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Test the convergence of $\sum_{n=1}^{n=\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$. Test the convergence of
$$
\sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)
$$
$$
\frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
|
Note if the two series $\displaystyle \sum_{n=1}^\infty u_n$ and $\displaystyle \sum_{n=1}^\infty v_n$ are convergent then the series $\displaystyle \sum_{n=1}^\infty (u_n+v_n)$ is also convergent.
Now, we know that the Riemann series $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2} $ is convergent and we have
$$\frac{n^2}{2^n}=_\infty o(\frac{1}{n^2})$$
hence the series $\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n}$ is also convergent by comparison with the convergent Riemann series.
|
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"url": "https://math.stackexchange.com/questions/359294",
"timestamp": "2023-03-29T00:00:00",
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|
Minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$ Is there an "easy" way to find the minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$ without the help of any computer programme?
If I knew $\sin(\pi/8)=\frac{\sqrt{2-\sqrt2}}{2}$ then it would obviously be easy to find it, but how would i evaluate $\sin(\pi/8)$ in the first place?
And $\cos(\pi/9)$ does not seem to have a nice alternate form. So I see no point to start.
|
This is an old question, but it doesn't seem to have a correct answer with full details.
Let $\omega = e^{i\pi/9}$ and $\zeta = e^{i\pi/8}$, so $\cos(\pi/9) = \mathrm{Re}(\omega)$ and $\sin(\pi/8) = \mathrm{Im}(\zeta)$, and note that $\omega$ is a primitive 18th root of unity and $\zeta$ is a primitive 16th root of unity. The conjugates of $\cos(\pi/9)$ must be the real parts of the other primitive 18th roots of unity, namely $\cos(5\pi/9)$ and $\cos(7\pi/9)$, and therefore $\cos(\pi/9)$ has degree three. Similarly $\sin(\pi/8)$ has degree $4$, with conjugates $-{\sin(\pi/8)}$ and $\pm{\sin(3\pi/8)}$.
To find the minimum polynomial for $\cos(\pi/9)$, we start by finding the minimum polynomial for $\omega=e^{i\pi/9}$. The roots of this polynomial must be the primitive 18th roots of unity, and with a little thought we can see that the desired cyclotomic polynomial is
$$
\frac{(x^{18}-1)(x^3-1)}{(x^9-1)(x^6-1)} \;=\; \frac{x^9+1}{x^3+1} \;=\; x^6 - x^3 + 1.
$$
Thus $\omega^6 - \omega^3 + 1 = 0$. Dividing through by $\omega^3$ gives the relation
$$
\omega^3 + \omega^{-3} - 1 \;=\; 0.
$$
Let $\alpha = 2\cos(\pi/9)$. Then
$$
\alpha \;=\; \omega + \omega^{-1}
\qquad\text{and}\qquad
\alpha^3 \;=\; \omega^3 + 3\omega + 3\omega^{-1} + \omega^{-3}
$$
so
$$
\alpha^3 - 3\alpha - 1 \;=\; 0.
$$
Then the minimum polynomial for $\cos(\pi/9)$ is $(2x)^3 - 3(2x) - 1$, or equivalently
$$
\fbox{$x^3 \,-\, \frac{3}{4} x \,-\, \frac{1}{8}$}.
$$
We can do something similar for $\sin(\pi/8)$. First, since $\zeta = e^{i\pi/8}$ is a primitive 16th root of unity, the minimum polynomial for $\zeta$ is
$$
\frac{x^{16}-1}{x^8-1} \;=\; x^8 + 1.
$$
Thus $\zeta^8 + 1 = 0$. Dividing through by $\zeta^4$ gives the equation
$$
\zeta^4 + \zeta^{-4} \;=\; 0.
$$
Let $\beta = 2i\sin(\pi/8)$. Then $\beta = \zeta - \zeta^{-1}$, so
$$
\beta^2 \;=\; \zeta^2 - 2 + \zeta^{-2}
\qquad\text{and}\qquad
\beta^4 \;=\; \zeta^4 - 4\zeta^2 + 6 - 4\zeta^{-2} + \zeta^{-4}.
$$
Then
$$
\beta^4 + 4\beta^2 + 2 \;=\; 0
$$
so the minimum polynomial for $\sin(\pi/8)$ is $(2ix)^4 + 4(2ix)^2 + 2$, or equivalently
$$
\fbox{$x^4 - x^2 + \frac{1}{8}$}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\circ}}<2$,then $a=1$
then $$\dfrac{b^2}{16}(1-\cos{(2c)})+\dfrac{b}{4}\sin{c}=1-\sin{50^{0}}$$
so $b=4$
then we have $\sin{c}-\cos{(2c)}=-\sin{50^{0}}$
then my question: How can prove this $c$ must equality 10?
Thank you everyone: yesterday,when I go to bed, I have consider this:let $f(c)=\sin{c}-\cos{(2c)}$.then we have $f(10)=\sin{10}-\cos{20}=\cos{80}-\cos{20}=2\sin{\dfrac{80-20}{2}}\sin{\dfrac{80+20}{2}}=\sin{50}$, by other hand, we have $f'(c)=\cos{c}+2\sin{2c}>0,0<c<\dfrac{\pi}{2}$,so if we $f(c)=f(10)$,we must $c=10$
|
We have the following (working in degrees):
$$\cos 20 - \cos 80 = \cos(50-30) - \cos(50+30) = 2 \sin 50 \sin 30 = \sin 50$$
Thus we have that
$$1 - 2\sin^2 10 - \sin 10 = \sin 50$$
(using $\cos 20 = 1 - 2 \sin^2 10$ and $\cos 80 = \sin (90 - 80) = \sin 10$)
And so
$$9 - 8 \sin 50 = 9 - 8(1 - 2\sin^2 10 - \sin 10) = 1 + 8\sin 10 + 16\sin^2 10 = (1 + 4 \sin 10)^2$$
Thus $a=1, b=4, c=10$.
|
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|
Find radius and interval of convergence for $\sum_{n=1}^\infty$ $\frac{5^n}{n+2^n}x^n$ Find radius and interval of convergence for $\sum_{n=1}^\infty$ $\frac{5^n}{n+2^n}x^n$
So if i apply ratio test, I get $\lim_{x\to \infty} 5|x||\frac{n+2^n}{n+1+2^{n+1}}| $
Now need help to to check endpoints x= 2/5 and -2/5
If i put in x= 2/5 i get $\sum_{n=1}^\infty$ $\frac{2^n}{n+2^n}$ How do i check if this converges or not? Same with x = -2/5 $\sum_{n=1}^\infty$ $\frac{(-2)^n}{n+2^n}$
|
By ratio test
$$ 5|x||\frac{n+2^n}{n+1+2^{n+1}}|\sim_\infty 5|x||\frac{2^n}{2^{n+1}}|=\frac{5}{2}|x|<1\iff|x|<\frac{2}{5}$$
hence the radius is $\frac{2}{5}$.
Added
-For $x=\frac{2}{5}$, $$\frac{5^n}{n+2^n}x^n=\frac{2^n}{n+2^n}\to1\neq0$$
so the series is divergent.
-For $x=-\frac{2}{5}$,
$$\frac{5^n}{n+2^n}x^n=\frac{(-1)^n2^n}{n+2^n}$$
has not a limit so the series is also divergent.
|
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|
Prove that $4x-x^4 \leq 3, x \in \Bbb R$ How can I tackle the following inequality :
Prove that $4x-x^4 \leq 3$, where $x$ is any real number.
Can someone point me in the right direction?
|
We see that , $x^4-4x+3=x^4-2x^2+1+2x^2-4x+2=(x^2-1)^2+2(x-1)^2 \geq 0$ and so $-(x^4-4x+3)\leq 0 \implies -x^4+4x-3 \leq 0$ and
hence $4x-x^4 \leq 3$
|
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|
To find the logarithm of $1728$ to the base $2 \sqrt{3}$
Find the logarithm of: $1728$ to base $2\sqrt{3}$.
Let, $\log_{2\sqrt{3}} 1728 = y$, then
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
&= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should I do next to find the logarithm of $1728$?
|
Hint: by the very definition of logarithm
$$\log_{2\sqrt 3}1728=y\iff (2\sqrt 3)^y=1728=12^3=2^6\cdot 36\ldots$$
|
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|
Expressing $\cos\theta - \sqrt{3}\sin\theta = r\sin(\theta - \alpha)$ My book explains that $a\cos\theta + b\sin\theta$ is a sine (or cosine) graph with a particular amplitude/shift (i.e. $r\sin(\theta + \alpha)$) and shows me some steps to solve for $r$ and $\alpha$:
$$r\sin(\theta + \alpha) \equiv a\cos\theta + b\sin\theta$$
$$\Rightarrow r\sin\theta\cos\alpha + r\cos\theta\sin\alpha \equiv a\cos\theta + b\sin\theta$$
I see the basic trig identity $\sin(a + b) \equiv \sin a\cos b + \sin b\cos a$ is being used, and the coefficients $a$ and $b$ are easily identified as:
$$r\sin\alpha = a$$
$$r\cos\alpha = b$$
Then the book squares and adds the equations:
$$r^2(\sin^2\alpha + \cos^2\alpha) = a^2 + b^2 \Rightarrow r = \sqrt{a^2 + b^2}$$
I see that the basic identity $\sin^2\alpha + \cos^2\alpha \equiv 1$ is being used here.
All the examples in the book involve $r$ being a positive quantity.
An exercise asks me to express $\cos\theta - \sqrt{3}\sin\theta$ in the form $r\sin(\theta - \alpha)$. I tried this and ran into problems as the squaring and square rooting process only ever produces a positive $r$, but the answer is $-2\sin(\theta - \frac{1}{6}\pi)$.
My workings:
$\cos\theta - \sqrt{3}\sin\theta = r\sin(\theta - \alpha) = r\sin\theta\cos\alpha - r\sin\alpha\cos\theta$
$\Rightarrow r\cos\alpha = -\sqrt{3}$, $r\sin\alpha = -1$
$\Rightarrow r^2\cos^2\alpha = 3$, $r^2\sin^2\alpha = 1 => r^2(\cos^2\alpha + \sin^2\alpha) = 4$
$\Rightarrow r^2 = 4 \Rightarrow r = \pm2$.
My question is how should the process in the book be refined so one knows whether $r$ should be positive or negative?
(sorry if this is a bit long winded for such a basic question but I thought showing what I do and don't know might get me an answer targeted at my simple level!).
|
The answer is that it depends on the choice of $\alpha$. This is because $\sin (x + \pi) = - \sin x$ (also if you were using $\cos$ instead, $\cos (x + \pi) = - \cos x$). This means that it could be either positive or negative, but then $\alpha$ may need to be reduced by $\pi$
|
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|
The smallest three digit number that is equal to the sum of its digits plus twice the product of its digits How can I do the following?
Find the least three-digits number that is equal to the sum of its digits plus twice the product of its digit?
|
You are looking for integers $a$, $b$, and $c$, with $0\leq a,b,c\leq 9$ ($c\neq 0$), and
$$
a+10b+100c=a+b+c+2abc
$$
Simplifying, that becomes
$$
9b+99c=2abc
$$
We look for the smallest solution, so we want the smallest integer $c$ satisfying
$$
c=\frac{9b}{2ab-99}
$$
From this, we need $2ab\geq100$ or $ab\geq 50$. So the possible values of $a$ and $b$ are - $(a,b) = (6,9),(7,8),(7,9),(8,7),(8,8),(8,9),(9,6),(9,7),(9,8)$ or $(9,9)$.
For these, respectively, we have $2ab-99 = 9,13,27,13,29,45,9,27,45$ and $63$. Now, we can rule out those with $13$ and $29$, as $9b$ will not be divisible by those numbers. None of the valid values for $b$ are divisible by $5$, which also rules out $45$. We're down to $(a,b)=(6,9),(7,9),(9,6),$ $(9,7)$ and $(9,9)$. Five is easy enough to check. This gives:
$$
(6,9):\quad c=\frac{81}{9}=9\\
(7,9):\quad c=\frac{81}{27}=3\\
(9,6):\quad c=\frac{54}{9}=6\\
(9,7):\quad c=\frac{63}{27}\not\in\mathbb{Z}\\
(9,9):\quad c=\frac{81}{63}\not\in\mathbb{Z}\\
$$
It's quite obvious from here that the smallest is $c=3$, which gives our number, $397$.
|
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|
Can someone check the solution to this recurrence relation? Here's the recurrence relation: $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$
Here's the solution:Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply the recurrence by $z^n$ and sum over $n \ge 0$ you get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{1}{1 - 2 z} + \frac{z}{(1 - z)^2} + 3 \frac{1}{1 - z}
$$
This gives:
$$
\begin{align*}
A(z) &= \frac{1 - 4 z + 9 z^2 - 12 z^3 + 5 z^4}
{1 - 8 z + 24 z^2 - 34 z^3 + 23 z^4 - 6 z^5} \\
&= \frac{23}{8} \cdot \frac{1}{1 - 3 z}
- \frac{1}{1 - 2 z}
+ \frac{3}{8} \cdot \frac{1}{1 - z}
- \frac{3}{4} \cdot \frac{1}{(1 - z)^2}
- \frac{1}{2} \cdot \frac{1}{(1 - z)^3}
\end{align*}
$$
Expanding the geometric series, and also:
$$
(1 - z)^{-k} = \sum_{n \ge 0} (-1)^n \binom{-k}{n} z^n
= \sum_{n \ge 0} \binom{n + k - 1}{k - 1} z^n
$$
gives:
$$
a_n = \frac{23}{8} \cdot 3^n
- 2^n
+ \frac{3}{8}
- \frac{3}{4} \cdot \binom{n + 1}{1}
- \frac{1}{2} \cdot \binom{n + 2}{2}
= \frac{23}{8} \cdot 3^n - 2^n + \frac{3}{8}
- \frac{1}{6} (n^3 + 6 n^2 + 5 n)
$$
The problem is that when I check this with Wolfram, it has the solution of $a_n = -4(2^n) - n^2 / 4 - 5n / 2 + 1/8 + (39/8)(3^n)$. I just wanted to know if this was an error or what..thanks!
|
After the correction others have pointed out in your first line, you do the same z-transform/generating function stuff you did before, the only difference is now your two forcing terms are shifted by two.
\begin{align}
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^{n+2} + (n+2) + 3 \quad a_0 = 1, a_1 = 4
\end{align}
You really only need to figure the generating functions for the following term
$$
\sum_{n=0}^{\infty} 2^{n+2} z^n = 2^2\sum_{n=0}^{\infty} 2^{n} z^n = \frac{4}{1-2z}
$$
What's left is $n+2+3=n+5$, so your last term turns into a $5$ in the denominator, while the second to last stays the same. This gives you
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} - 3 A(z)
+ \frac{4}{1-2z} + \frac{z}{(1 - z)^2}+ \frac{5}{1 - z}
$$
At this point, you should be able to finish of the problem the same way as before: solve for $A(z)$, expand by partial fractions, then inverse transform. It seems that you need help with this step? I didn't do this by hand but used Mathematica quickly:
$$
A(z) = \frac{-12 z^4+24 z^3-14 z^2+4 z-1}{(z-1)^3 (2 z-1) (3 z-1)}
$$
$$
= -\frac{7}{4 (z-1)^2}+\frac{1}{2 (z-1)^3}+\frac{4}{2 z-1}-\frac{39}{8 (3 z-1)}-\frac{19}{8 (z-1)}
$$
This is in the same form as your second to last step, just with different coefficients. Match them up in the function forms of $z$ and you have your final answer.
|
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|
With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$ I know how to expand the binomial expansion but I have no idea how to do this second part.
Hence Express $$(3 - \sqrt 2)^4$$ in the form of $$P+Q\sqrt 2$$
Where P and Q are integers, and then how to state the values of P and Q.
|
Using the binomial theorem
$$
\begin{align*}
(3-\sqrt{2})^4 &= \sum_{k = 0}^4 \binom{4}{k} (3)^k (-\sqrt{2})^{4-k}\\
&= 3^0\binom{4}{0}(-\sqrt{2})^4 + (3^1) \binom{4}{1} (-\sqrt{2})^3 + (3^2)\binom{4}{2}(-\sqrt{2})^2 \\
&\quad + (3^3)\binom{4}{3}(-\sqrt{2}) + (3^4)\binom{4}{4}(-\sqrt{2})^0\\
&= (1)(1)(4) + (3)(4)(-2\sqrt{2}) + (9)(6)(2) + (27)(4)(-\sqrt{2}) + (81)(1)(1)\\
&= 193 - 132\sqrt{2}
\end{align*}
$$
|
{
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|
Finding the coefficient in the closed form of the generating function I try to solve the recursion $a_n=5a_{n-1}+5^n$ with $a_0=1$ with generating function, but I could not find the coefficient of $x^n$ in the closed form
\begin{eqnarray*}
g(x)&=&a_0+\sum^n_1a_nx^n\\
g(x)&=&1+\sum^n_1(5a_{n-1}+5^n)x^n\\
g(x)&=&1+5\sum^n_1a_{n-1}x^n + \sum^n_15^nx^n\\
g(x)&=&1+5x\sum^n_0a_{n-1}x^{n-1} + \sum^n_1(5x)^n\\
g(x)&=&1+5xg(x)+\frac{1}{1-5x}-1\\
g(x)(1-5x)&=&\frac{1}{1-5x}\\
g(x)&=&\frac{1}{(1-5x)^2}
\end{eqnarray*}
I tried the partial fraction and wish to find something in the form $\frac{A}{1-5x}+\frac{B}{1-5x}$, but it did not work out because the method of partial fraction required this in form of $\frac{A}{1-5x}+\frac{B}{(1-5x)^2}$, so it seems like partial fraction does not help. Can someone help.
|
You get the generating function:
$$
g(z) = \frac{1}{(1 - 5 z)^2}
= \sum_{n \ge 0} \binom{-2}{n} (-1)^n 5^n z^n
= \sum_{n \ge 0} \binom{n + 2 - 1}{2 - 1} 5^n z^n
= \sum_{n \ge 0} (n + 1) \cdot 5^n z^n
$$
so that:
$$
a_n = (n + 1) \cdot 5^n
$$
|
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|
Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
|
It may be of help to consider that $x^3 + y^3 = ( x + y ) \cdot ( x^2 - xy + y^2)$, so one could start by looking for values of $x$ and $y$ for which those factors are equal (and then for values where one factor "completes a square" with the other).
ADDENDUM: I had a little time to think more on this during my snowy walk home... We can show that if we set $x = y$ , the two factors are $2x \cdot x^2$, which only gets us the (2, 2, 4) family of solutions [apart from $(0,0,0)$].
In fact, if we set $y = kx$ ($k$ integral), the factors are $[k + 1]x \cdot [k^2 - k + 1]x^2$, so we have $ k + 1 = k^2 - k + 1 \Rightarrow k = 0, 2$ if we require the two factors $( x + y ) $ and $ ( x^2 - xy + y^2)$ to be equal. [Just noticed that $k = 2$ will produce, for instance, the $(1,2,3)$ family.] So almost all the solutions are ones for which one of these factors contains a divisor just once that also appears in the other factor just once. An example is $(4,8,24)$, for which the factors are $(4+8) \cdot (4^2 - 4 \cdot 8 + 8^2) = ( 2^2 \cdot 3 ) \cdot (3 \cdot 4^2)$ . This ought to help narrow the search somewhat.
|
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|
How can I solve these two equations to find theta? I am doing a projectile motion questions and I have to solve these simultaneous equations:
$$\frac{-5t^{2}+30t\sin\theta }{30t\cos\theta }=\frac{1}{\sqrt3}$$
$$\frac{-10t+30\sin\theta }{30\cos\theta }=-\sqrt3$$
I solved them but the solution is long and tedious and I got theta = $-60^\circ$ which is wrong.
Thank you in advance!
|
Subtracting the second from the first equation gives
$$\frac{t}{6\cos\theta} = \frac{4}{3}\sqrt{3}$$
or
$$t=8\sqrt{3}\cos\theta.$$
Plugging this back into the first equation gives
$$ \frac{-4\sqrt{3}\cos\theta+3\sin\theta}{3\cos\theta} = \frac{1}{3}\sqrt{3}$$
or
$$3\sin\theta = 5\sqrt{3}\cos\theta.$$
Thus finally we have
$$\tan\theta = \frac{5}{3}\sqrt{3}$$
and
$$ \frac{25}{3} = \tan(\theta)^2 = \cos(\theta)^{-2}-1,$$
thus
$$ t = \pm\frac{12}{7}\sqrt{7}.$$
|
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|
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this result can be generalized to other triples of
natural numbers.
Question. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$
For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$
$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$
and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$
A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.
For $(2)$ the very same idea yields
$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$
and
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$
I tried to solve this system for $a,b$ but since the solution is of the form
$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$
where $x$ satisfies the cubic equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach.
Is this problem solvable, at least partially?
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?
|
The solutions are of the form $\displaystyle(p, q)= \left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right)$, for any rational parameter $t$. To prove it, we start with $$\left(p+q\sqrt{r}\right)^{1/3}+\left(p-q\sqrt{r}\right)^{1/3}=n\tag{$\left(p,q,n,r\right)\in\mathbb{N}^{4}$}$$
and cube both sides using the identity $(a+b)^3=a^3+3ab(a+b)+b^3$ to, then, get $$\left(\frac{n^3-2p}{3n}\right)^3=p^2-rq^2,$$ which is a nicer form to work with. Keeping $n$ and $r$ fixed, we see that for every $p={1,2,3,\ldots}$ there is a solution $(p,q)$, where $\displaystyle q^2=\frac{1}{r}\left(p^2-\left(\frac{n^3-2p}{3n}\right)^3\right)$. When is this number a perfect square? Wolfram says it equals $$q^2 =\frac{(8p-n^3) (n^3+p)^2}{(3n)^2\cdot 3nr},$$ which reduces the question to when $\displaystyle \frac{8p-n^3}{3nr}$ is a perfect square, and you get solutions of the form $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right).$ Note that when $r=3$, this simplifies further to when $\displaystyle \frac{8p}{n}-n^2$ is a perfect square.
Now, we note that if $\displaystyle (p,q)=\left(p,\frac{n^3+p}{3n}\sqrt{\frac{8p-n^3}{3nr}}\right) \in\mathbb{Q}^2$, $\displaystyle\sqrt{\frac{8p-n^3}{3nr}}$ must be rational as well. Call this rational number $t$, our parameter. Then $8p=3t^2nr+n^3$. Substitute back to get $$(p,q)=\left(\frac{3t^2nr+n^3}{8},\,\frac{3n^2t+t^3r}{8}\right).$$ This generates expressions like $$\left(\frac{2589437}{8}+\frac{56351}{4}\sqrt{11}\right)^{1/3}+\left(\frac{2589437}{8}-\frac{56351}{4}\sqrt{11}\right)^{1/3}=137$$
$$\left(\frac{11155}{4}+\frac{6069}{4}\sqrt{3}\right)^{1/3}+\left(\frac{11155}{4}-\frac{6069}{4}\sqrt{3}\right)^{1/3}=23$$
for whichever $r$ you want, the first using $(r,t,n)=(11,2,137)$ and the second $(r,t,n)=(3,7,23)$.
|
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|
Prove that $\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1} \geq \frac12$
Let $a_1, a_2, a_3, \dots , a_n$ be positive real numbers whose sum is $1$. Prove that
$$\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \ldots +\frac{a_n^2}{a_n+a_1} \geq \frac12\,.$$
I thought maybe the Cauchy and QM inequalities would be helpful. But I can't see how to apply it. Another thought (might be unhelpful) is that the sum of the denominators on the left hand side is $2$ (the denominator on the right hand side). I would really appreciate any hints.
|
Thanks to Sanchez for giving me a hint to solve this. Here is a full solution.
By the Cauchy-Schwarz inequality we have:
$${\frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+ \cdots \frac{a_n^2}{a_n+a_1}=\frac{a_1^2}{(\sqrt{a_1+a_2})^2}+\frac{a_2^2}{(\sqrt{a_2+a_3})^2}+ \cdots+ \frac{a_n^2}{(\sqrt{a_n+a_1})^2} \geq \frac{1}{a_1+\cdots + a_n+a_1+ \cdots + a_n}\left(\frac{a_1 \cdot \sqrt{a_1+a_2}}{\sqrt{a_1+a_2}} + \frac{a_2 \cdot \sqrt{a_2+a_3}}{\sqrt{a_2+a_3}}+ \cdots + \frac{a_n \cdot \sqrt{a_n+a_1}}{\sqrt{a_n+a_1}}\right)\\=\frac{a_1+a_2+a_3+ \cdots a_n}{{2(a_1+a_2+a_3+ \cdots a_n)}}=\frac12}$$
as required. (We know that $a_1+a_2+a_3+ \cdots +a_n=1$)
|
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|
integrating fraction/completing the square Does anyone know how to integrate the following?
$\frac{dx}{9x^2 + 6x + 17}$
I have been trying for ages and cannot get an answer anywhere close to the answer I get on maple?
|
$$9x^2+6x+17=(3x)^2+2\cdot3x\cdot1+1^2+17-1=(3x+1)^2+4^2$$
Put $3x+1=4\tan\theta$ so that $(3x+1)^2+4^2=\cdots=16\sec^2\theta$ and $3dx=4\sec^2\theta d\theta$
So, $$\int \frac{dx}{9x^2+6x+17}=\int\frac{4\sec^2\theta d\theta}{3\cdot16\sec^2\theta}=\frac1{12}\int d\theta=\frac{\theta}{12}+C=\frac{\arctan\left(\frac{3x+1}4\right)}{12}+C$$
|
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|
Eigenvector of matrix of equal numbers For matrix the matrix
$$A = \begin{bmatrix}
3&1&1\\
1&3&1\\
1&1&3\\
\end{bmatrix}$$
with eigenvalues $\lambda_1=5$, $\lambda_2=2$, $\lambda_3=2$, I am trying to find the corresponding eigenvector corresponding to the eigenvalue 2. I got
$$(A - 2I_3) = \begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1\\
\end{bmatrix}$$
Reducing it (row reduced echelon form), I get:
$$\left[
\begin{array}
{ccc|c}
1&1&1&0\\
0&0&0&0\\
0&0&0&0\\
\end{array}\right]$$
Ending up with $x_1 + x_2 + x_3 = 0$. How would I find the eigenvector from there? Usually, I end up getting two equations and it's easy from there. How would you do it with one?
|
For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal) eigenvectors for the $n$ by $n$ case can be read from the columns of
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
In your case, the upper left 3 by 3 corner. If you want the result orthonormal you need to divide each column by a square root of something appropriate. I have displayed the 10 by 10 version, notice how the diagonal numbers go up to 9 = 10 - 1.
|
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|
How to count permutations with restrictions on how items are grouped I am trying to solve the following problem:
A town contains $4$ people who repair televisions. If $4$ sets break down, what is the probability that exactly $i$ of the repairers are called? Solve the problem for $i=1,2,3,4$.
For $i=1$, there are ${}_4P_1$ ways to assign $1$ person to $4$ televisions, so the probability is $\frac{{}_4P_1}{4^4}=\frac{1}{64}$.
For $i=4$, there are ${}_4P_4$ ways to assign $4$ people to $4$ televisions, so the probability is $\frac{{}_4P_4}{4^4}=\frac{3}{32}$.
I am having trouble with $i=2,3$. How should I go about these cases?
|
Conditions:
*
*The repairmen and TVs are distinct.
*Every TV owner calls ONLY one repairman, that is, the relationship between the set of TVs, $T$, and a set that contains $i$ of the repairmen, $Ri$, is functional, $f: T -> Ri$. Note that $|T|=4$ and $|Ri|=i$
*Every one of the $i$ repairmen gets at least one call, that is the range of $f$ is $Ri$.
The size of the sample space is $|S| = 4^4 = 256$.
$i = 2$:
There are $\binom{4}{2}=6$ ways to select two repairmen. Due to conditions #2 and #3, each repairman can get 1 OR 2 OR 3 calls, for the other will get the remaining calls; therefore, there are $\binom{4}{1} + \binom{4}{2} + \binom{4}{3} = 14$ ways for each repairman to get calls.
You can also argue as follows to obtain the same result: To satisfy conditions #1, #2 and #3, the number of calls can be divided between the two repairmen in one of the following ways: 1|3, 2|2, or 3|1; This is the same way we must partition the set of TVs. Using multinomial coefficients, these partitions can be counted in $\binom{4}{1,3}+\binom{4}{2,2}+\binom{4}{3,1}=14$ ways.
Hence $P\{i=2\} = (6*14)/256 = 84/256$.
$i = 3$:
There are $\binom{4}{3} = 4$ ways to select the repairmen. A repairman can get 1 OR 2 calls, for the other two repairmen will get the remaining calls.
If a repairman gets 1 (out of 4) call, the second repairman can get either 1 OR 2 (out of remaining 3) calls, and the third repairman gets the remaining calls. This can be counted in$\binom{4}{1}\binom{3}{1} + \binom{4}{1}\binom{3}{2}=24$ number of ways.
If a repairman gets 2 (out of 4) calls, the second repairman can only get 1 (out of 2) call, and the third repairman gets the remaining call. This can be counted in $\binom{4}{2}\binom{2}{1}=12$ number of ways.
Therefore, the total number of ways the repairmen are called is $24+12=36$.
The same result can be more succinctly obtained by multinomial coefficients: $\binom{4}{1,1,2}+\binom{4}{1,2,1}+\binom{4}{2,1,1}=36$.
Hence $P\{i=3\} = (4*36)/256 = 36/64$.
|
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|
Integral Question - $\int\frac{1}{\sqrt{x^2-x}}\,\mathrm dx$ Integral Question - $\displaystyle\int\frac{1}{\sqrt{x^2-x}}\,\mathrm dx$.
$$\int\frac{1}{\sqrt{x(x-1)}}\,\mathrm dx =\int \left(\frac{A}{\sqrt x} + \frac{B}{\sqrt{x-1}}\right)\,\mathrm dx$$
This is the right way to solve it?
Thanks!
|
The Partial Fraction Decomposition is for rational fraction only.
$$\int\frac{dx}{\sqrt{x^2-x}}=\int\frac{2dx}{\sqrt{4x^2-4x}}=\int\frac{2dx}{\sqrt{(2x-1)^2-1^2}}$$
Now, put $2x-1=\sec\theta$
EDIT: completing as requested
So,$2dx=\sec\theta\tan\theta d\theta$
$$\text{So,}\int\frac{2dx}{\sqrt{(2x-1)^2-1^2}}=\int \frac{\sec\theta\tan\theta d\theta}{\tan\theta}=\int \sec\theta d\theta =\ln|\sec\theta+\tan\theta|+C $$ (where $C$ is an arbitrary constant of indefinite integral )
$$=\ln\left|2x-1+\sqrt{(2x-1)^2-1}\right|+C=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C$$
Alternatively,using
$$\frac{dy}{\sqrt{y^2-a^2}}=\ln\left|y+\sqrt{y^2-a^2}\right|+C$$
$$\int\frac{dx}{\sqrt{x^2-x}}=\int\frac{dx}{\sqrt{\left(x-\frac12\right)^2-\left(\frac12\right)^2}}$$
$$=\ln\left|x-\frac12+\sqrt{x^2-x}\right|+C=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C-\ln2=\ln\left|2x-1+2\sqrt{x^2-x}\right|+C'$$ where $C'=C-\ln2$ another arbitrary constant
|
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|
Is there a pattern in figures whose perimeter is the same as their area? Here's what I've found so far.
*
*Circumference of a circle with identical circumference and area: $4\pi$
*Side length of a triangle with identical perimeter and area: $4\sqrt{3}$
And so on...
*
*Square: $4$
*Pentagon: $4\sqrt{5 - 2\sqrt{5}}$
*Hexagon: $\frac{4\sqrt{3}}{3}$
*Heptagon: $\frac{4\cos{\frac{3\pi}{14}}}{1 + \sin{\frac{3\pi}{14}}}$
*Octagon: $4(1+ \sqrt{2})$
Is there a pattern here with the 4s?
|
We can look at a regular $n$-gon of "radius" $r$, i.e., the convex hull of $r$ times the $n$-th roots of unity.
Connecting the vertices of the polygon to the origin gives you $n$ isosceles triangles of area $\frac{r^2}{2}\sin \frac{2\pi}{n}$. The total area of the $n$-gon is thus
$$A = \frac{nr^2}{2} \sin \frac{2\pi}{n}.$$
To calculate the length $l$ of the third side of these triangles, we can bisect the side to get two right triangles, and then
$$l = 2r\sin \frac{\pi}{n}$$
and the total perimeter is
$$P = 2r n \sin \frac{\pi}{n}.$$
Now we can solve for $r$:
$$\begin{align*}
\frac{nr^2}{2}\sin\frac{2\pi}{n} &= 2rn\sin \frac{\pi}{n}\\
r \sin \frac{2\pi}{n} &= 4\sin \frac{\pi}{n}\\
2r \sin \frac{\pi}{n} \cos \frac{\pi}{n} &= 4\sin \frac{\pi}{n}\\
r &= 2\sec \frac{\pi}{n}.
\end{align*}$$
Plugging this radius into the formula for perimeter gives you
$$P = 4n\tan \frac{\pi}{n}$$
and as you observed, this formula has a natural factor of 4 in it. And indeed,
$$\lim_{n\to\infty} 4n\tan\frac{\pi}{n} = 4\pi.$$
|
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|
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2}$$
following is my some nice methods,
use this inequality
$$\dfrac{x-y}{\ln{x}-\ln{y}}>\sqrt{xy},x>y$$
then we let
$x=2,y=1$
so $$\ln{2}<\dfrac{\sqrt{2}}{2}$$
solution 2:
since
$$\dfrac{1}{n+1}\le\dfrac{1}{2}\cdot\dfrac{3}{4}\cdots\dfrac{2n-1}{2n}$$
then
$$\ln{2}=\sum_{n=0}^{\infty}\dfrac{1}{(n+1)2^{n+1}}<\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\dfrac{1}{\sqrt{2}}$$
solution 3
since
$$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2=t(1-t)>0$$
so
$$\ln{2}=\int_{0}^{1}\dfrac{1}{t+1}dt<\int_{0}^{1}\left(\dfrac{1+\sqrt{2}}{t+1+\sqrt{2}}\right)^2dt=\dfrac{\sqrt{2}}{2}$$
solution 4:
$$\ln{2}=\dfrac{3}{4}-\dfrac{1}{4}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(2n+1)}<\dfrac{3}{4}-\dfrac{1}{4}\left(\dfrac{1}{1\times 2\times 3}-\dfrac{1}{2\times 3\times 5}\right)=\dfrac{7}{10}<\dfrac{\sqrt{2}}{2}$$
solution 5
$$\dfrac{1}{\sqrt{2}}-\ln{2}=\sum_{n=1}^{\infty}\dfrac{\sqrt{2}}{(4n^2-1)(17+2\sqrt{2})^n}>0$$
But $$\ln{2}>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$$ I can't have this nice solution
Thank you everyone can help.
|
Hmm, first I thought the following would allow to compute the proof mentally, but, well... although I find it a remarkable simplification I'll need the pocket calculator in the end. But let's see:
$$\ln(2) \gt (2 / 5)^{2 /5} = \left({16 \over 100 }\right)^{1/5} $$ We have also $$
\ln(2) = \ln \left( 1+1/3 \over 1-1/3 \right) = 2\left( {1\over3} + {1\over 3^3 \cdot 3} + {1 \over 3^5 \cdot 5}+ \cdots \right) \\ = {2 \over 3} \left( 1 + { 1 \over 9\cdot3 } + {1 \over 81 \cdot 5}+ \cdots ) \right) $$
Turn the factor 2/3 to the the rhs then
$$ 1 + 1/27 + 1/81/5+ 1/729/7 \cdots \gt \left( {3^5 \over2 \cdot 100 }\right) ^{1/5}= \left( 1+{43\over 200}\right)^{1/5} $$
Now in general we have for a fifth root
$$ (1+x)^{1/5} = 1 + x/5 - 2(x/5)^2 + 6(x/5)^3 - 21(x/5)^4 + 399/5 (x/5)^5 - \cdots $$
Thus we must evaluate
$$ 1 + 1/27 + 1/81/5+ 1/729/7 \cdots \gt 1 +43/1000 - 2(43/1000)^2 + \cdots $$
"In principle" this can be done with paper & pen only because the terms decrease quickly, and some adaptions of denominators are possible, however, that was too tidy for me. I found using a calculator (Pari/GP) that we must evaluate the lhs with 4 terms and the rhs with 5 terms (of course excluding the 1's) to get the decision - because after that the partial sums in the lhs still increase but in the rhs decrease.
|
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|
On the lengths of the cycles of random permutations A random permutation of $n$ from $n$ is chosen. What is the mathematical expectation of the sum of squares of lengths of its cycles?
|
The exponential generating function of permutations by the sum of the squares of the lengths of its cycles is
$$ G(z, u) =
\exp
\left(u z + \frac{1}{2} u^4 z^2 + \frac{1}{3} u^9 z^3 + \frac{1}{4} u^{16} z^4 \cdots\right)$$
which means that the probability generating function of the expectation that we are looking for is
$$ \left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1} \\=
\left.
\exp
\left(u z + \frac{1}{2} u^4 z^2 + \frac{1}{3} u^9 z^3 + \frac{1}{4} u^{16} z^4 \cdots\right)
\left(z + 2 u^3 z^2 + 3 u^8 z^3 + 4 u^{15} z^4 \cdots\right)
\right|_{u=1} \\=
\exp\log\frac{1}{1-z}\times \frac{z}{(1-z)^2} =
\frac{1}{1-z}\times \frac{z}{(1-z)^2} =
\frac{z}{(1-z)^3}.$$
The conclusion is that the expectation we are trying to compute is given by
$$[z^n] \left.\frac{\partial}{\partial u} G(z, u) \right|_{u=1} =
[z^n] \frac{z}{(1-z)^3} = \frac{1}{2} n (n+1).$$
|
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|
Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x $
2) via subsitution $ u = \cos x $
$ u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x $
3) using $ \sin 2x = 2 \sin x \cos x $
$ \int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x $
So, we have:
$$ \frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x $$
|
$$\frac{d\{f(x)+c\}}{dx}=f'(x)$$ for any arbitrary constant $c$
$$\implies \int f'(x)dx=f(x)+d $$ for any arbitrary constant $d$
So, in indefinite integral we can get answers which differ by some constant
|
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|
Integral Of $\int\sqrt{\frac{x}{x+1}}dx$ I want to solve this integral
$$\int\sqrt{\frac{x}{x+1}}dx$$
And think about:
1) $t=\frac{x}{x+1}$
2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$
Now I need your advice! Thanks!
|
If you know about Möbius transformations, from the change of variables
$$
t = \frac{x}{x+1}, \quad \Longrightarrow \quad x = \frac{t}{1-t}
$$
because
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}.
$$
If you don't know about Möbius transformations, then you can do this :
$$
t = \frac x{x+1} = 1 - \frac 1{x+1} \quad \Longrightarrow \quad 1-t = \frac 1{x+1} \quad \Longrightarrow \quad x = \frac 1{1-t} -1 = \frac t{1-t}.
$$This means that
$$
\left( \frac 1{x+1} - \frac x{(1+x)^2} \right) = \frac{x+1-x}{(x+1)^2} = \frac 1{(x+1)^2} = \frac{1}{\left( \frac t{1-t} + 1 \right)^2} = (1-t)^2.
$$
So now you can try to solve
$$
\int \frac{\sqrt t}{(1-t)^2} \, dt.
$$
Letting $u = \sqrt t$, $du = \frac 1{2 \sqrt t} dt$ which implies $2u du = dt$, hence now we have
$$
\int \frac{2u^2}{(1-u^2)^2} \, du
$$
to solve. Use partial fractions.
|
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|
How can I evaluate this given improper integral? How can I evaluate this integral:
$$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} \ ?$$
|
Try $y = (3-x)^{1/3}$.
$y^3 = 3-x$,
so $x = 3-y^3$
and $dx = -3 y^2 dy$.
Putting this in,
since $y$ goes from $3^{1/3}$ to $0$
as $x$ goes from $0$ to $3$,
$\begin{align}
\int_{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx}
&=\int_{3^{1/3}}^0 \frac{3-y^3}{y}(-3 y^2)dy\\
&=\int_0^{3^{1/3}} \frac{3-y^3}{y}(3 y^2)dy\\
&=3 \int_0^{3^{1/3}} y(3-y^3)dy\\
&=3 \int_0^{3^{1/3}} (3y-y^4)dy\\
&=3 (3y^2/2-y^5/5)\big|_0^{3^{1/3}}\\
&=3(3\cdot3^{2/3}/2 - 3^{5/3}/5)\\
&=3(3^{5/3}/2 - 3^{5/3}/5)\\
&=3\cdot 3^{5/3}(1/2-1/5)\\
&=3\cdot 3^{5/3}(3/10)\\
&=(9/10) 3^{5/3}\\
&=(27/10) 3^{2/3}\\
\end{align}
$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$ Please help me to find a closed form for the infinite product
$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.
|
For $x < 1$, we have the Taylor series expansion:
$$f(x):= \frac{-1}{4} \log \left(- \frac{x - x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$
Then
$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots
= \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots $$
$$= - \frac{1}{2} \log(1 - x^2).$$
Now let $x = e^{-2}$. Then
$$ \log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) =
\frac{1}{2^n} \log \left( \frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)
$$
$$= \frac{-4}{2^n} f(e^{-2^n}) = \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$
Hence summing over all $n \ge 1$, we see that, if the product is $P$, then
$$\log P = -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}})
= -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 - x^2),$$
and thus
$$P = \exp \log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved How are can the above recurrence relation be solved?
I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$
And then here:
$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$
And we are given that these: $T(0) = 2, T(1) = 3$
So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.
|
Use "generatingfunctionology" techniques. Define $G(z) = \sum_{n \ge 0} T(n) z^n$, write your recurrence as:
$$
T(n + 2) = T(n + 1) + T(n) + 3 n + 3
$$
Multiply by $z^n$, add for $n \ge 0$. Remember that:
$$
\sum_{n \ge 0} (n + 1) z^n = \frac{1}{(1 - z)^2}
$$
and you get:
$$
\frac{(G(z) - T(0) - T(1) z}{z^2}
= \frac{G(z) - T(0)}{z} + G(z) + 3 \cdot \frac{1}{(1 - z)^2}
$$
Solve for $G(z)$, expand in partial fractions:
$$
G(z) = \frac{2 - 3 z + z^2 + z^3}{1 - 3 z + 2 z^2 + z^3 - z^4}
= \frac{4 + 2 z}{1 - z - z^2} - \frac{1}{1 - z} - \frac{1}{(1 - z)^2}
$$
We have the last two terms, the first one is for Fibonacci numbers:
$$
T(n) = 4 F_{n + 1} + 2 F_n - 1 - (n + 1) = 4 F_{n + 1} + 2 F_n - n - 2
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$ I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks
|
HINT:
Using Partial Fraction Decomposition formula,
$$\frac{x^4+2x+4}{x^4-1}=1+\frac{ax+b}{x^2+1}+\frac c{x+1}+\frac d{x-1}$$ where $a,b,c,d$ are arbitrary constants to determined by equating the coefficients of the different powers of $x$ in
$$x^4+2x+4=x^4-1+(x^2-1)(ax+b)+c(x-1)(x^2+1)+d(x+1)(x^2+1)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} 1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{pmatrix};
A_5 = \begin{pmatrix} 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2\\
1& 2& 3& 3& 3\\
1& 2& 3& 4& 4\\
1& 2& 3& 4& 5
\end{pmatrix}$$
$$A_6 = \begin{pmatrix} 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4\\
1& 2& 3& 4& 5& 5\\
1& 2& 3& 4& 5& 6
\end{pmatrix};
A_7 = \begin{pmatrix} 1& 1& 1& 1& 1& 1& 1\\
1& 2& 2& 2& 2& 2& 2\\
1& 2& 3& 3& 3& 3& 3\\
1& 2& 3& 4& 4& 4& 4\\
1& 2& 3& 4& 5& 5& 5\\
1& 2& 3& 4& 5& 6& 6\\
1& 2& 3& 4& 5& 6& 7
\end{pmatrix}
$$
|
Recall that adding a multiple or subtracting a multiple of one row does not change the value of the determinant, see, for example ProofWiki.
Using this fact and Laplace expansion you get
$$|A_4|=
\begin{vmatrix}
1& 1& 1& 1\\
1& 2& 2& 2\\
1& 2& 3& 3\\
1& 2& 3& 4
\end{vmatrix}=
\begin{vmatrix}
1& 1& 1& 1\\
0& 1& 1& 1\\
0& 1& 2& 2\\
0& 1& 2& 3
\end{vmatrix}=
\begin{vmatrix}
1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{vmatrix}=|A_3|.$$
(We have subtracted the first row from the other rows.)
In the same way you get $|A_{n+1}|=|A_n|$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$? Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$
|
$$
\begin{align}
n\cot^{-1}(n)-1
&=n\tan^{-1}\left(\frac1n\right)-1\\
&=n\int_0^{1/n}\frac{\mathrm{d}x}{1+x^2}-1\\
&=-n\int_0^{1/n}\frac{x^2\,\mathrm{d}x}{1+x^2}\\
&=-\int_0^1\frac{x^2\,\mathrm{d}x}{n^2+x^2}\tag{1}
\end{align}
$$
Using formula $(9)$ from this answer and substituting $z\mapsto ix$, we get
$$
\sum_{n=1}^\infty\frac{1}{n^2+x^2}=\frac{\pi\coth(\pi x)}{2x}-\frac{1}{2x^2}\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty(n\cot^{-1}(n)-1)
&=\frac12\int_0^1(1-\pi x\coth(\pi x))\,\mathrm{d}x\\
&=\frac12\int_0^1\left(1-\pi x\left(1+\frac{2e^{-2\pi x}}{1-e^{-2\pi x}}\right)\right)\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\int_0^1\frac{xe^{-2\pi x}}{1-e^{-2\pi x}}\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\int_0^1x\left(\sum_{n=1}^\infty e^{-2\pi nx}\right)\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\sum_{n=1}^\infty\left(\color{#C00000}{\frac1{(2\pi n)^2}}-\left(\color{#00A000}{\frac1{2\pi n}}+\color{#0000FF}{\frac1{(2\pi n)^2}}\right)e^{-2\pi n}\right)\\
&=\frac{2-\pi}{4}-\color{#C00000}{\frac\pi{24}}-\color{#00A000}{\frac12\log\left(1-e^{-2\pi}\right)}+\color{#0000FF}{\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)}\\
&=\frac12+\frac{17\pi}{24}-\frac12\log\left(e^{2\pi}-1\right)+\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)\tag{3}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}} \cr
& = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|) \cr
& = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C \cr
& = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$
The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!
|
I am sure there are more tidy ways to do this but as an alternative...
Why not do $\int \ln(2x + 1)dx$ using:
$v^\prime = 1 \Rightarrow v = x$ and $u = \ln(2x+1)\Rightarrow u^\prime=\frac{2}{2x+1}$
Therefore,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{2x}{2x+1}dx$$
Then make the substitution $u=2x + 1$ to yield,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{1}{2}\frac{u-1}{u}du$$
Thus,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}u - ln(u) + c$$
Then,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}(2x-1) - \ln(2x-1) + c$$
$$= \ln(2x-1)(x-1) - \frac{1}{2}(2x-1) + c$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range of a function Find the range of
$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.
|
$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$
$$f(a)= \dfrac{(a-b)(a-c)}{(a-b)(a-c)}=1$$
$$f(b)= \frac{(b-c)(b-a)}{(b-c)(b-a)}=1$$
$$f(c)= \frac{(c-a)(c-b)}{(c-a)(c-b)}=1$$
Now what is $f'(x)?$, $f'(x)=\dfrac{2x-(a+b)}{(c-a)(c-b)}+\dfrac{2x-(b+c)}{(a-b)(a-c)}+\dfrac{2x-(c+a)}{(b-c)(b-a)}$. Now watch the house of cards fall. Then you have a constant function, and you are aware of that constant.
|
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"timestamp": "2023-03-29T00:00:00",
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|
real mechanism behind addition, subtraction, multiplication, division we all know the basic rules for operations of addition, subtraction, multiplication and division.
but what i don't know is why these rules (of addition, subtraction, multiplication and division) works.
as if we have been given algorithm to do these operations but not explained how they were derived.
if anyone can explain please expain .
|
$$
\begin{array}{cccccc}
& & 4 & 3 \\
& \times & 7 & 9 \\
\hline & 3 & 8 & 7 \\
3 & 0 & 1 \\
\hline 3 & 3 & 9 & 7
\end{array}
$$
Why is this algorithm, taught in elementary school, justified?
It's because multiplication distributes over addition:
$$
9\cdot(43) = 9\cdot(4\cdot10 \quad+\quad 3) = (9\cdot4)(10)\quad+\quad(9\cdot3)
$$
Hence
$$
\begin{array}{ccc}
& 2 & 7 \\
3 & 6 \\
\hline
3 & 8 & 7
\end{array}
$$
Similarly one multiplies $43$ by $7$, getting $301$, but the $301$ is moved one place to the left because it is $70$ rather than $7$ by which $43$ was to be multiplies; thus one multiplies by $10$ by moving one place to the left.
A think and explanation of long division might be more involved. Or might not$\ldots$
Later edit: $1352/43$. There's a quotient and a remainder: $1352 = 31\cdot 43 + 19$. The quotient is $31$ and the remainder is $19$. So $\dfrac{1352}{43} = 31 + \dfrac{19}{43}$. The remainder must be less than $43$ so the fraction must be less than $1$. So if we move the decimal point over to get a one-digit quotient, we have
$$
\frac{135.2}{43} = 3.1 +\text{less than $0.1$} = 3 + \text{less than $1$}
$$
Since $43$ goes $3+\text{fraction}$ times into $135.2$, and $43\cdot3$ must be an integer, we have $43$ goes $3+\text{some fraction}$ times into $135$. In particular,
$$
135 = 43\cdot3 + 6.\quad \text{The quotient is $3$ and the remainder is $6$.}
$$
Now look at the usual algorithm:
$$
\begin{array}{cccccccccc}
& & & & 3 \\
\hline
43 & ) & 1 & 3 & 5 & 2 \\
& & 1 & 2 & 9 \\
\hline & & & & 6
\end{array}
$$
Why should we now bring down the $2$ and then divide the resulting $62$ by $43$?
$$
\frac{135.2}{43} = \frac{135}{43} + \frac{0.2}{43} = 3 + \underbrace{\frac{6}{43} + \frac{0.2}{43}} = 3 + \frac{6.2}{43}.
$$
So
$$
\frac{1352}{43} = 30 + \frac{62}{43}.
$$
So that's why.
|
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|
Generating function: Find a closed form of $\sum_{k=0}^n (-3)^k(k+1)$ Find the closed form of $\sum_{k=0}^n (-3)^k(k+1)$.
So the generating function would be: $$A(x)=1-6x+18x^2-108x^3...$$ So what I did notice is that its closed form is perhaps some variation of $1\over {1+x}$ but I didn't manage to find a general formula.
Thanks for any help in advance!
|
Let $f(x) = \sum_{k=0}^n x^{k+1}
= \frac{x-x^{n+2}}{1-x}
$.
Then $f'(x) = \sum_{k=0}^n (k+1)x^k$,
so $f'(-3) = \sum_{k=0}^n (k+1)(-3)^k$.
So
$\begin{align}
f'(x) &= \frac{(x-x^{n+2})'(1-x) - (1-x)'(x-x^{n+2})}{(1-x)^2}\\
&= \frac{(1-x)(1-(n+2)x^{n+1})+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}-x+(n+2)x^{n+2}+x-x^{n+2}}{(1-x)^2}\\
&= \frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(1-x)^2}\\
\end{align}
$
Putting $x = -3$,
$\begin{align}
f'(-3)
&= \frac{1-(n+2)(-3)^{n+1}+(n+1)(-3)^{n+2}}{4^2}\\
&= \frac{1-(-3)^{n+1}((n+2)+(n+1)3)}{4^2}\\
&= \frac{1-(-3)^{n+1}(4n+5)}{16}\\
\end{align}
$
As a check (since I did this off the top of my head):
If $n=0$,
$f'(-3) = \frac{1-(-3)^{1}(5)}{16}
=\frac{1+15}{16}
=1
$
and $1*(-3)^0 = 1$.
If $n=1$,
$f'(-3) = \frac{1-(-3)^{2}(4+5)}{16}
=\frac{1-9*9}{16}
=-5
$
and $1+2*(-3) = -5$.
If $n=2$,
$f'(-3) = \frac{1-(-3)^{3}(8+5)}{16}
=\frac{1+27*13}{16}
=22
$
and $-5+3*(-3)^2 = -5+27 = 22$.
|
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|
is $(x+1)^4-x^4$ non-prime for all natural positive integers $x$ Looking at difference between two neighbouring positive integers raised to the power 4, I found that all differences for integer neighbours up to $(999,1000)$ are non-prime.
Does this goes for all positive integers?
And can someone please prove?
|
Hint : you can write $(x+1)^4-x^4=((x+1)^2-x^2)((x+1)^2+x^2)=(x+1-x)(x+1+x)((x+1)^2+x^2)=(2x+1)((x+1)^2+x^2)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
|
sorry about to answer my Question but Zarrax's way and her comment lead me to the answer
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}=\frac{3L}{27}-\frac{4}{27}$$
$$L=\frac{3L}{27}-\frac{4}{27}$$
$$L=-\frac{1}{6}$$
$$2L=-\frac{1}{3}=\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
thanks alot Zarrax
thanks for every one
|
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}
|
Finding the limit, multiplication by the conjugate I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$
I tried and tried... friends of mine tried as well and we don't know how to get out of:
$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$
(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)
How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.
|
You had the right idea: the issue is in your simplification of the numerator:
$$\begin{align}
(2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) & = 2^2 - \left(\sqrt{(3 + x)}\right)^2 \\ \\
& = 4 - (3 + x) \\ \\
& = 4 - 3 - x \\ \\
& = 1 - x = -(x - 1)
\end{align}$$
That gives you $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} & = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}}
& = -\frac 14
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/400838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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|
Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$
Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$
How do this transformation was done?
Thanks!
|
Look at $n^3 - (n-1)^3 = 3n^2 - 3n + 1$. Now substitute.
|
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|
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d
Is there any help?
|
I do not know how to solve it, but $a=-3,b=4,c=2,d=-1$
|
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|
curl of what yields $(0,s^{-1},0)$ in cylindrical coordinates? In cylindrical coordinates $(s,\theta,z)$, what function $\mathbf{A}$ has the property $$\nabla\times \mathbf{A} = (0, \frac{1}{s} , 0) $$
I know generally that $$\nabla\times \mathbf{A} = \left(\frac{1}{s}\frac{\partial A_3}{\partial \theta}-\frac{\partial A_2}{\partial z}, \frac{\partial A_1}{\partial z}-\frac{\partial A_3}{\partial s} , \frac{1}{s} \left( \frac{\partial}{\partial z}(s\cdot A_2) - \frac{\partial A_1}{\partial \theta}\right)\right) $$
Is there some better way of solving for $\mathbf{A}$ other than a lot of ugly equations?
|
There is a path integral formula to construct the inverse of the exterior derivative operator for closed differential forms, for curl in three dimensions:
$$
\newcommand{\A}{\mathbf{A}}\newcommand{\ps}{\boldsymbol{\psi}} \newcommand{\p}{\mathbf{p}} \mathscr{R}(\ps) = -(\p - \p_0)\times \int^1_0 \ps\big(\p_0 + t(\p- \p_0)\big)t\,dt,\tag{1}
$$
where $\p = (x,y,z)$ and $\ps$ has to be divergence free. We have that $\mathscr{R}(\ps) $ satisfies:
$$
\nabla \times \mathscr{R}(\ps) = \ps.
$$
The vector field you gave, $(0,1/s,0) = (0,\dfrac{1}{\sqrt{x^2+y^2}},0)$, is not divergence free, there does not exist any vector potential $\A$ satisfying:
$$
\nabla \times \A = (0,\frac{1}{s},0).
$$
However, we can just modify a little bit to make your question a well-posed problem, we can move $1/s$ to the $z$-component, and it is divergence free:
$$
\nabla \cdot \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right) = 0,
$$
and
$$
\nabla \times \A = (0,0,\frac{1}{s})
$$
has a solution (not unique though).
Let $\ps$ be the above vector, and in (1) we take $\p_0$ as the origin we have:
\begin{align}
\A = &\mathscr{R}(\ps) = -\p\times \int^1_0 \ps(t\p)t\,dt
\\
&= -(x,y,z)\times \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right)
\\
&= \left(-\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}},0\right).\tag{2}
\end{align}
We can check:
\begin{align}
\nabla \times\A &= \left(0,0, \frac{\partial }{\partial x}\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) + \frac{\partial }{\partial y}\Big(\frac{y}{\sqrt{x^2+y^2}}\Big)\right)
\\
&= \left(0,0, \frac{\sqrt{x^2+y^2} - x\frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2} + \frac{\sqrt{x^2+y^2} - y\frac{y}{\sqrt{x^2+y^2}}}{x^2+y^2}\right)
\\
& = \left(0,0,\frac{1}{\sqrt{x^2+y^2}}\right).
\end{align}
Or you can rewrite (2) in cylindrical coordinates:
$$
\A = (-\sin\theta, \cos\theta,0),
$$
and use the curl formula you gave.
|
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"url": "https://math.stackexchange.com/questions/403316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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|
Elementary Row Matrices Let $A$ =
$$
\begin{align}
\begin{bmatrix}
-4 & 3\\
1 & 0
\end{bmatrix}
\end{align}
$$
Find $2 \times 2$ elementary matrices $E_1$,$E_2$,$E_3$ such that $A$ = $E_1 E_2 E_3$
I figured out the operations which need to be performed which are;
$E_1$ = $R_2 \leftrightarrow R_1$
$E_2$ = $R_2$ = $R_2$ + $4R_1$
$E_3$ = $R_2$ * $\frac{1}{3}$
My question is how would I go about writing the elementary matrices? The solution says that they are;
$E_1$ =
$
\begin{align}
\begin{bmatrix}
1 & -4\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_2$ =
$
\begin{align}
\begin{bmatrix}
3 & 0\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_3$ =
$
\begin{align}
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
\end{align}
$
|
Note that the solutions are not unique. With your elementary row operations, we have
$$
\pmatrix{1&0\\ 0&\tfrac13}\pmatrix{1&0\\ 4&1}\pmatrix{0&1\\ 1&0}A = I_2.
$$
Therefore, by performing the reverse row operations (and also in reverse order) on $I_2$, we get
$$
A = \pmatrix{0&1\\ 1&0}\pmatrix{1&0\\ -4&1}\pmatrix{1&0\\ 0&3}I_2.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/404316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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|
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{1}=-1,\hspace{5mm} s_{2}=\frac18,\hspace{5mm} s_{3}=-\frac{215}{216},\hspace{5mm} s_{4}=\frac{155}{1728},\hspace{5mm} \text{for all} \quad k>4, \\ s_{k} &=\frac1{k^3(2k-3)}\left(\left(-4k^4+18k^3-25k^2+12k-2\right)s_{k-1}+\left(12k^3-39k^2+38k-10\right)s_{k-2} \right.\\
& \hspace{5mm} \left. +\left(4k^4-18k^3+25k^2-10k\right)s_{k-3}\\+\left(2k^4-15k^3+39k^2-40k+12\right)s_{k-4}\right),
\end{align}
but it could not express $S$ or $s_k$ in a closed form.
Can you suggest any ideas how to calculate $S$?
|
let $$y=\sum_{n=1}^{\infty}H^2_{n}x^n$$
then we have
$$y=x+xy+\ln^2{(1-x)}+\int_{0}^{x}\dfrac{\ln{(1-t)}}{t}dt$$
so
$$y=\dfrac{\ln^2{(1-x)}}{1-x}+\sum_{n=1}^{\infty}\left(1+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}\right)x^n$$
then you can use:Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Simple divisibility problem
Suppose $a$ and $b$ are distinct integers and $n\mid a^n-b^n$. Prove that $n\mid\frac{a^n-b^n}{a-b}$.
Here is how I do it: Write $a=b+d$, then $a^n-b^n=(b+d)^n-b^n=\binom{n}{1}b^{n-1}d+\binom{n}{2}b^{n-2}d^2+ \cdots + \binom{n}{n-1}bd^{n-1}+d^n \equiv 0 \pmod{n}$. Since $n| \binom{n}{k}$ for $k=1,2,\cdots,n-1$, $n|d^n$, which implies $n|d^{n-1}$. Therefore $(a^n-b^n)/(a-b)=\binom{n}{1}b^{n-1}+\binom{n}{2}b^{n-2}d+ \cdots + \binom{n}{n-1}bd^{n-2}+d^{n-1} \equiv 0 \pmod{n}$ and we are done. I wanna ask why $n|d^n$ will implies $n|d^{n-1}$ and that if there are any other possible solutions, thanks in advance.
|
Write down the unique prime factorization of $n = p_1^{k_1} \cdot p_2^{k_2} \cdot \ldots \cdot p_l^{k_l}$. Then $n \mid d^n$ implies $p_1 \cdot p_2 \cdot \ldots \cdot p_l \mid d$. Since $k_i < n$ for all $i \in \{1, \ldots , l\}$, we have $n \mid p_1^{n-1} \cdot p_2^{n-1} \cdot \ldots \cdot p_l^{n-1}$ and with $p_1^{n-1} \cdot p_2^{n-1} \cdot \ldots \cdot p_l^{n-1} \mid d^{n-1}$ our desired result $n \mid d^{n-1}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx$ This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$
where $\arctan{x}=\tan^{-1}{x}$
mark: this integral is my favorite one. Thanks to whoever has nice methods.
I have proved the following:
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$
where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
I also have got a few by-products
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$
where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$
and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$
and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant.
|
FWIW, here's Maple:
> f:= arctan(x)/(1+x)*ln((1+x^2)/2);
> int(f, x=-1..1);
$$ {\frac {7}{64}}\,{\pi }^{3}-{\frac {5}{16}}\,\pi \, \left( \ln
\left( 2 \right) \right) ^{2}-\ln \left( 2 \right) {\it Catalan}+1/
2\, \left( \ln \left( 1-i \right) \right) ^{2}\pi -1/2\,i \left(
\ln \left( 1+i \right) \right) ^{2}\ln \left( 2 \right) +\ln
\left( 1+i \right) {\it Catalan}+1/2\, \left( \ln \left( 1+i
\right) \right) ^{2}\pi -i{\it polylog} \left( 3,-i \right) +i{\it
polylog} \left( 3,i \right) +1/2\,i \left( \ln \left( 1-i \right)
\right) ^{2}\ln \left( 2 \right) -1/2\,i \left( \ln \left( 1-i
\right) \right) ^{3}-1/48\,i\ln \left( 1+i \right) {\pi }^{2}+1/2\,
i \left( \ln \left( 1+i \right) \right) ^{3}+\ln \left( 1-i
\right) {\it Catalan}+1/48\,i\ln \left( 1-i \right) {\pi }^{2}
$$
> simplify(%);
$$ \frac{\pi^3}{96}
$$
I don't know if that qualifies as "nice", but it's certainly easy.
|
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|
How to prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?
This should be easy, but I'm not seeing how. A hint would be appreciated.
|
Putting everything together, suppose $x,y>0$. Then $0\le2\sqrt{xy}$. Hence:
$$
\sqrt{x+y} = \sqrt{x+0+y}
\le \sqrt{x+2\sqrt{xy}+y}
= \sqrt{(\sqrt{x}+\sqrt{y})^2}
= \sqrt{x}+\sqrt{y}
$$
as desired. Note that this relied on the fact that $f(x)=\sqrt{x}$ is monotonically increasing.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Product of nilpotent matrices. Let $A$ and $B$ be $n \times n$ complex matrices and
let $[A,B] = AB - BA$.
Question: If $A , B$ and $[A,B]$ are all nilpotent matrices,
is it necessarily true that $\operatorname{trace}(AB) = 0$?
If,in fact, $[A,B] = 0$, then we can take $A$ and $B$ to be strictly upper triangular matrices so that the answer would be yes in this very special case.
|
The answer is no.
Take $A=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right)$, $B=XAX^{-1} = \left(\begin{array}{cccc} -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 1 & -1 & 1\end{array}\right)$, where we chose $X=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right)$.
Then $[A,B]= \left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 \end{array}\right)$ is nilpotent, but we have $AB=\left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right)$ and $\operatorname{trace}(AB)=-1\not=0$.
|
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|
Finding another way of doing this integral $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ Problem :
Integrate : $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
I have the solution : We can substitute $\sqrt{x}= \cos^2t$ and proceeding further,
I got the the answer which is $-2\sqrt{1-x}+\cos^{-1}\sqrt{x}+\sqrt{x-x^2}+C$
Can we do this problem some other way as well.. Thanks..
|
You could also try $\sqrt{x}=tan\theta$.
$dx=2tan\theta sec^2\theta d\theta$
$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\frac{1-tan\theta}{1+tan\theta}}$
Let $tan\theta=y$. Then $dy=sec^2\theta$. Substituting, we have
$\int{2y\sqrt{\frac{1-y}{1+y}}dy}=\int{\frac{2y(1-y)}{\sqrt{1-y^2}}dy}=\int{\frac{2y}{\sqrt{1-y^2}}dy}-\int{\frac{2y^2}{\sqrt{1-y^2}}dy}=I_{1}+I_{2}$
$I_{1}=-ln[\sqrt{1-y^2}]$; $I_{2}$: Let $y=sin\phi$. $dy=cos\phi d\phi$. $I_{2}=\int{2sin^2\phi d\phi}=\phi-\frac{sin2\phi}{2}=sin^{-1}y-y\sqrt{1-y^2}$
$I_{1}+I_{2}=y\sqrt{1-y^2} -ln[\sqrt{1-y^2}]-sin^{-1}y+C$. Now substiute $y=\sqrt{x}$.
|
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|
Square roots of complex numbers I know that the square root of a number x, expressed as $\displaystyle\sqrt{x}$, is the number y such that $y^2$ equals x. But is there any simple way to calculate this with complex numbers? How?
|
Write $(x+iy)^2=a+ib$, thus $x^2-y^2=a$ and $2xy=b$.
Now, $-x^2y^2=-b^2/4$, so $x^2$ and $-y^2$ are the solutions of $t^2-at-b^2/4=0$.
Then $\Delta=a^2+b^2$, and $t=\frac{a \pm \sqrt{a^2+b^2}}{2}$.
Obviously, since $x^2 \geq 0$ and $-y^2 \leq 0$, we have
$$x^2 = \frac{\sqrt{a^2+b^2}+a}{2}$$
$$y^2 = \frac{\sqrt{a^2+b^2}-a}{2}$$
Then, using the fact that $xy$ has the same sign than $b$, we have the two square roots of $a+ib$,
$$x+iy = \pm \left( \sqrt{\frac{\sqrt{a^2+b^2}+a}{2}} + i \ \mathrm{sign}(b) \sqrt{\frac{\sqrt{a^2+b^2}-a}{2}}\right)$$
with $\mathrm{sign}(x)=1$ if $x\ge0$ and $-1$ otherwise.
|
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|
In how many ways can you split a string of length n such that every substring has length at least m? Suppose you have a string of length 7 (abcdefg) and you want to split this string in substrings of length at least 2.
The full enumation of the possibilities is the following:
ab/cd/efg
ab/cde/fg
abc/de/fg
abc/defg
abcd/efg
abcde/fg
abcdefg
And perhaps some others.
In general, if we have a string of length n and we want to split the string in substrings of length at least m, in how many ways can we achieve this?
|
We can model the situation with generating functions. In order to do so we consider binary strings consisting of $0s$ and $1s$. Let $$0^\star=\{\varepsilon,0,00,000,\ldots\}$$ denote all strings containing only $0$s with length $\geq0$. The empty string is denoted with $\varepsilon$. The corresponding generating function is $$x^0+x^1+x^2+\ldots=\frac{1}{1-x},$$ with the exponent of $w^n$ marking the length $n$ of a string of $0s$ and the coefficient of $x^n$ marking the number of strings of length $n$.
We encode substrings of length at least $m(\geq 2)$ as strings starting with $1$ followed by at least $m-1$ zeros. So, each substring has the form
\begin{align*}
(1)(0^{m-1})(0^\star)
\end{align*}
with generating function
\begin{align*}
\frac{x^m}{1-x}
\end{align*}
$$ $$
Since each string consists of one or more substrings, the strings can be encoded as
\begin{align*}
(10^{m-1}0^\star)(10^{m-1}0^\star)^\star
\end{align*}
with generating function
\begin{align*}
\frac{\frac{x^m}{1-x}}{1-\frac{x^m}{1-x}}=\frac{x^m}{1-x-x^m}\qquad\qquad m\geq 2
\end{align*}
$$ $$
OPs example with $m=2$ results in a generating function for the Fibonacci Numbers
\begin{align*}
\frac{x^2}{1-x-x^2}=x^2+x^3+2x^4+3x^5+5x^6+8x^7+\mathcal{O}(x^8)
\end{align*}
So, the coefficient of $x^7$ is equal to $8$ corresponding to eight different strings of length $7$ with substrings of length at least $2$. These strings are
\begin{align*}
&1010100\quad\quad(2,2,3)\\
&1010010\quad\quad(2,3,2)\\
&1001010\quad\quad(3,2,2)\\
&1010000\quad\quad(2,5)\\
&1001000\quad\quad(3,4)\\
&1000100\quad\quad(4,3)\\
&1000010\quad\quad(5,2)\\
&1000000\quad\quad(7)
\end{align*}
|
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For any positive integer $n$, is it possible to find a nonzero integer $p$ so that $p^2$ is the sum of $i$ nonzero squares for all $1 \leq i \leq n$? I need to prove this result for something I am working on:
For any positive integer $n$, is it possible to find a nonzero integer $p$ so that $p^2$ is the sum of $i$ nonzero squares for all $1 \leq i \leq n$?
What I'd like to do is as follows. First find $a,b,c$ nonzero integers so that $a^2 + b^2 = c^2$. Then find a nonzero integers $d$ and $e$ so that $c^2 + d^2 = e^2$. Then find nonzero integers $f$ and $g$ so that $e^2 + f^2 = g^2$. Note that then
$$
g^2 = e^2 + f^2 = c^2 + d^2 + f^2 = a^2 + b^2 + d^2 + f^2
$$
and I'd like to continue in this way. However, I am wondering if this construction is even possible? Am I always guaranteed that such a string of integers exist? If so, how do I prove this rigorously? Another thought of mine is use Euclid's Formula. For example, let $m = 3$ and $n = 1$. Then
$$
\begin{align*}
a &= 3^2 - 1^2 = 8\\
b &= 2\cdot 3 \cdot 1 = 6\\
c &= 3^2 + 1^2 = 10.
\end{align*}
$$
Furthermore, $c = 2\cdot 5 \cdot 1$, so letting $m = 5$ and $n = 1$ gives
$$
\begin{align*}
d &= 5^2 - 1^2 = 24\\
e &= 5^2 + 1^2 = 26.
\end{align*}
$$
And again, $e = 2\cdot 13 \cdot 1$, so letting $m = 13$ and $n = 1$ gives
$$\begin{align*}
f &= 13^2 - 1^2 = 168\\
g &= 13^2 + 1^2 = 170.
\end{align*}$$
Then
$$\begin{align*}
g^2 = 170^2 = 26^2 + 168^2 = 10^2 + 24^2 + 168^2 = 6^2 + 8^2 + 24^2 + 168^2
\end{align*}
$$
Going one more step we can write $170 = 2\cdot 17 \cdot 5$ so $m = 17$ and $n = 5$ gives
$$\begin{align*}
h &= 17^2 - 5^2 = 264\\
i &= 17^2 + 5^2 = 314,
\end{align*}$$
in which $314 = 2\cdot 157 \cdot 1$. It seems I may be on the right track, but I'm not sure if this is true in general. Maybe it can be proven that this process can be continued arbitrarily given we start with two distinct primes $m$ and $n$ in the first step? Any thoughts or references or anything would be wonderful. Thanks in advance!
|
Yes, your process can always be continued if you start with $m_1>n_1$ both odd.
$$
a_1 = m_1^2-n_1^2 \equiv 0 \pmod{4} \\
b_1 = 2m_1n_1 \equiv 2 \pmod{4} \\
c_1 = m_1^2+n_1^2 \equiv 2 \pmod{4}
$$
So $c_1/2$ is odd and we can factor it as $m_2\cdot n_2$ with $m_2>n_2$ both odd, take $m_2=c_1/2, n_2=1$ if necessary or any other factorization. Then for $k>1$
$$
a_k=m_k^2-n_k^2 \\
b_k = 2m_k n_k = c_{n-1} \\
c_k = m_k^2+n_k^2 \equiv 2 \pmod{4}
$$
and $c_k/2$ is odd for every $k$ and we can repeat.
|
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|
How do I solve such logarithm I understand that
$\log_b n = x \iff b^x = n$
But all examples I see is with values that I naturally know how to calculate (like $2^x = 8, x=3$)
What if I don't? For example, how do I solve for $x$ when:
$$\log_{1.03} 2 = x\quad ?$$
$$\log_{8} 33 = x\quad ?$$
|
The logarithm $\log_{b} (x)$ can be computed from the logarithms of $x$ and $b$ with respect to a positive base $k$ using the following formula:
$$\log_{b} (x) = \frac{\log_{k} (x)}{\log_{k} (b)}.$$
So your examples can be solved in the following way with a calculator:
$$x = \log_{1.03} (2) = \frac{\log_{10} (2)}{\log_{10} (1.03)} =
\frac{0.301}{0.013} = 23.450, $$
$$x = \log_{8} (33) = \frac{\log_{10} (33)}{\log_{10} (8)} =
\frac{1.519}{0.903} = 1.681.$$
If you know that $b$ and $x$ are both powers of some $k$, then you can evaluate the logarithm without a calculator by the power identity of logarithms, e.g.,
$$x = \log_{81} (27) = \frac{\log_{3} (27)}{\log_{3} (81)} =
\frac{\log_{3} (3^3)}{\log_{3} (3^4)} = \frac{3 \cdot \log_{3} (3)}{4 \cdot \log_{3} (3)} =
\frac{3}{4}.$$
|
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|
Given that $x = 4\sin \left( {2y + 6} \right)$ find dy/dx in terms of x My attempt:
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr} $
$\eqalign{
& x = 4\sin \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = \left( 2 \right)\left( 4 \right)\cos \left( {2y + 6} \right) \cr
& {{dx} \over {dy}} = 8\cos \left( {2y + 6} \right) \cr
& {{dy} \over {dx}} = {1 \over {8\cos \left( {2y + 6} \right)}} \cr
& {\cos ^2}\left( {2y + 6} \right) + {\sin ^2}\left( {2y + 6} \right) = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) + {\left( {{x \over 4}} \right)^2} = 1 \cr
& {\cos ^2}\left( {2y + 6} \right) = 1 - {{{x^2}} \over {16}} \cr
& \cos \left( {2y + 6} \right) = \sqrt {{{16 - {x^2}} \over {16}}} \cr
& \cos \left( {2y + 6} \right) = {{\sqrt {16 - {x^2}} } \over 4} \cr
& {{dy} \over {dx}} = {1 \over {2\sqrt {16 - {x^2}} }} = 1 \cr} $
Okay I've got it right, but the official answer confuses me, it says:
${{dy} \over {dx}} = {1 \over {8cos\left( {\arcsin \left( {{x \over 4}} \right)} \right)}} = \left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
This is the part i'm struggling to get my head around, although I arrive at the same answer.
Okay
$\arcsin {x \over 4} = 2y + 6$
but how does the answer then go :
$\left( {\left( \pm \right){1 \over {2\sqrt {\left( {16 - {x^2}} \right)} }}} \right)$
is there a shortcut or trick I overlooked?
I think I need some sleep, thanks...
|
To find $\frac{dy}{dx}$ in terms of $x$ for $x=4 \sin(2y+6)$, I wouldn't bother with implicit differentiation. Instead:
First, solve for $y$,
$$\frac{x}{4} = \sin(2y+6)$$
$$\arcsin(\frac{x}{4}) = 2y + 6$$
$$y = \frac{\arcsin(\frac{x}{4})-6}{2} = \frac{1}{2} \arcsin{\frac{x}{4}} - 3.$$
Now differentiate with respect to $x$,
$$\frac{dy}{dx} = \frac{1}{2} \frac{1}{\sqrt{16-x^2}} = \frac{1}{2 \sqrt{16-x^2}}.$$
|
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|
How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
|
As $\sin3x=3\sin x-4\sin^3x$
$$\sin^6\alpha=(\sin^3\alpha)^2=\left(\frac{3\sin\alpha-\sin3\alpha}4\right)^2$$
$$=\frac{9\sin^2\alpha+\sin^23\alpha-3(2\sin\alpha\sin2\alpha)}{16}$$
$$=\frac{9(1-\cos2\alpha)+1-\cos6\alpha-6(\cos\alpha-\cos3\alpha)}{32}$$
using $2\sin A\sin B=\cos(A-B)-\cos(A+B)$ and $\cos3x=1-2\sin^2x$
Now, we find $\int_o^\pi\cos mx dx=\frac{\sin mx}m\big |_0^\pi=\frac{\sin m\pi}m$ which becomes $0$ for any integer $m$
So, the definite integral for all the cosine ratios in the given integral will vanish leaving behind $\frac{9+1}{32}=\frac5{16}$ to be integrated resulting in $$\frac5{16}\cdot \pi$$
Alternatively,
$$I=\int_0^\pi\sin^6\alpha d\alpha=\int_0^{\frac\pi2}\sin^6\alpha d\alpha+\int_{\frac\pi2}^\pi\sin^6\alpha d\alpha$$
Now,putting $\beta=\alpha-\frac\pi2$ $$\int_{\frac\pi2}^\pi\sin^6\alpha d\alpha=\int_0^{\frac\pi2}\cos^6\beta d\beta=\int_0^{\frac\pi2}\cos^6\alpha d\alpha$$
So, $$I=\int_0^\frac\pi2(\sin^6\alpha+ \cos^6\alpha )d\alpha$$
Now,
$$\sin^6\alpha+ \cos^6\alpha=(\sin^2\alpha+\cos^2\alpha)^3-3(\sin^2\alpha\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)$$
$$=1-\frac34(\sin^22\alpha)=1-\frac{3(1-\cos4\alpha)}8=\frac{5+3\cos4\alpha}8$$
|
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|
Number of ways to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball
How many ways are there to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball?
My approach is $ \binom{5}{3} 3! $ + $ \binom{2}{2} \binom{3}{2}2!$ which is equal to $66$?
|
$$\sum_{r_1+r_2+r_3=5,r_i\geq1}\binom{5}{r_1}\binom{5-r_1}{r_2}\binom{5-r_1-r_2}{r_3}=$$
$$=\sum_{r_1+r_2+r_3=5,r_i\geq1}\frac{5!}{r_1!r_2!r_3!}=\frac{5!}{1!1!3!}+\frac{5!}{1!3!1!}+\frac{5!}{3!1!1!}+\frac{5!}{1!2!2!}+\frac{5!}{2!1!2!}+\frac{5!}{2!2!1!}=$$
$$=3(20+30)=150$$
|
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|
Find the volume using triple integrals Using triple integrals and Cartesian coordinates, find the volume of the solid bounded by
$$ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $$
and the coordinate planes $x=0, y=0,z=0$
My take
I have set the parameters to $$ 0\le x \le a$$ $$0\le y \le b\left( 1 - \frac{x}{a} \right)$$ $$0\le z \le c \left( 1 - \frac{y}{b} -\frac{x}{a} \right) $$ and evaluated $$ \int_0^{a} \int_0^{b\left( 1 - \frac{x}{a}\right)} \int_0^{c \left( 1 - \frac{y}{b} -\frac{x}{a} \right)} 1 dzdydx$$ and gotten $0$ as my final answer but the actal answer is $\frac{abc}{6}$
Never mind, I found the mistake I was making, just a simple integral mistake but it was the right procedure. Thank you for viewing! :)
|
Here is an alternative computation using a single variable integral that confirms your result. The following figure represents the given pyramid.
The equations of the lines situated on the planes $y=0$ and $z=0$ are:
$$y=0,\qquad\frac{x}{a}+\frac{z}{c}=1\Leftrightarrow z=\left( 1-\frac{x}{a}\right) c,$$
$$z=0,\qquad\frac{x}{a}+\frac{y}{c}=1\Leftrightarrow y=\left( 1-\frac{x}{a}\right) b.$$
The intersection of the pyramid with the plane perpendicular to the $x$-axis in $x$ is a right triangle with catheti $\left( 1-\frac{x}{a}\right) c$ and $\left( 1-\frac{x}{a}\right) b$, whose area $A(x)$ is given by
$$A(x)=\frac{1}{2}\left( 1-\frac{x}{a}\right) b\left( 1-\frac{x}{a}\right) c=\frac{bc}{2}\left( 1-\frac{x}{a}\right) ^{2}.$$
Hence the volume is given by the integration of the area $A(x)$ from $x=0$ to $x=a$
$$\begin{eqnarray*}
V &=&\int_{0}^{a}A(x)dx \\
&=&\frac{bc}{2}\int_{0}^{a}\left( 1-\frac{x}{a}\right) ^{2}dx \\
&=&\frac{bc}{2}\left( a-\frac{2}{a}\frac{a^{2}}{2}+\frac{1}{a^{2}}\frac{a^{3}
}{3}\right) \\
&=&\frac{abc}{6}.
\end{eqnarray*}$$
|
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Trying to prove that $\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$ How could one prove that: $$\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$$
This is about as far as I got:
$$\prod_{k=1}^j \frac{2 k}{j+k-1} = \frac{2^j j!}{(j)_j} \implies$$
$$\sum_{j=2}^\infty\frac{2^j j!}{(j)_j} = \frac{2^2 2!}{(2)_2} + \frac{2^3 3!}{(3)_3} + \frac{2^4 4!}{(4)_4}+\cdots \implies$$
$$?$$
where $(x)_n$ denotes the Pochhammer symbol.
Maybe reduction isn't the way to go?
|
This is going to be a little out of the blue, but here goes.
Consider the function
$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
$f(x)$ has a Maclurin expansion as follows:
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$
Differentiating, we get
$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle \binom{2 n}{n}} x^{2 n}$$
Evaluate at $x=1/\sqrt{2}$:
$$f'\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{2}+2 = \sum_{n=0}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} $$
Thus we have established that
$$\sum_{n=2}^{\infty} \frac{2^{n}}{\displaystyle \binom{2 n}{n}} = \frac{\pi}{2}$$
Now consider the original sum:
$$\begin{align}\sum_{n=2}^{\infty} \prod_{k=1}^n \frac{2 k}{n+k-1}&= \sum_{n=2}^{\infty}\frac{2^n n!}{n (n+1) \cdots (2 n-1)}\\ &=\sum_{n=2}^{\infty}\frac{2^n n! (n-1)!}{(2 n-1)!}\\ &= 2 \sum_{n=2}^{\infty}\frac{2^n}{\displaystyle \binom{2 n}{n}} \\&= 2 \frac{\pi}{2} \\ &= \pi \end{align} $$
QED
|
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Prove that $\log _5 7 < \sqrt 2.$
Prove that $\log _5 7 < \sqrt 2.$
Trial : Here $\log _5 7 < \sqrt 2 \implies 5^\sqrt 2 <7.$ But I don't know how to prove this. Please help.
|
Observe that:
$$ \begin{align*}
\log_5 7 &= \dfrac{3}{3}\log_5 7 \\
&= \dfrac{1}{3}\log_5 7^3 \\
&= \dfrac{1}{3}\log_5 343 \\
&< \dfrac{1}{3}\log_5 625\\
&= \dfrac{1}{3}\log_5 5^4\\
&= \dfrac{1}{3}(4)\\
&= \sqrt{\dfrac{16}{9}}\\
&< \sqrt{\dfrac{18}{9}}\\
&= \sqrt{2}\\
\end{align*} $$
as desired.
|
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The dimension of the vector space of all trace-zero symmetric matrices Find the dimension of the vector space of all symmetric matrices of order $n \times n$ (real entries) and trace equal to zero.
|
$$V_F=\{A\in M_M(F)\:A^T=A , Trace(A=0)\}=$$ forall $A\in V_F$ we have $trac(A)=0$ and $a_{ij}=a_{ji}$ such that $A=[a_{ij}]_{n \times n}$
$$A=\begin{pmatrix}
a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\
a_{12} & a_{22} & a_2^2 & \cdots & a_2^n \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
a_{1n} & a_{2n} & a_{3n} & \cdots & (-a_{11}-...-a_{n-1 n-1})
\end{pmatrix}=a_{11}\begin{pmatrix}
1 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -1
\end{pmatrix}+a_{22}\begin{pmatrix}
0 & 0 & 0 & \cdots & 0 \\
0 & 1 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -1
\end{pmatrix}+...+a_{n-1n-1}\begin{pmatrix}
0 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -1
\end{pmatrix}+a_{12}\begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 0
\end{pmatrix}+a_{13}\begin{pmatrix}
0 & 0 & 1 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots &
\end{pmatrix}+....+a_{1n}\begin{pmatrix}
0 & 0 & 0 & \cdots & 1 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 0
\end{pmatrix}+........+a_{n-1n-1}\begin{pmatrix}
0 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots \\0 & 0 & 0 & \cdots & 1 \\
0 & 0 & 0 & \cdots & 0
\end{pmatrix}\Rightarrow$$$$V_{F}=\{\sum_{i=1}^{n-1}a_{i}E_{ii}+\sum_{i\gt j}c_jE_{ij}\}=<E_{11},E_{22},...,E_{n-1\times n-1},E_{12,....E_{1n},.....,E_{n-1\times n}}>=<X>$$ such that $ X =\{E_{11},E_{22},...,E_{n-1\times n-1},E_{12},....E_{1n},.....,E_{n-1\times n}\}$ then $X$ generate $V_F $ and easily we can prove $ X$ is linear independent totally X is basis for $V_F$ and $$|x|=\frac{(n-1)(n+2)}{2}$$
|
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|
Proving that a graph is self complementary I've been given the following adjacency matrix:
$$\left(\begin{array}{cccccccc} 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 &
1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 &
1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 &
1 \\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 &
0 \\ \end{array}\right)$$
and I need to prove that it is a self complementary graph. All this must be done by hand so I found the complementary graph but I don't know how to continue.
Do I need to solve
$$Ax=b$$
or
$$x^{t}Ax=b$$
is $x$ a matrix defined by 8 columns and 8 rows?
|
The blue graph has the big square 2-4-6-8 with little triangles 2-3-4, 4-5,6, 6-7-8, and 8-1-2, and also the little triangle vertices which are opposite relative to the big square are joined, i.e. 1-5 and 3-7.
The red graph has a similar description: there is a big square 1-3-5-7, and triangles 1-6-3, 3-8-5, 5-2-7, and 7-4-2, and also the triangle vertices which are opposite relative to the big square are joined, i.e. 2-6 and 8-4.
This means one can set up an isomorphism by corresponding the vertices as
Blue: $[1,2,3,4,5,6,7,8],$
Red: $[4,1,6,3,8,5,2,7].$
|
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Maximum of a trigonometric function without derivatives I know that I can find the maximum of this function by using derivatives but is there an other way of finding the maximum that does not involve derivatives? Maybe use a well-known inequality or identity?
$f(x)=\sin(2x)+2\sin(x)$
|
The idea is to use $\sin^2 x + \cos^2x = 1$ to reduce to dealing with only 1 trigonometric function, and then proceed as a standard 1-variable inequality.
We wish to find the maximum of $f(x) = \sin 2x + 2 \sin x = 2 \sin x ( 1 + \cos x )$.
It is clear that we may assume $\sin x \geq 0, \cos x \geq 0$, to maximize this product.
Let's consider $$[f(x)]^2 = 4 \sin^2 x ( 1 + \cos x)^2 = 4 (1-\cos^2x ) ( 1 + \cos x)^2 = 4 ( 1 - \cos x ) ( 1 + \cos x)^3 $$
By AM-GM, applied to $ 3(1-\cos x) , (1 + \cos x), (1 + \cos x), (1 + \cos x)$, we get that
$ \sqrt[4]{3 ( 1 - \cos x ) ( 1 + \cos x)^3} \leq \frac{ 6}{4}$, or that $ ( 1 - \cos x ) ( 1 + \cos x)^3 \leq \frac{27}{16}$.
Hence, $ [f(x)]^2 \leq \frac{27}{4}$ so $f(x) \leq \frac{3 \sqrt{3} } {2} $.
It remains to verify that equality can occur, which it does at $3(1-\cos x) = 1 + \cos x$ of $\cos x = \frac{1}{2}$.
|
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Divisibility by Quadratics $b^2+ba+1\mid a^2+ab+1\Rightarrow\ a=b$
The natural numbers $a$ and $b$ are such $a^2+ab+1$ is divisible by $b^2+ba+1$. Prove that $a = b$.
I tried to algebraically manipulate it as follows:
$(b^2 + ba + 1)k = a^2 + ab + 1$
$[b(a + b) + 1]k = a(a + b) + 1$
$kb(a + b) + k = a(a + b) + 1$
$k - 1 = (a - kb)(a + b)$
I'm stuck here. What should I do next? A case-by-case analysis of possible congruencies would be too tedious and inelegant.
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If $a^2+ab+1$ is divisible by $b^2+ba+1$, then so is $(a^2+ab+1)-(b^2+ba+1)=a^2-b^2$.
Note that $a+b$ and $b^2+ba+1$ are relatively prime. So $b^2+ab+1$ divides $a-b$. Now you should be able to finish, using considerations of size.
|
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|
Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (examples of my working below):
*
*U-Substitution.
*Guess and Check.
*Reversing the question (multiplying the answer out).
Here is my working for each case so far.
(1) U-Substitution.
I tried a simpler case via u-substitution.
Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.
(2) Guess and Check.
To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:
$(x^2 \pm y^2)(x^2 \pm y^2)$.
But I'm at a loss as to how to get the final solution's middle term from guessing and checking.
(3) Reversing the question
Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.
The original answer:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$
$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$
$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$
$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$
The $3y^3x$ terms cancel out, and we are left with:
$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.
The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).
|
Note that the answer is not unique. There are four linear factors, and you can get different quadratic factors by grouping them the other way.
For instance:
$$
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= x^4 + 2x^2y^2 + y^4 - 13x^2y^2 \\
&= (x^2+y^2)^2-(\sqrt{13}xy)^2 \\
&= (x^2+y^2-\sqrt{13}xy)(x^2+y^2+\sqrt{13}xy).
\end{align*}
$$
I admit that I got the given answer first (by the same method as this answer)
, because 9 looks more like a square than 13.
|
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|
Limit on the expression containing sides of a triangle To find the bounds of the expression $\frac{(a+b+c)^2}{ab+bc+ca}$, when a ,b, c are the sides of the triangle.
I could disintegrate the given expression as $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} + 2$$ and in case of equilateral triangle, the limit is 3.
Now how to proceed further?
|
Without loss of generality, $a \leqslant b \leqslant c = 1$. Thus it remains to see that
$$1 \leqslant \frac{1+a^2+b^2}{a + b + ab} \leqslant 2.$$
For the left inequality, we rewrite
$$a + b + ab \leqslant 1 + a^2 + b^2 \iff 0 \leqslant 1 - a - b + ab + a^2 -2ab + b^2 = (1-a)\cdot(1-b) + (b-a)^2,$$
which is evident.
For the right inequality,
$$1+a^2+b^2 \leqslant 2(a+b) + 2ab \iff 1 + (b-a)^2 \leqslant 2(a+b),$$
which is seen to be true since by assumption - that $a \leqslant b\leqslant 1$ be the sides of a triangle - $0 \leqslant b-a \leqslant 1 \leqslant a+b$.
letting $a \to 0$ resp. $a \to 1$ (both imply $b \to 1$) shows that the bounds are sharp.
|
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|
Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
This is similar to the series $$\frac{1}{4} + \frac{1\cdot 3}{4 \cdot 6} + \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \dotsb,$$which one can show is convergent by Raabe's test, as follows: $$\frac{a_{n+1}}{a_n}= \frac{2n-1}{2n+2}= \frac{2n + 2 - 3}{2n+2}=1 - \frac{3}{2(n+1)}.$$
However, in the series I am looking at, $\frac{a_{n+1}}{a_n}=\frac{(2n-1)^2}{(2n)^2}= 1 - \frac{4n-1}{4n^2}$, so Raabe's test doesn't work (this calculation also happens to show that the ratio and root tests will not work).
Any ideas for this one? It seems the only thing left is comparison...
|
There are several methods as commented already, but I will put mine for what is worth.
Here's my answer:
Let
$$
a_n=\frac 1 2 \cdot \frac 3 4 \cdots \frac{2n-1}{2n}$$
Then
$$
a_n^2=\frac 1 2 \frac 1 2 \frac 3 4 \frac 3 4 \cdots \frac{2n-1}{2n}\frac{2n-1}{2n}$$
$$a_n^2 > \frac 1 2 \frac 1 2 \frac 2 3 \frac 3 4 \cdots \frac{2n-2}{2n-1}\frac{2n-1}{2n}=\frac 1 {4n}$$
Thus, $\sum a_n$ diverges.
|
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|
Finding the relation between function x,y,z - trigo problem Problem :
For $\displaystyle 0 < \theta < \frac{\pi}{2}$ if
$$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$
then
options are :
(a) $xyz = xz+y$
(b) $xyz = xy+z$
(c) $xy^2 =y^2+x$
I have solution of this however I have one doubt in that, it is mention that :
$\displaystyle x=\sum^{\infty}_{n =0} \cos^{2n}\theta = \frac{1}{1-\cos^2\theta}$ (How it is derived... or what about this result.). Please guide on this... thanks.
|
$$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$
$$x=1+{\cos^2\theta}+{\cos^4\theta}+\cdots\infty \;terms$$
this is infinite geometric series :
$S_{\infty}=\dfrac {a}{1-r}\;\;\,|r|<1$ ,here $\cos\theta<1, when\; 0<\theta<\dfrac \pi2 \;so\; \cos^2\theta<1$
$$x=\dfrac {1}{1-\cos^2\theta}\implies \dfrac {1}{\sin^2\theta} \implies x=\csc^2\theta$$ here $a=1$ and r=$\cos^2\theta$
similarly for y $a=1$ and $r=\sin^2\theta$
$$y=\dfrac {1}{1-\sin^2\theta}\implies \dfrac {1}{\cos^2\theta}\implies y=\sec^2\theta$$
and $$z=\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$
so option (b) $$xy+z=\csc^2\theta\cdot\sec^2\theta+\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$
$$xy+z=\dfrac {\csc^2\theta\cdot\sec^2\theta}{1-\cos^2\theta\cdot\sin^2\theta}$$
$$xy+z=xyz$$
|
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|
How many times does the function $y=e^x $ meet $y=x^2$? As you know $y=e^x$ and $y= x^2$ meet once on $x<0$.
But I want to know whether or not they meet on $x>0$.
Since $\lim_{x\rightarrow \infty } e^x/x^2=\infty$, if they meet once on $x>0$, they
must meet again.
To summarize, my question is whether or not they meet on $x>0$.
Thank you in advance.
|
Note that $$e^x - x^2 = \left(e^{x/2}-x\right)\left(e^{x/2}+x\right)$$
For $x\ge 0$, the second factor is strictly positive, so consider the first factor:
$$\begin{align}e^{x/2}-x &= \left(1+\frac{x}{2}+\frac{x^2}{8} + \frac{x^3}{48} + \cdots\right) - x \\
&= 1 - \frac{x}{2}+\frac{x^2}{8} + \frac{x^3}{48}+\cdots \\[6pt]
&= \frac{1}{2} + \frac{1}{8}\left( 4-4x+x^2 \right) + \frac{x^3}{48} + \cdots \\[6pt]
&= \frac{1}{2} + \frac{1}{8} \left( 2-x \right)^2 + \frac{x^3}{48} + \cdots \\[6pt]
&\ge \frac{1}{2}
\end{align}$$
Thus, that first factor is also strictly positive, whence $e^x - x^2$ is as well. QED
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|
Integral of $\int \frac{dx}{\sqrt{x^2 -9}}$ $$\int \frac{dx}{\sqrt{x^2 -9}}$$
$x = 3 \sec \theta \implies dx = 3 \sec\theta \tan\theta d\theta$
$$\begin{align} \int \frac{dx}{\sqrt{x^2 -9}} & = \frac{1}{3}\int \frac{3 \sec\theta \tan\theta d\theta}{\tan\theta} \\ \\ & = \int \sec\theta d\theta \\ \\ & = \ln | \sec\theta + \tan\theta| + c\end{align}$$
$x = 3\sec \theta \implies \sec\theta = \frac {x}{3}$
$\tan\theta = \frac{\sqrt{x^2 - 9}}{x}$
I have have confused x with 3 but I cannot get the proper answer which is
$$\ln | x + \sqrt{x^2 - 9}| + c$$
I always get $\dfrac{x}{3}$ or $\dfrac{\sqrt{x^2 - 9}}{x}$ or some variation of that, I can't eliminate them to get their answer.
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What you should end with is
$\sec\theta = \dfrac x3\quad$ and $\quad\tan \theta = \dfrac{\sqrt{ x^2 - 9}}{3}$.
Then you have $$\begin{align} \log \Big| \frac 13\left(x + \sqrt{x^2 - 9}\right)\Big| + C & = \log|x +\sqrt{x^2 - 9}| -\log 3 + C \\ \\ & = \log|x + \sqrt{x^2 - 9}| + C'\end{align}$$
|
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|
what is the explicit form of this iterativ formular I am not sure, if there is an explicit form, but if there is, how do I get it?
This is the formula:
$$c_n=\frac{1-n \cdot c_{n-1}}{\lambda}$$
where $\lambda \in \mathbb{R}$ and $n \in \mathbb{N}$
I already tried some forms for c via trail and error, but I couldn't find a solution ...
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We have $$\begin{align} c_{10} & = \tfrac 1\lambda - \tfrac{10}\lambda c_9 \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} (1-9c_8) \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^2} c_8 \\
& = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}c_7 \\
& = \ldots = \tfrac 1\lambda - \tfrac{10}{\lambda^2} + \tfrac{10\cdot 9}{\lambda^3} -\tfrac{10\cdot 9\cdot 8}{\lambda^4}+\tfrac{10\cdot 9\cdot 8\cdot 7}{\lambda^5} - \ldots + (-1)^{10} \frac{10!}{\lambda^{11}}\cdot c_0 \end{align}$$
So if I have not done any mistakes, the solution is
$$c_n = \sum_{k=0}^{n-1} \left((-1)^k \frac{n!}{(n-k)!\cdot \lambda^{k+1}} \right) + (-1)^n \frac{n!}{\lambda^{n+1}} c_0$$
|
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|
Volume using disk method $y = 2\sqrt{x} \quad y=x$ $y = 2\sqrt{x}$
$y=x$
about $x=-2$
I know that $y=x$ is on the outside and they meet at 4.
I need these in terms of y since I rotate about y.
$x = y$
$x = \frac{y^2}{4}$
$$\pi \int_0^4 (y - (-2))^2 - \left(\frac{y^2}{4} - (-2)\right)^2 dy$$
$$\pi \int_0^4 (y + 2)^2 - \left(\frac{y^2}{4} + 2\right)^2 dy$$
$$\pi \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4y+4\right) dy$$
$$\pi \int_0^4 y^2 - \frac{y^4}{16} dy$$
$$\pi \left(\frac{y^3}{3} - \frac{y^5}{16*5}\right)$$
Gives me $\dfrac{128\pi}{15}$
Where did I go wrong?
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The issue appears to be in your second expansion. $$\begin{align}\left(\frac{y^2}4-(-2)\right)^2 &= \left(\frac{y^2}4+2\right)^2\\ &= \left(\frac{y^2}4\right)^2+2\left(\frac{y^2}4\right)(2)+(2)^2\\ &= \frac{y^4}{16}+y^2+4.\end{align}$$
Can you get the rest of the way from there?
|
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|
Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are particularly sought.
As an example of the utility of this result,
set $x = \frac12$.
Then
this shows that
$2 > (1+\frac1{2n})^n$
or
$2^{1/n} > 1+\frac1{2n}$
.
This is an example
of what I call a
contra-Bernoulli inequality
(CBI)
which gives an upper bound to
$(1+y)^n$
as opposed to Bernoulli's inequality,
which gives a lower bound to
$(1+y)^n$
of $1+ny$.
Note that
any CBI of the form
$(1+y)^n < 1+c y$
for $n \ge 2$
requires that $y$ is bounded,
since
$(1+y)^n > 1+y^n$
so
$1+cy > 1+y^n$
or
$cy > y^n$
or
$y < c^{1/(n-1)}$.
|
For the first part, we only need the binomial theorem:
$$\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k} = \sum_{k=0}^n \frac{\prod_{j=1}^k(n+1-j)}{k!n^k}x^k \leqslant \sum_{k=0}^n \frac{x^k}{k!} \leqslant \sum_{k=0}^n x^k < \frac{1}{1-x}.$$
For the second, we observe that
$$\frac{1}{1-x} \leqslant 1 + cx$$
for $0 < x < \frac{c-1}{c}$, since
$$(1-x)(1+cx) -1 = (c-1)x - cx^2$$
has zeros in $x = 0$ and $x = \frac{c-1}{c}$, and is positive between the zeros, since the coefficient of $x^2$ is negative.
|
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Find all values of a for which the equation $x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$ possesses at least two distinct negative roots Find all values of a for which the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0 $$ possesses at least two distinct negative roots.
I am able to prove that all roots would be negative .How to proceed after this.
|
This is a symmetric (the coefficients) polynomial. We may want to take advantage of that. We can divide by $x^2$ and get $$(x+\frac{1}{x})^2+(a-1)(x+\frac{1}{x})-1=0$$
From there we can solve for $x+\frac{1}{x}$ and get $$x+\frac{1}{x}=\frac{-(a-1)\pm\sqrt{(a-1)^2+4}}{2}.$$
This is a quadratic equation. Solve for $x$ and get the roots.
|
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|
An identity which applies to all of the natural numbers Prove that any natural number n can be written as $$n=a^2+b^2-c^2$$ where $a,b,c$ are also natural.
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I like this approach to the solution of this equation.
If we consider the Diophantine equation: $qX^2+Y^2=Z^2+j$
If the root is a : $a=\sqrt{\frac{j}{q}}$
We use the solutions of Pell's equation: $p^2-(q+1)s^2=1$
Solutions can be written:
$X=2s(s\pm{p})L\pm{ap^2}+2aps\pm{a(q+1)s^2}=bL+af$
$Y=(p^2\pm2ps+(1-q)s^2)L\pm{ap^2}+2aps\pm{a(q+1)s^2}=cL+af$
$Z=(p^2\pm2ps+(q+1)s^2)L\pm{ap^2}+2a(q+1)ps\pm{a(q+1)s^2}=fL+at$
$L$ - any integer number given by us
number: $b,c,f,t$ - are solutions of the following equations
$qb^2+c^2=f^2$
$t^2-(q+1)f^2=\pm{q}$
If we take the solutions of Pell's equation: $p^2-(q+1)s^2=k$
number $b,c$ - are solutions of the equation: $qb^2+c^2=f^2$
wherein: $c-b=k$
number $t,f$ - solutions of the equation: $t^2-(q+1)f^2=\pm{qk^2}$
These formulas allow us to find some solutions of Pell's equation using solutions of simpler equations. At least there will be another opportunity to find a solution to this equation. Later draw solutions with other factors.
All of numbers can be any character.In Equation: $qX^2+Y^2=Z^2+a$
If the ratio is factored so: $a=(b-c)(b+c)$
Then we use the solutions of Pell's equation: $p^2-fs^2=\pm1$
where: $f=(q+1)k^2-2kt-(q-1)t^2$
Then the solutions are of the form:
$X=2(ck-bt)ps+2(bk^2-(b+c)kt+ct^2)s^2$
$Y=bp^2+2c(k-t)ps-(b(q-1)k^2+2(b-qc)kt+b(q-1)t^2)s^2$
$Z=cp^2+2b(k-t)ps+(c(q+1)k^2-2(bq+c)kt+c(q+1)t^2)s^2$
All of numbers can be any character.
|
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|
Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $
Problem: Find $\displaystyle\int \frac {\tan 2x} {\sqrt {\cos^6 x +\sin^6 x}} dx $
Solution: $\tan 2x= \dfrac{2\tan x}{1-\tan^2 x}$
Also I can take $\cos^6x$ common from $\sqrt {\cos^6x +\sin^6x}$
I don't know whether it is good approach to the question
Please help
|
HINT:
$$\cos^6x+\sin^6x=(\cos^2x+\sin^2x)^3-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$
$$=1-3\cos^2x\sin^2x=1-\frac34(\sin2x)^2$$
$$=1-3\cos^2x\sin^2x=1-
\frac34(\sin2x)^2=1-\frac34(1-\cos^22x)=\frac{1+3\cos^22x}4$$
Use $\cos2x=u$
|
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|
If $x,y,z \in \Bbb{R}$ such that $x+y+z=4$ and $x^2+y^2+z^2=6$, then show that $x,y,z \in [2/3,2]$. If $x,y,z\in\mathbb{R}$ such that $$x+y+z=4,\quad x^2+y^2+z^2=6;$$then show that the each of $x,y,z$ lie in the closed interval $[2/3,2]$.
I have been able to solve using $2(y^2+z^2)\geq(y+z)^2$.
Is there any another method to solve it.
|
The mean of the numbers is $\frac{4}{3}$, while the mean of their squares is $2$,
that is
$$E(x)= \bar x =\frac{1}{n}\sum x= \frac{4}{3} \\
E(x^2) = \frac{1}{n}\sum x^2 = 2$$
($n$ is $3$)
Therefore, the variance is
$$\frac{1}{n}\sum(x - \bar x)^2 = E(x^2) - E(x)^2 = 2 - (\frac{4}{3})^2 = \frac{2}{9}$$
In other words,
$$\frac{1}{3}\left( (x-\frac{4}{3})^2 + (y - \frac{4}{3})^2 + (z- \frac{4}{3})^2\right) = \frac{2}{9}$$
This is the average quadratic deviation from the mean.
Rewrite it as
$$ (x-\frac{4}{3})^2 + (y - \frac{4}{3})^2 + (z- \frac{4}{3})^2 = \frac{2}{3}$$
With $a = x-\frac{4}{3}$, $b=y - \frac{4}{3}$, $c= z- \frac{4}{3}$, we have
$$a+b+c=0\\
a^2 + b^2 + c^2 = \frac{2}{3}$$
Now the question is how large can be the coordinate of a point $(a,b,c)$ on the circle of radius $r=\sqrt{\frac{2}{3}}$ described by the above equations. It turns out that the largest value for a coordinate is $\sqrt{\frac{2}{3}} \cdot r=\frac{2}{3} $. Hence the numbers $x$, $y$, $z$ differ from $\frac{4}{3}$ by at most $\frac{2}{3}$.
|
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Need help using the alternative formula to find the derivative at $x = c$ Don't know if I'm simplifying wrong or if it is a simple mistake but I cannot get the answer of 4. Please help and include exact step by step details. Thank you.
$$f(x)=x^3 + 2x^2 + 1,\;\;\;c= -2$$
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Using the alternative definition of the derivative, given what you posted in a comment, we'll start with the approximation of the derivative given by $\lim_{x \to -2}\dfrac{f(x) - f(c)}{x - c}$
\begin{align*}\require{cancel}
\lim_{x \to -2} \dfrac{f(x) - f(c)}{x - c}
&=\lim_{x\to -2} \dfrac{x^3 + 2x^2 + 1 - ((-2)^3 + 2(-2)^2 + 1)}{x - (-2)}\\
&=\lim_{x \to -2} \dfrac{x^3 + 2x^2 + 1 -(-8 + 2\cdot 4 + 1)}{x + 2}\\
&=\lim_{x \to -2} \dfrac{x^3 + 2x^2 + 1 - (1)}{x + 2}\\
&= \lim_{x \to -2} \dfrac{x^2({x + 2)}}{{x+2}}\\
&= \lim_{x \to -2} \dfrac{x^2\cancel{(x + 2)}}{\cancel{x+2}} \\
&= \lim_{x \to -2} x^2 = (-2)^2 = 4
\end{align*}
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|
using Gauss' algorithm (for linear congruences) for A > B To solve $Bx \equiv A \pmod{m}$, use Gauss' algorithm.
The algorithm works perfectly when $A < B$. For example, to solve $6x \equiv 5 \pmod{11}$: $$x \equiv \frac{5}{6} \equiv \frac{5(2)}{6(2)} \equiv \frac{10}{12} \equiv \frac{10}{1}$$
so $x \equiv 10$
But when $A > B$, I run into problems. For example, trying to solve $7x \equiv 13 \pmod{100}$: $$x \equiv \frac{13}{7} \equiv \frac{13(15)}{7(15)} \equiv \frac{195}{105} \equiv \frac{95}{5} \equiv \frac{95(21)}{5(21)} \equiv \frac{1995}{105} \equiv \frac{95}{5}$$ and it continues to be $\frac{95}{5}$. Am I missing a step?
PS: I was applying the algorithm on random linear congruence problems I could find. The second example comes from http://www.johndcook.com/blog/2008/12/10/solving-linear-congruences/, which says the answer is $x \equiv 59$.
update
This answer answered my question. It explains that Gauss' algorithm works only on prime modulo.
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Frank, you can use Gauss's Algorithm even if modulo is not prime. The only thing you need to take care is that multiplier should be co-prime to modulo.
Just keep multiplying denominator by a number so that denominator is near 100 till denominator become 1. However, the multiplier must be co-prime to 100.
$$\frac{13}{7} \pmod {100} \equiv \frac{13 × 29}{7 × 29} \pmod {100} \equiv \frac{-23}{3} \pmod {100}$$
$$ \equiv \frac {-23-100}{3} \pmod {100} \equiv \frac {-123}{3} \pmod {100} \equiv {-41} \pmod {100}$$
$$ \equiv {59} \pmod {100}$$
|
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|
Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem :
In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$
My approach :
Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get:
\begin{equation}
a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i})
\end{equation}
By using projection formula which states: $b=a\cos C +c \cos A $
(i) will become :
$$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here?
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Which means that $c \sin(B+C)=a \sin(A+B)$
$$c\sin(B+C)-a\sin(A+B)=O$$
|
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|
"Question: Show that $n^5 - n$ is divisible by 30; for all natural n"
Show that $n^5 - n$ is divisible by $30;$ $\forall n\in \mathbb{N}$
I tried to solve this three-way. And all stopped at some point.
I) By induction:
testing for $0$, $1$ and $2$ It is clearly true.
As a hypothesis, we have $30|n^5-n\Rightarrow n^5-n=30k$.
Therefore, the thesis would $30|(n+1)^5-n-1\Rightarrow (n+1)^5-n-1=30j$. By theorem binomial $(n+1)^5-n-1=n^5+5n^4+10n^3+10n^2+5n+1-n-1$ $=30k+5n^4+10n^3+10n^2+5n.$
It was a little messy and not given to proceed.
II)I tried by Fermat's Little Theorem:
$$30|n^5-n\Rightarrow 5\cdot3\cdot2|n^5-n $$
Analyzing each case, we have; clearly $5|n^5-n$ (Fermat's Little Theorem). Now cases $3|n^5-n$ and $2|n^5-n$ I could not develop.
III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that
$$gcd(n^5-n,30)=30$$
Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics
What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD?
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Method $1:$
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$
Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer
Method $2:$
Using Fermat's Little Theorem, $n^p\equiv n\pmod p$ for prime $p$
$\implies 5$ divides $n^5-n$
$n^5-n=n(n^2-1)(n^2+1)=(n^3-n)(n^2+1)$
As $3$ divides $n^3-n\implies 3$ divides $n^5-n$
$n^5-n=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$
As $2$ divides $n^2-n\implies 2$ divides $n^5-n$
$\implies $lcm $(2,3,5)$ divides $n^5-n$
|
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Maclaurin series of $f(x) = e^x \sin x$ $$f(x) = e^x \sin x$$
I tried applying the given formula in my book but it didn't work.
The maclaurin for $e^x$ is given as $\displaystyle \sum \frac{x^n}{n!}$ and $\sin x$ $\displaystyle \sum \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
I attempted to multiply them together, failed teh books answer. I tried inputting values for them and then multiplying and that fails as well. Why?
$\displaystyle \frac{x^n}{n!} \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
Fails
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Since you are looking for the Maclaurin Series of $ f(x):=\exp(x)\sin(x) $ you can start by looking at $f(0)$ and higher derivatives of $f$ at $x=0$. Let $f^{(n)}$ denote the $n$-th derivative of $f$:
$$
\begin{alignat}{2}
f(0) &= \exp(0)\sin(0) &&{}= 0 \\
f^{(1)}(0) &= \exp(0)\sin(0)+ \exp(0)\cos(0) &&{}= 1 \\
f^{(2)}(0) &= 2\exp(0)\cos(0) &&{}= 2 \\
f^{(3)}(0) &= 2\exp(0)\cos(0) - 2\exp(0)\sin(0) &&{}= 2 \\
f^{(4)}(0) &= -4\exp(0)\sin(0) &&{}= 0 \\
f^{(5)}(0) &= -4\exp(0)\sin(0) - 4\exp(0)\cos(0) &&{}= -4 \\
f^{(6)}(0) &= -8\exp(0)\cos(0) &&{}= -8 \\
f^{(6)}(0) &= -8\exp(0)\cos(0) + 8\exp(0)\sin(0) &&{}= -8 \\
\dots
\end{alignat}
$$
One can now identify the pattern: Define a real sequence $(a_n)_{n \in \mathbb{N}_0}$ with terms $a_n:= f^{(n)}(0)$. If one now consults the On-Line Encyclopedia of Integer Sequences, one can find an explicit, real, non-recursive expression for former sequence, namely:
$a_n = 2^{\frac{n}{2}}\sin(\frac{n\pi}{4})$
Knowing this, one can write the Maclaurin Series as follows:
$$\begin{align}
\exp(x)\sin(x)
&= \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\\
&= \sum_{n=0}^\infty \frac{a_n}{n!}x^n \\
&= \sum_{n=0}^\infty \frac{2^{\frac{n}{2}}\sin(\frac{n\pi}{4})}{n!}x^n \\
&= x + x^2 + \frac{1}{3}x^3 - \frac{1}{30}x^5 - \frac{1}{90}x^6 - \frac{1}{630}x^7 + \frac{1}{22680}x^9 + \frac{1}{113400}x^{10} + \dots \\
\end{align}
$$
|
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How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$
$\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P.
Therefore the sum of the series can be written as :
$\Rightarrow \frac{(3)^3}{(A+B+C)}$
Is this correct? Please suggest.
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The exact expression for $\displaystyle H_n:=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is not known, but you can estimate $H_n$ as below
Let us consider the area under the curve $\displaystyle \frac{1}{x}$ when $x$ varies from $1$ to $n$.
Now note that $\displaystyle H_{n}-\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n-1}$ is an overestimation of this area by rectangles. See below
And $\displaystyle H_n-1=\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{n} $ is an underestimation of the area. See below
(source: uark.edu)
Hence $$\large H_n-1<\int_{1}^n\frac{1}{x}dx<H_n-\frac{1}{n}\\
\Rightarrow \ln n+\frac{1}{n}<H_n<\ln n+1$$
Also, Euler discovered this beautiful property of harmonic number $H_n$ that $$\large \lim_{n\rightarrow \infty}\left(H_n-\ln n\right)=\gamma\approx 0.57721566490153286060651209008240243104215933593992…$$ $\gamma$ is called the Euler-Mascheroni constant.
|
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Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\cos2\alpha\sin\alpha$ can be expressed in three ways:
$$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha - \sin\alpha$$
$$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha - 2\sin^3\alpha$$
I tried adding these, but nothing came close to the required answer.
So then I tried calculating $\sin4\alpha$ (from the required answer):
$$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$
$$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$
Still looking at the answer, I calculated $\cos4\alpha$
$$\cos4\alpha = 1- \sin^2(2\alpha)$$
If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$
then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$
and $$\cos4\alpha = 1 - 4\sin^2\alpha\cos^2\alpha$$
and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$
I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.
|
Hint
Take the real part of this complex number
$$-ie^{i\alpha}e^{2i\alpha}=-ie^{-i\alpha}e^{4i\alpha}$$
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|
Condition for collinearity of points $(a, a^3), (b, b^3), and (c, c^3)$ The following is a statement I have been trying to prove (while solving problem 1.4.26 in Algorithms (4th edition) by Robert Sedgewick).
Show that three points $(a, a^3), (b, b^3), and
(c, c^3)$ are collinear if and only if $a + b + c = 0$.
I am aware that the rule for collinearity of three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is -
$x_1(y_2- y_3 ) + x_2(y_3 - y_1 ) + x_3(y_1 - y_2 ) = 0$
OR
$(x_2 - x_1) : (x_3 - x_2) = (y_2 - y_1) : (y_3 - y_2)$
But I cannot factor out the term $(a+b+c)$ when I apply either of these rules to the points $(a, a^3), (b, b^3), and (c, c^3)$.
Can someone help me with this proof?
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We need $$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}=0$$
Now,
$$\det\begin{pmatrix} a^3 & a & 1 \\ b^3 & b & 1 \\ c^3 & c &1\end{pmatrix}$$
$$=\det\begin{pmatrix} a^3-b^3 & a-b & 0 \\ b^3 & b & 1 \\ c^3-b^3 & c-b &0\end{pmatrix}$$
$$=-(a-b)(b-c)\det\begin{pmatrix} a^2+ab+b^2 & 1 & 0 \\ b^3 & b & 1 \\ c^2+cb+b^2 &1&0\end{pmatrix}$$
$$=-(a-b)(b-c)\{a^2+ab+b^2-(c^2+cb+b^2)\}$$
$$=-(a-b)(b-c)(a^2+ab-bc-c^2)$$
$$=-(a-b)(b-c)\{(a+c)(a-c)+b(a-c)\}$$
$$=(a-b)(b-c)(c-a)(a+b+c)$$
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$\frac1a+\frac1b+\frac1c=0 \implies a^2+b^2+c^2=(a+b+c)^2$? How to prove that $a^2+b^2+c^2=(a+b+c)^2$ given that $\frac1a+\frac1b+\frac1c=0$?
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$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc=a^{2}+b^{2}+c^{2}+2abc\big(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\big)=a^{2}+b^{2}+c^{2}$
since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$.
|
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Checking Independence for $\left\{\dfrac{1}{x^2+x-6},\dfrac{1}{x^2-5x+6},\dfrac{1}{x^2-9}\right\}$?
Is this subset of $V=\mathbb{F[0,1]}$ linearly independent, where the subset of $V$ is
$\left\{\dfrac{1}{x^2+x-6},\dfrac{1}{x^2-5x+6},\dfrac{1}{x^2-9}\right\}$?
My work:
If this subset is independent, then $\alpha=\beta=\gamma=0$ must be the only solution for $\alpha\dfrac{1}{x^2+x-6}+\beta\dfrac{1}{x^2-5x+6}+\gamma\dfrac{1}{x^2-9}=0\\ \implies\dfrac{\alpha}{x^2+x-6}+\dfrac{\beta}{x^2-5x+6}+\dfrac{\gamma}{x^2-9}=0\\\implies \dfrac{\alpha}{(x+3)(x-2)}+\dfrac{\beta}{(x-2)(x-3)}+\dfrac{\gamma}{(x+3)(x-3)}=0\\$
I'm note sure how to proceed from this step (assuming what I've done so far is correct). A hint would be appreciated. Thanks.
|
Hint: checking that this set is linearly independent is equivalent to checking if the set
$$\{x-3, x-2, x+3\}$$ is linearly independent.
|
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Diophantine equation $x^3=a^2+b^2+c^2$ Does anyone know if a formula exists to obtain all solutions of the above Diophantine equation? All numbers integers.
Addendum: After seeing the answer from Tito Piezas III, I reconsidered the above equation and came up with an original solution that applies to two consecutive cubes when one of the cubes is equal to two squares. Solutions for three consecutive cubes exist, when the middle cube is equal to two squares, but I regret to say that I could not find a general solution for the third cube.
$[(a^2+b^2) +1]^3= [a(a^2+b^2)+a]^2+[(a^2)b+b+b^3]^2+[a^2+b^2+1]^2$, all numbers positive integers and $a>b$.
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For any positive integer $k>2$, the kth power of the sum of two squares is the sum of three squares,
$$x^2+y^2+z^2 = (a^2+b^2)^k$$
thus,
$$a^2(a^2+b^2)^2 + (2ab^2)^2 + (a^2-b^2)^2b^2 = (a^2+b^2)^3$$
$$(a^4-b^4)^2 + (4a^2b^2)^2 + (2ab(a^2-b^2))^2 = (a^2+b^2)^4$$
$$a^2(a^2+b^2)^4 + (4a^3b^2-4ab^4)^2 + (a^4b-6a^2b^3+b^5)^2 = (a^2+b^2)^5$$
and so on. Similarly, for $k\geq2$, the kth power of the sum of three squares is itself the sum of three squares. Thus,
$$(a^2-b^2-c^2)^2 + (2ab)^2 + (2ac)^2 = (a^2+b^2+c^2)^2$$
$$(a^3-3ab^2-3ac^2)^2 + b^2(-3a^2+b^2+c^2)^2 + c^2(-3a^2+b^2+c^2)^2 = (a^2+b^2+c^2)^3$$
and so on.
Note: For $k>2$, this does not give all solutions, but at least you have formulas. For more details, see Sums of Three Squares.
|
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How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
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Note that
$${n^3 - 1 \over n^3 + 1} = {n - 1 \over n + 1}{n^2 + n + 1 \over n^2 - n + 1}$$
Also note that
$$(n-1)^2 + (n - 1) + 1 = n^2 - n + 1$$
So the infinite product in question is really the product of two telescoping products.
|
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|
Differentiation of logarithmic functions using the chain rule What's the derivative of $x^2(\ln(x^2))$?
I'm having a really hard time with logarithmic differentiation. Can someone help rationalize it for me?
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Hints:
*
*$\bigl( f \cdot g \bigr)' = f \cdot g' + g \cdot f'$.
Using this we see that if we see that
\begin{align*}y &= x^{2} \cdot \log(x^{2})\\
\implies y' &= x^{2} \cdot \frac{d}{dx}(\log(x^{2})) + \log(x^{2}) \cdot \frac{d}{dx} (2x) \\ &= x^{2} \cdot \frac{d}{dx} (2\cdot \log(x)) + \log(x^{2}) \times 2 \\
&=2 \cdot x^{2} \cdot \frac{1}{x} \cdot \log(x) + 4 \cdot \log(x)
\\ &= 2x \cdot \log(x) + 4 \cdot \log(x)
\end{align*}
|
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Proving an equality involving binomial coefficients and summations Question:
$$\sum_{k=0}^{n}\left ( -1 \right )^{k}\binom{2n}{k}\binom{2n-k}{2n-2k}=\sum_{2n}^{k=0}\binom{2n}{k}^{2}\left ( \frac{1+\sqrt{5}}{2} \right )^{2n-k}\left ( \frac{1-\sqrt{5}}{2} \right )^{k}$$
Attempt:
It looks like I need to start with a quadratic function like $x^{2}-x-1$ since the solutions of this function are $\frac{1\pm \sqrt{5}}{2}$. Is that the correct approach?
How do I get the LHS using this function?
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Suppose we seek to verify that
$$\sum_{k=0}^n (-1)^k {2n\choose k} {2n-k\choose 2n-2k}
= \sum_{k=0}^{2n} {2n\choose k}^2
\left(\frac{1+\sqrt{5}}{2}\right)^{2n-k}
\left(\frac{1-\sqrt{5}}{2}\right)^{k}.$$
For the LHS introduce
$${2n-k\choose 2n-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} (1+z)^{2n-k} \; dz.$$
Observe that when $2n\ge k\gt n$ the integral vanishes so we may use
it to control the range, extending the sum in $k$ on the LHS to $2n$
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n}
\sum_{k=0}^{2n} (-1)^k {2n\choose k}
\frac{z^{2k}}{(1+z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n}
\left(1-\frac{z^2}{1+z}\right)^{2n}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}}
(1+z-z^2)^{2n} \; dz.$$
This means the LHS is
$$[z^{2n}] (1+z-z^2)^{2n}.$$
On the other hand we may re-write the RHS as
$$\sum_{k=0}^{2n} {2n\choose k} {2n\choose 2n-k}
\left(\frac{1+\sqrt{5}}{2}\right)^{2n-k}
\left(\frac{1-\sqrt{5}}{2}\right)^{k}.$$
By inspection this is
$$[w^{2n}]
\left.(1+z)^{2n}\right|_{z=w(1+\sqrt{5})/2}
\left.(1+z)^{2n}\right|_{z=w(1-\sqrt{5})/2}
\\ = [w^{2n}]
\left(1+\frac{1+\sqrt{5}}{2}w\right)^{2n}
\left(1+\frac{1-\sqrt{5}}{2}w\right)^{2n}
\\ = [w^{2n}]
(1 + w + (1-5)/4 \times w^2)^{2n}
\\ = [w^{2n}] (1+w-w^2)^{2n}.$$
We have equality for the LHS and the RHS, QED.
|
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|
Identifying the nature of the eigenvalues I wish somebody could help me in this one. We have to choose one of the $4$ options.
Let $a,b,c$ be positive real numbers such that $b^2+c^2<a<1$. Consider the $3 \times 3$ matrix
$$A=\begin{bmatrix}
1 & b & c \\
b & a & 0 \\
c & 0 & 1 \\
\end{bmatrix}.$$
*
*All the eigenvalues of $A$ are negative real numbers.
*All the eigenvalues of $A$ are positive real numbers.
*$A$ can have a positive as well as a negative eigenvalue.
*Eigenvalues of $A$ can be non-real complex numbers.
Now, $\det(A-\lambda I)=0$, so
$$\begin{vmatrix}
1-\lambda & b & c \\
b & a-\lambda & 0 \\
c & 0 & 1-\lambda \\
\end{vmatrix}=0$$
$\implies(1-\lambda)(a-a\lambda -\lambda +\lambda^2)-b(b-b\lambda)-c(ac-c\lambda)=0$
$\implies a-a\lambda -\lambda +\lambda^2-a\lambda+a\lambda^2 +\lambda^2 -\lambda^3-b^2+b^2\lambda-ac^2+c^2\lambda=0$
$\implies-\lambda^3+\lambda^2(2+a)+\lambda(-2a-1+b^2+c^2)+a-b^2-ac^2=0$
I am stuck here, don't know how to proceed. Need your help, please.
|
Notice that the matrix is symmetric (so, we can exclude option 4 immediately), and the question basically asks if the matrix is negative definite (all eigenvalues are negative), positive definite (all eigenvalues are positive), or indefinite (we have both negative and positive eigenvalues).
Let's check the leading principal minoras:
\begin{align*}
\det A_{11} &= \det \begin{bmatrix} 1 \end{bmatrix} = 1 > 0, \\
\det A_{12} &= \det \begin{bmatrix} 1 & b \\ b & a \end{bmatrix} = a - b^2 > c^2 > 0, \\
\det A_{33} &= \det A = \det \begin{bmatrix} 1 & b & c \\ b & a & 0 \\ c & 0 & 1 \end{bmatrix} = a - ac^2 - b^2 > c^2 - ac^2 = c^2 (1-a) > 0,
\end{align*}
So, all of these are positive and we can conclude that the matrix $A$ is positive definite, i.e., option 2 is correct.
|
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|
Solve the equation : $x^2 − 6 |x − 2| − 28 = 0$ The following is an absolute value quadratic equation that I want to solve:
$$x^2 − 6 |x − 2| − 28 = 0$$
Here is what I did :
$x^2 − 6 |x − 2| − 28 = 0$
$x^2 − 6 |x − 2| − 28 = 0$
$-6|x-2|=28-x^2$
$6|x-2|=x^2-28$
$6x-12=x^2-28$ or $28-x^2$
(Is this step correct ?)
Solving this two quadratic equations I get the answers $x=-2,8,-10,4$
But when I actually substitute these in the original equation I see that only $x=8,-10 $ satisfy and other two solutions are invalid . Why so?
|
HINT:
First of all, let $x=a+ib$ where $a,b$ are real
So, we have $$(a+ib)^2-28=6|a+ib-2|\implies a^2-b^2-28+2ab i=6\sqrt{(a-2)^2+b^2}$$
Equating the imaginary parts, we have $ab=0$
If $a=0,-(b^2+28)=6\sqrt{4+b^2}>0$ which is impossible
So, $b$ must be $0\implies x$ must be real
We know for real $m,$ $$|m|=\begin{cases} +m &\mbox{if } m\ge0 \\
-m & \mbox{if } m<0 \end{cases} $$
$$\text{So, }|x-2|=x-2\text{ if } x-2\ge0 \iff x\ge2$$
In that case, we have $$x^2-6(x-2)-28=0\implies x^2-6x-16=0\implies x=8\text{ or } -2$$
But we have $x\ge2,$ so we need to discard the solution $x=-2$
Similarly for $|x-2|=-(x-2)$ if $x-2<0\iff x<2$
|
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|
Double Integral Question on unit square Hints on solving following double integral will be appreciated.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$
|
Just split the integral as
$$ \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
= \int_0^1 \int_0^1 \frac{x^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x
- \int_0^1 \int_0^1 \frac{y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x. $$
Now, just use standard integration techniques to evaluate the integrals. For instance
$$ \int_{0}^{1}\frac{1}{(x^2+y^2)^2}dy=\frac{xy}{x^2 + y^2} + \frac{\arctan(y/x))}{2x}\Big|_{0}^{1} =\dots. $$
|
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|
equilateral triangle; $3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$ In equilateral triangle ABC of side length d, if P is an internal point with PA = a, PB = b, and PC = c, the following pleasingly symmetrical relationship holds:
$3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.$
Please prove this identity.
source: http://www.qbyte.org/puzzles/p117s.html
|
A Cayley-Menger determinant gives the volume of "tetrahedron" $PABC$:
$$288\; V^2 = \left|\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & d^2 & d^2 & a^2 \\
1 & d^2 & 0 & d^2 & b^2 \\
1 & d^2 & d^2 & 0 & c^2 \\
1 & a^2 & b^2 & c^2 & 0
\end{array}\right| = d^2 \left( \left( a^2+b^2+c^2+d^2 \right)^2-3\left(a^4+b^4+c^4+d^4\right) \right)$$
Since the tetrahedron is flat, its volume is zero.
|
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|
proof for the Ramanujan's formula ? I found this formula in a textbook in which the proof to the formula was not given
Ramanujam's formula
$$\sqrt{1 +n\sqrt{1 +(n+1)\sqrt{1 + (n+2)\sqrt{1 + (n+3)\sqrt{1 +....\infty}}}}} = n+1$$
Its a great equation andhow do you prove this. its a bit difficult for me and tried different methods to solve this. Iam an undergraduate and I want you to elaborate the method of solving if its complex.
|
More general formulation can be gotten:
$F(x) = \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt{\mathrm{\cdots}}}}}$
$$F(x)^2 = ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}} $$
Which can be simplified to:
$$F(x)^2 = ax+(n+a)^2 +xF(x+n) \tag1$$
Let's assume that $F(x) = mx+k $
$deg(F(x))$ cannot be more than 1 otherwise left side degree will be bigger than right side
in Equation 1
$$F(x)^2 = ax+(n+a)^2 +xF(x+n) $$
$$( mx+k)^2=ax+(n+a)^2+x(m(x+n)+k) $$
$$ m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$$
$m^2=m$
$m=1$ or $m=0$
$2mk=mn+k+a$
If $m=1$ then
$2k=n+a+k$
$k=n+a$
And It is also confirming constant term of $ m^2x^2+2mkx+k^2=ax+(n+a)^2+mx^2+(mn+k)x$ :
$$k^2=(n+a)^2$$
$F(x) = mx+k $
$F(x) = x + n + a $
$$ x + n + a= \sqrt{ax+(n+a)^2 +x\sqrt{a(x+n)+(n+a)^2+(x+n) \sqrt{\mathrm{\cdots}}}} $$
So, setting $a=0$, $n=1$,
$$ x+1=\sqrt{1 +x\sqrt{1 +(x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)\sqrt{1 +....}}}}} $$
Please see Infinitely nested radicals in the wiki page as reference.
|
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|
Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$. Let $n$ be a positive integer. Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$.
I tried rewriting $\lceil(\sqrt{3}+1)^{2n}\rceil$ as $m*2^{n+1}$ for some m, but couldn't get anywhere.
|
Hint: The value is equal to $( \sqrt{3} + 1)^{2n} + (\sqrt{3}-1 )^{2n}$
Hint: Expanding the square, we get $ (4 + 2\sqrt{3})^n + (4-2\sqrt{3})^n$
This term is clearly a multiple of $2^n$, which we can factor out.
Hint: $(2 + \sqrt{3})^n + (2-\sqrt{3})^n$ is even, by the Binomial Theorem expansion.
|
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|
Evaluate $\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{\mathrm dx}{1+x^2}$ $$\int \limits _{0}^{\infty}\ln\left({x+\frac{1}{x}}\right)\cdot\frac{dx}{1+x^2}$$
we are asked to solve this definite integral so here's what i did
$$\int \limits_{0}^{\infty}\ln \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2} = \int\limits_{0}^{\infty}\ln \left(x^2+1\right)\cdot\frac{dx}{1+x^2} - \int \limits_{0}^{\infty}\ln (x).\frac{dx}{1+x^2}$$ now how to proceed after that ? should i intregrate both seperate functions by substituting $(x^2+1)$ and what should i susbtitute in other integral , by-parts integration is making troubles when substituting $\infty$ , now what to do ?
|
As you have done the integral can be fairly easy evaluated by splitting it into two easier integrals.
$$
\int_{0}^{\infty}\log \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2}\mathrm{d}x =
\int_{0}^{\infty}\frac{\log\left(x^2+1\right)}{1+x^2}\,\mathrm{d}x -
\int_{0}^{\infty}\frac{\log x}{1+x^2}\,\mathrm{d}x
$$
For the last integral we have that
$$
\int_0^\infty R(x) \log x = 0
$$
given that $R(x)$ is a rational function satisfying $R(x)=R(1/x)/x^2$. I will leave it to you to checki that $R(x) = 1/(1+x^2)$ satisfies this functional relation. Alternatively one can see that the last integral is zero by splitting the integral by splitting it into $[0,1]\cap[1,\infty)$
$$
\int_0^1 \frac{\log x}{1+x^2}\mathrm{d}x + \int_1^\infty \frac{\log x}{1+x^2}\mathrm{d}x
$$
and using $x = 1/u$ in the last integral. Now we have that
$$
\int_{0}^{\infty}\log \left({\frac{x^2 +1}{x}}\right)\cdot\frac{dx}{1+x^2}\mathrm{d}x
=
\int_{0}^{\infty}\frac{\log\left(x^2+1\right)}{1+x^2}\,\mathrm{d}x
$$
which can be evaluated by introducing a variable $\alpha$, and differentiating under the integral sign. By looking at the following integral
$$
I(\alpha) := \int_{0}^{\infty}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x
$$
we see that our integral equals $I(1)$. Now by differentiating by $\alpha$ we obtain
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}\alpha}I(\alpha) & = \frac{\mathrm{d}}{\mathrm{d}\alpha} \int_{0}^{\infty}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x \\
& = \int_{0}^{\infty} \frac{\partial}{\partial \alpha}\frac{\log\left(1 + \alpha^2x^2\right)}{1+x^2}\,\mathrm{d}x \\
& = \int_0^\infty \frac{2a x^2}{(1+\alpha^2x^2)(1+x^2)}\,\mathrm{d}x \\
& = \frac{2\alpha}{\alpha^2-1} \int_0^\infty \frac{1}{1+x^2} - \frac{1}{1 +\alpha^2x^2} \,\mathrm{d}x \\
& = \frac{\pi}{a+1}
\end{align}
Integrating both sides now gives
$$
I(\alpha) = \pi \log(1+\alpha) + \mathcal{C}
$$
But since $I(\alpha)=0$, then $\mathcal{C}=0$. We can now "finaly" conclude that
$$
\int_0^\infty \left( \alpha^2 x + \frac{1}{x}\right) \frac{\mathrm{d}x}{1+x^2} = \pi \log(1+\alpha)
$$
Your integral is now evaluated by simply plugging in $\alpha = 1$. The switching of the derivation is legal since we have that
$$
\frac{\log(1+\alpha x^2)}{1+x^2} < \frac{\log(\alpha x^2+\alpha x^2)}{1+x^2} \, \forall \, x>1
$$
and the last integral converges (why?). By similar means we have that $\pi/(1+\alpha)$ converges as well.
|
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|
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
|
We know $(A+B)\mid(A^n+B^n)$ for $n$ odd. What about with three terms? Compute
$$\mod A+B+C:\quad A^3+B^3+C^3\equiv-(B+C)^3+B^3+C^3\equiv-3BC(B+C)\equiv 3ABC.$$
So $(A+B+C)\mid(A^3+B^3+C^3-3ABC)=f(A,B,C)$. Further
$$f(A,B,C)=f(A,\omega B,\bar{\omega}C)=f(A,\bar{\omega}B,\omega C)=\rm etc.$$
by inspection so both $A+\omega B+\bar{\omega}C$ and $A+\bar{\omega}B+\omega C$ are also factors.
This argument exploits divisibility properties and inherent symmetry. It is generalized by the first proof (using matrix operations) mentioned in my other answer to compute $\Phi(G)$ for $G$ abelian.
|
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|
Summation of series Write down the sum of $\displaystyle \sum_1^{2N} n^3$ in terms of $N$, and hence find:
$1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3$ in terms of $N$, simplifying your answer.
I found this to be $n^2(2n+1)^2$ but the next part is not making sense to me.
Why is the general term of this sum $-(2N)^3$, where it doesn't work for N=0 etc?
thanks
|
HINT:
As
$$
\begin{align}
& \sum_{1\le r\le n}r^3=\frac{n^2(n+1)^2}4 \\[10pt]
& 1^3 - 2^3 + 3^3 - 4^3 + \cdots - (2N)^3 \\[10pt]
& =\sum_{1\le r\le 2N}r^3-2\sum_{1\le r\le N}(2r)^3 \\[10pt]
& =\sum_{1\le r\le 2N}r^3-2\cdot8\sum_{1\le r\le N}r^3 \\[10pt]
& =\frac{(2N)^2(2N+1)^2}4-16\frac{N^2(N+1)^2}4 \\[10pt]
& =N^2\{(2N+1)^2-4(N+1)^2\} \\[10pt]
& =-N^2(4N+3)
\end{align}
$$
|
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|
Find all $x,y,z \in \mathbb{Q}.x^2+y^2+z^2+x+y+z=1$ Find all $x,y,z \in \mathbb{Q}$
where.$x^2+y^2+z^2+x+y+z=1$
|
Assume $(x,y,z)$ is sastifying the equation.
We can multiply $K=N^2(N \in \mathbb{Z}^+)$ to the both sides of equation.
And the both sides are integers.
Set $N=2^{h}s$,$s$ is odd.
Then $K((2x+1)^2+(2y+1)^2+(2z+1)^2)=7N^2$
The $LHS$ is sum of 3 squares.
From A production by Gauss, $a^2+b^2+c^2=m,(a,b,c,m,\alpha,t \in\mathbb{N})$$\Leftrightarrow m\ne 4^{\alpha}(8t+7)$
Because,$s^2 \equiv 1\pmod{8}\Rightarrow 7s^2 \equiv 7 \pmod{8}$
Which is impossible!
|
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|
Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x)
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:
*
*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?
*Are there trigonometric identities that I could have used to simplify either derivative?
|
The equation $\cos x+\sin x=0$ is really the same as $\cos x-\sin x=0$ once you make the transformation $x\mapsto x+\frac{\pi}2$, because $\cos (x+\frac{\pi}2)=-\sin x$ and $\sin (x+\frac{\pi}2)=\cos(x)$. The latter can be solved by resorting to $\cos ^2+\sin ^2=1$.
Edit: We can derive $\cos (x+\frac{\pi}2)=-\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\cos (x+y)=\cos x\cos y-\sin x\sin y$ and $\sin (x+y)=\cos x\sin y+\cos y\sin x$. These can either be proved geometrically, or by resorting to some definition of $\sin$ and $\cos$, for example $\sin x=(e^{ix}-e^{-ix})/2i, \cos x=(e^{ix}-e^{-ix})/2$
|
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|
Prove that:$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geqslant a^n+b^n+c^n$ Find $n\in\mathbb{N}^+$
For all Positive real numbers $a,b,c$ sastifying $a+b+c=3$
$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geqslant a^n+b^n+c^n$
|
For an upper bound, consider $(a,b,c)=\left(\frac{3}{2},\frac{3}{4},\frac{3}{4}\right)$. (I thought of this myself, but I'm going to make it Community Wiki as this is reflected in the AoPS page.)
We need to solve for the smallest $n$ such that:
$\left(\frac{2}{3}\right)^n+2\left(\frac{4}{3}\right)^n<\left(\frac{3}{2}\right)^n+2\left(\frac{3}{4}\right)^n$.
Intuitively, a smallest $n$ should exist as $2\left(\frac{4}{3}\right)^n<<\left(\frac{3}{2}\right)^n$ for sufficiently large $n$.
A quick check on Wolfram Alpha gives $n>5.63493$.
As such, this inequality can only hold for $n=1,2,3,4,5$, since $n$ is an integer.
As a side note, the way I see how this works is trying things of the form $(a,b,c)=(3-2x,x,x)$, where $0<x<1$, as we want to make something on the right-hand side of the inequality to become very big.
The essential condition required to get a bound for $n$ is $\frac{1}{x}<3-2x$, or $2x^2-3x+1<0$. Factorising, we have $(2x-1)(x-1)<0$, which means $1>x>\frac{1}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.