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Proof polynomial has only one real root. I need to prove that this polynomial equation:
$$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2}).$$
has only one root. That it has one real root is obvious because it is of odd degree. But Descartes rules here fails to bound the number of roots to one.
|
Consider
$$p(x) = x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a$$
First, we note that if $x< 0$, each term is negative, hence there are no negative roots. Also, $p(0) = -a < 0$. Further,
$$p'(x) = 5x^4-4(3-a)x^3+3(3-2a)x^2-2ax+2a$$
So it is sufficient to show that $p'(x) > 0$ for $x > 0, \; a \in (0, \frac12)$.
For this, note that by AM-GM inequality, $\frac12 ax^2+2a \ge 2ax$, so it is sufficient to show that:
$5x^4+\frac12(18-13a)x^2 > 4(3-a)x^3$. By AM-GM we again have:
$$5x^4+\tfrac12(18-13a)x^2 \ge 2\sqrt{\frac{5(18-13a)}2}x^3$$
So it is enough to show $5(18-13a) > 8(3-a)^2 \iff 8a^2+17a < 18$, which is true for $a \in (0, \frac12)$.
|
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|
if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
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Here is an exotic solution based on geometry.
Let $\mathcal{M}$ and $\mathcal{S}$ be surfaces defined by
\begin{align*}
\mathcal{M} : abc = 1
\quad \text{and} \quad
\mathcal{S} : a^{2} + b^{2} + c^{2} = a + b + c.
\end{align*}
Then we have the following observations:
*
*$\mathcal{M}$ lies outside the sphere of radius $\sqrt{3}$. Indeed, if $X = (a, b, c) \in \mathcal{M}$, then the square-distance from the origin $O$ satisfies
$$ \overline{OX}^{2} = a^{2} + b^{2} + c^{2} \geq 3\sqrt[3]{a^{2}b^{2}c^{2}} = 3. $$
*$\mathcal{S}$ is contained in the sphere of radius $\sqrt{3}$. Indeed, $\mathcal{S}$ is the sphere of radius $\frac{\sqrt{3}}{2}$ centered at the point $P = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$, hence if $X = (a, b, c) \in \mathcal{S}$ then by the triangle inequality
$$ \overline{OX} \leq \overline{OP} + \overline{PX} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}.$$
Combining two facts, we find that $\mathcal{M}$ lies outside of $\mathcal{S}$ (possible except for the tangent point). Therefore
\begin{align*}
abc = 1
& \quad \Longrightarrow \quad (a, b, c) \in \mathcal{M} \\
& \quad \Longrightarrow \quad (a, b, c) \text{ lies outside } \mathcal{S} \\
& \quad \Longrightarrow \quad \left(a - \tfrac{1}{2}\right)^{2} + \left(b - \tfrac{1}{2}\right)^{2} + \left(c - \tfrac{1}{2}\right)^{2} \geq \tfrac{3}{4} \\
& \quad \Longrightarrow \quad a^{2} + b^{2} + c^{2} \geq a + b + c.
\end{align*}
|
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|
How was this solution found? Consider an empty spherical bowl of radius $r$. I was trying to find the height to which I would need to fill the bowl with water so that it would be one quarter full (in terms of volume).
The total volume is $\frac{4}{3}\pi r^3$ and the volume filled be water up to a height $h$ is $\pi r h^2 - \frac{1}{3}\pi h^3$. For the bowl to be one quarter full (in terms of volume), $h$ needs to satisfy $h^3 - 3rh^2 + r^3 = 0$.
I couldn't solve this by hand and so used Maple, expecting only a numerical solution. Astonishingly, Maple was able to give two beautiful solutions. The positive one is such that
$$ \color{blue}{\frac{h}{r} = \sqrt{3}\sin\frac{\pi}{9}-\cos\frac{\pi}{9}+1}$$
I've tried making the substitution $h = \rho\mathrm{e}^{\mathrm{i}\theta}$ into $h^3 - 3rh^2 + r^3 = 0$ and then equating real and imaginary parts. Taking the imaginary parts gives $4\rho\cos^2\theta - 6r\cos\theta - \rho = 0$. This seemed promising, but solving doesn't give anything useful.
I can see that the solution could have been found by solving a general cubic and then simplifying the ghastly expressions by applying de Moivre's formula.
*
*Is there an elegant and simple way of arriving at the result by hand?
*Under what conditions are such nice solutions possible?
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After some reading around, I think that I have found the solution.
Starting with $h^3-3rh^2+r^3=0$, we make the substitution $h=x+r$ to get the reduced cubic $x^3 -3r^2x-r^3$, or equivalently $x^3 = 3r^2x+r^3$.
Next, we make the substitution $x=2r\cos\theta$ which yields $8r^3\cos^3\theta = 6r^3\cos\theta + r^3$. Since $r>0$ in the context of the original problem, we have $8\cos^3\theta = 6\cos\theta + 1$, or equivalently
$$4\cos^3\theta - 3\cos\theta = \frac{1}{2}$$
At this point we apply the identity $\cos3\theta \equiv 4\cos^3\theta - 3\cos\theta$ to give $\displaystyle{\cos3\theta = \frac{1}{2}}$. This gives
$$3\theta = \pm\frac{\pi}{3},\pm\frac{5\pi}{3},\pm\frac{7\pi}{3},\pm\frac{11\pi}{3},\cdots$$
$$\theta = \pm\frac{\pi}{9},\pm\frac{5\pi}{9},\pm\frac{7\pi}{9},\pm\frac{11\pi}{9},\cdots$$
Since $x=2r\cos\theta$ and $h=x+r$, the solutions are
$$\color{blue}{h = 2r\cos\frac{\pi}{9}+r, \ 2r\cos\frac{5\pi}{9}+r, \ 2r\cos\frac{7\pi}{9}+r}$$
The negative solutions for $\theta$ are not used because $\cos\theta \equiv \cos(-\theta)$. The later solutions for $\theta$ are not used because, due to cosine's periodicity, they start to repeat the same answers.
Note that these three solutions take the same values as the ones in the OP.
|
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|
Use the integration formula $\frac{1}{a}\arctan\frac{x}{a}$ to solve $\frac{1}{2} \int_{-1}^1 \mathrm{ \frac{dx}{1+\sqrt{2}x+x^2} }\, $ As question states, I am trying to figure out how to use the integration formula to solve the integral. My issue is that the integral isn't of the form $\frac{dx}{a^2+x^2}$
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Hint: $$x^2\pm\sqrt{2}x+1 = (x\pm 1/\sqrt{2})^2+1/2$$
details:
$$\int_{-1}^1 \frac {dx}{x^2+\sqrt{2}x+1}
= \int_{-1}^1 \frac {dx}{(x+ 1/\sqrt{2})^2+1/2}
= \int_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
\frac {du}{u^2+1/2}
\\
= \left[1/\sqrt{2} \arctan \sqrt{2} u
\right]_{-1+1/\sqrt{2}}^{1+1/\sqrt{2}}
$$
|
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|
How to find the indefinite integral $\int \frac{dx}{1+x^{n}}$? How to find the indefinite integral $$\int \frac{dx}{1+x^{n}}$$ where n is a positive integer?
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I just wanted to comment to give the general formula for this explicitly in case you have interest, but I can't comment so I must post an answer. For $n=2q-1,$ with $q\in\mathbb{N},$ and $m<n$ both natural numbers,
$$\int\frac{x^{m-1}}{x^n+1}dx=\frac{\left(-1\right)^{m-1}}{n}\log\left(x+1\right)\\-\frac{1}{n}\sum\limits_{k=1}^{\frac{n-1}{2}}\cos\frac{(2k-1)m\pi}{n}\log\left(x^2-2x\cos\frac{(2k-1)\pi}{n}+1\right)\\+\frac
{2}{n}\sum\limits_{k=1}^{\frac{n-1}{2}}\sin\frac{(2k-1)m\pi}{n}\tan^{-1}\frac{x-\cos\frac{(2k-1)\pi}{n}}{\sin\frac{(2k-1)\pi}{n}}.$$ If $n=2q,$
$$\int\frac{x^{m-1}}{x^n+1}dx=-\frac{1}{n}\sum\limits_{k=1}^{n/2}\cos\frac{(2k-1)m\pi}{n}\log\left(x^2-2x\cos\frac{(2k-1)\pi}{n}+1\right)\\+\frac
{2}{n}\sum\limits_{k=1}^{n/2}\sin\frac{(2k-1)m\pi}{n}\tan^{-1}\frac{x-\cos\frac{(2k-1)\pi}{n}}{\sin\frac{(2k-1)\pi}{n}}.$$
Method of proof:
Partial fractions are obtained by $$\frac{x^{m-1}}{x^n+1}=\frac{1}{n}\sum\limits_{\alpha}\frac{\alpha^m}{x-\alpha},$$ where $\alpha^n+1=0.$ If $n$ is odd it is clear the first $\alpha$ is $-1.$ The rest of the terms must be from
$$\frac{e^{\frac{(2k-1)m\pi}{n}i}}{x-e^{\frac{(2k-1)\pi}{n}i}}+\frac{e^{-\frac{(2k-1)m\pi}{n}i}}{x-e^{-\frac{(2k-1)\pi}{n}i}}=\frac{2\cos\frac{(2k-1)m\pi}{n}\left(x-\cos\frac{(2k-1)\pi}{n}\right)-2\sin\frac{(2k-1)m\pi}{n}\sin\frac{(2k-1)\pi}{n}}{x^2-2x\cos\frac{(2k-1)\pi}{n}+1}$$ and integrate from here; $$\int\frac{dx}{x^2+r^2}=\frac{1}{r}\tan^{-1}\frac{x}{r}.$$
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|
Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$
I can see that 3 cancels out after 1/3, but what next? I can't go further.
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\begin{align*}
1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+
\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\ldots
&=\sum_{n=0}^{\infty}
\frac{(2n-1)!!}{3^{n} n!} \\
&=\sum_{n=0}^{\infty}
\frac{(-\frac{1}{2})(-\frac{3}{2})\ldots (-\frac{2n-1}{2})}
{3^{n} n!} (-2)^{n} \\
&=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}
\frac{ \left( -\frac{2}{3} \right)^{n}}
{n!} \\
&= \left[ 1+\left( -\frac{2}{3} \right) \right]^{-\frac{1}{2}} \\
&= \sqrt{3}
\end{align*}
|
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p odd prime. Prove that if $a\equiv b\pmod p$ then $a^p\equiv b^p\pmod p^2$. Then show $x^5+y^5=z^5$ has no integer solutions with $5\not\mid xyz$ Question:
Let $p$ be an odd prime. Prove that if $a\equiv b \pmod p$ then $a^p\equiv b^p \pmod p^2$. Then show the Diophantine equation $x^5+y^5=z^5$ has no integer solutions with $5\not\mid xyz$.
My working so far:
$a\equiv b \pmod p\Longrightarrow a=b+kp$ for some $k\in\mathbb{Z}$ so
$$ a^p = (b+kp)^p = \sum_{i=0}^p {p\choose i}(kp)^i ~b^{p-i} = b^p + k~p^2~b^{p-1} + \sum_{i=2}^p {p\choose i}(kp)^i ~b^{p-i} \equiv b^p\pmod{p^2}.$$
How can I use this to prove the latter statement?
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Write the Fermat equation of exponent $5$ in the form $x^5+y^5+z^5=0$.
We have $x^5\equiv x \mod 5$, $y^5\equiv y \mod 5$ and $z^5\equiv z \mod 5$, hence $x+y+z\equiv 0\mod 5$. Without loss of generality we may assume, because $5\nmid xyz$, that $x\equiv y\mod 5$, hence $x^5\equiv y^5 \mod 25$, and $−z^5 ≡ x^5 + y^5 ≡ 2 x^5 \mod 25$. However, the equation $x ≡ y \mod 5$ also implies that
$−z ≡ x + y ≡ 2 x \mod 5$ and $−z^5 ≡ 2^5 x^5 ≡ 32 x^5 \mod 25$.
Combining the two results and dividing both sides by $x^5$ yields a contradiction
$2 ≡ 32 \mod 25$.
|
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Help with telescoping sum $\sum_{i=3}^n \frac{1}{i(i+3)} $ How can I use the telescoping technique to compute the following sum?
I'm having issues getting started. I know the basic steps but I don't know how to perform them. I know I have to separate the fraction into A and B. After that I have to perform the sum but I'm not sure what comes next.
$$\sum_{i=3}^n \frac{1}{i(i+3)} $$
Thanks for any help!
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$\displaystyle\sum_{i=3}^n \dfrac{1}{i(i+3)}=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+3}\right)=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}+\dfrac{1}{i+1}-\dfrac{1}{i+2}+\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$
$\displaystyle=\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i}-\dfrac{1}{i+1}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+1}-\dfrac{1}{i+2}\right)+\dfrac{1}{3}\sum_{i=3}^n \left(\dfrac{1}{i+2}-\dfrac{1}{i+3}\right)$
$=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{n+1}+\dfrac{1}{4}-\dfrac{1}{n+2}+\dfrac{1}{5}-\dfrac{1}{n+3}\right)$
$=\dfrac{(n-2)(47n^2+196n+189)}{180(n+1)(n+2)(n+3)}$
|
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|
How to prove this inequation? $$
1+\frac{2}{3n-2}\leqslant \sqrt[n]{3}\leqslant 1+\frac{2}{n}, n\in \mathbb{Z}^{+}
$$
How to prove this inequation?
|
For the LHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $\frac13$, we get
$$\begin{align}
&\left(1 + \frac{2}{3n-2}\right)^{-1}= 1 - \frac{2}{3n} = \frac{1}{n}\left( (n-1)\times 1 + \frac{1}{3}\right) \ge \frac{1}{\sqrt[n]{3}}\\
\implies &
1 + \frac{2}{3n-2} \le \sqrt[n]{3}
\end{align}$$
For the RHS, apply $\text{AM} \ge \text{GM}$ to $n$ numbers with $n-1$ copies of $1$ and one copy of $3$, we get
$$\sqrt[n]{3} \le \frac{1}{n}\left( (n-1)\times 1 + 3 \right) = 1 + \frac{2}{n}$$
|
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A little more integration. Help integrating
$$\int\cos^{-1}(a\tan\theta)\ d\theta$$
I understand that Wolfram gives a solution, but I'd like to know the steps. I haven't been able to rewrite the equation into anything helpful.
|
Mathematica outputs $$\theta \cos ^{-1}(a \tan (\theta))+\frac{1}{4} \left(4 \theta \sin ^{-1}(a \tan (\theta))+i \left(2 i \left(\text{Li}_2\left(\left(\sqrt{a^2+1}-a\right) e^{-i \sin ^{-1}(a \tan (\theta))}\right)+\text{Li}_2\left(-\left(a+\sqrt{a^2+1}\right) e^{-i \sin ^{-1}(a \tan (\theta))}\right)\right)-2 i \left(\text{Li}_2\left(\left(a-\sqrt{a^2+1}\right) e^{-i \sin ^{-1}(a \tan (\theta))}\right)+\text{Li}_2\left(\left(a+\sqrt{a^2+1}\right) e^{-i \sin ^{-1}(a \tan (\theta))}\right)\right)+\left(-2 \sin ^{-1}(a \tan (\theta))-4 \sin ^{-1}\left(\frac{\sqrt{1+i a}}{\sqrt{2}}\right)+\pi \right) \log \left(1-\left(\sqrt{a^2+1}+a\right) e^{-i \sin ^{-1}(a \tan (\theta))}\right)-\left(-2 \sin ^{-1}(a \tan (\theta))+4 \sin ^{-1}\left(\frac{\sqrt{1-i a}}{\sqrt{2}}\right)+\pi \right) \log \left(e^{-i \sin ^{-1}(a \tan (\theta))} \left(-\sqrt{a^2+1}+e^{i \sin ^{-1}(a \tan (\theta))}+a\right)\right)+\left(-2 \sin ^{-1}(a \tan (\theta))+4 \sin ^{-1}\left(\frac{\sqrt{1+i a}}{\sqrt{2}}\right)+\pi \right) \log \left(e^{-i \sin ^{-1}(a \tan (\theta))} \left(\sqrt{a^2+1}+e^{i \sin ^{-1}(a \tan (\theta))}-a\right)\right)-\left(-2 \sin ^{-1}(a \tan (\theta))-4 \sin ^{-1}\left(\frac{\sqrt{1-i a}}{\sqrt{2}}\right)+\pi \right) \log \left(e^{-i \sin ^{-1}(a \tan (\theta))} \left(\sqrt{a^2+1}+e^{i \sin ^{-1}(a \tan (\theta))}+a\right)\right)-8 i \sin ^{-1}\left(\frac{\sqrt{1-i a}}{\sqrt{2}}\right) \tan ^{-1}\left(\frac{(a-i) \cot \left(\frac{1}{4} \left(2 \sin ^{-1}(a \tan (\theta))+\pi \right)\right)}{\sqrt{a^2+1}}\right)+8 i \sin ^{-1}\left(\frac{\sqrt{1+i a}}{\sqrt{2}}\right) \tan ^{-1}\left(\frac{(a+i) \cot \left(\frac{1}{4} \left(2 \sin ^{-1}(a \tan (\theta))+\pi \right)\right)}{\sqrt{a^2+1}}\right)+2 \log (a+i a \tan (\theta)) \sin ^{-1}(a \tan (\theta))+\log (a+i a \tan (\theta)) \left(\pi -2 \sin ^{-1}(a \tan (\theta))\right)-2 \log (a-i a \tan (\theta)) \sin ^{-1}(a \tan (\theta))-\log (a-i a \tan (\theta)) \left(\pi -2 \sin ^{-1}(a \tan (\theta))\right)\right)\right)$$
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Prove the inequality.Let a, b and c be nonnegative real numbers. Let $a$, $b$ and $c$ be nonnegative real numbers. Prove that $a^4+b^4+c^2\ge 8^{½}abc$
|
Using AM,GM Inequality
$$a^4+b^4\ge 2a^2b^2$$ which can be demonstrated as $$a^4+b^4=(a^2-b^2)^2+2a^2b^2\ge 2a^2b^2$$ as the square of any real number is $\displaystyle\not<0$
Again similarly, $$2a^2b^2+c^2\ge 2\sqrt{2a^2b^2\cdot c^2}$$
|
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Find $T(1)$, $T(x)$ and $T(x^{2})$ and $T(ax^2+bx+c)$ Let $$T:P_{3}\rightarrow P_{3}$$ be a linear transformation such that
$$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1.$$
Find $$T(1)$$, $$T(x)$$ and $$T(x^{2})$$
and $$T(ax^{2}+bx+c)$$.
I will be completely honest and say that I have no idea where to even begin with this problem.
Thank you for the help!
|
Here's one way:
given that
$T(2x^{2})=2x^{2}+3x, T(\frac{1}{2}x+2)=2x^{2}+4x-3, T(2x^{2}-1)=3x-1, \tag{0}$
start with $T(2x^2)$. We have, by linearity,
$T(2x^2) = 2T(x^2), \tag{1}$
and since we are given
$T(2x^2) = 2x^2 + 3x, \tag{2}$
we see by (1) that
$2T(x^2) = 2x^2 + 3x, \tag{3}$
or
$T(x^2) = x^2 + \dfrac{3}{2}x; \tag{4}$
that covers $T(x^2)$; next, we use
$T(2x^{2}-1)=3x-1, \tag{5}$
together with
$T(2x^2 - 1) = 2T(x^2) -T(1), \tag{6}$
which again follows from the linearity of $T$, to obtain
$2T(x^2) - T(1) = 3x - 1, \tag{7}$
and if we insert the value of $T(x^2)$ from (4) we see that
$2(x^2 + \dfrac{3}{2}x) - T(1) = 3x -1, \tag{8}$
which may be solved for $T(1)$:
$T(1) = 2x^2 + 1. \tag{9}$
Finally, we employ
$T(\dfrac{1}{2}x+2)=2x^{2}+4x-3 \tag{10}$
and the linearity of $T$ once again to see that
$\dfrac{1}{2}T(x) + 2T(1) = 2x^{2}+4x-3; \tag{11}$
substituting $T(1)$ from (9) yields
$\dfrac{1}{2}T(x) + 2(2x^2 + 1) = 2x^2+4x-3, \tag{12}$
and some algebraic fiddling around with (12) further yields
$T(x) = -4x^2 + 8x - 10. \tag{13}$
We have now found $T(1)$, $T(x)$ and $T(x^2)$; finding $T(ax^2 + bx + c)$ is simply one more application of linearity:
$T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1)$
$= a(x^2 + \dfrac{3}{2}x) + b(-4x^2 + 8x - 10) + c(2x^2 + 1)$
$=(a - 4b + 2c)x^2 + (\dfrac{3}{2}a + 8b)x + (c - 10b). \tag{14}$
The above shows how it can be done in this specific case, in which it is easy to successively isolate $T(1)$, $T(x)$, $T(x^2)$ due to the form of the givens in (0); in the more general case, of which this is but a specific instance, a general linear system in the "variables" $T(1)$, $T(x)$, $T(x^2)$ could be set up and solved using standard linear algebraic techniques. It boils down to a system of three variables in three unkowns.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I did:
the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:
$d(x,y,z)=\sqrt{x^2+y^2+z^2}$
$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$
$C_2(x,y,z)=x-2z$
$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:
$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$
Let's derive and solve when derivatives are zero:
$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$
$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$
$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$
$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$
$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$
Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?
|
The circle can be parametrized as
$$ x = 2 + \frac{4}{\sqrt 5} \sin \theta, \; y = 2 + 2 \cos \theta, \; z = 1 + \frac{2}{\sqrt 5} \sin \theta. $$
The squared distance of such a point from the origin is
$$ f(\theta) = 13 + 4 \sqrt 5 \sin \theta + 8 \cos \theta. $$
Derivative is
$$ f'(\theta ) = 4 \sqrt 5 \cos \theta - 8 \sin \theta. $$
So, the two extrema occur where $\tan \theta = \frac{\sqrt 5}{2};$ from $1 + \tan^2 \theta = \sec^2 \theta$ we get $$\sec^2 \theta = \frac{9}{4}, \cos^2 \theta = \frac{4}{9},$$ and
$$ \cos \theta = \pm \frac{2}{3}, \; \sin \theta = \pm \frac{\sqrt 5}{3}, $$ with matching $\pm.$
The farthest point from the origin is
$$ \left( \frac{10}{3}, \; \frac{10}{3}, \; \frac{5}{3} \right), $$
the nearest
$$ \left( \frac{2}{3}, \; \frac{2}{3}, \; \frac{1}{3} \right). $$
The center of the circle is at $(2,2,1),$ and the plane goes through the origin, so all we really did was find the two points on the given sphere along that line, $$ x = 2t,y=2t,z=t. $$
|
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|
Generating Functions for collection of balls There are 10000 identical red balls, 10000 identical yellow balls
and 10000 identical green balls. In how many different ways can we
select 2005 balls so that the number of red balls is even or the
number of yellow balls is odd?
in this question, being even or odd affects to what. As a result, the numbers of balls with different colors are same. I did not get it.
|
You are asking for generating functions.
First of all, the total number of balls available is irrelevant, as long as it is enough to fill all slots. For simplicity, take infinite number of balls.
*
*Red balls even: $1 + z^2 + \ldots = \frac{1}{1 - z^2}$
*Yellow balls odd: $z + z^3 + \ldots = \frac{z}{1 - z^2}$
*Any number of green balls: $1 + z + z^2 + \ldots = \frac{1}{1 - z}$
Now you have to consider (red even) + (yellow odd) - (red even and yellow odd), as the last case was considered twice in the first two collections:
*
*Red balls even, others arbitrary:
$\frac{1}{1 - z^2} \cdot \frac{1}{(1 - z)^2}$
*Yellow balls odd, others arbitrary:
$\frac{z}{1 - z^2} \cdot \frac{1}{(1 - z)^2}$
*Red even, yellow odd:
$\frac{z}{(1 - z^2)^2} \cdot \frac{1}{1 - z}$
In summary, you are interested in:
$$
[z^{2005}] \left(
\frac{1}{(1 - z^2) (1 - z)^2}
+ \frac{z}{(1 - z^2) (1 - z)^2}
- \frac{z}{(1 - z^2)^2 (1 - z)}
\right)
$$
You can get the result by expanding in partial fractions and handle the resulting terms:
$$
- \frac{1}{16 (1 + z)}
- \frac{1}{8 (1 + z)^2}
- \frac{1}{16 (1 - z)}
+ \frac{4}{5 (1 - z)^3}
$$
Geometric series or the generalized binomial theorem with negative integer exponent finishes this off:
\begin{align}
(1 + u)^{-m} &= \sum_{k \ge 0} \binom{-m}{k} u^k \\
&= \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
\end{align}
|
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|
Calculate $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ I am trying to calculate:
$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $\ln(1-x)$ then,
\begin{align*}
=&-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\\
=&-(x-x^2)-\frac{(x-x^2)^2}{2}-...\\
=&-x(1-x)+\frac{x^2}{2}(1-x)^2-\frac{x^3}{3}(1-x)^3\\
\text{thus the pattern is:}\\
=&\frac{x^n(1-x)^n}{n}
\end{align*}
Am I right?
Then our Integral would be: $$\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$\underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
\begin{align*}
\frac{n(n-1)...1}{(n+1)(n+2)...(2n)}\int_0^1 x^{2n} dx
\end{align*}
This shows a pattern:
\begin{align*}
=&\frac{(n!)^2}{(2n)!} (\frac{1}{2n+1})\\
=& \frac{(n!)^2}{(2n+1)!}
\end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
|
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2\int_{-1}^1\frac{\ln\frac{3+s^2}{4}}{1-s^2}ds.\tag{1}$$
The antiderivative of any expression of the type $\displaystyle\frac{\ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$\displaystyle \int\frac{\ln(a-x)}{x+b}dx=\mathrm{Li}_2\left(\frac{a-x}{a+b}\right)+\ln(a-x)\ln\frac{x+b}{a+b}.\tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
\begin{align}I=&-\int_{-1}^1\frac{2s}{3+s^2}\ln\frac{1+s}{1-s}ds=
4\Re\int_{-1}^1\frac{\ln(1-s)}{s+i\sqrt3}ds.
\end{align}
Applying (2), this reduces to
$$I=-4\Re\,\mathrm{Li}_2\left(e^{i\pi/3}\right)=-\frac{\pi^2}{9},$$
where at the last step we have used that for $z\in(0,1)$ one has
$$\Re\,\mathrm{Li}_2\left(e^{2i\pi z}\right)=\pi^2\left(z^2-z+\frac16\right).$$
|
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|
Prove determinant of $n \times n$ matrix is $(a+(n-1)b)(a-b)^{n-1}$? Prove $\det(A)$ is $(a+(n-1)b)(a-b)^{n-1}$ where $A$ is $n \times n$ matrix with $a$'s on diagonal and all other elements $b$, off diagonal.
|
For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal but not orthonormal) eigenvectors for the $n$ by $n$ case can be read from the columns of
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 2 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 3 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 4 & -1 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 5 & -1 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 6 & -1 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 7 & -1 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 8 & -1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 9
\end{array}
\right).
$$
Note: I made this up. My own self.
|
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|
The product of three consecutive natural numbers is divisible by $6$ Please give me feedback for my answer to this question.
Prove or find a counterexample: The product of any three consecutive
natural numbers is divisible by $6$
My answer: True. Suppose $n$ is a natural number, such that the $3$ consecutive natural numbers is $n, n+1, n+2$. Then,
$$\begin{align}\frac{n(n+1)(n+2)}{6}
&= \frac{1(1+1)(1+2)}{6} \\
&= \frac 66 \\
&= 1\end{align}$$
Thus, $n(n+1)(n+2)$ is divisble by 6.
|
You have proved that $1\times2\times3$ is divisible by six, not that the product of any 3 consecutive natural numbers is divisible by $6$.
If a number is divisible by $6$, then it must be divisible by both $2$ and $3$. Your product is $$n(n+1)(n+2)$$ so you could try showing that at least one of $n$, $n+1$ or $n+2$ is a multiple of 3, and at least one is even.
|
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|
Evaluate $\int x \sqrt{1 - x^4} \,\mathrm{d}x$ I have the following question
$$\int x \sqrt{1 - x^4} \,\mathrm{d}x$$
I know we have to use trig. substitution for this and therefore, I did the following by letting $x = \sin \theta$ and $dx = \cos \theta \,\mathrm{d}\theta$
\begin{align}
&\int x \sqrt{1-x^4} \,\mathrm{d}x \\
&=\int \sin \theta \cos \theta\sqrt{1 - (\sin \theta)^4} \, \mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(1-\sin^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
&=\int \sin \theta \cos \theta \sqrt{(\cos^2 \theta)(1+\sin^2\theta)} \,\mathrm{d}\theta \\
\end{align}
Now, I'm confused. How do I proceed?
Thanks!
EDIT: Taking from the answer, I have a (nearly) full solution below for future users.
Instead of letting $x = \sin \theta$. We'll let $x^2 = \sin \theta$ and this will greatly simplify everything. Since, $\mathrm{d}x = \frac{\cos \theta}{2x} \,\mathrm{d\theta}$
\begin{align}
&\int x \sqrt{1-x^4} \quad \mathrm{d}x \\
&=\frac{1}{2}\int \cos \theta\sqrt{1 - (\sin x)^2} \quad \mathrm{d}\theta \\
&= \frac{1}{2} \int \cos \theta \cos \theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \int 1 + \cos 2\theta \quad \mathrm{d}\theta \\
&= \frac{1}{4} \left(\theta + \frac{\sin2\theta}{2} \right) \\
&= \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} \\
\end{align}
After this, you only have to put $\sin \theta$ back in terms of $x$ and you're done!
|
Try another substitution: $x^2 = \sin (u)$.
We have $2x dx = \cos (u) du$ so $dx = \frac{\cos (u)}{2x} du$
So now we have $\frac{1}{2} \displaystyle \int \cos(u) \sqrt{1 - \sin^2 (u)} du$
And I think you can do the rest.
|
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|
Show that the follow function is Riemann integrable on $[0 , 2]$, and use te definition to find $\int_0^2f.$ Show that the follow function is Riemann integrable on $[0 , 2]$, and use te definition to find $\int_0^2f.$
$$
f(x) =
\left\{
\begin{array}{c}
-1, &0 \le x < 1 \\
2, &1 \le x \le 2
\end{array}
\right.
$$
Take $\epsilon > 0$, consider the partition $P = \{0, 1 - \frac{\epsilon}{2}, 1 + \frac{\epsilon}{2}, 2\}$ where $\epsilon$ is sufficiently small.
Note: $\Delta x_1 = 1 - \frac{\epsilon}{2}, \Delta x_2 = \epsilon, \Delta x_3 = 1 - \frac{\epsilon}{2}$
$L(P,f) = m_1\Delta x_1 + m_2\Delta x_2 + m_3\Delta x_3$
$L(P,f) = (-1)\Delta x_1 + (-1)\Delta x_2 + (2)\Delta x_3$
$L(P,f) = (-1)(1 - \frac{\epsilon}{2}) + (-1)(\epsilon) + (2)(1 - \frac{\epsilon}{2}) = 1 + \frac{\epsilon}{2} - 2\epsilon$
$U(P,f) = M_1\Delta x_1 + M_2\Delta x_2 + M_3\Delta x_3$
$U(P,f) = (-1)\Delta x_1 + (2)\Delta x_2 + (2)\Delta x_3$
$U(P, f) = (-1)(1 - \frac{\epsilon}{2}) + (2)(\epsilon) (2)(1 - \frac{\epsilon}{2}) = 1 + \epsilon + \frac{\epsilon}{2}$
$U(P,f) - L(P,f) = [1 + \epsilon + \frac{\epsilon}{2}] - [1 + \frac{\epsilon}{2} - 2\epsilon] = 3\epsilon$
Change to $\frac{\epsilon}{8}$
Then we get $L(P,f) = 1 - \frac{3\epsilon}{8}$ and $U(P,f) = 1 + \frac{3\epsilon}{8}$
$U(P,f) - L(P,f) = \frac{3\epsilon}{4}$ which is $< \epsilon$
Thus $f$ is integrable on $[0,2]$
$1 - \frac{\epsilon}{8} \le \int_\underline 0^2 f\le \int_0^\overline 2 f \le 1 + \frac{\epsilon}{8}$
Therefore $\int_0^2 f = 1$
Did I do this correctly?
|
Yes your proof looks rigorous and correct. You can also use the fact that a function with finitely many discontinuities in an interval satisfies that the integral of the function is equal to the sum of integrals of the function over the subintervals in-between the discontinuities, which is basically what you just proved. In your case you would get $\int_0^2 f = \int_0^1f + \int_1^2f= -1 + 2 = 1$.
|
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|
Kind of basic combinatorical problems and (exponential) generating functions I have a pretty straightforward combinatorical problem which is an exercise to one paper about generating functions.
*
*How many ways are there to get a sum of 14 when 4 distinguishable dice are rolled?
So, one die has numbers 1..6 and as dice are distinguishable then we should use exponential generating functions (we count sequences of rolled dice), because $3,4,3,4$ differs from $3,3,4,4$. So, we end up with answer
$$[\frac{x^{14}}{14!}](x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!})^4$$
How can we nicely calculate the coefficient of $\frac{x^{14}}{14!}$? I don't want to do this brutally, because next task is
2 Find generating function for the number of ways a sum of n can occur when rolling a die an infinite (or at least n) number of times.
I'd appreciate some help on such problems and how to "wrap" such limited exponential series.
|
As ShreevatsaR pointed out it's sufficient to consider ordinary generating functions, since they already take into account that $3,4,3,4$ and $3,3,4,4$ are different. The first is coded as the coefficient of $x^3x^4x^3x^4$, while the second as the coefficient of $x^3x^3x^4x^4$ when considering the ogf $(x^1+\cdots+x^6)^4$.
Therefore we get for the first part
\begin{align*}
[x^{14}]&(x^1+\cdots+x^6)^4\\
&=[x^{14}]x^4(1+\cdots+x^5)^4\\
&=[x^{10}](1+\cdots+x^5)^4\\
&=[x^{10}]\left(\frac{1-x^6}{1-x}\right)^4\\
&=[x^{10}](1-4x^6+6x^{12}-4x^{18}+x^{24})\sum_{k\geq0}\binom{-4}{k}(-x)^k\\
&=([x^{10}]-4[x^{4}])\sum_{k\geq0}\binom{k+3}{k}x^k\\
&=\binom{13}{10}-4\binom{7}{4}\\
&=146
\end{align*}
For the second part we observe that each roll contributes at least $1$ to the value $n$. We can therefore restrict ourselves to the bracketed formulation: A die will be rolled at least $n$ number of times since all further rolls will not contribute to $n$.
An ordinary generating function in this case is
$$x^n\left(\frac{1-x^6}{1-x}\right)^n$$
Added 2014-04-19: Supplement - Using exponential generating functions instead.
This is admittedly a rather simple minded attempt to answer the question of Chris from the comment below: What do we count, if we would have used exponential generating functions here? Please feel free, to provide better examples with four dice, if you like.
If we use exponential instead of ordinary generating functions, we could imagine that we also have the pips of the faces of the dies distinguishable. Let's assume we have four magic dice $(M_1,M_2,M_3,M_4)$ (M for magic) and we are asking what's the contribution of a roll $(3,4,3,4) \rightarrow 14$ with respect to
$$\left[\frac{x^{14}}{14!}\right](x+\frac{x^2}{2}+\cdots+\frac{x^6}{6!})^4$$
The contribution is the multinomial coefficient
$$\left[\frac{x^{14}}{14!}\right]\frac{x^3}{3!}\frac{x^4}{4!}\frac{x^3}{3!}\frac{x^4}{4!}=\binom{14}{3,4,3,4}=\frac{14!}{3!^{2}4!^{2}}=4204200$$
and the explanation: In this case (namely roll with $14$ pips) the magic dice can choose for the resulting pips from $14$ different colors. $M_1$ chooses $3$ colors, $M_2$ chooses $4$ from the remaining $11$ colors, $M_3$ takes $3$ from the remaining $7$ and the rest of $4$ colors is used by $M_4$. So, we have
$$\binom{14}{3}\binom{11}{4}\binom{7}{3}\binom{4}{4} = \binom{14}{3,4,3,4}$$
different possibilities to see the magic dice colorized with $14$ different colors.
|
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|
Minimal polynomial: is $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$? I was wondering about the minimal polynomial of real number
$$u=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$$
over field $\mathbb{Q}$.
As you can see here, I worked out that $u$ is a root of monic rational polynomial $x^3+3x-4$. This is not irreducible:
$$x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)$$
and the second, quadratic, factor has complex roots $\frac{-1\pm i\sqrt{15}}{2}$.
Can I claim that the minimal polynomial of $u$ over $\mathbb{Q}$ is $x-1$?
In other words: does this prove that $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1$?
In case it does, how can we demonstrate $u=1$ working on the roots in a way different from how I did in the link?
Edit:
$\sqrt[3]{2-\sqrt{5}}$ is meant to be the real cube root of $2-\sqrt{5}$.
|
Here's another way of looking at this through a reverse lens:
Let's solve $u^3+3u-4=0$ by Cardano's method, putting $u=x+y$.
Then $(x+y)^3-3xy(x+y)-(x^3+y^3)=0$ and we require:
$$x^3+y^3=4$$ and $$-3xy = 3 \text { so that }xy=x^3y^3=-1$$
Then we note that $x^3$ and $y^3$ are roots of the quadratic $$z^2-4z-1=0$$So that $$z=\frac{4\pm\sqrt{16+4}}{2}=2\pm \sqrt 5$$
Whence $$u=\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}$$ where the cube roots have to be compatible with the constraint $xy=-1$. There are three possibilities and three roots of the cubic. Since $u=1$ is a root of the cubic, one choice will give the root $u=1$.
Since the cubic has only one real root, choosing the real cube roots will give $u=1$.
|
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|
Calculate the integral using Jacobian The problem asks you to calculate the following integral using Jacobian
$$\int{4x^2+y^2}dA$$ and it tells you to substitute $y+2x=u,y-2x=v,0<u<2,-2<v<0$, and the Jacobian is $\cfrac{1}{4}$.
Could anyone here help me out? Thank you.
|
First, we determine $u$ and $v$ in term of $x$ and $y$. Using elimination/ substitution, it's easy to find that $x=\frac{1}{4}(u-v)$ and $y=\frac{1}{2}(u+v)$. Just adding and subtract $u$ and $v$. Then
\begin{align}
\int(4x^2+y^2)\,dA&=\int\int(4x^2+y^2)\,dx\,dy\\
&=\int_{-2}^0\int_0^2\left(4\left(\frac{1}{4}(u-v)\right)^2+\left(\frac{1}{2}(u+v)\right)^2\right)|J|\,du\,dv\\
&=\int_{-2}^0\int_0^2\frac{1}{4}\left((u-v)^2+(u+v)^2\right)\frac{1}{4}\,du\,dv\\
&=\frac{1}{16}\int_{-2}^0\int_0^2\left(2u^2+2v^2\right)\,du\,dv\\
&=\frac{1}{16}\int_{-2}^0\int_0^22\left(u^2+v^2\right)\,du\,dv\\
&=\frac{1}{8}\int_{-2}^0\left[\frac{1}{3}u^3+v^2u\right]_0^2\,dv\\
&=\frac{1}{8}\int_{-2}^0\left(\frac{8}{3}+2v^2\right)\,dv\\
&=\frac{1}{8}\left[\frac{8}{3}v+\frac{2}{3}v^3\right]_{-2}^0\\
&=\frac{1}{8}\left(\frac{16}{3}+\frac{16}{3}\right)\\
&=\frac{4}{3}
\end{align}
|
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|
Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ I'm looking for a way to prove
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$
I know that
$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$
so maybe I could simplify the above more?
|
Because it wasn't explicitly stated, I probably should state that $E_{2m}$ are the Euler numbers.
Let's integrate the function $$ f(z) = \frac{\pi \csc (\pi z)}{(2z+1)^{2m+1}}, \quad m \in \mathbb{N}_{\ge 0},$$
around a square contour with vertices at $\pm (N+\frac{1}{2}) \pm i(N+\frac{1}{2})$ , where $N$ is a positive integer.
Letting $N$ go to infinity throught the positive integers, the integral vanishes (see here), and we end up with
$$ \begin{align} 0 &= \sum_{n=-\infty}^{\infty} \text{Res}[f(z),n] + \text{Res} \left[f(z),- \frac{1}{2} \right] \\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} + \text{Res}\left[f(z),-\frac{1}{2} \right]. \end{align}$$
Notice that
$$\text{Res} \left[ f(z),- \frac{1}{2} \right] = \text{Res} \left[f \left(z - \frac{1}{2}\right),0 \right] = \text{Res} \left[-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}},0 \right] .$$
Using the Maclaurin series of sec(z), that is, $$ \sec (z) = \sum_{n=0}^{\infty} \frac{(-1)^n E_{2n}}{(2n)!}z^{2n} \, , \quad |x| < \frac{\pi}{2},$$
we have
$$-\frac{\pi \sec (\pi z)}{(2z)^{2m+1}} = -\frac{\pi}{2^{2m+1}} \sum_{n=0}^{\infty} \frac{(-1)^{k} E_{2n}}{(2n)!} \pi^{2n} z^{2n-2m-1} ,$$
from which we can conclude that
$$\text{Res} \left[-\frac{\pi \sec \pi z}{(2z)^{2m+1}},0 \right] = -\frac{\pi (-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m} .$$
Therefore,
$$2 \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2m+1}} = \frac{(-1)^{m} E_{2m}}{2^{2m+1} (2m)!} \pi^{2m+1}\, , $$
and the result follows.
EDIT:
Another approach is to use the partial fractions expansion
$$ \sec(z) = \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1) \pi}{\left(\frac{2n+1}{2} \right)^{2} \pi^{2}-z^{2}}. $$
For $|z| < \frac{\pi}{2}$, we have
$$ \begin{align}\sec(z) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} (2n+1)\pi}{\left(\frac{2n+1}{2} \right)^{2}\pi^{2}} \frac{1}{1-\left( \frac{2z}{(2n+1)\pi}\right)^{2}} \\ &=\frac{4}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+1} \sum_{k=0}^{\infty} \frac{(2z)^{2k}}{\left((2n+1) \pi\right)^{2k}} \\ &= \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} . \end{align} $$
But we know that
$$ \sec(z) = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k} .$$
Therefore,
$$ \frac{4}{\pi} \sum_{k=0}^{\infty} \frac{2^{2k}z^{2k}}{\pi^{2k}} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2k+1}} = \sum_{k=0}^{\infty} \frac{(-1)^k E_{2k}}{(2k)!}z^{2k},$$
and the result follows by comparing the $k$th coefficient of both series.
|
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|
Proving the area of an equilateral triangle How do you prove that How do you prove that for any equilateral triangle with side length s, area is $\frac{s^2 √3}{4}$ ? I tried using an equilateral triangle in a square, but I keep coming up with a $2x^2√3$ , as shown below. What am I doing wrong?
I started with the following:
The area of the full square is:
$ 2x * 2x = 4x^2$
To find the area of the triangle, I will subtract the non-triangle parts from the square.
The part shaded green is:
$ (2x - x√3) * 2x = 4 x^2-2x^2√3$
The parts shaded blue are:
$ \frac{x * ( x√3)}{2} + \frac{x * ( x√3)}{2} = x^2√3$
Adding blue and green:
$(x^2√3) + (4 x^2-2x^2√3) = 4 x^2-x^2√3 $
Subtract blue and green from whole square:
$(4x^2) -(4 x^2-x^2√3) = x^2√3$
Multiply by 2 because I am referring to the $2x$ side, not half of it ($x$):
Final answer: $2 * (x^2√3) = 2x^2√3$
And of course, $ 2x^2√3 \neq \frac{x^2 √3}{4}$
|
Let's consider a rectangle wit height $x\sqrt{3}$ and length $x$ such that the right angled triangle with legs as the sides forms half of the equilateral triangle (as shown in the picture in the question). Then, the area of the right angled triangle is $\dfrac{1}{2}x^2\sqrt{3}$. Two times this is the area of the equilateral triangle. This gives $x^2\sqrt{3}$. Now, $x$ is half of the side $s$ of the equilateral triangle, i.e., $x=s/2$, implying that the area is $\dfrac{s^2}{4}\sqrt{3}$.
|
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|
What is the average value of $y=x^2\sqrt{x^3+1}$ on the interval $(0,2)$ I know the average value formula is $$\frac{1}{b-a}\int_a^b f(x)\;dx$$
I have no problem plugging in 0 and 2 for a and b respectively. I think i'm struggling actually taking the integral.
|
here is an alternative method to substitution (actually its the same method, just arranged a bit differently).
$$\frac{1}{2-0}\int^2_0 x^2\sqrt{x^3+1}dx$$
$$=\frac{1}{2}\int^2_0 x^2\sqrt{x^3+1} dx\cdot\frac{d(x^3+1)/dx}{d(x^3+1)/dx}$$
$$=\frac{1}{2}\int^2_0 x^2\sqrt{x^3+1}\cdot\frac{d(x^3+1)}{3x^2}$$
$$=\frac{1}{6}\int^2_0\sqrt{x^3+1} d(x^3+1)$$
now we substitute $u=x^3+1$ and $du=d(x^3+1)$
$$=\frac{1}{6}\int^9_1\sqrt{u} du=\frac{1}{6}\bigg(\frac{2}{3} u^{3/2} \bigg)\bigg|_1^9$$
|
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|
Find the minimum value of this expression with absolute values The expression is
$$|x-3| + |x-1| + |x| + |x+2| + |x+4|$$
I know that the minimum values for this expression is when x = 0 but is there any algebraic way to find this out? I did it on the calculator
|
Let $f$ the function defined by the equation $f(x)=|x-3|+|x-1|+|x|+|x+2|+|x+4|$ for all $x\in \mathbb{R}$. $$x<-4\Longrightarrow f(x)=-(x-3)-(x-1)-x-(x+2)-(x+4)=-5x-2>18$$
$$-4\le x <-2 \Longrightarrow f(x)=-(x-3)-(x-1)-x-(x+2)+(x+4) = -3x+6>12$$
$$-2\le x <0 \Longrightarrow f(x)=-(x-3)-(x-1)-x+(x+2)+(x+4) = -x+10>10$$
$$0\le x < 1 \Longrightarrow f(x)=-(x-3)-(x-1)+x+(x+2)+(x+4) = x+10\ge 10$$
$$1\le x < 3 \Longrightarrow f(x)=-(x-3)+(x-1)+x+(x+2)+(x+4) = 3x+8\ge 11$$
$$x\ge 3 \Longrightarrow f(x)=(x-3)+(x-1)+x+(x+2)+(x+4) = 5x+2\ge 17$$
It follows $f(x)\ge 10=f(0)$ for all $x\in \mathbb{R}$.
|
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|
How to solve this equation $2\cos(\frac {x^2+x}{6})=2^x+2^{-x}$ How do I solve for $x$ from this equation?
$$2\cos\frac {x^2+x}{6}=2^x+2^{-x}$$
|
$-2\le2\cos(\frac{x^2 + x}{6})\le 2$ and $2\le2^x + 2^{-x}$.
Therefore the only possible solution is when both equal $2$.
|
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|
using substitution wrongly Solving integral, first way:
$$\int \frac{du}{u^2-9}=-\int \frac{du}{9-u^2}$$
$$u={3\sin v}$$
$$du=3\cos vdv$$
$$-\int \frac{3\cos vdv }{9-9\sin^{2}v}=-\frac 13\int\frac{dv}{\cos v}=-\frac 13\ln\left(\sec v+\tan v\right)=-\frac13 \ln \frac {\frac u3}{\sqrt{1-\frac{u^2}{9}}}$$
$$-\frac13 \ln \frac{u}{\sqrt{9-u^2}}=-\frac16 \ln \frac {u^2}{9-u^2}=\frac16 \ln \frac{9-u^2}{u^2}$$
second way:
$$\int \frac{du}{u^2-9}=\frac 16\int \frac{du}{u-3}-\frac16 \int \frac{du}{u+3}=\frac16 \ln \frac {u-3}{u+3}$$
what's wrong with first way?
Thanks all. $1$ from $\sec$ is missed, must be $1+\frac u3$
|
The first term should be
$$
\begin{align}
-\frac 13\ln\left|\sec v+\tan v\right|+C&=-\frac 13\ln\left|\frac{3}{\sqrt{9-u^2}}+\frac{u}{\sqrt{9-u^2}}\right|+C\\
&=-\frac 13\ln\left|\frac{3+u}{\sqrt{9-u^2}}\right|+C\\
&=-\frac 13\ln\left|\sqrt{\frac{(3+u)(3+u)}{(3-u)(3+u)}}\right|+C\\
&=\frac 16\ln\left|\frac{3-u}{3+u}\right|+C.
\end{align}
$$
|
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|
Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$ Prove that $\sin\frac{\pi}{14}$ is a root of $8x^3 - 4x^2 - 4x + 1=0$.
I have no clue how to proceed and tried to prove that the whole equation becomes $0$ when $\sin\frac{\pi}{14}$ is placed in place of $x$ but couldn't do anything further. I think the symbols might be different but can be the same. If it is correct, please help me to solve this; if the equation is wrong, then please modify it and solve it.
|
Like my other answers $$\sin\frac\pi{14}=\cos\left(\frac\pi2-\frac\pi{14}\right)=\cos\frac{3\pi}7$$
Using this, $$\cos\frac{\pi}7-\cos\frac{2\pi}7+\cos\frac{3\pi}7=0$$
Now $\displaystyle\cos\frac{\pi}7=\cos\left(\pi-\frac{6\pi}7\right)=-\cos\left(2\cdot\frac{3\pi}7\right)$ (use Double Angle formula $\displaystyle\cos2A=2\cos^2A-1$)
$\displaystyle\cos\frac{2\pi}7=-\cos\left(\pi+\frac{2\pi}7\right)=-\cos\left(3\cdot\frac{3\pi}7\right)$ (use $\displaystyle\cos3A=4\cos^3A-3\cos A$ formula)
|
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|
Probability, Uniform Distribution. Suppose $A$ is a random number chosen uniformly from the interval $(-5,6)$,what is the probability that the quadratic equation $x^2 + Ax + 1 = 0$, has no real root?
This is my approach,
Using the quadratic equation we know,
$b^2 - 4ac < 0$ (has no real roots)
Therefore solving we get $A^2 - 4 < 0$,
hence $-2 < A < 2$
Now we know the $p.d.f$ of a uniform distribution is as follows.
Therefore finally we get,
$P(-2<x<2)= 1/(6+5)$ or $1/11$
Now knowing this is a uniform distribution we can convert the above into a $c.d.f$
using integration therefore arriving at the formula on Wikipedia. However since it's uniform distribution why can't we just multiply $1/11$ by $5$ ? Since $[-2,2]$ has $5$ distinct values. Using the formula for $c.d.f$ on Wikipedia I get $4/11$.
|
The first part looks fine - $x^2 + Ax + 1 = 0$ having no real solution means $(x + A/2)^2 = A^2/4 - 1$ < 0, which implies $|A| < 2$.
But then you're done! Since $A$ is uniformy distributed on $(-5,6)$, and since $(-2,2)$ lies fully within that interval, the probabily is simply the ratio of the interval lengths, i.e. $$
P(A \in [-2,2]) = \frac{2 - (-2)}{6 - (-5)} = \frac{4}{11} \text{.}
$$
You can also derive that using the CDF, of course. The CDF of the uniform distribution on $(-5,6)$ is $$
F(x) = \begin{cases}
0 &\text{if $x < -5$} \\
\frac{x+5}{11} &\text{if $x \in [-5,6]$} \\
1 &\text{if $x \geq 6$.} \\
\end{cases}
$$
Therefore, $$\begin{eqnarray}
P(A \in (-2,2)) &=& P((X \leq 2) \setminus (X < -2)) = P(X \leq 2) - P(X \leq -2) \\
&=& \frac{2+5}{11} - \frac{-2+5}{11} = \frac{4}{11} \text{.}
\end{eqnarray}$$
|
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|
Lagrange multiplier - space probe i am stuck on this question which uses the Lagrange multiplier. I am trying to construct the equations using the partial derivatives but the $x$'s and $y$'s cancel. can anyone help?
A space probe in the shape of the ellipsoid
$x^2 + y^2 + 3z^2 = 3$
enters a planet's atmosphere and begins to heat up. The temperature on its surface is found
to be
$T(x, y, z) = x^2 + 2y^2 + 6z$:
Use the method of Lagrange multipliers to find the hottest points on the probe's surface.
|
Set $\Lambda \colon\mathbb R^4\to \mathbb R, (x,y,z,\lambda)\mapsto x^2+2y^2+6z+\lambda (x^2+y^2+3z^2-3)$.
Let $(x,y,z,\lambda)\in \mathbb R^4$.
The following holds:
$$\begin{cases} \Lambda _x(x,y,z,\lambda)&=2x+2\lambda x\\ \Lambda _y(x,y,z,\lambda)&=4y+2\lambda y\\ \Lambda_z(x,y,z, \lambda)&=6+6\lambda _z\\ \Lambda _\lambda(x,y,z,\lambda)&=x^2+y^2+3z^2-3.\end{cases}$$
Suppose $$\begin{cases} 0&=2x+2\lambda x\\ 0&=4y+2\lambda y\\ 0&=6+6\lambda z\\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
then $$\begin{cases} 0&=(1+\lambda)x\\ 0&=(2+\lambda)y\\ \lambda&=-\dfrac1 z \land z\neq 0\neq \lambda \\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
which implies $$\begin{cases} \lambda =-1\lor x=0\\ \lambda =-2 \lor y=0\\ \lambda=-\dfrac1 z \land z\neq 0\neq \lambda \\ x^2+y^2+3z^2-3=0.\end{cases}$$
$\boxed{\text{Case }\lambda =-1}$
It follows that $y=0$ and $z=1$. Thus $x^2+0+3-3=0$ and $x=0$, yielding the critical point $\color{blue}{(0,0,1)}$.
$\boxed{\text{Case }x=0}$
*
*$\boxed{\text{Sub case }\lambda =-2}$ It comes $z=\dfrac 1 2$ and $x^2+y^2-\dfrac 9 4=0$, that is $y=\pm\dfrac 3 2$, yielding the critical points $\color{blue}{\left(0, \dfrac 3 2, \dfrac 1 2\right)}$ and $\color{blue}{\left(0, -\dfrac 3 2, \dfrac 1 2\right)}$.
*$\boxed{\text{Sub case }y=0}$ It comes $0+0+3z~2-3=0$, that is, $z=\pm 1$, yielding the critical points $\color{blue}{(0,0,-1)}$ and $\color{blue}{(0,0,1)}$.
Now just check where it is hotter.
|
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|
Closed form of $ \int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx$ Hello I am trying to solve an incredible integral given by
$$
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx=\pi \ln\bigg[\frac{1}{2}\left(\cos^2\alpha +\sqrt{\cos^4 \alpha +\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}\right)\bigg],\qquad \alpha > \beta >0.
$$
I defined
$$
I\equiv
\int_0^{\pi/2}\ln\big[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\big]dx
$$
and using $\cos^2 x=1-\sin^2 x$ but obtained a more complicated expression. Usually it is easier to work with a closed form of Log Sine so I was trying this. I am not really sure how else to approach this at all.
The result looks like very nice and similar to something we all know :)
I am looking for real or complex methods to solve this problem. I am not sure of what substitutions to make but maybe we could work in hyperbolic space
|
With the shorthands $r=\frac{\sin^2\alpha-\sin^2\beta}{2\sin\beta}$, $s=\frac{\sin^2\alpha+\sin^2\beta}{2\sin\beta}$ and the known integral
$$\int_{0}^{\pi/2} \ln(p + q \cos^2x)dx
= \pi\ln\frac{\sqrt p+\sqrt{p+q}}2
$$evaluate
\begin{align}
I=& \int_0^{\pi/2}\ln\left[1-\cos^2 x(\sin^2\alpha-\sin^2\beta \sin^2 x)\right]dx\\
=& \int_0^{\pi/2}\ln\left[\left(\sqrt{1+r^2}+r+\sin\beta \cos^2x\right)
\left(\sqrt{1+r^2}-r-\sin\beta \cos^2x\right)\right]dx\\
=& \>\pi\ln\frac{\sqrt{\sqrt{1+r^2}+r}+\sqrt{\sqrt{1+r^2}+s} }2
+\pi\ln\frac{\sqrt{\sqrt{1+r^2}-r}+\sqrt{\sqrt{1+r^2}-s} }2
\end{align}
Combine the two terms with
$\sqrt{a+\sqrt c}+\sqrt{a-\sqrt c} =\sqrt{2a+2\sqrt{a^2-c}}$ to obtain
$$
I=\>\pi\ln \frac{1+\cos\alpha+\sqrt{(1+\cos \alpha)^2+\sin^2\beta}}4$$
which can also be expressed as
$$I=\pi \ln\frac{\cos^2\frac\alpha 2+\sqrt{\cos^4 \frac\alpha 2+\cos^2\frac{\beta}{2} \sin^2 \frac{\beta}{2}}}2
$$
|
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|
Find the remainder when $15!$ is divided by $31$. Find the remainder when $15!$ is divided by $31$. I know I have to apply Wilsons theorem but i am a little confused how.
|
This might be hitting a fly with a brick but.....
Primes less than or equal to 15: 2,3,5,7,11,13.
How many multiples of 2,4,8 are less than or equal to 15:7,3,1
How many multiples of 3,9:5,1
How many multiples of 5: 3
How many multiples of 7: 2.
11,13 > 15/2.
So $15! = 2^{7+3+1}3^{5+1}5^37^211\cdot13=2^{11}3^65^27^211\cdot13$
$2^{11} = (2^5)^2*2 = 32^2*2 \equiv 2 \mod 31$
$3^6 = (3^3)^2 = 27^2 \equiv (-4)^2 \equiv 16 \equiv -15 \mod 31$
$5^3 = 25*5 \equiv -6*5 \equiv -30 \equiv 1 \mod 31$
So $2^{11}3^65^3 \equiv -30 \equiv 1 \mod 31$
So $15! \equiv 7^2*11*13 \mod 31$.
$7^2 = 49 \equiv 18 \equiv 36/2 \equiv 5/2 \mod 31$
$11 \equiv -20 \mod 31$
$13 \equiv 26/2 \equiv -5/2 \mod 31$
So $15! \equiv 5/2*-5/2*-20 \equiv 25*5 \equiv -6*5 \equiv -30 \equiv 1 \mod 31$
|
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|
Prove or disprove inequality $a^2+b^2+c^2\ge a^rb^{2-r}+b^rc^{2-r}+c^ra^{2-r}$. If $a$, $b$ and $c$ are real numbers greater than $0$ and $r$ is a real number with $0 \le r \le 2$. Does inequality $$a^2+b^2+c^2\ge a^rb^{2-r}+b^rc^{2-r}+c^ra^{2-r}$$ hold?
|
Hint: Use Rearrangement Inequality, noting $a^r, b^r, c^r$ and $a^{2-r}, b^{2-r}, c^{2-r}$ are similarly ordered.
|
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|
Factor a quadratic equation to get two binomials I'm wrestling with this quadratic and trying to figure out how to factor it:
$$3x^2 - 5x + 2 = 0$$
I know that the product of the last terms of the binomial for an equation equals the third term of the polynomial. Also, the sum of the products of those two numbers should be the middle (second) term of the polynomial.
But what two numbers multiply to give $2$ and added together produce $-5$? How do you use the $3$ in this process?
|
Let me offer a different approach. Try to complete the square, by adding and subtracting suitable terms. $3x^2$ begs for an extra $x^2$ to be added to it, so there we go:
$$
\begin{aligned}
3x^2−5x+2 &= 4x^2 -4x + 1 - x^2 - x + 1 \\
&= (2x-1)^2 -(x^2- 2x+1) - 3x + 2 \\
&= (2x-1)^2 -(x-1)^2 - (3x - 2) \quad \color{red}{\mbox{<-- $a^2-b^2=(a-b)(a+b)$}}\\
&= (2x-1+x-1)(2x-1-x+1) - (3x-2) \\
&= (3x-2)x-(3x-2) \quad \color{red}{\mbox{<-- factor out $(3x-2)$}}\\
&= (3x-2)(x-1)
\end{aligned}
$$
|
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|
Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ ... Problem : Let f be a continuous function defined on [-2009,2009] such that f(x) is irrational for each $x \in [-2009,2009]$ and $f(0) =2+\sqrt{3}+\sqrt{5}$ Prove that the equation $f(2009)x^2 +2f(0)x +f(2009)=0$ has only rational roots.
Solution :
Let us assume the function f(x) be $2+(\sqrt{3}+\sqrt{5})(x+1)$ which satisfies $f(0) =2+\sqrt{3}+\sqrt{5}$
now $f(2009)x^2 =(2010((\sqrt{3}+\sqrt{5})x^2$ .....(1)
Also $2f(0)x =2(2+\sqrt{3}+\sqrt{5})x$ ...(2)
f$(2009)=(2010)(\sqrt{3}+\sqrt{5})$ ......(3)
Now solving for $f(2009)x^2 +2f(0)x +f(2009)=0$ by putting the values of $f(2009)x^2 , 2f(0)x ; f(2009)$ from 1,2 & 3 I am getting roots which are irrational please suggest some other method for this ... thanks....
|
A continuous function only taking irrational values is constant, hence $f(2009)x^2 + 2f(0)x + f(2009)=0$ simplifies to $x^2+2x+1=0$ or $(x+1)^2=0$.
|
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|
Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
|
You can simply solve the inequality $\frac{1}{\sqrt{1+\frac{1}{n}}+1}>\frac{1-2\epsilon}{2}$and find pick N = $\frac{(1-2\epsilon)}{8\epsilon^2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
determining the limit tending to infinity of n times squared roots I'm having troubles showing that
$$
\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}-\frac23n\sqrt{n}}{\sqrt{n}} = \frac12.
$$
I have tried this but I am not able to solve this.
|
Applied Stolz-Cesaro Theorem:
$$
\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}-\frac23n\sqrt{n}}{\sqrt{n}} = \lim_{n\rightarrow \infty}\frac{\sqrt{n+1}-\frac{2}{3}(n+1)\sqrt{n+1}+\frac{2}{3}n\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{2n\sqrt{n}-(2n-1)\sqrt{n+1}}{\sqrt{n+1}-\sqrt{n}}=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{4n^3-(2n-1)^2(n+1)}{2n\sqrt{n}+(2n-1)\sqrt{n+1}}(\sqrt{n+1}+\sqrt{n})=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{(3n-1)(\sqrt{n+1}+\sqrt{n})}{2n\sqrt{n}+(2n-1)\sqrt{n+1}}=\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Understanding substitution to compute primitive To compute the primitive
$$\int x^2 \sqrt[3]{x^3+3}\ dx$$
I am trying this:
$t = x^3+3$
$dt = 3x^2dx$
but
$\int x^2 \sqrt[3]{x^3+3}\ dx=\ ?$
How to continue from here?
|
Make the substitution $t=x^3+3$. Then $dt=3x^2dx$, and we have
$$
\frac{1}{3}\int t^{1/3}dt = \frac{1}{3}\left(\frac{3}{4}t^{4/3}\right) + C=\frac{1}{4}\left( x^3+3\right)^{4/3} +C.
$$
|
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|
Find the axes of the ellipse What are the equations of the major and minor axes of the ellipse $x^2+2y^2-2xy-1=0$. The centre of the ellipse is $(0,0)$ but the axes are tilted (with respect to $x-y$ axes). I don't know how to find those.
|
You take an ellipse of the form $a x^2 + b y^2 + c x y + d x + e y + f = 0$ and match coefficients with the equation of an ellipse centered about $(x_c,y_c)$ at an oblique angle $\theta$
$$ \left( \frac{ (x-x_c) \cos\theta + (y-y_c) \sin\theta}{R_x} \right)^2
+ \left( \frac{ -(x-x_c) \sin\theta + (y-y_c) \cos\theta}{R_y} \right)^2 - 1 = 0 $$
where $R_x$ and $R_y$ are the major and minor radii. From here we deduce that the coefficient $c$ is critical in defining the angle as you find that (by matching $x y$) $$ \sin(2 \theta) \left( \frac{1}{R_y^2} - \frac{1}{R_x^2} \right) + c = 0$$
You already have $(x_c,y_c) = (0,0)$ and $x^2+2 y^2 - 2 x y - 1 =0$ so I propose to change the coordinates to $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}$$ to get
$$ -(x'^2+4 x' y' - y'^2) \cos^2 \theta - 2 (x'^2 - x' y' -y'^2) \sin\theta \cos\theta + (2 x'^2 +2 x' y' + y'^2)-1 = 0$$
The correct choice of $\theta$ should convert the above to $\frac{x'^2}{R_x^2} + \frac{y'^2}{R_y^2}-1 =0$. If you match the $x' y'$ coefficients you get
$$ \left. 2 \cos(2 \theta) - \sin(2 \theta) = 0 \right\} \left. \tan( 2\theta) = 2 \right\} \theta = 31.71° $$
and
$$ R_x = \frac{\sqrt{5}}{2} + \frac{1}{2} = 1.618 \\ R_y = \frac{\sqrt{5}}{2} - \frac{1}{2} = 0.618 $$
Curious that these are equal to the golden ratio
|
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|
Let $z=\ln \tan\frac xy.$ What is $z_x$ and what is $z_y$? Let $$z=\ln \tan\frac xy.$$ What is $z_x$ and what is $z_y$?
Thanks ahead:)
What I have tried:
$$z_x=\frac{1}{\tan \frac xy} \frac{1}{1+(\frac xy)^2} \frac 1y=\frac {y}{\tan \frac xy (x^2+y^2)}$$
$$z_y=\frac {-x}{\tan \frac xy (x^2+y^2)}.$$
I am not sure I am right. Help, help, help...
|
As pointed out in comments, you misread $\tan$ as $\tan^{-1}$.
The answers are:
$$
\begin{array}{c}
z_x = \frac{1}{\tan\frac{x}{y}}\frac{1}{y}\sec^2\frac{x}{y}=\frac{\sec\frac{x}{y}\csc\frac{x}{y}}{y} \\
z_y = -\frac{1}{\tan\frac{x}{y}}\frac{x}{y^2}\sec^2\frac{x}{y}=\frac{-x \sec\frac{x}{y}\csc\frac{x}{y}}{y^2}
\end{array}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$ Hi I am trying to prove this result $$
I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1.
$$
Thanks. Since $x\in[0,1]
$ we can write$$
I=\sum_{n=0}^\infty \int_0^1 \log x (1+x^2) x^{c-2+2cn}dx.
$$
since $\sum_{n=0}^\infty x^n= (1-x)^{-1} , |x| < 1.$ Simplifying
$$
I=\sum_n \int_0^1 \log x\, x^{c-2+2n} dx+\sum_n \int_0^1 \log x \, x^{c+2cn}\, dx.
$$
Now we can write
$$
I=-\frac{1}{4}\psi_1 \left(\frac{c+1}{2}\right)-\frac{1}{4} \psi_1\left( \frac{3+c}{2}\right)
$$
where I summed the results of the integrals using general result of $\int_0^1 \log x \, x^n \, dx=-\frac{1}{(n+1)^2}$. Howeever this is not the result $\sec^2...$. Thanks
The function $\psi$ is the polygamma function which is defined in general by
$$
\psi_m(z)=\frac{d^{m+1}}{dz^{m+1}} \log \Gamma(z)
$$
and $\Gamma(z)=(z-1)!$, for this case $m=1$.
|
Another derivation that uses only very elementary complex analysis and no special functions:
(all integrals are to be regarded as Cauchy PV integrals if necessary)
By substituting $x\mapsto 1/x$, we find that the original integral equals
$$\int_1^{\infty} (-\log x) \frac{(1+1/x^2)x^{2-c}}{1-x^{-2c}}\frac{dx}{x^2} = \int_1^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx$$
Adding both copies of the integral, the problem is reduced to showing that
$$\int_0^{\infty} \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}} \,dx = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$
We will do this by the method that is by now well-known: let $I(\mu) = \int_0^{\infty} \frac{(1+x^2)x^{c-2+\mu}}{1-x^{2c}} \,dx$, then we are looking for $I'(0)$.
We have (remember we are taking PV everywhere)$$\int_0^{\infty} \frac{x^a}{1-x^b} \,dx = \frac{\pi}{b}\cot\left(\pi\frac{a+1}{b}\right)$$
which can be easily derived using residues. (Integrate around a pizza slice contour.)
This immediately gives $$I(\mu) = \frac{\pi}{2c} \left[\cot\left(\pi\frac{c-1+\mu}{2c}\right) + \cot\left(\pi\frac{c+1+\mu}{2c}\right) \right]$$
and hence
$$ I'(\mu) = -\frac{\pi^2}{4c^2}\left[\csc^2\left(\frac{\pi}{2c} (c-1+\mu)\right) +\csc^2\left(\frac{\pi}{2c}(c+1+\mu)\right)\right]$$
$$ \implies I'(0) = -\frac{\pi^2}{2c^2}\sec^2\left(\frac{\pi}{2c}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Beautiful Indefinite Integrals. These are some of the integrals with beautiful solutions I came across-
$$\int \frac{x^2}{(x\sin x+\cos x)^2} dx$$
$$\int\frac {1}{\sin^3x+\cos^3x} dx$$
$$\int \frac{1}{x^4+1}dx$$
I'd love if you share some of the ones you came across.
|
\begin{align}
I_1 & = \int \sqrt{ \sqrt{ x + 2\sqrt{2x-4} } +
\sqrt{ x - 2\sqrt{2x-4} } } \,\mathrm{d}x \, , \quad x>4\\
I_2 & = \int \log( \log x) + \frac{2}{\log x} - \frac{1}{(\log x)^2} \mathrm{d}x \\
I_4 & = \int (1 + 2x^2) e^{x^2}\, \mathrm{d}x \\
I_5 & = \int \frac{\sqrt{x+\sqrt{x^2+1\,}\,}\,}{\sqrt{x^2+1\,}\,} \mathrm{d}x \\
I_6 & = \int \frac{2^x 3^x}{9^x - 4^x} \,\mathrm{d}x
\end{align}
\begin{align*}
I_7
= \int \left( \frac{\arctan x}{x - \arctan x}\right)^2 \mathrm{d}x
= \frac{1 + x \arctan x}{\arctan x - x}
= \frac{1}{\tan (\beta - \tan \beta)}\,,
\end{align*}
where $x = \tan \tan \beta$ or $\beta = \arctan (\arctan x)$.
$$
I_6 = \int \frac{x^2+2x+1+ (3x+1)\sqrt{x+\ln x}}{x\,\sqrt{x+\ln x}(x+\sqrt{x+\ln x})}\mathrm{d}x
= 2 (\sqrt{x+\ln x} + \ln(x+\sqrt{x+\ln x})) + C
$$
I have a bunch more of these here, see p.68 for instance. (click on the problems for solution)
|
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|
Partial fraction integral Question:
$\int \dfrac{5 }{(x+1) (x^2 + 4) } dx $
Thought process:
I'm treating it as a partial fraction since it certainly looks like one.
I cannot seem to solve it besides looking at it in the "partial fraction" way.
My work:
1) Focus on the fraction part first ignoring the $\int $ and $dx$ for the moment.
Multiply $(x+1) (x^2+4)$ on both sides of the equation and get:
$5$ = $\dfrac{A }{(x+1) } $ + $\dfrac{Bx+C }{(x^2+4) } $
Note: the x^2 + 4 is irreducible which explains the Bx+c as the numerator.
$ 5 = A(x^2+ 4) + Bx + C(x+1) $
I tried x = -1 which knocks out C:
$5 = A5 + - B $
I also tried x = 0 which knocks out B
$5 = A4 + C$
A is a lot harder to knock out since the squared changes the picked value to be positive.
I decided to add the two found equation together and get
$10 = A9 - B + C $
Now I am officially stuck at this step.
|
Your equation is incorrect--you have some algebra mistakes that are causing you to not get a solution (which is obvious unless you start plugging in values to $x$):
$$
\frac{5}{(x + 1)(x^2 + 4)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 4}
$$
Now cross multiply to get the numerator:
\begin{align}
5 =& A(x^2 + 4) + (Bx + C)(x + 1) = Ax^2 + 4A + Bx^2 + Bx + Cx + C \\
=& (A + B)x^2 + (B + C)x + (4A + C)
\end{align}
This yields exactly three equations for three unkowns. You don't plug in values, you match coefficients: $5 = 0*x^2 + 0 * x + 5$ therefore you have:
\begin{align}
(A + B)x^2 =& 0*x^2& &\longrightarrow &A + B = 0 \\
(B + C)x =& 0*x&& \longrightarrow &B + C = 0 \\
4A + C =& 5 && \longrightarrow & 4A + C = 5 \\
\end{align}
Solving those equations yields:
\begin{align}
A + B = 0 &\longrightarrow A = -B \\
B + C = 0 & \longrightarrow C = -B = A\\
4A + C = 5 &\longrightarrow 4A + A = 5 \longrightarrow A = 1, B = -1, C = 1\\
\end{align}
Solve those equations and you will get the correct partial fraction decomposition--no need to plug in arbitrary values of $x$.
$$
\frac{5}{(x + 1)(x^2 + 4)} = \frac{1}{x + 1} + \frac{1 - x}{x^2 + 4} = \frac{1}{x + 1} + \frac{1}{x^2 + 4} - \frac{x}{x^2 + 4}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the two points where the shortest distance occurs on two lines Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance
between the lines AB and CD, given $\vec{AB} = \begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
,\vec{CD} = \begin{pmatrix}
0\\
1\\
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}
$ the normal vector $n=\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix}
$
and the shortest distance is $\frac{3}{\sqrt{17}}$. I figured all this out from 4 given points but don't know how to find points P and Q. Please help, I'm stuck...
|
If the normal vector is not provided in the question, you can still solve the question.
Any point $P$ on line $\vec{AB}$ given $u$ can be represented as
$\begin{pmatrix}
1-2u \\
2u\\
2+u
\end{pmatrix}$
, whilst point any point $Q$ on line $\vec{CD}$ given $v$ can be represented as
$\begin{pmatrix}
2v \\
1-v\\
1-2v
\end{pmatrix}$,
$\vec{PQ}=$
$\begin{pmatrix}
-1+2u+2v \\
1-2u-v\\
-1-u-2v
\end{pmatrix}$
You can interpret the shortest distance between $\vec{AB}$ and $\vec{CD}$ at points $P$ and $Q$ as a separate line that is perpendicular to both $\vec{AB},\vec{CD}$ at the same time, similar to how the shortest distance from a point to a line functions:
$$
\vec{PQ} \perp \vec{AB} \\
\vec{PQ} \perp \vec{CD}
$$
The dot product between two perpendicular vectors is 0, which gives:
$$
\vec{PQ} \cdot \vec{AB} =
\begin{pmatrix}
-1+2u+2v \\
1-2u-v\\
-1-u-2v
\end{pmatrix}
\cdot
\begin{pmatrix}
-2 \\
2\\
1
\end{pmatrix} =
0 \\
\vec{PQ} \cdot \vec{CD} =
\begin{pmatrix}
-1+2u+2v \\
1-2u-v\\
-1-u-2v
\end{pmatrix}
\cdot
\begin{pmatrix}
2 \\
-1\\
-2
\end{pmatrix} =
0
$$
Which gives the following system of equations
$$
\begin{aligned}
3-9u-8v = 0\\
-1+8u+9v=0
\end{aligned}
$$
Solving for $u,v$ gives $u=\frac{19}{17},v=-\frac{15}{17}$, which can be substituted back to calculate the coordinates of $P$ and $Q$, giving $P=(\frac{38}{17}, \frac{53}{17}, -\frac{21}{17}), Q=(-\frac{30}{17}, \frac{32}{17}, \frac{47}{17})$
|
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|
Calculate $\displaystyle \lim_{x \to \frac{3}{2}} \frac{2x^2-3x}{|2x-3|}$ I know that $\displaystyle \lim_{x \to \frac{3}{2}} \frac{2x^2-3x}{|2x-3|}$ does not exist, because the lateral limits are different and I also know that the absolute value on the denominator has something to do with it. But I can´t get my mind around it.
Can some one explain it to me slowly?
|
The key is to realize that absolute value is a piecewise defined function in disguise. Recall that $|u| = u$ if $u > 0$ and $|u| = -u$ if $u < 0.$ In your case, $u = 2x – 3,$ so $u > 0$ corresponds to $x > \frac{3}{2}$ (solve $2x – 3 > 0)$ and $u < 0$ corresponds to $x < \frac{3}{2}$ (solve $2x – 3 < 0).$
Thus, we want to separately consider the right and left limits, making use of the fact that $|2x – 3|$ can be rewritten without the use of absolute value bars when we stay on one side of $\frac{3}{2}.$ Specifically, when we're on the right side of $\frac{3}{2}$ we have $2x – 3 > 0$ and hence $|2x – 3| = (2x – 3);$ and when we're on the left side of $\frac{3}{2}$ we have $2x – 3 < 0$ and hence $|2x – 3| = -(2x – 3).$
Therefore,
$$ \lim_{x \to {\frac{3}{2}}^{+}} \frac{x(2x – 3)}{|2x - 3|} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{+}} \frac{x(2x – 3)}{(2x – 3)} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{+}} \frac{x}{1} \;\; = \;\; \frac{3}{2} $$
and
$$ \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{|2x - 3|} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{-(2x – 3)} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x}{-1} \;\; = \;\; -\frac{3}{2} $$
|
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|
log and poisson-like integral Here is a fun looking one some may enjoy.
Show that:
$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$
|
Starting from
$$
{\rm Log}(1+x e^{ia})=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n}x^n
$$
we see that
$$
\int_0^1{\rm Log}(1+x e^{ia})\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}e^{ina}}{n^2}
$$
Taking real parts we get
$$
\int_0^1 \log|1+x e^{ia}|\cdot \frac{1}{x}dx=\sum_{n=1}^\infty\frac{(-1)^{n-1}\cos(na)}{n^2}
$$
applying this to $a+\pi$ instead of $a$ we obtain also
$$
\int_0^1 \log|1-x e^{ia}|\cdot \frac{1}{x}dx=-\sum_{n=1}^\infty\frac{\cos(na)}{n^2}
$$
Subtracting these two formulas:
$$
\int_0^1 \log\frac{|1+x e^{ia}|}{|1-x e^{ia}|}\cdot \frac{1}{x}dx=
\sum_{n=1}^\infty\frac{((-1)^{n-1}+1)\cos(na)}{n^2}
=\sum_{n=0}^\infty\frac{2\cos((2n+1)a)}{(2n+1)^2}
$$
or
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\sum_{n=0}^\infty\frac{4\cos((2n+1)a)}{(2n+1)^2}\tag{1}
$$
On the other hand if $f$ is the $2\pi$-periodic even function that coincides with $a\mapsto \frac{\pi^2}{2}-\pi a$ on $[0,\pi]$ then it is straightforward to check that the Fourier series expansion of $f$ coincides with the right side of $(1)$. So, we have shown that
$$
\int_0^1 \log\left(\frac{1+2x\cos(a)+x^2}{ 1-2x \cos(a)+x^2 }\right)\cdot \frac{1}{x}dx
=\frac{\pi^2}{2}-\pi |a|
$$
for $a\in[-\pi,\pi]$.$\qquad\square$
|
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"url": "https://math.stackexchange.com/questions/808144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Any purely geometric solution to this problem?
What is the largest possible area of a rectangle(in square units) inscribed in the triangle shown in the picture above?
|
Let $\triangle ABC$ be the given triangle with $AB = 10$, $AC = 17$, and $BC = 21$. Choose an arbitrary point $M$ on the side $AB$, and let $x = AM$. Let $N$ be the point on $AC$ such that $MN \parallel BC$, and points $P$, and $Q$ on $BC$ such that $NP \perp BC$, and $MQ \perp BC$. Thus $MNPQ$ is a rectangle. Let $h_a$ be the length of the altitude from $A$. Note that $h_a$ is constant. The we have the followings proportions:
$\dfrac{BM}{BA} = \dfrac{MQ}{h_a} \to \dfrac{10-x}{10} = \dfrac{MQ}{h_a} \to MQ = \dfrac{h_a}{10}\cdot (10-x)$. Also:
$\dfrac{AM}{AB} = \dfrac{MN}{BC} \to \dfrac{x}{10} = \dfrac{MN}{21} \to MN = \dfrac{21}{10}\cdot x$.
Let $S(x)$ be the area of the rectangle $MNPQ$, then:
$S(x) = MN\cdot MQ = \dfrac{21h_a}{100}\cdot x(10 - x)$. Since:
$x + (10 - x) = 10$, a constant, $S$ achieves a maximum value when $x = 10 - x$, or $x = 5$.
From this we can solve for $MN =\dfrac{21}{2}$, and $MQ = \dfrac{h_a}{2}$. Thus:
$S_{max} = \dfrac{21h_a}{4} = \dfrac{S_{\triangle ABC}}{2}$
Note: $h_a$ can be calculated using Heron formula.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that $3x^2+2x\sin(x) + x^2\cos(x) > 0$ for all $x\neq 0$ I got this question:
Show that for all $x\neq 0$, $3x^2+2x\sin(x) + x^2\cos(x) > 0$
I tried to show it but got stuck.
|
Another approach:
$$
3x^2+2xsin(x)+x^2cos(x) = x(3x+\sqrt{2+x^2}sin(\alpha+x))
$$
where, $\alpha = tan^{-1}(x/2)$
Since, $ x> 0$, we need only prove: $3x-\sqrt{2+x^2}>0$, for $x>0$
or $3x>\sqrt{2+x^2}$, squaring both sides, and rearranging terms, $x>\frac{1}{2}$
Also, for $0<x<\frac{1}{2}<\frac{\pi}{2}$, all terms in the expression are positive.
*
* When $x>\frac{1}{2}>0$ the value as proved from the reduced expression is positive
* When $0<x<\frac{1}{2}$, all values are positive and so the expression is positive.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find the prime $p$ such $p\mid\sum_{k=1}^{p+2}\frac{T_{k}}{k+1}$
let $k\in N^{+}$, and such
$$T_{k}=\sum_{i=1}^{k}\dfrac{1}{i\cdot 2^i}$$
Find all prime number $p$, such that
$$p\mid\sum_{k=1}^{p+2}\dfrac{T_{k}}{k+1}$$
I think this problem maybe use this Abel transformation:see http://en.wikipedia.org/wiki/Summation_by_parts
let $$a_{n}=T_{n},b_{n}=\dfrac{1}{n+1},H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
then
$$\sum_{k=1}^{p+2}\dfrac{T_{k}}{k+1}=T_{p+2}\cdot H_{p+2}-\sum_{k=1}^{p+1}H_{k}\dfrac{1}{k\cdot 2^k}$$
then I can't Continue
|
What I get from partial summation is:
$$\begin{eqnarray*}I_p=\sum_{k=1}^{p+2}\frac{1}{k+1}\sum_{i=1}^{k}\frac{1}{i\,2^i}&=&(H_{p+3}-1)T_{p+2}-\sum_{k=1}^{p+1}\frac{H_{k+1}-1}{(k+1)2^{k+1}}\\&=&H_{p+3}T_{p+2}-\sum_{k=1}^{p+2}\frac{H_k}{k\,2^k}.\tag{1}\end{eqnarray*}$$
Now, it is well-known that $H_{p-1}\equiv 0\pmod{p^2}$, and by this paper of Zhi-Wei Sun we know that:
$$\sum_{k=1}^{p-1}\frac{H_k}{k\, 2^k}\equiv 0\pmod{p},\tag{2}$$
so $p\,|\,I_p$ is equivalent to:
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{H_p}{p\,2^p}-\frac{H_{p+1}}{(p+1)\,2^{p+1}}-\frac{H_{p+2}}{(p+2)\,2^{p+2}},$$
or:
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{H_p}{2p}-\frac{H_{p+1}}{4}-\frac{H_{p+2}}{16},$$
$$ p\; | \left(\frac{1}{p}+\frac{1}{p+1}+\frac{1}{p+2}+\frac{1}{p+3}\right)T_{p+2}-\frac{1}{2p^2}-\frac{H_{p+1}}{4}-\frac{H_{p+2}}{16},$$
$$ p\; | \left(\frac{1}{p}+\frac{11}{6}\right)\left(T_{p-1}+\frac{1}{2p}+\frac{5}{16}\right)-\left(\frac{1}{2p^2}+\frac{5}{16p}+\frac{3}{32}\right),$$
$$ p\; | \frac{11}{6}\left(T_{p-1}+\frac{1}{2p}\right)+\frac{T_{p-1}}{p}+\frac{23}{48},\tag{3}$$
hence to solve the problem it suffices to study the residue classes of $T_{p-1}\pmod{p^2}$.
We must have:
$$ p^2 \mid (48+88p) T_{p-1}+(23p+44), $$
from which
$$ 12 T_{p-1} + 11 \equiv 0 \pmod{p}$$
follows. The calculation of $T_{p-1}\pmod{p}$ occurs in the problem N7 of the IMO-2011 shortlist.
It is shown that:
$$ T_{p-1}\equiv \frac{1-\frac{1}{2^{p-1}}}{p}\pmod{p},$$
so the only primes for which we may have divisibility are the ones for which:
$$ 2^{p-1}+\frac{11}{12}p 2^{p-1}\equiv 1\pmod{p^2}.$$
Anyway, we require a closed expression for $T_{p-1}\pmod{p^2}$ in order to really close the problem.
I bet that a key element is the following identity:
$$\frac{1}{p}\binom{p}{k}\equiv (-1)^{k-1}\frac{1+pH_{k-1}}{k}\pmod{p^2},\tag{4}$$
from which we get:
$$ T_{p-1}\equiv\frac{1-\frac{1}{2^{p-1}}}{p}+p\cdot\sum_{k=1}^{p-1}\frac{1}{k^2 2^k}\pmod{p^2}.\tag{5}$$
|
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|
Find volume of sphere $ x^2+y^2+z^2=9$ bounded by planes $z=0$ and $z=2$ using double integral Find volume of sphere $x^2+y^2+z^2=9 $ bounded by planes $z=0$ and $z=2$ using double integral
I tried to take the total volume of the bigger hemisphere but i get zero, i managed to take the volume of the smaller hemisphere only.
|
We need to use the fact that $x^2+y^2=r^2$, so we can convert to polar coordinates. That is, $x^2+y^2+z^2=9$ implies $z = \pm \sqrt{9-(x^2+y^2)}=\pm\sqrt{9-r^2}$.
However, since the volume of the solid we want is above $z=0$, we only need to consider the top half of the sphere, that is, $z=+\sqrt{9-r^2}$.
Now here's the trick:
Find the volume of the half-sphere of radius $3$ (and you can figure out why the radius is $3$, and subtract the double integral below (which you should evaluate):
\begin{align}
\int_0^{2\pi} \int_2^{\sqrt{9-r^2}} r \, dr \, d\theta
\end{align}
So we can find the volume:
$$V=\underbrace{\frac{1}{2}\left(\frac{4}{3} \pi r^3 \right)}_{\text{half sphere}} - \underbrace{\int_0^{2\pi} \int_2^{\sqrt{9-r^2}} r \, dr \, d\theta}_{\text{double integral to subtract}}$$
Basically, we should try to employ as much geometry as possible, to simplify things whenever we can.
|
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|
A system of equations Given three equations $x^2+y^2+xy=a$, $y^2+z^2+yz=b$ and $x^2+z^2+xz=c$, how can I solve for $x,y$ and $z$ in terms of $a,b$ and $c$?
|
If we multiply second equation by $-1$ and add to first we have:
$x^2+y^2+xy-y^2-z^2-yz=(x+y+z)(x-z)=a-b$
Next do the same with other equation:
$(x+y+z)(y-z)=a-c$
$(x+y+z)(y-x)=b-c$
Now if $a \neq c$ we can divide first by second and get:
$\frac{x-z}{y-z}=\frac{a-b}{a-c}$, so:
$(x-z)(a-c)=(a-b)(y-z)$. It's linear.
Next do the same with the other equations and we get a system of three linear equation which is easy to solve. There are also cases $a=c$, $a=b$ or $b=c$, but it's easier than in general case.
|
{
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"url": "https://math.stackexchange.com/questions/819655",
"timestamp": "2023-03-29T00:00:00",
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|
How prove this interesting identity $(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$
let $0<x_{1}<x_{2}<x_{3}$, and there exsit $a$ such
$$\begin{cases}
y_{1}=x_{1}-\ln{x_{1}}=\dfrac{x^2_{1}}{ax_{1}+\ln{x_{1}}}\\
y_{2}=x_{2}-\ln{x_{2}}=\dfrac{x^2_{2}}{ax_{2}+\ln{x_{2}}}\\
y_{3}=x_{3}-\ln{x_{3}}=\dfrac{x^2_{3}}{ax_{3}+\ln{x_{3}}}
\end{cases}$$
show that:
$$(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$$
My try: since
$$(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$$
$$\Longleftrightarrow \dfrac{x^4_{1}x^2_{2}x^2_{3}}{(ax_{1}+\ln{x_{1}})(ax_{2}+\ln{x_{2}})(ax_{3}+\ln{x_{3}})}=x^2_{1}\cdot x_{2}\cdot x_{3}$$
$$\Longleftrightarrow (ax_{1}+\ln{x_{1}})(ax_{2}+\ln{x_{2}})(ax_{3}+\ln{x_{3}})=x^2_{1}\cdot x_{2}\cdot x_{3}$$
since
$$\ln{x_{1}}=x_{1}-y_{1},\ln{x_{2}}=x_{2}-y_{2},\ln{x_{3}}=x_{3}-y_{3}$$
then
$$\Longleftrightarrow [(a+1)x_{1}-y_{1}][(a+1)x_{2}-y_{2}][(a+1)x_{3}-y_{3}]=x^2_{1}\cdot x_{2}\cdot x_{3}$$
then I can't Continue ,This problem is from china's college entrance examination simulation.
and this problem not have solution
Thank you
|
Note $x_i>0$, and since $e^x \geq 1+x$ we have $y_i=x_i-\ln x_i \geq 1$. We have
$$y_i=x_i-\ln x_i=\frac{x_i^2}{ax_i+\ln x_i}=\frac{x_i^2}{ax_i+x_i-y_i}=\frac{x_i^2}{(a+1)x_i-y_i}$$
$$(a+1)x_iy_i-y_i^2=x_i^2$$
Let $a+1=2b$, so $2bx_iy_i-y_i^2=x_i^2$.
$$(x_i-by_i)^2=(b^2-1)y_i^2$$
$$x_i=\left(b \pm \sqrt{b^2-1}\right)y_i$$
Lemma: The equation $\frac{x}{x-\ln x}=c$ has exactly one positive real root if $0<c \leq 1$, and at most two positive real roots if $1<c$.
Proof: Let $f(x)=\frac{x}{x-\ln x}$. We see $f'(x)=\frac{1-\ln x}{(x-\ln x)^2}$, so $f$ is strictly increasing on $(0, e]$, has a maximum at $x=e$, and is strictly decreasing on $[e, \infty)$. Furthermore $f(e)=\frac{e}{e-1}>1$, $\lim_{x \to 0}{f(x)}=0$ and $\lim_{x \to \infty}{f(x)}=1$.
It follows that $f$ restricted to $(0, e]$ is a strictly increasing bijection to $(0, \frac{e}{e-1}]$ and $f$ restricted to $[e, \infty)$ is a strictly decreasing bijection to $(1, \frac{e}{e-1}]$. Now the desired conclusion follows easily.
Case 1: $x_i=(b-\sqrt{b^2-1})y_i$.
Since $b-\sqrt{b^2-1} \leq 1$, we have $x_i \leq y_i=x_i-\ln x_i$ so $x_i \leq 1$.
Note $\frac{x_i}{x_i-\ln x_i}=b-\sqrt{b^2-1}$. By the lemma applied to $0<c=b-\sqrt{b^2-1} \leq 1$, we have at most one possible solution for $x_i$.
Case 2: $x_i=(b+\sqrt{b^2-1})y_i$.
Since $b+\sqrt{b^2-1} \geq 1$, we have $x_i \geq y_i=x_i-\ln x_i$ so $x_i \geq 1$.
Note $\frac{x_i}{x_i-\ln x_i}=b+\sqrt{b^2-1}$. By the lemma applied to $1<c=b+\sqrt{b^2-1}$, we have at most two possible solutions for $x_i$.
Since $0<x_1<x_2<x_3$, we must have $x_1$ satisfying Case 1 and $x_2, x_3$ satisfying case 2.
Thus
$$x_1=(b-\sqrt{b^2-1})y_1$$
$$x_2=(b+\sqrt{b^2-1})y_2$$
$$x_3=(b+\sqrt{b^2-1})y_3$$
Thus
$$x_1^2x_2x_3=(b-\sqrt{b^2-1})^2(b+\sqrt{b^2-1})(b+\sqrt{b^2-1})y_1^2y_2y_3=y_1^2y_2y_3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determinant of specially structured block matrix How do you compute the determinant of the following $nd \times nd$ block matrix?
$$M = \begin{bmatrix}A+B & A & A & \dots & A & A\\ A & A+B & A & \dots & A & A\\ A & A & A+B & \dots & A & A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ A & A & A & \dots & A+B & A\\ A & A & A & \dots & A & A+B\end{bmatrix}$$
where $A$ and $B$ are $d \times d$ matrices.
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Let us assume that $B$ is invertible. Write
$$\begin{array}{rl} M &= \begin{bmatrix} A + B & A & \ldots & A\\ A & A + B & \ldots & A\\ \vdots & \vdots & \ddots & \vdots\\A & A & \ldots & A + B \end{bmatrix}\\\\ &= \begin{bmatrix} B & O_d & \ldots & O_d\\ O_d & B & \ldots & O_d\\ \vdots & \vdots & \ddots & \vdots\\ O_d & O_d & \ldots & B \end{bmatrix} + \begin{bmatrix} A \\ A\\ \vdots \\ A\end{bmatrix} \begin{bmatrix} I_d \\ I_d\\ \vdots \\ I_d\end{bmatrix}^T\\\\ &= (I_n \otimes B) + (1_n \otimes A) (1_n \otimes I_d)^T\\\\ &= (I_n \otimes B) \left(I_{nd} + (I_n \otimes B^{-1}) (1_n \otimes A) (1_n \otimes I_d)^T\right)\end{array}$$
Using the Weinstein-Aronszajn determinant identity,
$$\begin{array}{rl} \det (M) &= \det\left((I_n \otimes B) \left(I_{nd} + (I_n \otimes B^{-1}) (1_n \otimes A) (1_n \otimes I_d)^T\right)\right)\\\\ &= \det(I_n \otimes B) \cdot \det \left( I_{nd} + (I_n \otimes B^{-1}) (1_n \otimes A) (1_n \otimes I_d)^T\right)\\\\ &= \det(I_n \otimes B) \cdot \det \left( I_{d} + (1_n \otimes I_d)^T (I_n \otimes B^{-1}) (1_n \otimes A) \right)\\\\ &= \left(\det(B)\right)^n \cdot \det \left( I_{d} + n B^{-1} A \right)\\\\ &= \det(B^n) \cdot \det \left( I_{d} + n B^{-1} A \right)\\\\ &= \det \left( B^n + n B^{n-1} A \right)\end{array}$$
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{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for $n$ and $m$ integers: $ 3^mn \mid \sum\limits_{k=0}^{m} {\binom{3m}{3k}}(3n-1)^k$ If $m$ and $n$ are positive integers, with $m$ odd, then prove that:
$$3^mn \mid \sum\limits_{k=0}^m \binom{3m}{3k} (3n-1)^k$$
Proving divisibility by $3n$, we look at $\sum\limits_{k=0}^m (-1)^k\binom{3m}{3k}$. My idea is to substitute cube roots of unity in binomial expansion of $(1+x)^{3m}$. But, how do I get the additional divisibility by $3^{m-1}$?
Perhaps we can use some variation of $\displaystyle \binom{3m}{3k} \equiv \binom{m}{k} \pmod {3^{2+2\operatorname{ord}_3 m}}$, or induction on $m$ henceforth. ($\operatorname{ord_{3}}n$ is the heighest power of $3$ in $n$).
Edit: I add my approach (induction on $m$)
We have the identity $$(1+x)^{3m}+(1+\omega x)^{3m}+(1+\omega^2 x)^{3m} = 3\sum\limits_{k=1}^{m}\binom{3m}{3k}x^{3k}$$
Let, $\omega_1 = 1+x$, $\omega_2 = 1+\omega x$ and $\omega_3 = 1+\omega^2 x$, then $\omega_i$ are the roots of $P(t) = t^3 - 3t^2 + 3t -(1+x^3)$.
Denoting, $S_k = \sum\limits_{j=1}^{3}\omega_j^k$, then it satisfies the recurssion: $S_{k+3} - 3S_{k+2} + 3S_{k+1}-(1+x^3)S_k = 0, \forall \, x \in \mathbb{N}$
Setting $x = \sqrt[3]{3n-1} \implies x^3+1 = 3n$.
The recurssion becomes $S_{k+3} = 3S_{k+2} - 3S_{k+1} + 3nS_k$.
A small calculation shows $S_{k+7}= 63nS_{k+2} + 9(n^2-3n+3)S_{k+1} + 27n(2n+1)S_k$.
Where, $S_{2+1} = 9n$, so it follows from induaction that $3^{2k+2}n \mid S_{6k+3}$, i.e. $3^{m+1}n \mid S_{3m}$, where $m=2k+1$.
I am also very interested in a solution that uses a combinatorial approach to the problem(if possible), any ideas with computing the residues $\binom{3m}{3k} \pmod{3^m}$ to solve the problem, or any other approach. Thank you.
|
Let us denote
$$A_m = \sum_{k=0}^m \binom{3m}{3k}(3n-1)^k.$$
Writing $z = \sqrt[3]{3n-1}$ and $\rho = e^{2\pi i/3}$, we find
$$3A_m = (1+z)^{3m} + (1+\rho z)^{3m} + (1+ \rho^2 z)^{3m}.$$
The sequence given by
$$3u_k = (1+z)^k + (1+\rho z)^k + (1+\rho^2 z)^k$$
satisfies the linear recurrence
$$u_{k+3} = 3 u_{k+2} - 3 u_{k+1} + 3n u_k\tag{1}$$
since
$$\bigl(X - (1+z)\bigr)\bigl(X-(1+\rho z)\bigr)\bigl(X-(1+\rho^2 z)\bigr) = (X-1)^3 - z^3 = X^3 - 3X^2 + 3X - 3n.$$
We have $u_0 = u_1 = u_2 = 1$ and want to prove
$$3^{2r+1}n \mid u_{6r+3}.$$
It is probably easiest to replace $n$ with an indeterminate $Y$ and consider
$$u_k(Y) = \frac{1}{3}\left((1+\sqrt[3]{3Y-1})^k + (1+\rho\sqrt[3]{3Y-1})^k + (1+\rho^2\sqrt[3]{3Y-1})^k\right)$$
with the recurrence
$$u_{k+3}(Y) = 3 u_{k+2}(Y) - 3 u_{k+1}(Y) + 3Y u_k(Y).$$
An easy induction then shows that $u_k(Y) \equiv 0 \pmod{Y}$ if and only if $k\equiv 3\pmod{6}$, and that all coefficients of $u_k(Y)$ are divisible by $3^{\lfloor k/3\rfloor}$.
|
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|
Integrating $(3x + 1) / x^\frac{1}{2}$ I am trying to integrate the following equation, but my answer is different from the textbook and I cannot see where I am going wrong:
\begin{align} \int_1^2\frac{ 3x + 1}{x^{1/2}}dx
&= \int_1^2 \frac{3x}{x^{1/2}}dx + \int_1^2 \frac{1}{x^{1/2}}dx \\
&= \int_1^2 3x^{1/2}dx + \int_1^2 x^{-1/2}dx\\
&= 2x^{3/2}\vert_1^2 + 2x^{1/2}\vert_1^2
\end{align}
The book has an answer $6\sqrt{2} - 4$, but according to what I have just done the answer is roughly $7.07 - 4 = 3.07$.
|
$$\begin{align} \int_1^2 3x^{1/2} \,dx + \int_1^2 x^{-1/2}\,dx & = \frac{3x^{3/2}}{{\frac 32}} + \frac{x^{1/2}}{1/2}\Big|_1^2\\ \\
& = 2 x^{3/2} + 2x^{1/2}\Big|_1^2 \\ \\
& = 4\sqrt 2 + 2\sqrt 2 - (2 + 2)\\ \\ & = 6\sqrt 2 - 4\end{align}$$
|
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|
Product of Gamma functions II What is the value of the product of Gamma functions
\begin{align}
\prod_{k=1}^{10} \Gamma\left(\frac{k}{10}\right)
\end{align}
and can it be shown that
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) \approx \frac{\pi^{9}}{54}
\end{align}
and
\begin{align}
\prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501}\right) \pi^{18}.
\end{align}
|
Since @robjohn has done an excellent job at providing an answer in detail I will add my comments as an additional solution. The results presented here follow the numbering in robjohn's work.
In robjohn's equation (12) the factor
\begin{align}
\frac{9! \ \pi^{9}}{2 \cdot 5^{10}}
\end{align}
has been provided. Numerically this can be seen as $.018579456\cdots \ \pi^{9}$. By comparing this to that of $1/54 = .0185185\cdots$ one can make a good approximation by
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{\pi^{9}}{54}. \tag{14}
\end{align}
Another possible value is $20/1077 = .0185701021\cdots$ which leads to the statement
\begin{align}
\prod_{k=1}^{20} \Gamma\left(\frac{k}{10}\right) = \frac{9! \ \pi^{9}}{2 \cdot 5^{10}} \approx \frac{20 \ \pi^{9}}{1077}. \tag{15}
\end{align}
From equation (13) the factor is
\begin{align}
\frac{2^{18} \ (10)!(20)!(30)!}{10^{62}} = 6.138858832\cdots = 6 + .138858832\cdots .
\end{align}
Since $625/4501 = .1388580315\cdots$ then to a fair approximation it can be stated
\begin{align}
\prod_{k=1}^{40} \Gamma\left(\frac{k}{10}\right) \approx \left( 6 + \frac{625}{4501} \right) \pi^{18}. \tag{16}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$
My approach :
$= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$
Now how to solve this I am not getting any clue on this please help thanks.
|
Break $n^2$ into two parts: $\underbrace{n(n-1)} + \underbrace{{}\quad n\quad {}}$.
The first part appears in the second derivative of $x^n$ and the second part in the first derivative:
\begin{align}
\sum_{n=1}^\infty n^2 x^n & = \sum_{n=1}^\infty x^2\Big(n(n-1) x^{n-2} \Big) + x\Big(nx^{n-1}\Big) \\[10pt]
& = \left(x^2 \frac{d^2}{dx^2} \sum_{n=1}^\infty x^n\right) + \left( x\frac{d}{dx}\sum_{n=1}^\infty x^n \right) \\[10pt]
& = \left( x^2 \frac{d^2}{dx^2} \frac{x}{1-x}\right) + \left( x\frac{d}{dx} \frac{x}{1-x} \right) \\[10pt]
& = \cdots\cdots
\end{align}
(Finally at the end, put $x=1/6$.)
|
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|
Is this fourier even? $$
f(x) =
\begin{cases}
\frac{\pi}{4}-\frac{x}{2} & [0,\pi] \\
-\frac{3\pi}{4}+\frac{x}{2}, & (\pi,2\pi)
\end{cases}
$$
Is it right to compute only $a_n \text{ and } a_0$ coefficient for fourier series because $f(x)$ is even for fourier? How can I proove it since $f(x)!=f(-x)$
|
For $2\pi$-periodic function $f(x)$ defined as:
$$f(x) =
\begin{cases}
\frac{\pi}{4}-\frac{x}{2} & x\in [0,\pi] \\
-\frac{3\pi}{4}+\frac{x}{2}, & x\in(\pi,2\pi)
\end{cases}$$
We can calculate
$$f(2\pi-x) =
\begin{cases}
\frac{\pi}{4}-\frac{2\pi-x}{2} & x\in [\pi,2\pi] \\
-\frac{3\pi}{4}+\frac{2\pi-x}{2}, & x\in (0,\pi)
\end{cases}$$
$$ =
\begin{cases}
\frac{-3\pi}{4}+\frac{x}{2} & x\in [\pi,2\pi] \\
+\frac{1\pi}{4}-\frac{x}{2}, & x\in (0,\pi)
\end{cases}$$
Thus we proved
$$f(2\pi-x)=f(x)\text{ (1)}$$.
Since for $2\pi$-periodic function $f(-x)$ we have
$$f(-x)= f((-x)+2\pi)=f(2\pi-x)\text{ (2)}$$
Combining (1) and (2), we proved
$$f(x)= f(-x)$$.
Thus the coefficients $b_n$ that go with $\sin(2nx)$ are zero.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How would I factor $a^3+b^3+c^3-6abc$ How would I factor the polynomial $a^3+b^3+c^3-6abc$? The values are homogenous, so so must be the factors. I don't know where to go from there.
|
$$a^3+b^3+c^3-6abc= (a+b)^3-3ab(a+b)+c^3-6abc
\\(a+b+c)^3-3(a+b)c(a+b+c) -3ab(a+b)-6abc
\\=(a+b+c)^3 -3(a+b)c(a+b+c)-3ab(a+b)-(3+3)abc
\\=(a+b+c)^3-3c((a+b)(a+b+c)) -3ab(a+b+c) $$
now factor $$ (a+b+c) $$ and go on
$$(a+b+c)((a+b+c)^2 -3c(a+b) -3ab) )
\\=(a+b+c)(a^2+b^2+c^2+2ab+2ac+2bc-3ab-3bc-3ac)
\\=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
$$
|
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|
Prove this inequality: If $|x+3|< 0.5$, show that $|4x+13| < 3$ If $|x + 3| < 0.5$, show that $|4x + 13| < 3$
This is what I've got so far:
$|4x + 13| = |(x + 3) + (3x + 10)|$
by the Triangle Inequality:
$|(x + 3) + (3x + 10)| \le |x + 3| + |3x + 10|$
Now I continue to apply the Triangle Inequality to reach:
$|(x + 3) + (3x + 10)| \le |x + 3| + |x + 3| + |x + 3| + |x + 3| + |1|$
So I come to the conclusion: $|4x + 13| \le 4|x + 3| + 1$
Since $|x + 3| < 0.5$, then $4|x + 3| + 1 < 4\cdot 0.5 + 1 = 3$
Then $|4x + 13| < 3$
Please take a look and let me know if there are errors, if so, enlighten me.
Thank you :)
|
You can also remove the absolute value and work with a two-sided equation. It's a little simpler, as it doesn't involve the triangle inequality.
$$-0.5<x+3<0.5$$
$$-2<4(x+3)<2$$
$$-2<4x+12<2$$
$$-1<4x+13<3$$
Therefore:
$$|4x+13|<3$$
|
{
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|
Find the coefficients $a, b, c$ and $d$ so that the curve shown in the accompanying figure is the graph of the equation.
Find the coefficients $a, b, c$ and $d$ so that the curve shown in the accompanying figure is the graph of the equation $y = ax^3 + bx^2 + cx + d$.
I have no clue how to solve this. This looks like nothing from any example in my book. How do I solve this? Where do I even start?
|
$\begin{vmatrix}
x^3&x^2&x&1&y\\
0^3&0^2&0&1&10\\
1^3&1^2&1&1&7\\
3^3&3^2&3&1&-11\\
4^3&4^2&4&1&-14\\
\end{vmatrix}=
\begin{vmatrix}
x^3&x^2&x&1&y\\
0&0&0&1&10\\
1&1&1&1&7\\
27&9&3&1&-11\\
64&16&4&1&-14\\
\end{vmatrix}=
-72x^3+432x^2-144x-720+72y=
x^3-6x^2+2x+10-y=0$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Is there a simpler/better proof of this simple trigonometric property? The sine function has the following nice property : for any
$x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of
$x,y,x+y$ is $0$ modulo $2\pi$.
I sketch below my current proof of it, which I find somewhat
unsatisfying. Does anyone know a better proof ?
Let $\xi=\sin(x)$ and $\eta=\sin(y)$. Then
$\sin(x+y)=\xi\cos(y)+\eta\cos(x)$, so
$$
\begin{array}{lcl}
\sin^2(x+y)&=&\xi^2(1-\eta^2)+\eta^2(1-\xi^2)+2\xi\eta\cos(x)\cos(y) \\
&=& \xi^2+\eta^2-2\eta^2\xi^2+2\xi\eta\cos(x)\cos(y)
\end{array} \tag{1}$$
Whence
$$
\big(\sin^2(x+y)-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2=
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \tag{2}
$$
If $\sin(x)+\sin(y)=\sin(x+y)$, it follows then that $A(\xi,\eta)=0$
where
$$
\begin{array}{lcl}
A(\xi,\eta)&=&\big((\xi+\eta)^2-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta-2\eta^2\xi^2\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta\big(1-\eta\xi\big)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-\eta\xi\big)^2-(1-\xi^2)(1-\eta^2) \Bigg) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-2\eta\xi+\eta^2\xi^2\big)-\big(1-\xi^2-\eta^2+\eta^2\xi^2\big) \Bigg) \\
&=& 4(\xi\eta)^2 \big(\xi-\eta\big)^2 \\
\end{array} \tag{3}$$
So one of $\xi,\eta,\xi-\eta$ must be zero. A little more case analysis then shows
that in the end one of $x,y,x+y$ must be
zero modulo $2\pi$ as wished.
|
We have $\DeclareMathOperator{\Ima}{Im}$
$$\begin{align}
\sin (x+y) - \sin x - \sin y &= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy}\right)\\
&= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy} + 1\right)\\
&= \Ima \left(\left(e^{ix}-1\right)\left(e^{iy}-1\right)\right)\\
&= \Ima \left(e^{i(x+y)/2}\left(e^{ix/2} - e^{-ix/2}\right)\left(e^{iy/2} - e^{-iy/2}\right)\right)\\
&= \Ima \left(e^{i(x+y)/2} \left(2i\sin \tfrac{x}{2}\right)\left(2i\sin\tfrac{y}{2}\right) \right)\\
&= - 4 \sin \frac{x+y}{2}\sin \frac{x}{2} \sin \frac{y}{2}.
\end{align}$$
That product is $0$ if and only if at least one of the factors is $0$.
|
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|
How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c=3$, show that :$$f(a,b,c)=\dfrac{a}{a+bc}+\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\ge\dfrac{3}{2}$$
solution can see:inequality
I found this three answer all is not true,
1、such as dear @Macavity, in fact
$$\sum_{cyc}\dfrac{a^2}{a^2+abc}\ge\dfrac{16}{\sum_{cyc}(a^2+abc)}$$
the Right not
$\dfrac{16}{4+\sum_{cyc}abc}$
2、and the @ante.ceperic is also not true.in fact
$$a^2+b^2+c^2+d^2+abc+bcd+cda+dab\le 8$$ is not true with $a+b+c+d=4$
such let $a=3$
|
Let's denote the left-hand expression by L.
Then, by Cauchy-Schwarz:
$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)]L \geq (a + b + c + d)^2 = 16$
Now, we have to show
$[a(a + bc) + b(b + cd) + c(c + da) + d(d+ab)] \leq 8$ and we are over.
Let's write it down like:
$A(a, b, c, d) = a^2 + b^2 + c^2 + d^2 + ac(b+d) + bd(a+c) \leq 8$
Notice that we get equality for $a = c, b = d$. I'll try this tactic:
if I prove that $A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d)$ and $A(a, b, c, d) \leq A(a, \frac{b + d}{2}, c, \frac{b + d}{2})$ (these two statements are analogous, so I'll just try to show the first) we will have this chain:
$A(a, b, c, d) \leq A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) \leq A(\frac{a + c}{2}, \frac{b + d}{2}, \frac{a+c}{2}, \frac{b + d}{2}) = 8$.
Let's evaluate:
$A(\frac{a + c}{2}, b, \frac{a+c}{2}, d) - A(a, b, c, d) =
\\ = 2(\frac{a + c}{2})^2 - a^2 - c^2 +(b+d)[(\frac{a + c}{2})^2 - ac]
\\ = -2(\frac{a - c}{2})^2 + [4 - (a+c)](\frac{a - c}{2})^2
\\ = (\frac{a - c}{2})^2[2 - (a+c)]$
This works just if $a + c \leq 2$. If it's not true, $b + d \leq 2$ is, so we can start with the other substitution, and make at least one step in our chain. So, we can assume $a + c \leq 2$ and now we deal with:
$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd \leq 8$
where $2a + b + d = 4, a \leq 1$.
In the next few calculations, I'm just using AM-GM on $bd$ and $b + d = 4 - 2a$.
$2a^2 + b^2 + d^2 + a^2(b + d) + 2abd
\\ = 2a^2 + (b+d)^2 + a^2(b + d) + 2(a-1)bd
\\ = 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)bd
\\ \leq 2a^2 + 4(2-a)^2 + 2a^2(2 - a) + 2(a - 1)(2-a)^2$
After tidying this up a little bit, it is easy to see that for $a = 0$ and $a = 1$ upper expression equals 8. Differentiate it twice to show that it's convex, and we are finished.
|
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|
Equation $1+x^8y^4+x^4y^8-x^2y^4-x^6y^6-x^4y^2=0$ How to prove that the following equation: $$1+x^8y^4+x^4y^8-x^2y^4-x^6y^6-x^4y^2=0$$
has for solution(in real numbers): $|x|=|y|=1~$ only.
Any hint would be appreciated.
|
Note that the proposed equation is equivalent to
$$
(1-x^2y^4)^2+(1-x^4y^2)^2+x^4y^4(x^2-y^2)^2=0
$$
So, any solution satisfies $x^2y^4=x^4y^2=1$ and $xy(x^2-y^2)=0$. This implies clearly that $x^2=y^2=1$.$\qquad\square$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$a,b \in \mathbb N $ , $b$ odd $\implies$ $ \dfrac{2a^2-1}{b^2+2} \notin \mathbb Z $? If $a,b$ are positive integers and $b$ is odd , then is it ever possible that $ \dfrac{2a^2-1}{b^2+2} $ is an integer ?
|
Since $b$ is odd, $b^2 + 2\equiv 3\pmod{8}$. Thus $2$ is not a quadratic residue mod $(b^2 + 2)$. Hence $2a^2\not\equiv 1\pmod{b^2 + 2}$, from which the result follows.
|
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|
How do I evaluate the integral $\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$? I have no idea how to start, it looks like integration by parts won't work.
$$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$
If someone could shed some light on this I'd be very thankful.
|
I try to tackle the problem by differentiation under integral sign.
Using the famous result:
$$
\int_0^{\infty} \frac{\cos a x}{x^2+t^2} dx=\frac{\pi}{2 t} e^{-a t}
$$
where $a,t>0.$
we get the result of the integral
$$
\int_0^{\infty} \frac{\cos a x}{1+t x^2} d x=\frac{\pi}{2} \frac{1}{\sqrt{t} e^{\frac{a}{\sqrt t}}}
$$
Differentiating both sides w.r.t $t$ twice at $t=1$ yields
$$
\int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x=\frac{\pi\left(3-5 a+a^2\right)}{16 e^a}
$$
Differentiating both sides w.r.t $a$ once yields
$$
\frac{\partial}{\partial a} \int_0^{\infty} \frac{x^4 \cos (a x)}{\left(1+x^2\right)^3} d x =\frac{\pi}{16} \frac{\partial}{\partial a}\left(\frac{3-5 a+a^2}{e^a}\right)
$$
Putting $a=1$ yields
$$
-\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{16}\left(-\frac{2}{e}\right)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{x^5 \sin x}{\left(1+x^2\right)^3}=\frac{\pi}{8e}}$$
|
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|
Find $zw, \frac{z}{w},\frac{1}{z}$ for $ z=2\sqrt{3}-2i, w=-1+i$ Find $zw, \frac{z}{w},\frac{1}{z}$ for $ z=2\sqrt{3}-2i, w=-1+i$
I went wrong somewhere, this is what I have so far (this is in polar):
$z=4\left(\cos\left(\frac{11\pi}{6}\right)+\sin\left(\frac{11\pi}{6}\right)\right) $
$w=\sqrt2\left(\cos\left(\frac{7\pi}{4}\right)+\sin\left(\frac{7\pi}{4}\right)\right) $
Now my setup should be:
$zw=4\sqrt2\left(\cos\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)+\sin\left(\frac{11\pi}{6}+\frac{7\pi}{4}\right)\right) $
The common denominator is $12$ so
$zw=4\sqrt2\left(\cos\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)+\sin\left(\frac{22\pi}{12}+\frac{21\pi}{12}\right)\right) $
which then should equal out to
$zw=4\sqrt2\left(\cos\left(\frac{43\pi}{12}\right)+\sin\left(\frac{43\pi}{12}\right)\right) $
The answer in the book says:
$zw=4\sqrt2\left(\cos\left(\frac{7\pi}{12}\right)+\sin\left(\frac{7\pi}{12}\right)\right) $
Where did I go wrong?
I haven't even tried the other problems yet.
|
First, notice that the argument of $w$ is $\frac{3 \pi}{4}$, not $\frac{7\pi}{4}$. And you forgot to put the "$i$"'s together with the sines. Just a little distraction. The other argument, and the absolute values are ok. You had setup everything else correctly. Using the right value above, we get: $$\begin{align}zw &=4\sqrt2\left(\cos\left(\frac{11\pi}{6}+\frac{3\pi}{4}\right)+i\sin\left(\frac{11\pi}{6}+\frac{3\pi}{4}\right)\right) \\ &= 4\sqrt2\left(\cos\left(\frac{22\pi + 9\pi}{12}\right)+i\sin\left(\frac{22\pi + 9\pi}{12}\right)\right) \\ &= 4\sqrt2\left(\cos\left(\frac{31\pi}{12}\right)+i\sin\left(\frac{31\pi}{12}\right)\right)\end{align}$$
Normally, we would stop here, but, we can always reduce the argument so it is between $0$ and $2\pi$, and use that $\sin$ and $\cos$ have both period $2\pi$. Notice that: $$\frac{31\pi}{12} = \frac{7\pi}{12} + \frac{24\pi}{12} = \frac{7\pi}{12} + 2\pi$$
This way, we obtain: $$zw = 4\sqrt2\left(\cos\left(\frac{7\pi}{12}\right)+i\sin\left(\frac{7\pi}{12}\right)\right)$$
as desired. Ok?
|
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|
Derivative of $\frac{1}{2s}-\frac{5}{4s^{3}}$
$r=\dfrac{1}{2s}-\dfrac{5}{4s^3}$
$r=\dfrac{1}{2}s^{-1}-\dfrac{5}{4}s^{-3}$
$r^{\prime}=-\dfrac{1}{2}s^{-2}-\dfrac{5}{4}(-3)s^{-4}$
$r^{\prime}=-\dfrac{1}{2s^{2}}+\dfrac{15}{4s^{4}}$
Is this correct? Because I solved it again a different way by using the quotient rule and I got a different answer. Why wouldn't you use the quotient rule here if there are fractions? I found a common denominator and then used the quotient rule and got a different answer. Thanks.
|
Finding a common denominator and using the quotient rule we have
$$r=\frac{1}{2s}-\frac{5}{4s^3}$$
$$r=\frac{2s^2-5}{4s^3}$$
$$r'=\frac{4s^3\cdot4s-(2s^2-5)\cdot12s^2}{16s^6}=\frac{16s^4-(24s^4-60s^2)}{16s^6}=\frac{60s^2-8s^4}{16s^6}$$
$$r'=-\frac{1}{2s^2}+\frac{15}{4s^4}$$
As you calculated.
|
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|
Fibonacci Calculation using a larger matrix So the formula to generate the fibonacci sequence in matrix form is:
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
So, is there a way to generalize it to a larger matrix, a 4x4 for example?
|
$$
\begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n & 0 & 0 \\
F_n & F_{n-1} & 0 & 0 \\
0 & 0 & F_{2n+1} & F_{2n} \\
0 & 0 & F_{2n} & F_{2n-1} \\
\end{pmatrix}
$$
The point is you're asking for something very arbitrary
|
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|
Express $\log_5 288$ in terms of decimal logarithms $\log 2$ and $\log 3$ Assuming $a=\log 2$ and $b=\log 3$ (log is the base 10 logarithm). I have to find $\log_5 288$. How can I do this?
I've tried transforming $\log2$ to $\frac{\log_5 2}{\log_5 10}$ and same for $b$. Then it's $$\frac{5\log_5 2+2\log_5 3}{\log_5 10}=\frac{\log_5 288}{\log_5 10}=5\log2+2\log5=5a+3b$$ Is that correct?
|
\begin{align}
\log_5 288 & = \frac{\log_{10} 288}{\log_{10} 5} = \frac{\log_{10}(2\cdot2\cdot2\cdot2\cdot2\cdot3\cdot3)}{\log_{10} (10/2)} \\[15pt]
& = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 + }{\log_{10} 10 - \log_{10} 2} \\[15pt]
& = \frac{\log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}2 + \log_{10}3 + \log_{10}3 + }{1 - \log_{10} 2}
\end{align}
|
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|
Two diophantine equations with lots of unknowns Is it possible (tractable) to determine if the following system of equations has any nontrivial solutions (ie, none of the unknowns are zero) in the domain of integers?
$$A^2 + B^2=C^2 D^2$$
$$2 C^4 + 2 D^4 = E^2 + F^2$$
|
for the second one, take $C > D > 0,$ then
$$ E = C^2 - D^2, \; \; \; F = C^2 + D^2 $$
If you wanted a system, take any $C,D \equiv 1 \pmod 4$ distinct primes, such as $5,13.$ We get the Pythagorean triple $16^2 + 63^2 = 65^2 = 5^2 13^2.$ Then $2 \cdot 5^4 + 2 \cdot 13^4 = (13^2 - 5^2)^2 + (13^2 + 5^2)^2 = 144^2 + 194^2.$
|
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|
Generating function satisfying a second degree equation I got this problem in an exercise list:
Let $G(x)$ be the generating function of the numeric sequence $(C_n; n \geq 0)$ satisfying the recurrence equation:
$$C_n = \sum_{k=0}^{n-1}C_kC_{n-k-1}, C_0=C_1=1.$$
Show that $xG(x)^2 - G(x)+1=0$ and conclude that
$$G(x)=\frac{1-\sqrt{1-4x}}{2x}.$$
Then, show that
$$G(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n.$$
What I tried to do:
Since $G(x)=\sum_{n=0}^{\infty}C_nx^n$,we have that
$$xG(x)^2 - G(x)+1=x\left(\sum_{n=0}^{\infty}C_nx^n\right)^2-\sum_{n=0}^{\infty}C_nx^n+1$$
$$=\left(\sum_{n=0}^{\infty}C_nx^n\right)\left(\sum_{n=0}^{\infty}C_nx^{n+1}-1\right)+1$$
$$=x\left(\sum_{n=0}^{\infty}C_nx^n\right)\left(\sum_{n=1}^{\infty}C_nx^n\right)+1$$
But I really don't know how to conclude that this thing is 0.
Besides that, let's assume just for a second that I was able to prove that $xG(x)^2 - G(x)+1=0$.
From the equation above, we get that
$$G(x)=\frac{1+\sqrt{1-4x}}{2x}$$
or
$$G(x)=\frac{1-\sqrt{1-4x}}{2x}$$
How do I know which is the right one?
Now, let's pretend again that I fully understood the problem until this point, and that I know why $G(x)=\dfrac{1-\sqrt{1-4x}}{2x}$, how can I conclude that $G(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n?$
Hints, solutions, everything here is welcome... Thanks!
|
You can tell that the minus sign is correct, since you know that $G(x)$ is a power series, and if you used the plus sign, the numerator would start $1+1+...$, so that $G(x)$ would have a term $\frac{1}{x}$. To see that $G(x) = \sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n$, note that
$$(1+x)^{1/2} = \sum_0^\infty \dbinom{1/2}{k}x^k.$$
Now, $\dbinom{1/2}{k} = (-1)^k2^{1-2k}\frac{1}{k}\dbinom{2k-2}{k-1}$ for $k>0$, as you can see by just expanding the series for the left-hand side and simplifying. Thus
$$(1+x)^{1/2} = \dbinom{1/2}{0} + \sum_{k=1}^\infty \dbinom{1/2}{k}x^k
= 1 + \sum_{k=1}^\infty (-1)^{k-1}2^{1-2k}\frac{1}{k}\dbinom{2k-2}{k-1}x^k.$$
Now replace $x$ by $-4x$ to get
$$(1-4x)^{1/2}
= 1 + \sum_{k=1}^\infty (-1)^{k-1}2^{1-2k}\frac{1}{k}\dbinom{2k-2}{k-1}(-1)^k2^{2k}x^k
= 1 + \sum_{k=1}^\infty (-2)\frac{1}{k}\dbinom{2k-2}{k-1}x^k.$$
Thus
$$(1-4x)^{1/2}-1 = \sum_{k=1}^\infty (-2)\frac{1}{k}\dbinom{2k-2}{k-1}x^k
= \sum_{k=0}^\infty (-1)\frac{1}{k+1}\dbinom{2k}{k}x^k.$$
Now multiply through by $-\frac{1}{2}$ to get
$$\frac{1-\sqrt{1-4x}}{2} = \sum_{k=0}^\infty \frac{1}{k+1}\dbinom{2k}{k}x^k.$$
|
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|
Orthogonal Projection of $v$ on sub-space $U$
What is the orthogonal projection of $v = (1,5,-10)$ on the sub-space $U = Sp\{(5,-2,1),(1,2,-1)\}$?
Well, I managed to compute it on the way I've been taught by the book, but it doesn't seem to work. Any chance you guys can solve it? If needed, I can provide the way I tried to solve it.
|
Hint:
1) Compute an orthonormal base $u_1,u_2$ of $U$ using Gramm-SChmidt
2) Consider the projector $p_U(x) = \sum_{i=1}^2 <u_i,x>u_i$
3) Compute $p_U(v)$
So let's do it:
Note that $$\langle \begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}, \begin{pmatrix}5\\ -2 \\ 1 \end{pmatrix}\rangle = 0 \quad \text{ and }\quad \|\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}\|_2 = \sqrt{6}, \ \|\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\|_2 = \sqrt{30}.$$
So we may choose
$$ u_1 = \frac{1}{\sqrt{30}}\begin{pmatrix} 5\\ -2 \\ 1 \end{pmatrix}\quad \text{ and }\quad u_2=\frac{1}{\sqrt{6}}\begin{pmatrix} 1\\ 2 \\ -1 \end{pmatrix}$$
and thus $p_U(v)$, the projection of $v$ on $U$ is given by
$$p_U(v) = \langle u_1,v\rangle u_1 +\langle u_2,v\rangle u_2= \frac{-15}{\sqrt{30}}u_1+\frac{1}{\sqrt{6}}u_2 =
\begin{pmatrix}\frac{-75}{30}+\frac{1}{6} \\ \frac{30}{30}+\frac{2}{6} \\ \frac{-15}{30}-\frac{1}{6}\end{pmatrix}=\frac{1}{3}\begin{pmatrix}-7 \\ 4\\ -2\end{pmatrix}.$$
|
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|
Convergence N'th Harmonic number minus the Natural Logarithm of N. I was hoping if someone could show me the proof of exactly why this converges to the Euler–Mascheroni constant.
|
The sequence,
$$1+\frac{1}{2}+\cdots +\frac{1}{n}- \ln n $$ converges.
Consider the series
$$\sum\limits_{n=1}^{\infty} \frac{1}{n}-\ln\left( 1+\frac{1}{n} \right).$$
We show that this series converges. We use the inequality,
$$\ln (1+x) < x \qquad \text{for} -1<x <\infty.$$
First, all of its terms are positive, since
$$\ln\left( 1+\frac{1}{n} \right) <\frac{1}{n}.$$
We now make a comparison with the telescoping series, $\sum\limits_{n=1}^{\infty} \frac{1}{n(n+1)}$.
We have,
\begin{equation*}
\begin{split}
-\ln\left( 1+\frac{1}{n} \right)&=\ln \left( \frac{n}{n+1} \right)\\
&=\ln \left( 1-\frac{1}{n+1} \right)\\
&<-\frac{1}{n+1}\\
\end{split}
\end{equation*}
So,
$$\frac{1}{n}-\ln\left( 1+\frac{1}{n} \right) < \frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}.$$
Therefore the series $$\sum\limits_{n=1}^{\infty} \frac{1}{n}-\ln\left( 1+\frac{1}{n} \right)$$
converges.
The $n$th partial sum of this series is
$$1+\frac{1}{2}+\cdots +\frac{1}{n}- \ln (n+1) $$
and this differs from our original sequence by a term of $\ln (n+1) -\ln n= \ln\left( 1+\frac{1}{n} \right)$ but this goes to zero as $n$ goes to infinity.
|
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|
How find this integral $I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$ Find this integral
$$I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$$
since
$$(ax+by)^2=a^2x^2+b^2y^2+2abxy$$
so
$$I=I_{1}+I_{2}+I_{3}$$
where
$$I_{1}=\iiint_{x^2+y^2+z^2\le 1}ax^2\,dV=a\iiint_{x^2+y^2+z^2\le 1}x^2\,dV$$
since
$$\iiint_{x^2+y^2+z^2\le 1}x^2\,dV=\iiint_{x^2+y^2+z^2\le 1}y^2\,dV$$
maybe have other methods? Thank you
|
Expanding out, your integral is
$$a^2 \iiint_{x^2+y^2+z^2\le 1}x^2\,dV - 2ab \iiint_{x^2+y^2+z^2\le 1}xy\,dV + b^2 \iiint_{x^2+y^2+z^2\le 1}y^2\,dV$$
The middle term is zero since for fixed $y$ and $z$ the integrand is odd in $x$ and the domain of integration is a symmetric interval in $x$. Furthermore, by symmetry the first and third integrals are the same, so your answer will be
$$ (a^2 + b^2) \iiint_{x^2+y^2+z^2\le 1}x^2\,dV $$
The integrand is constant in $y$ and $z$, so if we integrate in those variables first, we introduce a factor given by the area of the $yz$-cross section for fixed $x$, namely $\pi(1 - x^2)$ (This is because the cross section is the circle $y^2 + z^2 = 1 - x^2$, of radius $\sqrt{1 - x^2}$).
So the answer is
$$(a^2 + b^2) \int_{-1}^1 \pi(1 - x^2)x^2\,dx$$
$$= \pi(a^2 + b^2)\int_{-1}^1 (x^2 - x^4)\,dx$$
$$= \pi(a^2 + b^2) \bigg({x^3 \over 3} - {x^5 \over 5}\bigg)\bigg|_{x = -1}^{x=1}$$
$$= {4 \over 15}\pi(a^2 + b^2)$$
|
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|
How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
|
$\Large \frac{2^{n+1}}{n+1}=\frac{4+2^{n}}{3}$
this equation tells us that $n+1>0 \Rightarrow n>-1 $((i)we can't divide by zero (ii) RHS is +ve so LHS must also be +ve)
$\Large \frac{2^{n+1}}{n+1}=\frac{8+2^{n+1}}{6}=\frac{8}{6}+\frac{2^{n+1}}{6}$(multiply and divide RHS by $2$)
$\Large \frac{2^{n+1}}{n+1}-\frac{2^{n+1}}{6}=\frac{8}{6}$(difference of two terms is +ve)
this equation tells us that
$\Large \frac{2^{n+1}}{n+1} > \frac{2^{n+1}}{6}\Rightarrow n+1<6 \Rightarrow n<5$
So the integer values to check for the answer are $0,1,2,3,4$
|
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|
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
y^2=z+1-y\\
z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$ $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..
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No solutions exist for real, distinct $(x,y,z)$.
Writing $f(x)=x^2+x-1$, the existence of such a solution would mean that $f$ has a real point of least period 3; that is, $f^3(x)=x$ for some real $x$ with $f(x)\neq1$ i.e. $x\neq \pm 1$. Sarkovsky's theorem then implies that $f$ has points of arbitrary least period. In particular, there would exist $x\in\mathbb{R}-\{-1,1\}$ such that $f^2(x)=x$. But $$f^2(x) - x = (x^2+x-1)^2+(x^2+x-1)-1-x = x^4+2x^3-2x-1 = (x-1)(x+1)^3.$$ and so there are no points of least period 2. Hence $f$ has no real points of least period 3.
P.S.: I'm actually only using the simplest case of Sarkovsky's theorem, namely that 3-cycles imply period 2-cycles. But the full theorem implies that the same conclusion follows for every least period other than one (that is, $f$ has fixed points but not cycles.) This means that we would be no no better off if we increased the number of variables and equations beyond 3: we would still find no nontrivial real solutions. (Though this can be proven much more simply by the same RMS-GM inequality stated by cirpis in his answer...)
|
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|
If $a + b + c = 0$ prove that If $a + b + c = 0$, prove that
1)$$
\sum_{\text{cyc}}{\frac{4bc - a^2}{bc + 2a^2}} = 3
$$
2)$$
\prod_{\text{cyc}}{\frac{4bc - a^2}{bc + 2a^2}} = 1
$$
There is a solution that uses two cubic equations. First is of form $x^3+px-q=0$ that has the roots $a,b$ and $c$. After he is forming another cubic equation with the roots the components of the sum of the product. I can't figure out how they've come to the equation
$y^3-3y^2-3\frac{5p^3-27q^2}{4p^3+27q^2}y-1=0$
|
From $\displaystyle x^3+px-q=0, abc=q, ab+bc+ca=p$
and $\displaystyle a^3+pa-q=0\ \ \ \ (1)\iff a^3=q-pa\ \ \ \ (2)$
Assuming $q\ne0,$
$\displaystyle y=\frac{4bc-a^2}{bc+2a^2}=\frac{4abc-a^3}{abc+2a^3}$
$\displaystyle y=\frac{4q-a^3}{q+2a^3}$
Express $a^3$ in terms of $y$ and compare the value of $a^3$ with that of $(2)$ to express $a$ in terms of $y$
Replace the value of $a$ in $(1)$
|
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|
How to express a trigonometic equation in $\sin 2\theta $ and $\cos 2\theta $? How do I express the given equation in $\sin 2\theta $ and $\cos 2\theta $ in terms of x?
$x + 3 = 7\sin \theta $ with $\frac{\pi }{2}{\text{ < }}\theta {\text{ < }}\pi $
for $\sin 2\theta $ i got $\frac{{( - 2x + 6)\sqrt { - {x^2} - 6x + 40} }}{7}$
for $\cos 2\theta $ i got $\frac{{ - 12x - 49}}{{49}}$
i think im missing something, can someone double check
|
HINT: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$
$\cos^2(\theta) + \sin^2(\theta) = 1$
$\frac{\pi}{2} < \theta < \pi$
$\cos(\theta) < 0$
|
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|
How to calculate $\int_{-\infty}^\infty\frac{x^2+2x}{x^4+x^2+1}dx$? I want to calculate the following integral: $$I:=\displaystyle\int_{-\infty}^\infty\underbrace{\frac{x^2+2x}{x^4+x^2+1}}_{=:f(x)}dx$$ Of course, I could try to determine $\int f(x)\;dx$ in terms of integration by parts. However, I don't think that's the way one should do this. So, what's the trick to calculate $I$?
|
Notice $$\frac{x^2+2x}{\color{blue}{x^4+x^2+1}}
= \frac{x(x+2)}{\color{blue}{(x^2+1)^2-x^2}} =
\frac12\left(\frac{x+2}{x^2-x+1} - \frac{x+2}{x^2+x+1}\right)\\
= \frac12\left(\frac{(x-\frac12)+\frac52}{(x-\frac12)^2+\frac34}
- \frac{(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34}
\right)
$$
Plug this into original integral will split it into two pieces.
Change variables to $y = x \mp \frac12$ for the two new integrals.
After throwing away terms that will get cancel out due to symmetry, i.e. the $y$ term in the numerators, we get
$$\int_{-\infty}^\infty \frac{x^2+2x}{x^4+x^2+1}
=\frac12\left(\frac52-\frac32\right)\int_{-\infty}^{\infty} \frac{dy}{y^2+\frac34} =
\frac12\frac{\pi}{\sqrt{\frac34}} = \frac{\pi}{\sqrt{3}}$$
|
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|
Line not intersecting circle, maximum value of expression involving radius If line $y+x=2$ do not intersect any member of circles $x^2 + y^2 -ax = 0$ at two distinct points where a is parameter, then maximum value of $|a + 4|$.
My try:
Since the line does not intersect the circle at 2 distinct points therefore the distance of line from center of circle is greater than or equal to the radius of the circle
Circle is $x^2 + y^2 -ax = 0$ therefore its radius is $a$ and center is $(a, 0)$
Distance of line $x^2 + y^2 -ax = 0$ from center of circle is given by $\frac{|a - 2|}{\sqrt2}$
Therefore we get the inequality $$\frac{|a - 2|}{\sqrt2} \ge a$$
But I can't find a way to solve it further to obtain the maximum value of |a + 4|
|
Use substitution to solve the system of equations:
\begin{align*}
\begin{cases}
x^2 + y^2 - ax = 0 \\
y + x = 2
\end{cases}
&\implies x^2 + (2 - x)^2 - ax = 0 \\
&\implies x^2 + (4 - 4x + x^2) - ax = 0 \\
&\implies 2x^2 + (-a - 4)x + 4 = 0 \\
\end{align*}
But since there is at most one intersection point, we know that this equation must have at most one solution so that the discriminant of the LHS is either negative or zero:
\begin{align*}
(-a - 4)^2 - 4(2)(4) &\leq 0 \\
(-(a + 4))^2 - 32 &\leq 0 \\
(a + 4)^2 &\leq 32 \\
\sqrt{(a + 4)^2} &\leq \sqrt{32} \\
|a + 4| &\leq 4\sqrt{2}
\end{align*}
|
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|
How to solve a system of three nonlinear equation in a simple way Given the system:
$$
\begin{cases}
x^2y^2+x^2z^2=axyz & \\
y^2z^2+y^2x^2=bxyz &\\
z^2x^2+z^2y^2=cxyz
\end{cases}
$$
The solution could be gotten in a very tedious way. Is it possible to solve it considering some symmetry property of the system if any? Thanks in advance.
|
If is clear we cannot have some $a,b,c$ positive while the rest negative. Otherwise,
$xyz = 0$ and we only have trivial solution.
For simplicity, I will assume $a, b, c > 0$.
If any one of $x,y,z$ vanishes, say $x = 0$, it is clear the equations implies $y z = 0$. This means if one of $x,y,z$ vanish, at least two of $x,y,z$ vanish. If two of $x,y,z$ vanish, then the third one is arbitrary.
To eliminate this sort of trivial solutions, we will first assume $x, y, z > 0$.
If we divide everything by $x^2 y^2 z^2$, we have
$$\begin{cases}
\frac{1}{z^2} + \frac{1}{y^2} &= \frac{a}{xyz}\\
\frac{1}{x^2} + \frac{1}{z^2} &= \frac{b}{xyz}\\
\frac{1}{y^2} + \frac{1}{x^2} &= \frac{c}{xyz}
\end{cases}
\quad\iff\quad
\begin{cases}
\frac{1}{x^2} &= \frac{b+c-a}{2xyz}\\
\frac{1}{y^2} &= \frac{a+c-b}{2xyz}\\
\frac{1}{z^2} &= \frac{a+b-c}{2xyz}
\end{cases}
\tag{*1}
$$
This implies
$$\begin{align}
\frac{1}{(xyz)^2} &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)\frac{1}{(xyz)^3}\\
\iff\quad\quad xyz &= \left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)
\end{align}
$$
Substitute this back into $(*1)$ immediately give us
$$\begin{cases}
x &= x_0 \stackrel{def}{=} \sqrt{\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}\\
y &= y_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+b-c}{2}\right)}\\
z &= z_0 \stackrel{def}{=} \sqrt{\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)}
\end{cases}\tag{*2}$$
What happens if some or all of the $x,y,z$ are negative? If one look back the original
set of equations, the LHS is positive. Under the assume that $a, b, c > 0$, we find $xyz$ is positive. This means in general, the set of non-trivial solutions of the original
equation is given by that in $(*2)$ with an even number of $x,y,z$ flipped sign. i.e.
$$(x,y,z) = (x_0,y_0,z_0), (x_0,-y_0,-z_0), (-x_0,y_0,-z_0) \text{ or } (-x_0,-y_0,z_0)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Square of $7+\sum_{k=1}^n6\times10^k$ If we build a number as follow:
$$N=7+\sum_{k=1}^n6\times10^k$$
we find:
$$N^2=9+\sum_{k=1}^n8\times10^k+\sum_{j=n+1}^{2n+1
}4\times10^{j}$$
that means for example:
$67^2=4489$,
$667^2=444889$,
$6667^2=44448889$
and so on. How can we prove that given an arbitrary $n$ and a number $N$ built as above, the square of $N$ can be obtained by th equation for $N^2$? Thanks
|
We have
$$N=1+6\sum_{k=0}^n10^k=1+6\cdot \frac{10^{n+1}-1}9=\frac{2\cdot 10^{n+1}-1}3.$$
Hence
$$N^2=\frac{4\cdot 10^{2n+2}-4\cdot 2^{n+1}+1}{9}=4\cdot 10^{n+1}\cdot \frac{10^{n+1}-1}{9} +8\cdot \frac{10^{n+1}-1}{9}+1$$
|
{
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|
Where find point $M$ on diagonal of cuboid $ABCDPQRS$ such that the sum $MA+MD$ is minimal? Let a rectangular cuboid of bases $ABCD$ i $PQRS$ such that $AB=1 , BC=2 , AP=3$ . How find on the diagonal $BS$ of the cuboid a point $M$ such that the sum $MA+MD$ is minimal?
|
Let $A = (0,0,0)$, $B = (1,0,0)$, $C = (1,2,0)$, $R = (1,2,3)$, $D = (0,2,0)$, $P = (0,0,3)$, $Q = (1,0,3)$, $S = (0,2,3)$. Point $M$ can be expressed as: $t(1,0,0) + (1-t)(0,2,3) = (t,2-2t,3-3t)$, $t \in [0,1]$. Thus: $MA + MD = \sqrt{(t-0)^2 +(2-2t-0)^2 + (3-3t-0)^2} + \sqrt{(t-0)^2 +(2-2t-2)^2 + (3-3t-0)^2} = \sqrt{t^2 + 13(1-t)^2} + \sqrt{5t^2 + 9(1-t)^2} = f(t)$. We can use calculus to find $f_{min}$. You can continue. Once you find critical value $t_0$ which minimizes $f$, then use this same value $t_0$ to find the point $M$.
|
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|
Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $ Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$
I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a multiple of $6$.
|
This technique is known as the Roots of Unity Filter. See this related question.
Note that $(1+x)^{n} = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}x^k$. Let $\omega = e^{i2\pi/3}$. Then, we have:
$(1+1)^{n} = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}1^k$
$(1+\omega)^{n} = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}\omega^k$
$(1+\omega^2)^{n} = \displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}\omega^{2k}$
Add these three equations together to get $\displaystyle\sum_{k = 0}^{n}\dbinom{n}{k}(1+\omega^k+\omega^{2k}) = 2^n+(1+\omega)^n+(1+\omega^2)^n$
You can see that $1+\omega^k+\omega^{2k} = 3$ if $k$ is a multiple of $3$ and $0$ otherwise.
Thus, $\displaystyle\sum_{m = 0}^{\lfloor n/3 \rfloor}\dbinom{n}{3m} = \dfrac{1}{3}\left[2^n+(1+\omega)^n+(1+\omega^2)^n\right]$
Now, simplify this.
|
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|
Find optimal least square solution to the normal equation What is the optimal solution for $\beta_1$ and $\beta_2$ in the following normal equation:
$$\beta _{ 1 }\sum _{ i=1 }^{ n }{ { x }_{ i } } +\beta _{ 0 }=\sum _{ i=1 }^{ n }{ { y }_{ i } } $$
EDIT
Suppose you are given a set of data $(x_i,y_i)$ with y = $\beta_1x+\beta_0$ and $\beta_1,\beta_0 \in \Bbb R$ are the parameters we want to determine.
A good criteria to find parameter values is to find $\beta_1$ and $\beta_0$ such that the residual sum of squares is minimum. In other words, we find the
values $\beta_0$, $\beta_1$ that minimize:
$$\sum _{ i=1 }^{ n }{ ({ \beta }_{ 1 }{ x }_{ i }+{ \beta }_{ 0 }-y_{ i })^{ 2 } } $$ which I wrote as
$$\left \|\ A\begin{pmatrix} \beta _{ 1 } \\ \beta _{ 0 } \end{pmatrix}-b\ \right \|^2$$
with $A=\begin{pmatrix} { x }_{ 1 } & 1 \\ { x }_{ 2 } & 1 \\ ... & ... \\ { x }_{ n } & 1 \end{pmatrix}$ and $b = \begin{pmatrix} y_{ 1 } \\ { y }_{ 2 } \\ ... \\ y_{ n } \end{pmatrix}$
From there, you can extract the normal equation
$$\beta _{ 1 }\sum _{ i=1 }^{ n }{ { x }_{ i } } +\beta _{ 0 }=\sum _{ i=1 }^{ n }{ { y }_{ i } } $$
Now equation is, how do you find the optimal solution for $\beta_1$ and $\beta_0$?
|
Given a sequence of measurements, $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$, and a trial function
$$
y(x) = \beta_{0} + \beta_{1} x,
$$
the linear system is
$$
\begin{align}
\mathbf{A} \beta &= y \\
\left[
\begin{array}{cc}
1 & x_{1} \\
1 & x_{2} \\
\vdots & \vdots \\
1 & x_{m}
\end{array}
\right]
%
\left[
\begin{array}{c}
\beta_{0} \\
\beta_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{m}
\end{array}
\right].
%
\end{align}
%
$$
Here are the normal equations expressed in terms of column vectors:
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} \beta &= \mathbf{A}^{*} y \\
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array}
\right]
%
\left[
\begin{array}{c}
\beta_{0} \\
\beta_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array}
\right].
%
\end{align}
%
$$
The solution is
$$
\beta
= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} b
= \frac{1}{
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( x \cdot x \right) -
\left( \mathbf{1} \cdot x \right)^{2}}
%
\left[
\begin{array}{rr}
x \cdot x & -\mathbf{1} \cdot x \\
-\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
\end{array}
\right]
%
\left[
\begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array}
\right].
$$
|
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|
If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c^2(a^{n-2})=a^{n-2}(b^2+c^2)=a^{n-2}\times a^2=a^n \implies b^n+c^n<a^n$
But I stuck in part (b). Can I prove it in a way I proved part (a)?
|
Like you did in (a), we can do as fellows: when $0<n<2$
$$a^n=\frac{a^2}{a^{2-n}}=\frac{b^2}{a^{2-n}}+\frac{c^2}{a^{2-n}}<\frac{b^2}{b^{2-n}}+\frac{c^2}{c^{2-n}}=b^n+c^n.$$
|
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|
Find the value of this infinitely nested radical (it appears to obtain multiple values)
Find the value of $$\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\cdots}}}}$$
This is not as simple as it looks for one reason - there are $2$ real solutions to the equation $$x=\sqrt{1-\sqrt{\frac{17}{16}-x}}\implies\begin{cases}x_1=0.5\\x_2\approx 0.073\end{cases}$$
You can see that for yourself on Wolfram Alpha.
How can we know the real value of this infinite radical?
|
Solution technique:
Let
$$L = \sqrt{1 - \sqrt{\frac{17}{6} - \sqrt{1 - \sqrt{\frac{17}{6} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }} } $$
The natural method of solution is to observe that
$$L^2 = 1 - \sqrt{\frac{17}{6} - \sqrt{1 - \sqrt{\frac{17}{6} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }}$$
So...
$$ L^2 - 1= -\sqrt{\frac{17}{16} - \sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }}$$
Therefore
$$ (L^2 - 1)^2 = \frac{17}{16} - \sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }$$
Implying:
$$(L^2 - 1)^2 - \frac{17}{16} = -\sqrt{1 - \sqrt{\frac{17}{16} -\sqrt{1 - \sqrt{\frac{17}{16} -... } } } }$$
And therefore we substitute
$$(L^2 - 1)^2 - \frac{17}{16} = -L$$
Now this equation has 4 possible solutions each expression for $L$ behaves algebraically the same as this nested radical.
The reason is because if you recall from algebra there are two possible $\sqrt{x}$ for any number (a positive and negative root). Now we need to de-nest 2 roots before we can exploit our substitution so what ends up happening is that that are really four possible infinite-radicals (depending on the types of roots we alernated with) that could've yielded this $L$. Your job is to then approximate YOUR problem to a good amount and see which of the 4 solutions to
$$(L^2 - 1)^2 - \frac{17}{16} = -L$$
Corresponds and even then, we may have more than one converging value if our nested radical trends in a way analogous to
$$1 - 1 + 1 - 1 + 1 - 1...$$
Whose sum can be seen as 1 or 0 depending on how one chooses to enumerate its values.
|
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|
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
|
Here is a geometric way of looking at this problem. I will change the angle variable to $t.$ We need to find $\tan t$ if $\sin t + \cos t = 1/2.$ I will use $x = \cos t, y = \sin t$ so we need to solve $$x^2 + y^2 = 1, x + y = 1/2$$ for $x, y.$ that is, to find the two points where the line cuts the unit circle. By eliminating one variable, say $x,$ gives
$$1 = x^2 +(1/2-x)^2= 2x^2 - 1/2x + 1/4$$ which is a quadratic equation $$8x^2 - 4x -3 = 0 $$ in $x.$ Now use the quadratic formulae to find $x,$ then $y = 1/2 -x$ and $\tan t = y/x ={1\over 2x } - 1$ can then be computed. Solving the quadratic equation for
$${1 \over x} = {-4 \pm 2\sqrt{28} \over 6} = {2(-1 \pm \sqrt 7)/3},\
{1 \over 2x} -1= {-4 \pm 2\sqrt{28} \over 6} = {(-4 \pm \sqrt 7) \over 3}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find all $n$ for which $2^n \ge (n+1)^2$
Find all of the elements of $X= \{ n \in \mathbb N: 2^n \ge (n+1)^2\}$
Could someone give me a hint to nudge me in the right direction?
|
$\color{green}{2^0=1\ge(0+1)^2=1}$ ?*
$\color{red}{2^1=2\ge(1+1)^2=4}$ ?
$\color{red}{2^2=4\ge(2+1)^2=9}$ ?
$\color{red}{2^3=8\ge(3+1)^2=16}$ ?
$\color{red}{2^4=16\ge(4+1)^2=25}$ ?
$\color{red}{2^5=32\ge(5+1)^2=36}$ ?
$\color{green}{2^6=64\ge(6+1)^2=49}$ ?
$\color{green}{2^7=128\ge(7+1)^2=64}$ ?
$\color{green}{2^8=256\ge(8+1)^2=91}$ ?
$\color{green}{2^9=512\ge(9+1)^2=100}$ ?
...
*whether $0$ is considered natural or not is a matter of convention.
|
{
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|
How do you use the BBP Formula to calculate the nth digit of π? I know what the Bailey-Borweim-Plouffe Formula (BBP Formula) is—it's $\pi = \sum_{k = 0}^{\infty}\left[ \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6} \right) \right]$— but how exactly do I use it to calculate a given digit of pi?
|
The basic idea depends on the following easy result:
The $d+n$-th digit of a real number $\alpha$ is obtained by computing the $n$-th digit of the fractional part of $b^d \alpha$, in base $b$ . (fractional part denoted by $\lbrace \rbrace$.)
For instance:
if you want to find the $13$-th digit of $\pi$ in base $2$, you must calculate the fractional part of $2^{12} \pi$ in base $2$.
$\lbrace2^{12} \pi\rbrace=0.\color{red} 1\color{blue} {111011}..._2$
hence $13$-th digit of $\pi $ is $\color{red} 1$.
$\pi=11.001001000011\color{red} 1\color{blue} {111011010101}..._2$
Now if we want to compute the $n+1$-th hexadecimal digit of $\pi$
we only need to calculate $\lbrace 16^{n} \pi\rbrace$
you can do this by using $BBP$ formula
$$\pi = \sum_{k = 0}^{\infty} \frac{1}{16^k} \left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6} \right) $$
$$16^{n} \pi= \sum_{k = 0}^{\infty} \left( \frac{4 \cdot 16^{n-k}}{8k + 1} - \frac{2\cdot 16^{n-k}}{8k + 4} - \frac{ 16^{n-k}}{8k + 5} - \frac{16^{n-k}}{8k + 6} \right)$$
and
$$\lbrace 16^{n} \pi\rbrace=\bigg\lbrace\sum_{k = 0}^{\infty} \left( \frac{4 \cdot 16^{n-k}}{8k + 1} - \frac{2\cdot 16^{n-k}}{8k + 4} - \frac{ 16^{n-k}}{8k + 5} - \frac{16^{n-k}}{8k + 6} \right)\bigg \rbrace$$
Now let $S_j=\sum_{k=0}^{\infty} \frac{1}{16^k(8k+1)}$ then
$$\color{blue}{\lbrace 16^{n} \pi\rbrace=\lbrace 4\lbrace 16^{n} S_1\rbrace-2\lbrace 16^{n} S_4\rbrace-\lbrace 16^{n} S_5\rbrace-\lbrace 16^{n} S_6\rbrace \rbrace}$$
using the $S_j$ notation
$$\lbrace 16^{n} S_j\rbrace=\bigg \lbrace \bigg \lbrace\sum_{k=0}^{n}\frac{16^{n-k}}{8k+j} \bigg \rbrace+\sum_{k=n+1}^{\infty}\frac{16^{n-k}}{8k+j}\bigg \rbrace $$
$$=\bigg \lbrace \bigg \lbrace\sum_{k=0}^{n}\frac{16^{n-k} \mod {8k+j} }{8k+j} \bigg \rbrace+\sum_{k=n+1}^{\infty}\frac{16^{n-k}}{8k+j}\bigg \rbrace$$
Now compute $\lbrace 16^{n} S_j\rbrace$ for $j=1,4,5,6$, combine these four result, then discarding integer parts.
The resulting fraction, when expressed in hexadecimal notation, gives the hex digit of $\pi$ in position $n+1$ .
|
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|
Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\right)\right]$$
$$\lim_{x\to-\infty} {x^2 - (x^2 + 2x)\over x - \sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2x\over x - \sqrt{x^2+2x}}$$
divide by highest power of denominator
$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x - ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2\over 1 - \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$
but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.
|
Hint :
Set $x=-y$, then we will have
$$
\lim_{y\to\infty} \left(\sqrt{y^2 - 2y}-y\right)
$$
Now, repeat the process as you have done (multiplying by its conjugate and dividing by the highest power of denominator).
|
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|
Find the smallest positive number $p$ for which the equation $\cos(p\sin x)=\sin(p \cos x)$ has a solution $x\in[0,2\pi].$ Find the smallest positive number $p$ for which the equation
$\cos(p\sin{x})=\sin(p\cos{x})$ has a solution $x$ belonging $[0,2\pi]$.
I am not able to solve this problem.
Please help me.
|
It must be that $p\sin{x}+p\cos{x}=\dfrac{\pi}{2}$ with, $p=\dfrac{\pi}{2(\sin{x}+\cos{x})}$.
So, to minimize $p$, $\sin{x}+\cos{x}$ must be maximized.
$\sin{x}+\cos{x}=\sqrt{2} \sin\left(x+\dfrac{\pi}{4}\right)$, which is maximized when $\sin\left(x+\dfrac{\pi}{4}\right)=1$ at $x=\dfrac{\pi}{4}, \dfrac{7\pi}{4}$.
Hence, $p=\dfrac{\pi}{2\sin\left(\dfrac{\pi}{2}\right)}$.
$p=\dfrac{\pi}{2\sqrt2}$.
Hence, $x=\dfrac{\pi}{4}, \dfrac{7\pi}{4}$ in the interval $[0, 2\pi]$.
|
{
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|
Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
|
Hint:
$$\frac{A}{x+3} + \frac{B}{x+1} = \frac{A(x+1) + B(x+3)}{(x+1)(x+3)} = \frac{(A+B)x + (A+3B)}{x^2+4x+3} = \frac{2x}{x^2+4x+3}$$
What must the values of $A$ and $B$ so that $A+B = 2$ and $A + 3B = 0$?
|
{
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|
Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$ Evaluate Limits
$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
\begin{align*}
\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\stackrel{LHR}{=}\\
&= \lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\right)\right)}{\ln \left(\cos \left(3x\right)\right)}\right)&\\
&=\lim _{x\to \:0}\frac{\left(\ln \left(\cos \left(2x\right)\right)\right)'}{\left(\ln \left(\cos \left(3x\right)\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\sin \left(2x\right)\cos \left(3x\right)}{3\cos \left(2x\right)\sin \left(3x\right)}\right)\\
&\stackrel{LHR}{=}\lim _{x\to \:0}\frac{\left(2\sin \left(2x\right)\cos \left(3x\right)\right)'}{\left(3\cos \left(2x\right)\sin \left(3x\right)\right)'}\\
&=\lim _{x\to \:0}\left(\frac{4\cos \left(2x\right)\cos \left(3x\right)-6\sin \left(3x\right)\sin \left(2x\right)}{9\cos \left(3x\right)\cos \left(2x\right)-6\sin \left(2x\right)\sin \left(3x\right)}\right)\\
&=\lim _{x\to \:0}\left(\frac{2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)}{3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)}\right)\\
&=\frac{\lim _{x\to \:0}\left(2\left(2\cos \left(2x\right)\cos \left(3x\right)-3\sin \left(3x\right)\sin \left(2x\right)\right)\right)}{\lim _{x\to \:0}\left(3\left(3\cos \left(2x\right)\cos \left(3x\right)-2\sin \left(3x\right)\sin \left(2x\right)\right)\right)}=\dfrac{4}{9}
\end{align*}
*
*Could we do it in others ways?
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Once you apply L'Hopital's Rule once, and simplify you get $\displaystyle \lim_{x \to 0}\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}} = \lim_{x \to 0}\dfrac{2\tan 2x}{3 \tan 3x}$.
Now, applying L'Hopital's Rule one more time gives you $\displaystyle\lim_{x \to 0}\dfrac{4\sec^2 2x}{9 \sec^2 3x} = \dfrac{4}{9}$.
Alternatively, $\displaystyle \lim_{x \to 0}\frac{-\frac{2\sin \left(2x\right)}{\cos \left(2x\right)}}{-\frac{3\sin \left(3x\right)}{\cos \left(3x\right)}} = \lim_{x \to 0}\dfrac{2}{3} \cdot \dfrac{\cos 3x}{\cos 2x} \cdot \dfrac{\frac{\sin 2x}{x}}{\frac{\sin 3x}{x}} = \lim_{x \to 0}\dfrac{4}{9} \cdot \dfrac{\cos 3x}{\cos 2x} \cdot \dfrac{\frac{\sin 2x}{2x}}{\frac{\sin 3x}{3x}}$,
which is easily broken up into well known limits.
Note: More often than not, it is better to simplify a limit before applying L'Hopital's Rule.
|
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"url": "https://math.stackexchange.com/questions/883053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
|
Let $k$ be a positive integer and let $x$ be such that $x \in [k-1,k]$. You have
$$
n+k-1 \leq x +n
$$ and
$$
\int_{k-1}^k \frac{1}{x+n} dx \leq \int_{k-1}^k \frac{1}{n+k-1} dx =\frac{1}{n+k-1}
$$ thus, summing from $k=1$ to $n+1$, $n\geq1$, we get
$$
\int_{0}^{n+1} \frac{1}{x+n} dx \leq \sum_{k=1}^{n+1}\frac{1}{n+k-1}=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}
$$ or
$$
\ln \left(2+\frac{1}{n}\right) \leq \frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}
$$
But $$ e^{7/12} \leq 2 < 2+\frac{1}{n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/883670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ I'm looking for an asymptotic equivalent of
$$\sum_{0 < k \le n} \frac{2^k}{k}$$
as $n \to \infty$. A plausible candidate seems to be $\frac{2^{n+1}}{n+1}$ (WolframAlpha plot, and the intuitive similarity with $\sum_{k \le n} 2^k = 2^{n+1}$ is also appealing), but my usual tricks seem powerless here.
I've tried:
*
*Interpreting $x^k/k$ as the primitive of $x^{k-1}$ and setting $x=2$
*Replacing $2^k/k$ with $\int_{x=0}^2 x^{k-1}$
*Reordering the sum's terms to expose log-resembling sub-sums like $\sum_k 1/k$
*Finding lower and upper bounds asymptotically equivalent to $\frac{2^{n+1}}{n+1}$ -- the lower bound is easy ($\sum\limits_{k \le n} \frac{2^k}{k} \ge \sum\limits_{k \le n} \frac{2^k}{n+1} \ge \frac{\sum\limits_{k \le n} 2^k}{n+1} \ge \frac{2^{n+1}}{n+1}$), but the upper bound seems trickier (I couldn't think of a sequence $\varepsilon_k \in o(2^k/k)$ that would make it easy to estimate $\sum_{0 < k \le n} 2^k/k - \varepsilon_k$)
*... and a few others, to no avail
Any hints?
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The Euler-Maclaurin summation formula is useful for approximating sums and often reveals the asymptotic behavior with only a few terms. This problem is an interesting application because the precise asymptotic behavior requires summing an infinite number of terms with Bernoulli numbers as coefficients - the terms that are typically neglected.
Using the Euler-Maclaurin summation formula with $f(x) = 2^x/x$
$$\sum_{k=1}^{n}\frac{2^k}{k} = C+\int_{1}^{n}f(x)\,dx + \frac1{2}f(n)+\frac{B_2}{2!}f'(n) + \frac{B_4}{4!}f'''(n)+ \ldots.$$
The integral is the exponential integral which behaves asymptotically as $Ei(x) \sim e^x/x$ as $x \rightarrow \infty:$
$$\int_{1}^{n}f(x)\,dx=\int_{1}^{n}\frac{2^x}{x}\,dx= Ei(n\log2)-Ei(\log2) \sim \frac{e^{n\log 2}}{n\log 2}.$$
Taking the odd-order derivatives we find a pattern
$$f'(x) = \frac{2^x}{x}\left(\log2-\frac1{x}\right)\sim\frac{2^x}{x}\log2\\
f'''(x) = \frac{2^x}{x}\left[\left(\log2-\frac1{x}\right)^3+O(x^{-2})\right]\sim\frac{2^x}{x}(\log 2)^3\\ \ldots$$
Hence,
$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^n}{n}\left[\frac1{\log 2}+\frac1{2}+\frac1{\log 2}\left(\frac{B_2}{2!}(\log 2)^2 + \frac{B_4}{4!}(\log 2)^4+ \ldots\right)\right]\,\,(*)$$
The trailing terms can be summed exactly. The generating function for the Bernoulli numbers is $x/(e^x-1)$ where
$$\frac{x}{e^x-1} = \sum_{k=0}^{\infty}\frac{B_kx^k}{k!}.$$
The first two Bernoulli numbers are $B_0 = 1$ and $B_1 = -1/2$. They are zero for odd $n \geq 3$.
Hence,
$$\sum_{k=2}^{\infty}\frac{B_k(\log 2)^k}{k!} = \frac{\log 2}{e^{\log 2}-1}-1 + \frac{\log 2}{2}= \log 2-1 + \frac{\log 2}{2}.$$
Substituting into $(*)$ we get
$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^n}{n}\left[\frac1{\log 2}+\frac1{2}+\frac1{\log 2}\left(\log 2-1 + \frac{\log 2}{2}\right)\right]=\frac{2^n}{n}2,$$
and
$$\sum_{k=1}^{n}\frac{2^k}{k} \sim \frac{2^{n+1}}{n}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/888354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
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|
Solve $(1+z)^8=(1-z)^8$ My guess is to write this as $$\left(\frac{1+z}{1-z}\right)^8=1.$$ We can then find 8 possibilities for $\frac{1+z}{1-z}$, namely $\cos(k\pi/4)+i\sin(k\pi/4)$, $k=1,\ldots,8$. For each $k$ we can then deduce 2 equations by putting $z=x+iy$, for example for $k=1$ we get: $$\frac{1+x+iy}{1-x-iy}=\frac12 (1+i).$$ Now we find two equations for $x$ and $y$, by noting both the real and imaginary parts of the equations should be equal, and can thus find $z$.
However, doing this for $k=1,\ldots,8$ seems somewhat cumbersome. Does anyone know of a faster way to find the $z$? Thans in advance.
|
This reduces to:
$$1+8x+28x^2+56x^3+70x^4+56x^5+28x^6+8x^7+x^8=1-8x+28x^2-56x^3+70x^4-56x^5+28x^6-8x^7+x^8$$Which becomes $$16x+112x^3+112x^5+16x^7=0$$
Whence $x=0$ or $$x^6+7x^4+7x^2+1=0$$
Write $y=x^2$ to obtain $$y^3+7y^2+7y+1=0$$
We have the obvious solution $y=-1$ and $$y^2+6y+1=0$$ for the others
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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|
Solve the equation $(x^2-9)+\sqrt{2-x}=0$ Solve the equation $(x^2-9)+\sqrt{2-x}=0$
*
*$(x+3)(x-3)+\sqrt{2-x}=0$
Conditions: $x\neq\pm3 \wedge x\leq2$
*$(x+3)(x-3) = -\sqrt{2-x}$
*$(x+3)^2(x-3)^2 = 2-x$
*$x^4-18x^2+81 = 2-x$
*$x^4-18x^2+x+79 = 0$
I'm positive this isn't going to the right direction.
Please help.
|
Consider making a sign diagram. You will get your answer if you fill in (x+3)(x-3) and ...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/892816",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 0
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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.