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Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$ Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$
I used the R method to simplify it -
$ \sqrt{80} \cos (x-26.565) $
Minimum value of that =
$ \sqrt{80} \cos (x-26.565) = - \sqrt{80}$
$ \cos (x-26.565) = -1$
This cosine value lies in the 2nd and 3rd quadrant
letting $x-26.565 = y$
y reference angle = $ \cos^-1 (-1) = 180$
2nd quadrant - $180 - y (ref) = 0 $
3rd quadrant - $180 + y(ref) = 360$
Therefore , $x = 26.565, 386.565$
Why am I wrong ? The minimum value is $206.6$
|
Since
$$
8^2+4^2=80
$$
you know that $8\cos x+4\sin x=\sqrt{80}\cos(x-\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\pi/2$:
\begin{align}
8&=\sqrt{80}\cos\alpha\\
4&=\sqrt{80}\sin\alpha
\end{align}
Thus the angle $\alpha$ is in the first quadrant and so
$$
\alpha=\arcsin\frac{4}{\sqrt{80}}=\arcsin\frac{1}{\sqrt{5}}
$$
In degrees this is $26.565$. In radians it is $0.464$ (rounding to three decimal digits).
The point where the minimum value $-\sqrt{80}$ is reached is when $x-\alpha$ is the straight angle. In degrees the value of $x$ is $180+26.565=206.565$.
In radians it is $\pi+0.464=3.605$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Condition for two cubic equations to have two common roots The equations $x^3-x^2+bx+c=0$ and $x^3+cx^2+bx-d=0$ have two common roots.The question is to show that $b^2=d$
I couldn't get how to approach this problem.Any help would be appreciated. thanks
|
Let $\alpha,\beta,\gamma$ be the roots of $x^3-x^2+bx+c$.
Also, let $\alpha,\beta,\omega$ be the roots of $x^3+cx^2+bx-d$.
By Vieta's formulas, we get
$$\alpha+\beta+\gamma=1,\qquad \alpha+\beta+\omega=-c\tag1$$
Since we get
$$\alpha^3-\alpha^2+b\alpha+c=0$$
$$\alpha^3+c\alpha^2+b\alpha-d=0$$
subtracting the latter from the former gives
$$(-1-c)\alpha^2+c+d=0$$
Similarly, we get
$$(-1-c)\beta^2+c+d=0$$
So, we see that $\alpha,\beta$ are the roots of $(-1-c)x^2+c+d$.
By Vieta's formulas, we get
$$\alpha+\beta=0\tag2$$
From $(1)(2)$, we have $\gamma=1$ and $\omega=-c$ from which
$$1^3-1^2+b\cdot 1+c=0,\qquad (-c)^3+c(-c)^2+b(-c)-d=0$$
follow.
Now, eliminating $c$ gives $b^2=d$.
|
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|
Evaluate the Integral $ \iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz$ I am attempting to evaluate the following triple integral using spherical coordinated:
$$
\iiint_W\frac{1}{\sqrt{x^2+y^2+z^2}}dxdydz \\
W = \{(x,y,z) \in \mathbb{R}^3 : 1 \le x^2+y^2+z^2 \le 9; 0 \le x \le y, z \ge 0\}
$$
This tells me that I have the following regions:
$$
x^2 + y^2 + z^2 = 1 \rightarrow \text{Sphere of radius 1}\\
x^2 + y^2 + z^2 = 9 \rightarrow \text{Sphere of radius 3} \\
x=y \rightarrow \text{line in the xy plane}
$$
I have also determined that:
$$
1\le r \le 3
$$
But I am not sure how to determine which part of the intersection of the regions I am to integrate and to solve for $\theta$ and $\phi$. How do I proceed?
|
$x\geq0, y\geq0, z\geq0$ show $0\leq\theta\leq\dfrac{\pi}{2}$ and $0\leq\phi\leq\dfrac{\pi}{2}$. Also $x\leq y$ states $r\cos\theta\sin\phi\leq r\sin\theta\sin\phi$ or $\tan\theta\geq1$ so $\dfrac{\pi}{4}\leq\theta\leq\dfrac{\pi}{2}$ then
$$\int _{\frac{\pi }{4}}^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _1^3\frac{r^2 \sin (\varphi )}{r}drd\varphi d\theta=\color{blue}{\pi}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$ $$L=\lim_{x\to0}\frac{\ln((1+x)^{1+x})}{x^2}-\frac1x$$
I am not sure of my answer. Please help me.
$$L=\lim_{x\to0}\frac{1+x}x\frac{\ln(1+x)}x-\frac1x=\lim_{x\to0}\frac{1+x-1}x=1$$
|
You are missing a factor of $\frac12$, so I recommend you be careful and write your steps out.
Using a MacLaurin Expansion: $\ln(1+x) = x - \frac{x^2}{2} + O(x^3)$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{\ln(1+x)}{x} - \frac1{x} = $$
$$\lim_{x \to 0} \frac{1+x}{x} \frac{x - \frac{x^2}2 + O(x^3)}{x} - \frac{1}{x} =$$
$$\lim_{x \to 0} \frac{(1+x)(1 - \frac{x}2 + O(x^2)) - 1}{x}=$$
$$\lim_{x \to 0} \frac{(1 - \frac{x}2 + O(x^2)) + (x - O(x^2)) - 1}{x}$$
$$ \lim_{x \to 0} \frac{\frac{x}2 + O(x^2)}{x} =$$
$$\lim_{x \to 0} (\frac12 + O(x)) =\frac12$$
|
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|
How to Simplify $ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)} $
Simplify $$ \frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)}.$$
The solution is : $-\cot(x)$
I tried to: $$\frac{\sin(2x)\cos(x)+\cos(2x)\sin(x)+\sin^3(x)}{\cos(2x)\cos(x)+\sin(2x)\sin(x)-\cos^3(x)}.$$
|
With one step further you have
$$\sin(3x) = - 4 \sin^3 x + 3 \sin x~~~and~~~\cos 3x = -3\cos x+4\cos^3x$$that is
$$\sin(3x) +\sin^3 x = 3\sin x( 1-\sin^2 x ) = 3\sin x\cos^2x ~~~$$and$$~~~\cos 3x-\cos^3x = -3\cos x( 1-\cos^2x)= -3\cos x\sin^2x$$
Then $$\frac{\sin(3x)+\sin^3(x)}{\cos(3x)-\cos^3(x)} =\frac{3\sin x\cos^2x}{-3\cos x\sin^2x} =-\cot x.$$
See here:
http://2000clicks.com/mathhelp/geometrytrigequivcos3xetc.aspx
|
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|
Time and Work. How much work does C do per hour?
A, B and C need a certain unique time to do a certain work. C needs 1
hours less than A to complete the work. Working together, they require
30 minutes to complete 50% of the job. The work also gets completed if
A and B start working together and A leaves after 1 hour and B works
for a further 3 hours. How much work does C do per hour?
(A)16.66%
(B)33.33%
(C)50%
(D)66.66%
My attempt:
Let the total work be 100 units.
Let the work done by A,B and C be a units/hour,b units/hour,c units/hour respectively.
Let the time taken by A alone to complete the work be t hours.
ATQ:
\begin{align*}
(a+b+c) \cdot \frac{1}{2} & =50 \tag{1}\\
(a+b) \cdot 1+b \cdot 3 & =100 \tag{2}\\
c \cdot (t-1)& =100 \tag{3}\\
at & =100 \tag{4}
\end{align*}
Please help me solve these equations. When I am solving it is getting cumbersome.
Also if someone tells us some other way of solving, that would be helpful as well. Thanks.
|
Let $A$, $B$ and $C$ be the rates measured in an amount of work per hour $\left(\frac{work}{hour}\right)$ at which workers $A$, $B$ and $C$ accomplish work. If $C$ is the amount of work worker $C$ can do in one hour and $1$ ($100\%$) is the amount of work of the entire job, then $t$ is the time it takes him to bring it to completion: $t=\frac{1}{C}\cdot\frac{\text{work}}{\text{work/hour}}$. We also know that it takes worker $A$ one hour longer to complete the job than it takes worker $C$: $A(t+1)=1 \implies A\left(\frac{1}{C}+1\right)=1$. Another condition that we have is that three of them together require $30$ (or $\frac{1}{2}$ of an hour, don't forget that time is masured in hours here) minutes to complete $50\%$ of the job: $\frac{1}{2}(A+B+C)=\frac{1}{2} \implies (A+B+C)=1$. It is aslo stated that the work also gets completed if worker $A$ and worker $B$ start working together and worker $A$ leaves after $1$ hour and worker $B$ works for a further $3$ hours: $1\cdot(A+B)+3B=1 \implies A+4B=1$. We're all set to go now. Our conditions:
$$
A\left(\frac{1}{C}+1\right)=1\\
A+B+C=1\\
A+4B=1
$$
Solution:
$$
A\cdot\left(\frac{1}{C}+1\right)=1 \implies A=\frac{C}{1+C},\\
A+4B=1 \implies B=\frac{1-A}{4} \implies B=\frac{1}{4(1+C)},\\
A+B+C=1 \implies \frac{C}{1+C}+\frac{1}{4(1+C)}+C=1 \implies C^2+C-1=0 \implies \\
C_{1}=-\frac{3}{2}\text{ (discarded)}\\
C_{2}=\frac{1}{2}=\frac{1\text{ work}}{2\text{ hour}}≡50\%
$$
Answer: $C$ does $50\%$ of work per hour.
|
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|
Minimum values of the sequence $\{n\sqrt{2}\}$ I have been studying the sequence
$$\{n\sqrt{2}\}$$
where $\{x\}:= x-\lfloor x\rfloor$ is the "fractional part" function. I am particularly interested in the values of $n$ for which $\{n\sqrt{2}\}$ has an extremely small value - that is, when $n\sqrt 2$ is extremely close to (but greater than) its nearest integer. Of course, for integer $n$, this always takes on values between (but never including) $0$ and $1$.
I have defined the sequence $M_n$ as follows:
*
*$M_1=1$
*$M_{n+1}$ is the smallest positive integer such that $\{M_{n+1}\sqrt 2\}\lt \{M_n\sqrt 2\}$
The first few terms of this sequence are
$$1,3,5,17,29,99,169,...$$
And after a quick trip to the OEIS, I have found that this sequence is equal to OEIS entry A079496, as written in the comments. It also provides the recurrence
$$a_0=a_1=1$$
$$a_{2n+1}=2a_{2n}-a_{2n-1}$$
$$a_{2n}=4a_{2n-1}-a_{2n-2}$$
How can I prove that the sequence $M_n$ satisfies this recurrence, using the definition I wrote for $M_n$?
Thanks!
EDIT: Here is the closed-form explicit formula for $a_n$, for anyone who wants to know:
$$a_{2n}=\frac{(3+2\sqrt 2)^n+(3-2\sqrt 2)^n}{2}$$
$$a_{2n+1}=\frac{(1+\sqrt 2)(3+2\sqrt 2)^n-(1-\sqrt 2)(3-2\sqrt 2)^n}{2\sqrt 2}$$
|
This is about Pell type equations. Absolutely complete detail would be a bit long. I have posted many times about solving $x^2 - n y^2 = T,$ where $T$ is some target number, and $n$ is positive not a square.
Your sequence is this: given positive integer $x,$ let $v = \lfloor x \sqrt 2 \rfloor.$ Your numbers will be
$$ 2 x^2 - v^2 \in \{ 1,2 \}. $$
Furthermore, the ratios of consecutive terms are bounded: if $w$ is a number with worse $2 w^2 - w_0^2,$ where $w_0 = \lfloor w \sqrt 2 \rfloor,$ then there is one of your numbers $t$ with $t > w/(2 + \sqrt 2).$
Alright, the positive values of $2x^2 - y^2$ are $1,2,4,7,8,9, 14 \ldots$ where we are allowing common factors of $x,y.$
If $2x^2 - v^2 = 4,$ then both $x,v$ are even, we can take $t = x/2,$ and the fractional part of $t \sqrt 2$ is half the fractional part of $x \sqrt 2,$ meaning $x$ cannot be part of your $M$ sequence. Now that I think of it, the same applies to any number divisible by $4$ or by $9.$ As a result, if we calculated $2x^2 - v^2 = 7$ as a special case (we know how to find all representations) we could then continue with $2 x^2 - v^2 \geq 14.$
It appears we need just the one special case. After this, we may take
$$ 2 x^2 - v^2 \geq 7, $$ with
$$ \{ x \sqrt 2 \} \geq \frac{7}{x \sqrt 2 + v} \approx \frac{7}{2x \sqrt 2 }\approx \frac{2.4748737}{x } . $$
$$ t > \frac{x}{2 + \sqrt 2}, $$
$$ 2 t^2 - w^2 \in \{ 1,2 \}. $$
$$ (t \sqrt 2 - w)(t \sqrt 2 + w) \leq 2. $$
$$ t \sqrt 2 - w \leq \frac{2}{t \sqrt 2 + w}. $$
$$ \{ t \sqrt 2 \} \leq \frac{2}{t \sqrt 2 + w} \approx \frac{2}{2t \sqrt 2 } \approx \frac{1}{t \sqrt 2 } < \frac{2 + \sqrt 2}{x \sqrt 2 } = \frac{1 + \sqrt 2}{x } \approx \frac{2.1421356}{x } . $$
This is smaller than
$$ \{ x \sqrt 2 \} \geq \frac{7}{x \sqrt 2 + v} \approx \frac{7}{2x \sqrt 2 }\approx \frac{2.4748737}{x } . $$
A more careful analysis is possible, correction terms all over, but the heart of it is the observation that the floor of $t \sqrt 2$ is extremely close to the real number itself, when $0 < 2t^2 - w^2 \leq 2$.
Here is the Conway topograph for the negative of your form, $x^2 - 2 y^2.$ Your numbers show up representing negative number, (4,3) gives $-2$, then (7,5) gives $-1,$ then (24,17) give $-2,$ (41,29) gives $-1$
Let's see, it will take an hour or so, but I can draw a topograph for $2x^2 - v^2$ that shows numbers represented up to $17,$ as numbers such as $4,8,9$ are not squarefree, and are not primitively represented (we get $\gcd(x,v) \neq 1$). Will, 9:19 Pacific time
Done. Note that the green coordinate pairs have a motion that gives the same value of $2x^2 - y^2,$ namely
$$ (x,y) \mapsto (3x+2y, 4x+3y). $$ For example, $ (1,1) \mapsto (5, 7 ) $ and $ (3,4) \mapsto (17, 24 ). $
Let's see, Cayley Hamilton for the coefficient matrix
$$
\left(
\begin{array}{cc}
3 & 2 \\
4 & 3
\end{array}
\right)
$$
gives the linear recurrence,
$$x_{n+4} = 6 x_{n+2} - x_n $$
if we combine the $x$ values in a single list. We need to separate odd index and even index because we are combining $2x^2-v^2 = 1$ and $2x^2 - v^2 = 2$ in one list of $x$ values
There is a book that Allen Hatcher makes available online, here is his diagram for $x^2 - 2 y^2,$ but without the green coordinates that I like to include. The topograph diagram was introduced by J. H. Conway
|
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|
Arclength Parameterization of the Trefoil Knot I would like to find an arclength parameterization of the trefoil knot
The parameterizations I can find are:
$(sin(t) + 2sin(2t),$ $cos(t) - 2cos(2t),$ $sin(3t))$
and
$((2+cos(3t))cos(2t),$ $(2+cos(3t))sin(2t),$ $sin(3t))$
for $t \in [0,2\pi)$
writing $t = t(\theta)$, the magnitude of the derivatives wrt. $\theta$ of these two parameterizations are:
$| \frac{d}{d\theta} (sin(t) + 2sin(2t),$ $cos(t) - 2cos(2t),$ $sin(3t)) |$
= $| (cos(t) + 4cos(2t),$ $-sin(t) + 4cos(2t),$ $3cos(3t))\frac{dt}{d\theta} |$
= $\sqrt{cos^2(t) + sin^2(t) + 16[cos^2(2t) + sin^2(2t)] + 8[cos(t)cos(2t) - sin(t)sin(2t)] + 9cos^2(3t)}\frac{dt}{d\theta}$
= $\sqrt{17 + 8cos(3t) + 9cos^2(3t)}\frac{dt}{d\theta}$
and similarly
$|\frac{d}{d\theta}((2+cos(3t))cos(2t),$ $(2+cos(3t))sin(2t),$ $sin(3t))|$
= $\sqrt{25 + 16cos(3t) + 4cos^2(3t)}\frac{dt}{d\theta}$
I need a solution to, for example
$\int \sqrt{17 + 8cos(3t) + 9cos^2(3t)}dt$ so that I can get $t$ in terms of $\theta$
Wolfram alpha doesn't like either of these integrals, and I can see no way to solve them.
There are two types of answers to this question: one would solve one of these integrals, another would tell me how to change the parameterization, reasonably, so that the integral, and the obtained formula for t, is solvable.
I guess the third is to tell me that this question isn't solvable like this.
My thoughts on the latter: Below, I make the integral solvable by changing the parameterization, but I can't solve for $t(\theta)$
The freedom in the parameterizations can be expressed as:
$(sin(t) + Asin(2t),$ $cos(t) - Acos(2t),$ $Bsin(3t))$
where $A > 1$ and $B > 0$
which gives integrand:
$\sqrt{1 + 4A^2 + 4Acos(3t) + 9B^2cos^2(3t)}$
To eliminate the $\sqrt{}$ we want some factorization:
$(1 + 4A^2 + 4Acos(3t) + 9B^2cos^2(3t)) = (3Bcos(3t) + \lambda)^2$
where $\lambda^2 = 1 + 4A^2$ and $6B\lambda = 4A$
so $\lambda = \frac{2A}{3B} $
so $\frac{4A^2}{9B^2} = 1 + 4A^2$
ie. $4A^2\frac{9B^2 - 1}{9B^2} = -1$ or $4A^2 = \frac{9B^2}{1 - 9B^2} > 4$
so $9B^2 > 4/5$ and I will choose $9B^2 = 9/10$ giving:
$B = \frac{1}{\sqrt{10}}$
$A = 3/2$
$\lambda = \sqrt{10}$
ie. the trefoil $(sin(t) + \frac{3}{2}sin(2t),$ $cos(t) - \frac{3}{2}cos(2t),$ $\frac{1}{\sqrt{10}}sin(3t))$
has arclength parameterization given by $t(\theta)$ where:
$\int (\frac{3}{\sqrt{10}} cos(3t) + \sqrt{10}) dt = \theta$
$\frac{sin(3t)}{\sqrt{10}} + t\sqrt{10} = \theta$
But I want $t$ in terms of $\theta$
|
Since both integrands here are square roots of quadratic polynomials in $\cos3t$, they are reducible to elliptic integrals through the method I detailed here. There will be an extra algebraic/trigonometric term here, but it simplifies quite a bit because of preceding substitutions. I did all of the work in Mathematica and my notebook can be found here.
The end results for $0\le y\le\frac\pi3$ (all other values of $y$ follow by symmetry) are
$$\int_0^y\sqrt{9\cos^23t+8\cos3t+17}\,dt=L+\sqrt g\left(n\left(\frac{F(\varphi,m)-E(\varphi,m)}m\right)+\frac{16}{137}(1-m)\Pi(n,\varphi,m)\right)\\
\cos\varphi=\frac{Au-1}{A-u},u=\cos3y,L=\frac13\left(\frac{\sqrt{(1-u^2)(9u^2+8u+17)}}{u-A}+\frac43\tan^{-1}\sqrt{\frac{1-u^2}{u^2+8u/9+17/9}}\right)\\
A=\frac{-13-3\sqrt{17}}4,g=\frac{46359+11297\sqrt{17}}{192},m=\frac12-\frac2{3\sqrt{17}},n=\frac12-\frac{13}{6\sqrt{17}}$$
$$\int_0^y\sqrt{4\cos^23t+16\cos3t+25}\,dt=L+\sqrt g\left(n\left(\frac{F(\varphi,m)-E(\varphi,m)}m\right)+\frac{16}9(1-m)\Pi(n,\varphi,m)\right)\\
\cos\varphi=\frac{Au-1}{A-u},u=\cos3y,L=\frac13\left(\frac{\sqrt{(1-u^2)(4u^2+16u+25)}}{u-A}+4\tan^{-1}\sqrt{\frac{1-u^2}{u^2+4u+25/4}}\right)\\
A=\frac{-29-3\sqrt{65}}{16},g=\frac{-195+99\sqrt{65}}{512},m=\frac12-\frac7{2\sqrt{65}},n=\frac12-\frac{29}{6\sqrt{65}}$$
where all elliptic integral arguments follow Mathematica/mpmath conventions.
|
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|
How can i solve $5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$? How can i solve it?
$$5^{2x-\frac{1}{3}x^2} < 5^{2-2x} * (5^\frac{1}{3})^{x^2}+24$$
I don't have idea how to solve it..
|
Taking uldek's comment in the form of an answer.
Let $A=5^{2x-{x^2}/3}$. Note that $A>0$ for any $x$.
Notice that A appears as is on the LHS.
Notice that A is hidden on the RHS, as $5^{2-2x}*(5^{1/3})^{x^2}=5^2*5^{-2x+{x^2}/3}=5^2*A^{-1}$.
Now the inequality is $A < 25*A^{-1} +24$;
Take all the terms on the same side:
$A -24 - 25*A^{-1} < 0$
As $A>0$, you can multiply both sides by A without changing the sense of the inequality:
$A^2 - 24A - 25 < 0$
Factor:
$(A-25)(A+1)<0$
Draw a little table where you show the sign of the product:
$$
\begin{array}{c|ccc}
A & \text{under -1} & \text{between} & \text{over 25} \\
\hline
A+1 & - & + & + \\
A-25 & - & - & + \\
(A-25)(A+1) & + & - & + \\
\end{array}
$$
So, that the inequation is negative implies $-1<A<25$
Remember that $A>0$ no matter what, so we solve:
$0<A<25$
$5^{2x-{x^2}/3}<25$
We notice that $25=5^2$
$5^{2x-{x^2}/3}<5^2$
By definition of what $5^r$ is (that is : $5^r=e^{r*ln(5)}$), and the exponential function being a strictly increasing function, so $e^X<e^Y$ is equivalent to $X<Y$.
Put another way, "you take the logarithm" of the inequality - logarithm being also a strictly increasing function.
$2x-{x^2}/3<2$
Put all the terms on the same side of the inequality (this time, I am choosing the right hand side):
${x^2}/3-2x+2>0$
Multiplying by 3 > 0 keeps the sense of the inequality:
${x^2}-6x+6>0$
Factor... yes, discuss $b^2-4ac$, that stuff...
$(x-(3+\sqrt{3}))(x-(3-\sqrt{3}))>0$
Discuss the sign of the product...
$$
\begin{array}{c|ccc}
x & \text{under $3-\sqrt{3}$} & \text{between} & \text{over $3+\sqrt{3}$} \\
\hline
x-(3-\sqrt{3}) & - & + & + \\
x-(3+\sqrt{3}) & - & - & + \\
(x-(3+\sqrt{3}))(x-(3-\sqrt{3})) & + & - & + \\
\end{array}
$$
You can now write down the conclusion.
Hope this helps!
(and I hope that no typo sneaked in my text)
|
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|
Solving the integral I'm trying to calculate the following integral :$\int \sqrt x (2-3x^2)^2dx$, but somehow I can't get it right, here's what I did:
First I expanded the expression:
\begin{align}&\int \sqrt x (4-12x^2 + 9x^4)
\\&= \int(4\sqrt x-12x^2\sqrt x + 9x^4\sqrt x) \end{align}
Then I evaluated each term individually:
\begin{align}&4 \int \sqrt x - 12 \int x ^\frac{5}{2} + 9 \int x^\frac{9}{2} \\&
= \frac{4x^\frac{3}{2} \cdot 2}{3} - \frac{12x^\frac{7}{2} \cdot 2}{7} + \frac{9x^\frac{11}{2} \cdot 2}{11}\end{align}
Which gives
$$\frac{8x^\frac{3}{2}}{3} - \frac{24x^\frac{7}{2}}{7} + \frac{18x^\frac{11}{2}}{11}$$
What's wrong with my solution?
|
You just forget the constant (and the minus sign as commented), other than that, your answer is correct.
We can verify the solution by differentiating the solution.
\begin{align}\frac{d}{dx}\left(\frac{8x^\frac{3}{2}}{3} - \frac{24x^\frac{7}{2}}{7} + \frac{18x^\frac{11}{2}}{11}+c\right) &=\left( \frac32\right)\frac{8x^\frac12}{3}-\left( \frac72\right)\frac{24x^\frac52}{7}+\left(\frac{11}2\right) \frac{18x^\frac92}{11}\\&= 4x^\frac12-12x^\frac52+9x^\frac92\\&= \sqrt x (4-12x^2 + 9x^4) \end{align}
|
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"url": "https://math.stackexchange.com/questions/2571027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
system of equations with 3 variables in denominator I'm having problems with this system of equations:
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2}$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3}$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}$$
I've tried substitution method but that took me nowhere, I'm not sure I can solve it and would appreciate if anyone showed me the proper method to solve this problem.
|
I hope you mean the following.
$$\frac{3}{x+y}+\frac{2}{x-z}=\frac{3}{2},$$
$$\frac{1}{x+y}-\frac{10}{y-z}=\frac{7}{3},$$
$$\frac{3}{x-z}+\frac{5}{y-z}=-\frac{1}{4}.$$
If so, let $\frac{1}{x+y}=c$, $\frac{1}{x-z}=b$ and $\frac{1}{y-z}=a$.
Thus,
$$3c+2b=\frac{3}{2},$$ $$c-10a=\frac{7}{3}$$ and
$$3b+5a=-\frac{1}{4}.$$
The last two equations give $$6b+c=\frac{11}{6},$$ which with first gives
$$c=\frac{1}{3},$$ $$b=\frac{1}{4}$$ and from here we'll get $$a=-\frac{1}{5}.$$
Thus, $$x+y=3,$$ $$x-z=4$$ and $$y-z=-5,$$ which gives
$$(x,y,z)=(6,-3,2).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
note that $$\frac{1-2x}{2x+3}\geq 0$$ is hold if $$1-2x\geq 0$$ and $$2x+3>0$$ or
$$1-2x\le 0$$ and $$2x+3<0$$ you have to solve These two cases
|
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|
How to algebraically solve $\frac{1}{x} < 5$ inequality to obtain two solutions? Given the inequality:
$\frac{1}{x} < 5$
In order to find a solution, I would normally multiply both sides by $x$:
$1 < 5x$
Then I would divide by $5$
$\frac{1}{5} < x$
To obtain the solution: $x > \frac{1}{5}$.
Now, the thing is, the solutions are actually two: $x > \frac{1}{5}$ and $x < 0$
How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ($x < 0$).
Thanks!
|
$$\frac { 1 }{ x } <5\quad \Rightarrow \frac { 1 }{ x } -5<0\quad \Rightarrow \quad \frac { 1-5x }{ x } <0\quad \Rightarrow \frac { x\left( 1-5x \right) }{ { x }^{ 2 } } <0\\ x\left( 5x-1 \right) >0\quad \Rightarrow x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 5 } ,+\infty \right) $$
|
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|
If $\vert{z_{1}+ \cdots + z_{n}}\vert$ = $\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert$ then $z_{j} = c_{j}z_{1}$ Could someone give me just one suggestion to solve this problem?
Let $z_{1}, \ldots z_{n} \in \mathbb{C}$, with $z_{1}\neq 0$. Prove that if
$\vert{z_{1}+ \cdots + z_{n}}\vert$ =
$\vert{z_{1}}\vert + \cdots + \vert{z_{n}}\vert$ then $z_{j} = c_{j}z_{1}$, where
$c_{j}\geqslant 0 $, $j=1,\ldots n $
|
Here's the base case.
If
$|(a+bi)+(c+di)|
=|a+bi|+|c+di|
$
then,
$\sqrt{(a+c)^2+(b+d)^2}
=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}
$.
Squaring,
$(a+c)^2+(b+d)^2
=a^2+b^2+c^2+d^2+2\sqrt{(a^2+b^2)(c^2+d^2)}
$
or
$2ac+2bd
=2\sqrt{(a^2+b^2)(c^2+d^2)}
$.
Dividing by 2
and squaring again,
$a^2c^2+2abcd+b^2d^2
=a^2c^2+a^2d^2+b^2c^2+b^2d^2
$
or
$0
=a^2d^2+b^2c^2-2abcd
=(ad-bc)^2
$
so
$ad=bc$.
Now it becomes the cases
depending on
which, if any,
of $b$ abd $d$ are zero.
If $b\ne 0, d\ne 0$,
then
$a/b = c/d$.
Letting
$r = a/b$,
then
$a = rb, c = rd$
so
$a+ib
=b(r+i)
$
and
$c+id
= d(r+i)
$
so
$c+id
= (d/b)b(r+i)
=(d/b)(a+ib)
$.
I'll let you work out the
other cases.
|
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|
Find the MSE of the MLE
Let $X_1,\ldots,X_n$ be i.i.d exponentially distributed r.v's with pdf:
$$ f_\theta(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$$
such that $x\ge 0$.
I have found that the MLE is given by $$ \hat{\theta}(X)= \frac{1}{n}\sum_{i=1}^n X_i.$$
Note that $\hat{\theta}$ is unbiased as
$$ E(\hat{\theta}) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \theta $$
since the r.v's are i.i.d.
But then I am having difficulties computing the MSE as
$$ MSE = E_{\theta}\left( (\theta-\hat{\theta})^2 \right) $$
$$= E(\hat{\theta^2})+\theta^2-2\theta E(\hat{\theta})$$
$$ =E(\hat{\theta^2}) -\theta^2$$
So then $E(\hat{\theta^2}) = E\left(\frac{1}{n^2}\sum_{i=1}^n X_i^2\right) =E(X^2) = \int_0^\infty f_\theta dx = ... = \frac{2\theta^2}{n^2} $
$ \Rightarrow MSE = \frac{2\theta^2}{n^2} - \theta^2$
Supposedly the final answer should be $\frac{\theta^2}{n}$ but im not getting this at all...
|
It's not true that
$$ E(\hat{\theta^2}) = E\left(\frac{1}{n^2}\sum_{i=1}^n X_i^2\right).$$
Rather, one has
\begin{align}
E(\hat{\theta^2}) &= E\left[\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\right] \\
&= E\left[\frac{1}{n^2}\sum_{i,j} X_iX_j \right] \\
&= \frac{1}{n^2}E\left[\sum_{i=1}^n X_i^2 + \sum_{i\ne j} X_iX_j \right] \\
&= \frac{1}{n^2}E\left[\sum_{i=1}^n X_i^2 \right] + \frac{1}{n^2}E\left[\sum_{i\ne j} X_iX_j \right] \\
&= \frac{1}{n^2} \sum_{i=1}^nE\left[ X_i^2 \right] + \frac{1}{n^2} \sum_{i\ne j} E\left[X_iX_j \right] \\
&= \frac{1}{n^2} \sum_{i=1}^nE\left[ X^2 \right] + \frac{1}{n^2} \sum_{i\ne j} E[X_i]E[X_j] \\
&= \frac{1}{n^2} \sum_{i=1}^n 2\theta^2 + \frac{1}{n^2} \sum_{i\ne j} \theta^2 \\
&= \frac{1}{n^2} n2\theta^2 + \frac{1}{n^2} (n^2 - n)\theta^2 \\
&= \frac{\theta^2}{n^2} (n^2 - n + 2n) \\
&= \theta^2 + \frac{\theta^2}{n}.
\end{align}
|
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|
How to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}$ using AM-GM? In this question Prove that, if $g(x)$ is concave, for $S = {x : g(x) > 0}$, $f(x) = 1/g(x)$ is convex over $S$. , in the proof of Math536,
how to prove $2\le\frac{g(x)}{g(y)}+\frac{g(y)}{g(x)}?$
This is the proof of Math536:
$$\begin{align}
1 &= (a+(1-a))^2 \\
&= a^2 + 2a(1-a) + (1-a)^2 \\
&\le a^2 + a(1-a)\left(\frac{g(x)}{g(y)} + \frac{g(y)}{g(x)}\right) + (1-a)^2 \\
&= (ag(x)+(1-a)g(y)) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&\le g(ax+(1-a)y) \left(\frac{a}{g(x)} + \frac{1-a}{g(y)}\right) \\
&= \frac{af(x) + (1-a)f(y)}{f(ax+(1-a)y)}
\end{align}$$
He said we should use the AM-GM inequality which is $\forall a_i\in\mathbb R^+: \frac{a_1+...+a_n}{n}\ge (a_1\dots a_n)^{1/n}.$
I tried to find a similarity with the LHS of the AM-GM inequality:
$$g(x)f(y)+f(x)g(y)\ge\sqrt{g(x)g(y)f(x)f(y)}$$
I honestly don't know how will I get the $2$ in the inequality?
Can someone help me please?
|
Don't worry about functions.
For any two $x,y$ then $\frac {g(x)}{g(y)} = k$ will be a positive number, and $\frac {g(y)}{g(x)} = \frac 1k$.
So by AM-GM. $\frac {k + \frac 1k}2 \ge \sqrt {k*\frac 1k}=1$.
And that's it.
....
In fact this basically is the proof of the AM-GM.
$(k - 1)^2 \ge 0$ so
$k^2 + 1 \ge 2k$ so
$k + \frac 1k \ge 2$.
Or in general:
$(\sqrt a - \sqrt b)^2 \ge 0$ so
$a + b \ge 2\sqrt{ab}$ so
$\frac {a+b}2 \ge \sqrt{ab}$.
|
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|
Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have: $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
So we have a system of three Diophantine equations:
$$n=a^2+b^2$$
and $$n=c^2+d^2+e^2$$
and $$n=f^4+g^4+h^4+i^4$$
where, $n,a,b,c,d,e,f,g,h,i \in \mathbb N$.
Is there a finite number of these numbers?
Edit : Also, it is $$2018=35^2+26^2+8^2+7^2+2^2$$ a sum of five squares.
And of $$2018=11^3+7^3+7^3+1^3$$ four cubes.
|
There are an infinite number of such integers.
One way to show this is to start from the fact that there are an infinite number of Pythagorean triples $a^2 + b^2 = c^2$. Given any such triple, let $N$ be given by:
$$N = (ab)^4 + (bc)^4 + (ac)^4 + c^4$$
By an identity of Fauquembergue (1) we have (given $a^2 + b^2 = c^2$):
$$(ab)^4 + (bc)^4 + (ac)^4 = (a^4 + a^2b^2 + b^4)^2$$
Hence:
$$N = (a^4 + a^2b^2 + b^4)^2 + (c^2)^2$$
Furthermore:
$$(c^2)^2 = (a^2 + b^2)^2 = a^4 + 2a^2b^2 + b^4 = (a^2 - b^2)^2 + (2ab)^2$$
Hence:
$$N = (a^4 + a^2b^2 + b^4)^2 + (a^2 - b^2)^2 + (2ab)^2$$
Reference:
1) Dickson L E History of the Theory of Numbers Vol 2 Ch XXII p 658
|
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|
Find $\lim\limits_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$ $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$$
I'm surprisingly struggling with this limit, could you give me a hint how to handle it (no L'Hospital and no prior knowledge what the limit value is)?
Generally what are some basic methods to handle $\infty \cdot 0$?
|
$$\lim _{ n\rightarrow \infty } n\left( \cos \left( \frac { 1 }{ \sqrt { n } } \right) -1 \right) =\lim _{ n\rightarrow \infty } n\left( \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) - } } \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) - } } \right) =\\ =\lim _{ n\rightarrow \infty } n\left( -2\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n } } \right) } \right) =-\frac { 1 }{ 2 } \lim _{ n\rightarrow \infty }{ \left( \frac { \sin { \left( \frac { 1 }{ 2\sqrt { n } } \right) } }{ \frac { 1 }{ 2\sqrt { n } } } \right) } ^{ 2 }=-\frac { 1 }{ 2 } $$
|
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|
Translate and rotate the coordinate axes to put the quadric in standard position, $2xy+2xz+2yz-6x-6y-4z=-9$
Referring to the image attached. Is my working correct?
The answer gives the coefficient on the other side of the equation is -1. Please show where went wrong.
Thanks.
|
Let $A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix} \, .
$
To write the quadric in standard form, we must write it with respect to an orthonormal set of eigenvectors, which amounts to orthogonally diagonalizing $A$.
As mentioned in the comments, you need to apply Gram-Schmidt orthogonalization to obtain an orthonormal basis for the eigenspace of the eigenvalue $-1$.
Diagonalizing $A$, we find that $P^{-1} A P = D$ where
$$
D =
\left(\begin{array}{rrr}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{array}\right)
\qquad
P =
\left(\begin{array}{rrr}
1 & 1 & 0 \\
1 & 0 & 1 \\
1 & -1 & -1
\end{array}\right) \, .
$$
Let $w_i$ be the $i^\text{th}$ column of $P$. Note that the last two columns $w_2, w_3$ are not orthogonal. Applying Gram-Schmidt, we set $v_2 = w_2$ and compute
$$
v_3 = w_3 - \operatorname{proj}{v_2}(w_3) = w_3 - \frac{w_3 \cdot v_2}{v_2 \cdot v_2} v_2 = w_3 - \frac{1}{2} v_2 = \begin{pmatrix} -1/2\\ 1\\ -1/2 \end{pmatrix} \, .
$$
Normalizing the vectors, we obtain the orthogonal matrix
$$
Q =
\left(\begin{array}{rrr}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\
\frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}
\end{array}\right) \, .
$$
To deal with the degree $1$ terms, we compute
$$
\left(-6,\,-6,\,-4\right)
\left(\begin{array}{rrr}
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}\\
\frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{2}}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}}
\end{array}\right)
= \left(-\frac{16}{\sqrt{3}}, -\sqrt{2},\,-\frac{\sqrt{2}}{\sqrt{3}}\right) \, .
$$
Thus the transformed quadric is $0 = 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9$ and completing the square yields
\begin{align*}
0 &= 2 x^2 - y^2 - z^2 - \frac{16}{\sqrt{3}} x - \sqrt{2}y -\frac{\sqrt{2}}{\sqrt{3}} z + 9\\
&= 2\left(\left(x - \frac{4}{\sqrt{3}} \right)^2 - \frac{16}{3} \right) - \left(\left(y + \frac{\sqrt{2}}{2} \right)^2 - \frac{1}{2} \right) - \left(\left(z + \frac{1}{\sqrt{6}} \right)^2 - \frac{1}{6} \right) + 9\\
&= 2 u^2 - v^2 - w^2 - \frac{32}{3} + \frac{1}{2} + \frac{1}{6} + 9 = 2 u^2 - v^2 - w^2 - 1 \, .
\end{align*}
|
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|
Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
|
This is really the same as aid78's method.
$$(3+4i)e^{ix}=(3\cos x-4\sin x)+i(3\sin x+4\cos x)
=2+iA$$
for $A=3\sin x+4\cos x$. As $|(3+4i)e^{ix}|^2=3^2+4^2=25$
then $25=4+A^2$, so $A=\pm\sqrt{21}$ (both are possible).
|
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|
Sum of the infinite series The series is $$\frac{5}{1\cdot2}\cdot\frac{1}{3}+\frac{7}{2\cdot3}\cdot\frac{1}{3^2}+\frac{9}{3\cdot4}\cdot\frac{1}{3^3}+\frac{11}{4\cdot5}\cdot\frac{1}{3^4}+\cdots$$
This is my attempt:
$$T_n=\frac{2n+3}{n(n+1)}\cdot\frac{1}{3^n}$$
Assuming $$\frac{2n+3}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$$
we find $A=3,B=-1.$ Putting these values in $T_n$ we get,
$$T_n=\frac{1}{n}\cdot\frac{1}{3^{n-1}}-\frac{1}{n+1}\cdot\frac{1}{3^{n}}$$
How do I find the sum of the series from here $?$
|
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=1}^{\infty} \frac{1}{n+1}.\frac{1}{3^{n}}$$
$$Sum=\sum_{n=1}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}-\sum_{n=2}^{\infty} \frac{1}{n}.\frac{1}{3^{n-1}}$$
$$Sum=\sum_{n=1}^{1} \frac{1}{n}.\frac{1}{3^{n-1}}=1$$
|
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|
In which interval lies the minimum? Let $a_1,a_2,a_3,a_4$ be real numbers such that $a_1+a_2+a_3+a_4 =0 $ and $a_1^2+a_2^2+a_3^2+a_4^2=1$.
Then in what interval does the smallest possible value of the following expression lies?
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$$
Here lies the actual question with the options.
|
Let $a_1=0$, $a_2=\frac{1}{\sqrt2},$ $a_3=0$ and $a_4=-\frac{1}{\sqrt2}.$
Hence, we get a value $2$.
We'll prove that it's a minimal value.
Indeed, let $a_1=a$, $a_2=b$, $a_3=c$.
Thus, $a_4=-a-b-c$, $a^2+b^2+c^2+(a+b+c)^2=1$ and we need to prove that
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\geq2$$ or
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\geq2\left(a^2+b^2+c^2+(a+b+c)^2\right)$$ or
$$(a+c)^2\geq0.$$
Done!
We'll find a maximal value for the collection.
Let $a_i=(-1)^{i-1}\frac{1}{2}.$
Hence, we get a value $4$.
We'll prove that it's a maximal value.
Indeed, let $a_1=a$, $a_2=b$, $a_3=c$.
Thus, $a_4=-a-b-c$, $a^2+b^2+c^2+(a+b+c)^2=1$ and we need to prove that
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\leq4$$ or
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\leq4\left(a^2+b^2+c^2+(a+b+c)^2\right)$$ or $$(a+b)^2+(b+c)^2\geq0.$$
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the derivative of $\frac{2}{x^2}$ directly from definition I have to find the derivative of $\frac{2}{x^2}$ using only the definition. I have tried to do this several times but I always get the wrong answer - could you tell me where I made an error?
$$\frac{d}{dx}[\frac{2}{x^2}] = \lim_{h \to 0} \frac{\frac{2}{(x+h)^2}- \frac{2}{x^2}}{h} = \lim_{h \to 0} \frac{2x^2-2(x+h)^2}{x^2(x+h)^2h} = \lim_{h \to 0} \frac{-4xh -2h^2}{x^4h + 2x^3h^2 + x^2h^3} = \lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$
As you see, I keep getting $x^4$ instead of $x^3$ but I simply cannot see where I made a mistake - any suggestions?
|
You missed an $x$$$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4}{x^4}$$ It should have been $$\lim_{h \to 0} \frac{-4x -2h}{x^4+2x^3h +x^2h^2} = \frac{-4x}{x^4}= \frac{-4}{x^3}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
When are these eigenvalues non-negative? I'm trying to find a pair of real numbers $(a,b)$ which ensure that some matrix is strictly positive semi-definite.The eigenvalues of this matrix are $$\lambda=1 + a \pm \sqrt{(c+b)^2+2(x'-ax)^2}$$ and $$\lambda=1 - a \pm \sqrt{(c-b)^2+2(x'-ax)^2}.$$
I therefore need one of the eigenvalues to be zero and the rest non-negative.
For a fixed $c,x,x' \in \mathbb R$, what is the best way to determine some $a,b$ exist? If not, are there conditions on $x,x',c$ so that $a,b$ exist? Here I know that $0\leq x,x' \leq 1/\sqrt2$ and $0 \leq c \leq 1$.
|
I guess you know if we put $a=m+n$, $b=m-n$, $t=x’-ax$, $a=x$, $b=x’$ then we obtain this problem with additional conditions that $0\le a,b\le 1/\sqrt{2}$ and one of $\lambda$’s is $0$. Nevertheless, these conditions turn us to a very different way.
Let’s start. The required $a$ and $b$ exist iff all the following conditions hold
$|a|\le 1$
$(1 + a)^2\ge (c+b)^2+2(x'-ax)^2$
$(1 – a)^2 \ge (c-b)^2+2(x'-ax)^2$
and one of the inequalities is an equality.
Let $a$ is fixed. Put $d=\sqrt{2}|x’-ax|$. We need $d\le\min\{1-a,1+a\}$. Then
$\sqrt{(1 + a)^2-d^2}\ge |b+c|$
$\sqrt{(1 - a)^2-d^2}\ge |b-c|$
and one of the inequalities is an equality. That is
$-\sqrt{(1 + a)^2-d^2}\le b+c\le \sqrt{(1 + a)^2-d^2}$
$-\sqrt{(1 - a)^2-d^2}\le b-c\le \sqrt{(1 - a)^2-d^2}$
and one of the inequalities is an equality. That is
$-\sqrt{(1 + a)^2-d^2}-c\le b\le \sqrt{(1 + a)^2-d^2}-c$
$-\sqrt{(1 - a)^2-d^2}+c\le b\le \sqrt{(1 - a)^2-d^2}+c$.
and one of the inequalities is an equality.
Such $b$ exists iff one of endpoints of the segments $[-\sqrt{(1 + a)^2-d^2}-c,\sqrt{(1 + a)^2-d^2}-c]$ and $[-\sqrt{(1 - a)^2-d^2}+c,\sqrt{(1 - a)^2-d^2}+c]$ belongs to the other, that is when the segments intersect. This holds iff the distance between their midpoints is not greater than a half of the sum of their lengths, that is when
$$2c\le \sqrt{(1 + a)^2-d^2}+\sqrt{(1 - a)^2-d^2}\equiv f(a).$$
For the sake of simplicity we introduce variables $y=\sqrt{2}x$ and $y’=\sqrt{2}x’$ (so $0\le y,y’\le 1$). Then $d^2=(y’-ay)^2$ and
$$f(a)=\sqrt{1+2a+a^2-y’^2+2ayy’-a^2y^2}+\sqrt{1-2a+a^2-y’^2+2ayy’-a^2y^2}.$$
From the above we see that the required $a$ and $b$ exist iff there exists $a$ such that $f(a)\ge 2c$ and $d\le\min\{1-a,1+a\}$. We claim that it suffices to look for such not-negative $a$. Indeed, if the needed conditions are satisfied then when we replace $a$ by $|a|$ we obtain that the new expression under the first (resp., the second) radical will become the old expression under the second (resp., the first) radical plus $4|a|yy’$, so $f(|a|)\ge f(a)$. The condition $$|d|\le \min\{1-|a|,1+|a|\}=\min\{1-a,1+a\}$$ will be also kept, because $$|(y’-|a|y|)|\le|y’|+|a||y|=|y’-ay|.$$
So now we have $0\le a\le 1$ and $|y’-ay|\le 1-a$. The last condition determines our maximal possible value $a_\max$ of $a$, namely, $a_\max=\frac{1+y’}{2}$ if $y=1$ and $a_\max=\min\left\{\frac {1+y’}{1+y},\frac {1-y’}{1-y}\right\}$, if $y<1$.
If $a_\max=0$ (which holds iff $y<1$ and $y’=1$) then the required $a$ and $b$ exist iff $f(0)=0\ge 2c$, that is iff $c=0$. Further we assume that $a_\max>0$. The function $f(a)$ is continuous at the closed segment $[0,a_\max]$, so it attains its maximum at some point $a_0\in [0,a_\max]$. Moreover, since the function $f(a)$ is differentiable on the interval $(0, a_\max)$, we have $a_0=0$ (remark that $f(0)=2\sqrt{1-y’^2}$), or $a_0=a_\max$, or $f’(a_0)=0$. Routine calculations simplify the last condition to
$-1+a+yy’-ay^2\le 0$ and $(ay-y’)(ay^2y’-ay’+y-yy’^2)=0$.
If $ay-y’=0$ then $d=0$, so $f(a)$ is defined and equals $2$, so the required $a$ and $b$ exist for any $0\le c\le 1$. The equality $ay-y’=0$ can be satisfied by some $0\le a\le 1$ iff $y\ge y’$. So further we assume that $y<y’<1$. Then $a_\max=\frac {1-y’}{1-y}$ and $f(a_\max)=2\sqrt{\frac{1-y’}{1-y}}.$
Let $ay^2y’-ay’+y-yy’^2=0$.
If $y’-y’y^2=0$ then $y-yy’^2=0$ and we have no conditions for $a$, but these equalities hold iff $y=y’=0$ (then $f(a)=2$) or $y=y’=1$ (then $f(a)=2\sqrt{a}$, $a_\max=1$ and $f(a_\max)=2$). Anyway, the required $a$ and $b$ exist for any $0\le c\le 1$. By the way, these cases were already excluded.
If $y’-y’y^2>0$ then $a=\frac{y-yy’^2}{y’-y’y^2}$. Since $y<y’$, $a<a_\max$ and $-1+a+yy’-ay^2<0$, so this $a$ is an admissible value. We have $f(a)=2\sqrt{\frac{1-y’^2}{1-y^2}}$. It is easy to check that $f(a)>f(0)$ and $f(a)>f(a_\max)$.
The summary. The required $a$ and $b$ exist iff $(1-2x^2)c^2\le (1-2x’^2)$.
|
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|
Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive members but I couldn't manage to show something. It is obvious that $$\frac{n^n}{n!}$$ is increasing but that don't help us much (I think)
|
The ratio of the consecutive terms of the sequence $a_n=\frac{n}{\sqrt[n]{n!}}$ is greater than $1$.
Therefore, $\frac{n}{\sqrt[n]{n!}}$ is increasing.
$$
\begin{align}
\left(\frac{a_{n+1}}{a_n}\right)^{n(n+1)}
&=\left[\frac{\frac{\color{#C00}{n+1}}{\color{#090}{\sqrt[n+1]{(n+1)!}}}}{\frac{\color{#C00}{n}}{\color{#090}{\sqrt[n]{n!}}}}\right]^{n(n+1)}\tag1\\
&=\left(\color{#C00}{\frac{n+1}n}\right)^{n(n+1)}\color{#090}{\frac{n!^{n+1}}{(n+1)!^n}}\tag2\\[6pt]
&=\left(\frac{n+1}n\right)^{n(n+1)}\frac{n!}{(n+1)^n}\tag3\\[6pt]
&=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}\tag4\\[15pt]
&\ge2\tag5
\end{align}
$$
Explanation:
$(2)$: distribute the exponent $n(n+1)$ over the ratios
$(3)$: cancel $n!^n$ in the numerator and denominator of the right term
$(4)$: move $\left(\frac{n+1}n\right)^n$ from the left term to the right term
$(5)$: inequality $(10)$ shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing
$\phantom{\text{(5):}}$ since $b_1=2$, $b_n\ge2$
Bernoulli's Inequality shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing:
$$
\begin{align}
\frac{b_{n+1}}{b_n}
&=\frac{\color{#C00}{\left(\frac{n+2}{n+1}\right)^{(n+1)^2}}\color{#090}{\frac{(n+1)!}{(n+1)^{n+1}}}}{\color{#C00}{\left(\frac{n+1}n\right)^{n^2}}\color{#090}{\frac{n!}{n^n}}}\tag6\\
&=\color{#C00}{\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{2n+1}}\color{#090}{\left(\frac{n}{n+1}\right)^n}\tag7\\
&=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{n+1}\tag8\\[6pt]
&\ge\left(1-\frac1{n+1}\right)^{n+1}\left(\frac{n+1}n\right)^{n+1}\tag9\\[12pt]
&=1\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: multiply numerator and denominator by $\left(\frac{n+1}n\right)^{2n+1}$ and combine terms
$\phantom{1}(8)$: $\frac{n(n+2)}{(n+1)^2}=1-\frac1{(n+1)^2}$
$\phantom{1}(9)$: Bernoulli's Inequality
$(10)$: $1-\frac1{n+1}=\frac{n}{n+1}$
|
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|
Numerical radius of a pair of operators in Hilbert spaces Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=\displaystyle\sup_{\|x\|=1}\left(|\langle Cx,x \rangle|^2+|\langle Dx,x \rangle|^2\right)^{1/2}.$$
Moreover, the following inequality holds:
$$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}\leq w_e(C,D)\leq \|C^*C+D^*D\|^{1/2}.$$
I want to show that the constants $\frac{\sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ (operator on $(\mathbb{C}^2,\|\cdot\|)$). Hence, I get
$w_e(C,D)=\sqrt{2}$ and $\|C^*C+D^*D\|^{1/2}=\sqrt{2}.$
I want to find $(C,D)$ such that
$$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}= w_e(C,D).$$
For a single operator, we have the following theorem:
Do you think that if $\text{Im}(C)\perp \text{Im}(C^*)$ and $\text{Im}(D)\perp \text{Im}(D^*)$ we have
$$\frac{\sqrt{2}}{4}\|C^*C+D^*D\|^{1/2}= w_e(C,D)\,?$$
Thank you in advance.
|
For your first question, consider
$$ C = D = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ \frac{\sqrt{2}}{4}\| 2\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}\| = \frac{\sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = \sqrt{2} \sup_{\|x\|=1} |\langle \begin{pmatrix}0 \\ x_1 \end{pmatrix}, \begin{pmatrix} x_1\\x_2 \end{pmatrix}\rangle| = \sqrt{2} \sup_{\|x\|=1} |x_1x_2| = \frac{\sqrt{2}}{2}.$$
For your second question: Consider
$$ C = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, D = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},$$
then $Im(C) \perp Im(C^\ast)$ and $Im(D) \perp Im(D^\ast)$ but
$$ \frac{\sqrt{2}}{4} \| CC^\ast + DD^\ast\|^{1/2} = \frac{\sqrt{2}}{4}, \quad w_e(C,D)) = \frac{1}{\sqrt{2}}.$$
|
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|
How to prove that this triangle is equilateral? Question: If $\cos A +\cos B +\cos C=\frac{3}{2}$, prove that the triangle is equilateral.
My attempt: According to cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$, $\cos B=\frac{c^2+a^2-b^2}{2ca}$ and $\cos C=\frac{a^2+b^2-c^2}{2ab}$.
$\cos A +\cos B +\cos C=\frac{3}{2}$
$\therefore\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\implies\frac{ab^2+ac^2-a^3+bc^2+ba^2-b^3+ca^2+cb^2-c^3}{2abc}=\frac{3}{2}$
I have tried simplifying the given equation using cosine rule but could not get far. Please help.
|
I Assume that all angles are acute. Now its wise to use concavity of $\cos$ to say (from Jensen's Inequality)
$$\cos(A)+\cos(B) + \cos(C) \ge 3\cos(\frac{A+B+C}{3}) = 3 \cos(\frac{\pi}{3})=\frac{3}{2}$$
equality occurs for $A=B=C = \pi / 3$.
I don't know how to handle obtuse triangles with this..,
|
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|
sum $\displaystyle \sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$ I have the following series:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{n^2+n-2}$$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$$\sum _{n=2}^{\infty}\frac{(-1)^n}{(n+2)(n-1)}$$
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?
|
$$\sum_{k=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}=\frac{1}{3}\sum_{k=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$
$$=\frac{1}{3}\left(1-\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{7}+...\right)=$$
$$=-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)+\frac{2}{3}\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}=\frac{2}{3}\ln2-\frac{5}{18}.$$
|
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|
Simple ODE Analytical Solution Question: $\frac{dx}{dt} = ax + bu$ my question is in regards to the following formula:
\begin{align}
\frac{dx}{dt}=ax+bu
\end{align}
where x(0)=0 & u=constant and a & b are scalar
The answer to this or at least what I was given as the answer is the following:
\begin{align}
x=-\frac{bu}{a}(1-e^{at)}
\end{align}
Now I understand that I need to manipulate the formula and integrate both sides. Utilizing the integration table I perform the following:
Since u=constant we can then say bu=b therefore:
\begin{align}
\frac {dx}{ax+b}= {dt}
\end{align}
\begin{align}
\int \frac {dx}{ax+b}= \int{dt}
\end{align}
\begin{align}
\frac{1}{a}ln|ax+b| = t + C
\end{align}
\begin{align}
e^{ln|ax+b|} = e^{(t + C)a}
\end{align}
\begin{align}
ax+b = e^{at+Ct}
\end{align}
\begin{align}
x = -\frac{b}{a} \frac{e^{at}e^{Ct}}{a}
\end{align}
I do not see where I am going wrong in the calcs above. I would appreciate any help in identifying where I went wrong.
Thank you
|
$$\begin{align}
x = -\frac{b}{a} \frac{e^{at}e^{Ct}}{a}
\end{align}$$
This line isn't correct, it should be:
$$\begin{align}
x = -\frac{b}{a} +\frac{e^{(t+C)a}}{a}
\end{align}$$
|
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|
Calculating limit $\lim\limits_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$ for an unknown function. Given that $f(x)$ is a continuous function and satisfies $f'(x)>0$ on $(-\infty,\infty)$ and $f''(x)=2 \forall x \in(0,\infty)$.We need to find the limit
$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$$
Now the numerator is tending to infinity so denominator must also go to infinity else limit won't exist.So I tried the L'Hospitals rule and it became$$\lim_{x\to\infty}\frac{6x+\frac{6x}{(x^2+1)^2}-4f''(x)}{f'(x)}$$The numerator is still infinity so once again applying L'Hospitals rule (assuming denominator must still be infinity) we get
$$\lim_{x\to\infty}\frac{6+\frac{6(x^2+1)^2-6x×2(x^2+1)×2x}{(x^2+1)^4}+0}{f''(x)}$$
Now putting $f''(x)=2$ we get
$$3+\lim_{x\to\infty}\frac{3(x^2+1)^2-12x^2(x^2+1)^2}{(x^2+1)^4}$$
Collecting the coefficients of $x^4$ from numerator and denominator we get the limit to be$3-9=-6$ but the answer is not -6. Is applying LHospital wrong?Help.Thanks.
|
Since $f''(x)=2~~for ~~x>0$, Therefore $f$ has the form $f(x)=x^2+bx +c~~~for~~~x>0$
Since the limit is at $+\infty$ it suffices to consider $x>0$
$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)} = \lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-8x-4b}{x^2+bx +c} =3$$
|
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|
Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$ What is the sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots?$$
I know how to check if a series is convergent or not.Is there any technique to find out the sum of a series like this where each term increases by a pattern?
|
My solution is adapted from another similar solution.
\begin{align}
\sum_{n=1}^{\infty}\frac{4\cdot7\cdot\cdots\cdot(3n+1)}{n!10^n}
&=\sum_{n=1}^{\infty}\frac{\frac43\cdot\frac73\cdot\cdots\cdot\frac{3n+1}3}{n!\left(10/3\right)^n}\\
&=\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}\left(\frac{3}{10}\right)^n\\
&=\left[\sum_{n=1}^{\infty}\binom{\frac{3n+1}{3}}{n}x^n\right]_{x=3/10}\\
&=\left[\sum_{n=1}^{\infty}\binom{-\frac{4}{3}}{n}(-x)^{n}\right]_{x=3/10}\\
&=\left[ \left(\left(1-x\right)^{-4/3}-1\right) \right]_{x=3/10}\\
&=\sqrt[3]{\left(\frac{10}{7}\right)^4}-1
\end{align}
|
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|
An alternative way to find the sum of this series? $\displaystyle \frac{4}{20}$+$\displaystyle \frac{4.7}{20.30}$+$\displaystyle \frac{4.7.10}{20.30.40}$+...
Now I have tried to solve this in a usual way, first find the nth term $t_n$.
$t_n$= $\displaystyle \frac{1}{10}$($\displaystyle \frac{1+3}{2}$) + $\displaystyle \frac{1}{10^2}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$) + ...+ $\displaystyle \frac{1}{10^n}$($\displaystyle \frac{1+3}{2}$)($\displaystyle \frac{1+6}{3}$)...($\displaystyle \frac{1+3n}{n+1}$)
=$\displaystyle \frac{1}{10^n}\prod$(1+$\displaystyle \frac{2r}{r+1}$) , $r=1,2,..,n$
=$\displaystyle \prod$($\displaystyle \frac{3}{10}-\displaystyle \frac{1}{5(r+1)}$)
thus, $t_n=$ (x-$\displaystyle \frac{a}{2}$)(x-$\displaystyle \frac{a}{3}$)...(x-$\displaystyle \frac{a}{n+1}$), x=$\displaystyle \frac{3}{10}$, a=$\displaystyle \frac{1}{5}$
Now to calculate $S_n$, I have to find the product $t_n$, and then take sum over it. But this seems to be a very tedious job. Is there any elegant method(may be using the expansions of any analytic functions) to do this?
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over 10^{n}\pars{n + 1}!}} =
\sum_{n = 2}^{\infty}{3^{n - 1}
\prod_{k = 1}^{n - 1}\pars{k + 1/3} \over 10^{n - 1}\,n!}
\\[5mm] = &\
{10 \over 3}\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,
{\Gamma\pars{4/3 + \bracks{n - 1}}/\Gamma\pars{4/3} \over n!}
\\[5mm] = &\
{10 \over 3}\,{\pars{-2/3}! \over \Gamma\pars{4/3}}
\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,
{\pars{n - 2/3}! \over n!\pars{-2/3}!}
\\[5mm] = &\
{10 \over 3}\,{\Gamma\pars{1/3} \over \pars{1/3}\Gamma\pars{1/3}}
\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}\,{n - 2/3 \choose n}
\\[5mm] = &\
10\sum_{n = 2}^{\infty}\pars{3 \over 10}^{n}
\bracks{{-1/3 \choose n}\pars{-1}^{n}}
\\[5mm] = &\
10\bracks{%
\sum_{n = 0}^{\infty}{-1/3 \choose n}\pars{-\,{3 \over 10}}^{n}
-\ \overbrace{-1/3 \choose 0}^{\ds{= 1}}\ -\
\overbrace{-1/3 \choose 1}^{\ds{= -\,{1 \over 3}}}\
\pars{-\,{3 \over 10}}}
\\[5mm] = &\
10\braces{\bracks{1 + \pars{-\,{3 \over 10}}}^{-1/3} - 1 - {1 \over 10}}
=
\bbx{10\pars{10 \over 7}^{1/3} - 11}
\\[5mm] \approx &\ 0.2625
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$
But know, I don't know how to continue for get what that rquired.
I will be happy for help or hint.
After beautiful ideas for solving this question, someone know how to do it with induction too?
|
$1+7+7^2+7^3=400$, so $$\begin{array}{rcl}7+7^2+7^3+\cdots+7^{4k}&=&(1+7+7^2+7^3)(7+7^5+7^9+\cdots+7^{4k-3})\\&=&100\cdot 4(7+7^5+7^9+\cdots+7^{4k-3})\end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2611490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.
|
$$2(xy+xz+yz)=(x+y+z)^2-(x^2+y^2+z^2)=0,$$ which says $xy+xz+yz=0$.
Thus, $$1=(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+xz+yz)-3xyz=1-3xyz$$ and we are done!
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determinant equal to zero : elements are angles Let $\alpha, \beta, \gamma$ be internal angles of an arbitrary triangle.
I want to show that $$\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=0$$
We have the following:
\begin{align*}&\det \begin{pmatrix}\cos \beta & \cos \alpha&-1 \\ \cos\gamma & -1 & \cos\alpha \\ -1 & \cos\gamma & \cos\beta\end{pmatrix}=\cos \beta \cdot \det \begin{pmatrix} -1 & \cos\alpha \\ \cos\gamma & \cos\beta\end{pmatrix}-\cos \alpha \cdot \det \begin{pmatrix} \cos\gamma & \cos\alpha \\ -1 & \cos\beta\end{pmatrix}+(-1)\cdot \det \begin{pmatrix} \cos\gamma & -1 \\ -1 & \cos\gamma \end{pmatrix} \\ & = \cos \beta \cdot \left (-\cos \beta-\cos\gamma\cdot \cos\alpha\right )-\cos \alpha \cdot \left (\cos\gamma\cdot \cos\beta-(-1)\cdot \cos\alpha\right )-\left (\cos^2 \gamma -(-1)^2\right ) \\ & =-\cos^2 \beta-\cos\gamma\cdot \cos\alpha\cdot \cos \beta-\cos\alpha\cdot \cos\gamma\cdot \cos\beta- \cos^2\alpha-\cos^2 \gamma +1 \\ & =1-2\cdot \cos\alpha\cdot \cos\beta\cdot \cos\gamma - \cos^2\alpha-\cos^2 \beta-\cos^2 \gamma \end{align*}
Do we use here the Law of cosines?
\begin{align*}&c^2=a^2+b^2-2ab\cdot \cos \gamma \Rightarrow \cos \gamma=\frac{a^2+b^2-c^2}{2ab} \\ &b^2=a^2+c^2-2ac\cdot \cos \beta \Rightarrow \cos \beta=\frac{a^2+c^2-b^2}{2ac} \\ &a^2=b^2+c^2-2bc\cdot \cos \alpha \Rightarrow \cos\alpha=\frac{b^2+c^2-a^2}{2bc}\end{align*}
Or how could we continue?
|
If a point $P$ has trilinear coordinates $[\cos B,\cos A,-1]$ it lies on the height from vertex $C$, since the circumcenter and the orthocenter are isogonal conjugates and the distances of the circumcenter from the $BC$ and $AC$ sides are exactly given by $R\cos A$ and $R\cos B$. The claim turns out to be equivalent to the collinearity of three points at infinity, or, by duality, to the concurrency of the heights of a triangle. In a simpler way,
$$ \begin{pmatrix}c & b & a\end{pmatrix}\begin{pmatrix}\cos B & \cos A & -1 \\ \cos C & -1 & \cos A \\ -1 & \cos C & \cos B\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
hence the determinant of the given matrix is trivially zero.
|
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"question_score": "2",
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|
Find the range of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$
Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$
MY Approach: By $n$th term divergence test,
$$\lim_{n\rightarrow\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}=0 \Longleftrightarrow |4x-12|<1\Longleftrightarrow\frac{11}{4}<x<\frac{13}{4},$$
not in the options.
Edit
|
Using Cauchy root test we have:$$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{1+\frac{2}{n}}}|4x-12|\frac{1}{(n^2+1)^{\frac{1}{n}}}=\frac{1}{3}|4x-12|$$then for $|4x-12|<3$ the series converges and for $|4x-12|>3$ it diverges. Also for $|4x-12|=3$ we have:$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{1}{n^{2}+1}$$or$$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{3^{n+2}}\frac{1}{n^{2}+1}$$which converges in both cases. Then the domain is $|4x-12|\le 3\to2.25\le x\le 3.75$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Let $f\left(x\right)=\begin{cases} 0 & -1\leq x\leq0\\ x^4 & 0
Question Let $f\left(x\right)=\begin{cases}
0 &\text{if } {-}1\leq x\leq0\\
x^4 &\text{if } 0<x\leq1
\end{cases}$. If $f(x) = \sum_{k=0}^{n}$$\frac{f^{(k)}\left(0\right)}{k!}x^{k}+\frac{f^{(n+1)}\left(\xi\right)}{\left(n+1\right)!}x^{n+1}$
is the Taylor's formula for $f$ about $x=0$ with maximum possible value of $n$, then the value of $\xi$ for $0 <x\leq1$ is?
I know
I don't know how to apply the theorem on this question
Any and all help will be appreciated
|
First, let us find the value of $n$. Your function $f$ is only 3 times differentiable at $x = 0$, because
\begin{align}
f'(x) &=
\begin{cases}
0 &\text{if } {-}1 \leq x \leq 0 \\
4x^3 &\text{if } 0 < x < 1
\end{cases}
&
f''(x) &=
\begin{cases}
0 &\text{if } {-}1 \leq x \leq 0 \\
12x^2 &\text{if } 0 < x < 1
\end{cases}
&
f'''(x) &=
\begin{cases}
0 &\text{if } {-}1 \leq x \leq 0 \\
24x &\text{if } 0 < x < 1
\end{cases}
\end{align}
and clearly $f'''$ is not differentiable at $x = 0$.
Therefore, $n = 2$ (it is the maximum value for which the Taylor formula in the OP is well-defined). Note that $f(0) = f'(0) = f''(0) = 0$.
So, the Taylor formula for $f$ about $x = 0$ (more precisely, for all $0 < x \leq 1$) is:
\begin{align}
f(x) = f(0) + \frac{f'(0)}{1}x + \frac{f''(0)}{2}x^2 + \frac{f'''(\xi)}{6}x^3 = 4 \,\xi\, x^3
\end{align}
where $0 < \xi < 1$. Since $f(x) = x^4$ for all $ 0 < x \leq 1$, then $$\xi = \frac{x}{4} \ .$$ Note that the condition $0 < \xi < 1$ is fulfilled by the above solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation of parabola which touches a line and coordinate axis
Equation of parabola which touches $y=x$ line at $(1,1)$ and touches $x$ axis at $(1,0)$
Try: let focus of parabola be $S(p,q)$ and equation of directrix be $y=mx+c$ and a point $P(x,y)$ on parabola.
The definition of parabola $$(x-p)^2+(y-q)^2=\displaystyle \frac{mx-y+c}{\sqrt{1+m^2}}$$
Parabola passes through $(1,0)$ and $(1,1)$
So $$(1-p)^2+q^2=\frac{m-c}{\sqrt{1+m^2}}$$
and $$(1-p)^2+(1-q)^2=\frac{m-1+c}{\sqrt{1+m^2}}$$
Could some help me how to solve it with shorter way. Thanks
|
When you render the parabola in the form
$(ax^2+bxy+cy^2)+(dx+ey)+f=0$
the quadratic terms make a completed square ($b^2-4ac=0$). Exploiting this we may use a simplified form, noting the coefficient of $x^2$ must be nonzero since the parabola does not have a horizontal symmetry axis:
$(x+ay)^2+(bx+cy)+d=0$; $x+ay$ and $bx+cy$ must be linearly independent for a true parabola.
Now a tangent point to the $x$ axis at $(1,0)$ means if we put $y=0$ there must be a double root at $x=1$. So
$x^2+bx+d=0$, $(x-1)^2=x^2-2x+1=0$
Then $b=-2, d=1$.
Now apply the tangency point at $(1,1$). There must be a double root at $x=1$ (the point of tangency) when $y=x$ (the tangent line) so
$(x+ax)^2+(bx+cx)+d=(1+a)^2x^2+(c-2)x+1=0$
$x^2-2x+1=0$
We must have $c-2=-2$, so $c=0$; and $(1+a)^2=1$ so $a\in\{0,-2\}$. We then have two candidate equations:
$x^2-2x+1=0$ but this is just the line $x=1$; fails.
$(x-2y)^2-2x+1=0$, this is really a parabola because $x-2y$ and $x$ are linearly independent; good.
|
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|
Find k for Positive Definite Quadratic Form I have two quadratic forms and I need to find $k$ (different $k$ for each possibly) that makes them positive definite. Here are the two:
*
*$Q(y)=5y_1^2+y_2^2+ky_3^2+4y_1y_2=2y_1y_3-2y_2y_3$
*$Q(y)=ky_1^2+ky_2^2+ky_3^2+2y_1y_2+2y_1y_3-2y_2y_3$
What I would like to do is ensure that each expression is always positive for our chosen range of $k$, but I am not supposed to use matrices. Only the quadratic form definition.
My attempt: I was trying to factor the equations in an attempt to get some square terms (which are always positive) and then some other terms that could determine k, but I was unable to work out the algebra.
|
For the first we need to find a value of $k$ for which
$$5a^2+b^2+kc^2+4ab-2ac-2bc\geq0$$ or
$$kc^2-2(a+b)c+5a^2+b^2+4ab\geq0$$ for which we need
$k>0$ and since it's a quadratic inequality of $c$, we need also $\Delta\leq0$, which is $$(a+b)^2-k(5a^2+b^2+4ab)\leq0$$ or
$$(5k-1)a^2+2(2k-1)ab+(k-1)b^2\geq0,$$ for which we need
$$(2k-1)^2-(5k-1)(k-1)\leq0$$ or
$$k^2-2k\geq0,$$ which gives $$k\geq2.$$
The work with the second form is the same.
I got that all $k\geq2$ they are valid.
By the way, for the second there is an easy solution:
For $k\geq2$ we obtain:
$$k(a^2+b^2+c^2)+2ab+2ac-2bc=$$
$$=(k-2)(a^2+b^2+c^2)+(a+b)^2+(a+c)^2+(b-c)^2\geq0.$$
I used the following.
For $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)=$$
$$=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right).$$ If we want that $ax^2+bx+c\geq0$ for all real $x$ then we need $a>0$ and $\Delta\leq0.$
|
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|
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sqrt{n})(\sqrt{n+\sqrt{n}}+\sqrt{n})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}+\sqrt{n}\sqrt{n+\sqrt{n}}-\sqrt{n}\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) \\
&= \lim_{n \to \infty}\left( \frac{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}-n)}{\frac{1}{\sqrt{n}}(\sqrt{n+\sqrt{n}}+\sqrt{n})} \right) \\
&= \lim_{n \to \infty}\left( \frac{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}-\frac{n}{\sqrt{n}})}{(\sqrt{\frac{n}{n}+\frac{\sqrt{n}}{n}}+ \frac{\sqrt{n}}{\sqrt{n}})} \right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{n}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\frac{\sqrt{n}\sqrt{n}}{\sqrt{n}}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{\sqrt{1}-\sqrt{n}}{2}\right) \\
&= \lim_{n \to \infty}\left( \frac{1}{2} - \frac{\sqrt{n}}{2} \right) \\
\end{align}$
Which is wrong.
Where could be my mistake?
|
The third line which is
$$\lim_{n \to \infty}\left( \frac{\sqrt{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
should be
$$\lim_{n \to \infty}\left( \frac{{n+\sqrt{n}}-n}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \right) $$
|
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|
Integration by Parts - $\int \frac{x}{(x+1)^2} dx$ I have been trying to evaluate the following integral by using integration by parts, but I contine to yield the incorrect answer.
$$\int \frac{x}{(x+1)^2} dx$$
I choose $u=x$, $dv=\frac{1}{(x+1)^2}dx => du=dx$, $v=\frac{-1}{x+1}$.
The integration by parts formula, $\int udv = uv - \int vdu$ yields
$$\int \frac{x}{(x+1)^2} dx = \frac{-x}{x+1} - \int \frac{-1}{x+1}dx = \frac{-x}{x+1}+ ln(x+1)+C$$
However, the integral should be
$$\frac{1}{x+1} + ln(x+1) + C$$
Where did I go wrong? This is my first ask on math stack exchange, so please be kind.
|
$$g'= (\frac{-x}{x+1}+ ln(x+1)+C)'= \frac {-1(x+1)+x}{(x+1)^2}+\frac 1 {x+1}=\frac x {(x+1)^2}$$
Your answer is correct..
|
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|
let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$ now find the $\text{min}(\dfrac{x}{z})=?$
Let $x,y ,z \in \mathbb{R^+}$ such that $x \geq y \geq z$ and $x^2+y^2+z^2 \geq 2xy+2xz+2yz$. Find $\text{min}\bigg(\dfrac{x}{z}\bigg)= \ ?$
I found $(x-y+z-2\sqrt{xz})(x-y+z+\sqrt{xz})\geq 0$
but now what do I do?
|
Since $$2(xy+xz+yz)-x^2-y^2-z^2)=$$
$$=(\sqrt{x}+\sqrt{y}+\sqrt{z})(\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{x}-\sqrt{y}+\sqrt{z})(\sqrt{y}+\sqrt{z}-\sqrt{x})\leq0,$$
We obtain $$\sqrt{y}+\sqrt{z}-\sqrt{x}\leq0.$$
Thus, $$\frac{x}{z}\geq\frac{(\sqrt{y}+\sqrt{z})^2}{z}\geq4.$$
The equality occurs for $y=z$, which says that we got a minimal value.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2630801",
"timestamp": "2023-03-29T00:00:00",
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|
Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. I have started by assuming that $x + \sqrt{3} = \frac{a}{b}$ and substituting $x = \frac{a}{b} - \sqrt{3}$ into $x^2 + \sqrt{3}$. It led me to $\frac{a^2}{b^2} + 3 + \sqrt{3} \cdot \frac{b - 2a}{b}$. I don't know how to continue.
|
Well, since $x^2 + \sqrt{3}$ is assumed to be rational, then your work implies that
$$b - 2a = 0.$$
Make sure that you understand why this is the case. Once you have this, $2a = b$ and
$$x = \frac 1 2 - \sqrt{3}$$
is the only solution.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How can I integrate $\sqrt{ \frac{x}{ax^3 + b}}$ analytically? How can I calculate $$\int^1_0\sqrt{\frac{x}{ax^3 + b}} \mathrm dx$$ analytically?
I searched my integral table, but I haven't found the solution.
But using WolframAlpha, I could find the analytic result like this,
$$\int \sqrt{\frac{x}{ax^3 + b}} \mathrm dx = \frac{
2 \sqrt{\frac{x}{ax^3+b}} \sqrt{ax^3+b} \log{(\sqrt{a}\sqrt{ax^3 +b} + ax^{3/2})}}{3\sqrt{a}\sqrt{x}} + C \\= \frac{
2\log{(\sqrt{a}\sqrt{ax^3 +b} + ax^{3/2})}}{3\sqrt{a}} + C$$
How can I obtain this result?
|
Hint...substitute $u^2=x^3$ and you have a standard $\operatorname{arsinh}$ integral
|
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|
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by parts implies $$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{(x^2+a^2)-a^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{1}{(x^2+a^2)^{n}}-2na^2\int\frac{dx}{(x^2+a^2)^{n+1}}$$ but I don't think I'm doing this correctly since the power of the denominator $(x^2+a^2)$ is not decreasing.
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Let $a\neq0$, $n\in\mathbb{N}$ and define, $$I_n:=\int{\frac{dx}{(x^2+a^2)^n}}$$
So, If we use integral by parts, we have $$I_n = \frac{x}{(x^2+a^2)^n}+2n\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}\qquad(*)$$
On the other hand, is clear that $$I_n=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2\int{\frac{dx}{(x^2+a^2)^{n+1}}}=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2I_{n+1}\qquad(**)$$
Now using $(*)$ and $(**)$, follows that $$I_n=\frac{x^2}{(x^2+a^2)^{n}}+2n(I_n-a^2I_{n+1})\implies I_{n+1}=\frac{1}{2a^{2}n}\left(\frac{x}{(x^2+a^2)^n}-(1-2n)I_n\right)$$
In particular, if we replace $n$ by $n-1$, we have
$$I_n=\frac{1}{2a^{2}(n-1)}\left(\frac{x}{(x^2+a^2)^{n-1}}-(3-2n)I_{n-1}\right)$$
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Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the first element of every individual term makes a sequence $1, 2, 4, 7...$. $\therefore$ the first element of each term is $(1+\frac{n(n-1)}2)^3$ and this the last element is $(\frac{n(n-1)}2 + 1 + (n-1))^3$. But I don't understand what to do next.
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HINT
Note that $$a_n= \sum_1^{T(n)} k^3-\sum_1^{T(n-1)} k^3$$
where $T(n)=\frac{n(n+1)}{2}$ are triangular numbers
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If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that $x^2+y^2+z^2+2xyz=1$
If $\sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)=\dfrac {\pi}{2}$, prove that:
$$x^2+y^2+z^2+2xyz=1$$
My Attempt:
$$\sin^{-1} (x) + \sin^{-1} (y) + \sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2}+y\sqrt {1-x^2})+\sin^{-1} (z)=\dfrac {\pi}{2}$$
$$\sin^{-1} (x\sqrt {1-y^2} + y\sqrt {1-x^2})=\dfrac {\pi}{2} - \sin^{-1} (z)$$
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I want to show a slight variation of Aretino's nice proof that does not presuppose knowledge of the identity we want to prove, but rather derives it from the given equation. We need the identity $\cos(\sin^{-1}(a)) = \sqrt{1-a^2}$ as well as
$$ \sin(a+b+c) = \sum\limits_{cyc} \Big(\sin(a)\cos(b)\cos(c)\Big)-\sin(a)\sin(b)\sin(c) $$
Thus by applying $\sin$ on both sides of $\frac \pi2 = \sin^{-1} (x) + \sin^{-1} (y)+ \sin^{-1} (z)$ we deduce
$$ 1 = \sum_{cyc} \Big( x\sqrt{1-y^2}\sqrt{1-z^2}\Big)- xyz $$
On the other hand if we solve the original equation for any of the variables, say, $x$ we obtain:
$$ x = \sqrt{1-y^2}\sqrt{1-z^2} - yz $$
Which we can plug into the former equation to eliminate the roots:
$$ 1 = \sum_{cyc} \Big( x(x+yz)\Big)- xyz = x^2+y^2+z^2 +2xyz$$
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FInd the limit without the l’Hospital’s rule FInd the limit without the l’Hospital’s rule:
$$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x}-\sqrt{x+2}}$$
I have tried multiplying the equation with:
$\frac{\sqrt{2x}+\sqrt{x+2}}{\sqrt{2x}+\sqrt{x+2}}$ and got stuck with
$\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{2x-(x+2)}$
$\implies$ $\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{x-2}$
$\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}$ and got stuck with
$\frac{3^2-(x^2+5)}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$
$\implies$ $\frac{14-x^2}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$
I tried to solve the denominator in the second one (which is the same as the nominator in the first but with different signs) which ended up with:
$\frac{14-x^2}{3\sqrt{2x}+\sqrt{2x(x^2+5)}-3\sqrt{x+2}-\sqrt{(x+2)(x^2+5)}}$
What can I do after that?
The answer with the l’Hospital’s rule (using Symbolab) $-\frac{8}{3}$
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Your method is correct, but there a sign error in the numerator of your first fraction. Doing what you were, it should have been:
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}\color{red}+\sqrt{x+2})}{2x-(x+2)}$$
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}+\sqrt{x+2})}{2x-(x+2)}\cdot\color{blue}{\frac{(3+\sqrt{x^2+5})}{(3+\sqrt{x^2+5})}}$$
You can do it fairly easily now...
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Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0$$
Now what do I do?
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The function $\phi(x):=|x+a|-|x+2|$ is $\ \equiv a-2$ when $x\gg1$ and is $\ \equiv2-a$ when $x\ll-1$. On the other hand $|x|\gg1$ implies ${\rm sgn}\bigl(b|x+5|\bigr)={\rm sgn}(b)$ when $b\ne0$, which is forbidden for an odd function. It follows that necessarily $b=0$.
The graph of $\phi$ has corners at $x=-a$ and at $x=-2$. In order to make $\phi$ odd we can choose $a=2$, which results in $f(x)=\phi(x)\equiv0$, or we can choose $a=-2$ in the hope of obtaining a nontrivial solution of the problem. This second choice results in
$$f(x)=\phi(x)=|x-2|-|x+2|=|(-x)+2|-|x+2|\ ,$$
which is obviously odd.
It follows that the possible values of $a+b$ are $\pm2$.
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Unable to prove that $n^5-n$ is a multiple of 30. It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.
Basically, the issue is how to show that :
$(n-1)(n)(n+1)(n^2+1)$ for $n$ being odd or even, is a multiple of $30$.
Let us take two cases: (a)$n$ is odd, i.e. of the form $2k+1$, for a suitable integer $k$.
So, $(2k)(2k+1)(2k+2)(2k+1)$, as $n^2+1$ will also be an odd number.
This leads to a final expression of the form: $2k+2$.
But, not able to prove its being a multiple of $30$.
(b) Need not take even case, as the final even form is proved by one term being even; & still not able to prove it being a multiple of $30$.
So, try by smaller values, and that would not serve the cause until use induction (strong case might be needed, as need consider all lower value for the case of $k=1$).
Am I correct, & should proceed by strong induction?
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You can rearrange:
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2-4+5)=n(n-1)(n+1)(n^2-4)+5n(n-1)(n+1)=$$
$$(n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1).$$
Both terms are divisible by $2,3,5$, hence by $30.$
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Find the functions satisfying $ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } $
Find all the functions $ f : \mathbb R \to\mathbb R $ such that
$$ \frac { f ( a b ) } { a b } = \frac { f ( a + b ) } { a + b } \cdot \frac { f ( a - b ) } { a - b } \text . $$
I think $ f ( x ) = x $, but I don't know how to get it.
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First of all, note that you have $ f : \mathbb R \to \mathbb R $, and in particular $ 0 $ is in the domain of $ f $. But the functional equation
$$ \frac { f ( x y ) } { x y } = \frac { f ( x + y ) } { x + y } \cdot \frac { f ( x - y ) } { x - y } \tag 0 \label 0 $$
can only hold when $ x , y \ne 0 $ and $ x \ne \pm y $. That means \eqref{0} doesn't tell you anything about $ f ( 0 ) $, and given any solution $ f $ of \eqref{0}, you can change the value of $ f ( 0 ) $, and still get a solution; i.e. any function $ \tilde f : \mathbb R \to \mathbb R $ with $ \tilde f ( x ) = f ( x ) $ for all $ x \ne 0 $ is a solution. Thus the better question is to ask for all $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{0} for all $ x , y \in \mathbb R \setminus \{ 0 \} $ with $ x \ne \pm y $. Then, defining $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ with $ g ( x ) = \frac { f ( x ) } x $, \eqref{0} gives
$$ g ( x y ) = g ( x + y ) g ( x - y ) \tag 1 \label 1 $$
for all nonzero $ x $ and $ y $ with $ x \ne \pm y $. Conversely, any $ g : \mathbb R \setminus \{ 0 \} \to \mathbb R $ satisfying \eqref{1} gives rise to an $ f : \mathbb R \setminus \{ 0 \} \to \mathbb R $ defined with $ f ( x ) = x g ( x ) $, which satisfies \eqref{0}. So we will try to solve this equivalent problem. It's straightforward to check that the constant functions $ g ( x ) = 0 $ and $ g ( x ) = 1 $ satisfy \eqref{1} (equivalently, the constant zero function and the identity function both satisfy \eqref{0}). We will show that these are the only solutions.
Setting $ y = 1 $ and then $ y = - 1 $ in \eqref{1} we get
$$ g ( x ) = g ( x + 1 ) g ( x - 1 ) = g ( x - 1 ) g ( x + 1 ) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 1 $. Similarly, substituting $ \frac x 2 $ for $ x $ in \eqref{1} and once setting $ y = 2 $ and once $ y = - 2 $, we get
$$ g ( x ) = g \left( \frac x 2 + 2 \right) g \left( \frac x 2 - 2 \right) = g \left( \frac x 2 - 2 \right) g \left( \frac x 2 + 2 \right) = g ( - x ) $$
for all nonzero $ x $ with $ x \ne \pm 4 $. Thus we have
$$ g ( - x ) = g ( x ) \tag 2 \label 2 $$
for all nonzero $ x $.
Now, suppose there is a nonzero $ a $ with $ g ( a ) = 0 $. Without loss of generality, we can assume that $ a > 0 $, because if $ a < 0 $, by \eqref{2} we have $ g ( - a ) = 0 $ while at the same time $ - a > 0 $. Letting $ x = y + a $ in \eqref{1} we get $ g \big( y ( y + a ) \big) = 0 $ for all positive $ y $ (note that since $ a > 0 $, thus $ 0 \ne y + a \ne \pm y $). This means that we have $ g ( x ) = 0 $ for all positive $ x $, as when $ x > 0 $ there is a positive $ y $, namely $ y = \frac { - a + \sqrt { a ^ 2 + 4 x } } 2 $, such that $ y ( y + a ) = x $. Using \eqref{2}, this can be generalized for negative $ x $ as well, which means that $ g $ is the constant zero function, one of the mentioned solutions.
Finally, consider the other case where for all nonzero $ x $, $ g ( x ) $ is not equal to $ 0 $. Suppose that $ y $ is negative. Then we have $ 0 \ne \frac y { y - 1 } \ne \pm y $, and we can let $ x = \frac y { y - 1 } $ in \eqref{1}. In that case, we have $ x y = x + y = \frac { y ^ 2 } { y - 1 } $, and by the fact $ g \left( \frac { y ^ 2 } { y - 1 } \right) $ is not equal to $ 0 $, we find out that $ g ( x - y ) = g \left( \frac { y ^ 2 - 2 y } { y - 1 } \right) = 1 $. This means that we have $ g ( x ) = 1 $ for all negative $ x $, since for $ x < 0 $ there is a negative $ y $, namely $ y = \frac { x + 2 - \sqrt { x ^ 2 + 4 } } 2 $, such that $ \frac { y ^ 2 - 2 y } { y - 1 } = x $. By \eqref{2}, that is also true for positive $ x $, and $ g $ is the constant one function, the other mentioned solution.
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$T(n) = T(n-1) + T(n-2) + n \in \Theta(2^n)$ I've tried enclosing $T(n)$ between two functions that represent the lower bound and upper bound, respectively, and prove that both of them are in $\Theta(2^n)$
I've noticed a good candidate for the upper bound would be $S(n) = S(n-1) + S(n-1) + n = 2S(n-1) + n = 4S(n-2) + 2(n-1) + n = 2^{n-1} + 2^{n-2} \cdot 2 + 2^{n-3} \cdot 3 + ... + 2^2(n-2) + 2(n-1) + n$
$2S(n) = 2^{n} + 2^{n-1} \cdot 2 + 2^{n-2} \cdot 3 + ... + 2^3(n-2) + 2^2(n-1) + 2n$
$2S(n) - S(n) = 2^n + 2^{n-1} + 2^{n-2} + 2^{n-3} + ... + 2^3 + 2^2 + 2 + n$ (and we multiply by 2 again)
$2S(n) = 2^{n+1} + 2^n + 2^{n-1} + ... + 2^4 + 2^3 + 2^2 + 2n$
$2S(n) - S(n) = 2^{n+1} + 2 + n \in \Theta(2^{n+1})$
As for the lower bound part, I tried choosing $I(n) = 2I(n-2) + n$ but haven't reached any valid result since it decreases exponentially faster. Any tip would be appreciated.
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Note that $$T(n)+n+3=T(n-1)+n+2+T(n-2)+n+1$$ hence $S(n)=T(n)+n+3$ solves the recursion $$S(n)=S(n-1)+S(n-2)$$ The characteristic equation reads $$x^2=x+1$$ hence, except for specific initial conditions, $S(n)=\Theta(\lambda^n)$ with $$\lambda=\frac{1+\sqrt5}2\approx1.618$$ Since $\lambda>1$, one gets $$T(n)\in\Theta(\lambda^n)$$ in particular, $$T(n)\notin\Theta(2^n)$$
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I need to find the general nth degree polynomial expansion of $x(x+1)(x+2)\dots(x+n)$ in order to solve an integral. Expanding $x(x+1)(x+2)\dots(x+n)$ has been challenging for me.
Say $n=5$, you will notice many patterns like $x^n+(n-1)x^{n-1}+\dotsb$ et cetera, but does there exist a general polynomial expansion for any $n$? This has been driving me nuts. I need a whiteboard.
The integral is $$\frac{1}{x(x+1)(x+2)\dots(x+n)}$$ and will be easy once I expand the polynomial. I am trying to avoid partial fraction decomposition.
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Let us find the partial fraction decomposition of the rational function
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)}, \quad n = 0,1,2,\ldots$$
using the Heaviside cover-up method.
As we have $n + 1$ distinct linear factors we can write the rational function as
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \frac{A_0}{x} + \frac{A_1}{x + 1} + \frac{A_2}{x + 2} + \cdots + \frac{A_n}{x + n}.$$
To find each of the unknown constants $A_0$ through to $A_n$ we "cover-up" the first linear factor in the denominator on the left hand side then substitute in the value of $x$ that makes the factor covered up equal to zero which then gives the value for $A_0$. In the first case $x = 0$. We then repeat this process for each linear factor in the denominator. Doing so we find
\begin{align*}
A_0 &= \frac{1}{1.2.\ldots n} = \frac{1}{0! n!}\\
A_1 &= \frac{1}{(-1).1.2\ldots (n - 1)} = -\frac{1}{1!(n - 1)!}\\
A_2 &= \frac{1}{(-2)(-1).1.2\ldots (n - 2)} = \frac{1}{2!(n - 2)!}\\
& \vdots\\
A_n &= \frac{1}{(-1)(-n + 1)(-n + 2) \cdots (-1)} = \frac{(-1)^n}{n! 0!}
\end{align*}
So we see the $k$th constant $A_k$ where $0 \leqslant k \leqslant n$ will be given by
$$A_k = \frac{(-1)^k}{(n - k)! k!}, \quad 0 \leqslant k \leqslant n.$$
The partial fraction decomposition for the rational function will therefore be
$$\frac{1}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \frac{1}{x + k}.$$
Now the integral you are after can be readily found. Here we have
$$\int \frac{dx}{x(x + 1)(x + 2) \cdots (x + n)} = \sum_{k = 0}^n \frac{(-1)^k}{(n - k)! k!} \ln |x + k| + C.$$
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Complex integral $\int^{2\pi}_0 e^{i\tan^{-1}\frac{r\sin(t-t_0)}{1+r\cos(t-t_0)}-i\tan^{-1}\frac{1-r \cos( t-t_0)}{r\sin(t-t_0)}}dt$ Is there a way to evaluate the following integral by simplifying the exponent?
$$\int^{2\pi}_0 \exp \left(i\tan^{-1}\frac{r\sin(\phi-\phi_0)}{1+r\cos(\phi-\phi_0)}-i\tan^{-1}\frac{1-r \cos( \phi-\phi_0)}{r\sin(\phi-\phi_0)}\right) \, d\phi$$
where $r$ and $\phi_0$ are constants.
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$$
\arctan u + \arctan v = \arctan \frac{u+v}{1-uv} \tag 1
$$
\begin{align}
& \arctan \frac A {1+B} - \arctan\frac{1-B} A = \arctan \frac{\frac A{1+B} - \frac{1-B} A}{1 + \frac{A(1-B)}{A(1+B)}} \\[10pt]
= {} & \arctan\frac{A^2 - (1-B^2)}{A(1+B) + A(1-B)} = \arctan \frac{A^2+B^2 - 1}{2A}
\end{align}
In your case $A^2+B^2 = r^2$ and $2A = 2r\sin(\varphi - \varphi_\xi).$
But you need to think through the question of when $(1)$ holds, since "arctan" is multiple-valued unless the domain of the tangent function is restricted.
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$\tan(\theta) = \frac{a}{b}$ s.t. $\gcd(a,b) = 1$, $a+b$ is odd. Show $\theta$ can be trisected iff $a^2 + b^2$ is a perfect cube I've been stuck on this problem for quite some time.
Let tan$(\theta) = \dfrac{a}{b}$ s.t. $a,b \in \mathbb{Z}$, $b \neq 0$, $\gcd(a,b) = 1,$ and $a+b$ is odd. Show $\theta$ can be trisected iff $a^2 + b^2$ is a perfect cube.
What I've come up with so far is:
Let $\phi = \dfrac{\theta}{3}$, then $$\tan(\theta)=\frac{3\tan(\phi)-\tan^3(\phi)}{1-3\tan^2(\phi)}.$$ Thus $\tan(\phi)$ is a zero of $$f(t) = t^3 -3t^2\tan(\theta)-3t+\tan(\theta)$$ and thus a zero of
$$g(t) = bt^3-3t^2a-3bt+a$$
I know that if I can show that $g$ is reducible iff $a^2 + b^2$ is a perfect cube, then I'm done, but I'm unsure of how to show this.
The book gave a hint of considering that $\mathbb{Z}[i]$ is a UFD, which I don't know how to use in relation to this problem.
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Since $R=\mathbb Z[i]$ is a UFD, you can factorize $a+bi$ as a product of primes raised to certain exponents. Classifying exponents according to their value modulo $3$ (which can be $0,1$ or $2$), you can write $a+bi=x^3yz^2$ with $x,y,z\in R$ and $y$ and $z$ are coprime in $R$.
Suppose that the norm $N(a+bi)=a^2+b^2$ is a perfect cube, say $a^2+b^2=w^3$ with $w\in {\mathbb N}$. We then have the equality (in $\mathbb N$) $|x|^3 |y| |z|^2=w^3$, and $|y|$
and $|z|$ are coprime. This is possible only if $|y|=|z|=1$, so $y$ and $z$ are units in $R$, but then $y,z$ are in $\lbrace \pm 1,\pm i \rbrace$ whence $a+bi=(xy^3z^6)^2$, so that $a+bi$ is a perfect cube in $R$. The implication in the other side being trivial, we have :
$$ a^2+b^2 \text{ is a perfect cube in } {\mathbb Z} \Leftrightarrow
a+bi \text{ is a perfect cube in } R \tag{1} $$.
Then, $\theta$ is trisectable iff $\frac{a}{b}=\frac{t^3-3t}{3t^2-1}$ for some $t\in{\mathbb Q}$, iff $\frac{a}{b}=\frac{c^3-3cd^2}{3dc^2-d^3}$ for some coprime $c,d\in{\mathbb Z}$ with $d\neq 0$ and $3c^2-d^2\neq 0$ (put $t=\frac{c}{d}$). Let $C=c^3-3cd^2$ and $D=3dc^2-d^3$. I claim that
$$ \text{ if } p>2 \text{ is a prime in } {\mathbb Z}, \text{ it cannot divide both }
C \text { and } D \tag{2}$$
For suppose that this were the case, and that $p$ divides both $C=c(c^2-3d^2)$ and $D=d(3c^2-d^2)$. If $p$ divides $c$, then $p$ cannot divide $d$ and so must divide $3c^2-d^2$, and hence divides $d^2$, contradiction. So $p$ cannot divide $c$, and similarly $p$ cannot divide $d$. So $p$ divides $u=c^2-3d^2$ and $v=3c^2-d^2$, and hence divides $u+3v=10c^2$, so that $p$ divides $10$. So $p=5$. But then $-3 \equiv \big(\frac{c}{d}\big)^2 [{\sf{mod}}\ 5]$ is a square modulo $5$ which is absurd. Next, we can show that
$$ \text{ if } p=2,\ p \text{ cannot divide both }
C \text { and } D \tag{3}$$
For suppose that this were the case. Then $c$ and $d$ are both odd, and it follows that $C$ and $D$ are both congruent to $2$ modulo $4$, and hence $\frac{C}{2}$ and $\frac{D}{2}$ are coprime by (2). Then $\frac{a}{b}=\frac{C/2}{D/2}$ forces $(a,b)=(\varepsilon\frac{C}{2},\varepsilon\frac{D}{2})$ for some $\varepsilon=\pm 1$, contradicting the hypothesis that $a+b$ is odd.
From (2) and (3) we deduce that $C$ and $D$ are coprime. Then $\frac{a}{b}=\frac{C}{D}$ forces $(a,b)=(\varepsilon C,\varepsilon D)$ for some $\varepsilon=\pm 1$. Replacing $(c,d)$ with $(-c,-d)$, we may assume without loss that $\varepsilon= 1$.
So $\theta$ is trisectable iff $a=c^3-3cd^2, b=3dc^2-d^3$ for some some coprime $c,d\in{\mathbb Z}$ with $d\neq 0$ and $3c^2-d^2\neq 0$. But this is exactly equivalent to
$a+bi=(c+di)^3$, and we are then done by (1).
|
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|
Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $
For positive real $a, b$, $$ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right).$$
I know the inequality $ (a+b)^2 \le 2(a^2 +b^2 ) $, but this is not useful for me. Then I think the function $$f(x) = \frac{ x^2 } {\sqrt{x^2 +1 }},$$
and the inequality is $$f(a+b) <2 ( f(a) + f(b)).$$
But I meet a wall and I don't know if this inequality is true. I want some hints. Thank you.
|
The inequality you know gives you
$$\frac{(a + b)^2}{\sqrt{(a + b)^2 + 1}} \le 2\left(\frac {a^2}{\sqrt{(a + b)^2 + 1}} + \frac{b^2}{\sqrt{(a + b)^2 + 1}}\right).$$
Now just remember that $(a + b)^2 > a^2$ as well, so
$$\frac{1}{\sqrt{(a + b)^2 + 1}} < \frac{1}{\sqrt{a^2 + 1}}$$
and so on.
|
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|
Suppose $a,b,c \neq 0$. Show that $\det(1+a,1,1),(1,1+b,1),(1,1,1+c) = abc\left(1 + \frac1a + \frac1b + \frac1c\right)$ Suppose $a,b,c \neq 0$ .Show that the \begin{eqnarray*} \det \begin{bmatrix} 1+a &1 &1 \\1 &1+b & 1 \\ 1 & 1 & 1+c \end{bmatrix} = abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right). \end{eqnarray*}
I've use cofactor expansion and reduced to get the form
$bc + ac + ab + abc = det$. Choosing a variety of number to plug in for $a,b,$ and $c$, I know I'm on the right track. How can I reduce to get to $$bc + ac + ab + abc = abc\left(1 + \frac1a + \frac1b + \frac1c\right)$$
|
$$abc(1 + {1\over a} + {1\over b} + {1\over c})= abc \cdot 1 + abc \cdot {1\over a}+ abc \cdot {1\over b}+ abc \cdot {1\over c}= abc + bc + ac + ab$$
|
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|
$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar{a}\cdot d) \cdot e$, then $(G, \cdot)$ is a group
Let $G$ be a non-empty set, $"\cdot"$ be a binary operation on $G$ and a single operation $x \mapsto \bar{x}$ such that $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar{a}\cdot d) \cdot e, \forall a,b,c,d,e \in G.$$
Prove that $(G, \cdot)$ is a group.
For $b = c = d = e = a \implies a = (\bar{a} \cdot a ) \cdot a, \forall a \in G$.
For $d = a, e = b \implies b = (\bar{a} \cdot a) \cdot b, \forall a,b \in G$.
Swapping $a$ and $b$ in the hypothesis equation and making $d = b, e = a$ we obtain that $b = (\bar{a} \cdot a) \cdot b$ and $a = (\bar{b} \cdot b) \cdot a$.
I think I need to show that $\bar{\bar{a}} = a, \forall a \in G$ and that $a \cdot \bar{a} = \bar{a} \cdot a = e$ (the identity element), but I don't know how. It would also help If I could prove that the binary operation is associative.
|
For all $a,b,c,d,e \in G$, $$a\cdot(b \cdot c) = d \cdot (e \cdot c) \implies b = (\bar a\cdot d) \cdot e. \tag 1$$
Lemma 1: For all $x \in G$, we have $\bar x \cdot x$ is a left identity.
Proof: If we take $a=c=d=x$ and $b=e=y$ in $(1)$ we obtain,
$$x\cdot(y \cdot x) = x \cdot (y \cdot x) \implies y = (\bar x\cdot x) \cdot y,$$
and $x\cdot(y \cdot x) = x \cdot (y \cdot x)$ is true for all $x,y \in G$, so $y = (\bar x\cdot x) \cdot y$ for all $x,y \in G$. $\qquad \blacksquare$
Lemma 2: For all $x,y,z \in G$, if $x \cdot z = y \cdot z$ then $x=y$, i.e. the operation $\cdot$ is right cancellative.
Proof: If we take $a=b=d=x$, $c=z$ and $e=y$ in $(1)$ we obtain,
$$x\cdot(x \cdot z) = x \cdot (y \cdot z) \implies x = (\bar x\cdot x) \cdot y,$$
and if $x \cdot z = y \cdot z$ then $x\cdot(x \cdot z) = x \cdot (y \cdot z)$, so $x = (\bar x\cdot x) \cdot y = y$. $\qquad \blacksquare$
Lemma 3: For all $x,y \in G$, we have $\bar x \cdot x = \bar y \cdot y$.
Proof: From Lemma 1, we know $\bar x \cdot x$ and $\bar y \cdot y$ are both left identities, so $(\bar x \cdot x) \cdot x = (\bar y \cdot y) \cdot x$, so from Lemma 2 it follows that $\bar x \cdot x = \bar y \cdot y$. $\qquad \blacksquare$
Defintion: Let $1$ denote the element of $G$ that is equal to $\bar x \cdot x$ for all $x \in G$.
Lemma 4: $\bar 1 = 1$.
Proof: By the definition of $1$, since $1 \in G$ we have $1 = \bar 1 \cdot 1$, and from Lemma 1 we know that $1$ is a left identity so $1 = 1 \cdot 1$. Hence $\bar 1 \cdot 1 = 1 \cdot 1$ so, from Lemma 2, we know $\bar 1 = 1$. $\qquad \blacksquare$
Lemma 5: $1$ is a right identity.
Proof: If we take $a=e=1$ and $b=c=d=x$ in $(1)$ we obtain,
$$1\cdot(x \cdot x) = x \cdot (1 \cdot x) \implies x = (\bar 1\cdot x) \cdot 1,$$
and $1\cdot(x \cdot x) = x \cdot (1 \cdot x)$ is true for all $x \in G$ since $1$ is a left identity, so $$x = (\bar 1\cdot x) \cdot 1 = (1\cdot x) \cdot 1 = x \cdot 1 \mbox{ for all } x \in G. \qquad \blacksquare$$
Lemma 6: For all $a,b,d,e \in G$,
$$a\cdot b = d \cdot e \implies b = (\bar a\cdot d) \cdot e. \tag 2$$
Proof: If we take $c=1$ in $(1)$ we obtain,
$$a\cdot(b \cdot 1) = d \cdot (e \cdot 1) \implies b = (\bar a\cdot d) \cdot e,$$
and since $1$ is a right identity
$$a\cdot b = d \cdot e \implies a\cdot(b \cdot 1) = d \cdot (e \cdot 1)$$
so the result follows. $\qquad \blacksquare$
Lemma 7: For all $x,y \in G$, we have $\bar x \cdot (x \cdot y) = y$.
Proof: If we take $a=x$, $b=y$, $d=x \cdot y$ and $e = 1$ in $(2)$ we obtain,
$$x\cdot y = (x \cdot y) \cdot 1 \implies y = (\bar x\cdot (x \cdot y)) \cdot 1,$$
and $x\cdot y = (x \cdot y) \cdot 1$ since $1$ is a right identity, so
$$y = (\bar x\cdot (x \cdot y)) \cdot 1 = \bar x\cdot (x \cdot y). \qquad \blacksquare$$
Lemma 8: For all $x \in G$, we have $\bar{\bar x} = x$.
Proof: If we take $a=\bar x$, $b=x$ and $d=e=1$ in $(2)$ we obtain,
$$\bar x \cdot x = 1 \cdot 1 \implies x = (\bar{\bar x}\cdot 1) \cdot 1,$$
and $\bar x \cdot x = 1 = 1 \cdot 1$ for all $x \in G$, so for all $x \in G$, $$x = (\bar{\bar x}\cdot 1) \cdot 1 = \bar{\bar x}. \qquad \blacksquare$$
Lemma 9: The operation $\cdot$ is associative.
If we take $a = \bar x$, $b = x \cdot (y \cdot z)$, $d = y$ and $e = z$ in $(2)$ we obtain
$$\bar x\cdot (x \cdot (y \cdot z)) = y \cdot z \implies x \cdot (y \cdot z) = (\bar{\bar x}\cdot y) \cdot z,$$
and $\bar x\cdot (x \cdot (y \cdot z)) = y \cdot z$ is true for all $x,y,z\in G$ by Lemma 7, so for all $x,y,z\in G$ we have
$$x \cdot (y \cdot z) = (\bar{\bar x}\cdot y) \cdot z = (x\cdot y) \cdot z,$$
using Lemma 8. $\qquad \blacksquare$
|
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|
How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}$$
$$\bigg(\frac{x}{8\sqrt{x^2+8}}\bigg)'=\frac{1}{8}\cdot\frac{\sqrt{x^2+8}-\frac{x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{\frac{x^2+8-x^2}{\sqrt{x^2+8}}}{x^2+8}=\frac{1}{8}\cdot\frac{8}{\sqrt{(x^2+8)^3}}=\frac{1}{\sqrt{(x^2+8)^3}}$$
It there a way to extract the solution from these types of integrals which argument is born from the easy quotient rule?
|
substitute $$x=2\sqrt{2}\tan(u)$$ then we get
$$dx=2\sqrt{2}\sec^2(u)du$$
|
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|
If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is? If $2\tan^2x - 5\sec x = 1$ has exactly $7$ distinct solutions for $x\in[0,\frac{n\pi}{2}]$, $n\in N$, then the greatest value of $n$ is?
My attempt:
Solving the above quadratic equation, we get $\cos x = \frac{1}{3}$
The general solution of the equation is given by $\cos x = 2n\pi \pm \cos^{-1}\frac{1}{3}$
For having $7$ distinct solutions, $n$ can have value = 0,1,2,3
So, from here we can conclude that $n$ is anything but greater than $6$. So, according to the options given in the questions, the greatest value of $n$ should be $13$. But the answer given is $14$. Can anyone justify?
|
By replacing $\tan^2 x=\sec^2x-1$ we have$$4\sec^2x-5\sec x=5$$which yields to $$\sec x=\dfrac{5\pm\sqrt{25+80}}{8}=\dfrac{5\pm\sqrt{105}}{8}$$where only $\sec x=\dfrac{5+\sqrt{105}}{8}$ is acceptable. Also the equation $\sec x=p>1$ has exactly two roots in $[0,2\pi)$ so for having 7 distinct roots we must be in the interval $[0,3\times 2\pi+\pi=14\dfrac{\pi}{2})$ which yields to $n<14$ or $n=13$. An illusion is as following
|
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|
Summation of primes
Let $P(n)=\displaystyle\sum^{n}_{i=1}p_i$, where $p_i=i^{th}$ prime $\geq 2$. Does exist $k\in \mathbb{N}$, such that, $P(k)$ and $P(k+1)$, both are squares?
Please provide any clue.
|
No, there is not such $k$.
If such integer $k$ exists then there are positive integers $a$ and $b$ with $a>b$ such that
$$p_{k+1}=P(k+1)-P(k)=a^2-b^2=(a+b)(a-b)\implies a-b=1,\; a+b=p_{k+1}.$$
Hence $b=(p_{k+1}-1)/2$ and
$$\sum_{i=1}^{k}p_i=P(k)=b^2=\left(\frac{p_{k+1}-1}{2}\right)^2.$$
Now we show that the above equality does not hold for any positive integer $k$.
For $n=1,2,3$, we easily verify that
$$\sum_{i=1}^{n}p_i>\left(\frac{p_{n+1}-1}{2}\right)^2.$$
Moreover, we show by induction that for $n\geq 4$,
$$\sum_{i=1}^{n}p_i<\left(\frac{p_{n+1}-1}{2}\right)^2.$$
For $n=4$ it holds: $2+3+5+7=17<25=\left(\frac{11-1}{2}\right)^2$.
Inductive step: we have that
$$\sum_{i=1}^{n+1}p_i=\sum_{i=1}^{n}p_i+p_{n+1}<\left(\frac{p_{n+1}-1}{2}\right)^2+p_{n+1}=\left(\frac{p_{n+1}+1}{2}\right)^2\leq \left(\frac{p_{n+2}-1}{2}\right)^2$$
where the first (strict) inequality is the inductive hypothesis and the last inequality holds because $p_{n+2}-p_{n+1}\geq 2$.
Hence
$$\sum_{i=1}^{n+1}p_i< \left(\frac{p_{n+2}-1}{2}\right)^2$$
and we are done.
|
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|
Prove polynomial identity $x^k-1=(x-1)(x^{k-1}+x^{k-2}+\ldots+1).$ I am confused, as not clear except by multiplying both terms on the r.h.s, and showing that all cancel out except the two on the l.h.s., as below:
$(x)(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)= x^k -1$
|
It's like this: $(x-1)(x^{k-1}+x^{k-2}+\ldots+1)$
$=x(x^{k-1}+x^{k-2}+\ldots+1) - (x^{k-1}+x^{k-2}+\ldots+1)$
$=x^{k-1+1}+x^{k-2+1}+x^{k-3+1}+x^{k-4+1}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x-1$
$=x^{k}+x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x-x^{k-1}-x^{k-2}-x^{k-3}-...-x^2-x-1$
$=x^{k}+(x^{k-1}+x^{k-2}+x^{k-3}...+x^3+x^2+x)-(x^{k-1}+x^{k-2}+x^{k-3}+...+x^2+x)-1$
$=x^k-1$
You can also prove it by reversing the solution above.
|
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|
Find the kernel of a 4x4 matrix $$
\begin{pmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
$$
and for $x$ I find $x = \alpha + 2\beta$, whereas $y = -2\alpha -3\beta$
Therefore,
$$
\begin{pmatrix}
\alpha + 2\beta\\
-2\alpha - 3\beta\\
\alpha\\
\beta\\
\end{pmatrix}
$$
When we take outside alpha and beta:
we get two vectors:
$$
\begin{pmatrix}
1\\
-2\\
1\\
0\\
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
2\\
-3\\
0\\
1\\
\end{pmatrix}
$$
which are linearly independent and form a basis of this $ker(M)$
Could you please confirm with me whether you get the same result? Thank you.
|
Yes it is correct, in case of doubt you can check it directly by simple multiplication for RREF matrix
$$\begin{bmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}\begin{bmatrix}
1\\
-2\\
1\\
0\\
\end{bmatrix}=0$$
$$\begin{bmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}\begin{bmatrix}
2\\
-3\\
0\\
1\\
\end{bmatrix}
=0
$$
or/and also for the original matrix
$$\begin{bmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{bmatrix}
\begin{bmatrix}
1\\
-2\\
1\\
0\\
\end{bmatrix}=0$$
$$\begin{bmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{bmatrix}
\begin{bmatrix}
2\\
-3\\
0\\
1\\
\end{bmatrix}
=0
$$
Note that also after this check, for the correctness of the result it is crucial that RREF is calculated properly. In case of doubt on it we can evaluate $rank(M)$ with others methods or follow the nice suggestion give by Atmos to be sure that $dim(Ker M)=2$ and thus that what we have found is a basis for $Ker(M)$.
|
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|
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2k+1) + 7 \\
\implies & (2k+1) = (4k^2+4k+1) - 4k-2+7-1 \\
\implies & 2k+1 = 4k^2+1-2+7-1 \\
\implies & 2k = 4k^2 + 4 \\
\implies & 2(2k^2-k+2)
\end{align}
Now, this is even but I wanted to prove that this is odd(the contradiction). Can you some help me figure out my mistake?
Thank you.
|
HINT
Note that
*
*$n^2-2n+7=n^2-2n+1+6=(n-1)^2+6$ is even $\iff$ $n-1$ is even
and
*
*$n+1=(n-1)+2$
|
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|
Binomial Coefficient deck of cards probability question A regular deck of 52 playing cards has 13 ranks in 4 suits. The ranks of Jack, Queen, King and Ace of each suit are top cards. Suppose you are randomly dealt seven cards. What is the probability of getting three top cards in the same suit and any four cards in another suit (but all
four in one suit)?
This is what I was thinking of calculating but am completely unsure:
$$
{ \begin{pmatrix}
4 \\
3 \\
\end{pmatrix}
× \begin{pmatrix}
39 \\
4 \\
\end{pmatrix} × \begin{pmatrix}
9 \\
4 \\
\end{pmatrix} \over
\begin{pmatrix}
52 \\
7 \\
\end{pmatrix} } = 0.30986...$$
Any help appreciated. Thanks.
|
There are $\binom{4}{1}$ ways to choose the suit from which the three top cards are drawn, $\binom{4}{3}$ ways to choose three top cards of that suit, $\binom{3}{1}$ ways to choose a suit from which four cards are drawn, and $\binom{13}{4}$ ways to choose four cards of that suit. Hence, the number of favorable cases is
$$\binom{4}{1}\binom{4}{3}\binom{3}{1}\binom{13}{4}$$
Since there are $\binom{52}{7}$ ways to select seven cards from the deck, the desired probability is
$$\frac{\dbinom{4}{1}\dbinom{4}{3}\dbinom{3}{1}\dbinom{13}{4}}{\dbinom{52}{7}}$$
|
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|
Evaluation of $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration I don't succed to evaluate this sum $\sum_{n=1}^{+\infty}\frac{n}{n^4+4}$ using integration because using fraction method is too hard for me and i have got that is equal $\frac 3 8$ , Is there any way to use integral for Evaluation ?
|
It's telescopic. By Sophie Germain's identity $$n^4+4 = (n^2+2)^2-(2n)^2 = (n^2-2n+2)(n^2+2n+2),$$
so
$$ \frac{1}{n^2-2n+2}-\frac{1}{n^2+2n+2} = \frac{1}{f(n)}-\frac{1}{f(n+2)} = \frac{4n}{n^4+4} $$
and
$$ \sum_{n\geq 1}\frac{n}{n^4+4}=\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{f(n)}-\frac{1}{f(n+2)}\right) = \frac{1}{4}\left(\frac{1}{f(1)}+\frac{1}{f(2)}\right)=\frac{3}{8}. $$
Through the inverse Laplace transform:
$$\begin{eqnarray*} \sum_{n\geq 1}\frac{n}{n^4+4}&=&\lim_{\varepsilon\to 0^+}\frac{1}{2}\sum_{n\geq 1}\int_{0}^{+\infty}\sin(s)\sinh(s)e^{-(n+\varepsilon)s}\,ds\\&=&\frac{1}{4}\lim_{\varepsilon\to 0^+}\int_{0}^{+\infty}(1+e^{-s})\sin(s)e^{-\varepsilon s}\,ds\\&=&\frac{1}{4}\left(1+\frac{1}{2}\right)\end{eqnarray*}$$
we reach the same conclusion.
|
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|
Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please give me a hint?
|
From the original equation, we get:
$$abc=(a+b+c)(ab+bc+ca)$$
which is equivalent to
$$a^2(b+c)+bc(b+c)+ab(b+c)+ca(b+c)=0$$
$$\implies(b+c)(a+c)(a+b)=0$$
Then, obviously any one of the following must hold:
$$a=-b\\b=-c\\c=-a$$
In any case we can prove the equation $$\frac{1}{a^n+b^n+c^n}=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}$$
with odd $n$.
Since if we take $a=-b$
we get
$$\frac{1}{(-b)^n+b^n+c^n}=\frac{1}{(-b)^n}+\frac{1}{b^n}+\frac{1}{c^n}$$
which is equivalent to
$$\frac{1}{c^n}=\frac{1}{c^n}$$
and this is true.....
|
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|
find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$ $A=\begin{bmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{bmatrix}$
here after row reduction
$\begin{bmatrix}1&0&2&1\\0&1&-1&2\\0&0&1&2\\0&0&0&0\end{bmatrix}$
clearly determinant is zero
but how can I find vector $v$ , $v \in R^4$ that satisfy $Av\neq0$ and $A^2v=0$
$A^2=\begin{bmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{bmatrix}$
so I try to row reduce $A^2$ using gauss elimination
$A^2=\begin{bmatrix}1&0&\frac{3}{2}&0\\0&1&\frac{-5}{3}&\frac{2}{3}\\0&0&0&0\\0&0&0&0\end{bmatrix}$
so $x_3$ and $x_4$ is free variable, but in this question I'm not sure what I need to find and should I find Eigen value first?
|
No need for row reduction.
From $$A^2=\begin{pmatrix}6&3&4&2\\0&0&0&0\\4&0&6&0\\2&0&3&0\end{pmatrix}$$ you can immediately see that $A^2 e_2 = 3e_1$ and $A^2e_4 = 2e_1$.
Therefore $$A^2e_2 = \frac32 A^2 e_4 \implies 0 = A^2\left(e_2 - \frac32 e_4\right) = A^2 \begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix}$$
so $\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} \in \ker A^2$. On the other hand, we have that $$A \begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} =\begin{pmatrix}1&0&2&1\\0&-1&1&-2\\2&1&1&0\\1&1&0&1\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} = \begin{pmatrix}-\frac32 \\ 2 \\ 1 \\ -\frac12\end{pmatrix} \ne 0$$
so $\begin{pmatrix}0 \\ 1 \\ 0 \\ -\frac32\end{pmatrix} \notin \ker A$.
|
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|
How many $k$ satisfy the equation $(p \cdot k)^2 \equiv 0 \pmod{p^n}$ where $k < p^n$ and $p$ is prime How many $k$ satisfy the equation $(p \cdot k)^2 \equiv 0 \pmod{p^n}$ where $k < p^n$ and $p$ is prime?
I attach very simple code:
#include <stdio.h>
#include <math.h>
unsigned long long int ipow( unsigned long long int base, unsigned long long int exp )
{
unsigned long long int result = 1ULL;
while( exp )
{
if ( exp & 1 )
{
result *= (unsigned long long int)base;
}
exp >>= 1;
base *= base;
}
return result;
}
int main()
{
long long int p = 3;
long int j;
for(j=2; j < 25; j++)
{
long long int limit = ipow(p, j);
long long int count = 0;
long long int i;
for(i=0; i < limit; i+=p)
{
if( (i*i) % limit == 0 )
count++;
}
printf("n=%ld count=%lld\n", j, count);
}
return 0;
}
For $p=3$ it return:
n=2 count=3
n=3 count=3
n=4 count=9
n=5 count=9
n=6 count=27
n=7 count=27
n=8 count=81
n=9 count=81
n=10 count=243
n=11 count=243
n=12 count=729
n=13 count=729
n=14 count=2187
n=15 count=2187
n=16 count=6561
n=17 count=6561
n=18 count=19683
n=19 count=19683
n=20 count=51432
n=21 count=17144
n=22 count=17146
I am interested in the case for very large $n$ (and $p$).
|
I believe your program is wrong, for example for $p=3$, $n=3$ and $k<3^3$
$$3^3 \mid (3 \cdot 3 \cdot 1)^2$$
$$3^3 \mid (3 \cdot 3 \cdot 2)^2$$
$$...$$
$$3^3 \mid (3 \cdot 3 \cdot 8)^2$$
where $k=3\cdot 8 < 3^3$, so there are $8$ cases, not $3$.
Generally (using Euclid lemma)
$$p^n \mid p^2 k^2 \Rightarrow p^{n-2} \mid k^2$$
from $p \mid k^2 \Rightarrow p^2 \mid k^2$ we have
*
*if $n-2$ - even then $p^{n-2}\mid k^2$
*if $n-2$ - odd then $p^{n-2+1}\mid k^2$
which is equivalent to $p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor }\mid k^2$. As a result
$$p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \leq k_1^2=p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \cdot 1^2 < p^{2n}$$
$$p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \leq k_2^2=p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \cdot 2^2 < p^{2n}$$
$$...$$
$$p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \leq k_q^2=p^{2 \left \lfloor \frac{n-2+1}{2} \right \rfloor } \cdot q^2 < p^{2n}$$
or
$$1 \leq 1^2 < p^{2n - 2 \left \lfloor \frac{n-2+1}{2} \right \rfloor }$$
$$1 \leq 2^2 < p^{2n - 2 \left \lfloor \frac{n-2+1}{2} \right \rfloor }$$
$$...$$
$$1 \leq q^2 < p^{2n - 2 \left \lfloor \frac{n-2+1}{2} \right \rfloor }$$
and there are
$$p^{n - \left \lfloor \frac{n-2+1}{2} \right \rfloor }-1$$ such $k$'s.
|
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|
Solutions to diophantinte equation $x^4+2y^4=z^2$ It is well known that the equation $x^4+y^4=z^2$ has no non-trivial solutions. The same holds also for the equation $x^4+2y^4=z^4$. What about the equation
$x^4+2y^4=z^2$?
|
The classic book Diophantine Analysis by R. D. Carmichael says on page 17 that the equation $x^4+2y^4=z^2$ has no non-trivial solutions. Here is the original text, slightly edited:
*The equation $x^4+2y^4=z^2$ is impossible in integers $x$, $y$, $z$, all of which
are different from zero.
Suggestion. This may be proved by the method of infinite descent. Begin by writing $z$ in the form
$
z = x^2 + \frac{2py^2}{q},
$
where $p$ and $q$ are relatively prime integers, and thence show that $x^2=q^2-2p^2$,
$y^2=2pq$, provided that $x$, $y$, $z$ are prime each to each.
|
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|
Extrema of a funcion of $x$ depending on $n$. I have this function:
$$f(x)=\frac{x^{3}+3(\sqrt{n}-2)x^{2}-24\sqrt{n}x-2}{n^{2}}$$
I have to find the critical point of this funcion depending on $n\ge1\in \mathbb N$.
So I have computed the first derivative
$$f'(x)=\frac{3x^{2}+6(\sqrt{n}-2)x-24 \sqrt{n}}{n^{2}}$$
So the problem is, since in this case $n^{2}\ge1$ is always positive and $\neq 0$, studying:
$$3x^{2}+6(\sqrt{n}-2)x-24 \sqrt{n}=0$$
I've tried and tried to solve this equation without finding the correct answer. Could someone help me? Thank you :).
|
You have obtained a quadratic equation $ax^2+bx+c=0$ then, as a standard method, we can apply the formula for the roots
$$x_{1,2}=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
to $$x^2+(2\sqrt n-4)x-8\sqrt n=0$$
and obtain
$$x_{1,2}=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{-(2\sqrt n-4)\pm \sqrt{4n+16-16 \sqrt n + 32\sqrt n}}{2}=\frac{-(2\sqrt n-4)\pm (2\sqrt n+4)}{2}= 4, \quad-2\sqrt n$$
As an alternative, this is the way indicated by Lord Shark the Unknown, use that
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2=0$$
and thus
*
*$x_1+x_2=-2\sqrt n+4$
*$x_1x_2=-8\sqrt n$
and solve from here.
|
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|
Find a pattern and prove it by mathematical induction: $1 = 1$
$3 + 5 = 8$
$7 + 9 + 11 = 27$
$13 + 15 + 17 + 19 = 64$
Etc...
I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic
|
Hint: $$ [1+2\cdot( 1+ 2+\ldots +(n-1))]\cdot n + 2(1+2+\ldots +(n-1)) $$
You can then use Gauss formula for $1+ 2+\ldots +(n-1)$.
As in $n=4$:
$$13 + (13 + 2) + (13 +4) + (13 +6) = 13\cdot 4 + 2\cdot(1+2+3)$$
and
$$ 13 = 1 + 2\cdot( 1+ 2 + 3)$$
|
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|
Summation of binomial coefficients upon $(r+2)$
$$\sum^{50}_{r=0}(-1)^r \dfrac{\dbinom {50}r}{r+2}= ?$$
Attempt:
$(1-x)^{50}= \sum (-1)^r \dbinom{50}r x^{n-r}$
Integrating both sides and then placing limits 0 to 1:
$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{52-r}$
So the answer should be $\dfrac{1}{51}$
But answer given is $1/(51\times 52)$
Where have I gone wrong?
Edit:
I understood my mistake from the comment.
Now I have:
$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{51-r}$
Is it possible to complete it from here?
|
OP's expression is a special case of a formula for the reciprocal binomial coefficient using the Beta function
\begin{align*}
\color{blue}{\binom{n}{k}^{-1}}&\color{blue}{=(n+1)\int_0^1 z^k(1-z)^{n-k}\,dz}\\
&=(n+1)\int_0^1z^k\sum_{r=0}^{n-k}\binom{n-k}{r}(-z)^r\,dz\\
&=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}(-1)^r\int_0^1z^{k+r}\,dz\\
&\,\,\color{blue}{=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}}\tag{1}
\end{align*}
We obtain from (1)
\begin{align*}
\sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}=\frac{1}{(n+1)\binom{n}{k}}\tag{2}
\end{align*}
Putting $n=51$ and $k=1$ in (2) we derive OP's formula
\begin{align*}
\color{blue}{\sum_{r=0}^{50}\binom{50}{r}\frac{(-1)^r}{r+2}}=\frac{1}{52\binom{51}{1}}\color{blue}{=\frac{1}{52\cdot 51}}
\end{align*}
|
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|
Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$ Question: Sketch the region in the plane consisting of all points (x,y) such that $2xy\le |x-y|\le x^2+y^2$
My attempt:
If $x>y$ then
$|x-y|=x-y$
Thus now,
$2xy\le x-y\le x^2+y^2$
If $x-y\geq 2xy$ then let $x-y=k$ ($k$ is some real number)
then $y=x-k \tag{1}$
Thus the straight line (1) has a negative slope.
After this how to proceed?
|
$$2xy\le |x-y|\le x^2+y^2$$
First, consider the inequality on the right.
Depending on whether $x$ or $y$ is the larger of the two the second inequality resolves into the two inequalities
$$ \left(x-\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\ge\left(\frac{\sqrt{2}}{2}\right)^2 $$
or
$$ \left(x+\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\ge\left(\frac{\sqrt{2}}{2}\right)^2 $$
So the solution set must lie outside the interiors of these two circles.
The first inequality resolves into the two inequalities
$$ 2xy\le x-y \text{ given that } x\ge y$$
$$ 2xy\le y-x\text{ given that } x\le y $$
The first inequality can be re-written as
\begin{eqnarray}
2xy-x+y &\le& 0\\
xy-\frac{1}{2}x+\frac{1}{2}y-\frac{1}{4} &\le& -\frac{1}{4}\\
\left(x+\frac{1}{2}\right)\left(y-\frac{1}{2}\right) &\le& -\frac{1}{4} \text{ and }y\le x
\end{eqnarray}
The graph of this inequality consists of all points lying on or below the right branch of the hyperbola.
Similiarly, the second inequality can be re-formed as
$$ \left(x-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \le -\frac{1}{4} \text{ and }y\ge x$$
The graph of this inequality consists of all points lying on or above the left branch of this hyperbola.
The darker region in the following is the solution set.
|
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|
A difficult improper integral
What is the value of the $$\int_{1}^{\infty}\frac{dx}{ x^2(e^x+1)}?$$
I got $2$ as a result. Wolfram has given $0.11111$ as a result. I am confused.
Is it correct? Or incorrect ? Please tell me...
Thanks in advance
|
We are dealing with
$$ \int_{1}^{+\infty}\frac{1-\tanh(x/2)}{2x^2}\,dx=\frac{1}{2}-\frac{1}{4}\int_{1/2}^{+\infty}\frac{\tanh(z)}{z^2}\,dz=\frac{1}{2}-\int_{1/2}^{+\infty}\sum_{n\geq 0}\frac{2\,dz}{z(4z^2+(2n+1)^2\pi^2)}$$
or
$$ \frac{1}{2}-\sum_{n\geq 0}\frac{\log(1+(2n+1)^2 \pi^2)}{(2n+1)^2 \pi^2}=\frac{1}{2}+\frac{\gamma}{4}+\frac{\log 2}{3}-3\log A-\sum_{n\geq 0}\frac{\log\left(1+\frac{1}{(2n+1)^2\pi^2}\right)}{(2n+1)^2 \pi^2} $$
where the last series can be numerically approximated by exploiting the Padé relation $\log(1+z^2)\approx\frac{z^2}{1+\frac{z^2}{2}}$ in a neighborhood of the origin. This leads to
$$ \int_{1}^{+\infty}\frac{dx}{x^2(e^x+1)}\,dx\approx \frac{1}{12} \left(3+3\gamma+4\log 2-36\log A+6 \sqrt{2}\tanh\frac{1}{2\sqrt{2}}\right)=\color{green}{0.11916}9\ldots$$
|
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|
Finding the limit: $L = \lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}$ I'm trying to solve the following question. Have tried taking $\log$ on both sides, but didn't get very far. Seems difficult to apply L'Hopital's rule.
$L = \lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}$
Any help is much appreciated.
|
$$\prod_{r=3}^{n}\frac{r^3-8}{r^3+8}=\prod_{r=3}^{n}\frac{r-2}{r+2}\prod_{r=3}^{n}\frac{(r+1)^2+3}{(r-1)^2+3}=\frac{\prod_{k=1}^{n-2}k}{\prod_{k=5}^{n+2}k}\cdot\frac{\prod_{h=4}^{n+1}(h^2+3)}{\prod_{h=2}^{n-1}(h^2+3)}$$
for any $n\geq 7$ can be written as
$$ \frac{4!}{(n+2)(n+1)n(n-1)}\cdot\frac{((n+1)^2+3)(n^2+3)}{(2^2+3)(3^2+3)} $$
whose limit as $n\to +\infty$ is given by
$$ \frac{4!}{(2^2+3)(3^2+3)}=\frac{24}{7\cdot 12}=\color{red}{\frac{2}{7}}.$$
|
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|
$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ without 'Hopital Rule Compute $$\lim_{x\rightarrow 0} \frac{(x+2)\cdot \ln(1+x)-2x}{x^3}$$ without L'Hopital Rule.
I proved before that that $$\lim_{x\rightarrow 0} \frac{ \ln(1+x)-x}{x^2}=-\frac{1}{2}$$
I tried to use it and I have to compute $$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$$
This limit looks easier to compute.
|
Use that
$$
2\ln(1+x)=2x-x^2+2\frac{x^3}{3}+o\left(x^3\right)
$$
Hence
$$\frac{2\ln(1+x)+x^2-2x}{x^3}=\frac{2x^3+o\left(x^3\right)}{3x^3}\underset{x \rightarrow 0}{\rightarrow}\frac{2}{3}$$
|
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|
prove that if $p$ is a prime number, then $\sqrt{p}$ is an irrational number.
*
*$\sqrt{p}$ is rational.
*$\sqrt{p}=\frac{a}{b}$, where $a,b$ are integers with $\gcd(a,b)=1$.
*$a^2=b^2p$.
Since $p$ divides $a^2$, $p$ divides $a$.
*
*$a=kp$.
*$a^2=k^2p^2=b^2p$
*$p=\frac{b^2}{k^2}\Rightarrow\sqrt{p}=\frac{b}{k}\Rightarrow\frac{a}{b}=\frac{b}{k}$
*$b^2=ak\Rightarrow b=\sqrt{ak}$.
*$\sqrt{p}=\frac{a}{b}=\frac{a}{\sqrt{ak}}=\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\sqrt{k}}$ which has $\gcd(a,b)=\sqrt{a}\neq 1$ , a contradiction.
($a\neq 1$ because the only number that divides 1 is 1, but 1 is not a prime from $a=kp$)
Hence, $p$ is irrational. Is this legit?
|
By way of contradiction, assume $\sqrt{p}$ is rational. Then there exist $a, b \in \mathbb{Z}$ with $b\neq 0$ such that $\sqrt{p} = \frac{a}{b}$. Without loss of generality, we may assume $\text{gcd}(a,b) \neq 1$
We can make this assumption, because we still lose no generality.
Now using $\text{gcd} (a,b) = d \neq 1$. Then we can write $a = d \cdot a'$ and $b = d \cdot b'$, for some relatively prime integers $a'$ and $b'$.
Hence
$$
\sqrt{p} = \frac{a}{b} = \frac{d a'}{d b'} = \frac{a'}{b'},
$$
So we have shown that $\sqrt{p}$ is a ratio of two relatively prime integers.
|
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|
Solving a cubic modular congruence I need to solve this congruence:
$$x(x^2-1) \equiv 34 \pmod {35}$$
I noticed that the left hand side is divisible by $3$, namely, $x(x-1)(x+1) \equiv0 \pmod 3$. Also, I thought that using the Chinese remainder theorem might work here. However, I can't find a good way to apply this theorem to single $x$'s.
Any hints will be most appreciated.
|
Well $x(x+1)(x-1)$ being the product of three consecutive numbers is pretty cute if we are going to take modulo low terms.
So by CRT we want to solve
$x(x+1)(x-1) \equiv 4\equiv -1 \pmod 5$ and if we have $x\equiv 0,1,4$ we have $x(x+1)(x-1)\equiv 0 \pmod 5$ so we must have $x\equiv 2,3$ so $x(x+1)(x-1) \equiv$ either $1*2*(-2) \equiv 1$ or $2*(-2)*(-1)\equiv -1$. So $x \equiv 3 \pmod 5$.
Similar if $(x-1)x(x+1) \equiv 6\equiv -1 \pmod 7$. We can't have $x \equiv 0, \pm 1$ so $x \equiv \pm 2, \pm 3\pmod 7$. And $(x-1)x(x+1) \equiv 1*2*3;2*3*(-3);3*(-3)*(-2);(-3)*(-2)*(-1)\equiv -1, 3, -3, 1$ so $x \equiv 2 \pmod 7$.
So by CRT $x\equiv 23 \pmod {35}$.
Can't help but wonder if we could have been more efficient.
$x^3- x + 1\equiv 0 \pmod 5$. As $x^4 \equiv 1 \pmod 5$ when $x \ne 0$ then
$\frac 1x - x + 1 \equiv 0 \pmod 5$ so
$x^2 - x - 1 \equiv 0 \pmod 5$ so $x \equiv \frac {1 \pm \sqrt{1 +4}}{2} \equiv \frac {1 \pm 0}{2} = \frac 12\equiv 3 \pmod 5$ but I'd hardly call that efficient.
And $x^6\equiv 1 \pmod 7$ for $x \ne 0$ so $x^3 \equiv 1 \pmod 8 \implies x^3 \equiv \pm 1$. So $x^3 - x + 1\equiv -x;-x + 2 \equiv 0\pmod 7$ means that $x\equiv 2\pmod 7$ as $x\not \equiv 0 \pmod 7$.
True... but definitely not efficient.
|
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|
$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11}=-ie^{
\frac{i2 \pi k}{11}}$$
so,$$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=-i \sum_{k=1}^{10}e^{
\frac{i2 \pi k}{11}}=-i \times e^{
\frac{i2 \pi }{11}} \times \frac{{\{e^{
\frac{i2 \pi }{11}}\}}^{10} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
$$=-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}$$
now how can i proceed and simplify the result?
Answer $=i$
|
$$S-i=-i\cdot (\sum_{k=1}^{10}({e^{\frac {2\pi i}{11}}})^k+1)=-i\cdot (\frac{1-({e^{\frac{2\pi i}{11}}) ^{11}}}{1-e^{\frac{2\pi i}{11}}})=-i\cdot 0=0$$.
|
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|
Find the smallest distance between point and ellipsoid
Find the smallest distance between the points on the ellipsoid $x^2+2y^2+z^2=16$ and the point $(0,0,1)$.
Answer:
Let $(x,y,z)$ be the closest point of the ellipsoid. The distance is
$$\sqrt{x^2+y^2+(z-1)^2}$$
Let $f(x,y,z)=x^2+y^2+(z-1)^2$. Construct the Lagrangian:
$$L(x,y,z)=x^2+y^2+(z-1)^2+\lambda (x^2+2y^2+z^2-16)$$
For smallest distance,
$$L_x=0 \Rightarrow x(\lambda+1)=0\\
L_y=0 \Rightarrow y(2 \lambda+1)=0\\
L_z=0 \Rightarrow z(\lambda+1)=1$$
The 3rd equation implies that $\lambda\ne-1$. Thus we have $x=0$. Now 2nd and 3rd equation solution gives $y=0$ or $\lambda=-\frac12$ and $z=\frac1{1+\lambda}$.
But I can not find the final solution .
Help me out
|
If $\lambda=-\frac12$, $z=2$ and from the original constraint $x^2+2y^2+z^2=16$ (where $x=0$ as shown prior) we get $y=\pm\sqrt 6$. In this case, the distance is $d=\sqrt{0^2+\sqrt6^2+(2-1)^2}=\sqrt7$.
If $y=0$, from the original constraint we have $z=4\lor z=-4$, which give $d=\sqrt{0^2+0^2+(4-1)^2}=3$ and $d=\sqrt{0^2+0^2+(-4-1)^2}=5$ respectively.
$\sqrt7$ is thus the shortest distance from $(0,0,1)$ to the ellipsoid, the closest points being $(0,\pm\sqrt6,2)$.
|
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|
Find $y$ for $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$ where $y$ is a positive real number Here is the answer in the textbook with which I disagree / don't understand:
$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$
$$y^{y^{\frac{3}{2}}} = (y^{\frac{3}{2}})^y = y^{\frac{3}{2}y}$$
Comparing indices, we get:
$$y^{\frac{3}{2}} = \frac{3}{2}y$$
from $y^{\frac{1}{2}} = {\frac{3}{2}}$, we get: $$y = {\frac{9}{4}}$$
Here what I came up with:
$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$
$$(\sqrt[y]{y})^y = (y^{\frac{1}{y}})^y = y$$
$$y^{\sqrt[y]{y}} = y \Rightarrow \sqrt[y]{y} = 1$$
And I get stuck somewhere around: $y = 1^y$
|
For positive $y$ you have $(\sqrt[y]{y})^y = y$
so $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y\implies y^{\sqrt[y]{y}} = y \implies y^{\sqrt[y]{y}-1} = 1$
so either $y=1$ or $\sqrt[y]{y}-1=0$
in both cases implying $y=1$
Alternatively if the original question is actually $y^{y \sqrt{y}} = (y \sqrt{y})^y$
Then this indeed gives $y^{y^{3/2}}=y^{3y/2}$
so either $y=1$ or $y^{3/2}=\frac{3y}{2}$
i.e. either $y=1$ or $y^{1/2}=\frac{3}{2}$
i.e. either $y=1$ or $y=\frac{9}{4}$
|
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|
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?
|
Starting from
$$
\left\{ \matrix{
{{x^{\,2} } \over 9} + {{y^{\,2} } \over 4} = 1 \hfill \cr
A = 4xy \hfill \cr} \right.
$$
which gives the intercepts between the ellipse and a hyperbola,
you could have simply substitute $y=A/(4x)$, to get
$$
{{x^{\,2} } \over 9} + {{A^{\,2} } \over {64x^{\,2} }} = 1
$$
i.e.
$$
64x^{\,4} - 9 \cdot 64x^{\,2} + 9A^{\,2} = 0
$$
put the discriminant to be null, and get $A$
$$
\Delta = \left( {9 \cdot 64} \right)^{\,2} - 4 \cdot 64 \cdot 9A^{\,2} = 0\quad \Rightarrow \quad A^{\,2} = 9 \cdot 16\quad \Rightarrow \quad A = 12
$$
and since the discriminant is null, from the above it follows
$$
x^{\,2} = {{9 \cdot 64} \over {2 \cdot 64}}\quad \Rightarrow \quad x = {3 \over 2}\sqrt 2 \quad \Rightarrow \quad y = A/4x = \sqrt 2
$$
|
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|
Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
|
They have to be divisible by $15$, $2$, and $3$, or divisible by $30$, but from your method, the divisors can be divisible by $10$. What I mean mathematically is:
$$30(2^2\cdot3^3\cdot5^2\cdot7\cdot11)\ne 2079000$$
Any even number must be divisible by $2$, and any number divisible by $15$ and $2$ must be divisible by $30$; as $15$ and $2$ are relatively prime numbers.
Can you find the amount of numbers which are divisible by $30$ and divide into $2079000$?
Hint: You have to perform an operation with $2079000$ and $30$.
|
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|
How I can solve $s(n)=n+s(n-1)$ by iteration method? $$
s(n)=
\begin{cases}
0, \text{if $n=0$}\\
n+s(n-1), \text{if $n>0$}\\
\end{cases}
$$
Using the relation
\begin{align}
s(n) &= n+s(n-1) \\
&= n+n-1+s(n-2) \\
&= n+n-1+n-2+s(n-3) \\
&= \dots \\
&= n+n-1+n-2+\dots+1 \\
&= \dfrac{n^2+n}2 \\
&= \theta(n^2)
\end{align}
Is it right?
|
I like to turn
recurrences to
telescoping sums
whenever possible.
From
$s(n) = n+s(n-1)
$
I get
$s(n)-s(n-1) = n$.
Summing both sides from $1$ to $m$,
$\sum_{n=1}^m s(n)-s(n-1) = \sum_{n=1}^m n $.
The left side telescopes into
$s(m)-s(0)$
and the right side is just
$\dfrac{m(m+1)}{2}
$.
Note that if
the first value is $s(1)$,
just make the sum from $2$
instead of $1$.
|
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|
Haggling problem Adam is trying to sell Bob a bike for $a$ dollars.
Bob does not agree on the price $b$ ($0 < b < a$).
Adam does not agree on this price but does lower his initial price to $\dfrac{a+b}{2}$ dollars.
Bob responds by offering $\dfrac{b+\frac{a+b}{2}}{2}$ dollars.
They continue haggling this way, each time taking the average of the previous two amounts.
On what price will they converge? Express this price in terms of the two initially proposed amounts $a$ and $b$.
|
*
*We first write the series of bids: Start with $a$ and then
$$
\tfrac{b}{1} \to \tfrac{a+b}{2} \to \tfrac{2b+(a+b)}{4}=\tfrac{a+3b}{4} \to \tfrac{2(a+b)+(a+3b)}{8}=\tfrac{3a+5b}{8} \to \tfrac{2(a+3b)+(3a+5b)}{16}=\tfrac{5a+11b}{16}\to\dotsb \tag{*}
$$
*The denominators in (*) are the series $\{ 2^0, 2^1, 2^2, 2^3, 2^4,\dotsc\}$.
*The nominators in (*) are as follows:
$$
0a+1b \to 1a+1b \to 1a+3b \to 3a+5b \to 5a+11b \to 11a+21b \to \dotsb
$$
*Let us rewrite the coefficients of $a$ and $b$ as sequences:
$$
\begin{align*}
\text{coefficients of $a$}: \{a_n\mid n\geq0\}=\{ 0, 1, 1, 3, 5, 11, 21, 43, \dotsc \} \\
\text{coefficients of $b$}: \{b_n\mid n\geq0\}=\{ 1, 1, 3, 5, 11, 21, 43, 85, \dotsc \}
\end{align*}
$$
*Two sequences are the same: $a_n=s_n$ and $b_n=s_{n+1}$ for $n\geq0$, where
$$
s_0=0, \quad s_1=1, \quad s_n = 2s_{n-2} + s_{n-1} \quad\text{for $n\geq2$}
$$
Note that $a_n+b_n=s_n+s_{n+1}=2^n$ for all $n\geq0$. Moreover,
$$
s_n = 2s_{n-2} + s_{n-1} = 2(2^{n-2}-s_{n-1})+s_{n-1} = 2^{n-1}-s_{n-1} \quad\text{for $n\geq2$}
$$
so that
$$
\frac{1}{2}\left(\frac{s_{n-1}}{2^{n-1}}\right) + \frac{s_n}{2^n}=\frac{s_{n-1} + s_n}{2^n} = \frac{2^{n-1}}{2^n}=\frac{1}{2} \tag{**}
$$
*If the sequence $\dfrac{s_n}{2^n}$ converges to $p$, then the equation (**) follows that $\frac{1}{2}p+p=\frac{1}{2}$ so that
$$
\lim_{n\to\infty} \frac{s_n}{2^n} = p = \frac{1}{3} \tag{***}
$$
*Let us rewrite the bids in (*) as an equation:
$$
\frac{a_na+b_nb}{2^n} = \frac{s_na+s_{n+1}b}{2^n} = \frac{s_na+(2^n-s_n)b}{2^n} = \frac{s_n}{2^n}(a-b) + b \quad\text{for $n\geq0$}
$$
On what price will they converge? Express this price in terms of the two initially proposed amounts a and b.
Now from the limit in (***), the final answer is:
$$
\lim_{n\to\infty} \left[ \frac{s_n}{2^n}(a-b) + b \right] = \frac{1}{3}(a-b)+b
$$
|
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|
Number of arrangements of four blue, three green, and two red balls in which no two blue balls are adjacent Find the number of ways of arranging four blue, three green and two red balls such that no two blue balls are adjacent.
I have done this question and got the answer which is
$$\binom{6}{4} \cdot \frac{5!}{2!\cdot3!}=150$$
using the concept of arranging $4$ blue balls in the six vacant spaces among the five other balls.
How can this problem be solved by excluding combinations which consist of $4$ blue balls together, $3$ blue balls together, or two blue balls together?
|
There are $4 + 3 + 2 = 9$ positions to fill. We fill four of them with blue balls, three of the remaining five positions with green balls, and the remaining two positions with red balls. Therefore, the number of possible arrangements of the balls is
$$\binom{9}{4}\binom{5}{3}\binom{2}{2} = \frac{9!}{4!5!} \cdot \frac{5!}{3!2!} \cdot \frac{2!}{2!0!} = \frac{9!}{4!3!2!}$$
From these, we wish to exclude those arrangements with one or more pairs of adjacent blue balls. Since there are only four blue balls, at most three such pairs can be formed.
A pair of adjacent blue balls: We have eight objects to arrange: $bb$, $b$, $b$, $g$, $g$, $g$, $r$, $r$. We choose three of those eight positions for the green balls, two of the remaining five positions for the red balls, two of the remaining three positions for the blue balls, and place the block of two blue balls in the remaining position. Hence, there are
$$\binom{8}{3}\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{8!}{3!2!2!}$$
such arrangements.
If we subtract the number of arrangements with a pair of adjacent blue balls from the total, we will have subtracted too much since we will have subtracted those arrangements with two pairs of adjacent blue balls twice, once for each way of designating one of those pairs as the pair of adjacent blue balls. Since we only want to subtract such arrangements once, we must add them back.
Two pairs of adjacent blue balls: This can occur in two ways.
*
*There are two disjoint pairs of adjacent blue balls.
*There are two overlapping pairs of adjacent blue balls, meaning that there are three consecutive blue balls.
Two disjoint pairs of adjacent blue balls: We have seven objects to arrange: $bb$, $bb$, $g$, $g$, $g$, $r$, $r$. We choose two of the seven positions for the blocks of blue balls, three of the remaining five positions for the blocks of green balls, and then fill the remaining the two positions with the red balls. Hence, there are
$$\binom{7}{2}\binom{5}{3}\binom{2}{2} = \frac{7!}{2!3!2!}$$
such arrangements.
Two overlapping pairs of adjacent blue balls: We again have seven objects to arrange: $bbb$, $b$, $g$, $g$, $g$, $r$, $r$. We choose three of the seven positions for the green balls, two of the remaining four positions for the red balls, and arrange the remaining two distinct objects in the remaining two positions. Hence, there are
$$\binom{7}{3}\binom{4}{2}2! = \frac{7!}{3!4!} \cdot \frac{4!}{2!2!} \cdot 2! = \frac{7!}{3!2!}$$
such arrangements.
If we subtract the number of arrangements containing a pair of adjacent blue balls and add the number of arrangements containing two pairs of adjacent blue balls, we will have not subtracted arrangements containing three pairs of adjacent blue balls at all. This is because we subtract three times when we subtract arrangements containing a pair of adjacent blue balls, once for each way we could designate one of those pairs as the pair of adjacent blue balls, and add them three times when we add arrangements containing two pairs of adjacent blue balls, once for each of the $\binom{3}{2}$ ways we could designate two of the three pairs as the pairs of adjacent blue balls. Hence, we must subtract the number of arrangements with three pairs of adjacent blue balls from the total.
Three pairs of adjacent blue balls: This means all four blue balls are consecutive. Hence, we have six objects to arrange: $bbbb$, $g$, $g$, $g$, $r$, $r$. We choose three of the six positions for the green balls, two of the remaining three positions for the red balls, and fill the final position with the block of blue balls. Hence, there are
$$\binom{6}{3}\binom{3}{2}\binom{1}{1} = \frac{6!}{3!2!}$$
such arrangements.
By the Inclusion-Exclusion Principle, the number of arrangements of four blue, three green, and two red balls in which no two blue balls are consecutive is
$$\frac{9!}{4!3!2!} - \frac{8!}{3!2!2!} + \frac{7!}{2!3!2!} + \frac{7!}{3!2!} - \frac{6!}{3!2!} = 150$$
which agrees with your result.
|
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|
Find a basis for $W$ and identify $\dim(W)$. Let $\mathbf{R}^{3\times3}$ be the linear space of all $3 \times 3$ matrices.
Let $W$ be the set of all symmetric $3\times3$ matrices. Then
$W$ is a linear subspace of $\mathbf{R}^{3×3}$. Find a basis for $W$ and identify $\dim(W)$.
Would the $\dim(W)$ be $3$ since it is a $3\times3$ matrix. Is that correct? How do you go about finding the basis? From my understanding, if the span of $W$ are linearly independent then they would be the basis of the $3\times3$ matrix.
Please help!
:) Thanks
|
The $3\times3$ symmetric matrices are those of the form$$\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}.$$They form a $6$-dimensional space. A basis of this space is$$\left\{\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix},\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},\begin{pmatrix}0&0&0\\0&0&1\\0&1&0\end{pmatrix},\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix}\right\}.$$Can you prove it?
|
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|
Number of 4x4 matrices such sum of all rows and columns is zero. The number of matrices $A = [a_{ji}], 1 \le i,j \le 4$ such that $a_{ij} = +- 1$ and $\sum_{i = 1}^{4}a_{ij} = \sum_{j=1}^{4}a_{ij} = 0$ is...
I solved it by first selecting two values of “i” which is $\binom{4}{2}$ and two values of “j” similarly. Now if I fill thes four positions by 1, the value of remaining positions also get fixed because sum of all rows and columns is zero. So answer should be 36 $(\binom{4}{2}\times \binom{4}{2})$but the correct answer is 90.
Why this method is wrong and what is the correct way of solving this?
|
You are both over-counting and under-counting!
First, you are over-counting:
Suppose you first pick $i=1,2$ and $j=1,2$. So, $a_{11}$, $a_{12}$, $a_{21}$, and $a_{22}$ are all set to $1$. But that forces $a_{33}$, $a_{34}$, $a_{43}$, and $a_{44}$ to be a $1$ as well.
OK, but now start with $i=3,4$ and $j=3,4$. Then $a_{33}$, $a_{34}$, $a_{43}$, and $a_{44}$ are a $1$. But that forces $a_{11}$, $a_{12}$, $a_{21}$, and $a_{22} $ to $1$. So, you get the exact same matrix.
But, you are also under-counting, since it is not true that there must be two row with the same entries being a $1$. Consider:
$$\begin{bmatrix}
1& 1 &-1 & -1 \\
-1& 1 &1 & -1 \\
-1& -1 &1 & 1 \\
1& -1 &-1 & 1 \\
\end{bmatrix}$$
OK, here's how you can get them all.
First, pick two entries from the first column that are a $1$
Then, for the second row, you have the following possibilities:
$A$. Pick the exact same entries. That can be done in one way only, and it forces the rest of the matrix, so $ 1$ possibility
$B$. Pick the 'opposite' entries (example, if you picked entries $1$ and $2$ for row $1$, pick $3$ and $4$ for row $2$). That can also only be done in one way. Now, notice that for the third row, since rows $1$ and $2$ do not share any entries, you can still pick any two entries from row $3$, so $6$ possibilities for that one. Of course, after row $3$ is done, row $4$ is forced, so option $B$ leads to a total of $6$ possibilities
C. Row $2$ shares one entry with row $1$, which can be done in $4$ ways ($2$ choices for the shared entry, and $2$ for the other two entries. (Example: if the entries for row $1$ are $1$ and $2$, then for row $2$ we can pick $1$ and $3$, $1$ and $4$, $2$ and $3$, and $2$ and $4$)). Then for row $3$, you cannot use the shared entry from rows $1$ and $2$, but you have to pick the one non-shared entry, and then one out of the remaining $2$ entries, so $2$ options there (Example, if you picked entries $1$ and $2$ for row $1$, and entries $1$ and $3$ for row $2$, then the entries for row $3$ can only be $2$ and $4$ or $3$ and $4$)
And again, since row $4$ is now fixed, option $C$ gives $4\cdot 2 = 8$ possibilities.
Finally, since there were $6$ ways to pick the first two entries for row $1$, that gives $6 \cdot(1 + 6 + 8) = 90$ possibilities
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing a function grows like a certain power of $n$ Consider $f(n) = \frac{1}{n} {n \choose (n+1)/2} \frac{1}{2^n}$ for odd $n$. I suspect that $\sum_{n \geq 1} n^{\beta}f(n)= \infty$ for $\beta \geq 1/2$. For this purpose, it would be enough to show that $f(n) \sim n^{-3/2}$. How can I show this last statement?
|
Letting $n=2m+1, m\in\mathbb{N}$ we obtain
\begin{align*}
\color{blue}{f(2m+1)}&=\frac{1}{2^{2m+1}(2m+1)}\binom{2m+1}{m+1}\\
&=\frac{1}{2^{2m+1}(m+1)}\binom{2m}{m}\tag{1}\\
&\,\,\color{blue}{=\frac{1}{2^{2m+1}}C_m}\tag{2}
\end{align*}
with $C_m$ the $m$-th Catalan-number.
In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
The asymptotic behaviour of the Catalan numbers $C_m$ can be shown using Stirling's formula and is according to this MSE answer
\begin{align*}
C_m\sim\frac{4^m}{\sqrt{\pi}m^{\frac{3}{2}}}
\end{align*}
We conclude together with (2)
\begin{align*}
\color{blue}{f(2m+1)\sim \frac{1}{2\sqrt{\pi}m^{\frac{3}{2}}}}
\end{align*}
Note: Since $f(n)$ is given for odd $n$ only, we do not know anything about the behaviour of $f$ for even arguments.
|
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|
Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose function $f$ and $h.$
|
For $0< x< 1$ it holds: $1+x^2<1+x+x^2<3$. Therefore
$$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$
|
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|
Power Series Representation of a Function, (differentiation) Suppose that $f(x) =\frac{x^2+x}{(1-x)^3}$. What would the power series representation for this function?
I have found the power series representation for $\frac{1}{(1-x)^3}$ which is $\sum_{n=2}^{\infty} \frac{n(n-1)x^{n-2}}{2}$. Would it be allowed to multiply the series by $x^2+x$ to get $\left(x^2+x\right) \sum_{n=2}^{\infty} \frac{n(n-1)x^{n-2}}{2}$?
|
This is indeed correct and the result is then given by $\sum_{n=0}^{\infty}n^2x^n$, as stated in M. Iwaniuk's comment.
$$\left(x^2+x\right)\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=x^2\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}+x\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=\sum_{n=2}^{\infty}\frac{n(n-1)x^{n}}{2}+\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-1}}{2}=\sum_{n=2}^{\infty}\frac{n(n-1)x^{n}}{2}+\sum_{n=1}^{\infty}\frac{(n+1)nx^{n}}{2}=x+\sum_{n=2}^{\infty}\frac{n(n-1)+n(n+1)}{2}x^{n}=\sum_{n=0}^{\infty}n^2x^n$$,
as $0x^0=0$ and $x=1^2x^1$
|
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|
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
|
It is a matrix such that $P^4= P^2$ and some additional relations.
We see that
$$ 0 = P^4 - P^2 = P^2(P^2 - I) = P^2(P-I)(P+I)$$
so the minimal polynomial $\mu_P(x)$ divides $x^2(x-1)(x+1)$.
However, the minimal polynomial is not any of $$x, x^2, x-1, x+1, \underbrace{x(x-1)}_{P^2 \ne P}, \underbrace{x(x+1)}_{P^2 \ne -P}, \underbrace{x^2(x-1)}_{P^3 \ne P^2}, \underbrace{x^2(x-1)}_{P^3 \ne -P^2}, \underbrace{(x-1)(x+1)}_{P^2 \ne I}, \underbrace{x(x-1)(x+1)}_{P^3 \ne P}$$ so it has to be precisely $x^2(x-1)(x+1)$.
The possibilities for the Jordan normal form of $P$ are:
$$\pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0}, \pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1}, \pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1}$$
meaning $5 \times 5$ matrices have the above properties if and only if they are similar to one of the three matrices above.
|
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|
Computing a series rised from definite integral of a floor function I interested to compute $$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx.$$
So, I did like this:
$$\int_0^1\left[\frac{1}{\sqrt{x}}\right]dx=\sum_{n=1}^\infty
\int_{1/(n+1)^2}^{1/n^2} ndx=\\
\sum_{n=1}^{\infty}\frac{2n+1}{n(n+1)^2}=\\
=\sum_{n=1}^\infty \frac{2}{(n+1)^2}+\sum_{n=1}^\infty \frac{1}{n(n+1)^2}=\frac{\pi^2}{3}-2+\sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$
Now, how to compute the last series? Is it computible or it relates to $\zeta(3)$?
Thanks.
|
$\sum_{n=1}^\infty \frac{1}{n(n+1)^2}=\sum_{n=1}^\infty \frac{n+1-n}{n(n+1)^2}=\sum_{n=1}^\infty \frac{1}{n(n+1)}-\sum_{n=1}^\infty \frac{1}{(n+1)^2}=1-\sum_{n=1}^\infty \frac{1}{(n+1)^2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Stuck on this square root conjugate problem I have to multiply the following:
$((x+h)\sqrt{x+h} - x\sqrt{x}) * ((x+h)\sqrt{x+h} + x\sqrt{x})$
I tried to use $(a - b)(a + b) = a^2 + b^2$ so:
$((x+h)\sqrt{x+h} - x\sqrt{x})((x+h)\sqrt{x+h} + x\sqrt{x}) = (x^2 + h^2)(x + h) - x^2(x)$
and then:
$x^3 + x^2h + h^2x + h^3 - x^3 $
and then:
$x^2h + h^2x + h^3$
But the book is showing the answer as $3x^2h + 3xh^2 + h^3$
I don't know where they are getting those $3$'s from. Can someone please tell me what I'm doing wrong?
|
Notice that $$\begin{align} (x+h)^3 &= x^3 + h^3 + 3xh(x+h) \\ &= x^3 + h^3 + 3x^2h + 3xh^2.\end{align}$$ Therefore, the book's answer is equivalent to $$(x+h)^3 - x^3 = (x+h)^2(x+h) - x^2(x),\tag1$$ but the answer you obtained was $(x^2 + h^2)(x+h) - x^2(x)$, and $x^2 + h^2\neq (x+h)^2$. From this observation, it is clear that you have calculated it incorrectly.
Since $(a+b)(a-b) = a^2 - b^2$ then substitute $a = (x+h)\sqrt{x+h}$ and $b = x\sqrt{x}$. Now, by letting $b = f(x)$, it follows that $a = f(x+h)$, and so if $b^2 = x^2(x)$ then $$a^2 = (x+h)^2(x+h)\neq(x^2+h^2)(x+h),$$ and Eq. $(1)$ is consequently implied.
|
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|
$5$ points on ellipse I want to take five points and locate them on ellipse so that the distance between the point $1$ and point $2$, the point $2$ and point $3$, the point $3$ and point $4$, the point $4$ and point $5$, the point $5$ and point $1$ were equal to each other. I think it will look like a polygon with equal side lengths, but with different angles.
Can you help me to figure out how to find the distance among those points and the coordinates of those points if the minor and major axes are $b$ and $a$ respectively?
|
Equation of the ellipse :
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$
with given values $a,b$.
They are many solutions, depending where we put the first point on the ellipse.
If you want only one solution, one of the simplest is obtained in putting the first point A on $(x=a,y=0)$.
See Figure below. With AB=BC=CD=DE=EA=$L$, at beginning $L$ is unknown. To find it, the equations are :
Point A$(x_1,y_1)$ :
$$y_1=\sqrt{1-\left(\frac{x}{a}\right)^2} \quad;\quad y_1^2+(a-x_1)^2=L^2$$
$$x_1=a\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2} \tag 1$$
$$y_1=b\sqrt{1-\left(\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}\right)^2} \tag 2$$
Point B$(x_2,y_2)$ :
$$y_2=\frac{L}{2} \tag 3$$
$$x_2=a\sqrt{1-\left(\frac{L}{2b}\right)^2} \tag 4$$
Condition to have BC=$L$ : $\quad (x_1-x_2)^2+(y_1-y_2)^2=L^2$
$$\left(a\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}-a\sqrt{1-\left(\frac{L}{2b}\right)^2}\right)^2+ +\left(b\sqrt{1-\left(\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}\right)^2}-a\sqrt{1-\left(\frac{L}{2b}\right)^2}\right)^2=L^2$$
this equation contains only the unknown $L$.
It can be simplified by normalization, with $x=\frac{L}{a}$ , $\beta=\frac{b}{a}$. This leads to an equation with parameter $\beta$ and unknown $x$. But the benefit is not large because transforming the equation with radicals leads to a polynomial equation of degree higher than 5.
Thus, there is no closed form for the unknown $L$. Nevertheless, one can easily solve it for $L$ with numerical calculus. Then the coordinates of the points are computed with Eqs.$(2-4)$. An example is shown on the figure below.
|
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Algebraic Proof on equivalence $$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}\equiv{a+b+c}$$
Demonstrate the identities above considering a, b & c real numbers and distinct from each other.
I'm stucked on this problem. Take a look on what I did:
$x=(a-b)\therefore -x=(b-a)$; $y=(a-c);-y=(c-a)$; $z=(b-c);-z=(c-b)$.
___________________________________________________________________$$\frac{a^3}{xy}+\frac{b^3}{(-x)z}+\frac{c^3}{(-y)(-z)}$$
$$\frac{a^3z-b^3y+c^3x}{xyz}$$
$$\frac{a^3(b-c)+b^3(a-c)+c^3(a-b)}{xyz}$$
$$\frac{a^3(b-c)+b^3(\color{red}{-b}+a-c\color{red}{+b})+c^3(a-b)}{xyz}$$
$$\frac{a^3b+ab^3-a^3c+ac^3-b^3c-bc^3}{xyz}$$
$$\frac{ab(a^2+b^2)-ac(a^2-c^2)-bc(b^2+c^2)}{xyz}$$
after this step I was thinking about doing the same technique on the second term, $-ac(\color{red}{b^2}+a^2-c^2\color{red}{-b^2})$, but after doing some more algebraic factorization I got stucked, could you help me to demonstrate?
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Put the LHS over a common denominator, getting
$$\frac{a^3(b-c)}{(a-c)(b-c)(a-b)}-\frac{b^3(a-c)}{(a-c)(b-c)(a-b)}+\frac{c^3(a-b)}{(a-c)(b-c)(a-b)}=$$
$$=\frac{a^3b-a^3c-ab^3+ac^3+b^3c-bc^3}{(a-c)(b-c)(a-b)}=\frac{(a+b+c)(a-c)(b-c)(a-b)}{(a-c)(b-c)(a-b)}=a+b+c$$
|
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parabola locus problem If $Q_1$ and $Q_2$ be the angle made by tangents to the axis of $y^2=4x$ from point $P$ and if $Q_1+Q_2=45^{\circ}$ then locus of $P$ is
for options see here
|
Let $P(X,Y)$ be the pole.
The equation of polar or equivalently the chord:
$$Yy-2(x+X)=0 \tag{1}$$
Substitute $x=\dfrac{y^2}{4}$ into $(1)$,
\begin{align}
y^2-2Yy+2X &= 0 \\
y_1+y_2 &= 2Y \\
y_1 y_2 &= 4X \\
m_1 &= \frac{2}{y_1} \tag{$2yy'=4$} \\
m_2 &= \frac{2}{y_2} \\
\frac{m_1+m_2}{1-m_1 m_2} &= \frac{2(y_1+y_2)}{y_1 y_2-4} \\
\tan 45^{\circ} &= \frac{4Y}{4X-4} \\
Y &= X-1
\end{align}
|
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If $\gcd(m,15)=\gcd(n,15)=1,$ show that $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$. If $\gcd(m,15)=\gcd(n,15)=1$ show $15\mid (m^4+n^4)$ or $15\mid (m^4-n^4)$.
This is what I have so far but I'm stuck on essentially the last step.
Proof: Assume $\gcd(m,15)=\gcd(n,15)=1$. By the Euler's Phi Function: If $\gcd(a,m)=1$ then $a^{\phi (m)}\equiv 1 \pmod m$.
$\phi(15)=\phi(5)\phi(3)=(5-1)(3-1)=8$
Hence, $m^8\equiv 1 \pmod{15}$ and $n^8\equiv 1 \pmod{15}$.
Then, $m^8\equiv 1 \pmod{15}\iff 15\mid m^8-1$. Similarly $15\mid n^8-1$.
Divisibility property sates if $a\mid b$ and $a\mid c$ then $a\mid b\pm c$.
So $15\mid (m^8-1)-(n^8-1) \Rightarrow 15\mid m^8-n^8$
So I believe I'm right up to this point, please correct if I'm not. I know that $m^8-n^8$ factors to $(m^4+n^4)(m^4-n^4)$ however since $15$ is not a prime number I can't assume $15\mid m^4+n^4$ or $15\mid m^4-n^4$.
Please help on these last few steps. Thanks.
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Since $\gcd(m,5) = \gcd(n,5)=1$ we have by Fermat little theorem:
$$ m^4\equiv 1\equiv n^4 \pmod 5$$ so $5\mid m^4-n^4$.
On the other hand since $\gcd(m,3) = \gcd(n,3)=1$ we have again by Fermat little theorem:
$$ m^2\equiv 1\equiv n^2 \pmod 3$$ so $3\mid m^2-n^2$ and so $3\mid m^4-n^4$
|
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|
Finite Complex Exponential Geometric Series with Negative Exponents So I want to know if I am doing my manipulations correctly. I have the following expression:
$$ X(\omega) = A\frac{1 - e^{-j \frac{1}{2} \omega N}}{1 - e^{j
\frac{1}{2} \omega}} $$
Using the geometric series formula:
$$ \displaystyle\sum\limits_{n=0}^{N-1} A r^{n} = \frac{1 - r^{N}}{1 - r}$$
Can I rewrite the first expression as:
$$ X(\omega) = A\frac{1 - \big(e^{j \frac{1}{2} \omega }\big)^{-N}}{1 - e^{j
\frac{1}{2} \omega}} $$
$$ X(\omega) = A \displaystyle\sum\limits_{n=0}^{-(N-1)} e^{j \frac{1}{2} \omega n} $$
$$ X(\omega) = A \displaystyle\sum\limits_{n=0}^{N-1} e^{-j \frac{1}{2} \omega n} $$
Somehow I don't think this is right, because if we reuse the geometric series expansion, the denominator will be different. But, I can't see to convince myself of it.
Or this is manipulation correct?
|
Addendum
Let me try and clear your doubts about summing definition and handling.
There are fundamentally three ways to express a sum.
a) over a set
e.g.
$$
\sum\limits_{k\, \in \,\left\{ {1,2,3} \right\}} k = 1 + 2 + 3 = 2 + 1 + 3 = \ldots
$$
b) under a condition on the index
e.g.
$$
\sum\limits_{1\, \le \,k\; \le \,3} k = 1 + 2 + 3 = \ldots \quad \Rightarrow \quad \sum\limits_{1\, \le \,k\; \le \, - 3} k = \sum\limits_\emptyset k = 0
$$
or
$$
\sum\limits_{\,k\;\backslash \,4} k = 1 + 2 + 4
$$
etc.
c) as antidelta
that is
$$
\eqalign{
& f(x) = \Delta _{\,x} F(x) = F(x + 1) - F(x) = \quad \Rightarrow \cr
& \Rightarrow \quad F(x) = \Delta _{\,x} ^{\,\left( { - 1} \right)} f(x) = \sum\nolimits_x {f(x)} \quad \Rightarrow \cr
& \Rightarrow \quad \sum\nolimits_a^b {f(k)} = F(b) - F(a) \cr}
$$
So, concerning the geometric sum
when the exponent is positive, e.g. $3$, then
$$
{{1 - r^{\,3} } \over {1 - r}} = \sum\limits_{k\, \in \,\left\{ {0,1,2} \right\}} {r^{\,k} } = \sum\limits_{0\, \le \,k\; \le \,2} {r^{\,k} } = \sum\limits_{0\, \le \,k\; < \,3} {r^{\,k} } = \sum\nolimits_0^3 {r^{\,k} }
$$
the three definitions coincide.
But when it is negative
$$
\eqalign{
& {{1 - r^{\, - 3} } \over {1 - r}}\quad \ne \quad \sum\limits_{k\, \in \,\left\{ {0, - 1, - 2} \right\}} {r^{\,k} } = {{1 - \left( {1/r} \right)^{\,3} } \over {1 - \left( {1/r} \right)}} \cr
& \quad \quad \quad \;\; \ne \quad \sum\limits_{0\, \le \,k\; \le \, - 2} {r^{\,k} } = 0 \cr}
$$
while the antidelta gives
$$
\eqalign{
& {{1 - r^{\, - 3} } \over {1 - r}} = \quad \sum\nolimits_0^{ - 3} {r^{\,k} } = - {{r^{\, - 3} } \over {1 - r}} + {{r^{\,0} } \over {1 - r}} = \cr
& = - \sum\nolimits_{ - 3}^0 {r^{\,k} } = \sum\limits_{ - 3\, \le \,k\; < \,0} {r^{\,k} } = \sum\limits_{ - 3\, \le \,k\; \le \, - 1} {r^{\,k} } = \cr
& = \sum\limits_{1\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} } = \sum\limits_{0\, \le \,k\; \le \,3} {\left( {1/r} \right)^{\,k} } - 1 = {{1 - \left( {1/r} \right)^{\,4} } \over {1 - \left( {1/r} \right)}} - 1 = \cr
& = {{r\left( {r^{\,4} - 1} \right)} \over {r^{\,4} \left( {r - 1} \right)}} - 1 = {{r^{\,3} - 1} \over {r^{\,3} \left( {r - 1} \right)}} \cr}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2707952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.