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Quadrilateral inscribed in semicircle. How to find BC using sin ABD? AB = 3, BD = 5 , tanABD = 0.75
BC is diameter
Question : How to find BC using sin ABD?
I can find sin cos ABD and lenght AD.
I can find BC using Ptolemy's theorem.
|
Draw the other diagonal $AC$. Since $BC$ is a diameter of the circumcircle of the quad $BADC$, $$\angle\, BAC = \angle \, BDC = 90^{\circ}$$
Let $P$ be the intersection point of the diagonals $AC$ and $BD$. Then triangle $ABP$ is right-angled, so
$$\frac{AP}{AB} = \frac{AP}{3} = \tan(\angle\, ABP) = \tan(\angle\, ABD) = \frac{3}{4}$$ so
$AP = \frac{9}{4}$. By Pythagoras' theorem for $ABP$
$$BP = \sqrt{AB^2 + AP^2} = \sqrt{3^2 + \frac{9^2}{4^2}} = \frac{15}{4}$$ Then $DP = BD - BP = 5 - \frac{15}{4} = \frac{5}{4}$
In the cyclic quad $BADC$ the triangles $ABP$ and $DCP$ are simiplar, right-angled triangles, therefore
$$\frac{DP}{CD} = \frac{5}{4 \, CD} = \tan(\angle \, DCP) = \tan(\angle \, ABP) = \frac{3}{4}$$ so
$$CD = \frac{5}{3}$$ Apply Pythagors' theorem to the right-angled triangle $BCD$
$$BC = \sqrt{BD^2 + CD^2} = \sqrt{5^2 + \frac{5^2}{3^2}} = \frac{5}{3}\sqrt{10}$$
Finally, the triangles $ADP$ and $BCP$ are similar ($BADC$ is cyclic), so
$$\frac{AD}{BC} = \frac{AP}{BP}$$
$${AD} = \frac{AP}{BP}{BC} = \frac{9\cdot 4}{4 \cdot 15} \cdot \frac{5}{3}\sqrt{10} = \sqrt{10}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3606170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find all points order 3 on an elliptic curve Let $P=(x,y)\in E(\mathbb{F}_p)$ by a Weierstrass equation $y^2=x^3+ax+b$. Show that $3P=\mathcal{O}$ iff $3x^4+6ax^2+12bx-a^2=0$.
I derived that every point in $\{P\in E(\mathbb{F}_p)|3P=\mathcal{O}\}$ is a root of the above equation, then no further clue. I also tried to from $y^2=x^3+ax+b$ to yield $3x^4+6ax^2+12bx-a^2=0$, but not sure is the right approach because I can't find a way to do it.
Any help or hints will be great, thanks.
|
$\textit{Proof:}$
First, note that $3P=\infty$ iff $2P=-P$. Then let $P=\mathbb{F}_p$. If $P$ is the finite point $(x,y)$, then $-P=(x,-y)$; iff $P=\infty$, then $2P=3P=\infty$. If $y=0$, then $2P=\infty\neq-P$. Otherwise, we can calculate $2P=(x', y')$ by
$$\lambda=\frac{3x^2+a}{2y}, x'=\lambda-2x, y'=\lambda(x-x')-y$$
So $2P=-P$ iff $x'=x$ and $y'=y$. Now let's consider:
\begin{align*}
3x&=\lambda^2\\
3x&=\frac{(3x^2+a)^2}{4y^2} \\
3x&=\frac{3x^2+a)^2}{4(x^3+ax+b)}\\
12x(x^3+ax+b)&=(3x^2+a)^2\\
12x^4+12ax^2+12bx&=9x^4+6ax^2+a^2\\
3x^4+6ax^2+12bx-a^2&=0
\end{align*}
And vice versa for the other direction of the proof. So in conclusion, $P=(x,y)\in E(\mathbb{F}_p)$ satisfies $3P=\infty$ iff $3x^4+6ax^2+12bx-a^2=0$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3607389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1-c<1.$$
Thus by AM-GM inequality we have $$\frac{(1-a)+(1-b)+(1-c)}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{(1-a)(1-b)(1-c)}\ge 27.$$
Also by AM-GM inequality we have $$\frac{a+b+c}{3}\ge (abc)^\frac{1}{3}\\ \implies \frac{2}{3}\ge (abc)^\frac{1}{3}\\\implies abc\le \frac{8}{27}\\\implies \frac{1}{abc}\ge \frac{27}{8}.$$
This implies that, we have $$\frac{1}{abc(1-a)(1-b)(1-c)}\ge \frac{27^2}{8}\\ \implies \frac{(abc)^2}{abc(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}\\\implies \frac{abc}{(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}.$$
How to proceed after this?
|
Hint: The inequality is:
$$\frac{a}2\cdot\frac{b}2\cdot\frac{c}2 \geqslant (1-a)\cdot(1-b)\cdot(1-c)$$
which follows from Karamata's inequality as $\left(1-a, 1-b, 1-c\right) \succ \left(\dfrac{a}2, \dfrac{b}2, \dfrac{c}2\right)$, and $t\mapsto \log t$ is concave.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}{4}$
Now I know that values of a and b but how do I get x and y?
Note:
The textbook solution says that if we consider a=1 and b=-2 then x and y are the roots of
$$t^2+t-2=0$$
But shouldn't it be
$$t^2-t-2=0$$
from Vieta's formulas?
I would like a clarification for this or any other solutions would be fine too.
|
$$ax^2+bx+c=0$$ so $$x_1=\frac{-b + \sqrt{b^2-4ac}}{2a} \; and \; x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$$
From this we have $x_1x_2=\frac{c}{a}$ and $x_1+x_2=-\frac{b}{a}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of the complex power series $\sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n$ I have to study the convergence of
$$S(z)=\displaystyle \sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n,\quad \quad z\in \mathbb{C}$$
I have defined $c_n=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}$ and I calculated, first of all, the convergence radius $R$ of the series. I have observed that
$$\dfrac{|c_{n+1}|}{|c_n|}=\dfrac{2n+1}{2n+2}\rightarrow 1,\quad n\to \infty,$$
Therefore, $R=1$ and the convergence theorem of the power series ensures that
*
*$S(z)$ converges absolutely (and punctually) for all $z\in \mathbb{D}(0,1)$.
*$S(z)$ converges uniformly on all $\mathbb{D}(0,1)$-compacts.
*$S(z)$ diverge for all $z$ with $|z|>1$.
Now I was trying to see what happens to the $\mathbb{D}(0,1)$-frontier points. Let $z$ with $|z|=1$ and $b_n$ be the general term for $S(z)$. So,
$$|b_n|=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}|z|^n=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}=\dfrac{\frac{(2n)!}{2^n\cdot n!}}{2^n\cdot n!}=\dfrac{(2n)!}{4^n\cdot (n!)^2}\sim \dfrac{\sqrt{4\pi n}}{2\pi n}\to 0,\quad n\to \infty$$
applying Stirling. I don't know how to continue the study of the series on the border. I was trying with $z=e^{i\theta}$ and the criteria of Dirichlet and Abel, but I can't get anywhere.
|
Using
$$
\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}} = \frac{2}{\pi }\int_0^{\pi /2} {\sin ^{2n} tdt}
$$
and summation together with analytic continuation, we obtain
$$
\left( {\frac{1}{{\sqrt {1 - z} }} -1= } \right)\sum\limits_{n =1}^{N - 1} {\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}}z^n } + z^N \frac{2}{\pi }\int_0^{\pi /2} {\frac{{\sin ^{2N} t}}{{1 - z\sin ^2 t}}dt}
$$
for all $z \in \mathbb{C}\setminus \left[ {1, + \infty } \right)$ and $N\geq 1$. Note that for $0<s<1$,
$$
\left| 1 - zs \right|^2 = \left| z \right|^2 s^2 - 2(\Re z)s + 1 \ge \begin{cases} 1 & \text{ if } \;
\Re z \le 0, \\ \frac{(\Im z)^2}{ \left| z \right|^2} & \text{ if } \;
0<\Re z \le \left| z \right|^2, \\ \left|1-z\right|^2 & \text{ if } \; \Re z > \left| z \right|^2.\end{cases}
$$
Consequently, the remainder term in absolute value is at most
$$
\left| {\frac{{1 \cdot 3 \cdots (2N - 1)}}{{2 \cdot 4 \cdots 2N}}z^N } \right| \times \begin{cases} 1 & \text{ if } \;
\Re z \le 0, \\ \left|\frac{z}{\Im z}\right| & \text{ if } \;
0<\Re z \le \left| z \right|^2, \\ \frac{1}{|1-z|} & \text{ if } \; \Re z > \left| z \right|^2.\end{cases}
$$
On the unit circle,
$$
\left| {\frac{{1 \cdot 3 \cdots (2N - 1)}}{{2 \cdot 4 \cdots 2N}}z^N } \right| < \frac{1}{{\sqrt {\pi N} }}
$$
by the known inequality for the central binomials. Hence, on the unit circle
$$
\left| {\sum\limits_{n = N}^\infty {\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}}z^n } } \right| \le \frac{1}{{\sqrt {\pi N} }} \times \begin{cases} 1 & \text{ if } \;
\Re z \le 0,\,|z|=1, \\ \frac{1}{|\Im z|} & \text{ if } \;
0<\Re z \le 1,\,|z|=1.\end{cases}
$$
This shows that the series is convergent at every point of the circle except perhaps at $z=1$. Clearly, it cannot converge at $z=1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $ without using L'Hopital's rule. Let $ n $ be a positive integer greater than $ 1 $.
Find : $$ \lim\limits_{x\to 0}{\frac{1-\sqrt{1+x}\sqrt[3]{1-x}\cdots\sqrt[2n+1]{1-x}}{x}} $$
Without L'Hopital's rule or series expansion.
Here is What I did to solve the problem.
\begin{aligned}\displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}-\prod\limits_{i=2}^{k}{\sqrt[i]{1+\left(-1\right)^{i}x}}}{x}}}\\&=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{1-\sqrt[k]{1+\left(-1\right)^{k}x}}{x}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\lim_{x\to 0}{\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{\sum\limits_{j=0}^{k-1}{\sqrt[k]{1+\left(-1\right)^{k}x}^{j}}}\displaystyle\prod\limits_{i=2}^{k-1}{\sqrt[i]{1+\left(-1\right)^{i}x}}}}\\ &=\displaystyle\sum_{k=2}^{2n+1}{\displaystyle\frac{\left(-1\right)^{k+1}}{k}} \\ &=\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k+1}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}\\ &=-1+\displaystyle\sum_{k=0}^{n}{\displaystyle\frac{1}{2k+1}}+\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{2k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ &=-1+\displaystyle\sum_{k=1}^{2n+1}{\displaystyle\frac{1}{k}}-\displaystyle\sum_{k=1}^{n}{\displaystyle\frac{1}{k}}\\ \displaystyle\lim_{x\to 0}{\displaystyle\frac{1-\prod\limits_{k=2}^{2n+1}{\sqrt[k]{1+\left(-1\right)^{k}x}}}{x}}&=H_{2n+1}-H_{n}-1 \end{aligned}
What's your approach to solve the problem ?
|
By the binomial formula $$(1+(-1)^kx)^{1/k}=1+\frac{(-1)^k}kx+O(x^2),$$ so that
$$\begin{align}\prod_{k=2}^{2n+1}(1+(-1)^kx)^{1/k}
&=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx+o(x)\right)\\
&=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}kx\right)+o(x)\\
&=1+x\sum_{k=2}^{2n+1}\frac{(-1)^k}k+o(x)
\end{align}$$
so that the desired limit is $$-\sum_{k=2}^{2n+1}\frac{(-1)^k}k$$
This differs from your answer. Have I made a mistake?
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
How to solve the following radical equation?
$$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$
I'm looking for an easy way to solve it.
|
If $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$
Then:
$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}}}}}=x$
In general:
$x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\ldots}}}}$
Which means that:
$2-(x^2-2)^2=x$
$x^4-4x^2+4+x-2=0$
$x^4-x^3+x^3-x^2-3x^2+3x-2x+2=0$
$(x-1)(x^3+x^2-3x-2)=0$
$(x-1)(x^3+2x^2-x^2-x-2x-2)=0$
$(x-1)(x+2)(x^2-x-2)=0$
$(x-1)(x+2)\left(x - \dfrac{1 \pm \sqrt{5}}{2}\right)$
So the 'possible' answers for x would be :
$\left(1,-2 ,\phi,\dfrac{1}{\phi} \right)$
All are too small to be the answer of the original question except for $\phi$, the golden ratio. So that's the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3615926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Reduction of ${}_{4}F_{3}(\cdots; 1)$ It is suspected that
$$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{8}{\pi} - \frac{4 \, \sqrt{2} \, \Gamma^2\left(\frac{3}{4}\right)}{\Gamma^3\left(\frac{1}{2}\right)} = 1.020959562466698996....$$
The ${}_{4}F_{3}$ is nicely set to use Clausen's formula,
$${}_{4}F_{3}(a, b, c, d; e, f, g; 1) = \frac{(2 a)_{|d|} (a b)_{|d|} (a + b)_{|d|} }{(a)_{|d|} (b)_{|d|} (2a + 2b)_{|d|} }$$
where $\frac{1}{2} = a + b + c -d$, $e = a + b + \frac{1}{2}$, $a + f = d + 1 = b + g$. The condition that is not satisfied is that $d$ should be a negative integer.
Another attempt made was to use
$$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \, \int_{0}^{1} t^{-1/4} (1-t)^{-1/4} \, {}_{3}F_{2}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}; 1, \frac{3}{2}; t \right) \, dt$$
but switching around the ${}_{3}F_{2}$ seems to just cycle the result and not provide a reduction.
Another method is to use
$$ {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) = 2 \, {}_{4}F_{3}\left(\frac{1}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{1}{2}, \frac{3}{2}; 1 \right) - {}_{4}F_{3}\left(\frac{5}{4}, \frac{1}{4}, \frac{3}{4}, \frac{3}{4}; 1 , \frac{3}{2}, \frac{3}{2}; 1 \right) $$
but doesn't seem to be able to fit a known reduction like Clausen's identity.
Might there be a better way to show this identity is true?
|
Maple says:
convert(hypergeom([1/4,1/4,3/4,3/4],[1,3/2,3/2],1), `2F1`);
$$ \frac{8}{\pi}+\frac{8}{\pi}{\it EllipticK} \left( i \right) -\frac{8}{\pi}{\it EllipticE}(i)
$$
This could also be written as
$$ \frac{8}{\pi} - \frac{8}{\pi} \int_0^1 \frac{t^2\; dt}{\sqrt{1-t^4}}$$
and Maple says that is indeed
$$ \frac{8}{\pi} - \frac{2}{\pi} B(1/2, 3/4) = \frac{8}{\pi} - 4\,{\frac { \left( \Gamma \left( 3/4 \right) \right) ^{2}\sqrt {2}}{{
\pi}^{3/2}}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What value does the area of the region enclosed by $\sum_{k=1}^m (x^k+y^k)=2$ converge to as $m\rightarrow\infty$? Find the area enclosed by the curve
$$\sum_{k=1}^m (x^k+y^k)=2$$
where $m$ is even and $m \rightarrow\infty$.
That is for a sufficiently large even number $m$, we have:
$$x^1+y^1+x^2+y^2+x^3+y^3+\dots+x^m+y^m=2$$
See the graph of the curve ($m=250$): the $250$ here is to illustrate the region, but the required area to be calculated when $m$ is even that is tending to infinity.
WHAT I DID:
I tried to convert the equation of the curve, which has an infinite series, into a function of $x$ where $x>0 ,y>0$.
After very long and difficult steps, I got the area $(A)$ enclosed by that curve as:
$$A=\int_0^{\frac{12}{7}}\frac{7x-12}{4x-7}dx=3-\frac{1}{8}\ln(7)\approx 2.75676 \text{ unit}^2$$.
I. Is my answer right?
II. How can we solve it easily? I used $5$ pages to solve it.
Any help would be really appreciated. THANKS!
|
Doing some obvious stuff.
$\begin{array}\\
2
&=\sum_{k=1}^m (x^k+y^k)\\
&=\sum_{k=1}^m x^k+\sum_{k=1}^m y^k\\
&=\dfrac{x-x^{m+1}}{x-1}+\dfrac{y-y^{m+1}}{y-1}\\
&\to \dfrac{x}{x-1}+\dfrac{y}{y-1}
\qquad\text{as }m \to \infty\\
&= \dfrac{x-1+1}{x-1}+\dfrac{y-1+1}{y-1}\\
&= 1+\dfrac{1}{x-1}+1+\dfrac{1}{y-1}\\
\text{so}\\
0
&= \dfrac{1}{x-1}+\dfrac{1}{y-1}\\
&= \dfrac{(y-1)+(x-1)}{(x-1)(y-1)}\\
&= \dfrac{x+y-2}{(x-1)(y-1)}\\
&\text{just a line}\\
2
&=\dfrac{x-x^{m+1}}{x-1}+\dfrac{y-y^{m+1}}{y-1}\\
&=\dfrac{x(1-x^{m})}{x-1}+\dfrac{y(1-y^{m})}{y-1}\\
&=\dfrac{(y-1)(x-x^{m+1})+(x-1)(y-y^{m+1})}{(x-1)(y-1)}\\
&=\dfrac{(y-1)x-(y-1)x^{m+1}+(x-1)y-(x-1)y^{m+1}}{(x-1)(y-1)}\\
&=\dfrac{2xy-x-y-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
\text{so}\\
0
&=\dfrac{2xy-x-y-(y-1)x^{m+1}-(x-1)y^{m+1}-2(x-1)(y-1)}{(x-1)(y-1)}\\
&=\dfrac{2xy-x-y-2(xy-x-y+1)-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
&=\dfrac{x+y-2-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
\text{or}\\
0
&=x+y-2-(y-1)x^{m+1}-(x-1)y^{m+1}\\
&=(x-1)+(y-1)-(y-1)x^{m+1}-(x-1)y^{m+1}\\
\end{array}
$
Don't know how much this helps,
but here it is.
|
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|
Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
|
Let
*
*$f(x) = x^4 + px^3 + qx^2 + rx + 1$.
*$\Lambda = \{ a, b, c, d \}$ be the set of roots of $f(x)$.
*$\Omega = \left\{ \frac{\pm 1 \pm i}{\sqrt{2}} \right\}$ be the set of roots of $x^4 + 1$.
We have
$$\begin{align}\mathcal{P}\stackrel{def}{=}\prod_{\lambda \in \Lambda}(\lambda^4 + 1)
&= \prod_{\lambda \in \Lambda}\prod_{\omega \in \Omega}(\lambda - \omega)
= \prod_{\omega \in \Omega}\prod_{\lambda _in \Lambda}(\omega - \lambda)
= \prod_{\omega \in \Omega}f(\omega)\\
&= \prod_{\omega \in \Omega}(p \omega^3 + q \omega^2 + r\omega)
\end{align}
$$
Since $\prod_{\omega \in \Omega} \omega^2 = (-1)^2 = 1$ and $\omega^{-1} = \bar{\omega}$ for all $\omega \in \Omega$, we find
$$\begin{align}
\mathcal{P} = & \prod_{\omega \in \Omega}(q + p \omega + r \bar\omega)\\
= & \phantom{+}
\left(q + \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q + \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
& \times
\left(q - \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q - \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
= & \left(q^2 + p^2 + r^2 + \sqrt{2}q(p+r)\right)
\left(q^2 + p^2 + r^2 - \sqrt{2}q(p+r)\right)\\
= & (q^2 + p^2 + r^2)^2 - 2q^2(p+r)^2\\
= & (p^2 + \color{red}{r}^2)^2 + q^4 - 4q^2 pr
\end{align}$$
Please note that there is a typo in the formula you want to prove.
|
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|
Stuck trying to sum this series I actually got this question from a book - TMH. Here is the question:
Prove that $\cot^2 \frac{\pi}{2n+1}\cdot \cot^2 \frac{\pi}{2n+1}\cdot \ldots \cdot \cot^2 \frac{\pi}{2n+1}$ are the roots of the equation
$$
x^n - \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} x^{n-1} +\frac{\binom{2n+1}{5}}{\binom{2n+1}{1}}x^{n-2}-\cdots = 0,
$$and hence show that $\lim\limits_{n\to\infty} \sum_{r=1}^{n} r^{-2}=\frac{\pi^2}{6}$.
I noticed the following, but couldn't solve later.
$$
\cot ^{2} \frac{\pi}{2 n+1}=\tan ^{2} \frac{n \pi}{2 n+1}
$$
|
You may know the following formula that you will find for example here in paragraph "Sine, cosine, tangent of multiple angles"
$$\displaystyle \tan(n\theta )=\frac {\sum _{k{\text{ odd}=2q+1}}(-1)^{q}{n \choose k}(\tan\theta) ^{2q+1}}{\sum _{k{\text{ even}=2q}}(-1)^{q}{n \choose k}(\tan\theta)^{2q} }\tag{1}$$
Let us take $2n+1$ instead of $n$ in (1) :
$$\displaystyle \tan((2n+1)\theta )=\frac {\sum _{q=0}^n(-1)^{q}{2n+1 \choose 2q+1}(\tan\theta) ^{2q+1}}{\sum _{q=0}^{n}(-1)^{q}{2n+1 \choose 2q}(\tan\theta)^{2q} }\tag{2}$$
Therefore, we have
$$\tan((2n+1)\theta)=0 \ \ \iff \ \ \sum _{q=0}^n (-1)^{q}{2n+1 \choose 2q+1}x^q = 0\ \ \text{with} \ \ x:=\tan(\theta)^2\tag{3}$$
Two remarks :
1) The second equation in (3) is exactly your equation (multiplied by the common denominator) ?
2) This equation has been simplified by $x$ assumed $\ne 0$.
The first equation in (3) is equivalent to :
$$(2n+1) \theta=k\dfrac{\pi}{2} \ \ \text{for any integer} \ k \tag{4}$$
Therefore, $\theta$ can take the values :
$$\theta=\dfrac{k}{2n+1}\dfrac{\pi}{2}\tag{5}$$
Plugging (5) into relationship $x=(\tan\theta)^2$, we get
$$x=\tan(\dfrac{k}{2n+1}\dfrac{\pi}{2})^2=(\cot(\dfrac{\pi}{2}-\dfrac{k}{2n+1}\dfrac{\pi}{2}))^2=(\cot(\dfrac{2n+1-k}{2n+1}\dfrac{\pi}{2}))^2$$
as desired.
|
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|
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
|
Without loss of generality, we can assume $a\leq b\leq c$. Let's do some remarks. I will suppose $d> 3$, the first cases are easy to check by hands and they will coincides with the solutions in $a=1$.
If $a,b,c\geq 3$ then $3$ divides $a!+b!+c!$, but $3$ does not divide $2^d$. Hence $a=1,2$. (the case $a=0$ is equal to case $a=1$; in the solutions you can replace $0$ with $1$)
Case $a=2$
We have $2! + b! + c! = 2^d$ that is $1+\frac{b!}{2}+\frac{c!}{2}=2^{d-1}$.
If $b,c\geq 4$ then the LHS is odd and the RHS is even. Then $b$ has to be $2$ or $3$.
If $b=2$ we have $c!=2^{d}-4 = 4(2^{d-2}-1)$. So we need $c\geq 4$ in order to have a factor $2^2$ in the LHS. But now the LHS have a factor $2^3$ in its factorization, and the RHS don't, a contradiction.
If $b=3$ we have $c!=2^d-8=8(2^{d-3}-1)$. As above we need $c\geq 4$ in order to have a factor $2^3$ in the LHS, but if $c\geq 6$ we have a factor $2^4$ in the LHS factorization and the RHS don't. So $c$ can be only $4$ or $5$.
With these considerations, the solutions are:
$$(a,b,c,d) = (2,3,4,5), (2,3,5,7)$$
Case $a=1$
We have $1+b!+c! = 2^d$ that is $b!+c! = 2^d-1$. The RHS is odd, so $b!+c!$ has to be odd. For, we need $b!$ odd and $c!$ even (because $b\leq c$). Hence, the unique case is $b=1$.
So now we have $c! = 2^{d}-2 = 2(2^{d-1}-1)$, and using the same argument used in the case above, we will find that $c$ can be only $2$ or $3$.
With these considerations the solutions are:
$$
(a,b,c,d) = (1,1,2,2), (1,1,3,3)
$$
Edit: Thanks to Gareth Ma for his remark (case $a=1$).
|
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|
Solve the ODE $2y'' - y' = 2\cos(3t)$ using laplace transform
$2y'' - y' = 2\cos(3t)$
\begin{align*}
2\mathcal{L}[y''] - \mathcal{L}[y'] &= 2\mathcal{L}[\cos3t] \\
2\bigg[-y(0) - sy(0) + s^2Y(s) \bigg] + y(0) - sY(s) &= \frac{2s}{s^2+3^2} \\
\end{align*}
This works out to $$Y(s)(s^2-s) = \frac{2s}{s^2+9} + 3 - 2s$$ and $$Y(s) = \frac{2s^3 + 3s^2-16s+27}{(2s^2-s)(s^2+9)} $$
I get the following partial fraction decomposition, but here is where I think I am messing up and getting the wrong coefficients
\begin{align*}
\frac{2s^3 + 3s^2-16s+27}{(2s^2-s)(s^2+9)} &= \frac{A}{s} + \frac{B}{2s-1} + \frac{Cs + D}{s^2+9} \\
2s^3+3s^2-16s + 27 &= 2As^3 + 18As - As^2 - 9A + Bs^3 + 9Bs + 2Cs^3 - Cs^2 + 2Ds^2 - Ds \\
2s^3+3s^2-16s + 27 &= s^3(2A+B+2C) + s^2(-A-C+2D) + s(18A+9B-D) - 9A
\end{align*}
Here is the corresponding system of linear equations
\begin{align*}
2 &= 2A + B + 2C \\
3 &= -A - C + 2D \\
-16 &= 18A + 9B - D \\
27 &= -9A
\end{align*}
I got $A = -3, B = \frac{136}{35}, C = \frac{212}{35}, D = \frac{106}{35}$ which seems way off because the correct answer with coefficients is $y(t) = \frac{-4}{37}\cos3t - \frac{2}{111}\sin(3t) - 4 + \frac{152}{37}e^{t/2}$
I think something might be off even before partial fractions because $A=-3$ would imply that the corresponding laplace transform is $-3 = \frac{-3}{s}$, which doesn't even appear in the answer.
edit: Sorry I forgot the initial conditions! $y(0) = -1, y'(0) = =1$
edit: additional work on partial fractions
\begin{align*}
\frac{4s+2}{(s^2+9)(2s-1)} &=\color{blue}{2(2s+1)} \left ( \frac{A}{s^2+9} + \frac{B}{4s^2-1} \right ) \\
&=-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right) \\
1= 4As^2-A + Bs^2 + 9B \\ \vdots \\
&=-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right)
\end{align*}
I get following system of linear equations to solve for A and B and it isn't clear how the final result is reached...
\begin{align*}
0 &= 4A + B\\
1 &= -A + 9B
\end{align*}
\begin{align*}
A &= -\dfrac 1 {37}\\
B &= \dfrac 4 {37}
\end{align*}
|
You should get:
$$Y(s)(2s^2-s)+y(0)(1-2s))-2y'(0)=\dfrac {2s}{s^2+9}$$
Maybe you are given initial conditions ?
You wrote :
$$Y(s)(\color{red}{2s^2}-s) = \frac{2s}{s^2+9} + 3 - 2s$$
You should keep it this way for simplicity:
$$Y(s) = \frac{2}{(s^2+9)(2s-1)} + \dfrac { 3 - 2s}{s({2s}-1)}$$
It's easier to decompose in simple fractions.
$$Y(s) = \frac{2}{(s^2+9)(2s-1)} - \dfrac { 3}{s} + \dfrac {4}{{2s}-1}$$
$$Y(s) =-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right) - \dfrac { 3}{s} + \dfrac {4}{{2s}-1}$$
$$Y(s) =-\dfrac {2}{37}\left ( \dfrac{(2s+1)}{s^2+9} \right)+ \dfrac{156}{37(2s-1)} - \dfrac { 3}{s} $$
Apply inverse Laplace Transform.
$$\boxed {y(t)=-\dfrac {4}{37} \cos (3t)-\dfrac {2}{111} \sin(3t) -3 +\dfrac {78}{37}e^{t/2}}$$
|
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|
Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove this afterwards, but failed. I would love to see another proof to this Problem.
-A proof-: we know that $\displaystyle\sum_{i=1}^{n}\cos{a_i}= \sum_{k=1}^{n}\cos{(a+(k-1)x)}=\frac{\cos{\frac{a+a_n}{2}}\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}=T(x)\quad(1)$
(Here's the source for $(1)$).
Also, $\displaystyle\sum_{i=1}^{n}a^2_i=\sum_{k=1}^{n}(a+(k-1)x)^2\\=\displaystyle\sum_{k=1}^{n}(a^2+2ax(k-1)+x^2(k-1)^2)\\=\displaystyle\sum_{k=1}^{n}a^2+2ax\sum_{k=1}^{n}(k-1)+x^2\sum_{k=1}^{n}(k-1)^2\\=na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\quad(2)$
We consider the function $f(x)=\frac{x^2}{2}+\cos{x}\implies f''(x)=1-\cos{x}\geq{0}$ therefore $f(x)$ is a concave function. From Jensens inequality for concave functions:$ \displaystyle\sum_{i=1}^{n}f(a_i)\geq{nf\left(\frac{\sum_{i=1}^{n}a_i}{n}\right)}\iff\displaystyle\sum_{i=1}^{n}\left(\frac{a^2_i}{2}+\cos{a_i}\right)\geq n\Big(\frac{1}{2}\left(\frac{\frac{n}{2}(a+a_n)}{n}\right)^2+\cos{\frac{\frac{n}{2}(a+a_n)}{n}}\Big)\iff\frac{1}{2}\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}\cos{a_i}\geq \frac{n}{2}\left(\frac{a+a_n}{2}\right)^2+n\cos{\frac{a+a_n}{2}}\overset{(1),(2)}{\iff}\frac{1}{2}\left(na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\right)+T(x)\geqslant \frac{n\left(2a+(n-1)x\right)^2}{8}+n\cos{\frac{a+a_n}{2}}\overset{\ldots}{\iff} \frac{x^2n(n^2-1)}{3}\geqslant n\cos{\frac{a+a_n}{2}}-T(x)\overset{(1)}{\iff} \frac{x^2n(n^2-1)}{3}\geq \cos{\frac{a+a_n}{2}}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\iff \frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\quad(3) $
$Lemma.$ For every dinstict $x,x\in\mathbb{R}_{\neq kπ}$ there exists at least one value of $a$ such that $\cos{(a+\frac{(n-1)x}{2}})=1\quad(5)$
Proof of the lemma: $(5)\iff (n-1)x=2-2a+4kπ\iff a=2kπ+1-\frac{(n-1)x}{2}$.
Hence, for every $x,x\in\mathbb{R}_{\neq k\pi}$
we have $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$
*
*To me it looks like $n-\frac{x^2n(n^2-1)}{6}$ is a good approximation function of $\frac{\sin{nx}}{\sin{x}}$ for small values of $x$ (or for small values of $x+ z\pi$, $z\in\mathbb{Z}$, it depends on $n$, because the second one is periodic with period multiple of $\pi$).
|
Let $f(n,x)=n-\frac{(n-1)n(n+1)}{6}x^2$ the RHS of the required inequality. Wlog we can ssume $n \ge 3$ as for $n=1,2$ the inequality is trivial ($n=2$ reduces to $2\sin^2(\frac{x}{2}) \le \frac{x^2}{2}$ which follows from $|\sin x| \le |x|$).
Using the well known inequality $|\frac{\sin nx}{\sin x}| \le n$, the required inequality is non-trivial and needs to be proven only for $n-\frac{(n-1)n(n+1)}{6}x^2 \ge -n$ or $x^2 \le \frac{12}{n^2-1}$ which means $x^2 \le \frac{3}{2}$ when $n \ge 3$ so definitely $|x| < \frac{\pi}{2}$ so $\cos x >0$.
Note that $\frac{\sin(n+1)x}{\sin x}-f(n+1,x)=(\frac{\sin(n)x}{\sin x}-f(n,x))\cos x-2\sin^2(\frac{x}{2})f(n,x)-2\sin^2(\frac{nx}{2})+\frac{n(n+1)}{2}x^2$
So assuming by induction $\frac{\sin(n)x}{\sin x}-f(n,x) \ge 0$ (starting at $n=2$ as $2x^2 \ge 3$ hence $\cos x>0$ when $n+1 \ge 3$ while the inequality is trivial otherwise as seen above) and noticing that $-2\sin^2(\frac{nx}{2}) \ge -\frac{n^2x^2}{2}$. while if $f(n,x) \ge 0$ then $f(n,x) \le n$ so in any case $-2\sin^2(\frac{x}{2})f(n,x) \ge -\frac{n}{2}x^2$, we get:
$\frac{\sin(n+1)x}{\sin x}-f(n+1,x) \ge 0$ and we are done!
|
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|
Evaluation of $\int^{1}_{0}\frac{x^4}{1+x^8}dx$
How can I Integrate $\displaystyle \int^{1}_{0}\frac{x^4}{1+x^8}dx$?
I have searched in that forum and get the result for $\displaystyle \int^{\infty}_{-\infty}\frac{x^4}{1+x^8}dx$ or for $\displaystyle \int^{\infty}_{0}\frac{x^4}{1+x^8}dx$
Although i know the formula $\displaystyle \int^{\infty}_{0}\frac{x^{m-1}}{1+x^n}dx=\frac{\pi}{n}\cdot \csc\frac{\pi m}{n}$
But did not understand How can i use in my original Integration
Help me please Thanks
|
Note
$$\int^{1}_{0}\frac{x^4}{1+x^8}dx
=\frac1{2\sqrt2}\int_0^1 \left(\frac{x^2}{x^4-\sqrt2x^2+1}
- \frac{x^2}{x^4+\sqrt2x^2+1}\right)dx $$
where
$$\begin{align}
I(a)& =\int_0^1 \frac{x^2}{x^4+ax^2+1}dx \\
&=\frac12\int_0^1 \frac{x^2+1}{x^4+ax^2+1}dx -
\frac12\int_0^1 \frac{1-x^2}{x^4+ax^2+1}dx \\
&=\left(\frac1{2\sqrt{2+a}}\tan^{-1}\frac{x-\frac1x}{\sqrt{2+a}}
- \frac1{2\sqrt{2-a}}\coth^{-1}\frac{x+\frac1x}{\sqrt{2-a}}
\right)_0^1 \\
&=\frac\pi{4\sqrt{2+a}}
- \frac1{2\sqrt{2-a}}\coth^{-1}\frac{2}{\sqrt{2-a}} \\
\end{align}$$
Thus,
$$\begin{align}
&\int^{1}_{0}\frac{x^4}{1+x^8}dx
=\frac1{2\sqrt2}\left[I(-\sqrt2)-I(\sqrt2)\right]\\
=&\frac{\sqrt2}{8\sqrt{2-\sqrt2}}\left( \frac\pi2
+ \coth^{-1}\frac{2}{\sqrt{2-\sqrt2}} \right)
-\frac{\sqrt2}{8\sqrt{2+\sqrt2}}\left( \frac\pi2+\coth^{-1}\frac{2}{\sqrt{2+\sqrt2}}\right) \\
\end{align}$$
|
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|
Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
|
Note that we can write
$$\begin{align}
\int_{-\infty}^\infty \frac{x^2}{(x^2+a^2)(x^2+b^2)}\,dx&=\frac12\int_{-\infty}^\infty \frac{(x^2+b^2)+(x^2+a^2)}{(x^2+a^2)(x^2+b^2)}\,dx-\frac{1}2\int_{-\infty}^\infty\frac{a^2+b^2}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac12\int_{-\infty}^\infty \frac1{x^2+a^2}\,dx+\frac12\int_{-\infty}^\infty \frac1{x^2+b^2}\,dx\\\\
&-\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\\\\
&=\frac\pi {2a}+\frac\pi {2b} -\frac{a^2+b^2}2\int_{-\infty}^\infty\frac{1}{(x^2+a^2)(x^2+b^2)}\,dx\tag1
\end{align}$$
Now apply Parseval to the integral on the right-hand side of $(1)$ with $f(x)=\frac{1}{x^2+a^2}$ and $g(x)=\frac1{x^2+b^2}$ and $F(k)=\frac{\pi}{|a|}e^{-|a|k}$ and $G(k)=\frac\pi{|b|}e^{-|b|k}$.
|
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|
Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$ Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$
My Attempt :
$$(y-px)^2 (1=p^2)=a^2p^2$$
$$y-px=\frac {ap}{\sqrt {1+p^2}}$$
$$y=px+\frac {ap}{\sqrt {1+p^2}} .....(1)$$
This is Clairaut's Equation so we differentiate both sides with respect to $x$
$$\frac {dy}{dx} = p + x\cdot \frac {dp}{dx} + \frac {\sqrt {1+p^2} \cdot a \cdot \frac {dp}{dx} + ap\cdot \frac {1}{2\sqrt {1+p^2}} \cdot 2p \cdot \frac {dp}{dx}}{(1+p^2)}$$
$$p=p+x+\frac {a\sqrt {1+p^2} + \frac {ap^2}{\sqrt {1+p^2}}}{(1+p^2)} \cdot \frac {dp}{dx} $$
$$(x+\frac { a+2ap^2 }{(1+p^2)^{\frac {3}{2}}}) \cdot \frac {dp}{dx}=0$$
Either,
$$\frac {dp}{dx}=0$$
$$\textrm {so p}=c$$
Using $p=c$ in $(1)$ we get:
$$y=cx+\frac {ac}{\sqrt {1+c^2}}$$ which is the required general solution.
How do I evaluate the singular solution ?
|
Both sign variants of the square roots give solutions, you can include this by using also the solutions where $a$ is replaced by $-a$.
In $y=px+f(p)$ the derived equation is $0=y''(x+f'(y'))$ so that any solution will be composed of segments where one of the factors is zero. You already found the lines $y'=p=c$.
linear solution family for $|a|=2$, blue lines for $a=2$ and red lines for $a=-2$
The other factor gives with
$$
f'(p)=\frac{a}{\sqrt{1+p^2}}-\frac{ap^2}{\sqrt{1+p^2}^3}=\frac{a}{\sqrt{1+p^2}^3}
$$
the equation
$$
\sqrt{1+p^2}=-\sqrt[3]{\frac{a}{x}}\implies p=\pm\sqrt{\left(\frac{a}{x}\right)^{2/3}-1},~~ax<0
$$
which can be inserted into the original equation
\begin{align}
y(x)&=p\left(x-a\sqrt[3]{\frac{x}{a}}\right)
\\
&=\pm\sqrt{\left(\frac{a}{x}\right)^{2/3}-1}\left(x-a^{2/3}x^{1/3}\right)
\end{align}
This last formula does not depend on the sign of $a$. Switching the parts to have a unique formula above and below the $x$ axis results in
$$
y(x)=\pm \left(a^{2/3}-x^{2/3}\right)^{3/2},~~|x|\le|a|.
$$
|
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|
y=x^3sinx,what's the 6th derivative of y at x = pi/6? question 1： $y=x^3\sin x$, what's $y^{\left(6\right)}\left(0\right)$ ?
I have solved it use Taylor's Formula in the follow way.
step1:
$y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$
and x_0 = 0 for the question1, So $y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(0\right)}{n!}\left(x\right)^n$
step2:
$y=x^3\sin x=x^3\left(x-\frac{1}{3!}x^3+\ldots \right)=x^4-\frac{1}{6}x^6+\ldots$
[PS: $\sin x=\displaystyle \sum _{n=0}^{\infty }\:\frac{\left(-1\right)^n}{\left(2n+1\right)!}x^{2n+1}$]
step3:
$\dfrac{y^{\left(6\right)}\left(0\right)}{6!}\:=\:-\dfrac{1}{6}. \quad $
So $y^{\left(6\right)}\left(0\right)=-120$
The above method is so useful. But for question2 I find that I can't use the method above.Is there a way to use Taylor's formula to solve this problem, such as substitution? I know leibniz formula can solve it but I want to know if Taylor's formula can be made better in this case?
Question2: $y=x^3\sin x$, what's $y^{\left(6\right)}\left(\dfrac{\pi }{6}\right)$?
|
Thanks for @InterstellarProbe's helping me solve this problem.
$y=x^3sinx$
step1:
Let : $x=u+\frac{\pi }{6}$
then
$y=\sum _{n=0}^{\infty }\:\frac{y^{\left(n\right)}\left(0\right)}{n!}u^n$
and
$y=\left(u+\frac{\pi }{6}\right)^3sin\left(u+\frac{\pi }{6}\right)
=\left(u^3+\frac{\pi }{2}u^2+\frac{\pi^2 }{12}u+\frac{\pi \:^3}{6^3}\right)\left(sin\left(u\right)cos\left(\frac{\pi }{6}\right)+cos\left(u\right)sin\left(\frac{\pi }{6}\right)\right)
=\left(u^3+\frac{\pi }{2}u^2+\frac{\pi^2 }{12}u+\frac{\pi ^3}{6^3}\right)\left(\frac{\sqrt{3}}{2}\left(u-\frac{1}{3!}u^3+\frac{1}{5!}u^5-\frac{1}{7!}u^7+...\right)+\frac{1}{2}\left(1-\frac{1}{2!}u^2+\frac{1}{4!}u^4-\frac{1}{6!}u^6+...\right)\right)\:$
step2:
$\frac{y^{\left(6\right)}\left(0\right)}{6!}u^6=u^3\cdot \left(\frac{\sqrt{3}}{2}\cdot \left(-\frac{1}{6}\right)u^3\right)+\frac{\pi }{6}u^2\cdot \left(\frac{1}{2}\cdot \frac{1}{4!}u^4\right)+\frac{\pi^2 }{12}u\cdot \left(\frac{\sqrt{3}}{2}\cdot \frac{1}{5!}u^5\right)+\frac{\pi ^3}{6^3}\cdot \left((\frac{1}{2})(-\frac{1}{6!})u^6\right)$
So
$\frac{y^{\left(6\right)}\left(0\right)}{6!}=-\frac{\sqrt{3}}{12}+\frac{\pi }{4\cdot 4!}+\frac{\sqrt{3}\pi }{24\cdot 5!}-\frac{\pi ^3}{2*6!\cdot 6^3}$
and the anwser is
$y^{\left(6\right)}\left(0\right)=\left(-\frac{\sqrt{3}}{12}+\frac{\pi \:}{4\cdot \:4!}+\frac{\sqrt{3}\pi^2 \:}{24\cdot \:5!}-\frac{\pi \:^3}{2*6!\cdot \:6^3}\right)\cdot 6!$
|
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|
Simplifying $(1+\cos2x)^2+(-2\sin2x)^2=1+2\cos2x+\cos^22x+4\sin^22x$ I'm supposed to find the arc length of this polar curve : $r=1+\cos2x$ where $0\leq x \leq \pi/4$.
I know that I have to use this formula $s=\displaystyle\int_{0}^{\pi/4} \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2}
dx$ My teacher always tells us to simplify the expression under the square root before taking the integral. But I'm stuck.
$(1+\cos2x)^2+(-2\sin2x)^2=1+2\cos2x+\cos^22x+4\sin^22x$
I have tried to use half-angle formulas and everything, but the expression does not get easier. Does someone have any tips?
|
For $r(x) = 1+\cos(2x)$, you have that your expression reads
$$ \sqrt{r^2+\bigg(\frac{dr}{dx}\bigg)^2} = \cos(x) \sqrt{10 - 6 \cos(2 x)} = \cos(x) \sqrt{4 + 12 \sin(x)^2} $$
The integral now can be done by using $y=\sin(x)$ and $dy = \cos(x) dx$. You should obtain
$$
s = \sqrt{5}/2 + \tanh^{-1} \left( \sqrt{3/5} \right) /\sqrt{3}
$$
Edit: to simplify use the "double-angle formulae"
$(1+\cos2x)^2+(-2\sin2x)^2 = 1 +2 \cos 2x+ \cos^2 2x + 4 \sin^2 2x \\
= 2 + 2 \cos 2x+ 3 \sin^2 2x \\
= 4 \cos^2 x + 3 \sin^2 2x \\
= 4 \cos^2 x + 12 \sin^2 x \cos^2 x$.
This sequence of passages is not unique, other intuitions are possible.
|
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|
Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$ Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$
My Attempt:
The given equation is
$$(D^2+4)y=x\sin^2 x$$
It's auxiliary equation is
$$m^2+4=0$$
$$m^2=-4$$
$$m=\pm 2i$$
$$\textrm {Complementary Function (C.F)}=c_1 \cos (2x) + c_2 \sin (2x)$$
Now, the particular integral is given by
$$\textrm {P.I.}=\frac {x\sin^2 x}{D^2+4}$$
$$=\frac {x}{D^2+4}\cdot \frac {1-\cos (2x)}{2}$$
$$=\frac {x}{2(D^2+4)} - \frac {x\cdot \cos (2x)}{2(D^2+4)}$$
$$=\frac {1}{2} \cdot \frac {1}{4} \cdot (1+\frac {D^2}{4})^{-1} \cdot x - \frac {1}{2}(x\cdot \frac {\cos (2x)}{D^2+4} - \frac {2D \cos (2x)}{(D^2+4)^2})$$
$$=\frac {x}{8} - \frac {1}{2} (x\cdot \frac {x \cos (2x)}{2D} - \frac {2D \cos (2x)}{(D^2+4)^2} )$$
How do I solve further?
|
Starting from here:
$$y_p=\frac {x\sin^2 x}{D^2+4}$$
$$2y_p=\frac {x(1-\cos (2 x))}{D^2+4}$$
$$2y_p=\frac {x }{D^2+4}-\frac {x\cos (2 x)}{D^2+4}$$
$$2y_p=\dfrac x 4-(x-\dfrac {2D}{D^2+4})\frac {\cos(2 x)}{D^2+4}$$
More simply:
$$2y_p=\dfrac x 4-x\frac {\cos(2 x)}{D^2+4}$$
$$2y_p=\dfrac x 4-\Re \{x\frac {e^{2 ix}}{D^2+4}\}$$
$$2y_p=\dfrac x 4-\Re \{xe^{2 ix}\frac {1}{D(D+4i)}\}$$
$$2y_p=\dfrac x 4-\Re \{ \dfrac {xe^{2 ix}}{4i}({\frac 1D -\dfrac 1{D+4i}})\}$$
$$2y_p=\dfrac x 4-\Re \{\dfrac {xe^{2 ix}}{4i}({x -\dfrac 1{4i}})\}$$
Now apply Euler's formula to get the final result:
$$\boxed {y_p=\dfrac x 8-\frac {x^2}{16}\sin (2x)-\dfrac {x}{32} \cos(2x)}$$
|
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|
Computing the index of fixed point for the system of equations $x'=x^2$ and $y'=-y$. Consider the system of equations $$x' = x^2$$ $$y'=-y.$$ Compute the index of fixed point at the origin both visually and computationally, or just write down the integral.
$\textbf{Solution:}$ To find the fixed point, integrate both equations $$\int f'(x) = \int x^2 dx \text{ and } \int f'(y) = \int -y dy.$$ Therefore, $f(x) = \frac{x^3}{3} + C \text{ and } f(y) = -\frac{y^2}{2} + C.$ Next, to find the fixed point index, let $x=0, \text{ and } y=0$ so $f(0) = C.$ Therefore, $f(x) = \frac{x^3}{3} + 0$ and $f(y) =-\frac{y^2}{2} + 0.$ So the fixed point of $f(x) = 0$ and $f(y) = 0.$
So the index for the first function is $I(f(x),x_0) = (\frac{x^3}{3}, 0)$ and for our second function it is $I(f(y),y_0) = (-\frac{y^2}{2},0).$ Can someone please help me complete the problem?
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To get fixed points, we need $x'=0$ and $y'=0$. So, the fixed points are $x^2 = 0$ and $-y=0$ so $x=0$ and $y=0$. Thus, $(0,0)$ is a fixed point.
Let $f=x^2$ and $g=-y$, then $f_x = 2x, f_y=0, g_x = 0,$ and $g_y=-1.$ Therefore the Jacobian is $$J = \begin{pmatrix} f_x & g_x \\ f_y & g_y \end{pmatrix}$$ $$J_{(0,0)} = \begin{pmatrix} 2x & 0 \\ 0 & -1 \end{pmatrix}_{(0,0)} = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}.$$
The eigenvalues at $(0,0)$ are $|J-I\lambda| = 0.$ $$\begin{pmatrix} 0-\lambda & 0 \\ 0 & -1-\lambda \end{pmatrix} = 0 \text{ so } \lambda = 0, -1.$$ The system is marginally stable if it has one or more distinct poles on the imaginary axis and any remaining poles have negative real part.
Now, taking the integrals, $x' = x^2, \frac{dx}{x^2} = dt$ $$\implies \int x^{-2}dx = t + C$$ $$\implies -\frac{1}{x} = t + C.$$ Now, $y' = -y , \frac{dy}{dt} = -y \implies \frac{dy}{y} = \int -dt = \log y = -t + \log C$. So, $\frac{y}{C} = e^{-t}$ and $y= Ce^{-t}.$
|
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|
MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$
After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??
|
Using $\sum $ to denote cyclic sums,
\begin{align}
\sum \sqrt{a^2-ab+b^2} \;+ 9\sqrt[3]{abc} &\leqslant \sum \left(\sqrt{a^2-ab+b^2}+3\sqrt{ab}\right) \tag{*}\\
&\leqslant \sum 2(a+b) \tag{**}\\
&=4\sum a
\end{align}
Notes: * AM-GM inequality ** Jensen's inequality with the concave function $t \mapsto\sqrt{t}$
|
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|
Use induction to prove that $3 |(7^n −1)$ for every natural number n. This is what I have so far and now I'm getting lost.
Proof - We prove by induction. Let $P(n)$ be the statement "$3|(7^n - 1)$". Since $3|6$, we see that $P(1)$ holds. Suppose that is true for $n = k$. We must show the result is true for $n = k + 1$.
Consider,
$$ 7^{(k+1)} - 1 = (7-1)(7^k + 7^{(k-1)} + \cdots + 1)
= 6(7^k + 7^{(k-1)} + \cdots + 1)$$
I do not know if I am doing it correctly or if I assumed my goal.
|
Your proof is valid if we assume that we know that $a^{m+1} - b^{m+1}=(a^m + ba^{m-1} + ... + b^{m-1}a + b^m)$.
But if we do know that then your proof is one line: $7^n - 1=(7-1)(7^{m-1}+...+1) = 6*(7^{m-1}+...+1)$ and no need for induction at all.
So I'm assuming you aren't supposed to know that result. But could you prove the result?
To do it without the result we can do
Induction step: Assume $6|7^k-1$
then $7^{k+1}-1 = 7*7^k- 1 = (6+1)7^k - 1=$
$6*7^k + (7^k -1)$ and $6|6*7^k$ and $6|7^k-1$ so $6|6*7^k + (7^k -1)$
....
or you could just prove the result
You can prove it by induction:
$a^1-b^1 = (a-b)$
Induction step:
Assume $a^k - b^k = (a-b)(a^{k-1} + a^{k-2}b + ... +ab^{k-2} + b^{k-1})$
Then $a^{k+1} -b^{k+1} =(a*a^{k} - a*b^k)+(a*b^k- b*b^k)=$
$a(a^k-b^k) +(a-b)b^k =$
$(a-b)a(a^{k-1} + a^{k-2}b + ... + ab^{k-2} + b^{k-1}) + (a-b)b^k=$
$(a-b)[a(a^{k-1} + a^{k-2}b + ... + ab^{k-2} + b^{k-1}) + b^k]=$
$(a-b)[(a^k + a^{k-1}b + .... +a^2b^{k-2} + ab^{k-1}) + b^k]=$
$(a-b)(a^k + a^{k-1}b + .... +a^2b^{k-2} + ab^{k-1} + b^k$
Now the result is proven you can use it.
.....
|
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|
If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to
Let an angle between the vectors $a$ and $b$ be $\frac{2\pi}{3}$. If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to $\frac23/\frac25/\frac13/\frac15$?
My attempt: $b^2=4a^2$. Also, $$a^2+\frac{|a||b|}{2}-\frac{x|a||b|}{2}-b^2=0$$$$\implies a^2+\frac{|a|*2|a|}{2}-\frac{x|a|*2|a|}{2}-4a^2=0$$$$\implies 1+1-x-4=0$$$$\implies x=-2$$ But it doesn't match with the options.
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Remember, $a$ and $b$ are vectors, not scalars, so $a^2 = 4b^2$ doesn't make a lot of sense. Much of your working suffers from the same issue, where it is unclear what is a vector, what is a scalar, and hence what can even be added together.
Our information is:
*
*The angle between the two vectors is $\frac{2\pi}{3}$,
*$|b| = 2|a|$,
*$a + xb \perp a - b$.
All of these pieces of information are relevant, including the first piece, which you seem to be neglecting. From 1, I get
$$a \cdot b = |a| |b| \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}|a||b|.$$
Using 2, we get
$$a \cdot b = -|a|^2.$$
Using 3,
\begin{align*}
0 &= (a + xb) \cdot (a - b) \\
&= |a|^2 + (x - 1)(a \cdot b) - x|b|^2 \\
&= |a|^2 - (x - 1)|a|^2 - 4x|a|^2 \\
&= |a|^2(2 - 5x).
\end{align*}
Note $a \neq 0$, as we need an angle to be defined. We must therefore have $x = \frac{2}{5}$.
|
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|
If $\omega$ is a cube root of unity $\not = 1$ then find the minimum value of $|a+b\omega +c\omega^2|$, where $a,b,c$ are integers but not all equal. Let $z=a+b\omega + c\omega^2$
$$z=a+b\omega -c (1+\omega)$$
$$z=a-c+\omega (b-c)$$
Therefore $$|a-c+\omega (b-c)| \ge ||a-c|-|b-c||$$
How should I proceed?
|
Note that:
$$|a+b\omega+c{\omega}^2|^2=(a+b\omega+c{\omega}^2)(a+b{\omega}^2+c\omega)$$
This is because $\bar{\omega}={\omega}^2$, and $\bar{{\omega}^2}=\omega$. If you multiply out the terms, you'll get $a^2+b^2+c^2-ab-bc-ca$, which is equal to $\frac {(a-b)^2+(b-c)^2+(c-a)^2}{2}$. Since $a,b,c$ are not all equal, two or all three of these square terms cannot be $0$. Hence for minimum let any one of them be $0$ and let the remaining variable differ by $1$. You will get final answer $1$.
|
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|
if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and when $P(x)=x^4+2x^3+2x^2+2x+1=(x+1)^2(x^2+1)$ but if the $P(x)$ have four real roots,I can't it.
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Clearly, $x=0$ is not a root. WLOG, assume that $|b| \ge |a|$ (otherwise, letting $x = \frac{1}{y}$, we have
$y^4 + by^3 + 2y^2 + ay + 1 = 0$; then swap $a$ and $b$).
Let $x_1 \le x_2 \le x_3 \le x_4$ be the four real roots. We split into two cases.
1) If $x_1x_4 > 0$, then either $x_i>0, \forall i$ or $x_i < 0, \forall i$. By AM-GM, we have
$|a| = |x_1+x_2+x_3+x_4| \ge 4\sqrt[4]{x_1x_2x_3x_4} = 4$ and hence $a^2 + b^2 \ge 2a^2 \ge 32$.
2) If $x_1x_4 < 0$, then $x_1 \le x_2 < 0 < x_3 \le x_4$ since $x_1x_2x_3x_4 = 1$.
From $x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4 = 2$, we have
\begin{align}
x_1x_2 + x_3x_4 &= 2 - x_2x_3 - x_4x_1 - x_1x_3 - x_2x_4\\
&\ge 2 + 4\sqrt[4]{(-x_2x_3)\cdot (-x_4x_1) \cdot (-x_1x_3) \cdot (-x_2x_4)}\\
&\ge 6.
\end{align}
Then, we have
\begin{align}
a^2 &= (x_1 + x_2 + x_3 + x_4)^2 \\
&= x_1^2 + x_2^2 + x_3^2 + x_4^2 + 2 (x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1 + x_1x_3 + x_2x_4)\\
&\ge 2x_1x_2 + 2x_3x_4 + 4\\
&\ge 16.
\end{align}
Thus, $a^2 + b^2 \ge 2a^2 \ge 32$.
We are done.
|
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|
Prove $\sum_{n=1}^{\infty}((n+\frac{1}{2})\ln(1+\frac{1}{n})-1)=1-\ln(\sqrt{2\pi})$ I am looking for a derivation of the following sum:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$
My current derivation(s) uses the zeta function at negative integers (and or Stirling approximation/ the derivative of $\zeta'(0)$). I want to avoid those.
How I got an answer was via regularization of
$$-\sum_{i=1}^{\infty}\frac{\zeta(-i)}{i}=\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+ \frac{1}{n}\right)-1\bigg)$$
My own other try was rewriting it via:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=\sum_{k=2}^{\infty} \zeta(k)(-1)^k \bigg(\frac{1}{k+1}-\frac{1}{2k}\bigg)$$
If this works I am already happy. If there's another simple way I'd love to hear it as well.
|
One has
\begin{align*}
& \sum\limits_{n = 1}^\infty {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - t}}{{n + t}}dt} } = \sum\limits_{n = 1}^\infty {\int_0^1 {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{n + t}}dt} } \\ & = \sum\limits_{n = 1}^\infty {\int_{n - 1}^n {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} } = \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + 1}}dt} .
\end{align*}
Now, by the Euler--Maclaurin formula,
$$
\log k! = \left( {k + \frac{1}{2}} \right)\log k- k + C + \int_0^{ + \infty } {\frac{{\frac{1}{2} - (t - \left\lfloor t \right\rfloor )}}{{t + k}}dt}
$$
with some constant $C$. It can be shown that the integral is $\mathcal{O}(k^{-1})$ and so by Stirling's formula (or the Wallis product), $C=\frac{1}{2}\log (2\pi )$. Thus
\begin{align*}\sum\limits_{n = 1}^\infty {\left[ {\left( {n + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{n}} \right) - 1} \right]} & = \log 1! - \left( {\left( {1 + \frac{1}{2}} \right)\log 1 - 1 + \frac{1}{2}\log (2\pi )} \right) \\ &= 1 - \frac{1}{2}\log (2\pi ).
\end{align*}
|
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How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$ Then $3,6,9$ are divisible by $3$ so we remove it. So we are left with $1, 2, 4, 5, 7, 8, 10, 11$. Clearly the $7th$ number not divisible by $3$ is $10$. So answer is $10$. It turns out the answer is:
$$k+⌊\tfrac{k-1}{n-1}⌋$$
I want the mathematical intuition behind this formula.
|
==== old answer ===
Assume the answer is $M$.
Then there are $k$ numbers less than or equal to $M$ that are not divisible by $n$.
So $k = f(M)$ where $f(M)$ is the function that tells you how many number less than or equal to $M$ are not divisible by $n$.
So if we can "undo" $f(M)$ to get $f^{-1}(k) = M$ we can solve for $M$.
Now there is a unique number $m$ so that $M = m*n + r$ where $0\le r < n$. That's just division.
In other words, $m*n \le M < (m+1)n$. But as $n$ doesn't divide $M$ we have.
$m*n < M < (m+1)n$.
And the numbers $n, 2n, 3n,.....,mn$ are all divisible by $n$. And all the rest of the numbers are not.
So there are $M- m = k$ numbers that are not divisible $n$ that are equal or less than $M$.
So what is $m$?
So $mn < M < (m+1)n$ so $m < \frac Mn < m+1$ so $m = \lfloor \frac Mn \rfloor$.
And we have $M - \lfloor \frac Mn \rfloor = k$
So now we must solve for $M$ in terms of $n$ and $k$.
Let's go.....
$M - \lfloor \frac Mn \rfloor = k$
$\lfloor \frac Mn \rfloor = M-k$
$M-k < \frac Mn < M-k +1$ (We know $n\not \mid M$ so we have strict inequalities)
$Mn-kn < M < Mn -kn + n$
$-kn < M-Mn < -kn +n$
$kn- n < Mn-M < kn$
$(k-1)\frac n{n-1} < M < k\frac n{n-1}$
Now $\frac n{n-1} < 1$ so there is (apparently) exactly one integer, $M$ so that $(k-1)\frac n{n-1} < M$ and that $k\frac n{n-1} > M$.
$k\frac n{n-1} = k\frac {(n-1)+1}{n-1} = k(1 + \frac 1{n-1})= k + \frac k{n-1}$
And $(k-1)\frac n{n-1} = (k-1)(1 + \frac 1{n-1}) = k-1 + \frac {k-1}{n-1}=k + \frac{k-n}{n-1}$.
So we have the only integer $M$ so that $k + \frac {k-n}{n-1} < M < k + \frac {k}{n-1}$.
If $\frac k{n-1}$ is an integer then $k+\frac k{n-1}$ is too big.
But if $h \le k-1$ and $\frac h{n-1}$ is an integer then $M \ge k + \frac h{n-1}$.
So $M$ is the $k + $ the largest integer less than or equal to $\frac {k-1}{n-1}$
So $M = k + \lfloor \frac {k-1}{n-1} \rfloor$.
=====
Actually that last bit was pretty confusing.
Go back to $(k-1)\frac n{n-1} < M < k\frac n{n-1}$
Now lets jump ahead and consider $k + \lfloor \frac {k-1}{n-1}\rfloor$
Let $h = \lfloor \frac {k-1}{n-1}\rfloor$.
So $h\le \frac {k-1}{n-1}< \frac {k}{n-1}$
And $\frac {k-1}{n-1} - 1=\frac {k-n}{n-1} < h$.
So $k + \frac {k-n}{n-1}< k + \lfloor \frac {k-1}{n-1}\rfloor <k+\frac {k}{n-1}$
But also $(k-1)\frac n{n-1} < M < k\frac n{n-1}$.
But there is at most one integer in that range. So as both $M$ and $k + \lfloor \frac {k-1}{n-1}\rfloor$ are both integers, they must be equal.
|
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|
Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate thrice but it's very lenghty.
|
Let $n = 15$ and $P(x) = \frac{x^n-1}{x-1} = \sum\limits_{k=0}^{n-1} x^k$. Last part of your question suggest you already know:
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = - \frac12 \left.\frac{d^2}{dx^2} \frac{P'(x)}{P(x)}\right|_{x=1} = -\frac12\left.\frac{d^3}{dx^3}\log P(x)\right|_{x=1}$$
To evaluate the derivative, change variable to $t = \log x$ and let $D$ be the operator $\frac{d}{dt}$, we have
$$-2\mathcal{S} = \left.\left(x^3\frac{d^3}{d x^3}\right)\log P(x)\right|_{x=1}
= D(D-1)(D-2)\left.\log P(e^t)\right|_{t=0}\tag{*1}
$$
Notice
$$\log P(e^t) = \log\frac{e^{nt} - 1}{e^t-1} = \log n + f(nt) - f(t)
\quad\text{ where }\quad f(t) = \log\frac{e^t - 1}{t}$$
We just need to figure out the Taylor expansion of $f(t)$ up to $O(t^4)$. Since
$$f(t) = \log\left( e^{\frac{t}{2}} \frac{\sinh(\frac{t}{2})}{\frac{t}{2}}\right)
= \frac{t}{2} + \log\left( 1 + \frac{t^2}{3! 2^2} + O(t^4)\right)
= \frac{t}{2} + \frac{t^2}{24} + O(t^4)$$
We have
$$Df(t) = \frac12 + \frac{t}{12} + O(t^3)$$
and hence
$$Df(t)|_{t=0} = \frac12,\quad D^2f(t)|_{t=0} = \frac{1}{12}\quad\text{ and }\quad D^3f(t)|_{t=0} = 0$$
Substitute this in $(*1)$, we get
$$-2\mathcal{S} = \bigg[(D^2 - 3D + 2)D(f(nt) - f(t))\bigg]_{t=0} = -\frac{3}{12}(n^2-1) + \frac{2}{2}(n-1)$$
Similplify this give us
$$\sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = \mathcal{S} = \frac{(n-3)(n-1)}{8}$$
For $n = 15$, this reduces to $\displaystyle\;\frac{(15-3)(15-1)}{8} = 21$ as first pointed out by @user64494 in comment.
|
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|
Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$ Prove that $\frac{1}{2020} < \frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times ... \times \frac{2019}{2020} < \frac{1}{44}$
I have proven the first half of the inequality, which is the expression is less than $\frac{1}{2020}$ by using the following method:
$\frac{1}{2} \times \frac{3}{4} \times \cdot \times \frac{2017}{2018} \frac{2019}{2020}$ and $\frac{2}{3} \times \frac{4}{5} \times \cdot \times \frac{2018}{2019}$ are both less than $1$, so their product = $\frac{1}{2019} \times \frac{2019}{2020}$ = $\frac{1}{2020}$ is smaller than both of them. Therefore $\frac{1}{2} \times \frac{3}{4} \times \cdot \times \frac{2017}{2018} \frac{2019}{2020}>\frac{1}{2020}$
However, I am unable to use the same idea to prove the other half of the inequality. Any ideas? Thank you in advance.
|
Because $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2019}{2020}=\sqrt{\frac{1\cdot3}{2^2}\cdot\frac{3\cdot5}{4^2}\cdot\frac{5\cdot7}{6^2}\cdot...\cdot\frac{2017\cdot2019}{2018^2}\cdot\frac{2019}{2020^2}}<$$
$$<\sqrt{\frac{2019}{2020^2}}<\frac{1}{\sqrt{2020}}<\frac{1}{44}.$$
|
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|
Integrate $\int \frac {dx}{x\sqrt{x^2-49}}\,$ I'm trying to integrate the following:
$$\int \frac {dx}{x\sqrt{x^2-49}}\,$$
using the substitution $x=7\cosh(t)$
This is as far as I've gotten:
$\int \frac {dx}{x\sqrt{x^2-49}}\,$ = $\int \frac {7\sinh(t)dt}{7\cosh(t)7\sinh(t)}\,$ = $\int \frac {dt}{7\cosh(t)}\,$ = $\int \frac {\cosh(t)dt}{7\cosh^2(t)}\,$ = $\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\,$
Let $u=\sinh(t)$, $du=\cosh(t)dt$
$$\int \frac {\cosh(t)dt}{7(1+\sinh^2(t))}\,
=\int \frac {du}{7(1+u^2)}\,$$
$$=\frac {1}{7}\arctan(u)+C=\frac{1}{7} \arctan(\sinh(t))+C$$
This is as far as I have been able to get. Somehow from here I need to get to
$$-\frac{1}{7} \arctan(\frac{7}{\sqrt{x^2-49}})+C$$
Can someone please show me how to finish this integration problem off? I would appreciate it so so much.
|
Note
$$\frac{1}{7} \arctan(\sinh t)
=\frac{1}{7} \arctan\sqrt{\cosh^2t -1}\\
=\frac{1}{7} \arctan\sqrt{\frac{x^2}{49}-1}
=\frac{1}{7} \arctan\frac{\sqrt{x^2-49}}7\\
= \frac{1}{7} \text{arccot } \frac7{\sqrt{x^2-49}}
= \frac{1}{7}(\frac\pi2- \text{arctan} \frac7{\sqrt{x^2-49}})\\
=- \frac{1}{7}\text{arctan} \frac7{\sqrt{x^2-49}}+C
$$
|
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|
Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$$
*About minimal value:By $\sqrt x +\sqrt y \ge \sqrt{x+y}$
So $T\ge \sqrt{5+1+1}=\sqrt 7$
I think the minimal value has a little wrong. Pls help me.
|
Hint. That's equivalent to extremising $$\sqrt{6-5x}+\sqrt {1+5x}$$ on the interval $[0,1].$ Can you now continue?
|
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|
Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theorem, by parametrizing the sphere and by algebraic manipulating the function but I haven't been able to come up with the solution.
Could anybody help me?
Thank you in advance.
|
Here's a more routine answer. First, to find critical points in the interior, compute the gradient:
$$
\nabla f = (4x^3-4yz,4y^3-4xz,4z^3-4xy)
$$Certainly the origin is a critical point, and if any of the variables equal zero, the other two do as well. Otherwise, we have
$$
\begin{cases}
x^3-yz &=0\\
y^3-xz &=0\\
z^3-xy &=0\\
\end{cases}
$$
$$
\begin{cases}
x^4 &=xyz\\
y^4 &= xyz\\
z^4 &=xyz\\
\end{cases}
$$Since we are assuming all variables are now positive, this only admits the symmetric solution (because $x^4=y^4$ so we can infer $x=y$ and so on).
Now along the boundary. Let $f$ be the objective function and $g$ the constraint. No harm in substituting $\lambda\mapsto 2\lambda$ to simplify the arithmetic:
$$
\begin{cases}
4x^3-4yz &=4\lambda x\\
4y^3-4xz &=4\lambda y\\
4z^3-4xy &=4\lambda z\\
\end{cases}
$$Do the same symmetry trick as before:
$$
\begin{cases}
x^4-\lambda x^2 &= xyz\\
y^4-\lambda y^2 &= xyz\\
z^4-\lambda z^2 &= xyz\\
\end{cases}
$$First, suppose $z=0$. Then either we have $(x,y)=(0,0)$, $(x,y)=(\sqrt{\lambda},0)$, or $(\sqrt{\lambda},\sqrt{\lambda})$; with $x^2+y^2+z^2=9$, these reduce to $(x,y)=(3,0)$ or $(3/\sqrt{2},3/\sqrt{2})$ (this holds up to permutation). If $xyz\neq 0$, to be quite honest, I used a computer algebra system because the equations become horrific. I couldn't find any other solutions where all variables were positive.
All that remains is to check our candidate points.
*
*$(x,y,z)=(0,0,0)$. $f(0,0,0)=0$
*$(x,y,z)=(x,x,x)$. $f(x,x,x)=3x^4-4x^3,$ $0\leq x\leq \sqrt{3}$; this has a minimum at $x=1$, corresponding to $f(1,1,1)=-1$, and a maximum of $27-12\sqrt{3}$ at $x=\sqrt{3}$
*$(x,y,z)=(3,0,0)$. $f(3,0,0)=81$.
*$(x,y,z)=(3/\sqrt{2},3/\sqrt{2},0)$. $f(3/\sqrt{2},3/\sqrt{2},0)=81/2$.
In conclusion, $f$ has a maximum at $(3,0,0)$ (cyclic) and a minimum at $(1,1,1$).
|
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|
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I just thought of using $uvw$, but I am not allowed to use it during my exam. Because of the equality I cannot use AMGM, Cauchy-Schwarz, etc.
I tried to use Mixing-Variables, but I failed. Please help.
|
There is also a straight-forward solution with Lagrange multipliers. It is more convenient to examine the (continuous) function
$$
f(x, y, z) = (2-x)(2-y)(2-z)
$$
on the (compact) set $K = \{ (x, y, z) \in \Bbb R^3 \mid x^2+y^2+z^2 = 3 \}$ first, without the restriction to positive arguments.
$f$ attains both minimum and maximum on $K$, and the Lagrange method gives that at the extremal points
$$
-(2-y)(2-z) = \lambda 2x \\
-(2-x)(2-z) = \lambda 2y \\
-(2-x)(2-y) = \lambda 2z
$$
for some $\lambda \in \Bbb R$. We see that $\lambda$ cannot be zero, so that
$$
x(2-x) = y(2-y) = z(2-z) =: \alpha \, ,
$$
i.e. $x, y, z$ are all solutions of the quadratic equation $t^2 - 2t + \alpha = 0$. It follows that any two of them are equal or add to $2$. So we have two possible cases:
*
*$x=y=z$. Then necessarily $x=y=z = \pm 1$.
*Up to a permutation, $x=y=2-z$. Substituting that into $x^2+y^2+z^2$ gives
$$
3 = 2x^2 + (2-x)^2 = 3x^2 -4x + 4 \iff (3x-1)(x-1) = 0 \, .
$$
If $x=1$ then (again) $x=y=z=\pm 1$. Otherwise $(x, y, z) = (1/3, 1/3, 5/3)$.
So the only possible extremal values of $f$ on $K$ are
$$
\begin{align}
f(1, 1, 1) &= 1 \, \\
f(-1, -1, -1) &= 27 \, \\
f(1/3, 1/3, 5/3) &= 25/27 \, ,
\end{align}
$$
and we conclude that $25/27$ is the minimum.
This is also the mimimum for the given problem because it is attained at a point with $x, y, z > 0$.
|
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|
Computing $\int_Q \frac{xy}{x^2+y^2}dxdy$ I'm asked to compute the following integral:
$$\int_Q \frac{xy}{x^2+y^2}dxdy \qquad Q=[0,1]^2$$
Solution:
First I'm going to study if the integral is convergent. To do this, we notice that
$$\int_Q \frac{xy}{x^2+y^2}dxdy < \infty \iff \int_S \frac{xy}{x^2+y^2}dxdy<\infty$$
where $S=\{(x,y) \in \mathbb{R}^2 | x\geq0, y\geq 0, x^2+y^2\leq1\}$ and this is clear because the difference between $\int_Q f(x,y)dxdy$ and $\int_S f(x,y)dxdy$ is a proper integral.
Computing we have:
$$\int_S \frac{xy}{x^2+y^2}dxdy=\lim_{\epsilon \to 0}\int_{\epsilon}^1\int_0^{\frac{\pi}{2}}\rho^3\sin\theta\cos\theta d\rho d\theta=\frac{1}{8}$$
and so I'm granted the convergence.
Now I have to compute the real integral, as we have assured that it's convergent.
$$\int_Q \frac{xy}{x^2+y^2}dxdy=\int_0^1 \int_0^1 \frac{xy}{x^2+y^2}dxdy= \int_0^1 \frac{y}{2} \int_0^1 \frac{2x}{x^2+y^2}dxdy = \frac{1}{2} \int_0^1 y(log(1+y^2)-log(y^2))dy =$$
$$ = \frac{\log2}{2}-\frac{1}{2}\lim_{\epsilon \to 0}\int_{\epsilon}^1y\log(y^2)dy= \frac{\log2}{2}$$
I checked the result and it's correct, but I'm asking for a review of the process: did I do anything wrong?
|
As you wrote it, it looks like you claim
$$ I = \frac{1}{2} \int_0^1 y \ln (y^2) dy = 0.$$
But this is not the case:
\begin{align*}
\frac{1}{2} \int_0^1 y \ln(y^2) dy &= \frac 1 4 \int_{0}^1 \ln(y^2)2y dy \\
&= \frac 1 4 \int_0^1 \ln(u) du.
\end{align*}
Using the fact the $u \ln u - u $ is a primitive of $\ln u$ you get
$$ I = \frac 1 4 \big(- 1 - (0 - 0)\big) = \frac {-1}{4}.$$
Same goes for
$$J = \frac 1 2 \int_0^1 y \ln (1 + y^2) dy \neq \frac {\ln(2)} {2}.$$
but rather
\begin{align*}
\frac{1}{2} \int_0^1 y \ln (1 + y^2) &= \frac{1}{4} \int_0^1 \ln(1+y^2) 2ydy \\
&= \frac{1}{4} \int_1^2 \ln(u) du \\
&= \frac{1}{4} (2 \ln 2 - 2 - (1\cdot\ln 1 - 1)) \\
&= \frac 1 4 (2 \ln 2 -1) \\
&= \frac{\ln(2)}{2} - \frac{1}{4}.
\end{align*}
So the integral is
$$J - I = \left(\frac{\ln(2)}{2} -\frac{1}{4}\right) - \frac{-1}{4} =\frac{\ln(2)}{2}.$$
Concerning the first part it seems easier to me to do the following
$$ \left\vert \int_Q f(x,y)dxdy \right\vert \leq \int_Q \vert f(x,y)\vert dxdy \leq \int_{D_R^+} \vert f(x,y)\vert dxdy$$
where $D_R^+$ is the positive par of a disc of radius $R$ with $R$ such that $Q \subset D_R^+.$
Using polar coordinates you will have the following inside the integral
$$ \left \vert \frac{r^2 \cos \sin}{R^2} r \right \vert\leq r \leq R$$ and
$ \int_Q f(x,y) dxdy < \infty.$
|
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|
Boundaries of a triple integral between $x^2+y^2=2z$ and $x+y=z$ I am calculating the integral $\iiint_V z\,dV$ between $x^2+y^2=2z$ and $x+y=z$, but I am not sure how to find boundaries for this case. I tried with cylindrical coordinates, but I was unable to solve the problem.
I would also like to know if there is any general procedure to solve this kind of problem.
|
Here I present a way to solve the integral. Maybe there's an easier way to integrate, because now you get some troubling terms in the last integral.
Since we have a function of $z$, we would like to integrate over $z$ first. The boundaries of $z$ evaluates to
$$\frac{x^2 + y^2}2 \le z \le x + y$$
since that's the only bounded volume. Therefore we get
$$\int_V z \ dV = \int_{\pi(V)} \left[\int_{\frac{x^2 + y^2}2}^{x + y} z \ dz \right] dA = \int_{\pi(V)} f(x, y) \ dA,$$
where $\pi(V)$ is the projection of $V$ on the $xy$-plane. The boundaries for $\pi(V)$ is where $\frac{x^2 + y^2}2 = x + y$. Solving for $x$ we get $x = 1 \pm \sqrt{1 + 2 y - y^2}$. Therefore $x$ in $\pi(V)$ is therefore contained by
$$1 - \sqrt{1 + 2 y - y^2} \le x \le 1 + \sqrt{1 + 2 y - y^2}.$$
and we get
$$\int_{\pi(V)} f(x, y) \ dA = \int_{Y} \left[\int_{1 - \sqrt{1 + 2 y - y^2}}^{1 + \sqrt{1 + 2 y - y^2}} f(x, y) \ dx\right]dy$$
where $Y$ is the projection of $\pi(V)$ on the $y$-axis. Now the last step is to solve where $Y$ is located. Well, solve where $1 - \sqrt{1 + 2 y - y^2} = 1 + \sqrt{1 + 2 y - y^2}$. Here $y$ is bounded by
$$1 - \sqrt2 \le y \le 1 + \sqrt2,$$
so $Y = [1 - \sqrt2, 1 + \sqrt2]$.
|
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|
The area of a triangle determined by two diagonals at a vertex of a regular heptagon
In a circle of diameter 7, a regular heptagon is drawn inside of it. Then, we shade a triangular region as shown:
What’s the exact value of the shaded region, without using trigonometric constants?
My attempt
I tried to solve it with the circumradius theorem : $A=(abc)/(4R)$, where $a$, $b$, and $c$ are the three sides, and $R$ is the circumradius of the triangle. However, I needed to find the exact value of $\cos(5\pi/14)$, $\cos(4\pi/7)$, and $\sin(5\pi/14)$. Finally, I found an explicit formula for this particular triangle, but the proof was missing.
You can find the formula in Wikipedia's "Heptagonal triangle" entry.
|
It boils down to
$$I= \sin\frac\pi7 \sin\frac{2\pi}7 \sin\frac{3\pi}7= \frac{\sqrt7}8
$$
Let $a =\frac\pi7$ and evaluate
\begin{align}
I^2 & = (1-\cos^2 a ) (1-\cos^2 2a ) (1-\cos^2 3a )\\
& =\frac18 (1-\cos 2a ) (1-\cos 4a ) (1-\cos 8a )\\
& =\frac18 (1-\cos 2a \cos 4a \cos 8a )\tag1\\
\end{align}
where the following equality via $2\cos x\cos y= \cos(x+y)+ \cos(x-y)$ is used
\begin{align}
\cos 2a \cos 4a + \cos 4a \cos 8a+ \cos 8a \cos 2a
=\cos 2a +\cos 4a + \cos 8a\\
\end{align}
Note that
$$\cos 2a \cos 4a \cos 8a= \frac{\sin4a \cos 4a \cos 8a}{2\sin 2a}
= \frac{\sin16a}{8\sin 2a}=\frac18
$$
Thus, (1) yields
$$I = \sqrt{\frac18(1-\frac18)}= \frac{\sqrt7}8$$
|
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|
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
& =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
|
To continue from
$\tag 1 \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) = \dfrac{(ad+bc)^2}{abcd}$
set
$\quad u = ad$
and
$\quad v = bc$
Then substituting in the rhs of $\text{(1)}$, we have
$\quad \dfrac{(u+v)^2}{uv} \ge 4 \text{ iff } (u-v)^2 \ge 0$
Note that if we let $a,b,c,d \in \Bbb R$ satisfy $abcd \gt 0$ then
$\quad \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \ge 4$
and if $abcd \lt 0$ then
$\quad \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \le 4$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$ For $x,y,z\geqq 0$ and $x+y+z=1.$ Prove that$:$
$$2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \leqq \left(x^2+y^2+z^2+3xyz\right)^2.$$
Let $y=\hbox{mid} \{x,y,z\}$ and
$\text{P}= \left[\left(x^2+y^2+z^2\right)(x+y+z)+3xyz\right]^2$
$-2\left[x^2+y^2+z^2+(x+y+z)^2\right]\left[x^3y+y^3z+z^3x+xyz(x+y+z)\right]$
Ji Chen gave the following expression:
$\text{P}=\left[x^3-x^2y-xy^2-y^3+z\left(x^2-y^2+z^2-yz+zx+xy\right)\right]^2$
$+4z(x-y)(y-z)(y+z)\left(x^2+y^2+z^2+yz+zx+xy\right)\geqq 0,$
Result by SBM (me)$:$
$\text{P}=\Big[{x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}+z \left( {x}^{2}+2\,xz-2\,yz+{z}^{2}
\right) \Big]^2$
$+z\left( x-y
\right) \left( y-z \right) \cdot \text{M} \geqq 0,$
where $\text{M}=\left( 2\,{x}^{3}+2\,{x}^{2}y+6\,{x}^{2}z+2\,x{y}^{2}+9\,xyz+7\,x{z}^
{2}+2\,{y}^{3}+7\,{y}^{2}z+5\,y{z}^{2}+6\,{z}^{3} \right) $
|
Also, BW helps.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that:
$$\left(\sum_{cyc}(x^3+x^2y+x^2z+xyz)\right)^2\geq4\sum_{cyc}(x^2+xy)\sum_{cyc}(x^3y+x^2yz)$$ or
$$16(u^2-uv+v^2)x^4+8(3u^3-5u^2v+4uv^2+3v^3)x^3+$$
$$+(17u^4-34u^3v+3u^2v^2+38uv^3+17v^4)x^2+$$
$$+2(3u^5-6u^4v-3u^3v^2+5u^2v^3+8uv^4+3v^5)x+$$
$$+(u^3-u^2v-uv^2-v^3)^2\geq0,$$ which is obvious.
|
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|
A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any more?
|
Once we have proved that all of $x,y,z$ are, if the gcd is 1, cubes themselves, we arrive at, using
$$ \delta = \frac{-1 + i \sqrt 3}{2} $$
so that $\delta^2 + \delta + 1 = 0$ and $\delta^3=1,$
$$ ( u^3 + v^3 + w^3)^3 - 27 u^3 v^3 w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right) $$
Let me post a clean proof that, if $\gcd(x,y,z) = 1$ and $(x+y+z)^3 = 27 xyz,$ then actually $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1,$ therefore $x,y,z$ are all cubes.
We demand, by dividing through if necessary, that $\gcd(x,y,z) = 1.$ Let us ASSUME that there are a pair that are not coprime, let those be $y,z.$
Next, let $g = \gcd(y,z)$ and ASSUME $g > 1.$ Next we get $y = g \beta$ and $z = g \gamma,$ with $\gcd(\alpha, \beta) = 1.$ The important part is that, from the hypothesis $\gcd(x,y,z) = 1,$ actually
$$ \gcd(x,g) = 1. $$
Pick some new letters,
$$ \color{magenta}{mx + ng = 1.} $$
The diophantine equation becomes
$$ 27x g^2 \beta \gamma = (x + g \beta + g \gamma)^3 = x^3 + g \cdot (\mbox{stuff}) $$
Thus we have
$$ g | x^3 \; . \; \; $$
Well, $$g | (m x^3 + ng x^2) = x^2 (mx+ng) = x^2$$
so $g | x^2.$
Well, $$g | (m x^2 + ng x) = x (mx+ng) = x$$
so $g | x.$
Well, $$g | (m x + ng ) = 1$$
so $g | 1.$ Then $$ g = 1 $$
which contradicts the assumption that there is a pair of coprime variables among $x,y,z.$ The contradiction proves that $\gcd(x,y) = \gcd(y,z) = \gcd(z,x) = 1.$ As we have $xyz = t^3$ for some integer $t,$ we find that $x,y,z$ are all cubes. Taking $x = u^3, y = v^3, z=w^3,$ the identity
$$ ( u^3 + v^3 + w^3)^3 - 27 u^3 v^3 w^3 = \left( \color{red}{u^3 + v^3 + w^3 - 3uvw} \right) \left( \color{blue}{u^3 + v^3 + w^3 - 3 \delta uvw} \right) \left( \color{green}{u^3 + v^3 + w^3 - 3 \delta^2 uvw} \right) $$
says that either $x=y=z$ or $x+y+z=0.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for an integer $x \ge 7$, it follows that $x\# > x^2+x$ Is the following argument sufficient to show that for an integer $x \ge 7, x\# > x^2 + x$.
Please let me know if I made a mistake or if there is a more straight forward way to make the same argument.
Let:
*
*$p_n$ be the $n$th prime
*$p\#$ be the primorial for $p$
Here's the argument by induction:
(1) Base case: $p_4=7$
For $7 \le x < 14, 7\# = 210 > x^2+x$ since:
$$7^2 + 7 < 8^2 + 8 < 9^2 + 9 < 10^2 + 10 < 11^2 + 11 < 12^2 + 12 < 13^2 + 13 = 182 < 210$$
(2) Assume up to some prime $p_n \ge 7$ that for $p_n \le x < 2p_n, p_n\# > x^2+x$.
(3) From Bertrand's Postulate, $p_n < p_{n+1} < 2p_n$
(4) $p_{n+1}\# > p_{n+1}[(2p_n-1)^2 + 2p_n-1] = p_{n+1}(2p_n)^2 - 2p_{n+1}p_n$
(5) Since $p_{n+1} \ge 11$, it follows from Bertrand's Postulate:
$$p_{n+1}(2p_n)^2 - 2p_{n+1}p_n > p_{n+1}(p_{n+1})^2 - 2p_{n+1}p_n > 9(p_{n+1})^2$$
(6) It follows:
$$p_{n+1}\# > (3p_{n+1})^2 > (2p_{n+1})^2 - 2p_{n+1} = (2p_{n+1} - 1)^2 + 2p_{n+1} - 1$$
(7) For any integer $x \ge 7$, let $p_n$ be the highest prime less than or equal to $x$.
(8) If $x$ is prime, from the above, $x\# > x^2 + x$
(9) If $x$ is not prime, from Bertrand's Postulate, it follows:
$$x\# = p_n\# > (2p_n-1)^2 + 2p_n - 1 \ge x^2 + x$$
|
What you've done looks correct. However, one small thing to note in your step $(1)$ base case, you only need to show $10^2 + 10 = 110 \lt 210$ since $11$ is prime and, as such, will be handled later by induction as $p_5$.
Also, I believe a somewhat simpler way to proceed after your step $(5)$ is to then show, for all $p_{n+1} \le x \lt 2p_{n+1}$, that
$$\begin{equation}\begin{aligned}
p_{n+1}\# & \gt 9p_{n+1}^2 \\
& = 4p_{n+1}^2 + 5p_{n+1}^2 \\
& \gt (2p_{n+1})^2 + 2p_{n+1} \\
& \gt x^2 + x
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
As you stated in your step $(3)$, Bertrand's postulate shows there's a prime $p_{n+1} \lt p_{n+2} \lt 2p_{n+1}$. Thus, for all $p_{n+1} \le x \lt p_{n+2}$, you have $x\# = p_{n+1}\#$, with \eqref{eq1A} showing
$$x\# \gt x^2 + x \tag{2}\label{eq2A}$$
|
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|
Solutions to $\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2$. $$\sqrt{3 - \tan^2(\frac{3x}{2})} \cdot \sin x - \cos x = 2.$$
How can such an equation be solved? I don't know where to start?
My attempt: $$(3 - \frac{\sin^2 3 y}{\cos^2 3 y}) \cdot 2 \cos y \sin y = (3 - 2 \sin^2 y)^2$$, where $y = \frac{x}{2}$. This does not simplify anything at all, I do not understand what to do next
|
By transferring $\cos x$ to the RHS and squaring, we have
$$(3 - \tan^2\tfrac32x)\sin ^2x=(2+\cos x)^2.$$
Using the fact that $\cos^2x+\sin^2x=1$, some rearrangement gives
$$(\tan\tfrac32x\sin x)^2+(1+2\cos x)^2=0.$$
A real solution to this equation would simultaneously require, modulo $2\pi$, that $x=0$ and $x=\pm\frac23\pi$. Since these requirements are inconsistent, the original equation has no solution.
|
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|
Prove that for integer n greater than 2 and is coprime to 10, the decimal expansion of $\frac{1}{n}$ repeats with period of ${e_n(10)}$ Prove that for integer n greater than 2 and is coprime to 10, the decimal expansion of $\frac{1}{n}$ repeats with period of ${e_n(10)}$ (the order of 10 mod n). I tried computing $\frac{1}{10^{e_n(10)}-1}$ but got stuck. Any insights?
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Note: $\frac 1{a-1} = \frac a{a(a-1)} = \frac {a-1}{a(a-1)} + \frac 1{a(a-1)} = \frac 1a + \frac1{a(a-1)}=\frac 1a + \frac 1a(\frac 1a + \frac 1{a(a-1)})=\frac 1a + \frac 1{a^2} + \frac 1{a^2(a-1)}=....$.
And by induction $\frac 1{a-1} = \frac 1a + \frac 1{a^2} + \frac 1{a^3} + ....$
So $\frac 1{10^m -1} = \frac 1{10^m} + \frac 1{10^m(10^m-1)}$.
ANd by induction:
$\frac 1{10^m-1} = \frac 1{10^m} + \frac 1{10^{2m}} + \frac 1{10^{3m}} + ....=$
$\sum\limits_{k=1}^\infty \frac 1{10^{mk}}$.
So...
$10^{e(n)} \equiv 1 \pmod n$ so
$10^{e(n)} = 1 + Kn$ for some $K$. (And obviously $K < 10^{e(n)}$)
So $\frac 1n = K\frac 1{10^{e(n)} -1}=$
$K\sum\limits_{k=1}^\infty \frac 1{10^{e(n)k}}=\sum\limits_{k=1}^\infty \frac K{10^{e(n)k}}$
And as $K$ has $10^{e(n)}$ digits or fewer this is a repeating decimal with a period of $e(n)$.
|
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|
I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ I am trying to understand each step in order to get from $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the other factorings. I thought the $x$ could be factored out. I'm confused.
I know I can just insert $a^2-b^2$ into the difference of squares formula like so $(a-b)(a+b)$ but I am practicing factoring. I'm just curious to see each and every step of the factoring.
|
If $x^2 - 4$ can be factored then there exist $a, b \in \Bbb R$ such that
$\quad x^2 - 4 = (x-a) (x-b)$
Now multiplying out the rhs and collecting terms,
$\quad (x-a) (x-b) = x^2 - (a+b)x + ab$
So we need to solve the system of equations
$\tag 1 a + b = 0$
$\tag 2 ab = -4$
From equation $\text{(1)}$ we have
$\quad a = -b$
and plugging that into equation $\text{(2)}$,
$\quad -b^2 = -4$
We have two solutions for $b$.
If $b = 2$ then $a = -2$ and so
$\tag {ANS} x^2 - 4 = (x+2) (x-2)$
|
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|
How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then the original inequality is equal to
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \sqrt{3(ab+bc+ca)}+2(a+b+c)
\end{align*}
By using Cauchy's inequality,we can get
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \frac{1}{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2
\end{align*}
So if we can proof (Since the original equality is true ,then the following equality must be true)
\begin{align*}
(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\geq 2\sqrt{3(ab+bc+ca)}+4(a+b+c)
\end{align*}
or
\begin{align*}
\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)\tag{*}
\end{align*}
then the problem is done.But I can't prove (*).
|
Solution of AOPS member (don't remember name)
Setting
$$x=\sqrt{(a+b)(a+c)}-a, \quad y=\sqrt{(b+c)(b+a)}-b, \quad z=\sqrt{(c+a)(c+b)}-c,$$
we have
$$ab+bc+ca=xy+yz+zx,$$
Inequality become
$$x+y+z \geqslant \sqrt{3(xy+yz+zx)}.$$
Done.
|
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|
Continuity of multivariable function $\frac{xy}{\sqrt{16-x^2-y^2}}$ How can I define where is the following function continuous? The function being: $$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$$
Is it enough with getting the domain by doing $16-x^2-y^2>0$. How does it change if instead of $xy$ I have $\frac{y}{\sqrt{16-x^2-y^2}}$ or $\frac{x^2+y^2}{\sqrt{16-x^2-y^2}}$?
Thanks for the help.
|
For $\sqrt{x}$ to be defined, we must have $x\geq 0$. For $\frac{1}{\sqrt{x}}$ to be defined, we must have $\sqrt{x}\neq 0$, i.e. $x\neq 0$. Hence, for the function
$f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$ to be defined, we must have $16-x^2-y^2>0$, just as you said.
Now, $xy$ is a continuous function, and $\sqrt{16-x^2-y^2}$ is a nonzero (in fact, positive) continuous function when $16-x^2-y^2>0$. Therefore, the quotient of these two continuous functions, $f(x,y) = \frac{xy}{\sqrt{16-x^2-y^2}}$, is a continuous function when $16-x^2-y^2>0$.
The situation is the same for $\frac{y}{\sqrt{16-x^2-y^2}}$ or $\frac{x^2+y^2}{\sqrt{16-x^2-y^2}}$, since $y$ and $x^2+y^2$ are still continuous functions.
|
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|
Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$ I want to calculate the following improper integral:
$$\int_A \frac{z}{y^2}\\
A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$$
First I noticed that the conditions imply $x^2<2x\rightarrow x<2;1<x^2<y^2\rightarrow y>1$, and thus $B=A\cap \{y>2\}=\{z\in(0,2),x\in(1,2), x^2+z^2<2x\}$. Thus this part of the integral is fairly easy:
$$\int_B\frac{z}{y^2}=\int_2^\infty\int_1^2\int_0^\sqrt{2x-x^2}\frac{z}{y^2}dzdxdy=\int_2^\infty\frac1{y^2}dy\int_1^2\frac{2x-x^2}{2}dx=\frac{1}{6}$$
We are now left with a "proper" integral (i.e. the region on which we are integration is finite, and the integrand is bounded):
$$\int_{A\cap\{y\in(1,2)\}}\frac{z}{y^2}$$
I tried to split the domain in two regions: $A'=\{x\in[1,2);y\in(x,\sqrt{2x});z\in [0,\sqrt{y^2-x^2})\}, A''=\{x\in[1,2);y\in[\sqrt{2x},2];z\in[0,\sqrt{2x-x^2})$
Is my approach correct (I'm not sure, since the computations that follow bring me to an uncorrect result)? Is there any other approach to computing this integral, perhaps less messy?
|
The conditions $z\ge 0$, $x\ge 1$, $y\ge 0$, $x^2+z^2 \lt 2x$, and $x^2+z^2 \lt y^2$ define the intersection of two solids. Solid one is the quarter of the cylinder $(x-1)^2 + z^2 = 1$ to the right of the plane $x=1$ and above the plane $z=0$. Solid two is the portion of the cone $x^2 + z^2 = y^2$ to the right of the plane $x=1$ and above the plane $z=0$. For $y\ge 2$, the cone completely contains the cylinder, so $A$ is just the cylinder in this portion of $A$. The integral is indeed $1/6$ over this subregion $A_1$.
You can split the remainder of $A$ into two subregions $A_2$ and $A_3$. The first is the subregion where the boundary of the cone is intersecting the slice of the plane $x=1$ between $z=0$ and $z=1$. The second is the subregion where the boundary of the cone is intersecting the upper boundary of the cylinder between $x=1$ and $x=2$. The integral over $A_2$ is
$$
\begin{align}
\int_{A_2} z/y^2 &= \int_1^{\sqrt{2}} \frac{1}{y^2} \int_0^{\sqrt{y^2-1}} z
\int_1^{\sqrt{y^2-z^2}} \,dx\,dz\,dy \\
&= \int_1^{\sqrt{2}}\frac{1}{y^2}\int_0^{\sqrt{y^2-1}} (z\sqrt{y^2-z^2} - z) \, dz\, dy \\
&= \int_1^{\sqrt{2}} \frac{1}{y^2} \left.\left(-(y^2-z^2)^{3/2}/3 - z^2/2\right)
\right|_{z=0}^{z=\sqrt{y^2-1}} \,dy \\
&= \int_1^{\sqrt{2}} \frac{1}{y^2}\left(\frac{1}{6} - \frac{y^2}{2} + \frac{y^3}{3}\right) \,dy \\
&= \left. -\frac{1}{6y} - \frac{y}{2} + \frac{y^2}{6}\right|_1^{\sqrt{2}} \\
&= \frac{5}{6} - \frac{7}{12}\sqrt{2}.
\end{align}
$$
The integral over $A_3$ is
$$
\int_{A_3} z/y^2 = \int_{\sqrt{2}}^{2}\frac{1}{y^2}\int_0^1 z
\int_1^{\min(\sqrt{1-z^2}+1, \sqrt{y^2-z^2})} \,dx\,dz\,dy. \\
$$
The cylinder boundary intersects the cone boundary at $z = y\sqrt{1-y^2/4}$. This means that $\sqrt{y^2-z^2} \le \sqrt{1-z^2}+1$ for $0\le z \le y\sqrt{1-y^2/4}$ and $\sqrt{1-z^2}+1 \le \sqrt{y^2-z^2}$ for $y\sqrt{1-y^2/4} \le z \le 1$, so we can write
$$
\int_0^1 z \left({\min(\sqrt{1-z^2}+1, \sqrt{y^2-z^2})} - 1\right) \, dz
$$
as the sum
$$
\int_0^{y\sqrt{1-y^2/4}} \left(z\sqrt{y^2-z^2} - z\right) \, dz +
\int_{y\sqrt{1-y^2/4}}^1 z\sqrt{1-z^2} \, dz.
$$
This sum of two integrals simplifies to
$$
-\frac{y^4}{8} + \frac{y^3}{3} - \frac{1}{3}.
$$
So
$$
\begin{align}
\int_{A_3} z/y^2 &= \int_{\sqrt{2}}^2 -\frac{y^2}{8} + \frac{y}{3} - \frac{1}{3y^2} \, dy \\
&= \frac{1}{6} - \frac{\sqrt{2}}{12}.
\end{align}
$$
So the integral over the entire region $A$ is
$$
\left(\frac{1}{6}\right) + \left(\frac{5}{6} - \frac{7}{12}\sqrt{2}\right) +
\left(\frac{1}{6} - \frac{1}{12}\sqrt{2}\right) =
\frac{7}{6} - \frac{2}{3}\sqrt{2}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof of "equations cannot have an integral solutions" using modular arithmetic. We a got an equation $$15x^2 -7y^2 =9$$ and we have to prove that this equation cannot have an integral solution. So, this what we will do:
$7y^2 = 15x^2 -9$, and since $3| 15x^2 -9 \implies 3|7y^2$. As 3 cannot divide 7 therefore, it must divide $y^2$ and hence it must divide $y$. So, let $y=3m$. Let's substitute this value of $y$ into our original equation:
$$
15x^2 - 63m^2 = 9 \\
5x^2 = 3+21m^2
$$
Since, 3 divides RHS, so it would divide LHS, that is $3|5x^2$, but 3 doesn't divide 5 so it must divide $x^2$ and hence it must divide $x$, so let $x= 3n$. Let's substitute this value in the last equation :
$$
45n^2 = 3+21m^2\\
15n^2= 1+7m^2
$$
That is, $ 7m^2 +1 \equiv 0 \mod 3$ but thats not possible because $$m \equiv 0 \implies m^2 \equiv 0 \\
m\equiv 1 \implies m^2 \equiv 1 \\
m \equiv 2 \implies m^2 \equiv 4 \implies m^2 \equiv 1 $$ (all mod 3)
And
$$
7m^2 \equiv 0 ~~~~~~~~~~~~~~~~~~~~~~(i)\\
7m^2 \equiv 7 \implies 7m^2 \equiv 1~~~~~~~~~~~~~~~~~~~~~~~~(ii) $$
But we know $$
1 \equiv 1 \mod 3 \\
\text {Adding congruence relations of (i) and (ii) to the above one, we get}\\
7m^2 +1 \equiv 1 \\
7m^2 +1 \equiv 2 $$
Hence, $7m^2 +1 \equiv \mod 3$ is not possible. Therefore, the equation doesn't have any integral solutions (and I really don't know why does this show that our original equation doesn't have any integral solutions).
Let's consider this equation $$3x^3 +y^3 =6$$
We want to prove that this equation doesn't have any integral solution. So, we've $y^3 = 6-3x^3$, that means $3| y^3$ which implies $y= 3m$. Let's substitute this into our original equation,
$$
3x^3 +27m^3 = 6 \\
x^3 + 9m^3 = 2\\
9m^3 = 2-x^3
$$
That means, $2-x^3 \equiv 0 \mod 3$. So, we have
$$
x \equiv 0 \implies -x^3 \equiv 0 \\
x \equiv 1 \implies -x^3 \equiv -1 \\
x \equiv 2 \implies x^3 \equiv 8 \implies x^3 \equiv 2 \implies -x^3 \equiv -2 ~~~~~~~~~~(iii)$$
(all are mod 3)
And we know
$$
2 \equiv 2 \mod 3$$
Now, adding the congruence relation (iii) with the above one we get
$$
2-x^3 \equiv 0 \mod 3$$ as wanted. But this doesn't prove that our equation have integral solutions (someone said that) and of course, our equation doesn't have any integral solutions but why we didn't reach any contradiction in the way we reached in the first proof.
So, my question is why in the first proof when reached a contradiction we concluded "no integral solutions are possible" but in the second proof everything agreed with one-another but we didn't conclude "integral solutions are possible", Why?
Please explain me.
|
Given
$$15x^2 -7y^2 =9\quad\implies\quad x = ± \frac{\sqrt{7 y^2 + 9}}{\sqrt{15}}\quad\land\quad y = \sqrt{\frac{3}{7}} \sqrt{5 x^2 - 3}$$
we can see that there are no rational solutions, let alone integers.
|
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|
Derivation of natural log inequality While looking at a proof of this inequality
$(1+\frac{1}{x})^x < e < (1+\frac{1}{x})^{(x+1)}$
The authors take the natural log of on both sides and get
$xln(1+\frac{1}{x}) < 1 < (x+1)ln(1+\frac{1}{x})$
But then they write this last inequality as follows (which I don't understand)
$\frac{1}{1+x} <ln(1+\frac{1}{x}) < 1/x $
How can you re-write the second inequality as the third inequality?
|
Let's start from the known equation
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
This is, in fact, two different results:
$$
x \ln\left( 1+ \frac{1}{x} \right) < 1 \qquad \text{and} \qquad 1 < (x+1) \ln\left( 1+ \frac{1}{x} \right)
$$
When you divide both sides of the first equation by $x > 0$ and both sides of the second equation by $(x+1)$, you get
$$
\ln\left( 1+ \frac{1}{x} \right) < \frac{1}{x} \qquad \text{and} \qquad \frac{1}{x+1} <\ln\left( 1+ \frac{1}{x} \right)
$$
Can you see it now?
|
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|
Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0
$$
But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?
|
The conversion to polar coordinates has not failed. Note what happens when $r=\tan^2(\theta)\sec(\theta)$, $|\theta| > \pi/2$. Then,
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=\frac{\sin^4(\theta)}{2\sin^4(\theta)}=\frac12$$
And of course, if $\theta =0$ or $\theta =\pi/2$, then
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=0$$
EXPLANATION: So what is happening here?
Note that $x^6+y^4\ge 2\sqrt{x^6y^4}=2|x|^3y^2$, with equality for $y^2=|x|^3$. Therefore, we have
$$\left|\frac{x^3y^2}{x^6+y^4}\right|\le \frac12$$
with equality when $y^2=|x|^3$.
If we transform to polar coordinates, this analysis is identical to
$$\left|\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}\right|\le \left|\frac{r\cos^3(\theta)\sin^2(\theta)}{2\sqrt{r^2\cos^6(\theta)\sin^4(\theta)}}\right|=\frac12$$
with equality when $r=\frac{\sin^2(\theta)}{|\cos^3(\theta)|}=\tan^2(\theta)|\sec(\theta)|$.
|
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|
If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I can not find a way to decompose of combine.
Also, how do you prove that $x^4+y^4+z^2 \ge xyz \sqrt{8}$?
I got $x^4+y^4+z^2\geq3\sqrt[3]{x^4*y^4*z^2}=3xy\sqrt[3]{xyz^2}.$
I do not know what to do from here.
|
For the first inequality, using the substitution of $ a = \frac{y}{x}, b = \frac{z}{y},c = \frac{x}{z}$, (which is a good approach to nomarlize these inequalities), we WTS (see Michaels' solution)
$$ \sum y \times \frac{1}{ (x+y+z) - y } \geq \sum x \times \frac{ 1}{(x+y+z) - y }.$$
This is just rearrangement inequality, since $\{ y\}$ and $\{ \frac{ 1}{ (x+y+z) - y } \}$ are similarly ordered.
|
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|
Given $S =\{ x-y \mid (x,y) \in \Bbb R , x^2+y^2=1\}$, find $\max S$ My solution:
$x^2+y^2=1$ is the equation of a circle.
Let $x-y=k$. That becomes the equation of a line. Since $(x-y) \in S$,
the point $(x,y)$ satisfies both the circle and the line equation. So we know that the graph of the circle and the line have at least $1$ intersection. Now here's the part I have a question about. I drew the graph of the circle and visualised the line equation (slope$=1$) and one can easily conclude that the maximum and minimum values of $k$ occur when the line is a tangent to the circle.
So after concluding this, rest is just calculation. ( The answer is $\sqrt2$, by the way).
My questions are:
*
*Can we prove that(without visualization) the maximum and minimum values of $k$ occur when the line is tangent to the circle?
2.Is there any other method to solve such a question?
|
Well,... let $y = \pm \sqrt {1-x^2}$ and let $d_1(x) = x-y= x-\sqrt{1-x^2}$ when $y\ge 0$ and $d_2(x) =x-y= x+\sqrt{1-x^2}$ when $y < 0$.
Then $d_1'(x) = 1 -\frac 12\frac 1{\sqrt{1-x^2}}*(-2x)= 1+\frac x{\sqrt{1-x^2}}$
And $d_2'(x) = 1-\frac {x{\sqrt{1-x^2}}$.
Solving for $d_1'(x) = 0$ and $d_2'(x)=0$ we get:
$d_1'(x) = 1+\frac x{\sqrt{1-x^2}}=0$
$x = -\sqrt{1-x^2}$
$x^2 = 1-x^2; x\le 0$
$x = -\frac 1{\sqrt 2}$.
$d_1'(x) = 1-\frac x{\sqrt{1-x^2}}=0$
And $x=\sqrt{1-x^2}$
$x^2 = 1-x^2; x\ge 0$
$x = \frac 1{\sqrt 2}$. are the two critical points.
Could do second derivative test to determine if max/min but it's easier and practical to plug the values in.
At $x =-\frac 1{\sqrt 2};y \ge 0;x\le 0$ we have $x^2 + y^2 =1$ so $y=\frac 1{\sqrt 2}$ and $x-y = -\sqrt 2$.
And at $x = \frac 1{\sqrt 2}; y \le 0; x \ge 0$ we have $x^2 + y^2 =1$ so $y=-\frac 1{\sqrt 2}$ and $x-y = \sqrt 2$.
so $\sqrt 2$ is the max $x-y$ can be and $-\sqrt 2$ is the min.
|
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|
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$. For $x > 1$, $x^2(x^2+1) > x(x^2+1)$, hence $P>Q$.
The part where I have the problem is I can't prove this for the range $0 < x < 1$ without the help of a graphing calculator. Can anyone help?
What I've done in this region so far is:
*
*Prove that $P$ and $Q$ is always increasing in this region,
*The range for $P$ starts from $1 < P < 3$, and
*The range for $Q$ starts from $0 < Q < 2$.
The only thing I need to prove now is that $P$ and $Q$ will not intersect at $0 < x < 1$, but I can't prove this part.
|
In other words you want to show that $$x^4-x^3+x^2-x+1>0$$ identically.
This is clearly true if $x=0.$
So for $x\ne 0$ factor out $x^2>0$ to get $$x^2\left(x^2-x+1-1/x+1/x^2\right),$$ and then we only need consider the expression in parentheses. This may be written as $$x^2+1/x^2-(x+1/x)+1,$$ and now we set $u=x+1/x,$ to obtain $$u^2-u-1,$$ whose roots are $$\frac12(1\pm \sqrt 5),$$ so that we only need focus on the interval between these roots. That is this quantity is not positive for $$\frac{1-\sqrt 5}{2}< x+\frac 1x<\frac{1+\sqrt 5}{2}.$$
We now show this to have no solution by considering when the roots of $x+1/x=a$ are not real for some given real value of $a.$ That is, we consider the equation $$x^2-ax+1=0,$$ whose discriminant $a^2-4<0\implies |a|<2.$
We only need show that the roots of $u^2-u-1=0$ are less than $2$ in magnitude, and the proof is complete. But this is easily shown.
|
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|
What went wrong in the evaluation of $\int \frac{1}{3-2\sin(x)}dx$? I tried to evaluate the following integral:
$$\int \frac{1}{3-2\sin(x)}dx\,\, $$ with universal substitution, using the fact that
$t(x):=\tan\left(\frac{x}{2}\right)$
$\sin(x)=\frac{2t(x)}{1+t(x)^2}$ and $t'(x)=\frac{1+t(x)^2}{2}$
$\displaystyle\int \frac{1}{3-2\sin(x)}dx=$$\displaystyle\int \frac{1}{3-2(\frac{2t(x)}{1+t(x)^2})}dx=\displaystyle\int \frac{1}{3-\frac{4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1}{\frac{3+3t(x)^2-4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1+t(x)^2}{3+3t(x)^2-4t(x)}dx$
Now I substituted: $u:=t(x) \Longrightarrow dx= \frac{du}{t'(x)}=\frac{2du}{1+t(x)^2}$
So:
$\displaystyle\int \frac{1+u^2}{3+3u^2-4u}\frac{2}{1+u^2}du=\displaystyle\int \frac{2}{3+3u^2-4u}du=2\displaystyle\int \frac{1}{3u^2-4u+3}du=2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du$
Here I used the formular:
$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right]$
So:
$2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{(\sqrt{3}u-\frac{2}{\sqrt{3}})\sqrt{3}}{\sqrt{5}}\right)\right]=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3u-2}{\sqrt{5}}\right)\right]$
Resubstituting:
$2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$
Here is the issue:
Wolfram tells:
$\displaystyle\int \frac{1}{3-2\sin(x)}dx=2\left[\frac{1}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$
I cannot figure out where I went wrong.. like.. its only one factor..
could someone maybe show me what went wrong? Thank you :)
|
Use the fact that, if $\color{blue}{\int f(x)dx = F(x)}$, then $\color{blue}{\int f(ax+b)dx = \frac{1}{a} F(x)}$
So, you'd have
$$\frac{1}{\sqrt3}\cdot\left[\frac{2\sqrt3}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)\right] =\frac{2}{\sqrt5}\arctan\left(\frac{3u-2}{\sqrt5}\right)$$
|
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|
Does the following integral converge? $ \int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx $ Does the following integral converge?
$$
\int\limits_0^\pi\frac{\sin x}{\sqrt{x}}\ dx
$$
I haven't solved such problems for a while. So, I would really appreciate it if someone gave me a hint.
Or maybe my solution is correct?
$$
\sin x\sim x\Rightarrow\frac{\sin x}{\sqrt{x}}\sim\sqrt{x}
$$
$$
\int\limits_0^\pi\sqrt{x}\ dx\ \ \text{is convergent}
$$
Therefore, the initial integral is convergent as well.
|
Just for the fun of it !
The problem of the convergence being solved, there are analytical solution for this kind of integrals (and antiderivatives; have a look here.
Since @Von Neumann wrote an answer where complex numbers do appear, I wondered what would give the $1,400$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
$$\int\frac{\sin (x)}{\sqrt{x}}\, dx \sim \int \frac{16 (\pi -x) \sqrt{x}}{5 \pi ^2-4 (\pi -x) x} \,dx=$$ and then the integral is$$-8 \sqrt{\pi }+2 i \sqrt{(-2-4 i) \pi } \cot
^{-1}\left(\sqrt{-\frac{1}{2}-i}\right)-(4+3 i) \sqrt{\left(-\frac{2}{5}+\frac{4
i}{5}\right) \pi } \cot ^{-1}\left(\sqrt{-\frac{1}{2}-i}\right)-2 i \sqrt{(-2+4 i)
\pi } \cot ^{-1}\left(\sqrt{-\frac{1}{2}+i}\right)-(4-3 i)
\sqrt{\left(-\frac{2}{5}-\frac{4 i}{5}\right) \pi } \cot
^{-1}\left(\sqrt{-\frac{1}{2}+i}\right)$$ which is $\approx 1.78995$ while the "exact" value is $1.78966$.
Edit
Another amazing approximation is
$$\sin(x)=\pi \sum_{n=1}^\infty a_n \Big[\left(1-\frac x \pi\right)\frac x \pi\Big]^n$$ where coefficients $a_n$ make the sequence
$$\left\{1,1,2-\frac{\pi ^2}{6},5-\frac{\pi ^2}{2},14-\frac{3 \pi ^2}{2}+\frac{\pi
^4}{120},42-\frac{14 \pi ^2}{3}+\frac{\pi ^4}{24},132-15 \pi ^2+\frac{\pi
^4}{6}-\frac{\pi ^6}{5040}\right\} $$
This makes the integration very easy
$$\int\limits_0^\pi\frac{\sin (x)}{\sqrt{x}}\, dx=\pi ^2\sum_{n=1}^\infty \frac{\Gamma (2 n+1)}{4^n \,\Gamma \left(2 n+\frac{3}{2}\right)}\,a_n$$
Using the $a_n$'s given in the table, the definite integral is then
$$\frac{4 \pi ^{3/2} \left(46190338425-595324620 \pi ^2+1781520 \pi ^4-704 \pi
^6\right)}{503889568875}$$ which is $1.789662938921$ while the exact value is $1.789662938968$
|
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|
Solving $\sin^2x + 3\sin x\cos x + 2\cos^2x=0$, for $0\leq x\leq 2\pi$
Solve
$$\sin^2x + 3\sin x\cos x + 2\cos^2x=0$$
for $0\leq x\leq 2\pi$.
My answers are
$$x=2.03, 5.18 \qquad\text{or}\qquad
x=\frac{3\pi}{4},\frac{7\pi}{4} \qquad\text{or}\qquad
x=\frac{\pi}{2}, \frac{3\pi}{2},$$
but the answer states $x=2.03, 5.18$ or $x=3\pi/4,7\pi/4$ only.
I got $x=\pi/2, 3\pi/2$ from $(\cos x)^2=0$,
where it is a factor in one of my steps:
$$\cos^2x\left(\tan^2x+3\tan x+2\right)=0.$$
|
Use that $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
\right) \right) ^{2}}}
$$ and
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left(
\tan \left( x/2 \right) \right) ^{2}}}
$$
|
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|
Prove: $\int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = \frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$ I'm trying to prove
$$I := \int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = \frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$$
which is around $2.5063328837$.
Using the $u$-substitution $u = \sqrt{-1 + \sqrt{1 + (\tan{x})^{2/3}}}$, we have
\begin{align*}
du
&= \frac{1}{6}\frac{1}{\sqrt{1 + (\tan{x})^{2/3}}}\frac{1}{\sqrt{-1 + \sqrt{1 + (\tan{x})^{2/3}}}} (\tan{x})^{-1/3} \sec^2{x}\,dx \\
&= \frac{1}{6 u (u^2 + 1)} \frac{1}{\sqrt{(u^2 + 1)^2 - 1}} \left(((u^2 + 1)^2 - 1)^3 + 1\right) \,dx \\
&= \frac{1}{6 u^2 (u^2 + 1)} \frac{1}{\sqrt{u^2 + 2}} \left(u^6 (u^2 + 2)^3 + 1\right) \,dx
\end{align*}
and thus,
$$I = \int_{0}^{\pi/2}{\sqrt{1 + \sqrt{1 + (\tan{x})^{2/3}}}\,dx} = 6 \int_{0}^{\infty}{\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \,du}$$
So we have the integral of a rational function, where the roots of the denominator can easily be found in closed form; hence using the method of residues, we can get an answer in closed form. Yet, I've done this and it's still not so obvious that the answer miraculously simplifies to $\frac{\pi}{2} (3^{1/4} + 3^{3/4} - 2)$.
My question is, seeing that the answer is so nice, is there a more clever way to solve this integral? I've tried different substitutions, integration by parts, the Feynman trick by putting a parameter in place of the exponent $2/3$, etc., but they all seem to go nowhere.
Thanks!
|
The explanation for simplicity of the result follows the same reasoning as in my answer here.
You obtained
$$\frac{I}{6} = \int_{0}^{\infty}{\frac{u^2 (u^2 + 1) (u^2 + 2)}{u^6 (u^2 + 2)^3 + 1} \,du}$$
The denominator factors over $\mathbb{Q}(i\sqrt[4]{3})$ as $$u^6 (u^2 + 2)^3 + 1 = (u^2+1)^2 h(x) \overline{h}(x)$$
where
$$h(x) = u^4+\sqrt[4]{3} i u^3+3^{3/4} i u^3-2 \sqrt{3} u^2-u^2-2 i \sqrt[4]{3} u+1$$
Note that $h$ has a very special distribution of zeroes: all of its zeroes lie in lower half plane (this has to checked numerically). Let $p(x) = u^2 (u^2+2)$, then residue theorem says
$$\frac{I}{{12\pi i}} = - \sum\limits_{\alpha :h(\alpha ) = 0} {\frac{{p(\alpha )}}{{h'(\alpha )\overline{h}(\alpha )({\alpha ^2} + 1)}}} - \frac{{p( - i)}}{{( - 2i)h( - i)\overline{h}( - i)}} $$
First term of RHS is symmetric in roots of $h(x) \in \mathbb{Q}(i\sqrt[4]{3})[x]$, so it is in $\mathbb{Q}(i\sqrt[4]{3})$. Second term is in $\mathbb{Q}(i,i\sqrt[4]{3})$. Therefore
$$I/\pi \in \mathbb{Q}(i,i\sqrt[4]{3}) \cap \mathbb{R} = \mathbb{Q}(\sqrt[4]{3})$$
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|
Solving $\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$
I need some tips with solving this algebraic equation
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$
I've tried subtracting: $$\frac{8x}{x+2}$$
and also setting $$\sqrt x = t$$
Where "t" is a substitution to make things simpler.
This is how it ends up:
$$x + 2 = 6\sqrt x - \frac{8x}{x+2}$$
or
$$t^2 + 2 = 6t - \frac{8t^2}{t^2+2}$$
Unfortunately, I couldn't find a way from here without getting polynomials of the fourth degree or equations with $$x\sqrt x$$
I'd just like to clarify that I'm not looking for the solution here. I'd just like it if I could have some pointers or tips on where to go from here, or even if I did something wrong.
Thanks in advance!!!
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Middle term splitting will always remain my favorite method for factorising polynomials!
\begin{align*}
&\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x\\
\iff &\dfrac{(x+2)^2+8x}{x+2}=6\sqrt x\\
\iff &(x+2)^2-6\sqrt x(x+2)+8x=0\\
\iff &(x+2)^2-4\sqrt x(x+2)-2\sqrt x(x+2)+8x=0\\
\iff &(x+2-4\sqrt x)(x+2-2\sqrt x)=0
\end{align*}
Now apply the quadratic formula.
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Suppose that $x$ and $y$ are real numbers. Prove that if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. Not a duplicate of
Prove that if $x \neq 0$, then if $ y = \frac{3x^2+2y}{x^2+2}$ then $y=3$
Prove that for any real numbers $x$ and $y$ if $x \neq 0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.
This is exercise $3.2.10$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose that $x$ and $y$ are real numbers. Prove that if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$.
Here is my proof:
Proof. We will prove the contrapositive. Suppose $y=\frac{3x^2+2y}{x^2+2}$ and $y\neq3$. Suppose $x=0$. Then substituting $x=0$ into $y=\frac{3x^2+2y}{x^2+2}$ we obtain $y-y=0$ which means that $y$ can be any number and in particular $y=3$ which contradicts the assumption that $y\neq 3$. Thus $x\neq 0$. Therefore if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. $Q.E.D.$
Is my proof valid$?$
Edit:
I was reviewing the material today and I noticed a fatal error in the above proof. I am not allowed to assume $y\neq3$ and conclude $y=3$. So the above proof is certainly not valid.
Proof. Suppose $x\neq0$. Suppose $y=\frac{3x^2+2y}{x^2+2}$. Simplifying $y=\frac{3x^2+2y}{x^2+2}$ we obtain $(y-3)x^2=0$. Since $x\neq 0$ and $(y-3)x^2=0$, then $y-3=0$ which is equivalent to $y=3$. Thus if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. Therefore if $x\neq0$, then if $y=\frac{3x^2+2y}{x^2+2}$ then $y=3$. $Q.E.D.$
I think this one should be valid.
Thanks for your attention.
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You could have said this:
if $y=\dfrac{3x^2+2y}{x^2+2}$, then $y(x^2+2)=3x^2+2y$, so $(y-3)x^2=0$, so $x=0$ or $y=3$.
|
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|
Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}})+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{63}{-16}+\tan^{-1}\frac{63}{16}$$
$$=-\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{63}{16}$$
$$=0$$
But the answer is given as $\pi$. What is my mistake?
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Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
As $\dfrac{12}5\cdot\dfrac{3}4>1,$
$$\tan^{-1}\dfrac{12}5+\tan^{-1}\dfrac{3}4=\pi+\tan^{-1}\left(\dfrac{\dfrac{12}5+\dfrac{3}4}{1-\dfrac{12}5\cdot\dfrac{3}4}\right)=?$$
$$\text{and }\tan^{-1}(-y)=-\tan^{-1}y$$
Alternatively,
$$(5+12i)(4+3i)(16+63i)=(36i-16)(36i+16)=-(36)^2-16^2$$
Apply argument in both sides.
Now as $0<\tan^{-1}\dfrac{12}5,\tan^{-1}\dfrac34,\tan^{-1}\dfrac{63}{16}<\dfrac\pi2,$
$$0<\tan^{-1}\dfrac{12}5+\tan^{-1}\dfrac34+\tan^{-1}\dfrac{63}{16}<\dfrac{3\pi}2$$
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|
What maths rule allows this expression with powers to be rewritten as below? I have been reading through a programming book and the author asked to calculate the following expression in the program.
$$4*5^3+6*5^2+7*5+8$$
I approached this by expanding the expression like so:
$$(4*5*5*5)+(6*5*5)+(7*5)+8$$
$$500+150+35+8 = 693$$
In his solution he states that the expression can be rewritten as:
$$((4*5+6)*5+7)*5+8$$
$$(26*5+7)*5+8$$
$$(130+7)*5+8$$
$$(137*5)+8$$
$$685+8 = 693$$
Which produces the correct answer, but there is no explanation as to why this works. I wasn't aware the expression could be rewritten like this and the only pattern I can see is that the five to the power of x is decreasing by one each time.
Is there a technical name for this rule? I am curious to know why this works.
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The main rule used is the distributivity rule which states that for real $a,b,c$, $$a(b+c)=ab + ac.$$
This rule can easily be used on more than $2$ summands, so say you have a real number $a$ and $n$ real numbers $b_1,\dots, b_n$.
Then,
$$a(b_1+\cdots + b_n) = ab_1 + ab_2 + \cdots + ab_n$$
This means that
$$\begin{align}4\cdot 5^3 + 6\cdot 5^2 + 7\cdot 5 + 8 &= (4\cdot 5^2\cdot 5 + 6\cdot 5\cdot 5 + 7\cdot 5) + 8 \\&= (4\cdot 5^2+6\cdot 5 + 7)\cdot 5 + 8,\end{align}$$
where the second row follows from the first precisely because of the rule written above (and the fact that multiplication is commutative, which is why I can have $5$ on the right side instead of the left).
Using the same rule again on the expression in parentheses, you can see that
$$4\cdot 5^2 + 6\cdot 5 + 7 = 4\cdot 5\cdot 5 + 6\cdot 5 + 7 = (4\cdot 5 + 6)\cdot 5 + 7$$
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How to show $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0$ have five different roots? I have to prove this fact when determining the elementary divisors of a matrix. The numerical solution confirms that this equation indeed has five different roots (one real root, two pairs of conjugate complex roots), but how to show it theoretically? I looked into its derivatives but it didn't solve the problem.
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To tell if $f$ has any multiple zeros, take the GCD of $f$ and $f'$. (Any multiple zero of $f$ is also a zero of $f'$.)
In this case
$$
\gcd(f,f') = 1 ,
$$
so there are no multiple zeros, so there are $5$ simple zeros.
Euclidean algorithm computation.
$f(x) = x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$,
$f'(x) = 5x^4+20x^3+12x^2+6x+2$.
Divide $x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1$ by $5x^4+20x^3+12x^2+6x+2$:
Remainder is $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$.
Divide $5x^4+20x^3+12x^2+6x+2$
by $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$.
Remainder is ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$.
Divide $-{\frac {12\,{x}^{3}}{5}}-\frac35\,{x}^{2}+\frac25\,x+\frac35$
by ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$.
Remainder is $-{\frac {116000\,x}{152881}}-{\frac{216624}{152881}}$.
Divide ${\frac {391\,{x}^{2}}{48}}+{\frac {83\,x}{8}}+{\frac{107}{16}}$
by $-{\frac {116000\,x}{152881}}-{\frac{216624}{152881}}$.
Remainder is the constant $\frac{13220690237}{841000000}$.
Conclusion: $f$ has no multiple zero. If $a$ were a multiple zero, then $x-a$ divides both $f$ and $f'$, and therefore all the remainders in the calculation. But of course $x-a$ cannot divide a constant.
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If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
$= 7\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
So, $a^2+b^2+c^2\ge \frac72$
Therefore, the minimum value of $a^2+b^2+c^2$ is $\frac72$
I want to know are there any other method to find the minimum value of
$a^2+b^2+c^2$ such as using inequalities and calculus by assuming function $f(x,y,z)=x^2+y^2+z^2$.
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Think geometrically: The equation $3x+2y+z=7$ describes a plane in $\mathbb{R}^3$ with normal vector proportional to $(3,2,1)$. The closest point to the origin is therefore of the form $(a,b,c)=(3r,2r,r)$. From $3a+2b+c=7$ we have $9r+4r+r=7$, so $r=1/2$, and thus $a^2+b^2+c^2=(9+4+1)/4=7/2$.
(Rohan Nuckchady posted this approach as a comment beneath the OP. I either missed it, or was composing my answer when it appeared.)
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If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.
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The solution is due to @Calvin Lin.
Problem: Find the best constant $C$
such that $x^2+y^2 - 1 \ge C(1-x)(1-y)$ holds for all
$x, y \ge 0$ with $x^3+y^3 = 1$.
Solution: The best constant is $C = 2^{4/3} + 2^{2/3} + 2$. Let us prove it.
Let $w = x + y$. Since $x^3+y^3 \le (x+y)^3 \le 4(x^3 + y^3)$, we have $1\le w \le \sqrt[3]{4}$.
We have
\begin{align}
&(x+y)[x^2+y^2 - 1 - C(1-x)(1-y)]\\
=\ & x^3+y^3 - (C-1)xy(x+y) + C(x+y)^2 - (C+1)(x+y)\\
=\ & x^3+y^3 - \frac{C-1}{3}[(x+y)^3-x^3-y^3] + C(x+y)^2 - (C+1)(x+y)\\
=\ & 1 - \frac{C-1}{3}[(x+y)^3-1] + C(x+y)^2 - (C+1)(x+y)\\
=\ & \frac{1}{3}(w-1)^2[2 + w - C(w-1)].
\end{align}
From $2 + w - C(w-1) \ge 0$ for $1\le w \le \sqrt[3]{4}$,
we have
$$C \le \inf_{1 < w \le \sqrt[3]{4}} \frac{2+w}{w-1} = 2^{4/3} + 2^{2/3} + 2.$$
On the other hand, when $C = 2^{4/3} + 2^{2/3} + 2$, we have
$$2 + w - C(w-1) = (2^{4/3} + 2^{2/3} + 1)(\sqrt[3]{4} - w) \ge 0$$
for $1\le w \le \sqrt[3]{4}$. We are done.
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Taylor series for 1/(1-x). I am trying to write a Taylor series for $$ f(x)=\frac{1}{1-x}, \ x<1 \ .$$
In most sources, it is said, that this function can be written as a Taylor series, if $$ \left| x \right|<1. $$
However, I don't get the same condition for x.
Because $$ f^{\left(n\right)}\left(x\right)=\frac{n!}{\left(1-x\right)^{n+1}},$$ the remainder term (or its abolute value) is $$ \left|\frac{x^{n+1}}{\left(1-c\right)^{n+2}}\right|=\frac{\left|x\right|^{n+1}}{\left|1-c\right|^{n+2}}{,}\ \mathrm{where} \ 0\le c\le x<1\ \mathrm{or}\ x\le c\le0.$$
If $$ 0 \le c \le x<1, $$ then
$$ 0\ge-c\ge-x>-1\ \Leftrightarrow\ 1\ge1-c\ge1-x>0\ \Leftrightarrow\ 0<\left(1-x\right)^{n+2}\le\left(1-c\right)^{n+2}\le1\ \Leftrightarrow\ 1\le\frac{1}{\left(1-c\right)^{n+2}}\le\frac{1}{\left(1-x\right)^{n+2}} \Leftrightarrow\ \left| x^{n+1} \right|\le \left| \frac{x^{n+1}}{\left(1-c\right)^{n+2}} \right| \le \left| \frac{x^{n+1}}{\left(1-x\right)^{n+2}} \right|=\begin{cases}
e^{\ln\frac{x}{1-x}\cdot n+\left(\ln x-2\ln\left(1-x\right)\right)}\ , \ 0<x<1&\\
0\ , \ x=0&
\end{cases} .$$
When $$ n \rightarrow \infty, $$ for remainder to converge to zero, x has to be such that $$ 0 \le x < \frac{1}{2}.$$
Then the lower bound of the remainder $$ \left|x^{n+1} \right| \rightarrow 0 $$ and the upper bound $$ \left| \frac{x^{n+1}}{\left(1-x\right)^{n+2}}\right| \rightarrow 0 .$$
If $$ x \le c \le 0 ,$$ then
$$ \ -x\ge-c\ge0\ \Leftrightarrow\ 1-x\ge1-c\ge1\ \Leftrightarrow\ \left(1-x\right)^{n+2}\ge\left(1-c\right)^{n+2}\ge1\ \Leftrightarrow\ \frac{1}{\left(1-x\right)^{n+2}}\le\frac{1}{\left(1-c\right)^{n+2}}\le1\ \Leftrightarrow\ \left|x\right|^{n+1}\ge\left|\frac{x^{n+1}}{\left(1-c\right)^{n+2}}\right|\ge\left|\frac{x^{n+1}}{\left(1-x\right)^{n+2}}\right|=\frac{\left|x\right|^{n+1}}{\left|1-x\right|^{n+2}}=\begin{cases}
e^{\ln\left|\frac{x}{1-x}\right|\cdot n+\left(\ln\left|x\right|-2\ln\left|1-x\right|\right){,}\, x<0}&\\
0\ , \ x=0&
\end{cases} . $$
When $$ n \rightarrow \infty, $$ for remainder to converge to zero, x has to be such that $$ -1 < x \le 0.$$
Then the upper bound of the remainder $$\left|x\right|^{n+1} \rightarrow 0 $$ and the lower bound $$ \frac{\left|x\right|^{n+1}}{\left|1-x\right|^{n+2}} \rightarrow 0. $$
Thus I would say, that function f can be written as Taylor series only when $$ -1 < x < \frac{1}{2} . $$ Isn't this true?
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It is a well-known and indisputable (even elementary) fact that for all $x\ne1$ and all natural $n$
$$\sum_{k=0}^n x^k=\frac1{1-x}-\frac{x{^{n+1}}}{1-x},$$
implying that $\forall|x|<1$,
$$\lim_{n\to\infty}\sum_{k=0}^n x^k=\frac1{1-x}.$$
This falsifies your claim.
Needless to say,
$$\left.\left(\frac1{1-x}\right)^{(k)}\right|_{x=0}=k!$$
Note that the remainder can be evaluated exactly as
$$\sum_{k=n+1}^\infty x^k=\frac{x^{n+1}}{1-x}.$$
Comparing to the Taylor-Lagrange expression of the remainder,
$$\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\frac{x^{n+1}}{(1-c)^{n+2}},$$
we have
$$c=1-\sqrt[n-2]{1-x}\le1.$$
This formula shows that a partial sum plus the remainder yield a correct evaluation of the function for all $x<1$ and cannot work for $x\ge1$ (because of the singularity).
In fact, the study of the Taylor-Lagrange remainder cannot tell you about convergence.
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Find maximum point of $f(x,y,z) = 8x^2 +4yz -16z +600$ with one restriction I need to find the critical points of $$f(x,y,z) = 8x^2 +4yz -16z +600$$ restricted by $4x^2+y^2+4z^2=16$.
I constructed the lagrangian function $$L(x, y, z, \lambda ) = 8x^2 +4yz -16z +600 - \lambda (4x^2+y^2+4z^2-16) $$
but I'm very confused about how to determine those points. I know I need to make a system with all the first derivatives of $L$ equaled to $0$. I did it but every time I try to solve it I get different solutions. How can I get the points?
Thanks.
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Using Lagrange Multipliers, we wish to find the points $(x, y, z)$ such that $\nabla f(x, y, z) = \lambda \nabla g(x, y, z),$ where $g(x, y, z) = 4x^2 + y^2 + 4z^2 = 16$ and $\lambda$ is some constant. Observe that the gradients are given by $\nabla f = \langle f_x, f_y, f_z \rangle = \langle 16x, 4z, 4y - 16 \rangle$ and $\nabla g = \langle 8x, 2y, 8z \rangle,$ hence we must solve the following $4 \times 4$ system of equations. $$\begin{cases} 16x = 8 \lambda x \\ 4z = 2 \lambda y \\ 4y - 16 = 8 \lambda z \\ 4x^2 + y^2 + 4z^2 = 16\end{cases}$$ Using the first equation, we have that $(16 - 8 \lambda)x = 0,$ from which it follows that $\lambda = 2$ or $x = 0.$ Using the second equation, we have that $2z = \lambda y$ so that $4z^2 = \lambda^2 y^2.$ Using the third equation, we have that $4y = 8 \lambda z + 16 = 4 \lambda^2 y + 16$ (by the second equation) so that $(4 \lambda^2 - 4)y + 16 = 0.$ Using the fourth equation, we have that $4x^2 + (\lambda^2 + 1) y^2 = 16$ (by the second equation). $$\begin{cases} 4z^2 = \lambda^2 y^2 \\ (4 \lambda^2 - 4)y + 16 = 0 \\ 4x^2 + (\lambda^2 + 1) y^2 = 16 \end{cases}$$
Given that $\lambda = 2,$ we may solve for $y$ in the second equation; solve for $z$ in the first equation; and solve for $x$ in the third equation. On the other hand, if $x = 0,$ then we are dealing with the functions $f(0, y, z) = 4yz - 16z + 600 = h(y, z)$ and $g(0, y, z) = y^2 + 4z^2 = k(y, z) = 16,$ and we can use Lagrange Multipliers to find the critical points of $h(y, z)$ with respect to the constraint $k(y, z) = 16.$ Explicitly, we have that $\langle 4z, 4y - 16 \rangle = \nabla h(y, z) = \mu \nabla k(y, z) = \mu \langle 2y, 8z \rangle$ so that $$\begin{cases} 4z = 2 \mu y \\ 4y - 16 = 8 \mu z \end{cases}$$ is the relevant system of equations. Considering that $\mu$ is nonzero, we can eliminate it by taking $$16 \mu z^2 = (2z)(8 \mu z) = (2z)(4y - 16) = (\mu y)(4y - 16) = \mu (4y^2 - 16y)$$ and cancelling a factor of $\mu$ from the left- and right-hand sides. We have therefore that $16z^2 = 4y^2 - 16y.$ Use the fact that $16z^2 = 4(16 - y^2)$ to solve for $y.$
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Integral $\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx$ For $a>0$ and $D>0$, I want to evaluate the integral
$$\int_{-D}^D \frac{1}{x-ia} \log \frac{(x-2D)^2+a^2}{x^2+a^2} dx.$$
Here $i$ is the imaginary unit. This integral arises when evaluating a Feynman diagram in physics.
|
It does not seem to be the most pleasant integral. Focusing on the antoderivative, let first $x=a y$ and $k=\frac{2D}a$to make
$$I=\int \frac{1}{x-ia} \log \left(\frac{(x-2D)^2+a^2}{x^2+a^2}\right)\,dx=\int\frac{1}{y-i}\log \left(\frac{\left(y-k\right)^2+1}{y^2+1}\right)\,dy$$ Deloping the logarithm of the ratio, you then face two integrals looking like
$$J_k=\int \frac{1}{y-i}\log \left({\left(y-k\right)^2+1}{}\right)\,dy$$
$$J_k=\text{Li}_2\left(\frac{k-y-i}{k-2
i}\right)+\text{Li}_2\left(\frac{k-y+i}{k}\right)+\log (y-i) \log
\left((k-y)^2+1\right)-\log (k) \log (-k+y-i)+\left(\log \left(\frac{y-i}{k-2
i}\right)-\log (y-i)\right) \log (-k+y+i)$$
$$J_0=\text{Li}_2\left(-\frac{1}{2} i (y+i)\right)+\left(\log \left(y^2+1\right)-\log
(y+i)\right) \log (y-i)-\frac{1}{2} \log ^2(y-i)+$$ $$\log \left(\frac{1}{2} (1+i
y)\right) \log (y+i)$$
The definite integral will not be very nice looking.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3746761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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|
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.
|
Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ be points on the $xy$-plane. Then the points that divide the line segment $\overline{AB}$ in the ratio $m:n$ are
$$ P = \left( \frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \right), \quad P = \left( \frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n} \right). $$
The point $P$ is the internal divisor and the point $Q$ the external divisor.
We want $x \in \mathbb R$ such that
$$ 3 \left|x-\frac{5}{3}\right| > 2 \left|x-\left(-\frac{3}{2}\right)\right|. $$
Take $A=\left(-\frac{3}{2},0\right)$ and $B=\left(\frac{5}{3},0\right)$, and look for points $P$ and $Q$ that divide the line segment $\overline{AB}$ in the ratio $3:2$.
The formulae above give
$$ P = \left(\frac{(3 \cdot \frac{5}{3})+(2 \cdot -\frac{3}{2})}{3+2},0 \right) = \left(\frac{2}{5},0 \right), \quad Q = \left(\frac{(3 \cdot \frac{5}{3})-(2 \cdot -\frac{3}{2})}{3-2},0 \right) = (8,0). $$
Therefore,
$$ 3 \left|x-\frac{5}{3}\right| > 2 \left|x-\left(-\frac{3}{2}\right)\right| \Longleftrightarrow x>8 \:\:\text{or}\:\: x<\frac{2}{5}. $$
|
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"url": "https://math.stackexchange.com/questions/3747907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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|
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}
{
x
}
}
$$
Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
x
}
\right]
}
{
x
}
}
$$
Multiplying and dividing by the right terms:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}+\sin^2{x}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Grouping the $\sin{x}$ in the first exponential function:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$
We should end up with:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
=
\frac
{
\exp{\left[1(1)(1)\right]}
-
\exp{\left[1(1)\right]}
}
{
0
}
= \frac{e-e}{0}
= \frac{0}{0}
$$
Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.
|
Hint: Write the function of interest as
$$
(1+\sin x)^\frac{1}{x}\,\frac{e^{\frac{1}{x}\log(1+\frac{\sin^2 x}{1+\sin x})}-1}{x}
$$
and remember that
$$
\lim_{x\to0}\frac{\sin x}{x}=1\,,\qquad \lim_{x\to0}\frac{\log(1+x)}{x}=1\,,\qquad \lim_{x\to0}\frac{e^{x}-1}{x}=1\,.
$$
You should find that the limit is $e$.
|
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"url": "https://math.stackexchange.com/questions/3748637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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|
Integer solutions of $2a+2b-ab\gt 0$ Let $a\in\mathbb{N}_{\ge 3}$ and $b\in\mathbb{N}_{\ge 3}$. What are the solutions of the Diophantine inequality
$$2a+2b-ab\gt 0?$$
By guessing, I found 5 solutions:
$$\text{1)}\, a=3,\, b=3$$
$$\text{2)}\, a=3,\, b=4$$
$$\text{3)}\, a=4,\, b=3$$
$$\text{4)}\, a=5,\, b=3$$
$$\text{5)}\, a=3,\, b=5.$$
Are these all the solutions? How could I find all the solutions rigorously?
|
$2a+2b-ab>0$ is equivalent to $(a-2)(b-2)<4$, or to $(a-2)(b-2) \le 3$. If $a \ge 3$ and $b \ge 3$, then $(a-2)(b-2) \ge 1$. Thus, $(a-2)(b-2) \in \{1,2,3\}$.
$\bullet$ If $(a-2)(b-2)=1$, then $a-2=b-2=1$, so that $(a,b)=(3,3)$.
$\bullet$ If $(a-2)(b-2)=2$, then $\{a-2,b-2\}=\{1,2\}$, so that $\{a,b\}=\{3,4\}$.
$\bullet$ If $(a-2)(b-2)=3$, then $\{a-2,b-2\}=\{1,3\}$, so that $\{a,b\}=\{3,5\}$.
Therefore, we have the five solutions $(a,b)=(3,3), (3,4), (4,3), (3,5),\:\text{or}\: (5,3)$. $\blacksquare$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3749995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
System of quadratic homogeneous Diophantine equations Is there a method for determining if a system of quadratic diophantine equations has any solutions?
My specific question is:
\begin{align*}
a^2+b^2&=c^2\\[4pt]
2a^2+b^2&=d^2\\
\end{align*}
I want to know if there are any positive integers $a, b, c, d$ which satisfy both equations.
The question came from my student and I cannot answer her in class.
|
Suppose $(a,b,c,d)$ is a solution to the system
$$
\left\lbrace
\begin{align*}
&a^2+b^2=c^2\\[4pt]
&2a^2+b^2=d^2\\[4pt]
\end{align*}
\right.
$$
where $a,b,c,d$ are positive integers, and where $d$ is as small as possible.
If $a,b$ have a common factor $g > 1$, then $c$ and $d$ must also be divisible by $g$, hence
$$\left(\frac{a}{g},\frac{b}{g},\frac{c}{g},\frac{d}{g}\right)$$
would also be a solution, contrary to the assumed minimality of $d$.
Hence $\gcd(a,b)=1$.
Suppose $b$ is even.
Then from $2a^2+b^2=d^2$, it follows that $d$ is even.
But then from $2a^2=d^2-b^2$, it follows that $2a^2$ is a multiple of $4$, hence $a$ is even, contrary to $\gcd(a,b)=1$.
Hence $b$ is odd.
Since $\gcd(a,b)=1$ and $b$ is odd, using the standard form for primitive Pythagorean triples, it follows that the equation
$$a^2+b^2=c^2$$
holds if and only if
$$
\left\lbrace
\begin{align*}
a&=2st\\[4pt]
b&=s^2-t^2\\[4pt]
c&=s^2+t^2\\[4pt]
\end{align*}
\right.
$$
for some positive integers $s,t$ such that
*
*$s > t$.$\\[4pt]$
*$\gcd(s,t)=1$.$\\[4pt]$
*One of $s,t$ is even (and the other is odd).
Then the equation $2a^2+b^2=d^2$ can be recast as
$$
2(2st)^2+(s^2-t^2)^2=d^2
$$
or equivalently
$$
s^4+6s^2t^2+t^4=d^2
$$
hence, letting $x={\Large{\frac{t}{s}}}$ and $y={\Large{\frac{d}{s}}}$, it follows that the equation
$$y^2=x^4+6x^2+1$$
has a rational point $(x,y)$ with $0 < x < 1$.
That's as far as I can go.
I suspect there is no such rational point, but if so, I doubt that it can be proved in an elementary way.
Using more advanced methods, perhaps there is a way to convert the equation $y^2=x^4+6x^2+1$ to the equation of an elliptic curve in such a way that the non-existence of qualifying rational points on the curve $y^2=x^4+6x^2+1$ can be inferred from the non-existence of qualifying rational points on the elliptic curve.
|
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"url": "https://math.stackexchange.com/questions/3750610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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|
Differentiate $\left(x^6-2x^2\right) \ln\left(x\right) \sin\left(x\right)$ Differentiate
$$\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)$$
with respect to $x$
My work so far
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)\right]}}$
$=\class{steps-node}{\cssId{steps-node-3}{\class{steps-node}{\cssId{steps-node-2}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^6-2x^2\right]}}\cdot\ln\left(x\right)\sin\left(x\right)}}+\class{steps-node}{\cssId{steps-node-5}{\left(x^6-2x^2\right)\cdot\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\ln\left(x\right)\right]}}\cdot\sin\left(x\right)}}+\class{steps-node}{\cssId{steps-node-7}{\left(x^6-2x^2\right)\ln\left(x\right)\cdot\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\sin\left(x\right)\right]}}}}$
$=\class{steps-node}{\cssId{steps-node-8}{\left(\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^6\right]}}-2\cdot\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^2\right]}}\right)}}\ln\left(x\right)\sin\left(x\right)+\left(x^6-2x^2\right)\cdot\class{steps-node}{\cssId{steps-node-11}{\dfrac{1}{x}}}\sin\left(x\right)+\left(x^6-2x^2\right)\ln\left(x\right)\class{steps-node}{\cssId{steps-node-12}{\cos\left(x\right)}}$
$=\left(\class{steps-node}{\cssId{steps-node-13}{6}}\class{steps-node}{\cssId{steps-node-14}{x^5}}-2\cdot\class{steps-node}{\cssId{steps-node-15}{2}}\class{steps-node}{\cssId{steps-node-16}{x}}\right)\ln\left(x\right)\sin\left(x\right)+\dfrac{\left(x^6-2x^2\right)\sin\left(x\right)}{x}+\left(x^6-2x^2\right)\ln\left(x\right)\cos\left(x\right)$
$=\left(6x^5-4x\right)\ln\left(x\right)\sin\left(x\right)+\dfrac{\left(x^6-2x^2\right)\sin\left(x\right)}{x}+\left(x^6-2x^2\right)\cos\left(x\right)\ln\left(x\right)$
$=x\left(\left(\left(6x^4-4\right)\ln\left(x\right)+x^4-2\right)\sin\left(x\right)+\left(x^5-2x\right)\cos\left(x\right)\ln\left(x\right)\right)$
I've had great difficulty in solving this. Was my method correct?
Also, would there be any shortcuts in solving this, or is this method the best way of getting the solution?
|
Your method is correct. If you have an derivative of a function that is the product of three other functions, it looks like this:
$$h(x)=f(x)\cdot g(x)\cdot y(x)$$
$$h'(x)=f'(x)\cdot g(x)\cdot y(x)+f(x)\cdot g'(x)\cdot y(x)+f(x)\cdot g(x)\cdot y'(x)$$
This derivative is just particularly algebraically intense, and at the end of the day I would leave it as you have on line 3.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3751546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding the closed form of $\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:\mathrm{d}x$ I solved a similar case which is also a very well known integral
$$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x=\frac{\pi }{4}\ln \left(2\right)+G$$
My teacher gave me a hint which was splitting the integral at the point $1$,
$$\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x+\int _1^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x=\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x+\int _0^1\frac{\ln \left(\frac{1+x}{x}\right)}{1+x^2}\:\mathrm{d}x$$
$$2\int _0^1\frac{\ln \left(1+x\right)}{1+x^2}\:\mathrm{d}x-\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:\mathrm{d}x=\frac{\pi }{4}\ln \left(2\right)+G$$
I used the values for each integral since they are very well known.
My question is, can this integral be generalized for $a>0$?, in other words can similar tools help me calculate
$$\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:\mathrm{d}x$$
|
You can evaluate this integral with Feynman's trick,
$$I\left(a\right)=\int _0^{\infty }\frac{\ln \left(1+ax\right)}{1+x^2}\:dx$$
$$I'\left(a\right)=\int _0^{\infty }\frac{x}{\left(1+x^2\right)\left(1+ax\right)}\:dx=\frac{1}{1+a^2}\int _0^{\infty }\left(\frac{x+a}{1+x^2}-\frac{a}{1+ax}\right)\:dx$$
$$=\frac{1}{1+a^2}\:\left(\frac{1}{2}\ln \left(1+x^2\right)+a\arctan \left(x\right)-\ln \left(1+ax\right)\right)\Biggr|^{\infty }_0=\frac{1}{1+a^2}\:\left(\frac{a\pi \:}{2}-\ln \left(a\right)\right)$$
To find $I\left(a\right)$ we have to integrate again with convenient bounds,
$$\int _0^aI'\left(a\right)\:da=\:\frac{\pi }{2}\int _0^a\frac{a}{1+a^2}\:da-\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$$
$$I\left(a\right)=\:\frac{\pi }{4}\ln \left(1+a^2\right)-\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$$
To solve $\displaystyle\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da$ first IBP.
$$\int _0^a\frac{\ln \left(a\right)}{1+a^2}\:da=\ln \left(a\right)\arctan \left(a\right)-\int _0^a\frac{\arctan \left(a\right)}{a}\:da=\ln \left(a\right)\arctan \left(a\right)-\text{Ti}_2\left(a\right)$$
Plugging that back we conclude that
$$\boxed{I\left(a\right)=\:\frac{\pi }{4}\ln \left(1+a^2\right)-\ln \left(a\right)\arctan \left(a\right)+\text{Ti}_2\left(a\right)}$$
Where $\text{Ti}_2\left(a\right)$ is the Inverse Tangent Integral.
The integral you evaluated can be proved with this,
$$I\left(1\right)=\int _0^{\infty }\frac{\ln \left(1+x\right)}{1+x^2}\:dx=\frac{\pi }{4}\ln \left(2\right)-\ln \left(1\right)\arctan \left(1\right)+\text{Ti}_2\left(1\right)$$
$$=\frac{\pi }{4}\ln \left(2\right)+G$$
Here $G$ denotes the Catalan's constant.
|
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"url": "https://math.stackexchange.com/questions/3751665",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
$$x^2 -xz+z^2 +y^2-yz+z^2+2\sqrt{(x^2-xz+z^2)(y^2-yz+z^2)} - x^2 - xy - y^2 \geq 0$$
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2$$
Squaring both sides yields
$$4(x^2-xz+z^2)(y^2-yz+z^2)\geq(xy+xz+yz-2z^2)^2$$
After some expanding it becomes
$$(xy-xz-yz)^2 \geq 0$$
which completes the proof.
However, I want to ask whether the squaring in the $3^{rd}$ - $4^{th}$step is problematic. Since it is possible for $xy + xz+ yz -2z^2$ to be negative. Squaring both sides will invert the sign. Do I have to analyse the positive and negative case here seperately?
|
You got that we need to prove that:
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2.$$
Now, by C-S $$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}=\sqrt{(2x^2-2xz+2z^2)(2y^2-2yz+2z^2)}=$$
$$=\sqrt{((x-z)^2+x^2+z^2)(z^2+y^2+(y-z)^2)}\geq$$
$$\geq(x-z)z+xy+z(y-z)=xy+xz+yz-2z^2$$ and we are done!
Now we see that the starting inequality is true for any reals $x$, $y$ and $z$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
|
Here is another easier way to integrate as follows
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac14\frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac14\int \frac{x}{\sqrt{4x^2+9}}dx-\frac14\int \frac{9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac1{32}\int \frac{d(4x^2+9)}{\sqrt{4x^2+9}}-\frac9{32}\int \frac{d(4x^2+9)}{(4x^2+9)^{3/2}}$$
$$=\frac1{32}2\sqrt{4x^2+9}-\frac9{32}\frac{-2}{\sqrt{4x^2+9}}+C$$
$$=\bbox[15px,#ffd,border:1px solid green ]{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve Complex Equation $z^3 = 4\bar{z}$ I'm trying to solve for all z values where $z^3 = 4\bar{z}$.
I tried using $z^3 = |z|(\cos(3\theta)+i\sin(3\theta)$ and that $|z| = \sqrt{x^2+y^2}$ so:
$$z^3 = \sqrt{x^2+y^2}(\cos(3\theta)+\sin(3\theta))$$ and $$4\bar z = 4x-4iy = 4r\cos(\theta)-i4r\sin(\theta)$$
but I have no idea where to go from there.
|
If $z^3=4\overline z$, then $z^4=4z\overline z=4|z|^2$. So, $|z|^4=|z^4|=4|z|^2$, and therefore $z=0$ or $|z|=2$. So, unless $z=0$, $z$ can be written as $2(\cos\theta+i\sin\theta)$, in which case$$z^3=4\overline z\iff8\bigl(\cos(3\theta)+i\sin(3\theta)\bigr)=8(\cos\theta-i\sin\theta).$$Can you take it from here?
|
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|
Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $ 1+\frac{x^2}{6} $ [what I got] then think again!". My attempt was:
When $ x $ is small, $ \sin x \approx x $ so
$$ \frac{\sin^2{x}}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} = \frac{x^2}{x^2 \sqrt{1-\frac{x^2}{3}}} = \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} $$
Then using the binomial series approximation,
$$ \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6} $$
...and so it looks like I've fallen into whatever trap the question set.
Where is my error?
|
You can also note the the singularity at $x=0$ is removable and that $f$ is in fact at least 4 times differentiable. Taylor's expansion gives you the answer...
$$
f(x)=f(0)+f'(0)x + \frac 12 f''(0) x^2 + O(x^3)
$$
where
$$
f(0)=\lim_{x\to 0}f(0) = 1, \quad f'(0) = \lim_{x\to 0}f'(x)=0, \quad f''(0)=\lim_{x\to 0}f''(x)= -\frac 13, \quad f'''(0)=0
$$
yielding
$$
f(x)=1-\frac 16 x^2 + O(x^4)\approx 1-\frac 16 x^2 (\textrm{for small } x ).
$$
|
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|
How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the entire denominator, and set $u$ equal to each expression in the radical.
However, all my attempts have just made the integral more complex without an obvious way to simplify. Can someone provide insight please? Thank you.
|
$$\int \frac{1}{\sqrt{2x}-\sqrt{x+4}}\ dx=\int \frac{(\sqrt{2x}+\sqrt{x+4})}{(\sqrt{2x}-\sqrt{x+4})(\sqrt{2x}+\sqrt{x+4})}\ dx=$$
$$=\int \frac{\sqrt{2x}+\sqrt{x+4}}{x-4}\ dx$$
$$=\int \frac{\sqrt{2x}\ dx}{x-4} + \int \frac{\sqrt{x+4}\ dx}{x-4}$$
$$=\int \frac{\sqrt{2}\ xd(\sqrt{x})}{x-4} + \int \frac{2(x+4)d(\sqrt{x+4})}{x-4}$$
$$=\int \frac{(2\sqrt{2}(x-4)+8\sqrt2)d(\sqrt{x})}{x-4} + \int \frac{(2(x-4)+16)d(\sqrt{x+4})}{x-4}$$
$$=2\sqrt{2} \int d(\sqrt{x})+8\sqrt2\int \frac{d(\sqrt{x})}{(\sqrt{x})^2-2^2} + 2\int d(\sqrt{x+4})+16\int \frac{d(\sqrt{x+4})}{(\sqrt{x+4})^2-(2\sqrt2)^2}$$
$$=2\sqrt2\sqrt{x}+8\sqrt{2}\frac{1}{2\cdot 2}\ln\left|\frac{\sqrt x-2}{\sqrt{x}+2}\right|+2\sqrt{x+4}+16\frac{1}{2\cdot 2\sqrt2}\ln\left|\frac{\sqrt{x+4}-2\sqrt2}{\sqrt{x+4}+2\sqrt2}\right|$$
$$=2\sqrt{2x}+2\sqrt{2}\ln\left|\frac{\sqrt x-2}{\sqrt{x}+2}\right|+2\sqrt{x+4}+2\sqrt2\ln\left|\frac{\sqrt{x+4}-2\sqrt2}{\sqrt{x+4}+2\sqrt2}\right|+C$$
|
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Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$. Let $a,b$ be coprime integers. I am trying to find all integers $x,y$ such that:
$$x^3+5y^3=(a^3+5b^3)^3$$
What I have tried:
$$5y^3=(a^3+5b^3-x)[(a^3+5b^3)^2+(a^2+5b^3)x+x^2 ]$$
There are 2 Cases depending which factor of $5y^3$ is divisible by $5$:
Case 1: $5$ divides the first factor:
$$a^3+5b^3-x=5y_0^{3-n}y_1^3$$
$$(a^3+5b^3)^2+(a^2+5b^3)x+x^2 =y_0^ny_2^3$$
where $$y=y_0y_1y_2$$ $$(y_1,y_2)=1$$ $$n=\{0,1,2,3\}$$
It follows $$x=a^3+5b^3-5y_0^{3-n}y_1^3$$
We substitute $x$
$$(a^3+5b^3)^2+(a^2+5b^3)(a^3+5b^3-5y_0^{3-n}y_1^3)+(a^3+5b^3-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
For simplicity, we write $u=a^3+5b^3$:
$$u^2+u(u-5y_0^{3-n}y_1^3)+(u-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
$$u^2+u^2-5y_0^{3-n}y_1^3u+u^2-10uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
$$3u^2-15uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
What do I do at this point? Any input will be appreciated. Thanks.
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As noted in the other answer, this is a question about rational points on elliptic curves:
Claim: There are no solutions to $x^{3}+y^{3} = 5$ with $x,y \in \mathbb{Q}$.
This follows from the following Proposition:
Proposition: Set $R := \mathbb{Z}[t]/(t^{2}+t+1)$. Let $A,B,C$ be elements of $R_{\mathbb{Q}}$ such that \begin{align*} A+B+C = 0 \quad\text{ and }\quad ABC = 5w^{3} \end{align*} for some $w \in R_{\mathbb{Q}}$. Then two of $A,B,C$ must be equal.
Proof of Proposition: See this post by Paul Monsky (with "2" instead of "5"); I write some more details here. It seems this argument is called "3-descent" and is usually attributed to Sylvester, Pepin, Lucas in the literature. I think we can replace "5" with any prime $p \in \mathbb{Z}$ such that $p$ remains prime in $R$.
We will use without proof that $R$ is a UFD. Let $\xi \in R$ be the class of $t$. There is a norm map \begin{align*} |\;\;| : R \to \mathbb{Z} \end{align*} sending \begin{align*} a+b\xi \mapsto \det(a\begin{bmatrix} 1 & \\ & 1 \end{bmatrix} + b\begin{bmatrix} & -1 \\ 1 & -1 \end{bmatrix}) = a^{2}-ab+b^{2} = (a+b\xi)(a+b\xi^{2}) \end{align*} for all $a,b\in \mathbb{Z}$. (We always have $a^{2}-ab+b^{2} = (a-\frac{1}{2}b)^{2}+\frac{3}{4} b^{2} \ge 0$.) The norm is multiplicative, satisfies the triangle inequality, and $a+b\xi$ is a unit of $R$ if and only if $|a+b\xi|$ is a unit of $\mathbb{Z}$ (i.e. $|a+b\xi| = 1$).
If there's a counterexample to the Proposition, there's one with $A,B,C$ in $R$ by clearing denominators (then $w \in R$ also); take such a counterexample with \begin{align*} d = \max(|A|,|B|,|C|) \end{align*} as small as possible. Here necessarily $d > 1$ (otherwise each $A,B,C$ would be either $0$ or a unit); in fact $d > 2$ since $2$ is not a norm from $R$ (the equation $(2a-b)^{2}+3b^{2} = 8$ has no integer solutions). Since $A+B+C = 0$ and $(A,B,C)$ was assumed minimal, we have that $A,B,C$ are pairwise coprime. Note that $5$ is a prime in $R$ since $R/(5)R \simeq \mathbb{F}_{5}[t]/(t^{2}+t+1)$ is a field. Note also that the only units of $R$ are $\pm \xi^{i}$ for $i \in \{0,1,2\}$. By unique factorization in $R$, there exist $r,s,t \in R$ such that \begin{align*} A = 5\xi^{i}r^{3} \quad\text{ and }\quad B = \xi^{j}s^{3} \quad\text{ and }\quad C = \xi^{k}t^{3} \end{align*} for some $i,j,k \in \{0,1,2\}$. Here $w^{3} = \xi^{i+j+k}(rst)^{3}$ so $i+j+k \equiv 0 \pmod{3}$; we may thus assume that $i=j=k=0$, so that \begin{align*} A = 5r^{3} \quad\text{ and }\quad B = s^{3} \quad\text{ and }\quad C = t^{3} \end{align*} in $R$. Consider \begin{align*} A' := s+t \quad,\quad B' := \xi^{1}s+\xi^{2}t \quad,\quad C' := \xi^{2}s+\xi^{1}t \end{align*} in $R$; then $A'+B'+C' = 0$ and $A'B'C' = s^{3}+t^{3} = -5r^{3}$; it remains to show that they are distinct. We have $\xi^{1}s+\xi^{2}t \ne \xi^{2}s+\xi^{1}t$ since $\xi^{1}-\xi^{2}$ is a nonzerodivisor in $R$ (since $t-1$ does not divide $t^{2}+t+1$) and $s,t$ are distinct (since $B,C$ are distinct). We have $s+t \ne \xi^{1}s+\xi^{2}t$ (since otherwise $(1-\xi)s = (\xi^{2}-1)t$ implies $-s = (\xi+1)t$ implies $s = \xi^{2}t$ but $s,t$ are coprime in $R$). By symmetry, we have $s+t \ne \xi^{2}s+\xi^{1}t$.
We have $|\xi^{i}s+\xi^{j}t| \le |s|+|t| = |B|^{1/3}+|C|^{1/3} \le 2d^{1/3}$, so $(A',B',C')$ is a new counterexample with $\max(|A'|,|B'|,|C'|) \le 2d^{1/3} < d$.
|
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|
Is it possible to justify these approximations about prime numbers? A recently closed question asked for a possible closed form of the infinite summation
$$f(a)=\sum _{i=1}^{\infty } a^{-p_i}$$ for which I already proposed a first simple but totally empirical approximation.
Since we quickly face very small numbers, I tried to find approximations of
$$g(a)=\Big[\sum _{i=1}^{\infty } a^{-p_i}\Big]^{-1} \qquad \text{and} \qquad h(a)=\Big[\sum _{i=1}^{\infty } (-1)^{i-1} a^{-p_i}\Big]^{-1}$$
All calculations where done with integer values of $a$ for the range $2 \leq a \leq 1000$.
What I obtained is
$$\color{blue}{g(a)\sim\frac{(a-1) (2a^3+2a-1)}{2 a^2}}\qquad \text{and} \qquad \color{blue}{h(a)\sim\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}}$$
If the corresponding curve fits were done, in both cases we should have $R^2 > 0.999999999$.
For the investigated values of $a$,
$$\text{Round}\left[\frac{(a-1) (2a^3+2a-1)}{2 a^2}-{g(a)}\right]=0$$
$$\text{Round}\left[\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}-{h(a)}\right]=0$$
Not being very used to work with prime numbers, is there any way to justify, even partly, these approximations ?
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These estimates are correct within a reasonable degree of accuracy. Below is the explanation for $f(a)$; the case for $h(a)$ can be dealt similarly. We have
$$
f(a) = \frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{a^5} + O\bigg(\frac{1}{a^7}\bigg)
$$
whereas
$$
\frac{2a^2}{(a-1)(2a^3 + 2a - 1)} = \frac{1}{a^2} + \frac{1}{a^3} + \frac{1}{2a^5} + \frac{3}{2a^6} + O\bigg(\frac{1}{a^7}\bigg).
$$
Hence,
$$
f(a) = \frac{2a^2}{(a-1)(2a^3 + 2a - 1)} + O\bigg(\frac{1}{a^5}\bigg)
$$
For large values of $a$ the error would obviously be negligible, since it grows no faster than a constant times $a^{-5}$. So this may or may not be a good estimate depending upon weather you are satisfied with the magnitude of the error term $O(a^{-5})$.
The best possible estimate of the form $\dfrac{Ax^2}{(x-1)(Bx^3 + Cx^2 + Dx + E)}$ is obtained by the Laurent series expansion of about the point $x = \infty$ and equating the coefficient of smallest non prime powers to zero which gives $A = B = D = 1, C = 0,E = -1$.
Hence we have,
$$
f(a) = \frac{a^2}{(a-1)(a^3 + a - 1)} + O\bigg(\frac{1}{a^6}\bigg)
$$
which reduces the error by a factor of $a$.
Update 21-Jul-2020: However, using basic properties of primes we can get remarkably sharper estimates. Since every primes $\ge 5$ are of the form $6k \pm 1$, by summing up the geometric sequences $a^{-6k-1} + a^{-6k+1}$ for $k = 1,2,\ldots, \infty$ and adding $a^{-2} + a^{-3}$, and taking advantage of the fact the the density of primes among the first few numbers of these form is high we get
$$
f(a) = \frac{a^7 + a^6 + a^4 + a^2 -a - 1}{a^3(a^6 - 1)} + O\bigg(\frac{1}{a^{25}}\bigg)
$$
|
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What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i do next?
|
Okay, so you have shown the expression to be equal to $20\cdot 2^n\cdot 25^n+18\cdot3^n\cdot4^n=20\cdot 50^n+18\cdot12^n$ and as @mwt as shown in comments , write $20=19+1$ and $18=19-1$ to get the expression equal to $19(50^n+12^n)+50^n-12^n$. Now we know that $a^n-b^n$ is divisible by $a-b$ for any natural number $n$. If you don't know that, you can prove it by factorizing $a^n-b^n$.
So $50^n-12^n$ is divisble by $38$ and so $19$ divides the whole expression for any natural number $n$.
|
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|
Contradiction on an Inequality The problem on which I have a problem is this-
Let a, b, c be non-negative real numbers. Prove that $$ \sum_{cyc} {a^2-bc \over 2a^2+b^2+c^2} \ge 0 $$
While solving and after some resolution, we get $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le 3 $$
And by C-S, we have,
$$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le \sum_{cyc} {2(a^2+b^2) \over a^2+b^2+2c^2} $$
It rests to prove that
$$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} \le {3 \over 2} $$
By cyclic substitutions of $x$ for $b^2+c^2$, we get,
$$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} = \sum_{cyc} {(a^2+b^2) \over (c^2+a^2)+(b^2+c^2)} = \sum_{cyc} {z \over x+y} \le {3 \over 2} $$
But by the Nesbitt's Inequality,
$$ \sum_{cyc} {z \over x+y} \ge {3 \over 2} !$$
Can anybody explain me where's the mistake and the correction?
Thanks!
|
Your first step was not so strong, which gave a wrong inequality.
SOS helps here:
We need to prove that $$\sum_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{2a^2+b^2+c^2}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a+c}{2a^2+b^2+c^2}-\frac{b+c}{2b^2+a^2+c^2}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(c^2-(a+b)c+a^2-ab+b^2)(2c^2+a^2+b^2)\geq0,$$ which is true because $$c^2-(a+b)c+a^2-ab+b^2\geq c^2-(a+b)c+\frac{1}{4}(a+b)^2=\frac{1}{4}(2c-a-b)^2\geq0.$$
|
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Finding $\frac{\cot\gamma}{\cot \alpha+\cot\beta}$, given $a^2+b^2=2019c^2$ This is a question that appeared in the $2018$ Southeast Asian Mathematical Olympiad:
In a triangle with sides $a,b,c$ opposite angles $\alpha,\beta,\gamma$, it is known that $$a^2+b^2=2019c^2$$ Find $$\frac{\cot\gamma}{\cot\alpha+\cot\beta} $$
Well, by the Sine Law we have $$\sin^2\alpha+\sin^2\beta=2019\sin^2\gamma$$ and by the Cosine Law, $$\cos\gamma=\frac{a^2+b^2-c^2}{2ab} = \frac{1009c^2}{ab}$$ I’m stuck here. I tried to convert everything in our target expression to sines and cosines, but that makes the expression more complicated. I guess we can use the fact that $\cot\gamma=-\cot(\alpha+\beta)$.
How can you tackle this question? (also, apparently there are no worked solutions online)
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Here is the complete solution from AoPS.
Continuing from Aqua's answer, we know that $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Furthermore, as you have found, $\cos C = \frac{1009c^2}{ab}$. Substituting these values in gives:
$$\frac{\cot\gamma}{\cot \alpha+\cot\beta} = \frac{\cos\gamma \sin \alpha \sin \beta}{\sin ^2\gamma} = \frac{\frac{1009c^2}{ab} \cdot \frac{a}{2R} \cdot \frac{b}{2R}}{\frac{c^2}{4R}} = \frac{\frac{1009c^2}{ab} \cdot ab}{c^2} = \frac{1009c^2}{c^2} = \boxed{1009}.$$
|
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|
What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\
&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\
&=j_{1}+j_{2}+j_{3}\\
\end{align*}
$$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
Wait for a review to find solutions to this
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{a,b,c \in \mathbb{R}_{\ >\ 0}}$:
\begin{align}
\mrm{f}\pars{a,b,c} & \equiv
\int_{0}^{\infty}{\ln\pars{a^{2} + b^{2}x^{2}} \over c^{2} + x^{2}}\,\dd x =
\Re\int_{-\infty}^{\infty}{\ln\pars{a + bx\ic} \over
\pars{x + c\ic}\pars{x - c\ic}}\,\dd x
\end{align}
With the change
$\ds{\pars{x = -\,{s - a \over b}\,\ic \implies s = a + bx\ic}}$:
\begin{align}
\mrm{f}\pars{a,b,c} & =
-\,\Im\int_{a - \infty\ic}^{a + \infty\ic}{b\ln\pars{s} \over
\bracks{s - \pars{a - bc}}\bracks{s - \pars{a + bc}}}\,\dd s
\\[5mm] & =
-\,\Im\bracks{-2\pi\ic\,{b\ln\pars{a + bc} \over 2bc}} =
\bbx{\pi\,{\ln\pars{a + bc} \over c}} \\ &
\end{align}
$$
\left\{\begin{array}{rcccccl}
\ds{j_{1}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{1 + 4x^{2}} \over 9 + x^{2}}\,\dd x}
& \ds{=} & \ds{\mrm{f}\pars{1,2,3}} & \ds{=} & \ds{{\pi \over 3}\ln\pars{7}}
\\[2mm]
\ds{j_{2}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 1 + 9x^{2}}\,\dd x}
& \ds{=} & \ds{{1 \over 9}\,\mrm{f}\pars{2,1,{1 \over 3}}} & \ds{=} & \ds{{\pi \over 3}\,\ln\pars{7 \over 3}}
\\[2mm]
\ds{j_{3}} & \ds{=} &
\ds{\int_{0}^{\infty}{\ln\pars{4 + x^{2}} \over 9 + x^{2}}\,\dd x}
& \ds{=} & \ds{\mrm{f}\pars{2,1,3}} & \ds{=}
& \ds{{\pi \over 3}\,\ln\pars{5}}
\end{array}\right.
$$
\begin{align}
\mbox{} &
\\
j_{1} + j_{2} + j{3} & =
\bbx{{1 \over 3}\,\pi\ln\pars{245 \over 3}} \approx 4.6104 \\ &
\end{align}
|
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"url": "https://math.stackexchange.com/questions/3768690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Combination to find integers satisfying a condition
Let $n$ and $k$ be positive integers such that $n\ge\frac{k(k+1)}{2}$.
The number of solutions $(x_1,x_2,\dots,x_{k})$, with $x_1\ge1$, $x_2\ge2$,..., $x_{k}\ge k$ for all integers satisfying $x_1+x_2+\dots+x_{k}=n$ is?
I substituted the last equation in the first inequality.
$$x_1+x_2+\dots+x_{k}\ge\frac{k(k+1)}{2}.$$
Took $x_1$ as $1+ t_1$, $x_2$ as $2+t_2$...where $t_i\ge 0$. On simplifying by using sum of k numbers, I end with with
$t_1+t_2+\dots+t_{k} \ge0$. Since $x_1,x_2$... are in increasing order, and sum of all $t$ values is $0$, I conclude that this is only possible when $t=0$. Therefore only one solution is possible when $LHS = RHS$. The inequality is not valid.
But the answer is $\frac{1}{2}(2n-k^2+k-2)$. What am I missing here?
|
No, the sequence $x_1,x_2,\dots,x_{k}$ is not necessarily increasing, so your conclusion is not correct.
Let $t_k=x_k-k\geq 0$, then we have to count the number of non-negative integer solutions of
$$t_1+t_2+\dots+t_k=n-\frac{k(k+1)}{2}$$
which is, by Stars-and-Bars, given by
$$\binom{n-\frac{k(k+1)}{2}+k-1}{k-1}=\binom{\frac{2n-k^2+k-2}{2}}{k-1}.$$
P.S. My answer is correct. Probably there is a typo in your book. The integer $\binom{\frac{2n-k^2+k-2}{2}}{k-1}$ is precisely the coefficient of $t^n$ in
$$ (t + t^2 +t^3+ \dots)(t^2 + t^3 +t^4+\dots)\dots (t^k + t^{k+1}+t^{k+2}+\dots),$$
that is
$$[t^n]\prod_{j=1}^k\frac{t^j}{1-t}=[t^{n-\frac{k(k+1)}{2}}](1-t)^{-k}
=(-1)^{{n-\frac{k(k+1)}{2}}}\binom{-k}{n-\frac{k(k+1)}{2}}=\binom{n-\frac{k(k+1)}{2}+k-1}{k-1}.$$
Numerical example. Take $n=8$ and $k=3$, then it is easy to see that
$$ (t + t^2 +t^3+ \dots)(t^2 + t^3+ t^4 +\dots)(t^3 + t^{4}+t^{5}+\dots)
=t^6+3t^7+6t^8+\dots$$
and the coefficient of $t^8$ is $6$:
$$\binom{\frac{16-9+3-2}{2}}{3-1}=\binom{4}{2}=6.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3768789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here
|
Substitute $x = \sin\theta$ and let's restrict our attention to $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ since that covers the whole range of $\sin\theta$:
$$\sin^{-1}\left(\frac{\sin\theta+\cos\theta}{\sqrt{2}}\right) = \frac{\pi}{4}+\sin^{-1}(\sin\theta)$$
$$\implies \sin^{-1}\left(\sin\left(\theta+\frac{\pi}{4}\right)\right) = \frac{\pi}{4}+\theta$$
However, the equality is only true when $$\left|\theta+\frac{\pi}{4}\right|\leq \frac{\pi}{2} \implies -\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}$$
Taking the intersection of this set with our original domain, we have that
$$-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{4} \implies -1 \leq x \leq \frac{1}{\sqrt{2}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}$$
Using $$\frac{n!}{2^n}=\frac{4!}{2^4}\frac{5}{2}\frac{6}{2}\cdots \cdots >1\Longrightarrow \frac{n!}{2^n}>1$$
So $$\frac{2^n}{n!}\leq 1\Longrightarrow \frac{2^n\cdot n^5}{n!}<n^5$$
How do i solve it Help me please, Thanks
|
$$\frac{2^n\cdot n^5}{n!} = \frac{6^n}{n!} \cdot \left(\frac{2}{3}\right)^n<\frac{6^n}{n!} $$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
Consider
$P(x)= ax^9+bx^8+1$ and $\alpha$ one of the numbers$\frac{1 \pm \sqrt 5}{2}.$
We have equalities:
$\alpha^2$ = $\alpha$ + 1, $\alpha^4$ = 3$\alpha$ + 2,
$\alpha^8$ = 21$\alpha$ + 13, $\alpha^9$ = 34$\alpha$ + 21.
(The coefficients to the right of the equations are terms of the Fibonacci sequence)
From $P(\alpha) = 0$ results $a(34\alpha + 21) +b(21\alpha + 13) + 1 =0$ and then
$(34a + 21b)\alpha + 21a + 13b + 1 = 0$
If $a$ and $b$ are rational numbers then $$ a = 21, b = -34.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.
|
Guide:
Case $1$: If $z=1$. Check what happens here.
Case $2$: If $z \ne 1$, then $x+y \ne 0$,
$$x^2-xy+y^2=1+z=1+(1-(x+y))$$
$$x^2+y^2-xy+x+y = 2$$
$$(x^2-xy+x)+(y^2+y)=2$$
$$(x^2-x(y-1))+(y^2+y)=2$$
$$\left(x - \frac{y-1}2\right)^2-\left(\frac{y-1}2 \right)^2 + (y^2+y)=2$$
$$(2x-y+1)^2 -(y^2-2y+1) + 4y^2+4y=8$$
$$(2x-y+1)^2 + 3y^2+6y-1=8$$
$$(2x-y+1)^2+3(y^2+2y+1)=12$$
$$(2x-y+1)^2+3(y+1)^2=12$$
Hence we have $|y+1| \in \{1,2\}$.
*
*If $|y+1|=1$, then $|2x-y+1|=3.$
*If $|y+1|=2$, then $|2x-y+1|=0.$
I will leave the rest as an exercise.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor/Maclaurin series of $\arctan(e^x - 1)$ Right, so this keeps bugging me, and I'm probably stuck in some tunnel trying the same thing over and over again.
Give the Maclaurin series of the function $\arctan(e^x - 1)$. up to terms of degree three. Since I try to be as lazy as possible and differentiating this thing sounds like a lot of work, I want to cheat!
Knowing that the Maclaurin series for $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} + ....$ and the Maclaurin series for $e^x - 1 = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ..... $
So yeah, substition to the resque, $\arctan(e^x - 1)$, use the known series and substite in the known series for $e^x - 1$:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3] - \frac{[1 + x]}{3}^3 + \frac{[1]}{5!}^5$$
Something along the lines of:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3] - (\frac{1 + 3x + 3x^2 + x^3}{3})$$
Okay, lets all get 6's:
$$[\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3] - (\frac{2}{6} + \frac{18x}{6} + \frac{18x^2}{6} + \frac{2x^3}{6})$$
Great... Lets do some algebruh
$$\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3 - \frac{18x}{6} - \frac{18x^2}{6} - + \frac{1}{6}x^3$$
Join some like terms
$$\frac{6}{6} - \frac{2}{6} + \frac{x}{6} - \frac{18x}{6} + \frac{3}{6}x^2 - \frac{18x^2}{6} + \frac{1}{6}x^3 - \frac{2}{6}x^3$$
This is so far removed from any of the possible answers and I've spent lots of time on this one, so it's time to get some real help.
Please help me :)
|
The correct expansion should be $e^x-1=\sum_{i=1}^\infty \frac{x^i}{i!}$.
Hence substituting it, we get
\begin{align}
\arctan (e^x - 1) &\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{(x+\frac{x^2}{2}+\frac{x^3}{6})^3}{3} \\
&\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{x^3}{3}\\
&= x+\frac{x^2}2 - \frac{x^3}6
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
|
Another way.
We need to prove that $$\prod_{cyc}(3+3a^2)\geq8\cdot27$$ or
$$\prod_{cyc}(ab+ac+bc+3a^2)\geq8(ab+ac+bc)^3$$ or
$$\sum_{sym}(3a^4b^2+a^3b^3+3a^4bc-6a^3b^2c-a^2b^2c^2)\geq0,$$ which is true by Muirhead because
$(4,2,0\succ(3,2,1),$ $(4,1,1)\succ(3,2,1)$ and $(3,3,0)\succ(2,2,2).$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$$
and :
\begin{align*}
\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\ \mathrm{d}x&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{1}{2n+1}\displaystyle\int_0^{\sqrt{2}-1}x^{2n}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=S\\
\end{align*}
So :
\begin{align*}
S&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{2x}\left(\log(1+x)-\log(1-x)\right)\mathrm{d}x\\
&=\frac{1}{2}(J_1-J_2)
\end{align*}
Where:
\begin{align*}
J_1&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1+x)}{x}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\displaystyle\int_0^{\sqrt{2}-1}x^n\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
And:
\begin{align*}
J_2&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1-x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}-\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n+1}dx\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]_0^{\sqrt{2}-1}\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
Finally we find :
$$S=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}((-1)^n+1)}{(n+1)^2}\right)$$
But I could not find a way to calculate $S$.
Any help please? and thank's in advance.
|
No need a sledgehammer to crack a nut.
\begin{align}
J&=\int_0^{\sqrt{2}-1} \frac{\text{arctanh } x}{x}\,dx\\&=-\frac{1}{2}\int_0^{\sqrt{2}-1}\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}\,dx\\
&\overset{y=\frac{1-x}{1+x}}=-\int_{\sqrt{2}-1}^1 \frac{\ln y}{1-y^2}dy\\
&=\int_0^{\sqrt{2}-1} \frac{\ln y}{1-y^2}dy-\int_0^1 \frac{\ln y}{1-y^2}dy\\
&\overset{\text{IBP}}=\frac{1}{2}\left[\ln\left(\frac{1+y}{1-y}\right)\ln y\right]_0^{\sqrt{2}-1}+\frac{1}{2}\int_0^{\sqrt{2}-1} \frac{\ln\left(\frac{1-y}{1+y}\right)}{y}dy-\int_0^1 \frac{\ln y}{1-y^2}dy\\
&=-\frac{1}{2}\ln^2\left(\sqrt{2}-1\right)-J-\int_0^1 \frac{\ln y}{1-y^2}dy\\
\end{align}
And,
\begin{align}\int_0^1 \frac{\ln y}{1-y^2}dy&=\int_0^1 \frac{\ln y}{1-y}dy-\int_0^1 \frac{u\ln u}{1-u^2}du\\
&\overset{y=u^2}=\int_0^1 \frac{\ln y}{1-y}dy-\frac{1}{4}\int_0^1 \frac{\ln y}{1-y}dy\\
&=\frac{3}{4}\int_0^1 \frac{\ln y}{1-y}dy\\
&=\frac{3}{4}\times -\frac{\pi^2}{6}\\
&=-\frac{\pi^2}{8}
\end{align}
Therefore,
$\displaystyle \boxed{J=\frac{\pi^2}{16}-\frac{1}{4}\ln^2\left(\sqrt{2}-1\right)}$
NB:
I assume that $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$
and, observe that if $\alpha=\sqrt{2}-1$ then $\displaystyle \dfrac{1-\alpha}{1+\alpha}=\alpha$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\begin{align*}
&2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \\
\Rightarrow \quad & 2^{-x} = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} \\
\Rightarrow \quad & 2^{-x}-1 = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1 \\
\Rightarrow \quad & \dfrac{2^{-x}-1}{x} = \frac{\frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1}{x} \\
\Rightarrow \quad & \frac{2^{-x}-1}{x} = \frac{1-\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}{x\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}
\end{align*}
From this part I don't know how to continue
|
$$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!} $$
Proof:
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-1x} = 1 +\sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1 = \sum_{n=1}^{\infty} \dfrac{(-1)^n(x\ln(2))^n}{n!} $$
$$\Rightarrow 2^{-x} -1 = \sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}(x\ln(2))^{n+1}}{(n+1)!} $$
$$\Rightarrow \dfrac{2^{-x} -1}{x} = \sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^n\ln(2)^{n+1}}{(n+1)!} $$
|
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|
convergence of $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1}\left\{\frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ (new methods) The series $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\}\right]$ is said to converge when $0<s<1,$ which seems impossible, for
$s-1<0,$ $\frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}<0$, and so $$\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\} \right]>\sum_{n=1}^\infty \frac{1}{n^s} \quad (>0),$$ which diverges.
So is the series really convergent?
PS: I would prefer a method without using integral.
And this series is an example of a series whose ratio of nearby items tends to 1.
|
In $\sum_{n=1}^\infty \left[\frac{1}{n^s}+\frac{1}{s-1} \left\{ \frac{1}{{(n+1)}^{s-1}}-\frac{1}{n^{s-1}}\right\} \right]$, thinking of the subtraction may result in $\frac{1}{n^\sigma}$ where $\sigma\geq1>s$, I add up the fractions and then get
$$
\frac{u_{n+1}}{u_n}=\frac{1-s-(\frac{n+1}{n+2})^s(n+2)+n+1}{1-s-(\frac{n}{n+1})^s(n+1)+n}(\frac{n}{n+1})^s
=\frac{1-s-(\frac{1}{1+\frac{1}{1+n}})^s(n+2)+n+1}{1-s-(\frac{1}{1+\frac{1}{n}})^s(n+1)+n}(\frac{1}{1+\frac{1}{n}})^s
.$$
Then using $(1+x)^{-s}=1-sx+s(s+1)x^2+O(x^3)$ (for we need to multiply O(n), we need to expand to 2nd order) and changing the denominator to nominator by
$\frac{1}{-s^2+s(s+1)\frac{1}{n}}=-s^2(1-\frac{s(s+1)}{s^2}\frac{1}{n}),$
we have the ratio equals
$$
(-s^2\frac{n}{n+1}-s(s+1)\frac{n}{(n+1)^2})(1-s\frac{1}{n}+s(s+1)\frac{1}{n^2})\frac{(-)}{s^2}(1-(1+\frac{1}{s})\frac{1}{n}).
$$
Then by $\frac{n}{(n+1)^2}=\frac{1}{n(1+\frac{2}{n}+\frac{1}{n^2})}=\frac{1}{n}(1-(\frac{2}{n}+\frac{1}{n^2})+O((\frac{2}{n}+\frac{1}{n^2})^2))=\frac{1}{n}+O(\frac{1}{n^2})$, we have the ratio equals
$$
(1+\frac{1}{s}\frac{1}{n})(1-s\frac{1}{n})(1-(1+\frac{1}{s})\frac{1}{n})+O(\frac{1}{n^2})=1+\frac{-1-s}{n}+O(\frac{1}{n^2}).
$$
Since $-1-s<-1$, we have the series converge absolutely.
(This result also supports the guessing that for $u_n=1/n^\lambda$, $\frac{u_{n+1}}{u_n}$ equals $1-\lambda/n$.
My thought is to try using the ratio test. But even if so, I can try first to simplify $u_n$ to $\alpha_0+\frac{\alpha_1}{n}+O(\frac{1}{n^2}) $ (or 2nd order expansion, etc.) before I do the division, which will make the calculation easier. But the above calculation also verify however and in whatever order we do the calculation, the expansion of a function (here, of 1/n) won't change. Such perfect consistence is so nice, considering how different the paths the result is gotten via.
|
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|
The integral of $ \int_{0}^{1} 2^{x^2 +x} dx$ So, What I tried was,
$$ I(b) = \int_{0}^{1} 2^{x^2 + x +b} dx$$
And hence,
$$ I'(b) = \ln(2) I$$
Hence,
$$ I = C_o 2^{b}$$
or,
$$ C2^{b} = \int_{0}^{1} 2^{x^2 +x + b} dx$$
Now I'm trying to find an easy 'b' to evaluate the right side integral at, so as to figure out my constant. However I'm not sure how to find that 'b'. A guess was to take $ b= \frac{1}{4}$ however that was not a fruitful substitution
|
If you do not want to use special functions, build a series expansion around $x=\frac 12$ to get for the integrand
$$2^{3/4} \left(1+2 L \left(x-\frac{1}{2}\right)+L (2 L+1) \left(x-\frac{1}{2}\right)^2+\frac{2}{3}
L^2 (2 L+3) \left(x-\frac{1}{2}\right)^3+\frac{1}{6} L^2 (4 L^2+12L+3)
\left(x-\frac{1}{2}\right)^4+L^3\left(\frac{4 L^2}{15}+\frac{4 L}{3}+1\right)
\left(x-\frac{1}{2}\right)^5+O\left(\left(x-\frac{1}{2}\right)^6\right) \right)$$ where $L=\log(2)$.
Integrate termwise and an approximation is
$$\int_{0}^{1} 2^{x^2 +x} dx \sim \frac{480+40 L+83 L^2+12 L^3+4 L^4 } {240 \sqrt[4]{2} }\approx 1.93589$$ while the exact solution is $1.93749$.
For sure, adding terms will improve the accuracy.
|
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|
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}= \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)}$$
but I don't know how to reduce this further.
|
Near $0$, you have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x} = \frac{1 + \frac{x^2}{2} + o(x^3)}{x + \frac{x^3}{6} + o(x^4)} - \frac{1}{x} = \frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\left( 1 + \frac{x^2}{6} + o(x^3) \right)^{-1}-1\right]$$ $$ =\frac{1}{x}\left[\left( 1 + \frac{x^2}{2} + o(x^3) \right)\left( 1 - \frac{x^2}{6} + o(x^2) \right)-1\right] =\frac{1}{x}\left( \frac{x^2}{3} + o(x^2) \right) = \frac{x}{3} + o(x) $$
So the limit is $0$.
|
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|
Finding $ \mathop {\lim }\limits_{n \to \infty } \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$ I tried to find the limit $$\mathop {\lim }\limits_{n \to \infty } \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$$, but I couldn't find a conclusive argument.
By assuming that the sequence $\displaystyle x_n = \ln (n) \cdot \int_0^1 \ln ( n^{-t} + 1 ) \, \mathrm dt$ is convergent to $k$, I managed to do the following:
It is well known that $T_{2q,0} (x) \le \ln(1+x) \le T_{2q+1,0} (x) \; , \; \forall x>0 , \forall q \in \mathbb{N} ^{*}$. (I used the Taylor polynom here)
Using this inequality, we obtain a bounding for $x_n$. By making $n \to \infty$, we obtain:
$$\sum\limits_{p = 1}^{2q} { \frac{ (-1)^{p+1} }{ p^2 } } \le k \le \sum\limits_{p = 1}^{2q+1} { \frac{ (-1)^{p+1} }{ p^2 } } \, , \, \forall q \in \mathbb{N} ^{*} $$
Then, by making $q \to \infty$, we have that $ \sum\limits_{p = 1}^{ \infty } { \frac{ (-1)^{p+1} }{ p^2 } } = \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } - 2 \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ (2p)^2 } } = \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } - \frac{1}{2} \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } = \frac{1}{2} \cdot \sum\limits_{p = 1}^{ \infty } { \frac{ 1 }{ p^2 } } = \frac{ \pi ^2}{12} $, which implies that $k = \frac{ \pi ^2 }{12}$.
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Substitute $x=t\ln(n)$, then the integral becomes
$$\lim_{n\to\infty}\int_0^{\ln(n)}\ln\left(1+e^{-x}\right)\,dx=\int_0^{\infty}\ln\left(1+e^{-x}\right)\,dx.$$
Now
$$\int_0^\infty \ln(1+e^{-x})dx=\int_0^\infty\left(e^{-x}-\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}-\frac{e^{-4x}}{4}+\cdots\right)dx $$
$$=\int_0^\infty e^{-x}dx-\frac{1}{2}\int_0^\infty e^{-2x}dx+\frac{1}{3}\int_0^\infty e^{-3x}dx-\frac{1}{4}\int_0^\infty e^{-4x}dx+\cdots$$
$$=1-\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}+\cdots $$
$$=\eta(2)=\frac{\pi^2}{12},$$
where $\eta(s)$ is the Dirichlet eta function. The last equality follows from the Basel Problem by the amazing
$$\eta(2)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}.$$
|
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|
Given $U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$, find $\frac{100U_{10}-1}{U_8}$
If
$$U_n=\int_0^\frac{\pi}{2} x\sin^n x dx$$
Find $\frac{100U_{10}-1}{U_8}$
Answer: $90$
My Attempt:
I tried applying Integration By Parts, and when that failed, I tried the substitution $x\rightarrow \frac{\pi}{2} -x$ , only to establish a (probably useless) relationship between the $U_n$ and its cosine counterpart:
$$U_n=\frac{\pi}{2}\int_0^{\frac{\pi}{2}} \cos^n \ dx-\int_0^{\frac{\pi}{2}}x\cos ^n x\ dx$$
Any help would be appreciated!
|
Note
$$U_{n-2} -U_n =\int_0^\frac{\pi}{2} x\sin^{n-2}\cos^2x dx
\overset{IBP}=\frac1{n-1} \int_0^\frac{\pi}{2} x\cos xd(\sin^{n-1}x)\\
= \frac1{n-1}\left( U_n -\frac1n \int_0^\frac{\pi}{2} d(\sin^{n}x)\right)= \frac1{n-1}\left(U_n - \frac1n\right)
$$
Thus, $U_n = \frac{n-1}nU_{n-2} + \frac1{n^2}$ and
$$\frac{100U_{10}-1}{U_8} = \frac{ 100(\frac9{10}U_8+\frac1{100})-1}{U_8}=90$$
|
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Correct result for this integral $\int \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}}\, dx$ Wolfram|Alpha and its CAS, Wolfram Mathematica are, as far as I know, the only website and software that give the correct solution to this integral,
$$ f(x) = \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}} $$
$$ F(x) = \int f(x)\, dx$$
because deriving the function given as result and using the FullSimplify function of Mathematica we get to the original function that we wanted to integrate.
This is the solution:
$$ F(x) = \frac{1}{5} (-8) \sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1} \sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2} \left(\sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2}-2\right) \sqrt{\sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2}+2} \csc \left(5 \sqrt{x}+4\right) + C $$
Fricas finds another expression for the integral that I'm curious about:
$$ F(x) = \frac{1}{5} (-8) \sqrt{2} \sqrt{\sqrt[4]{2} \sqrt{\sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}}+2} \left(\sqrt{2} \cos \left(5 \sqrt{x}+4\right)-2 \sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1} \left(\sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}\right)+2 \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}}\right) \csc \left(5 \sqrt{x}+4\right) + C$$
However, in this video, an incorrect result is given although the integration process seems correct. As above, you know that the result is incorrect since deriving the resulting function doesn't result in the original function we wanted to integrate.
Is the expression given by Fricas equivalent to the one given by Wolfram|Alpha and Mathematica?
Final update: According to Mathematica, the expression given by FriCAS is not equal to the correct solution given by Mathematica and Wolfram|Alpha. This means it is not the solution to this problem.
Even though I have a Pro Premium subscription, the step-by-step solution is not available for this input. What are the steps taken by Mathematica/Wolfram|Alpha to get to the given result?
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These are the steps to get to the given result:
$$f(x)=\dfrac1{\sqrt x}\sqrt{2+\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}
=\dfrac1{\sqrt x}\,\dfrac{\sqrt{2-\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}}$$
$$=\dfrac1{\sqrt x}\,\dfrac{\sqrt{2-2\cos(5\sqrt x+4)\large\mathstrut}}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}$$
$$=\dfrac1{\sqrt x}\,\dfrac{\sqrt{4-4\cos^2(5\sqrt x+4)\large\mathstrut}}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}$$
$$=\dfrac25\,\dfrac{2\cdot\dfrac5{2\sqrt x}\,|\sin(5\sqrt x+4)|}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}$$
$$=-\dfrac{4}5\,\dfrac{\operatorname{sign}(\sin(5\sqrt x+4))\cdot(2+2\cos(5\sqrt x+4))'}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}\;2\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}$$
$$=-\dfrac{8}5\,\dfrac{\operatorname{sign}(\sin(5\sqrt x+4))\cdot\left(2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}\right)'}{\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\;2\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}$$
$$=\dfrac{16}5\,\dfrac{\operatorname{sign}(\sin(5\sqrt x+4))\cdot\left(2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}\right)'}{2\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}}$$
$$=\dfrac{16}5\,\operatorname{sign}(\sin(5\sqrt x+4))\cdot\left(\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}\right)'.$$
Now we can easily verify that the function
$$F(x)=\dfrac{16}5\,\operatorname{sign}(\sin(5\sqrt x+4))\cdot\sqrt{2-\sqrt{2+\sqrt{2+2\cos(5\sqrt x+4)\large\mathstrut}}}+C$$
corresponds to the given one in the OP.
|
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$2$ fair coins, $1$ biased coin $-$ probability of getting a $3^{\text{rd}}$ head after $2$ consecutive heads? I solved this problem but my solution is apparently incorrect.
Let $A$ and $B$ denote the event that we chose the fair and biased (both sides are heads) coin, respectively. $P(A) = \frac{2}{3}$, $P(B) = \frac{1}
{3}$. Let $HH$ denote event of getting $2$ heads.
Then
\begin{align}
P(A|HH) = \frac{P(HH|A)P(A)}{P(HH)} \\
= \frac{\frac{1}{4}\cdot\frac{2}{3}}{\frac{1}{4}\cdot\frac{2}{3} + 1\cdot\frac{1}{3}} \\
= \frac{1}{3} \\
P(B|HH) = \frac{P(HH|B)P(B)}{P(HH)} \\
= \frac{\frac{1}{3}}{\frac{1}{4} \cdot \frac{2}{3} + 1\cdot\frac{1}{3}} \\
= \frac{2}{3} \\
\end{align}
Let $3H$ denote the event of getting a $3^{\text{rd}}$ head. We can essentially treat $A$ as $1$ single fair coin chosen with probability $\frac{1}{3}$ and $B$ as $1$ single biased coin chosen with probability $\frac{2}{3}$.
\begin{align}
P(3H) = P(3H|A)P(A) + P(3H|B)P(B) \\
= \frac{1}{2} \cdot \frac{1}{3} + \frac{2}{3} = \frac{5}{6}
\end{align}
But apparently the solution is $\frac{3}{4}$. Where did I go wrong in my thought process?
|
You need to be precise describing the experiment. It looks like you are saying that (1) a single coin is chosen at random from three coins and that coin is flipped three times, and (2) two of the three coins are fair, and the third always comes up heads. And you want to know, given that the first two flips were heads, what is the probability that the third flip is also a head?
If that description is right, then your Bayesian analysis is also correct. The probability that you selected the biased coin is $2/3$, and given that, the probability of a third head is $(1/2)(1/3) + (1)(2/3)=(5/6)$. So, if the "correct" answer is $3/4$, your description of the experiment must be wrong. An alternate experiment might be to choose a coin at random, flip it, and replace it, three times. In that case, the probability of a third head would be just $(1/2)(2/3) + (1)(1/3) = (2/3)$. So that's not right either.
Yet another guess would be that the experiment is to flip each coin once, in a random order. In that case, you have either flipped the two fair coins, or a fair coin and then the biased coin, or the biased coin and then a fair coin.
$$
P(2H\vert AB)=P(2H\vert BA)=1/2; \qquad P(2H\vert AA)=1/4.
$$
$$
P(AB\vert 2H)=P(BA\vert 2H)=\frac{P(2H\vert AB)P(AB)}{P(2H\vert AB)P(AB) + P(2H\vert BA)P(BA) + P(2H\vert AA)P(AA)}=\frac{(1/2)(1/3)}{(1/2)(1/3)+(1/2)(1/3)+(1/4)(1/3)}=\frac{2}{5}.
$$
$$
P(AA\vert 2H)=\frac{1}{5}.$$
In which case, $P(3H\vert 2H)=(4/5)(1/2)+(1/5)=(3/5)$. Which is still not right. So I'm out of guesses :).
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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.