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Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2n^2-2-2n^2-3}{2(2n^2+3)}\right| \\
&= \left|\frac{-5}{2(2n^2+3)}\right|\\
&= \frac{5}{2(2n^2+3)} \\
&<\frac{5}{4n^2},
\end{align}$$
$$\frac{5}{4n^2}<\epsilon \iff \frac{1}{n^2}<\frac{4 \epsilon}{5} \iff n>\sqrt{\frac{5}{4\epsilon}}$$
We choose $n_0=\left[\sqrt{\frac{5}{4\epsilon}} \right]+1$, Then $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$.
Let $(x_n)=\frac{1}{\ln(n+1)}$ for $n \in \mathbb{N}$.
a) Use the definition of the limit to show that $\lim(x_n)=0$.
$|\frac{1}{\ln(n+1)}-0|=\frac{1}{\ln(n+1)}<\epsilon \Leftrightarrow ln(n+1) > \epsilon \Leftrightarrow n> e^{\epsilon} -1$
We choose $n_0=\left[ e^\epsilon -1 \right]+1$, Then $\lim(x_n)=0$.
b) Find specific value of $n_0 (\epsilon)$ as required in definition of limit for $\epsilon=\frac{1}{2}$.
$n_0=\left[\sqrt{e}-1\right]+1$
Is that true, please?
|
If you are using the definition of a limit at infinity, you should include a few more references to the definition in the proof:
Prove: $\lim_{n \to \infty} \left(\frac{n^2-1}{2n^2+3}\right)=\frac{1}{2}$
Proof: Let $\epsilon>0$. Show that there is a positive integer $n_0$ such that if $n>n_0$ then $\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right|<\epsilon$
Then proceed with the steps which you have given.
|
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|
Evaluating $\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}$ What is the value of $$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}?$$
What I have tried:
$$\implies\frac{2013^2(2013-2\cdot2014)+2014^2(3\cdot 2013-2014)+1}{2013\cdot 2014}$$
$$\implies\frac{2013^2(-2015)+2014^2(4025)+1}{2013\cdot 2014}$$
I'm not sure what to do next...
Help is appreciated!
Furthermore, if you are nice, could you also help me on this problem(Transferring bases of numbers.) too?
Thanks!
Max0815
|
HINT:
Let $a=2013$ and $b=2014$. You have:
$$\frac{a^3 - 2a^2b + 3ab^2 - b^3 + 1}{ab}$$
Recall that $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ and use that to simplify.
|
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|
Evaluate the integral $\int_2^\infty \frac{1}{\sqrt x}log(\frac{x+1}{x-1})\,dx$
I actually need to check whether the following series is convergent
$\sum_{n=2}^\infty \frac{1}{\sqrt n}log(\frac{n+1}{n-1})$.
The question specifically asks to use Cauchy's integral method, but I am not able to evaluate the integral nor able to find the answer through comparison test.
Any help is appreciated.
|
$$\int_2^\infty \log\left(\frac{x+1}{x-1}\right)\frac{dx}{\sqrt{x}}=2\sqrt{x}\log\left(\frac{x+1}{x-1} \right)|_2^\infty-2\int_2^\infty\sqrt{x}\left(\frac{1}{x+1}-\frac{1}{x-1}\right)dx\\=-\log(9)\sqrt{2}-2\int_2^\infty \frac{\sqrt{x}}{x+1}-\frac{\sqrt{x}}{x-1}dx$$ Let $u^2=x$, $2udu=dx$.
$$-2\int_2^\infty \frac{\sqrt{x}}{x+1}-\frac{\sqrt{x}}{x-1}dx=-4\int_\sqrt{2}^\infty \frac{u^2}{u^2+1}-\frac{u^2}{u^2-1}du$$ We can evaluate this integral in two pieces due to its linearity and compute the indefinite integrals.
$$\int \frac{u^2}{u^2+1}du=\int\frac{u^2 +1 -1}{u^2+1}du=u-\tan^{-1}(u)$$
$$\int \frac{u^2}{u^2-1}du=\int\frac{u^2 -1 +1}{u^2 -1}du=u+\frac{1}{2}\int \frac{1}{u-1}-\frac{1}{u+1}du=u+\frac{1}{2}\log\left(\frac{u-1}{u+1}\right)$$ Combining and taking the respective limits, we get $$-\log(9)\sqrt{2}-4\left(u-\tan^{-1}(u)-u-\frac{1}{2}\log\left(\frac{u-1}{u+1}\right)\right)\bigg|_\sqrt{2}^\infty\\=-\log(9)\sqrt{2}+2\pi-4\tan^{-1}(\sqrt{2})+2\log\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$$
|
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|
Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$
Compute $\sum_{k=1}^{25} (\frac{1}{k}-\frac{1}{k+4})$
I know that some of the terms will cancel each other. Have it been $k+1$ instead of $k+4$, I could have easily see the pattern in which the terms cancel each other. I don't know what would I do if it was say $k+6$ or bigger.
How can I identify the pattern in these type of problems?
|
You may write
$$\frac{1}{k} - \frac{1}{k+4} = \left( \frac{1}{k} - \frac{1}{k+1} \right) + \left( \frac{1}{k+1} - \frac{1}{k+2} \right)+ \left( \frac{1}{k+2} - \frac{1}{k+3} \right) + \left( \frac{1}{k+3} - \frac{1}{k+4} \right)$$
Now, apply telescoping and you get for the sum
$$\left( \frac{1}{1} - \frac{1}{26} \right) + \left( \frac{1}{2} - \frac{1}{27} \right)+ \left( \frac{1}{3} - \frac{1}{28} \right) + \left( \frac{1}{4} - \frac{1}{29} \right)$$
|
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|
Finding $\lim_{n\to\infty}\frac{\log(1^1 +2^2 +\cdots+ n^n)}{\sqrt{n^4 + 2n^3\log(n)}-\sqrt{n^4-n^3}}$ I have the limit and I tried to find the limit but I am stuck after few steps:
$\lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)}{\sqrt{n^4 + 2n^3\log(n)}-\sqrt{n^4-n^3}} $
$= \lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)(\sqrt{n^4 + 2n^3\log(n)}+\sqrt{n^4-n^3})}{n^4 + 2n^3\log(n)-n^4-n^3} $
$=\lim_{n\to\infty}\dfrac{n^2\log(1^1 +2^2 + \cdots + n^n)\left(\sqrt{1+\frac{ 2\log(n)}{n}}+\sqrt{1-\frac{1}{n}}\right)}{n^3 (2\log(n)-1)}$
$=\lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)\left(\sqrt{1+\frac{ 2\log(n)}{n}}+\sqrt{1-\frac{1}{n}}\right)}{n (2\log(n)-1)} \,\quad?
$
What can I do after to find the limit?
|
Try to use the inequality
$$
n^{n} < 1^{1} + 2^{2} + \cdots + n^{n} < n^{n} + n^{n} + \cdots + n^{n} = n^{n+1}
$$
and apply the squeeze theorem.
|
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|
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
|
Note that
$$x^4-2x^3+x-2 = x^3(x-2)+1(x-2)=(x-2)(x^3+1)=\color{red}{(x-2)(x+1)}(x^2-x+1).$$
Regarding the taking a factor out part, we use
$$ab+ac = a(b+c)$$
to factor $a$ out of the terms. In your case, it is $ab+ac+ad = a(b+c+d)$.
|
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|
Test $\sum_{n=1}^\infty [n^4 \sin^2 (\frac{2n}{3n^3 - 2n^2 + 5})]^n$ for convergence Any hints regarding this question would be appreciated - is the ratio test the best place to start?
|
Hint:
$$a_n^{1/n} = n^4 \sin^2 \bigg( \frac{2n}{3n^3-2n^2+5} \bigg) \sim n^4 \bigg(\frac{2n}{3n^3-2n^2+5} \bigg)^2 = \frac{4n^6}{(3n^3-2n^2+5)^2}$$
|
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|
Finding all real $x$ such that $1+\sum_{j=1}^n\sin{\frac{j\pi x}{n+1}} = 0$, where $n=18$.
The task is to find all $ x \in \mathbb R $ such that
$$ 1 + \sum_{j=1}^n \sin{\frac{j\pi x}{n + 1}} = 0, \qquad n = 18 $$
What I have tried
Using the following formulas:
$$1. \sin{x} + \sin{y} = 2\sin{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$2. \cos{x} + \cos{y} = 2\cos{\frac{x + y}{2}}\cos{\frac{x - y}{2}} $$
$$3. \sin{x}\sin{y} = \frac{1}{2} (\cos{(x - y)} - \cos{(x + y)}) $$
First approach
$$ 1 + \sin{\frac{\pi x}{19}} + \sin{\frac{18\pi x}{19}} + \sin{\frac{2 \pi x}{19}} + \sin{\frac{17\pi x}{19}} + ... + \sin{\frac{9 \pi x}{19}} + \sin{\frac{10 \pi x}{19}} = 0 $$
Using formula 1:
$$ 1 + 2\sin{\frac{\pi x}{2}} (\cos{\frac{17\pi x}{19}} + \cos{\frac{15\pi x}{19}} + ... + \cos{\frac{\pi x}{19}})$$
Skipping some steps, we get:
$$ 1 + 2\sin{\frac{\pi x}{2}}\cos{\frac{9\pi x}{38}}(2\cos{\frac{3\pi x}{38} + 1})(2\cos{\frac{\pi x}{38}} + 1) = 0 $$
From here I couldn't find a way to go further
Second approach
Multiply both sides by $ \sin{\frac{\pi x}{38}} $:
$$ \sin{\frac{\pi x}{38}} (1 + \sin{\frac{\pi x}{19}} + \sin{\frac{2\pi x}{19}} + \sin{\frac{3 \pi x}{19}} + ... + \sin{\frac{17 \pi x}{19}} + \sin{\frac{18 \pi x}{19}}) = 0 $$
Using formula number 3
$$ \sin{\frac{\pi x}{38}} + \frac{1}{2}(\cos{\frac{\pi x}{38}} - \cos{\frac{3\pi x}{38}} + \cos{\frac{3\pi x}{38}} - \cos{\frac{5\pi x}{38}} + ... + \cos{\frac{35\pi x}{38}} - \cos{\frac{37\pi x}{38}}) = 0$$
$$ \sin{\frac{\pi x}{38}} - \sin{\frac{\pi x}{2}}\sin{\frac{18\pi x}{38}} = 0 $$
From here I couldn't find any way to go further
Any hints will be appreciated!
EDIT:
The upper-most formula had a mistake, but now it is up to date.
PS
The problem is supposed to be solvable assuming you have high-school level math (i.e only simple trigonometric formulas are allowed, no complex numbers, no derivatives for the sine and cosine functions).
Also, this equation on Wolfram Alpha shows the following solution:
$ x = \frac{19}{2}(4n - 1), n \in \mathbb Z $
Update 2: Using @Doug M's answer, I managed to conclude that ($ \sin{\frac{\pi x}{19}} = 1 $ and $ \cos{\frac{\pi x}{19}} = 0 $), which equalates to $ \frac{\pi x}{19} = \frac{\pi}{2} + 2\pi n, n \in \mathbb Z $ is one of the solutions to the equation
|
You want to find $x$ s.t. $-(n+1) = \sum_{j=0}^{n} \sin (j\pi x)$.
If $x\not = 1$, then the RHS is the imaginary part of
$$
X = \sum_{j=0}^n (\cos(j\pi x) + i\sin(j\pi x))= \sum_{j=0}^n \left( \cos(\pi x) + i\sin(\pi x) \right)^j = \frac{1 - \left( \cos(\pi x) + i\sin(\pi x) \right)^{n+1}}{1 - \left( \cos(\pi x) + i\sin(\pi x) \right)}
$$
$$
X = \frac{1 - (\cos \pi + i \sin \pi)^{x(n+1)}}{1 - (\cos \pi + i \sin \pi)^x} = \frac{1 - (-1)^{x(n+1)}}{1 - (-1)^x}
$$
Its imaginary part is of course $0$, so it doesn't depend on $x$. Just check what happens for $x = 1$.
|
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Find the particular solution of the differential equation $\sqrt{x} + \sqrt{y}y' =0$, with $y(1) = 9$.
I get:
$\sqrt{y}dy = -\sqrt{x}dx \Rightarrow \frac{2}{3}y^{\frac{3}{2}} = -\frac{2}{3}x^{\frac{3}{2}}+C$.
What should I do from here?
|
Solve it by separation of variables because it's a separable differential equation:
$$
\sqrt{x} + \sqrt{y}y' =0\\
\sqrt{y}\frac{dy}{dx}=-\sqrt{x}\\
\int\sqrt{y}\frac{dy}{dx}\,dx=-\int\sqrt{x}\,dx\\
\int\sqrt{y}\,dy=-\int\sqrt{x}\,dx\\
\frac{2\sqrt{y^3}}{3}+C_1=-\frac{2\sqrt{x^3}}{3}+C_2\\
\sqrt{y^3}=-\sqrt{x^3}+C\\
y=\sqrt[3]{(C-\sqrt{x^3})^2}
$$
Now that we have found the general solution, we can use the initial condition $y(1)=9$ to find a particular solution:
$$
y(1)=9\implies \sqrt[3]{(C-\sqrt{1^3})^2}=9\implies C=28
$$
Your final answer is $y=\sqrt[3]{(28-\sqrt{x^3})^2}$.
|
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Let $x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}$; then the value of $(2x-1)^2$ equals... Let $$x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}};$$ then the value of $(2x-1)^2$ equals...
I don't how to start this question. Please help.
|
If you only want to know the value that the continued fraction converges to, you use a simple technique:
$$
x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}=1+\frac{1}{2+\frac{1}{x}}
$$
With some manipulation you could come up with the value of $x$, but you want $(2x-1)^2:$
$$
x\left(2+\frac{1}{x} \right)=\left(2+\frac{1}{x} \right)+1
$$
$$
2x+1=3+\frac{1}{x}
$$
Multiply everything by $x$, since we know it's not zero:
$$
2x^2-2x-1=0
$$
Complete the square that we want by multiplying by $2$:
$$
4x^2-4x-2=4x^2-4x+1-3=(2x-1)^2-3=0
$$
Hence the answer is 3.
|
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|
Find the value of expression $\frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt3 \sin250^{\circ}}$.
Find the value of expression $\frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt3 \sin250^{\circ}}$.
Though this question seems easy, I just seem to get stuck somewhere. This is what I have tried so far:
$$\frac{1}{\cos70^{\circ}}-\frac{1}{\sqrt3\sin70^{\circ}}$$
$$=\frac{\sqrt3\sin70^{\circ}-\cos70^{\circ}}{\sqrt3\sin70^{\circ}\cos70^{\circ}}$$
From here, I do not know how to continue. Pls help. Thank you :)
|
$$\frac{\sqrt{3}\sin 70^\circ-\cos 70^\circ }{\sqrt{3}\sin 70^\circ \cos 70^\circ}=\frac{\frac{\sqrt{3}}{2}\sin 70^\circ-\frac{1}{2}\cos 70^\circ }{\frac{\sqrt{3}}{2}\sin 70^\circ \cos 70^\circ}=$$
$$=\frac{\sin 70^\circ \cos 30^\circ -\sin 30^\circ\cos 70^\circ }{\frac{\sqrt{3}}{4}(2\sin 70^\circ \cos 70^\circ)}=\frac{\sin (70^\circ - 30^\circ)}{\frac{\sqrt{3}}{4}(\sin 140^\circ)}=\frac{\sin 40^\circ}{\frac{\sqrt{3}}{4}(\sin 40^\circ)}=\frac{4}{\sqrt{3}}.$$
|
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For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that? For three positive numbers verifying $xyz=2+x+y+z$, is there an upper bound on $x+y+z$? And, if so, which is that?
I've managed to find only a lower bound, from $ xyz=2+x+y+z \le \frac{(x+y+z)^3}{27} $, from which I got $x+y+z \ge 6 $
|
No, there is no upper bound. We solve for $x$ and get
$$ x = \frac{2+y+z}{yz - 1} $$
Pick $z=\frac{2}{y}$, then we get
$$ x = 2 + y + \frac{2}{y} $$
and hence for this choice
$$ x+y+z = 2 + 2y + \frac{4}{y} $$
Choosing $y$ large will get you a large number for $x+y+z$.
Added: To make it more fun. Let us consider the same question when $x,y,z\geq 2$. For that case we get indeed a bound. Wlog we may assume that $2\leq x\leq y \leq z$. Then we get
$$ y = \frac{2+x+ z}{xz -1} \leq \frac{2+2z}{2 z-1} \leq \frac{6}{3} =2 $$
and the way we get $x=2$. Thus, we get
$$ z = \frac{2+x+y}{xy -1} = \frac{6}{3} =2 $$
Thus, $(x,y,z)=(2,2,2)$ is the only solution and hence, $x+y+z=6$.
Let us now consider the case, where we only allow positive integer solutions. By the argument above, we only have to consider the case where $x=1$ otherwise we automatically get $x=y=z=2$. We narrow down the possibilites for $y$. We have
$$ y = \frac{2+x+z}{xz -1} = \frac{3+z}{z-1} =:f(z)$$
Note that $f$ is decreasing on $(1;\infty)$. We get
$$ f(2)=5, f(3)=3 $$
The first case cannot happen, as we assumed $y\leq z$. Hence, $y\in \{ 1, 2, 3\}$. Now we simply test which combinations solve our equation.
For $y=1$ we get
$$ z = 4 + z $$
which has no solution.
For $y=2$ we get
$$ 2z = 5 + z $$
hence, we get the valide solution $(1; 2; 5)$.
For $y=3$ we get
$$ 3z = 6 + z $$
which yields $z=3$ and hence another solution, namely $(1; 3; 3)$.
Therefore, we have that - up to permuting the entries - that we only have the solutions
$$ (2;2 ;2), \ (1; 2;5), \ (1;3; 3) $$
Thus, we get that $x+y+z\leq 8$ if we assume that $x,y,z$ are positive integers.
|
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|
Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Least common denominator is $ab$.
Multiplying each part to get the LCD in the denominator on both sides I get:
$$\frac{a}{b} \cdot \frac{a}{a} - \frac{b}{a} \cdot \frac{b}{b}$$
$$= \frac{a^2 - b^2}{ab}$$
Multiplying this expression by the reciprocal in the original problem:
$$\frac{a^2 - b^2}{ab} \cdot \frac{ab}{a+b}$$
$$= \frac{ab(a^2-b^2)}{ab(a+b)}$$
Cancel out common factor $ab$:
$$\frac{a^2 - b^2}{a + b}$$
Where did I go wrong and how can I arrive at $a - b$?
|
Not that $$\frac{a}{b}-\frac{b}{a}=\frac{(a-b)(a+b)}{ab}$$ so we get
$$\frac{(a-b)(a+b)ab}{(a+b)ab}=a-b$$
|
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|
Ramanujan's radical and how we define an infinite nested radical I know it is true that we have
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
The argument is to break the nested radical into something like
$$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
However, I am not convinced. I can do something like
$$4 = \sqrt{16}=\sqrt{1+2\sqrt{56.25}}=\sqrt{1+2\sqrt{1+3\sqrt{\frac{48841}{144}}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$$
Something must be wrong and the reason behind should be a misunderstanding of how we define infinite nested radical in the form of
$$ \sqrt{a_{0}+a_{1}\sqrt{a_{2}+a_{3}\sqrt{a_{4}+a_{5}\sqrt{a_{6}+\cdots}}}} $$
I researched for a while but all I could find was computation tricks but not a strict definition. Really need help here. Thanks.
|
$4=\sqrt{16}=\sqrt{1+3\sqrt{25}}=\sqrt{1+3\sqrt{1+4\sqrt{36}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{49}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}}}=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{81}}}}}}$
$=\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+\cdots}}}}}}}$
$5=\sqrt{25}=\sqrt{1+4\sqrt{36}}=\sqrt{1+4\sqrt{1+5\sqrt{49}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{64}}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{81}}}}}=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{100}}}}}}$
$=\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+8\sqrt{1+9\sqrt{1+\cdots}}}}}}}$
$\vdots$
$n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+(n+3)\sqrt{1+(n+4)\sqrt{1+\cdots}}}}}}}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3119631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
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|
Find envelope of $x \sin \theta - y \cos \theta + z = a \theta$, where $\theta$ is a parameter. I have tried to find envelope for
$$x \sin \theta - y \cos \theta + z = a \theta$$
First I find derivative w.r.t. $\theta$
$$F(\theta)=x \sin \theta - y \cos \theta + z - a \theta = 0$$
$$\frac{\partial F(\theta)}{\partial \theta}=y \sin \theta + x \cos \theta - a = 0$$
Then by solving these two above equation in order to eliminate parameter $\theta$, I find values of $\sin \theta$ and $\cos \theta$
$$\sin \theta = \frac{ax\theta + ay - xz}{x^2 + y^2}$$
and
$$\cos \theta = \frac{ax - ay\theta + yz}{x^2 + y^2}$$
Finally to find value of $\theta$ I squares and add above two equations then I get a quadratic equation in $\theta$ as
$$a^2\theta^2 - 2az\theta + a^2 + z^2 - x^2 - y^2=0$$
Then solving this for $\theta$ by applying quadratic formula, I get condition for real values of $\theta$ i.e.
$\theta$ is real only if
$$x^2+y^2 \ge a^2$$
Now here is my question. Either $x^2+y^2 \ge a^2$ is required envelope or something more to do?
Because $x^2+y^2 \ge a^2$ is an equality in which we have eliminated parameter $\theta$.
|
Hint:
You get rid of the annoying trigonometric functions by
$$(z-a\theta)^2+a^2=x^2+y^2.$$
From this you draw $\theta$, which you plug in one of the initial equations. Not a really nice expression.
Alternatively, in cylindrical coordinates $(\phi,\rho)$ we rewrite
$$\begin{cases}\rho\sin(\phi-\theta)=z-a\theta,\\\rho\cos(\phi-\theta)=a.\end{cases}$$
Then
$$\phi=\theta+\arccos\frac a\rho=\frac{z\pm\sqrt{\rho^2-a^2}}a+\arccos\frac a\rho.$$
|
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|
Proving the identity $\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$ by considering a more general form I came across this exercise:
Prove that
$$\tan x+2\tan2x+4\tan4x+8\cot8x=\cot x$$
Proving this seems tedious but doable, I think, by exploiting double angle identities several times, and presumably several terms on the left hand side would vanish or otherwise reduce to $\cot x$.
I started to wonder if the pattern holds, and several plots for the first few powers of $2$ seem to suggest so. I thought perhaps it would be easier to prove the more general statement:
For $n\in\{0,1,2,3,\ldots\}$, prove that
$$2^{n+1}\cot(2^{n+1}x)+\sum_{k=0}^n2^k\tan(2^kx)=\cot x$$
Presented this way, a proof by induction seems to be the smart way to do it.
Base case: Trivial, we have
$$\tan x+2\cot2x=\frac{\sin x}{\cos x}+\frac{2\cos2x}{\sin2x}=\frac{\cos^2x}{\sin x\cos x}=\cot x$$
Induction hypothesis: Assume that
$$2^{N+1}\cot(2^{N+1}x)+\sum_{k=0}^N2^k\tan(2^kx)=\cot x$$
Inductive step: For $n=N+1$, we have
$$\begin{align*}
2^{N+2}\cot(2^{N+2}x)+\sum_{k=0}^{N+1}2^k\tan(2^kx)&=2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)+\sum_{k=0}^N2^k\tan(2^kx)\\[1ex]
&=2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)-2^{N+1}\cot(2^{N+1}x)+\cot x
\end{align*}$$
To complete the proof, we need to show
$$2^{N+2}\cot(2^{N+2}x)+2^{N+1}\tan(2^{N+1}x)-2^{N+1}\cot(2^{N+1}x)=0$$
I noticed that if I ignore the common factor of $2^{N+1}$ and make the substitution $y=2^{N+1}x$, this reduces to the base case,
$$2^{N+1}\left(2\cot2y+\tan y-\cot y\right)=0$$
and this appears to complete the proof, and the original statement is true.
First question: Is the substitution a valid step in proving the identity?
Second question: Is there a nifty way to prove the special case for $n=2$?
|
Let $\displaystyle S =\sum^{2}_{k=0}2^k\tan(2^k x)+8\cot (8x).$
Then $\displaystyle \int Sdx=-\sum^{2}_{k=0}\ln\bigg(\cos(2^k x\bigg)+\ln(\sin 8x)$
Using $\displaystyle \prod^{n-1}_{r=0}\cos(2^r x)=\frac{1}{2^n}\frac{\sin(2^nx)}{\sin x}.$
$\displaystyle \int Sdx=-\ln\bigg(\frac{1}{2^3}\frac{\sin 8x}{\sin x}\bigg)+\ln(\sin 8x)$
$$\displaystyle \frac{d}{dx}\int Sdx=\frac{d}{dx}\bigg[\ln(\sin x)+\ln(8)\bigg]$$
$$S=\sum^{2}_{k=0}2^k\tan(2^k x)+8\cot (8x)=\cot x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...
|
Geometrically, the solution set is the set of points outside of the circle $x^2+y^2=1$ but inside the superellipse $x^4+y^4=1$. There are an infinite number of such points.
|
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|
Integral $\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$
Evaluate $$\int^{1}_{1/2}\frac{\ln(x)}{1-x}dx$$
Try: Let $$ I =\int^{1}_{1/2}\frac{\ln x}{1-x}dx=\int^{1}_{1/2}\sum^{\infty}_{k=0}x^k\ln(x)dx$$
$$I =\sum^{\infty}_{k=0}\int^{1}_{1/2}x^k\ln(x)dx$$
Using integration by parts, gives: $$I =\sum^{\infty}_{k=0}\frac{\ln(x)x^{k+1}}{k+1}\bigg|^{1}_{1/2}-\sum^{\infty}_{k=0}\frac{x^{k+1}}{(k+1)^2}\bigg|^{1}_{1/2}$$
Can someone help me to solve it?
|
Here we will address the integral:
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx
\end{equation}
We first observe that:
\begin{equation}
\frac{\partial}{\partial a} x^a = x^a \ln(x) \Longrightarrow \ln(x) = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a
\end{equation}
Thus,
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\ln(x)}{1 - x}\:dx = \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx
\end{equation}
Which by Leibniz's Integral Rule becomes:
\begin{equation}
I = \int_{\frac{1}{2}}^1 \frac{\lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} x^a }{1 - x}\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx
\end{equation}
We now employ the Taylor Series for $\frac{1}{1 - x}$ (which is convergent on the bounds of the integral):
\begin{equation}
\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^n
\end{equation}
Thus our integral becomes:
\begin{align}
I &= \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 \frac{x^a}{1 - x}\:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \int_{\frac{1}{2}}^1 x^a\sum_{n = 0}^{\infty} x^n \:dx = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \int_{\frac{1}{2}}^1 x^{a + n} \:dx \nonumber \\
&= \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{x^{a + n + 1}}{a + n + 1}\right]_{\frac{1}{2}}^1 = \lim_{a \rightarrow 0^+} \frac{\partial}{\partial a} \sum_{n = 0}^{\infty} \left[ \frac{1}{a + n + 1} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\
&= \lim_{a \rightarrow 0^+} \sum_{n = 0}^{\infty} \left[ \frac{-1}{\left(a + n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{a + n + 1}}{\left(a + n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{a + n + 1}}{a + n + 1}\right] \nonumber \\
&= \sum_{n = 0}^{\infty} \left[\frac{-1}{\left(n + 1\right)^2} - \frac{\left(\frac{1}{2}\right)^{n + 1}}{\left( n + 1\right)^2} + \frac{\ln\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{n + 1}}{n + 1}\right] \nonumber \\
&= -\sum_{n = 0}^{\infty} \frac{1}{(n + 1)^2} - \frac{1}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{(n + 1)^2} + \frac{\ln\left(\frac{1}{2}\right)}{2}\sum_{n = 0}^{\infty} \frac{\left(\frac{1}{2}\right)^n}{n + 1} \nonumber \\
&= -\Phi\left(1,1,2 \right) - \frac{1}{2}\Phi\left(\frac{1}{2},1,2 \right) + \ln(\sqrt{2})\Phi\left(\frac{1}{2},1,1 \right)
\end{align}
Where $\Phi(\cdot, \cdot, \cdot, \cdot)$ is the Lerch Transcedent Function.
|
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|
What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x+2(x-1)^2 \ln(1-x)+3}{(1-x)^2}$$
$$f'''(x) = \dfrac{2x^2-5x+1}{(x-1)^3}$$
$$f^{(n)}(x) = \ ?$$
|
As you noticed,
$$
f^{(3)}(x)=\frac{2x^{2}-5x+1}{\left(x-1\right)^{3}}.
$$
Wolfram claims that
$$
f^{(3+n)}(x)=n!\frac{n^{2}+n\left(x+2\right)-2x^{2}+5x-1}{\left(x-1\right)^{3+n}}.
$$
To convince yourself that this is indeed the case, fix $n\geq0$ and differentiate the expression for $f^{(3+n)}$ to get the expression for $f^{(3+n+1)}$.
|
{
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"url": "https://math.stackexchange.com/questions/3126890",
"timestamp": "2023-03-29T00:00:00",
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|
Taylor series, Convergence I need to find the radius of convergence (R) of the Taylor series expansion of the following function $f(x) =\frac{x^3 -2x+1}{x+7}$, $x_0=5$
The Taylor series of $f(x) = \frac{29}{3}+\frac{\frac{95}{18}}{1!}\left(x-5\right)+\frac{\frac{175}{108}}{2!}\left(x-5\right)^2+\frac{\frac{41}{432}}{3!}\left(x-5\right)^3+\frac{-\frac{41}{1296}}{4!}\left(x-5\right)^4+...$
But, in order to find the R, I need TS collect in the form of $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n$. From this, I can extract $a_n$ term and use it to get $R = \frac{1}{\lim_{n\to \infty} \lvert \frac{a_{n+1}}{a_n}\rvert}$
Does anyone has an idea how to write TS in this $\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n$ form?
Or are there any other possible ways to solve it?
Thanks
|
For simplicity, I let $x=y+5$ to make
$$\frac{x^3 -2x+1}{x+7}=\frac{y^3+15 y^2+73 y+116}{y+12}$$ and using the long division
$$\frac{y^3+15 y^2+73 y+116}{y+12}=\frac{29}{3}+\frac{95 y}{18}+\frac{175 y^2}{216}+\frac{41 y^3}{216 (y+12)}$$
Focusing on the last term and using the classical expansion, we then have
$$\frac{ y^3}{ (y+12)}=-\sum_{n=3}^\infty (-1)^n 12^{2-n} y^n$$
If you want to make life easier, , let, for the time being, $y=12 t$ to make
$$\frac{41 y^3}{216 (y+12)}=\frac{82 }{3 } \frac{ t^3}{ (t+1)}$$
|
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|
What is the value of $2x+3y$ if $x+y=7$ and $x^2-y^2=21$? If $x+y=7$ and $x^2-y^2=21$, then what is $2x+3y$?
I solved it like this:
\begin{align}
y& =7-x \\
x^2-(7-x)^2-21&=0 \\
x^2-49+14x-x^2-21&=0 \\
14x&=70 \\
x&=5
\end{align}
Then I solved for $y$ and I got $2$. I plugged in the values of $x$ and $y$ to $2x+3y$, and I got $16$.
Is this correct?
|
Quicker way is to note that
$$
21=(x-y)(x+y)=7(x-y)
$$
so
$$
x-y=3
$$
Combined with $x+y=7$, we get that $2x=10$ i.e. $x=5$ and $y=2$.
So your answer is correct.
|
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|
What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtain
\begin{equation}
x^2+(1-x^2)+(1+x)+(1-x)
>4 (v^2)^{\frac{1}{4}}
\implies \frac{9}{16} \ge v.
\end{equation}
Therefore the max value is $\frac{9}{16}$.
Method 2
$$
v=x (1-x^2)
\implies 2v^2= 2x^2 (1-x^2)^2
$$
With the AM-GM-inequality
$$2x^2+(1-x^2)+(1-x^2)
>3 (2v^2)^{\frac{1}{3}}
\implies \frac{2}{(27)^{\frac{1}{2}}} > v.
$$
|
Hint:
$$\dfrac{ax+b(1+x)+c(1-x)}3\ge ?$$
for $a,b,c>0$
Set $a+b-c=0\ \ \ \ (1)$
The equality will occur if $ax=b(x+1)=c(x-1)=k$(say)
Put the values of $a,b,c$ in$(1)$
|
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|
Evaluate $\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy$ [Background]: I'm trying to find the volume of the region bounded by the $xy$-plane, the cone $z^2=x^2+y^2$ and the cylinder $(x-1)^2+y^2=1$.
[Attempt]: I tried to use the polar coordinate:
\begin{align*}
\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy &= \int_0^{2\pi}\int_0^1 \sqrt{(1+r\cos{\theta})^2+(r\sin{\theta})^2}rdrd\theta\\
&= \int_0^{2\pi}\int_0^1 \sqrt{1+2r\cos\theta+r^2}r drd\theta
\end{align*}
Then I cannot continue. Could you give me a hint?
|
By using the usual polar coordinates $x=r\cos(\theta)$ and $y=r\sin(\theta)$ then the domain is
$$1\geq (x-1)^2+y^2=(r\cos(\theta)-1)^2+r^2\sin^2(\theta)\Leftrightarrow r\leq 2\cos(\theta),$$
and the integral becomes
$$\iint_{\{(x,y)\mid (x-1)^2+y^2\leq 1\}}\sqrt{x^2+y^2} dxdy=\int_{\theta=-\pi/2}^{\pi/2} \int_{r=0}^{2\cos(\theta)}r^2\,dr\,d\theta.$$
Can you take it from here?
|
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|
maximum value of $ab$ If $a,b\in R$ and $\displaystyle a^2+b^2=1+\frac{2ab}{a-b}$ and $\sqrt{a-b}=a^2+5b$, what is the maximum of $ab$?
Here is what I tried:
$a=r\cos\alpha$ and $b=r\sin \alpha$
$\displaystyle r^2=1+\frac{r\sin 2\alpha}{\cos \alpha-\sin \alpha}$
and $\sqrt{r(\cos \alpha-\sin \alpha)}=r^2\cos^2\alpha+5r\sin \alpha$
|
The domain gives $a-b>0$ and from the first equation we obtain:
$$(a-b)^2+2ab=1+\frac{2ab}{a-b}$$ or
$$(a-b)^2-1+2ab\left(1-\frac{1}{a-b}\right)=0$$ or
$$(a-b-1)\left(a-b+1+\frac{2ab}{a-b}\right)=0$$ or
$$a=b+1.$$
Can you end it now?
I got the following answer: $42$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$ Since I don't have the answer to this one, I want to make sure I've done this correctly.
Show that
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
Since
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
we have
$$\tan\left(\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)\right) = x$$
Applying the Pythagorean theorem, we learn that $b = 1$, so we have $\tan(x/1) = x$.
It may be a bit short, but I really want to be sure I have this right before I continue on. :)
|
$\tan \arcsin \frac{x}{\sqrt{x^2+1}} = \frac{\sin}{\cos} \arcsin \frac{x}{\sqrt{1+x^2}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}}$ and we know that $\cos^2 x+\sin^2 x=1 $ so $\cos x= \sqrt{1-\sin^2 x}$
so $ \frac{\frac{x}{\sqrt{x^2+1}}}{\cos \arcsin \frac{x}{\sqrt{x^2+1}}} = \frac{\frac{x}{\sqrt{x^2+1}}}{\sqrt{1-\sin^2 \arcsin \frac{x}{\sqrt{1+x^2}}}}=\frac{\frac{x}{\sqrt{x^2+1}}}{\sqrt{1-\frac{x^2}{1+x^2}}} = \frac{x}{\sqrt{1+x^2} \sqrt{\frac{1}{x^2+1}}} = x$
|
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|
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is odd
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv -1 \mod 8$
$7^n +1 \equiv 0 \mod 8$
Therefore, $7^n+1$ is divisible by $8$ if $n$ is odd.
*
*If $n$ is even
$7 \equiv -1 \mod 8$
$7^n \equiv (-1)^n \mod 8$
$7^n \equiv 1 \mod 8$
$7^n +1 \equiv 2 \mod 8$
Therefore, the remainder of the division of $7^n+1$ is $2$.
Is that true, please?
|
Just to be different:
$(a-1)(a^{n-1} + a^{n-2} + ..... + a + 1)=(a^n - 1)$.
$(a+1)(a^{n-1} -a^{n-2} + ........ \mp a \pm 1) = (a^n \pm 1)$.
with $n$ even yielding $(a+1)(a^{n-1} -a^{n-2} + ........ + a - 1)=a^n -1$
and $n$ odd yielding $(a+1)(a^{n-1} -a^{n-2} + ...... -a +1) = a^n+1$
So we $a = 7$ we get $7+1=8|a^n+1$ if $n$ is odd. And $7+1=8|a^n-1$ if $n$ is even. And as $a^n-1 = (a^n -1) + 2$, $a^n + 1$ will have remainder $2$ if $n$ is even.
But that's just to be different.
What you did was shorter and better. This just shows why it's no surprise.
|
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|
Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$ =
$\int \tan^{\frac{-5}{2}}\theta \sec\theta$
Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(
Update with more work after initial answers:
$\int \frac{\cos\theta}{\sin^2\theta}$ $u = \sin\theta, du = \cos\theta d\theta$
I found $\sin^{-1}\theta = \frac{\sqrt{x^2-1}}{x}$
$= \int \frac{du}{u^2} = \frac{1}{ \frac{1}{3}u^3} =
\frac{1}{3\sin^3\theta}
= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3$
|
If you do $x=\sec\theta\,\mathrm d\theta$ and $\mathrm dx=\sec(\theta)\tan(\theta)\,\mathrm d\theta$, what you get is$$\int\frac{\sec(\theta)\tan(\theta)}{\tan^3\theta}\,\mathrm d\theta=\int\frac{\cos^2\theta}{\sin^3\theta}\,\mathrm d\theta=\int\frac{\cos^2(\theta)\sin(\theta)}{\bigl(1-\cos^2(\theta)\bigr)^2}\,\mathrm d\theta.$$Now, do $\cos\theta=u$ and $\sin\theta\,\mathrm d\theta=\mathrm du$.
|
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|
minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$
How do I solve it ? Help me please
|
In
$$(2\cos\alpha+3\sin\alpha)\sin\beta+4\cos\beta$$ the parenthesised factor takes values in $[-\sqrt{2^2+3^2},\sqrt{2^2+3^2}]$. Using the largest value, the minimum of $$\sqrt13\sin\beta+4\cos\beta$$ is
$$-\sqrt{13+4^2}.$$
Justification:
$$a\cos t+b\sin t$$ is the scalar product of $(a,b)$ with the unit rotating vector $(\cos\theta,\sin\theta)$, which takes the extreme values $\pm\|(a,b)\|=\pm\sqrt{a^2+b^2}$.
We use this property twice.
|
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|
Computing $m-n$
When $P(x+4)$ is divided by $P(x)$, the remainder is $3x+m$. When
$P(x)$ is divided by $P(x+4)$, the remainder is $nx-6$. Compute $m-n$
Here I wrote down the equations as follows
$$P(x+4) = P(x)Q_1(x)+3x+m$$
$$P(x) = P(x+4)Q_2(x) + nx-6$$
However, these will not give me anything useful. Could you assist me with what I'm missing?
Regards
|
firstly, we write :
$m=P(x+4)-P(x)Q_1(x)-3x,$
$n=[P(x)-P(x+4)Q_2(x)+6]/x=(-3)[P(x)-P(x+4)Q_2(x)+6]/[m-P(x+4)+P(x)Q_1(x)]$
pick $n=\pm{3}$ by irreducible and improve above formula to :
$P(x)(Q_1(x)-1)+P(x+4)(Q_2(x)-1)=6-m,$$P(x)(Q_1(x)+1)-P(x+4)(Q_2(x)+1)=-6-m$
also note that:
$P(x)Q_1(x)-P(x)=P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6,$$P(x)Q_1(x)+P(x)=P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6$
then we can get:
$P(x+4)-3x-m-P(x+4)Q_2(x)-nx+6+P(x+4)(Q_2(x)-1)=6-m$
$P(x+4)-3x-m+P(x+4)Q_2(x)+nx-6-P(x+4)(Q_2(x)+1)=-6-m$
therefore, $m=\pm{6},n=\mp{3}$
|
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|
Evaluate integral $\int \sin^4(t)\cos^3(t)dt$ $$\int \sin^4(t)\cos^3(t)dt = \int \sin^4(t)(1-\sin^2(t))\cos(t) dt $$
$$u = \sin(t) \\ du = \cos(t)dt$$
$$ \int \sin^4(t)\cos^3(t)dt = \int u^4(1-u^2) du \\
= u^4 - u^6 = \frac{1}{5}u^5 - \frac{1}{7}u^7 + C \\
= \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) + C $$
This seems like a simple enough trig substitution integral problem to me. However, when I check my answer with wolphram alpha, it gives:
This looks like a simplified version of my answer, but it is not entirely clear to me how it gets reduced down. The furthest I can get is this:
$$ \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) = \sin^5(t) \bigg( \frac{1}{5} - \frac{1}{7} sin^2(t) \bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{1}{5} - \frac{1}{7} \cdot \frac{1}{2} (1 - \cos 2x) \bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{1}{5} - \frac{1}{14} - \frac{1}{14}\cos 2x\bigg) \\
= \frac{1}{5}\sin^5(t) \bigg(\frac{14}{90} - \frac{5}{90} - \frac{5}{90} \cos 2x \bigg)$$
and I feel like my simplification is not really going anywhere meaningful...
|
I think your answer is all right since $$\cos 2 \theta =1-2 \sin^2 \theta$$
You see : $$\dfrac{1}{5} \sin^5 (x )+ \dfrac{1}{7} \sin^7 (x) =\dfrac{1}{70} \sin^5( x) (14+10 \sin^2 (x)) =\dfrac{1}{70} (9+5 \cos (2x))$$
|
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|
An interesting indefinite integral Here's an interesting integral that I'm struggling with.
$$ \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$
MY EFFORTS:
I'm very very close in reaching a closed form.
\begin{align}
&= \frac12\int \frac{2\ dx}{\sqrt{\left(x^2+\dfrac{1}{x^2}\right) + 4\left(x+\dfrac{1}{x}\right)-6}} \\
&= \frac12\int \frac{\left(1 - \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x+\dfrac{1}{x}\right)^2 + 4\left(x+\dfrac{1}{x}\right)-8}} + \frac12\int \frac{\left(1 + \dfrac{1}{x^2}\right)\ dx}{\sqrt{\left(x-\dfrac{1}{x}\right)^2 + 4\sqrt{\left(x-\dfrac{1}{x}\right)^2+4}-4}} \\
&= \frac12 \int \frac{du}{\sqrt{u^2+4u-8}} + \underbrace{\frac12 \int \frac{dt}{\sqrt{t^2 + 4\sqrt{t^2+4}-4}}}_{\text{How to solve this?}}
\end{align}
As you all can see, I'm able to break integral by 2 parts and I'm struck in evaluating 2nd part. The first part can easily be solved by completing the square and trigonometric substitutions.
NOTE: In my effort pic. the variable of integration is x and should not be confused with letter n.
([1]: https://i.stack.imgur.com/k2Vhu.jpg)
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I don't think that your effort leads to a simpler integral that the original one. They are two radicals in your last integral. This makes it more difficult to integrate. So, you are not close to solve it.
HINT :
$$x^4+4x^3-6x^2+x^4+1= (x-r_1)(x-r_2)(x-r_3)(x-r_4)$$
$r_1=\frac12\left(1+\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$
$r_2=\frac12\left(1-\sqrt{2}+\sqrt{7-4\sqrt{2}} \right)$
$r_3=\frac12\left(1+\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$
$r_4=\frac12\left(1-\sqrt{2}-\sqrt{7-4\sqrt{2}} \right)$
$$I= \int \frac{x\ dx}{\sqrt{x^4 + 4x^3 - 6x^2 + 4x + 1}} $$
It is easier to integrate on this form :
$$I= \int \frac{x\ dx}{\sqrt{(x-r_1)(x-r_2)(x-r_3)(x-r_4)}} $$
The result involves an elliptic integral as a term of a complicated formula. From WolframAlpha :
https://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt((x-r_1)(x-r_2)(x-r_3)(x-r_4))
NOTE :
If the integral comes from an academic exercise, there is probably a typo in it, because solving it requires an hight level of knowledge of special functions. I will not loose time with a problem suspected of mistake in the wording.
Please check your calculus which leads to the integral. Eventually edit the original version of the problem in your question.
|
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|
Let $A=[1 2 0 1 ]$. Find all $2×2$ matrices B, $B≠O_2$ and $B≠I_2$ such that $AB=BA$.
Let $A=[1 2 0 1 ]$. Find all $2×2$ matrices B, $B≠O_2$ and $B≠I_2$ such that $AB=BA$
Explain your answer.
I know two ways to find B
A) As det of A is 1, it is invertible. So A = B (inverse)
B) Or by assigning a,b,c,d as four elements of B matrix and then solving for AB=BA
But the question says to find all 2×2 matrices B. Is there any way i can find all possible matrices for B such that AB=BA?
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So $$A = \begin{pmatrix}1&2 \\ 0&1 \end{pmatrix}$$
If $$B = \begin{pmatrix}a&b \\ c&d \end{pmatrix}$$
From $$\begin{pmatrix}1&2 \\ 0&1 \end{pmatrix} \begin{pmatrix}a&b \\ c&d \end{pmatrix} = \begin{pmatrix}a&b \\ c&d \end{pmatrix} \begin{pmatrix}1&2 \\ 0&1 \end{pmatrix}$$
we have $$ a+2c = a\implies c=0$$
and $$b+2d = 2a+b \implies a=d$$
so $$B = \begin{pmatrix}a&b \\ 0&a \end{pmatrix}$$
Now you have two more equations, but from them you do not get any new information so we are done.
|
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Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin x }{x}$$
I am getting somewhere from here, but I don't know how to continue. Please help!
The answer is 1
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From the comments, I knew the solution to this question. So, I am just writing it out as an answer.
For the case where $x \ne 0$,
$$(x+\frac{1}{2})^2 + \frac{7}{4} = - \frac{\sin x}{x}$$
Since, LHS is at least $\frac{7}{4}$
From the comment of Eric Yau, $0<\frac{\sin x}{x}<1$ and hence, $-1 < \frac{\sin x}{x} < 0$
Hence for $ x \ne 0 $, there is no solution for x.
However, clearly by substituting 0, $x=0$ is definitely a root of this equation.
Hence, $x=0$ is the only solution for this equation and thus, there is only one root for this equation.
|
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Solve for $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\frac{\sqrt 3}2$ Find out the value of $x$ in $x\sqrt{1-4x^2}+2x\sqrt{1-x^2}=\dfrac{\sqrt 3}2$.
I tried squaring both the sides but it only makes it more complicated. Is there any other way?
|
Put $2x = \cos(\alpha), x = \cos(\beta) \implies \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha) = \dfrac{\sqrt{3}}{2}\implies \sin(\alpha+\beta) = \dfrac{\sqrt{3}}{2}\implies \alpha+\beta=\dfrac{\pi}{3}, \dfrac{2\pi}{3}\implies \beta = \dfrac{\pi}{3} - \alpha\implies x = \cos(\dfrac{\pi}{3} -\alpha)= \dfrac{1}{2}(2x)+\dfrac{\sqrt{3}}{2}\sin(\alpha)\implies \sin(\alpha) = 0 \implies \cos(\alpha) = 1\implies x = \dfrac{1}{2}$. The other case that $\alpha+\beta = \dfrac{2\pi}{3}$, I leave it for you to finish.
|
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How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follows:
$f(p) = \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$
where $p$ is a prime number and $[...]$ are Iverson brackets.
Here are the first few values of $f(p)$:
$f(3) = \frac{1}{(3^2 - 1)} \times \frac {(3^2-3)}{3}$
$f(3) = \frac{1}{8} \times \frac{6}{3}$
$f(3) = 0.25$
$f(5) = \frac{1}{(5^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} ]$
$f(5) = \frac{1}{24} \times [ \frac{6}{3} +\frac{22}{5} ]$
$f(5) = 0.26667$
$f(7) = \frac{1}{(7^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}]$
$f(7) = \frac{1}{48} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}]$
$f(7) = 0.273214286$
$f(11) = \frac{1}{(11^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}+\frac{(11^2-3)}{11}]$
$f(11) = \frac{1}{120} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}+\frac{118}{11}]$
$f(11) = 0.198679654$
How to prove that $f(p)$ is always less than 0.3 for all prime numbers $p$?
I wrote a program to calculate $f(p)$ for the first $1000$ primes up to $p = 7,927$ and graphically it appears to approach $0$ with a maximum value of $0.2732$ at $p=7$. From the graph, it looks like it should be easy, but for some reason, I cannot figure it out.
Notice that $f(3) < f(5) < f(7)$, but $f(7) > f(11)$ so a proof by induction may not work.
|
$$
\begin{align}
\sum_{q=3}^p\frac{q^2-3}{q}
&=\int_{3^-}^{p^+}\frac{x^2-3}{x}\,\mathrm{d}\pi(x)\\
&=\left[\frac{p^2-3}{p}\pi(p)-2\right]-\int_{3^-}^{p^+}\pi(x)\left(1+\frac3{x^2}\right)\mathrm{d}x\\[3pt]
&=\frac{p^2}{2\log(p)}+\frac{p^2}{4\log(p)^2}+\frac{p^2}{4\log(p)^3}+O\!\left(\frac{p^2}{\log(p)^4}\right)
\end{align}
$$
Thus, we have
$$
\frac1{p^2-1}\sum_{q=3}^p\frac{q^2-3}{q}\sim\frac1{2\log(p)}+\frac1{4\log(p)^2}+\frac1{4\log(p)^3}
$$
whose plot looks like
|
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Prove that $\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$ Prove that $$\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$$
I have tried to solve this on my own, and I want to check my solution.
My steps:
Set $x=\left\lfloor \frac {n+3}{2}\right\rfloor $
then, for an integer x
$$ \frac {n+3}{2}=x+\epsilon , 0 \le \epsilon \lt 1$$
$$ n=2x+2 \epsilon-3 $$
We substitute in the right hand side and get
$$ \left\lceil \frac {2x+2\epsilon -1}{2} \right\rceil = \left\lceil x + \epsilon - \frac {1}{2}\right\rceil $$
Now, sine x is an integer we can write the last statement as $$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil $$
Now , I played around with the inequality. I have subtracted a half from all the sides so I got
$$ -\frac {1}{2} \le \epsilon -\frac {1}{2} \lt \frac {1}{2} $$
I have drawn the number line and found that the inequality turns to
$$ -1 \lt \epsilon -\frac {1}{2} \le 0 $$
(Since we are dealing with integers)
Now, we see that $\left\lceil \epsilon - \frac {1}{2}\right\rceil = 0$
So, we proved that$$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil = x $$
Which the same as the left hand side.
Is my procedure correct??
This is the solution that I have found
|
If $n$ is an integer, then $\epsilon$ is either $0$ or $\frac12$ and hence
$$ \bigg\lceil\epsilon-\frac12\bigg\rceil=0. $$
Otherwise, the identity doesn't hold. For example, $n=4.3$ and then
$$ \bigg\lfloor\frac{4.3+3}{2}\bigg\rfloor=\lfloor3.65\rfloor=3, \bigg\lceil\frac{4.3+2}{2}\bigg\rceil=\lceil3.25\rceil=4. $$
|
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Prove $x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$ Let $x,y,z>0$ and $n\in\mathbb{N}$, prove that:
$$x^n(x-y)(x-z)+y^n(y-x)(y-z)+z^n(z-x)(z-y)>0$$
Some idea? I have tried to develop the multiplication and certain assumptions buy I can't see it clearly. Note that this is not true if $x=y=z$ or $x=y=z=0$.
|
Well. It's NOT true. If $x = y = z > 0$ then $x^n(x-y)(x-z) + y^n(y-x)(y-z) + z^n(z-x)(z-y) = 0$.
So you can not prove it.
But maybe we can prove it is we stipulate that at least two of the terms are distinct.
Note that we can label the three variables whatever we want and by symmetry we might as well, and we can assume without loss of generality that $0 < x \le y \le z$. (we can just relabel the variables so that is true.)
Case 1: $x=y < z$
Then $x^n(x-z)(x-y) + y^n(y-z)(y-x) + z^n(z-y)(z-x) = 0+0 + z^n(z-y)^2 > 0$ as $(z-y)^2 > 0, z^n > 0$.
Case 2: $x < y=z$
Then $x^n(x-y)(x-z) + y^n(y-z)(y-x) + z^n(z-x)(z-y) = x^n(x-y)^2 + 0 + 0> 0$ as $x^n, (x-y)^2 > 0$.
Case 3: $x < y < z$.
Then $x^n(x-y)(x-z) > 0$ and $y^n(y-z)(y-x) < 0$ and $z^n(z-x)(z-y) > 0$.
So we need to prove that $x^n(y-x)(z-x) + z^n(z-x)(z-y) > y^n(z-y)(y-x)$. (Note: all terms are positive.)
That will be true if and only if $\frac {x^n}{z-y} + \frac {z^2}{y-x} > \frac {y^n}{z-x}$
So it if we can prove: $\frac {x^n}{z-y} + \frac {z^2}{y-x} > \frac {y^n}{z-x}$ we will be done.
$z-y < z-x$ and $y-x < z-x$ so
$\frac {x^n}{z-y} + \frac {z^n}{y-x} >$
$ \frac {x^n}{z-x} + \frac{z^n}{z-x} > $
$\frac {z^n}{z-x} >$
$\frac {y^n}{z-x}$.
And that's it.
|
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Minimum value of PA+PB+AB If $P(2,1) $ and $A $ and $B $ lie on $x$ axis and $y=x $ respectively,
then find the minimum value of $PA+PB+AB $ .
If $A$ was given , I could have worked geometrically, by using that image of $P$ and point $A$ and $B$ should be collinear.
But here vary two things , $A$ and $B$....
|
Another way.
By Minkowski we obtain:
$$PA+PB+AB=\sqrt{(x-2)^2+1}+\sqrt{(y-x)^2+y^2}+\sqrt{(1-y)^2+(2-y)^2}\geq$$
$$\geq\sqrt{(x-2+y-x+1-y)^2+(1+y+2-y)^2}=\sqrt{10}.$$
The equality occurs for $A\left(\frac{5}{3},0\right)$ and $B\left(\frac{5}{4},\frac{5}{4}\right),$ which says that $\sqrt{10}$ is a minimal value.
|
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Trouble calculating probability Machine generates one random integer in range ${[0;40)}$ on every spin.
You should choose 5 numbers in that range.
Then the machine will spit out 5 numbers (numbers are independent of each other).
what is the probability that you will get exactly two numbers correct?
My logic:
You should get two of them right. chance of that is: $r = { \left( 1 \over 40 \right)^ 2 }$
You should get 3 of wrong. Chance of that is: $w = { \left( 39 \over 40 \right)^3 }$
As order doesn't matter answer should be: $$ans = { rw \over 2!3!}$$
Simulator tells me I'm wrong. Where is my logic flawed?
P.s. Machine can spit out duplicates
|
If the machine is choosing with replacement, but the person is choosing five distinct numbers, and if success means two distinct numbers the person chose are picked by the machine, then the problem is much more difficult:
Probability machine chooses five distinct numbers:
$$\dfrac{40!}{35!40^5}$$
Probability machine chooses four distinct numbers (choose the four distinct numbers the machine selected, choose the one number that is repeated twice, permute the multiset to find all possible orders these numbers could be chosen, divide by the total number of ways to choose five numbers):
$$\dfrac{\dbinom{40}{4}\dbinom{4}{1}\dfrac{5!}{1!1!1!2!}}{40^5}$$
Probability machine chooses three distinct numbers (choose the three distinct numbers the machine selected, either one number is repeated three times or two numbers are repeated twice each):
$$\dfrac{\dbinom{40}{3}\dbinom{3}{1}\left(\dfrac{5!}{3!1!1!}+\dfrac{5!}{2!2!1!}\right)}{40^5}$$
Probability machine chooses two distinct numbers (choose the two distinct numbers the machine selected. You can have one number four times and the other once or one number three times and the other twice.):
$$\dfrac{\dbinom{40}{2}\dbinom{2}{1}\left(\dfrac{5!}{4!1!}+\dfrac{5!}{3!2!}\right)}{40^5}$$
So, the probability of you matching exactly two numbers:
$$\begin{align*} & \dfrac{40!}{35!40^5}\cdot \dfrac{\dbinom{5}{2}\dbinom{35}{3}}{\dbinom{40}{5}}\\ + & \dfrac{\dbinom{40}{4}\dbinom{4}{1}\dfrac{5!}{2!}}{40^5}\cdot \dfrac{\dbinom{4}{2}\dbinom{36}{3}}{\dbinom{40}{5}} \\ + & \dfrac{\dbinom{40}{3}\dbinom{3}{1}\left(\dfrac{5!}{3!}+\dfrac{5!}{(2!)^2}\right)}{40^5}\cdot \dfrac{\dbinom{3}{2}\dbinom{37}{3}}{\dbinom{40}{5}} \\ + & \dfrac{\dbinom{40}{2}\dbinom{2}{1}\left(\dfrac{5!}{4!}+\dbinom{5}{3}\right)}{40^5}\cdot \dfrac{\dbinom{2}{2}\dbinom{38}{3}}{\dbinom{40}{5}} \approx 0.09116015625\end{align*}$$
Note: It is possible for the machine to choose one distinct number, but there is a zero probability that you wind up with two matching numbers, so I ignored that case. To show that my numbers work out though, you can test that the following holds:
$$\dfrac{40!}{35!}+\dbinom{40}{4}\dbinom{4}{1}\dfrac{5!}{2!1!1!1!}+\dbinom{40}{3}\dbinom{3}{1}\left(\dfrac{5!}{3!1!1!}+\dfrac{5!}{2!2!1!}\right)+\dbinom{40}{2}\dbinom{2}{1}\left(\dfrac{5!}{4!1!}+\dfrac{5!}{3!2!}\right)+\dbinom{40}{1} = 40^5$$
(I verified this is true myself using Wolframalpha).
|
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|
Finding intercalates within a (reduced) latin square I need confirmation on if my intuition on finding intercalates is correct, suppose we have the following reduced latin square
\begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 4 & 1 & 3 \\
3 & 1 & 4 & 2\\
4 & 3 & 2 & 1
\end{bmatrix}
we know from a theorem that for a latin square of order n(even), there is at most $\frac{n^2(n-1)}{4}$ intercalates; in this case 12.
I am not sure if two 2 x 2 sub matrixes can be considered the same intercalate if they are flipped and rotated versions of each other. From the square above we get: (e.g R1 = row 1)
R1,R2
\begin{bmatrix}
1 & 2 & | 3&4\\
2 & 4 & |1&3
\end{bmatrix}
R1,R3
\begin{bmatrix}
1 & 2 & | 3&4\\
3 & 1 & |4&2
\end{bmatrix}
R1,R4\begin{bmatrix}
1 & 2 & | 3&4\\
4 & 3 & |2&1
\end{bmatrix}
R2,R3
\begin{bmatrix}
2 & 4 & | 1&3\\
3 & 1 & |4&2
\end{bmatrix}
R2,R4
\begin{bmatrix}
2 & 4 & | 1&3\\
4 & 3 & |2&1
\end{bmatrix}
R3,R4
\begin{bmatrix}
3 & 1 & | 4&2\\
4 & 3 & |2&1
\end{bmatrix}
And only keeping ones that are not flips and rotations of each other (I keep 1 copy). These should be the only intercalates right? So 6 of them.
\begin{bmatrix}
1&2&|3&4&|&3&4&|&1&2&|1&2&|&2&4&\\
2&4&|1&3&|&4&2&|&3&1&|4&3&|&3&1&
\end{bmatrix}
|
None of these are intercalates. As defined in this paper (for example), we're looking for two rows $i,j$ and two columns $x,y$ such that $L_{i,x} = L_{j,y}$ and $L_{j,x} = L_{i,y}$. There are four of these in your example:
*
*$\{i,j\} =\{1,4\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}1 & 4 \\ 4 & 1\end{bmatrix}$$
*$\{i,j\} =\{2,3\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}4 & 1 \\ 1 & 4\end{bmatrix}$$
*$\{i,j\} =\{1,4\}$ and $\{x,y\} = \{2,3\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$
*$\{i,j\} =\{2,3\}$ and $\{x,y\} = \{1,4\}$: $$\begin{bmatrix}2 & 3 \\ 3 & 2\end{bmatrix}$$
|
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|
A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money can be expressed as an irreducible fraction p/q.
What I did was--
WLOG let us assume the correct number be 1. Total number of outcomes can be calculated this way
(0,1)(0,2)(0,3)(0,4)(0,5)(0,6)(0,7)(0,8)(0,9)
(1)
(2,0)(2,1)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)
(3,0)(3,1)(3,2)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)
(4,0)(4,1)(4,2)(4,3)(4,5)(4,6)(4,7)(4,8)(4,9)
(5,0)(5,1)(5,2)(5,3)(5,4)(5,6)(5,7)(5,8)(5,9)
(6,0)(5,1)(5,3)(5,4)(5,6)(5,7)(5,7)(5,8)(5,9)
(7,0)(7,1)(7,2)(7,3)(7,4)(7,5)(7,6)(7,8)(7,9)
(8,0)(8,1)(8,2)(8,3)(8,4)(8,5)(8,6)(8,7)(8,9)
(9,0)(9,1)(9,2)(9,3)(9,4)(9,5)(9,6)(9,7)(9,8)
So total number of outcomes is 82 and total number of favorable outcomes is 10 so my answer is 10/82 or 5/41. But it is not correct. Please help me understand this.....
|
Alex tries two numbers. Then the probability that the correct number belongs to this pair is simply
$$\frac{2}{10} = 20 \%$$
|
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|
Finding the value of $x-y$. Given two equations \begin{align} & x^4+y^4=\dfrac {-7}{9} \\ & x^3-y^3=3. \end{align} From these equations find the value of $(x-y).$
I have just factorize $x^3-y^3=(x-y)(x^2+xy+y^2)=3$. But don't know how to proceed from here. One way to find the values of $x$ and $y$ and then compute possible values of $x-y$. But that is a too long way. I think there is any simple way to find the value of $x-y$. Please help me to solve this.
|
Here is a hint to get you started:
$$\frac{-7}9= (x^2+y^2)^2-2(xy)^2$$$$3=(x-y)(x^2+y^2+xy)=(x-y)^3+3(xy)(x-y)$$ If we substitute $a=x^2+y^2$ and $b=xy$ and $c=x-y$, we get $$\frac{-7}9=a^2-2b^2$$$$3=c\cdot(a+b)=c^3+3bc$$
Now, you have three equations and three variables. Can you figure it out?
|
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|
Can $a(n) = \frac{n}{n+1}$ be written recursively? Take the sequence $$\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \dots$$
Algebraically it can be written as $$a(n) = \frac{n}{n + 1}$$
Can you write this as a recursive function as well?
A pattern I have noticed:
*
*Take $A_{n-1}$ and then inverse it. All you have to do is add two to the denominator. However, it is the denominator increase that causes a problem here.
I am currently in Algebra II Honors and learning sequences
|
\begin{align*}
a_{n+1} &= \frac{n+1}{n+2} \\
&= \frac{n+2-1}{n+2} \\
&= 1 - \frac{1}{n+2} \text{, so } \\
1 - a_{n+1} &= \frac{1}{n+2} \text{, } \\
\frac{1}{1 - a_{n+1}} &= n+2 &[\text{and so } \frac{1}{1 - a_n} = n+1]\\
&= n+1+1 \\
&= \frac{1}{1- a_n} +1 \\
&= \frac{1}{1- a_n} + \frac{1-a_n}{1-a_n} \\
&= \frac{2-a_n}{1- a_n} \text{, then } \\
1 - a_{n+1} &= \frac{1-a_n}{2- a_n} \text{, and finally } \\
a_{n+1} &= 1 - \frac{1-a_n}{2- a_n} \\
&= \frac{2-a_n}{2- a_n} - \frac{1-a_n}{2- a_n} \\
&= \frac{1}{2- a_n} \text{.}
\end{align*}
|
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|
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
|
Construct the Groebner basis of your system, with the variables ordered $x$, $y$:
$$ \mathrm{GB}(\{3x+2y-36, 5x+4y-64\}) = \{y-6, x-8\} $$
and read out the solution. (If we reverse the variable order, we get the same basis, but in reversed order.) Under the hood, this is performing Gaussian elimination for this problem. However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.
Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$. A sequence of reductions (similar to the Euclidean algorithm for GCDs): \begin{align*}
(5,4,-64) - (3,2,-36) &= (2,2,-28) \\
(3,2,-36) - (2,2,-28) &= (1,0,-8) \tag{1} \\
(2,2,-28) - 2(1,0,-8) &= (0,2,-12) \tag{2} \\
\end{align*}
From (1), we have $x=8$. From (2), $2y = 12$, so $y = 6$. (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components. One implementation of the LLL algorithm, terminates with the shorter vectors $\{(-1,2,4), (-2,2,4)\}$, but we would continue to manipulate these to get the desired zeroes.)
|
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|
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two terms and performing subtraction we will get it) so is $nt_n$. Now can we prove it is monotonically decreasing? If so then $\lim_{n \to \infty} nt_n=\lim_{n \to \infty}[(\frac{1}{2+1/n}-\frac{1}{2+2/n})+(\frac{1}{2+3/n}-\frac{1}{2+4/n})+\cdots +(\frac{1}{4-1/n}-\frac{1}{4})]$ and each of these terms will go to zero so is the limit.
Second attempt: I was trying to use Riemann summation $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}=
\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}+\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}-2[\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots \frac{1}{4n}]\Rightarrow \lim \frac 1n [nt_n]=\int_0^2\frac{dx}{2+x}-\int_0^1\frac{dx}{1+x}=\ln4-\ln 2-\ln 2=0$
So $\lim t_n=0$
So what will happen with $\lim nt_n$
Edit: As I got the answer is not $0$ because of the flaw. So can we have different approaches even with Riemann Sum to have the answer?
|
Another approach
$$t_n=\sum_{k=1}^n\left(\frac{1}{2n+2k-1}-\frac1{2n+2k}\right)=\sum_{k=1}^n\frac1{(2n+2k-1)(2n+2k)}$$
and then
$$nt_n=\frac1{n}\sum_{k=1}^n\frac1{(2+(2k-1)/n)(2+2k/n)}$$
which is a Riemann sum for
$$\int_0^1\frac{dx}{(2+2x)^2}=\frac18.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
|
You can use
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)$,
$(ab+ac+bc)^2=a^2+b^2+c^2+2(ab+ac+bc)$,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,
$(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$
|
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|
Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.
Step by Step procedure to get the answer.
Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.
The distance d between the two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$ is given by the formula
$$
d= \sqrt{ (x_2-x_1)^2 + (y_2 - y_1)^2 }
$$
So we get the distance $AB = 2 \sqrt{2}$
. Similarly we have to find the distance between BC & AC using distance formula where we get the equation in the form of x & y even though the distance is $2\sqrt{2}$ because it is equilateral triangle. But the correct solution is not able to get. Please help.
|
Distance between two given points will be the side of the triangle $a=\sqrt{(1+\sqrt{3})^2+(\sqrt{3}-1)^2}=2\sqrt{2}$. Now you just need to draw two circles of radius $a$ and with centers at the known vertices. The equations of the circles are:
$(x-1)^2+(y+1)^2=8$
$(x+\sqrt{3})^2+(y+\sqrt{3})^2=8$
The circles will intersect at two points and to find the coordinates you need to solve this system:$$ \begin{cases}
(x-1)^2+(y+1)^2=(x+\sqrt{3})^2+(y+\sqrt{3})^2 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
y(2-2\sqrt{3})=x(2+2\sqrt{3})+4 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
y=\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}} \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$Finally, we get a quadratic equation for $x$: $$x^2-2x+1+\Big(\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}}+1\Big)^2=8 $$ The solutions are: $x=-1, x=2-\sqrt{3}$. We already have $y$ expressed via $x$ so just plug the $x$ values in that equation and find $y$.
|
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|
Find the minimum of $\space\frac{1}{x}+\frac{1}{y}+c\cdot xy\space$ subject to $\space x+y-c=0$ Let $f(x,y):\mathbb{D}\rightarrow\mathbb{R}$ be the function:
$$f(x,y)=\frac{1}{x}+\frac{1}{y}+c\cdot xy\space\space|\space\space c\in(0,\sqrt[4]8)\text{ $\space$constant}$$
$$\mathbb{D}=\{(x,y)\space|\space x>0,\space y>0\}$$
Find the point $P$ where $f$ gets its minimum value, subject to the equality:
$$x+y-c=0$$
I tried:
1) Lagrange multipliers. Unfortunately, they don't seem to help since I get equations I cannot solve (5th degree).
2) Substituting $y=c-x$ to $f$ in order to solve $\frac{d}{dx}f(x,c-x)=0$. That wasn't helpful either, from the same reasons.
Final Solution: The final solution should be $(\frac{c}{2},\frac{c}{2})$; However I could not figure out how to find it by myself.
Thanks!
|
$\begin{array}\\
f(x,y)
&=\dfrac{1}{x}+\dfrac{1}{y}+c xy\\
&=\dfrac{c}{x(c-x)}+c x(c-x)\\
&=\dfrac{c+cx^2(c-x)^2}{x(c-x)}\\
&=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\
f_x(x,y)
&= \dfrac{c^3 x^2 - 4 c^2 x^3 + 5 c x^4 - c - 2 x^5 + 2 x}{x^2 (c - x)^2}
\quad\text{according to Wolfy}\\
&= \dfrac{(c - 2 x) (c x - x^2 - 1) (c x - x^2 + 1)}{x^2 (c - x)^2}
\quad\text{also according to Wolfy}\\
\end{array}
$
so
$f$ is extreme at
$x=\frac{c}{2}$
or
$x^2-cx =\pm 1$
or
$x^2-cx+c^2/4 =c^2/4\pm 1$
or
$(x-c/2)^2
=c^2/4\pm 1$.
If
$(x-c/2)^2
=c^2/4 -1
$
then,
since
$0 \le c \le \sqrt[4]{8}
$,
$0 \le c^2 \le \sqrt{8}
=2\sqrt{2}
$
so
$c^2/4
\le \dfrac{\sqrt{2}}{2}
\lt 1$
so this can not be.
If
$(x-c/2)^2
=c^2/4 +1
$
then
$x
=\dfrac{c}{2}\pm\sqrt{\dfrac{c^2}{4} +1}
$.
Since $x > 0$
we must have
$x
=\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1}
$.
If
$x = \dfrac{c}{2}$,
$x = y$
so
$x(c-x)
=\dfrac{c^2}{4}
$
so
$\begin{array}\\
f(x, y)
&=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\
&=c\dfrac{1+c^4/16}{c^2/4}\\
&=\dfrac{4(1+c^4/16)}{c}\\
&=\dfrac{4}{c}+\dfrac{c^3}{4}\\
\end{array}
$
If
$x
=\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1}
$,
$y
=c-x
=\dfrac{c}{2}-\sqrt{\dfrac{c^2}{4} +1}
\lt 0
$
which is not allowed.
Therefore
the extreme value is at
$x=y=\dfrac{c}{2}$
where the value is
$\dfrac{4}{c}+\dfrac{c^3}{4}
$.
|
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|
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{1-2y}\cdot\frac{1}{(1-\frac{1}{4y^2})^2}=$$ $$=-\frac{1}{2}\sum(2y)^n\sum(\frac{1}{4y^2})^n$$That is why I have $$f^{(3)}=3!\frac{-1}{2}\cdot2\cdot\frac{1}{4}=\frac{1}{4}$$Why is it not correct solution?
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$\begin{array}\\
f(x)
&=\dfrac{(1+x)^4}{(1+2x)^3(1-2x)^2}\\
&=\dfrac{(1+x)^4(1-2x)}{(1+2x)^3(1-2x)^3}\\
&=\dfrac{(1+x)^4(1-2x)}{(1-4x^2)^3}\\
&=(1+4x+6x^2+4x^3+x^4)(1-2x)(1-4x^2)^{-3}\\
&=(1+2x-2x^2+...)\sum_{k=0}^{\infty} \binom{k+2}{2})(4x^2)^k
\qquad\text{generalized binomial theorem}\\
&=(1+2x-2x^2+...)(1+3(4x^2)+6(4x^2)^2+...)\\
&=(1+2x-2x^2+...)(1+12x^2+96x^4+...)\\
&=1+2x+10x^2+...\\
\end{array}
$
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"url": "https://math.stackexchange.com/questions/3188040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1 - 6n^3 -6n -24}{2n^3+2n+8}\right| < \left|\frac{-n -25}{2n^3+2n+8}\right| = \left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right|$$
Since $|-x| = |x|$:
$$\left|-\left(\frac{n +25}{2n^3+2n+8}\right)\right| = \left|\frac{n +25}{2n^3+2n+8}\right|<\varepsilon $$
This is the part which I have trouble with. Here, I always end up with my minimum-required index to be smaller than zero, which seems peculiar to me:
$$\left|\frac{n +25}{2n^3+2n+8}\right| < |n+25| = n+ 25 < \varepsilon = 1/2$$
From here, I will get that my $N$ will be less than zero, which means that all the elements of the sequence are inside of my given epsilon environment, but I know that's not true since $a_1 = 5/6 < 3 - \varepsilon $, so what I did do wrong? Is it because I completely removed the denominator? If so, why does that break the inequality?
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Note that for $n\geq 1$ we have $n+25\leq n+25n=26n$ and $2n^3+2n+8\geq 2n^3$ so you have that $$\left|\frac{n+25}{2n^3+2n+8}\right|\leq \frac{26n}{2n^3}=\frac{13}{n^2}$$
Can you take it from here?
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
System of quadratic equations with three variables (generic form) Try solve a system of equation like this one.
\begin{cases}
(O_x -A_x)^2+(O_y-A_y)^2+(O_z-Az)^2=x^2 \\
(O_x -B_x)^2+(O_y-B_y)^2+(O_z-Bz)^2=y^2 \\
(O_x -C_x)^2+(O_y-C_y)^2+(O_z-Cz)^2=z^2
\end{cases}
Is there any way to express $O_x, O_y, O_z$ in terms of $A_x, A_y, A_z, B_x, B_y, B_z,C_x,C_y,C_z, x, y, z$?
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OK lets check this out I'll replace $x \ y \ z$ with $a\ b\ c$ to prevent name collision with function independent variables.
\begin{cases}
(O_x -A_x)^2+(O_y-A_y)^2+(O_z-A_z)^2=a^2 \ ①\\
(O_x -B_x)^2+(O_y-B_y)^2+(O_z-B_z)^2=b^2 \ ②\\
(O_x -C_x)^2+(O_y-C_y)^2+(O_z-C_z)^2=c^2 \ ③
\end{cases}
$④=①-②$ and $⑤=②-③$ we get
\begin{cases}
2(A_x-B_x)O_x+2(A_y-B_y)O_y+2(A_z-B_z)O_z-A_x^2-A_y^2-A_z^2+B_x^2+B_y^2+B_z^2+a^2-b^2=0;\ ④ \\
2(B_y-C_y)O_y+2(B_z-C_z)O_z+2(B_x-C_x)O_x-B_y^2-B_z^2-B_x^2+C_y^2+C_z^2+C_x^2+b^2-c^2=0;\ ⑤
\end{cases}
eliminate $O_y$ in $④$ and $O_z$ in $⑤$
\begin{cases}
O_y=\frac{(B_z-C_z)(A_x^2+A_y^2+A_z^2-a^2)/2+(C_z-A_z)(B_x^2+B_y^2+B_z^2-b^2)/2+(A_z-B_z)(C_x^2+C_y^2+C_z^2-c^2)/2-(A_x(B_z-C_z)+B_x(C_z-A_z)+C_x(A_z-B_z))*O_x}{A_y(B_z-C_z)+B_y(C_z-A_z)+C_y(A_z-B_z)} \\ \\
O_z=\frac{(A_y-B_y)(C_z^2+C_x^2+C_y^2-c^2)/2+(B_y-C_y)(A_z^2+A_x^2+A_y^2-a^2)/2+(C_y-A_y)(B_z^2+B_x^2+B_y^2-b^2)/2-(C_x(A_y-B_y)+A_x*(B_y-C_y)+B_x(C_y-A_y))O_x}{C_z(A_y-B_y)+A_z(B_y-C_y)+B_z(C_y-A_y)}
\end{cases}
name
$A_\sharp =(A_x^2+A_y^2+A_z^2-a^2)/2\quad B_\sharp =(B_x^2+B_y^2+B_z^2-b^2)/2\quad
C_\sharp =(C_x^2+C_y^2+C_z^2-c^2)/2\\
DET_{xy} =\begin{vmatrix}A_x&B_x&C_x\\1&1&1\\A_y& B_y&C_y\end{vmatrix}\quad
DET_{xz} =\begin{vmatrix}A_x&B_x&C_x\\1&1&1\\A_z& B_z&C_z\end{vmatrix}\quad
DET_{yz} =\begin{vmatrix}A_y&B_y&C_y\\1&1&1\\A_z& B_z&C_z\end{vmatrix}\\
DET_{\sharp x} =\begin{vmatrix}A_\sharp&B_\sharp&C_\sharp\\1&1&1\\A_x& B_x&C_x\end{vmatrix}\quad
DET_{\sharp y} =\begin{vmatrix}A_\sharp&B_\sharp&C_\sharp\\1&1&1\\A_y& B_y&C_y\end{vmatrix}\quad
DET_{\sharp z} =\begin{vmatrix}A_\sharp&B_\sharp&C_\sharp\\1&1&1\\A_z& B_z&C_z\end{vmatrix}\quad$
$$ \bbox[,border:2px solid red]
{
\begin{aligned}
&\qquad \text{ An interesting property to notice is}\\
& DET_{xz}DET_{\sharp y}-DET_{xy}DET_{\sharp z}=DET_{yz}DET{\sharp x}\\
&\qquad \quad also \quad DET_{xy}=-DET_{yx}
\end{aligned}
}$$
renaming $O_y$ and $O_z$ to
\begin{cases}
O_y=\frac{(B_z-C_z)*A_\sharp+(C_z-A_z)*B_\sharp+(A_z-B_z)*C_\sharp+DET_{xz}*O_x}{-DET_{yz}} \\ \\
O_z=\frac{(A_y-B_y)*C_\sharp+(B_y-C_y)*A_\sharp-(C_y-A_y)*B_\sharp+DET_{xy}*O_x}{DET_{yz}}
\end{cases}
and further simplifies to
\begin{cases}
O_y=\frac{\ DET_{\sharp z}-DET_{xz}*O_x}{DET_{yz}} \quad ⑥\\
O_z=\frac{-DET_{\sharp y}+DET_{xy}*O_x}{DET_{yz}} \quad ⑦
\end{cases}
Substitute $O_x$ and $O_y$ into $①$
$O_x^2-2A_xX+A_x^2-a^2
+\bigl[(\frac{DET_{\sharp z}-DET_{xz}O_x}{DET_{yz}})^2-2(\frac{DET_{\sharp z}-DET_{xz}O_x}{DET_{yz}})A_y+A_y^2\bigr]
+\bigl[(\frac{-DET_{\sharp y}+DET_{xy}O_x}{DET_{yz}})^2-2(\frac{-DET_{\sharp y}+DET_{xy}O_x}{DET_{yz}})A_z+A_z^2\bigr]=0$
and further reduce to
$(DET_{xy}^2+DET_{yz}^2+DET_{xz}^2)
-2(DET_{yz}^2A_x+DET_{xz}(DET_{\sharp z}-DET_{yz}Ay)+DET_{xy} (DET_{\sharp y}+DET_{yz}A_z))O_x
+DET_{yz}^2(A_x^2-a^2) +(DET_{\sharp z}-DET_{yz}Ay)^2+(DET_{\sharp y-DET_{yz}Az})^2=0$
then we can solve $O_x$ using "Quadratic formula"
$
Ox=\frac{DET_{yz}^2A_x+DET_{xz}(DET_{\sharp z}-DET_{yz}A_y)+DET_{xy}(DET_{\sharp y}+DET_{yz}A_z)
\pm \sqrt{
-DET_{yz}^2(DET_{\sharp z}-DET_{yz}A_y)(DET_{\sharp z}-DET_{yz}A_y-2DET_{xz}A_x)
-DET_{yz}^2(DET_{\sharp y}+DET_{yz}A_z)(DET_{\sharp y}+DET_{yz}A_z-2DET_{xy}A_x)
-(DET_{xz}(DET_{\sharp y}+DET_{yz}A_z)-DET_{xy}(DET_{\sharp z}-DET_{yz}A_y))^2
-DET_{yz}^2(DET_{xz}^2+DET_{xy}^2)A_x^2
+DET_{yz}^2(DET_{xy}^2+DET_{yz}^2+DET_{xz}^2)a^2}}
{DET_{xy}^2+DET_{yz}^2+DET_{xz}^2} $
substitude $O_x$ into $⑥\ ⑦$ we get $O_y\ O_z$
or interestingly because $O_x O_y O_z$ are in symmetrical position all you have to do is to swap all $x \leftrightarrow y$ or $x \leftrightarrow z$ to get O_y and O_z
$
Oy=\frac{DET_{xz}^2A_y+DET_{yz}(DET_{\sharp z}-DET_{xz}A_x)+DET_{yx}(DET_{\sharp x}+DET_{xz}A_z)
\pm \sqrt{
-DET_{xz}^2(DET_{\sharp z}-DET_{xz}A_x)(DET_{\sharp z}-DET_{xz}A_x-2DET_{yz}A_y)
-DET_{xz}^2(DET_{\sharp x}+DET_{xz}A_z)(DET_{\sharp x}+DET_{xz}A_z-2DET_{yx}A_y)
-(DET_{yz}(DET_{\sharp x}+DET_{xz}A_z)-DET_{yx}(DET_{\sharp z}-DET_{xz}A_x))^2
-DET_{xz}^2(DET_{yz}^2+DET_{yx}^2)A_y^2
+DET_{xz}^2(DET_{yx}^2+DET_{xz}^2+DET_{yz}^2)a^2}}
{DET_{yx}^2+DET_{xz}^2+DET_{yz}^2} $
$
Oz=\frac{DET_{yx}^2A_z+DET_{zx}(DET_{\sharp x}-DET_{yx}A_y)+DET_{zy}(DET_{\sharp y}+DET_{yx}A_x)
\pm \sqrt{
-DET_{yx}^2(DET_{\sharp x}-DET_{yx}A_y)(DET_{\sharp x}-DET_{yx}A_y-2DET_{zx}A_z)
-DET_{yx}^2(DET_{\sharp y}+DET_{yx}A_x)(DET_{\sharp y}+DET_{yx}A_x-2DET_{zy}A_z)
-(DET_{zx}(DET_{\sharp y}+DET_{yx}A_x)-DET_{zy}(DET_{\sharp x}-DET_{yx}A_y))^2
-DET_{yx}^2(DET_{zx}^2+DET_{zy}^2)A_z^2
+DET_{yx}^2(DET_{zy}^2+DET_{yx}^2+DET_{zx}^2)a^2}}
{DET_{zy}^2+DET_{yx}^2+DET_{zx}^2} $
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to show that $ab+bc+ca\le \frac34$ Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$
I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ Using AM-GM in $a+b,b+c$ & $c+a$, I get $a+b+c\ge \frac32$. Any hint will be thankful.
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$$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's
$$\sum_{cyc}(a^2b+a^2c-2abc)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(a^2b+a^2c-2abc)=\sum_{cyc}(a^2c+b^2c-2abc)\geq\sum_{cyc}(2\sqrt{a^2c\cdot b^2c}-2abc)=0.$$
Thus, it's enough to prove that
$$(a+b+c)^2\geq3(ab+ac+bc),$$ which is true by AM-GM again:
$$(a+b+c)^2=\sum_{cyc}(a^2+2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2+4ab)\geq\frac{1}{2}\sum_{cyc}(2ab+4ab)=3\sum_{cyc}ab.$$
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{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Evaluating $\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx$ $$\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx$$
I started like this:
$$\int_{0}^{\frac{\pi}{4}}\dfrac{x\sin(x)}{\cos^{3}(x)}\,dx=\int_{0}^{\frac{\pi}{4}}\dfrac{1}{\cos^{2}(x)}\cdot \dfrac{\sin(x)}{\cos(x)}\cdot (x)\,dx=\int_{0}^{\frac{\pi}{4}}(x\cdot \tan(x))\cdot (\tan'(x)) \,dx$$
How to continue ?
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An alternative solution.
Observe that:
\begin{align}\dfrac{\partial}{\partial x}\left(\frac{1}{2\cos^2 x}\right)=\frac{\sin x}{\cos^3 x}\end{align}
Therefore,
\begin{align}\int_{0}^{\frac{\pi}{4}}\frac{x\sin(x)}{\cos^{3}(x)}\,dx&=\left[\frac{x}{2\cos^2 x}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{2}\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos^2 x}\,dx\\
&=\frac{\pi}{4}-\frac{1}{2}\Big[\tan x\Big]_{0}^{\frac{\pi}{4}}\\
&=\boxed{\frac{\pi-2}{4}}
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3202482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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|
Cant find an eigenvector for an eigenvalue For the matrix
$$
\begin{pmatrix}
1 & 2 & 3 \\
0 & 3 & 4 \\
0 & 0 & 5 \\
\end{pmatrix}
$$
I know that $5, 2+\sqrt3, 2-\sqrt3$ are eigenvalues. I am trying to find an eigenvector for $2+\sqrt3$ using $(A-\lambda I)V=0$. But this gives me:
$$
\begin{pmatrix}
-1-\sqrt3 & 2 & 3 \\
0 & 1-\sqrt3 & 4 \\
0 & 0 & 3-\sqrt3 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}=\begin{pmatrix}
0 \\
0 \\
0 \\
\end{pmatrix}
$$
Which implies $x=y=z=0$. But this isnt possible as an eigenvector cannot be a $0$ vector.
What am I doing wrong?
NOTE: Thank you all, I see it now.
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Indeed, as Jose has rightly pointed out, the eigenvalues are the diagonal entries of a triangular matrix:
$$ \det(A - \lambda I) = 0$$
$$ \begin{vmatrix} 1-\lambda & 2 & 3 \\ 0 & 3-\lambda & 4 \\ 0 & 0 & 5- \lambda \end{vmatrix} = 0$$
$$ (1-\lambda) \begin{vmatrix} 3-\lambda & 4 \\ 0 & 5-\lambda \end{vmatrix} - 2\begin{vmatrix} 0 & 4 \\ 0 & 5-\lambda \end{vmatrix} + 3\begin{vmatrix} 0 & 3 - \lambda \\ 0 & 0 \end{vmatrix} = 0 $$
$$ (1 - \lambda)(3 - \lambda)(5 - \lambda) = 0$$
Its one reason I like linear algebra-the very neat and mesmerising results.
|
{
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"url": "https://math.stackexchange.com/questions/3203807",
"timestamp": "2023-03-29T00:00:00",
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|
Find $\sum_{i=10}^n (\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10})$=? Evaluate:
$$\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right)$$
I tried this and my result come:
$$n\binom{30}{10}\binom{20}{1}+ (n-1)\binom{30}{11}\binom{20}{2} + (n-2)\binom{30}{12}\binom{20}{3} + ...... +(n-(n-1))\binom{30}{n-1}\binom{20}{(n-1)-10} + (n-(n))\binom{30}{n}\binom{20}{n-10}$$
I'm not getting what to do further.
|
Note
$$
\sum_{i=10}^n \left(\sum_{r=10}^i \binom{30}{r}\binom{20}{r-10}\right) = \\
\sum_{i=10}^n \left(\sum_{k=0}^{i-10 } \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{n-10} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right)= \\
\sum_{j=0}^{20} \left(\sum_{k=0}^{j} \binom{30}{k+10}\binom{20}{k}\right) = \\
\sum_{j=0}^{20} \left(\sum_{k=0}^{j} \binom{30}{20-k}\binom{20}{20-k}\right)
$$
Observe that you can rearrange $\sum_{j=0}^{20}\sum_{k=0}^{j} = \sum_{k=0}^{20} \sum_{j=k}^{20}$ which can be used:
$$
\cdots =
\sum_{k=0}^{20} \left(\sum_{j=k}^{20} \binom{30}{20-k}\binom{20}{20-k}\right) = \\
\sum_{k=0}^{20} (20-k) \binom{30}{20-k}\binom{20}{20-k} = \\
30 \sum_{k=0}^{19} \binom{29}{19-k}\binom{20}{20-k} = \\
30 \cdot 18851684897584 = 565550546927520
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $x = 2$ is a root of $\det\left[\begin{smallmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{smallmatrix}\right]=0$, find other two roots
If $x = 2$ is a root of equation
$$ \begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0 $$
Then find the other two roots.
I solved it and got a cubic equation, and then I divided it by $(x-2)$ to get the other two roots. But this is a long method to do.
Please help me with some shorter approach to this question.
|
$$
\begin{align*}
\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} & =
\begin{vmatrix}
x-2 & 3x-6 & 2-x \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 3 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix}
\\
& = (x-2)
\begin{vmatrix}
1 & 0 & 0 \\
2 & -3x-6 & x-1\\
-3 & 2x+9 & x-1
\end{vmatrix}
\\
& = (x-2)(x-1)
\begin{vmatrix}
1 & 0 & 0 \\
2 & -3x-6 & 1\\
-3 & 2x+9 & 1
\end{vmatrix}
\\
& = (x-2)(x-1)
\begin{vmatrix}
1 & 0 & 0 \\
5 & -5x-15 & 0\\
-3 & 2x+9 & 1
\end{vmatrix}
\\
& = (x-2)(x-1)(-5x-15) \\
& = -5(x-2)(x-1)(x+3) \\
\end{align*}
$$
Therefore, the other roots are $1$ and $-3$.
|
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"url": "https://math.stackexchange.com/questions/3209154",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.
|
Hint:
$\frac{\tan x- \sin x}{x^3}=\frac{1}{ \cos x} \cdot \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2}.$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$x^3-8x^2+30x-20=0$ has roots $a$, $b$, $c$. Find the equation with roots $a+2$, $b+2$, $c+2$ I have the equation $$x^3-8x^2+30x-20=0$$ let's call the roots $a,b,c$
It's easy to find the equation with roots $(a+2)(b+2)(c+2)$
these are the steps in my books
so the equation is $$ aX^3+ bX^2 + cX + D = 0 $$
$$ x^3 -14x^2 +74x -120 = 0$$
my problem is, I can't understand why divide by $-2$ not $2$
how do I do this division with long division? like what am I dividing by if the symbols were here?!
|
Alternatively: for the roots $a,b,c$ of $x^3-8x^2+30x-20=0$, by Vieta's we have:
$$a+b+c=8\\
ab+bc+ca=30\\
abc=20$$
Now we want the roots $a+2,b+2,c+2$ of $X^3+kX^2+lX+m=0$. Hence:
$$a+2+b+2+c+2=a+b+c+6=14=-k\\
(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2)=\\
(ab+bc+ca)+4(a+b+c)+12=74=l\\
(a+2)(b+2)(c+2)=\\
abc+2(ab+bc+ca)+4(a+b+c)+8=120=-m \Rightarrow \\
x^3-14x^2+74x-120=0.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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}
|
Find $1 \le a < b \le n$ such that $a\cdot b + a + b = n\cdot (n+1)/2$
Find $1 \le a < b \le n$ such that $$a\cdot b + a + b = \frac{n\cdot (n+1)}2$$
Is there a more efficient way than picking $a$ or $b$ and trying all values between $1$ and $n$ ?
|
$(a+1)(b+1)=\dfrac{n^2+n+2}{2}$.
If $a$ and $b$ are restricted to integers, consider the factors of $\dfrac{n^2+n+2}{2}$.
|
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|
Having the identity: $\int_0^1 \frac{x^{2n}}{1+x^2} dx= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}$, how do I square the fraction inside the sum? I have the integral:
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} dx &= \int_0^1 x^{2n} \sum_{k=0}^{\infty}(-1)^kx^{2k} dx\\
&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}
\end{align}
For $n=0$ we have $\pi/4$. So, for some positive integer $n$, the integral gives something in terms of $\pi/4$.
I wish the fraction on right hand to have a squared denominator inside the sum, like this:
$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2n+2k+1)^2} $$
Because for a positive integer $n$ it gives something in terms of the Catalan constant $(1-1/3^2+1/5^2-\cdots)$.
one way to do this is the following (due to Beukers):
\begin{align}
\int_0^1 \frac{x^{2n+y}}{1+x^2} dx &= \int_0^1x^{2n+y} \sum_{k=0}^{\infty}(-1)^k x^{2k} dx\\
&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1+y}
\end{align}
Differentiate inside the integral with respect to $y$:
\begin{align}
& \int_0^1 \frac{\partial}{\partial y} \frac{x^{2n+y}}{1+x^2} dx
= \sum_{k=0}^{\infty} \frac{\partial}{\partial y} \frac{(-1)^k}{n+k+1+y} \\
& \int_0^1 \frac{x^{2n+y}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1+y)^2}
\end{align}
for $y=0$
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1)^2}
\end{align}
The problem is the $\ln(x)$ because it makes the integral way too larger than before. If we chose some polynomial $(P_{2n}(x))$ with integer coefficients and even powers:
$$\int_0^1 P_{2n}(x)\ln (x)/(1+x^2)dx \gg \int_0^1 P_{2n}(x)/(1+x^2)dx$$
for some integer $n$. By the sign $\gg$ I mean that the integral is very "sensitivity" for values of $n$, even if the difference between the integrals are $0.01$ this already makes a big difference.
Does anyone have another idea?
|
HINT: Manipulate this identity involving the Digamma Function $$\sum_{k=0}^\infty \frac{(-1)^k}{zk+1}= \frac{1}{2z}\left( \psi_0(\frac{1}{2}+\frac{1}{2z})-\psi_0(\frac{1}{2})\right)$$
to get $$\frac{1}{4}\left(\psi_0(\frac{n}{2}+\frac{3}{4})-\psi_0(\frac{n}{2}+\frac{1}{4})\right)$$ for the value of the integral.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p-1)(p+1) $$
We know that (for example from here) that this is dividable by $2$ and by $3$ so by $6$
Let consider $5$ cases:
$$\exists_k p=5k \rightarrow \mbox{false because p is prime}$$
$$\exists_k p=5k+1 \rightarrow p^2 - 1 = 5k(5k+2) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
$$\exists_k p=5k+2 \rightarrow p^2 - 1 = (5k+1)(5k+3) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+3 \rightarrow p^2 - 1 = (5k+2)(5k+4) \rightarrow \mbox{ ??? }$$
$$\exists_k p=5k+4 \rightarrow p^2 - 1 = (5k+3)(5k+5) \rightarrow \mbox{ dividable by 6 and by 5 so we have rest 0}$$
I have stucked with $???$ cases...
|
$\ 5< p$ prime $\,\Rightarrow\, \begin{align} p&\equiv\pm1\ \ \ \ \ \pmod{\!6}\\ p&\equiv \pm1,\pm2\!\!\!\pmod{\!5}\end{align}$
$\,\Rightarrow\,\begin{align}\color{#0a0}{p^2}&\: \color{#0a0}{\equiv\, 1 \pmod{\!6}}\\ p^2&\equiv \color{#c00}{\pm1}\!\!\!\!\pmod{\!5}\end{align}\!$
$\overset{\rm CRT}\iff p^2\equiv 1,19\pmod{\!30}$
because $\ p^2\!-\!1\bmod 30 = 6\left[\color{#0a0}{\dfrac{p^2\!-\!1}{6}}\bmod 5\right] = 6\left[\dfrac{\color{#c00}{0,3}}{1}\bmod 5\right] = 0,18.\ \ \ \small\rm QED $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing to volume of ball intersection with cone and another ball I am trying to calculate the integral $\ \int\int\int_V z dV $ where $\ V $ is the volume inside the ball $\ x^2 + y^2 + (z-2)^2 = 4 $ and also inside the cone $\ z^2 = x^2 + y^2 $ and outside the ball $\ x^2 + y^2 + z^2 = 4 $ .
So my attempt it to set the limits of $\ z $ integral from the lower ball to the higher ball. so $\ z $ should start at $\ \sqrt{4-x^2-y^2} $ and ends at $\ \sqrt{4-x^2-y^2} + 2 $ and the project of the intersection of cone with the ball is a circle with radius of $\ \sqrt{2} $ and so $ x $ will go from $\ - \sqrt{2-y^2} $ to $ \sqrt{2-y^2} $ and $\ y $ from $\ -\sqrt{2} $ to $ \sqrt{2} $
I don't know how to convert though. What should be $\ \theta, \phi , r $ ? trying to use the formula the outcome doesn't seem reasonable.
Thanks
|
$x = rsin\theta cos\phi$
$y = rsin\theta sin\phi$
$z = rcos\theta$
$x^2+y^2+z^2 = r^2$
$dxdydz = r^2sin\theta \ dr \ d\theta \ d\phi$
Case1:
$x^2+y^2+z^2 = r^2 = 4$
$$r = 2$$
Case2:
$x^2+y^2+(z-2)^2 = x^2+y^2+z^2 - 4z +4 = 4$
$r^2 - 4rcos\theta = 0$
$$r = 4cos\theta$$
(refer the graph for $r\not=0$)
$\theta$ varies from $0$ to $\pi/4$
and
$\phi$ varies from $0$ to $2\pi$
Solution:
$ I = \int^{2\pi}_{\phi = 0} \ \int^{\pi/4}_{\theta = 0}\ \int^{4\cos\theta}_{r=2} r^3\cos\theta \ \sin\theta dr \ d\theta \ d\phi$
$I = [\int^{2\pi}_{\phi = 0}d\phi]\ \int^{\pi/4}_{\theta = 0}\bigg[\frac{r^4}{4}\bigg]^{4cos\theta}_2 sin\theta cos\theta d\theta$
$I = [2\pi].\frac{1}{4}\bigg[\int^{\pi/4}_{\theta = 0}\big[4^4cos^4\theta.cos\theta.sin\theta d\theta] + \int^{\pi/4}_{\theta = 0}\big[(-2^4)sin\theta.cos\theta.d\theta\big]\bigg]$
$I = \frac{\pi}{2}\bigg[\int^{\pi/4}_{\theta = 0}\big[4^4cos^5\theta d(- cos\theta)] + \int^{\pi/4}_{\theta = 0}\big[(-2^3)sin2\theta d\theta\big]\bigg]$
$I = \frac{\pi}{2}\bigg[\big[\frac{- 4^4cos^6\theta}{6}]^{\pi/4}_0 + 2^3\big[\frac{\cos 2\theta}{2}\big]^{\pi/4}_0\bigg]$
$I = \frac{\pi}{2}\bigg[\frac{-128}{3}(\frac{1}{8} -1) + 4(0 -1)\bigg]$
$I = \frac{\pi}{2}\big[\frac{112}{3} - 4 \big]$
$I = \frac{\pi}{2}(\frac{100}{3})$
Thus, $$I = \frac{50\pi}{3} $$
Proof of the value:
$x = rsin\theta cos\phi$
$y = rsin\theta sin\phi$
$z = rcos\theta$
Computing partial derivatives,
$x_r = sin\theta cos\phi \ , x_{\theta} = rcos\theta cos\phi \ , x_{\phi} = -rsin\theta sin \phi$
$y_r = sin\theta sin\phi \ , y_{\theta} = rcos\theta sin\phi \ , y_{\phi} = rsin\theta cos \phi$
$z_r = cos\theta \ , z_{\theta} = -rsin\theta , z_{\phi} = 0$
$J = \
\begin{vmatrix}
sin\theta cos\phi & rcos\theta cos\phi & -rsin\theta sin \phi \\
sin\theta sin\phi & rcos\theta sin\phi & rsin\theta cos \phi \\
cos\theta & -rsin\theta & 0
\end{vmatrix} $
Expanding along third row,
$J = cos\theta[r^2cos\theta sin\theta cos^2\phi \ + r^2cos\theta sin\theta sin^2\phi] + rsin\theta[rsin^2\theta cos^2\phi + rsin^2\theta sin^2\phi]$
$J = r^2sin\theta[cos^2\theta(cos^2\phi+sin^2\phi)] + r^2sin\theta[sin^2\theta(cos^2\phi+sin^2\phi)]$
$J = r^2sin\theta[cos^2\theta+sin^2\theta] = r^2sin\theta$
$$dxdydz = |J|dr d\theta d\phi = r^2sin\theta dr d\theta d\phi$$
|
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|
Solving differential equation with power series $y'' - xy' - y = 0$ I need a work check
$$y'' -xy' - y = 0$$ given $y(0) = 1$ and $y'(0) = 0$
so we have:
$$
\begin{split}
y &= \sum_{n=0}^\infty C_nx^n\\
y' &= \sum_{n=1}^\infty nC_nx^{n-1}\\
y' &= \sum_{n=2}^\infty n(n-1)C_nx^{n-2}
\end{split}
$$
so making a substitution into the original differential equation:
$$y' = \sum_{n=2}^\infty n(n-1)C_nx^{n-2} - \sum_{n=0}^\infty nC_nx^{n-1} - \sum_{n=0}^\infty C_nx^{n} = 0$$
$$y' = \sum_{n=0}^\infty (n+2)(n+1)C_{n+2}x^{n} - \sum_{n=0}^\infty (n+1)C_nx^n = 0$$
$$C_{n+2} = \frac{C_n}{n+2}$$
so first a few terms:
$C_0 = C_0$ and $C_1 = C_1$ and $C_2 = \frac{C_0}{2}$ and $C_3 = \frac{C_1}{3}$ and $C_4 = \frac{C_2}{4} = \frac{C_0}{4 \cdot 2}$ and $c_5 = \frac{c_3}{5} = \frac{C_1}{5 \cdot 3}$
even terms are: $\frac{C_0}{2^n \cdot n!}$
odd terms are: $\frac{C_1}{1 \cdot 3 \cdot 5 \cdot ...(2n+1)}$
and so $$y = \sum_{n=0}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$
Where do I go from here? if I do $y(0)$, I don't get 1. How do I take the derivative of this to check the initial condition $y'(0) = 0$
|
You're given initial conditions so you must define $c_0$ and $c_1$ to satisfy these initial conditions.
Notice that you can write your series as
$$y(x) =c_0+ \sum_{n=1}^\infty \frac{c_0}{2^n \cdot n!}x^{2n} + \sum_{n=0}^\infty \frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n+1}$$
where I have simply taken the first term of the first series out front. Now it is clear, when $x=0$, both series are zero, so you are left with
$$y(0)=c_0=1$$
so $c_0=1$. Now, take the derivative with respect to $x$ of your entire function,
$$y'(x)=\sum_{n=1}^\infty \frac{2n}{2^n \cdot n!}x^{2n-1} + \sum_{n=0}^\infty (2n+1)\frac{c_1}{1 \cdot 3 \cdot 5 \cdot (2n+1)} x^{2n}$$
Now notice again when $x$ is zero, the only surviving term is the zeroeth term of the second series, giving you
$$y'(0)=\frac{c_1}{15}=0$$
and thus your entire series is
$$y = \sum_{n=0}^\infty \frac{1}{2^n \cdot n!}x^{2n}$$
|
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|
How to show the following problem of determinant If the matrices $A,B\in M_{3}(\Bbb{Z})$ are singular and $AB=BA$, show that the number $$\det(A^3+B^3)+\det(A^3-B^3)$$ is the double of a perfect cube.
I have considered the polynomial $$\det(A+xB)=\det A+mx+nx^2+x^3\det B$$
From this how I can show
|
Following Anurag A's suggestion in the comments, let $\omega$ be a primitive third root of unity (so $\omega^2+\omega + 1=0$).
Then observe that since $A$ and $B$ commute, we have $$A^3+B^3 = (A+B)(A+\omega B)(A+\omega^2B)$$ and $$A^3-B^3 = (A-B)(A-\omega B)(A-\omega^2 B).$$
Then using that $\det(A)=\det(B)=0$, we have that $p(x)=\det(A+xB)= x(m+nx)$ for some integers $n$ and $m$ from the formula that you've written in the question.
Thus
$$\det(A^3+B^3)+\det(A^3-B^3) = p(1)p(\omega)p(\omega^2) +p(-1)p(-\omega)p(-\omega^2) $$
$$= (m+n)(m+n\omega)(m+n\omega^2) - (m-n)(m-n\omega)(m-n\omega^2) = m^3+n^3 - (m^3-n^3) = 2n^3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
show this inequality with $xy+yz+zx=3$ let $x,y,z>0$ and such $xy+yz+zx=3$,show that
$$\dfrac{x}{x^3+y^2+1}+\dfrac{y}{y^3+z^2+1}+\dfrac{z}{z^3+x^2+1}\le 1$$
To prove this inequality,I want use following Cauchy-Schwarz inequality
$$(x^3+y^2+1)(\frac{1}{x}+1+z^2)\ge (x+y+z)^2$$
$$\dfrac{x}{x^3+y^2+1}\le \dfrac{1+x+xz^2}{(x+y+z)^2}$$
we have
$$\sum\dfrac{x}{x^3+y^2+1}\le\dfrac{3+x+y+z+xz^2+yx^2+zy^2}{(x+y+z)^2}$$
it suffices to prove that
$$x+y+z+xz^2+yx^2+zy^2\le x^2+y^2+z^2+3$$
|
By Muirhead
$$\sum_{cyc}\frac{x}{x^3+y^2+1}\leq\sum_{cyc}\frac{x}{x^2+y^2+x}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{x}{x^2+y^2+x}\leq1$$ or
$$\sum_{cyc}\left(x^4y^2+x^4z^2+\frac{2}{3}x^2y^2z^2-x^3y-x^2yz-\frac{2}{3}xyz\right)\geq0$$ or
$$\sum_{cyc}(3x^4y^2+3x^4z^2+2x^2y^2z^2-x^4y^2-x^4yz-x^3y^2z-x^3y^2z-x^3z^2y-x^2y^2z^2-2xyz)\geq0$$ or
$$\sum_{cyc}(2x^4y^2+3x^4z^2-x^4yz-2x^3y^2z-x^3z^2y+x^2y^2z^2-2xyz)\geq0$$ or
$$\sum_{cyc}(x^4z^2-x^3y^2z)+2\sum_{cyc}(x^4y^2+x^4z^2-2x^4yz)+$$
$$+xyz\sum_{cyc}(x^3-x^2y-x^2z+xyz)+2xyz\sum_{cyc}(x^3-1)\geq0,$$
which is true by AM-GM and Schur.
|
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|
How to find the sum: $\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) $ $$S_n=\sum_{n=0}^{\infty} \frac{1}{r-s}\left(\frac{1}{n+s+1}-\frac{1}{n+r+1}\right) = \frac{1}{r-s}\left( \frac{1}{s+1}+\frac{1}{s+2}\cdots+\frac{1}{r}\right)$$
I fail to see how this:
$$S_n = \frac{1}{r-s}\left( \frac{1}{s+1}-\frac{1}{r-1}+\frac{1}{s+2}-\frac{1}{r+2}+\cdots\right)$$
converges to the right answer. Can someone please give a detailed explanation?
$r,s$ are integers and $r>s>0$
|
Since $r > s$,
$$\begin{array}{rl} & \frac{1}{n+s+1} & - \frac{1}{n+r+1}\\
= & \frac{1}{n+s+1} &+ \color{red}{\frac{1}{n+s+2}} &+ \cdots &+ \color{blue}{\frac{1}{n+r}}\\
& & - \color{red}{\frac{1}{n+s+2}} &- \cdots &- \color{blue}{\frac{1}{n+r}} &- \frac{1}{n+r+1}\end{array}$$
we have
$$\begin{align}\frac{1}{n+s+1} - \frac{1}{n+r+1}
&= \sum_{k=1}^{r-s} \frac{1}{n+s+k} - \sum_{k=1}^{r-s} \frac{1}{(n+1)+s+k}\\
&= U(n) - U(n+1)\end{align}
$$
where $U(n) = \sum\limits_{k=1}^{r-s}\frac{1}{n+s+k}$. The sum at hand is a telescoping sum. The partial sums has the form
$$\begin{align}
\sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right)
= & \frac{1}{r-s}\sum_{n=0}^p \left(U(n) - U(n+1)\right)\\
= & \frac{1}{r-s}\left(U(0) - U(p+1)\right)\end{align}$$
Since $U(p+1)$ is a finite sum of $r-s$ terms and each terms converge to $0$ as $p\to \infty$, we find
$$\lim_{p\to\infty} U(p+1) = 0$$
As a result,
$$\begin{align}
& \sum_{n=0}^\infty \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right) \\
= & \lim_{p\to\infty} \sum_{n=0}^p \frac{1}{r-s}\left(\frac{1}{n+s+1} - \frac{1}{n+r+1}\right)\\
= & \lim_{p\to\infty}\frac{1}{r-s}\left(U(0) - U(p+1)\right)\\
= & \frac{1}{r-s}U(0)\\
= & \frac{1}{r-s}\sum_{k=1}^{r-s}\frac{1}{s+k}\\
= & \frac{1}{r-s}\left(\frac{1}{s+1} + \frac{1}{s+2} + \cdots + \frac{1}{r}\right)
\end{align}$$
|
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|
Can $\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}$ be simplified? I am trying to simplify the following but I cannot.
$$
\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha}
$$
Can it be simplified?
Edit
My last result is
$$
- \frac{\cos \alpha \left( 1 + \sin \alpha \cos \alpha\right)}
{\sin^2 \alpha \left(1+\sin \alpha\right)}
$$
I am wondering it might be a wrong question given by my student's teacher.
|
First, let us state of the fundamentals.
$$\csc \alpha = \frac{1}{\sin \alpha};\ \tan \alpha = \frac{\sin \alpha}{\cos \alpha};\ \cot \alpha = \frac{\cos\alpha}{\sin \alpha};\ \sec \alpha = \frac{1}{\cos \alpha};\ \sin^2\alpha + \cos^2\alpha = 1;$$
So,
$$\frac{\csc \alpha +\cos \alpha}{\cos \alpha - \tan \alpha - \sec \alpha} \cdot 1= \frac{\frac{1}{\sin \alpha} + \cos \alpha}{\cos \alpha - \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\cos\alpha}} \cdot\frac{\cos \alpha}{\cos \alpha} = \frac{\cot \alpha + \cos^2\alpha}{\cos^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{1-\sin^2\alpha - \sin\alpha - 1} = \frac{\cot\alpha + 1 -\sin^2\alpha}{\left(0-\sin\alpha\right)\cdot\left(1+\sin\alpha\right)}\cdot\frac{\sin\alpha}{\sin\alpha} = \frac{\cos\alpha + \sin\alpha - \sin^3\alpha}{- \sin^2\alpha\left(1+\sin\alpha\right)} = -\frac{\cos\alpha + \sin\alpha\cos^2\alpha}{\sin^2\alpha\left(1+\sin\alpha\right)} = - \frac{\cos\alpha\left(1 + \sin\alpha\cos\alpha\right)}{\sin^2\alpha\left(1+\sin\alpha\right)}$$
Alright, a great number of further simplifications are possible from here, but what do you think will eliminate the denominator and bring the order down to one without changing the inner $\alpha$. There are 4 terms and various manipulations left.
You'll see this is actually the simplest you can get it to. It is not trivial to prove this though.
|
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|
Solve $\frac{2 f'}{(f-1)^2}=1$ with initial condition $f(0)=c$ I'm trying to solve this by writing it as the derivative of a log of a polynomial but I can't make it work. Any hints?
|
With
$g = f - 1, \; g(0) = f(0) - 1 = c - 1, \tag 1$
we have
$g' = f'; \tag 2$'
thus,
$\dfrac{2f'}{(f - 1)^2} = \dfrac{2g'}{g^2} = 1, \tag 3$
whence
$g^{-2}g' = \dfrac{1}{2}, \tag 4$
or
$(-g^{-1})' = \dfrac{1}{2}; \tag 5$
we integrate 'twixt $0$ and $x$:
$-g^{-1}(x) - (-g^{-1}(0)) = \displaystyle \int_0^x (-g^{-1}(s))\; ds = \int_0^x \dfrac{1}{2} \; ds = \dfrac{1}{2}x, \tag 6$
and perform a few simple algebraic maneuvers using
$g^{-1}(0) = (c - 1)^{-1}: \tag 7$
$-g^{-1}(x) + g^{-1}(0) = \dfrac{1}{2}x; \tag 8$
$-g^{-1}(x) + (c - 1)^{-1} = \dfrac{1}{2}x; \tag 9$
$g^{-1}(x) = (c - 1)^{-1} - \dfrac{1}{2}x = \dfrac{1}{c - 1} - \dfrac{1}{2}x$ $= \dfrac{2}{2(c - 1)} - \dfrac{(c - 1)x}{2(c - 1)} = \dfrac{2 - (c - 1)x}{2(c - 1)}; \tag{10}$
$g(x) = \dfrac{2(c - 1)}{2 - (c - 1)x} = 2(c - 1)(2 - (c - 1)x)^{-1}. \tag{11}$
We Check:
Differentiating (11) yields
$g'(x) = -2(c - 1)(2 - (c - 1)x)^{-2} (-(c - 1))$
$= 2(c - 1)^2(2 - (c - 1)x)^{-2} = 2(((c - 1)(2 - (c - 1)x)^{-1})^2 = \dfrac{g^2}{2}, \tag{12}$
in accord with (3). It is also a simple matter to verify that
$g(0) = c - 1. \tag{13}$
Finally, since $g(x)$ is the solution to (4),
$f(x) = g(x) + 1 = 2(c - 1)(2 - (c - 1)x)^{-1} + 1 \tag{14}$
satisfies (3) with
$f(0) = g(0) + 1 = c. \tag{15}$
|
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=1.
\end{aligned}
\end{equation}
Please give me some ideas,thank you!
|
With the help of reuns,I can complete the proof.
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{2k^2}-\frac{1}{k+1}\right)\\&=\frac{1}{2}.
\end{aligned}
\end{equation}
Since
$$\frac{1}{n}=\sum_{k=n}^\infty \frac{1}{k(k+1)}<\sum_{k=n}^\infty \frac{1}{k^2}<\sum_{k=n}^\infty \frac{1}{k(k-1)}=\frac{1}{n-1}$$
and
$$\frac{1}{n^3}\leq\sum_{k=n}^\infty\frac{1}{k^3}\leq\frac{1}{n}\sum_{k=n}^\infty\frac{1}{k^2}<\frac{1}{n(n-1)}.$$
Thanks to Martin R for reminding.
The following part is by Adam Latosiński, I just modified the typographical errors.Thanks for his help!
We still need to show that $\lim\limits_{x\rightarrow 0} \frac{F(x)}{x} = \frac{1}{2}$ also when we approach by a sequence with $x\neq\frac{1}{n}$. Let $n=[ 1/x]$, so that $\frac{1}{n+1}<x\le\frac{1}{n}$, that is $n\le\frac{1}{x}<{n+1}$. We have
\begin{align} \left|\frac{F(1/n)}{1/n} - \frac{F(x)}{x}\right| &\le \left|\frac{F(1/n)}{1/n} - \frac{F(1/n)}{x}\right| + \left|\frac{F(1/n)}{x} - \frac{F(x)}{x}\right| \\
&\le F(1/n) \left|n-\frac{1}{x}\right| + (n+1) |F(1/n)-F(x)| \\
&\le F(1/n) + (n+1) \int_{x}^{\frac{1}{n}} \Big(\frac{1}{t}- \left[\frac{1}{t}\right]\Big) dt \\
&\le F(1/n) + (n+1) \int_{x}^{\frac{1}{n}} 1 \,dt \\
&\le F(1/n) + (n+1)(\frac{1}{n}-\frac{1}{n+1}) \rightarrow 0\end{align}
which proves that $$ \lim_{x\rightarrow 0} \frac{F(x)}{x} = \lim_{n\rightarrow\infty} nF(\frac{1}{n}) = \frac12$$
|
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Anders, Bodil, Cecilia, and David shall receive 4 oranges. In how many ways is this possible if Anders should have at least one? Anders, Bodil, Cecilia, and David shall receive 4 oranges. In how many ways is this possible if Anders should have atleast one?
Correct answer: 29
My solution:
How many solutions are there to
$x_{1} + x_{2} + x_{3} + x_{4} = 4$
where
$1 \leq x_{1}$
$0 \leq x_{2}$
$0 \leq x_{3}$
$0 \leq x_{4}$
Substituting
$y_{1} = x_{1} - 1$
$y_{2} = x_{2} - 0$
$y_{3} = x_{3} - 0$
$y_{4} = x_{4} - 0$
satisfying
$$y_{1} + y_{2} + y_{3} + y_{4} = 3 $$
$$ y = (3,0,0,0) \rightarrow 4$$
$$y = (2,1,0,0) \rightarrow 12$$
$$ y = (1,1,1,0) \rightarrow 4$$
My answer: 4 + 12 + 4 = 20
I'm basically trying to use the same method as "RMWGNE96" did here: How many solutions are there to $x_{1} + x_{2} + x_{3} + x_{4} = 15$
|
Your solution is correct.
If we ignore the restriction that Anders must receive at least one orange, the number of ways we could distribute the oranges is the number of solutions of the equation
$$a + b + c + d = 4$$
in the nonnegative integers. A particular solution of this equation corresponds to the placement of three addition signs in a row of four ones. For instance,
$$1 1 + 1 + 1 +$$
corresponds to the solution $a = 2$, $b = 1$, $c = 1$, $d = 0$, while
$$1 + 1 + 1 + 1$$
corresponds to the solution $a = b = c = d = 1$. The number of such solutions is the number of ways we can place three addition signs in a row of four ones, which is
$$\binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = 35$$
since we must choose which three of the seven positions required for four ones and three addition signs will be filled with addition signs.
From these, we subtract the number of ways of distributing the four oranges so that Anders receives none, which is the number of solutions of the equation
$$b + c + d = 4$$
in the nonnegative integers, which is
$$\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15$$
Hence, the number of ways the four oranges can be distributed to the four people if Anders receives at least one is
$$\binom{7}{3} - \binom{6}{2} = 35 - 15 = 20$$
as you found.
|
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|
Does dividing a common factor out from numerator and denominator of a rational function create a new function with different domain? Suppose a function is defined as $ f(x) =\frac{x^2 - 9}{x-3}. $
If we divide the common factor $ x-3 $ from both the numerator and denominator :-
$$ \frac{x^2 - 9}{x - 3} \\ = \frac{(x+3)(x-3)}{x-3} \\ = x+3 $$
1) Is $ x+3 $ a different function from the original $ \frac{x^2 - 9}{x - 3} $ ?
2) Since $ \frac{x^2 - 9}{x - 3} $ does not have 3 in its domain, when we simplify it to become $x + 3$, does it now have 3 in its domain ?
|
Two functions are equal if and only if they have the exact same domain, and the exact value at each element of the domain. (If you like your functions to also have a specified codomain, then you also require the functions to have the exact same codomain).
“Cancelling” a common factor includes the implicit assertion that this common factor is not equal to $0$. So in your example, when you cancel $x-3$, you should also be saying under your breath “and $x\neq 3$”, so the correct simplification would be
that $\frac{x^2-9}{x-3}$ is equal to the function $x+3,\ x\neq 3$.
(So, yes, the function $f(x) = \frac{x^2-9}{x-3}$ and the function $g(x) = x+3$ are different functions, because they have different domain; on the other hand, $f(x)$ is equal to the function $h(x) = x+3$, $x\neq 3$. )
However, cancelling does not always lead to a different function. For example, the function
$$f(x) = \frac{x^2-5x+6}{x^2-6x+9}= \frac{(x-3)(x-2)}{(x-3)^2}$$
is equal to the function
$$g(x) = \frac{x-2}{x-3}$$
because they have the exact same domain (all $x\neq 3$), and take the same value at every point of the domain.
(This is similar to the fact that the function
$$f(x) = \frac{1}{\quad\frac{1}{x}\quad}$$
and the function $g(x) = x$ are different functions, because they have different domains.)
|
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Why do we take the positive square root only from $\sqrt{a^2-x^2}$ when integrating using trig substitutions? $$\int\frac{\mathrm dx}{x^2 \sqrt{16-x^2}}$$
when substituting $\,x=4\sin\theta ,\;\mathrm dx=4\cos\theta\, \mathrm d\theta,\,$ it becomes
$$\int\frac{4\cos\theta\, \mathrm d\theta}{4^2\sin^2\theta\sqrt{16-4^2\sin^2\theta}}$$
then
$$\frac{1}{4}\int\frac{\cos\theta \,\mathrm d\theta}{\sin^2\theta\times4\sqrt{1-\sin^2\theta}}$$
The question is why did we take the positive root of 16 and took it out? as $\sqrt{16} \;is \pm 4 $
|
You wrote that “$\sqrt{16}$ is $\pm4$”. This is wrong. Yes, the number $16$ has two square roots: $4$ and $-4$. But $\sqrt{16}$ stands for the non-negative root of $16$, which is $4$. For the same reason, $\sqrt{16-4^2\sin^2\theta}=4\sqrt{1-\sin^2\theta}$.
|
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|
Computing the matrix powers of a non-diagonalizable matrix Define
\begin{equation}
A = \begin{pmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{pmatrix}.
\end{equation}
Note that the sum of the dimensions of the eigenspaces of $A$ is only two. $A$ is thus not diagonalizable. How can we compute $A^n$?
|
Here is a different way using a rather classical trick, converting the issue into a binomial expansion. Indeed, we can write :
$$A=\frac12(I+B) \ \text{where} \ B:=\begin{pmatrix}0&1&0 \\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}$$
where matrix $B$ has the following particularity
$$B^n=C \ \text{for all} \ n>1 \ \text{where} \ C:=\begin{pmatrix}0&1/2&1/2\\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}$$
Therefore
$$A^n = \dfrac{1}{2^n}\left(I+\binom{n}{1}B+\binom{n}{2}B^2+\binom{n}{3}B^3+\cdots+\binom{n}{n}B^n\right)$$
$$A^n = \dfrac{1}{2^n}\left(I+nB+\binom{n}{2}C+\binom{n}{3}C+\cdots+\binom{n}{n}C\right)\tag{1}$$
As is well known, $\sum_{k=0}^n \binom{n}{k}=2^n$, reducing (1) to :
$$A^n = \dfrac{1}{2^n}\left(I+nB+(2^n-n-1)C\right)$$
It suffices now to replace $B$ and $C$ by their expression
$$A^n = \dfrac{1}{2^n}\left(\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+n\begin{pmatrix}0&1&0 \\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}+(2^n-n-1)\begin{pmatrix}0&1/2&1/2\\0&1/2&1/2\\0&1/2&1/2\end{pmatrix}\right)$$
to get the result (coinciding with yours !).
|
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If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.
Given $x + y + z = 2$, prove
$$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$
I have tried to solve this using AM $\geq$ GM inequality.
Attempt :
$$\frac{\dfrac{1-x}{x} + \dfrac{1-y}{y} + \dfrac{1-z}{z}}{3}\geq\left(\frac{(1-x)(1-y)(1-z)}{xyz}\right)^{1/3}$$
What should I do to calculate the value of $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$?
Doubt:
*
*Let $a, b, c, d$ all be positive numbers.
*Let $a > c$ and $b > d$.
*Can we say that $\dfrac {a}{b} > \dfrac{c}{d}$?
Problem number 2:
$x,y,z$ are unequal positive quantities, prove that
$(1+x^3)(1+y^3)(1+z^3) > (1+xyz)^3$
My attempt:
$ \frac{x^3+y^3+z^3}{3} > xyz$
$ => (1+x^3)+(1+y^3)+(1+z^3) > 3(1+xyz)$
now by cubing the both sides we get,
$ => \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > (1+xyz)^3$ ---(eqn. 1)
$ \frac{(1+x^3)+(1+y^3)+(1+z^3)}{3} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}^{\frac{1}{3}}$
$=> \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}$ ---(eqn. 2)
So, now if we do : $\frac{(eqn. 1)}{(eqn. 2)}$
then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} > (1+xyz)^3$
and this is the correct result for all $x,y,z$ greater than zero.
But if we do $\frac{(eqn. 2)}{(eqn. 1)}$ then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} < (1+xyz)^3$
So, my question is, how to prove it correctly without doing the division operation, cause when we divide the two equations then we get two separate results.
|
$\dfrac ab-\dfrac cd=\dfrac{ad-bc}{be}$ will be $\ge0$
only if $ad-bc>0\iff \dfrac ab>\dfrac cd$
Alternatively
Let $a=c+p,b=d+q,p,q>0$
$\dfrac ab-\dfrac cd=\dfrac{c+p}{d+q}-\dfrac cd=\dfrac{d(c+p)-c(d+q)}{...}$
$=\dfrac{dp-cq}{...}$
will be $>0$
only if $dp-cq>0\iff \dfrac dc>\dfrac qp$
as $c,d,p,q>0$
|
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Evaluating $\lim\limits_{n \to \infty} \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx$ Can you please verify my proof below?
My solution
Notice that $f(x):=\ln^n(x)$ and $g(x):=\dfrac{1}{1+x^2}$ is continuous over $[0,1]$, and $g(x)>0$. As per the first integral mean value theorem, on can have
$$0\leq \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx=\ln^n(1+\xi)\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}\ln^n(1+\xi)\leq \frac{\pi}{4}\ln^n(1+1)\to 0,$$
which implies $$\lim\limits_{n \to \infty} \int_0^1 \frac{\ln^n (1+x)}{1+x^2}dx=0.$$
|
It's correct, but you can get the same bound easier, without using MVT:
For $0\le x \le 1$ we have $0 \le \ln(1+x) \le \ln 2$, so $0\le \frac{\ln^n(1+x)}{1+x^2} \le \frac{\ln^n 2}{1+x^2}$ and
$$ 0\le \int_0^1 \frac{\ln^n(1+x)}{1+x^2} dx \le \int_0^1 \frac{\ln^n 2}{1+x^2} dx=\frac{\pi}{4} \ln^n 2$$
|
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For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equiv 3\pmod 4$
Let $n = k+1$ such that
$${(k+1)^2 + 2(k+1)}\equiv {k^2 + 2k + 1 + 2k + 2}\pmod 4$$
Then I substitute $k^2 + 2k$ with both $0$ or $3$ from the earlier assumption.
So we have
$$(0 + 2k + 3) \equiv (2k + 3) \pmod 4$$
Or,
$$(3 + 2k + 3) \equiv (2k + 6) \pmod 4$$
Where do I go from here?
|
You only need one equation.
$\begin{array}\\
(n+2)^2+2(n+2)
-(n^2+2n)
&=n^2+4n+4+2n+4
-(n^2+2n)\\
&=4n+8\\
\end{array}
$
which is divisible by $4$,
so whatever remainder $n$ has
mod 4, $n+2$ also has.
Since $n=0 \implies n^2+2n = 0$,
all even numbers have a
remainder of 0 mod 4.
Since $n=1 \implies n^2+2n = 3$,
all odd numbers have a
remainder of 3 mod 4.
|
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Is $x = \frac{a}{b} = \frac{b}{a/3}$ a continued fraction? How to solve for $x$? I believe following is a continued fraction. I'm stumped on how to solve for x
$x = \frac{a}{b} = \frac{b}{a/3}$
I know it can be re-written as
$x = \frac{a}{b} = \frac{3b}{a}$
$x = a^2 = 3b^2$
I'm unsure where to go from here.
Below are the choices
$x=9$
$x=\frac{1}{3}$
$x=3$
$x=\frac{1}{\sqrt3}$
$x={\sqrt3}$
|
The part where you wrote $x=a^2=3b^2$ is not correct.
From $$ \frac{a}{b} = \frac{b}{a/3} \implies a^2/3 = b^2 \implies a^2=3b^2$$
So $a = \sqrt{3}b$ and thus $$x = \frac{a}{b} = \sqrt{3}$$
|
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as perfect squares to get the least value? a Hint is much appreciated. Thanks!
|
Using Cauchy Schwarz, we can write $\displaystyle (a + b + c) \left(\frac{1}{a} +\frac{1}{b} + \frac{1}{c} \right) \geq 9$. Thus, we have,
$$ \frac{a+b+c}{a} +\frac{a+b+c}{b} + \frac{a+b+c}{c} \geq 9 \implies 1+ \frac{b+c}{a} + 1 + \frac{a+c}{b} + 1 + \frac{a+b}{c} \geq 9$$
Hence, $$ \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6 $$
|
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$\frac{x}{4} = \frac{y}{x} = \frac{7}{y}$ Could someone please explain to me what's wrong here?
$\cfrac{x}{4} = \cfrac{y}{x} = \cfrac{7}{y}$
So $$\cfrac{x^2}{4} = y \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ y = \sqrt{7x}$$
Hence $$\cfrac{x^2}{4} =\sqrt{7x}$$
With a solution of $x = 2.57$
But $$\cfrac{2.57^2}{4} \neq \sqrt{7 (2.57)}$$
|
$x\neq 2.57$. You want to solve for $x^2/4 = \sqrt{7x}$, which corresponds to solving the polynomial equation $x^4/16 - 7x = 0$, which has a real and nontrivial zero at $\sqrt[3]{112}$. Hence, $x = \sqrt[3]{112}$ and $y = (112)^{2/3} / 4$. You can check that this in fact, is a solution to your system.
|
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$
I have not come up with any ideas to solve the problem yet. I will probably in the near future but right now, I can't.
|
SOS helps!
Let $a=b=c=\frac{1}{3\sqrt2}.$
Thus, we get a value $\frac{1}{2\sqrt2}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{a^2+b^2}$$ or
$$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{a}{2}\right)\geq\frac{1}{4}\sum_{cyc}\left(\sqrt{2(a^2+b^2)}-a-b\right)$$ or
$$\sum_{cyc}\frac{a(a-b-(c-a))}{b+c}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or
$$\sum_{cyc}(a-b)\left(\frac{a}{b+c}-\frac{b}{a+c}\right)\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or
$$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ and since by C-S $$\sqrt{2(a^2+b^2)}=\sqrt{(1^2+1^2)(a^2+b^2)}\geq a+b,$$ it's enough to prove that:
$$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{4}\sum_{cyc}\frac{(a-b)^2}{a+b}$$ or
$$\sum_{cyc}(a-b)^2(4(a+b+c)(a+b)-(a+c)(b+c))\geq0,$$ for which it's enough to prove that
$$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq0.$$
Now, let $a\geq b\geq c$.
Thus,
$$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq(a-c)^2(a^2+c^2-b^2)+(b-c)^2(b^2+c^2-a^2)\geq$$
$$\geq(b-c)^2(a^2-b^2)+(b-c)^2(b^2-a^2)=0$$ and we are done!
|
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|
Calculation of a series using series definition I have to calculate the series $$\sum_{k=2}^\infty \left(\frac 1k-\frac 1{k+2}\right)$$
Using the definition: $$L = \lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=0}^na_k$$
Obviously $\lim_{n\to\infty} (\frac 1n-\frac 1{n+2})=0$, but I don't think that this is the right way to calculate the value of the series.
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As suggested from the comment by JMoravitz, you should write down the series explicitly and see if you can see something interesting:
$$ \sum_{2}^\infty \big( \frac{1}{k} - \frac{1}{k+2} \big) = \bigg(\frac{1}{2} - \frac{1}{4} \bigg) + \bigg(\frac{1}{3} - \frac{1}{5} \bigg) + \bigg(\frac{1}{4} - \frac{1}{6} \bigg) + \bigg(\frac{1}{5} - \frac{1}{7} \bigg) + \bigg(\frac{1}{6} - \frac{1}{8} \bigg) + \cdots + \bigg(\frac{1}{n} - \frac{1}{n+2} \bigg) + \bigg(\frac{1}{n+1} - \frac{1}{n+3} \bigg) + \bigg(\frac{1}{n+2} - \frac{1}{n+4} \bigg) + \bigg(\frac{1}{n+3} - \frac{1}{n+5} \bigg) + \cdots $$
From here, you can see which terms are cancelling with one another and which one is left.... Essentially you should see that all the terms will cancel with each other except the $\frac{1}{2}$ and $\frac{1}{3}$. Therefore, the answer is $\frac{5}{6}$.
$\textbf{Another, Different way}$ that uses limit like you wanted to is to find the partial sum! You first compute, $S_2 = \frac{1}{2} - \frac{1}{4} $, $S_3 = S_2 + \bigg( \frac{1}{3} - \frac{1}{5} \bigg)$ and so on... What you will find (and you can prove this fact by induction) is that
$$ S_n = \dfrac{5n^2 + 3n -8}{6(n+1)(n+2)}$$
Now, $$ \lim_{n \to \infty} \dfrac{5n^2 + 3n -8}{6n^2 + 18n + 12} = \frac{5}{6} $$
|
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|
Evaluate $\lim_{x\to 1}\frac{x^{x}-x^{x^2}}{(1-x)^2}$ $$L=\lim_{x\to 1}\frac{x^{x}-x^{x^2}}{(1-x)^2}$$
Any hints on how to approach this problem in the first place? The answer should be $-1$.
I tried adding and subtracting $1$, then evaluating
$$L_1=\lim_{x\to 1}\frac{x^{x}-1}{(1-x)^2}$$ $$and$$ $$L_2=\lim_{x\to 1}\frac{x^{x^2}-1}{(1-x)^2}$$
But, unfortunately after applying L'Hospital to both of them, I'd got to
$L_1-L_2$ which is the equivalent of $\infty-\infty$.
I also tried applying L'Hospital to $L$
$$\lim_{x\to 1}\frac{x^{x}(lnx+1)-x^{x^2}(2xlnx+x)}{-2(1-x)}$$
Which weirdly doesn't give the answer.
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We have \begin{align} \frac{x^x - x^{x^2}}{(x-1)^2} &= x^x \cdot \frac{1 - x^{x(x-1)}}{(x-1)^2} = \\
&= x^x \cdot \frac{1-e^{x(x-1)\ln x}}{(x-1)^2} = \\
&= x^x \cdot\frac{1-e^{x(x-1)\ln x}}{x(x-1)\ln x}\cdot\frac{x\ln x}{x-1} \end{align}
We calculate the limits separately: $$ \lim_{x \rightarrow 1} x^x = 1^1 = 1$$
$$ \lim_{x \rightarrow 1} \frac{1-e^{x(x-1)\ln x}}{x(x-1)\ln x} = \lim_{z \rightarrow 0}\frac{1-e^z}{z} = -1$$
$$ \lim_{x \rightarrow 1} \frac{x\ln x}{x-1} =^H \lim_{x \rightarrow 1} \frac{1 + \ln x}{1} = 1$$
so
$$ \lim_{x \rightarrow 1} \frac{x^x - x^{x^2}}{(x-1)^2} = -1$$
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|
Solve integer equation $2^m.m^2=9n^2-12n+19$
Problem: Find $m,n\in \mathbb{N}^*$ satisfies: $2^m.m^2=9n^2-12n+19$.
This is my attempt:
We have $9n^2-12n+19\equiv 1 \pmod 3$, so: $2^{m}m^2\equiv 1 \pmod 3\tag{1}$
In addition, we have: $$m^2\equiv 0\text{ or }1 \pmod 3 $$
So $(1)\implies m\equiv \pm 1\pmod 3$.
Suppose that: $m=3k+1(k\in \mathbb{N})$.
We have: $$2^{m}m^2=2^{3k+1}(3k+1)^2=8^{k}(3k+1)^22\equiv 2^{k+1}(3k+1)^{2}\pmod 3\\ \implies k\equiv 1\pmod 2$$
I can only come here!!
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First of all, let's complete the square on the RHS and write it as $(3n-2)^2+15$. Next, suppose $m$ is odd. Then $2^mm^2$ will be $\equiv 2\pmod 4$ (can you see why?) but the RHS — as a sum of a square and a number $\equiv 3\pmod 4$ — will be congruent to either $3$ or $0$, so they can't be equivalent.
Now, suppose $m$ is even; then $m=2r$, say, and so the LHS $m^22^m = (2r)^22^{2r} = \left(r2^{r+1}\right)^2$. This means we can write $a^2=b^2+15$, where $a=r2^{r+1}$ and $b=3n-2$. This can be written perhaps more suggestively as $a^2-b^2=15$; can you take it from there?
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|
Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$
Compute
$$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$
Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+(1)(x^2-x))}{x^2-x}
-\frac{\ln(1+(0)(x^2-x))}{x^2-x}\right]dx\\
&=\int_{0}^{1} \frac{\ln(1+x^2-x)}{x^2-x}dx
\end{split}
$$
And here I can't find any substitution to solve this integral.
Can anyone help me?
By the way, I also used Simpson's 3/8 method to find the approximation and got $1.063$.
But I want to find it using Calculus.
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$$I=\int_0^1 \frac{\ln(1-x+x^2)}{x(x-1)}dx=-\int_0^1 \frac{\ln(1-x(1-x))}{x}dx-\int_0^1\frac{\ln(1-x(x-1))}{1-x}dx$$
The substitution $1-x\to x $ in the second integral reveals that:
$$\int_0^1\frac{\ln(1-x(1-x))}{1-x}dx=\int_0^1\frac{\ln(1-x(1-x))}{x}dx$$
$$\Rightarrow I=-2\int_0^1\frac{\ln(1-x(1-x))}{x}dx=2\int_0^1\frac{\ln(1+x)-\ln(1+x^3)}{x}dx$$
But we are lucky again, because: $$\int_0^1\frac{\ln(1+x^3)}{x}dx\overset{x=t^{1/3}}=\frac13\int_0^1 \frac{\ln(1+t)}{t^{1/3}}\,t^{1/3-1}dt\overset{t=x}=\frac13\int_0^1\frac{\ln(1+x)}{x}dx$$
Also we can use this equality which will lead to:
$$ I=\frac43 \int_0^1 \frac{\ln(1+x)}{x}dx=-\frac23 \int_0^1 \frac{\ln x}{1-x}dx=-\frac23 \sum_{n=0}^\infty \int_0^1 x^n\ln x dx=\frac23 \sum_{n=1}^\infty \frac{1}{n^2}$$
Combining with heropup's answer we can deduce that $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
I guess we can add one more to this big list Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem).
|
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|
Evaluation of $\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}$ I trying to verify the following:
$$
\int_{0}^{1}{\int_{0}^{1}{\frac{\ln \left( 1-x \right)-\ln \left( 1-y \right)}{x-y-1}dx}dy}=\frac{{{\pi }^{2}}}{12}+{{\ln }^{2}}\left( 2 \right)
$$
I thought the solution may be similar to this problem but i can't handle the integral in a similar way?!?!?!
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With
$$\int\frac{\ln(x+z)}{x-a}dx = \text{Li}_2\left(\frac{x+z}{a+z}\right) - \ln(x+z)\,\text{Li}_1\left(\frac{x+z}{a+z}\right) + C $$
we get
$$\int\limits_0^1\frac{\ln(1-x)}{x-y-1}dx = \int\limits_0^{-1}\frac{\ln(x+1)}{x+y+1}dx = -\text{Li}_2\left(-\frac{1}{y}\right)$$
and the other part is:
$$\int\limits_0^1\frac{-\ln(1-y)}{x-y-1}dx = -\ln(1-y)\ln\left(\frac{y}{y+1}\right)$$
Now we need
$$\int -\text{Li}_2\left(-\frac{1}{y}\right)dy = -y\,\text{Li}_2\left(-\frac{1}{y}\right) + y\ln\left(1+\frac{1}{y}\right) + \ln(y+1) + C$$ and
$$\int -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy = \\ 2\,\text{Li}_2\left(\frac{1-y}{2}\right) + \text{Li}_2(y) + (y + (1-y)\ln(1-y))\ln\left(\frac{y}{y+1}\right) \\ + 2\ln(1-y)\ln\left(\frac{y+1}{2}\right) - \ln(y+1) + C$$
which leads to:
$$\int\limits_0^1\int\limits_0^1 \frac{\ln(1-x) - \ln(1-y)}{x-y-1} dxdy = \int\limits_0^1 -\text{Li}_2\left(-\frac{1}{y}\right)dy + \int\limits_0^1 -\ln(1-y)\ln\left(\frac{y}{y+1}\right)dy \\ = \left(\frac{\pi^2}{12} + \ln 4\right) + \left((\ln 2)^2 - \ln 4\right) = \frac{\pi^2}{12} + (\ln 2)^2$$
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|
Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to rewrite it as
$$ \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$
Then say $$\frac{x+y+z}{yzx}= a^2.$$
We get
$$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$
And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$
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Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,
$-\dfrac\pi2\le \tan^{-1}a\le\dfrac\pi2$
Now $\sqrt b\ge0$ for real $\sqrt b$
$\implies0\le\tan^{-1}\sqrt b\le\dfrac\pi2$
So, here the sum will lie in $\in[0,3\pi/2]$
Now the sum will be $=0$ only if each term under is individually $=0$
i.e. if $x+y+z=0$
Otherwise the sum will be $\ne0$
Also, the general value of the sum is $n\pi$ where $n$ is an integer
So, for $x+y+z\ne0,n=1$
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Solve $(y^2 + xy)(x^2 - x + 1) = 3x - 1$ over the integers.
Solve $$(y^2 + xy)(x^2 - x + 1) = 3x - 1$$ over the integers.
There are many solutions to this problem, and perhaps I chose the worst one possible. I hope that someone could come up with a better answer.
This problem is adapted from a recent competition (which is different than all of the "recent competitions" that I have mentioned before.
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After getting $(x^2-x+1)\mid(3x-1)$, and bounding $-2\leq x\leq 3$ as a result, you should first go back and check the condition $(x^2-x+1)\mid(3x-1)$:
*
*$x=-2$: $x^2-x+1=7$, $3x-1=-7\quad\checkmark$
*$x=-1$: $x^2-x+1=3$ so doesn't divide $3x-1$.
*$x=0,1$: $x^2-x+1=1\quad\checkmark$
*$x=2$: $x^2-x+1=3$, so doesn't divide $3x-1$.
*$x=3$: $x^2-x+1=7$ does not divide $3x-1=8$.
then find $y$.
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|
Calculate the maximum value of $\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$ where $x + 2y + 3z = 2$
$x$, $y$ and $z$ are positives such that $x + 2y + 3z = 2$. Calculate the maximum value of $$ \sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$
This problem is brought to you by a recent competition. There should be different answers that are more creative than the one I have provided. Oh well...
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We have that
$$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$
$$ = \sqrt{\frac{3yz}{3yz + (2 - 2y - 3z)}} + \sqrt{\frac{3zx}{3zx + 2(2 - 3z - x)}} + \sqrt{\frac{xy}{xy + (2 - x - 2y)}}$$
$$ = \sqrt{\frac{3yz}{(1 - y)(2 - 3z)}} + \sqrt{\frac{3zx}{(2 - 3z)(2 - x)}} + \sqrt{\frac{xy}{(2 - x)(1 - y)}}$$
$$\le \frac{1}{2}\left(\frac{2y}{2 - 3z} + \frac{3z}{2 - 2y}\right) + \frac{1}{2}\left(\frac{3z}{2 - x} + \frac{x}{2 - 3z}\right) + \frac{1}{2}\left(\frac{x}{2 - 2y} + \frac{2y}{2 - x}\right)$$
$$ = \frac{1}{2}\left(\frac{2y + 3z}{2 - x} + \frac{3z + x}{2 - 2y} + \frac{x + 2y}{2 - 3z}\right)$$
$$ = \frac{1}{2}\left[(x + 2y + 3z - 2)\left(\frac{1}{2 - x} + \frac{1}{2 - 2y} + \frac{1}{2 - 3z}\right) - 3\right] = \frac{3}{2}$$
The equality sign occurs when $\left\{ \begin{align} x(2 - x) = 2y(2 - 2y) = 3z(2 - 3z)\\ x + 2y + 3z = 2\end{align} \right.$
$\iff \left\{ \begin{align} (x - 1)^2 = (2y - 1)^2 = (3z - 1)^2\\ x + 2y + 3z = 2 \end{align} \right.$$\implies x = 2y = 3z = \dfrac{x + 2y + 3z}{1 + 1 + 1} = \dfrac{2}{3}$
$\iff (x, y, z) = \left(\dfrac{2}{3}, \dfrac{1}{3}, \dfrac{2}{9}\right)$.
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|
Showing that two curves do not intersect I want to show that
$$x+1 \neq (x^3(x+2))^{1/4} + \sqrt{x+1-\sqrt{x^2+2x}}$$
for any real $x>0$.
There are two approaches I've taken: showing they are equal and arriving at a contradiction (but this hasn't worked) and computing the derivative of the RHS and showing it is strictly less than $1$ everywhere. Its derivative is given by
$$g(x)=\frac{1-\frac{2x+2}{2\sqrt{x^2+2x}}}{2\sqrt{-\sqrt{x^2+2x}+x+1}}+\frac{x^3+3x^2\left(x+2\right)}{4\left(x^3\left(x+2\right)\right)^\frac{3}{4}}$$
I'm not sure how to show it is less than $1$. One approach may be to show that $g(x)$ is (strictly) increasing and $\lim_\limits{x\to \infty}g(x)=1$. Nothing has worked yet. Thank you for any help!
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Suppose that there exists an $x\gt 0$ such that
$$x+1 - \sqrt{x+1-\sqrt{x^2+2x}}= (x^3(x+2))^{1/4} $$
Squaring the both sides gives
$$(x+1)^2-2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}+x+1-\sqrt{x^2+2x}=x\sqrt{x^2+2x},$$
i.e.
$$(x+1)^2+x+1-(x+1)\sqrt{x^2+2x}=2(x+1)\sqrt{x+1-\sqrt{x^2+2x}}$$
Let $x+1=y\ (\gt 1)$. Then, we have
$$y^2+y-y\sqrt{y^2-1}=2y\sqrt{y-\sqrt{y^2-1}}$$
Dividing the both sides by $y$ gives
$$y+1-\sqrt{y^2-1}=2\sqrt{y-\sqrt{y^2-1}}$$
Squaring the both sides gives
$$(y+1)^2-2(y+1)\sqrt{y^2-1}+y^2-1=4\left(y-\sqrt{y^2-1}\right),$$
i.e.
$$y^2+2y+1+y^2-1-2(y+1)\sqrt{y^2-1}=4y-4\sqrt{y^2-1},$$
i.e.
$$2y^2+2y-4y=(2y+2-4)\sqrt{y^2-1},$$
i.e.
$$2y(y-1)=2(y-1)\sqrt{y^2-1}$$
Dividing the both sides by $2(y-1)$ gives
$$y=\sqrt{y^2-1}$$
which is impossible.
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|
Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.
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A shorter approach:
Generally for numbers $m=\sqrt{n+a}+\sqrt{n+\sqrt{n+a}}$ we must have:
$n+a=b^2$
$b^2+b-a=c^2$
If this system of equations, with given a, has integer solutions for n then m is integer.For example $a=3$; $m=\sqrt{n+3}+\sqrt{n+\sqrt{n+3}}$ we have:
$n+3=b^2$
$b^2+b-3=c^2$
It can be seen that with $c=3$ we have:
$b^2+b-3=3^2=9$; or $b^2+b-12=0$ which gives $b=3$ and $b=-4$. With $b=3$ we get $n=6$ and $m=6$, with $b=-4$ we get $n+3=16$ or $n=13$ which gives $m=4+\sqrt{17}$ which is not integer.
In question $a=2$ and we can search and see that reasonable value of c can only be zero. Other values give irrational numbers for one of terms $\sqrt{n+2}$ or $\sqrt{n+\sqrt{n+2}}$. Hence we have:
$$b^2+b-2=0$$
Which gives:
$b=1$; $a= 1$; $n=-1$; $m= 1$
$b=-2$;$a= 4$; $n=2$; $m=4$
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|
Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$
Investigate convergence of $\int_0^{\pi} \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} dx$ for $\alpha \in \mathbb R$
My try:
I want to use Direct comparison test so: $$0\le \frac{\sin x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}}=\frac{\sin(\pi-x)}{x^{\frac{3}{2}}(\pi-x)^{\alpha}} \le \frac{\pi-x}{x^{\frac{3}{2}}(\pi-x)^{\alpha}}=\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}$$
However in this moment I don't know what I can do with $\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}$ because for $x \in [1,\pi)$ I have $\frac{1}{x^{\frac{3}{2}}(\pi-x)^{\alpha-1}}\le \frac{1}{(\pi-x)^{\alpha-1}}$ but for $x\in (0,1]$ it is not right.Have you got any ideas?
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Divide the integral $$ \int_0^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx = \int_0^{\pi/2} \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx + \int_{\pi/2}^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx $$
For $x\in[0,\pi/2]$ we have $$ \frac{2}{\pi} \le \frac{\sin x}{x} \le 1$$
$$ 0 < M_1(\alpha) = \min\{\frac{1}{\pi^\alpha},\frac{1}{(\pi/2)^\alpha}\} \le \frac{1}{(\pi-x)^\alpha} \le \max\{\frac{1}{\pi^\alpha},\frac{1}{(\pi/2)^\alpha}\} = M_2(\alpha) < \infty$$
$$ \frac{2M_1(\alpha)}{\pi}\frac{1}{x^\frac12} \le \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} \le \frac{M_2(\alpha)}{x^\frac12 }$$
so
$$ \int_0^{\pi/2} \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx \text{ is convergent} \Leftrightarrow \int_0^{\pi/2} \frac{1}{x^\frac12} dx \text{ is convergent} \Leftrightarrow \alpha \in \mathbb R$$
since
$$\int_0^{\pi/2} \frac{1}{x^\frac12} dx = \Big[ 2x^\frac12\Big] \Big|_{x=0}^{x=\pi/2} = \sqrt{2\pi}$$
is convergent, then this part of integral is convergent for all $\alpha\in\mathbb R$.
For $x\in[\pi/2,\pi]$ we have $$ \frac{2}{\pi} \le \frac{\sin x}{\pi -x} = \frac{\sin(\pi-x)}{\pi-x} \le 1$$
$$ \frac{1}{\pi^\frac32} \le \frac{1}{x^\frac32} \le \frac{1}{(\pi/2)^\frac32}$$
$$ \frac{2}{\pi^\frac52}\frac{1}{ (\pi-x)^{\alpha-1}} \le \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} \le \frac{1}{(\pi/2)^\frac32}\frac{1}{(\pi-x)^{\alpha-1}}$$
so
$$ \int_{\pi/2}^\pi \frac{\sin x}{x^\frac32 (\pi-x)^\alpha} dx \text{ is convergent} \Leftrightarrow \int_{\pi/2}^\pi \frac{1}{(\pi-x)^{\alpha-1}} dx \text{ is convergent} $$
and we have $$ \int_{\pi/2}^\pi \frac{1}{(\pi-x)^{\alpha-1}} dx = \Big[\frac{-1}{2-\alpha}(\pi-x)^{2-\alpha}\Big]\Big|_{x=\pi/2}^{x=\pi}$$
which is finite iff $\alpha<2$.
In other words we use the comparison $$ \sin x < \frac{\pi}{2} - |x-\frac{\pi}{2}| = \left\{\begin{array}{ll} x & \text{for } 0 \le x \le \frac{\pi}{2}\\ \pi - x & \text{for } \frac{\pi}{2} \le x \le\pi\end{array}\right. $$
|
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|
Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$ This problem was already solved here (in different closed form).
But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $
Where $\displaystyle \operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $ is the the trilogarithm.
|
The following solution is by Cornel Valean:
\begin{gather*}
\int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\textrm{d}x\overset{x= \frac1y}{=}\int_0^\infty\frac{\ln^2\left(\frac{y}{1+y}\right)}{1+y^2}\textrm{d}y\\
\overset{\frac{y}{1+y}=x}{=}\int_0^1\frac{\ln^2(x)}{x^2+(1-x)^2}\textrm{d}x\\
\left\{\text{write $\frac{1}{x^2+(1-x)^2}=\Im \frac{1+i}{1-(1+i)x}$}\right\}\\
=\Im \int_0^1\frac{(1+i)\ln^2(x)}{1-(1+i)x}\textrm{d}x\\
=2\ \Im\{\operatorname{Li}_{3}(1+i)\}.
\end{gather*}
|
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|
Constraints on the roots to a quadratic equation If both roots of the quadratic equation
$$2x^2 +kx -(k+1)=0$$
are greater than $1$, then $k$ lies in what interval?
I tried to solve this using using different graphical and algebraic method but i seem to miss a crucial insight which is stopping me from getting the answer. Please help here.
|
Let $x_1,x_2$ be the two roots of your equation, so $x_1 \geq 1$, $x_2 \geq 1$.
In general, given a quadratic equation $ax^2+bx+c=0$ you have that $\frac{c}{a}$ is the product of the two roots and that $-\frac{b}{a}$ is the sum of the two roots.
In our case $x_1+x_2=-\frac{k}{2}$ and $x_1 x_2=-\frac{k+1}{2}$ and $x_1+x_2 \geq 2$ and $x_1x_2 \geq 1$.
We get $-\frac{k}{2} \geq 2$ or equivalently $k \leq -4$ and $-\frac{k+1}{2}\geq 1$ or equivalently $k \leq -3$. Together they give you $k \leq -4$.
You have to check one last thing: you equation must have two real roots, so $\Delta \geq 0$. Here $\Delta=k^2+4\cdot 2 (k+1)=k^2+8k+8$. Now $k^2+8k+8 \geq 0$ and we have that $k^2+8k+8=(k-2\sqrt{2}+4)(k+2\sqrt{2}+4)$ so $k \leq -2\sqrt{2}-4$ or $k \geq 2\sqrt{2}-4$.
Now we can combine our conditions and the final answer is $k \in (-\infty, -2\sqrt{2}-4]$
|
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|
How to convert the following Quadratic optimization problem to a linear one? I have an optimization problem
$$Min : \ 3x_{11} + 5x_{12} + 4x_{21} + 3 x_{22} - (10x_{11}x_{22} + 2x_{12}x_{21})$$
subject to the following constraints :
$$ x_{11} + x_{12} = 1$$
$$ x_{21} + x_{22} = 1$$
where $x_{11}. x_{12}, x_{21}, x_{22}$ are Integer Variables taking values either $0$ or $1$.
How do I convert the objective function to a linear one?
|
The general linearization of a product of binary variables is overkill here. Instead, first substitute $x_{12} = 1 - x_{11}$ and $x_{22} = 1 - x_{21}$ to obtain objective function
\begin{align}
&3x_{11} + 5x_{12} + 4x_{21} + 3 x_{22} - 10x_{11}x_{22} - 2x_{12}x_{21}\\
&=3x_{11} + 5(1 - x_{11}) + 4x_{21} + 3 (1 - x_{21}) - 10x_{11}(1 - x_{21}) - 2(1 - x_{11})x_{21}\\
&=3x_{11} + 5 - 5 x_{11} + 4x_{21} + 3 - 3 x_{21} - 10x_{11} +10x_{11} x_{21} - 2x_{21} +2 x_{11} x_{21}\\
&=8 - 12x_{11} - x_{21} +12x_{11} x_{21}.
\end{align}
Now introduce $y \ge 0$ to represent $x_{11} x_{21}$. Because we are minimizing, and the objective coefficient is $12>0$, we need only enforce $y\ge x_{11} x_{21}$, which we can do with a single linear constraint $y\ge x_{11} + x_{21} - 1$. In summary, the problem is to minimize $$8 - 12x_{11} - x_{21} +12y$$ subject to:
\begin{align}
y&\ge x_{11} + x_{21} - 1\\
x_{11} &\in \{0,1\}\\
x_{12} &\in \{0,1\}\\
y &\ge 0
\end{align}
The optimal solution is $(x_{11},x_{21},y)=(1,0,0)$, which yields objective value $-4$ and corresponds to $(x_{11},x_{12},x_{21},x_{22})=(1,0,0,1)$ in the original space.
|
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Reasoning about $x(x+2)$ relatively prime to $\dfrac{p\#}{w}$ Let:
*
*$p_k$ be the $k$th prime so that $p_3 = 5$
*$p\#$ be the primorial of $p$.
*$\text{gcd}(a,b)$ be the greatest common divisor of integer $a$ and integer $b$.
*$w \ge 1$ be an integer that divides $p\#$
Does it follow that for all integers $k \ge 3$ and all integers $m$, there exists an integer $x$ where:
*
*$m < x \le m + \dfrac{p_k\#}{w}$
*$\text{gcd}\left(x[x+2],\dfrac{p_k\#}{w}\right)=1$
Here is the inductive argument:
(1) Base Case: $k=3$
For $p=p_3=5$ where $p\# = 30$, we need only consider $w \in \{1,2,3,5,6,10,15,30\}$
*
*For $w=30$, $x=m+1$.
*For $w=15$, $x = m + 1 + c$ where $m \equiv c \pmod 2$.
*For $w=10$, $x = m + 1 + \dfrac{c^3-3c+2}{2}$ where $m \equiv c \pmod 3$
*For $w =6$, $x = m + 2 - d$ where $m \equiv c \pmod 5$ and $c \equiv d \pmod 2$
*For $w = 5$, if $c < 5$, $x = m + 5 - c$ else $x = m+6$ where $m \equiv c \pmod 6$
*For $w = 3$, if $c < 7$, $x = m + 7 - c$ else $x = m + 11 - c$ where $m \equiv c \pmod {10}$
*For $w = 2$, if $c < 11$, $x = m + 11 - c$ else $x = m + 17 - c$ where $m \equiv c \pmod {15}$
*For $w = 1$, if $c < 29$, $x = m + 29 - c$ else $x = m + 30$ where $m \equiv c \pmod {30}$
(2) Assume up to some $k \ge 3$, for all integer $m, w | p_k\#$, there exists $x$ where $m < x \le m + \dfrac{p_k\#}{w}$ and $\text{gcd}\left(x[x+2],\dfrac{p_k\#}{w}\right)=1$
(3) For all $m,w | p_{k+1}\#$, there exists $x$ where $m < x \le m + \dfrac{p_{k+1}\#}{w}$ and $\text{gcd}\left(x[x+2],\dfrac{p_{k+1}\#}{w}\right)=1$
*
*If $w$ divides $p_k\#$:
*
*There exists $x_1, x_2, \dots x_{p_{k+1}}$ where $m+(i-1)\dfrac{p_k\#}{w} < x_i \le m+i\dfrac{p_k\#}{w}$ from step (2).
*Now, $x_1, x_1+\dfrac{p_k\#}{w}, \dots, x_1+(p_{k+1}-1)\dfrac{p_k\#}{w}$ and $x_1+2, x_1+2 + \dfrac{p_k\#}{w}, \dots, x_1+2 + (p_{k+1}-1)\dfrac{p_k\#}{w}$ form complete residue systems modulo $p_{k+1}$.
*So, it follows that at most two of $x_1, \dots x_{p_{k+1}}$ don't have the property where $m < x_i \le m + \dfrac{p_{k+1}\#}{w}$ and $\text{gcd}\left(x[x+2],\dfrac{p_{k+1}\#}{w}\right)$
*
*If $w$ does not divide $p_k\#$:
*
*In this case, $p_{k+1}$ divides $w$ and there exists $w' = \dfrac{w}{p_{k+1}}$ such that $\dfrac{p_{k+1}\#}{w} = \dfrac{p_k\#}{w'}$
*In this case, the $x$ that solves for $\dfrac{p_k\#}{w'}$ also solves for $\dfrac{p_{k+1}\#}{w}$
Is this argument valid? Did I make a mistake?
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We can solve for any integer $A$, not just for primorial divided by something, and proof is clearly easier.
Here, we should state that, for any positive number $m$, there is a number $x$ in $[m, m+A)$ which $gcd(x(x+2), A) = 1$.
The necessary/sufficient condition for $gcd(x(x+2), A) = 1$ is, $gcd(x, A)=1$ and $gcd(x+2, A) = 1$.
It is obvious that the GCD will be $1$ when $x \ mod \ A = A - 1$, because $(x+2) \ mod \ A = 1$ will also hold, and it is clear that both GCD is $1$.
Additionally, for any positive number $m$, there is a number $x$ in $[m, m+A)$ that $x \ mod \ A = A-1$, because all possible $x \ mod \ A$ will appear in $[m, m+A)$.
|
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Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ . Given three positive numbers $a,\,b,\,c$ . Prove that $$(\!abc+ a+ b+ c\!)^{3}\geqslant 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!).$$
My own problem is given a solution, and I'm looking forward to seeing a nicer one(s), thank you !
Solution. Without loss of generality, we can suppose $(a- 1)(b- 1)\geqslant 0 \rightarrow 1+ ab\geqslant a+ b$. Then
\begin{align*} (abc+ a+ b+ c)^{3}&= \left ( c(1+ ab)+ a+ b \right )^{2}\left ( c(1+ ab)+ a+ b \right )\\& \geqslant 4\,c(1+ ab)(a+ b)\left ( c(a+ b)+ (a+ b) \right )\\&= 2c(a+ b)^{2}(2+ 2ab)(1+ c)\\& \end{align*}
Again, by using a.m.-g.m.-inequality, we have $$2c\left(a+ b\right)^{2}\left( 2+ 2ab\right)\left( 1+ c\right)\geqslant 8abc\left(1+ a\right)\left(1+ b\right)\left(1+ c\right)$$
That is q.e.d!
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Hint: By AM-GM,
$$(abc+a) + (b+c) \geqslant 2a\sqrt{bc}+2\sqrt{bc} = 2\sqrt{bc}(1+a)$$
Now multiply three such inequalities (after cyclical shift).
|
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Given $a,b,c,d\in\mathbb{Z}_{+}$ with $a\geqq b>c>d\geqq 0$ and $ac+bd=(\!b+d+a-c\!)(\!b+d-a+c\!)$. Prove $ab+cd$ is not prime. Given positve integers $a, b, c, d$. For $a\geqq\!b>\!c>\!d\geqq\!0$ and $ac+\!bd= (\!b+ d+ a- c\!)(\!b+ d- a+ c\!)$
Prove $ab+ cd$ is not a prime number.
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
We have
$$(\!ab+ cd\!)(\!ab- bc- cd\!)\!= (\!b^{2}- c^{2}\!)(\!c^{2}- ca+ a^{2}\!)+ c^{2}\left (\!ac+ bd- (\!b+ d- a+ c\!)(\!b+ d+ a- c\!)\!\right )\!=$$
$$= (b^{2}- c^{2})(c^{2}- ca+ a^{2})$$
where $(\!2a+ b- c+ 2d\!)(ab- bc- cd)= (\!b- c\!)\left (\!a^{2}+ b^{2}+ c^{2}+ d^{2}+ bc+ 2ad+ (\!b- c\!)(\!a+ d\!)\!\right )+$
$$+ (b+ c)\left (ac+ bd- (b+ d- a+ c)(b+ d+ a- c) \right )=$$
$$= (b- c)\left (a^{2}+ b^{2}+ c^{2}+ d^{2}+ bc+ 2ad+ (b- c)(a+ d) \right )> 0$$
$$\therefore\,ab- bc- cd> 0$$
$$(a+ b- c+ d)\left ( (b^{2}- c^{2})- (ab- bc- cd) \right )= (b- c)(c- d)(d+ a- c)-$$
$$- b\left ( ac+ bd- (b+ d- a+ c)(b+ d+ a- c) \right )=$$
$$= (b- c)(c- d)(d+ a- c)> 0\,\therefore\,b^{2}- c^{2}> ab- bc- cd$$
$$(c^{2}- ca+ a^{2})- (ab- bc- cd)= cd+ c^{2}- c(a- b)+ (a- b)^{2}+ b(a- b)> 0$$
$$\therefore\,c^{2}- ca+ a^{2}> ab- bc- cd$$
q.e.d
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We prove that $a^nb^m+c^md^n$ is composite for odd $n$ and arbitrary $m.$
$ a + b - c + d | ac + bd \rightarrow a + b - c + d |(a + b)(a + d)$
$ \therefore \exists k \in N$ ; $ (a + b)(a + d) = k(a + b - c + d)$
Assume that $ gcd(a + d,a + b - c + d) = 1 \implies a + b = l(a + b - c + d)$, where $l$ is a divisor of $ k.$
If $ l = 1\implies c = d$, a contradiction.
If $ l \geq 2 \rightarrow a + b \geq 2(a + b - c + d)$, so $2c \geq a + b + 2d$ which is again a contradiction.
Hence, $ \gcd(a + d,a + b - c + d) > 1.$
Let $ \gcd(a + d,a + b - c + d) = z > 1$
so $ z|a + d ,z|a + b - c + d \implies z | b - c.$
Thus $ a \equiv - d (\mod z)$ and $ b \equiv c (\mod z).$
Finally \begin{cases} a^n \equiv - d^n (\mod z)\\
b^m \equiv c^m (\mod z)
\end{cases}
Therefore, $ a^nb^m + c^md^n \equiv 0 (\mod z)$
|
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Integral by the residue theorem
Check that the value of
$$I(\alpha)=28\int_{-\infty}^\infty \frac{1-x^2}{2+3x^2+2\alpha^2x^4}\text{d}x$$
is
$$I(\alpha)=14\pi\left(1-\frac{1}{\alpha}\right)\sqrt{\frac{2}{4\alpha+3}}$$
for some $\alpha>0$.
I've tried to apply the Residue Theorem to this integral with the contour $C_R\cup C$, where $C_R$ is the semicircle of radius $R$ centered at the origin
(with initial point $(R,0)$ and final point $(-R,0)$) and $C$ is the segment $(-R,R)$ (with initial point $(-R,0)$ and final point $(R,0)$).
$\textbf{Lemma}$. Suppose $f(z)$ is defined in the upper half-plane. If there is an $a > 1$ and $M > 0$ such that $|f(z)| <\frac{M}{|z|^a}$ for large $|z|$. Then,
$$\lim_{R\to \infty} \int_{C_R} f(z)\text{d}z=0$$
The roots of the polynomial $p(x)=2+3x^2+2\alpha^2x^4$ are
$$x_1=-\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_2=\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$
$$x_3=-\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_4=\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$
By the previous lemma and the residue theorem, we have that
$$I(\alpha)=\int_{-\infty}^\infty \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=\lim_{R\to \infty} \int_C \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=2\pi i \sum_{i=1}^n\text{Res}(f,x_i)$$
However, I don't know how to discuss what roots are in the upper half-plane due to the parameter $\alpha$.
Can anyone continue from here or give some hints to follow?
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You are on the right track, and only needs a bit of extra input to get to the answer. Assume for a moment that $\alpha > 3/4$. Then the zeros of $2+3t+3\alpha^2 t^2 = 0$ are
$$ t = \frac{-3 \pm i\sqrt{16\alpha^2-9}}{4a^2} = \left( \frac{\sqrt{4\alpha-3} \pm i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \right)^2. $$
Now, among the square roots of these values, the only ones with positive imaginary parts are
$$ x_{(1)} = \frac{\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \qquad \text{and} \qquad x_{(2)} = \frac{-\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha}. $$
Then, after some painful algebra,
$$ I(\alpha) = 2\pi i \sum_{k=1}^{2} \operatorname{Res}(f, x_{(k)}) = 14\pi \left(1 - \frac{1}{\alpha}\right) \sqrt{\frac{2}{4\alpha+3}}. $$
At this point, this equality is established only for $\alpha > 3/4$. But since both sides are analytic near $(0, \infty)$, it extends to all of $(0, \alpha)$ by the principle of analytic continuation.
|
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Stuck on this - In ABC right triangle AC= $2+\sqrt{3}$ and BC = $3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$ In ABC right triangle $AC= 2+\sqrt{3}$ and $BC = 3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$
Here's my strategy of solving this, I'm not sure if it's correct, if you find my explanation hard to understand you can just ignore and write the solution in your own way, thanks.
1. first area of main triangle, we know AC and CB so it'll be easy to calculate that,
2. to find the radius, we'll reflect triangle ABC on the left side of the circle, turning it into circle inscribed in isosceles triangle, and find it with the formula
3. to find the area of $AMD$ I'll subtract the area of sector $OMD$, triangle $OAD$ and triangle $CDB$ from triangle $AMD$,
4. $DBC$ is an isosceles triangle, so $CB=DB$, then to find the area, I split it into 2 right triangles(it becomes 90 30 60 triangle) and find its height. So we got the Area of $DBC$
5. Now similarly $OAD$ is isosceles, $OD=OC=radius$ of the circle which we "found"
also, split this in 2 to get right triangles and then calculate with Pythagorean theorem to find the height so we get Area of $OMD$ too, maybe we could find angles with trigonometry? I don't know that, and if we get the angle of $DOA$ we could find the sector $OMD$ as well and subtract it to the main triangle so we get the area of $AMD$.
|
In a video by Mathematics professor Michael Penn, this is solved first by finding out the hypotenuse of the triangle and writing it in the form $a+b\sqrt{3}$, where $a, b \in \mathbb{R}$. Then, he finds the area of the triangle ($AOD$) and the sector ($EOD$) as shown in Figure 1 and subtracts the area of the sector from the triangle to find the shaded area.
Figure 1
However, the shaded area can also be found out using calculus.
We need to find the equation of the line $AB$ and the circle. Before that, we need to find the radius of the circle.
Figure 2
In $\Delta ABC$, let $\angle ABC = \theta$. Then,
\begin{align*}
\tan \theta &= \frac{AC}{CB}=\frac{1}{\sqrt{3}} \\
\implies \theta &= \frac{\pi}{6}
\end{align*}
Let us consider $\Delta COB$ and $\Delta DOB$. It can be easily shown that $\Delta COB \cong \Delta DOB$. Hence, $OB$ bisects $\angle ABC$.
$$
\therefore \angle CBO = \frac{\pi}{12}
$$
In $\Delta CBO$,
\begin{align*}
\tan{\frac{\pi}{12}} &= \frac{r}{3+2\sqrt{3}} \\
\therefore r &= \sqrt{3}
\end{align*}
Now, consider a cartesian coordinate system with point $C$ as the origin, as shown in Figure 2. Equation of line $AB$ is given by,
\begin{equation}
y=f(x)=\frac{(12+7\sqrt{3})-x(2+\sqrt{3})}{(3+2\sqrt{3})} \tag{1}\label{eq:1}
\end{equation}
The circle is centred at $(0, \sqrt{3})$. The equation of the circle is,
\begin{equation}
y=g(x)=\sqrt{3} \pm \sqrt{3-x^2} \tag{2}\label{eq:2}
\end{equation}
Here, the $\pm$ sign joins the upper half of the circle with the lower half to create the desired circle. But our purpose can be served with the upper half of the circle itself, i.e., the semicircle (The blue semicircular curve as shown in Figure 3. So, we can replace the $\pm$ with $+$ to reduce calculational labour.
Figure 3
If we solve for $x$ and $y$ in equations \eqref{eq:1} and \eqref{eq:2}, we will get the point $D(x, y)$ at which line $AB$ touches the circle (or semicircle).
We get $x=\frac{\sqrt{3}}{2}$ and $y=\frac{3}{2}+\sqrt{3}$. To compute the shaded area, we only need $x$. We will integrate $f(x)$ and $g(x)$ from $x=0$ to $x=\sqrt{3}/2$
Area under the line AB from $x=0$ to $x=\sqrt{3}/2$ is,
\begin{align}
I_{1} &= \int_{0}^{\sqrt{3}/2} f(x)dx \\
&= \frac{1}{3+2\sqrt{3}}\left(\frac{45\sqrt{3}+78}{8}\right) \\
&= \frac{7\sqrt{3}+12}{8} ~~\text{ sq. unit} \tag{3}\label{eq:3}
\end{align}
Also, area under the upper semicircular curve from $x=0$ to $x=\sqrt{3}/2$ is,
\begin{align}
I_{2} &= \int_{0}^{\sqrt{3}/2} g(x)dx \\
&= \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4} ~~\text{ sq. unit} \tag{4}\label{eq:4}
\end{align}
Finally, the required shaded area is given by,
\begin{equation}
I_{1} - I_{2} = \frac{7\sqrt{3}+12}{8} - \frac{3\sqrt{3}+12}{8} + \frac{\pi}{4}
\end{equation}
\begin{equation}
\boxed{\therefore ~~\text{Reqd. Area} = \left( \frac{\sqrt{3}}{2} - \frac{\pi}{4} \right) ~~\text{ sq. unit}} \tag{5}\label{eq:5}
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3272856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
}
|
Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get
\begin{bmatrix}
0&0 &0 \\
-7&7 &7 \\
0&0 &0
\end{bmatrix}
So we have geom multi. = alg. multi, so our matrix is complete.
Take $v_1 =
\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
$
Similarly, solve $\det(A-9\lambda)$ to get
$
\begin{bmatrix}
-7&0 &0 \\
-7&0 &7 \\
0&0 &-7
\end{bmatrix}
$
So take $v_2 =
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
$
So the general solution should be $x(t) = C_1e^{2t}\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+C_2e^{2t} \begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+ C_3e^{9t}\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$
However, according to the back of the book solution, this is incorrect. What am I missing here? Thank you.
|
The system $$x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}$$
is simply
$$ x_1'=2x_1$$
$$ x_2'=-7x_1+9x_2+7x_3$$
$$ x_3'=2x_3$$
Therefore $$x_1=c_1e^{2t}$$ and $$x_3=c_2 e^{2t}$$
Solving for $$x_2'=9x_2+7(c_2-c_1)e^{2t}$$
we get $$x_2=c_3e^{9t}+(c_1-c_2)e^{2t}$$
Thus we have $$x(t) =
c_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} +
c_2 e^{2t} \begin{pmatrix} 0\\-1 \\ 1 \end{pmatrix} +
c_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3273970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.