
I can easily see why $\begin{pmatrix}3 \\ 1\end{pmatrix}$ is an eigenvector for the eigenvalue $\lambda_2$.
But I am struggling to understand why $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is an eigenvector.
Could someone help me to understand why?
| For $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \end{pmatrix}\Rightarrow x_1=0\Rightarrow v_2=\begin{pmatrix}0 \\ x_2 \end{pmatrix}=x_2 \begin{pmatrix}0 \\ 1 \end{pmatrix}
$$
So your second eigenvector is $\begin{pmatrix}0 \\ 1 \end{pmatrix}$
| {
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Prove that $\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}$
If $0 < x,y < \frac {\pi}{2}$, prove that:
$$
\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}
$$
My attempt. First, I tried to change this inequality:
$$
\frac {\frac{x}{\sin x} + \frac{y}{\sin y}}{2} < \frac{1}{\cos \frac{x+y}{2}}
$$
Then,it's easy to know:
$$
LHS = \frac{1}{2} \left ( \frac {x}{2\sin \frac {x}{2} \cos \frac{x}{2}} + \frac{y}{2 \sin \frac{y}{2} \cos \frac{y}{2}} \right ) < \frac{1}{2} \left ( \frac{1}{\cos^{2} \frac{x}{2}} +\frac{1}{\cos^{2} \frac{y}{2}} \right )
$$
$$
RHS > \frac{1}{\cos \frac{x}{2} \cos \frac{y}{2}}
$$
How to solve it next? It seems this way is wrong.
| Lemma 1. $f(x)=\frac{x}{\sin x}$ is a positive, increasing and convex function on $I=\left(0,\frac{\pi}{2}\right)$.
Let us assume that $\mu = \frac{x+y}{2}\in I$ is fixed and $x\leq y$. Let us set $\delta=\frac{y-x}{2}$. By Lemma 1,
$$ \sup_{\substack{0\leq \delta < \min\left(\mu,\frac{\pi}{2}-\mu\right)\\ }}\left(\frac{\mu-\delta}{\sin(\mu-\delta)}+\frac{\mu+\delta}{\sin(\mu+\delta)}\right)$$
is achieved at the right endpoint of the range for $\delta$. If $\mu\leq\frac{\pi}{4}$ such supremum equals $1+\frac{2\mu}{\sin(2\mu)}$.
If $\mu\geq\frac{\pi}{4}$ such supremum equals $\frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}$. The inequality
$$ \frac{\pi}{2}+\frac{2\mu-\pi/2}{\sin(2\mu-\pi/2)}\leq \frac{2}{\cos\mu} $$
over the interval $\left[\frac{\pi}{4},\frac{\pi}{2}\right)$ is very loose, hence the problem boils down to showing that
$$ 1+\frac{2\mu}{\sin(2\mu)} \leq \frac{2}{\cos\mu} $$
holds over the interval $\left(0,\frac{\pi}{4}\right)$. By multiplying both sides by $\sin\mu\cos\mu$ we get that the inequality is equivalent to
$$ \sin\mu\cos\mu + \mu \leq 2\sin\mu\tag{E} $$
which follows from the termwise integration of $1+\cos(2\mu)\leq 2\cos\mu$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$x^{x^2} + x^{x^8} =?$ Given $x^{x^4} = 4$ If x is any complex number such that $x^{x^4} = 4$ , then find all the possible values of :
$x^{x^2} + x^{x^8}$
First, I used laws of exponents to give $18$ as answer. However , I realised that I've misused it. Further I used logarithms which yielded
$$ x^{x^2} + x^{x^8} = 4^{1/x^2} + 4^{x^2} $$
After this, I am stuck and can't proceed further. By hit and trial, $x = \sqrt{2}$ seems to be one of the solution. Any help is appreciated.
| This answer only works (based on the original question) on the set of real numbers.
Hint:
\begin{align}
x^{x^4}=4&\implies \left(x^4\right)^{x^4}=4^4\\
&\implies x^4=4\\
&\implies x=\pm\sqrt 2=\pm 2^{\frac 12}.\end{align}
Justification about the step $$\left(x^4\right)^{x^4}=4^4\implies x^4=4$$
We know that if $x\ge0$ and $x,y\in\mathbb R$ then the equation $ye^y=x$ has exactly one solution $W_0(x)$. Hence, we have
$$x^4\ln x^4=\ln x^4 e^{\ln x^4}=4\ln 4\ge 0$$
So, the fact $x^4=4$ is an only possible solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find range of values of $a,b,c$ such that $ax^2+bx+c$ satisfies given conditions Let $y = f(x) = ax^2+bx+c$ where $a\neq0$. Find the range of values $a,b$ and $c$ such that it satisfies the following:
*
*$0 \le f(x) \le 1 \quad \forall \space x \in [0,1]$
What I have found so far?
*
*For $x=0$, we get $0 \le c \le 1$
*For $x=1$, we get $0 \le a+b+c \le 1$
*If roots exist, then they should lie outside the interval $[0,1]$.
Therefore, ${{-b - \sqrt{D}}\over{2a}} < 0$ and ${{-b + \sqrt{D}}\over{2a}} >1$. I tried to solve these but didn't get any fruitful results.
EDIT 1: As explained in comments by @insipidintegrator, this is only for distinct roots.
*Vertex of quadratic $f(x)$ is $\big( \frac{-b}{2a}, -\frac{D}{4a}\big)$. If the x-coordinate of the vertex lies between 0 and 1, then y-coordinate should also be between 0 and 1. But I have no idea how to proceed from here.
I want to find some lower/upper limits for $a,b,c$ or any kind of relation between them. Any solution or hints for solving this problem is greatly appreciated.
| We may obtain the necessary and sufficient conditions for
$$a, b, c \in \mathbb{R}, ~ 0 \le ax^2 + bx + c \le 1, \forall x \in [0, 1].$$
Fact 1: Let $A, B, C \in \mathbb{R}$. Then
$$Ax^2 + Bx + C \ge 0, ~ x \ge 0
\iff A \ge 0, ~ C \ge 0, ~ B \ge -\sqrt{4AC}.$$
(The proof is given at the end.)
Fact 2: Let $A, B, C \in \mathbb{R}$. Then
$$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1] \iff
\left\{\begin{array}{l}
A + B + C \ge 0\\[5pt]
C \ge 0\\[5pt]
B + 2C \ge - \sqrt{4(A + B + C)C}.
\end{array}
\right.
$$
(The proof is given at the end.)
Now, using Fact 2, we have
$$ax^2 + bx + c \ge 0, ~ x \in [0, 1]
\iff
\left\{\begin{array}{l}
a + b + c \ge 0\\[5pt]
c \ge 0\\[5pt]
b + 2c \ge - \sqrt{4(a + b + c)c}
\end{array}
\right.$$
and
$$ - ax^2 - bx - c + 1 \ge 0,~ x \in [0, 1]$$
$$\iff
\left\{\begin{array}{l}
-a - b - c + 1 \ge 0\\[5pt]
- c + 1 \ge 0\\[5pt]
- b + 2(-c + 1) \ge - \sqrt{4(-a - b - c + 1)(-c + 1)}.
\end{array}
\right.$$
Thus, we have
$$0 \le ax^2 + bx + c \le 1, ~ x\in [0, 1]$$
$$\iff \left\{\begin{array}{l}
0 \le a + b + c \le 1\\[5pt]
0 \le c \le 1\\[5pt]
b + 2c \ge - \sqrt{4(a + b + c)c}\\[5pt]
b + 2c \le 2 + \sqrt{4(-a - b - c + 1)(-c + 1)}.
\end{array}
\right.$$
$\phantom{2}$
Proof of Fact 1:
“$\Longleftarrow$”: $\quad$ For $x\ge 0$, we have $Ax^2+Bx+C\ge Ax^2 - \sqrt{4AC}\ x + C = (x\sqrt{A} - \sqrt{C})^2\ge 0$.
“$\Longrightarrow$”: $\quad$ Clearly, $A \ge 0$ and $C \ge 0$.
If $B < - \sqrt{4AC}$ and $AC > 0$,
then $Ax_0^2 + Bx_0 + C < 0$ where $x_0 = \sqrt{\frac{C}{A}} > 0$. Contradiction.
If $B < - \sqrt{4AC}$ and $AC = 0$,
there exists $x_1 > 0$ such that $Ax_1^2 + Bx_1 + C < 0$ (easy). Contradiction.
We are done.
$\phantom{2}$
Proof of Fact 2:
By Fact 1, it suffices to prove that
$$Ax^2 + Bx + C \ge 0, ~ x\in [0, 1]
\iff (A + B + C)s^2 + (B + 2C)s + C \ge 0, ~ s \ge 0.$$
“$\Longleftarrow$”: $\quad$ Clearly, $A+B+C \ge 0$.
If $x = 1$, then $Ax^2 + Bx + C = A + B + C \ge 0$.
If $x\in [0, 1)$, letting $s = \frac{x}{1 - x} \ge 0$, we have
$x = \frac{s}{1+s}$ and
$$Ax^2 + Bx + C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0.$$
“$\Longrightarrow$”: $\quad$ For $s\ge 0$, letting $x = \frac{s}{1+s}\in [0,1]$, we have
$$Ax^2+Bx+C = \frac{1}{(1+s)^2}[(A+B+C)s^2+(B+2C)s+C] \ge 0$$
which results in
$(A+B+C)s^2+(B+2C)s+C\ge 0$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\sum_{k=0}^n(-1)^{\frac{k(k+1)}2}k$ I asked this question a few days ago, where I noticed $$\left\lfloor\frac{n}4\right\rfloor+\left\lfloor\frac{n+1}4\right\rfloor-\left\lfloor\frac{n+2}4\right\rfloor-\left\lfloor\frac{n+3}4\right\rfloor=\cos\left(\frac{n\pi}{2}\right)-1,\quad n\in\mathbb N$$
We can also write $\cos\left(\frac{n\pi}{2}\right)-1=\sum_{k=1}^n(-1)^{\frac{k(k+1)}2}$, and so I decided to try and evaluate $$\sum_{k=1}^n(-1)^{\frac{k(k+1)}2}k$$
to perhaps arrive at a another equivalence between trigonometric and floor functions(I do realise that we are only using trivial values of the cosine, but nevertheless find it rather intriguing).
I have arrived at a trigonometric relation as follows:
Since $\sum_{k=1}^nka_k=n(a_1+a_2+\dots+ a_n)-\sum_{k=1}^n(a_1+a_2+\dots+a_{k-1})$, we have (for $a_k=(-1)^{\frac{k(k+1)}2}$ )
$$\begin{align}\sum_{k=1}^nk(-1)^{\frac{k(k+1)}2}&=n\left(\cos\left(\frac{n\pi}{2}\right)-1\right)-\sum_{k=1}^n\left(\cos\left(\frac{k\pi}{2}\right)-1\right)\\
&=n\cos\frac{n\pi}{2}-\frac{\sin(\frac{n\pi}4)}{\sin\frac\pi4}\cdot\cos\frac{(n-1)\pi}4\\
&=n\cos\frac{n\pi}{2}-\frac{1}{\sqrt2}\cdot\sin\frac{(2n-1)\pi}4-\frac12
\end{align}$$
While not as concise as I had hoped, it is still a simple trigonometric equation. My problem however, lies in somehow generating a floor function evaluation for the series. Could somebody please help? Thanks in advance!
| We can write the series as $$\sum_{k=1}^n(-1)^{\frac{k(k+1)} 2}k=\sum_{4j\le n}4j+\sum_{4j-1\le n}(4j-1)-\sum_{4j-2\le n}(4j-2)-\sum_{4j-3\le n}(4j-3)$$
Since $$\sum_{4j\le n}4j=\left(4+8+\dots+4\left\lfloor\frac n4\right\rfloor\right)=\frac12\left\lfloor\frac n4\right\rfloor\left(2\cdot4+\left(\left\lfloor\frac n4\right\rfloor-1\right)\cdot4\right)$$
$$\implies\sum_{4j\le n}4j=2\left\lfloor\frac n4\right\rfloor^2+2\left\lfloor\frac n4\right\rfloor$$
We can use this result to evaluate the remaining terms as :
$$\sum_{4j-1\le n}(4j-1)=\sum_{4j\le n+1}4j-\sum_{4j\le n+1}1=2\left\lfloor\frac {n+1}4\right\rfloor^2+2\left\lfloor\frac {n+1}4\right\rfloor-\left\lfloor\frac {n+1}4\right\rfloor$$
$$\implies\sum_{4j-1\le n}(4j-1)=2\left\lfloor\frac {n+1}4\right\rfloor^2+\left\lfloor\frac {n+1}4\right\rfloor$$
Similarly,
$$\sum_{4j-2\le n}(4j-2)=2\left\lfloor\frac {n+2}4\right\rfloor^2,\sum_{4j-3\le n}(4j-3)=2\left\lfloor\frac {n+3}4\right\rfloor^2-\left\lfloor\frac {n+3}4\right\rfloor$$
Thus,
$$\sum_{k=1}^n(-1)^{\frac{k(k+1)} 2}k=2\left(\left\lfloor\frac {n}4\right\rfloor^2+\left\lfloor\frac {n+1}4\right\rfloor^2-\left\lfloor\frac {n+2}4\right\rfloor^2-\left\lfloor\frac {n+3}4\right\rfloor^2\right)\\+2\left\lfloor\frac {n}4\right\rfloor+\left\lfloor\frac {n+1}4\right\rfloor+\left\lfloor\frac {n+3}4\right\rfloor$$
$$\bbox[5px,border:2px solid #C0A000]{\begin{align} n\cos\frac{n\pi}{2}-\frac{1}{\sqrt2}\cdot\sin\frac{(2n-1)\pi}4-\frac12=&2\left(\left\lfloor\frac {n}4\right\rfloor^2 +\left\lfloor\frac {n+1}4\right\rfloor^2-\left\lfloor\frac {n+2}4\right\rfloor^2-\left\lfloor\frac {n+3}4\right\rfloor^2\right)\\&+2\left\lfloor\frac {n}4\right\rfloor+\left\lfloor\frac {n+1}4\right\rfloor+\left\lfloor\frac {n+3}4\right\rfloor \end{align}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rewrite this equation using cartesian coordiantes $x$ and $y$ In this question the $x$ and $y$ coordinates are given, however I do not know what to substitute in
$x=5t-2\;\quad y=-5t+7$
So far I have rearranged for $t$
$x+2=5t$
$\dfrac{x+2}5=t$
therefore would the next step by $y=-5\left(\dfrac{x+2}5\right)+7$ ?
| $x=5t-2$
$x+2=5t$
$\dfrac{x+2}5=t$
$y=-5t+7$
$y=-5\left(\dfrac{x+2}5\right)+7$
$y=-(x+2)+7$
$y=-x-2+7$
$y=-x+5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| HINT:
We have
$$x^2+7x+(4-2^y)=0\\
\Delta_x=2^{y+2}+33=z^2\\
$$
Let, $y=2k$, then
$$\left(z-2^{k+1}\right)\left(z+2^{k+1}\right)=33$$
and so
$$
\begin{cases}z- 2^{k+1}=\pm 1,3,11,33\\
z+ 2^{k+1}=\pm 33,11,3,1\end{cases}
$$
Then, let $y=2k-1$ we have
$$2^{2k+1}+1=(z^2+1)-33$$
Since, $2^{2k+1}+1 \mod 3=0$, putting $z=3n\pm 1 $ and $z=3n$ we get $z^2+1\mod 3 \neq 0$. Thus, this case is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as follow
$z = r{e^{i\theta }}$& $\overline z = r{e^{ - i\theta }}$
$r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = i\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right) \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = {e^{i\frac{\pi }{2}}}\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right)$
$ \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = \left( {r{e^{ - i\left( {\theta - \frac{\pi }{2}} \right)}} + {r^2}{e^{i\left( {2\theta + \frac{\pi }{2}} \right)}}} \right)$
$ \Rightarrow r\left( {\cos \theta - i\sin \theta } \right) - {r^2}\left( {\cos 2\theta + i\sin 2\theta } \right) = \left( {r\left( {\sin \theta + i\cos \theta } \right) + {r^2}\left( { - \sin 2\theta + i\cos 2\theta } \right)} \right)$
$ \Rightarrow r\cos \theta - {r^2}\cos 2\theta - r\sin \theta + {r^2}\sin 2\theta - i\left( {r\sin \theta - {r^2}\sin 2\theta - r\cos \theta - {r^2}\cos 2\theta } \right) = 0$
Not able to proceed further
| We have
$$\bar{z} - z^2 = i(\bar{z} + z^2)$$
and multiplying by $i$
$$-\bar{z} - z^2 = i(\bar{z} -z^2)$$
then summing
$$z^2=-i\bar z\iff r^2e^{i2\theta}=re^{i\left(-\theta+\frac 3 2\pi\right)}$$
from which we can conclude that $r=0$ or $r=1$ with
$$2\theta=-\theta+\frac 3 2\pi+2k\pi \iff \theta =\frac \pi 2 +\frac 2 3 k\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve $xdx-ydy=y^2(x^2-y^2)dy$ Question:
$$xdx-ydy=y^2(x^2-y^2)dy$$
I'm having trouble matching the solution in the book, which is $\frac{1}{2}\ln(x^2-y^2)=\frac{1}{3}y^3+C$.
I'm getting an integral that requires the incomplete gamma function.
My attempt:
Rewrite the equation:
$x + (-(x^2 - y^2) y^2 - y)\frac{dy}{dx} = 0$
This is not an exact equation, but I found the integrating factor: $μ(y) = e^{-(2 y^3)/3}$
Multiply both sides of $x + \frac{dy}{dx} (-(x^2 - y^2) y^2 - y) = 0$ by $μ(y):$
$xe^{-\frac{2}{3}y^3} + (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y) \frac{dy}{dx} = 0$
Let $R(x,y) =xe^{-\frac{2}{3}y^3} $ and $S(x,y) = (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y)$. So I want to seek $f(x,y)$ such that $\frac{\partial f(x,y)}{x} = R(x,y)$ and $\frac{\partial f(x,y)}{y} = S(x,y)$
Integrating w.r.t $x$:
$f(x,y) = \int xe^{-\frac{2}{3}y^3}\,dx = \frac{1}{2}x^2 e^{-\frac{2}{3}y^3} + g(y)$
$\frac{dg(y)}{dy} = e^{-\frac{2}{3}y} y (y^3 - 1)$
Integrating w.r.t $y$:
$g(y) = \int e^{-\frac{2}{3}y} y (y^3 - 1) dy = ?$
I'm stuck here.
| Your way is correct.
https://www.wolframalpha.com/input?i=Integral+e%5E%7B-2%2F3y%5E3%7D%28y%5E4-y%29dy
I saw this trick:
$$\frac{d(x^2-y^2)}{x^2-y^2}=2y^2dy$$
$$\ln(x^2-y^2)=\frac{2}{3}y^3+2c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches)
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.
I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?
|
$3x+4y=15$ represents a straight line and $\sqrt{x^2+y^2}$ represents the distance of the point $(x,y)$ from the origin. So the question is basically telling us to find the minimum distance of any point lying on the line $3x+4y=15$, from the origin.
This shortest distance must be the perpendicular distance from the origin to the line.
The perpendicular distance of a point $(h,k)$ from the line $ax+by+c=0$ is $\Bigg|\dfrac{ah+bk+c}{\sqrt{a^2+b^2}}\Bigg|$. Replacing $(h,k)$ with $(0,0)$ and the line with $3x+4y-15=0$ gives us the minimum value as 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
| $$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\dfrac{(\sin(\dfrac{3\pi}{4}+\alpha))^2}{(\cos(\dfrac{3\pi}{4}+\alpha))^2}=\dfrac{(\sin\dfrac{3\pi}{4}\cos\alpha+\cos\dfrac{3\pi}{4}\sin\alpha)^2}{(\cos\dfrac{3\pi}{4}\cos\alpha-\sin\dfrac{3\pi}{4}\sin\alpha)^2}=\dfrac{(\dfrac{1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}{(\dfrac{-1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)^2}=\dfrac{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha-\sin\alpha\cos\alpha}{\dfrac{1}{2}\cos^2\alpha+\dfrac{1}{2}\sin^2\alpha+\sin\alpha\cos\alpha}=\dfrac{1-\sin2\alpha}{1+\sin2\alpha}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Find all values of a so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2 My goal is to find all values of "a" so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2
The correct answer is: $a = -6$ and $a = 2$
I tried solving it by doing this:
$x^2 - ax + y^2 +2y=a$
$x^2 - ax + (y+1)^2-1=a$
$(x - \frac a2)^2 - (\frac a2)^2 + (y+1)^2-1=a$
$(x - \frac a2)^2 - {a^2\over 4} + (y+1)^2-1=a$
$(x - \frac a2)^2 + (y+1)^2=a + {a^2\over 4} + 1$
$(x - \frac a2)^2 + (y+1)^2={a^2+4a + 4\over 4}$
We want the radius to be 2 so set this ${a^2+4a + 4\over 4}$ equal to 2
${a^2+4a + 4\over 4}=2$
$a^2+4a + 4=8$
$a^2+4a -4=0$
Solve for a:
$a=-2 \pm \sqrt{4+4}$
$a=-2 \pm \sqrt{8}$
This is not correct as you can see. I don't understand what I do wrong, I'm not sure if there is one of those tiny mistakes somewhere in my solving process or if I'm completely wrong from the beginning. Thanks in advance.
| $\frac{a^2+4a+4}{4}$ is not a radius.
Actually, it is the square of radius.
So, you should solve $\frac{a^2+4a+4}{4}=2^2$
And its solution is a=-6, a=2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| Suggestion
In case$-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$ simply means $a \ge 8$.
After re-writing the function in vertex form and plugging in some valid values of a (like 8, 10 etc) will give a graph on quadrant I for x in (0, 2). This means the function is greater than 0 for all x in that domain under that restriction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 6
} |
Prove that $\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$ Prove that $$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$$ for any number $x$.
My attempt:
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \\ \iff \frac{x}{1+x^2} \geq -\frac{1}{2} \land \frac{x}{1+x^2} \leq \frac{1}{2}$$ $$\iff (x+1)^2 \geq 0 \land (x-1)^2 \geq 0$$ the last two inequalities are obviously true, which concludes my proof attempt.
Not sure if this is a correct way to prove the inequality, also it's clearly not very elegant.
Could someone please verify my solution, and maybe suggest a more elegant or efficient approach?
| Your solution looks fine, as an alternative, by a single inequality, we have
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff \left|1+x^2\right|\ge 2|x|
\iff x^4-2x^2+1\ge 0\iff (x^2-1)^2 \ge 0$$
or also
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff 1+x^2\ge 2|x|
\iff x^2-2|x|^2+1\ge 0\iff (|x|-1)^2 \ge 0$$
Another way by AM-GM
$$\frac{1+x^2}{2}\ge \sqrt{x^2}=|x|$$
Another way, by $x=\tan \theta$ we have
$$\left|\frac{x}{1+x^2}\right|= \left|\frac{\tan \theta}{1+\tan^2 \theta}\right|=\frac12|\sin 2\theta|\le \frac12$$
Another way, by rearrangement
$$\left|\frac{1+x^2}{x}\right|=\frac{1+x^2}{|x|}=\frac1{|x|}\cdot 1+|x|\cdot 1\ge \frac1{|x|}\cdot |x|+1\cdot 1= 2$$
Another one
$$\left|\frac{x}{1+x^2}\right|\le \frac12 \iff \frac{2|x|}{(|x|-1)^2+2|x|}\le 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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