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Find the least positive value of alpha for the following trignometric equation Given question is : $$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)$$
We are to find the least positive value of $\alpha$ for which above equation holds.
The way I did is as, $$\sin5\alpha \cos\theta + \cos5\alpha \sin\theta = \cos\theta \cos3\alpha + \sin\theta \sin3\alpha$$
Now for this to be true $$\sin5\alpha = \cos3\alpha$$ and $$\cos5\alpha = \sin3\alpha$$
How do I find the value of $\alpha$ that satisfies the above criteria?
|
My suggestion would be:
$$\sin(5\alpha + \theta) = \cos(\theta - 3\alpha)=\sin(\pi/2-\theta+3\alpha)$$
so,
$$\sin(5\alpha + \theta) -\sin(\pi/2-\theta+3\alpha)=0$$
$$2\sin\left(\frac{(5\alpha + \theta)-(\pi/2-\theta+3\alpha)}{2}\right)\cdot \cos\left(\frac{(5\alpha + \theta)+(\pi/2-\theta+3\alpha)}{2}\right)=0$$
$$2\sin\left(\alpha + \theta-\frac{\pi}{4}\right)\cdot \cos\left(4\alpha+\frac{\pi}{4}\right)=0$$
So your full solution is:
$$\cos\left(4\alpha+\frac{\pi}{4}\right)=0\to 4\alpha+\frac{\pi}{4}=\frac{\pi}{2}+k\pi\to \alpha=\frac{\pi}{16}+\frac{k\pi}{4}$$
or
$$\sin\left(\alpha + \theta-\frac{\pi}{4}\right)=0\to \alpha + \theta-\frac{\pi}{4}=k\pi\to \alpha=\frac{\pi}{4}-\theta+k\pi$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2357466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculating permutations for a specific password policy The security researcher Troy Hunt posted an example of an obscure password policy and I've been trying to figure out how to calculate the possible permutations
(Source: https://twitter.com/troyhunt/status/885243624933400577)
The rules are:
*
*The password must contain $9$ numbers (and only $9$ numbers)
*It must include at least $4$ different numbers
*It cannot include the same number more than three times
I understand the basic permutations will be $10^9$ ($0-9$ nine times) $= 1,000,000,000$
What I don't understand is how you factor in the reduction in permutations by enforcing $4$ different numbers and limiting repeats to $3$.
|
A brute force solution:
Nine numbers: Each number appears exactly once. We choose nine of the ten digits, then arrange the selected digits.
$$\binom{10}{9}9!$$
Eight numbers: We must use of the ten digits twice. We choose that digit, choose two of the nine locations for that number, choose seven of the other nine numbers, then arrange them in the seven open positions.
$$10\binom{9}{2}\binom{9}{7}7!$$
Seven numbers: There are two ways of partitioning $9$ into seven positive integers.
\begin{align*}
9 & = 3 + 1 + 1 + 1 + 1 + 1 + 1\\
& = 2 + 2 + 1 + 1 + 1 + 1 + 1
\end{align*}
One number appears three times, and six appear once each: We choose which of the ten digits appears three times, choose three of the nine positions for the numbers, select six of the other nine numbers, and arrange them in the six open positions.
$$10\binom{9}{3}\binom{9}{6}6!$$
Two digits appear twice, five numbers appear once each: We choose which two of the ten digits will appear twice, choose two of the nine positions for the smaller of those numbers, choose two of the seven remaining open positions for the larger of those numbers, choose five of the eight remaining numbers, and arrange them in the five open positions.
$$\binom{10}{2}\binom{9}{2}\binom{7}{2}\binom{8}{5}5!$$
Six numbers: We partition $9$ into six positive integers.
\begin{align*}
9 & = \color{red}{4 + 1 + 1 + 1 + 1 + 1}\\
& = 3 + 2 + 1 + 1 + 1 + 1\\
& = 2 + 2 + 2 + 1 + 1 + 1
\end{align*}
The partition shown in red is prohibited since no number may appear more than three times.
One number appears three times, another number appears twice, and four numbers appear once each: We choose the number that appears three times, choose three of the nine positions for that number, choose which of the nine remaining numbers appears twice, choose two of the six remaining positions for that numbers, choose four of the eight remaining numbers, and arrange them in the four remaining positions.
$$10\binom{9}{3} \cdot 9\binom{6}{2}\binom{8}{4}4!$$
Three numbers appear twice each, and three numbers appear once each: We choose three numbers to appear twice each, choose two of the nine positions for the smallest of those numbers, choose two of the seven remaining positions for the next smallest of those numbers, choose two of the five remaining positions for the largest of those numbers, choose three of the seven remaining numbers to appear once each, and arrange them in the the three remaining positions.
$$\binom{10}{3}\binom{9}{2}\binom{7}{2}\binom{5}{2}\binom{7}{3}3!$$
We can partition $9$ into exactly five numbers in the following ways:
\begin{align*}
9 & = \color{red}{5 + 1 + 1 + 1 + 1}\\
& = \color{red}{4 + 2 + 1 + 1 + 1}\\
& = 3 + 3 + 1 + 1 + 1\\
& = 3 + 2 + 2 + 1 + 1\\
& = 2 + 2 + 2 + 2 + 1
\end{align*}
We can partition $9$ into exactly four numbers in the following ways:
\begin{align*}
9 & = \color{red}{6 + 1 + 1 + 1}\\
& = \color{red}{5 + 2 + 1 + 1}\\
& = \color{red}{4 + 3 + 1 + 1}\\
& = \color{red}{4 + 2 + 2 + 1}\\
& = 3 + 3 + 2 + 1\\
& = 3 + 2 + 2 + 2
\end{align*}
The partitions shown in red are prohibited by the constraint that no digit can appear more than thrice.
Can you continue?
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many $6$ digit numbers can you make with the numbers $\{1, 2, 3, 4, 5\}$ so that the digit $2$ appears at least 3 times? I can't seem to understand.
I thought about it this way. We take a first example:
$222aaa$ for each blank space we have $5$ positions, so in this case the answer would be $125$ ($5 \times 5 \times 5$) numbers.
Now the position of the twos can change so we calculate the numbers for that taking this example:
$222aaa$ and the possibilities for this are the number of ways you can arrange $6$ digits ($6!$) and divide by repetitions so divided by $2 \cdot (3!)$ so answer $= 20$
So final answer should be $125 \cdot 20 = 2500$. But this answer is wrong and I don't understand why.
|
How many six-digit numbers can be formed using numbers from the set $\{1, 2, 3, 4, 5\}$ with replacement if the digit $2$ must appear at least three times?
There are $\binom{6}{k}$ ways of choosing exactly $k$ positions for the $2$'s and $4^{6 - k}$ ways to fill the remaining $6 - k$ positions with a number different from $4$. The number of six-digit numbers that can be formed using numbers from the set $\{1, 2, 3, 4, 5\}$ in which the digit $2$ appears at least three times can be found by adding the number of outcomes in which the digit $2$ appears exactly three times, exactly four times, exactly five times, and exactly six times
$$\sum_{k = 3}^{6} \binom{6}{k}4^{6 - k} = \binom{6}{3}4^3 + \binom{6}{4}4^2 + \binom{6}{5}4^1 + \binom{6}{6}4^0 = 1545$$
(as Dionel Jaime found) or by subtracting the number of outcomes in which the digit $2$ appears fewer than three times from the total number of words that can be formed from the five numbers in the set when those numbers are used with replacement
$$5^6 - \sum_{k = 0}^{2} \binom{6}{k}4^{6 - k} = 5^6 - \left[\binom{6}{0}4^6 + \binom{6}{1}4^5 + \binom{6}{2}4^4\right] = 1545$$
Where did you make your mistake?
By designating three positions for the $2$'s and then filling the remaining three positions with any of the five numbers in the set, you counted cases in which $2$ appears more than three times multiple times.
You counted cases in which the digit $2$ appears four times four times, once for each of the $\binom{4}{3}$ ways you could designate three of the four $2$'s as your three $2$'s. To see this, observe that you count the number $232422$ four times:
$$\color{blue}{2}3\color{blue}{2}4\color{blue}{2}2$$
$$\color{blue}{2}3\color{blue}{2}42\color{blue}{2}$$
$$\color{blue}{2}324\color{blue}{22}$$
$$23\color{blue}{2}4\color{blue}{22}$$
You counted cases in which the digit $2$ appears five times ten times, once for each of the $\binom{5}{3}$ ways you could designate three of your five $2$'s as your three $2$'s. You counted cases in which the digit $2$ appears six times twenty times, once for each of the $\binom{6}{3}$ ways you could designate three of your six $2$'s as your three $2$'s. Notice that
$$\binom{6}{3}4^3 + \binom{4}{3}\binom{6}{4}4^2 + \binom{5}{3}\binom{6}{5}4^1 + \binom{6}{3}\binom{6}{6}4^0 = 2500$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).
There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!
|
You may use Lagrange multipliers method to show that we need $a=b=c$. The optimisation problem would be to minimise $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, having the constraint $a+b+c=t , t\leq \frac{3}{2}$.
$$F=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\lambda(a+b+c-t)$$
Then, we need to find the right $t$, for which the minimum is achieved. Knowing that $a=b=c$, the problem is simplified to minimizing
$$3(a+\frac{1}{a})$$
As mentioned in comments, it is a convex problem and it achieves its minimum at $a=1$. However, the constraint $a+b+c\leq\frac{3}{2}$ does not allow $a=1$. So, because of convexity, you should decrease $a$ until you satisfy the constraint.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2360813",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove that a length is equal to the inradius of a triangle $D$ is the midpoint of the side $BC$ of the triangle $ABC$. The line joining $D$ and the incentre $I$ of the triangle intersects altitude $AA'$ at the point $P$.
Prove that the length of $AP$ is equal to the radius of the incircle of the triangle..
|
Let $E$ be a tangency point with side $BC$ and let $c>b$.
Hence,
$$DE=BE-BD=\frac{a+c-b}{2}-\frac{a}{2}=\frac{c-b}{2},$$
$$BA_1=c\cos\beta=c\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2a}.$$
Thus, $$DA_1=BA_1-BD=\frac{a^2+c^2-b^2}{2a}-\frac{a}{2}=\frac{c^2-b^2}{2a}.$$
Now, since $\Delta DIE\sim\Delta DPA_1,$ we obtain:
$$\frac{PA_1}{IE}=\frac{DA_1}{AE}$$ or
$$\frac{PA_1}{r}=\frac{\frac{c^2-b^2}{2a}}{\frac{c-b}{2}}$$ or
$$PA_1=\frac{r(b+c)}{a}.$$
In another hand, $$S_{\Delta ABC}=pr=\frac{(a+b+c)r}{2}=\frac{1}{2}ah_a.$$
Thus, $$AA_1=h_a=\frac{(a+b+c)r}{a},$$
which gives $$AP=AA_1-PA_1=\frac{(a+b+c)r}{a}-\frac{(b+c)r}{a}=r$$
and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2361398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$x^7+y^7+z^7$ is divisible by $7^3$, then $x+y+z$ is divisible by $7^2$
Let $x, y, z$ be positive integers, and $7 \nmid xyz$. If $7^3|x^7+y^7+z^7$, show that $7^2|x+y+z$.
by Fermat's little theorem, $x^7 \equiv x \pmod7$, then $x^7+y^7+z^7\equiv x+y+z \equiv 0 $ (mod 7)
so we have $7 | (x+y+z)$.
what should I do next?
|
HINT:
If $x+y\equiv0\pmod7,$ we are done
Otherwise
We can write $z=7a-x-y$
$$x^7+y^7+z^7=x^7+y^7+(7a-x-y)^7$$
$$(7a-x-y)^7\equiv-(x+y)^7+7(7a)(x+y)^6\pmod{7^3}$$
$$\implies-(x+y)^7+7(7a)(x+y)^6\equiv0\pmod{p^3}$$
As $7\nmid(x+y),$ $$x+y\equiv-49a\pmod{7^3}$$
which is impossible as $7\nmid(x+y)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2361499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Express $355$ as a sum of three squares. PROBLEM
Find all the ways to express $355$ as a sum of three squares.
MY ATTEMPT
By Legendre's three-square theorem, since $355$ is not of the form $n = {4^a}(8b+7)$ (for $a, b \in \mathbb{Z}$), then $355$ can be expressed as a sum of three squares.
WLOG, we can restrict to positive integral solutions in
$$x^2 + y^2 + z^2 = 355,$$
where $x \leq y \leq z$.
WolframAlpha gives
$$x = 7, y = 9, z = 15,$$
and
$$x = 3, y = 11, z = 15.$$
QUESTION
Are these all of the solutions?
|
I do not consider any order between $x,y,z$ here.
$x^2+y^2+z^2\equiv 1\pmod 3$ and since $(0,1,2)\xrightarrow{x^2}(0,1,1)$ then we can only be in the $0+0+1$ configuration, so two numbers are divisible by $3$.
$x=3a,\ y=3b$ gives $9(a^2+b^2)+z^2=355\iff z^2\equiv 4\pmod 9\iff z\equiv 2,7\pmod 9$
Since $19^2=361>355$ then only $z=2,7,11,16$ are possible.
For $a^2,b^2$ we are limited to values $0,1,4,9,16,25,36$ in the table below and it is easy to find them manually
$\begin{array}{|c|c|c|c|}\hline
z & \frac{355-z^2}9 & a^2+b^2 & x,y\\\hline
2 & 39 & \varnothing \\
7 & 34 & 9+25 & 9,15\\
11 & 26 & 1+25 & 3,15\\
16 & 11 & \varnothing \\\hline
\end{array}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2361586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\int (x^2+1)/(x^4+1)\ dx$ I have Divided the numerator and Denominator by $x^2$
to get $\dfrac{1+x^{-2}}{x^2+x^{-2}}$ then changed it into $(1+(x^{-2}))/[(x-x^{-1})^2 +2]$ then took $x-(1/x)$ as $u$ and Differentiated it with respect to $x$ to get $dx=du/(1+x^{-2})$ Finally I got this expression:
$$
\int\frac{x^2+1}{x^4+1} \, dx = \int (u^2+2)^{-1} \, du
$$
After this I need help!
|
Since $x^4+1=x^4+2x^2+1-2x^2=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$, we obtain
$$\frac{x^2+1}{x^4+1}=\frac{1}{2}\left(\frac{1}{x^2+\sqrt2x+1}+\frac{1}{x^2-\sqrt2x+1}\right).$$
Now, use $\int\frac{1}{1+x^2}dx=\arctan{x}+C$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2361853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$
Prove that \begin{align}\begin{vmatrix}
1 & a &bc \\
1& b & ac\\
1&c & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\
1& b&b^2 \\
1& b & c^2
\end{vmatrix}\\&=(c-a)(b-a)(c-b)\begin{vmatrix}
1 & a & a^2\\
0& 1 &b+a \\
0& 0 & 1
\end{vmatrix}\\\\
&=(c-a)(b-a)(c-b)\end{align}
I got the last two equalities but I did't get first two. I need help with first two
|
From the last column, we remove the common factor $abc$;
Then we multiply the first line by $a$, the second line by $b$ and the last line by $c$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2362387",
"timestamp": "2023-03-29T00:00:00",
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|
Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere?
|
Rotating the parabola $y=x^2$ by $\theta$ clockwise gives $v=u^2$, where
$$\left(u\atop v\right)=\left(\cos\theta\quad-\sin\theta\atop\sin\theta\quad\;\;\;\cos\theta\right)\left(x\atop y\right)$$
i.e.
$$x\sin \theta+y\cos\theta=(x\cos\theta-y\sin\theta)^2$$
Putting $\theta=\frac\pi 4$ gives
$$\frac 1{\sqrt2}(x+y)=\left(\frac 1{\sqrt2}(x-y)\right)^2\\
\sqrt2(x+y)=(x-y)^2$$
which when expanded is
$$x^2+y^2-2xy-\sqrt2 x-\sqrt2 y=0$$
|
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"url": "https://math.stackexchange.com/questions/2363075",
"timestamp": "2023-03-29T00:00:00",
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|
$33=\left\{a^3+b^3+c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$ I was just wondering, if I had an equation like:
$$33 = \left\{a^3 + b^3 + c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$$
What are the values of $(a, b \land c)$. Is there a way of proving or disproving that such integers of $a$, $b$, and $c$ exist to satisfy this equation? Recently the equation:
$$74 = a^3 + b^3 + c^3$$
Was solved, and there proved to be integers $a$, $b$, and $c$ that satisfied this equation. Could you please help for $33$? Thanks :)
UPDATE: I did not realise this was an open question and that we need hefty computers to solve an equation like this, but thank you people for helping me out. I guess if $(a, b \lor c) \to \pm \infty$ then it is only a matter of trial and error before we find a solution. After looking at Numberphile, I realised how this question was not just any "ordinary problem".
|
Note in your case, the max value of $a,b,c$ is achieved when both others are zero, and hence is $33^{1/3} \approx 3.2$, so since they must be integers, you have $a,b,c \in \{0,1,2,3\}$.
Now it is easy to find all combinations of these in 4 variables. Without loss of generality, assume $a \ge b \ge c$, and you get the following list of combinations (I used Excel but you can enumerate any way you like):
$$
\begin{array}{ccc|c}
a & b & c & a^3+b^3+c^3 \\
\hline
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 \\
1 & 1 & 0 & 2 \\
1 & 1 & 1 & 3 \\
2 & 0 & 0 & 8 \\
2 & 1 & 0 & 9 \\
2 & 1 & 1 & 10 \\
2 & 2 & 0 & 16 \\
2 & 2 & 1 & 17 \\
2 & 2 & 2 & 24 \\
3 & 0 & 0 & 27 \\
3 & 1 & 0 & 28 \\
3 & 1 & 1 & 29 \\
3 & 2 & 0 & 35 \\
3 & 2 & 1 & 36 \\
3 & 2 & 2 & 43 \\
3 & 3 & 0 & 54 \\
3 & 3 & 1 & 55 \\
3 & 3 & 2 & 62 \\
3 & 3 & 3 & 81 \\
\end{array}
$$
It is clear that no answer results in 33...
UPDATE
As @Ture points out in the comments below, doing this for $74$ shows that there are no solutions to $a^3+b^3+c^3=74$ for non-negative $a,b,c$ either. This shows conclusively that you cannot restrict $a,b,c$ thus, and the result OP quotes likely features at least one of $a,b,c$ less than zero.
|
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|
Improper Integral Involving A Probable Partial Fraction Conversion I was having trouble evaluating the following integral:$$\int^{+\infty}_{-\infty} \frac{dx}{(x^2\pm ax+a^2)(x^2\pm bx+b^2)}$$
I have tried approaches that involve converting into partial fractions using thumb rule and without using it.I also tried the online integral calculator. No luck whatsoever. The answer according to the book is:
$$\frac{2\pi}{\sqrt{3}}.\frac{a+b}{ab(a^2+ab+b^2)}$$
|
We have
$$\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} = \frac{cx+d}{x^2+ax+a^2} + \frac{ex+f}{x^2+bx+b^2}.$$
Solving for $c,d,e,f$ one should find
\begin{align*}
\frac{1}{(x^2+ax+a^2)(x^2+bx+b^2)} &= \frac{1}{(a-b)(a^2+ab+b^2)}\left(
\frac{x-b}{x^2+ax+a^2} + \textrm{something similar}\right).
\end{align*}
Now use the hint in the answer by @Dr. Sonnhard Graubner.
Addendum:
Letting $x=t-a/2$ we find
$$\frac{x-b}{x^2+ax+a^2} = \frac{t-(b+a/2)}{t^2+3a^2/4}.$$
The integral
\begin{align*}
\int_{-\infty}^\infty \frac{dt}{t^2+3a^2/4} \tag{1}
\end{align*}
is a standard integral.
The integral
\begin{align*}
\int_{-\infty}^\infty \frac{t dt}{t^2+3a^2/4} \tag{2}
\end{align*}
is divergent.
Letting $x=t-b/2$ in the "something similar" above we find terms like (1) and (2) above.
Call them (3) and (4), respectively.
Combining (2) and (4) we find we must integrate
$$\frac{1}{t^2+3a^2/4}-\frac{1}{t^2+3b^2/4}
= (\mathrm{const})\frac{t}{(t^2+3a^2/4)(t^2+3b^2/4)}.$$
This function is integrable on $(-\infty,\infty)$ and odd and therefore integrates to zero.
|
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|
Committee selection:Where is my mistake?
There are $5$ boys and $6$ girls. A committee of $4$ is to be selected so that it must consist at least one boy and at least one girl?
I consider $1$ boy can be chosen in $5$ ways.
Also, consider $1$ girl can be chosen in $6$ ways.
So, others $2$ can be chosen in $9C2$ ways.
So, required answer is $5 * 6 * 9C2$
But this is the wrong answer. Where is my mistake?
|
Let's denote the girls $\{g_1,g_2,\ldots,g_6\}$ and the boys $\{b_1,b_2,\ldots,b_5\}$.
You count some configurations more than once. The selection $g_1,g_2,b_1,b_2$ is counted four times namely as
\begin{align*}
(g_1,b_1),(g_2,b_2)\\
(g_1,b_2),(g_2,b_1)\\
(g_2,b_1),(g_1,b_2)\\
(g_2,b_2),(g_1,b_1)
\end{align*}
One way to count the number of admissible selections is: We select either one girl and three boys, or two girls and two boys or three girls and one boy giving
\begin{align*}
\binom{5}{1}\binom{6}{3}+\binom{5}{2}\binom{6}{2}+\binom{5}{3}\binom{6}{1}&=20\cdot 5+ 10\cdot 15+ 10\cdot 6\\
&=\color{blue}{310}
\end{align*}
Another way:
We count all selections of $4$ people and subtract the selections containing only girls and only boys giving
\begin{align*}
\binom{11}{4}-\binom{5}{4}-\binom{6}{4}&=330-5-15\\
&=\color{blue}{310}
\end{align*}
|
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|
GRE Practice Question I need a calculus refresher! This question was one of the lowest percent correct on the practice (27%).
A curve in the xy-plane is given by
\begin{align*}
x &= t^2+2t \\
y &= 3t^4+4t^3
\end{align*}
for all $t > 0.$ The value of $\frac{d^2y}{dx^2}$ at the point $(8,80)$ is
(a) 4, (b) 24, (c) 32, (d) 96, (e) 192
The first thing we can do is figure out what the $t-$value is at the point $(8,80)$. Using the first equation we have $8 = t^2 + 2t.$ By solving the quadratic we get $t = 2$ and $t = -4$, but only the first solution is in the domain. We can check to see that $t = 2$ satisfies the second equation.
Then comes the part I'm not sure about. Can someone explain how you determine the derivative (first and second) of $x$ with respect to $y$? I think I remember $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ because it sort of cancels like a fraction.
|
Solve the quadratic equation $t^2+2t-x=0$
to find $t=-1+\sqrt{1+x}$ and substitute into $y$ to get:
$$y=3(-1+\sqrt{1+x})^4+4(-1+\sqrt{1+x})^3.$$
Thus:
$$y'=12(-1+\sqrt{1+x})^3\cdot \frac{1}{2\sqrt{1+x}}+12(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}.$$
$$y''=-\frac{6}{2(1+x)^{3/2}}\cdot((-1+\sqrt{1+x})^3+(-1+\sqrt{1+x})^2)+\frac{6}{\sqrt{1+x}}\cdot (3(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}+2(-1+\sqrt{1+x})\cdot \frac{1}{2\sqrt{1+x}}).$$
$$y''(8)=-\frac{6}{2\cdot 27}(8+4)+2(2+\frac{2}{3})=4.$$
|
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|
$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$
Find the value of $$\cos^2 76^{\circ} + \cos^2 16^{\circ} -\cos 76^{\circ} \cos 16^{\circ}$$
I did it like this
$$\cos^2 76^{\circ}+\cos^2 16^{\circ} = \cos(76^{\circ}+16^{\circ}) \, \cos(76^{\circ}-16^{\circ}).$$
So the expression is $$\cos 92^{\circ} \cos 60^{\circ}-\cos 76^{\circ} \, \cos16^{\circ}.$$
I couldn't simplify after that.
|
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2(A+60^\circ)+\cos^2A=1+\cos(A+60^\circ+A)\cos(A+60^\circ-A)=1+\dfrac{\cos(2A+60^\circ)}2$$
Using Werner's formula,
$$\cos(A+60^\circ)\cos A=\dfrac{\cos(A+60^\circ+A)+\cos(A+60^\circ-A)}2=?$$
Can you recognize $A$ here?
|
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|
Quartic with 3 distinct roots Consider a quartic equation
$$x^4 – kx^3 + 11x^2 – kx + 1 = 0$$
The value of $k$ so that given equation has three real and distinct solutions can be equal to - i think that for any four degree to have 3 roots, its $f'(x)$ should have two zero and by the same way its $f''(x)$ should have one zero which means that the determinant of $f''(x)$ should be zero but this is wrong since the correct answer to this question is $\frac{13}{2}$.
|
Hint
Observe that $0$ is not a root of this equation and since three real roots exist, therefore fourth must be real as well and there must be a repeated root.
Start with:
\begin{align*}
x^4 – kx^3 + 11x^2 – kx + 1 & = 0\\
x^2 – kx + 11 – \frac{k}{x} + \frac{1}{x^2} & = 0\\
\left(x+\frac{1}{x}\right)^2-k\left(x+\frac{1}{x}\right)+9 & =0\\
t^2-kt+9 & =0 && \left(\text{where } t=x+\frac{1}{x}\right)
\end{align*}
Now can you proceed from here to impose conditions for real roots?
|
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|
Understanding a Symmetry Argument
A rectangle has dimensions $a$ units by $b$ units with $a > b$. A diagonal divides the rectangle into two triangles. A square, with sides parallel to those of the rectangle, is inscribed in each triangle. Find the distance between the vertices (of the squares) that lie in the interior of the rectangle.
In the solution, we find that one square has an 'interior' vertex $(\frac{ab}{a+b},\frac{ab}{a+b})$, because the vertex must lie on both $y=x$ and $y = -\frac{b}{a}x + b$. Then the author asserts that symmetry permits us to conclude that the coordinates of the other vertex of the other inscribed square are $(\frac{a^2}{a+b}.\frac{b^2}{a+b})$. I have drawn several pictures trying to see the symmetry, but I can't seem to discern it, so I left no other option but a (relatively) brute-force method. Perhaps someone with a more geometric eye could kindly point out the symmetry being appealed to.
|
We can draw a picture of this situation as follows:
We need to find $c,d,e,f$ to know the interior vertices of the squares.
Can you see that this is rotationally symmetric, if we spin it around by $180^\circ$, it will be exactly the same problem?
This means that the distance between $(0,b)$ and $(c,f)$ is equal to the distance between $(e,d)$ and $(a,0)$. Equally, the distance from $(c,f)$ to $(a,0)$ is equal to the distance from $(e,d)$ to $(0,b)$.
We can see that $$(c,f)=\left(\frac{ab}{a+b}, \frac{ab}{a+b}\right)$$ by the definition of a square, and then we can find the distance from here to $(a,0)$:
\begin{align}D_1&=\sqrt{\left(a-\frac{ab}{a+b}\right)^2+\left(0-\frac{ab}{a+b}\right)^2}\\
&=\sqrt{\left(\frac{a^2}{a+b}\right)^2+\left(-\frac{ab}{a+b}\right)^2}\\
&=\sqrt{\frac{a^4}{(a+b)^2}+\frac{a^2b^2}{(a+b)^2}}\\
&=\frac{a\sqrt{a^2+b^2}}{a+b}\end{align}
Now we know that the distance from $(e,d)$ to $(0,b)$ is equal to this:
\begin{align}\sqrt{e^2+(d-b)^2}&=\frac{a\sqrt{a^2+b^2}}{a+b}\end{align}
We can repeat this with the other distances to get \begin{align}D_2&=\sqrt{\left(\frac {ab}{a+b}-0\right)^2+\left(\frac{ab}{a+b}-b\right)^2}\\
&=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}
and therefore \begin{align}\sqrt{(e-a)^2+d^2}&=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}
Now we have two simultaneous equations in $e,d$ to solve:
\begin{align}\sqrt{e^2+(d-b)^2}&=\frac{a\sqrt{a^2+b^2}}{a+b}\\
\sqrt{(e-a)^2+d^2}&=\frac{b\sqrt{a^2+b^2}}{a+b}\end{align}
If we solve these while remembering that $0<e<a$ and $0<d<b$ then we find that \begin{align}e&=\frac{a^2}{a+b}\\
d&=\frac{b^2}{a+b}\end{align}
|
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|
Limit $\lim_{x \to 2^{-}} \left ( \frac{1}{\sqrt[3]{x^{2} -3x+2}} + \frac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$ $$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )$$
I've tried using the $A^3-B^3$ identity, but that doesn't help. Also, I tried multiplying every fraction with $\sqrt[3]{A^2}$ to get rid of the roots in the denominator, but that doesn't help either. Can someone suggest a solution? Thanks.
|
Let $x-2=t$.
Hence,
$$\lim_{x \to 2^{-}} \left ( \dfrac{1}{\sqrt[3]{x^{2} -3x+2}} + \dfrac{1}{\sqrt[3]{x^{2} -5x+6}} \right )=$$
$$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t}}\left(\frac{1}{\sqrt[3]{t+1}}+\frac{1}{\sqrt[3]{t-1}}\right)=$$
$$=\lim_{t\rightarrow0^-}\frac{1}{\sqrt[3]{t+1}\sqrt[3]{t-1}}\lim_{t\rightarrow0^-}\frac{\sqrt[3]{t+1}+\sqrt[3]{t-1}}{\sqrt[3]t}=$$
$$=-\lim_{t\rightarrow0^-}\frac{\sqrt[3]{t+1}+\sqrt[3]{t-1}}{\sqrt[3]t}=$$
$$=-\lim_{t\rightarrow0^-}\frac{2t}{\sqrt[3]t\left(\sqrt[3]{(t+1)^2}-\sqrt[3]{t^2-1}+\sqrt[3]{(t-1)^2}\right)}=$$
$$=-\lim_{t\rightarrow0^-}\frac{2\sqrt[3]{t^2}}{3}=0.$$
I used the following formula $a^3+b^3=(a+b)(a^2-ab+b^2)$,
where $a=\sqrt[3]{t+1}$ and $b=\sqrt[3]{t-1}$.
By this formula we obtain: $$\sqrt[3]{t+1}+\sqrt[3]{t-1}=\frac{t+1+t-1}{\sqrt[3]{t+1)^2}-\sqrt[3]{(t+1)(t-1)}+\sqrt[3]{(t+1)^2}}$$
|
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|
Find derivative of $u = \sin(y^2 + u)$ using implicit differentiation Find $\frac{du}{dy}$, when $u = \sin(y^2 + u)$, using chain rule and expanding I get:
$1 = (2y\frac{du}{dy} + 1) \cos(y^2 + u)$
$2y \cos(y^2 + u)\frac{du}{dy} + \cos(y^2 + u) = 1$
$2y \cos(y^2 + u)\frac{du}{dy} = 1- \cos(y^2 + u)$
$\frac{du}{dy} = \frac{1 - \cos(y^2 + u)}{2y \cos(y^2 + u)}$
Since this was wrong, I massaged it a bit more to:
$\frac{du}{dy} = \frac{\sec(y^2 + u) - 1}{2y}$
Still wrong... But what am I doing wrong, wolfram alpha seems to agree with me...
|
Taking derivative towards $u$ w.r.t.$y$ is not $1$, but should be $\frac{d u}{d y}$
Thus, it instead should be:
$$\frac{du}{dy}=\cos(y^2 + u)(2y + \frac{du}{dy})$$
$$\frac{du}{dy}=\frac{2y\cos(y^2+u)}{1-\cos(y^2 + u)}$$
|
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A game concerning Sudoku We (me and you) start with a blank $9\times 9$ blank square (as an empty Sudoku) and I fill the first three rows legally according to Sudoku rules. Is it always possible that you complete the Sudoku with this given three rows?
|
Suppose you filled the first three rows according to Sudoku rules, let a1, a2, a3 be the 3x1 vectors in the top-left block; b1, b2, b3 be the next three 3x1 vectors in the top-middle block; and c1, c2, c3 be the last three 3x1 vectors in the top-right block.
Then consider:
\begin{array}{|ccc|ccc|ccc|}
\hline | & | & | & | & | & | & | & | & |\\
{\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}}\\
| & | & | & | & | & | & | & | & |\\
\hline & & & & & & & & \\
& ? & & & ? & & & ? & \\
& & & & & & & & \\
\hline & & & & & & & & \\
& ? & & & ? & & & ? & \\
& & & & & & & & \\
\hline \end{array} and fill it with\begin{array}{|ccc|ccc|ccc|}
\hline | & | & | & | & | & | & | & | & |\\
{\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}}\\
| & | & | & | & | & | & | & | & |\\
\hline | & | & | & | & | & | & | & | & |\\
{\vec{a}_{2}} & {\vec{a}_{3}} & {\vec{a}_{1}} & {\vec{b}_{2}} & {\vec{b}_{3}} & {\vec{b}_{1}} & {\vec{c}_{2}} & {\vec{c}_{3}} & {\vec{c}_{1}}\\
| & | & | & | & | & | & | & | & |\\
\hline | & | & | & | & | & | & | & | & |\\
{\vec{a}_{3}} & {\vec{a}_{1}} & {\vec{a}_{2}} & {\vec{b}_{3}} & {\vec{b}_{1}} & {\vec{b}_{2}} & {\vec{c}_{3}} & {\vec{c}_{1}} & {\vec{c}_{2}}\\
| & | & | & | & | & | & | & | & |
\\\hline \end{array}
Note the rest of the blocks are just permutations of the columns of the blocks above. It is done in a way so that in each block, we have all 1-9 numbers, since the first three rows are Sudoku-satisfied. And each column will also have 1-9, because we used the top three blocks to make the columns. Lastly, the rows are all satisfied with 1-9, because they are just the same rows as the first three rows except permuted again.
By way of an example, consider this:
\begin{array}{|ccc|ccc|ccc|}
\hline 5 & 3 & 4 & 6 & 7 & 8 & 9 & 1 & 2\\
6 & 7 & 2 & 1 & 9 & 5 & 3 & 4 & 8\\
1 & 9 & 8 & 3 & 4 & 2 & 5 & 6 & 7\\
\hline & & & & & & & & \\
& ? & & & ? & & & ? & \\
& & & & & & & & \\
\hline & & & & & & & & \\
& ? & & & ? & & & ? & \\
& & & & & & & & \\
\hline \end{array}
We will fill the remaining blocks by permuting the columns of the 3x3 blocks directly above it cyclically:
\begin{array}{|ccc|ccc|ccc|}
\hline 5 & 3 & 4 & 6 & 7 & 8 & 9 & 1 & 2\\
6 & 7 & 2 & 1 & 9 & 5 & 3 & 4 & 8\\
1 & 9 & 8 & 3 & 4 & 2 & 5 & 6 & 7\\
\hline 3 & 4 & 5 & 7 & 8 & 6 & 1 & 2 & 9\\
7 & 2 & 6 & 9 & 5 & 1 & 4 & 8 & 3\\
9 & 8 & 1 & 4 & 2 & 3 & 6 & 7 & 5\\
\hline 4 & 5 & 3 & 8 & 6 & 7 & 2 & 9 & 1\\
2 & 6 & 7 & 5 & 1 & 9 & 8 & 3 & 4\\
8 & 1 & 9 & 2 & 3 & 4 & 7 & 5 & 6
\\\hline \end{array}
|
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Guessing a number Find the number $n$ which has exactly $3$ prime divisors: $3$, $5$, and $7$, and has $4$ divisors that are powers of $5$. Moreover, $n$ has as many divisors that are powers of $3$ as there are divisors that are powers of $7$. Finally, $24$ divisors of $n$ are multiples of $3$.
What I understood is that $5$, $25$, $125$ and $625$ are the divisors of $n$. But do $7$ and $4$ have the same power in the prime factorization of $n$?
|
"$24$ divisors of $n$ are multiples of $3$". So, $n/3$ itself has $24$ divisors which are each one third of a divisor of $n$.
To have four power of $5$ divisors and $m$ divisors apiece that are powers of $3$ and powers of $7$, the number must have the form $n=3^{m-1}×5^3×7^{m-1}$ counting $1=p^0$ as a power of each prime factor. Then $n/3=3^{m-2}×5^3×7^{m-1}$ with $4m(m-1)$ divisors (add one to each exponent and multiply the augmented exponents). To match this with $24$ set $m=3$ therefore $n=3^2×5^3×7^2=55 125$.
|
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Factoring limits Couldn't find this limit someone help me?
$$\lim_{x\rightarrow0} \frac{ (1+x)^{1/3} - (1-x)^{1/3}}{x}$$
I tried to take $x^{1/3}$ common from above expression
|
Let $A=(1+x)^{1/3}$ and $B=(1-x)^{1/3}$ then
$$2x=A^3-B^3=(A-B)(A^2+AB+B^2)$$
Hence, as $x\to 0$, we have that $A\to 1$, $B\to 1$ and
$$\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}x=\frac{A-B}{x}=\frac{2}{A^2+AB+B^2}\to\frac{2}{3}.$$
|
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|
Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer. Question:
Prove $ \sqrt{2k}$ is irrational where $ k$ is an odd integer.
My attempt:
Proof by contradiction:
Now, assume $ \sqrt{2k}$ is rational. Then, $ \sqrt{2k} = \frac{a}{b}$ where $a,b \in Z$, $b$ not equal $0$ and $a,b$ have no common factors.
$ \sqrt{2k} = \frac{a}{b} \implies 2k = \frac{a^{2}}{b^{2}} \implies (b^{2})(2k) = a^{2} \implies 2|a^{2} \implies 2|a, \ since \ 2 \ is\ prime \implies \exists c \in Z$ such that $ a = 2c$
Then, $ (2k)(b^{2}) = a^{2} \implies (2k)(b^{2}) = 4c^{2} \implies kb^{2} = 2c^{2} \implies 2|kb^{2} \implies 2 | b^{2} \implies 2|b$, since $2$ is prime.
So, $ 2|a$ and $ 2|b$ , a contradiction.
|
You are almost there, suppose that $a$ and $b$ are relatively prime,
write $a^2=b^2(4m+2)=2b^2(2m+1)$, you deduce that $2$ divides $a^2$, and $a$, write $a=2a_1$, you have $2b^2(2m+1)=4a_1^2$, you deduce that $b^2(2m+1)=2a_1^2$. This implies that $2$ divides $b$, contradiction since $a$ and $b$ are relatively prime.
|
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Is $\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\sqrt{\phi }\right)\right|$ a radical? In this post, Reshetnikov gave the enormous even $80$-deg equation satisfied by,
$$x=\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\color{blue}{\sqrt{\phi }}\right)\right|=1.2054797\dots\tag1$$
with golden ratio $\phi$ and absolute value $|u|$, though mentioned he was unsure if it has a solvable Galois group.
After some experimentation, it turns out an even $40$-deg equation is a satisfied by the analogous,
$$y=\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\,\color{blue}\phi\right)\right|=1.162132\dots\tag2$$
If we define $z=\big(\frac{4y}5\big)^2$, then we just have the $20$-deg,
$$2^{16} - 20480 z^3 - 32000 z^4 - 24576 z^5 - 25600 z^6 - 20000 z^7 -
13065 z^8 - 6000 z^9 - 309 z^{10} + 2800 z^{11} + 2500 z^{12} +
160 z^{13} + 375 z^{14} - 96 z^{15} - 100 z^{16} - 5 z^{18} + z^{20}=0$$
This is more manageable, and Magma says this has permutation group $G=2^5 \cdot3^2 \cdot 5^2 = 7200$ which is unsolvable.
Q: If $(2)$ has an unsolvable group, does that imply the same for $(1)$ as well?
|
The scaled Bring quintic
$$x^5-\frac5nx+\frac4n=0\tag1$$
is solved by,
$$x =\frac45\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};n\right)$$
while the decic,
$$y^4(n y^2-5) (n y^2+5)^2 + \frac{16^2}n = 16 y^3 (2 n y^2+5)\tag2$$
is solved by,
$$y=\left(\frac45\right)^2\,\left|\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};n\right)\right|^2$$
with absolute value $|u|$. Eq $(2)$ then explains the $80$-deg since the argument $n=\sqrt{\phi}$ with golden ratio is a quartic root so $10\times4\times2 = 80$. Eliminating $n$ between $(1)$ and $(2)$, easily done using Mathematica, we get a solvable polynomial relationship for $x,y$, implying if one is a radical (or not), then so is the other.
However, the complete radical parameterization to $(1)$ is known and given by the Blair-Spearman quintic,
$$x^5+\frac{5u^4(4v+3)}{v^2+1}x+\frac{4u^5(2v+1)(4v+3)}{v^2+1}=0\tag3$$
Equating coefficients between $(1)$ and $(3)$, then eliminating $v$, we get the sextic in $u$,
$$n=\frac{5u^2+2u+1}{4(2-u)u^5}\tag4$$
Thus, for any radical $n$ such that $u$ is radical, then $x$ is also a radical.
Example. If $n=81$, then $u=1/3$ so we know,
$$x =\frac45\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};81\right)\approx 0.42429 - 0.239932 i$$
is a radical quintic root of $x^5-\frac5{81}x+\frac4{81}=0$.
The problem then for Reshetnikov's $80$-deg is to establish if $u$ is a radical in,
$$\sqrt{\phi}=\frac{5u^2+2u+1}{4(2-u)u^5}\tag5$$
But eq $(5)$ is a $24$-deg with integer coefficients and Magma says this has permutation group $G$ of order $207360000 = 2^{12} \cdot 3^4 \cdot 5^4$. Therefore it is not solvable, hence the $80$-deg root is not a radical.
|
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|
Show that $\frac{d^n}{dx^n}(\frac{\log x}{x})=(-1)^n\frac{n!}{x^{n+1}}(\log x-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{n})$ Show that $\frac{d^n}{dx^n}\left(\frac{\log x}{x}\right)=(-1)^n\frac{n!}{x^{n+1}}\left(\log x-1-\frac{1}{2}-\frac{1}{3}-\ldots -\frac{1}{n}\right)$
I am not allowed to use induction. I do not know how to approach this. Any help would be appreciated.
|
Another proof
Let $f(x) = \frac{\log(x)}{x}$, then for $x, y > 0$
$$f(x y) = \frac{\log(x)+ \log(y)}{x y} = \frac{f(x)}{y} + \frac{f(y)}{x}$$
Derivating $n$ times with respect to $y$ yields
$$x^n f^{(n)}(x y) = f(x) \left(\frac{1}{y}\right)^{(n)} + \frac{f^{(n)}(y)}{x}$$
Now replacing $y$ by $1$ yields
$$x^n f^{(n)}(x) = f(x) (-1)^n n! + \frac{f^{(n)}(1)}{x}$$
hence
$$f^{(n)}(x) = (-1)^n n! \frac{\log(x)}{x^{n+1}} + \frac{f^{(n)}(1)}{x^{n+1}}$$
The value of $f^{(n)}(1)$ is missing, it can be obtained by the Taylor series of
$$f(1+h) = \frac{\log(1+h)}{1+h}$$
used by Professor Vector, but there is another way: rewrite the above formula as
$$f^{(n)}(x) = (-1)^{n}n! \frac{\log(x) - s_n}{x^{n+1}}$$
We have $s_0=0$ and we claim that $s_{n+1} - s_n = \frac{1}{n+1}$ from which it follows that
$$s_n = s_n - s_0 = \sum_{k=0}^{n-1}(s_{k+1}-s_k) = \sum_{k=0}^{n-1}\frac{1}{k+1} = 1 + \frac{1}{2} +\cdots+\frac{1}{n}$$
To prove our claim, we compute
$$(-1)^{n+1}(n+1)! \frac{\log(x) - s_{n+1}}{x^{n+2}} = f^{n+1}(x)
= (f^{(n)}(x))^\prime
=(-1)^{n}n! \frac{1 -(n+1)(\log(x) - s_n)}{x^{n+2}}$$
It follows that $s_{n+1} = \frac{1}{n+1} + s_n$, QED
|
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|
Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM and CS inequality but failed to get a result. Please give me some hint in how to solve this question.
|
If $a=3$ and $b=-1$ and $c=-2$ we get a value $1$.
But
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=-\frac{abc(a+b+c)\sum\limits_{cyc}(a^2-ab)}{\prod\limits_{cyc}(2a^2+bc)}=0.$$
Thus, $P=1$ for all $a+b+c=0$ and $\prod\limits_{cyc}(2a^2+bc)\neq0$
and we are done!
|
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|
Find all solutions to $x^3+(x+1)^3+ \dots + (x+15)^3=y^3$
Find all pairs of integers $(x, y)$ such that
$$x^3+(x+1)^3+ \dots + (x+15)^3=y^3$$
What I have tried so far:
The coefficient of $x^3$ is $16$ in the left hand side. It is not useful then to trying bound LHS between, for example, $(ax+b)^3$ and $(ax+c)^3$ and then say that $ax+b<y<ax+c$.
I also tried to use modulo a prime. But it seems unlikely to bound variables this way.
EDIT : Though, it can be factored as $(2x+15)(x^2+15x+120)=(y/2)^3$. LSH factors are almost co-prime and we can say that $x^2+15x+120=3z^3$ or $x^2+15x+120=5z^3$. These are still too difficult to solve!
Any ideas?
|
You are looking for $(x, y) \in \mathbb{Z}^2$ for which
$$ \sum_{i= 0}^{15} (x+i)^3 = y^3, \tag{0}$$
that is,
$$ \sum_{i= 0}^{15} \left( x^3 + 3 i x^2 + 3i^2 x + i^3 \right) = y^3, $$
that is,
$$ 16 x^3 + 3 \frac{15 (15+1)}{2} x^2 + 3 \frac{15 (15+1)(2 \times 15 + 1)}{6} + \left( \frac{15 (15+1)}{2} \right)^2 = y^3, $$
that is,
$$ 16 x^3 + 360 x^2 + 3720 x + 14400 = y^3, $$
which can be written as
$$ 8 (2x^3 + 45 x^2 + 465 x + 1800) = y^3 $$
Can you get anywhere from here?
|
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|
Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $[x,y,z]$ and applying the property of orthogonal matrix.
In each approach I got a different solution, which is not correct.
Even I suspect there is an error in given problem.
Please help me in finding the third row.
Adding cross product calculation below.
\begin{vmatrix}
\bar{\imath}&\bar{\jmath}&\bar{k}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}
\end{vmatrix}
\begin{align}
&=\bar{\imath}(0)-\bar{\jmath}\left(-\frac{1}{3\sqrt{2}}\right)+\bar{k}\left(\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\frac{1}{\sqrt{\frac{1}{18}+\frac{1}{18}}}\left(0,\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)
\end{align}
|
Let this matrix be denoted as:
\begin{equation}
\mathbf{Q}=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
x & y & z
\end{array}\right]
\end{equation}
From the definition of the orthogonal matrix $\mathbf{Q}\,\mathbf{Q}^T = \mathbf{I}$:
\begin{equation}
\left[\begin{array}{ccc}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3}\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}\\
x & y & z
\end{array}\right]
\left[\begin{array}{ccc}
\frac{1}{3} & 0 & x\\
\frac{2}{3} & \frac{1}{\sqrt{2}} & y\\
\frac{2}{3} & -\frac{1}{\sqrt{2}} & z
\end{array}\right]
=
\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]
\end{equation}
we get three equations:
\begin{align}
&\frac{1}{3}x + \frac{2}{3}y + \frac{2}{3}z = 0\\
&\frac{1}{\sqrt{2}}y - \frac{1}{\sqrt{2}}z = 0\\
&x^2 + y^2 + z^2 = 1
\end{align}
By solving this you get two solutions $x_1 = -\frac{4}{\sqrt{18}}, y_1 = \frac{1}{\sqrt{18}}, z_1 = \frac{1}{\sqrt{18}}$ or $x_2 = \frac{4}{\sqrt{18}}, y_2 = -\frac{1}{\sqrt{18}}, z_2 = -\frac{1}{\sqrt{18}}$.
|
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|
Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
|
There is a shortcut way to evaluate $$p(x):=(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4)\\ - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3).$$
It is clear that this is a polynomial of at most second degree, as all cubic term will cancel out*.
Then we perform the easy evaluations
$$p(1)=-6,\ p(2)=-2,\ p(3)=2,\ p(4)=6$$ and clearly $$p(x)=4x-10.$$
*Actually we can also find the quadratic coefficient to be $$-2-3-4+1+3+4+2+1+4-1-2-3=0,$$ but we needn't use this fact, direct evaluation is anyway required to get the linear terms.
|
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|
Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\frac{x}{\sqrt{2}})^2$$
and then expand out, and complete the square. But I seem to get an error. Perhaps there is a smart way to do this?
|
One might recall (from the proof of Kepler's first law, for instance) that
$$ \rho(\theta) = \frac{\frac{b^2}{a}}{1+\frac{c}{a}\cos\theta} $$
is the polar equation of an ellipse (with semi-axis $b<a$ and $c=\sqrt{a^2-b^2}$) with respect to a focus. The associated cartesian equation clearly is $\frac{(x+c)^2}{a^2}+\frac{y^2}{b^2}=1$. In a similar way the cartesian equation associated to
$$ \rho(\theta) = \frac{\frac{b^2}{a}}{1-\frac{c}{a}\cos\theta} $$
clearly is $\frac{(x-c)^2}{a^2}+\frac{y^2}{b^2}$. So, if the polar equation is $\rho(\theta)=\frac{A}{1-\frac{1}{\sqrt{2}}\cos\theta}$, we have $a^2=2b^2$ and $\frac{b^2}{a}=A$ and the cartesian equation is given by
$$ \frac{\left(x-\sqrt{2}A\right)^2}{(2A)^2}+\frac{y^2}{(\sqrt{2} A)^2}=1. $$
|
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|
smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$
My attempt,
$$|S_n-S_\infty|<0.001$$
$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$
$$|2[1-(-\frac{1}2)^n]-2|<0.001$$
How to proceed? Thanks in advance.
|
$$|2[1-(-\frac{1}2)^n]-2|<0.001\\
|-2(\frac{-1}2)^n|<0.001\\
+2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\
+2.\frac{1}{2^n}<\frac{1}{1000}\\
\frac{2^n}{2}>1000\\
2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$
|
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|
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.
The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1-
\dfrac{x^2}{a^2}\right) \tag{1}.$$
Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$
We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.
Substituting back in $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.
The given answer is $3A$.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?
|
We perform an orthogonal projection to map the ellipse to the unit circle.
Let the maximal area of our isosceles triangle be $\mathcal{A}$ which we wish to find, and let $\triangle ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\triangle ABC$ is the projection of the triangle with area $\mathcal{A}$. Note that since we wish to maximise the area, $\triangle ABC$ is simply an equilateral triangle with area $\frac{3\sqrt{3}}{4}$. Hence, by preservation of area ratios,
$$\begin{align*} \frac{\mathcal{A}}{\text{ Area of ellipse}} &=\frac{[ABC]}{\text{ Area of circle}} \\ \implies \mathcal{A} &=\frac{3\sqrt{3}}{4\pi}\cdot \pi ab \\ &= \frac{3ab\sqrt{3}}{4} \end{align*}$$which is our answer.
|
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|
Maclaurin Series $ \int \frac{\sin(x)}{5x} $ I am supposed to evaluate the indefinite integral as an infinite series centered at $ x=0 $ and give the first five non-zero terms of the series.
$ \int \frac{\sin(x)}{5x} $
Here is what I have done so far:
$ g(x) = \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} $
$ \int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{5x} $
$ \frac{1}{5}\int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} x^{2n} $
$ \frac{1}{5} \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{2n+1} $
For $ n=0 $, $ \frac{1}{5}(x)$
$ n=1 $, $ \frac{1}{5}(\frac{-1}{3!} \frac{x^3}{3})$
$ n=2 $, $ \frac{1}{5}(\frac{1}{5!} \frac{x^5}{5})$
$ n=3 $, $ \frac{1}{5}(\frac{-1}{7!} \frac{x^7}{7})$
$ n=4 $, $ \frac{1}{5}(\frac{1}{9!} \frac{x^9}{9})$
And sum them.
I've done this a few different ways and am getting the same result. Can anyone shed some light on where I am going wrong?
|
Your result is perfectly good (before the expansion of the factorials, what I cannot check).
Forgetting the signs, the denominators are (as you wrote)
$$5 \qquad 5\times 3!\times 3=90\qquad 5\times 5!\times 5=3000 \qquad 5\times 7!\times 7=176400$$ $$\qquad 5\times 9!\times 9=16329600$$ Making then $$\frac{x}{5}-\frac{x^3}{90}+\frac{x^5}{3000}-\frac{x^7}{176400}+\frac{x^9}{16329600}
+O\left(x^{11}\right)$$
|
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|
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:
A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$
I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.
|
$$\cot\alpha = -\frac{7}{24} \implies \tan\alpha = -\frac{24}{7}$$
Since $1 + \tan^2\alpha = \sec^2\alpha$,
$$\sec^2\alpha = 1 + \left(-\frac{24}{7}\right)^2 = 1 + \frac{576}{49} = \frac{625}{49} \implies |\sec\alpha| = \frac{25}{7}$$
Observe that $450^\circ < \alpha < 540^\circ \implies \cos\alpha < 0$. Hence,
$$\sec\alpha = -\frac{25}{7} \implies \cos\alpha = -\frac{7}{25}$$
Since $450^\circ < \alpha < 540^\circ$, $225^\circ < \frac{\alpha}{2} < 270^\circ$. Thus,
$\cos(\frac{\alpha}{2}) < 0$. Therefore,
$$\cos\left(\frac{\alpha}{2}\right) = -\sqrt{\frac{1 + \cos\alpha}{2}} = -\sqrt{\frac{1 - \frac{7}{25}}{2}} = -\sqrt{\frac{\frac{18}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$
|
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|
Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove
$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$
I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$.
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
|
All of them fit in this infinite sequence of inequalities:
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots \ge 0 \ \forall x \in \mathbb{R} \\
\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} -\cdots > 0 \ \forall x>0 \\
\frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots > 0 \ \forall x \in \mathbb{R^*}\\
\cdots \cdots \\
$$
It's enough to show the first sum is $\ge 0$ and $>0$ for $0<|x|< \epsilon$, and notice that the derivative of any of those sums is the previous sum. The first inequality follows from the identity of power series
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots= 2 \left( \sum_{k \ge 0}\,(-1)^k\frac{x^{2k+1}}{2^{2k+1}(2k+1)!}\right)^2$$ which one can check directly.
|
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|
Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\lambda(x^2+y^2+z^2-2x+2y+6z+9)=0$$
Differentiated partially wrt $x$, $y$ and $z$, and equated to $0$.
I get the following equations
$$(2\lambda+2)x-y-z=2\lambda$$
$$-x+(2\lambda+2)y-z=-2\lambda$$
$$-x-y+(2\lambda+2)z=6\lambda$$
In matrix form:
$$\begin{bmatrix} 2\lambda+2 & -1 & -1 \\-1 & 2\lambda+2 & -1 \\ -1 & -1 & 2\lambda+2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}2 \\ -2 \\ -6 \end{bmatrix} $$
When I try to find $x$, $y$ and $z$ from these equations, it gets complex. Is there an easier way to ascertain their values?
|
I get the following equations
$$2x-y-z=\lambda(2x-2) \\-x+2y-z=\lambda(2y+2) \\-x-y+2z=\lambda(2z+6) $$
or equivalently
$$\begin{bmatrix} 2-2\lambda & -1 & -1 \\-1 & 2-2\lambda & -1 \\ -1 & -1 & 2-2\lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}-2 \\ 2 \\ 6 \end{bmatrix} $$
If the matrix is invertible, there is only one solution $(x,y,z)$. Otherwise, its determinant is zero, and that gives you the values of $\lambda$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m},$$ where $m\in \mathbb{Z}.$
So $$1728e^{-2\pi i+2\pi m}=1728cos(-2\pi)=1728.$$
Is it correct?
|
$(\sqrt{3}-3i)^6=27(1-\sqrt{3}i)^6=27\times 2^6\Big(\dfrac{1}{2}-\dfrac{\sqrt3}{2}i\Big)^6=27\times64\space\omega^6=27\times 64$
$\omega$ is a non-real solution of the equation $x^6=1$
|
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|
How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
|
Consider three cases:.
i) $\bf x\leq -1$. Then $x\leq -1$, $2^x< 1$, and the equation becomes
$\frac{1}{2}2^{-x} - 2^x = 1- 2^x + 1$, that is $2^{-x}=4$ which implies that $x=-2\leq -1$.
ii) $\bf -1<x<0$. Then $x>-1$, $2^x< 1$, and the equation becomes $2^x=2\cdot 2^{x} - 2^x = 1- 2^x + 1=2-2^x$, that is $2^x=1$ which is impossible for $-1<x<0$.
iii) $\bf 0\leq x$. Then $x>-1$, $2^x\geq 1$, and the equation becomes $2^x=2^x$ which holds for all $x\geq 0$.
Hence the complete set of solutions is $\{-2\}\cup [0,+\infty)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many positive integer cubes are divisors of $1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!$? What I have tried:
$$N= 1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$
|
Now, for each prime factor, how many possible exponents are there?
For example, for the prime factor $2$, the exponent can be $0,3,6,9,12,15,18$ or $21$.
|
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|
Determining if a vector is in the row space I am just trying to determine if the vector $[0, 7, 4]$ belongs in the row space of the matrix $$A = \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}
$$
What I have done so far is created an augmented matrix like so
$$ \left[
\begin{array}{ccc|c}
1 & 3 & 1 & 0 \\
2 & -1 & -5 & 7 \\
0 & 4 & 4 & 4 \\
\end{array}
\right]
$$
(naive gaussian) reducing to
$$ \left[
\begin{array}{ccc|c}
1 & 3 & 1 & 0 \\
0 & -7 & -7 & 7 \\
0 & 0 & 0 & 32/7 \\
\end{array}
\right]
$$
returning inconsistent, i.e. not existing in the row space. However apparently it does in fact belong in the row space, so clearly I have gone about this the wrong way. Is someone able to correct this?
|
Your working is fine.
Consider the matrix$$ \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}$$
Minus $2$ times of row $1$ plus row $2$ gives us row $3$, the row space is spanned by the first two rows.
Suppose $$a\begin{bmatrix} 1 & 2 & 0 \end{bmatrix} + b\begin{bmatrix} 3 & -1 & 4 \end{bmatrix}=\begin{bmatrix} 0 & 7 & 4 \end{bmatrix}$$
From the third coordinate, $b=1$.
Hence $a+3=0$ and $2a-1=7$ of which we can see a contradiction.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?
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Result: $\sin x\geq x-\dfrac{x^2}{2}\dots(*)$
Now $\sin \dfrac{2\pi}{9}=\dfrac{1}{\sqrt2}(\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})$
To show $\sin \dfrac{2\pi}{9}<\sqrt{\dfrac{3}{7}}\Leftrightarrow (\cos\dfrac{5\pi}{180}-\sin\dfrac{5\pi}{180})<\sqrt{\dfrac{6}{7}}\Leftrightarrow\sin\dfrac{\pi}{18}>\dfrac{1}{7}$
Using $(*)$, you get $\sin\dfrac{\pi}{18}>\Big(\dfrac{\pi}{18}\Big)-\dfrac{1}{2}\Big(\dfrac{\pi}{18}\Big)^2>\dfrac{1}{7}$
(You can see this calculation here, or you can do by hand by taking $\pi\sim 3.14$)
Hence $$\sin(40^\circ)<\sqrt{\frac{3}7}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this the only solution to the problem? Find all mutually non-equivalent $A(p,q)$ so that $(Aq\rightarrow \overline{p}) = (p+A)$ is true regardless of $p$ and $q$, where $\overline{x}$ is negation, $xy$ is conjunction, $x+y$ disjunction, $x\rightarrow y$ implication and $x=y$ equivalence.
\begin{matrix}p&q&A&(Aq\rightarrow \overline{p}) = (p+A)\\
0&0&0&0\\
0&0&1&1\\
0&1&0&0\\
0&1&1&1\\
1&0&0&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{matrix}
From the truth table we see that: $((Aq\rightarrow \overline{p}) = (p+A)) = \overline{p}\overline{q}A + \overline{p}qA + p\overline{q}\overline{A} + p\overline{q}A + pq\overline{A}$
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A} = 1$$
\begin{align}
A&=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)}=\\
&=\overline{p}+\overline{p}+\overline{q}=\\
&=\overline{p}+\overline{q}
\end{align}
Cheking result
$$((\overline{p}+\overline{q})q\rightarrow \overline{p})=p+(\overline{p}+\overline{q})$$
For $p=0$ and $q=0$
$$
((1+1)0\rightarrow 1)=0+(1+1)\\
(0\rightarrow 1) = 1 \\
1=1 \\
$$
For $p=0$, $q=1$
$$
((1+0)1\rightarrow 1)=0+(1+0)\\
(1\rightarrow 1)=1 \\
1=1 \\
$$
For $p=1$, $q=0$
$$
((0+1)0\rightarrow 0)=1+(0+1)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
For $p=1$, $q=1$
$$
((0+0)1\rightarrow 0)=1+(0+0)\\
(0\rightarrow 0)=1 \\
1=1 \\
$$
This is a correct solution, but is it the only one?
I ask because the question uses the word all, and though that may not mean anything, I'd just like to check in case I missed something.
|
It seems to me you are correct up to here:
$$\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
= 1.$$
It does not follow that
$A=\overline{p}\overline{q}+\overline{p}q+\overline{(pq)}
= \overline p + \overline q,$ however.
Yes, it is true that
$$
(A = \overline p + \overline q)\rightarrow
(\overline{p}\overline{q}A + \overline{p}qA + p\overline{q} + pq\overline{A}
= 1)
$$
for all possible truth assignments,
but the converse is not true.
The truth table shows that $A$ is one of two possible non-equivalent expressions, because for just one assignment to $p$ and $q,$
namely, $p=1, q=0,$ you have two choices for the value of $A.$
In particular, try the substitution $A = \overline p.$
With that substitution,
$Aq\rightarrow \overline p$ becomes
$\overline pq\rightarrow \overline p$, a tautology,
and $p+A$ becomes
$p+\overline p$, also a tautology,
so $(Aq\rightarrow \overline p)=(p+A).$
You can find this solution (rather than having to guess it) by selecting rows $2$, $4$, $5$, and $7$ from the truth table.
|
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|
Smallest diameter of a balanced subset of the Hamming cube Let $\{0,1\}^n$ be the Hamming cube with the Hamming metric. It's a metric space of diameter $n$.
Let's call a set $B\subset \{0,1\}^n$ balanced if its center of mass is the center of the cube; that is, the average of all vectors contained in $B$ is
$(1/2,\dots,1/2)$. What is the smallest possible diameter of a nonempty balanced subset of $\{0,1\}^n$?
Partial results
Let $d_n$ be the aforementioned smallest diameter, and let $B$ be a balanced set that realizes it. Without loss of generality, $B$ contains the zero vector $(0,\dots,0)$. To be balanced, it must also contain some vector with more $1$s than $0$s. Hence
$$
d_n \ge \left\lceil \frac{n+1}{2} \right\rceil \tag{1}
$$
The Cartesian product of two balanced sets is balanced, and its diameter is the sum of diameters of its factors. Therefore, the sequence $(d_n)$ is subadditive:
$$
d_{m+n} \le d_m+d_n \tag2
$$
(which in particular implies that $d_n/n$ has a limit.)
The values of $d_n$ I know so far:
$$
\begin{array}{|c|c|}
\hline
n & d_n \\
\hline
1 & 1 \\
2 & 2 \\
3 & 2 \\
4 & 3 \\
5 & 4 \\
6 & 4 \\
\hline
\end{array}
$$
Most of the table is obtained from $(1)$-$(2)$ with the help of the example
$$
\begin{pmatrix}
0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0
\end{pmatrix}
$$
where the columns form a balanced set of diameter $2$, showing $d_3\le 2$. (By the way, this is a tetrahedron inscribed in a cube.)
For $d_5$, the bounds $(1)$-$(2)$ give $3\le d_5\le 4$. To see that the diameter cannot be $3$, note that WLOG a balanced set of diameter $3$ contains the following columns:
$$
\begin{pmatrix}
0 & 1 \\
0 & 1 \\
0 & 1 \\
0 & 0 \\
0 & 0
\end{pmatrix}
$$
Since the total number of $0$s and $1$s in the last two rows must be the same, we need a column ending with two $1$s:
$$
\begin{pmatrix}
0 & 1 & * \\
0 & 1 & * \\
0 & 1 & *\\
0 & 0 & 1 \\
0 & 0 & 1
\end{pmatrix}
$$
but no matter how the asterisks are filled, the diameter is greater than $3$.
|
(This would probably be better as a comment than an answer but it looks like I can't comment yet.)
I think I can slightly improve on your partial results, with $d_7=4$ and (hence) $d_8=5$. (Since whenever $d_{2k-1} = k$ your bounds $(1)$-$(2)$ give $d_{2k}=k+1$.)
For the first claim, it's enough to find a balanced set with diameter equal to $4$, and it appears that
$$\left\{
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}
\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}
\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 1 \end{pmatrix}
\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}
\begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \\ 1 \\ 0 \\ 1 \end{pmatrix}
\begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}
\right\}$$
is exactly such a set.
|
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|
Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?
|
Like my answers in
Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$
In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$
let $S=\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
$2S=\left(\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right)\sin\dfrac C2$
As $\dfrac{A+B}2=\dfrac\pi2-\dfrac C2,$
$2S=\left(\cos\dfrac{A-B}2-\sin\dfrac C2\right)\sin\dfrac C2$
$\iff\sin^2\dfrac C2-\sin\dfrac C2\cos\dfrac{A-B}2+2S=0$
As $\sin\dfrac C2$ is real, the discriminant must be $\ge0$
i.e., $$\cos^2\dfrac{A-B}2-8S\ge0\iff8S\le\cos^2\dfrac{A-B}2\le1$$
So, $8S\le1\implies S\le\frac{1}{8}$
the equality occurs if $\cos\dfrac{A-B}2=1\iff\cos(A-B)=1\iff A=B$ as $0<A,B<\pi$
and hence $\sin\dfrac C2=\dfrac12\implies C=\dfrac\pi3$ as $0<C<\pi$
$\implies A=B=\dfrac{A+B}2=\dfrac\pi3=C$
|
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|
Find polynomials : $ xP(x-1)=(x-11)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ xP(x-1)=(x-11)P(x)$$
(Attempted work has been moved to answer)
|
Substitute $x=0$, we have $P(0)=0$ and substitute $x=11$, we have $P(10)=0$,
so $P(x)$ has $0$ and $10$ as its roots, i.e., $\exists Q(x)$, $P(x)=x(x-10)Q(x)$
Since $ xP(x-1)=(x-11)P(x)$, so $x(x-1)(x-11)Q(x-1)=(x-11)x(x-10)Q(x)$
thus, $(x-1)Q(x-1) = (x-10)Q(x)$ ---[2]
Similarly, substitute $x=1$ and $x=10$ in [2], $P(x)$ has $1$ and $9$ as its roots.
$\exists R(x)$, $Q(x)=(x-1)(x-9)R(x)$ substitute in [2], we have
$R(x-1)(x-1)(x-2)(x-10)=(x-10)(x-1)(x-9)R(x)$
so $R(x-1)(x-2)=(x-9)R(x)$
Finally,we have $A(x)=A(x+1)$, $\forall x \in \mathbb{R}$
Since $A$ is continuous function so $A(x) = c$, $\forall x \in \mathbb{R}$
Therefore, $P(x)=c(x)(x-1)(x-2)\ldots(x-10)$, where $c$ is constant.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
|
Let $K>1$ be a constant. Let $y = K^x$. Then $\ln y = x \ln K$. So
$\dfrac{y'}{y} = \ln K$. Hence $$\frac{d}{dx}K^x = \ln(K) K^x \tag 1$$.
As a reality check, note that this gives you $\frac{d}{dx}e^x = e^x$.
Using L'Hospital's rule, we see that
$\displaystyle \lim_{x \to \infty}\dfrac{x}{K^x} =
\lim_{x \to \infty}\dfrac{1}{\ln(K)K^x} = 0$. Using induction, we see that
$$\displaystyle \lim_{x \to \infty}\dfrac{x^n}{K^x} = 0 \tag 2$$
\begin{align}
\lim_{x \to \infty}\frac{7x^4+x^2\, 3^x+2}{x^3+x\,4^x+1}
&= \lim_{x \to \infty}\frac
{\dfrac{7x^3}{4^x} + \dfrac{x}{\left(\frac 43\right)^x} + \dfrac{2}{4^x}}
{\dfrac{x^2}{4^x} +1 + \dfrac{1}{x 4^x}} = 0
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be easy to implement in a computer program.
For instance, suppose
$$\begin{eqnarray}
F_0&=&0\\
F_1&=&1\\
F_{n+1}&=&F_n+F_{n-1},
\end{eqnarray}$$
so that $F_n$ is the $n^{th}$ term in the Fibonacci sequence. If we set
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix}$$
we see that
$$A^1=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix},$$
and it can be shown that
$$
A^n =
\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}.$$
This example is "interesting" in that it provides a novel way to compute the Fibonacci sequence. It is also relatively easy to implement a simple program to verify the above.
Other examples like this will be much appreciated.
|
Rather than give a new example, I'd like to present a way to expand on your own example. You mention that
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} =
\begin{pmatrix}
F_2 & F_1 \\ F_1 & F_0
\end{pmatrix},\quad
A^n =
\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix}.$$
Though, as @Henrik mentions, the matrix multiplication simply consists of doing the "usual" addition of Fibonacci numbers.
However, when you see a matrix power, e.g. $A^n$, you think of diagonalisation to efficiently compute this matrix power.
Any decent mathematical software should be able to compute for you that
$$A=\begin{pmatrix}
1 & 1 \\ 1 & 0
\end{pmatrix} = \begin{pmatrix}
-\frac{1}{\varphi} & \varphi \\ 1 & 1
\end{pmatrix}\begin{pmatrix}
-\frac{1}{\varphi} & 0 \\ 0 & \varphi
\end{pmatrix}\begin{pmatrix}
-\frac{1}{\sqrt{5}} & \frac{\varphi}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{1}{\varphi\sqrt{5}}
\end{pmatrix} = P\Lambda P^{-1},$$
where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. Therefore,
$$A^n = P\Lambda^n P^{-1} = \begin{pmatrix}
-\frac{1}{\varphi} & \varphi \\ 1 & 1
\end{pmatrix}\begin{pmatrix}
\left(-\varphi\right)^{-n} & 0 \\ 0 & \varphi^n
\end{pmatrix}\begin{pmatrix}
-\frac{1}{\sqrt{5}} & \frac{\varphi}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{1}{\varphi\sqrt{5}}
\end{pmatrix}.$$
Working out the product of these three matrices is really easy, so we get:
$$\begin{pmatrix}
F_{n+1} & F_{n} \\ F_{n} & F_{n-1}
\end{pmatrix} = A^n = \begin{pmatrix}
\frac{\varphi^{n+1} - \left(-\varphi\right)^{-(n+1)}}{\sqrt{5}} & \frac{\varphi^{n} - \left(-\varphi\right)^{-n}}{\sqrt{5}} \\ \frac{\varphi^{n} - \left(-\varphi\right)^{-n}}{\sqrt{5}} & \frac{\varphi^{n-1} - \left(-\varphi\right)^{-(n-1)}}{\sqrt{5}}
\end{pmatrix}.$$
In other words, through the use of diagonalisation we have proven that
$$F_n = \frac{\varphi^{n} - \left(-\varphi\right)^{-n}}{\sqrt{5}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "65",
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|
Show $f_{2n} = f_{n}^2 + 2f_{n-1}f_{n}$ where $f_n$ is the $n^{th}$ fibonacci number Note that for this problem $f_0 = 0, f_1 = 1, f_n = f_{n-1} + f_{n-2}$
My work so far is to write down the base case and the induction hypothesis. Then the problem becomes showing:
$f_{2n} = f_{n}^2 + 2f_{n-1}f_{n} \implies f_{2(n+1)} = f_{n+1}^2 + 2f_{n}f_{n+1}$
I've done this so far but beyond here I'm stuck. I don't know what to do with $f_{2n+1}$
$f_{2(n+1)} = f_{2n + 1} + f_{2n} = f_{2n+1} + f_n^2 + 2f_{n-1}f_{n} $
|
Alternatively, using:
$$f_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right),$$
we get:
$$f_{2n}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n}-\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\right),$$
$$f_n^2=\frac15 \left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n}-2(-1)^n+\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\right),$$
$$2f_{n-1}f_n=2\cdot \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right) \cdot \frac{1}{\sqrt{5}} \left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)=$$
$$\frac{1}{5}\left((\sqrt{5}-1)\left(\frac{1+\sqrt{5}}{2}\right)^{2n}+2(-1)^n-(\sqrt{5}+1)\left(\frac{1-\sqrt{5}}{2}\right)^{2n}\right).$$
Indeed:
$$2f_{n-1}f_n+f_n^2=f_{2n}.$$
|
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|
Logarithmic inequality with substitution I have this problem that I can solve halfway, but I'm struggling to find the interval for the solution.The inequality is this: $$\log_3(4^x+1)+\log_{4^x+1}(3)>2.5.$$
Now here is the method how I tried to solve this inequality:
$\log_3(4^x+1)+ \frac{1}{\log_{3}(4^x+1)}>2.5$
Substitute $\log_3(4^x+1)$ with $u$:
$\log_3(4^x+1)=u$
$u+ \frac{1}{u}>2.5$ multiply both side with $u$
$u^2-2.5u+1>0$
Now solving for $u$ I get $u_1=1$ and $u_2=2$
Next: going back to the substitution:
$\log _3(4^x+1)=u$
How do I proceed from now on assuming my calculations are right? How do I find the intervals?
|
$log_3(4^x+1)+log_{4^x+1}(3)>2.5$
$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$
Before I go any further, let me establish something. Assuming x is finite a real number, $4^x$>0. This means $4^x+1$>1. ln($4^x$+1)>0 if $4^x$+1 is greater than 1, which I have just established it is. Now, continuing from where we left off:
$\frac{ln(4^x+1)}{ln(3)}+\frac{ln(3)}{ln(4^x+1)}>2.5$
Multiply both sides by $ln(4^x+1)$
$\frac{ln²(4^x+1)}{ln(3)}+ln(3)>2.5ln(4^x+1)$
Since $ln(4^x+1)>0$, we did not have to change the > sign.
$\frac{ln²(4^x+1)}{ln(3)}-2.5ln(4^x+1)+ln(3)>0$
Let's assign u to be ln(4$^x$+1)
$\frac{u²}{ln(3)}-\frac{5u}2+ln(3)>0$
If you graph this quadratic function out, you'll see the value is greater than zero when u is less than the first quadratic root or greater than the second. So, to finish up this inequality, let's solve for the roots of this quadratic:
$\frac{u²}{ln(3)}-\frac{5u}2+ln(3)=0$
$\frac{-B±\sqrt{B²-4AC}}{2A}$
A=$\frac{1}{ln(3)}$ B=$-\frac{5}2 $ C=ln(3)
$\frac{-(-\frac{5}2)±\sqrt{(-\frac{5}2)²-4(\frac{1}{ln(3)})(ln(3))}}{2\frac{1}{ln(3)}}$
$\frac{\frac{5}2±\sqrt{\frac{25}4-4}}{2\frac{1}{ln(3)}}$
$\frac{ln(3)(\frac{5}2±\sqrt{\frac{25}4-\frac{16}4})}2$
$\frac{ln(3)(\frac{5}2±\sqrt{\frac{9}4})}2$
$\frac{ln(3)(5±3)}4$
$\frac{ln(3)(8)}4$ or $\frac{ln(3)(2)}4$
2ln(3) or $\frac{ln(3)}2$
since ln(3)>0, the greater of the two would be 2ln(3).
u<$\frac{ln(3)}2$ or u>2ln(3)
As mentioned before, u=ln($4^x$+1)
$ln(4^x+1)<\frac{ln(3)}2 or ln(4^x+1)>2ln(3)$
$ln(4^x+1)<ln(\sqrt{3}) or ln(4^x+1)>ln(9)$
As e$^x$ is an increasing function over all values x is real, we can put e to the power of both sides of both equations.
$4^x+1<\sqrt{3} or 4^x+1>9$
$4^x<\sqrt{3}-1 or 4^x>8$
As log$_4$(x) is also increasing over all x is real, we can apply the function $log_4(x)$ over both sides of both equations.
$x<log_4(\sqrt{3}-1) or x>log_4(8)$
$x<log_4(\sqrt{3}-1) or x>\frac{3}2$
|
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|
Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$ I was reading a solution for an analysis problem and they argued that $$\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$$ for $x\geq 1$ and $n\geq 1$, $x$ a real number and $n$ a natural number. Why is that?
|
We have $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(x)+\ln(2)}{x^2}$ such that $n \in \mathbb{N} \geq 1$ and $x \in \mathbb{R} \geq 1$,
Using the fact that $\ln x+\ln y = \ln(x y)$ we get that $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(2x)}{x^2}$.
We will assume that $n\geq 3$ so $1+x^2 \ln n$ and $x^2$ are positive numbers so we multiply by them does not change the inequality.
We arrive at $x^2 \ln(1+ n x) \leq (1+x^2 \ln n)(1+\ln (2x))$
Which is just $x^2 \ln (1+n x) \leq 1+x^2 \ln n+\ln(2x) + x^2 \ln(2x) \ln n$
Now since $1+\ln(2x)$ is positive for all $x\geq 1$ we can treat them as $0$ to strengthen the inequality and get that $x^2 \ln (1+n x) \leq x^2 \ln n + x^2 \ln(2x) \ln n$ Now divide by $x^2 \ln n$
We get that $\frac{\ln(1+n x)}{\ln n} \leq 1+\ln(2x)$ which is just $\frac{\ln(1+n x)-\ln n+\ln n}{\ln n} \leq 1+\ln(2x)$
Using the fact that $\ln x-\ln y = \ln(x/y)$ we get that $\frac{\ln(\frac{1+n x}{n})}{\ln n }+1 \leq 1+\ln(2x)$
Which is just $\frac{\ln(\frac{1}{n}+x)}{\ln n}\leq \ln(2x)$
Since $n\geq 3$ its easy to see that $\ln n \geq 1$ so $\ln(\frac{1}{n}+x) \geq \frac{\ln(\frac{1}{n}+x)}{\ln n}$
So to strengthen the inequality we get that $\ln(\frac{1}{n}+x) \leq \ln(2x)$
And since $\ln$ function is increasing then the inequality is true whenever $\frac{1}{n}+x \leq 2x$ which is just $\frac{1}{n} \leq x$ and since $x\geq 1$ and $n\geq 3$ so $\frac{1}{n} \leq \frac{1}{3}$ the we get that $\frac{1}{n} \leq \frac{1}{3} \leq 1 \leq x$ which is always true.
Now we are left with cases $n=1,2$.
For $n=1$ we get that $\ln(1+x) \leq \frac{1+\ln(2x)}{x^2}$ which is obviously false for all $x\geq 2$
For $n=2$ we get that $\frac{\ln(1+x)}{1+x^2 \ln 2} \leq \frac{1+\ln(2x)}{x^2}$
Which is also false for all number $x\geq6$ because the terms the play the major rule are $x^2$ and $x^2 \ln 2 \approx 0.693 x^2$ and since $x^2$ is much bigger than $0.693 x^2$ so the inequality obviously will be false from some point on.
thus concluding that the proof is true whenever $n\geq3$ and $x\geq1$.
|
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|
Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}$
Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}.$
I'm currently trying to finish a task that requires me to use mathematical induction to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$. This is the first time that the divider is not a simple integer and I'm having trouble trying to prove it.
This is what I managed to do:
First step:
For $n=1$:
$2^3+1=9$
$3^2=9$
$L=P$
Second step: I assume that for some numbers $n>=1$ $2^{3^n}+1$ is divisible by $3^{n+1}$
Third step: (induction hypothesis):
$2^{3^{n+1}}+1$ is divisible by $3^{n+1}$
I'm getting stuck here, trying to do some algebra and simplify the dividend:
$2^{3^{n+1}}+1 = 2^{3^n*3}+1$
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You should use Wilsons theorem: It says that if $(m,n)=1$ then $m^{\phi(n)}\equiv$ 1 (mod $n$ ), where $\phi(n)$ is the number of natural numbers less than $n$ co-prime to $n$
Using this it follows that $m^{2\cdot 3^n} \equiv 1$ ( mod $3^{n+1}$ ) if $(m,3) = 1$
Now whats left is that $5$ is not a square mod 3 (one can check this easily)
Next it follows $5$ is not a square mod $3^{n+1}$ cause if it were one immediately gets $5$ is a square mod 3
So now we have $5^{2 \cdot 3^n} \equiv $ 1 (mod $3^{n+1}$)
since $5$ is not a square mod $3^{n+1}$ raising it to an odd power also yields a number which is not a square mod $3^{n+1}$ so we get $5^{3^{n}}$ is not a square mod $3^{n+1}$
Since $5^{2 \cdot 3^{n}} \equiv 1 $( mod $3^{n+1}$) we get that
$5^{3^n} \equiv \pm 1 $ ( mod $3^{n+1}$ ) Now also it cannot equal 1 cause that would imply that $5^{3^n}$ is a square mod $3^{n+1}$ which we have shown is not true.
At last we conclude that $5^{3^n} \equiv$ -1 (mod $3^{n+1}$) so we get the result:
$5^{3^n} + 1 \equiv $ 0 (mod $3^{n+1}$)
If you really want induction then as was said cube and one has:
$5^{3^n} + [3\cdot 5^{n-1}(5^{n-1} +1)] + 1 \equiv $ 0 (mod $3^{n+1}$)
Since by the inductive hypothesis we also have the part in brackets is divisible by $3^{n+1}$ the result follows.
|
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|
$a^2=ab+b^2+b+5$ has no integer solutions Prove that the equation $a^2=ab+b^2+b+5$ has no integer solutions.
My attempt:
$4a^2=4ab+4b^2+4b+20$
$4a^2-4ab+b^2 = 5b^2+4b+20$
$5(4a^2-4ab+b^2) = 5(5b^2+4b+20)$
$5(2a-b)^2=(5b+2)^2+96$
Let $2a-b = x$, $\;5b+2=y$
we get $\;5x^2-y^2=96$.
Please suggest on how to proceed.
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$$5x^2-y^2 = 84 \implies 2x^2-y^2 \equiv 0 \pmod{3}$$
which implies $3\mid x$ and $3\mid y$. But $9\nmid 84$.
Your original equation should be equivalent to $$5x^2-y^2 = 96$$
where $x=2a-b, y=5b+2$. But the above method works fine for $96$ too.
|
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|
Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:
Use counting arguments to prove these identities:
$$1. \quad\begin{pmatrix}
n \\
0\\
\end{pmatrix}^2 + \begin{pmatrix} n \\ 1\\ \end{pmatrix}^2 +\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 + \ldots+\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
$$2. \quad\ \begin{pmatrix}
n \\
1\\
\end{pmatrix}^2 + 2\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 +3\begin{pmatrix} n \\ 3\\ \end{pmatrix}^2 + \ldots+n\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \frac n2\begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
Use $(1+x)^{2n}$ to prove Identity $1$.
Use Identity $1$ to prove Identity $2$.
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Part 1:
$$\begin{align}
(1+x)^{2n}&=\sum_{m=0}^{2n}\binom {2n}mx^m\tag{1}\\
(1+x)^{2n}&=(1+x)^n(1+x)^n\\
&=\sum_{r=0}^n\binom nr x^r\sum_{s=0}^n\binom ns x^s\\
&=\sum_{r=0}^n\sum_{s=0}^n\binom nr\binom nsx^{r+s}\\
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\tag{2}\\
[x^m](2)=[x^m](1):\qquad
\sum_{r=0}^m\binom nr\binom n{m-r}&=\binom {2n}m\\
\text{Put }m=n:\qquad
\sum_{r=0}^n\binom nr\binom n{n-r}&=\binom {2n}n\\
\sum_{r=0}^n\binom nr^2&=\binom {2n}n\color{red}\blacksquare
\end{align}$$
Part 2:
$$\begin{align}
(1)=(2):\qquad
\sum_{m=0}^{2n}\binom {2n}mx^m&=\sum_{m=0|}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}x^m\\
\frac d{dx}:\qquad
\sum_{m=0}^{2n}\binom {2n}m mx^{m-1}
&=\sum_{m=0}^{2n}\sum_{r=0}^m\binom nr\binom n{m-r}mx^{m-1}\\
[x^{m-1}]:\qquad \qquad
m\binom {2n}m&=\sum_{r=0}^m m\binom nr\binom n{m-r}\\
\text{Put }m=n:\qquad
n\binom {2n}n
&=\sum_{r=0}^n n\binom nr\binom n{n-r}\\
&=\sum_{r=0}^n n\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+(n-r)\binom nr^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r=0}^n (n-r)\binom n{n-r}^2\\
&=\sum_{r=0}^n r\binom nr^2+\sum_{r'=0}^n r'\binom n{r'}^2
&&(r'=n-r)\\
&=2\sum_{r=0}^n r\binom nr^2\\
\sum_{r=0}^n r\binom nr^2
&=\frac n2\binom {2n}n\color{red}\blacksquare\end{align}$$
|
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Solving a System of Two Differential Equations and getting the wrong answer Problem:
Solve the following system of differential equations.
\begin{eqnarray*}
5x' + y' - 3x + y &=& 0 \\
4x' + y' - 3x &=& -3t \\
\end{eqnarray*}
Answer:
To solve this, we use the operator method.
\begin{eqnarray*}
(5D-3)x + (D+1)y &=& 0 \\
(D-3)x + Dy &=& -3t \\
D(5D-3)x + D(D+1)y &=& 0 \\
(D+1)(D-3)x + D(D+1)y &=& (D+1)(-3t) = -3t - 3 \\
D(5D-3)x - (D+1)(D-3)x &=& 3t + 3 \\
(5D^2 - 3D)x - (D^2 - 2D + 3)x &=& 3t + 3 \\
(4D^2 - D - 3)x &=& 3t + 3 \\
\end{eqnarray*}
Now to solve this equation, we need to find both the complementary solution and the particular solution. We call the complementary solution $x_c$ and the particular solution $x_p$. To find the complementary solution, we
set up the characteristic equation.
\begin{eqnarray*}
4m^2 - m - 3 &=& 0 \\
m &=& \frac{1 \pm \sqrt{1 - 4(4)(-3)} }{2(4)} = \frac{1 \pm \sqrt{1+ 48} }{8} \\
m &=& \frac{1 \pm 7 }{8} \\
m &=& 1 \,\, \text{or} \,\, m = -\frac{3}{4} \\
x_c &=& c_1 e^{x} + c_2 e^{-\frac{3}{4}} \\
\end{eqnarray*}
Now we need to find $x_p$.
\begin{eqnarray*}
x_p &=& At + B \\
x'_p &=& A \\
x''_p &=& 0 \\
4(0) - A - 3(At + B) &=& 3t + 3 \\
-3A &=& 3 \\
A &=& -1 \\
-A - 3B &=& 3 \\
-(-1) - 3B &=& 3 \\
-3B &=& 3 - 1 = 2\\
B &=& -\frac{2}{3} \\
x_p &=& -t -\frac{2}{3} \\
\end{eqnarray*}
The books answer is:
\begin{eqnarray*}
x &=& c_1 e^{t} + c_2 e ^{3t} + t + \frac{7}{3} \\
y &=& - c_1 e^{t} - 3c_2e^{3t} + 3t + 1 \\
\end{eqnarray*}
Since my answer for $x$ is going to be different from the book, I conclude that I went wrong some where. Where did I go wrong?
Bob
|
This is the screen copy of your answer with the mistake pointed out :
NOTE :
I got :
\begin{eqnarray*}
x &=& c_1 e^{t} + c_2 e ^{3t} + t + \frac{7}{3} \\
y &=& - c_1 e^{t} - 3c_2e^{3t} + 3t - 1 \\
\end{eqnarray*}
which isn't exactly the expected result.
|
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|
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
|
Because $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n=$$
$$=20(50^n-12^n)+38\cdot12^n$$ and since
$$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1}),$$
we are done!
|
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|
What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am having trouble in dealing with increasing number of multiples in each step. Kindly Guide in How to proceed
|
$$a_0=1, \quad \text{and }a_n=\frac{x^{2n}}{2}\left(\prod_{k=0}^{n-1} \frac{2k+1}{2k+4}\right) \text{ for } n\ge1.$$
|
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|
Find $\tan^{-1} (i\sqrt{2})$. Problem: Find $\tan^{-1} (i\sqrt{2})$.
My attempt: We must find $z \in \mathbb{C}$ such that $\tan z = i\sqrt{2}$.
$$\tan z = \frac{\sin z}{\cos z} = i\frac{e^{-iz} - e^{iz}}{e^{iz} + e^{-iz}}$$
Let $u = e^{iz}$. Then $$i\frac{u^{-1}-u}{u+u^{-1}} = i\sqrt{2}$$ so $$1-u^2 = \sqrt{2}(u^2 + 1)$$
and
$$u^2 = \frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Then we have
$$e^{iz} = \sqrt{\frac{1-\sqrt{2}}{1+\sqrt{2}}}$$
Taking the natural logarithm of both sides gives
$$z = \frac{1}{2i}\ln\frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Since $\ln$ is multi-valued and $\frac{1-\sqrt{2}}{1+\sqrt{2}}+0i=\frac{1-\sqrt{2}}{1+\sqrt{2}}e^{i2n\pi},$
we have
$$\frac{1}{2i}\ln{\frac{1-\sqrt{2}}{1+\sqrt{2}}}=\frac{-i}{2}\texttt{Ln}\frac{1-\sqrt{2}}{1+\sqrt{2}}+ n\pi$$
That's how far I got. The book lists the final answer as $$\pi/2 + n\pi -i\texttt{Ln}(\sqrt{2}-1)$$
|
A much easier solution is as follows:
$$
\tan^{-1}(z)=-i\tanh^{-1}(iz)\\
\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}\\
\begin{align}
\tan^{-1}(i\sqrt{2})
&=-i\tanh^{-1}(-\sqrt{2})=i\tanh^{-1}(\sqrt{2})\\
&=\frac{i}{2}[\ln(1+\sqrt{2})-\ln(1-\sqrt{2})]\\
&=\frac{i}{2}[\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-i(\pi+2n\pi)]\\
&=\frac{\pi}{2}+n\pi+\frac{i}{2}\ln\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}\\
&=\frac{\pi}{2}+n\pi-i\ln(\sqrt{2}-1)
\end{align}
$$
since
$$
\ln(1-\sqrt{2})=\ln[(\sqrt{2}-1)e^{i(\pi+2n\pi)}]\\
\ln\frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}=-\ln[(\sqrt{2}-1)^2]=-2\ln(\sqrt{2}-1)
$$
EDIT: here is another interpretation of the result that gives a slighly different result. [This result agrees with Matlab if you type >atan(i*sqrt(2)).]
$$
\begin{align}
\tan(i\sqrt{2})
&=\frac{i}{2}\ln\frac{1+\sqrt{2}}{1-\sqrt{2}}\\
&=\frac{i}{2}\ln\left(\frac{1+\sqrt{2}}{1-\sqrt{2}}\frac{1+\sqrt{2}}{1-\sqrt{2}}\right)\\
&=\frac{i}{2}(\ln(-1)-2\ln(\sqrt{2}-1))\\
&=-\frac{\pi}{2}-n\pi-i\ln(\sqrt{2}-1)
\end{align}
$$
I just thought this was curious and you can imagine my angst when the first result disagreed with Matlab.
|
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|
Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \geqslant 5.$$
This is an assignment I got from a teacher. I've been baffled by the negative sign in the denominator.
Could you please provide insight as to how would one resolve this inequality please? (Do not solve yet, I've yet to turn in this assignment, I just want to get some tips)
|
Start with
\begin{eqnarray*}
(2y^2z+2z^2x+2x^2y-yz^2-zx^2-xy^2-3xyz)^2 \geq 0
\end{eqnarray*}
This can be expanded out to give
\begin{eqnarray*}
4 \sum_{cyc} y^2 z^4 -4 \sum_{cyc} y^3 z^3 +\sum_{cyc} y^4 z^2 -4 \sum_{cyc} x y z^4 +\\14 \sum_{cyc} xy^2 z^3 -10 \sum_{cyc} x y^3 z^2 -3x^2 y^2 z^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
(2x-y)^2(y-z)^2(z-x)^2+(x-y)^2(2y-z)^2(z-x)^2+(x-y)^2(y-z)^2(2z-x)^2 \geq 5(x-y)^2(y-z)^2(z-x)^2
\end{eqnarray*}
Now divide by $(x-y)^2(y-z)^2(z-x)^2$ and we have
\begin{eqnarray*}
\frac{(2x-y)^2}{(x-y)^2}+\frac{(2y-z)^2}{(y-z)^2}+\frac{(2z-x)^2}{(z-x)^2} \geq 5.
\end{eqnarray*}
Edit:
I worked my way backwards from the result, using the "reduce" algebra package. I then guessed a square that would "consume" some of the terms & got lucky !
(2*x-y)^2*(y-z)^2*(z-x)^2+(x-y)^2*(2*y-z)^2*(z-x)^2+(x-y)^2*(y-z)^2*(2*z-x)^2-5*(x-y)^2*(y-z)^2*(z-x)^2;
(2*(y^2*z+z^2*x+x^2*y)-(yz^2+zx^2+x*y^2)-3*xyz)^2;
|
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|
Evaluate $\int \frac{dx}{(x^2-x+1)(\sqrt{x^2+x+1})}$ Evaluate $$I=\int \frac{dx}{(x^2-x+1)(\sqrt{x^2+x+1})}$$
My Try:
we have $x^2-x+1=(x+w)(x+w^2)$ where $w$ is complex cube root of unity
I have splitted $I$ as
$$I=AI_1+BI_2$$ where $A,B$ are some constants
$$I_1=\int \frac{dx}{(x+w)\sqrt{x^2+x+1}}$$
By taylor's series $$x^2+x+1=P(x+w)^2+Q(x+w)+R=(x+w)^2 \left(P+\frac{Q}{x+w}+\frac{R}{(x+w)^2}\right)$$ for some constants complex $P,Q,R$
hence
$$I_1=\int \frac{\frac{dx}{(x+w)^2}}{\sqrt{ \left(P+\frac{Q}{x+w}+\frac{R}{(x+w)^2}\right)}}=\int \frac{dt}{\sqrt{Rt^2+Qt+P}}$$ which is a standard Integral. Similar analysis for $I_2$.
Any other approach?
|
Substituting
$$ x = \frac{\alpha + \beta t}{1 +t}$$
we have:
$$ x^2- x+1 = \frac{(\alpha +\beta t)^2-(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$
$$ x^2+x +1 = \frac{(\alpha +\beta t)^2+(1+t)(\alpha +\beta t)+(1+t)^2}{(1+t)^2}$$
Numbers $ \alpha, \beta $ we define like, that coefficients at $ t $ are zero.
Hence
$$ 2\alpha \beta - \alpha - \beta + 2 =0,\ \ 2\alpha \beta +\alpha +\beta +2=0.$$
$$ \alpha = 1, \ \ \beta = -1.$$
We have
$$ x = \frac{1-t}{1+t}; \ \ dx = \frac{-2dt}{(1+t)^2};$$
$$ x^2 -x +1 = \frac{3t^2 +2}{1 + t^2};$$
$$\sqrt{x^2+x +1} =\frac{\sqrt{t^2+3}}{1+t}; \ \ 1+t>0. $$
$$I =-2\int \frac{(t+1)dt}{(3t^2+1)\sqrt{t^2+3}}= -2\int \frac{tdt}{(3t^2+1)\sqrt{t^2+3}} - 2\int \frac{dt}{(3t^2+1)\sqrt{t^2+3}} = I_{1} + I_{2}.$$
$$ I_{1} = -2\int \frac{tdt}{(3t^2+1)\sqrt{t^2+3}} = (\sqrt{t^2+3}= u) = 2\int \frac{du}{8 -3u^2} = \frac{1}{2\sqrt{6}}\ln \left| \frac{2\sqrt{2}+\sqrt{3}u}{2\sqrt{2}-\sqrt{3}u}\right| = \frac{1}{\sqrt{6}}\ln\left|\frac{2\sqrt{2}+\sqrt{3(t^2+3)}}{2\sqrt{2}- \sqrt{3(t^2+3)}}\right|=\frac{1}{\sqrt{6}}\ln\left|\frac{(1+x)\sqrt{2}+\sqrt{3(x^2+x+1)}}{\sqrt{x^2-x+1}}\right|.$$
$$ I_{2}=-2\int\frac{dt}{(3t^2+1)\sqrt{t^2+3}}=(\frac{t}{\sqrt{t^2+3}}=z)=-2 \int\frac{dz}{8z^2 +1} =-\frac{1}{\sqrt{2}}\arctan\left(\frac{2\sqrt{z}}{1}\right)= -\frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{2}(1-x)}{\sqrt{x^2+x +1}}\right).$$
In the end
$$ I =\frac{1}{\sqrt{6}}\ln\left|\frac{(1+x)\sqrt{2}+\sqrt{3(x^2-x+1)}}{\sqrt{x^2-x+1}}\right| -\frac{1}{\sqrt{2}}\arctan\left(\frac{\sqrt{2}(1-x)}{\sqrt{x^2+x +1}}\right) + C.$$
|
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|
Minimise a sum of unknown function.
Let $f(x)$ be a real valued function such that $f(x) > 0\ \ \forall x
\in \Bbb R$. $f(x)$ is symmetrical about $x = 2$ and $x = 4$. If
$\displaystyle S = \sum^{50}_{r= 1} f(r + 4)$ and
$\displaystyle\prod^{50}_{r = 1}f(r) = 2^{50}$. Find the minimum value
of $S$.
$$\begin{align}S &= \sum^{46}_{r = 1} f(r + 4) + f(51) + f(52) + f(53) + f(54)\\ &= \sum^{50}_{r=1} f(r) + f(51) + f(52) + f(53) + f(54) - \left(\ f(1) + f(2) + f(3) + f(4)\ \right)\\&\ge 50 \times\left(\prod^{50}_{r=1} f(r)\right)^{1/50} + f(51) + f(52) + f(53) + f(54) - \left(\ f(1) + f(2) + f(3) + f(4)\ \right)\\&= 100 + f(51) + f(52) + f(53) + f(54) - \left(\ f(1) + f(2) + f(3) + f(4)\ \right) \end{align}$$
Now I don't have any idea how to eliminate the remaining unknowns from the above equation. I suspect that $f(51) + f(52) + f(53) + f(54) =\ f(1) + f(2) + f(3) + f(4)$ and $100$ is the answer. I guess we have to use $f(4+ x) = f(4 -x)$. Any idea how to eliminate these unknowns ?
|
From the symmetry at $x=4$, $f(5) = f(3)$ and from the symmetry at $x=2$, $f(3)=f(1)$. Therefore, $f(1)=f(3)=f(5)$. Similarly, we can show that for $\forall r:$
$$f(r+2)=f(r)$$
Consequently,
$$S=25 (f(1)+f(2))$$
Also, note that $\prod_{r=1}^{50}f(r)=(f(1)f(2))^{25}=2^{50}$. Thus,
$$f(1)f(2)=4$$
Therefore, $S$ is minimum when $f(1)=f(2)=2$, and thus $\min{(S)}=100$.
|
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|
Need help with a proofs question.. Q : If $x_1$ and $x_2$ are two solutions to quadratic equation $ax^2+bx+c = 0$, then show that
$$x_1 + x_2 = -\frac{b}{a}\qquad\text{and}\qquad x_1x_2 = \frac{c}{a}.$$
From this question I gathered that the discriminant $b^2-4ac > 0$
I also tried to use the AGM and I got the inequality of $$\frac{c}{a} < \frac{b^2}{4a^2}.$$ But I don't know if I'm on the right track or not. What should I do?
|
when you try to prove any thing try to think in the simple way first and start with what you already know.
$$x_1 = \frac{- b + \sqrt{b^2 - 4ac}}{2 a}$$
$$x_2 = \frac{- b - \sqrt{b^2 - 4ac}}{2 a}$$
Thus
$$x_1 + x_2 = \frac{-2b}{2a} = -\frac{b}{a};$$
the square root part cancel out because the had different signs.
\begin{align}
x_1 x_2
&= \frac{(- b + \sqrt{b^2 - 4ac})}{2 a} \frac{(- b - \sqrt{b^2 - 4ac})}{2 a} \\
&= \frac{b^2 - b^2 + 4 a c}{4 a^2} \\
&= \frac{c}{a}.\end{align}
|
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|
Integrate $\int \frac{1}{\sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}}dx$ For a physics assignment we are analyzing an anharmonic oscillator whose force is given by: $$ F =-kx + \frac{kx^3}{a^2},$$ where $k$ and $a$ are the spring constant and an arbitrary positive constant respectively. One of the questions is to find $x(t)$ for a particular initial velocity of $${v_0}^2 = \frac{ka^2}{2m}.$$ Separating variables I find that $${v(x)} = \sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}.$$ So then $x(t)$ is given by the integral in the title which I cannot solve, $$\int \frac{1}{\sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}}dx.$$ Any help would be much appreciated.
Edit: My professor suggested using the identity $$\frac{2a}{a^2 -b^2} = \frac{1}{a-b} + \frac{1}{a+b},$$ and that the problem then becomes 'logarithimically easy'.
|
Just observe $$\int \frac{1}{\sqrt{\frac{ka^2}{2m}-\frac{k}{m}x^2\:+\frac{kx^4}{2ma^2}}}dx = \int \frac{1}{\sqrt{\frac{k}{2ma^2}(x^2-a^2)^2}} dx$$
and then using substitution like $x = a \sin \theta$ to solve the integral ?
|
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|
Eigenvalues of $A^n$, $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$
Let $A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$ and $\alpha_n$ and $\beta_n$ denote the two eigenvalues of $A^n$ such that $|\alpha_n|\geq |\beta_n|$. Then
*
*$\alpha_n\rightarrow \infty$ as $n\to\infty$
*$\beta_n\to 0$ as $n\to\infty$
*$\beta_n$ is positive if $n$ is even.
*$\beta_n$ is negative if $n$ is odd.
$F_n$ is $n$-th Fibonacci sequence, with $F_{-1}=0,F_0=1,F_1=1$
I found $A^n=\begin{bmatrix}F_n&F_{n-1}\\F_{n-1}&F_{n-2}\end{bmatrix}$
Eigenvalues are $\dfrac{F_n+F_{n-2}\pm\sqrt{(F_n-F_{n-2})^2+4F^2_{n-1}}}{2}$
Is there any result I need to know, because it is a MSQ(Multiple selection question) and meant to solve within 4-5 minutes.
One other thing, I found (which may not be important here):
$\begin{bmatrix}F_n&F_{n-1}\\F_{n-1}&F_{n-2}\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}=\begin{bmatrix}F_{n+1}&0\\0&F_n\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$
|
It can be easily noticed that for all $n\geq 3$, $A^n=A^{n-1}+A^{n-2}$ and so $\alpha_n=\alpha_{n-1}+\alpha_{n-2}$ and $\beta_n=\beta_{n-1}+\beta_{n-2}$. Also note that
$$\alpha_1=\dfrac{1+\sqrt{5}}{2} \quad \alpha_2=\dfrac{3+\sqrt{5}}{2} \qquad \beta_1=\dfrac{1-\sqrt{5}}{2} \quad \beta_2=\dfrac{3-\sqrt{5}}{2} $$ Now characteristic equation of recurring relation (See: Wikipedia) $\alpha_n=\alpha_{n-1}+\alpha_{n-2}$ is $x^2-x-1=0$, which gives us
$$\alpha_n = a\left( \frac{1+\sqrt{5}}{2} \right)^n +b\left( \frac{1-\sqrt{5}}{2} \right)^n. $$
Taking $n=1$ we have $a=1, b=0$ and so we get $$ \alpha_n = \left( \frac{1+\sqrt{5}}{2} \right)^n=\alpha_1^n $$
Similarly we have $$\beta_n = \left( \frac{1-\sqrt{5}}{2} \right)^n = \beta_1^n$$.
As $\alpha_1>2$, we have $\alpha_n\to\infty$, as $n\to\infty$. Again that $\beta_1\in(-1,0)$, so we have $\beta_n\to 0$, as $n\to \infty$. Also $\beta_n$ is positive if $n$ is even and $\beta_n$ is negative if $n$ is odd.
Answer: $(1), (2), (3), (4)$.
|
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|
Number of permutations of $9$ people of three nationalities in which no two people of the same nationality are adjacent $9$ different people must be put in a row.
Three of them are of nationality $X$, three are of $Y$, and the remaining three are of $Z$.
In how many combinations there will be no two people of the same nationality sitting next to each other?
My understanding:
I couldn't find a pattern to follow if few are already occupied. Like, if the first seat is occupied by a person of nationality $X$ then the next seat can be occupied by $6$ other people from $Y$ or $Z$, and then the tree grows where I can't seem to have a control on.
Any hints will be appreciated.
|
There are $9!$ ways to arrange nine individuals. From these, we must exclude those seating arrangements in which two people from the same country are seated in adjacent seats.
One pair of adjacent people: We have three ways to choose the nationality of the pair. We have $\binom{3}{2}$ ways to choose the people of that nationality who form the pair. We have $2$ ways to arrange the people within the pair. Finally, we have eight objects to arrange, the pair and the other seven people, so there are
$$\binom{3}{1}\binom{3}{2}2!8!$$
such seating arrangements.
Two pairs of adjacent people: There are two possibilities, both pairs are from the same nationality (so the three people of that nationality sit in consecutive seats) or they are from different nationalities.
Both pairs are from the same nationality: We have three ways to choose the nationality that sits together and $3!$ ways to arrange the people of that nationality in a row. That leaves us with seven objects to arrange, the block of three people from one nationality and the other six people. Hence, we have
$$\binom{3}{1}3!7!$$
such seating arrangements.
Two pairs from different nationalities: We have three ways to choose the nationalities from which the pairs are selected. In each case, we have $\binom{3}{2}$ ways to select two people of that nationality to sit together and $2$ ways to arrange them within the pair. That leaves us with seven objects to arrange, the two pairs and the five other people. Hence, there are
$$\binom{3}{2}^32!^27!$$
such seating arrangements.
Three pairs of adjacent people: Again, we have two cases to consider. Either we have two pairs of one nationality and one pair from a different nationality or three pairs of different nationalities.
Two pairs from one nationality and one pair from a different nationality: We have three ways to select the nationality from which two pairs of adjacent people come, $3!$ ways to arrange the people of that nationality, two ways to choose the nationality of the remaining pair, $\binom{3}{2}$ ways to choose the members of that nationality who sit in adjacent seats, and $2$ ways to arrange those people within the pair. That leaves us with six objects to arrange, the block of three people of one nationality, the pair of another nationality, and the other four people. Hence, there are
$$\binom{3}{1}3!\binom{2}{1}\binom{3}{2}2!6!$$
such seating arrangements.
Three pairs from different nationalities: We must choose two people from each nationality who sit in adjacent seats and arrange the two people within each pair. This leaves us with six objects to arrange, the three pairs and the other three people. Hence, we have
$$\binom{3}{2}^32!^36!$$
such seating arrangements.
Four pairs of adjacent people of the same nationality: Again, we have two cases. Either there are two nationalities with two pairs of adjacent people or there is one nationality with two pairs of adjacent people and one pair of adjacent people from each of the other nationalities.
Two nationalities with two pairs of adjacent people: We choose two of the the three nationalities. Within each such nationality, all three people must be adjacent, so they can be arranged in $3!$ ways within their blocks. That leaves us with five objects to arrange, the two blocks of three people of one nationality and the other three people. Hence, there are
$$\binom{3}{2}3!^25!$$
such seating arrangements.
One nationality with two pairs of adjacent people and two other nationalities with one pair of adjacent people: We have three ways of choosing the nationality from which two pairs of adjacent people are drawn and $3!$ ways of arranging the people of that nationality. For each of the other two nationalities, we have $\binom{3}{2}$ ways to choose two people of the same nationality to sit in adjacent seats and $2$ ways to arrange them within the pair. This leaves us with five objects to arrange, the block of three people of one nationality, the two pairs, and the other two people. Hence, there are
$$\binom{3}{1}3!\binom{3}{2}^22!^25!$$
such seating arrangements.
Five pairs of adjacent people: We must have two nationalities with two pairs of adjacent people and one nationality with one pair of adjacent people. There are $\binom{3}{2}$ ways to choose the nationalities with two pairs of adjacent people. Since the three people of those nationalities must be adjacent, there are $3!$ ways to arrange the people of each of those nationalities. There are $\binom{3}{2}$ ways to choose two people from the third nationality who sit together and $2$ ways to arrange them within the pair. This leaves us with four objects to arrange, the two blocks of three people, the pair, and the remaining individual. Hence, we have
$$\binom{3}{2}3!^2\binom{3}{2}2!4!$$
such seating arrangements.
Six pairs of adjacent people: We must have two pairs of adjacent people from each nationality. Hence, the three people of each nationality must be adjacent. We have $3!$ ways to arrange the nationalities and $3!$ ways to arrange the block of three people within each nationality. Hence, there are
$$3!^4$$
such seating arrangements.
By the Inclusion-Exclusion Principle, there are
$$9! - \binom{3}{1}\binom{3}{2}2!8! + \binom{3}{1}3!7! + \binom{3}{2}^32!^27! - \binom{3}{1}3!\binom{2}{1}\binom{3}{2}2!6! - \binom{3}{2}^32!^36! + \binom{3}{2}3!^25! + \binom{3}{1}3!\binom{3}{2}^22!^25! - \binom{3}{2}3!^2\binom{3}{2}2!4! + 3!^4$$
|
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|
$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$ Find the $x$ :
$$\lfloor \sqrt[3]{|8x|}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200$$
$$x>0 , \to \lfloor \sqrt[3]{8x}\rfloor +\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor =200 \\ \lfloor \sqrt[3]{8x}\rfloor \in \mathbb{Z}+\lfloor \sqrt[3]{\lfloor 8x \rfloor}\rfloor \in \mathbb{Z}=200$$
now what ?
|
We can write
$$
\left\{ \matrix{
0 < x\quad \Rightarrow \quad \left\lfloor {\root 3 \of {8x} } \right\rfloor + \left\lfloor {\root 3 \of {\left\lfloor {8x} \right\rfloor } } \right\rfloor = 200 \hfill \cr
x = - y < 0\quad \Rightarrow \quad \left\lfloor {\root 3 \of {8y} } \right\rfloor + \left\lfloor {\root 3 \of {\left\lfloor { - 8y} \right\rfloor } } \right\rfloor = 200 \hfill \cr} \right.
$$
Now we have a fundamental property of the floor of a function which says
$$
\eqalign{
& \left\{ \matrix{
f(x){\rm continuous}{\rm , monotone (strictly) increasing} \hfill \cr
f(x) = {\rm integer}\quad \Rightarrow \quad x = {\rm integer} \hfill \cr} \right.\quad \Rightarrow \cr
& \Rightarrow \quad \left\{ \matrix{
\left\lfloor {f\left( {\left\lfloor x \right\rfloor } \right)} \right\rfloor = \left\lfloor {f\left( x \right)} \right\rfloor \hfill \cr
\left\lceil {f\left( {\left\lceil x \right\rceil } \right)} \right\rceil = \left\lfloor {f\left( x \right)} \right\rfloor \hfill \cr} \right. \cr}
$$
which can be easily proved considering that
$$
f\left( x \right) = \left\lfloor {f(x)} \right\rfloor + \left\{ {f(x)} \right\} = f\left( {\left\lfloor x \right\rfloor + \left\{ x \right\}} \right)
$$
That it is in particular the case with root function. So, for the case $0<x$
$$
\eqalign{
& 200 = \left\lfloor {\root 3 \of {8x} } \right\rfloor + \left\lfloor {\root 3 \of {\left\lfloor {8x} \right\rfloor } } \right\rfloor = \left\lfloor {\root 3 \of {8x} } \right\rfloor + \left\lfloor {\root 3 \of {8x} } \right\rfloor \cr
& 100 = \left\lfloor {\root 3 \of {8x} } \right\rfloor \cr
& 100 \le 2\root 3 \of x < 101 \cr
& 50^{\,3} \le x < \left( {50.5} \right)^{\,3} \cr}
$$
While the case $x<0$ does not provide any solution, since
$$
\eqalign{
& 200 = \left\lfloor {\root 3 \of {8y} } \right\rfloor + \left\lfloor {\root 3 \of {\left\lfloor { - 8y} \right\rfloor } } \right\rfloor = \cr
& = \left\lfloor {\root 3 \of {8y} } \right\rfloor + \left\lfloor {\root 3 \of { - \left\lceil {8y} \right\rceil } } \right\rfloor = \cr
& = \left\lfloor {\root 3 \of {8y} } \right\rfloor - \left\lceil {\root 3 \of {\left\lceil {8y} \right\rceil } } \right\rceil = \cr
& = \left\lfloor {\root 3 \of {8y} } \right\rfloor - \left\lceil {\root 3 \of {8y} } \right\rceil \cr}
$$
which would require
$$
- 1 \le 200 = \left\lfloor {\root 3 \of {8y} } \right\rfloor - \left\lceil {\root 3 \of {8y} } \right\rceil \le 0
$$
|
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Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls?
1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?
I am getting different answer from this solution manual. My argument is the following:
In order for us to have an arrangement where there are exactly $2$ cells with $3$ balls, we can follow the following procedure. We first decide which $3$ balls we want to put them together among all $7$ balls, and then decide to which cell we want to put them in. This gives us $\binom{7}{3} \binom{7}{1}$. Now, among the left $4$ balls, we choose $3$ balls to put them together, and choose one cell among the remaining $6$ empty cells. This gives us $\binom{4}{3} \binom{6}{1}$. Finally, we are left with one ball, and we have $5$ choices regarding where to put it. Putting them together, we have
$$P(X_3 = 2) = \frac{\binom{7}{3} \binom{7}{1} \binom{4}{3} \binom{6}{1} 5}{7^7}$$
But the solution manual says the answer should be
$$\frac{\binom{7}{2}\binom{7}{3}\binom{4}{3}5}{7^7}$$
Who is wrong and why?
|
The manual is right. The difference between your solution and the books solution is a factor of $2.$ That is ${7\choose1}{6\choose1} = 2{7\choose 2}$
You have placed 3 balls in one of 7 cells and 3 balls in one of 6 remaining. But you have double counted at this step.
|
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|
Turkevicius inequality From Zdravko Cvetkovski's Inequalities — Theorems, Techniques and Selected:
Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that:
$$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$
Please suggest how to prove this inequality using basic methods. No logarithms.
|
Let $a\geq b\geq c\geq d$.
Thus, $$d(d-a)(d-b)(d+2c)+d(d-a)(d+2b)(d-c)+d(d+2a)(d-b)(d-c)\geq0$$ or
$$d^4-(ab+ac+bc)d^2+2abcd\geq0.$$
Thus, it's enough to prove that
$$a^4+b^4+c^4\geq(a^2+b^2+c^2-ab-ac-bc)d^2+a^2b^2+a^2c^2+b^2c^2$$ or
$$\sum_{cyc}(a^2-b^2)^2\geq\sum_{cyc}(a-b)^2d^2$$ or
$$\sum_{cyc}(a-b)^2((a+b)^2-d^2)\geq0,$$
which is obvious.
Here $\sum\limits_{cyc}$ means $\sum\limits_{a\rightarrow b\rightarrow c\rightarrow a}$.
Also, we can use the BW.
Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.
Thus, $u$, $b$ and $w$ are non-negatives and
$$\sum_{cyc}a^4+2abcd-\frac{1}{6}\sum_{sym}a^2b^2=$$
$$\sum_{cyc}(3u^2-2uv)a^2+\sum_{cyc}\left(4u^3-u^2v-u^2w+\frac{2}{3}uvw\right)a+\sum_{cyc}(u^4-u^2v^2)\geq0.$$
Also there are proofs by EV Method and by SOS.
|
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|
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
|
For odd $n$, $(-1)^{n-1}=1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=1^2+(-2^2+3^2)+(-4^2+5^2)+\cdots+(-(n-1)^2+n^2)\\
=1+\color{blue}{(-2+3)}(2+3)+\color{blue}{(-4+5)}(4+5)+\cdots+\color{blue}{(-(n-1)+n)}((n-1)+n)\\
=\color{blue}{1\cdot}\left(1+2+3+4+5+\cdots+(n-1)+n\right)\\
=\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
For even $n$, $(-1)^{n-1}=-1$, so
$$1^2-2^2+3^2-4^2+\cdots+n^2\\
=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+\cdots+((n-1)^2-n^2)\\
=\color{orange}{(1-2)}(1+2)+\color{orange}{(3-4)}(3+4)+\color{orange}{(5-6)}(5+6)+\cdots+\color{orange}{((n-1)-n)}((n-1)+n)\\
=\color{orange}-(1+2+3+4+5+\cdots+(n-1)+n)\\
=-\frac {n(n+1)}2\\
=(-1)^{n-1}\frac {n(n+1)}2$$
Hence, for both odd and even $n$,
$$1^2-2^2+3^2-4^2+\cdots+n^2=(-1)^{n-1}\frac {n(n+1)}2$$
|
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|
Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gives
$$\cos{y}=1-\sin^2{\left(\arcsin{\frac{1}{\sqrt{10}}}\right)}=1-\frac{1}{10}=\frac{9}{10}$$
So I have reduced the problem to the following
$$\sin{x}\cdot \frac{9}{10}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Here I'm stuck. I cant just divide this expression by $(\sin{x}\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\sin(\arctan{2})$ and $\cos{\arctan2}.$ I also tried rewriting $\cos(\arctan{2})$ as $1-\sin^2(\arctan{2}),$ but to no avail.
Any suggestions on
*
*how to proceed from where I left;
*how tocompute this by means more effective;
*both of the above.
|
Hint:
$$\sin x =\sqrt{1-\cos^2 x} =\sqrt{1-\frac{1}{1+\tan^2 x}} $$
$$\cos x =\sqrt{\frac{1}{1+\tan^2 x}} $$
Since
$$ 1+\tan^2 x =1+\frac{\sin x^2}{\cos^2 x} = \frac{\cos^2x+\sin x^2}{\cos^2 x} =\frac{1}{\cos^2 x}$$
|
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|
How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\[5 pt]
+\cdots \qquad\quad \vdots\qquad\qquad\\[5 pt]
+ \left(\frac{1}{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{2016}\right)^2\\
+ \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)\;
\end{align}
I can not find a good way to solve it. Any ideas?
Edit: That is, with no dots, $$S_{2016}=\sum_{k=1}^{2016}\left(\sum_{n=k}^{2016}\frac1n\right)^2+\sum_{k=1}^{2016}\frac1k$$
|
We show that for any positive integer $n$
$$
S_n:=\sum_{j=1}^n\left(\sum_{k=j}^n\frac{1}{k}\right)^2+\sum_{k=1}^n\frac{1}{k}=\sum_{j=1}^{n} (H_n-H_{j-1})^2 +H_n=2n.$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$.
We have that
\begin{align*}
S_n&=nH_n^2-2H_n\sum_{j=1}^{n}H_{j-1}+\sum_{j=1}^{n}H_{j-1}^2+H_n\\
&=
nH_n^2-2H_n((n+1)H_{n}-n-H_n)\\
&\quad+((n+1)\,H_n^2-(2n+1)\,H_n+2n-H_n^2)+H_n\\
&=2n
\end{align*}
where we used
$$\sum_{j=1}^{n}H_j=(n+1)\,H_n-n,\quad \sum_{j=1}^n H_j^2=(n+1)\,H_n^2-(2n+1)\,H_n+2n$$
(see Sum of Squares of Harmonic Numbers).
|
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|
Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?
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The RHS should be $$1^{ 2 }+3^{ 2 }+...+\left( 2n-1 \right) ^{ 2 }+\left( 2n+1 \right) ^{ 2 }=\underset { \left( n\left( 2n-1 \right) \left( 2n+1 \right) \right) /3 }{ \underbrace { 1^{ 2 }+3^{ 2 }+...+(2n-1)^{ 2 } } } +{ \left( 2n+1 \right) }^{ 2 }=\frac { n\left( 2n-1 \right) \left( 2n+1 \right) }{ 3 } +{ \left( 2n+1 \right) }^{ 2 }=\\ =\frac { n\left( 2n-1 \right) \left( 2n+1 \right) +3{ \left( 2n+1 \right) }^{ 2 } }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }-n+6n+3 \right) }{ 3 } =\frac { \left( 2n+1 \right) \left( 2{ n }^{ 2 }+5n+3 \right) }{ 3 } =\frac { \left( n+1 \right) \left( 2n+1 \right) \left( 2n+3 \right) }{ 3 } $$
|
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Prove $3^n$ > $n^4$ if $n\geq 8$ As the title states, I am tasked with proving $3^n$ > $n^4$ if $n\geq 8$
The base case is trivial to prove. It is obvious that $3^8 > 8^4$ since
$3^8 = 6561$ and $8^4 = 4096$, and $6561 > 4096$, thus the base case $n = 8$ is true.
Now for the Induction Hypothesis (IH) we let $n = m$ which then says $3^{m} > m^{4}$
I know that $3^{m+1} > (m+1)^{4}$ is what we are trying to show. So when you expand this you get $$3*3^{m} > m^{4} + 4m^{3} + 6m^{2} + 4m + 1$$ (Binomial Expansion Theorem is how you get the right hand side of the above inequality).
But this is where I'm stuck. I know that $3^{m} > m^{4}$ via the IH but idk what else to say or where to go. Any help would be appreciated. Thanks!
|
Pulling it together into a compact induction proof:
Base case: $3^8 = 9^4>8^4$
Inductive hypothesis: $3^k >k^4$ for some $k\ge 8$
$\begin{align}
\text{Observe }(k+1)^4 &= k^4 + 4k^3 + 6k^2 + 4k + 1 \\
&< k^4 + (4+6+4+1)k^3 \\
&< k^4 + 16k^3 \\
& \le 3k^4
\end{align}$
Then $3^{k+1}=3\cdot3^k>3\cdot k^4> (k+1)^4$ as required
|
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|
better upper bound of Cauchy - Schwarz inequality for 4 variables Given $a,b,c,d,x,y,z,w \geq 0$.
By the Cauchy - Schwarz inequality we have that:
\begin{equation}
\label{1}\tag{1}
\left( ax + by + cz + dw \right) ^{2} \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) .
\end{equation}
Also, by the Cauchy - Schwarz inequality we also obtain:
\begin{equation}
\label{2}\tag{2}
\left( ax + by + cz + dw \right) ^{2} \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} .
\end{equation}
In general it is better to use \eqref{1} or \eqref{2} as I am looking for the tightest estimate as possible. That is, generally we have
\begin{align}
\label{3}\tag{3}
\left( ax + by + cz + dw \right) ^{2} & \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) \\
& \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2}
\end{align}
or
\begin{align}
\label{4}\tag{4}
\left( ax + by + cz + dw \right) ^{2} & \leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2} \\
& \leq \left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right) .
\end{align}
Edit: I forgot one assumption that $d = kc$ for some $k > 0$ and also $x,y,z,w > 0$ so that there is no trivial answer. The question is when will \eqref{3} and \eqref{4} corresponding to the choice of $k$.
|
Both your inequalities are wrong.
$$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right)
\leq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2}
$$ is wrong for $a=y=c=w\rightarrow0^+$.
$$\left( a^{2} + b^{2} + c^{2} + d^{2} \right) \left( x^{2} + y^{2} + z^{2} + w^{2} \right)
\geq 4 a^{2} x^{2} + 4 b^{2} y^{2} + 4 c^{2} z^{2} + 4 d^{2} w^{2}
$$
is wrong for $b=c=d=y=z=w\rightarrow0^+$.
|
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$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a math olympics. I tried to use known Newton identities and other symmetric polynomial results but without success (perhaps a wrong approach). Sorry if it is a duplicate. Hints and answers are always welcomed.
Edit: There is a problem with the original statement of the question in the original source. Under these assumptions it is impossible to have $a, b, c$ with value 2, as spotted in the comments and proved by the answers below
|
Let me try. Note that $2(ab+bc+ca) = a^2+b^2+c^2-(a+b+c)^2 = 2$, then $ab+bc+ca=1$.
We have $$(a-3)^3 + (b-2)^3+(c-2)^3 = 0$$
$$(a-3)^3+(b+c-4)(b^2+c^2+4-2b-2c-bc) = 0$$
$$(a-3)^3 + (-2-a)(6-a^2+4-2(2-a)-(1-a(2-a)))=0$$
$$(a-3)^3-(2+a)(-2a^2+4a+5)=0$$
$$3a^3-9a^2+14a-37=0$$
$$3(a-1)^3+5(a-1)-29=0$$
Solve this equation, you get $a = 2$ is not the root.
If $b=2$, then $a+c=0$, $a^2+c^2 = 2$, so $a=1$, $c=-1$ but it doesn't satisfy the first equation.
Similarly for $c=2$.
|
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|
A simple factoring problem $3x^3 - x^2 -12x + 4$ I'm stuck on a simple factoring problem of
$$3x^3 - x^2 -12x + 4$$
I keep coming up with
$$x^2(3x - 1) -4(3x - 1) = (3x - 1) (x^2 - 4) = 3x^3 -12x - x^2 + 4 =
3x^3 - x^2 - 12x + 4$$
but it doesn't seem to be right, can anyone point out my mistake?
|
From $x^2(3x - 1) -4(3x - 1)$, factor out $(3x-1)$ to get $(x^2-4)(3x-1)$.
Knowing that $a^2-b^2=(a+b)(a-b)$, we get $(x-2)(x+2)(3x-1)$.
|
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|
Prove that there exist no natural number k such that $3^k+5^k$ is a square of an integer number? How to prove that there exist no natural number $k$ such that $3^k+5^k$ is a square of an integer number
|
An elementary solution without modular arithmetic. :)
If $k$ is odd, then write
$$a^2 = 3^k+5^k= 8(3^{k-1}-3^{k-2}5+...+5^{k-1})$$
Since expression in bracket is odd (we have odd odd numbers) we have $8|a^2$ but $16\not|a^2$. A contradiction.
Say $k$ is even. Then $k=2n$ so $(a-3^n)(a+3^n)= 5^{2n}$, so $a-3^n= 5^{x}$ and $a+3^n= 5^{y}$ for some non negative integer $x$ and $y$ (we see that $y>x$ and $x+y=2n$) thus:
$$ 2\cdot 3^n = 5^y-5^x= 5^x(5^{y-x}-1)$$
so $x=0$. So $2\cdot 3^n = 5^{2n}-1$ which is impossible since that left side is smaller then right side for each $n$ (easy to prove with induction).
|
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|
Why does the angle between a Pair of Straight Lines depend only on the Homogeneous Part? The most general form of a quadratic is: $ax^2 + by^2 + 2gx + 2fy + c + 2hxy= 0 $ and that for a homogeneous second degree equation is : $ax^2 + by^2 + 2hxy=0$
I derived the formula $$\tan \theta = \left|\dfrac{2\sqrt{h^2 - ab}}{a+b}\right|$$
but later, author claimed that the same formula is applicable for the general form too.
But I (and he too) derived it for the homogeneous case only. How can he claim this then?
|
For $2^\circ$ homogeneous equation $ax^2+2hxy+by^2=0$, we say $m_1+m_2=-\dfrac{2h}{b}$ where $m_1$ is the slope of first line and $m_2$ is the slope of second line.
Proof:
$$ax^2+2hxy+by^2=0$$
$$a+2h\dfrac{y}{x}+b\dfrac{y^2}{x^2}=0$$
$$bm^2+2hm+a=0$$
$$m_1+m_2=-\dfrac{2h}{b}$$
But if we talk about general $2^\circ$ equation $ax^2+2hxy+by^2+2gx+2fy+c=0$, then also we can say $m_1+m_2=-\dfrac{2h}{b}$
Here is the proof:
$ax^2+2hxy+by^2+2gx+2fy+c=0$ can be written as
$$\left(x+\dfrac{hy-y\sqrt{h^2-ab}}{a}+c_1\right)\left(x+\dfrac{hy+y\sqrt{h^2-ab}}{a}+c_2\right)=0$$
Slope of first line is
$$m_1=\dfrac{-a}{h-\sqrt{h^2-ab}}=-\dfrac{h+\sqrt{h^2+ab}}{b}$$
$$m_2=\dfrac{-a}{h+\sqrt{h^2-ab}}=-\dfrac{h-\sqrt{h^2+ab}}{b}$$
$$m_1+m_2=-\dfrac{2h}{b}$$
|
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|
Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.
|
Hint :$$a=\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}}=b+c \\a^3=b^3+c^3+3bc(b+c)\\
a^3=18+5\sqrt{13}+18-5\sqrt{13}+3\sqrt[3]{18+5\sqrt{13}}.\sqrt[3]{18-5\sqrt{13}}(a)\\a^3=36+3\sqrt[3]{324-325}a\\a^3=36-3a$$solve for a
$$a^3+3a-36=(a-3)(a^2+3a+12)=0 \to a=3$$
|
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|
Limit as $x\to -\infty$ for function having square root
Evaluate
$$\lim_{x\to -\infty} \frac{\sqrt{3x^2+2}}{x-2}.$$
My work: when I solved
$$\lim_{x\to -\infty} \frac{\sqrt{3+2/x^2}}{1-2/x}$$
I got answer $\sqrt{3}$
but if we take $|x|$ then I got $-\sqrt{3}$
what should be answer ?
$\sqrt{3}$ or $-\sqrt{3}$ ?
|
Note that $\sqrt{x^2}=|x|$ and, for $x<0$, $|x|=-x$. Therefore, as $x\to -\infty$,
$$\frac{\sqrt{3x^2+2}}{x-2}=\frac{\sqrt{x^2}\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=\frac{|x|\sqrt{3+\frac{2}{x^2}}}{x(1-\frac{2}{x})}=-\frac{\sqrt{3+\frac{2}{x^2}}}{1-\frac{2}{x}}\to -\sqrt{3}.$$
|
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|
Find the number of solutions of x+y+z=17? Find the number of solutions of $x+y+z=17$ where $2\le x\le 5, 3\le y \le 6, 4\le z\le7$.
My approach:
The number of solutions with the indicated constraints is the coefficient of $x^{17}$ in the expansion of
($x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$
I have changed the above polynomial to
$x^9(1+x+x^2+x^3)^3$
Now $x^{17}=x^9*x^8$
So I must now find the coefficient of $x^8$ in the expansion of ($1+x+x^2+x^3)^3$
$x_1+x_2+x_3=8$
This is equal to $C(3+8-1,8)=45$
However it takes into values of $x,y,z$ greater than $3$.
So I must subtract those combinations where either of $x_1,x_2,x_3$ is greater than $3$.
Let's suppose $x_1\ge 4$.Then
$x_1+x_2+x_3=4$
Solutions=$15$.
Similarly for $x_2,x_3$ we get $15$ solutions each.
Total=$45$.
Now we must consider the case when more than one of $x_1+x_2+x_3\ge 4$.
For this we have $3$ solutions.
Total solutions=$48$.
Now I must subtract these from original solutions of $45$. This gives answer $=-3$.
But the correct answer is $3$.
What is wrong in this approach?
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If $x,y,z \in \mathbb{Z}$ and then the constraints you had; then consider the following case where $x,y,z$ are the most they can be, i.e. $5$, $6$ and $7$:
$$5 + 6 + 7 = 18$$
$18$ is one more than $17$, which means that our solutions will be when only $x$ or only $y$ or only $z$ is one less than their maximum. We have $3$ terms, which means there must be $3$ solutions:
$$(5-1) + 6 + 7 = 17$$
$$5 + (6-1) + 7 = 17$$
$$5 + 6 + (7-1) = 17$$
|
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|
How to get the same answer for integral $\int\frac{dz}{\sqrt{1+z^2}}$ as in textbook I have some answer:
$$\int\frac{dz}{\sqrt{1+z^2}}=\ln ( z + \sqrt {1 + z^2})+C$$
But I can't get the same expression. So here what I've done and got stuck. How can I get the same answer?
\begin{align}
& \int\frac{dz}{\sqrt{1+z^2}}=\left\{z=\tan t, \ dz=\frac{dt}{\cos^2 t} \right\} = \int\frac{dt}{\cos^2 t\sqrt{1+\tan^2 t}} \\[10pt]
= {} & \left\{\sqrt{1+\tan^2 t}=\frac{1}{\cos t}\right\}=\int\frac{\cos t \ dt}{\cos^2 t} \\[10pt]
= {} & \left\{\cos t \ dt= d\sin t, \cos^2 t=1-\sin^2 t\right\}=\int\frac{d\sin t}{1-\sin^2 t} \\[10pt]
= {} &\{\sin t=u\}=\int\frac{d u}{1-u^2}=-\int\frac{du}{u^2-1}=-\frac{1}{2} \int\left(\frac{1}{u-1}-\frac{1}{1+u}\right)\,du \\[10pt]
= {} &-\frac{1}{2}(\ln |u-1|-\ln|u+1|)+C
\end{align}
From here I don't understand how to get the answer above. Need help
|
I'm not sure about how obvious this seems, but you can just make the direct substitution $z=x+\sqrt{1+x^2}$ in$$\int\frac {\mathrm dx}{\sqrt{1+x^2}}$$
Because$$\frac {\mathrm dz}{\mathrm dx}=1+\frac {x}{\sqrt{1+x^2}}=\frac {z}{\sqrt{1+x^2}}$$
This can be noted because the derivative of $\sqrt{1+x^2}$ is $x/\sqrt{1+x^2}$, and if you add one, you get$$1+\frac x{\sqrt{1+x^2}}=\frac {x+\sqrt{1+x^2}}{\sqrt{1+x^2}}$$which is just the derivative of the original function. Therefore, by substitution, we have the integral as$$\begin{align*}\int\frac {\mathrm dx}{\sqrt{1+x^2}} & =\int\frac {\mathrm dz}{\sqrt{1+x^2}}\frac {\mathrm dx}{\mathrm dz}\\ & =\int\frac {\mathrm dz}{\sqrt{1+x^2}}\frac {\sqrt{1+x^2}}{z}\\ & =\log z+C\end{align*}$$
|
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|
Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
-1 \\
-1\\
3\\
\end{pmatrix}$ , $T$ $\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix}$ = $\begin{pmatrix}
4 \\
-3 \\
-1\\
\end{pmatrix}$ then $T$ $\begin{pmatrix}
-5\\
4 \\
4 \\
\end{pmatrix}$ = $\begin{pmatrix}
? \\
? \\
? \\
\end{pmatrix}$
I do not know where to start with this question, I tried doing RREF of the transformation numbers but I feel that is wrong.
|
Define $\vec e_i = (0, \dots, 1, 0, \dots)$ where the 1 is in the $i^\text{th}$ position. Then note that your question is
$$ T(-5\vec e_1 + 4 \vec e_2 + 4 \vec e_3) = -5\pmatrix{3 \\ 1 \\ 4} + 4\pmatrix{-1 \\ -1 \\ 3} + 4\pmatrix{4 \\ -3 \\ -1} = \pmatrix{-3 \\ -21 \\ -12}.$$
|
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|
Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$ and these numbers are roots of $x^3 - 8x + b = 0$ Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$, that they are roots of $x^3 - 8x + b = 0$
I find that $a+b=5x(-x+3) $
Also I tried to solve second equation to get roots, depending on b, but my approach was unsuccessful.
Any help is appreciated!
|
Let $P(x) = x^3-5x^2+7x-a$ and $Q(x) = x^3-8x+b$ and $\gamma, \delta$ be their common roots.
We can decompose them as
$$\begin{cases}
P(x) &= (x-\alpha)M(x),\\
Q(x) &= (x-\beta)M(x)
\end{cases}\quad\text{ where }\quad M(x) = (x-\gamma)(x-\delta)
$$
Subtract them, we get
$$5x^2-15x+b+a = Q(x)-P(x) = (\alpha-\beta)M(x)$$
By comparing the coefficients of $x^2$, we find $\alpha - \beta = 5$ and
$$M(x) = x^2 - 3x + \frac{a+b}{5}$$
Notice
$$(2-\alpha)M(x) = P(x) - (x-2)M(x) = \frac15((b+a-5)x + (3a-2b))$$
If $\alpha \ne 2$, the LHS is a polynomial of degree $2$ while the RHS is a polynomial of degree at most $1$. This is impossible, so $\alpha = 2$ and
$$(b+a-5)x + (3a-2b) = 0
\quad\implies\quad
\begin{cases}
b+a-5 &= 0\\
3a-2b &= 0
\end{cases}
\quad\implies\quad
(a,b) = (2,3)
$$
|
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|
An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.
|
Reduce the rational degrees with successive substitutions as follows
$$\begin{align}
\int_{-\infty}^{\infty}\frac{(x^{4}+1)^{2}}{x^{12}+1}dx=&\ 2\int_{0}^{\infty}\frac{x^{4}+1}{x^{8}-x^{4}+1}\overset{x\to\frac1x}{dx}=\int_{0}^{\infty}\frac{(x^{4}+1)(x^2+1)}{x^{8}-x^{4}+1}
\overset{ x-\frac1x\to x}{ dx} \\
=& \ 2\int_0^\infty \frac{x^2+2}{x^4+4x^2+1} \overset{x\to\frac1x}{dx}= 3\int_{0}^{\infty}\frac{x^2+1}{x^4+4x^2+1}
\overset{x-\frac1x\to x}{ dx} \\
=&\ 6\int_0^\infty \frac1{x^2+6}dx
=\sqrt{\frac32}\pi
\end{align}$$
|
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|
What is the minimum value of? On positive reals if $3x+4y+7z=1$ what is the minimum value of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}?$ I have tried using the arithmetic mean and harmonic mean inequality but I failed. Not good at inequalities though. Please help.
|
From $3x+4y+7z=1$ I got $z= \frac{1}{7} (-3 x-4 y+1)$ and plugged in
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
gives $$f(x,y)=\frac{7}{1-3 x-4 y}+\frac{1}{x}+\frac{1}{y}$$
$$f'_x=\frac{21}{(1-3 x-4 y)^2}-\frac{1}{x^2};\;f'_y=\frac{28}{(1-3 x-4 y)^2}-\frac{1}{y^2}$$
$f'_x=0$ gives $21x^2=(1-3x-4y)^2$ and $f_y=0\to 28y^2=(1-3x-4y)^2$
so we have $21x^2=28y^2$ which for positive reals means $y=\frac{\sqrt 3}{2}\,x$
furthermore we have from the first equation
$x\sqrt{21}=1-3x-4y$ which after substituting becomes
$x\sqrt{21}=1-3x-2\sqrt{ 3}x$
$$x=\frac{1}{3+2 \sqrt{3}+\sqrt{21}};\;y=\frac{\sqrt{3}}{2 \left(3+2 \sqrt{3}+\sqrt{21}\right)};\;z=\frac{1}{7+2 \sqrt{7}+\sqrt{21}}$$
So minimum of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is
$2 \left(7+2 \sqrt{3}+2 \sqrt{7}+\sqrt{21}\right)\approx 40.676$
|
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|
Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}=-3$$
$$\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}=-3\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}$$
$$\frac{x}{x-\sqrt{x^2+3x}}=\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
To prepare the product $\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$ for solution by l'Hopital's rule, write it as $\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
Is it correct to use L'Hopital here like Wolfram did?
|
$$\lim_{x\rightarrow-\infty}\left(\sqrt{x^2+3x}+x\right)$$
Multiply by the conjugate
\begin{aligned}
\sqrt{x^2+3x}+x &= \left(\sqrt{x^2+3x}+x\right) \cdot \left(\frac{\sqrt{x^2+3x}-x}{\sqrt{x^2+3x}-x}\right)\\
&= \frac{\left(\sqrt{x^2+3x}\right)^2-x^2}{\sqrt{x^2+3x}-x}\\
&= \frac{x^2+3x-x^2}{\sqrt{x^2+3x}-x}\\
&=\frac{3x}{\sqrt{x^2+3x}-x}
\end{aligned}
Also
\begin{aligned}
\lim_{x\rightarrow-\infty}\left(\frac{3x}{\sqrt{x^2+3x}-x}\right)&= 3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2+3x}-x}\right)\\
&=3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2\left(1+\frac{3}{x}\right)}-x}\right)\\
&=3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}-x}\right)
\end{aligned}
Let $x \to -\infty \implies \sqrt{x^2} = -x$
\begin{aligned}
3\cdot\lim_{x\rightarrow-\infty}\left(\frac{x}{-x\sqrt{1+\frac{3}{x}}-x}\right)
\end{aligned}
Divide by $x$
\begin{aligned}
3\cdot\lim_{x\rightarrow-\infty}\left(\frac{\frac{x}{x}}{-\frac{x\sqrt{1+\frac{3}{x}}}{x}-\frac{x}{x}}\right) = 3\cdot\lim_{x\rightarrow-\infty}\left( \frac{1}{-\sqrt{1+\frac{3}{x}}-1}\right)
\end{aligned}
\begin{aligned}
3\cdot \frac{\lim _{x\to -\infty }\left(1\right)}{\lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}-1\right)} =3\cdot \frac{1}{-2} = -\frac{3}{2}
\end{aligned}
So
$$\lim_{x\rightarrow-\infty}\left(\sqrt{x^2+3x}+x\right) = -\frac{3}{2}$$
Note:
\begin{aligned}
\lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}-1\right) = \lim _{x\to -\infty }\left(-\sqrt{1+\frac{3}{x}}\right)-\lim _{x\to -\infty }\left(1\right)
= 1 - (-1) =2
\end{aligned}
|
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|
Solve $32x^2 -y^2 = 448$ I am trying to find all integer solutions to the following equation:
$$32x^2 - y^2 = 448$$
This is what I have tried so far:
The equation describes a hyperbola, and so I try the usual trick of intersecting the curve with a line of rational slope to find rational solutions first.
Knowing the point (4,8) satisfies the equation, I solve the following system:
$$\left\{
\begin{array}{c}
32x^2 - y^2 = 448 \\
y = m(x - 4) + 8 \\
\end{array}
\right.$$
After a bunch of algebra, I get:
$$x = \frac{-4m^2+16m-128}{32-m^2}$$
$$y = \frac{8m^2-256m+256}{32-m^2}$$
Finally, substituting $m = \frac{u}{v}$, I get:
$$x = \frac{-4u^2+16uv-128v^2}{32v^2-u^2}$$
$$y = \frac{8u^2-256uv+256v^2}{32v^2-u^2}$$
Cool, with any choice of $u$ and $v$, I get a rational solution.
But since cancelling the denominators does not work, I do not know how to continue to get integer solutions only.
Is this perhaps not the right way to go?
Any help would be much appreciated.
|
Note that you can simplify and substitute in the following way:
$$\begin{array}{lll}
y^2 = 2^5(x^2-14)&&\text{substitute } y = 2^2z \\
z^2 = 2(x^2-14)&&\text{substitute } z = 2a\\
2a^2 = x^2-14&& \\
2(a^2+7) = x^2&&\text{substitute } x = 2b\\
a^2+7 = 2b^2&&\text{substitute } a = 2c+1\\
2(c^2+c + 2) = b^2&&\text{substitute } b = 2d\\
2d^2 = c^2 + c + 2&&\\
2(d-1)(d+1) = c(c+1)&&\\
\end{array}$$
so that $y = 16c+8$ and $x=4d$. This might work better for your approach.
|
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|
Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm looking at thow to solve for the actual length. It looks like a very complex integral so I'm assuming I made a mistep or theres some easy reduction I can make.
After determining the area of an incredibly small section of the function:
$$ds = \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}$$
$$\frac{dx}{dy} = \frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2} $$
$$ds = \sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}$$
This leaves me with the integral
$$\int{\sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}} dy$$
I do still have to calculate from 1 - 25 but I like to plug in after I solve my integral. Anyway, I can't tell how to solve this, but I have a feeling I need to play with the squared term. Perhaps a substitution or maybe the reciprocals simplify into something. If anyone has any tips I appreciate it!
EDIT: Problem solved! (I think)
In the answer to the question I can make the simplification
$$\int{\sqrt{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)^2}} dy$$
$$\int{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)} dy$$
$$\frac{1}{2}\left(\int{y^{\frac{5}{2}}}dy + \int{y^{\frac{-5}{2}}}dy\right)$$
$$\frac{y^{\frac{7}{2}}}{7} + \frac{1}{-3y^{\frac{3}{2}}}$$
We plug in our bounds here and the answer is $F(25) - F(1)$
|
note that $$\left(\frac{-1}{2}y^{-5/2}+\frac{1}{2}y^{5/2}\right)^2+1=1/4\,{\frac { \left( y+1 \right) ^{2} \left( {y}^{4}-{y}^{3}+{y}^{2}-y
+1 \right) ^{2}}{{y}^{5}}}
$$
|
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|
find the equation of the cone $z=\sqrt{x^2+y^2}$ in spherical coordinates I have the following...
$$z=\sqrt{x^2+y^2}$$
I need to write this as an equation in spherical coordinates.
I know that $p^2 = x^2+y^2+z^2$
and that...
$$x = p\sin\phi \cos\theta$$
$$y=p\sin\phi \sin\theta$$
$$z = p\cos\theta$$
The answer is $\phi = \pi/4$
How do you get t0 this answer? I tried doing
$$p^2-z^2 = x^2+y^2$$
$$z^2 = p^2-z^2$$
$$z = p/2$$
But I really do not know what I am doing. How do I solve this?
|
The idea is to plug in the values of $x$, $y$ and $z$ in $$z = \sqrt{x^2+y^2}.$$
Specifically, by using the given expressions, we get
$$p \cos \phi = \sqrt{p^2\sin^2\phi \cos^2 \theta + p^2\sin^2\theta \sin^2 \phi}$$ $$p \cos\phi = \sqrt{p^2\sin^2 \phi \ (\sin^2 \theta + \cos^2 \theta)} $$
$$p \cos\phi = p \sin \phi$$
$$\cos \phi = \sin \phi$$
$$\phi = \pi/4.$$
|
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|
A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt{\cos2y}}\frac{dy}{\sqrt{cosy}} $$
put $$x=tany$$
But how to calculate this integral?
|
Try $x = \sin u$, so that $dx = \cos(u)\, du$.
Then use the fact that: $$\sqrt{1+\sqrt{1-\sin^2 u}} = \sqrt{1+\cos u} =\sqrt{2} |\cos\tfrac{u}{2}|$$
The integral converts to:
$$\begin{align}
I &= \int_{0}^{\tfrac{\pi}{2}} \frac{\sqrt{2} \cos\tfrac{u}{2} \, \cos u}{1+\sin^2 u} du \\
\end{align}$$
|
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|
Does $A,B>0$ imply that the projection onto the positive subspace of $A-B$ is smaller than $A+B$ for positive semidefinite matrices? I am trying to find a counterexample for the following matrix inequality which I suspect not to hold:
$$P^+(A-B)\leq A+B$$
where $A,B$ are positive semidefinite matrices and $P^+(X)$ the projector onto the positive subspace of $X$.
|
Let $A=\frac{1}{10}\begin{bmatrix} 2&4\\4&8\end{bmatrix},$ $B=\frac{1}{10}\begin{bmatrix}9&-3\\-3&1\end{bmatrix}.$ Then $A-B=\frac{1}{10}\begin{bmatrix}-7&7\\7&7\end{bmatrix},$ which has eigenvalues $\pm \sqrt{\frac{49}{50}},$ with corresponding eigenvectors $\begin{bmatrix}1\\1\pm\sqrt{2}\end{bmatrix}.$ Thus $P^{+}(A-B)=\frac{\sqrt{49/50}}{6+2\sqrt{2}}\begin{bmatrix}1&1+\sqrt{2}\\1+\sqrt{2}&5+2\sqrt{2}\end{bmatrix}.$ Letting $x=\frac{1}{\sqrt{6+2\sqrt{2}}}\begin{bmatrix}1\\1+\sqrt{2}\end{bmatrix},$ we see that $x^{*}P^{+}(A-B)x=\sqrt{\frac{49}{50}}\approx 0.9899,$ while $x^{*}(A+B)x=\frac{4+2\sqrt{2}}{6+2\sqrt{2}}\approx 0.7735.$ Therefore, $P^{+}(A-B)\not\leq A+B$ in this case.
Admittedly, $A$ and $B$ were both positive semidefinite here, but a slight modification yields a counterexample with $A'$ and $B'$ positive definite. Letting $\varepsilon>0,$ observe that $A+\varepsilon I,$ $B+\varepsilon I$ are positive definite, $(A+\varepsilon I)-(B+\varepsilon I)=A-B,$ and $x^{*}(A+B+2\varepsilon I)x=x^{*}(A+B)x+2\varepsilon,$ so for $\varepsilon<0.1,$ we still have $x^{*}(A'+B')x<x^{*}P^{+}(A'-B')x.$
Several factors are important to making the example above work. They both have one large eigenvalue and one $0$ (or close to $0$) eigenvalue, and the eigenvectors corresponding to the large eigenvalues are nearly orthogonal ($\frac{\langle [1,2],[3,-1]\rangle}{\sqrt{5}\sqrt{10}}=\frac{1}{5\sqrt{2}}$). Since $A$ and $B$ are roughly the same size, this means that the eigenvectors of $A-B$ are changed essentially as much as possible from the large eigenvector of either $A$ or $B$. This means that the eigenvector of $A-B$ corresponding to the positive eigenvalue is relatively far from either of the large eigenvectors of $A$ or $B$, which means that when we compute $x^{*}P^{+}(A-B)x$ for the largest eigenvector $x$, scaled to unit length, we obtain the largest eigenvalue of $P^{+}(A-B),$ but when we compute $x^{*}Ax$ (or $x^{*}Bx$), we are getting an average of the largest eigenvalue, $1$, and the smallest eigenvalue, $0$ (or $\varepsilon,$ if we consider the second example). It then turns out that the sum of these two ends up being less than the positive eigenvalue of $A-B$ (here is where a bit of tinkering might be necessary to make sure this happens as desired.
In higher dimensions, I would suggest considering matrices that are again of the form $uu^{T}+\varepsilon I$ for a unit vector $u,$ and with the vectors $u$ close to being orthogonal (note that if they are orthogonal, the matrices $A$ and $B$ will be simultaneously diagonalizable, and the counterexample will be impossible). Then $A-B=uu^{T}-vv^{T}$ will have one positive and one negative eigenvalue, and the positive unit eigenvector should satisfy $x^{*}P^{+}(A-B)x>x^{*}(A+B)x$ when $\varepsilon$ is sufficiently small (this obviously requires proof).
|
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}$, or is that not enough?
|
if $\sqrt{a+b} = \sqrt a + \sqrt b$ for some $a,b \in \mathbb{R^{\geq 0}}$, then $a+b = a + 2\sqrt{a}\sqrt{b} + b$, then $\sqrt{a}\sqrt{b} = 0$. So $a =0$ or $b=0$.
|
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|
Evaluate The Arc Length
Let $$\alpha(t)=(\alpha \cos^3t,\alpha \sin^3 t)$$ Evaluate the arc length when $\alpha>0$ and $t\in[0,2\pi]$
$$\alpha'(t)=(-3\alpha \cos^2{t} \sin{t} ,3\alpha \sin^2t\ \cos t)$$
$$||\alpha'(t)||=\sqrt{(-3\alpha \cos^2{t} \sin{t})^2+(3\alpha \sin^2t\ \cos t)^2}=\sqrt{9\alpha^2\cos^4{t} \sin^2{t}+9\alpha^2 \sin^4t\ cos^2t}=\sqrt{9\alpha^2\cos^2t\sin^2t(\cos^2t+\sin^2t)}=\sqrt{9\alpha^2\cos^4{t} \sin^2{t}+9\alpha^2 \sin^4t\cos^2t}=\sqrt{9\alpha^2\cos^2t\sin^2t}=3\alpha \cos t \sin t$$
So we have:
$$\int_{0}^{2\pi}||\alpha'(t)||dt=\int_{0}^{2\pi}(3\alpha \cos t \sin t)\,dt=3\alpha \int_{0}^{2\pi} \cos t \sin t\, dt$$
$u=\sin t$ and $du=\cos t\, dt$
$$3\alpha \int_{0}^{2\pi} udu=3\alpha\frac{u^2}{2}|_{0}^{2\pi}=\frac{3}{2}\alpha{\sin t^2}|_{0}^{2\pi}=\frac{3}{2}(0-0)=0$$
But the answer is $6\alpha$ where did I get it wrong?
|
Here another way to do this, this time in the complex plane. Let the astroid be given by
$$z=a(\cos^2 t+i~b\sin^3 t),\quad t\in[0,2\pi]$$
The arc length is given by
$$
\begin{align}
s
&=\int_0^{2\pi}|\dot z|~dt\\
&=4\int_0^{\pi/2}|\dot z|~dt,\quad \text{by symmetry}\\
&=4a\int_0^{\pi/2}|-3\cos^2 t~\sin t+3\sin^2 t~\cos t|~dt\\
&=4a\int_0^{\pi/2}\sqrt{9\cos^4 t~\sin^2 t+9\sin^4 t~\cos^2 t}~dt\\
&=4a\int_0^{\pi/2}\sqrt{\cos^2 t~\sin^2 t}~dt\\
&=12a\int_0^{\pi/2} \sin t~d(\sin t)\\
&=6a \sin^2 t\biggr|_0^{\pi/2}\\
&=6a
\end{align}
$$
|
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|
not understanding identity theorem for polynomials
Suppose $f(x)$ is a polynomial such that $f(x) = a_nx^n +
a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots +a_1 x + a_0$ and there are
at least $n+1$ different values of $x$ for which $f(x) =0$.
(a) If $a_n\neq 0$, then what does the Fundamental Theorem of Algebra
tell us about the number of different values of $x$ for which $f(x) =
0$?
(b) Why must we have $f(x) = 0$ for all values of $x$?
It is with the answer to question (b) that I have the most difficulty. The answer to (a) and (b) is:
If any of the $a_i$ are nonzero, then $\deg f \le n$. Therefore, by
the Fundamental Theorem of Algebra, if any of the $a_i$ are nonzero,
then $f(x)$ cannot have more than $n$ roots. However, we are told that
there are at least $n+1$ values of $x$ for which $f(x) = 0$. So, it is
impossible for any of the $a_i$ to be nonzero (since then the
Fundamental Theorem of Algebra would force $f(x)$ to have no more than
$n$ roots). Therefore, all the $a_i$ must be 0, which means $f(x) = 0$
for all values of $x$. □
I'm having difficulty understanding what is meant with any of the $a_i$ are nonzero. What does this mean? That none of them are 0?
As far as I'm understanding, $f(x) = 0$, like, literally.
Then, there is the following, related bit:
Suppose that $f(x)$ is a quadratic polynomial such that $f(2) = 4$,
$f(3) = 9$, and $f(4) = 16$. Prove that $f(x)=x^2$.
By inspection, we note that $f(x) = x^2$ satisfies $f(2) = 4$, $f(3) =
9$, and $f(4) = 16$. However, maybe there are other quadratics $f(x)$
such that $f(2) = 4$, $f(3) = 9$, and $f(4) = 16$. How can we tell if
$f(x)= x^2$ is the only possible one?
We let $g(x) = f(x) - x^2$ and focus on $g(x)$, because we know that
2, 3, and 4 are all roots of $g(x)$. Therefore, the Factor Theorem
tells us that $x-2$, $x-3$, and $x-4$ are all factors of $g(x)$. So,
we have $$ g(x) = (x - 2)(x - 3)(x - 4)q(x), $$for some polynomial
$q(x)$. This tells us that either the degree of $g(x)$ is at least 3,
or $q(x)=g(x)= 0$ for all values of $x$. However, because $g(x) = f(x)
-x^2$ and $f(x)$ is a quadratic, we know that the degree of $g(x)$ cannot be larger than 2. Therefore, we must have $q(x) = g(x)= 0$ for
all $x$. Finally, since $f(x) = g(x) + x^2$, we know that $f(x) = x^2$
for all $x$.
But why is $g(x) = f(x) - x^2$? I mean, I know why, but why not any other polynomial? In other words, I'm not convinced; I'm still thinking that there might be some polynomial out there which satisfies the conditions.
For example:
$f(1) = 1$, $f(2) = 8$, $f(3) = 27$, and $f(4) = 64$.
This looks like it might be $x^3$, but actually, it is $f(x) = x^3 + (x - 1)(x - 2)(x - 3)(x - 4) \left( 16,\!000,\!000x^{427} 473.15x^{101} - \frac{\pi}{\sqrt{2}} x^{23} + x^5 + 99 \right)$
I could have gone and done:
$g(x) = f(x) - x^3$ with $f(x) = x^3$. Then, $g(x) = (x - 1)(x - 2)(x - 3)(x - 4)q(x)$. Since the degree of $g(x)$ cannot be larger than 3, we must have that $q(x) = g(x) = 0$. Finally, we have $f(x) = g(x) + x^3$, and we know that $g(x) = 0$, and so $f(x) = x^3$.
But apparently it doesn't proof anything because there's a polynomial out there that produces the same result! So what did I proof?
|
A nonzero polynomial of degree $d$ has at most $d$ roots. And the expression
$$\sum_{k=0}^n a_kx^k$$
is a polynomial of degree at most $n$. Hence if it has $n+1$ roots, it must be identically $0$.
Then if there are two polynomials of degree at most $n$, let $p(x)$ and $q(x)$, achieving the same value at $n+1$ points, by the above theorem, the polynomial $p(x)-q(x)$ has $n+1$ roots and so it must be identically $0$. In other words, $p(x)=q(x)$.
|
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|
Minimizing $\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$ for nonnegative parameters with $a+b+c+d=4$
Minimize
$$\frac{d}{a^3+4}+\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}$$
where $a$, $b$, $c$, $d$ are nonnegative and $a+b+c+d = 4$.
I know the minimum is $2/3$ when, say, $a=b=2$. I found this using calculus, which is very computational.
Does anybody have an algebraic solution. I am unable to find one. Thank you!
Source: https://brilliant.org/problems/volcanic-inequality/?ref_id=1413038
|
For $(a,b,c,d)=(2,2,0,0)$ we get a value $\frac{2}{3}$.
We'll prove that it's a minimal value.
Indeed, by AM-GM
$$\sum_{cyc}\frac{a}{b^3+4}=\frac{1}{4}\sum_{cyc}\left(a-\frac{ab^3}{b^3+4}\right)=\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{2b^3+8}\right)\geq$$
$$\geq\frac{1}{4}\sum_{cyc}\left(a-\frac{2ab^3}{3\sqrt[3]{b^6\cdot8}}\right)=\frac{1}{12}\sum_{cyc}(3a-ab).$$
Thus, it remains to prove that
$$\sum_{cyc}(3a-ab)\geq8$$ or
$$a+b+c+d\geq ab+bc+cd+da$$ or
$$(a+b+c+d)^2\geq4(a+c)(b+d),$$
which is AM-GM again.
Done!
|
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|
How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?
|
Using $ \sin^2 \alpha + \cos^2 \alpha =1$
\begin{eqnarray*}
\sin^4 \alpha + \cos^4 \alpha =(\sin^2 \alpha + \cos^2 \alpha)^2 -2\sin^2 \alpha \cos^2 \alpha = 1-2\sin^2 \alpha \cos^2 \alpha \\
\sin^6 \alpha + \cos^6 \alpha =(\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha -\sin^2 \alpha \cos^2 \alpha+ \cos^4 \alpha) = 1-3\sin^2 \alpha \cos^2 \alpha. \\
\end{eqnarray*}
So your equation can be simplified to
\begin{eqnarray*}
R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha ) \\
=1-3\sin^2 \alpha \cos^2 \alpha +K(1-2\sin^2 \alpha \cos^2 \alpha) \\
=1+K-(2K+3)\sin^2 \alpha \cos^2 \alpha \\
\end{eqnarray*}
So it is independent of $\alpha$ when $\color{blue}{2K+3=0}$. (Giving the value $K=\color{red}{-\frac{3}{2}}$)
|
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|
Prove identity in a triangle I want to show that if $ABC$ is a triangle then
$$\sin^2(A/2)+ \sin^2(B/2) + \sin^2(C/2) =1-2\sin(A/2) \sin(B/2) \sin(C/2)$$
Well I eventually got it after much algebra, but I am looking for a shorter solution, or maybe even a geometric one?
|
As $\dfrac A2+\dfrac B2=\dfrac\pi2-\dfrac C2,\sin\dfrac C2=\cos\dfrac{A+B}2$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2$$
$$=1-\left(\cos^2\dfrac A2-\sin^2\dfrac B2\right)+\sin^2\dfrac C2$$
$$=1-\cos\dfrac{A+B}2\cos\dfrac{A-B}2+\cos^2\dfrac{A+B}2$$
$$=1-\cos\dfrac{A+B}2\left(\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right)$$
Can you take it from here?
|
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|
Divide $(x+1)^2$ by $x-1$ without using longdivision. Dividing a polynomial of degree $n$ with a polynomial of degree $n-1$ gives a polynomial of degree $1$. So,
$$\frac{(x+1)^2}{x-1}=ax+b\Leftrightarrow x^2+2x+1=ax^2+(b-a)x-b$$
Gives $a=1, \quad a-b=2, \quad -b=1$. So $a=1$ and $b=-1.$ The result I get is that $$\frac{(x+1)^2}{x-1}=x-1.$$ Which is far from the correct answer. I can't spot my mistake. I feel that the more I sit and study, the worse at math I become.
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Write $(x+1)^2 = (x-1)(ax+b)+c$ and plug in $x=1$, you get $4=0+c$ so $c=4$. Now
$$ (x+1)^2-4 = x^2+2x-3 = (x-1)(x+3)$$ so $a= 1$ and $b=3$.
|
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|
Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would be appreciated.
|
It's $$5(x^3-1)^2=(\sqrt3(2x^3-x)-1)^2$$ and from here
$$(5(x^3-1)^2-3(2x^3-x)^2-1)^2=12(2x^3-x)^2,$$
which is
$$(7x^6-12x^4+10x^3+3x^2-4)^2-12x^2(2x^2-1)^2=0$$ or
$$49x^{12}-168x^{10}+140x^9+186x^8-240x^7-76x^6+60x^5+153x^4-80x^3-36x^2+16=0.$$
|
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|
How can I find the points of intersection of $f(x)=\sin(\sqrt{x^2+4})$ and $g(x)=4-x^2$ analytically How can I find the points of intersection of $f(x)=\sin(\sqrt{x^2+4})$ and $g(x)=4-x^2$ analytically without having to resort to a graph? Thank you very much
I know that $-1\leq\sin(\sqrt{x^2+4})\leq 1$, with which $-1\leq4-x^2\leq 1$ and $-1\leq (2-x)(2+x)\leq 1$, but I do not know what else to do, could you help me?
|
Just to add a few things after Jam's answer.
Sooner or later, you will learn than, better than with Taylor expansions, functions can be approximated using Padé approximants which, built around $u=u_0$, are in the form of $$f(u)=\frac{\sum_{i=0}^m a_i (u-u_0)^i } {1+\sum_{i=1}^n b_i (u-u_0)^i}$$ Using, for simplicity, $m=1$, this gives as an approximate solution $$u_{(n)}=u_0-\frac{a_0^{(n)}}{a_1^{(n)}}$$ (remember that $a_0$ and $a_1$ depend on the degree $n$ used for the denominator).
Applying the method to $f(u)=\sin(u)+u^2-8$ with $u_0=\pi$, we could obtain
$$\left(
\begin{array}{ccc}
n & u_{(n)} & \text{approx} \\
0 & \frac{8-\pi +\pi ^2}{2 \pi -1} & 2.78771 \\
1 & \frac{-8+25 \pi -3 \pi ^2+\pi ^3}{9-4 \pi +3 \pi ^2} & 2.76231 \\
2 & \frac{432-230 \pi +318 \pi ^2-52 \pi ^3+6 \pi ^4+\pi ^5}{-38+228 \pi -76 \pi
^2+24 \pi ^3+\pi ^4} & 2.76208 \\
3 & \frac{-1824+13380 \pi -5628 \pi ^2+3612 \pi ^3-600 \pi ^4+24 \pi ^5+18 \pi
^6}{2436-2208 \pi +3828 \pi ^2-1104 \pi ^3+168 \pi ^4+24 \pi ^5} & 2.76218
\end{array}
\right)$$
|
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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.