Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Line through two given skew lines and origin Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connecting the two lines on the perpendicular from the required line to the given two lines is zero. Hence
\begin{equation}
\begin{vmatrix}
1&2&3\\
2&3&4\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
\begin{equation}
\begin{vmatrix}
-2&3&4\\
4&3&2\\
a&b&c\\
\end{vmatrix} = 0
\end{equation}
This gives me two equations
\begin{equation}
a-2b+c=0
\end{equation}
\begin{equation}
3a-10b+9c=0
\end{equation}
I get two different ratios for the direction cosines (a,b,c) as
\begin{equation}
(5,3,1),(2,3,1)
\end{equation}
Both these direction cosines give me a line intersecting only one given line.
Is my approach wrong or does this mean that there is no line that intersects both given lines and also goes through the origin?
Highly appreciate your help on this.Thanks in advance.
| You are correct that there is no such line other than a mistake you made in calculation. Here is another approach I used to validate.
An intersection point on line $1$ is $P (2s+1, 3s+2, 4s+3)$ and on line $2$, $Q (4t-2, 3t + 3, 2t + 4)$.
Now as the the intersecting line passes through origin $O$, we must have, $\vec {OP} = \lambda \ \vec {OQ}$ where $\lambda$ is a scalar constant and we must have $\lambda \ne 0$.
So, $(2s+1, 3s+2, 4s+3) = \lambda (4t-2, 3t + 3, 2t + 4)$
From first two, $s = \cfrac{18 \lambda - 5}{6}, t = \cfrac{12\lambda-1}{6\lambda}$
Plugging $\lambda$ into the third component and solving, $\lambda = 0$. Hence there is no such line that intersects both lines and passes through the origin.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$ How to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$
I tried using the sandwich theorem by taking logarithm of the limit and using the expansion
$$ x - (x^2)/2<\ln(1+x)<x$$
but I was getting different from answer from the two sides. The correct answer is $e^{1/3}$.
| Using the fact that $\log(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$ for small $x$, we find
\begin{align*}
\left( {\prod\limits_{r = 1}^n {\left( {1 + \frac{r}{{n^2 }}} \right)} } \right)^n & = \exp \left( {n\sum\limits_{r = 1}^n {\log \left( {1 + \frac{r}{{n^2 }}} \right)} } \right) \\ &= \exp \left( {n\left( {\sum\limits_{r = 1}^n {\left( {\frac{r}{{n^2 }} - \frac{1}{2}\frac{{r^2 }}{{n^4 }} + \mathcal{O}\!\left( {\frac{{r^3 }}{{n^6 }}} \right)} \right)} } \right)} \right)
\\ &
= \exp \left( {\frac{1}{n}\sum\limits_{r = 1}^n r - \frac{1}{{2n^3 }}\sum\limits_{r = 1}^n {r^2 } + \mathcal{O}\!\left( {\frac{1}{{n^5 }}} \right)\sum\limits_{r = 1}^n {r^3 } } \right) \\ &= \exp \left( {\frac{{n + 1}}{2} - \frac{{(n + 1)(2n + 1)}}{{12n^2 }} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right)
\\ &
= \exp \left( {\frac{n}{2} + \frac{1}{3} + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right).
\end{align*}
The result now follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing $ n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2}}\sin{\frac{\pi(n-2(k-1))}{n+2}}\sec^2{\frac{\pi(n-2(k-1))}{2n+4}}$ for natural $n$ Good afternoon,
I am a little confused and intrigued by this finite summation formula I came up with.
If $n$ is a natural number then
$$
n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2}}\sin{\frac{\pi(n-2(k-1))}{n+2}}\sec^2{\frac{\pi(n-2(k-1))}{2n+4}}
$$
This is kind of obtuse to look at so here are some examples:
$n=1$:
$$
1=\sin{\frac{2\pi}{3}}\sin{\frac{\pi}{3}}\sec^2{\frac{\pi}{6}}
$$
$n=3$:
$$
3=\sin{\frac{2\pi}{5}}\sin{\frac{3\pi}{5}}\sec^2{\frac{3\pi}{10}}+\sin{\frac{4\pi}{5}}\sin{\frac{\pi}{5}}\sec^2{\frac{2\pi}{10}}
$$
$n=5$:
$$
5=\sin{\frac{2\pi}{7}}\sin{\frac{5\pi}{7}}\sec^2{\frac{5\pi}{14}}+\sin{\frac{4\pi}{7}}\sin{\frac{3\pi}{7}}\sec^2{\frac{3\pi}{14}}+
\sin{\frac{6\pi}{7}}\sin{\frac{\pi}{7}}\sec^2{\frac{\pi}{14}}
$$
Does this look familiar to anybody? I have no idea what to compare this to? Is this new or even vaguely interesting?
I managed to prove that this is true for odd values of n. It seems to hold for even $n$ too (if you sum over $\left \lfloor\frac{m+1}{2}\right \rfloor$) but I don't know how to prove it. Thanks for any help!
| We have, $$\theta=\frac {\pi(n+2-2k)}{2n+4}$$
Then $\frac {2k\pi}{n+2}=\pi-2\theta $
Thus, the series is equivalent to:
$$S=\sum_{k=1}^{\frac {n+1}{2}} \sin (\pi-2\theta)\sin 2\theta \sec^2 \theta $$
Using $\sin 2\theta=2\cos \theta \sin \theta$ and $\sin (\pi-2\theta)=\sin 2\theta$, we get:
$$S=\sum 4\sin^2 \theta =\sum 4 \cos^2\left(\frac {k\pi}{n+2}\right)$$
Now note the identity:
$$4\cos^2 x=2\cos 2x+2$$
Hence we get:
$$S=\sum_{k=1}^{\left \lfloor \frac {n+1}{2} \right \rfloor} \left(2+2\cos \frac {2k\pi}{n+2}\right) {\tag 1}$$
It is already a well-known result that $$\sum_{k=1}^{n} \cos(a+(k-1)d)=\frac {\cos\left(a+\frac {(n-1)d}{2}\right) \sin \frac {nd}{2}}{\sin \frac d2} {\tag 2}$$
Thus, using $(2)$ in $(1)$ and putting $n=2m$, we arrive at the result for even $n$, as shown. Note that if $n=2m$ then $\left \lfloor \frac {n+1}{2} \right \rfloor=m$. Hence:
$$S=\left(\sum_{k=1}^m 2\right)+2\left(\sum_{k=1}^m \cos \frac {2k\pi}{2m+2}\right)$$
The first term obviously evaluates to $2m=n$, and if you note that the second term is a summation of $\cos$ in AP, and hence use $(2)$, you'll see that it evaluates to $0$.
Thus $$S=n+0=n$$
This completes the proof of the even case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $| f(p + q) – f(q)| \le \dfrac pq$ for all $p$ and $q\in \mathbb Q$ & $q\ne 0$, show that $\sum_{i=1}^k| f(2^k ) – f(2^i ) |\le \dfrac{k(k – 1)}2$ The following question is taken from the practice set of JEE exam.
If $| f(p + q) – f(q)| \le \dfrac pq$ for all $p$ and $q \in \mathbb Q$ & $q \ne 0$, show that $\sum_{i=1}^k| f(2^k ) – f(2^i ) |\le \dfrac{k(k – 1)}2$
$| f(2^k ) – f(2^1 ) |\le \dfrac{2^k-2}{2}=2^{k-1}-1$
$| f(2^k ) – f(2^2 ) |\le \dfrac{2^k-2^2}{2^2}=2^{k-2}-1$
.
.
.
$| f(2^k ) – f(2^{k-1} ) |\le \dfrac{2^k-2^{k-1}}{2^{k-1}}=2^1-1$
Adding all this, we get the required LHS$\le(2+2^2+...2^{k-1})-(k-1)=2(2^{k-1}-1)-k+1=2^k-k-1$
How to get the desired RHS?
| After the hints given by @Amit_Rajaraman and @leoli1, I am attempting the solution:
$|f(2^k)-f(2^i)|\le|f(2^k)-f(2^{k-1})|+|f(2^{k-1})-f(2^{k-2})|+...+|f(2^{i+1})-f(2^i)|$
$|f(2^k)-f(2^i)|\le\dfrac{2^k-2^{k-1}}{2^{k-1}}+\dfrac{2^{k-1}-2^{k-2}}{2^{k-2}}+...+\dfrac{2^{i+1}-2^i}{2^i}$
$|f(2^k)-f(2^i)|\le(2-1)+(2-1)+...(k-i)\text{times}=k-i$
So, $\sum_{i=1}^k|f(2^k)-f(2^i)|\le k\cdot k-\dfrac{k(k+1)}2=\dfrac{k(2k-k-1)}2=\dfrac{k(k-1)}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n(n+1)(n+2)$ where $n\in N$ and $n+1=k^2$ ($k\in N$) is divisible by $60$. My Attempt
Let $N=n(n+1)(n+2)$
Now $n=k^2-1 \implies N=(k^2-1)k^2(k^2+1)$
Now if $k=2m \implies N=(4m^2-1)4m^2(4m^2+1)$ which is divisible by $3$ and $4 \implies$ It is divisible by $12$.
Now if $k=2m+1 \implies N=(4m^2+4m)(4m^2+4m+1)(4m^2+4m+2)$ which is divisible by $8$.
This is as far as I got. How do I prove its divisible by $60$? Sorry if the formatting isn't the best.
| To conclude you have to show that it is divisible by $5$ so you have to see what the possible remainders of $n+1$ when dividing by $5$ are, and conclude that at least one of $n,n+1$ and $n+2$ is a multiple of $5$.
If $n+1$ is a square then $n+1\equiv -1,0$ or $1\bmod 5$. It follows at least one among $n,n+1,n+2$ is $0 \bmod 5$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there an analytical solution to $\int_{0^{-}}^{t} d{s} \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)$? I am trying to solve the integral
\begin{align}
\int_{0^{-}}^{t} d{s} \ \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)
\end{align}
where $T$ is Owen's T function. I have been trying to substitute some integral definition of the Owen T function, e.g.
\begin{align}
T(h, a)
& =\frac{1}{2 \pi} \int_{0}^{a} d{x} \ \frac{e^{-\frac{1}{2} h^{2}\left(1+x^{2}\right)}}{1+x^{2}} \label{eq: core def OwenT}\\
& =\frac{1}{2\sqrt{2 \pi}} \int_{h}^{\infty} d{\xi} \ e^{-\xi^{2}/2} \ \operatorname{Erf}\left(\frac{a \xi}{\sqrt{2}}\right)
\end{align}
which I found in [1] and [2]. However, this yields integrals over Error functions. Some sources one can check for integrals over Error function are [3], [4] and [5]. Nevertheless, so far, no luck. Is there someone who knows a way to tackle integrals over Owen's T functions?
| Let
$$ f \colon \mathbb{R} \times (0,\infty) \to (0,\infty) \, , \, f(x,t) = \int \limits_0^t \operatorname{T} \left(\frac{x}{\sqrt{2 t}},\sqrt{\frac{s}{2t-s}}\right) \, \mathrm{d} s \, .$$
The substitution $s = t u$ yields $f(x,t) = t g\left(\frac{x}{2\sqrt{t}}\right)$ with $g \colon \mathbb{R} \to (0,\infty)$ given by
\begin{align}
g(y) &= \int \limits_0^1 \operatorname{T} \left(\sqrt{2} y, \sqrt{\frac{u}{2-u}}\right) \, \mathrm{d} u = \frac{1}{2 \pi} \int \limits_0^1 \int \limits_0^{\sqrt{\frac{u}{2-u}}} \frac{\mathrm{e}^{- y^2 (1 + v^2)}}{1+v^2} \, \mathrm{d} v \, \mathrm{d} u \\
&\!\!\!\!\overset{\text{Tonelli}}{=} \frac{1}{2\pi} \int \limits_0^1 \int \limits_{\frac{2v^2}{1+v^2}}^1 \frac{\mathrm{e}^{- y^2 (1 + v^2)}}{1+v^2} \, \mathrm{d} u \, \mathrm{d} v = \frac{1}{2\pi} \int \limits_0^1 \frac{1-v^2}{(1+v^2)^2} \, \mathrm{e}^{- y^2 (1 + v^2)} \, \mathrm{d} v \\
&\!\overset{\text{IBP}}{=} \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \int \limits_0^1 \frac{v^2}{1+v^2} \, \mathrm{e}^{- y^2 (1 + v^2)} \, \mathrm{d} v\right] \\
&= \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \left(\frac{\sqrt{\pi}}{2y} \mathrm{e}^{-y^2} \operatorname{erf}(y)-2 \pi \operatorname{T} (\sqrt{2} y,1)\right)\right] \\
&= \frac{1}{2\pi} \left[\frac{1}{2} \mathrm{e}^{-2y^2} + 2 y^2 \left(\frac{\sqrt{\pi}}{2y} \mathrm{e}^{-y^2} \operatorname{erf}(y)-\frac{\pi}{2} \operatorname{erfc}(y) \left(1 - \frac{1}{2} \operatorname{erfc}(y)\right)\right)\right] \\
&= \frac{\mathrm{e}^{-2 y^2}}{4 \pi} + \frac{y \, \mathrm{e}^{-y^2}}{2 \sqrt{\pi}} \operatorname{erf}(y) - \frac{y^2}{4} \left(1 - \operatorname{erf}^2(y)\right) = \frac{\left(y \operatorname{erf}(y) + \frac{\mathrm{e}^{-y^2}}{\sqrt{\pi}}\right)^2 - y^2}{4} \, .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$
Let $a$ and $b$ be real numbers. Find the minimum of $$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$$
The problem is from an inequality book. While I was doing the problem myself, I got the following results:
From Minkowski's inequality, we have
$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}\\ \geq \sqrt{(a+a-1+a+a-1)^2+(b+b-1+b+b-1)^2}\\ =\sqrt{(4a-2)^2+(4b-2)^2}\\ \geq \sqrt{8(2a-1)(2b-1)} \ \ \ \ \ \ \text{[From AM-GM]}$
Equality holds when $4a-2=4b-2\implies a=b$. And the minimum $\sqrt{8(2a-1)(2b-1)}=0$ when $a=b=\frac 12$.
Here, everything seemed right to me. But when I plugged in $a=b=\frac 12$ in the original expression, I didn't get the minimum $0$ rather I got $2\sqrt 2$.
When I checked the solution in the book, a geometric solution was given and the minimum was indeed $2\sqrt 2$. So, I couldn't find where my mistake is. It will be helpful for me if someone can find that. Other algebraic solutions are also welcome.
I'm not good at inequalities, so pardon silly mistakes. Thanks.
| Assuming you got the minimum $0$ and you wanna ask what is your mistake.
Then each term reduces to $\sqrt{\dfrac{1}{2}}$
So $\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}} = \sqrt{2}$
Just to help you help here is another helpful solution-
In the cartesian system of coordinates consider the points:
$O(0,0)$, the origin;
$A(x,y);B(x,y-1);C(x-1,y-1);D(x-1,y)$.
$A,B,C,D$ are the vertices of a square with the side $1$.
Results: $AC=BD=\sqrt2$.
Denote $M\left(x-\dfrac{1}{2},y-\dfrac{1}{2}\right)=AC\cap BD$ the center of the square.
$\sqrt{x^2+y^2}=OA; \sqrt{x^2+(y-1)^2}=OB; \sqrt{(x-1)^2+(y-1)^2}=OC; \sqrt{(x-1)^2+y^2}=OD$.
Using the triangle inequality:
$\sqrt{x^2+y^2}+\sqrt{(x-1)^2+(y-1)^2}=OA+OC\ge AC=\sqrt2$, with equality for $O\in[AC]$.
$\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}=OB+OD\ge BD=\sqrt2$, with equality for $O\in[BD]$.
Results:
$\sqrt{x^2+y^2}+\sqrt{(x-1)^2+(y-1)^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-1)^2+y^2}\ge2\sqrt2$, with equality for
$O\equiv M\Longleftrightarrow x=y=\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\underbrace{\sum_{x = 1}^{r} \ldots \sum_{x = 1}^{r}}_{\text{n times}} x = ?$ Let $f(n,r) = \sum_{x = 1}^{r} f(n - 1,x)$ with $f(0,x) = x$.
We get
$$f(n,r) = \underbrace{\sum_{x_{n-1} = 1}^{x_n = r} \ldots \sum_{x = 1}^{x_1}}_{\text{n times}}x\text{ .}$$
This sum popped up when I was trying to find a formula for the numbers in Pascal's triangle and I have no idea about solving it.
I tried figuring out the first sum $f(1,r)$ and it is the triangular number $\frac{r(r + 1)}{2}$ and tried figuring out the second sum $f(2,r)$ in terms of the first by naming $\frac{r(r + 1)}{2}$ as $F(r)$:
$
\begin{eqnarray}
f(2,r)&=& \sum_{x_1 = 1}^{r}\color{blue}{\sum_{x = 1}^{x_1} x} \\
&=& \sum_{x_1 = 1}^{r} \color{blue}{\frac{x_1(x_1+1)}{2}} \\
&=& \frac{1}{2} \sum_{x_1 = 1}^{r} x_1(x_1 + 1) \\
&=& \frac{1}{2}(2 + 6 + 12 + \ldots + r(r + 1)) \\
&=& \frac{1}{2}((2 + (2 + 2 \cdot 2) + \ldots + r(r + 1)) \\
&=& \color{green}{\frac{1}{2}}(\color{green}{2}(1) + \color{green}{2}(1 + 2) + \ldots + \color{green}{2}(1 + 2 + \ldots + r)) \\
&=& (1) + (1 + 2) + \ldots + (1 + 2 + \ldots + r) \\
&=& \color{blue}{r(1 + 2 + \ldots + r)}\color{purple}{- ((r-0) + (r - 1) + \ldots + 2)} \\
&=& \color{blue}{\frac{r^2(r + 1)}{2}} \color{purple}{- r(r - 1) + \frac{(r - 1)(r - 2)}{2}} \\
&=& rF(r) - 2F(r - 1) + F(r - 2) \\
&=& rF(r) - 2(F(r) - r) + (F(r) - (r + (r - 1))) \\
&=& rF(r) - 2F(r) + 2r + F(r) - 2r + 1 \\
&=& rF(r) - F(r) + 1 \\
&=& (r - 1)F(r) + 1 \end{eqnarray}
$
After getting a formula for $f(n,r)$ in terms of $f(1,r)$, you swap $r$ for $x$ to get a sum $f(n + 1,x)$ that runs from $1$ to $r$, then you find a formula for $f(n + 1,r)$ in terms of $f(1,r)$ and swap to get $f(n+2,r)$ and so on, at least this was how the sum was defined in the context of my work on the formula for the numbers of Pascal's triangle.
For example:
$$\begin{eqnarray} f(1,r) &=& \sum_{x = 1}^{r} x \\
&=& \frac{r(r + 1)}{2} \\
&=& F(r) \\ \\
f(2,r) &=& \sum_{x = 1}^{r} F(x) \\
&=& (r - 1)F(r) + 1 \\
&=& G\circ F(r) \\ \\
f(3,r) &=& \sum_{x = 1}^{r} G\circ F(x) \end{eqnarray}$$
I don't know if their exists such a formula in terms of the first sum for any number of sums we take, I stopped to see if there is any easier way to solve it.
| There is a somewhat combinatorial approach to this problem that uses hockey-stick identity.
Note that $f(0,x)=\binom{x}{1}$. We have that
$$f(1,r)=\sum_{x=1}^r \binom{x}{1}$$
Using hockey-stick identity, this is
$$f(1,r)=\binom{r+1}{2}$$
We are motivated to proceed by induction. We hypothesize that $f(n,r)=\binom{r+n}{n+1}$. We have already established our base case. To prove our inductive step, we want to prove, given $f(n-1,r)=\binom{r+n-1}{n}$, that
$$f(n,r)=\binom{r+n}{n+1}$$
Since we have the recursion
$$f(n,r)=\sum_{x=1}^r \binom{x+n-1}{n}$$
By hockey-stick identity, this simplifies to
$$\binom{r+n}{n+1}$$
Hence, our proof by induction is complete.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify this equation? I'm trying to simplify equation
$$n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil$$
into
$$n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor$$
where $x$ is an integer. Indeed, both equations seem to output the same result when testing for some values. Thus far, I've done:
\begin{align*}
&n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil \\
\implies &\sqrt{2x +\frac14} - \frac12 \leq n < \sqrt{2x +\frac14} + \frac12
\end{align*}
The left side easily reduces as follows:
\begin{align*}
&\sqrt{2x +\frac14} - \frac12 \leq n \\
\implies &\sqrt{2x} - \frac12 < n
\end{align*}
but I'm having trouble with the right side which I would wish to reduce as follows:
\begin{align*}
&n < \sqrt{2x +\frac14} + \frac12 \\
\implies &\texttt{???} \\
\implies &n \leq \sqrt{2x} + \frac12
\end{align*}
so that I'm able to obtain my end result.
\begin{align*}
&\sqrt{2x} - \frac12 < n \leq \sqrt{2x} + \frac12\\
\implies &n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor
\end{align*}
| Is it possible that that there is an integer $k$ so that $2x < k^2 < 2x +\frac 14$?
As $x$ and $k^2$ are integers the answer to that is obviously no.
Let's get a little tighter. Is it possible there is "half-integer" $\frac k2$ where $k\in \mathbb Z$ and $2x < (\frac k2)^2 < 2x+\frac 14$.
That is $8x < k^2 < 8x + 1$. Again as $8x, k^2, 8x+1$ are all integers the answer is no.
So we either have $(k-\frac 12)^2 < 2x < 2x + \frac 14 < k^2$ (strict inequalitys as $2x+\frac 14$ and $(k-\frac 12)^2$ are not integers) for some integer $k$ or $k^2 \le 2x < 2x+\frac 14\le (k+\frac 12)^2$ for some integer $k$.
Case 1:
$(k-\frac 12)^2 < 2x < 2x +\frac 14 < k^2$.
Then $k-\frac 12 < \sqrt{2x + \frac 14} < k$
$k-1 < \sqrt{2x + \frac 14} < k-\frac 12$
And $\lceil \sqrt{2x +\frac 14} \rceil = k$
And $k-\frac 12 < \sqrt{2x} < k$
$k < \sqrt{2x} +\frac 12 < k + \frac 12$
And so $\lfloor \sqrt{2x} +\frac 12 \rfloor = k =\lceil \sqrt{2x +\frac 14} \rceil$
Case 2:
$k^2 \le 2x < 2x+\frac 14\le (k+\frac 12)^2$
$k < \sqrt{2x +\frac 14} \le k+\frac 12$
$k-\frac 12 < \sqrt{2x +\frac 14}-\frac 12 \le k$ so
$\lceil \sqrt{2x +\frac 14} \rceil = k$
And $k \le \sqrt {2x} < k+\frac 12$ so
$k+\frac 12 \le \sqrt{2x}+\frac 12 < k+1$ so
$\lfloor \sqrt{2x} +\frac 12 \rfloor = k =\lceil \sqrt{2x +\frac 14} \rceil$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
show this indentity $\sum_{k=1}^{2n-1}\frac{\sin{\frac{k^2\pi}{2n}}}{\sin{\frac{k\pi}{2n}}}=n$ let $n$ be postive integers.show that
$$\sum_{k=1}^{2n-1}\dfrac{\sin{\frac{k^2\pi}{2n}}}{\sin{\frac{k\pi}{2n}}}=n$$
Try:I can show $n$ is smaller number.
let $LHS=f(n)$.when $n=1$ it is clear
$$f(1)=\dfrac{\sin{\pi/2}}{\sin{\pi/2}}=1$$
when $n=2$ then
$$f(2)=\dfrac{\sin{\frac{\pi}{4}}}{\sin{\frac{\pi}{4}}}+\dfrac{\sin{\frac{4\pi}{4}}}{\sin{\frac{2\pi}{4}}}+\dfrac{\sin{\frac{9\pi}{4}}}{\sin{\frac{3\pi}{4}}}=1+0+1=2$$
when $n=3$,then
$$f(3)=\dfrac{\sin{\frac{\pi}{6}}}{\sin{\frac{\pi}{6}}}+\dfrac{\sin{\frac{4\pi}{6}}}{\sin{\frac{2\pi}{6}}}+\dfrac{\sin{\frac{9\pi}{6}}}{\sin{\frac{3\pi}{6}}}+\dfrac{\sin{\frac{16\pi}{6}}}{\sin{\frac{4\pi}{6}}}+\dfrac{\sin{\frac{25\pi}{6}}}{\sin{\frac{5\pi}{6}}}=1+1-1+1+1=3$$
$$\begin{align}f(4)&=\dfrac{\sin{\frac{\pi}{8}}}{\sin{\frac{\pi}{8}}}+\dfrac{\sin{\frac{4\pi}{8}}}{\sin{\frac{2\pi}{8}}}+\dfrac{\sin{\frac{9\pi}{8}}}{\sin{\frac{3\pi}{8}}}+\dfrac{\sin{\frac{16\pi}{8}}}{\sin{\frac{4\pi}{8}}}+\dfrac{\sin{\frac{25\pi}{8}}}{\sin{\frac{5\pi}{8}}}+\dfrac{\sin{\frac{36\pi}{8}}}{\sin{\frac{6\pi}{8}}}+\dfrac{\sin{\frac{49\pi}{8}}}{\sin{\frac{7\pi}{8}}}\\
&=1+\sqrt{2}-\tan{\frac{\pi}{8}}+0-\tan{\frac{\pi}{8}}+\sqrt{2}+1\\
&=4\end{align}$$
because use this well known $\tan{\frac{\pi}{8}}=\sqrt{2}-1$
| Let $\omega = e^{i\pi/2n}$. If we write $S$ for the sum, then from
$$ \frac{\sin(\pi k^2/2n)}{\sin(\pi k/2n)} = \frac{\omega^{k^2} - \omega^{-k^2}}{\omega^k - \omega^{-k}} = \omega^{k(-k+1)} \frac{(\omega^{2k})^k - 1}{\omega^{2k} - 1} = \sum_{j=0}^{k-1} \omega^{k(2j-k+1)}, $$
we may recast $S$ as the double sum
$$ S = \sum_{k=1}^{2n-1} \sum_{j=0}^{k-1} \omega^{k(2j-k+1)} = \sum_{(k,l) \in \mathcal{T}} \omega^{kl}, $$
where $l = 2j-k+1$ and the region $\mathcal{T}$ is defined by
$$\mathcal{T} = \{ (k, l) \in \mathbb{Z}^2 : 0 < |l| < k < 2n \text{ and } k + l \text{ is odd}\} .$$
The following figure illustrates the region $\mathcal{T}$ for $n = 8$:
Now the key observation is that the sum $S$ enjoys the following symmetry:
$$ S = \sum_{(l, k) \in \mathcal{T}} \omega^{kl} = \sum_{(-k,-l) \in \mathcal{T}} \omega^{kl} = \sum_{(-l,-k) \in \mathcal{T}} \omega^{kl} $$
Moreover, if we write
\begin{align*}
\mathcal{T}_1 &= \mathcal{T}, &
\mathcal{T}_2 &= \{(l, k) : (k, l) \in \mathcal{T}\}, \\
\mathcal{T}_3 &= \{(-k, -l) : (k, l) \in \mathcal{T}\}, &
\mathcal{T}_4 &= \{(-l, -k) : (k, l) \in \mathcal{T}\},
\end{align*}
then $\mathcal{T}_1, \ldots, \mathcal{T}_4$ are disjoint and
$$ \mathcal{T}_1 \cup \mathcal{T}_2 \cup \mathcal{T}_3 \cup \mathcal{T}_4
= \{(k, l) : |k|, |l| < 2n \text{ and } k + l \text{ is odd} \}. $$
From this, we may write
$$ 4S = Z - 2B, $$
where
$$ Z = \sum_{\substack{k, l \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd}}} \omega^{kl}
\qquad\text{and}\qquad
B = \sum_{\substack{k = 2n \\ l \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd}}} \omega^{kl}
= \sum_{\substack{k \in \mathbb{Z}/4n\mathbb{Z} \\ l = 2n \\ k+l \text{ odd}}} \omega^{kl}. $$
The figure below illustrates the regions $\mathcal{T}_k$'s as well as the points $(k, l)$ contributing the sum for $2B$ when $n = 5$.
Then the claim will follow once we prove:
Claim. $Z = 0$ and $B = -2n$.
The value of $B$ is easier to compute. Indeed, using $\omega^{2n} = -1$, we get
$$ B
= \sum_{\substack{|k| < 2n \\ k \text{ odd}}} \omega^{2nk}
= \sum_{\substack{|k| < 2n \\ k \text{ odd}}} (-1)^{k}
= -2n. $$
So we shift our focus to the value of $Z$. Substituting $l \mapsto l + 2r$ for $r = 0, 1, \dots, 2n-1$, we find that
$$ Z
= \frac{1}{2n} \sum_{r=0}^{2n-1} Z
= \frac{1}{2n} \sum_{r=0}^{2n-1} \sum_{\substack{k, l \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd}}} \omega^{k(l + 2r)}
= \sum_{\substack{k, l \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd}}} \omega^{kl} \left( \frac{1}{2n} \sum_{r=0}^{2n-1} \omega^{2kr} \right). $$
Then by using the identity
$$ \frac{1}{2n} \sum_{r=0}^{2n-1} \omega^{2kr}
= \begin{cases}
1, & \text{if $k \equiv 0$ (mod $2n$),} \\
0, & \text{otherwise,}
\end{cases}$$
it follows that
$$ Z
= \sum_{\substack{k, l \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd} \\ \text{$k \equiv 0$ (mod $2n$)}}} \omega^{kl}
= \sum_{l\in\mathbb{Z}/4n\mathbb{Z}} \sum_{\substack{k \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd} \\ \text{$k \equiv 0$ (mod $2n$)}}} \omega^{kl}. $$
Now the inner sum can be easily evaluated for each given $l\in\mathbb{Z}/4n\mathbb{Z}$, yielding
$$ \sum_{\substack{k \in \mathbb{Z}/4n\mathbb{Z} \\ k+l \text{ odd} \\ \text{$k \equiv 0$ (mod $2n$)}}} \omega^{kl}
= \begin{cases}
0, & \text{if $l$ is even,} \\
1 + (-1)^l = 0, & \text{if $l$ is odd.}
\end{cases} $$
Therefore $Z = 0$ and the desired claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$
Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$
I noted that
(A) ${}^{20}{C_{10}} - 110$, if $n=10$
(B) ${}^{22}{C_{11}} - 110$, if $n=11$
(C) ${}^{18}{C_{9}} - 100$, if $n=10$
(D) ${}^{20}{C_{10}} - 121$, if $n=11$
My basic approach is as follows:
$\sum\limits_{r = 1}^{n - 1} {\frac{{n\left( {n + 1} \right) - r\left( {2r + 3} \right) - 1}}{{{{\left( {r + 1} \right)}^2}}}.{}^{n + r}{C_r}.{}^n{C_r}} = $
$T = \frac{1}{{{{\left( {r + 1} \right)}^2}}}.{}^{n + r}{C_r}.{}^n{C_r}$
$T = \frac{1}{{{{\left( {r + 1} \right)}^2}}}.\frac{{\left( {n + r} \right)!}}{{r!.n!}}.\frac{{n!}}{{r!\left( {n - r} \right)!}} \Rightarrow T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times \frac{{\left( {n + r + 2} \right)\left( {n + r + 1} \right)\left( {n + r} \right)!}}{{\left( {n - r} \right)!\left( {2r + 2} \right)!}}.\frac{{2r + 2!}}{{r + 1!.r + 1!}}$
$T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times \frac{{\left( {n + r + 2} \right)!}}{{\left( {n - r} \right)!\left( {2r + 2} \right)!}}.\frac{{2r + 2!}}{{r + 1!.r + 1!}} \Rightarrow T = \left( {\frac{1}{{n + r + 1}} - \frac{1}{{n + r + 2}}} \right) \times {}^{n + r + 2}{C_{2r + 2}}.{}^{2r + 2}{C_{r + 1}}$
Not able to proceed.
| Note that
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&=\frac{n^2+n-2r^2-3r-1}{(r+1)^2}
\\&=\frac{n^2-r^2+n-r-(r^2+2r+1)}{(r+1)^2}\\
&=\frac{(n+r+1)(n-r)}{(r+1)^2}-1\end{align}$$
which implies
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&\binom{n+r}{r}\binom{n}{r}\\
&=\frac{n+r+1}{r+1}\binom{n+r}{r}\frac{n-r}{r+1}\binom{n}{r}-\binom{n+r}{r}\binom{n}{r}\\
&=\binom{n+r+1}{r+1}\binom{n}{r+1}-\binom{n+r}{r}\binom{n}{r}\\
&=b_{r+1}-{b_r}\end{align}$$
where $b_r=\binom{n+r}{r}\binom{n}{r}.$
Therefore the sum turns out to be telescopic:
$$\begin{align}\sum_{r=1}^{n-1}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\binom{n+r}{r}\binom{n}{r}&=\sum_{r=1}^{n-1}(b_{r+1}-{b_r})\\
&=b_{n}-b_1=\binom{2n}{n}-(n+1)n.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Roots of polynomials with repeated roots Let $a$ and $b$ be real numbers. Consider the cubic equation $$x^3+2bx^2-ax^2-b^2=0$$
(i) Show that if $x=1$ is a solution of the cubic then $$ -1+\sqrt{2}\leq b\leq1+\sqrt{2} $$
(ii) Show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic
I need some assistance on (ii). In order to show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic. I attempted to factorise the cubic. If $x=1$ is a solution, then the cubic may be expressed as follows with no remainder. $x^3+2bx^2-a^2x-b^2 \equiv (x-1)(Ax^2+Bx+C)$. Comparing coefficients, $A = 1, C = b^2, B = a^2+b^2$ $\therefore (x-1)(x^2+(a^2+b^2)x+b^2)$. Finally, if we sub $x=1$ into the resulting quadratic, we get $1+a^2+b^2+b^2 > 0$ for all real values of $a,b$. Therefore, $x=1$ cannot be a repeated root.
I have been told that $x^2+(a^2+b^2)x+b^2$ is not the quadratic equation for this and in fact it's $x^2+(2b+1)x+b^2$
$(x-1)(x^2+(2b+1)x+b^2) = x^3+2bx^2+(b^2-2b-1)x-b^2$. The coefficient of the $x$ term is correct, but the $b^2-2b-1 \not= -a^2$ therefore this cannot be right aswell.
My solution:
$(x-1)(x^2+(a^2+b^2)x+b^2) = x^3+(a^2+b^2-1)x^2-a^2x-b^2$
So my solution seems to get the correct coefficient for $x$ term, but the incorrect $a^2+b^2-1$ coefficient for the $x^2$ term.
What is going on here? Has math been broken?
| By polynomial long division, you get that$$x^3+2bx^2-ax^2-b^2=(x-1)^2(x+2b+2)+(-a^2+4b+3)x-(b^2+2b+2).$$So, if $1$ was a repeated root of $x^3+2bx^2-ax^2-b^2$, that is, if$$(x-1)^2\mid x^3+2bx^2-ax^2-b^2,$$the polynomial $(-a^2+4b+3)x-(b^2+2b+2)$ would be the null polynomial. But, for every $b\in\Bbb R$, $b^2+2b+2=(b+1)^2+1>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \cdot n +n^2-t^2=2\cdot t \cdot n +n^2$$
Also, we can build every possible square $w_n$ greater than $p$ in this way:
$$w_n=\left\lfloor\sqrt{p}\right\rfloor^2+2\cdot \left\lfloor\sqrt{p}\right\rfloor \cdot n +n^2 = \left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2$$
For example, let $p=5$. Follows that: $w_1=\left(1+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=9$, $w_2=\left(2+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=16$ and so on.
Now, using this idea and applying to the first equation:
$$9x^2=\left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2-p$$
I am not allowed to apply Pell's equation because $9=3^2$ and calcultaing $\Delta$ in $x$ doesn't help anymore.
Another approach is based on Pell's equation. I thought to express $9x^2=8x^2+x^2$ and then:
$$8x^2+x^2+p=y^2\leftrightarrow 8x^2+p=y^2-x^2\leftrightarrow (y^2-x^2)-8x^2=p \leftrightarrow u^2-8x^2=p$$
But then, in order to generate all the solutions, I have to guess the first one (or one of them) that is pretty complicated for big $p$.
So, how can we do that? Are there any other solutions?
Thanks.
| An experimental approach:
Here I find one solution for this equation which a family of solutions can be based on:
$9x^2=y^2-p=(y-\sqrt p)(y+\sqrt p)$
we can can construct following system of equations:
$\begin{cases}y-\sqrt p=9\\y+\sqrt p=x^2\end{cases}$
Subtracting first equation from second one we get:
$2\sqrt p=x^2-9=(x-3)(x+3)$
Suppose:
$x-3=2\Rightarrow x=5$
$x+3=\sqrt p$
subtracting first fro, second we get:
$\sqrt p-2=6\Rightarrow \sqrt p=8\rightarrow p=64$
Now we have:
$9x^2=y^2-64=(y-8)(y+8)$
Let $y-8=9\rightarrow y=17$
$x^2=y+8=25$
hence :
$x=5$, $y=17$ and $p=64$
$9\times5^2+64=17^2\space\space\space\space\space (1)$
$10^{12}\div 64=15625\times10^6>10^9$
So we can construct following relation by multiplying both sides of the relation (1) by $15625\times 10^6=(125\times 10^3)^2$:
$9(5\times 125000)^2+15625\times 10^6=(17\times125000)^2$
that is one solution of equation under condition $p=15625000000>10^9$ is:
$x=625\times10^3$
$y=2125\times 10^3$
This experiment shows that only with particular values of p the equation may have integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
A simple general expression for a definite integral Let
$$I_k = \int_0^{\frac{\pi}{2}} \sin x \cos^{2k} x \sqrt{1 + \sin x} \, dx,$$
where $k = 0,1,2,\ldots$. I wish to find a simple general expression for $I_k$ in terms of $k$. Simple here is the operative word.
Making a tangent half-angle substitution of $t = \tan \frac{x}{2}$ produces
$$I_k = 4\int_0^1 \frac{t(1 + t)^{2k + 1} (1 - t)^{2k}}{(1 + t^2)^{2k + \frac{5}{2}}} \, dt.$$
Trying to guess its general form one notices:
\begin{align*}
k = 0 : \quad I_0 &= 4\int_0^1 \frac{t(1 + t)}{(1 + t^2)^{\frac{5}{2}}} \, dt = 4 \left [\frac{(t - 1)(t^2 + t + 1)}{3(t^2 + 1)^{\frac{3}{2}}} \right ]_0^1 = \frac{4}{3}\\
k = 1 : \quad I_1 &= 4\int_0^1 \frac{t(1 + t)^3(1 - t)^2}{(1 + t^2)^{\frac{9}{2}}} \, dt = 4 \left [\frac{(t - 1)^3(11t^4 + 33t^3 + 52t^2 + 33t + 11)}{105(t^2 + 1)^{\frac{7}{2}}} \right ]_0^1 = \frac{44}{105}\\
k = 2 : \quad I_2 &= 4 \int_0^1 \frac{t(1 + t)^5(1 - t)^4}{(1 + t^2)^{\frac{13}{2}}} \, dt = 4\left [\frac{(1 - t)^5(211t^6 + 1055t^5 + 2593 t^4 + 3370t^2 + 1055t + 211)}{3465(t^2 + 1)^{\frac{11}{2}}} \right ]_0^1 = \frac{844}{3465}
\end{align*}
So it appears
$$I_k = 4 \left [\frac{(t - 1)^{2k + 1}p_k(t)}{a_k(t^2 + 1)^{2k + \frac{3}{2}}} \right ]_0^1,$$
where $a_k$ is a positive integer and $p_k(t)$ is a symmetric polynomial of degree $(2k + 2)$. But what is $a_k$ and $p_k(t)$?
Alternatively, a closed-form expression in terms of the hypergeometric function can be found. It is
$$I_k = \frac{2}{3\sqrt{2}} {}_2 F_1 \left (\frac{3}{2},-2k;\frac{5}{2};1 \right ) + \frac{1}{2^{2k + \frac{3}{2}}(2k+1)} {}_2F_1 \left (2k + 1,2k + \frac{5}{2};2k + 2; \frac{1}{2} \right ),$$
but this is hardly simple since ultimately $I_k$ is just a positive rational number. Perhaps this expression can be simplified in some way.
So my question is, can a simple expression for $I_k$ be found?
| Too long for a comment.
I did some back-of-the-envelope computation and obtained the following conjecture:
$$ I_k = 2 \sum_{n=0}^{k+1} \sum_{l=n}^{\left\lfloor\frac{n+k+1}{2}\right\rfloor} \frac{(-1)^{n}}{2n+2k+1} \binom{n+k+1}{2l}\binom{l}{n}. $$
The idea is that the following identity is likely to hold for $|z| < 1$:
$$ \sum_{k=0}^{\infty} I_k z^{2k} = \frac{\alpha\operatorname{arctanh}(\alpha) + \beta \operatorname{arctanh}(\beta)}{z^2}, \quad \text{where} \quad \alpha, \beta = \sqrt{z\bigl(z \pm i\sqrt{1-z^2}\bigr)} $$
(I did not compute this result by myself, rather jammed it to my Mathematica and simplified the result.)
Expanding the right-hand side and reading out the coefficient of $z^{2k}$ gives the above formula. Here is a Mathematica code for generating the above sequence:
Table[2 Sum[(-1)^n/(2 n + 2 k + 1) Binomial[n + k + 1, 2 l] Binomial[l, n], {n, 0, k + 1}, {l, n, (n + k + 1)/2}], {k, 0, 5}]
The following compares the various representations for $I_k$'s:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integrate $ \int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} \, \mathrm dx $ I tried writing $\tan^2 x+2\tan x+2$ as $(\tan x+1)^2+1$ and letting $\tan\theta = \tan x+1$ yields
$$ \int \frac{2\tan\theta-1}{\tan^2\theta-2\tan\theta+2}\,\sec\theta\,\mathrm d\theta $$
Then letting $ t = \tan(\frac{\theta}{2})$ yields
$$ \int\frac{t^2+4t-1}{t^4+2t^3-2t+1}\,\mathrm dt $$
Obviously what's next is factoring out the denominator and doing partial fraction decomposition, but this seems very hard.
I wanted to know if there's an easier method.
| Let $t=\tan x$ to rewrite the integral as
$$ I=\int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} dx
=\int \frac{2t+1}{(1+t^2)\sqrt{t^2+2t+2}} dt\tag1
$$
Note that
$$
\bigg( \tan^{-1} \frac{at-a^{-1}}{\sqrt{t^2+2t+2}}\bigg)’
= \frac{a^{-1} t +ab }{(t^2+b)\sqrt{t^2+2t+2}},\>\>\>\>\>
b= \frac{1+2a^2}{a^2+a^4}$$
Set $b=1$ to obtain $a^4-a^2-1=0$, which leads to $a= \sqrt{\phi}$, with $\phi=\frac{1+\sqrt5}2$ the golden ratio. Hence
$$
g(t) = \frac{\phi^{-1/2} t + \phi^{1/2} }{(t^2+1)\sqrt{t^2+2t+2}} =\bigg( \tan^{-1} \frac{\phi^{1/2}t-\phi^{-1/2}}{\sqrt{t^2+2t+2}}\bigg)’
$$
and similarly
$$
h(t) =\frac{\phi^{-1/2}- \phi^{1/2} t }{(t^2+1)\sqrt{t^2+2t+2}} =\bigg( \tanh^{-1} \frac{\phi^{-1/2}t+\phi^{1/2}}{\sqrt{t^2+2t+2}}\bigg)’
$$
Then, the integrand in (1) can be decomposed as
$$
\frac{2t+1}{(1+t^2)\sqrt{t^2+2t+2}}
= \phi^{1/2} g(t) - \phi^{-1/2} h(t)
$$
resulting in
$$I= \phi^{1/2} \tan^{-1} \frac{\phi^{1/2}t- \phi^{-1/2}}{\sqrt{t^2+2t+2}}
- \phi^{-1/2} \tanh^{-1} \frac{\phi^{-1/2}t+ \phi^{1/2} }{\sqrt{t^2+2t+2}}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1) &= \lim\limits_{x\rightarrow\infty} f(x) \\
&= \lim\limits_{x\rightarrow\infty}\frac{x^{\frac{1}{x}}-1}{\frac{\log x}{x}} \\
&= \lim\limits_{x\rightarrow\infty} \frac{x^{\frac{1}{x}-2}(1-\log x)}{\frac{x^2+\log x}{x^4}} \\
&= \lim x^{\frac{1}{x}}\lim x^2\lim\frac{1-\log x}{x^2+\log x} \\
&= 1 \cdot (+\infty)\cdot (0)\mbox{.}
\end{aligned}
This method does not work.
| Let $n=e^x.$ Then $n^{1/n}-1=$ $e^{x/e^x}-1=[x/e^x][1+(x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...]$
which lies between $x/e^x=(\log n)/n$ and $(x/e^x)(1+(x/e^x))=(\log n)/n+((\log n)/n)^2.$
Because if $x\ge 0$ then $0\le x/e^x<1$ so $$0\le (x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...\le$$ $$\le [(x/e^x)/2!][1+1/2^1+1/2^2+...]=x/e^x.$$
| {
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"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $a^a+b^b\ge a^b+b^a>1$ if $a,b> 0$. I succeeded in proving the second part of the inequality by showing that for every real number$x,y ∈ (0, 1)$, we have
$x^y≥ \frac x{x + y − xy}.$
By Bernoulli’s inequality we have
$x^{1−y}= (1 + x − 1)
^{1−y}
≤ 1 + (x − 1)(1 − y) = x + y − xy,$
$\implies$ $x^y ≥ \frac x{x + y − xy}.$
If $a ≥ 1$ or $b ≥ 1$ then the given inequality clearly holds.
So let $0 <a,b< 1.$ By the previous inequality we have
$a^b + b^a≥ \frac a{a + b− ab}
+ \frac b{a + b− ab}= \frac {a+b}
{a + b− ab}
> \frac {a+b}{a+b}= 1.$
But, I found some difficulty on how to prove the first part. Thanks in advance
| First, suppose $a>b>1$. Consider the function $f(x):=x^a-x^b$.
It suffices to show that $f$ is monotonically increasing on $[1,\infty[$.
But we can write $f(x)=x^b(x^{a-b}-1)$ which is the product of the two strictly positive strictly increasing functions on $]1,\infty[$. So $f$ is monotonically increasing in this case.
Second, Suppose $a>1>b$. Since $a>1$ then $a^{x}$ is strictly increasing and we have $a^a>a^b$. And since $b<1$ then $b^x$ is strictly decreasing and we have that $b^a<b^b$.
Third, Suppose $1>a>b$. Write $f(x):=x^a-x^b=x^a(1-x^{b-a})$ then $f$ is an increasing function on $]0,1[$ (since it is again the product of two strictly positive increasing strictly functions).
The rest follows by symmetry
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Bounds of an Expression
if $(a,b,c)>0$
and $abc(a+b+c)=3$
Then what can you say about the bounds of
$(a+b)(b+c)(c+a)$ ?
Hint:
think Geometrically!
My Approach
1.assumed $a\geq b \geq c$
2.Tried to think geometrically
3.used the Hadwiger-Finsler inequality
and modified it to get $(a+b+c) \geq \sqrt{4A\sqrt{3}+2(a^2+b^2+c^2)}$ (where A is the area of a triangle and a,b,c are it's sides)
4. I suspect that as we're to find bounds of $\Pi (a+b)$,we could try to find some vaues of (a+b)'s.
Could you please help me with this problem?
| The following inequality is true.
$$\sqrt[3]{\frac{(a+b)(a+c)(b+c)}{8}}\geq\sqrt[4]{\frac{abc(a+b+c)}{3}}.$$
For the proof use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc).$$
It gives a lower bound.
The upper bound is $+\infty$.
Can you end it now?
We need to prove that
$$\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^4\geq\left(\frac{abc(a+b+c)}{3}\right)^3.$$
Now, by using $(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$ and by AM-GM we obtain:
$$\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^4\geq\left(\frac{(a+b+c)(ab+ac+bc)}{9}\right)^4=$$
$$=\left(\frac{a+b+c}{3}\right)^3\cdot\frac{a+b+c}{3}\cdot\left(\frac{ab+ac+bc}{3}\right)^4\geq$$
$$\geq\left(\frac{a+b+c}{3}\right)^3\cdot\sqrt[3]{abc}\cdot\left(\sqrt[3]{a^2b^2c^2}\right)^4=\left(\frac{abc(a+b+c)}{3}\right)^3.$$
| {
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"url": "https://math.stackexchange.com/questions/4226354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the quadratic equation from given relatioship the quadratic equation whose roots are a and b where $a^2 +b^2=5$ and
$3(a^5+b^5)=11(a^3+b^3)$
What I Tried
$a^2 +b^2=5$ $(a+b)^2-2ab=5$ $(\text{sum of roots})^2 -2(\text{products of roots})=5$
$3(a^5+b^5)=11(a^3+b^3)$ $a^3(3a^2-11)=b^3(11-3b^2)$
| Hint: If $a+b = 0$, then you have $a = \pm \sqrt{\frac{5}{2}}, b = \mp \sqrt{\frac{5}{2}}$. From this you can find the equation. If $a + b \neq 0\implies 3(a^4-a^3b+a^2b^2-ab^3+b^4)=11(a^2-ab+b^2)= 11(5-ab)$. Using $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 = 25 - 2(ab)^2, a^3b + ab^3 = ab(a^2+b^2) = 5ab$. Thus you do have a quadratic equation in $ab$, and once you solve for $ab$, then you can solve for $a,b$ and you are done !
| {
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How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\cos\theta+1)^2}{(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)^2}$
But there were complicated calculations involved. Is there a simpler way to find the maximum of this function?
|
Step 1: Find the critical points of $f(\theta)$:
For this, it is necessary to find the derivative of the function and do $f'(\theta)=0$, such that:
\begin{align*}
f'(\theta)=&-\dfrac{2\cos^2\theta\sin\theta}{(cos\theta+1)(\sin\theta+1)^2} +\\
&+\dfrac{2\cos^2\theta}{(\cos\theta+1)(sin\theta+1)}+ \\
&+\dfrac{2\cos\theta\sin^2\theta}{(\cos\theta+1)^2(\sin\theta+1)} -\\
&-\dfrac{2\sin^2\theta}{(\cos\theta+1)(\sin\theta+1)}=0
\end{align*}
We can rewrite this:
\begin{equation}
f'(x)=\dfrac{2(\cos\theta-\sin\theta)(\cos\theta+\cos^2\theta+\sin\theta+\cos\theta\sin\theta+\sin^2\theta)}{(\cos\theta+1)^2(\sin\theta+1)^2}=0
\end{equation}
dividing by 2 both sides and reduce trigonometric functions:
\begin{equation}
\dfrac{3\cos\theta+4\cos(2\theta)+\cos(3\theta)-3\sin\theta +\sin(3\theta)}{4(\cos\theta+1)^2(\sin\theta+1)^2}=0
\end{equation}
multiplying both sides by 4 and assuming $(\cos\theta+1)\neq 0$ and $(\sin\theta+1)\neq 0$, we can treat the problem in another way, such that:
\begin{equation}
3\cos\theta+4\cos(2\theta)+\cos(3\theta)-3\sin\theta +\sin(3\theta)=0
\end{equation}
Show that this equation is equal to:
\begin{equation}
16\sqrt{2}\cos^2\left(\dfrac{\theta}{2}\right)\sin\left(-\theta+\dfrac{\pi}{4}\right)\sin^2\left( \dfrac{\theta}{2}+\dfrac{\pi}{4} \right)
\end{equation}
This implies that:
\begin{equation}
\cos^2\left(\dfrac{\theta}{2}\right)\sin\left(-\theta+\dfrac{\pi}{4}\right)\sin^2\left( \dfrac{\theta}{2}+\dfrac{\pi}{4} \right)=0
\end{equation}
We have 3 possibilities:
\begin{align}
\cos^2\left(\dfrac{\theta}{2}\right)&=0 \\
\sin\left(-\theta+\dfrac{\pi}{4}\right)&=0 \\
\sin^2\left( \dfrac{\theta}{2}+\dfrac{\pi}{4} \right)&=0
\end{align}
Solving this 3 equations, you find that:
$$\theta=\dfrac{\pi}{4}$$
Step 2: Determine the endpoints of the domain of $f(\theta)$
For $\theta=0^+$ and $\theta=\dfrac{\pi}{2}^{-}$, we have $f(\theta)=0$, this implies:
$$\fbox{$f\left(\dfrac{\pi}{4}\right)=6-4\sqrt{2}$}$$
Is the global maximum of $f(\theta)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exponent of Matrix A when A is not diagonalizable I want to compute the Matrix $A^{19}$ where
$$A = \begin{pmatrix}
6&1&0\\0&6&1\\0&0&6\end{pmatrix}$$
Since the only eigenvalue of the matrix is 6, it's not diagonalizable. How do you proceed to calculate the exponent matrix of A then?
| $ A = \begin{bmatrix} 6 && 1 && 0 \\ 0 && 6 && 1 \\ 0 && 0 && 6 \end{bmatrix} $
We can express $A^n$ as follows:
$A^n = \alpha_0 I + \alpha_1 A + \alpha_2 A^2 $
The above equation is satisfied by $\lambda = 6$, i.e.
$\lambda^n = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda^2 $
And since it is a repeated eigenvalue, we use derivatives of this last equation to obtain independent equations in the three unknowns $\alpha_0, \alpha_1, \alpha_2$, so we have
$ n \lambda^{n-1} = \alpha_1 + 2 \alpha_2 \lambda $
and
$ n(n-1) \lambda^{n-2} = 2 \alpha_2 $
Solving the above three equations for $\alpha_0, \alpha_1, \alpha_2$, we can evaluate the matrix $A^n$.
We have,
$\alpha_2 = (19)(9) (6)^{17} $
$\alpha_1 = 19 (6)^{18} - (19)(18) 6^{18} = - 323 (6)^{18}$
$\alpha_0 = (6)^{19} + (323) 6^{19} - (19)(9) 6^{19} = 153 (6)^{19}$
and we have
$A^2 = \begin{bmatrix} 6^2 && 12 && 2 \\ 0 && 6^2 && 12 \\ 0 && 0 && 6^2 \end{bmatrix} $
Putting it all together we can find the entries of $A^{19}$
$A^{19} = 153 (6)^{19} \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - 323 (6)^{18} \begin{bmatrix} 6 && 1 && 0 \\ 0 && 6 && 1 \\ 0 && 0 && 6 \end{bmatrix} + 171 (6)^{17} \begin{bmatrix} 6^2 && 12 && 2 \\ 0 && 6^2 && 12 \\ 0 && 0 && 6^2 \end{bmatrix} $
Which reduces to,
$A^{19} = \begin{bmatrix} 6^{19} && 19(6)^{18} && 57(6)^{18} \\ 0 && 6^{19} && 19 (6)^{18} \\ 0 && 0 && 6^{19} \end{bmatrix} $
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$ I was attempting this question: https://www.youtube.com/watch?v=600X-ZGNBbk
Which is the following:
$$\sum_{cyc (a,b,c)}{}\frac{a}{b^2+1} \geq \frac{3}{2}$$
Where $a+b+c = 3$ and $a,b,c > 0 $
The video itself used a solution where they added $3$ and then subtracted $a+b+c$ to rewrite the inequality into something more manageable. Though I understand this solution, it seems arbitrary to me - thinking to add 3 and then subtract the condition is somewhat out of the blue.
What I did was to first try to homogenize the inequality to get rid of the condition. This meant writing $1$ as $\frac{(a+b+c)^2}{9}$ and multiplying the numerator by $\frac{a+b+c}{3}$:
$$\sum_{cyc}^{}\frac{\frac{a(a+b+c)}{3}}{b^2+\frac{(a+b+c)^2}{9}} \geq \frac{3}{2}$$
$$\leftrightarrow $$
$$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$$
I'm fairly certain this inequality is true without the condition of $a+b+c=3$, but I can't see a way to prove it. Multiplying out seems like a lot of trouble; is there an easier way (ideally with Olympiad mathematics methods)?
| By AM-GM $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{1+b^2}\geq$$
$$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}\sum_{cyc}ab\geq3-\frac{3}{2}=\frac{3}{2}.$$
Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(10a^6+42a^5b+24a^5c-21a^4b^2+105a^4c^2+38a^3b^3)+$$
$$+\sum_{cyc}(84a^4bc-66a^3b^2c+78a^3c^2b-294a^2b^2c^2)\geq0,$$ which is true by AM-GM.
Indeed, $$66\sum_{cyc}a^4c^2=33\sum_{cyc}(a^4c^2+b^4a^2)\geq\sum_{cyc}66a^3b^2c$$ and by Muirhead
$$21\sum_{cyc}(a^5b+a^5c)\geq21\sum_{cyc}(a^4b^2+a^4c^2).$$ Thus, it's enough to prove that:
$$\sum_{cyc}(10a^6+21a^5b+3a^5c+60a^4c^2+38a^4b^3)+$$
$$+\sum_{cyc}(84a^4bc+78a^3c^2b-294a^2b^2c^2)\geq0,$$ which is AM-GM again.
I hope, now it's clear.
| {
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"timestamp": "2023-03-29T00:00:00",
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USAMO 1973 (Simultaneous Equations)
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$
Multiply equation I by 2 and subtract it from Equation II:
$$x^2-2x+y^2-2y+z^2-2z=-3$$
Complete the square for all three variables:
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
Since the RHS is zero, and the LHS has only perfect squares, by the trivial inequality, we must have:
$$(x-1)^2=0\Rightarrow x=1$$
$$(y-1)^2\Rightarrow y=1$$
$$(z-1)^2=0\Rightarrow z=1$$
Hence, the only solution for Equations I and II is
$$x=y=z=1$$
This satisfies Equation III as well. And hence, this is also the only solution to the overall system of equations.
1.
Would this be enough to get full marks?
Anything missing, or needed to be added?
2.
The complex roots is neither needed nor used anywhere. This was just used to (artificially?) increase the "complexity" of the problem. Is this correct.
| $$(x+y+z)^2=9=x^2+y^2+z^2+2(xy+yz+zx)\implies xy+yz+zx=3$$
$$(x+y+z)^3=27=6xyz+3(x+y+z)(x^2+y^2+z^2)-2(x^3+y^3+z^3)=6xyz+27-6\implies xyz=1$$
Hence by Viète's formulas $x,y,z$ are the roots of $t^3-3t^2+3t-1=(t-1)^3$, so $x=y=z=1$ is the only solution.
Note that Viète's formulas are valid even in the complex plane; the specification of "complex" in the question precludes using real-line tricks like the $x^2\ge0$ inequality in your attempt.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:
Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2).
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1},
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction
| As Kavi Rama Murthy and "user" point out, you don't need induction to show that $f(n + 1) \geq f(n)$. But (according to the usual definition of "increasing"), you need to show more than that; that for any $m$ and $n$ with $n \geq m$, $f(n) \geq f(m)$. And to derive that rigorously from $f(n + 1) \geq f(n)$, you do need induction (even though the inference is intuitively obvious).
But you can show $n \geq m \ \Rightarrow f(n) \geq f(m)$ without induction. GeometryLover and ilovemath deduce that from the premise that the sequence $n \mapsto \frac{1}{2n + 1}$ is decreasing. More directly, we have
\begin{align*}
f(n) - f(m)
&= \frac{n}{2n + 1} - \frac{m}{2m + 1}\\
&= \frac{n - m}{(2m + 1)(2n +1)}\\
&> 0,
\end{align*}
so $f(n) > f(m)$, whenever $n > m$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?
I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f(2)$ to use this condition and Bolzano's theorem. Unfortunately, I couldn't find a way to proceed.
| Let's start from $4a^2+3ab+2ac=a(4a+3b+2c)>0$.
Then we have
$8a^2+6ab+4ac>0$
Case 1: $-\frac{b}{2a}<2$
$\implies 8a^2+6ab+b^2 > b^2-4ac$
$\implies \frac{8a^2+6ab+b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$ , since $a^2 > 0, a \in \mathcal{R} \wedge a \neq 0$
$\implies 2+\frac{3b}{2a}+\frac{b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$
$\implies (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})(2+\frac{b}{2a}) > (1+\frac{b}{2a})(2+\frac{b}{2a}) > \frac{b^2-4ac}{4a^2}$, since $2+\frac{b}{2a}>0$
$\implies (2+\frac{b}{2a})^2 > \frac{b^2-4ac}{4a^2}$
$\implies (2+\frac{b}{2a})^2 - \left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2 > 0$
$\implies \left(2+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right)\left(2+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-2-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}\right)\left(-2-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}\right) > 0$
$\implies \left(-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}-2\right)\left(-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}-2\right) > 0$
$\implies -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} > 2$ (in which case there is no root in between 1 and 2,since both roots > 2)
OR $\;\;-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a} < 2$, in which case both the roots are less than 2. For the second case we need further analysis, can you take it from here?
Also, notice that $f(1)-(b+c)=a$ and $f(2)+(b+c)=4a+3b+2c$ have the same sign.
Case 2: $-\frac{b}{2a}\geq 2$
The roots are $-\frac{b}{2a}\pm\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}$, where the larger root $\geq$ 2, so that we may have at most one (the smaller) root in $(1,2)$, but it needs further analysis.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx\right)<\frac{1}{2(n+1)(n+2)}$$
We will be able to prove the above inequality
| Simply observe that for $x \in [0,1]$ and $n \in \mathbb Z^+$,
$$0 \le (1-x)x^n = (x+1)x^n - 2x^{n+1},$$ the first inequality arising from the fact that $x \ge 0$ and $1-x \ge 0$. Consequently,
$$0 \le \frac{x^{n+1}}{x+1} \le \frac{x^n}{2},$$ and integrating gives the desired inequality.
| {
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} |
Are $a=3, b=7$ the only solutions to $\sqrt{3}+\sqrt[3]{7}=\sqrt{a}+\sqrt[3]{b}$ for $a,b \in \mathbb{Q}$? So I had a lesson about calculating surds e.g. $\sqrt{5+2\sqrt{6}}$, then the teacher wrote the steps like that:
For some $a,b\in\mathbb{Q}_{\ge 0}$
$$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} \\
5+2\sqrt{6}=a+b+2\sqrt{ab} \\
$$
which directly implies $\begin{cases} a+b=5 \\ ab=6\end{cases}$. The teacher didn't know the proof and I eventually found a proof of it:
Rewrite the equation, we get:
$$5-a-b+2\sqrt{6}=2\sqrt{ab}$$
Then let $2c=5-a-b$, where $c$ should be rational. Then we continue the equation:
$$ 2c+2\sqrt{6}=2\sqrt{ab} \\ c+\sqrt{6}=\sqrt{ab} \\ (c+\sqrt{6})^2=ab \\ c^2+6-ab=-2c\sqrt{6}$$
As $c^2+6-ab$ is rational, $-2c\sqrt{6}$ is also rational, which implies $c=0$, i.e. $a+b=5$. Substituting it back to the original equation, we get $ab=6$. Q.E.D.
He found the proof quite interesting and asked a question as follows:
For rational numbers $a,b$, is $(3,7)$ the only solution to the equation $\sqrt{a}+\sqrt[3]{b}=\sqrt{3}+\sqrt[3]{7}$?
I try to prove this but it is harder than it looks. The cubic root is one of the annoying part of proving this. What I am able to prove is that $\sqrt{3}+\sqrt[3]{7}$ is irrational. Other than that, I have no idea. To prove this, I think it may probably consist of algebraic fields, which I totally have no idea. I would like to know if there are any ways to solve this as this problem is quite interesting. If you guys have any ideas, please feel free to send it out. Thanks :)
| By doing some calculations you can see that $\sqrt{3}+\sqrt[3]{7}$ is a root of the polynomial
$$
p(x) = x^6-9 x^4-14 x^3+27 x^2-126 x+22.
$$
Furthermore, $\sqrt{a}+\sqrt[3]{b}$ is a root of
$$
q_{a,b}(x) = x^6 -3 a x^4 -2 b x^3 +3 a^2 x^2 -6 a b x -a^3+b^2.
$$
The polynomial $p$ is irreducible. This means that it is the only monic degree 6 polynomial with rational coefficients of which $\sqrt{3}+\sqrt[3]{7}$ is a root. This means that if $a,b$ are rational such that $\sqrt{a}+\sqrt[3]{b}$ is equal to $\sqrt{3}+\sqrt[3]{7}$ that the polynomials $p$ and $q_{a,b}$ must be the same. Now you can look for example at the coefficients in front of the $x^3$ and $x^4$ terms to find that this implies that $a=3$ and $b=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't have an explicit closed-form.
An integral relation:
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=-\frac{3}{2}J(k)+\frac{\pi^2}{8k}
\left ( \operatorname{Li}_2\left ( \frac{1}{k} \right )
-\operatorname{Li}_2\left ( -\frac{1}{k} \right ) \right )
+\frac{\ln k}{8k} \ln^3\left ( \frac{k-1}{k+1} \right ),\qquad (k>1).$$
With some calculations, we followed that
$$\int_{0}^{\infty} \frac{\arctan^3x}{1+k^2x^2} \text{d}x
=\frac{\pi^4}{64k}
+\frac{3}{4k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1} \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1} \right )
\right ) +\frac{3\pi^2}{8k} \operatorname{Li}_2\left (- \frac{k-1}{k+1} \right ),\qquad(k>1).$$
Then the final result of $J(k)$ is
$${\color{Green}{J(k)
=\frac{\pi^2}{12k}
\left ( \operatorname{Li}_2\left ( \frac{1}{k} \right )
-\operatorname{Li}_2\left ( -\frac{1}{k} \right ) \right )
+\frac{\ln k}{12k} \ln^3\left ( \frac{k-1}{k+1} \right )
-\frac{\pi^4}{96k} -\frac{1}{2k}\left ( \operatorname{Li}_4\left ( \frac{k-1}{k+1} \right )
-\operatorname{Li}_4\left (- \frac{k-1}{k+1} \right )
\right ) -\frac{\pi^2}{4k} \operatorname{Li}_2\left (- \frac{k-1}{k+1} \right ),\qquad (k>1).}}$$
| A solution idea of calculating the integral by Cornel Ioan Valean
The post is extremely short since I have no time, but once you know what to do, all is trivial. So, what do to? Observe that any $J(2n,k)$ is half the real part of the integral over the positive real line.
$$\int_{0}^{1} \frac{\log(x)^{2n}\log\left(\frac{1-x}{1+x}\right)}{(x-1)^2-k^2(x+1)^2}\text{d}x=\frac{1}{2}\Re\biggr\{\int_{0}^{\infty} \frac{\log(x)^{2n}\log\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x\biggr\},$$
which is easy to reach by also using a simple substiution.
Once you have reached this point, exploit multiple integrals involving PV integrals as in the calculations of $J(s)$ here https://math.stackexchange.com/q/3488566.
The rest is easy, boring work to do.
End of story
| {
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"url": "https://math.stackexchange.com/questions/4247363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
$\sum_{i = 0}^{K - 1} {[\frac{1}{i + 1}\prod_{j=i+2}^{K}{(1 - \frac{1}{jq})}]} = q$ I've come across an interesting sum of products series. I know the sum, I can more or less see why it is so, but I'm stuck at a certain point of the proof.
Let $0 <q \leq 1$ be some fraction, $K$ some large positive integer.
$\sum_{i = 0}^{K - 1} {[\frac{1}{i + 1}\prod_{j=i+2}^{K}{(1 - \frac{1}{jq})}]} = q$
When I simulate it e.g. in R it seems legit:
K <- 1000
s <- 0
q <- 0.5
for (i in 0:(K - 1)) {
p <- 1
for (j in (i + 2): K) {
p <- p * (1 - 1 / (q *j))
}
s <- s + (1 / (i + 1)) * p
}
s
[1] 0.499996
And I can pretty much see how this happens but am stuck below:
$\sum_{i = 0}^{K - 1} {[\frac{1}{i + 1}\prod_{j=i+2}^{K}{(1 - \frac{1}{jq})}]} =$
$\frac{1}{1} \cdot (1 - \frac{1}{2q})(1 - \frac{1}{3q}) \cdots (1 - \frac{1}{Kq}) + \frac{1}{2} \cdot (1 - \frac{1}{3q}) \cdots (1 - \frac{1}{Kq}) + \cdots + \frac{1}{K} \cdot 1=$
$= \frac{1}{1} \cdot (\frac{2q - 1}{2q})(\frac{3q - 1}{3q}) \cdots (\frac{Kq - 1}{Kq}) + \frac{1}{2} \cdot (\frac{3q - 1}{3q}) \cdots (\frac{Kq - 1}{Kq}) + \cdots + \frac{1}{K} \cdot 1=$
$= \frac{(2q - 1)(3q - 1) \cdots (Kq - 1)}{\frac{K!}{0!}q^{K - 1}} + \frac{(3q - 1)(4q - 1) \cdots (Kq - 1)}{\frac{K!}{1!}q^{K - 2}} + \cdots + \frac{1}{\frac{K!}{(K - 1)!}q^{K - K}}$
(I can see the gist of all the $q^{K - i}$ disappearing but beyond that I'm lost)
| If you are familiar with Pochhammer symbols
$$\prod_{j=i+2}^{K}\left(1 - \frac{1}{jq}\right)=\frac{\left(i+2-\frac{1}{q}\right)_{K-i-1}}{(i+2)_{K-i-1}}$$
$$\sum_{i = 0}^{K - 1}\frac{\left(i+2-\frac{1}{q}\right)_{K-i-1}}{(i+2)_{K-i-1}}=q+\frac{q^2 \,\Gamma \left(K-\frac{1}{q}+1\right)}{\Gamma (K+1)\, \Gamma
\left(-\frac{1}{q}\right)}$$
If you make $q=\frac 12$ the result is then exactly $\frac 12 \forall K but this is the only case.
Close to $q=\frac 12$, a series expansion would give
$$\frac{1}{2}+\left(1+\frac{2}{K(K-1) }\right) \left(q-\frac{1}{2}\right)+O\left(\left(q-\frac{1}{2}\right)^2\right)$$
Now, if $K$ is large, using Stirling approximation
$$\frac{\Gamma \left(K-\frac{1}{q}+1\right)}{\Gamma (K+1)}=K^{-1/q}\Bigg[1+\frac{1-q}{2 K q^2}+O\left(\frac{1}{K^2}\right) \Bigg]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$.
Let $a,b,c,d$ be real numbers such that $a+b+c+d=0$ and $abcd=1$. Find
the minimum value of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$.
By $\text{Vieta}$'s theorem, $a,b,c,d$ are the roots of the equation $x^4+sx^2+tx+1=0$. Thus
\begin{align*}
&~~~~~~a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2\\&=(1-s)(a^2+b^2+c^2+d^2)-4\\
&=(1-s)[(a+b+c+d)^2-2(ab+ac+ad+bc+bd+cd)]-4\\
&=(1-s)(0-2s)-4=2(s^2-2s-4).
\end{align*}
Now, we only need find the range of $s$. But how to do?
| Let $f(a,b,c,d,\lambda,\mu)=\sum\limits_{cyc}(a^4+a^2)+\lambda(a+b+c+d)+\mu(abcd-1).$
Thus, in the minimal point we need $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial d}=\frac{\partial f}{\partial \lambda}=\frac{\partial f}{\partial a}=\frac{\partial f}{\partial \mu}=0.$$
Let $a$, $b$ and $c$ be different numbers.
Thus, since $$4a^3+2a+\lambda+\mu bcd=0$$ or
$$4a^4+2a^2+\lambda a+\mu abcd=0,$$ which with $$4b^4+2b^2+\lambda b+\mu abcd=0$$ gives
$$4a^4+2a^2+\lambda a=4b^4+2b^2+\lambda b$$ or
$$(a-b)(4(a^2+b^2)(a+b)+2(a+b)+\lambda)=0$$ or
$$4(a^2+b^2)(a+b)+2(a+b)+\lambda=0.$$
Similarly $$4(a^2+c^2)(a+c)+2(a+c)+\lambda=0$$ and $$4(c^2+b^2)(c+b)+2(c+b)+\lambda=0,$$ which gives
$$4(a^2+b^2)(a+b)+2(a+b)=4(a^2+c^2)(a+c)+2(a+c)$$ or
$$(b-c)\left(1+2\sum_{cyc}(a^2+ab)\right)=0$$ or
$$1+\sum_{cyc}(a+b)^2=0,$$ which is impossible.
We got a contradiction.
Thus, for any three variables at least two of them are equal.
Id est, it's enough to find a minimal value in the following cases.
*
*Three variables are equal.
Let $b=c=a$.
Thus, $d=-3a$ and $-3a^4=1,$ which is impossible;
*$a=b$ and $c=d$.
In this case $a+c=0$ and $a^2c^2=1,$ which gives $(a,b,c,d)=(1,1,-1,-1)$.
In the general we obtain any symmetric permutations of this, which gives a value $8$
and it's a minimal value because the maximal value does not exist.
Also, by AM-GM $$\sum_{cyc}(a^4+a^2)\geq4|abcd|+4\sqrt{|abcd|}=8,$$ and the equality occurs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression:
a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$
b) $$P=\frac{a^3}{2a+bc} + \frac{b^3}{2b+ac} + \frac{c^3}{2c+ab}, a+b+c=2$$
Only use AM-GM and Cauchy-Schwarz inequalities.
The problem here is that a, b, c are the 3 sides of the triangle so it has the triangle inequality so I can't find the equal condition.
Edit: My attempt:
a) I tried multiplying each fraction with the numerator variable, and then use Cauchy-Schwarz Engel's form and then it will become $$P\geq\frac{(2a+3b+4c)^2}{2ab+2bc+2ac-a^2-b^2-c^2}$$Let the RHS be Q. After that I use AM-GM with $2ab,2bc,2ac$ and then it will become $$Q\geq\frac{(2a+3b+4c)^2}{a^2+b^2+c^2}$$And then I'm stuck.
b) I tried dividing each fraction with the numerator, and then use Cauchy-Schwarz and get $$P\geq\frac{4}{6+\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}}$$And I'm stuck too.
| The first problem.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and by AM-GM we obtain:
$$\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}=\frac{4(y+z)}{2x}+\frac{9(x+z)}{2y}+\frac{8(x+y)}{z}=$$
$$=\frac{2y}{x}+\frac{9x}{2y}+\frac{2z}{x}+\frac{8x}{z}+\frac{9z}{2y}+\frac{8y}{z}\geq6+8+12=26.$$
The equality occurs for $2y=3z,$ $z=2x$ and $3z=4y,$ id est, occurs, which says that we got a minimal value.
The second problem.
By C-S Schur and Muirhead we obtain: $$\sum_{cyc}\frac{a^3}{2a+bc}=\sum_{cyc}\frac{a^4}{2a^2+abc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2+abc)}=$$
$$=\frac{2(a^2+b^2+c^2)^2}{(a+b+c)^2(a^2+b^2+c^2)+3abc(a+b+c)}=$$
$$=\frac{\sum\limits_{cyc}(2a^4+4a^2b^2)}{\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}=$$
$$=\frac{1}{2}+\frac{\sum\limits_{cyc}(3a^4-2a^3b-2a^3c+6a^2b^2-5a^2bc)}{2\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}=$$
$$=\frac{1}{2}+\frac{\sum\limits_{cyc}(3(a^4-a^3b-a^3c+a^2bc)+a^3b+a^3c+6a^2b^2-8a^2bc)}{2\sum\limits_{cyc}(a^4+2a^2b^2+2a^3b+2a^3c+5a^2bc)}\geq\frac{1}{2}.$$
The equality occurs for $a=b=c,$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the measure of the PQ segment in the figure below? Given a semi-circumference of diameter AB and center «O» and
let P and Q be points on this curve, we draw $QH\perp OB (H \in OB)$ .
Calculate "PQ", if AP = 2; QH = 6, $ m\measuredangle PAQ=3m\measuredangle QAB$
My progress:(hard question)
$\triangle BAQ: 6^2 = m.n\rightarrow 36 = mn\\
6q = a.b\\
\measuredangle QPB = \measuredangle QAB = x\\
\measuredangle PBQ = \measuredangle PAQ = 3x\\
\measuredangle ABP = \measuredangle AQP = \alpha\\
\measuredangle Q + \measuredangle A = 180^o\implies \boxed{4x+\alpha =90^o}\therefore \measuredangle A = 90-\alpha\\
\measuredangle B = 3x+\alpha = 3x+90-4x = 90^o -x\\
T.Ptolomeu: b.c =e.q+2a\\
\triangle BPA: c^2+4 = q^2\rightarrow q^2 - c^2 = 4 \\
\triangle QHA: b^2 = 36+n^2\rightarrow b^2-n^2= 36$
| Here is a solution using trigonometry. Say radius of the circle is $r$ and $O$ is the center of the circle.
$\angle BQH = x, \angle QOB = 2x$
So, $a = \frac{6}{\cos x}$
$\frac{a}{2} = OB \sin x \implies r = \frac{6}{\sin 2x}$
But in $\triangle ABP, AB = 2 r = \frac{2}{\cos 4x}$
Equating, $6 \cos 4x = \sin 2x$
$6 - 12 \sin^2 2x = \sin 2x$
Solving the quadratic, $\sin 2x = \frac{2}{3}$
$\cos 4x = \frac{1}{9}, r = 9$
$\cos 2x = \frac{\sqrt5}{3}, \cos x = \frac{\sqrt5 + 1}{2 \sqrt3}, \sin x = \frac{\sqrt5 - 1}{2 \sqrt3}$
So, $\sin 3x = \frac{7 + \sqrt5}{6 \sqrt3}$
$ \small PQ = 2 r \sin 3x = 7 \sqrt3 + \sqrt{15} \approx 16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$ Q:
Integrate $I=\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$
My Approach:
Put $$\tan^{-1}x=t\to x=\tan t$$
Also we have, $$\frac{dx}{1+x^{2}}=dt$$
We get, $$I=\int_{0}^{\frac{\pi}{2}}\frac{t}{\tan t}dt$$
I'm stuck here, how do I proceed further? I tried integration by parts but it doesn't seem to work out for me.
Edit: I tried again and I got the answer
| Another way using Feynmann trick
$$\frac d {da}\int_0^\infty \frac{\tan ^{-1}(a x)}{x \left(x^2+1\right)}\,dx=\int_0^\infty \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}$$
$$\frac{1}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac{a^2}{\left(a^2-1\right) \left(a^2x^2+1\right)}-\frac{1}{\left(a^2-1\right) \left(x^2+1\right)}$$
$$\int \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac{a \tan ^{-1}(a x)-\tan ^{-1}(x)}{a^2-1}$$
Assuming $a>0$
$$\int_0^\infty \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac \pi{2(1+a)}$$ Integrating from $0$ to $1$ then
$$\int_0^\infty \frac{\tan ^{-1}( x)}{x \left(x^2+1\right)}\,dx=\frac{1}{2} \pi \log (2)$$ and making it more general
$$\int_0^\infty \frac{\tan ^{-1}(k x)}{x \left(x^2+1\right)}\,dx=\frac{1}{2} \pi \log (k+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How to solve $ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0$ Hello friends please help me to solve this equation.
$ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0.$
Here is what I have done so far
$ \Rightarrow \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3\sqrt2(\frac{1}{\sqrt2}\sin( 2x) - \frac{1}{\sqrt2}\cos (2x) )+ \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x )=0$
$\Rightarrow 2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3(\frac{1}{\sqrt2}\sin(2x) - \frac{1}{\sqrt2}\cos (2x) )=0$
$\Rightarrow 3\cos(\frac{\pi}{4} + 2x) -2\cos(\frac{\pi}{4}+x) =0$
Please help me to continue this method and I'd there is an easier way please tell me it too.
| You changed a couple of $3x$ terms into $x$, so your attempt is faulty.
Consider instead that
$$
\sin 3x+\sin x=2\sin2x\cos x,\qquad
\cos 3x+\cos x=2\cos2x\cos x
$$
so your equation becomes
$$
3(\cos2x-\sin2x)-2\cos x(\cos2x-\sin2x)=0
$$
| {
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Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9$ as we only care $\pmod 9.$
So $(x+a)(x+b)(x+c)\equiv 1,2,0\mod 9.$
Note that $$\frac{\prod_{x=0}^8 (x+a)}{(a+2)(a+5)(a+8)}\cdot \frac{\prod_{x=0}^8(x+b)}{(b+2)(b+5)(b+8)}\cdot \frac{\prod_{x=0}^8(x+c)}{(c+2)(c+5)(c+8)}= [1\cdot 2\cdot 4\cdot 5\cdot 7\cdot 8]^3\equiv -1^3\equiv -1 $$
I am stuck. Solutions are appreciated.
| Perhaps a slight hint should do good, it seems (not entirely sure if this should work properly, but let's see how far insights take us).
As you saw in my comment, $x$ can either be $9k-1$ or an integer root of $x^2 - x + 1 - 9k = 0$. Let's take the first case.
For $x = 9k -1$, we see $9 \mid x^3 + 1$, so turning our focus onto $(x+a)(x+b)(x+c)$, on putting $x = 9k -1$, we get $(9k - 1 + a)(9k - 1 +b)(9k - 1 + c)$
From what I can see of this, if any of $a,b,c$ is $1\pmod{9}$, then surely we $9$ divides the expression. But also look at the specific condition that has to be satisfied by $x$ : it has to be $-1 \pmod{9}$ and not all integers are $-1 \pmod{9}$.
I guess that you can perhaps proceed...
Note: before downvotes, please hint out to me my errors, because I am still in the basics of modular arithmetic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Challenging series of a fraction with a cosine and a variable Is it possible to express the following series explicitly (e.g. as a polynomial in $\alpha$):
$$f(\alpha )=\sum_{k=1}^{\infty} \frac{\cos(\alpha2\pi k)}{(2\pi k)^2((2\alpha k)^2-1)^2} , $$
where $0\leq\alpha\leq1$ ?
Do you have any idea on how to approach this? Thank you so much!
| $$S_0(a)=\sum_{k=1}^{\infty} \frac{\cos(2\pi ka)}{(2\pi k)^2((2ka)^2-1)^2}=\frac{1}{4\pi^2}\frac{1}{16a^4}\Re \frac{1}{2}S(a)$$
where
$$S(a)=\sum_{\binom{k=-\infty}{k\neq0}}^{\infty}\frac{e^{2\pi ika}}{k^2(k-1/2a)^2(k+1/2a)^2} ,\,\,a\in(0;1), \,\,a\neq\frac{1}{2n} \text(n - integer)$$
To evaluates $S(a)$ we go in the complex plane and integrate the function $f(z)=\frac{2\pi i \,e^{2\pi iza}}{(e^{2\pi iz}-1)z^2(z-1/2a)^2(z+1/2a)^2} $ along a big circle with the radius $R\to\infty$.
The integrand declines fast enough at $|z|\to \infty $, so
$$\oint f(z)dz\to 0$$
On the other hand,
$$\oint f(z)dz =2\pi i\sum Res $$
We have inside the cirlcle the simple poles in the points $z=\pm1, \pm2,...$ , the pole of the third degree at $z=0$, and a couple of simple poles at $z=\pm1/2a$
The residues at $z=\pm1, \pm2,...$ give us the sum $S(a)$, therefore
$$S(a)=-Res_{z=0}f(z)-Res_{z=\pm1/2a}f(z)$$
The easiest way to evaluate $Res_{z=0}$ is to decompose the integrand at $z=0$ and evaluate the coefficient at $z^2$
$$f(z)=-2\pi i \,16a^4\frac{1+2\pi iaz-2\pi^2a^2z^2+...}{(1-1-2\pi iz+2\pi^2z^2-(2\pi iz)^3/3!+...)z^2(1-4a^2z^2)^2}$$
$$-Res_{z=0}f(z)=-(2a)^4(8a^2-2\pi^2a^2-\pi^2/3+2\pi^2a)$$
The residues evaluation at $z=\pm1/2a$ is straightforward:
$$-Res_{z=1/2a}f(z)=-\frac{8\pi ia^4}{e^{\pi i/a}-1}\Big(6a-2\pi ia+2\pi i\frac{e^{\pi i/a}}{e^{\pi i/a}-1}\Big)$$
$$-Res_{z=-1/2a}f(z)=-\frac{8\pi ia^4}{e^{-\pi i/a}-1}\Big(-6a-2\pi ia+2\pi i\frac{e^{-\pi i/a}}{e^{-\pi i/a}-1}\Big)$$
Taking all together,
$$S(a)=16a^4(\pi^2/3-\pi^2a+2\pi^2a^2-8a^2)-\frac{8\pi^2a^4}{\sin^2\frac{\pi}{2a}}-48\pi a^5\cot\frac{\pi}{2a}$$
Finally,
$$S_0=\frac{1}{4\pi^2}\frac{1}{16a^4}\frac{1}{2}S(a)=\frac{1}{24}-\frac{a}{8}+\frac{a^2}{4\pi^2}(\pi^2-4)-\frac{1}{16\sin^2\frac{\pi}{2a}}-\frac{3}{8\pi}\,a\cot\frac{\pi}{2a}$$
| {
"language": "en",
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"answer_id": 0
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Analytic expressions for two definite integrals containing trigonometric functions and a real parameter I encountered these two integrals in my research about the physics of Josephson junctions:
$$\int^{\pi}_{-\pi}\frac{dx}{\sqrt{1+\sqrt{1-D\sin^2\frac{x}{2}}}} $$
$$\int^{\pi}_{-\pi}\sqrt{1+\sqrt{1-D\sin^2\frac{x}{2}}} dx $$
where D is a real number between 0 and 1.
I wonder if there's an analytic expression for either of these two integrals. Mathematica doesn't seem to help.
|
Define the two functions $\mathcal{I}:(0,1)\rightarrow\mathbb{R}$ and $\mathcal{J}:(0,1)\rightarrow\mathbb{R}$ via the respective integrals,
$$\mathcal{I}{\left(D\right)}:=\int_{-\pi}^{\pi}\mathrm{d}\omega\,\frac{1}{\sqrt{1+\sqrt{1-D\sin^{2}{\left(\frac{\omega}{2}\right)}}}}$$
and
$$\mathcal{J}{\left(D\right)}:=\int_{-\pi}^{\pi}\mathrm{d}\omega\,\sqrt{1+\sqrt{1-D\sin^{2}{\left(\frac{\omega}{2}\right)}}}.$$
Suppose $a\in(0,1)$, and note that $0<\sqrt{1-a^{2}}<1$. Setting $b:=\sqrt{1-a^{2}}$, we have
$$\begin{align}
\mathcal{I}{\left(a^{2}\right)}
&=\int_{-\pi}^{\pi}\mathrm{d}\omega\,\frac{1}{\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\frac{\omega}{2}\right)}}}}\\
&=2\int_{0}^{\pi}\mathrm{d}\omega\,\frac{1}{\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\frac{\omega}{2}\right)}}}};~~~\small{even\,symmetry}\\
&=2\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{2}{\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\theta\right)}}}};~~~\small{\left[\omega=2\theta\right]}\\
&=4\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}\sqrt{1+\sqrt{1-a^{2}x^{2}}}};~~~\small{\left[\theta=\arcsin{\left(x\right)}\right]}\\
&=4\int_{0}^{a}\mathrm{d}y\,\frac{1}{\sqrt{a^{2}-y^{2}}\sqrt{1+\sqrt{1-y^{2}}}};~~~\small{\left[ax=y\right]}\\
&=4\int_{1}^{\sqrt{1-a^{2}}}\mathrm{d}t\,\frac{(-1)t}{\sqrt{1-t^{2}}}\cdot\frac{1}{\sqrt{a^{2}-1+t^{2}}\sqrt{1+t}};~~~\small{\left[y=\sqrt{1-t^{2}}\right]}\\
&=4\int_{\sqrt{1-a^{2}}}^{1}\mathrm{d}t\,\frac{t}{\sqrt{1+t}\sqrt{1-t^{2}}\sqrt{t^{2}-\left(1-a^{2}\right)}}\\
&=4\int_{b}^{1}\mathrm{d}t\,\frac{t}{\left(1+t\right)\sqrt{\left(1-t\right)\left(t^{2}-b^{2}\right)}},\\
\end{align}$$
and
$$\begin{align}
\mathcal{J}{\left(a^{2}\right)}
&=\int_{-\pi}^{\pi}\mathrm{d}\omega\,\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\frac{\omega}{2}\right)}}}\\
&=2\int_{0}^{\pi}\mathrm{d}\omega\,\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\frac{\omega}{2}\right)}}};~~~\small{even\,symmetry}\\
&=4\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\sqrt{1+\sqrt{1-a^{2}\sin^{2}{\left(\theta\right)}}};~~~\small{\left[\omega=2\theta\right]}\\
&=4\int_{0}^{1}\mathrm{d}x\,\frac{\sqrt{1+\sqrt{1-a^{2}x^{2}}}}{\sqrt{1-x^{2}}};~~~\small{\left[\theta=\arcsin{\left(x\right)}\right]}\\
&=\int_{0}^{a}\mathrm{d}y\,\frac{4\sqrt{1+\sqrt{1-y^{2}}}}{\sqrt{a^{2}-y^{2}}};~~~\small{\left[ax=y\right]}\\
&=\int_{1}^{\sqrt{1-a^{2}}}\mathrm{d}t\,\frac{(-1)t}{\sqrt{1-t^{2}}}\cdot\frac{4\sqrt{1+t}}{\sqrt{a^{2}-1+t^{2}}};~~~\small{\left[y=\sqrt{1-t^{2}}\right]}\\
&=\int_{\sqrt{1-a^{2}}}^{1}\mathrm{d}t\,\frac{t}{\sqrt{1-t^{2}}}\cdot\frac{4\sqrt{1+t}}{\sqrt{t^{2}-\left(1-a^{2}\right)}}\\
&=4\int_{b}^{1}\mathrm{d}t\,\frac{t}{\sqrt{\left(1-t\right)\left(t^{2}-b^{2}\right)}}.\\
\end{align}$$
Hence, both $\mathcal{I}$ and $\mathcal{J}$ are ultimately elliptic integrals and can be reduced to standard form by the usual algorithms.
| {
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What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $\triangle ABC$. (Answer:3)
My progress:
I made the drawing
Inradius = r
Circumradius = R
$r=\frac{a+c-b}{2}=\frac{ab}{a+b+c}\\
R = \frac{b}{2}\\
\frac{R}{r} = \frac{b}{a+c-b}$
| Using standard notations for side lengths,
Please note that in $\triangle ADB$, as $AI$ is the angle bisector,
$AD:c = 3:4 \implies AD = \frac{3c}{4}$
Similarly in $\triangle CBD, CD = \frac{3a}{4}$
So, $b = 2R = AD + CD = \frac{3}{4} (a + c)$
$a + c = \frac{8R}{3} \tag1$
Now use the fact that $\triangle ABC$ is a right triangle and hence,
$r = \frac{1}{2} (a + c - 2R)$
| {
"language": "en",
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Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
| Let $\sqrt {\log n}=u,\thinspace u≥0$, then we have
$$\begin{align}&\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n}{2}\\
\iff &\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n^2}{2n}\\
\iff &n≥2^{\sqrt{\log(n)}-1}\\
\iff &e^{u^2}≥2^{u-1}\\
\iff &e^{u^2}>e^{u-1}>2^{u-1}.\end{align}$$
This means, the equality is impossible.
Therefore, we have
$$\frac{n^2}{2^{\sqrt{\log(n)}}} > \frac{n}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int \frac{1}{a+b\sec(x)}dx$ I tried to find a closed-form expression for the integral
$$\int\frac{1}{a+b\sec(x)}\>dx$$ and, afterwards, set $a=0$, $b=1$ to recover the result for the simplified integral
$$\int\cos(x)\>dx=\sin(x)+C$$
The above integral was inspired by Américo Tavares’s rather illuminating solution to Ways to evaluate $\sec(t)$,
Here's the progress I've made which makes use of the integral
$$(\star)\int\frac{dx}{m^2+n^2x^2}=\frac{1}{mn}\arctan\left(\frac{nx}{m}\right)\quad m,n>0$$
Using $\cos(x)=2\cos^2(x/2)-1$, we get the following
$$I=\int\frac{dx}{a+b\sec(x)}=\int\frac{\cos(x)}{a\cos(x)+b}dx=\int\frac{2\cos^2(x/2)-1}{2a\cos^2(x/2)+b-a}dx\\=2\underbrace{\int\frac{dx}{2a+(b-a)\sec^2(x/2)}}_\color\red{(1)}-\underbrace{\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx}_\color\red{(2)}$$
$\color\red{(2)}$ becomes
$$\int\frac{\sec^2(x/2)}{2a+(b-a)\sec^2(x/2)}dx=\int\frac{\sec^2(x/2)}{a+b+(b-a)\tan^2(x/2)}dx\\=\frac{2}{\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)$$
from the substitution $u=\tan(x/2)$ and then applying $(\star)$.
For tackling $\color\red{(1)}$, the same u-substitution and applying partial fractions converts the integral into
$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=2\int\frac{du}{(a+b+(b-a)u^2)(1+u^2)}\\=2\left[\frac{a-b}{2a}\int\frac{du}{(a+b+(b-a)u^2)}+\frac{1}{2a}\int\frac{du}{1+u^2}\right]$$
Thus
$$\int\frac{dx}{2a+(b-a)\sec^2(x/2)}=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}u\right)+\frac{1}{a}\arctan(u)\\=\frac{a-b}{a\sqrt{b^2-a^2}}\arctan\left(\sqrt{\frac{b-a}{a+b}}\tan(x/2)\right)+\frac{1}{a}\arctan(\tan(x/2)))$$
However, the problem is that setting $a=0$ afterwards results in the denominator becoming zero. Also, I couldn't think another approach to this.
Any help?
| As @Hussain-Alqatari commented, using immediately the tangent half angle substitution, you have
$$I=\int \frac{dx}{a+b\sec(x)}=\int \frac{2(1-t^2)}{ (b-a)t^4+2 b t^2+(a+b)}\,dt$$ Assuming $b\neq a$
$$I=\frac 2{b-a}\int \frac {1-t^2}{(t^2+1)(t^2-\alpha)}\,dt \qquad \text{with} \qquad \alpha=\frac{a+b}{a-b}$$ Using partial fraction decomposition
$$ \frac {1-t^2}{(t^2+1)(t^2-\alpha)}=-\frac{2}{1+\alpha }\frac 1{t^2+1}+\frac{1-\alpha }{1+\alpha }\frac 1 {t^2-\alpha}$$ which is very simple
$$\color{blue}{I=\frac{2 }{a}\tan ^{-1}(t)-\frac {2b}{a \sqrt{a^2-b^2}}\tanh ^{-1} \left(\sqrt{\frac{a-b}{a+b}}t\right)}$$
| {
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"source": "stackexchange",
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Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$.
My work-
$\frac{10^n}{x}-1<\lfloor \frac{10^n}{x}\rfloor≤\frac{10^n}{x}\Rightarrow\frac{10^n}{x}-1<1989≤\frac{10^n}{x}\Rightarrow\frac{10^n}{1990}<x≤\frac{10^n}{1989}$
I am unable to proceed beyond this. Any help or other method is appreciated.
| $\frac{10^4}5=2000>1989$
$\Rightarrow x>5 $ for n=4
let $x=5.1$ for $10^4$ or 51 for $10^5$ we have:
$\frac{10^5}{51}=1960$
$\Rightarrow 5.1>x>5$ for $n=5$
let $x=50.1$ for $10^5$ or 501 for $10^6$ we have:
$\frac{10^6}{501}=1996$
$\Rightarrow 502>x>501$ for n=6
let $x=502$ we have:
$\frac {10^6}{502}=1992$
$\Rightarrow 503>x>502 $ for $n=6$ or $5020<x<5030$ for $10^7$
With few try we find $x=5027$ is the integer solution for $n=7$:
$\frac{10^7}{5027}=1989.25$
or:
$\lfloor{\frac {10^7}{5027}}\rfloor=1989$
| {
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"answer_count": 2,
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} |
Find $a$ for which $y\ge x^2+4a^2,x\ge y^2+4a^2$ has only one solution For what $a$ does this system of inequalities have only one solution?
$$
\begin{cases}y\ge x^2+4a^2\\x\ge y^2+4a^2\end{cases}
$$
So, I have tried getting $y$ from the first equation then putting it into second, but then I have two quadratic and $x$, which of these quadratics are the first term? I know that I have to put it into $\Delta=0$. But don't really get which of them are first, second etc.
Even tried subtracting second to first.
| We have,
$$\begin{align}&\begin{cases}y^2≥\left(x^2+4a^2\right)^2≥0\\y^2≤x-4a^2,\thinspace x≥4a^2\end{cases}\\\\
\implies &\left(x^2+4a^2\right)^2≤y^2≤x-4a^2\\
\implies &\left(x^2+4a^2\right)^2≤x-4a^2 \\
\implies &16a^4+a^2(8x^2+4)+(x^4-x)≤0\\
\implies &\left(4a^2+\left(x^2+\frac 12\right)\right)^2≤\left(x+\frac 12\right)^2\\
\implies &x^2-x+4a^2≤0\\
\implies &\left(x-\frac 12\right)^2≤\frac{1-16a^2}{4}\end{align}$$
Similary, we have
$$\begin{align}\left(y^2+4a^2\right)^2≤y-4a^2\\
\implies \left(y-\frac 12\right)^2≤\frac{1-16a^2}{4}\end{align}$$
We see that, if $x,y$ are unique then
$$16a^2=1\iff a=±\frac 14$$
which implies,
$$x=y=\frac 12.$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Integrating a two complex functions I need to evaluate the norm $||G(s)||_2$ where $G(s) = \frac{\alpha_1 s+\alpha_o}{s^2+\beta_1s+\beta_o }$ is a transfer function.
And
$||G(s)||^2_2$ = $\int_{-\infty}^{\infty} trace {G(j\omega) G^*(j\omega)} \,d\omega$
I am able to get to this point
$||G(s)||^2_2$ = $\int_{-\infty}^{\infty} \frac{\alpha_1^2\omega^2 + \alpha_0^2}{\omega^4+(\beta_1^2-2\beta_o)\omega^2+\beta_o^2} \,d\omega$
The above expression is analogous to $\int_{-\infty}^{\infty} \frac{ax^2 + b}{x^4+cx^2+d^2} \,dx$
I tired splitting the functions and use square method but it doesn't seem to let me anywhere. Any suggestions in solving this integral? Thanks
| I am assuming that all the constants are $\gt 0$.
We start by noticing that the integrand is even, thus
$$I =2 \int\limits_0^\infty\frac{\alpha_1^2x^2+\alpha_0^2}{x^4+(\beta_1^2-2\beta_0)x^2+\beta_0^2 }\, \mathrm dx $$
Now, using change of variables $x^4=\beta_0^2 t^4$, we get
$$\begin{align} I &= \frac2{\beta_0^{3/2}}\int\limits_0^\infty \frac{\alpha_1^2\beta_0t^2+\alpha_0^2}{t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\,\mathrm dt \\ &= \frac{2\alpha_1^2\beta_0}{\beta_0^{3/2}}\underbrace{\int\limits_0^\infty \frac{t^2}{t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\,\mathrm dt}_{t\mapsto\frac1t}+\frac{2\alpha_0^2}{\beta_0^{3/2}} \int\limits_0^\infty \frac{\mathrm dt} {t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1} \\ I &= \frac{2(\alpha_1^2\beta_0+\alpha_0^2)}{\beta_0^{3/2}} \int\limits_0^\infty \frac{\mathrm dt} {t^4+\frac{t^2}{\beta_0}(\beta_1^2-2\beta_0)+1}\end{align}$$
Now, again using $t\mapsto \frac1t$ and adding the result, we get
$$\begin{align}I &= \frac{\alpha_1^2\beta_0+\alpha_0^2}{\beta_0^{3/2}}
\int\limits_0^\infty\frac{1+\frac1{t^2}}{t^2+\frac1{t^2}+\frac{\beta_1^2-2\beta_0}{\beta_0}}\,\mathrm dt \\ &= \frac{\alpha_1^2\beta_0+\alpha_0^2}{\beta_0^{3/2}}
\int\limits_{-\infty}^\infty\frac{\mathrm dy}{y^2+\frac{\beta_1^2}{\beta_0} }\ , t-\frac1t=y \end{align}$$
Thus, we conclude that
$$\color{red}{\boxed{\boxed{\,\int\limits_{-\infty}^\infty\frac{\alpha_1^2x^2+\alpha_0^2}{x^4+x^2(\beta_1^2-2\beta_0)+\beta_0^2}\,\mathrm dx = \frac{\pi(\alpha_1^2\beta_0+\alpha_0^2)}{\beta_0\beta_1}}}}$$
| {
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How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}(!)$$
since by C-S inequality: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c$. But (!) is not true.
Anyone can help me give a hint to solve this nice problem?. Thanks!
| Also, we can use $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$$ and
$$\sqrt{2(a^2+b^2)}\leq\frac{3a^2+2ab+3b^2}{2(a+b)}.$$
By this way we can get something stronger:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt[3]{4(a^3+b^3)}+\sqrt[3]{4(b^3+c^3)}+\sqrt[3]{4(c^3+a^3)}.$$
The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt[4]{8(a^4+b^4)}+\sqrt[4]{8(b^4+c^4)}+\sqrt[4]{8(c^4+a^4)}.$$
| {
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$\int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+n\cdot h))\, \text{d}x = ?$ I would like to dissolve this expression:
\begin{align}
f(h) = \int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+nh)) \ \text{d} x
\end{align}
I have a guess by working out the first two expressions that the following appears.
\begin{align}
f(h) = -b_n \cdot h^{n+2}
\end{align}
And I noticed that for even $n$ the integral disappears. But the $b_n$ that appear seem very chaotic. I have listed them:
\begin{align}
b_n = \left(\frac{1}{6},0,\frac{9}{10},0,\frac{1375}{84},0,\frac{57281}{90},0,\frac{1891755}{44},\ldots\right)
\end{align}
I was also able to express this product term in gamma functions. But now I don't know if this makes it so much easier to calculate the expression.
\begin{align}
f(h) &= h^{n+1}\int_a^{a+nh} \frac{\Gamma\left[\frac{a - x}{h} + 1 + n\right]}{\Gamma\left[\frac{a-x}{h}\right]} \ \text{d} x \\
&= -h^{n+2} \int_0^{-n} \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
&= h^{n+2} \int_{-n}^0 \frac{\Gamma\left[u + 1 + n\right]}{\Gamma\left[u\right]} \ \text{d} u \\
\end{align}
| A closed form can be given in terms of the Gregory co-efficients. Note that we can write $$b_n=\int_{0}^{n}x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx$$
We have, $$\int_{ }^{ }x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx=\psi_{n+2}\left(x\right)(n+1)!\tag{1}$$
Where, $\psi_n(x)$ is the $n$-th Bernoulli polynomial of the second kind and hence,
$$b_{n}=\left(\psi_{n+2}\left(n\right)-\psi_{n+2}\left(0\right)\right)(n+1)!\tag{2}$$.
Some things to consider: the linked article states the following properties of the polynomials,
$$\begin{align}\psi_{n+2}(0)&=G_{n+2}=(-1)^{n+1}|G_{n+2}|\\\psi_{n+2}(n)&=-|G_{n+2}|\end{align}$$
Where $G_n$ are the Gregory coefficients, which appears in the expansion of $x/\ln(1+x)$ for $|x|<1$,
$$\begin{align}\frac{x}{\ln(1+x)}&=1+G_1x+G_2x^2+G_3x^3+ \cdots\\&=1+\frac{1}{2}x-\frac{1}{12}x^{2}+\frac{1}{24}x^{3}-....\end{align}$$
Substituting this in $(2)$,
$$b_n=\left(\left(-1\right)^{n}-1\right)\left(n+1\right)!\left|G_{n+2}\right|\tag{3}$$
We can see from $(3)$ that for $n$ being even, $b_n=0$
| {
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Find $\lim_{n\to \infty}\frac{-7n^5-6n^4+122n^3-15n^2+7n-2}{21n^5+12n^3-501}$ I am stuck with this question, I do not know what I'm doing wrong. I am dividing everything by $n^5$ and getting end result as $\frac{-7}{21}$ but the result in book answer says -$\frac{1}{3}$. Can someone advice? My calculation:
$${\displaystyle\lim_{n\to \infty}}\frac{-7n^5-6n^4+122n^3-15n^2+7n-2}{21n^5+12n^3-501}=\frac{n^5(-7-\frac{6n^4}{n^5}+\frac{122n^3}{n^5}-\frac{15n^2}{n^5}+\frac{7n}{n^5}-\frac{2}{n^5})}{n^5(21+\frac{12n^3}{n^5}-\frac{501}{n^5})}$$
So obviously here $n^5$ cancels out, $\frac{-7}{21}$ stays and all other values go to $0$.
| You are correct, infact we have
$$\frac{-7}{21} = \frac{-7}{7 \times 3}=\frac{-1}{3}$$
| {
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What's the measure of the segment $PQ$ in the triangle below? For reference:
NIn figure CQ is bisector, and
PE = EC, AC = b and BC = a. Calculate PQ.(Answer: $\frac{ab}{a+b}$)
My progress:
$(Bissector ~Th.):\triangle ABC:\\ \frac{BQ}{AQ}=\frac{a}{b} \\
By~ proportionality: \frac{BQ}{BQ+AQ}=\frac{a}{a+b}=\frac{BQ}{AB}$
$(Bissector~ Th.):\triangle CEP:\\ \frac{PQ}{EQ}=\frac{CP}{CE}\\
By~ proportionality: \frac{PQ}{PQ+EQ}=\frac{CP}{CE+CP}\implies \frac{PQ}{PE=EC} = \frac{CP}{CE+CP}$
$(Menelaus~ Th.): CEP-AB:\\: a \cdot PQ \cdot AE =BP \cdot EQ \cdot b\implies PQ = \frac{b}{a} \cdot.\frac{BP \cdot EQ}{AE}(I) \\ (Menelaus - Th.): \triangle ABC - PE:\\CE \cdot AQ \cdot BP = AE \cdot BQ \cdot CP \implies \frac{BP}{AE} = \frac{a}{b}\cdot\frac{CP}{CE}(II)\\
(II)in (I): PQ = \frac{CP}{CE} \cdot EQ \implies PQ = \frac{(a+PB) \cdot (b-PQ)}{(b-EA)}$
|
Draw segment $QR = QP$ then,
$\angle PRQ = \angle PCA \implies AC \parallel QR$. Also, $CQ$ is angle bisector of $\angle ACB$.
So, $ \displaystyle \frac{QR}{AC} = \frac{BQ}{BA} = \frac{BC}{BC+AC} = \frac{a}{a+b}$
$ \displaystyle \implies PQ = QR = \frac{ab}{a+b}$
| {
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Maximum eigenvalue of an adjacency matrix Let $G = (V,E)$ be an undirected graph without multiple edges.
Additionally, let $|V| = n + 1$, where $n \geq 3$.
$G$ is created in such a way, that $n$-first vertices create a cycle and $n+1$ vertex is connected only to the previous $n$-vertices.
Prove that the maximum eigenvalue of adjacency matrix generated by this graph is equal:
$$
\lambda_{0} = \sqrt{n+1} + 1
$$
Example. Let $n+1 = 10$. The $G$ looks like this - [1]. The adjacency matrix $M$ for $G$ is the following:
$$
M =
\left(\begin{array}{rrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0
\end{array}\right)
$$
The characteristics polynomial $W$ of $M$:
$$
W = \lambda^{10} - 18\lambda^8 - 18\lambda^7 + 72\lambda^6 + 90\lambda^5 - 93\lambda^4 - 126\lambda^3 + 27\lambda^2 + 34\lambda - 9
$$
Lastly, real roots of $W$ are the following:
$$
\begin{array}{lar}
\lambda_0 = 1 + \sqrt{10} \\
\lambda_1 = -1 \\
\lambda_2 = 1 - \sqrt{10} \\
\end{array}
$$
Of course, $\lambda_0$ satisfies the equality $\lambda_0 = \sqrt{n+1} + 1 = \sqrt{10} + 1 \approx 4,162277 $.
| Hint: using the symmetries in the graph, try out vectors that equal 1 on the first $n$ components and equal a variable $x$ on the last component. The eigenvalue equation gives you two equations and two unknowns; without giving too much away, you’ll end up with a quadratic that you can solve to get the desired eigenvalue.
To prove it is maximal, try using Perron-Frobenius and Courant-Fischer; what do you notice about the entries of the corresponding eigenvector?
| {
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Apply the Cauchy–Goursat theorem to compute the integral $\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})} dz$ where $C$ is the circle $|z|=2$. Apply the Cauchy–Goursat theorem to compute the integral
$$
\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})}dz,
$$
where $C$ is the circle $|z|=2$.
Solution:
$z^{4}(1-z^{2})=0 \Rightarrow$
$z=0,z=±1$
points where the funtion is not analytic.
| You need to find the residue at $0$ ,$1$ and $-1$.
For the residue at $0$ you need to calculate the coefficient of $\frac{1}{z}$ in the Laurent Series expansion around $0$.
To do that use :-
$\sinh(z)=\frac{e^{z}-e^{-z}}{2}=z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....$.
And $\frac{1}{1-z^{2}}=1+z^{2}+z^{4}+z^{6}+....$
Combining them you need to find coefficient of $\frac{1}{z}$ in:-
$$\frac{1}{z^{4}}(z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....)(1+z^{2}+z^{4}+z^{6}+....)=(z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....)(\frac{1}{z^{4}}+\frac{1}{z^{2}}+1+z^{2}+...)$$.
Clearly the coefficient is $\frac{1}{3!}+1=\frac{7}{6}$.
Now for the residues at $1$ and $-1$. You see that
$\displaystyle\frac{\frac{\sinh(z)}{z^{4}(1-z)}}{1+z}$ has a simple pole at $z=-1$ and $\displaystyle\frac{\frac{-\sinh(z)}{z^{4}(1+z)}}{z-1}$ has a simple pole at $z=1$. So their residies are respectively $\frac{\sinh(-1)}{2}$ and $\frac{-\sinh(1)}{2}$ .
So the value of the integral by Cauchy Residue Theorem is :-
$$2i\pi\left(\frac{7}{6}-\sinh(1)\right)$$
| {
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$\frac1{12}(\sin^2A+\sin^2B+\sin^2C)\leq \sin^2\frac12B\sin^2\frac12C+\sin^2\frac12C\sin^2\frac12A+\sin^2\frac12A\sin^2\frac12B$ for $A+B+C=180^\circ$
If $A+B+C=180^{\circ}$, prove that
$$\begin{align}
\tfrac{1}{12}(\sin^2 A+\sin^2 B+\sin ^2C)
&\leq
\sin^2\tfrac{1}{2}B\sin^2\tfrac{1}{2}C \\[0.5em]
&+\sin^2\tfrac{1}{2}C\sin^2\tfrac{1}{2}A \\[0.5em]
&+\sin^2\tfrac{1}{2}A\sin^2\tfrac{1}{2}B
\end{align}$$
This is probably not that difficult, but I have a mental block when it comes to inequalities.
I have been trying the identity,
$$\sin^2\frac{1}{2}A=\frac{1}{2}(1-\cos A)$$
and
$$\sin^2 A+\sin^2 B+\sin ^2C=2+2\cos A\cos B\cos C$$ which holds when
$A+B+C=180^{\circ}$.
What is the insight or trick I am missing in this problem ?
| Is known that
$$\sin^2u+\sin^2v=1-\dfrac{\cos2u+\cos2v}2=1-\cos(u-v)\cos(u+v).$$
Let
$$\alpha=\dfrac12\angle A,\quad \beta=\dfrac12\angle B.$$
Then
$$\delta=12(\sin^2\alpha\sin^2\beta+(\sin^2\alpha+\sin^2\beta) \cos^2(\alpha+\beta))-(\sin^22\alpha+\sin^22\beta+\sin^22(\alpha+\beta))$$
$$=4\big(3\sin^2\alpha\sin^2\beta-\sin^2\alpha\cos^2\alpha-\sin^2\beta\cos^2\beta$$
$$+(3\sin^2\alpha+3\sin^2\beta-\sin^2(\alpha+\beta))\cos^2(\alpha+\beta)\big)$$
$$=4\big(\sin^4\alpha+3\sin^2\alpha\sin^2\beta+\sin^4\beta-\sin^2\alpha-\sin^2\beta$$
$$+(3\sin^2\alpha+3\sin^2\beta-1+\cos^2(\alpha+\beta)) \cos^2(\alpha+\beta)\big)$$
$$=4\big((\sin^2\alpha+\sin^2\beta)^2+\sin^2\alpha\sin^2\beta-\sin^2\alpha-\sin^2\beta$$
$$+(3\sin^2\alpha+3\sin^2\beta-1+\cos^2(\alpha+\beta)) \cos^2(\alpha+\beta)\big)$$
$$=4\big((1-\cos(\alpha-\beta)\cos(\alpha+\beta))^2
-(1-\cos(\alpha-\beta)\cos(\alpha+\beta))\big)
+\big(\cos(\alpha-\beta)-\cos(\alpha+\beta)\big)^2$$
$$+4\big(2-3\cos(\alpha-\beta)\cos(\alpha+\beta)+\cos^2(\alpha+\beta)\big) \cos^2(\alpha+\beta),$$
$$\delta=\cos^2(\alpha-\beta)\big(1+4\cos^2(\alpha+\beta)\big)
-6\cos(\alpha-\beta)\cos(\alpha+\beta)\big(1+2\cos^2(\alpha+\beta)\big)$$
$$+\cos^2(\alpha+\beta)(9+4\cos^2(\alpha+\beta)).$$
Then
$$D=9\big(1+2\cos^2(\alpha+\beta)\big)^2-\big(1+4\cos^2(\alpha+\beta)\big)\big(9+4\cos^2(\alpha+\beta)\big)$$
$$=4\cos^2(\alpha+\beta)(5\cos^2(\alpha+\beta)-1).$$
If $\;\cos(\alpha+\beta)\in[0,\sqrt5\,^{-1}),\;$ then $\;D<0,\;\delta>0.\;$
Else
$$\delta=y^2(1+4x^2)-6y(1+2x^2)+9+4x^2,$$
where
$$x=\cos(\alpha+\beta)\in[\sqrt5^{-1},1],\quad y=\dfrac{\cos(\alpha-\beta)}{\cos(\alpha+\beta)}\in[1,x^{-1}].$$
If $\;\underline{x=\sqrt5^{-1}},\;$ then
$$\delta=\dfrac15(3y-7)^2\ge0.$$
If $\;\underline{x=y=1},\;$ then $\;\delta=0.\;$
In the inner stationary points
$$(2+8x^2)y-6(1+2x^2),\quad y=\dfrac{3+6x^2}{1+4x^2}>1.$$
Therefore,
$$\delta\ge 0.$$
Proved.
| {
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Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$ Eliminate $\theta$ from the equations.
$$4x=5\cos\theta -\cos 5\theta$$
$$4y=5\sin\theta -\sin 5\theta$$
Alternative forms are
$$x=5\cos^3\theta-4\cos^5\theta$$
$$y=5\sin^3\theta-4\sin^5\theta$$
and $$x=\cos^3\theta(3-2\cos2\theta)$$
$$y=\sin^3\theta(3+2\cos2\theta)$$
| These are the parametric equations of an epicycloid, where the large circle has radius $R=1$, and the smaller exterior rolling circle has radius $r=\frac{1}{4}$.
A possible simplification is writing
$$4a=4(x+i y) = 5 z - z^ 5\\
4b=4(x-i y) = \frac{5}{z} - \frac{1}{z^5}$$
where $z = \cos \theta + i \sin \theta$
and get a relation between $a$, $b$, that implies a relation between $x$ and $y$, see link. Or, we can just give the result provided by WA :
$$-81 - 45 x^2 + 365 x^4 - 15 x^6 - 480 x^8 + 256 x^{10} - 45 y^2 - 2395 x^2 y^2 - 45 x^4 y^2 - 1920 x^6 y^2 + 1280 x^8 y^2 + 365 y^4 - 45 x^2 y^4 - 2880 x^4 y^4 + 2560 x^6 y^4 - 15 y^6 - 1920 x^2 y^6 + 2560 x^4 y^6 - 480 y^8 + 1280 x^2 y^8 + 256 y^{10}=0$$
a curve of degree $10$
$4$ leaves" />
We can simplify the defining polynomial by writing it as a polynomial in $x^2 + y^2$ and $x^2 y^2$.
$\bf{Added:}$ Following the solution of @ACB: we have the polar representation ($x = r \cos \phi$, $y=r \sin \phi$)
$$\cos^2 2 \phi = \frac{(9 - 4 r^2)(16 r^4+ 3 r^2 + 6)^2}{3125 r^4}= f(r)$$
The function in $r>0$ on RHS is strictly decreasing on $(0, \infty)$. Indeed, we have
$$f'(r) = -24\, \frac{(-1 + r^2)^2 (9 + 16 r^2) (6 + 3 r^2 + 16 r^4)}{3125\, r^5}$$
Now, $f(1) = 1$, and $f(\frac{3}{2})=0$. We see that our curve is situated between the circles of radius $1$, and $\frac{3}{2} = 1 + 2\cdot \frac{1}{4}$, as it should.
So now we can write the curve in polar coordinates
$$r = f^{-1}(\cos^2 2\phi)$$
where $f^{-1}\colon [0,1]\to [1, \frac{3}{2}]$ is the inverse function of $f$.
| {
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Existence of integer coefficient polynomial such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ The original problem asked to "Show that there exists a polynomial $f(x)$ with rational coefficients such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$."
I was wondering if there exists an integer coefficient polynomial. Intuitively, I think the answer should be no.
| We know that $\sqrt2 + \sqrt3$ is a root of the polynomial $$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2-\sqrt3) = x^4 - 10x^2+1.$$
So in any polynomial $f$ of degree at least $4$, we can replace the leading term $a_n x^n$ by $a_n(10x^{n-2} - x^{n-4})$, reducing the degree while not changing the value of $f(\sqrt2 + \sqrt3)$. This means that it's enough to check all cubic polynomials.
If $x=\sqrt2+\sqrt3$, then out of $x^0 = 1$, $x^1 = \sqrt2 + \sqrt3$, $x^2 = 5 + 2\sqrt6$, and $x^3 = 11\sqrt2 + 9 \sqrt3$, only $x$ and $x^3$ contain any square roots of $2$. There is only one linear combination of $\sqrt2 + \sqrt3$ and $11\sqrt2 + 9\sqrt3$ that gives $\sqrt2$, and it does not have integer coefficients.
| {
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Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$
I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it results in a quintic polynomial in terms of $\cos x $. Wolfram alpha has simplified the left hand side all the way to $\frac{15}{16}\cos3x$ but I'm unable to think of any method to reach there.
| Use $\cos (A+B)$ formula to show for a natural number $n$,
$$\cos nx + \cos \left(nx+\frac{2n\pi}{3}\right) + \cos \left(nx+\frac{4n\pi}{3}\right) = \begin{cases} 0 , \quad \text{when n is coprime to 3} \\ 3\cos nx , \quad \text{when n is a multiple of 3} \end{cases}$$
Combining this with the identity from other answer,
$$\cos^5 \theta = \frac{1}{16}\cos 5\theta + \frac{5}{16}\cos 3\theta + \frac{10}{16}\cos \theta$$
it is easy to see that first and third terms, $n=1,5$, should sum to zero, while middle term remains to give
$$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) = \frac{15}{16}\cos 3x$$
| {
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Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$
$(x-r_1)(x-r_2)(x-r_3)= x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3 +r_2r_3)x-r_1r_2r_3$
$r_1+r_2+r_3=\frac{13}{a}$
$(2+i)(2-i)+r_3=\frac{13}{a}$
$5+r_3=\frac{13}{a}$
We also have:
$r_1r_2r_3=\frac{25}{a}$
$(2+i)(2-i)r_3=\frac{25}{a}$
$5r_3=\frac{25}{a}$
$r_3=\frac{5}{a}$
Going back to $5+r_3=\frac{13}{a}$
We have $5+\frac{5}{a}=\frac{13}{a}$
Thus $a=\frac{8}{5}$
The statement I am having trouble with is
"$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$"
Why do we know what $r_2$ is? I understand it's the conjugate and that $r_1r_2=5$ is a real number. But I don't understand why the coefficients being real implies that $r_2=2-i$
| Plug in $2+i$ into $p(x)$. Note that $(2+i)^2=3+4i$, $(2+i)^3=2+11i$. So $$p(2+i)= a*(2+11i) -13*(3+4i)+15a*(2+i)-25=
(32a-64)+(26a-52)i =(a-2)(32+26i)=0.$$ So $a=2$.This is much easier than what is in the OP and gives the right result. You do not need the assumption that $a$ is real.
| {
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Exponential diophantine equation $2^x+7^y=9^z$. The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions:
\begin{eqnarray*}
(x_1,y_1,z_1)&=&(3,0,1),\\
(x_2,y_2,z_2)&=&(1,1,1)\\
(x_3,y_3,z_3)&=&(5,2,2).\\
\end{eqnarray*}
There are no other natural solutions for $z≤10000$.
It seems that the equation $2^{x}+7^{y}=9^{z}$ has no other solutions in natural numbers. How can this be proven?
| here is a method I learned from Exponential Diophantine equation $7^y + 2 = 3^x$ and polished. I have answered many questions of the form $A^u - B^v = C$ for given positive integers $A,B,C$
This answer finishes $x=1.$
The following shows that, with $x = 1,$ the largest solution to $9^z - 7^y = 2 $ is $z=y=1.$ First, write $9^z - 9 = 7^y -7$ Next introduce non-negative integers $p,q$ with
$$ 9(9^p - 1) = 7 (7^q - 1), $$
ASSUME $p,q \geq 1$ and reach a contradiction. We alternate steps, each is a calculation.
$$ 7 | 9^p -1 \Longrightarrow \; \; 3 | p$$
$$ 9^3 - 1 = 2^3 \cdot 7 \cdot 13 $$
$$ 13 | 7^q -1 \Longrightarrow \; \; 12 | q$$
$$ 7^{12} - 1 = 2^5 \cdot 3^2 \cdot 5^2 \cdot 13 \cdot 19 \cdot 43 \cdot 181 $$
$$ 19 | 9^p -1 \Longrightarrow \; \; 9 | p$$
$$ 9^9 - 1 = 2^3 \cdot 7 \cdot 13 \cdot 19 \cdot 37 \cdot 757 $$
$$ 37 | 7^q -1 \Longrightarrow \; \; 9 | q$$
$$ 7^{9} - 1 = 2 \cdot 3^3 \cdot 19 \cdot 37 \cdot 1063 $$
This number divides $9 (9^p - 1). $ In particular, $27 | 9 (9^p - 1).$
This CONTRADICTS the assumption that $p > 0$
| {
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Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By considering the equation $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$
I have completed part (a). I am stuck on part (b) however.
My Workings:
By Factor formulae, $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{1}{2} (\cos(\frac{2\pi}{5}) + \cos(\frac{\pi}{5}))$
Considering roots of $z^5 = 1$. The sum of the roots equals 0.
So $\cos(\frac{2\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{8\pi}{5}) + \cos(\frac{10\pi}{5})=0$
$\cos(\frac{2\pi}{5}) = \cos(\frac{8\pi}{5})$, $\cos(\frac{4\pi}{5}) = \cos(\frac{6\pi}{5}) = -\cos(\frac{\pi}{5})$
So $2\cos(\frac{2\pi}{5})-2\cos(\frac{\pi}{5}) = -1$
$\cos(\frac{2\pi}{5})-\cos(\frac{\pi}{5}) = -\frac{1}{2}$
Been on stuck on this problem for an hour. Please help me.
| I don't know if this is what is intended, but here is where I went with this problem.
let $u = \cos \theta$
$u(16 u^4 - 20 u^2 + 5) = 0$
From the binomial theorem:
$u^2 = \frac {20 \pm \sqrt {80}}{32}\\
u^2 = \frac {5 \pm \sqrt {5}}{8}$
Yes, $u = 0$ is a solution associated with $\theta = \frac {\pi}{2}$ but that isn't particularly interesting.
$u^2 = \cos^2 \theta = \frac 12 (1+\cos 2\theta)$
$\cos 2\theta = \frac {5 \pm \sqrt {5}}{4} - 1\\
\cos 2\theta = \frac {1 \pm \sqrt {5}}{4}$
There are multiple values of $\theta$ such that $\cos 5\theta = 0$ the smallest of which is $\theta = \frac {\pi}{10}.$ This will be associated with the largest possible value of $\cos \theta$
$\cos \frac {\pi}{5} = \frac {1 + \sqrt {5}}{4}\\
\cos \frac {2\pi}{5} = 2\cos^2 \frac {\pi}{5} - 1 = 2 \frac {3 + \sqrt 5}{8} - 1\\
\cos \frac {2\pi}{5} = \frac {-1 + \sqrt 5}{4}$
$\frac 12 (\cos \frac {\pi}{5} + \cos \frac {2\pi}{5}) = \frac 12(\frac {1 + \sqrt {5}}{4} + \frac {-1 + \sqrt 5}{4}) = \frac {\sqrt 5}{4}$
| {
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Find the maximum value of $P= \frac2{1+a^2}-\frac2{b^2+1}+\frac3{c^2+1}$ The problem is, given that $a,b,c>0$ and $abc+a+c=b$, then find the maximum value of
\begin{align}
P= \frac2{1+a^2}-\frac2{b^2+1}+\frac3{c^2+1}
\end{align}
One thing I know for sure is that $a,c$ are symmetric, or so I thought.
In the condition, clearly $a,c$ are symmetric, and thus I guessed that if (a,b,c) satisfies P is maximum then (c,b,a) should also be one. But unfortunately, in $P$, the role of $a$ and $c$ are irreversible.
So how should I approach this problem?
Any help is appreciated!
| You have two ways to solve the problem without changing variables.
First way
From the constraint, we have, for example, $b=-\frac{a+c}{a c-1}$. Replacing in $P$ and simplifying
$$P=\frac{a^2 \left(3-2 c^2\right)+4 a c+2 c^2+3}{\left(a^2+1\right)
\left(c^2+1\right)}$$ Computing the partial derivatives
$$\frac{\partial P}{\partial a}=-\frac{4 c \left(a^2+2 a c-1\right)}{\left(a^2+1\right)^2 \left(c^2+1\right)}=0\tag 1$$
$$\frac{\partial P}{\partial c}=-\frac{2 \left(a \left(5 a c+2 c^2-2\right)+c\right)}{\left(a^2+1\right) \left(c^2+1\right)^2}=0\tag 2$$
From $(1)$, $c=\frac{1-a^2}{2 a}$. Plug in the numerator of $(2)$ to obtain
$$-4 a^3-2 a+\frac{2}{a}=0 \implies -\frac{2 \left(a^2+1\right) \left(2 a^2-1\right)}{a}=0\tag 3$$ So, taking into account the conditions
$$a=\frac{1}{\sqrt{2}} \implies c=\frac{1}{2 \sqrt{2}} \implies b=\sqrt{2}\implies
P=\frac{10}{3}$$
Second way
Using Lagrange multipliers, consider
$$F=P+\lambda(abc+a+c-b)$$
$$\frac{\partial F}{\partial a}=\lambda (b c+1)-\frac{4 a}{\left(a^2+1\right)^2}=0\tag 4$$
$$\frac{\partial F}{\partial b}=\lambda (a c-1)+\frac{4 b}{\left(b^2+1\right)^2}=0\tag 5$$
$$\frac{\partial F}{\partial c}=\lambda (a b+1)-\frac{6 c}{\left(c^2+1\right)^2}=0\tag 6$$
$$\frac{\partial F}{\partial \lambda}=abc+a+c-b=0\tag 7$$
But, for this case, it is quite tedious to solve equations $(4,5,6,7)$ but it leads to the same results with $\lambda=\frac{16 \sqrt{2}}{27}$.
| {
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how to express the $n$th power of the cosine as a series of cosines? Which is the correct way for expressing the $n$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n=
\frac{1}{2^n}\left( e^{j\theta}+e^{-j\theta} \right)^n=\frac{1}{2^n}\left( z+z^{-1} \right)^n.$
with $z=e^{j\theta}$.
Since the term $\left( z+z^{-1} \right)^n$ is the $n$th power of a binomial, I could express it using the binomial identity, thus
$\cos^n{(\theta)} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^k(z^{-1})^{n-k}
= \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^{2k-n}.$
If we expand the expression above, we obtain
$\cos^n{(\theta)}=\frac{1}{2^n}\Bigg ( z^{-n} + \binom{n}{1}z^{-(n-2)} + \binom{n}{2}z^{-(n-4)} + \dots + \binom{n}{2}z^{n-4} + \binom{n}{1}z^{(n-2)} + z^n \Bigg )$
which can be rewritten as
$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (z^{-n} + z^n) + \binom{n}{1} \left(z^{-(n-2)} + z^{(n-2)}\right) + \binom{n}{2}\left( z^{-(n-4)} + z^{(n-4)}\right) + \dots \Bigg ) $
Finally, since $z=e^{j\theta}$
$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (e^{-jn\theta} + e^{jn\theta}) + \binom{n}{1} \left(e^{j(n-2)\theta} + e^{-j(n-2)\theta}\right) + \binom{n}{2}\left( e^{j(n-4)\theta} + e^{-j(n-4)\theta}\right) + \dots \Bigg )$
By applying the Euler's formula once again, we obtain
$\cos^n{(\theta)}=\frac{2}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$
Unfortunately, if I plug n=2, I obtain
$\cos^2{(\theta)}= \cos{(2\theta)} + 1.$
instead of the well-known result
$\cos^2{(\theta)}= \frac{1}{2}(\cos{(2\theta)} + 1).$
a) Why my result is scaled by a factor of 2 ?
b) Is the correct general formula
$\cos^n{(\theta)}=\frac{1}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$
c)If so, why ?
| We start from
$$ s_n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} z^{2k-n} $$
Case 1: Let $n = 2m$
$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{2m} \binom{2m}{k} z^{2(k-m)} $$
$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{k=m+1}^{2m} \binom{2m}{k} z^{2(k-m)} $$
$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{l=0}^{m-1} \binom{2m}{2m - l} z^{2(m - l)} $$
$$ s_{n} = \frac{n!}{2^{n}(n/2)!^2} + \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} (z^{2k-n} + z^{n - 2k}) $$
$$ cos^{n}(\theta) = \frac{n!}{2^{n}(n/2)!^2} + \frac{2}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) $$
$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n/2-1} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}}\binom{n}{n/2} \cos((2(n/2)-n)\theta) + \frac{1}{2^{n}} \sum_{l=n/2+1}^{n} \binom{n}{l} \cos((2l-n)\theta) $$
$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$
Case 2: Let $n = 2m - 1$
$$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} $$
$$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k=m}^{2m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} $$
$$ s_{2m - 1} = \frac{1}{2^{2m - 1}} \sum_{k=0}^{m - 1} \binom{2m - 1}{k} z^{2(k-m) + 1} + \frac{1}{2^{2m - 1}} \sum_{k = 0}^{m} \binom{2m - 1}{k} z^{2(m-k) - 1} $$
$$ s_{n} = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} (z^{2k-n} + z^{n-2k}) $$
$$ cos^{n}(\theta) = \frac{2}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) $$
$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{(n - 1)/2} \binom{n}{k} cos((2k-n)\theta) + \frac{1}{2^{n}} \sum_{l=(n + 1)/2}^{n} \binom{n}{l} cos((2l-n)\theta) $$
$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$
Therefore, mixing the both cases we get
$$ cos^{n}(\theta) = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} cos((2k-n)\theta) $$
| {
"language": "en",
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About an inequality wich have a link with $\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2$ Hi it's a follow up of show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$:
Problem :
Let $a,x>0$ then (dis)prove :
$$\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\right)^{\frac{4}{5}}+\left(\frac{1}{2}x^{a}+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-a}+a}\right)^{\frac{4}{5}}\geq 2$$
If true I have tried to show it using the well-know formula :
Let $x$ be a real number then we have :
$$e^x\geq x+1$$
Applying to $a^x$ and $x^a$ .Unfortunately the inequality becomes false .
Also I have tried Tchebytchev's inequality but the application is similar to the use of Am-Gm and the inequality becomes wrong .
I found this refinement as a refinement of Am-Gm-Hm wich is a mix .
Edit :
It seems we have for $a\geq 1$ and $x\in(\frac{1}{a},a)$ :
$$g(x)=\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{-\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\right)^{-\frac{4}{5}}\cdot\left(\frac{1}{2}x^{a}+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-a}+a}\right)^{\frac{4}{5}}$$
And :
$$h(x)=\left(\frac{1}{2}a+\frac{1}{2}x^{-1}\right)^{-\frac{1}{5}}\cdot\left(\frac{2}{a^{-1}+x}\right)^{-\frac{4}{5}}\cdot\left(\frac{1}{2}x+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-1}+a}\right)^{\frac{4}{5}}$$
Then it seems we have :
$$h(x)< g(x)$$
Last edit 25/11/2021 correction 26/11/2021:
We need to show that for $a\leq 1$ and $x\ge 1$ :
$$2^{\frac{3}{5}}\cdot\frac{\left(\frac{a^{x}}{x}\right)^{\frac{4}{5}}}{\left(a^{x}+\frac{1}{x}\right)^{\frac{3}{5}}}+2^{\frac{3}{5}}\cdot\frac{\left(\frac{x^{a}}{a}\right)^{\frac{4}{5}}}{\left(x^{a}+\frac{1}{a}\right)^{\frac{3}{5}}}\geq 2 $$
Or :
$$x^{-\frac{5}{5}}r\left(a^{x}x\right)+a^{-\frac{5}{5}}r\left(x^{a}a\right)\geq 2$$
Where the function :
$$r\left(x\right)=2^{\frac{3}{5}}\cdot\frac{x^{\frac{4}{5}}}{\left(x+1\right)^{\frac{3}{5}}}$$
Is reciprocally convex on $(0,\infty)$ .
See this paper for further informations : http://artem.sobolev.name/posts/2021-05-02-reciprocal-convexity-to-reverse-the-jensen-inequality.html at ("defined by Merkle")
We get a lower bound :
$$\left(\frac{1}{x}+\frac{1}{a}\right)\cdot r\left(\frac{\left(\frac{1}{x}+\frac{1}{a}\right)}{\frac{1}{x}\cdot\frac{1}{\left(a^{x}x\right)}+\frac{1}{a}\left(\frac{1}{x^{a}\cdot a}\right)}\right)$$
It seems to be superior to two when $0.75\leq a\leq 1$ and $0<x\leq 6$
Question :
How to (dis)prove it ?
|
Your function looks like this on the reals:
So it is reasonable to expand to small $a$ and $x$ and to examine the curve where the value $2$ is taken. For example $a(x,2)$.
The problem is fundamental because $a^x$ and $x^a$ appear in the composite function. The function is rational and factored in both summands.
Special case is evident $a=x$.
From this visual approach, it is clear together with the fundamentality of the constituents of the composite function that only numerical methods are suitable and understandable to simplify and visualize this inequality.
The result is topping Your work in the very first steps.
An easy path is to approximate around the curve for small values of $a$ and $x$ and for the bigger ones. For small values the curve is for large enough values of the other variable almost a straight line as can be seen in the graph of the function. These approximations are a mathematical art of value to the world.
This is just a motivation to prefer visualization and approximation techniques from numerics and analysis as well as pure analysis.
Nevertheless here some progression:
And viewed from the direction of the surface height.
This is implicit a clear idea where the equality holds.
It presents crisp clear the boundaries where the given surface is 2. Outside the inequality is true, inside it is false.
A good idea is to approximate or the this boundary curve set.
| {
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How to show that the following sequence is monotinacly increasing? given $$a_{1}=1, \quad a_{n+1}=\frac{2\left(2 a_{n}+1\right)}{a_{n}+3}$$ I am trying to show $a_{n}\leq a_{n+1}$
I am trying to show in induction, so my assumption is that $a_{n}\leq a_{n+1}$ and I want to show $a_{n+1}\leq a_{n+2}$ the only thing I could get is that $$a_{n+2}=\frac{2\left(2 a_{n+1}+1\right)}{a_{n+1}+3}>\frac{2\left(2 a_{n}+1\right)}{a_{n+1}+3}=a_{n+1}\cdot \frac{(a_n +3)}{a_{n+1}+3}$$
| Notice that $a_n$ had better never be $-3$ or the recursion fails to be defined. We wish to show \begin{align*}
a_n &\leq a_{n+1} \\
&= \frac{2(2a_n+1)}{a_n+3} \\
&= \frac{4a_n+2}{a_n+3}
\end{align*}
so
\begin{align*}
\begin{cases}
4a_n + 2 \geq a_n(a_n+3) ,& a_n+3 > 0 \\
4a_n + 2 \leq a_n(a_n+3) ,& a_n+3 < 0
\end{cases} \\
\begin{cases}
4a_n + 2 \geq a_n^2+3a_n ,& a_n > -3 \\
4a_n + 2 \leq a_n^2+3a_n ,& a_n < -3
\end{cases} \\
\begin{cases}
0 \geq a_n^2-a_n-2 ,& a_n > -3 \\
0 \leq a_n^2-a_n-2 ,& a_n < -3
\end{cases} \\
\begin{cases}
0 \geq (a_n - 2)(a_n + 1) ,& a_n > -3 \\
0 \leq (a_n - 2)(a_n + 1) ,& a_n < -3
\end{cases} \\
\end{align*}
Drawing sample points from $(-\infty,-3)$, $(-3,-1)$, $(-1,2)$, and $(2,\infty)$, we find that this is satisfied for $a_n < 3$ and $a_n \in (-1,2)$.
We have $a_1 = 1$. To be monotonically increasing, we must have $a_n \in [1,2)$ for all $n \geq 1$. (In particular, although we get at least one step of increasing if $a_n < -3$, if any $a_n$ after $a_1$ is less than $a_1 = 1$, the sequence can't be increasing.) We have $\frac{\mathrm{d}}{\mathrm{d}x} \frac{2(2x+1)}{x+3} = \frac{10}{(x+3)^2} > 0$ (for $x \neq -3$), so is increasing on $[1,2)$. This means $\frac{2(2x+1)}{x+3}$ is lower bounded at $x = 1$ and is upper bounded at $x = 2$. We compute $\frac{2(2(1)+1)}{(1)+3} = \frac{3}{2}$ and $\frac{2(2(2)+1)}{(2)+3} = 2$, so $a_n \in [3/2,2)$ for all $n$. Therefore, the sequence is monotonically increasing.
| {
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If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$. If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$.
I am really sorry for the basic question, but I want to make things clear for me. By the AM-GM inequality we have $$\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2$$ Is this enough? I mean we know that the inequality holds only for positive numbers $a$ and $b$, but here we can have $a<0$ and $b<0$ and then $ab>0$ will also hold.
| I have followed another reasoning to prove the statement.
Case 1: $a=b$
$$\frac{a}{a}+\frac{a}{a}=2\quad \blacksquare$$
Case 2: $a\neq b$
Notice that $ab>0$ is the same as saying $a,b\neq 0$, and both have the same sign (which later becomes irrelevant).
$$\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{a^2+2ab+b^2-2ab}{ab}=\frac{(a+b)^2}{ab}-2\geq2$$
$$\frac{(a+b)^2}{ab}\geq4;\quad (a+b)^2\geq 4ab;\quad a^2+2ab+b^2\geq4ab;\quad a^2-2ab+b^2=(a-b)^2\geq0$$
which we know it is true regardless of the signs of $a$, $b$ since a squared number is always positive $\blacksquare$
| {
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What's the measure of the radius of the circle below? For reference: In the figure, calculate $R$. If : $\overset{\LARGE{\frown}}{AB}= 120°$, $CM = 1$ and $MB = 2$
(Answer: $\frac{3}{4}(\sqrt{3}+1$))
My progress:
Draw OP
Th.Chords:
$CM.MP = BM.AM \implies 1.[(R-1)+(R)] = 2.AM\\
\boxed{2AM = 2R-1}: \boxed{R=AM+ \frac{1}{2}}\\
\triangle AOB (isósceles):$\
Draw $AE$
$\implies$
$\triangle EAB(right):
AE^2+(2+AM)^2 = 4R^2\\
AE^2 +4+4AM + AM^2 = 4R^2\\
AE^2 + 4+8R-4 + R^2 - R+\frac{1}{4} = 4R^2\\
4AE^2+16+32R - 16+4R^2-4R+1 = 16R^2\\
4AE^2+28R-12R^2+1 = 0 \implies\\
AE^2 = 12R^2-28R-1\\$
...?
I have not found another relationship with AM
|
Extend $\overline{CMO}$ to meet the circle at E and Extend $\overline{AO}$ to meet the circle at $D$. As $\triangle ABD$ is a 30-60-90 triangle $AB=R\sqrt{3}$. Since $BM=2$ and $AM=R\sqrt{3}-2$.
Now, look at $\triangle ACM$ and $\triangle BEM$ they are similar.
$\frac{1}{2}=\frac{R\sqrt{3}-2}{2R-1}$
After solving for R you get,
$R=\frac{3}{2(\sqrt{3}-1)}$
by getting the decimal value for R,
$R=2.04$
The decimal value of the answer you have posted is also 2.04
Then the proof will be true
| {
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Given the PDF of $X$ which is right hand continuous, compute PDF of $Y$ = CDF$(x)$ Let x have a piecewise right hand continuous pdf, $f_X(x)$, which is defined as follows.
$$f_X(x) = \begin{cases}
\frac{1}{2}(x+2)^2 & -2 \leq x < -1 \\
-\frac{1}{2}x +\frac{1}{12}\delta (x+1) & -1 \leq x < 0 \\
\frac{1}{2}x & 0 \leq x < 1 \\
\frac{1}{2}(x-2)^2 + \frac{1}{12}\delta(x-1) & 1 \leq x \leq 2 \\
\end{cases}
$$
Let Y = $F_X(x)$. Compute $f_Y(y)$.
My approach: I have tried writing this:
$$
F_Y(y) = Pr\{Y \leq y\} = Pr \{F_X(x) \leq y\}
$$
But I honestly have no idea how to handle this piecewise function in this situation. So far, I've tried to compute the derivative of $F_Y(y)$ using:
$$
f_Y(y) = lim_{dy->0^+}\frac{Pr\{Y \leq y + dy \} - Pr\{Y \leq y \}}{dy}
$$
to no avail.
| To explain the solution step by step, a couple of drawings are needed but with this sketch I am sure you will be able to conclude by yourself.
First observe, that IF your X rv was continuous, $Y\sim U(0;1)$ by integral transform, being
$$F_Y(y)=F_X\left[F_X^{-1}(y) \right]=y$$
Your case is very similar but you have 2 discontinuity points in $F_X$ or, equivalently, you have 2 dirac impulses of $1/12$ each.
Immediately the result is that
$$f_Y(y)=\frac{1}{12}\delta\left( y-\frac{3}{12} \right)+\frac{1}{12}\delta\left( y-\frac{10}{12} \right)+\mathbb{I}_{\left(0;\frac{2}{12} \right)\cup \left(\frac{3}{12};\frac{9}{12} \right)\cup\left(\frac{10}{12};1 \right)}$$
or alternatively, using special function $\delta$ and $\text{Rect}$,
$$f_Y(y)= \text{Rect}\left( \frac{12y-1}{2} \right)+ \frac{1}{12}\delta\left( y-\frac{3}{12} \right)+ \text{Rect}\left( \frac{12y-6}{6} \right) +\frac{1}{12}\delta\left( y-\frac{10}{12} \right)+\text{Rect}\left( \frac{12y-11}{2} \right) $$
Observe also that $f_Y(y)$ is not a pdf, being it not absolutely continuous. In some branches as Signal Theory they define these function as "mixed densities"
To derive analytically this solution I suggest you
*
*First derive $F_X(x)$
$$F_X(x) =
\begin{cases}
0, & \text{if $x<-2$} \\
\frac{(x+2)^3}{6}, & \text{if $-2\le x<-1$} \\
\frac{1}{4}+\frac{1-x^2}{4}, & \text{if $-1\le x<0$} \\
\frac{1}{2}+\frac{x^2}{4}, & \text{if $0\le x<1$} \\
1+\frac{(x-2)^3}{6}, & \text{if $1\le x<2$} \\
1, & \text{if $x\ge 2$ }
\end{cases}$$
*
*Do a drawing of $F_X$
*Do a drawing of its inverse, $F_Y$
*Derive $f_Y$, considering that it is discrete in 2 points
| {
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"url": "https://math.stackexchange.com/questions/4310981",
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"question_score": "1",
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Playing tennis until one of the players wins 3 times (Binomial distribution) Question
Nadal and Federer playing tennis against each other.
Nadal's probability to win one match is $\frac{2}{3}$, independently from the previous results.
The two are playing until one of them wins 3 matches.
What is the probability for Nadal's victory?
My Take
Nadal's probability to win is Nadal's probability to win exactly 3 matches which is:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}=\frac{80}{243}
$$
The Book's Answer
According to the book I should take into consideration Nadal's chances of winning 3\4\5 times:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}+{5 \choose 4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^{5}=\frac{64}{81}
$$
I can't tell why that's the right answer.
As far as I understand the game should stop after Nadal gets 3 points, so why adding the probability of him winning 4\5 matches?
| The final score could be 3-2 for Nadal in which case we must have seen 5 experiments with 3 Nadal victories and 2 Federer ones. So $\binom{4}{2}=6$ ways to "place" two of those victories among the first $4$ games and then the final game must be Nadal's. So the probability of that happening is
$$6 \times (\frac{2}{3})^2 \times (\frac13)^2 \times \frac23 = \frac{48}{243}$$
To get 3-1 we need one Federer win among the first 3 and then a final Nadal one and so we get
$$3 \times (\frac23)^2 \frac13 \times \frac23 = \frac{24}{81} = \frac{72}{243}$$
For 3-0 we just have
$$(\frac23)^3 = \frac{8}{27}=\frac{72}{243}$$
So in total $$\frac{72+72+48}{243}= \frac{192}{243} = \frac{64}{81}$$
| {
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inhomogen linear equation system problem I need to find all the answers of this linear equation system.
$$\begin{vmatrix}
1 & 2 & 1 & 3 & 0 & =\beta \\
2&3&2&5&1&=0\\
2&1&1&4&1&=0\\
3&3&2&7&1&=1\\
\end{vmatrix} $$
I tried to solve it like that:
Row 4- row 3->
$$\begin{vmatrix}
1 & 2 & 1 & 3 & 0 & =\beta \\
2&3&2&5&1&=0\\
2&1&1&4&1&=0\\
1&2&1&3&0&=1\\
\end{vmatrix} $$
Row 1- row 4 ->
$$\begin{vmatrix}
0 & 0 & 0 & 0 & 0 & =\beta-1 \\
2&3&2&5&1&=0\\
2&1&1&4&1&=0\\
1&2&1&3&0&=1\\
\end{vmatrix} $$
and if I understand correctly, if a row is in the form of the first row here, all the variable are zero and equals something, the linear equation system don't have an answer.
but I checked that LES in online calculator, and apparently it does have an answer.
Am I getting something wrong here?
| As I wrote in the comment. From your work we conclude that the linear system might only possess a solution for $\beta = 1$. As it turns out the solutions for this case are of the form
$$\left(\begin{matrix}
1-2\cdot x_4 \\
2-x_4+x_5 \\
-4+x_4-2\cdot x_5 \\
x_4 \\
x_5
\end{matrix}\right)$$
for $x_4, x_5 \in \mathbb{R}$.
| {
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Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f$. Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f $.
1- $\langle 2\rangle $
2-$\langle 4\rangle $
3-$\langle 7\rangle $
4-$\langle 1\rangle $
I think first of all there is no group homomorphism with $f(1)=12$ because we know that every homomorphism between cyclic groups sends generators to generators . But in this case $12$ is not generator for $(\mathbb{Z}_{16}, +)$ because $(12,16) \neq 1$ .
| We have
$$\begin{align}
\color{red}{12+12}\pmod{16}&=f(1)+f(1)\\
&=f(1+1)\\
&=f(2)\\
&=\color{red}{8}\pmod{16},
\end{align}$$
$$\begin{align}
\color{green}{8+12}\pmod{16}&=f(2)+f(1)\\
&=f(2+1)\\
&=f(3)\\
&=\color{green}{4}\pmod{16},
\end{align}$$
and
$$\begin{align}
\color{blue}{4+12}\pmod{16}&=f(3)+f(1)\\
&=f(3+1)\\
&=f(4)\\
&=\color{blue}{0}\pmod{16}
\end{align}$$
Thus $\ker f=\langle 4\rangle$.
| {
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Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}, $$
where $n\in N.$
After trying some substitution and integration by parts, I still failed and got no idea for reducing the power n. After two days, the Leibniz Rule for high derivatives come to my mind.
Differentiating $I_1(a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\displaystyle \begin{array}{l}
\displaystyle \int_{0}^{\pi} \frac{(-1)^{n-1}(n-1) !}{(a-\cos x)^{n}} d x=\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right) \\ \displaystyle
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) \tag{*}\label{star}
\end{array}
$$
I am glad to see that the integration problem turn to be merely a differentiation problem.
Now I am going to find the $(n-1)^{th} $ derivative by Leibniz Rule.
First of all, differentiating $I_1(a)$ w.r.t. $a$ yields $$
\left(a^{2}-1\right) \frac{d y}{d a}+a y=0 \tag{1}\label{diffeq}
$$
Differentiating \eqref{diffeq} w.r.t. $a$ by $(n-1)$ times gets $$
\begin{array}{l}
\displaystyle \left(a^{2}-1\right) \frac{d^{n} y}{d a^{n}}+\left(\begin{array}{c}
n-1 \\
1
\end{array}\right)(2 a) \frac{d^{n-1} y}{d a^{n-1}}+2\left(\begin{array}{c}
n-1 \\
2
\end{array}\right) \frac{d^{n-2} y}{d a^{n-2}}+x \frac{d^{n-1} y}{d a^{n-1}}+(n-1) \frac{d^{n-2} y}{d a^{n-2}}=0
\end{array}
$$
Simplifying, $$
\left(a^{2}-1\right) y^{(n)}+(2 n-1) ay^{(n-1)}+(n-1)^{2} y^{(n-2)}=0 \tag{2}\label{diffrec}
$$
Initially, we have $ \displaystyle y^{(0)}=\frac{1}{\sqrt{a^{2}-1}}$ and $ \displaystyle y^{(1)}=-\frac{a}{\left(a^{2}-1\right)^{\frac{3}{2}}}.$
By \eqref{diffrec}, we get $$
y^{(2)}=\frac{2 a^{2}+1}{\left(a^{2}-1\right)^{\frac{5}{2}}}
$$ and $$
\displaystyle y^{(3)}=-\frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}}
$$
Plugging into \eqref{star} yields $$
\begin{aligned}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{3}} &=\frac{\pi}{2} y^{(2)}=\frac{\pi\left(2 a^{2}+1\right)}{2\left(a^{2}-1\right)^{\frac{5}{2}}} \\
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{4}} &=-\frac{\pi}{6} \cdot \frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}} =-\frac{\pi a\left(2 a^{2}+3\right)}{2\left(a^{2}-1\right)^{\frac{7}{2}}}
\end{aligned}
$$
Theoretically, we can proceed to find $I_n(a)$ for any $n\in N$ by the recurrence relation in $(2)$ .
By Mathematical Induction, we can further prove that the formula is $$
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{\pi P(a)}{\left(a^{2}-1\right)^{\frac{2 n-1}{2}}}
$$
for some polynomial $P(a)$ of degree $n-1$.
Last but not least, how to find the formula for $P(a)$? Would you help me?
| One way using complex analysis:
Since the integrand is even:
$$
I=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n}
$$
Since $a>1$, we can expand the integrand with the generalized binomial theorem:
$$ I = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2} \int_{-\pi}^{\pi} \sum_{j=0}^{\infty} \binom{-n}{j} a^{-n-j}(-1)^j\cos^j(x) dx $$
From Fubini-Tonelli we can interchange integral and series:
$$ I = \frac{1}{2} \sum_{j=0}^{\infty} \binom{-n}{j}a^{-n-j}(-1)^j\int_{-\pi}^{\pi} \cos^j(x) dx $$
Do the following substitution
$$ \cos x = \frac{z+z^{-1}}{2}$$
$$ dx = \frac{dz}{zi}$$
The integral is now a contour integral round the unit complex circle:
$$ I = \frac{1}{2} \sum_{j=0}^{\infty} \binom{-n}{j}\frac{a^{-n-j}(-1)^j}{2^j}\oint_{|z|=1} \frac{(z^2+1)^j}{z^{j+1} }dz$$
The integral has a pole at $z=0$. To find the residue, expand the integrand:
$$\frac{(z^2+1)^j}{z^{j+1} }= \sum_{k=0}^{j}\binom{j}{k} z^{2k-j-1}$$
The residue is the coefficient of $z^{2k-j-1}= z^{-1}$
Then
$$ 2k-j-1 = -1 \Longrightarrow k = \frac{j}{2}$$
Therefore, the residue exists if $j$ is $\textbf{divisible by}$ $2$:
$$\oint_{|z|=1} \frac{(z^2+1)^j}{z^{j+1} }dx = 2\pi i \operatorname{Res}\left(\frac{(z^2+1)^j}{z^{j+1}},0\right) = 2\pi i \binom{j}{\frac{j}{2}} $$
Hence, we have
$$ I = \pi\sum_{j=0}^{\infty} \binom{-n}{2j}\binom{2j}{j}\frac{a^{-n-2j}}{2^{2j}} $$
Note
$$\binom{-n}{2j} = \frac{(-n-2j+1)_{2j}}{(2j!}$$
where $(x)_{n} = x(x+1)\cdots(x+n-1)$ is the rising factorial (Pochhammer polynomial)
and
$$\binom{2j}{j} = \frac{(2j)!}{j!^2}$$
Hence
$$\binom{-n}{2j}\binom{2j}{j} = \frac{(-n-2j+1)_{2j}}{j!^2}$$
From the reflection formula for the Pochhammer polynomial:
$$(-x)_{m} = (-1)^m(x-m+1)_{m}$$
we have
$$(-n-2j+1)_{2j}= (n)_{2j}$$
and from the duplication formula for the degree of the Pochhammer polynomial:
$$(x)_{2m} = 4^m\left(\frac{x}{2}\right)\left(\frac{1+x}{2}\right)$$
we have
$$(n)_{2j} = 4^{j}\left(\frac{n}{2}\right)\left(\frac{n+1}{2}\right)$$
Hence
$$ I = \pi\sum_{j=0}^{\infty} \binom{-n}{2j}\binom{2j}{j}\frac{a^{-n-2j}}{2^{2j}} = \frac{\pi}{a^n}\sum_{j=0}^{\infty}\frac{\left(\frac{n}{2}\right)\left(\frac{n+1}{2}\right)}{(1)_{j}} \frac{\left(\frac{1}{a^2}\right)^j}{j!}$$
$$\Longrightarrow I = \frac{\pi}{a^n}{}_{2}F_{1}\left({\frac{n}{2},\frac{n+1}{2}\atop 1};\frac{1}{a^2}\right)$$
where ${}_{2}F_{1}({a,b\atop c};z)$ is the Guassian function
Now using the following forumula:
$${}_{2}F_{1}\left({a,a+\tfrac{1}{2}\atop c};z\right)=2^{c-1}z^{\frac{(1-c)}{2}}%
(1-z)^{-a+\left(\frac{c-1}{2}\right)}P^{1-c}_{2a-c}\left(\frac{1}{\sqrt{1-z}}\right) $$
where $P_{n}^{(\alpha,\beta)}$ is the Jacobi polynomial
we have
$$ I =\frac{\pi}{a^n}{}_{2}F_{1}\left({\frac{n}{2},\frac{n+1}{2}\atop 1};\frac{1}{a^2}\right)= \frac{\pi}{a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}P_{n-1}\left(\frac{a}{\sqrt{a^2-1}}\right)$$
where $P_{n}$ is the standard Legendre polynomial which satisfies the Rodrigues' formula:
$$ P_{n}(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2-1)^n$$
Finally, we can conclude
$$\boxed{\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{\pi}{2^{n-1}(n-1)!a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}\left( \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^{n-1} \right)_{x = \frac{a}{\sqrt{a^2-1}}}}$$
This is similar to the formula you found. However, Legendre polynomials also have an explicit formula:
$$P_{n} (x) = \sum_{j} (-1)^{\frac{j}{2}} \frac{(2n-j-1)!!}{j!!(n-j)!} x^{n-j} \quad j=0,2,4,...,n-\frac{1}{2}\pm \frac{1}{2}$$
(ends with $n$ in the case $n$ even and ends with $n-1$ in the case $n$ odd).
Hence
$$ \boxed{I =\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{\pi}{a^n}\left(1-\frac{1}{a^2}\right)^{-\frac{n}{2}}\sum_{j} (-1)^{\frac{j}{2}} \frac{(2n-j-3)!!}{j!!(n-1-j)!} \left(\frac{a}{\sqrt{a^2-1}}\right)^{n-1-j} \quad j=0,2,4,...,n-\frac{3}{2}\pm \frac{1}{2} \textrm{ ($n-1$ even or odd)}} $$
It turned out that the polynomial you are looking for is this Legendre polynomial.
| {
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Counting routes from home to office if you only return home if you realize you have forgotten something before you reach the office The following is how you can go to the office from home:
Home $\to$ Four Roads $\to$ Schools $\to$ Three Roads $\to$ University $\to$ Five Roads $\to$ Parks $\to$ Two Roads $\to$ Offices
You are forgetful minded. You may have forgotten something at home. You remember what you forgot at home, either at your school or university or in the park, and you go back to pick it up. Then the journey continues from the beginning again. You forget one thing at most during the day and when you reach the office you do not go back to take back what you have left. So how many different routes are possible for you?
I have tried this way:
Case 1: I remember what I forgot at school:
In this case, I can go to school in 4 ways and come back in $4$ ways and go back in $4$ ways. Then I can start the journey from there in $3 \cdot 5 \cdot 2$ ways.
Case 2: I remember what I forgot at university:
In this case, I can go to university in $4 \cdot 3$ ways and come back in $4 \cdot 3$ ways and go back in $4 \cdot 3$ ways. Then I can start the journey from there in $5 \cdot 2$ ways.
Case 3: I remember what I forgot at park:
In this case, I can go to park in $4 \cdot 3 \cdot 5$ ways and come back in $4 \cdot 3 \cdot 5$ ways and go back in $4 \cdot 3 \cdot 5$ ways. Then I can start the journey from there in $2$ ways.
Case 4: I go to office without remembering:
In this case, I can go to the office in $4 \cdot 3 \cdot 5 \cdot 2$ ways.
So, total ways would be $= (4 \cdot 3 + 3 \cdot 5 \cdot 2) + (4 \cdot 3 \cdot 3 + 5 \cdot 2) + (4 \cdot 3 \cdot 5 \cdot 3)+2 + 4 \cdot 3 \cdot 5 \cdot 3$ ways.
| Keep in mind that you add when two events are mutually exclusive, meaning that they cannot occur at the same time. However, when choices can be made independently (such as choosing how to travel from home to school, school to home, then from home to school to university to park to office), you should be multiplying.
Case 1: You have four ways of reaching the school. You then discover that you have forgotten something at home. For each of the four ways you could have reached the school, you have four ways of returning home. For each of the $4 \cdot 4$ ways you could have gone to school and then returned home, you then have four ways of reaching the school, three ways of traveling from the school to the university, five ways to travel from the university to the park, and two ways to travel from the park to the office. By the Multiplication Principle, there are
$$4 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$$
possible journeys if you realize that you have forgotten something once you reach the school.
Case 2: You should have $4 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$ possible journeys (travel to university via school, travel home from university via school, travel to office via school, university, and park).
Case 3: You should have $4 \cdot 3 \cdot 5 \cdot 5 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2$. Do you see why?
Case 4: You are correct that there are $4 \cdot 3 \cdot 5 \cdot 2$ possible journeys from home to the office via the school, university, and park if you do not realize that you have forgotten something until you reach the office, but you wrote the answer incorrectly when you added the four mutually exclusive cases.
Total: Since the four cases are mutually exclusive and exhaustive, you obtain the answer by applying the Addition Principle to the four cases, which yields
$$4 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 + 4 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 + 4 \cdot 3 \cdot 5 \cdot 5 \cdot 3 \cdot 4 \cdot 4 \cdot 3 \cdot 5 \cdot 2 + 4 \cdot 3 \cdot 5 \cdot 2$$
| {
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How do I know if the determinant is positive? I have the following problem.
Let $ A = \begin{pmatrix}a&b&c&d\\-b&a&d&-c\\-c&-d&a&b\\-d&c&-b&a\end{pmatrix}$
Find $\det(A)$.
I calculated $$AA^T = \begin{pmatrix}a^2+b^2+c^2+d^2&0&0&0\\0&a^2+b^2+c^2+d^2&0&0\\0&0&a^2+b^2+c^2+d^2&0\\0&0&0&a^2+b^2+c^2+d^2\end{pmatrix}$$ and got that $\det(A)^2 = (a^2+b^2+c^2+d^2)^4 \implies \det(A)=\pm(a^2+b^2+c^2+d^2)^2$
But I don't know how to determine the sign of the $\det(A)$. I want to do it without calculating the determinant in a different way
| the determinant is the sum of 4! terms, one of which is $a^4$, so we choose the + sign, not the - sign.
alternatively, if we don't know the Leibniz formula, sub a=1 and b=c=d=0, then A is the identity matrix, so $\det(A)=1$, so we choose the + sign.
| {
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Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$.
If there were such an inequality:
$ t_{1} \leq (xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \leq t_{2}$, it is obvious that $t_{1}=0$ is suitable. And if $x=y=z$, then we have:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2=3x^2 \cdot \dfrac{9}{4x^2}=\dfrac{27}{4}$. Next question is this value is the maximum or minimum?
If $x=1, y=1, z=0$, then we have $F(1,1,0)=(\dfrac{1}{2}+2)^2=\dfrac{25}{4}< \dfrac{27}{4}$, but if $x=1, y=1, z=n$, then we have $F(1,1,n)=(1+2n)(\dfrac{1}{2}+\dfrac{2}{1+n})^2>(1+2n)(\dfrac{1}{4}) \rightarrow +\infty$, which means, if I'm not mistaken, that this function has no extremum?
What to do with the case when $x,y,z$ are different I have not yet figured out.
Maybe in my case one of the following inequalities could be applied, but I'm stumped:
$1) \ xyz \geq(x+y-z)(x+z-y)(y+z-x)$
$2) \ x^3+y^3+z^3 \geq x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-3xyz$
$3) \ x^3+y^3+z^3 \geq 3xyz$.
Maybe someone can give an idea where to start with my case? Thanks!!
| Another way.
By my previous post we need to prove that:
$$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4},$$ which is true by Muirhead:
$$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2-\frac{25}{4}=$$
$$=\frac{\sum\limits_{sym}\left(4x^5y-x^4y^2-3x^3y^3+x^4yz+14x^3y^2z+\frac{19}{3}x^2y^2z^2\right)}{4\prod\limits_{cyc}(x+y)^2}\geq0.$$
| {
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Finding $\lim\limits_{n→∞}\frac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$
Find $\lim\limits_{n→∞}\dfrac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$.
Notice that I don't have $a>b$ or $b>a$, and also no $b≠0$ or $a≠0$, so I don't know if i even can use my solution.
My solution :
\begin{gather*}
\frac {a^n+b^n}{a^n-b^n} = \frac {a^n}{a^n-b^n} + \frac {b^n}{a^n-b^n}
= \frac{1}{1- \frac{b^n}{a^n}}-\frac {1}{\frac{a^n}{b^n}-1}\\
a>b \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=0,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=\infty\\
\frac{1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-0} + \frac{1}{\infty-1} = 1+\frac{1}{\infty}= 1+0=1\\
b>a \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=\infty,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=0\\
\frac {1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-\infty} + \frac {1}{0-1} = \frac{1}{-\infty}-1= 0-1=-1
\end{gather*}
Now I have two convergent solutions for different cases. What should I answer to the question mentioned above?
| Alternative approach:
First of all, I am interpreting the problem to represent that $n$ is restricted to the positive integers. This is because, apparently, it is permissible to have either $a$ or $b$ negative.
If (for example), $a = -1.5$, you run into problems evaluating $a^n$ for $n \in \Bbb{R^+}$. For example, is $(-1.5)^\pi$ a positive or negative number?
I avoid this by assuming that the problem intends that $n$ is restricted to $\Bbb{Z^+}.$
For fixed $a,b,~$ with $~n \in \Bbb{Z^+},~$ let $f(n) = \displaystyle \frac{a^n + b^n}{a^n - b^n}$.
$a = 0, b \neq 0 \implies f(n) = -1$, for all $n$, and for all $b$.
$b = 0 \implies f(n) = 1$, for all $n$, and for all $a$.
Therefore, without loss of generality, $a \neq 0 \neq b.$
$\underline{\textbf{Case 1:} ~|a| > |b|}$
Set $~r = \displaystyle \frac{b}{a} \implies -1 < r < 1 \implies \lim_{n \to \infty} r^n = 0$.
Then $f(n) = \displaystyle \frac{1 + r^n}{1 - r^n}$, which goes to $\displaystyle \frac{1 + 0}{1 - 0} = 1$.
Note that here, it is irrelevant whether $a$ is positive or negative, and irrelevant whether $b$ is positive or negative.
$\underline{\textbf{Case 2:} ~|a| < |b|}$
Set $~r = \displaystyle \frac{a}{b} \implies -1 < r < 1 \implies \lim_{n \to \infty} r^n = 0$.
Then $f(n) = \displaystyle \frac{r^n + 1}{r^n - 1}$, which goes to $\displaystyle \frac{0 + 1}{0 - 1} = -1$.
$\underline{\textbf{Case 3:} ~1 \leq |a| = |b|, ~a,b ~\text{are of opposite signs}}$
This is tricky. When $n$ is odd, the numerator of $f(n)$ is $0$, while the denominator is increasing in absolute value. So, the subsequence of just the terms corresponding to $n$ odd do go to $0$.
However, when $n$ is even, the numerator is increasing in absolute value, while the denominator is always $0$. When the denominator of $f(n)$ is $0$, the fraction represented by $f(n)$ is undefined.
Here, it is unnecessary to examine the behavior of the subsequence represented by $n$ even. The issue is the overall sequence, which is divergent, but does not go to infinity or go to minus infinity.
$\underline{\textbf{Case 4:} ~1 > |a| = |b|, ~a,b ~\text{are of opposite signs}}$
Superficially, this one may seem trickier than Case 3.
However, it is not trickier, because when $n$ is even, the denominator of $f(n)$ is flatly $0$, while the numerator is never $0$. When $n$ is odd, the reverse is true (i.e. numerator only is $0$).
Therefore, the evaluation is identical to Case 3: the sequence is divergent, going to neither infinity or minus infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining $A$ such that $\lim \limits_{x \to\infty }(\sqrt{Ax^2+2x}−5x)$ exists and is finite
Determine the value of the real number $A$ so that a finite limit exists, and then compute the limit:
$$\lim \limits_{x \to \infty}(\sqrt{Ax^2+2x}−5x)$$
Where would I start with this? I thought I could factor the highest degree variable from each term but that lets any real A work with it.
| $\begin{aligned} \sqrt{Ax^2+2x}-5x &= (\sqrt{Ax^2+2x}-5x)\frac{\sqrt{Ax^2+2x}+5x}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{Ax^2+2x-25x^2}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{(A-25)x+2}{\sqrt{A+\frac{2}{x}}+5} \end{aligned}$
Notice that, as $x$ goes to $\infty$, the denominator goes to $\sqrt{A}+5$ and the numerator goes to
*
*$+\infty$ if $A>25$
*$-\infty$ if $A<25$
*$2$ if $A = 25$
so $A=25$ is our only hope to obtain a finite result, which would be
$$\frac{2}{\sqrt{25}+5} = \frac{1}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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projection of $x^2+y^2+z^2-yz=1$ to xoy plane if the surface is projected to yoz plane, I known than the equation of the projection is $y^2+z^2-yz-1=0$. But what what about xoy plane, is the equation of project projection $x^2+y^2=1$?
| $$
x^2+y^2+z^2-yz=1
\\
z^2-yz+\left( x^2+y^2-1 \right) =0
\\
\Delta =-4x^2-3y^2+4\ge 0
\\
x^2+\frac{3}{4}y^2\le 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Vanier College Practice test - Prove that $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$ Source : Colorado.edu, Vanier College, Worksheet Logarithm Function, Question 8 (https://math.colorado.edu/math1300/resources/Exercises_LogarithmicFunction.pdf)
To be proved :
$\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$.
What I did, before getting stuck :
$n=\log_{\frac {1} {\sqrt b} }\sqrt x$
$\iff \bigg (\frac {1} {\sqrt b} \bigg)^n = \sqrt x $
$\iff \frac {n} { (\sqrt b)^n}= \sqrt x$
$\iff n = (\sqrt b)^n \sqrt x$
Whith the help of @jjagmath I continue the proof at the point I made a mistake :
$ \frac {1} {(\sqrt b)^n}= \sqrt x $
$ \iff 1 = \sqrt x (\sqrt b)^n$
$\iff \frac {1} {\sqrt x} = (\sqrt b)^n$
$ \iff x^{-1/2}= (\sqrt b)^n$
$ \iff n = \log_{\sqrt b} x^{- \frac {1} {2} }$
$ \iff n = - \frac {1} {2}\log_{\sqrt b} x$
$\iff n = - \frac {1}{2} \frac {\log_b x} {\log_b b^{1/2}}$ $\space \space \space$ by theChange of base formula.
$\iff n = - \frac {1}{2} \frac {\log_b x} {\frac {1} {2}}$
$ \iff n = - \log_b x$
| \begin{align}
\log_{\frac{1}{\sqrt{b}}}{\sqrt{x}} &= \frac{\log{\sqrt{x}}}{\log{\frac{1}{\sqrt{b}}}}\\
&= \frac{\log{\sqrt{x}}}{-\log{{\sqrt{b}}}} \\
&= \frac{\frac{1}{2}\log{{x}}}{-\frac{1}{2}\log{{{b}}}} \\
&= -\frac{\log{x}}{\log{b}} \\
&= -\log_b{x}
\end{align}
In total we use the following two rules:
\begin{align}
\log_ab &= \frac{\log{a}}{\log{b}}, \\
\log{(a^{c})} &= c\log{a}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need some help expressing this as a series, if possible Okay, so I'm working on a question, and after a bunch of simplification and manipulations, I get the following expression. Just wanted to ask if there's any way I can get this in the form of a series, with $n$ going from 0 to $\infty$. a is a random positive integer here.
$$
U(x) = Q\bigg[1 - \frac{a(a+1)}{2!}x^2 + \frac{a(a+1)(a+2)(a+3)}{4!}x^4 + ...\bigg]
+ R\bigg[x - \frac{(a-1)(a+2)}{3!}x^3 + \frac{(a-1)(a+2)(a-3)(a+4)}{5!}x^5 + ...\bigg]
$$
I'm having trouble trying to figure out a way to simplify this further, so I'm just curious if it's possible. If not, then that's okay too.
Thanks for taking the time to look at this!!
Editing to add the DE I was working on:
$$(1-x^2)U''(x) -2xU'(x) +a(a+1)U(x) = 0$$
| This is just for your curiosity
$U(x)$ is already expressed as a linear combination of two Taylor series (for which you give only the first terms but in which the patterns of the coefficients is quite clear). I suppose that this is the power series solution of a second oreder differential equation which could be intresting to know.
So, what I think is that the problem would better be to find the analytic functions to which correspond these Taylor series.
The first one is "amazing" since it seems that we have
$$f(x)=\sum_{n=0}^\infty (-1)^n \frac {c_n}{(2n)!} x^{2n}\quad \text{with} \quad c_n=\prod_{i=0}^{2n-1} (a+i)=a \,(a+1)_{2 n-1}$$ where appear Pochhammer symbols. This makes
$$f(x)=\, _2F_1\left(\frac{a}{2},\frac{a+1}{2};\frac{1}{2};-x^2\right)$$ where appears the gaussian hypergeometric function. But, guess what ?
$$\color{blue}{f(x)=\left(1+x^2\right)^{-\frac a2} \cos \left(a \tan ^{-1}(x)\right)}$$
For the second series, it seems to be much more complex since we have
$$g(x)=\sum_{n=0}^\infty (-1)^n \frac {d_n}{(2n+1)!} x^{2n+1}$$ with
$$d_n=\Bigg[\prod_{i=0}^{n} (a+2i)\Bigg]\,\, \Bigg[\prod_{i=0}^{n-1} (a+2i-1)\Bigg]=(-1)^{n+1}(a-1) 2^{2 n-1} \left(\frac{3-a}{2}\right)_{n-1} \left(\frac{a+2}{2}\right)_n$$
$$\color{blue}{g(x)=x \, _2F_1\left(\frac{1-a}{2},\frac{a+2}{2};\frac{3}{2};x^2\right)}$$ which, unfortunately, does not simplify but explains why was required a power series solution.
For integer values of $a$, we have quite nice and simple expressions
$$\left(
\begin{array}{cc}
a & \left(x^2+1\right)^a\,f(x) \\
1 & 1 \\
2 & 1-x^2 \\
3 & 1-3 x^2 \\
4 & x^4-6 x^2+1 \\
5 & 5 x^4-10 x^2+1 \\
6 & -x^6+15 x^4-15 x^2+1
\end{array}
\right)$$
For $g(x)$, it is a bit more complex but it write
$$g(x)= x Q(x)+ R(x) \tanh^{-1}(x)$$
$$\left(
\begin{array}{ccc}
a & Q(x) & R(x) \\
1 & 1 & 0 \\
2 & \frac{3}{4} & -\frac{3 x^2}{4}+\frac{1}{4} \\
3 & -\frac{5 x^2}{3}+1 & 0 \\
4 & -\frac{105 x^2}{64}+\frac{55}{64} & \frac{105 x^4}{64}-\frac{45
x^2}{32}+\frac{9}{64} \\
5 & \frac{21 x^4}{5}-\frac{14 x^2}{3}+1 & 0 \\
6 & \frac{1155 x^4}{256}-\frac{595 x^2}{128}+\frac{231}{256} & -\frac{1155
x^6}{256}+\frac{1575 x^4}{256}-\frac{525 x^2}{256}+\frac{25}{256}
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4333787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are two approaches I see, partial fractions, maybe setting $y=1+x$, but this is still alot of work. Or perform successive integration by parts. Both still involve a lot of calculation. The compact form of the answer makes me hope there is a nice way to achieve it. Guessing the ultimate form and then calculating the derivative is, to me, a less desirable method.
| Explore the embedded symmetry by denoting
$$f_{n,m}(x)= \frac{x^{n-m}+x^{n+m}}{(1+x)^{2n}}$$
which differentiates as
$$f’_{n,m}(x)
=\frac{(n-m)(x^{-m-1}-x^{m+1}) -2m( x^{-m}-x^{m})-(n+m)(x^{-m+1}-x^{m-1}) }{x^{-n}(1+x)^{2n+2}}
$$
In particular, with $n=8$ and $m=0,1,2$
\begin{align}
f’_{8,0}(x)&= \frac{16(x^7-x^9)}{(1+x)^{18}}\\
f’_{8,1}(x)&= \frac{-2(x^7-x^9) +7(x^6-x^{10})}{(1+x)^{18}}\\
f’_{8,2}(x)&
= \frac{-10(x^7-x^9)-4(x^6-x^{10})+ 6(x^5-x^{11})}{(1+x)^{18}}
\end{align}
which leads to
$$
\frac{x^5-x^{11}}{(1+x)^{18}}
= \frac16 f’_{8,2}(x)+\frac2{21} f’_{8,1}(x)+\frac{13}{112} f’_{8,0}(x)
$$
Integrate to obtain
\begin{align}
\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx
&= \frac16 f_{8,2}(x)+\frac2{21} f_{8,1}(x)+\frac{13}{112} f_{8,0}(x)\\
&= \frac{x^6+x^{10}}{6(1+x)^{16}} + \frac{2(x^7+x^{9})}{21(1+x)^{16}} + \frac{26x^8}{112(1+x)^{16}}\\
&=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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The minimal value of a fraction based on a focal chord of an ellipse I came across a very interesting olympiad problem. It goes as follows: suppose you have an ellipse given by $\frac{x^2}{16}+\frac{y^2}{9}=1$ and a line that goes through the point $A(\sqrt{7},0)$. This line also goes through the ellipse at two points $N$ and $M$. What is the minimum possible value of $\frac{AM\cdot AN}{NM}$? Here is a desmos link for visualization: https://www.desmos.com/calculator/mncpycyaj3 . My attempt: first I noted that $a=4$ and $b=3$ from the equation of the ellipse. Then I know that $a^2-b^2=c^2$, so $c=\sqrt{7}$. This immediately tells us that the line actually goes through the right focus $F_2$. So, the fraction of interest can be written as $\frac{AM\cdot AN}{NM}=\frac{F_2M\cdot F_2N}{F_2M+F_2N}$. Now I tried a few possible configurations of the line: 1) [a horizontal line] $N(4,0);M(-4,0)$, thus, $F_2N=\sqrt{(4-\sqrt{7})^2+(0-0)^2}=4-\sqrt{7}$ and $F_2M=\sqrt{(-4-\sqrt{7})^2+(0-0)^2}=4+\sqrt{7}$. So, $\frac{F_2M\cdot F_2N}{F_2M+F_2N}=\frac{(4-\sqrt{7})(4+\sqrt{7})}{8}=\frac{16-7}{8}=\frac{9}{8}=1.125$. 2) [a vertical line] $N\left(\sqrt{7},\frac{9}{4}\right);M\left(\sqrt{7},-\frac{9}{4}\right): F_2N=\sqrt{(\sqrt{7}-\sqrt{7})^2+\left(\frac{9}{4}-0\right)^2}=\frac{9}{4}=2.25=F_2M$, thus, $\frac{F_2M\cdot F_2N}{F_2M+F_2N}=\frac{2.25\cdot 2.25}{4.5}=1.125$. At this point I am thinking that this is some universal constant or something, but I have only checked the extreme cases. 3) [a line going through the highest point of the ellipse] $N(0,3)$, then the other point is found by $$y=\frac{y_M}{x_M-\sqrt{7}}(x-\sqrt{7});\space 3=-\frac{y_M}{x_M-\sqrt{7}}\sqrt{7}$$
But this point lies on the ellipse, $\frac{x_M^2}{16}+\frac{y_M^2}{9}=1$. Solving for $x_M$ and then for $y_M$ we get $M\left(\frac{32\sqrt{7}}{23},-\frac{27}{23}\right)$. Then $F_2N=\sqrt{(0-\sqrt{7})^2+(3-0)^2}=4$ and $F_2M=\sqrt{\left(\frac{32\sqrt{7}}{23}-\sqrt{7}\right)^2+\left(-\frac{27}{23}-0\right)^2}=\frac{36}{23}$;$\space \frac{F_2M\cdot F_2N}{F_2M+F_2N}=1.125$. My question is: is this fraction always equal to $1.125$ and if yes, then how can I prove this fact? I have tried some arbitrary points from desmos and plugged them in Wolframalpha, the outputs seem to be slightly bigger than 1.125, but I guess that is just due to rounding up of the points in desmos. And another thing I would like to know is: if it is indeed true then is this only a property of lines going through focuses or in general for any lines? Any hints are much appreciated.
| This is a focal chord, so the product of the distances $AN$ and $AM$ is a constant
$AN \cdot AM = (a - c) \cdot (a + c) = a^2 - c^2 = b^2$
Thus
$AN \cdot AM = 9 $
The ratio requested is thus $\dfrac{ 9 }{ NM } $
and the distance ${NM}$ is maximum when $N$ and $M$ are on the major axis, in which case ${NM} = 2a = 8$
Thus the minimum possible ratio is $\dfrac{9}{8} = 1.125$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_{0}^{\infty} \frac{4\pi}{16\pi^2 + x^2} \left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right) \, dx$ I’m trying to evaluate the following integral:
$$\int_{0}^{\infty} \frac{4\pi}{16\pi^2 + x^2} \left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right) \, dx$$
I’ve tried using contour integration by using a quarter-circle contour and going around the pole at $z=4\pi i$ with a semi-circular arc, however, I wasn’t able to evaluate the integrals along the imaginary axis.
I wasn’t able to come up with a real or complex method for evaluating this integral, so any help would be appreciated. I’m not sure if a closed-form exists.
| I'm afraid there is no a closed form for this integral.
$$I=\int_0^\infty \frac{4\pi}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{e^{-x}-1}) dx=4\pi\int_0^\infty \frac{dx}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{2}\coth\frac{x}{2}-\frac{1}{2}) dx$$
$$=-\frac{\pi}{4}+\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\Big(\frac{1}{t}-\coth t\Big)=-\frac{\pi}{4}-\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\frac{d}{dt}\Big(\ln\frac{\sinh t}{t}\Big)$$
Using the representation $\,\,\ln\frac{\sinh t}{t}=\sum_{k=1}^\infty\ln\Big(1+\frac{t^2}{\pi^2k^2}\Big)$ and making the change $x=t^2$
$$I=-\frac{\pi}{4}-\pi\int_0^\infty\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^N\frac{dx}{4\pi^2+x}\sum_{k=1}^\infty\frac{1}{\pi^2k^2+x}$$
$$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\sum_{k=1}^\infty\int_0^N\frac{dx}{4\pi^2+x}\frac{1}{\pi^2k^2+x}$$
$$=-\frac{\pi}{4}-\pi\lim_{N\to\infty}\int_0^Ndx\bigg(\frac{1}{3\pi^2}\Big(\frac{1}{\pi^2+x}-\frac{1}{4\pi^2+x}\Big)+\frac{1}{(4\pi^2+x)^2}\bigg)$$
$$-\pi\lim_{N\to\infty}\sum_{k=3}^\infty\int_0^N\frac{dx}{\pi^2(k^2-4)}\Big(\frac{1}{4\pi^2+x}-\frac{1}{\pi^2k^2+x}\Big)$$
Integrating and taking the limit
$$=-\frac{\pi}{4}-\frac{2\ln 2}{3\pi}-\frac{1}{4\pi}-\pi \sum_{k=3}^\infty\frac{1}{\pi^2(k^2-4)}\ln\frac{k^2}{4}$$
$$I(n)=-\frac{\pi}{4}+\frac{3\ln 2}{8\pi}-\frac{1}{4\pi}-\frac{2}{\pi}\sum_{k=3}^\infty\frac{\ln k}{k^2-4}\approx-1.36824...$$
It is not clear whether a closed form for the sum $\sum_{k=3}^\infty\frac{\ln k}{k^2-4}$ exists.
| {
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"url": "https://math.stackexchange.com/questions/4341440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
| Riemann integration for the area integrals is nasty as you need to find a way to enumerate the whole area.
I would suggest slightly different way, i.e. randomly picking a point from the triangle:
$$ f(x,y) = \sqrt{x^2-y^2} $$
$$ \int_D {f(x,y)} = A \cdot \frac{\sum_{n=1}^{k}{f(x_n,y_n)}}{k}, (x_n,y_n)\in D, k>1200$$, where $(x_n,y_n)$ is chosen randomly from the triangle and A is the area of the triangle. This is called "average-based integral", and it is slightly easier to use for areas like D than riemann integral.
Picking a point from a triangle with nice distribution can be done by picking a point from bounding box, and then filtering out random points that are outside of the triangle, as in:
$$ \{ (x,y) \in [x_0..x_1]\times [y_0..y_1] | (x,y) \in D \} $$
| {
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"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find the area of region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.
My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{BP}{PE}\\
\frac{S_{BPM}}{S_{BCE}}=\frac{BM.BP}{BC.BE}=\frac{BP}{2BE}\\
\frac{S_{ABM}}{S_{ABP}}=\frac{AB.AM}{AB.AP}\implies \frac{20}{S_{ABP}} = \frac{AM}{AP}\\
\frac{S_{ABC}}{S_{BCE}}=\frac{AC}{CE}$
...?
| The only data of this problem is $AB=3BM$ and area $S=40$ in the triangle $\triangle {ABC}$ in which $M$ is the midpoint of $BC$. There are a lot of such triangles whose shape can be quite different so either the problem is undetermined or there is a family of distinct triangles having area as a very nice invariant. We choose the angle $\beta=90^{\circ}$ and we must find the value $4$ according Math Lover's answer.
Put $\angle {ABC}=\beta=90^{\circ}, AB=6k, BC=2k$.
$$S_{\triangle{ABC}}=6k^2\sin(90^{\circ})=6k^2=40\Rightarrow k=\sqrt{\frac{20}{3}}\\S_{\triangle{BMP}}=\frac12k\cdot BP\sin45^{\circ}=\frac{k\cdot BP\sqrt2}{4}$$
In $\triangle{ABP} : \dfrac{BP}{\sin(\alpha_1)}=\dfrac{6k}{\sin(135^{\circ}+\alpha_1)}=\dfrac{12k}{\sqrt2(\cos(\alpha_1)-\sin(\alpha_1))}=\dfrac{12AM}{5\sqrt2}=\dfrac{6\sqrt2(AM)}{5}$
Then $BP=\sin(\alpha_1)\dfrac{6\sqrt2(AM)}{5}=\dfrac{k}{AM}\cdot\dfrac{6\sqrt2(AM)}{5}=\dfrac{6\sqrt2k}{5}$.
Consequently $$\\S_{\triangle{BMP}}=\frac{k\sqrt2}{4}\cdot\dfrac{6\sqrt2k}{5}=\frac{3k^2}{5}=\dfrac35\cdot\dfrac{20}{3}=\color{red}{\boxed4}$$
Thus, in the great variety of possible shapes for the triangle $\triangle{BMP}$ (we can change the angle $\beta$) the area of all is a pretty invariant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Make $2$ cubes out of $1729$ unit cubes, expected number of times you have to paint I'm trying to solve question 6 from the PuMac 2007 Combinatorics A competition:
Joe has $1729$ randomly oriented randomly arranged unit cubes, which are initially unpainted. He makes two cubes of sidelengths $9$ and $10$ or of sidelengths $1$ and $12$ (randomly chosen). These cubes are dipped into white paint. Then two more cubes of sidelengths $1$ and $12$ or $9$ and $10$ are formed from the same unit cubes, again randomly oriented and randomly arranged, and dipped into paint. Joe continues this process until every side of every unit cube is painted. After how many times of doing this is the expected number of painted faces closest to half of the total?
Here's what I got so far:
*
*${1\over2}$ chance of this happening: If you make two cubes of side lengths $9$ and $10$, then $16$ cubes will have $3$ faces sharing a vertex painted, $12(8 + 7) = 180$ cubes will have $2$ faces sharing an edge painted, $6(8^2 + 7^2) = 678$ cubes will have $1$ face painted, and the remaining $7^3 + 8^3 = 855$ cubes will have no faces painted.
*${1\over2}$ chance of this happening: If you make two cubes of side lengths $1$ and $12$, then $1$ cube will have all $6$ faces painted, $8$ cubes will have $3$ faces sharing a vertex painted, $12(10) = 120$ cubes will have $2$ faces sharing an edge painted, $6(10^2) = 600$ cubes will have $1$ face painted, and the remaining $10^3 = 1000$ cubes will have no faces painted.
But I'm stuck as this point, and don't know what to do next. Any help would be well-appreciated.
| Thankfully we are being asked about the expected value of this random variable, the random variable is a sum of simpler random variables, namely whether each unit face remains unpainted or not, so we can exploit linearity of expectation.
If we are able to find a good expression for the expected value after $k$ steps then we must simply find which $k$ yields the value closest to $6\times 1729/2$
First note all $6\times 1729$ of the faces are identical. The expected number of painted faces after $k$ steps is equal to the probability $p_k$ that a face remains unpainted after $k$ steps multiplied by $6\times 1729$.
Hence we are simply asked to find the value of $k$ for which $p_k$ is closest to $\frac{1}{2}$.
Now, since every step is independent we begin by finding the probability $q$ that a face is painted in a single step. Assume we are in the $9,10$ case and notice that each possible location and orientation of the unit cube of our face is equally likely, so the probability that it is on the outside (and hence gets painted) is $\frac{6\times9^2 + 6\times 10^2}{6\times1729}$. For the other case we get $\frac{6\times 1^2 + 6\times 12^2}{6\times 1729}$, so $q = \frac{1}{2}\frac{6\times9^2 + 6\times 10^2}{6\times 1729} + \frac{1}{2}\frac{6\times 1^2 + 6\times 12^2}{6\times 1729} = \frac{7^2 + 10^2 + 1^2 +12^2}{2\times 1729} = 21/247 \approx 0.085$
Since $p_k$ is the probability that a face is never colored we have $p_k = (1-q)^k$, so we must solve for $\frac{1}{2} = (1-q)^k$ which yields a solution of 7.8. The integer value that gives the closest result to $\frac{1}{2}$ is $8$.
| {
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$(\frac{a}{b}-1)^2+(\frac{b}{a}+1)^2\ge3$ I tried expanding the parenthesis and then grouping them back somehow but i couldn't get anywhere.
I also tried to apply known inequalities with no success. Can someone help me out?
| Substitute $x=\frac{a}{b}$.
\begin{align*}
\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}+1\right)^2 - 3 &= (x-1)^2+\left(1+\frac{1}{x}\right)^2-3 \\
&= \frac{x^2(x^2-2x+1)+(x^2+2x+1)-3x^2}{x^2} \\
&= \frac{x^4-2x^3-x^2+2x+1}{x^2} \\
&= \frac{(x^2-x-1)^2}{x^2} \\
&\geq 0.
\end{align*}
| {
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Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$ I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying
$$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$
I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities
$$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(t)$$
The new equation will be
$$f\Big(1-\frac{2}{\cos^2(t)}\Big)+f\Big(\frac{2}{\sin^2(t)}-1\Big)=$$
$$\cos^2(t)-\sin^2(t)$$
I think that this approach won't allow me the get the answer.
Any idea will be appreciated.
| Substitute in $x\mapsto\frac{x-3}{x+1}$. Then we have
$$f\left(\frac{x+3}{1-x}\right)+f\left(x\right)=\frac{x-3}{x+1}$$
Now substitute again $x\mapsto \frac{x-3}{x+1}$,
$$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{x+3}{1-x}$$
Now add these two to get
$$2f(x)+f\left(\frac{x+3}{1-x}\right)+f\left(\frac{x-3}{x+1}\right)=\frac{x-3}{x+1}+\frac{x+3}{1-x}$$
$$2f(x)=\frac{x-3}{x+1}+\frac{x+3}{1-x}-x$$
$$f(x)=\frac{1}{2}\left(\frac{x-3}{x+1}+\frac{x+3}{1-x}-x\right)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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A Recurrence Relations problem I need to bound
$$T(n) = 2T\left(\frac{n}{2}\right) + n\cdot \frac{\log \log n}{\log n}$$ as tight as possible.
Here is what I got:
Firstly, we introduce a change of variable. Let $n = 2^k$ while ignoring the rounding, then we have
$$
T(n) = T(2^k) =: g(k).
$$
Expand it further, we get
$$
\begin{split}
g(k) &= 2\cdot g(k-1) + 2^k\cdot \frac{\log \log 2^k}{\log 2^k}\\
&= 2\cdot g(k-1) + 2^k\cdot \frac{\log k}{k}\\
&= 2\cdot \left(2\cdot g(k-2) + 2^{k-1}\cdot \frac{\log k-1}{k-1}\right) + 2^k\cdot \frac{\log k}{k}\\
&= 2^2\cdot g(k-2) + 2^{k} \left(\frac{\log k-1}{k-1}\right) + 2^k\cdot \frac{\log k}{k}\\
&= 2^2\cdot g(k-2) + 2^{k} \left(\frac{\log k-1}{k-1} + \frac{\log k}{k}\right)\\
\end{split}
$$
By induction, we see that
$$
g(k) = 2^{k - 1} g(1) + 2^k\left(\sum\limits_{i=0}^{k-1} \frac{\log k-i}{k-i}\right)
= 2^{k - 1} + 2^k\left(\sum\limits_{i=0}^{k-1} \frac{\log k-i}{k-i}\right)
$$
since $g(1) = T(2) = 1$ by assumption. Let $k-i := j$, we further have
$$
g(k) = 2^{k - 1} + 2^k\sum\limits_{j = 1}^{k} \frac{\log j}{j}.
$$
Substitute $k$ back to $n$, we have
$$
T(n) = \frac{n}{2} + n\sum\limits_{k = 1}^{\log n} \frac{\log k}{k}.
$$
In order to bound the sum, we investigate $\frac{\log x}{x}$. We see that
$$
\left(\frac{\log x}{x}\right)^\prime = \frac{1 - \log x}{x^2},
$$
which is $0$ when $x = 2$, and positive in $(0, 2]$ while negative in $[2,+\infty)$ assuming base $2$. By taking the second derivative,
we see that it's negative at $2$, hence we conclude that it reaches its maximum at $2$ with the value $\frac{1}{2}$. Hence, we have
$$
T(n) = \frac{n}{2} + n\sum\limits_{k = 1}^{\log n} \frac{\log k}{k} \leq \frac{n}{2} + n \log n\cdot \frac{1}{2} = O(n\log n).
$$
As one can see, the main problem in this derivation is to bound the summation
$$
\sum\limits_{k = 1}^{\log n}\frac{\log k}{k},
$$
which I can't think of a better bound. It's clearly too large since I simply take the maximum of $\frac{\log x}{x}$ while $\frac{\log x}{x}$ is rapidly decreasing. Are there any formula I can use for improving this bound?
| Here I will present a bound for the sum
$$\sum_{j=1}^k \frac{\log j}{j}.$$
The basic idea is that the partial sum can be asymptotically approximated by the integral of the function $\frac{\log x}{x}$, which is easy to compute. More generally, if a function $f(x)$ is decreasing on $[a,\infty)$, then the following bounds hold.
$$f(b) + \int_a^b f(x)\ dx \leq \sum_{j=a}^b f(j) \leq f(a) + \int_a^b f(x)\ dx $$
This can be seen by drawing a picture and considering the left and right Riemann sums of step size $1$.
The function $(\log x)/x$ is zero at $x = 1$ and decreasing for $x \geq 3$, so we can approximate the sum like so:
\begin{align*}
\sum_{j=1}^k \frac{\log j}{j} &= \frac12 \log 2 + \sum_{j=3}^k \frac{\log j}{j}\\
&\leq \frac12 \log 2 + \frac13 \log 3 + \int_3^k \frac{\log x}{x}\ dx\\
&= \frac12 \log 2 + \frac13 \log 3 + \Big( \frac12\log^2 x \Big)\Big|_{x=3}^k\\
&= \frac12 \log 2 + \frac13 \log 3 + \Big(\frac12\log^2 k - \frac12\log^2 3\Big)\\
&\approx \frac12\log^2 k +0.1093\ldots\\
&\leq \frac12\log^2 k + 0.11
\end{align*}
Here I have assumed that the logarithm in use is the natural logarithm, but you can convert to another logarithm by dividing through by the appropriate change of base constant. At the very least, it shows that
$$\sum_{j=1}^k \frac{\log j}{j} = \mathcal O (\log^2 k)$$
which may be sufficient. Here is a plot of the asymptote.
| {
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"source": "stackexchange",
"question_score": "2",
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What is $\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1}$?
A bag contains $n$ white and $n$ black balls, all of equal size. Balls are drawn at random. Find the probability that there are both white and black balls in the draw and that the number of white balls is greater than that of black balls by $1$.
The total number of possible draws will be the number of subsets of a set with $2n$ elements (minus the empty set) since we have $2n$ balls, hence $2^{2n} - 1$. The number of draws satisfying the conditions there's both white and black in the draw and also the number of white being greater than that of black by $1$ involves computing the following sum:$$\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1},$$which I don't know how to do. Could anybody help? Thanks. I've tried playing around with paths on an $n \times n$ grid, also an $2n \times n$ grid to come up with a combinatorial proof, not to any success.
| A combinatoric proof : Let :
$$A = \{1, 2, \ldots, 2 k\}, B = \{1, \ldots, k\} \text{ and } C = \{k + 1, \ldots, 2 k\}$$
We need to choose $k - 1$ elements from $A$. There are two ways :
*
* First way : Choose $k - 1$ element directly for $A$ : In this case, we have then :
$$\binom{2 k}{k - 1}$$
ways.
* Second way : We can choose $r$ elements from $B$ and $k - 1 - r$ from $C$ for $r \in \{0, \ldots, k -1 \}$ : In this case we have :
$$\sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{k - 1 - r}$$
ways.
We deduce that :
$$\binom{2 k}{k - 1} = \sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{k - 1 - r} = \sum_{r = 0}^{k-1} \binom{k}{r} \binom{k}{r + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352574",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\dots$
In general, for $n \in \mathbb{N}$, let $A(n)$ be the same series as the above with sign alternates every $n$ terms.
I know that $A(1)=\log(2)$, and $A(2)=\frac{1}{4}(\pi+\log(4))$.
How can we evaluate $A(n)$ for $n \in \mathbb{N_{\ge3}}$?
| $A_n$ can be expressed as
$$
A_n=\sum_{k=1}^n \sum_{u=0}^\infty \frac{(-1)^u}{nu+k}.
$$
Now, from the Taylor series of $\frac{1}{1+x}$ for $|x|<1$ we have
$$
{
\frac{1}{1+x}=\sum_{u=0}^\infty (-1)^ux^u\implies
\\
\frac{1}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu}\implies
\\
\frac{x^{k-1}}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu+k-1}\implies
\\
\int_0^1\frac{x^{k-1}}{1+x^n}dx=\sum_{u=0}^\infty \frac{(-1)^u}{nu+k}
}
$$
hence
$$
A_n{=\sum_{k=1}^n \int_0^1\frac{x^{k-1}}{1+x^n}dx
\\=
\int_0^1\sum_{k=1}^n\frac{x^{k-1}}{1+x^n}dx
\\=
\int_0^1\frac{1-x^n}{1+x^n}\frac{1}{1-x}dx
\\=
\int_0^1\sum_{k=1}^n \frac{2u_k}{u_k-1}\frac{1}{x-u_k} dx
\\=
\sum_{k=1}^n\frac{2u_k}{u_k-1}\int_0^1 \frac{1}{x-u_k} dx
\\=
\sum_{k=1}^n\frac{2u_k}{u_k-1}[\ln(1-u_k)-\ln(-u_k)]
}
$$
where $u_k=\exp(i\pi\frac{2k+1}{n})$ is the $k$th root of $-1$ and $\ln$ is the main branch of complex logarithm.
Additional Remarks
The reason for interchanging integrals and summations, is the existence and boundedness of the integral values and the boundedness of summation scope.
| {
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"question_score": "3",
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Calculate the determinant of $a_{ij} = \frac{(1+x)^{i+j-1}-1}{i+j-1}$ There is a question asked by my classmate. Looking forward to some ideas, thanks.
Set $A=\{a_{ij}\}_{n\times n}$, where
$$a_{ij}=\frac{(1+x)^{i+j-1}-1}{i+j-1}.$$
Prove that $\det A=cx^{n^2}$ for some $c$.
I have tried to calculate it, but failed. I computed
$$\frac{(1+x)^{i+j-1}-1}{i+j-1}=\sum_{k=1}^{i+j-1}\frac{(i+j-2)!}{k!(i+j-1-k)!}x^k,$$
but I have no idea how to continue.
I know when $a_{ij}=\frac{1}{i+j-1}$, it is the Hilbert matrix, and we can get its determinant, but I don’t know how to calculate the above determinant. Are there some hints? Looking forward to your answer. Thanks!
| Firstly, we know there is a nondegenerate matrix $J(n)$ such that
$$(1,x,\dots,x^{n-1})=(1,(1+x),\dots,(1+x)^{n-1})J(n).$$
Then we have
$$
\begin{aligned}
&\begin{pmatrix}
1&x&\cdots&x^{n-1}\\
x&x^2&\cdots&x^{n}\\
\vdots&\vdots&\ddots&\vdots\\
x^{n-1}&x^n&\cdots&x^{2n-2}
\end{pmatrix}
=(1,x,\cdots,x^{n-1})^T(1,x,\cdots,x^{n-1})\\[7pt]&=
J(n)^T\begin{pmatrix}
1&1+x&\cdots&(1+x)^{n-1}\\
1+x&(1+x)^2&\cdots&(1+x)^{n}\\
\vdots&\vdots&\ddots&\vdots\\
(1+x)^{n-1}&(1+x)^n&\cdots&(1+x)^{2n-2}
\end{pmatrix}J(n).
\end{aligned}
$$
Hence, integrating respect to $x$ yields (Note that $J(n)$ does not depend on $x$)
$$
\begin{pmatrix}
\frac{x}{1}&\frac{x^2}{2}&\cdots&\frac{x^{n}}{n}\\
\frac{x^2}{2}&\frac{x^3}{3}&\cdots&\frac{x^{n+1}}{n+1}\\
\vdots&\vdots&\ddots&\vdots\\
\frac{x^{n}}{n}&\frac{x^{n+1}}{n+1}&\cdots&\frac{x^{2n-1}}{2n-1}
\end{pmatrix}=
J(n)^T\begin{pmatrix}
\frac{(1+x)-1}{1}&\frac{(1+x)^2-1}{2}&\cdots&\frac{(1+x)^{n}-1}{n}\\
\frac{(1+x)^2-1}{2}&\frac{(1+x)^3-1}{3}&\cdots&\frac{(1+x)^{n+1}-1}{n+1}\\
\vdots&\vdots&\ddots&\vdots\\
\frac{(1+x)^{n}-1}{n}&\frac{(1+x)^{n+1}-1}{n+1}&\cdots&\frac{(1+x)^{2n-1}-1}{2n-1}
\end{pmatrix}J(n).
$$
Then it is easy to see that
$$\det
\begin{pmatrix}
\frac{x}{1}&\frac{x^2}{2}&\cdots&\frac{x^{n}}{n}\\
\frac{x^2}{2}&\frac{x^3}{3}&\cdots&\frac{x^{n+1}}{n+1}\\
\vdots&\vdots&\ddots&\vdots\\
\frac{x^{n}}{n}&\frac{x^{n+1}}{n+1}&\cdots&\frac{x^{2n-1}}{2n-1}
\end{pmatrix}=ax^{n^2},$$
which implies $\det A=cx^{n^2}$.
| {
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I have a small problem with long division when dividing polynomials of the same degree I was doing long division with $x^2 + 1$, and $3x^2+5$. (the second polynomial is the quotient).
the problem I've found is related to dividing two polynomials of the same degree. Even if I know that the quotient is always a constant and the remainder is a polynomial of one degree less than the dividend, I still have problem when performing this long division, meaning:
$x^2+0+1\space /\space 3x^2+5$
x^2 is contained in 3x^2 3 times, so I write 3 in the quotient.
then, I perform multiplication between the quotient and the divisor. $3 * 3x^2 = 9x^2$, and $3*5 = 15$. I write them below the dividend.
Now, I subtract the dividend with the things I have below.
but, it's an infinite loop, because the degree doesn't change no matter how long I divide.
| You are confusing what you dividing into with what you are dividing by and you are taking the quotient, $3$, and multiply it by what you are dividing into; not what you are dividing by. You must multiply the quotient by what you are dividing by.
Question 1: $\frac {3x^2+ 5}{x^2 + 1}$ then we divide $x^2$ into $3x^2$ and get a quotient of $3$. So we multiply the denominator, $x^2 + 1$ by $3$ to get $3(x^2+1)=3x^2 + 3$. Then you subtract $(3x^2 + 5)-(3x^2 + 3) = 2$. Now you have the remainder.
So $\frac {3x^2 + 5}{x^2 + 1} = 3 + \frac 2{x^2 + 1}$.
Question 2: $\frac {x^2 + 1}{3x^2 + 5}$ then we divide $3x^2$ into $x^2$ and we get a quotient of $\frac 13$ (because $3x^2$ goes into $x^2$ a total of $\frac 13$ times). So we multiply the denominator, $3x^3 + 5$ by $\frac 13$ to got $\frac 13(3x^2 + 5) = x^2 + \frac 53$. Then we subtract $(x^2 + 1)-(x^2 + \frac 53)= -\frac 23$. Now we have a remainedr of $-\frac 23$.
So $\frac {x^2+1}{3x^2 + 5} = \frac 13 -\frac {\frac 23}{3x^2 + 5}$.
| {
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How to evaluate $\int_0^\infty \frac{\cos(ax)}{1+x^N}dx$ for $a,N\in\mathbb{R}$ and $N> 1$ Interesting challenge: Does anyone know how to evaluate the integral $$\int_0^\infty\frac{\cos(ax)}{1+x^N}\text{d}x$$ for $a,N\in\mathbb{R}$ and $N>1$?
The $N=2$ case is quite easily done using the Residue Theorem as well as for general $N>1$ for $a=0$. Both exploit symmetries when path integrating, which do not work in the general case.
Let me know if you have any ideas! Cheers!
| For even $N$ you can use the residue theorem as well. It suffices to consider $a>0$ and calculate the integral
$$\oint_{-\infty}^{\infty} \frac{e^{iax}}{1+x^N} \, {\rm d}x = \int_{0}^{\infty} \frac{2\cos(ax)}{1+x^N} \, {\rm d}x = 2\pi i \sum_{0 \leq k \leq \frac{N-1}{2}} {\rm Res}\left( \frac{e^{iax}}{1+x^N} \right)\Bigg|_{x=e^{\frac{i\pi+i2\pi k}{N}}} \\
=\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia e^{\frac{i\pi+i2\pi k}{N}} + \frac{i\pi+i2\pi k}{N}} = \frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{i\pi+i2\pi k}{N}}e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\
=\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} \Bigg[\cos\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right) \\ + i \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)\Bigg]e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\
=\frac{2\pi}{N} \sum_{k=0}^{N/2-1} \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)}$$
where the loop-integral on the LHS is closed in the upper half plane. The imaginary part is zero, since the LHS is real.
| {
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Show that the bissectors of $\angle BHC$ and $\angle BFC$ meet on $BC$ Given an acute triangle $\triangle ABC$ with orthocenter $H$. Let $D = BH \cap AC, E = CH \cap AB$ and $F = (AEDH) \cap (ABC) \neq A$. Show that the inner angle bisectors of $\angle BFC$ and $\angle BHC$ meet on the side $BC$.
It is easy to prove that the dotted ray passes through the midpoint of $BC$ and the antipode of $A$ in the $(ABC)$.
My first idea was to find $\angle CBF$ but that is a bit tough. So maybe drawing the perpendiculars from the meeting of the bissector of $\angle CHB$ with $BC$ to $FB$ and $FC$ may lead to something but it seems I need the forementioned angles.
|
With the ideas of sirous, we use barycentric calculus. The coordinates of $D$ and $E$ are
\begin{equation}
D\bigg(\frac{a^2+b^2-c^2}{2b^2},0,\frac{-a^2+b^2+c^2}{2b^2}\bigg),\quad E\bigg(\frac{a^2-b^2+c^2}{2c^2},\frac{-a^2+b^2+c^2}{2c^2},0\bigg)
\end{equation}
The circumcircle $(ABC)$ has equation $-a^{2}yz-b^{2}xz-c^{2}xy=0$ and $(AED)$ has $-a^{2}yz-b^{2}xz-c^{2}xy+\Big(\frac{a^2-b^2+c^2}{2}y+\frac{a^2+b^2-c^2}{2}z\Big)(x+y+z)=0$ whose intersection $F\neq A$ is
\begin{equation}
F\Big(a^2\big(a^2+b^2-c^2\big)\big(a^2-b^2+c^2\big):-\big(-b^2+c^2\big)\big(a^2+b^2-c^2\big)\big(-a^2+b^2+c^2\big):\big(-b^2+c^2\big)\big(a^2-b^2+c^2\big)\big(-a^2+b^2+c^2\big)\Big)
\end{equation}
The bisector of $BC$ is $y-z=\frac{-b^2+c^2}{a^2}x$ whose below intersection with $(ABC)$ is $U\big(-a^2:b(b+c):c(b+c)\big)$.
The lines $BC$ and $UF$ intersects on $G_1\Big(0:\big(a^2+b^2-c^2\big)c:\big(a^2-b^2+c^2\big)b\Big)$.
The orthocenter is $H\Big(\big(a^2+b^2-c^2\big)\big(c^2+a^2-b^2\big):\big(b^2+c^2-a^2\big)\big(a^2+b^2-c^2\big):\big(c^2+a^2-b^2\big)\big(b^2+c^2-a^2\big)\Big)$ and $(HBC)$ is $-a^{2}yz-b^{2}xz-c^{2}xy+\big(b^2+c^2-a^2\big)x(x+y+z)=0$ and the intersection of this circle with the bisector of $BC$ is $V\big(-a^2:a^2+bc-c^2:a^2+bc-b^2\big)$.
The lines $BC$ and $HV$ intersects on $G_2\Big(0:\big(a^2+b^2-c^2\big)c:\big(a^2-b^2+c^2\big)b\Big)$.
We have $G_1=G_2$ and the demostration is done.
| {
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"answer_count": 5,
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} |
Composition of piecewise functions - Strange result I'm trying to get $f(g(x))$, where:
$$
f(x)=
\begin{cases}
\sqrt{1-x} &\text{if } x \leq 1 \\
x &\text{if } x > 1
\end{cases}
$$
$$
g(x)=
\begin{cases}
x + 1 &\text{if } x \geq 0 \\
x^2 &\text{if } x < 0
\end{cases}
$$
I
followed these steps:
*
*$g(x \geq 0) = x + 1$ hence:
$$
f(g(x \geq 0)) =
\begin{cases}
\sqrt{1 - (x + 1)} = \sqrt{-x} &\text{if } 0 \leq x \leq 1 \\
x + 1 &\text{if } x > 1
\end{cases}
$$
but $\sqrt{-x}$ is not real! Is it correct? if not, what is the right result?
*$g(x < 0) = x^2$ hence:
$$
f(g(x < 0)) = \sqrt{1 - x^2} \quad \text{if } x < 0
$$
EDIT: steps to get the @JoséCarlosSantos solution
$
\boldsymbol{g(x) \leq 1}
$
$$
\begin{cases}
x + 1 \leq 1 \\
x^2 \leq 1
\end{cases}
$$
$$
\begin{cases}
x \leq 0 \\
-1 \leq x \leq 1
\end{cases}
$$
so $-1 \leq x \leq 0$.
$
\boldsymbol{g(x) > 1}
$
$$
\begin{cases}
x + 1 > 1 \\
x^2 > 1
\end{cases}
$$
$$
\begin{cases}
x > 0 \\
x < - 1 \text{ or } x > 1
\end{cases}
$$
this means that I can split it in two subsystems:
*
*$$ \begin{cases} x > 0 \\ x < - 1 \end{cases} $$ this system does not have any solution;
*$$
\begin{cases}
x > 0 \\
x > 1
\end{cases}
$$ the solution is $x > 1$.
Putting together the results of the two subsystems: $x > 1$. I made a mistake, the right result should be $x < - 1$ or $x > 0$.
| Since$$g(x)=\begin{cases}x+1&\text{ if }x\geqslant0\\x^2&\text{ if }x<0,\end{cases}$$you have$$g(x)\leqslant1\text{ if }x\in[-1,0]\quad\text{and}\quad g(x)>1\text{ if }x<-1\text{ or }x>0.$$So,\begin{align}f\bigl(g(x)\bigr)&=\begin{cases}\sqrt{1-g(x)}&\text{ if }x\in[-1,0]\\g(x)&\text{ if }x<-1\text{ or }x>0\end{cases}\\&=\begin{cases}\sqrt{1-x^2}&\text{ if }x\in[-1,0)\\0&\text{ if }x=0\\x^2&\text{ if }x<-1\\x+1&\text{ if }x>0.\end{cases}\end{align}
| {
"language": "en",
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} |
About integration of $\frac{1}{(1+a^2 x^2)(1+x^2)}$ in differentiation under integral sign Consider the function $F(a)=\int_0^{+\infty} \frac{\arctan(ax)}{x(1+x^2)} dx$. I assumed $a \ge 0$ because $F$ is odd, I applied the theorem of differentiation under the integral sign and arrive to $F'(a)=\int_0^{+\infty} \frac{1}{(1+a^2 x^2)(1+x^2)}dx$; while doing the partial fraction decomposition of $\frac{1}{(1+a^2 x^2)(1+x^2)}$, I get
$$\frac{1}{(1+a^2 x^2)(1+x^2)}=\frac{A}{1+a^2 x^2}+\frac{B}{1+x^2}=\frac{A+Ax^2+B+Ba^2 x^2}{(1+a^2 x^2)(1+x^2)}$$
$$=\frac{A+B+(A+Ba^2)x^2}{(1+a^2 x^2)(1+x^2)}$$
So
$$\begin{cases} A+B=1 \\ A+Ba^2=0 \end{cases} \iff \begin{cases} A=1-B \\ 1-B+Ba^2=0 \end{cases} \iff \begin{cases} A=1-B \\ B(1-a^2)=1 \end{cases} \iff \begin{cases} A=-\frac{a^2}{1-a^2} \\ B=\frac{1}{1-a^2} \\ a \ne 1\end{cases}$$
I assumed $a \ne 1$ because I divided by $1-a^2$ (and not $a \ne -1$ too, because I'm working with $a \ge 0$). So I obtain the same result of the question I linked above, that is $F'(a)=\frac{\pi}{2(1+a)}$ and so $F(a)=\frac{\pi}{2}\log(1+a)$ for $a \ge 0$, by oddity then for $a<0$ it is $F(a)=-\frac{\pi}{2} \log (1-a)$.
The question is: what about $a=1$ and $a=-1$? To obtain the antiderivative with the partial fraction decomposition I divided by $1-a^2$, so I believe that I can't use the same expression when $a=1$ (or $a=-1$ by oddity), however the value of the integral for $a=1$ or $a=-1$ is respectively $\frac{\pi}{2} \log 2$ and $-\frac{\pi}{2}\log 2$, so it is coherent with the result $F(a)=\text{sgn}(a)\frac{\pi}{2} \log(1+|a|)$ even if $a=1$ or $a=-1$. I don't understand why this works even if, at some step, I divided by something that could be $0$ and still get the right result. Moreover, I can't fix $a=1$ or $a=-1$ to do the partial fraction decomposition of $\frac{1}{(1+x^2)^2}$ because all the point of differentiation under the integral sign is having an parameter to fix appropriately at the end and get the desired value of the integral; so fixing it makes all this senseless (moreover, $1/(1+x^2)^2$ is already decomposed).
I searched other answers and no one seems to consider this possibility of dividing by $0$ when using partial fraction decomposition, so I assume that I am missing something about partial fractions decomposition theory or somewhere else. What am I missing?
Edit. I tried, failing, to see if $F(1)$ was integrable with some tricks typical of definite integrals, like trying to use the substitution $t=\frac{1}{x}$ and using the identity $\arctan\frac{1}{t}=\pi/2-\arctan t$ valid for $t>0$, but it doesn't seem to help. So I'm still stuck.
| The short idea of the story below is to use a further deformation parameter, $b$, and avoid the denominator $(a^2-1)$ which makes problems in $a=\pm1$ by having instead $(a^2-b^2)$. The deformation used is continuous in $b$. Everything transposes on the computational side.
In order to exchange the order of differentiating w.r.t. $a$ and integrating w.r.t. $x$ we really need a "dominated convergence" argument to estimate the expression under the integral, so let us do this first. We assume $a$ lives in some bounded interval like $[-M,M]$. Then
$$
\begin{aligned}
\int_0^\infty \frac{\arctan(ax)}{x(1+x^2)}\; dx
&=
a\int_0^\infty \frac{\arctan(ax)}{ax(1+x^2)}\; dx
\\
&\le
M\int_0^\infty \frac{1}{1+x^2}\; dx =\frac 12M\pi<\infty
\ .
\end{aligned}
$$
We have used $0\le u\le \tan u$ for $u$ from $0$ to $\pi/2$.
Similar dominance estimations can be used for the integrals below.
To get rid of the problems presented in the question for $a=\pm1$, we will introduce a new parameter, $b>0$.
We may and do assume that $a\ge 0$, so that only $a=1$ remains. So let us consider:
$$
\begin{aligned}
J(a,b)
&:=
\int_0^\infty \frac{\arctan(ax)}{x(1+b^2x^2)}\; dx\ ,
\text{ so that}
\\
J(0,b) &=0\ ,\\
J'_a(a,b)
&=
\int_0^\infty \frac{1}{(1+a^2x^2)(1+b^2x^2)}\; dx\ ,
\\
&=
\int_0^\infty
\frac 1{a^2-b^2}
\left(
\frac {a^2}{1+a^2x^2}
-
\frac{b^2}{1+b^2x^2}
\right)\;
dx
\\
&=\frac 12\pi\cdot\frac {a-b}{a^2-b^2}
\\
&=\frac 12\pi\cdot\frac 1{a+b}\ .
\end{aligned}
$$
The above computations are valid for all parameters $a,b$ with $a\ne b$. We restrict to the case $0\le a<b$, integrate and obtain the equality:
$$
J(a,b) := \int_0^\infty \frac{\arctan(ax)}{x(1+b^2x^2)}\; dx
=
\frac 12\pi(\log (a+b)-\log b)\ .
$$
Now we can easily cover the case $a=1$. Just set it so, and let $b>1$ converge to $1$, and use the continuity of $J(1,b)$ as function of $b$.
| {
"language": "en",
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"answer_count": 2,
"answer_id": 0
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Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{2x}{x^2+1}\right)=2\tan^{-1}(x)$$
(a)
We wish to show that these are equal for $-1\leq x \leq 1$.
For this domain $\displaystyle -1 \leq\frac{2x}{x^2+1} \leq 1 \implies -\frac{\pi}{2} \leq \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
$$\frac{2x}{x^2+1} = \sin(2\tan^{-1}(x))$$
The task is now to show $\sin(2\tan^{-1}(x))=\frac{2x}{x^2+1}$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
For $-1 \leq x \leq 1 \implies -\frac{\pi}{4}\leq\tan^{-1}(x)\leq\frac{\pi}{4} \implies -\frac{\sqrt{2}}{2}\leq \sin(\tan^{-1}(x)) \leq \frac{\sqrt{2}}{2}$
Also, $\frac{\sqrt{2}}{2}\leq\cos(\tan^{-1}(x)) \leq 1$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x)) \iff 2(\frac{x}{\sqrt{x^2+1}})(\frac{1}{\sqrt{x^2+1}})= \frac{2x}{x^2+1}$$
Which was to be shown.
(b)
$\displaystyle x \geq 1 \implies 0<\frac{2x}{1+x^2}\leq 1 \implies 0< \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
Hence,
We must show that $$\sin^{-1}(\frac{2x}{x^2+1}) = \pi - 2\tan^{-1}(x) \iff \frac{2x}{x^2+1} = \sin(\pi - 2\tan^{-1}(x))$$ for $x\geq 1$
$$\sin(\pi - 2\tan^{-1}(x))=\sin(\pi)\cos(2\tan^{-1}(x)) - \cos(\pi)\sin(2\tan^{-1}(x))=\sin(2\tan^{-1}(x))$$
$$\sin(2\tan^{-1}(x)) = 2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
Which was to be demonstrated.
Note: This problem didn't flow like I thought it would. I had imagined that during some of the intermediate steps, I would be presented with the option of choosing an $f(x)$ or trig function value that would only be true in one of the intervals. But no such situation presented itself. Did I do something wrong? Did I overlook something?
| Show that $f'(x)=\frac{2}{1+x^2}$, hence $f(x)$ and $2\arctan(x)$ differ only by a constant. Then substitute $x=0$ to show that this constant is $0$.
| {
"language": "en",
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"answer_id": 3
} |
Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
= 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
= 1 - \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
= \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.
| $$P((1-X)/(X+1)\leq t)=\begin{cases}
P(\{X\leq -1\}\cup\{X\geq (1-t)/(1+t)\})&t>-1\\
P(\{(1-t)/(1+t)\leq X\leq -1\}& t<-1
\end{cases}$$
So for $t>-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)+\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\\frac{3}{4}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}$$
while for $t<-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)-\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\-\frac{1}{4}-\frac{1}{\pi}\arctan\frac{1-t}{1+t}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way in exercise 8(a) to chapter 1.2.1:
Prove the following theorem of Nicomachus (A.D. c. 100) by induction: $1^3=1$, $2^3=3+5$, $3^3=7+9+11$, $4^3=13+15+17+19$, etc.
In the answers to exercises, the author gives the following formula: $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)=n^3$
My question is, how do I get to that formula from the sample sums given in the problem? It looks kind of odd, especially because the last summand doesn't give me any idea on how it is connected with first ones. Usually one is able to see this clearly, but not here.
I tried to get to that formula by the following set of thoughts:
To get $n^3$ one must sum up $n$ odd numbers, starting from the $(n-1)$th central polygonal number, meaning that a cube of number $n$ is made by summing up odd numbers, starting from $(n-1)$th central polygonal number to $n$th triangular number.
Since odd numbers form arithmetic progression with $a_1=1, d=2$, it's possible to use the following formula of summing up $p$th to $q$th member of this progression:
$$S_{p,q}=\dfrac{a_p+a_q}2\cdot(q-p+1)$$
We can simplify $\dfrac{a_p+a_q}2$, putting $a_p=2p-1$ and $a_q=2q-1$:
$$\dfrac{a_p+a_q}2=\dfrac{(2p-1)+(2q-1)}2=\dfrac{2p+2q-2}2=p+q-1$$
Thus, the formula of summing up $p$th to $q$th member of this progression is:
$$(p+q-1)(q-p+1)=(q+(p-1))(q-(p-1))=q^2-(p-1)^2$$
Substituting in $p=\dfrac{n(n-1)}{2}+1=\dfrac{n^2-n+2}{2}$, and $q=\dfrac{n(n+1)}{2}$, we get the formula:
$$\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}+1-1\right)^2=\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}\right)^2=\\
\dfrac{(n(n+1))^2-(n(n-1))^2}{4}=\dfrac{(n(n+1)-n(n-1))(n(n+1)+n(n-1))}{4}=\\
\dfrac{n^2(n+1-n+1)(n+1+n-1)}{4}=\dfrac{4n^3}{4}=n^3$$
This... kind of... proves the sum formula for any $n$, really. But it doesn't give out the formula in question, i.e. $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)$.
What is the correct way to get this formula? Any hints are greatly appreciated.
| $\sum_{k=0}^{n-1} (n^2 -n + 1 + 2k)$ = $\sum_{k=0}^{n-1} n^2 + \sum_{k=0}^{n-1} (-n + 1 + 2k)$, but the second sum is 0 because of the symmetry relative to 0:
$\sum_{k=0}^{n-1} (-n + 1 + 2k) = \sum_{2k\leq n-1} \left((-n+1+2k) + (n-1-2k)\right) =\sum 0$. What remains is $\sum_{k=0}^{n-1} n^2 = n\cdot n^2 = n^3$
| {
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"answer_count": 3,
"answer_id": 1
} |
Inverse of a 2x2 matrix with an example Need to find inverse of this matrix:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$
This is how it has been solved:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$ $
\begin {bmatrix}
x_1 & x_2\\
x_3 & x_4\\
\end {bmatrix}
$ = $
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$ - step 1
$
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$ $
\begin {bmatrix}
x_1 & x_2\\
x_3 & x_4\\
\end {bmatrix}
$ = $
\begin {bmatrix}
1 & -3/5\\
0 & 1\\
\end {bmatrix}
$ - step 2
From step 1 to step 2, -3/5 added to first row, second column of $
\begin {bmatrix}
1 & -3/5\\
0 & 1\\
\end {bmatrix}
$
Also, -3/5 to first row,second column of $
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {bmatrix}
$
My query is if it is correct to add or deduct a particular number from say first column, first row on the right side and on the left side and both leave the equation intact?
| One may find the inverse matrix by row operations on the augmented matrix:
$$
\left[\begin{array}{rr|rr}
1 & 3/5 & 1 &0 \\
0 & 1 & 0 &1 \\
\end{array}\right]\overset{R_1'=-3/5R2+R1}{\longrightarrow}
\left[\begin{array}{rr|rr}
1 & 0 & 1 &-3/5 \\
0 & 1 & 0 &1 \\
\end{array}\right],
$$
so $$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}^{-1}
=
\begin {bmatrix}
1 & -3/5\\
0 & 1\\
\end {bmatrix}.
$$
It is also useful to remember the general expression for the inverse of a 2x2 matrix:
$$
\begin {bmatrix}
a & b\\
c & d\\
\end {bmatrix}^{-1}=\frac{1}{ad-bc}
\begin {bmatrix}
d & -b\\
-c & a\\
\end {bmatrix}
$$
if the determinant $ad-bc\neq 0.$
| {
"language": "en",
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"answer_count": 3,
"answer_id": 0
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Newton's Method of Fluxions pg.8 paragraph $25$ question y'all. Newton has this problem. I'm using the free copy downloaded from the Library of Congress website, the original English. In paragraph 25 line 7 he states $y \times b - x = v$ . . . by which I guess he means $y(b-x)$. Then he says to write,$ v/(b-x) = y$. Then in the next line he gets that long equation:
$$v^3 + b^2v^2 - b^3x^3 + 3b^2x^4 - 3bx^5 + x^6 = 0$$
How the heck did he get this? Could it be a mistranslation from the Latin? Thanks.
Quote:
Source:
The method of fluxions and infinite series; with its application to the geometry of curve-lines by Sir Isaac Newton, published in 1736, p.8
| No, there is no mistake. He has the equation $y^3+\frac{b^2}{b-x}\cdot y^2-x^3=0$. At the next step we substitute $y$ by $\frac{v}{b-x}$
$$\left(\frac{v}{b-x}\right)^3+\frac{b^2}{b-x}\cdot \left(\frac{v}{b-x}\right)^2-x^3=0$$
$$\frac{v^3}{(b-x)^3}+\frac{b^2v^2}{(b-x)^3}-x^3=0$$
Expanding the last summand by $(b-x)^3$.
$$\frac{v^3}{(b-x)^3}+\frac{b^2v^2}{(b-x)^3}-\frac{x^3\cdot (b-x)^3}{(b-x)^3}=0$$
We can ommit the denominator, since the numerator has to be 0 only.
$$ v^3+b^2v^2-x^3\cdot (b-x)^3=0$$
To expand $(b-x)^3$ you can use the binomial theorem: $(b-x)^3=\sum\limits_{k=0}^3 \binom{3}{k}\cdot b^k\cdot (-x)^{3-k}$
$$v^3+b^2v^2-x^3b^3+3x^4b^2-3x^5b+x^6=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A better way to find the radius of the complex roots of $(z + 1)^5 = 32z^5$? I was gnawing on this problem today:
All the complex roots of $(z + 1)^5 = 32z^5,$ when plotted in the complex plane, lie on a circle. Find the radius of this circle.
I solved this by first dividing
$$
\left(\frac{z+1}{2z}\right)^5 = 1
$$
then using the roots of unity, and solving for complex $z$. Then, I did regression to determine the solution equation to be
$$
\left(x - \frac{1}{3}\right)^2 + y^2 = \left(\frac{2}{3}\right)^2
$$
Thus, the radius is $2/3$.
Gross. That is (in my opinion) an absolutely awful way to solve this, and not the intended way. I know that there exists a better, non-numerical solution to this problem; Could you please help me find it?
| Alternative approach.
The $5$th roots of $32$ are given by $2 \times k$, where
$k$ is one of the $5$ roots of $z^5 = 1.$
These roots are $~\displaystyle e^{i(2k\pi/5)} ~: ~k \in \{0,1,2,3,4\}.$
From the original problem, you have that
$$\left[\frac{z + 1}{z}\right]^5 = 32.$$
Therefore, $(z + 1) = z \times 2 \times e^{i(2k\pi/5)}.$
Therefore $1 = z \times \left[\left(2 \times e^{i(2k\pi/5)}\right) - 1\right].$
Therefore,
$$z = \frac{1}{\left(2 \times e^{i(2k\pi/5)}\right) - 1} = \frac{1}{\left[2\cos(2k\pi/5) - 1\right] + i \left[2\cos(2k\pi/5)\right]}. \tag1$$
Multiplying the numerator and denominator in (1) above by
$\displaystyle \left[2\cos(2k\pi/5) - 1\right] - i \left[2\sin(2k\pi/5)\right]$
gives
$$z = \frac{\left[2\cos(2k\pi/5) - 1\right] - i \left[2\sin(2k\pi/5)\right]}{5 - 4\cos(2k\pi/5)}.$$
Edit
For what it's worth, the $5$ roots of $z^5 - 1 = 0$ may be routinely converted into radical form by noting that
$$z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1).\tag2 $$
The 2nd factor in (2) above may be conquered by noting that $z = 0$ will not satisfy the equation.
Therefore, you have that
$$z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} = 0.$$
Then, you set $~\displaystyle w = z + \frac{1}{z}$.
Then, you note that $~\displaystyle w^2 = z^2 + \frac{1}{z^2} + 2.$
You end up with a quadratic equation in $w$ which will have $2$ roots, $w_1$ and $w_2$.
You then have $2$ separate quadratic equations :
$$z + \frac{1}{z} = w_1 ~~~\text{and}~~~ z + \frac{1}{z} = w_2.$$
You end up with the $4$ roots of the 2nd factor in (2) above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of ways to pick $3$ balls from a box with $8$ balls if $3$ of the balls are identical and the other $5$ are all different
A box contains 8 balls, of which 3 are identical and the remaining 5 are different from each other. 3 balls are to be picked out of the box; the order in which they are picked out does not matter. Find the number of different possible selections of 3 balls.
My Solution: I grouped them into cases
Case 1: All three balls are identical in which number of ways $= 1$
Case 2: Two of the balls are identical in which number of ways $= 5$
Case 3: One of the ball is from the $3$ identical ones in which number of ways $= 5 \cdot 4$ since there are $5$ ways to choose $1$ ball from the remaining $5$ balls and another $4$ ways to choose again from the remaining $4$ balls.
Case 4: I do not take from any of the $3$ identical balls so that leaves me with $5$ balls to choose $3$ from in which number of ways $= 5 \cdot 4 \cdot 3 = 60$.
Total number of ways $= 1+5+20+60 = 86$.
However, the answer given is $26$.
| $\mathbf{\text{Generating Function Approach:}}$
The generating function of identical objects : $(1+x+x^2+x^3)$
The generating function of distinct objects : $(\binom{5}{0}x^0+ \binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4 +\binom{5}{5}x^5)$
Then , find the coefficent of $x^3$ in the expansion of $$(1+x+x^2+x^3)\bigg(\binom{5}{0}x^0+ \binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4 +\binom{5}{5}x^5\bigg)$$
So , the answer is $26$
$\mathbf{\text{NOTE:}}$ Realize that the coefficents of generating functions determine th number of ways to select desired objects , for example , the coefficients of identical objects is always $1$ , because there is only one way to select identical objects , but when they are distinct , the coeffients are determined by combination.For example $\binom{5}{3}x^3$ means the number of selection of three distinct objects among $5$ distinct objects
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = \int \frac{2(u^{2}+1)}{(u+1)^{4}} du$$
$$= 2 \int \left(\frac{1}{(u+1)^{2}} - \frac{2}{(u+1)^{3}} + \frac{2}{(u+1)^{4}} \right)$$
$$= - \frac{2}{u+1} + \frac{2}{(u+1)^{2}} - \frac{4}{3(u+1)^{3}} + const.$$
$$= \frac{-2(3u^{2}+3u+2)}{3(u+1)^{3}} + const.$$
where
$$A = \sqrt{2x-x^{2}},~~ u = \sqrt{\frac{1-A}{1+A}}$$
However, Wolfram gives the following answer :
$$ - \frac{\sqrt{2x-x^{2}}~(x+1)}{3x^{2}} + const. $$
which is way more simple than mine. Is there any other way to integrate this? The $u$ in my answer is very complicated, I just couldn't change my answer into Wolfy's.
| If $x<0$ then $\sqrt{2x-x^{2}}$ does not make sense on $\mathbb{R}$ so I suppose that $x>0$, we have
\begin{align*}
\int\frac{1}{x^{2}\sqrt{2x-x^{2}}}{\rm d}x&=\int \frac{1}{x^{2}\sqrt{1-(x-1)^{2}}}{\rm d}x\\&\overset{x\mapsto x-1}{=}\int \frac{1}{(x+1)^{2}\sqrt{1-x^{2}}}{\rm d}x\\&\overset{x\mapsto \sin x}{=}\int \frac{1}{(1+\sin x)^{2}}{\rm d}x\\&\overset{x\mapsto \tan \frac{x}{2}}{=}\int \frac{2}{(x^{2}+1)\left(1+\frac{2x}{x^{2}+1}\right)^{2}}{\rm d}x\\&=2\int \frac{1+x^{2}}{(1+x)^{4}}{\rm d}x\\&=2\int \left(\frac{1}{(x+1)^{2}}-\frac{2}{(x+1)^{3}}+\frac{2}{(x+1)^{4}} \right){\rm d}x\\&=-\frac{2}{3}\left( \frac{3x^{2}+3x+2}{(x+1)^{3}}\right)+C\\&\overset{\text{returning}}{=}-\frac{\sqrt{2x-x^{2}}(x+1)}{3x^{2}}+C
\end{align*}
Details of "returning":
\begin{align*}
-\frac{2}{3}\left( \frac{3x^{2}+3x+2}{(x+1)^{3}}\right)+C&=\frac{\sec^{2}\frac{x}{2}(-3\sin x +\cos x -5)}{3\left(\tan \frac{x}{2}+1\right)^{3}}+C\\
&=\frac{(-3\sin \sin^{-1} x +\cos\sin^{-1}x -5)\sec^{2}\frac{1}{2}\sin^{-1}x)}{3(\tan \frac{1}{2}\sin^{-1}x +1)^{3}}+C\\
&=\frac{2(\sqrt{1-x^{2}}+1)^{2}(\sqrt{1-x^{2}}-3x-5)}{3(\sqrt{1-x^{2}}+x+1)^{3}}+C\\&=\frac{2(\sqrt{-(x-2)x}+1)^{2}(-3x+\sqrt{-(x-2)x}-2)}{3(x+\sqrt{-(x-2)x})^{3}}+C\\&=-\frac{\sqrt{2x-x^{2}}(x+1)}{3x^{2}}+C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.