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Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{3}}}\right)$. I can put the squareroot sign over the whole fraction, but that still doesn't really help me get an actual number for theta.
ALSO: It is a rule that I add $\pi$ to theta if $a < 0$. Why? Looking at the diagram of the triangle form by the complex vector $z$ and the real axis, the 'triangle' is in the second quadrant. Hence, the theta we find with $\theta = \tan^{-1}(\frac{b}{a})$ is the angle closes to the real axis on the left-side. However, we measure angle going counter-clockwise. Shouldn't we do the following computation to get the angle going clock-wise: $\pi - \tan^{-1}(\frac{b}{a})$?
|
Note that:
$$
\dfrac{\sqrt{2+\sqrt{3}}}{-\sqrt{2-\sqrt{3}}}=-\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=-\sqrt{\dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}=-(2+\sqrt{3})
$$
since $2+\sqrt{3}=\tan \left(\dfrac{5 \pi}{12}\right)$, and $\theta$ must be in the second quadrant we have $\theta =\left(\dfrac{7 \pi}{12}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\begin{align}
& \lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} = \\
& \lim_{x\to0}e^{\ln\left(\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{x^2}\ln\left(\left(\frac{\sin(x)}{x}\right)\right)} = \\
& e^{\lim_{x\to0}\frac{\ln\left(\left(\frac{\sin(x)}{x}\right)\right)}{x^2}} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \\
& e^{\lim_{x\to0}\frac{x}{2x \sin(x)}\cdot\frac{\cos(x)x -1\cdot\sin(x)}{x^2}} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\tan(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\tan(x)}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{1}{\cos^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(1 - \infty\right)} = \\
& e^{-\infty} = 0\\
\end{align}
$$
Edit #1:
Continuing after the mistake of the $\tan(x)$:
$$\begin{align}
& e^{\lim_{x\to0}\frac{1}{2}\cdot\frac{1}{\sin(x)}\cdot\left(\frac{\cos(x)}{x} - \frac{\sin(x)}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\cot(x)}{x} - \frac{1}{x^2}\right)} = \\
& e^{\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\frac{1}{\tan(x)}}{x} - \frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{1}{\tan(x)}}{x} - \lim_{x\to0}\frac{1}{x^2}\right)} \stackrel{\frac{0}{0}}{\stackrel{=}{L}} \quad\quad \text{(LHopital only for the left lim)} \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\frac{\frac{-1}{\cos^2(x)}}{\tan^2(x)}}{1} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{-1}{\sin^2(x)} - \lim_{x\to0}\frac{1}{x^2}\right)} = \\
& e^{\frac{1}{2}\cdot\left(-\infty - \infty\right)} = 0\\
\end{align}
$$
|
Using Taylor series:
$$\frac{\sin(x)}{x}=1- \frac{1}{3!}x^2+o(x^2).$$
So using this formula we get:
$$
\ln \left( \frac{\sin(x)}{x}\right)=\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right).
$$
Now we can use the Taylor expansion for the $\ln$ function so we get:
$$
\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right)= - \frac{1}{3!}x^2+o(x^2).
$$
Now divide all by $x^2$ and compute the limit while $x$ goes to zero:
$$
\frac{- \frac{1}{3!}x^2+o(x^2)}{x^2}
$$
as $x\rightarrow 0$, we obtain that
$$
\lim_{x \rightarrow 0} \ln \left( \frac{\sin(x)}{x}\right)=
-\frac{1}{6}.
$$
At the end
$$
\lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x} \right)^{\frac{1}{x^2}}=
e^{\lim_{x \rightarrow 0}\frac{\ln\frac{\sin(x)}{x}}{x^2}}=
e^{-\frac{1}{6}}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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|
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula:
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!
|
Here is a solution using complex numbers: Let $\omega = \exp(i\pi/2n)$. Then
\begin{align*}
\prod_{k=1}^{n} \left(1 + \cos \left( \tfrac{2k-1}{2n}\pi \right) \right)
&= \frac{1}{2^{n}} \prod_{k=1}^{n} \left(1 + 2\Re(\omega^{2k-1})+ 1 \right) \\
&= \frac{1}{2^{n}} \prod_{k=1}^{n} (1 + \omega^{2k-1})(1 + \omega^{-(2k-1)}).
\end{align*}
But notice that $\omega^{-(2n-1)}, \cdots, \omega^{2n-1}$ are $2n$ distinct zeros of the polynomial $ z^{2n} + 1$. Consequently we have
\begin{align*}
\prod_{k=1}^{n} \left(1 + \cos \left( \tfrac{2k-1}{2n}\pi \right) \right)
&= \frac{1}{2^{n}} \prod_{k=1}^{n} (-1 - \omega^{2k-1})(-1 - \omega^{-(2k-1)}) \\
&= \frac{1}{2^{n}} (z^{2n}+1)|_{z=-1} \\
&= \frac{1}{2^{n-1}}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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|
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws.
|
[\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{x{{\left( {x + 1} \right)}^{x + 1}}}}
{{{{\left( {x + 2} \right)}^{x + 2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{x}
{{x + 2}}.\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 1}}
{{x + 2}}} \right)^{x + 1}} = 1.\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1}
{{x + 2}}} \right)^{x + 1}} \hfill \\
= \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{1}
{{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{ - \frac{{x + 1}}
{{x + 2}}}} = {\left[ {\mathop {\lim }\limits_{x \to \infty } {{\left( {1 - \frac{1}
{{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{\mathop {\lim }\limits_{x \to \infty } - \frac{{x + 1}}
{{x + 2}}}} = {e^{ - 1}} = \frac{1}
{e} \hfill \\
\end{gathered} ]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
}
|
Partial differential equation (first order) I don't have ideas to solve the following PDE of the 1st order
$$
(x^2 - y^2 + 1)u_{x} + 2xyu_{y} = 0
$$
Could you give me a hint ?
Thanks,
R.
|
$(x^2-y^2+1)u_x+2xyu_y=0$
$\dfrac{x^2-y^2+1}{2xy}u_x+u_y=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dx}{dt}=\dfrac{x^2-y^2+1}{2xy}=\dfrac{x^2-t^2+1}{2xt}$
$2x\dfrac{dx}{dt}=\dfrac{x^2}{t}-\dfrac{t^2-1}{t}$
$\dfrac{d(x^2)}{dt}-\dfrac{x^2}{t}=-t+\dfrac{1}{t}$
$\dfrac{1}{t}\dfrac{d(x^2)}{dt}-\dfrac{x^2}{t^2}=-1+\dfrac{1}{t^2}$
$\dfrac{d}{dt}\left(\dfrac{x^2}{t}\right)=-1+\dfrac{1}{t^2}$
$\dfrac{x^2}{t}=x_0-t-\dfrac{1}{t}$
$x^2=x_0t-t^2-1=x_0y-y^2-1$
$\dfrac{du}{dt}=0$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)=f\left(\dfrac{x^2+y^2+1}{y}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1135346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+(x^2+2x-8)(6x^2+3)$
$(4x^4+4x^3+6x^2+6x+6x^4+12x^3-48x^2+3x^2+6x-24)$
$=10x^4+16x^3+39x^2+6x-18$
|
I would do it simply by expanding and then repeatedly using the Power Rule (using the Product Rule would be overkill I think):
$$
(2x^3+3x)(x-2)(x+4) = 2x^5+4x^4-13x^3+6x^2-24x.
$$
Now differentiate with respect to $x$, obtaining
$$
\frac{d}{dx}(2x^5+4x^4-13x^3+6x^2-24x) = 10x^4+16x^3-39x^2+12x-24.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1138666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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|
Trigonometrical limit $$\lim\limits_{x \to 0} \frac{x^4\cos(\frac{1}{x})} {\sin^3x},$$
I have a problem with this limit. I tried to use L'Hôpital's rule but it is not effective. Please help.
|
$$\lim\limits_{x → 0} \frac{x^4 \cos \left(x^{-1}\right)}{\sin ^3\left(x\right)} \times \frac{x^{-3}}{x^{-3}}$$
$$\lim\limits_{x → 0} \frac{x\cos \left(x^-1\right)}{x^{-3}\sin ^3\left(x\right)}$$
$$\lim\limits_{x → 0} \frac{ \frac{\cos \left(x^{-1}\right)}{x^{-1}} }{ \frac{\sin ^3\left(x\right)}{x^3} } $$
*
*Using $$\lim\limits_{x → 0} \frac{\sin \left(x\right)} {x}=1 \quad \textrm{ & } \lim\limits_{x → 0} \frac{\cos \left(1/x\right)} {1/x}=0$$
$$\therefore \frac{ \lim\limits_{x → 0}
\frac{\cos \left(x^{-1}\right)}{x^{-1}} }{ \lim\limits_{x → 0} \left( \frac{\sin \left(x\right)}{x} \right)^3 } =0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1138866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me that:
$$\frac{x}{(x-1)^2} = \frac{1}{(x-1)} + \frac{1}{(x-1)^2}$$
How to intuitively think about this partial fraction expansion? I've seen some examples but suddenly these conter intuitive examples opo out and I get confused. I can check that his is true but I couldn't find this expansion by myself
Then:
$$\int \frac{x}{(x-1)^2} dx = \int \frac{1}{(x-1)} + \frac{1}{(x-1)^2}dx = \ln (x-1) + (x-1)^{-1}$$
Then:
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+\frac{x}{(x-1)^2}dx = \frac{x^2}{2} + 2x + 3[\ln(x-1)+(x-1)^{-1}]$$ but wolfram alpha gives another answer. What I did wrong?
|
First: search for:
$$
\dfrac{x}{(x-1)^2}=\dfrac{1}{(x-1)^2}+\dfrac{A}{x-1}
$$
and find:
$$
x=1+Ax-A \iff x(1-A)=1-A
$$since this must be true $\forall x$ you must have $A-1=0$ and $A=1$.
Now integrates as you have done but be careful that you have a mistake in your work:
$$
\int \dfrac{1}{(x-1)^2} dx = - \dfrac{1}{x-1}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
}
|
Integral with un-intuitive U-Substitution On a recent midterm we were given the following integral and were expected to integrate it.
$$\int_{-27}^{-8} \frac{\mathrm dx}{x + x^{2/3}}$$
|
$$x+x^{\frac{2}{3}}=\\x^{\frac{3}{3}}+x^{\frac{2}{3}}=\\x^{\frac{2}{3}}(x^{\frac{1}{3}}+1)=\\\sqrt[3]{x^2}(\sqrt[3]{x}+1)$$now apply this$$u=\sqrt[3]{x}\\x=u^3\\dx=3u^2du\\\frac{1}{x+x^{\frac{2}{3}}}dx=\\\frac{1}{u^2(u+1)}*(3u^2du)=\\\frac{3du}{u+1}\\$$so $$\int\frac{1}{x+x^{\frac{2}{3}}}dx=\int \frac{3du}{u+1}=3ln|u+1|+c\\=3ln|\sqrt[3]{x}+1|+c$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1139635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?
|
Hint:
Use finite differences as was suggested earlier by @user314 and Robert Israel. They behave similar to derivatives.
With $\Delta [p](x) \colon = p(x+1) - p(x)$, we have for $p(x) = a_n x^n + \cdots $ a polynomial of degree $n$,
$$\Delta^n [p(x)] \equiv n ! \cdot a_n$$
Therefore, for our monic polynomial $p(x)= (x+a)(x+b)(x+c)(x+d)$ of degree $4$ we have
$$\Delta^4 [p] (x)= p(x+4) - \binom{4}{1}p(x+3) + \binom{4}{2} p(x+2) - \binom{4}{1} p(x+1) + p(x) \equiv 4! = 24$$
and so for $x=1$ we get
$$p(5) = 4\, p(4) - 6\, p(3) + 4 \,p(2) - p(1) + 24 = \,...$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
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|
What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$? Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come to the conclusion, using the pythagorean theorem, that the height of the trapezoid is $h=\sqrt { 1-\frac { x^{ 2 } }{ 4 } } $
The formula for the area of a trapezoid is $A=\frac { a+b }{ 2 } (h)$
I am basically solving for the $a$ here and so far I have:
$A=\frac { x+2 }{ 2 } (\sqrt { 1-\frac { x^{ 2 } }{ 4 } } )$
I simplified this in order to be able to easily take the derivative of it as such:
$$A=\frac { x+2 }{ 2 } (\sqrt { \frac { 4-x^{ 2 } }{ 4 } } )$$
$$A=\frac { x+2 }{ 2 } (\frac { \sqrt { 4-x^{ 2 } } }{ 2 } )$$
$$A=(\frac { 1 }{ 4 } )(x+2)(\sqrt { 4-x^{ 2 } } )$$
Then I took the derivative:
$$A'=\frac { 1 }{ 4 } [(1)(\sqrt { 4-x^{ 2 } } )+(x+2)((\frac { 1 }{ 2 } )(4-x^{ 2 })^{ -1/2 }(-2x)]$$
This all simplified to:
$$A'=\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } ]$$
Next, I set the derivative equal to zero in order to find the maximum point (I know that I can prove that it is actually a max point by taking the second derivative later)
$$\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } ]=0$$
$$\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } } } =0$$
$$\frac { 4-x^ 2-2x^ 2-4x }{ \sqrt { 4-x^ 2 } } =0$$
$$4-x^{ 2 }-2x^{ 2 }-4x=0$$
$$-3x^{ 2 }-4x+4=0$$
so I got:
$$x=-2\quad or\quad x=\frac { 2 }{ 3 }$$
$x=-2$ wouldn't make sense so I chose $x=\frac { 2 }{ 3 }$ and plugged it back into my original formula for the area of this trapezoid but my answer doesn't seem to match any of the multiple choice solutions. Where did I go wrong?
|
I also found the area equation for the trapezoid, but found $h$ in a different way.
The area of a trapezoid is $A=\dfrac{(b_1+b_2)H}{2}$. Base $1$ becomes $2$ units, Base $2$ is $2x$, and using the Pythagorean theorem you can find that $H=\sqrt{1-x^2}$. You then plug these values into the Area formula and find its derivative.
$A=(1+x)\sqrt{1-x^2}$, and thus $A'=\sqrt{1-x^2}+ (-x-x^2)(\sqrt{1-x^2})^{-1}$
You can then set the derivative equal to zero and solve for $x$.
$\sqrt{1-x^2} + (-x-x^2)(\sqrt{1-x^2})^{-1}=0$
Move $\sqrt{1-x^2}$ to the other side of the equation.
$(-x-x^2)(\sqrt{1-x^2})^{-1}=-\sqrt{1-x^2}$
Multiply both sides by $\sqrt{1-x^2}$.
$(-x-x^2)=-(1-x^2)$
$-x-x^2=x^2-1$
$x^2+x-1=0$
Use quadratic formula or other method to deduce that $x = -1, 0.5$. $x$ can't be negative as you can't have a negative measurement, so $x = 0.5$ is the only answer.
Plugging that back into the Area formula you get $\dfrac{3\sqrt{3}}{4}$ or $1.299 un^2$ as your biggest area.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1140339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-10 x-7 x^2+x^4}}+\frac{1+3x-2x^3}{\sqrt{5-2 x-3 x^2+x^4}}$$.
But failed to solve $f'(x)=0$ for finding $f(x)_{max}$.
I would be glad for your help.
Thanks.
|
here is geo solution which might be make things clear:
let $y=x^2,A(x,y),B(5,4),C(1,2)$,your problem become when $A$ moves on $y=x^2$, find max of $AB-AC$.
it is trivial $AB-AC <CB$, except when $A=D$ then $DB-DC=CB$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
|
Let $u=\sqrt{\frac{x}{x+1}}$. Then
$$
\begin{align}
\int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x
&=\int u\,\mathrm{d}\frac{u^2}{1-u^2}\\
&=\frac12\int u\,\mathrm{d}\left(\frac1{1-u}+\frac1{1+u}\right)\\
&=\frac12\int u\left(\frac1{(1-u)^2}-\frac1{(1+u)^2}\right)\,\mathrm{d}u\\
&=\frac12\int\left(\frac1{(1-u)^2}-\frac1{1-u}+\frac1{(1+u)^2}-\frac1{1+u}\right)\,\mathrm{d}u\\
&=\frac12\left(\frac1{1-u}+\log(1-u)-\frac1{1+u}-\log(1+u)\right)+C\\
&=\frac{u}{1-u^2}+\frac12\log\left(\frac{1-u}{1+u}\right)+C\\[6pt]
&=\sqrt{x(x+1)}+\log\left(\sqrt{x+1}-\sqrt{x}\right)+C
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Roll a die $n$ times ($n$ is a natural number). What is the probability that 1 and 6 are observed at least once? Note: This is a homework problem so I cannot accept solutions. I would like suggestions as to how to proceed.
I have that each trial of rolling a die is independent. So I can say:
Let $P(1)$ = Rolling a 1 and $P(6)$ = Rolling a 6. We want to find $P(1\cap 6)$. Since rolling a die and getting an outcome in a trial is independent of other trials: $P(1\cap6)=P(1)P(6) $
It's unclear to me how to find the probability that you roll a 1 if you roll $n$ times. Would it be $P(1)=1/6^n$
|
Let $t_{ab}(n)$ be the number of sequences of length $n$ containing $a$ and $b$ (assuming $a \ne b$).
Let $t_{a}(n)$ be the number of sequences of length $n$ containing $a$.
Let $t(n)$ be the number of sequences of length $n$.
The question is to investigate $t_{16}(n)$. Each sequence starts with a $1$, $6$, or something else, so:
$$t_{16}(n) = t_{1}(n - 1) + t_{6}(n - 1) + 4 ~ t_{16}(n-1) \tag{A}$$
Similarly, $t_{a}(n)$ either starts with $a$ or it doesn't:
$$t_{a}(n) = t(n-1) + 5 ~ t_{a}(n - 1) \tag{B}$$
Finally,
$$t(n) = 6^n \tag{C}$$
(C) is solved, so (B) becomes:
$$t_{a}(n) = 6^{n-1} + 5 ~ t_{a}(n - 1) \tag{B2}$$
A recursive equation which can be solved:
$$\begin{align}
%
-1~t_{a}(n) + 5~t_{a}(n-1) &= 6^{n-1} \\
%
-1~t_{a}(n + 1) + 5~t_{a}(n) &= 6\cdot 6^{n-1} \\
%
t_{a}(n + 1) &= 11~t_{a}(n) - 30~t_{a}(n-1) \\
%
\end{align}$$
Which is a linear recursive equation, and it solves to
$$t_{a}(n) = 6^n - 5^n \tag{B3}$$
So (A) becomes
$$t_{16}(n) = 2\left(6^{n-1} - 5^{n-1}\right) + 4 ~ t_{16}(n-1) \tag{A2}$$
Solving similarly:
$$\begin{align}
%
t_{16}(n) - 4 ~ t_{16}(n-1) &= 2~6^{n-1} - 2~5^{n-1} \\
%
t_{16}(n+1) - 4 ~ t_{16}(n) &= 12~6^{n-1} - 10~5^{n-1} \\
%
t_{16}(n+2) - 4 ~ t_{16}(n+1) &= 72~6^{n-1} - 50~5^{n-1} \\
\end{align}$$
Eliminating the exponents between the 3 equations, it yields:
$$t_{16}(n+2) = 15~t_{16}(n+1) - 74~t_{16}(n) + 120~t_{16}(n-1)$$
Another linear recursive equation, which solves to:
$$t_{16}(n) = 6^{n} - 2\cdot 5^{n} + 4^{n} \tag{A3}$$
And the resulting probability is then:
$$\boxed{\huge{\frac{6^{n} - 2\cdot 5^{n} + 4^{n}}{6^n}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
|
Also, $\cos x = \cos^2 (x/2) - \sin^2 (x/2)$.
Then letting $y=x/2$,
$$\frac{1-\cos x}{\sin x} = \frac{1 - \cos^2 y + \sin^2 y}{2 \sin y \cos y} = \frac{2 \sin^2 y}{2 \sin y \cos y} = \frac{\sin y}{\cos y} = \tan (x/2).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x^2 y''+(-2x-x^3)y'+5y=0$ Ok so for me I am having trouble solving this equation to get
$y=C_1y_1+C_2y_2$, but I'm having trouble dealing with the (-2x-x^3) part. Usually I would isolate $y''$, then make $y=x^m$, then go from there, but I'm having trouble dealing with the $(-2x-x^3)$ part. Thanks!
When I tried that approach, I got $[x^{3/2}x^{x^2/2}x^{(\sqrt{x^2+6x-11})/2}, x^{3/2}x^{x^2/2}x^{-(\sqrt{x^2+6x-11})/2}]$ form a basis, but it just seems too complicated when I need to solve
$y_1(1)=-4, y'_1(1)=5, y_2(1)=-5,y'_2(1)=5$
The point is to find the Wronskian, $w(x)=y_1(x)y'_2(x)-y_2(x)y'_1(x)$ for x>0
|
Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du}=4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}$
$\therefore x^2\left(4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}\right)+(-2x-x^3)2x\dfrac{dy}{du}+5y=0$
$4x^4\dfrac{d^2y}{du^2}-2x^2(x^2+1)\dfrac{dy}{du}+5y=0$
$4u^2\dfrac{d^2y}{du^2}-2u(u+1)\dfrac{dy}{du}+5y=0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Another method of proving $x^3+y^3=z^3$ has no integral solutions? The equation $$ax^2+by^2=z^3$$ has the following parametrization:
$$y=q(3ap^2-bq^2)$$
$$x=p(ap^2-3bq^2)$$
$$z=ap^2+bq^2$$ can we deduct from that the Diophantine equation $$x^3+y^3=z^3$$ has no nontrivial solutions by choosing $x=a$ and $y=b$?
|
That is easy to prove. The proof is rather long but simple.Please disregard any typos.
Case 1: ($p=-1$ and $q=1$)
\begin{eqnarray}
x &=& -x (x (-1)^2 - 3 y (1)^2) \\
2x &=& 3 y \\
\end{eqnarray}
Since $\gcd(x,y)=1$ then $x=y=z=0$
\begin{eqnarray}
y &=& 1 (3x (-1)^2 - y (1)^2) \\
2y &=& 3 x \\
\end{eqnarray}which implies as well $$x=y=z=0$$
Case 2: ($p \neq -1$ and $q \neq 1$)
In this case we have:
\begin{eqnarray}
x &=& \dfrac{3py q^2}{p^3+1 } \\
y &=& \frac{ 3qxp^2}{q^3-1}\\
\end{eqnarray}1) $ x \equiv 0 \pmod 3 $
$\gcd(x,y)=\gcd(x,q)=1$, therefore we can only have $$x=3p$$ $$yq^2=p^3+1$$ Similarly, since $3|x$ and $\gcd(y,x)=\gcd(y,p)=1$, then we must have:$$y=q$$ $$3xp^2=q^3-1$$
By replacing $x$ and $y$ by their respective values,we obtain:
$$
q^3 =p^3+1 $$
$$9p^3=q^3-1$$
Which yields:$$8p^3=0$$
Then, $p=0$ and $q=1$, which contradicts the case we are dealing with $q \neq 1$
2) $ y \equiv 0 \pmod 3 $
$\gcd(x,y)=\gcd(x,q)=1$, therefore we can only have $$x=p$$ $$3yq^2=p^3+1$$ Similarly, since $3|y$ and $\gcd(y,x)=\gcd(y,p)=1$, then we must have:$$y=3q$$ $$xp^2=q^3-1$$
By replacing $x$ and $y$ by their respective values,we obtain:
$$
9q^3 =p^3+1 $$
$$p^3=q^3-1$$
Which yields:$$8q^3=0$$
Then, $q=0$ and $p=-1$, which contradicts the case we are dealing with $p \neq {-1}$
3) The cases where $x=\pm 1$ or $y=\pm 1$ are easy to demonstrate. In every case, we have a product of 2 numbers which equal to $\pm 1$. Therefore, both must be each either $-1$ or $1$.
QED
|
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|
prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x - 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)
|
*
*$ \cos (x-60^\circ ) \times \cos ( x + 60^\circ ) = \frac{1}{2} ( \cos (2x) + \cos 120^\circ) = \frac{1}{2} \cos (2x) - \frac{1}{4}$
*$ (\frac{1}{2} \cos (2x) - \frac{1}{4}) \times \cos x = \frac{1}{4} ( \cos (3x) + \cos (x) ) - \frac{1}{4} \cos (x) = \frac{1}{4} \cos (3x) $
|
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"timestamp": "2023-03-29T00:00:00",
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|
List the elements of $\mathbb Z_2 \times \mathbb Z_3$ and write its operation table (the notation is additive). $\mathbb Z_2 \times \mathbb Z_3 = \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}.$
$$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 3) & (1, 1) & (1, 2) & (1, 3) \\
(0, 2) & (0, 2) & (0, 3) & (0, 4) & (1, 2) & (1, 3) & (1, 4) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (2, 0) & (2, 1) & (2, 3) \\
(1, 1) & (1, 1) & (1, 2) & (1, 3) & (2, 1) & (2, 2) & (2, 3) \\
(1, 2) & (1, 2) & (1, 3) & (1, 4) & (2, 2) & (2, 3) & (2, 4) \\
\end{array}
$$
Please, check my work.
|
$\mathbb Z_3=\{0,1,2\}$
\begin{array}{c|lcr}
+ & 0 &1 & 2\\
\hline
0 & 0 &1 & 2\\
1 & 1 &2 & 0\\
2 & 2 &0 & 1\\
\end{array}
$\mathbb{Z}_3\times\mathbb{Z}_2=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2) \\
(0, 1) & (0, 1) & (0, 2) & (0, 0) & (1, 1) & (1, 2) & (1, 0) \\
(0, 2) & (0, 2) & (0, 0) & (0, 1) & (1, 2) & (1, 0) & (1, 1) \\
(1, 0) & (1, 0) & (1, 1) & (1, 2) & (0, 0) & (0, 1) & (0, 2) \\
(1, 1) & (1, 1) & (1, 2) & (1, 0) & (0, 1) & (0, 2) & (0, 0) \\
(1, 2) & (1, 2) & (1, 0) & (1, 1) & (0, 2) & (0, 0) & (0, 1) \\
\end{array}
As a general idea, when you write the table group of an operation, each element should appear only once on each column and row. Of course that in $\mathbb{Z}_3, 4=1$ because $4\equiv 1\pmod 3$ but is preferable to keep the same notation for an element.
Edit: The operation on $\mathbb Z_2\times \mathbb Z_3$ is defined as follows: for $(a,b),(c,d)\in \mathbb Z_2\times \mathbb Z_3$, i.e. $a,c\in\mathbb Z_2, b,d\in\mathbb Z_3, (a,b)+(c,d)=(a+c,b+d)$. The idea when you write these elements is to keep the same writing always.
When you make a sum in $\mathbb Z_n$, the elements are classes of equivalence (in $\mathbb{Z}_n, 0=n=2n=\dots; 1=n+1=2n+1=\dots ; 2=n+2=2n+2=\dots$ and so on - basically whenever an n appears it is made $0$)and you choose to work with the smallest representants $\{0,1,\dots,n-1\}$, so if you have that for example $x+y=n+k$, you shall write $x+y=k$.
|
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|
Calculate $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$
Calculate the sum $$\displaystyle \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$$
where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, $(2n)!!=2\cdot 4 \cdots 2n$
Using Wolframalpha, the result is $\sqrt{2}-1$. But I don't have any idea to come up with that result.
Thanks a lot.
|
Since $(2n)!! = 2^n n!$ and $(2n - 1)!! = (2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$,
\begin{align} \sum_{n = 1}^\infty \frac{(2n-1)!!}{(2n)!!2^n} &= \sum_{n = 1}^\infty \frac{(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1}{2^n n!}\frac{1}{2^n}\\
& = \sum_{n = 1}^\infty \frac{\left(n - \frac{1}{2}\right)\left(n - \frac{3}{2}\right)\cdots \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}{n!}\frac{1}{2^n}\\
&= \sum_{n = 1}^\infty \frac{\left(-\frac{1}{2} - n + 1\right)\left(-\frac{1}{2} - n + 2\right)\cdots \left(-\frac{1}{2}\right)}{n!}\left(-\frac{1}{2}\right)^n\\
&= \sum_{n = 0}^\infty \binom{-1/2}{n}\left(-\frac{1}{2}\right)^n - 1\\
&=(1 - 1/2)^{-1/2} - 1 \\
&= \sqrt{2} - 1.
\end{align}
|
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|
What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's possible to color a maximum of $2$ boxes with red, and a maximum of $3$ colors with blue. Write the ordinary generating function associated with the problem and find the number of ways to color 10 boxes.
I've managed to find the generating function:
$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+\dots)^2=\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2\\ \frac{(1-x^3)(1-x^4)}{(1-x)^4}$$
But for calculating the number of colored boxes, I'd have:
$$\frac{(1-x^3)(1-x^4)}{(1-x)^4}=[1-x^3][1-x^4] \left[ \frac{1}{(1-x)} \right]^4$$
The expansion of $\left[\frac{1}{(1-x)}\right]^4$ is:
$$\left[\frac{1}{(1-x)}\right]^4=\sum_{j=0}^\infty {4+j-1 \choose j} y^k$$
But this doesn't seem too revealing. I don't know how to proceeed the counting in this exercise.
|
Already you have notice that$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+\dots)^2=\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2.$$ Also $$(1+x+x^2+\dots)^2=(1+2x+3x^2+\cdots+kx^{k-1}+\cdots)$$
$$(1+x+x^2)(1+x+x^2+x^3)=(1+2x+3x^2+3x^3+2x^4+x^5)$$
Therefore coefficient of $x^{10}$ in the product
$$(1+2x+3x^2+3x^3+2x^4+x^5)(1+2x+3x^2+\cdots+kx^{k-1}+\cdots)$$ is
$$11+(2\times10)+(3\times9)+(3\times8)+(2\times7)+6=102.$$
|
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|
If $0 \le a \lt b$ prove that $a^2 < b^2$ Below is how I prove it.
Case 1: $a = 0$
*
*$0^2 < b^2$ where $b$ is a positive number.
*$0 < b \times b$
*A positive number times a positive number is always positive.
*It is true.
Case 2: $a > 0$
*
*$a < b \Rightarrow a + x = b$
*$a^2 < b^2 \Rightarrow a^2 < (a+x)^2$
*$a^2 < (a+x) \times (a+x)$
*$a^2 < a^2 + 2ax + x^2$
*$0 < 2ax + x^2$
*It is true because $a, x$ are positive numbers.
I was wondering a) if my prove is correct and b) if there are other straightforward way to prove this?
|
You don't need cases. Just say:
$a^2 - b^2 = (a - b)(a + b)$.
Since $a \ge 0$ and $ b>0$, $a + b > 0$ and $a - b < 0$ since $a < b$, thus: $(a -b)(a +b) < 0$. So $a^2 - b^2 < 0$. Thus $a^2 < b^2$.
|
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|
Euler sequence, limit of an related sequence Study the convergence of the sequence $ \left( a_n\right)_{n\in\mathbb{N^*} }$ defined by
$$
a_{n}+e^{a_{n}}=\left( 1+\frac{1}{n}\right) ^{n}+1,~\forall n\in
\mathbb{N}
^{\ast}
$$
and find its limit. Moreover, deduce the values of limit
$$
\lim_{n\rightarrow\infty} n(1-a_n).
$$
I don't know how to obtain the condition of Bolzano-Weierstrass Theorem: a monotone and bounded sequence is convergent. The exponential function keeps me from proceeding; that's where I get stuck.
|
Since both the function $f(x) = x + e^x$ and the sequence $\left(1 + \frac{1}{n}\right)^n + 1$ are strictly increasing, it follows that the sequence $a_n$ must be strictly increasing; it is bounded because $\left(1 + \frac{1}{n}\right)^n + 1 \leq e + 1 = f(1)$ is bounded.
If $a = \lim a_n$, then $a$ must satisfy $a + e^a = 1 +e$, which implies that $a = 1$ (because $x + e^x$ is strictly increasing, hence injective).
For the second limit, let $g(x) = 1-x + e^{1-x}$ and $h(x) = \left(1 + \frac{1}{x}\right)^x$, so that $1 - a_n = g^{-1}(h(n))$. If we calculate
$$\lim_{x \to \infty} \frac{g^{-1}(h(x))}{1/x} = \lim_{x \to \infty} \frac{h'(x)}{g'(g^{-1}(h(x))) \cdot (-1/x^2)} = \lim_{x \to \infty} \frac{x^2 h'(x)}{-g'(g^{-1}(h(x))},$$
by L'Hôpital. Now $g'(y) = - (1 + e^{1-y})$ and $h'(x) = \left(1 + \frac{1}{x}\right)^x\left(\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right)$. Recalling that $g^{-1}(h(x)) \to 0$ as $x \to \infty$, we get that $-g'(g^{-1}(h(x)) \to e + 1$. Moreover,
$$x^2 h'(x) = \left(1 + \frac{1}{x}\right)^x \cdot x^2\left(\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}\right).$$
Looking good, since $\left(1 + \frac{1}{x}\right)^x \to e$. For the other term, another round of L'Hôpital yields
$$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right) - \frac{1}{x+1}}{1/x^2} = \lim_{x \to \infty} \frac{-1/(x(x+1)^2)}{-2/x^3} =\lim_{x \to \infty} \frac{x^3}{2x(x+1)^2} = \frac{1}{2}.$$
Collecting all the loose terms, we see that
$$\lim_{n \to \infty} n(1-a_n) =\lim_{x \to \infty} \frac{g^{-1}(h(x))}{1/x} = \frac{e}{2(e+1)}.$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0<r< \infty\}$ under the mapping $w=\frac{z}{z-1}$.
Here is what I got so far. First I got the reverse function
$$z= \frac{w}{w-1}=\frac{2u+2iv}{u-1+iv}$$
I did some algebra and I got $x=r \cos t=\frac{u^2 -u+v^2}{(u-1)^2 +v^2}$ and $y=r \sin t = \frac{v}{(u-1)^2 +v^2}$.
For here, I'm not sure how to use the given boundary to find the boundary of the image. I wonder if anyone would give me a hint please.
|
It's easy to see that the $x$-axis is mapped to itself, but note that the positive part is mapped to $(-\infty,0) \cup (1,\infty)$
To see what $re^{i\frac{\pi}{4}}$ is mapped to try to work with $f(re^{i\frac{\pi}{4}})$ and get a general formula of a curve in $r$. You should get a circle around $\frac{1}{2}-\frac{1}{2}i$ with radius $r=\frac{\sqrt{2}}{2}$
Then you should be able to figure out the image of the set.
Update:
You can show that $f(re^{i\frac{\pi}{4}}) = \frac{r^2-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1} - i\frac{r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1}$
So the image curve is represented as: $$h(r)=(\frac{r^2-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1},\frac{-r\frac{\sqrt{2}}{2}}{r^2-r\sqrt{2}+1})$$
Taking the derivative gives you the tangent to the curve:
$$
h'(r) = (\frac{-\frac{\sqrt{2}}{2}r^2 + 4r-\frac{\sqrt{2}}{2}}{(r^2-r\sqrt{2}+1)^2},\frac{\frac{\sqrt{2}}{2}r^2-\frac{\sqrt{2}}{2}}{(r^2-r\sqrt{2}+1)^2})
$$
Since you know it's a circle (Mobius), looking for the tangents parallel to the $x$-axis can give you $2$ points with distance $2r$ with the center in the middle. The tangent is parallel where the second term is zero which is at $r=1,-1$
So $h(1) = \frac{1}{2}-\frac{i}{2\sqrt{2}-2}, h(-1)=\frac{1}{2}+\frac{i}{2\sqrt{2}+2}$
The distance between these points is $\sqrt{2}$ and the point $z=\frac{1}{2}-\frac{1}{2}i$ is right in the middle.
So in total $f(re^{i\frac{\pi}{4}}) = \{z: |z-\frac{1}{2}+\frac{1}{2}i|=\frac{\sqrt{2}}{2}\}$
Putting it all together:
The $(0,\infty)$ is mapped to $(-\infty,0) \cup (1,\infty)$ but in the other direction, meaning points in the upper half plane are mapped to the lower half plane.
$re^{i\frac{\pi}{4}}, r \gt 0$ is mapped to the part in the circle $\{z: |z-\frac{1}{2}+\frac{1}{2}i|=\frac{\sqrt{2}}{2}\}$ from $0$ to $1$ going counter clockwise, i.e. in the lower half plane, as $r$ increases, therefore points to the right of $re^{i\frac{\pi}{4}}, r \gt 0$ are mapped to points outside the circle.
And so in total we have that the image is points in the lower half plane outside the circle around $\frac{1}{2}-\frac{1}{2}i$ with radius $\frac{\sqrt{2}}{2}$ but where the real part is not between $[0,1]$ or
$$\{z: |z-\frac{1}{2}+\frac{1}{2}i| \ge \frac{\sqrt{2}}{2}, \operatorname{Im}z \le 0, \operatorname{Re}z \notin [0,1] \}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1159067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Limit of $x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right)$ Find $$\lim_{x\to\infty}x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right).$$
I tried substituting $x=1/t$ with $t$ approaching $0$ but the term inside the bracket is not giving me ideas on how to compute the limit.
|
Recall that, as $u \to 0$, we have
$$
\begin{align}
\cos u& =1-\frac {u^2}{2}+\mathcal{O}(u^3)\\
\sin u& =u+\mathcal{O}(u^3)\\
\ln (1+u)&=u-\frac {u^2}{2}+\mathcal{O}(u^3)
\end{align}
$$ giving, as $x \to \infty$,
$$
\cos\frac{\pi}{x}=1-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)
$$
$$
\begin{align}
\log \left(\cos\frac{\pi}{x}\right)&=\log \left(1-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)\\\\
\log \left(\cos\frac{\pi}{x}\right)&=-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\\\\
\log \left(\sqrt{\cos\frac{\pi}{x}}\right)&=-\frac{\pi^2}{4x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)
\end{align}
$$ and
$$
\begin{align}
\sin \left(\log \left(\sqrt{\cos\frac{\pi}{x}}\right)\right)&=-\frac{\pi^2}{4x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\\\\
x^2\sin \left(\log \left(\sqrt{\cos\frac{\pi}{x}}\right)\right)&=-\frac{\pi^2}{4}+\mathcal{O}\left(\frac{1}{x}\right)
\end{align}
$$ giving $-\dfrac{\pi^2}{4}$ for the desired limit.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Summation of sine by considering the imaginary part of exp(ik*seta)
prove that
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac{\cos{\frac {1} {2}}\theta - \cos(n + {\frac 1 2}\theta)} {2\sin \frac{1} {2}\theta}
$$
I would like to solve this by considering the imaginary part of $\sum {\exp(ik\theta)}$.
This is an alternative method suggested by the paper.
$$
S_n = \sum\limits_{k=0}^n {e^{ik\theta}} = 1+e^{i\theta} + e^i\theta + \cdots +e^{in\theta}
$$
$$
= \frac{1 - e^{in\theta}} {1-e^{i\theta}}
$$
$$
=\frac{1-(\cos n\theta + i\sin n\theta)}{1-(\cos\theta + i\sin \theta)}
$$
Considering only the imaginary parts:
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac {\sin n\theta}{\sin \theta} = \frac {\sin ((n+{\frac 1 2})-\frac {1} {2})\theta}{{2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}}
$$
$$
= \frac {\sin (n + {\frac {1} {2}}) \theta \cos {\frac{1} {2}}\theta - \sin {\frac {1} {2}}\theta \cos (n+{\frac{1} {2}})\theta} {2\sin {\frac{1} {2}}\theta cos {\frac{1} {2}}\theta}
$$
Now I have all the parts required + some extra terms that I need to get rid of.
How do i do that?
Many thanks in advance.
|
Since
$$\sum_{k = 0}^n e^{ik\theta} = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}} = \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{-i(n+1)\theta/2} - e^{i(n+1)\theta/2}}{e^{-i\theta/2} - e^{i\theta/2}} = e^{in\theta/2} \frac{\sin [(n+1)\theta/2]}{\sin \theta/2},$$
Taking imaginary parts yields
\begin{align}\sum_{k = 0}^n \sin k\theta &= \frac{\sin(n\theta/2)\sin [(n+1)\theta/2]}{\sin(\theta/2)}\\
&= \frac{\cos\left(\frac{n\theta}{2} - \frac{(n+1)\theta}{2}\right) - \cos\left(\frac{n\theta}{2} + \frac{(n+1)\theta}{2}\right)}{2\sin(\theta/2)}\\
&= \frac{\cos(\theta/2)- \cos[(n + 1/2)\theta]}{2\sin(\theta/2)}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1168691",
"timestamp": "2023-03-29T00:00:00",
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|
Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$ Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$
|
We want to find integers $x$ that satisfy$$\begin{align}x&=7k\\x&=4l+2\\x&=3m+1\end{align}$$
Substitute $x=7k$ into the second equation and obtain
$$7k = 4l+2\\\implies 8k-k=4l+2\\\implies -k=4\heartsuit+2\\\implies k = 4n+2 $$
So we have $k=4n+2$, where $n$ is an integer. Then
$$x=7(4n+2) = 28n+14$$
For $x$ to satisfy the last equation, we must have
$$28n+14=3m+1\\\implies n+27n+14=3m+1\\\implies n = 3\spadesuit -1\\\implies n = 3s+2$$
This yields
$$x=28(3s+2)+14 = 84s+70$$
All in all, you want to find the integers $x=84s+70$ such that
$$\color{blue}{0\lt 84s+70\lt 1000}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1168841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to solve this sum problem?
For the first radical section.
$$\sqrt{1\times 2\times 3\times 4 + 1} - 1 = 1 + 3 + 1 - 1= 5 - 1 = 4$$
The second radical section.
$$(\sqrt{2\times 3\times 4\times 5 + 1}) = 4 = 4 + 6 + 1 - 4 = 7$$
The third radical (not shown)
$$\sqrt{3\times 4\times 5\times 6 + 1} - 9 = 9 + 9 + 1 - 9 = 10$$
If $x$ is considered. Then:
$$\sqrt{x(x+1)(x+2)(x+3) + 1} - x^2 = 3x + 1$$
I need to find:
$$\sum_{n=1}^{97} 3n + 1$$
$$=3\sum_{n=1}^{97} n + 97$$
$$= 3\cdot 97 \cdot 49 + 97 = 14 356$$
|
Split your sum up:
$$3\sum\limits_{n=1}^{97}n + \sum\limits_{n=1}^{97}1$$
$$3 {97 \choose 2} + 97$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1169429",
"timestamp": "2023-03-29T00:00:00",
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|
Finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ I'm having a difficult time finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ over $\mathbb{Q}$.
I know that $\sqrt{7}$ satisfies $f(x) = x^2 - 7$ and is irreducible over $\mathbb{Q}$. So $[\mathbb{Q}(\sqrt{7}) : \mathbb{Q}] = 2$. I also know that $\sqrt{5}$ satisfies $f(x) = x^2 - 5$ which is also irreducible in $\mathbb{Q}$. So $[\mathbb{Q(\sqrt{5})}: \mathbb{Q}] = 2$.
Then by some corollary which I forget the name of $[\mathbb{Q}(\sqrt{5}, \sqrt{7}) : \mathbb{Q}] \leq 4$.
I am unsure of what to do from here.
|
Some of the "hard part" done for you: It should be clear that
$\Bbb Q(\sqrt{5} + \sqrt{7}) \subseteq \Bbb Q(\sqrt{5},\sqrt{7})$ since $\sqrt{5} + \sqrt{7} \in \Bbb Q(\sqrt{5},\sqrt{7})$.
The other direction is a bit "trickier", note that:
$(\sqrt{5} + \sqrt{7})^3 = 5\sqrt{5} + 15\sqrt{7} + 21\sqrt{5} + 7\sqrt{7} = 26\sqrt{5} + 22\sqrt{7}$
Hence $(\sqrt{5} + \sqrt{7})^3 - 26(\sqrt{5} + \sqrt{7}) = -4\sqrt{7}$, that is, if we set $\alpha = \sqrt{5} + \sqrt{7}$, then:
$\sqrt{7} = -\frac{1}{4}(\alpha^3 - 26\alpha) \in \Bbb Q(\alpha) = \Bbb Q(\sqrt{5} + \sqrt{7})$.
In turn, this means $\sqrt{5} = \sqrt{5} + \sqrt{7} - \sqrt{7} \in \Bbb Q(\sqrt{5} + \sqrt{7})$, so that $\Bbb Q(\sqrt{5},\sqrt{7}) \subseteq \Bbb Q(\sqrt{5} + \sqrt{7})$
So we conclude $\Bbb Q(\sqrt{5},\sqrt{7}) = \Bbb Q(\sqrt{5} + \sqrt{7})$. This is useful.
Since we know the minimal polynomial of $\alpha$ is of degree at most $4$, it's natural to start with:
$\alpha^4 = 284 + 48\sqrt{35}$. Now $\alpha^2 = 12 + 2\sqrt{35}$, so:
$\alpha^4 - 24\alpha^2 = -4$, that is $\alpha$ is a root of $x^4 - 24x + 4$. Proving this is irreducible over $\Bbb Q$ then shows $[\Bbb Q(\sqrt{5},\sqrt{7}):\Bbb Q] = 4$.
One possible basis from this is $\{1,\alpha,\alpha^2,\alpha^3\}$, the linear independence of which is guaranteed by the minimal degree of our quartic. Spanning should be clear, from the degree of the extension, but also: any rational polynomial in $\alpha$ can be reduced to one of degree $< 4$, and from:
$\alpha^4 - 24\alpha^2 + 4 = 0$ we obtain: $\alpha\left(6\alpha - \dfrac{\alpha^3}{4}\right) = 1$, that is:
$\alpha^{-1} = 6\alpha - \dfrac{\alpha^3}{4}$, so any rational function of $\alpha$ (that is, $\Bbb Q(\alpha)$) can be expressed as a linear combination of our basis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1171268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Is $53$ expressible in this form? It seems as if prime numbers may always be expressed in the form $a\cdot 2^b+c \cdot 3^d$ for some nonnegative integers $b,d$ and $a,c\in \{-1,0,1\}$.
Examples:
$2=1\cdot 2^1+0\cdot 3^d$
$3=0\cdot 2^b+1\cdot 3^1$
$5=1\cdot 2^1+1\cdot 3^1$
$7=1\cdot 2^2+1\cdot 3^1$
$11=1\cdot 2^3+1\cdot 3^1$
$13=1\cdot 2^2+1\cdot 3^2$
$17=1\cdot 2^3+1\cdot 3^2$
$19=1\cdot 2^4+1\cdot 3^1$
$23=1\cdot 2^5+(-1)\cdot 3^2$
$29=1\cdot 2^5+(-1)\cdot 3^1$
$31=1\cdot 2^5+(-1)\cdot 3^0$
$37=1\cdot 2^6+(-1)\cdot 3^3$
$41=1\cdot 2^5+1\cdot 3^2$
$43=1\cdot 2^4+1\cdot 3^3$
$47=1\cdot 2^7+(-1)\cdot 3^4$
We hit a brick wall at $53$. Can anyone confirm if $53$ is/isn't expressible?
What about $n\in \mathbb{N}$ in general? Is it possible to always express $n$ in this form?
Thanks.
EDIT: I invented this question on my own, there are no sources.
Note: The reason I started with prime numbers is because I found it hard to find an expression for $6$, whereas prime numbers continued to be easy to find expressions for (easy until $53$, that is).
|
You can check that for any $k,l$ we have that
$$2^k \not\equiv 3^l+53 \pmod{117}$$
and that
$$2^k+53 \not\equiv 3^l\pmod{117}$$
Just enumerating the different powers.
Edit: If you consider the sets $Q_t = \{2^{k_1}3^{k_2}\dots p_t^{k_t}, k_1,\dots,k_t \ge 0\}$ and ask for integers not in $Q_t$ or sum or difference of elements in $Q_t$ then I found using this method that 103 is the smallest using primes up to 3 and that 583 is the smallest using primes up to 5 and I conjecture that 3737, 16579 and 41969 are the least integers using primes up to 7, 11 and 13 resp.
It seems probable that there is always an integer that can't be written in this way for any set of primes but I'm not sure how to prove it.
|
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|
Shortcuts for integrating $\int \sqrt{x^2-36} \,dx$ Is there any simple method to solve this integral
$$\int \sqrt{x^2-36} \,dx$$
Because my method is a bit complicated and long:
|
I think integration by parts is the shortest way here. Let $u = \sqrt{x^2 - 36}$ and $\mathrm dv = \mathrm dx$. Then,
$$\begin{align}
\int\sqrt{x^2 - 36}\,\mathrm dx &= x\sqrt{x^2 - 36} - \int\frac{x^2}{\sqrt{x^2 - 36}}\,\mathrm dx =\\[0.3em]
&= x\sqrt{x^2 - 36} - \int\frac{x^2 - 36 + 36}{\sqrt{x^2 - 36}}\,\mathrm dx =\\[0.3em]
&= x\sqrt{x^2 - 36} - \color{green}{\int\sqrt{x^2 - 36}\,\mathrm dx} - 36\int\frac1{\sqrt{x^2 - 36}}\,\mathrm dx=\\[0.3em]
&= \frac x2\sqrt{x^2 - 36} - 18\int\frac1{\sqrt{x^2 - 36}}\,\mathrm dx =\tag{$\star$}\\[0.5em]
&= \frac x2\sqrt{x^2 - 36} - 18\operatorname{arccosh}\frac x6 + C'
\end{align}$$
In step $(\star)$ we moved the highlighted term to the LHS and divided by $2$. Then we used the fact that
$$\frac{\mathrm d}{\mathrm dx}\operatorname{arccosh} x = \frac1{\sqrt{x^2 - 1}}.$$
You can also rewrite the result in another form, using $\operatorname{arccosh} x$ definition:
$$\operatorname{arccosh} x = \ln\left(x + \sqrt{x^2 - 1}\right);$$
in this case, we can write
$$\begin{align}
\int\sqrt{x^2 - 36}\,\mathrm dx &= \frac x2\sqrt{x^2 - 36} - 18\ln\left(\frac x6 + \frac16\sqrt{x^2 - 36}\right) + C' =\\[0.3em]
&= \boxed{\displaystyle\frac x2\sqrt{x^2 - 36} - 18\ln\left(x + \sqrt{x^2 - 36}\right) + C}
\end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/1174951",
"timestamp": "2023-03-29T00:00:00",
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|
Does every positive rational number appear once and exactly once in the sequence $\{f^n(0)\}$ , where $f(x):=\frac1{2 \lfloor x \rfloor -x+1} $ Consider the map $f:\mathbb Q^+ \to \mathbb Q^+$ defined as $f(x):=\dfrac1{2 \lfloor x \rfloor -x+1} , \forall x \in \mathbb Q^+$ ; then is the function
$g:\mathbb Z^+ \to \mathbb Q^+$ defined as $g(n):=f^n(1)=\underbrace{(f \circ f \circ \ldots f}_{n \text{ times}})(1)$ a bijection? That is, do the iterates of $f()$ hit all positive rationals?
|
As Henning noted in the comments,
$$\begin{align*}
f\left(\frac{p}q\right)&=\frac1{2\left\lfloor\frac{p}q\right\rfloor-\frac{p}q+1}\\
&=\frac{q}{2q\left(\frac{p-(p\bmod q)}q\right)-p+q}\\
&=\frac{q}{p+q-2(p\bmod q)}\;.
\end{align*}$$
If we let $p_0=q_0=1$ and define $p_n$ and $q_n$ by the recurrences
$$\begin{align*}
p_{n+1}&=q_n\\
q_{n+1}&=p_n+q_n-2(p_n\bmod{q_n})\;,
\end{align*}\tag{1}$$
clearly $g(n)=\frac{p_n}{q_n}$. Note that we can eliminate $q_n$ and reduce $(1)$ to
$$p_{n+1}=p_{n-1}+p_n-2(p_{n-1}\bmod{p_n})\;,$$
with initial conditions $p_0=p_1=1$.
An easy induction shows that $p_{2n}>p_{2n+1}$ for $n>0$, and $p_{2n+1}<p_{2n+2}$ for $n\ge 0$.
If $n>0$ is even, then by the induction hypothesis $p_{n-1}<p_n$, so $$p_{n+1}=p_{n-1}+p_n-2p_{n-1}=p_n-p_{n-1}<p_n\;.\tag{2}$$
If $n$ is odd, then $p_{n-1}>p_n$; let $p_{n-1}=kp_n+r$, where $k\ge 1$ and $0\le r<p_n$. Then $$p_{n+1}=p_{n-1}+p_n-2(p_{n-1}\bmod{p_n})=(k+1)p_n-r\ge 2p_n-r>p_n\;.$$
Let $r_0=1$, and define $r_n$ by the recurrence
$$\begin{align*}
r_{2n+1}&=r_n\\
r_{2n+2}&=r_n+r_{n+1}\;.
\end{align*}\tag{3}$$
Clearly $r_0=r_1=p_0=p_1=1$. Suppose that $r_k=p_k$ for $k\le 2n+1$. Then
$$\begin{align*}
p_{2n+2}&=p_{2n}+p_{2n+1}-2(p_{2n}\bmod{p_{2n+1}})\\
&=r_{2n}+r_{2n+1}-2(r_{2n}\bmod{r_{2n+1}})\\&=r_{n-1}+2r_n-2\big((r_{n-1}+r_n\big)\bmod{r_n}\big)\\
&=r_{n-1}+2r_n-2(r_{n-1}\bmod{r_n})\\
&=r_n+\big(p_{n-1}+p_n-2(p_{n-1}\bmod{p_n})\big)\\
&=r_n+p_{n+1}\\
&=r_n+r_{n+1}\\
&=r_{2n+2}\;,
\end{align*}$$
and $r_{2n+3}=r_{n+1}=r_{2n+2}-r_n=r_{2n+2}-r_{2n+1}=p_{2n+2}-p_{2n+1}=p_{2n+3}$ by (2). Thus, $r_n=p_n$ for all $n\ge 0$.
For $n\ge 1$ let $a_n=r_{n-1}=p_{n-1}$. Then $a_1=1$, and it follows from $(3)$ that
$$a_{2n}=r_{2n-1}=r_{n-1}=a_n$$
and
$$a_{2n+1}=r_{2n}=r_{n-1}+r_n=a_n+a_{n+1}$$
and hence that $\langle a_n:n\in\Bbb Z^+\rangle$ is the sequence defined in this question. In my answer to that question I showed that if $q_n=\frac{a_{n+1}}{a_n}$ for $n\in\Bbb Z^+$, the sequence $\langle q_n:n\in\Bbb Z^+\rangle$ is a bijection from $\Bbb Z^+$ to $\Bbb Q^+$, and the rational numbers $q_n$ are just the reciprocals of the numbers $g(n)$.
|
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|
Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give me some hints? Thanks in advance!
|
Let $(a+b\sqrt{2})^3=7+5\sqrt{2}$, then $(a-b\sqrt{2})^3=7-5\sqrt{2}$. This gives us
\begin{align}
a^3+6ab^2 & = 7\\
2b^3+3a^2b & = 5
\end{align}
From the two equations listed above we get.
$$\frac{a(a^2+6b^2)}{b(2b^2+3a^2)}=\frac{7}{5}.$$
Now let $x=\frac{a}{b}$. Then
$$5x(x^2+6)=7(3x^2+2).$$
Finally we get
$$5x^3-21x^2+30x-14=0$$
Observe that $x=1$ is an obvious solution of this equation. So we can rewrite this as:
$$(x-1)(5x^2-16x+14)=0.$$
The quadratic factor has only non-real roots. Thus $x=1$ is the only real solution. This gives $a=b=1$. Thus the given expression is equal to $2$.
|
{
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"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Lagrange Multipliers, maximize $f=xy$ restricted to $g=x^2+y^2=r^2$ So I have to solve the system of equations
$$\cases{\nabla f = \lambda \nabla g\\x^2+y^2 = r^2}.$$
Then $y=2\lambda x, x=2\lambda y$. Sorry if this is obvious, but how can I get $x$ and $y$ only as a function of $\lambda$? Otherwise I'm not able to find the values $\lambda$ to evaluate them and find a maximum.
|
Since $y = 2\lambda x$ and $x = 2\lambda y$, $x^2 + y^2 = 4\lambda^2(x^2 + y^2)$, i.e., $r^2 = 4\lambda^2 r^2$. Thus $4\lambda^2 = 1$, yielding $\lambda = \pm 1/2$. If $\lambda = 1/2$, then $y = x$, so from the condition $x^2 + y^2 = r^2$, we obtain $x = y = \pm r/\sqrt{2}$. If $\lambda = -1/2$, then $y = -x$, so the condition $x^2 + y^2 = r^2$ results in $x = r/\sqrt{2}$ and $y = -r/\sqrt{2}$ or $x = -r/\sqrt{2}$ and $y = r/\sqrt{2}$. The point $ (r/\sqrt{2}, r/\sqrt{2})$ (and $(-r/\sqrt{2}, -r/\sqrt{2})$) maximizes $xy$, with maximum $r^2/2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1181307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
|
As one can obviously see, $x=1$ is a root of the denominator, and thus, it can be divided by $(x-1)$. Using long division, we get $(x-1)(x^3+x^2+x-3)$
Again, we can see that $x=1$ is a solution, of the second factor, so, using long division again, we get $(x-1)^2(x^2+2x+3)$
$$\frac{-2x^2+6x+8}{(x-1)^2(x^2+2x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Ex+F}{(x^2+2x+3)}$$
Thus, we have
$$-2x^2+6x+8=A(x-1)(x^2+2x+3)+B(x^2+2x+3)+(Ex+F)(x-1)^2$$
$$-2x^2+6x+8=A(x^3+2x^2+3x-x^2-2x-3)+Bx^2+2Bx+3B+(Ex+F)(x^2-2x+1)$$
By identification, we find
$A=-1$, $B=2$ $E=1$ $F=-1$
$$\int \frac{-1}{x-1}\,dx =-\ln(x-1)$$
$$\int \frac{2}{(x-1)^2} \,dx =\int 2(x-1)^{-2} =-2(x-1)^{-1}\,dx$$
$$\int \frac{x-1}{x^2+2x+3}\,dx=\int \frac{x+1}{x^2+2x+3}\,dx -2\int \frac{1}{x^2+2x+3}\,dx =\frac{1}{2}\ln(x^2+2x+3)-2\int \frac{1}{2+(x+1)^2} \,dx =\frac{1}{2}\ln(x^2+2x+3)-\int \frac{1}{1+(\frac{x+1}{\sqrt{2}})^2} \,dx =\frac{1}{2}\ln(x^2+2x+3)-\sqrt{2}\arctan \frac{x+1}{\sqrt{2}}$$
Not sure if I made any mistake since it's pretty late and I'm dead tired, but I hope this gets the point.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying $c^n + c^{−n} = p(c + c^{−1})$. Let $c$ be a nonzero element of a field $F$ and let $n > 1$ be
an integer. Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying
$c^n + c^{−n} = p(c + c^{−1})$.
I made some particular cases, but not seems to have a form so I need some help, Thank you.
This is an exercise from the book: The Linear Algebra a Beginning Graduate Student Should Know by Golan.
|
Complete solution:
We can prove this by induction, starting with a base case of $n=0$. If $n=0$, then taking $p(X)=2$ is a solution to the problem. Suppose $n>0$ and that we have polynomials $p_k(X)\in F[X]$ such that $p_k(c+c^{-1})=c^k+c^{-k}$ for all $k<n$. Then notice that $(c+c^{-1})^n=c^n+nc^{n-1}c^{-1}+\left(\!\begin{array}{c}n\\2\end{array}\!\right)c^{n-2}c^{-2}+\cdots \left(\!\begin{array}{c}n\\n-2\end{array}\!\right)c^2c^{-(n-2)}+ncc^{-(n-1)}+c^{-n}.$ By noting that$\left(\!\begin{array}{c}n\\i\end{array}\!\right)=\left(\!\begin{array}{c}n\\n-i\end{array}\!\right)$, and by simplifying and gathering terms, we have that\begin{align*}(c+c^{-1})^n&=c^n+c^{-n}+n(c^{n-2}+c^{-(n-2)})+\cdots +\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)(c^{n-2\lfloor n/2\rfloor}+c^{-(n-2\lfloor n/2\rfloor)})\\&=c^n+c^{-n}+np_{n-2}(c+c^{-1})+\cdots+\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)p_{n-2\lfloor n/2 \rfloor}(c+c^{-1})\end{align*}if $n$ is odd and\begin{align*}(c+c^{-1})^n&=c^n+c^{-n}+n(c^{n-2}+c^{-(n-2)})+\cdots
+\left(\!\begin{array}{c}n\\n/2\end{array}\!\right)\\&=c^n+c^{-n}+np_{n-2}(c+c^{-1})+\cdots+\left(\!\begin{array}{c}n\\ n/2\end{array}\!\right)\end{align*}if $n$ is even. Therefore, taking $p_n(X)=X^n-np_{n-2}(X)-\left(\!\begin{array}{c}n\\2\end{array}\!\right)p_{n-4}(X)-\cdots -\left(\!\begin{array}{c}n\\\lfloor n/2\rfloor\end{array}\!\right)p_{1}(X)$ if $n$ is odd and $p_n(X)=X^n-np_{n-2}(X)-\left(\!\begin{array}{c}n\\2\end{array}\!\right)p_{n-4}(X)-\cdots -\left(\!\begin{array}{c}n\\ n/2\end{array}\!\right)$ if $n$ is even we have our desired polynomial for $n$. So we have completed the inductive step and proved that such a polynomial exists for all $n\geq 0$.
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|
Solve $\arg(-1/z)=-2\pi/3$ and $|1-\frac2z|=1$ $\arg(\frac{-1}z)=\frac{-2\pi}3$
what does this mean? how to i get the $\arg(z)$ from this.
I'm thinking of reciprocals.
$\left|1-\frac2z\right|=1$ how do i solve for this as well.. i'm confused when i negative sign appears
|
Note that:
$$\frac{1}{z} = \frac{x -i y}{x^2+y^2}$$
Now you have:
$$\arg\left(-\frac{1}{z}\right) = -\frac{2\pi}{3}\implies \arg \left(\frac{-x +i y}{x^2+y^2}\right) = \tan^{-1}{\left(-\frac{y}{x}\right)} = -\frac{2\pi}{3}$$
$$\implies \frac{y}{x} = \tan(\frac{2\pi}{3})$$
The other condition you can write as:
$$ \vert z - 2 \vert = \vert z \vert \implies (x-2)^2 + y^2 = x^2 + y^2$$
$$\implies 4 x = 4 \implies x = 1 \implies y = \tan(\frac{2\pi}{3}) = -\sqrt{3}.$$
Hence $z = 1 - i \sqrt{3}.$
|
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"url": "https://math.stackexchange.com/questions/1187763",
"timestamp": "2023-03-29T00:00:00",
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|
Coordinate transformation (or conversion) into yards Following is a soccer field with its dimensions.
There is a similar field, but I am capturing coordinates via mouse-movement. So, what (115,75) shows here, is 567.5,369 when I capture the coordinates when the mouse moves. How could I convert a general coordinate (x,y) to (x1,y1) where x1 and y1 are the coordinates in yards.
|
So you want an affine transformation $L\colon\Bbb R^2\to\Bbb R^2$ that maps
*
*$\begin{pmatrix} 0\\ 0\end{pmatrix}$ to $\begin{pmatrix} a_0\\ b_0\end{pmatrix}$
*$\begin{pmatrix} x_1\\ 0\end{pmatrix}$ to $\begin{pmatrix} a_1\\ b_1\end{pmatrix}$ where $x_1\neq 0$ (for you example, take $x_1=115$)
*$\begin{pmatrix} 0\\ y_2\end{pmatrix}$ to $\begin{pmatrix} a_2\\ b_2\end{pmatrix}$ where $y_2\neq 0$ (for you example, take $y_2=75$)
Since $L$ is supposed to be affine, it can be written in the form $L(x)=Mx+v$ where $$M=\begin{pmatrix}M_{1,1}&M_{1,2}\\ M_{2,1}& M_{2,2}\end{pmatrix}\in\Bbb R^{2\times 2}\qquad \text{ and }\qquad v=\begin{pmatrix} v_1\\ v_2\end{pmatrix}\in\Bbb R^2,$$
So,
*
*$\begin{pmatrix} a_0\\ b_0\end{pmatrix}=L\begin{pmatrix} 0\\ 0\end{pmatrix}=v\implies v = \begin{pmatrix} a_0\\ b_0\end{pmatrix}$
*$\begin{pmatrix} a_1\\ b_1\end{pmatrix}=L\begin{pmatrix} x_1\\ 0\end{pmatrix}=x_1\begin{pmatrix} M_{1,1}\\ M_{2,1}\end{pmatrix}+v\implies \begin{pmatrix} M_{1,1}\\ M_{2,1}\end{pmatrix}=\frac{1}{x_1}\begin{pmatrix} a_1\\ b_1\end{pmatrix}-\frac{1}{x_1}v=\frac{1}{x_1}\begin{pmatrix} a_1-a_0\\ b_1-b_0\end{pmatrix}$
*$\begin{pmatrix} a_2\\ b_2\end{pmatrix}=L\begin{pmatrix} 0\\ y_2\end{pmatrix}=y_2\begin{pmatrix} M_{1,2}\\ M_{2,2}\end{pmatrix}+v\implies \begin{pmatrix} M_{1,2}\\ M_{2,2}\end{pmatrix}=\frac{1}{y_2}\begin{pmatrix} a_2\\ b_2\end{pmatrix}-\frac{1}{y_2}v=\frac{1}{y_2}\begin{pmatrix} a_2-a_0\\ b_2-b_0\end{pmatrix}$
It follows that
$$L\begin{pmatrix}x_1\\ x_2\end{pmatrix}=\begin{pmatrix} \frac{a_1-a_0}{x_1}&\frac{a_2-a_0}{y_2}\\ \frac{b_1-a_0}{x_1}&\frac{b_2-a_0}{y_2}\end{pmatrix}\begin{pmatrix}x_1\\ x_2\end{pmatrix}+\begin{pmatrix}a_0\\ b_0\end{pmatrix}$$
is the transformation you are looking for.
|
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|
Geometric Interpretation of the Basel Problem? Does the Pi in the solution to the Basel problem have any geometric significance? Every time I see Pi, I have to think of a circle. I would love to see a nice intuitive picture connecting the Basel problem with geometrical figures. Anyone here been lucky enough to stumble upon such a thing?
|
The Montreal interpretation of the Basel problem
Suppose you randomly pick a street in the left window, one which is parallel to that window (red streets). Similarly, you pick a blue street in the right window:
Then, from this perspective the cross section between the two streets has 50% chance of being visible in the right window (the blue street is farther away than the red street).
Hence:
$$ \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6} $$
Proof
Okay, I assumed that in each window the elevation angle of the picked street was uniformly distributed between $0^\circ$ and $90^\circ$. This means that the distances are Cauchy distributed:
$$P_{\color{blue}{X}}(\color{blue}{x}) = \frac{2}{\pi (1+\color{blue}{x}^2)}$$
$$P_{\color{red}{Y}}(\color{red}{y}) = \frac{2}{\pi (1+\color{red}{y}^2)}$$
NB: The unit distance of $\color{blue}{x}$ and $\color{red}{y}$ is the viewer’s height above the ground.
The statement that the cross section is in 50% of the cases visible in the right window, is equivalent to saying that the ratio $\frac{\color{red}{y}}{\color{blue}{x}}$ has 50% chance being smaller than 1. Note that this ratio is the tangens of the azimuth where the cross section is located.
A ratio of two similar Cauchy distributions, is distributed as follows:
$$P_T(t) = \frac{4}{\pi^2}\frac{\log\left(\frac{1}{t}\right)}{(1 - t^2)}$$
Using $P_T(t<1)=0.5$ , we get:
$$ \begin{align}\frac12 &= \frac{4}{\pi^2}\int_0^1 \frac{\log\left(\frac{1}{t}\right)}{1 - t^2}dt \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty \int_0^1 \log\left(\frac{1}{t}\right) t^{2k}dt \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty \left.\frac{t^{2k+1}}{2k+1}\left(\frac{1}{2k+1} - \log(t)\right)\right|_0^1 \\ \frac{\pi^2}{8} &= \sum_{k=0}^\infty \frac1{(2k+1)^2} \\ \frac{\pi^2}{8} &= \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots \end{align}$$
The complete Basel sum
$$\displaystyle \begin{align} \sum_{k=1}^\infty \frac{1}{k^2} &= \frac{4}{3}\left(\sum_{k=1}^\infty \frac{1}{k^2} - \frac{1}{4}\sum_{k=1}^\infty \frac{1}{k^2} \right) \\ &= \frac{4}{3}\left(\sum_{k=1}^\infty \frac{1}{k^2} - \sum_{k=1}^\infty \frac{1}{(2k)^2} \right) \\ &= \frac{4}{3}\left(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \cdots\right) \\ \sum_{k=1}^\infty \frac{1}{k^2} &=\frac{\pi^2}{6} \\ & \qquad \qquad \blacksquare \end{align}$$
Based on a proof by Luigi Pace, as appeared in: ‘The American Mathematical Monthly’, August-September 2011, pp. 641-643
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|
1985 Putnam A1 Solution
I dont see what they mean by bijection of triples of subsets of $\{1, \ldots, 10\}$ and the $10\times3$ matrix with $0, 1$ entries?
How is that created?
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For example, suppose the three sets are
\begin{align}
A_1 & = \{1,5,6,8\} \\
A_2 & = \{1,2,3,4,10\} \\
A_3 & = \{2,4,5,7,9,10\}
\end{align}
(Note that $A_1\cup A_2\cup A_3=\{1,2,3,4,5,6,7,8,9,10\}$ and $A_1\cap A_2\cap A_3=\varnothing$.)
Then the $10\times3$ matrix is
$$
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}
$$
The first column of the matrix corresponds to the set $A_1$. It has a $1$ in the $1$st, $5$th, $6$th, and $8$th rows because the members of $A_1$ are $1,5,6,8$. Similarly the second and third columns correspond respectively to $A_2$ and $A_3$.
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
|
Build from inside:
$$
\begin{align}
\frac{d}{dx}\sqrt{7x}&=\frac{7}{2\sqrt{7x}}\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x}}&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{d}{dx}\sqrt{7x}\right)\\
&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{7}{2\sqrt{7x}}\right)\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x+\sqrt{7x}}}&=\frac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\cdot\left(7+\frac{d}{dx}\sqrt{7x+\sqrt{7x}}\right)
\end{align}
$$
where you can plug in the expression for $\frac{d}{dx}\sqrt{7x+\sqrt{7x}}$ found in the second to last line to finish the calculation to end up with
$$
y'=\frac{1}{2\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}\cdot\left(7+\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{7}{2\sqrt{7x}}\right)\right)
$$
and now it should be a childs play even finding the derivative of $\sqrt{7x+\sqrt{7x+\sqrt{7x+\sqrt{7x}}}}$ if you feel like it.
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"timestamp": "2023-03-29T00:00:00",
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|
Determine the coefficient for fourier sine series of f(x) = cos ${\pi x\over L}$ Determine the coefficient for Fourier sine series of f(x) = cos ${\pi x\over L}$
I know how to solve it but I get stuck at the end.
$$B_n = \int_{0}^{L} \cos\frac{\pi x}{L} \, \sin\frac{n\pi x}{L}$$
We use the identity $$\sin a \, \cos b=\frac{1}{2} \left(\sin(a+b)+ \sin(a−b) \right).$$
I solved it twice and I always end up with $$ \frac {-(n+1)-(n-1)}{(n^2 -1)\pi } (\cos{(n-1)\pi x\over L} + \cos{(n+1)\pi x\over L})$$
The solution in the book has the end result which is
0 if n odd
$\frac{4n}{\pi(n^2 -1)}$ if n even But I cant seem to figure out how even after i Substitue 0 and L
|
Consider
\begin{align}
B_{n} = \frac{2}{L} \int_{0}^{L} \cos\left( \frac{\pi x}{L} \right) \sin\left( \frac{n \pi x}{L} \right) \, dx.
\end{align}
For the case $n=1$ it is seen that
\begin{align}
B_{1} &= \frac{1}{L} \int_{0}^{L} \sin\left( \frac{2 \pi x}{L} \right) \, dx \\
&= \frac{1}{L} \left[ - \frac{L}{2 \pi} \cos\left( \frac{2 \pi x}{L} \right) \right]_{0}^{L} \\
&= 0.
\end{align}
Now, for $n > 1$ it is seen that
\begin{align}
B_{n} &= \frac{1}{L} \int_{0}^{L} \left(\sin\left(\frac{(n+1) \pi x}{L}\right) + \sin\left(\frac{(n-1)\pi x}{L} \right) \right) \, dx \\
&= \frac{-1}{L} \left[ \frac{L}{(n+1) \pi} \cos\left( \frac{(n+1)\pi x}{L} \right) + \frac{L}{(n-1) \pi} \cos\left( \frac{(n-1) \pi x}{L} \right) \right]_{0}^{L} \\
&= - \left[ \frac{\cos(n+1)\pi - 1}{(n+1) \pi } + \frac{\cos(n-1)\pi -1}{(n-1) \pi } \right] \\
&= \frac{(-1)^{n} +1}{\pi} \left[ \frac{1}{n+1} + \frac{1}{n-1} \right] \\
&= \frac{ 2 (1 + (-1)^{n}) \, n}{(n^{2} -1) \pi}.
\end{align}
The sine series then becomes
\begin{align}
\cos\left( \frac{\pi x}{L} \right) &= \sum_{n=1}^{\infty} B_{n} \, \sin\left( \frac{n \pi x}{L} \right) \\
&= B_{1} \sin\left( \frac{\pi x}{L} \right) + \sum_{n=2}^{\infty} B_{n} \, \sin\left( \frac{n \pi x}{L} \right) \\
&= \frac{2}{\pi} \sum_{n=2}^{\infty} \frac{(1+(-1)^{n}) \, n}{n^{2} -1} \, \sin\left( \frac{n \pi x}{L} \right) \\
&= \frac{8}{\pi} \sum_{n=1}^{\infty} \frac{n}{4 n^{2} -1} \, \sin\left( \frac{2 n \pi x}{L} \right).
\end{align}
|
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|
Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded.
Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded.
I'm approaching this starting with
$$ \left| \frac{2yx^3}{(x^2+y^2)^2} \right| = \left| \frac{2yx^3}{x^4+2x^2y^2+y^4} \right| \leq \left| \frac{2yx^3}{2x^2y^2} \right| = \left| \frac{x}{y}\right|.$$
However, this doesn't get me anywhere. What am I missing?
|
By AM-GM,
$$ |yx^3| = 27\left|y\cdot\frac{x}{3}\cdot\frac{x}{3}\cdot\frac{x}{3}\right| \leq 27\left(\frac{|x|+|y|}{4}\right)^4 = \frac{27}{64}(|x|+|y|)^4$$
while by AM-QM:
$$ \frac{x^2+y^2}{2} \geq \left(\frac{|x|+|y|}{2}\right)^2 $$
hence:
$$\left|\frac{2yx^3}{(x^2+y^2)^2}\right|\leq \frac{\frac{27}{32}}{\frac{1}{4}}=\frac{27}{8}.$$
An even sharper bound (thanks to Git Gud) is:
$$\left|\frac{2yx^3}{(x^2+y^2)^2}\right|\leq \left|\frac{2xy}{x^2+y^2}\right|\cdot \left|\frac{x^2}{x^2+y^2}\right|\leq 1,$$
but the sharpest possible bound is given by $\frac{3\sqrt{3}}{8}$, as shown in the comments.
|
{
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|
The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
|
\begin{array} \\
\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^2 + x} - \sqrt{x^2 + 1} &= \displaystyle \lim_{x \rightarrow \infty} \left(\sqrt{x^2 + x} - \sqrt{x^2 + 1}\right) \frac{\sqrt{x^2 + x} + \sqrt{x^2 + 1}}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \\
&= \displaystyle \lim_{x \rightarrow \infty} \frac{(x^2 + x) - (x^2 + 1)}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \\
&= \displaystyle \lim_{x \rightarrow \infty} \frac{x-1}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \cdot \frac{(1/x)}{(1/x)} \\
&= \displaystyle \lim_{x \rightarrow \infty} \frac{1-1/x}{\sqrt{1 + 1/x} + \sqrt{1 + 1/x^2}} \\
&= \displaystyle \frac{1}{2}
\end{array}
|
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|
Find a formula for $\sin(3a)$ and use to calculate $\sin(π/3)$ and $\cos(π/3)$? Problem:
Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$.
My attempt:
\begin{align}
\sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\
&= \sin(a + a)\cos(a) + \cos(a + a)\sin(a) \\
&= [\sin(a)\cos(a) + \cos(a)\sin(a)]\cos(a) + [\cos(a)\cos(a) - \sin(a)\sin(a)]\sin(a).
\end{align}
It can then be simplified to
$$2\sin(a)\cos^2(a) + \sin(a)\cos^2(a) - \sin^3(a) = 3\sin(a)\cos^2(a) - \sin^3(a).$$
My question is this: How am I supposed to use this formula to find $\sin(π/3)$ and $\cos(π/3)$?
|
hint: Put $x = \sin \left(\frac{\pi}{3}\right)$, then the equation $0=\sin\left(3\cdot\dfrac{\pi}{3}\right)= 3\sin\left(\frac{\pi}{3}\right) - 4\sin^3(\frac{\pi}{3})= 3x - 4x^3$ gives: $4x^3-3x = 0 \to x(4x^2-3) = 0$. Can you solve for $x$. Note that $x \neq 0$.
|
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|
Finding an isomorphism between rings Find an isomorphism between $T: \mathbb{F_3}[x]/(x^2+x+2) \to \mathbb{F_3}[x]/ (x^2 + 1)$. We have that both of these are fields. The answer states the map $T$ is given by $T(a+bx) = (a+b) + bx$ so $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Could someone explain how they got this?
|
There is a mechanical way to find isomorphisms like this. We know that such an isomorphism exists by the uniqueness of finite fields of a given order. Furthermore, such an isomorphism is completely determined by the image of $x$. Finally, we know that the image of $x$ can be represented by $a + bx$ such that $(a + bx)^2 + (a + bx) + 2 \in (x^2 + 1)$. Expanding, we get
\begin{align*}
(a + bx)^2 + (a + bx) + 2 &= (a^2 + a + 2) + (2a + 1)bx + b^2 x^2 \\
&= (a^2 + a + 2 - b^2) + (2a + 1)bx + b^2 (x^2 + 1).
\end{align*}
This gives the equations $a^2 + a + 2 - b^2 = 0$ and $(2a + 1)b = 0$. By solving for $a$ and $b$ in $\mathbb F_3$ we get $a = b = 1$ or $a = 1, b = 2$.
|
{
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|
Count the number of one pair hands in a standard deck In an attempt to answer this question, I tried the following solution:
*
*Let $A$ denote the number of one-pair hands
*Let $B$ denote the number of two-pair hands
*Let $C$ denote the number of three-of-a-kind hands
*Let $D$ denote the number of full-house hands
*Let $E$ denote the number of four-of-a-kind hands
*Let $F$ denote the total number of hands
Then:
*
*$A=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{10}{1}\cdot\binom{4}{2}\cdot\binom{4}{1}\cdot\binom{4}{1}\cdot\binom{4}{1}=\color{red}{6589440}$
*$B=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot\binom{4}{1}=247104$
*$C=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{11}{1}\cdot\binom{4}{3}\cdot\binom{4}{1}\cdot\binom{4}{1}=109824$
*$D=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{4}{3}\cdot\binom{4}{2}=3744$
*$E=\binom{13}{1}\cdot\binom{12}{1}\cdot\binom{4}{4}\cdot\binom{4}{1}=624$
*$F=\binom{52}{5}=\color{red}{2598960}$
The part marked red is obviously wrong, since the number of one-pair hands is larger than the total number of hands. I've been going around this in circles trying to find out where I went wrong, with no luck whatsoever. Now I'm starting to think that other parts are wrong as well.
Any idea where my mistake is?
Thanks
|
To choose a pair, you choose one of the thirteen kinds, choose two of the four suits from that kind, choose three of the remaining kinds, and choose one of the four suits from each of those kinds, which yields
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3 = 1,098,240$$
hands with one pair.
Say your hand is $\color{red}{4\heartsuit}, \color{red}{4\diamondsuit}, \color{red}{J\diamondsuit}, 7\clubsuit, 5\spadesuit$. Your mistake was that you did not take into account that there are $3! = 6$ orders in which you could select $\color{red}{J\diamondsuit}, 7\clubsuit, 5\spadesuit$. Since each such selection would leave you with the same hand, you must divide your answer by $6$.
|
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|
Proving a goal with an existential quantifier and making sure it covers all cases I'm trying to prove the following theorem:
Suppose x is a real number. Prove that if x $\neq$ 1 then there is a
real number y such that $\frac{y + 1}{y - 2}$ = x.
The logical structure of the sentence is:
x $\neq$ 1 $\implies$ $\exists$y($\frac{y + 1}{y - 2}$ = x)
I first suppose x $\neq$ 1 so my goal becomes $\exists$y($\frac{y + 1}{y - 2}$ = x).
How to Prove it: A Structured Approach says the following regarding goals with existential quantifiers:
To prove a goal of the form $\exists$x P(x): Try to find a value of x
for which you think P(x) will be true. Then start your proof with “Let
x = (the value you decided on)” and proceed to prove P(x) for this
value of x.
Let's say I chose y = 5. I can therefore let y = 5, conclude that $\frac{y + 1}{y - 2}$ = 2, and since my hypotheses say x $\neq$ 1 and 2 $\neq$ 1 finish my proof.
However this looks like I'm proving a theorem about a certain value of x instead or proving the theorem for all values of x except for 1. I feel I'm proving the following instead:
"Prove that if x = 2 then there is a real number y such that $\frac{y + 1}{y - 2}$ = x"
How should I deal with goals with existential quantifiers instead, to make sure I prove all the cases?
|
You should solve for $y$ in terms of $x$. Once you find it, you can then go back and choose $y$ to be this number and show that it works. Messing around, we see that:
\begin{align*}
\frac{y + 1}{y - 2} &= x \\
y + 1 &= xy - 2x \\
2x + 1 &= xy - y = y(x - 1) \\
y &= \frac{2x + 1}{x - 1}
\end{align*}
We're now ready to prove what we want.
Proof: Suppose that $x \neq 1$. Then consider $y = \frac{2x + 1}{x - 1}$ (which is a well-defined real number, since we're not dividing by zero). Observe that:
\begin{align*}
\frac{y + 1}{y - 2}
&= \frac{\frac{2x + 1}{x - 1} + 1}{\frac{2x + 1}{x - 1} - 2} \\
&= \frac{(2x + 1) + (x - 1)}{(2x + 1) - 2(x - 1)} \\
&= \frac{(2x + 1) + (x - 1)}{(2x + 1) + (-2x + 2)} \\
&= \frac{3x}{3} \\
&= x
\end{align*}
as desired. $~~\blacksquare$
|
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|
Limit of a two variable function $\frac{x^2\sin^2{y}}{2x^2+y^2}$ I am trying to find the limit as $(x,y) \rightarrow (0,0)$ of $\frac{x^2\sin^2{y}}{2x^2+y^2}$
I tried this:
$\frac{x^2}{2x^2+y^2}<1/2$ so using the squeeze theorem $\frac{x^2}{2x^2+y^2}<1/2 \sin^2{y}$
and the limit of the right handside is 0, so the limit of the function must be 0. But wolfram alpha says the limit doesnt exist. Where am I going wrong?
|
Since $$2x^2+y^2 \geq 2\sqrt{2}\,|xy|$$
it follows that:
$$\left|\frac{x^2\sin^2 y}{2x^2+y^2}\right|\leq\left|xy\cdot\frac{xy}{2x^2+y^2}\right|\leq\frac{|xy|}{2\sqrt{2}}$$
so the limit is zero.
|
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|
Solving $3\sin x - \cos x = 2$ for $x \in [0, 2\pi)$ Problem:
Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$
My attempt:
I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.
I've tried to find ways to rewrite the equation by using $\sin x / \cos x = \tan x$ which I expect is the "desired" method, but for some reason, my algebra is failing me, and I can't seem to eliminate the $\sin$ and $\cos$ terms.
I've tried squaring both sides, but given the coefficient on $\sin x$, I can't find a way to use that identity either.
So far I end up with $$9\sin^2x - 6\sin x\cos x + \cos^2x = 4$$
Any help appreciated!
|
When you arrive at
$$
9\sin^2x - 6\sin x\cos x + \cos^2x = 4
$$
just note that $4=4\sin^2x+4\cos^2x$ so you can rewrite your equation as
$$
5\sin^2x-6\sin x\cos x-3\cos^2x=0
$$
Since $\cos x=0$ is not a solution, divide by $\cos^2x$ and get
$$
5\tan^2x-6\tan x-3=0
$$
so
$$
\tan x=\frac{3\pm\sqrt{24}}{5}
$$
You need to exclude the extraneous solution, of course.
A different method that doesn't need $\tan(x/2)$ is to set $X=\cos x$, $Y=\sin x$ and transform the equation into
$$
\begin{cases}
3Y-X=2\\
X^2+Y^2=1
\end{cases}
$$
so $X=3Y-2$ and then
$$
(3Y-2)^2+Y^2=1
$$
so
$$
10Y^2-12Y+3=0
$$
which gives $Y=\dfrac{6\pm\sqrt{6}}{10}$ and $X=\dfrac{-2\pm\sqrt{6}}{10}$ (with the same choice of signs). Thus $\tan x=Y/X$ and so
$$
\tan x=\frac{3\pm2\sqrt{6}}{5}
$$
as it is easy to verify. Note that this method is guaranteed not to introduce extraneous solutions.
|
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|
What is the probability of getting at least one $3$ when rolling two standard dice? What is the probability of getting at least one $3$ when rolling two standard ($6$-sided) dice?
I was thinking the answer would be $11$ out of $36$, but the textbook says it is $10$ out of $36$.
Please help.
|
The correct answer is: $\frac {1}{6} \times \frac{5}{6} + \frac {5}{6} \times \frac{1}{6} + \frac {1}{6} \times \frac{1}{6} = \frac {11}{36}$
The test book is in error.
|
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|
Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)$
Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6
|
Since
$n^{3} + 11n = 6m$
for some integer $m,$
we have
$$(n+1)^{3} + 11(n+1) = 6m + 3n^{2} + 3n + 12 = 6m + 12 + 3(n^{2} + n).$$
It suffices to prove that $3(n^{2} + n)$ is a multiple of $6$. But, since if $n$ is odd then $n^{2} + n = 2m'$ for some integer $m'$ and if $n$ is even then of course $n^{2} + n = 2m''$ for some integer $m'',$ it follows that $6$ is indeed a multiple of $3(n^{2}+n),$ qed.
|
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|
Integrating over the two form Let $A=(0,1)^2$. Let $\alpha:A\to\Bbb R^3$ be given by the equation
$$\alpha(u,v)=(u,v,u^2+v^2+1)$$
Let $Y$ be the image set of $\alpha$. Evaluate the integral over $Y_\alpha$ of the 2-form
$x_2dx_2\land dx_3+x_1 x_3 dx_1\land dx_3$.
Can someone please give me a useful hint for solving this?
Thanks in advance!
|
Given parametrization:
$$\phi (u,v) = (u,v,{u^2} + {v^2} + 1)$$
Setting
$$x = u,y = v,z = {u^2} + {v^2} + 1$$
Then calculate differential resp. to parametrization:
$$\begin{gathered}
dx = du \hfill \\
dy = dv \hfill \\
dz = 2udu + 2vdv \hfill \\
\end{gathered} $$
Form in old coordinates:
$$\omega (x,y,z) = ydy \wedge dz + xzdx \wedge dz = (ydy + xzdx) \wedge dz$$
Form in new coordinates resp. surface, the world famous pull-back:
$$\begin{gathered}
{\phi ^ * }\omega (u,v) = (u({u^2} + {v^2} + 1)du + vdv) \wedge (2udu + 2vdv) \hfill \\
{\phi ^ * }\omega (u,v) = 2uv({u^2} + {v^2} + 1)du \wedge dv - 2uvdu \wedge dv \hfill \\
{\phi ^ * }\omega (u,v) = 2({u^3}v + u{v^3})du \wedge dv \hfill \\
\end{gathered}$$
Ready to integrate:$$2 \cdot \int\limits_0^1 {\int\limits_0^1 {({u^3}v + u{v^3})dudv} } = \int\limits_0^1 {(\frac{1}{2}v + {v^3})dv = } \frac{1}{2}$$
Another user (Callus) saw this:
$$\begin{gathered}
d\omega = d(xzdx \wedge dz) \hfill \\
= d(xz) \wedge dx \wedge dz \hfill \\
= (zdx + xdz) \wedge dx \wedge dz \hfill \\
= zdx \wedge dx \wedge dz + xdz \wedge dx \wedge dz = 0 \hfill \\
\end{gathered} $$
That means the form is closed. On the other hand: We know formula for exterior
derivative for wedge of forms:
$$d(\alpha \wedge \beta ) = d\alpha \wedge \beta + {( - 1)^k}\alpha \wedge d\beta$$
Here $\alpha$ is a k-form. Therefore $$\omega = y{\kern 1pt} dy \wedge dz + xz{\kern 1pt} dx \wedge dz$$
may be written like:
$$\omega = d(\frac{1}{2}{y^2} \wedge dz) + d(\frac{1}{2}{x^2} \wedge zdz) = d(\frac{1}{2}({y^2}dz + {x^2}z)dz)$$
Or, if we set:
$$\theta = \frac{1}{2}({x^2}z + {y^2})dz$$
we get for short:
$$\omega = d\theta $$
At least resp. local coordinates, $\omega$
is exact. It's the exterior derivative from another form $\theta$.
So, by Stoke's theorem, that means:
$$\int\limits_A {d\theta } = \int\limits_{\partial A} \theta = \int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz}$$
Let's evaluate the RHS:
If $z = {x^2} + {y^2} + 1$, so $dz = 2xdx + 2ydy$ and for RHS we get:
$$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz} = \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})xdx} + \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})ydy}$$
With counterclockwise orientation of boundary $\partial A = A\backslash (0,1) \times (0,1)$, we have to manage following four integrals:
$$\left. {\begin{array}{*{20}{c}}
{x \in [0,1]} \\
{y = 0}
\end{array}} \right\}\int\limits_0^1 {({x^5} + {x^3})dx} = \frac{5}{{12}}$$
$$\left. {\begin{array}{*{20}{c}}
{x = 1} \\
{y \in [0,1]}
\end{array}} \right\}2\int\limits_0^1 {({y^3} + y)dy} = \frac{3}{2}$$
$$\left. {\begin{array}{*{20}{c}}
{x \in [0,1]} \\
{y = 1}
\end{array}} \right\} - \int\limits_0^1 {({x^5} + 2{x^3} + x)dx} = - \frac{1}{6} - \frac{1}{2} - \frac{1}{2} = - \frac{7}{6}$$
$$\left. {\begin{array}{*{20}{c}}
{x = 0} \\
{y \in [0,1]}
\end{array}} \right\} - \int\limits_0^1 {{y^3}dy} = - \frac{1}{4}$$
Last two integrals got minus-sign resp. orientation. Summing up, we have:
$$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz} = \frac{5}{{12}} + \frac{3}{2} - \frac{{14}}{{12}} - \frac{3}{{12}} = \frac{6}{{12}} = \frac{1}{2}$$
|
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|
If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to find out relation but that will be long and tedious process , I want shorter approach to this problem . Btw I was given following options for this question
A) $a^2+c^2+ab=0$
B) $a^2-c^2+ab=0$
C) $a^2-c^2-ab=0$
D) $ap^2+bp+c=0$
Maybe we can relate something by looking at options?
|
Following the path you have taken, and correcting the algebra, one Gets
$$(\lambda X+D)(X^2+pX+1)=\lambda X^3+(\lambda p+D)X^2+(Dp+\lambda)X+D$$
And this is valid for all $X$. So by identifying the coefficients we get
$$\begin{cases} \lambda=a\\ \lambda p+D=0\\ Dp+\lambda=b\\ D=c \end{cases}$$
which reads
$$\begin{cases} a p+c=0\\ cp+a=b\end{cases}$$
and this leads to
$$-\frac{c^2}{a}+a=b$$
$$a^2-c^2-ab=0$$
|
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|
Dynamical system $x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...$
Consider the dynamical system
$$
x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...
$$
So by using the substitution $x_n = \cot(y_n)$, I have found:
$$
x_n = \cot(\cot^{-1} (x_0) \cdot 2^n )
$$
but the next part of the question says
Let us parametrize the initial condition $x_0$ by means of a unique angle $\theta \in (0, \pi)$ such that $x_0 = \cot(\theta)$. Show that, for every $p>1$, the choice
$$
\theta = \theta_p := \frac{\pi}{2^p - 1}
$$
yields a period-p solution.
So I know I have to show that $x_n = x_{n+p}$, that is:
$$
cot(\frac{\pi}{2^p - 1} \cdot 2^n) = cot(\frac{\pi}{2^p - 1} \cdot 2^{n+p})
$$
and so, using periodicity of $\cot$, that
$$
\frac{\pi}{2^p - 1} \cdot 2^n + k \pi= \frac{\pi}{2^p - 1} \cdot 2^{n+p}
$$
for some $k\in\mathbb{Z}$. Inductively, we have for $p = 1$ that $\theta_1 = \pi$ hence
$$
x_{n+1} = \cot (\pi \cdot 2^{n+1}) = \cot(\pi \cdot 2^n + \pi \cdot 2^n) = \cot(\pi \cdot 2^n) = x_n
$$
and so we have a period-1 solution. How do I show the inductive step for this proof?
|
For $x_0=\cot(\theta)$ we have that
$x_1=\frac{1}{2}\Big[\frac{\cos(\theta)}{\sin(\theta)}-\frac{\sin(\theta)}{\cos(\theta)}\Big]= \cot(2\theta)$
and inductively
$x_{n}=\cot(2^n\theta)$
Thus when $\theta=\frac{\pi}{2^p-1}$ we obtain
$x_{n+p}=\cot\big(\frac{2^{n+p}\pi}{2^p-1}\big)=\cot\Big(\frac{2^n(2^{p}-1)\pi}{2^p-1}+\frac{2^{n}\pi}{2^p-1}\Big)=\cot\Big(2^n\pi+\frac{2^{n}\pi}{2^p-1}\Big)=\cot\Big(\frac{2^{n}\pi}{2^p-1}\Big)=x_n$
since $2^n\pi$ ia an integer multiple of $\pi$.
|
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|
the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots + n^2 .$$
or, as another example if a = 0.01 .$$\frac{n (n + 0.01) (2 n + 0.01) }{6 \cdot 0.01} = 0.01^2 + 0.02^2 + \cdots + n^2 .$$
Now if we follow the same logic, I suppose if the difference between the closest numbers becomes smallest possible (a = 0.0...1), we will obtain
$$ \frac{n (n + 0.0..1) (2 n + 0.0..1)}{6 \cdot 0.0..1} = 0.0..1^2 + 0.0..2^2 + \cdots + n^2$$
So can conclude that
$$\frac{2n ^ 3}{6} = \frac{n ^ 3}{3} = \ (0.0..1 ^ 2 + 0.0..2 ^ 2 .. + n ^ 2) * {0.0..1}$$ If we take that, a = 0.0...1 and, b = $\frac {n}{0.0...1}$ and we rearrange the formula we will get this:
$$\frac{n ^ 3}{3} = 1^2a^3 + 2^2a^3 + \dots+b^2 a^3$$
following the same logic we prove that the formula holds for each degree (it is easy to check), so that we could generalize, If we take that degree call m, we will get
$$\frac{n ^ m}{m} = 1^{m-1}a^m + 2^{m-1}a^m + \dots+b^{m-1} a^m$$
so the question is this expression can be interpreted as a Riemann sum, and why?
|
I think this is correct, as you are essentially integrating $x^2$, to get the result of $\frac{x^3}3$. this, if you don't know, is adding together rectangles of increasingly smaller width (a) and heights which are the result of a function ($x^2$).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?
|
There are simpler ways. There is absolutely no preference between the numbers, so each number is equally likely to be the second one drawn. Since 3 of 5 numbers are odd, the probability is 3/5 that an odd number is drawn.
|
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|
Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{1}{2} \sec^2 (\frac{x}{2})\;dx.$ With algebra we get that $dx= \frac{2}{1+t^2}dt,$ $\sin x = \frac{2t}{1+t^2},$ and $\cos x = \frac{1-t^2}{1+t^2}.$ Therefore we get the following:
$$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4} = \int \frac{\frac{2}{1+t^2}}{(\sqrt{\frac{1-t^2}{1+t^2}}+\sqrt{\frac{2t}{1+t^2}})^4}dt = 2\int \frac{1}{1+t^2}\cdot \frac{(1+t^2)^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
$$
= 2\int \frac{1+t^2}{(\sqrt{1-t^2}+ \sqrt{2t})^4}dt
$$
I expanded the denominator, but only made things worse. According to wolfram alpha, the solution is in terms of elementary functions. Can you give me any clues or hints on how to evaluate this beast. Thank you!
|
To make the integral easier to evaluate. Write the integral as
$$ I = \int \frac{dx}{\cos^2(x)( 1+ \sqrt{\tan(x)} )^4} $$
and the use the substitution $\tan(x)=u^2$ which makes the integral falls a part
$$ I=2 \int \frac{u}{(1+u)^4} du .$$
You can use another substitution $1+u=t$ to finish the problem.
|
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|
Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
|
Set $y=x+2$. The equation rewrites as:
$$(y-1)^2+y^3+(y+1)^3==2\iff y(y^3+5y^2+7y^+2)=0$$
The second factor has an integer root: $-2$, hence the factorisation:
$$y^3+5y^2+7y +2=(y+2)(y^2+3y+1).$$
Finally, the roots (in $y$) are: $0, -2, \dfrac{-3\pm\sqrt 5}{2} $, whence the roots of the initial equation:
$$ \Bigl\{ -2,-4, \frac{-7\pm\sqrt 5}{2}\Bigr\}.$$
|
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|
Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2(n-1)+1)(n-1)$=$2n^2-3n+1$
for $k=n$
$a_n$= $\sum\limits_{k=1}^{2n} {(-1)^n \cdot n^2}$+$a_{n-1}$
$a_n$=${(-1)^n \cdot n^2}$+$2n^2-3n+1$
But that last equation seems somehow wrong to get $(2n+1)\cdot n$
|
$$a_n=\sum_{k=1}^{2(n-1)}(-1)^kk^2+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$$=a_{n-1}+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$=2n^2-3n+1+(2n)^2-(2n-1)^2=\cdots$
W/O using induction,
$$a_n=\sum_{k=1}^{2n}(-1)^kk^2=\sum_{k=1}^n[(2k)^2-(2k-1)^2]=4\sum_{k=1}^nk-\sum_{k=1}^n1=4\cdot\dfrac{n(n+1)}2-n$$
|
{
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|
Bonferroni Inequalities
Let $k$ and $m$ be positive integers with $k>m$.
Then the partial sums of
$$
1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}
$$
has alternating signs.
(The partial sums of the given sum are $P_1=1$, $P_2=1-\binom{k}{1}$, $P_3=1-\binom{k}{1}+\binom{k}{2}$, etc)
I arrived at the above problem while trying to show the follwoing (known as Bonferroni inequalities):
Let $A_1, \ldots, A_n$ be events of a probability space. For a subset $I$ of $\{1,\ldots, n\}$, write $A_I$ to denote $\bigcap_{j\in I}A_j$. Further, denote $\sum_{|I|=i}P(A_I)$ as $\sigma_i$. We agree by convention that $\sigma_0=1$.
Then the partial sums of
$$
P(A_1^c\cap A_2^c\cap \cdots\cap A_n^c)-\sigma_0+\sigma_1-\sigma_2\cdots
$$
have alternating signs.
|
$$\begin{align}
\require{cancel}
&1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}\\
&=1+\sum_{r=1}^{m}(-1)^k\binom kr\\
&=1+\sum_{r=1}^{m}(-1)^r\left[\binom {k-1}{r-1}+\binom {k-1}r\right]\\
&=\color{lightgrey}{\cancel{1}-\left[\cancel{{\binom {k-1}0}}+\bcancel{\binom {k-1}1}\right]
+\left[\bcancel{\binom {k-1}1}+\cancel{\binom {k-1}2}\right]
-\left[\cancel{\binom {k-1}2}+\bcancel{\binom {k-1}3}\right]
+\cdots
+(-1)^m\left[\bcancel{\binom {k-1}{m-1}}+\binom {k-1}m\right]}\\
&=(-1)^m\binom {k-1}m\qquad \blacksquare\\
\end{align}$$
Alternatively:
$$\begin{align}
&1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}\\
&=1+\sum_{r=1}^{m}(-1)^k\binom kr\\
&=1+\sum_{r=1}^{m}(-1)^r\left[\binom {k-1}{r-1}+\binom {k-1}r\right]\\
&=1+\sum_{r=0}^{m-1}(-1)^{r+1}\binom {k-1}{r}+\sum_{r=1}^{m}(-1)^r\binom {k-1}r\\
&=1\color{blue}{-\sum_{r=0}^{m-1}(-1)^r\binom {k-1}{r}}
\color{green}{+\sum_{r=1}^{m}(-1)^r\binom {k-1}r}\\
&=\cancel{1}\color{blue}{-\cancel{1}-\bcancel{\sum_{r=1}^{m-1}(-1)^r\binom {k-1}{r}}}
\color{green}{+\bcancel{\sum_{r=1}^{m-1}(-1)^r\binom {k-1}r}+(-1)^m\binom {k-1}m}\\
&=(-1)^m\binom {k-1}m\qquad \blacksquare\\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Scale invariant Image Moments - not scale variant? I came across a problem working with image moments [1].
It is stated that
$\eta_{ij} = \frac{\mu_{ji}}{\mu_{00}^{k}}$
where $k = 1 + \frac{i+j}{2}$
is scale invariant.
However, if I try to reproduce this, it does not appear scale invariant at all.
Consider a simple example:
In a binary image, we calculate $\eta_{20}$ of a 2x2 block of 4 pixels:
☐☐
☐☐
$\mu_{20} = 0.5^2 + 0.5^2 + (-0.5)^2 + (-0.5)^2 = 4 \cdot 0.25 = 1$
$\mu_{00} = 4$
$k = 1 + \frac{2+0}{2} = 1+1 = 2$
$\eta_{20} = \frac{1}{4^2} = \frac{1}{16} = 0.0625$
Now, let's scale this block by the factor two:
☐☐☐☐
☐☐☐☐
☐☐☐☐
☐☐☐☐
$\mu_{20} = 4 \cdot 1.5^2 + 4 \cdot 0.5^2 + 4 \cdot (-0.5)^2 + 4 \cdot (-1.5)^2 = 8 \cdot 2.25 + 8 \cdot 0.25 = 18 + 2 = 20$
$\mu_{00} = 16$
$k = 1 + \frac{2+0}{2} = 1+1 = 2$
$\eta_{20} = \frac{20}{16^2} = \frac{20}{256} = 0.078125$
Why do we have a different result after scaling the object if $\eta_{ij}$ is supposedly scale invariant? Is there any formal proof of the scale invariance of $\eta$?
[1] https://en.wikipedia.org/wiki/Image_moment
|
The invariance is exact in the continuous domain. Your problem here is how you scale, which is limited by the discrete grid.
Scaling by a factor 2 of your original 4 squares would result in 4 squares, centered at (-1,-1), (-1,1), (1,-1) and (1,1), each with a weight of 4:
$$
\mu_{20} = 4 \cdot 1^2 + 4 \cdot 1^2 + 4 \cdot (-1)^2 + 4 \cdot (-1)^2 = 16
$$
$$
\mu_{00} = 4 \cdot 4 = 16
$$
$$
\eta_{20} = \frac{16}{16^2} = 0.0625
$$
On the discrete grid, the invariances are approximate.
|
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|
How do you find the inflection point of this graph? The graph is this:
$$
\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}
$$
I know you can find second derivative and then solve for values that make it undefined or 0, but I was told apparently there is another faster way to get the inflection point. What is that method?
|
Let $$y=\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}$$
Then using the substitution, $u=x+1$, we obtain
$$y=\frac{u^3 - 4u^2 + 4u}{u^2 - 2u + 1}$$
This simplifies to $$y=\frac{u(u-2)^2}{(u-1)^2}$$
Differentiating with respect to $u$, we obtain
$$\frac{\partial y}{\partial u}=\frac{u^3-3u^2+4u-4}{(u-1)^3}$$
Since $$\frac{\partial u}{\partial x}=1$$
We have $$\frac{\partial y}{\partial x}=\frac{\partial y}{\partial u}\cdot\frac{\partial u}{\partial x}=\frac{\partial y}{\partial u}$$
Substituting back in $u=x+1$, we are left with
$$\frac{\partial y}{\partial x}=\frac{(x+1)^3-3(x+1)^2+4(x+1)-4}{x^3}$$
This simplifies to
$$\frac{\partial y}{\partial x}=\frac{(x^3+3x^2+3x+1)-(3x^2+6x+3)+(4x+4)-4}{x^3}=\frac{x^3+x-2}{x^3}$$
Computing the second derivative
$$\frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial x}(1+\frac{1}{x^2}-\frac{2}{x^3})=-\frac{2}{x^3}+\frac{6}{x^4}=\frac{6-2x}{x^4}$$
A point of inflection will occur when $\frac{\partial y}{\partial x}=0$, solving for this we get $6-2x=0$ gives $x=3$
The sufficient condition for points of inflection require that either side of the neighbourhood of $x=3$ have different signs.
Checking in the neighbourhood of $x=3$ we obtain a positive value for $\frac{\partial^2 y}{\partial x^2}$ when $x=2$ and a negative value for $\frac{\partial^2 y}{\partial x^2}$ when $x=4$.
Therefore $x=3$ is a point of inflection.
|
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|
Evaluate $\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$ This is a integration question from a previous calculus exam:
Evaluate $$\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$$
I rewrote it as $$\lim \limits_{b \to \infty} \int_0^b \left(\frac{x^{10}}{1+x^{14}} \right)^{2} \, dx,$$ where do I go on from there?
|
Hint:
$$\begin{align}\frac{1}{14}\int \frac{x^7 14x^{13}}{(1 + x^{14})^2}dx& \underbrace{=}_{\color{red}{\text{by parts}}}\frac{1}{14}\bigg( -x^7\frac{1}{1 + x^{14}} + 7\color{#05f}{\int \frac{x^6}{1+x^{14}}dx}\bigg) \\&= \frac{1}{14}\bigg(\arctan(x^7) -\frac{x^7}{1 + x^{14}}\bigg) + C\end{align}$$
For $$\color{#05f}{\int \frac{x^6}{1+x^{14}}dx = \frac{1}{7}\arctan (x^7) + C}$$ let $v = x^7$.
|
{
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|
Find a volume using a triple integral.. This is the problem:
Find the volumen of de solid bounded by $x^2+y^2=2$, $z=0$ and $x+2y+2z=2$.
I have set the parameters to:
$0 \leq z \leq \dfrac{2-x-y}{2}$
$ 0\leq y\leq 2-x$
$0 \leq x \leq 2$
and evaluated:
$
\displaystyle{ \int\limits_{0}^{2} \int\limits_{0}^{2-x} \int\limits_{0}^{\dfrac{2-x-y}{2}} dz\,dy\,dz}=\dfrac{4}{3}
$
Is there anything wrong?? Please help me... thank you so much...
|
There is no way in which a 2d-circle and a plane may bound a solid, so I assume you are trying to find the volume of the intersection between the cylinder $\{x^2+y^2\leq 2,z\geq 0\}$ and the half-space $x+2y+2z\leq 2$. It is worth to apply a rotation around the $z$-axis, by setting:
$$ u=\frac{x+2y}{\sqrt{5}},v=\frac{2x-y}{\sqrt{5}} $$
in order that the problem can be stated as: find the volume of the intersection between the cylinder $\{u^2+v^2\leq 2,z\geq 0\}$ and the half-space $\frac{\sqrt{5}}{2}u+z\leq 1$. Now the solid is symmetric with respect to the $uz$-plane.
Since $-\sqrt{2}\leq u\leq\sqrt{2}$, $z$ ranges between $0$ and $1+\frac{1}{2}\sqrt{10}$. Let us compute the area of a section $A_z$ according to the value of $z$ in the previous range:
$$ A_z = 2\int_{-\sqrt{2}}^{\frac{2}{\sqrt{5}}(1-z)}\sqrt{2-u^2}\,du =2\left.\left(w\sqrt{1-w^2}+\arcsin w\right)\right|_{-1}^{\sqrt{\frac{2}{5}}(1-z)}$$
giving:
$$ A_z = \pi + 2\left(\sqrt{\frac{2}{5}}(1-z)\sqrt{1-\frac{2}{5}(1-z)^2}+\arcsin\left(\sqrt{\frac{2}{5}}(1-z)\right)\right).$$
Now the volume is given by:
$$ V = \int_{0}^{1+\frac{1}{2}\sqrt{10}}A_z\,dz = \color{red}{\pi+\frac{4}{5}\sqrt{6}+2\arcsin\sqrt{\frac{2}{5}}}.$$
|
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|
Proving gcd($a,b$)lcm($a,b$) = $|ab|$
Let $a$ and $b$ be two integers. Prove that $$ dm = \left|ab\right| ,$$ where $d = \gcd\left(a,b\right)$ and $m = \operatorname{lcm}\left(a,b\right)$.
So I went about by saying that $a = p_1p_2...p_n$ where each $p_n$ is a prime. Same applies to $b = q_1q_2 ... q_c$. So then $m = (u_1u_2...)(p_1p_2...p_n)$ and $m = (t_1t_2...)(q_1q_2...q_c)$ since $a|m$ and $b|m$. $m$ has a unique factorization, so the primes $(u_1u_2...)(p_1p_2...p_n) = (t_1t_2...)(q_1q_2...q_n)$ and the gcd(a,b) = $(p_1p_2...p_n) \cap (q_1q_2...q_n)$ (I know this is not mathematically correct, so is there a correct way to express this?).
So $dm = |ab| \iff d= \frac{|ab|}{m}$. And by the definition above, $\frac{ab}{(t_1t_2...)b} = \frac{a}{(t_1t_2...)}$. And this is where I get stuck. Is my proof right? Am I going in the right direction?
Thanks
PS. I am trying to do this using only the prime factorization theorem and the definitions of the gcd and the lcm.
|
If $a=b=0$, then any integer is a common divisor of $a$ and $b$.
So, there is no greatest common divisor of $a$ and $b$.
So we don't consider the case in which $a=b=0$.
If $a=0$ and $b\neq 0$, then the set of the common divisors of $a$ and $b$ is equal to the set of divisors of $b$.
Obviously, the greatest divisor of $b$ is $|b|$.
So, $\gcd(a,b)=|b|$.
If $a=0$ and $b\neq 0$, the only common multiple of $a$ and $b$ is $0$ since the only multiple of $a$ is $0$ and $0$ is a multiple of $b$.
So, $\operatorname{lcm}(a,b)=0$.
Therefore, if $a=0$ and $b\neq 0$, then $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=|b|\cdot 0=|a\cdot b|$.
If $a\neq 0$ and $b=0$, then by symmetry, $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=|a|\cdot 0=|a\cdot b|$.
Let $a>0$ and $b>0$.
Obviously, $a\cdot b$ is a positive common multiple of $a$ and $b$.
So, the least common multiple of $a$ and $b$ exists, and it's positive.
Let $M$ be any common multiple of $a$ and $b$.
Let $M=q\cdot\operatorname{lcm}(a,b)+r,\,\, 0\leq r<\operatorname{lcm}(a,b)$.
Then, obviously, $r=M-q\cdot\operatorname{lcm}(a,b)$ is a common multiple of $a$ and $b$.
If $0<r<\operatorname{lcm}(a,b)$, then $r$ is smaller than $\operatorname{lcm}(a,b)$.
This is a contradiction.
So, $r=0$.
So, $M=q\cdot\operatorname{lcm}(a,b)$.
So, any common multiple of $a$ and $b$ is a multiple of $\operatorname{lcm}(a,b)$.
Since $a\cdot b$ is a positive common multiple of $a$ and $b$, so we can write $a\cdot b=D\cdot \operatorname{lcm}(a,b)$ for some positive integer $D$.
Let $\operatorname{lcm}(a,b)=a\cdot a^{'}$ and $\operatorname{lcm}(a,b)=b\cdot b^{'}$.
Then, $a\cdot b=D\cdot\operatorname{lcm}(a,b)=D\cdot a\cdot a^{'}=D\cdot b\cdot b^{'}$.
So, $b=D\cdot a^{'}$ and $a=D\cdot b^{'}$.
So, $D$ is a common divisor of $a$ and $b$.
Assume that there exists a common divisor $D^{'}$ of $a$ and $b$ which is larger than $D$.
Then, $M^{'}:=a\cdot\frac{b}{D^{'}}=\frac{a}{D^{'}}\cdot b$ is a positive common multiple of $a$ and $b$.
Then, $0<M^{'}=a\cdot\frac{b}{D^{'}}=\frac{a}{D^{'}}\cdot b<\frac{a\cdot b}{D}=\operatorname{lcm}(a,b)$.
This is a contradiction.
So, $D$ is the greatest common divisor of $a$ and $b$.
So, $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=a\cdot b=|a\cdot b|$.
If $a>0$ and $b<0$, then $\gcd(a, -b)\cdot\operatorname{lcm}(a,-b)=|a\cdot(-b)|=|a\cdot b|$.
Obviously, the set of the common divisors of $a$ and $b$ is equal to the set of the common divisors of $a$ and $-b$.
So, $\gcd(a,b)=\gcd(a,-b)$.
Obviously, the set of the common multiples of $a$ and $b$ is equal to the set of the common multiples of $a$ and $-b$.
So, $\operatorname{lcm}(a,b)=\operatorname{lcm}(a,-b)$.
So, $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=|a\cdot b|$.
If $a<0$ and $b>0$, then, by symmetry, $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=|a\cdot b|$.
If $a<0$ and $b<0$, then $\gcd(-a, -b)\cdot\operatorname{lcm}(-a,-b)=|(-a)\cdot(-b)|=|a\cdot b|$.
Obviously, the set of the common divisors of $a$ and $b$ is equal to the set of the common divisors of $-a$ and $-b$.
So, $\gcd(a,b)=\gcd(-a,-b)$.
Obviously, the set of the common multiples of $a$ and $b$ is equal to the set of the common multiples of $-a$ and $-b$.
So, $\operatorname{lcm}(a,b)=\operatorname{lcm}(-a,-b)$.
So, $\gcd(a, b)\cdot\operatorname{lcm}(a,b)=|a\cdot b|$.
|
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|
What does a positive definite matrix have to do with Cauchy-Schwarz Inequality? In my text book, Cauchy-Schwarz Inequality is extended to a positive definite matrix.
But I neither understand what the relationship between Cauchy-Schwarz Inequality and a positive definite matrix nor the sentence underlined in red, I am not a strong linear algebra person, I have just started studying it.
I'd like post the page here so that I might get a better explanation.
Thank you.
|
The sentence you underlined in red abbreviates these steps:
$$(\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{-1/2} \mathbf{d})^2 \leq (\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{1/2} \mathbf{b})(\mathbf{B}^{-1/2} \mathbf{d} \cdot \mathbf{B}^{-1/2} \mathbf{d}) $$
This is the Cauchy-Schwarz inequality applied as requested. Now we use the transpose-based definition of the dot product $(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a}^T \mathbf{b}$:
$$(\mathbf{b}^T (\mathbf{B}^{1/2})^T \mathbf{B}^{-1/2} \mathbf{d})^2 \leq (\mathbf{b}^T (\mathbf{B}^{1/2})^T \mathbf{B}^{1/2} \mathbf{b})
(\mathbf{d}^T (\mathbf{B}^{-1/2})^T \mathbf{B}^{-1/2} \mathbf{d})$$
Finally we take advantage of the the following observations:
*
*$\mathbf{B}$ is positive definite and therefore symmetric
*$\mathbf{B}^{1/2}$ is also positive definite and symmetric
*The inverse of a positive definite matrix is positive definite
As a result:
$$(\mathbf{B}^{1/2})^T \mathbf{B}^{-1/2} = \mathbf{B}^{1/2} \mathbf{B}^{-1/2} = \mathbf{I}$$
$$(\mathbf{B}^{1/2})^T \mathbf{B}^{1/2} = \mathbf{B}^{1/2} \mathbf{B}^{1/2} = \mathbf{B}$$
$$(\mathbf{B}^{-1/2})^T \mathbf{B}^{-1/2} = \mathbf{B}^{-1/2} \mathbf{B}^{-1/2} = \mathbf{B}^{-1}$$
Completing the proof.
You can see how this relates to the regular Cauchy-Schwarz inequality (for the dot product) by considering what it states when $\mathbf{B} = \mathbf{I}$ (Since $\mathbf{I}$ is a positive definite matrix). In this case we have
$$(\mathbf{b} \cdot \mathbf{d})^2 \leq (\mathbf{b} \cdot \mathbf{b}) (\mathbf{d} \cdot \mathbf{d})$$
Which is precisely the usual Cauchy-Schwarz inequality for the dot product.
|
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|
When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is even $x^2\equiv (1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
Thus, when $p\equiv 1\pmod{4}$, the congruence $x^2\equiv -1\pmod{p}$ is soluble.
Point of contention: I understand the general argument
I understand the relation
$(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\equiv (p-1)!\equiv -1\pmod{p}$
But I cant work out how $x^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
How is $((\frac{p-1}{2})!)^2\equiv(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))\pmod{p}$
|
For any $i$, $p-i\equiv -i\pmod{p}$, so the right-hand side of your final congruence is
\begin{equation*}
(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))
\equiv (1\cdot 2\cdot...\cdot r)((-1)\cdot(-2)\cdot...\cdot(-r))
\equiv r!(-1)^rr! \equiv (-1)^r(r!)^2\pmod{p}.
\end{equation*}
But since $p\equiv 1\pmod{4}$, it follows that $r = \frac{p-1}{2}$ is even, so this is just $(r!)^2 = \left(\left(\frac{p-1}{2}\right)!\right)^2\pmod{p}$.
|
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|
Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx $
but I am not sure how this was accomplished. Any ideas? I am more interested in the method than the answer.
|
when you get the $\int 1 - \frac{x-2}{x^2+x-2} dx$, you can use $x^2+x-2 = (x-1)(x+2)$.
then it become
$$
\int 1 - \frac{x+2-4}{(x+2)(x-1)}dx
$$
which equal to
$$
\int 1 - \frac{1}{x-1}-\frac{4}{(x+2)(x-1)}dx
$$
$$
\int 1 - \frac{1}{x-1}-4/3(\frac{1}{(x-1)}+\frac{1}{(x+2)})dx
$$
|
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|
Help with double simultaneous equations and roots I am in the course of a project, in which I need to solve these two simultaneous equations:
\begin{equation}
\sqrt{(1000-y)^2 + x^2} - \sqrt{y^2 + x^2} = 342.371
\end{equation}
\begin{equation}
\sqrt{(2000-x)^2 + y^2} - \sqrt{y^2 + x^2} = 961.674
\end{equation}
I know that the answers are y = 250, and x = 500, but for some reason, I cannot back it up with any calculations, because they always seem to get faulty. I think it has something to do with the square roots, but I was hoping for a useful hand out there.
Just for the sake of the argument, my methodology consists of:
- expanding the formulas
- cancel all equal terms in each formula
- write one variable in terms of another and...
- ... plug in the expression into the second equation.
But my problem already starts with the fact that when I start with the first equation and cancel all equal terms, I am left with an expression containing only one variable. (solving this variable results also in the wrong answer.)
Thanks again for the useful help.
Here my calculations for the first equation, which shall give me an expression to solve the second one:
\begin{equation}
((1000 - y)^2 + x^2) - (y^2 + x^2) = 342.371^2
\end{equation}
\begin{equation}
1000^2 - 2000y + y^2 + x^2 - y^2 - x^2 = 343.371^2
\end{equation}
\begin{equation}
1000^2 - 2000y = 343.371^2
\end{equation}
... and now I am left with an equation without an x in it... and solving this also does not result in the correct solution anyways.
|
$\sqrt{(1000-y)^2+x^2}-\sqrt{y^2+x^2}=Q$.
$\sqrt{(1000-y)^2+x^2}=\sqrt{y^2+x^2}+Q$.
$(1000-y)^2+x^2=y^2+x^2+2Q\sqrt{y^2+x^2}+Q^2$.
$(1000-y)^2+x^2-y^2-x^2-Q^2=2Q\sqrt{y^2+x^2}$.
$1000000-2000y-Q^2=2Q\sqrt{y^2+x^2}$.
Now square both sides to get an equation quadratic in $x$ and $y$. Then go through the same steps with the other equation to get a second quadratic in $x$ and $y$. Then solve the system of 2 equations in 2 unknowns.
|
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|
Dice roll probability, at least 9 total? If I have two dice with $6$ sides each, what is the probability of me rolling atleast $9$ total? I think I'm correct when thinking that the probability of rolling a $9$ is $\frac{4}{36}$, that is $11.1...\%$, but how do I go from here to calculate the "at least" part?
|
$\begin{array}{|c|c|c|c|}
\hline
&\overrightarrow{ D2} & \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6} \\ \hline D1\downarrow\\\hline
\color{blue}{1} &&2 &3 &4 & 5 & 6 & 7\\ \hline
\color{blue}{2}&& 3 & 4&5 & 6 & 7 & 8\\ \hline
\color{blue}{3} &&4 &5 &6 & 7 & 8 & \color{red}{9}\\ \hline
\color{blue}{4} &&5 & 6 & 7 & 8 &\color{red}{9} &\color{red}{10}\\ \hline
\color{blue}{5} &&6 &7 &8 &\color{red}{9} &\color{red}{10} &\color{red}{11} \\ \hline
\color{blue}{6} && 7 & 8 & \color{red}{9} &\color{red}{10} &\color{red}{11}&\color{red}{12}\\\hline
\end{array}$
Every cell containing a number in red, satisfies: $(D1+D2) \ge 9$
Assuming that each die is a fair die, the probability of obtaining any number from 1-6 on each of the two dice is $\frac{1}{6}$.
For example, the probability of obtaining $(D1,D2)=(1,1)$ is $\left(\frac{1}{6}\right)^2 = \left(\frac{1}{36}\right)$
Every individual outcome in the table is obtained with probability $\frac{1}{36}$ as each result is equally likely.
Since there are 36 (6 $\times$ 6 table) total outcomes, the probability will be $\frac{x}{36}$ or an equivalent fraction.
Try to find $x$ via a simple counting method (count how many numbers in red there are).
|
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|
What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$? What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$?
Let's try a several primes greater than 3...
If $p=5$, then we have $1^2 + 2^2 + 3^2 + 4^2 = 30$, so that $30\pmod{5} = 0$
If $p=7$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2= 91$, so that $91\pmod{7} = 0$
If $p=11$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2= 385$, so that $385\pmod{11} = 0$
So it seems like we will always get 0. So for primes greater than 3, we have the sum equal to a multiple of $p$.
I know that $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1)$.Well if $p>3$, then $p-1=2a$ for some integer $a$, so that $2p-1 = 4a +1$. Then we have
$$ 1^2 + 2^2 + 3^2 + \cdots + (p-1)^2 = \frac16p(p-1)(2p-1) = \frac16 p(2a)(4a+1)$$
If I could show that $(2a)(4a+1)$ is a multiple of 6, I'll be done, but I am having trouble showing that. Perhaps someone could provide a much needed tip, but don't figure this out for me, please.
|
Assuming the formula for sum of squares (which has to be an integer) we see that the 6 in the denominator has to divide the numerator: Now 6 is coprime to $p$ (for $p >3$). Now you can figure it out yourself.
|
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|
Question about matrix multiplication notation I have the following matrices:
$A=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix}$
; $B=\begin{pmatrix}
2 & 3\\
2 & 0 \\
0 & 3\\
\end{pmatrix}$
; $C=\begin{pmatrix}
6 & 3& 4 \\
6 & 6 & 0 \\
\end{pmatrix}$
;$u=\begin{pmatrix}2 \\ -5 \\ 4 \end{pmatrix}$
I want to calculate the following:
$\sum_j a_{ij}u_j$
$\sum_{jk} c_{ij}a_{jk}b_{kl}$
but I am not used to this kind of notation. Does the first one mean to multiply all $i$-rows of A with the $j$-column of u?
$\implies AB=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix} \cdot \begin{pmatrix}2 \\ -5 \\ 4 \end{pmatrix}=\begin{pmatrix}-3 \\ 4 \\ 0 \end{pmatrix}$?
I am not sure what the second one is trying to say? Maybe $C\cdot A\cdot B$?
|
Yes. Note that you can perform the usual rows-columns product of two matrices $AB$ only if the number of columns of $A$ is the same as the number of rows of $B$.
|
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|
Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$
|
Late answer since it was marked as duplicate somewhere else but the way I suggest was not among the answers.
Looking at the denominator of $\frac{x^2y}{x^2+y^3}$ it makes sense to consider a path of approaching $(0,0)$ close to the curve $x^2+y^3=0$.
So, consider for $x> 0$ and $\alpha > 0$ the path
$$(x,-\sqrt[3]{x^2 + x^{\alpha}})\Rightarrow \frac{x^2y}{x^2+y^3}=\frac{x^2\sqrt[3]{x^2 + x^{\alpha}}}{x^{\alpha}}\geq \frac{x^{2+\frac{\alpha}{3}}}{x^{\alpha}} = \frac{1}{x^{\frac{2}{3}\alpha-2}}$$
For $\frac{2}{3}\alpha-2 >0 \Leftrightarrow \alpha > 3$ we get
$$\frac{1}{x^{\frac{2}{3}\alpha-2}}\stackrel{x \to 0^+}{\longrightarrow}+\infty$$
So, approaching $(0,0)$ along such a path shows that the limit does not exist.
|
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|
Combinatorics Recurrence relation
Let $h_n$ be a number sequence where $h_n = 3h_{n-1} - 2h_{n-2}$ with $h_0 = 0$ and $h_1 = 1$. Compute the ordinary generating function of $h_n$ and then using the generating function compute a formula for $h_n$.
Does this start looks right?
We write the recurrence relation in form:
$$h_n - 3h_{(n-1)} + 2h_{(n-2)} = 0$$
Let $g(x) = h_0 + h_1 x + h_2 x^2+\ldots$ be the generating function for sequence $h_0, h_1,\ldots$.
We have $g(x) = h_0 + h_1 x + \ldots$ as well as $-3x\,g(x) = -3h_0 x -3 h_1 x^2 - \ldots$ and $2x^2\,g(x) = 2h_0 x^2 + 2h_1 x^3 +\ldots$.
Adding all three would give us :
$$(1 -3x + 2x^2) g(x) = h_0 + (-3h_0 + h_1) x + (2h_0 -3h_1 + h_2) x^2 + \ldots + (h_n -3h_{n-1} + h_{n-2}) x^n + \ldots$$
Since $h_n - 3h_{n-1} + 2 h_{n-2} = 0$ and since $h_0 = 0, h_1 = 1$.
So we know that $(1 - 3x + 2x^2) g(x)= 1 + x$ and that leads to:
$$ g(x) = \frac{1+x}{(1-2x)(1-x)}.$$
Does this look like a correct start?
|
Once you have:
$$ g(x)=\sum_{n\geq 0}h_n x^n = \frac{\color{red}{x}}{(1-2x)(1-x)} = \frac{1}{1-2x}-\frac{1}{1-x}\tag{1}$$
it is straightfoward to check that $\color{red}{h_n=2^n-1}.$
Just check your computations since the value in zero of $\frac{1+x}{(1-2x)(1-x)}$ is one, while you have $h_0=0$.
|
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|
Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& = \frac{1}{2} \frac{ \log(x)}{x^2+1} - \frac{1}{2}\int \left[(x^2+1)\frac{\frac{(x^2+1)^2}{x} - 4x(x^2+1) \log(x)}{(x^2+1)^4} \right ]dx\\
& = \frac{ \log(x)}{2(x^2+1)}-\frac{1}{2}\int \left [ \frac{x^2+1-4x^2 \log(x)}{x(x^2+1)^2} \right ] dx\\
& = \frac{ \log(x)}{2(x^2+1)} - \frac{1}{2}\int \frac{dx}{x(x^2+1)} + 2\int\frac{x \log(x)}{(x^2+1)^2}dx
\end{align}
Or this method doesn't work here, or I have done a mistake somewhere. However, I have also tried doing $u = x^2+1$ substitution, but this also didnt gave me any good results.Thank you.
|
You can evaluate using the residue theorem. In this case, by considering the contour integral
$$\oint_C dz \frac{z \log^2{z}}{(1+z^2)^2} $$
where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$, and letting $R \to \infty$ and $\epsilon \to 0$, we have
$$-i 4 \pi I_1 + 4 \pi^2 I_0 = i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log^2{z}}{(z+i)^2} \right ) \right ]_{z=e^{i \pi/2}} + i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log^2{z}}{(z-i)^2} \right ) \right ]_{z=e^{i 3 \pi/2}} = 2 \pi^2$$
where
$$I_j = \int_0^{\infty} dx \frac{x \log^j{x}}{(1+x^2)^2} $$
Now, for similar reasons,
$$-i 2 \pi I_0 = i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log{z}}{(z+i)^2} \right ) \right ]_{z=e^{i \pi/2}} + i 2 \pi \left [\frac{d}{dz} \left (\frac{z \log{z}}{(z-i)^2} \right ) \right ]_{z=e^{i 3 \pi/2}} = -i \pi$$
or $4 \pi^2 I_0 = 2 \pi^2$.
Therefore
$$I_1 = \int_0^{\infty} dx \frac{x \log{x}}{(1+x^2)^2} = 0$$
|
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|
Help in finding the sum of the series $$\sum_{n=1}^\infty \frac{1}{n^4+n^2+1}$$
I tried breaking into factors but it is not telescoping.
$$\frac {1}{(n^2+n+1)(n^2-n+1)} = \frac {1}{2n} \left(\frac {1}{n^2-n+1} - \frac {1}{n^2+n+1}\right)$$
|
Let $\omega=\exp\frac{2\pi i}{3}$. Then:
$$\sum_{n\geq 1}\frac{1}{n^4+n^2+1}=\frac{1}{i\sqrt{3}}\left(\sum_{n\geq 1}\frac{1}{n^2-\omega}-\sum_{n\geq 1}\frac{1}{n^2-\omega^2}\right)=\color{red}{-\frac{1}{2}+\frac{\pi\sqrt{3}}{6}\tanh\frac{\pi\sqrt{3}}{2}}$$
since we can deal with series like the ones appearing in the middle term through the logarithmic derivative of the Weierstrass product for the hyperbolic sine function. In particular, the identity:
$$ \sum_{n\geq 0}\frac{1}{n^2+z^2}=\frac{1+\pi z\coth(\pi z)}{2z^2}$$
is well-known.
|
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|
Every tetrahedron has a vertex whose adjacent edges can form a triangle
Prove that in every tetrahedron there exists a vertex $v$ such that the three edges incident at $v$ have lengths that can form a triangle.
It can be proved using a tedious casework: assume by contradiction that all vertex have edges that do not satisfy the criterion, WLOG $a+b \leq c$ and there are $3$ cases for vertex B... It works out in the end, but is quite lengthy and boring.
Is there a nicer way? A friend suggested using geometric vectors and thus proving equalities instead of the triangle inequality, but I can't wrap it up.
|
Lemma: Let $x,y,z>0$. Then there exists a triangle with sides $x,y,z$ if and only if $2x^2y^2+2y^2z^2+2z^2x^2>x^4+y^4+z^4$.
Proof of lemma: Observe that $2x^2y^2+2y^2z^2+2z^2x^2-x^4+y^4+z^4=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. If there exists a triangle with sides $x,y,z$, then each factor is positive, so their product is positive. Conversely, if there is no triangle with sides $x,y,z$ then exactly one factor is nonpositive, so whole product is nonpositive.
Let's get back to the problem. Using notations from your drawing we have:
$$2a^2b^2+2b^2e^2+2e^2a^2>a^4+b^4+e^4, \\
2a^2c^2+2c^2d^2+2d^2a^2>a^4+c^4+d^4, \\
2b^2c^2+2c^2f^2+2f^2b^2>b^4+c^4+f^4, \\
2d^2e^2+2e^2f^2+2f^2d^2>d^4+e^4+f^4. $$
Summing up and dividing by 2 gives $$a^2b^2+b^2e^2+e^2a^2+a^2c^2+c^2d^2+d^2a^2+b^2c^2+c^2f^2+f^2b^2+d^2e^2+e^2f^2+f^2d^2 > a^4+b^4+c^4+d^4+e^4+f^4. \qquad (\spadesuit)$$
Assuming the contrary of thesis, we get
$$2a^2b^2+2b^2c^2+2c^2a^2\le a^4+b^4+c^4, \\
2a^2d^2+2d^2e^2+2e^2a^2\le a^4+d^4+e^4, \\
2f^2c^2+2c^2d^2+2d^2f^2\le f^4+c^4+d^4, \\
2b^2e^2+2e^2f^2+2f^2b^2\le b^4+e^4+f^4. $$
Adding up and dividing by 2 yields
$$a^2b^2+b^2c^2+c^2a^2+a^2d^2+d^2e^2+e^2a^2+f^2c^2+c^2d^2+d^2f^2+b^2e^2+e^2f^2+f^2b^2 \le a^4+b^4+c^4+d^4+e^4+f^4,$$
which contradicts $(\spadesuit)$!
|
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|
Simplifying$\left|\frac{z-3}{z+3} \right|=2$ I want to graph the following, but simplifying is the question here:
$$\left|\frac{z-3}{z+3} \right|=2$$
Now I can do this : $$\frac{|z-3|}{|z+3|}=2 $$
$$|z-3|=2|z+3|$$
$$|x+iy-3|=2|x+iy+3|$$
What manipulation do I use here? I have the answer is a circle of radius $4$ centered at $(-5,0)$
|
$$|x+iy-3|^2=4|x+iy+3|^2$$
$$|x+iy-3|^2=(x-3)^2+y^2$$
$$4|x+iy+3|^2=4(x+3)^2+4y^2$$
So:
$$(x-3)^2+y^2=4(x+3)^2+4y^2$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2$$
$$0=3x^2+30x+27+3y^2$$
$$0=x^2+10x+9+y^2$$
$$0=(x+5)^2-16+y^2$$
$$16=(x+5)^2+y^2$$
It's an equation of circle of radius $4$ centered at $(-5,0)$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Non-standard question about random variables I am not sure which subbranch of mathematics this is, so I cannot give a precise tag. I am doing research, and this suddenly popped out of no where. So, please hear me out.
$x$ is a variable that takes on random values. $x$ is said to 'stable' if $|x| \leq 1$.
Here is an example: $x = \dfrac{cY+1}{2}$, where $Y \sim U(0,2)$. In this case, $c \in \mathbb{R}$ and $Y$ is a random variable. For what value of $c$ will $x$ become 'stable'?
Since we know that $\dfrac{1}{2}cY \sim U(0, c)$, then $\dfrac{1}{2}cY + \dfrac{1}{2} \sim U(\frac{1}{2}, c+\frac{1}{2})$. We want $x$ to be stable so $|c+\frac{1}{2}| \leq 1$, therefore, $c \in [-\frac{3}{2}, \frac{1}{2}]$.
Here is my question:
Let $Y \sim U(0,1)$. For what value of $c$ will make $x$ 'stable' if
$$x = \dfrac{\sqrt{(Y-\frac{2}{c})^2-4}}{2}?$$
|
You want to find $c$ such that $$-1\le \dfrac{\sqrt{(Y-2/c)^2-4}}{2} \le 1$$ (and in particular is real) for $0 \le Y \le 1$. Equivalently,
$$ 4 \le (Y - 2/c)^2 \le 8$$
for $0 \le Y \le 1$. So either $2 \le Y - 2/c \le 2 \sqrt{2}$ in that interval, or $-2\sqrt{2} \le Y - 2/c \le -2$. Now unfortunately,
$2\sqrt{2} - 2 < 1$, so this can't ever be true for all $Y \in [0,1]$.
EDIT: If complex numbers are acceptable, then you just need $0 \le (Y - 2/c)^2 \le 8$, so $-2\sqrt{2} \le Y - 2/c \le 2 \sqrt{2}$ for $0 \le Y \le 1$, and thus either $c \ge \sqrt{2}/2$ or $c \le -(2 + 4 \sqrt{2})/7$.
|
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|
What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
|
In terms of polar coordinate $(x,y) = (r\cos\theta,r\sin\theta)$, we have
$$\begin{align}
xdx + ydy
&=
r\cos\theta(\cos\theta dr - r\sin\theta d\theta) +
r\sin\theta(\sin\theta dr + r\cos\theta d\theta) = r dr\\
\frac{xdy - ydx}{x^2+y^2}
&=
\frac{r\cos\theta(\sin\theta dr + r\cos\theta d\theta) - r\sin\theta(\cos\theta dr - r\sin\theta d\theta)}{r^2} = d\theta
\end{align}$$
This leads to
$$dr^2 = 2rdr = 2d\theta
\;\;\implies\;\;
r^2 = 2\theta + C
\;\;\iff\;\;
x^2 + y^2 = 2\tan^{-1}\left(\frac{y}{x}\right) + C
$$
for suitably chosen constant $C$.
|
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|
How to solve $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$? I need to compute $$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}\ dx.$$ I tried it on wolfram but it timed out, maybe because I am on a mobile device. Any hint is appreciated.
|
Hint:
$$\begin{align}
I
&=\int\frac{x-1}{x^2\left(x+1\right)}\sqrt{\left(x^2+3x+1\right)\left(x^2-x+1\right)}\,\mathrm{d}x\\
&=\int\frac{y}{\left(\frac{1+y}{1-y}\right)^2}\sqrt{\left(\frac{5-y^2}{\left(1-y\right)^2}\right)\left(\frac{1+3y^2}{\left(1-y\right)^2}\right)}\,\frac{2\,\mathrm{d}y}{\left(1-y\right)^2};~~~\small{\left[\frac{x-1}{x+1}=y\right]}\\
&=\int\frac{2y}{\left(1+y\right)^2\left(1-y\right)^2}\sqrt{\left(5-y^2\right)\left(1+3y^2\right)}\,\mathrm{d}y\\
&=\int\frac{2y}{\left(1-y^2\right)^2}\sqrt{\left(5-y^2\right)\left(1+3y^2\right)}\,\mathrm{d}y\\
&=\int\frac{\sqrt{\left(5-z\right)\left(1+3z\right)}}{\left(1-z\right)^2}\,\mathrm{d}z;~~~\small{\left[y^2=z\right]}.\\
\end{align}$$
After that, you can complete the square under the radical and apply trigonometric substitutions to finish the integral.
|
{
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|
How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
|
It is true in general that $x^6-1=(x-1)(1+x+x^2+\cdots+x^5)$. Just substitute $x=10^4$.
|
{
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|
Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}Using the axioms, theorem, definitions of high school algebra concerning the real numbers, then prove the following:
Given $r>0$, find a $k>0$ such that:
$$\text{for all }x, y: \sqrt{(x-2)^2+(y-1)^2}<k\implies|xy-2|<r $$
I tried with several values given to $k$ and $r$ to find the relation between them.
Suppose then $r=1$ and we choose $k$ to be $\frac{1}{2}$
Now let us check if $\forall x,y$ such that $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|xy-2|<1$
However we know that $|x-2|\leq\sqrt{(x-2)^2+(y-1)^2}$
Therefore if $\sqrt{(x-2)^2+(y-1)^2}<\frac{1}{2}$ then $|x-2|<\frac{1}{2}$..........(1)
For the same reason $|y-1|<\frac{1}{2}$...................(2)
But, (1) implies $\frac{3}{2}<x<\frac{5}{2}$.......................(3)
Also (2) implies $\frac{1}{2}<y<\frac{3}{2}$....................(4)
And the question now is, do ALL the values of x and y satisfying (3) and (4), satisfy $|xy-2|<1$
No because for $x=\frac{16}{10}$ and $y=\frac{6}{10}$ the relation $|xy-2|<1$ is not satisfied.
And the question is, which is the proper relation between $r$ and $k$ so that the above inequality is satisfied for all the values of $x$ and $y$
|
The center of the circle is at $(2,1)$, which is also on the hyperbola $xy = 2$. So it makes sense to write $x = 2 + a$ and $y = 1 + b$. Then your problem can be restated as: Given $r > 0$, find a $k > 0$ such that if $\sqrt{a^2 + b^2} < k$, then $|a + 2b + ab| < r$.
If $\sqrt{a^2 + b^2} < k$, then both $|a|$ and $|b|$ are less than $k$, which implies $|a + 2b + ab| < 3k + k^2$. So your task is to find a $k$ such that $3k + k^2 < r$.
I think you can take it from here...
|
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|
Upper and Lower Darboux integral of a piecewise function $f(x)=x$ and $f(x)=0$. Let $0<a<b$. Find the upper and lower Darboux integrals for the function $$f(x)=x$$ if $x\in[a,b]\cap\mathbb{Q}$ and $$f(x)=0$$ if $x\in[a,b]-\mathbb{Q}$.
I am so lost on this problem. Any hints or solutions are greatly appreciated.
|
Any interval in any subdivision of $[a,b]$ must contain a rational and irrational point, since the intervals are not singletons nor empty. This is because the irrationals and rationals are both dense in $\mathbb{R}$. We then have that the supremum of $f$ on any non-empty, non-singleton closed interval $[x,y]$ is $\sup[x,y] = y$ and the infimum of $f$ on any interval is $0$.
So the upper sum of any uniform partition $P_n:a=t_0< t_1< ...<t_n=b$, with $t_k = a+\frac{k(b-a)}{n}$ must be $$U(f,P_n) =\sum_{k=1}^{n} (t_k-t_{k-1}) t_k=\sum^{n}_{k=1} (\frac{b-a}{n}) (t_k)=\sum^{n}_{k=1} (\frac{b-a}{n})(a+\frac{k(b-a)}{n}) = \sum^{n}_{k=1} (a(\frac{b-a}{n})+(\frac{b-a}{n})\frac{k(b-a)}{n}) = a(b-a) + \frac{(b-a)^2}{n^2}\sum^{n}_{k=1} k $$
$$= a(b-a) + \frac{(b-a)^2}{n^2}\frac{1}{2} (n)(n+1)= a(b-a) + \frac{1}{2} (b-a)^2(1+\frac{1}{n}) $$
since the supremum of any interval must be on the right hand side.
Now take the infimum of all the upper sums of the uniform partitions, and you'll note that it is, $$a(b-a) + \frac{1}{2} (b-a)^2(1) = ab-a^2 + \frac{1}{2}b^2-ab+\frac{1}{2}a^2 = \frac{1}{2}b^2-\frac{1}{2}a^2$$
So the upper integral is at most $$\frac{1}{2}b^2-\frac{1}{2}a^2$$.
If $Q = \{q_0,...,q_N\}$ is any partition of $[a,b]$, then
$$U(f, Q) = \sum_{i=1}^N q_i (q_i-q_{i-1}) = \sum_{i=1}^N q_i^2 -\sum_{i=1}^Nq_i q_{i-1} \geq \sum_{i=1}^N q_i^2 - \frac{1}{2} \sum_{i=1}^N (q_i^2 + q_{i-1}^2) $$$$= \frac{1}{2} \sum_{i=1}^N (q_i^2 - q_{i-1}^2) =\frac{1}{2}(\sum_{i=1}^N q_i^2 - \sum_{i=1}^N q_{i-1}^2) = \frac{1}{2}(b^2-a^2)$$
The inequality is because:
$$(x-y)^2 \geq 0 \implies x^2-2xy+y^2 \geq 0 \implies -2xy \geq -x^2-y^2 $$$$\implies 2xy \leq x^2 +y^2 \implies xy \leq \frac{1}{2}(x^2+y^2)$$
The last equality is because the squares cancel (telescoping sum) for all $q_i \neq a,b$.
So the upper integral is at least $\frac{1}{2}(b^2-a^2)$ since it is the greatest lower bound of the upper sums of all the partitions.
Hence the upper integral is $\frac{1}{2}(b^2-a^2)$.
The lower sum of any partition of $[a,b]$ must be $0$ as the infimum on any of the subintervals is $0$.
So the supremum of the lower sums over any partition must be $0$. Hence the lower integral is $0$.
Does that get you started?
|
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|
Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
|
By induction assume it is true for $n$.
Then, by induction,
$\sum_{k = 1}^{n+1} \frac{1}{k^{2}} =\sum_{k = 1}^{n} \frac{1}{k^{2}} +\frac{1}{(n+1)^2} < 2-\frac{1}{n} +\frac{1}{(n+1)^2}$.
So to finish we need $ 2-\frac{1}{n} +\frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$,
or $\frac{1}{n+1} \leq \frac{1}{n}-\frac{1}{(n+1)^2}$, which is true for $n>1$.
|
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|
Integral with residues $\int_0^\infty \tfrac{\sin^2(x)}{1+x^4}dx$ I am trying to calculate
$\displaystyle\int_0^\infty \dfrac{\sin^2(x)}{1+x^4}dx$
using method of residues. I have already seen this post, "Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem" And am integrating a quarter circle in the complex plane with a simple pole at $\exp(\frac{i\pi}{4})$. So if I put $z=e^{i\theta}, \sin^2(x)=-\frac{1}{4}(z-\frac{1}{z})$. Even so the residues are giving me difficulty. Any help would be great.
|
First note that the integrand is even, so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx.
$$
Furthermore $\sin^2 x = \dfrac{1-\cos 2x}{2}$ so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac14 \int_{-\infty}^\infty \frac{1-\cos 2x}{1+x^4}\,dx.
$$
Define a function
$$
f(z) = \frac{1-e^{2iz}}{1+z^4}.
$$
Then $\operatorname{Re}(f(x)) = \dfrac{1-\cos 2x}{1+x^4}$ for $x$ real. Let's compute $\int_C f(z)\,dz$ over the boundary of a large semi-disc: $C = \partial\{ z = x+iy : |z| < R, y = 0 \}$.
On the semi-circle (by the triangle inequality), we get
$$
|f(z)| \le \frac{1 + |e^{2iz}|}{|z|^4-1} \le \frac{2}{R^4-1}
$$
since $|e^{2i(x+iy)}| = |e^{-2y}| \le 1$ for $y \ge 0$.
The standard estimation lemma (ML-inequality) shows that
$$
\left| \int_{C_R^+} f(z)\,dz \right| \le \pi R \cdot \frac{2}{R^4-1} \to 0
$$
as $R \to \infty$. (Here $C_R^+$ is the semi-circle.)
Finally, the residue theorem shows that
\begin{align}
\int_C f(z)\,dz &= 2\pi i \big( \operatorname{Res}\limits_{\exp(i\pi/4)}(f(z)) +
\operatorname{Res}\limits_{\exp(3i\pi/4)}(f(z)) \big) \\
&= \frac{\pi\sqrt 2}{2}\,\big( 1- \exp(-\sqrt 2)(\sin \sqrt 2 + \cos \sqrt 2) \big).
\end{align}
(Tedious algebra omitted.)
Putting everything together, and taking the real part we get $1/4$ of the above, i.e.
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx = \frac{\pi\sqrt 2}{8}\,\big( 1- \exp(-\sqrt 2)(\sin \sqrt 2 + \cos \sqrt 2) \big).
$$
|
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|
Find the integral: $\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$ I have to find the following integral:
$$\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$$
My answer is $2x^2 -\ 1x + \frac{2\sqrt{27}}{3}$. Am I right?
|
$\bf{My \; Solution::}$ $\displaystyle \int (4x-1+3\sqrt{x})dx = 4\int xdx - \int 1 dx+3\int x^{\frac{1}{2}}dx = 4\cdot \frac{x^2}{2}-x+3\cdot \frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$
Above we have used the formula $\displaystyle \int x^ndx = \frac{x^{n+1}}{n+1}+\mathcal{C}\;,$ Where $n\neq -1$
|
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|
Evaluate $\lim _{n\to \infty }\int_1^2\:\frac{x^n}{x^n+1}dx$ We have $$I_n=\int _1^2\:\frac{x^n}{x^n+1}dx$$ and we need to find $\lim _{n\to \infty }I_n$. Have any ideea how we can evaluate this limit?
|
Note that $\frac{x^n}{x^n+1}=1-\frac{1}{x^n+1}$. Thus,
$$\int_1^2 \frac{x^n}{x^n+1}dx=1-\int_1^2 \frac{1}{1+x^n}dx$$
For $n>1$, the integral on the right-hand side satisfies the inequality
$$\left|\int_1^2 \frac{1}{1+x^n}dx\right|\le \int_1^2 x^{-n}dx=\frac{1-2^{1-n}}{n-1}$$
which clearly goes to zero as $n \to \infty$.
Putting all of this together we see that
$$1\ge \int_1^2 \frac{x^n}{1+x^n}dx=1-\int_1^2 \frac{1}{1+x^n}dx \ge 1-\int_1^2 x^{-n}dx=1-\frac{1-2^{1-n}}{n-1}$$
Therefore, by the "squeeze" theorem
$$\lim_{n \to \infty} \left(\int_1^2 \frac{x^n}{1+x^n}dx\right) =1$$
|
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|
Find with proof all solutions to $2^n = a! + b!$, where $a$, $b$, $n$ are positive integers and $a \leq b$. So far I have looked at $n=1$ with $a=1$ and $b=1$ which is $$2^1 = 1! + 1! = 2 $$
and $n=2$ with $a=2$ and $b=2$, $$2^2 = 2!+ 2! = 4$$
and finally $n=3$ with $a=3$ and $b=2$, $$2^3 = 3! + 2! = 8$$
I have also tried the positive integers up to $n=14$ and found that none of them work. I also tried random values such as $n=30$ and it doesn't seem to work either.
For example, $n=11$, $2^{11} = 2048$. $7! = 5040$ so this cannot be used. However, since $6! = 720$, the highest number we can make is $6! + 6! = 1440$.
However, exhaustively listing out the answers doesn't seem like a concrete enough proof to say that it cannot work for any other numbers.
Can anybody give a more concrete reason why this wouldn't work for positive integers above $n=3$?
|
*
*If $a,b \geq 3$, we have $3$ divides $a!$ and $3$ divides $b!$. Hence, $3$ divides $a!+b!$. However, $3$ doesn't divide $2^n$.
*$a \geq 3$, $b=1 \text{ or }2$. If $b=1$, then $a!+1$ is odd and cannot be $2^n$. If $b=2$, then $a!+2 = 2(a!/2+1)$, where $a!/2+1$ is odd for $a\geq 4$, and hence cannot equal $2^n$. For $a=3$, we see that $3!+2 = 8 = 2^3$.
*$a = 1 \text{ or }2$, $a = 1 \text{ or }2$. This we can check by brute-force and we see that $1!+1! = 2$ and $2! + 2! = 2^2$.
Hence, the only solutions are $(a,b,n) = (1,1,1)$, $(a,b,n) = (2,2,2)$, $(a,b,n) = (3,2,3)$ and $(a,b,n) = (2,3,3)$.
|
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Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\frac{x+y}{1-xy}=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan(a+b)$
$\implies a+b=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b} $
$\implies \arctan x+\arctan y = \arctan\frac{x+y}{1-xy}$
I know $\pi$ is the period of $\tan x$ so $xy\gt 1$ constraint must have something to do with this. But I am not able to figure out how exactly these period and $xy\gt 1$ are related. Any help is appreciated. Thanks!
|
Here is an insight given by a purely geometrical proof. Consider the Figure below.
Supposing $x>0$, the right-angled triangle $\triangle ABC$ has sides $\overline{AB} = 1$ and $\overline{BC} = x$, so that
$$\alpha = \arctan x.$$
Extend side $AB$ with a segment $\overline{BD}=\frac{x}{y}$, giving
$$\gamma = \arctan y.$$
Now draw from $D$ the line segment $DH$ perpendicular to $AC$ and note that $\triangle ADH \sim \triangle ABC$, with a scaling factor given by
$$\frac{\overline{AD}}{\overline{AC}} = \frac{x+y}{y\sqrt{1+x^2}}$$
yielding
$$\overline{DH} =\frac{x(x+y)}{y\sqrt{1+x^2}},$$
$$\overline{AH} = \frac{x+y}{y\sqrt{1+x^2}},$$
and
\begin{eqnarray}
\overline{CH} &=&\overline{AC}-\overline{AH}=\\
&=&\sqrt{1+x^2}- \frac{x+y}{y\sqrt{1+x^2}}=\\
&=& \boxed{\frac{x(xy-1)}{y\sqrt{1+x^2}}}.\tag{1}\label{boxed}
\end{eqnarray}
By definition finally we have
\begin{eqnarray}
\beta &=& \arctan\left( \frac{\overline{DH}}{\overline{CH}}\right)=\\
&=& \arctan\left(\frac{x+y}{xy-1}\right).
\end{eqnarray}
Which, once you plug in definitions of $\alpha$, $\beta$, and $\gamma$ in
\begin{equation}\alpha+\beta+\gamma=\pi,\end{equation}
leads to your identity.
When $x<0$ it is sufficient to to define $\overline{BC} = -x$ and then use symmetries of the $\arctan(\cdot)$ function.
The constraint, from this perspective, comes from the boxed equation \eqref{boxed}. If $xy<1$, that would lead to a negative number. As a matter of fact, this is because in this case $\beta$ would be obtuse and the altitude $DH$ would be external, yielding a slightly different formula for $\overline{DH}$ and thus $\beta$. You can figure out which one as an excersise.
|
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How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$
$$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$
$$\ln|\sec\theta + \tan\theta| + C $$
$$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$
My book says the answer should just be:
$$\ln\left({\sqrt{9+x^2}} + {x}\right) $$
I'm wondering where I went wrong with this?
|
The two results are equivalent, up to choice of constant.
\begin{align}
\ln\left({\sqrt{9+x^2}+x \over 3}\right)+C &= \ln\left(\sqrt{9+x^2} + x\right) - \ln(3) + C \\
&= \ln\left(\sqrt{9+x^2} + x\right) + C'
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Power series properties. The sum of two series. Let $\sum a_nz^n$ and $\sum b_nz^n$ power series with radiuses of convergences $R_1,R_2$ respectively. Suppose the radius of convergence of $(\sum a_n+b_n)z^n$ is $R$. Find an example in which $\infty>R>\min \{R_1,R_2\}$, given $R_1=R_2$.
What am I to do here? Previously I was required to show that $R=\{R_1,R_2\}$ (Given $R_1\ne R_2$, how does it matter?) and now that? Besides, $(\sum a_n+b_n)z^n=\sum a_nz^n+\sum b_nz^n$ and if $R$ is bigger the the minimum then one of the series must not converge, isn't that so? I would appreciate your help.
|
$$1+x+\frac{x^2}{2^2}+x^3 + \frac{x^4}{2^4} + x^5 + \frac{x^6}{2^6}+\cdots $$
and
$$1-x+\frac{x^2}{2^2}-x^3 + \frac{x^4}{2^4} - x^5 + \frac{x^6}{2^6}- \cdots$$
both have radius of convergence $1.$ But the sum of these series is
$$2+ \frac{x^2}{2} + \frac{x^4}{2^3}+ \frac{x^6}{2^5}+\cdots ,$$
which has radius of convergence $2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1262267",
"timestamp": "2023-03-29T00:00:00",
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|
How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$ How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$?
I try
$x=2r \cos \phi$,
$y=2r \sin \phi$,
$z=z$, but but probably not the way to go
|
Substitute $r^2 = x^2 + y^2$ and make $r^2$ the subject.
The inequalities are then
$4(1 - z^2) \geq r^2$ and
$(z+2)^2 \geq r^2$.
Now we compare the LHS of the two inequalities, to determine which gives a stricter upper bound on $r^2$.
$$(z+2)^2 - 4(1-z^2) = 5z^2 +4z$$
which has roots $z=-\frac45$ and $z=0$.
Thus the second inequality's LHS is smaller than the first only between these two values; therefore the second is stricter between these values and the first is stricter everywhere else.
You also want to ensure that your integral includes only values of $z$ for which the first inequality is possible; that is, for which its LHS is nonnegative. So $|z| \leq 1$.
Now the appropriate integral is $\pi(\int_{-1}^{-\frac45}4(1-z^2)dz + \int_{-\frac45}^0(z+2)^2dz + \int_0^14(1-z^2)dz)$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Computing $\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx$ I wish to compute
$$\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx, \quad a>0$$
but have no contour to work with. Does anyone have ideas on how to compute this integral?
|
\begin{align*}
\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \\
&= \frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^2 + 1}{x^6 + 1} \, dx+\frac{1}{a^{3/2}} \int_{1}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx\\
&= \frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^2 + 1}{x^6 + 1} \, dx+\frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^{-2} + 1}{x^{-6} + 1}x^{-2} \, dx \mbox{ (for the second:}x \rightarrow 1/x )\\
&= \frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^2 + 1}{x^4-x^2 +1} \, dx\\
&= \frac{1}{a^{3/2}} \int_{0}^{\frac{\pi}{2}} \frac{\sin^2t + 1}{\sin^4t-\sin ^2t +1} \, d\sin t =I\\
\end{align*}
Now by symmetry and $t \rightarrow \pi/2-t$ we obtain
\begin{align}
2I&=\frac{1}{a^{3/2}}\int_{0}^{\frac{\pi}{2}} \frac{\sin^2t + 1}{\sin^4t-\sin ^2t +1} \, \cos tdt+\frac{1}{a^{3/2}}\int_{0}^{\frac{\pi}{2}} \frac{\cos^2t + 1}{\cos^4t-\cos ^2t +1} \, \sin tdt\\
&=\frac{1}{a^{3/2}}\int_{0}^{\frac{\pi}{2}}\frac{\cos t +\sin t}{1-\cos t \sin t} \, dt\\
&=\frac{1}{a^{3/2}}\int_{0}^{\frac{\pi}{2}}\frac{(-2)(-\cos t -\sin t)}{1+(\cos t- \sin t)^2} \, dt\\
&=\frac{-2}{a^{3/2}}\arctan(\cos t- \sin t)\Big|_0^{\pi/2}\\
&=\frac{\pi}{a^{3/2}}
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1265905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
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|
Finding The Number Of Inversions In A Permutation
Let the be the following permutation:
$(1 5 4)(3 6)\in S_6$
How do I count the number of inversions to calculate the sign of the permutation?
$(1 5 4)(3 6)=(1 5)(1 4),(3 6)=3$ so it has an odd sign
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 2 & 6 & 1 & 4 & 3
\end{pmatrix}
before $5$ there are $0$ elements.
before $2$ there are $1$ elements (5).
before $6$ there are $0$ elements.
before $1$ there are $3$ elements (5,2,6).
before $4$ there are $2$ elements (5,6).
before $3$ there are $3$ elements (5,6,4).
Therefore there are $9$ inversions so it is $(-1)^9=-1$
Are both of the ways are ok?
|
You have not done anything wrong except in thinking the permutation is even. There are indeed 9 inversions as you have calculated.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1266702",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Power Series of $\frac{3}{(1-3x)^2}$ The problem is to find the power series of this function $$\frac{3}{(1-3x)^2}$$ centered at $x = 0$.
Normally you convert it into $\frac{1}{1-x}$ form. Since the denominator is squared do you multiply that out first then convert it into standard form?
|
Using the standard Taylor approach, compute the successive derivatives:
$$f(x)=\frac3{(1-3x)^2},f(0)=3$$
$$f'(x)=\frac{3\cdot3\cdot2(1-3x)}{(1-3x)^4}=\frac{3\cdot3\cdot2}{(1-3x)^3},f'(0)=3\cdot3\cdot2$$
$$f''(x)=\frac{3\cdot3^2\cdot3\cdot2(1-3x)^2}{(1-3x)^6}=\frac{3\cdot3^2\cdot3\cdot2}{(1-3x)^4},f''(0)=3\cdot3^2\cdot3\cdot2$$
$$f'''(x)=\frac{3\cdot3^3\cdot4\cdot3\cdot2(1-3x)^3}{(1-3x)^8}=\frac{3\cdot3^3\cdot4\cdot3\cdot2(1-3x)^3}{(1-3x)^5},f'''(0)=3\cdot3^3\cdot4\cdot3\cdot2$$
$$\cdots$$
$$f^{(n)}(0)=3\cdot3^n\cdot(n+1)!$$
So the series is$$\sum_{n=0}^\infty3\cdot3^n\cdot(n+1)\cdot x^n.$$
There is no essential difference with the form $1/(1-x)$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1266789",
"timestamp": "2023-03-29T00:00:00",
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|
Represent the transformation with respect to the standard basis
Consider a linear transformation from $R^2$ to $R^2$ defined via:
$$\left(\begin{matrix} 1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 1\end{matrix} \right)$$ and $$\left(\begin{matrix} -1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 2\end{matrix} \right)$$
Represent the transformation with respect to the standard basis
It has been some time I have tried to understand this problem, and I have seen solutions to problems similar to this and I still do not understand it. Those solutions talk about images of $e_1$ and $e_2$ (which are the standard vectors of the standard basis), but I don't know what the heck they are talking about.
|
A vector can be decomposed regarding a basis, esp. the standard basis:
$$
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
=
x
\left(
\begin{matrix}
1 \\
0
\end{matrix}
\right)
+
y
\left(
\begin{matrix}
0 \\
1
\end{matrix}
\right)
= x e_1 + y e_2
$$
so a linear transformation $A$ acting on a vector can be calculated by $A$ acting on the base vectors.
$$
A
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
=
x\,A
\left(
\begin{matrix}
1 \\
0
\end{matrix}
\right)
+
y\,A
\left(
\begin{matrix}
0 \\
1
\end{matrix}
\right)
=
x \, A e_1 + y \, A e_2
$$
The matrix of $A$ can be seen as
$$
A = \left((Ae_1) (Ae_2)\right)
$$
So from
$$
A
\left(
\begin{matrix}
1 \\
3
\end{matrix}
\right)
= 1\,A e_1 + 3\, A e_2 =
\left(
\begin{matrix}
3 \\
1
\end{matrix}
\right)
\quad
A
\left(
\begin{matrix}
-1 \\
3
\end{matrix}
\right)
= -1\,A e_1 + 3\, A e_2 =
\left(
\begin{matrix}
3 \\
2
\end{matrix}
\right)
$$
one can take the sum and difference:
$$
6\, A e_2 =
\left(
\begin{matrix}
6 \\
3
\end{matrix}
\right)
\quad
2\,A e_1 =
\left(
\begin{matrix}
0 \\
-1
\end{matrix}
\right)
$$
This gives
$$
A = \left((Ae_1) (Ae_2)\right) =
\left(
\begin{matrix}
0 & 1 \\
-1/2 & 1/2
\end{matrix}
\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Use Lagrange Interpolation polynomial to find this $\sum_{cyc}\frac{x^3}{(x^2-y^2)(x^2-z^2)}$
Let $x,y,z$ be the solutions of the equation $t^3-t^2+2t-3=0$. Find the sum
$$\dfrac{x^3}{(x^2-y^2)(x^2-z^2)}+\dfrac{y^3}{(y^2-x^2)(y^2-z^2)}+\dfrac{z^3}{(z^2-x^2)(z^2-y^2)}$$
How can I use the Lagrange
interpolation polynomial to solve this question?
I did a lot of calculations to find this answer
$$\dfrac{x^3}{(x^2-y^2)(x^2-z^2)}+\dfrac{y^3}{(y^2-x^2)(y^2-z^2)}+\dfrac{z^3}{(z^2-x^2)(z^2-y^2)}=\dfrac{xy+yz+xz}{(x+y+z)(xy+yz+xz)-xyz}=\dfrac{2}{1\cdot 2+3}=\dfrac{2}{5}$$
I think we can use the Lagrange polynomial method to solve this problem?
|
Yes, with some modification, we can use the Lagrange interpolation method. First, note that $(y+z)(z+x)(x+y)=(x+y+z)(yz+zx+xy)-xyz$. As $x+y+z=1$, $yz+zx+xy=2$, and $xyz=3$, $(y+z)(z+x)(x+y)=1\cdot 2-3=-1$. The required expression is then
$$\begin{align}
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}
=\left(\frac{1}{(y+z)(z+x)(x+y)}\right)^2\,\sum_\text{cyc}x^3(y+z)^2\left(\frac{(x+z)(x+y)}{\left(x-y\right)\left(x-z\right)}\right)\,.
\end{align}$$
As $y+z=1-x$, we get $x^3(y+z)^2=x^3(1-x)^2=x^3-2x^4+x^5$. Since $x^3-x^2+2x-3=0$, we obtain $x^3=x^2-2x+3$, $x^4=x^3-2x^2+3x=\left(x^2-2x+3\right)-2x^2+3x=-x^2+x+3$, and $x^5=x^4-2x^3+3x^2=\left(-x^2+x+3\right)-2\left(x^2-2x+3\right)+3x^2=5x-3$. Hence, $$x^3(y+z)^2=
\left(x^2-2x+3\right)-2\left(-x^2+x+3\right)+(5x-3)=3x^2+x-6\,.$$
Similarly, $y^3(z+x)^2=3y^2+y-6$ and $z^3(x+y)^2=3z^2+z-6$. That is, the quadratic polynomial $P(t):=3t^2+t-6$ interpolates the points $\left(x,x^3(y+z)^2\right)$, $\left(y,y^3(z+x)^2\right)$, and $\left(z,z^3(x+y)^2\right)$. Ergo,
$$P(t)=\sum_\text{cyc}\,P(x)\left(\frac{(t-y)(t-z)}{(x-y)(x-z)}\right)\,.$$ That is,
$$\begin{align}
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}
&=\left(-1\right)^2\,\sum_\text{cyc}P(x)\left(\frac{(x+z)(x+y)}{\left(x-y\right)\left(x-z\right)}\right)
\\
&=\sum_\text{cyc}\,P(x)\left(\left.\frac{(t-y)(t-z)}{(x-y)(x-z)}\right|_{t=x+y+z}\right)
\\
&=\big.P(t)\big|_{t=x+y+z}=P(x+y+z)=P(1)=3\cdot 1^2+1-6=-2\,.
\end{align}$$
Your result is wrong, by the way. (I have also verified my answer with Mathematica.)
In general, let $K$ be a field. The cubic $t^3+at^2+bt+c$ in $K[t]$, where $a,b,c\in K$, has three pairwise distinct roots $x,y,z$ such that the sum of any two of them is nonzero, then we have $ab- c\neq 0_K$, $a^2b^2-4b^2-4a^3c+18abc-27c^2\neq 0_K$, and
$$
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}=-\frac{b}{ab-c}=\frac{yz+zx+xy}{(x+y+z)(yz+zx+xy)-xyz}\,.$$
In this case, $\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)}=\frac{P(-a)}{(ab-c)^2}$, where $P(t):=-ct^2-(ac-b^2)t+bc$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
How can I find the indefinite integral of $\int\sin^3x \cos^3x dx $? I am looking to evaluate the indefinite integral
\begin{equation*}
\int\sin^3x \cos^3x dx.
\end{equation*}
I'm not sure if I started this right but I broke the terms up like this:
\begin{equation*}
\int\sin^2x \sin x \cos^2x \cos x
\end{equation*}
Edit: I got
\begin{equation*}
{1 \over 4}\cos^4x + {1 \over 6}\cos^6x + C
\end{equation*}
but my book says it should be sine not cosine
|
$$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\sin^3x(1-\sin^2x)\cos x dx \\
& (\sin x = t , \cos x dx = dt)\\
& = \frac{\sin^4x}{4} - \frac{\sin^6x}{6} + C
\end{align}$$
$$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\cos^3x(1-\cos^2x)\sin x dx \\
& (\cos x = t , \sin x dx = -dt)\\
& = -\frac{\cos^4x}{4} + \frac{\cos^6x}{6} + C
\end{align}$$
We can easily check that both the expressions differ by a constant by expanding one of them which can be adjusted with the $C$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.