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How to simplify this formula I have this equation: $\frac{\sqrt{(a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(-a+b+\sqrt{a^2+b^2-2ab\cos\gamma})(a-b+\sqrt{a^2+b^2-2ab\cos\gamma})(a+b-\sqrt{a^2+b^2-2ab\cos\gamma})}}{4}$
It calculates the area of a triangle using only 2 sides of a triangle $a, b$, and the angle of those sides, $\gamma$. It simply combines the law of cosines and Heron's Formula.
Is there any way to achieve the same function, and is it possible to simplify this equation?
| Let $c = \sqrt{a^2+b^2-2ab\cos\gamma}$ and $P_1,P_2,P_3,P_4$ be the 4 factors inside the square root. Notice
*
*$P_1 P_4 = (a+b+c)(a+b-c) = (a+b)^2 - c^2 = 2ab(1+\cos\gamma)$,
*$P_2 P_3 = (-a+b+c)(a-b+c) = c^2-(a-b)^2 = 2ab(1-\cos\gamma)$
We have
$$\frac{\sqrt{P_1P_2P_3P_4}}{4}
= \frac{\sqrt{(2ab)^2(1-\cos^2\gamma)}}{4}
= \frac12 ab\sin\gamma$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$? I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?
| $$\frac{1}{x^8(1+x^2)}=\frac{1+x^2-x^2}{x^8(1+x^2)}=\frac{1}{x^8}-\frac{1}{x^6(1+x^2)}=\cdots$$
and so on, finally you will get:
$$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
| Well, If $a,b,c$ form an arithmetic progression then there is a $d$ so that $a=b-d$ and $c=b+d$.
And to show that $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ form an arithmetic progression we have to show there is a $k$ so that
$k = (a^2+ac+c^2) - (a^2+ab+b^2)$ and $k= (b^2+bc+c^2) -(a^2+ac+c^2)$ or in other words that
$(a^2+ac+c^2) - (a^2+ab+b^2)= (b^2+bc+c^2) -(a^2+ac+c^2)$. So if we replace $a=b-d$ and $c = b+d$ we must show
$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)= (b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)$
Can we?
$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)=$
$((b-d)(b+d)+(b+d)^2) - ((b-d)b+b^2)=$
$((b-d)d+(b+d)^2) - (b^2)=$
$(b-d)d + 2bd+d^2=$
$3bd $.
And
$(b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)=$
$(b^2+b(b+d)) -((b-d)^2+(b-d)(b+d))=$
$(b^2) -((b-d)^2-d(b+d))=$
$(b^2) -(b^2-2bd-bd)=$
$3bd$.
Yep, they are equal all right.
========
To take your idea of $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$ into account:
We know $b = \frac{a+c}2$ and we need to prove
$a^2+ac+c^2= \frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2$ given that $b = \frac {a+c}2$
So
$\frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2=$
$\frac {a^2 + c^2 + b(a+c) + 2b^2}2=$
$\frac {a^2 +c^2 + \frac {a+c}2(a+c) + 2(\frac {a+c}2)^2}2=$
$\frac {a^2 + c^2 + \frac {(a+c)^2}2 + 2\frac {(a+c)^2}4}2=$
$\frac {a^2 + c^2 + (a+c)^2}2=$
$\frac {a^2 + c^2 + a^2 + 2ac + c^2}2 =$
$\frac {2a^2 + 2ac + 2c^2}2 =$
$a^2 + ac + c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triple Integral over a Superquadric How do I solve the following:
$\iiint _V (x^2+y^2+z^2) dxdydz $, where $V=\{x^{2n}+y^{2n}+z^{2n} \leq 1\}$ is a superquadric?
| By symmetry,
$$ \iiint _V (x^2+y^2+z^2) dxdydz=24\iiint_{V^+} x^2 dxdydz $$
where
$$ V^+=\{x^{2n}+y^{2n}+z^{2n} \leq 1, x\ge0,y\ge0,z\ge0\}. $$
Let
$$ x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi), x=r^{\frac1n}\sin^{\frac1n}(\theta)\cos^{\frac1n}(\phi),z=r^{\frac1n}\cos^{\frac1n}(\theta). $$
Then $V^+$ becomes
$$ \bar{V}^+=\{(r,\theta,\pi): 0\le r\le1, 0\le\theta,\phi\le\frac{\pi}{2}\}. $$
Now
$$ \det\frac{\partial(x,y,z)}{\partial r\partial \theta\partial\psi}=\frac{1}{n^3}r^{\frac{3}{n}-1} \sin ^{\frac{2}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta ) \sin
^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{1}{n}-1}(\phi ) $$
and then
\begin{eqnarray}
&&\iiint _V (x^2+y^2+z^2) dxdydz\\
&=&24\iiint_{V^+} x^2 dxdydz\\
&=&24\int_0^1\int_0^{\pi/2}\int_0^{\pi/2} r^{\frac2n}\sin^{\frac2n}\theta\cos^{\frac2n}\phi \det\frac{\partial(x,y,z)}{\partial r\partial \theta\partial\phi}drd\theta d\phi\\
&=&\frac{24}{n^3}\int_0^1\int_0^{\pi/2}\int_0^{\pi/2}r^{\frac{5}{n}-1} \sin ^{\frac{4}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta ) \sin
^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{3}{n}-1}(\phi )drd\theta d\phi\\
&=&\frac{24}{n^3}\int_0^1r^{\frac{5}{n}-1}dr\int_0^{\pi/2}\sin ^{\frac{4}{n}-1}(\theta ) \cos ^{\frac{1}{n}-1}(\theta )d\theta \int_0^{\pi/2} \sin
^{\frac{1}{n}-1}(\phi ) \cos ^{\frac{3}{n}-1}(\phi )d\phi\\
&=&\frac{24}{n^3}\cdot\frac{n}{5}\cdot \frac{\Gamma \left(\frac{2}{n}\right) \Gamma \left(\frac{1}{2n}\right)}{2 \Gamma
\left(\frac{5}{2n}\right)}\cdot\frac{\Gamma \left(\frac{1}{2n}\right) \Gamma \left(\frac{3}{2n}\right)}{2 \Gamma
\left(\frac{2}{n}\right)}\\
&=&\frac{6}{5n^2}\frac{ \Gamma^2 \left(\frac{1}{2n}\right)\Gamma \left(\frac{3}{2n}\right)}{\Gamma
\left(\frac{5}{2n}\right)}.
\end{eqnarray}
Here
$$ \int_0^{\pi/2}\sin^{-1+p}(t)\cos^{-1+q}(t)dt=\frac{\Gamma \left(\frac{p}{2}\right) \Gamma \left(\frac{q}{2}\right)}{2 \Gamma
\left(\frac{p+q}{2}\right)} $$
is used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$ For $n=5$, we get $\binom{10}{5}<4^4$. Asumme inductively that $\binom{2n}{n}<4^{n-1}$ for all positive integers $n\geq 5$. Then I need to prove that $\binom{2(n+1)}{n+1}<4^{n}$ for all positive integers $n\geq 5$. I've been looking around a bit, and I've found that the expression $\binom{2n}{n}$ is precisely a central binomial coefficient. Thus $\sum_{k=0}^{n+1}\binom{n+1}{k}^2=\binom{2(n+1)}{n+1}$. How can I use this to finish my proof?
| Using inductive argument, you can do something simpler.
$$\binom{2n+2}{n+1}=\frac{2(2n+1)}{n+1}\binom{2n}{n}$$
Now prove that
$$\frac{2(2n+1)}{n+1}<4.$$
Once the inequality is proven, we can multiply both sides by $\binom{2n}{n}$ and get,
$$\frac{2(2n+1)}{n+1}\binom{2n}{n}<4\binom{2n}{n}\to \binom{2n+2}{n+1}<4\binom{2n}{n}$$
and using the inductive hypothesis we get,
$$\binom{2n+2}{n+1}<4\binom{2n}{n}<4\cdot4^{n-1}=4^n.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$
Case $S(1)$ is true:
$$2((1)+2)\leq((1)+2)^{2}$$
$$6\leq9$$
Case $S(n)$ is true for all $n=1,2,...$
$$2(n+2)\leq(n+2)^{2}(i)$$
Case $S\left(n+1\right)$
$$2(n+3)\leq(n+3)^{2}(ii)$$
From (i)
$$2(n+2+1-1)\leq(n+2+1-1)^{2}$$
$$2(n+3)-2(1)\leq(n+3)^{2}-2(n+3)+1$$
$$2(n+3)\leq(n+3)^{2}-2(n+3)+3$$
$$2(n+3)\leq(n+3)^{2}-(2n+3)$$
$$2(2n+3)+3\leq(n+3)^{2}$$
Thus $(i)$ is true for all $n=1,2,...$
My question: How can I get the same expression as (ii)?
I got to $2(2n+3)+3\leq(n+3)^{2}$ but it's clearly wrong
| Recurrence :
$2(n+1)\leq(n+2)^2 $
For n=1
$4\leq 9$ that is correct
Suppose : $ 2(n+1)\leq(n+2)^2 $
Let's show $ 2(n+2)\leq(n+3)^2 $
$2(n+2)\leq(n+3)^2 $
$\Leftrightarrow $ $2(n+2)\leq ((n+2)+1)^2 $
$\Leftrightarrow $$2(n+2)\leq (n+2)^2 +2(n+2)+1$
$\Leftrightarrow $$0\leq(n+2)^2 +1$ that is correct
So:
$ 2(n+2)\leq(n+3)^2 $
Finally :
$ 2(n+1)\leq(n+2)^2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894498",
"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ $$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$
$$x=?$$
I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that, or someone will solve it in a better way. I'd be happy with that.
| Your way is fine with a solution for $x=-\frac12$, now we have that
$$3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1=x^2(3x ^ 2-x + 1)+(x-1)^2 > 0$$
since $3x ^ 2-x + 1>0$.
To find complex roots we can refer to Quartic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\prod_{i=1}^n(1+x_i)\leq \sum_{i=0}^n\frac{S^i}{i!}$, where $x_i\in\mathbb{R^+}$.
Let $x_1$, $x_2$, $\ldots$, $x_n$ be positive real numbers, and let $$S=x_1+x_2+\cdots+x_n.$$ Prove that $$(1+x_1)(1+x_2)\cdots(1+x_n)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots+\frac{S^n}{n!}.$$
My first thought is about using induction. For $n=1$, $LHS=1+x_1\leq1+S=RHS$.
Now I suppose that this inequality holds true for $n=k$; that is, $$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
And here I am stuck. I can't really see how I can find the relation between when $n=k$ and $n=k+1$.
Perhaps induction may not be the way? Any hints or suggestions will be much appreciated. Thanks.
| Induction, as you requested, works too. Just do it. Where are you stuck?
$$(1+x_1)(1+x_2)\cdots(1+x_k)\leq 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}.$$
Let $ S' = \sum_{i=1}^{k+1} x_i = S + x_{k+1}.$
$\prod_{i=1}^{k+1} (1 + x_i) \leq (1+x_{i+1}) (1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^k}{k!}) \\
= 1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!} $
Claim $ S^i ( S + (i+1) x_{i+1} ) \leq S' ^{i+1}$.
This is true because the sum of the terms on both sides are equal to $(i+1)S'$, and the terms are evenly distributed on the RHS.
Corollary $x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$
Now, do a sum of
$ 1 \leq 1 $
$x_{i+1} \frac{S^i}{i!} + \frac{S^{i+1}}{(i+1)!} \leq \frac{S'^{i+1}}{{i+1}!}$ for $ i = 1$ to $k$
$x_{i+1} \frac{ S^k}{k!} \leq \frac{ S'^{k+1} } { (k+1)!}$.
to conclude that
$1 + x_{i+1} + S + x_{i+1}S + \frac{S^2}{2!}S + x_{i+1}\frac{S^2}{2!} + \ldots + \frac{S^k}{k!} + x_{i+1} \frac{ S^k}{k!} \\
\leq 1 + S' + \frac{ S'^2}{2!} + \ldots + \frac{ S' ^{k+1} } { (k+1)!} $
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$ Solve in $\mathbb{R}$ the equation $3x+\sqrt[3]{x+1}+\sqrt[3]{(x+1)^2}=-1$
The way I attempted to the question is the following:
I state that $a=\sqrt[3]{x+1}$ hence we have that $3a^3-3+a+a^2=-1$. Hence $3a^3-2+a+a^2=0$, so $a^2(3a+1)+a-3=0$. And this is where I got stuck.
Could you please explain to me how to finish off the question with this train of thought and out of 10, what would be logical to assume that I would get for these workings out?
| $3a^3+a^2+a-2=0$
$(3a-2)(a^2+a+1)=0$
We can find this by using Rational Root theorem, or another way is by noting that $a^2+a+1$ is also a root because $a^2+a+1=0 \implies a^3=1$ (roots of unity) so
$$3a^3+a^2+a-2=3+a^2+a-2=a^2+a+1=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $ 2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx $ I've been trying to solve this this problem for so long but haven't had much luck, can someone please help me out?
The original question was:
Find the area of the part of the surface $z = \ln(x^2+y^2+2)$ that lies above the disk $x^2 + y^2 \le 1$ correct to four decimal places by expressing the area in terms of a single integral.
I set up the integral as:
$$
2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx.
$$
edit 1: Sorry for being unclear, I need a numerical value at the end, I've tried solving it by parts but it hasn't gotten me anywhere. I could just put it into a calculator and get an answer but I'm more curious about how to actually set it up and the steps to solve this problem.
| This is not a solid of revolution. The proble asks the area of the surface $z\log \left(x^2+y^2+2\right)$ inside the cylinder $x^2+y^2=1$
$z=\log \left(x^2+y^2+2\right)$ and $D:x^2+y^2\le 1$
$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$
$$z_x=\frac{2 x}{x^2+y^2+2};\;z_y=\frac{2 y}{x^2+y^2+2}$$
$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$
$$S=\iint\limits_{D}\sqrt{\frac{4 x^2}{\left(x^2+y^2+2\right)^2}+\frac{4 y^2}{\left(x^2+y^2+2\right)^2}+1}\,dA$$
$$S=\iint\limits_{D}\sqrt{1+\frac{4 \left(x^2+y^2\right)}{\left(x^2+y^2+2\right)^2}}\,dA$$
Changing to polar coordinates
$x=r\cos t;\;y=r\sin t$ and $Jacobian = r$ we get
$$S=\int _0^1\int _0^{2 \pi } r \sqrt{\frac{4 r^2}{\left(r^2+2\right)^2}+1}\,dtdr$$
$$S=\int _0^1\int _0^{2 \pi }\frac{r }{r^2+2}\sqrt{r^4+8 r^2+4}\,dtdr$$
$$S=\frac{1}{2} \pi \left(2 \sqrt{13}-4+\log \left(\frac{1}{162} \left(95 \sqrt{13}+343\right)\right)-4 \sqrt{2} \cot ^{-1}\left(\sqrt{2}\right)+4 \sqrt{2} \cot ^{-1}\left(\sqrt{26}\right)\right)\approx 3.5618$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of cube roots of two conjugate quadratic integers makes an integer. Consider the following expression:
$$(20-\sqrt{392})^{1/3}+(20+\sqrt{392})^{1/3}.$$
This equals $4$, but how can I show this?
Note that I do not want to make use of the following line of reasoning: 4 is a solution to $x^3-6x-40=0$, that this cubic has exactly one solution, and that all solutions to the cubic
$x^3+px +q = 0$ are given by
$$(-q/2-((q/2)^2+(p/3)^3)^{0.5})^{1/3}+(-q/2+((q/2)^2+(p/3)^3)^{0.5})^{1/3}.$$
I have noted the method set out here, How can I show that this complicated expression with square and cube roots reduces to the value 7? , but implementing it does not work very well.
I attempted this and got more complicated expressions than I started off with! It doesn't result in a unique solution for $a$ and $b$.
I do think that this is the right approach, though, i.e. making use of the fact that if $\sqrt c$ is irrational and $a+b\sqrt c=d+e \sqrt f$ then $a=d$ and $b=e$. (How do you prove this fact, by the way?)
| First note that $392=2^3\times7^2$ and so $\sqrt{392}=14\sqrt{2}$, and hence
$$\sqrt[3]{20+\sqrt{392}}=\sqrt[3]{20+14\sqrt{2}}.$$
Next, in the hope of finding a simple expression for the cube root, we compute
$$(20+14\sqrt{2})(20-14\sqrt{2})=20^2-2\times14^2=8,$$
which shows that
$$\sqrt[3]{20+14\sqrt{2}}\sqrt[3]{20-14\sqrt{2}}=\sqrt[3]{8}=2.$$
This suggests that perhaps there exist integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2},$$
and then certainly $a$ and $b$ should satisfy $a^2-2b^2=2$. An obvious choice is $a=2$ and $b=1$, and indeed
$$(2+\sqrt{2})^3=2^3+3\times2^2\times\sqrt{2}+3\times2\times\sqrt{2}^2+\sqrt{2}^3=20+14\sqrt{2},$$
and entirely analogously $(2-\sqrt{2})^3=20-14\sqrt{2}$. It follows that
$$\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=(2+\sqrt{2})+(2-\sqrt{2})=4.$$
Alternatively, we could directly try to find integers $a$ and $b$ such that
$$(a+b\sqrt{2})^3=20+14\sqrt{2}.$$
This boils down to the two simultaneous cubic equations
\begin{eqnarray*}
20&=&a^3+6ab^2&=&a(a^2+6b^2),\\
14&=&3a^2b+2b^3&=&b(3a^2+2b^2),
\end{eqnarray*}
from which it quickly follows that $a$ and $b$ are positive divisors of $20$ and $14$, respectively, and it is not hard to check that $a=2$, $b=1$ is the only solution.
| {
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sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?
| $$\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}=\frac{1\times 2}{2}+\frac{2\times 3}{2}+\frac{3\times 4}{2}+...+\frac{(n-1)\times n}{2}=\frac{1}{2}(1\times 2+2\times 3+3\times 4+...+(n-1)\times n)=\frac {1}{2}\times \frac{(n-1)n(n+1)}{3}=\frac{(n-1)n(n+1)}{6}$$
| {
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"url": "https://math.stackexchange.com/questions/3902467",
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"source": "stackexchange",
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Uniform convergence of $\frac{n^3x}{1+n^4x^2}$ I have been trying to prove the uniform convergence to f = 0 on [0,1] for the sequence of functions $\frac{n^3x}{1+n^4x^2}$.
So far i have attempted to prove pointwise converge as such;
Let $ N = 1/\epsilon x $
For every $n \ge N$ we have $ \vert\frac{n^3x}{1+n^4x^2} - 0 \vert = \frac{n^3x}{1+n^4x^2} < \frac{1}{Nx} = \frac{1}{(1/\epsilon x)x} = \epsilon $
Is this enough of a proof for pointwise convergence? and if so, how do i prove that the convergence is uniform (if it is in this case)?
| One strategy to check for uniform convergence is to find the maximum $M_n$ of $f_n(x) = \frac{n^3x}{1 + n^4 x^2}$ over $[0,1]$ using calculus, then check whether $M_n \to 0$ (keeping in mind that $0 < f_n(x) \leq M_n$ for all $x$).
That said, the same idea works with any upper bound for $f_n(x)$ that approaches $0$ as $n\to \infty$ and does not depend on $x$. For instance, one upper bound can be attained by combining an upper bound over $[0,1/n^2]$ and another over $[1/n^2,1]$.
With the first approach we find that
$$
f_n'(x) = \frac{n^3(1 + n^4x^2) - n^3x(2n^4x)}{[1 + n^4 x^2]} = 0 \implies\\
n^3(1 + n^4 x^2) = 2 n^3 \cdot n^4 x^2 \implies\\
(1 + n^4 x^2) = 2 n^4 x^2 \implies\\
1 = n^4 x^2 \implies\\
x = \pm 1/n^2.
$$
We can see that the critical point at $x = 1/n^2$ is indeed a maximum since $f_n'(0) > 0$ and $f_n'(1) < 0$. Thus, our maximum is
$$
M_n = f_n(1/n^2) = \frac{n^3 \cdot \frac 1{n^2}}{1 + n^4 \cdot \frac 1{n^4}} = \frac n2.
$$
So, we find in fact that $M_n$ does not converge to $0$, which means that $f_n$ fails to converge uniformly.
| {
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Is my $\epsilon$-$\delta$ calculation correct? I have to show that $\lim_{x \to 1} x^4-1 =0$. Here is how i have done it:
$\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < \epsilon \qquad$ and since we are close to 1, we can assume that the $\delta$-neighborhood of $c=1$ must be havea radius of max $\delta =1$ which implies that : $\mid x+1 \mid \le 2 \quad and \mid x^2+1 \mid \le 2 \quad \forall x \in V_{\delta}(c) \quad$
We now choose $\delta=min \left \{1,\frac{\epsilon}{4}\right\} \quad$ and we can conclude that if $\mid x-1 \mid < \delta$, it follows that $\mid x^4-1 \mid = \mid x-1 \mid\mid x+1 \mid\mid x^2+1 \mid < 4\frac{\epsilon}{4} =\epsilon$. Is this calculation correct? Do I miss something? Or some details?
| You want
$$
|x^2+1||x+1||x-1|<\varepsilon
$$
near $x=1$.
First step is to control the quantity $|x^2+1||x+1|$ near $x=1$. So first
restrict $x$ so that $|x-1|<1=\delta_1$. This means
$0<x<2$, and hence $|x+1|<3$ while $x^2+1<5$. Hence
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^2+1||x+1|<15,
$$
and thus
$$
|x-1|<1=\delta_1 \quad\Longrightarrow\quad |x^4-1|=|x^2+1||x+1||x-1|<15|x-1|,
$$
Now, we have
$$
|x-1|<\delta=\min\Big\{1,\frac{\varepsilon}{15}\Big\}\quad\Longrightarrow\quad
|x^4-1|<15|x-1|<15\cdot\frac{\varepsilon}{15}=\varepsilon.
$$
| {
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Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$ I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$
Then what is the value of $a/b$?
I tried evaluating the quantities on DESMOS to realise that: a=136 and b=17 which gives 8 as the answer. Any helpful insight about how to reach at these values !?
| Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$,
$$\tan(2n+1)x=\dfrac{\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}}{\binom{2n+1}0-\binom{2n+1}2t^2+\cdots+(-1)^n\binom{2n+1}{2n}t^{2n}}$$
where $t=\tan x$
So, the roots of $$\binom{2n+1}1t-\binom{2n+1}1t^3+\cdots+(-1)^n\binom{2n+1}{2n+1}t^{2n+1}=0$$ i.e., $$t^{2n+1}-\binom{2n+1}{2n-1}t^{2n-1}+\cdots+(-1)^n(2n+1)t=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=0,\pm1,\pm2,\cdots,\pm n$
As $r=0\implies \tan\dfrac{r\pi}{2n+1}=0$, the roots of $$t^{2n}-\binom{2n+1}{2n-1}t^{2n-2}+\cdots+(-1)^n(2n+1)=0$$ are $t_r=\tan\dfrac{r\pi}{2n+1}; r=\pm1,\pm2,\cdots,\pm n$
Writing $t_r^2=p_r,$
the roots of $$p^n-\binom{2n+1}{2n-1}p^{n-1}+\cdots+(-1)^n(2n+1)=0$$ are $p_r=\tan^2\dfrac{r\pi}{2n+1}; r=1,2,\cdots, n$
Use Vieta's formula
| {
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can I finish this solution by this method, if I can please tell me how $m^2+m=n^3+n^2+n$, find solutions in natural numbers. so this was the problem and answer is that it doesn't have solutions in natural numbers.
It was pretty hard to think about this problem, I wanted to analyze the ends
of the numbers, for example, if $m$ ends on 2, then $m^2$ ends on 4, so their sum ends on 6.
First of all, it is easy to prove that n is even. since the left side is even, then right side must be the even as well, so $n$ must be even. So I analyzed when n is even, at what number sum n^3+n^2+n ends on, for example if $n$ ends on 2 then $n^2$ ends on 4.
Then I analyzed when $n$ ends on 4,6,8. sum ends on 4 or 8. Then I analyzed ending of $m$ and $m^2$ then I got that $m^2+m$ doesn't end on 4 or on 8, it ends on 0 in 3 cases, when m ends on 4,5,9 $m+m^2$ ends on 0. In other cases $m$ doesn't end on 4 or on 8 so I supposed that $m+m^2$ doesn't equal $n+n^2+n^3$, for natural numbers, but unfortunately and I am very sad right now, I forgot to analyze when $n$ ends on 0. when $n$ ends on 0, it sum $n+n^2+n^3$ ends on 0 and unfortunately, I have to analyze 3 cases for $m$, when sum $m+m^2$ ends on 0.
So I really need your opinion about this can I finish this problem after analyzing ending 0 situations, please answer me it will be very helpful for me, I really want to defend this solution.
| Another approach:
$$m^2+m=\frac{m^3-1}{m-1}-1$$
$$n^3+n^2+n=\frac{n^4-1}{n-1}-1$$
Therefore:
$$\frac{m^3-1}{m-1}=\frac{n^4-1}{n-1}$$
Due to Fermat's little theorem:
$n^4-1 \equiv 0 \mod (5)$,$\rightarrow n^4-1=5t$
$m^3-1$ can be a multiple of 5 with some condition:
$m\equiv (0, 1, 2, 3, 4)\ mod (5)$
If $m \equiv 1 \ mod (5)$, then:
$m^3-1 \equiv 0 \ mod (5)$
But not all multiples of 5 are of the form $n^4-1$ unless $m=a^4$ such that $(a^4)^3-1= (a^3)^4-1$. I n this case $n=a^3$.But this does not satisfy the equation, for example let $a=2$ we have:
$m=2^4=16$, $n=2^3=8$
$m^3-1=2^{12}-1= 8^4-1$
But:
$m^2+m=16^2+16=272$
$n^3+n^2+n=8^3+8^2+8=584$
That is this equation can not have integer solution.
| {
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Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$?
If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$
We can write $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}=\left(a+\frac{a+b}{a}\right)^{2}+\left(b+\frac{a+b}{b}\right)^{2}=\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}$$
Using $Q.M\geq A.M\geq G.M$ we have
$$\sqrt{\frac{\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2}}{2}} \geq \frac{2+\frac{a}{b}+2+\frac{b}{a}}{2}=2+\frac{\frac{a}{b}+\frac{b}{a}}{2} \geqslant 2+1=3$$
So
$$\begin{array}{l}
\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{a}\right)^{2} \geqslant 18 \\
\Rightarrow \quad\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2} \geqslant 18
\end{array}$$
But in an alternate approach i got the correct minimum as $12.5$
| Another variation on the root-mean-square inequality (or Cauchy-Schwarz with $c = d = 1$) is:
$$\left(x + \frac{1}{x} \right)^2 + \left(1-x+\frac{1}{1-x} \right)^2 \ge \frac{\left(x + \frac{1}{x} +1-x+\frac{1}{1-x} \right)^2}{2} \tag{$0 \le x \le 1$}$$
$$= \frac{1}{2} \left(1 + \frac{1}{x} + \frac{1}{1-x} \right)^2 = \frac{1}{2} \left(1 + \frac{1}{x(1-x)} \right)^2.$$
Now this expression is minimised when $x(1-x)$ is maximised, which occurs halfway between the roots $x = 0,\ 1$. Thus the minimum value is $\frac{1}{2} \left(1 + \frac{1}{\frac{1}{2}(1-\frac{1}{2})} \right)^2 = \boxed{\frac{25}{2}}.$
| {
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find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem I'm trying to prove with squeeze theorem that the limit of the following series equals 1:
$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$
For the left side of the inequality I did:
$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} < \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$
For the right side, at first I did the following:
$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} < \frac{n\sqrt[n]{n}}{n}$$
But then I realized it wasn't true and that the direction of this inequality is the opposite.
Do you have any idea which series with limit 1 is bigger from the original series?
Thanks!
| First of all we have, for any $ n\in\mathbb{N}^{*} $, the following : $$ \sqrt[n]{n}=1+\frac{\ln{n}}{n}\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x} $$
Since : \begin{aligned}0\leq\int_{0}^{1}{n^{\frac{x}{n}}\,\mathrm{d}x}&\leq n^{\frac{1}{n}}\\ &\leq 2\end{aligned}
We have : \begin{aligned} 1\leq \sqrt[n]{n}\leq 1+\frac{2\ln{n}}{n}&=1+\frac{4\ln{\sqrt{n}}}{n}\\ &\leq 1+\frac{4\sqrt{n}}{n}= 1+\frac{8}{2\sqrt{n}}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq1+\frac{8}{\sqrt{n}+\sqrt{n-1}} \end{aligned}
That remains true for any $ n\in\mathbb{N}^{*} $, which means given $ n\in\mathbb{N}^{*} $, we have : \begin{aligned} 1\leq\frac{1}{n}\sum_{k=1}^{n}{\sqrt[k]{k}}&\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\frac{1}{\sqrt{k}+\sqrt{k-1}}} \\ &\leq 1+\frac{8}{n}\sum_{k=1}^{n}{\left(\sqrt{k}-\sqrt{k-1}\right)}\\ &\leq 1+\frac{8}{\sqrt{n}} \end{aligned}
Thus, using the squeezing theorem, the limit would be $ 1 \cdot$
| {
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find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$ I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$
Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$
Since $e^{x^4-2x^3+x^2-1}>0$ for all $x\in \Bbb R$ and $2x-1<0\iff x<\frac{1}{2}$. $f$ is decreasing on $(-\infty,\frac{1}{2})$.
Furthermore, $f$ is increasing on $(\frac{1}{2},\infty)$, $f$ is differentiable at $x=\frac{1}{2}$, and $f'(\frac{1}{2})=0$, $f$ attains its minimum value at $x=\frac{1}{2}$.
Am I right?
| Everything is fine ! A little bit more can be said:
$f$ is strictly decreasing on $(-\infty,\frac{1}{2}]$
and
$f$ is strictly increasing on $[\frac{1}{2}, \infty).$
| {
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Series expansion of $x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$ at $x=1$ In various texts it is stated that a good approximation of,
$x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$,
about $x=1$ is given by,
$\frac{4\sqrt{2}}{3} (x-1)^{3/2}$.
Plotting the graphs this is indeed the case, however, I am not sure how this approximation is derived. Entering the original equation in Wolfram Alpha it does list the above approximation as the first term in the Puiseux series expansion of the original function. I am not familiar with these and looking online I can't seem to find a way into calculating a Puiseux series for the relevant expression.
Any suggestions or pointers for how to go about this would be greatly appreciated :)
| The Approximation in the Question
I recognized this as the result of an integral that uses the trigonometric substitution $x=\sec(\theta)$, so I took the derivative of ${\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}$ and got $2{\textstyle\sqrt{t^2-1}}$. Thus,
$$
{\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}
=2\int_1^x{\textstyle\sqrt{t^2-1}}\,\mathrm{d}t\tag1
$$
For $t\approx1$, $t^2=(1+(t-1))^2\approx1+2(t-1)=2t-1$; therefore, $\sqrt{t^2-1}\approx\sqrt{2t-2}$. Thus,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\int_1^x\sqrt{2t-2}\,\mathrm{d}t\tag2\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}\tag3
\end{align}
$$
A Better Approximation
We can get a better approximation if we note that
$$
\begin{align}
\textstyle2\sqrt{t^2-1}
&=\textstyle2\sqrt{2(t-1)+(t-1)^2}\tag{4a}\\
&=\textstyle2\sqrt2\sqrt{t-1}\sqrt{1+\frac{t-1}2}\tag{4b}\\
&\approx\textstyle2\sqrt2\sqrt{t-1}\left(1+\frac{t-1}4\right)\tag{4c}\\
&=\textstyle2\sqrt2\sqrt{t-1}+\frac{\sqrt2}2\sqrt{t-1}^3\tag{4d}
\end{align}
$$
Therefore,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\sqrt2\int_1^x\sqrt{t-1}\,\mathrm{d}t+\frac{\sqrt2}2\int_1^x\sqrt{t-1}^3\,\mathrm{d}t\tag5\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}+\frac{\sqrt2}5(x-1)^{5/2}\tag6
\end{align}
$$
Graphs
| {
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Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
| Using Newton’s identities,
$$x_1^2+x_2^2+x_3^2= (x_1+x_2+x_3)^2- 2\sum x_ix_j=3^2-k=6$$
and therefore $k=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.
What I Tried: Here is a picture :-
I made an attempt solving this and found rather a weird conclusion.
We have that $A,P,C,R$ are collinear, and also $AP = CR$.
Let $DL = LS = BK = KQ = x$ . We have $AP = x\sqrt{2}$ . Since $PC$ = $(4 - 2\sqrt{2})$ , $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ .
Now, $PF = FC$ . So by Pythagoras Theorem on $\Delta PFC$ , $PF = FC = \sqrt{12 - 8 \sqrt{2}}$.
$AL = LS = \sqrt{12 - 8 \sqrt{2}} + 2x$ and $AS = (4 - 2\sqrt{2} + 2x\sqrt{2})$ . So now by Pythagoras Theorem :-
$$2\bigg(\sqrt{12 - 8 \sqrt{2}} + 2x\bigg)^2 = (4 - 2\sqrt{2} + 2x\sqrt{2})$$
$$\rightarrow 2\bigg(12 - 8\sqrt{2} + 4\sqrt{12 - 8\sqrt{2}}x + 4x^2\bigg) = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow 24 - 16\sqrt{2} + 8\sqrt{12 - 8\sqrt{2}}x + 8x^2 = 8x^2 + 16\sqrt{2}x - 16x - 16\sqrt{2} + 24$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}}x = 2\sqrt{2}x - 16x$$
$$\rightarrow \sqrt{12 - 8\sqrt{2}} = 2\sqrt{2} - 16$$
But this is obviously false.
So I literally ended up in a completely false equation, might be making a mistake in this complicated calculations, and I cannot find it. So can anyone tell me where I made the mistake, or I am solving this in the correct way or not?
| Let radius of circle be $r$. So side of square $a=2r$.
Let the circle touch $KR$ at $X$. So $KX=a/2=r$
Let $KQ=x$. So $PF=a-2x=2(r-x)$
Use power of point of $Q$, $$PQ\cdot QF=QX^2$$
| {
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Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
| Using algebra
Rewriting
$$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)$$ as
$$a_{n+1}=a_n-\left(a_n-\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)\right)=a_n-\frac{a_n^2-3}{2 a_n}$$ we can recognize the itegrative scheme for solving $a^2=3$; so the limit is $\sqrt 3$.
Considering the function $f(a)=a^2-3$, starting with $a_0=1$, we have
$f(1)=-2$ and $f''(1)=2$, $f(1)\times f''(1)=-4<0$, then, by Darboux theorem, we shall have one (and only one overshoot of the solution). This corresponds to $a_2$ and, later, we shall converge quadratically to the root.
$$\left(
\begin{array}{cc}
n & a_n \\
1 & 1.0000000000000000000 \\
2 & 2.0000000000000000000 \\
3 & 1.7500000000000000000 \\
4 & 1.7321428571428571429 \\
5 & 1.7320508100147275405 \\
6 & 1.7320508075688772953 \\
7 & 1.7320508075688772935
\end{array}
\right)$$
| {
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A question of Roots of unity
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
\cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
| The roots of unity for $z^9=1$ are $e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$ and their sum is equal to zero, i.e.
$$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9}
+ e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9}
+ e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9}
+ e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9}
$$
which leads to
$$\cos\frac{2\pi}{9}+
\cos\frac{4\pi}{9}+ \cos\frac{6\pi}{9} + \cos\frac{8\pi}{9} =- \frac{1}{2}$$
| {
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} |
Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
| When you know $a+b$, you can express $a^n+b^n$ in terms of powers of $c=ab$. A few cases
*
*$a^2+b^2=(a+b)^2-2ab$
*$a^3+b^3=(a+b)^3-3ab(a+b)$
What about $a^5+b^5$? The relations are obtained from the expansion of $(a+b)^5$:
\begin{align}
(a+b)^5&=a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)\\
&=a^5+b^5+5ab((a+b)^3-3ab(a+b))+10a^2b^2(a+b)\\
&=a^5+b^5+5ab(a+b)^3-5a^2b^2(a+b)
\end{align}
In your case you get
$$
a^5+b^5=32-40c+10c^2
$$
which yields
$$
32=32-40c+10c^2
$$
hence $c=0$ or $c=4$.
The first case yields $16-\sqrt{x}=0$; the second case yields
$$
256-x=4^5
$$
Alternative approach. Since $a=\sqrt[5]{16+\sqrt{x}}\ne0$, we can set $t=b/a$ and from $(a+b)^5=a^5+b^5$, we get
$$
f(t)=(t+1)^5-t^5-1=0
$$
We have $f'(t)=5(t+1)^4-5t^4$, which vanishes only at $t=-1/2$, and $f(-1/2)<0$. Thus $f$ only vanishes twice, but one root is negative. Since $0$ is a root, it is the only nonnegative one. Thus $b=0$.
| {
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"answer_count": 5,
"answer_id": 3
} |
How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$. How many invertible elements are there in $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$ . Justify your answer.
Hello, could someone explain to me how to find the invertible elements of a set as large as $\mathbb {Z_ {990}}$ and $\mathbb {Z_ {1060}}$
I thought about putting together the set and trying each one but it is very big and it can take forever
| For $k$ to be invertible in $\mathbb Z_n$ means there is an integer $a$ so that $ak \equiv 1 \pmod n$. And $ak\equiv 1\pmod n$ means there is an integer $b$ so that $ak = 1+bn$ which means $ak -bn = 1$.
By Bezouts Lemma that is possible if and only if $k,n$ are relatively prime.
So find all the numbers $k$ so that $k$ and $990$ are relatively prime.
As $990 = 2\cdot 3^2\cdot 5\cdot 11$ this will mean find all the integers less than $990$ that are not a mulitple of $2,3,5$ or $11$.
.....
There is a function that counts the number of elements that are relatively prime to $n$ and less than $n$. It is called the Euler Totient function.
$\phi(n) = $#of integers, $k: k< n; \gcd(k,n) = 1$.
There are some formulas I won't prove but:
*
*If $n = mp$ and $\gcd(m,p) = 1$ then $\phi(n) = \phi(m)\phi(n)$.
*If $p$ is prime $\phi(p) = p-1$.
*If $n =p^k$ and $p$ is prime $\phi(p^k)= p^{k-1}(p-1)$
From those we can conclude if If $n = \prod p_i^{k_i}$ then
$\phi(n) = \prod[p_i^{k_i-1}(p_k-1)] = n\prod\limits_{p\text{ prime};p|n}(1-\frac 1p)$.
So there are $\phi(990) = \phi(3^2)\phi(2)\phi(5)\phi(11) = 3^1\cdot 2\cdot 1\cdot 4\cdot 10 = 240$ such elements.
Or .... $990\prod (1-\frac 1p) = 990(1-\frac 12)(1-\frac 13)(1-\frac 15)(1-\frac 1{11})=$
$990\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$
$445\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$
$330\cdot\frac 45\cdot\frac {10}{11}=$
$264\cdot\frac {10}{11}= 260$
...or ....
$\require{cancel}$
$990\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {10}{11}=$
$990\cdot \frac 1{\cancel 2}\cdot\frac {\cancel 2}3\cdot\frac 4{\cancel 5}\cdot\frac {\cancel{10}2}{11}=$
$990 \cdot \frac 8{33}=$
$30\cdot 8 = 240$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$ There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent series: $\dfrac{1}{(z-1)^2}+\dfrac{1}{(z-1)^3}+\cdots$. Did I approach this correctly?
| *
*When the function is rational, the standard procedure is to start with finding its expansion into partial fractions. In the present case, you get $$f(z)=\dfrac{1}{z(z-1)}=\dfrac{1}{z-1}-\dfrac{1}{z}.$$ Hence, making use of your substitution $w=z-1$ (or $z=1+w$), we obtain
\begin{align}
\dfrac{1}{z-1}-\dfrac{1}{z}&=\frac{1}{w}-\frac{1}{w+1}=\frac{1}{w}-\frac{1}{w}\frac{1}{1-(-\dfrac{1}{w} ) }\\&=\frac{1}{z-1}-
\frac{1}{z-1}\frac{1}{1-(-\dfrac{1}{z-1})}.\tag 1
\end{align}
*And so, you just need to expand $\dfrac{1}{1-( -\dfrac{1}{w}
) }=\dfrac{1}{1-( -\dfrac{1}{z-1}) }$
for $|w|=|z-1|>1$:
$$\frac{1}{1-(-\dfrac{1}{z-1})}=\sum_{n\geq 0}\frac{(-1)^{n}}{(z-1)^{n}}=1+\sum_{n\geq 1}\frac{(-1)^{n}}{(z-1)^{n}}. \tag 2 $$
Please notice that this is an alternating series.
*Therefore, combining $(1)$ and $(2)$, for $|z-1|>1$ the original function may be expanded into
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-1}-\frac{1}{z-1}\sum_{n\geq 1}\frac{(-1)^{n}}{\left( z-1\right) ^{n}} \tag 3\\&=\sum_{n\geq 1}\frac{(-1)^{n+1}}{\left( z-1\right) ^{n+1}}=\sum_{n\geq 2}\frac{(-1)^{n}}{\left( z-1\right) ^{n}}\\&=\frac{1}{(z-1)^2}-\frac{1}{(z-1)^3}\pm\cdots\tag{4}
\end{align}
| {
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} |
Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$ I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
So far I have:
*
*Base case: true
*Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
here is where I get stuck on simplification of both left and right side so that they are equal
$\dfrac{3}{4}(3^n(2n-1)+1)+3^{n+1}+3^{n+1}n=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
Thanks for your help!
| Notice that
$$3^{n+1}(n+1) = 3 \cdot 3^n (n+1) = \frac{3}{4} \cdot 3^n \cdot 4(n+1)$$
Because of the 3/4, you can move this term into the parentheses and combine it with the $3^n(2n-1)+1$ part. And because of the $3^n$, you can move the $4(n+1)$ into the parentheses and combine with the $2n-1$ part, to get
$$ \frac{3}{4} \left( 3^n (2n-1) + 1 \right) + 3^{n+1}(n+1) = \frac{3}{4} \left( 3^n \left(6n+3 \right) + 1 \right) $$
Notice that $6n+3 = 3(2n+1)$, so in the parentheses you get $3^{n+1}(2n+1)+1$.
| {
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Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$ Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$
My attempt: multiply both sides by the denominators, and it is equivalent to prove:
$$
n\sum_{k=1}^n{\frac{1}{a_k}}-n\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n+1}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right) \\
\sum_{k=1}^n{\frac{1}{a_k}}-\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n\left( n+1 \right)}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right)\\ \sum_{k=1}^n{\left( \frac{1}{a_k}-\frac{1}{\frac{1}{k}+a_k} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\frac{1}{a_k}}\sum_{p=1}^n{\frac{1}{\frac{1}{p}+a_p}}
\\ \sum_{k=1}^n{\left( \frac{\frac{1}{k}}{a_k\left( \frac{1}{k}+a_k \right)} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\sum_{p=1}^n{\frac{1}{a_k\left( \frac{1}{p}+a_p \right)}}}
$$
But how to proceed further?
| I gave a solution years ago.
Denote $A = \sum_{k=1}^n \frac{1}{\frac{1}{k} + a_k}$,
$B = \sum_{k=1}^n \frac{1}{a_k}$.
The function $f(x) = \frac{1}{1 + \frac{1}{x}}, \ x > 0$ is concave. By Jensen's inequality, we have
\begin{align}
A &= f(\tfrac{1}{a_1}) + 2f(\tfrac{1}{2a_2})
+ 3(\tfrac{1}{3a_3}) + \cdots + n f(\tfrac{1}{na_n})\\
&\le \frac{n(n+1)}{2} f\Big( \frac{B}{\frac{n(n+1)}{2}}\Big)\\
&= \frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}.
\end{align}
Then we have
$$\frac{n}{A} - \frac{n}{B} \ge \frac{n}{\frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}} - \frac{n}{B}
= \frac{2}{n+1}\frac{B + \frac{n(n+1)}{2}}{B} - \frac{n}{B} = \frac{2}{n+1}.$$
We are done.
| {
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"answer_id": 0
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$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum. I need to find supremum of:
$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$
I found out that:
$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} = \frac{m^2(n^3 + 3n^2) + 2m(n^3+3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
$$\frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
And from inequality of means (AM - GM) I know that:
$$\frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} \leq \frac{1}{2}$$
Therefore I found a limitation of my set A. Now I need a sequence of elements belonging in $A$ that would be convergent to $\frac{1}{2}$. That is the part I can't do myself.
From graph I know that if I take $m = n$ then for $n \in [1 ; \infty)$ I get decreasing function and for that function $\frac{1}{2}$ is reached by $n < 1$. A maximal value of that function for $n \in [1 ; \infty)$ is equal to $\frac{12}{25}$ and is reached by $n = 1$. But that is not a solution.
There is no ANY fact that would legitimise value $\frac{12}{25}$ as the supremum of that set since there always might by something bigger if $n \neq m$.
What I need is a prove that in the supremum of set $A$: $m = n$ (I don't know why that would be the case) or that there is a bigger value reached for $n$ and $m$ that are not equal.
It might be important to add that I can not use integrals nor derivatives.
| We pick a sub-sequence of $A$ in the following fashion: $n=1, 2, \cdots, $, and $m$ depends on $n$ via $m=\lfloor \sqrt{a_n+1} \rfloor$.
Define $a_n = n^3+3n^2, b_m=m^2+2m=(m+1)^2-1=\lfloor\sqrt{a_n+1}+1 \rfloor^2-1$. Then $\frac{a_n b_m}{a_n^2+b_m^2} \in A$ and we will prove that $\lim_{n\to \infty} \frac{a_n b_m}{a_n^2+b_m^2} \to \frac 12.$
Note that $x-1 < \lfloor x \rfloor \le x, \forall x$, then
$$a_n=(\sqrt{a_n+1})^2-1 < \lfloor \sqrt{a_n+1}+1\rfloor^2-1 \le (\sqrt{a_n+1}+1)^2-1 \\
\implies 1 < \frac{b_m}{a_n} \le \frac{(\sqrt{a_n+1}+1)^2-1}{a_n} = 1 + \frac{1}{a_n}+\frac{2\sqrt{a_n+1}}{a_n}$$
Therefore $\lim_{n\to\infty} \frac{b_n}{a_n} = 1 \implies \lim_{n\to\infty} \frac{a_n b_n}{a_n^2+b_n^2} = \frac 12. \blacksquare$
| {
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How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:
Setting up for Integration by parts
$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1}$$ $$v^{'}=x^6 \implies v=\frac{1}{7}x^7$$
Applying Integration by parts
$$I_n=[\frac{1}{7}x^7(1+x^3)^n]^1_0 - \frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$
$$=\frac{2^n}{7}-\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$
However I have no idea how to bring the integral of $\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$ into the form of $\int^1_0{x^6(1+x^3)^{n-1}dx}$ so that I may substitute it as $I_{n-1}$.
How can I do it, so that I may find a relationship between $I_4$ and $I_3$ ?
| Notice that $$
I_n - I_{n - 1} = \int_0^1 {x^9 (1 + x^3 )^{n - 1} dx} .
$$
| {
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Stochastic Integration: Riemann Sums I am reading this book:
Introduction to Stochastic Differential Equations
On page 63, in the lemma in the middle of the page the author has the following
$\displaystyle \sum_{k=0}^{m_n - 1}W(\tau_k^n)(W(t_{k+1}^n) - W(t_k^n)) = \frac{W^2(T)}{2} - \frac{1}{2}\sum_{k=0}^{m_n - 1}(W(t_{k+1}^n) - W(t_k^n))^2 + \sum_{k=0}^{m_n - 1}(W(\tau_{k}^n) - W(t_k^n))^2 + \sum_{k=0}^{m_n - 1}(W(t_{k+1}^n) - W(\tau_k^n))(W(\tau_{k}^n) - W(t_k^n))$
$\tau_k$ is a midpoint in $[t_k, t_{k+1}]$
WLOG we can assume $W(t_0) = 0$ as it is a standard Brownian motion. Let $m_n = 2$ then we get the following:
NOTE: I will omit the exponent $n$ and $W(t_2) = W(T)$ as this is on $[0,T]$
The LHS becomes :
$W(\tau_0)(W(t_1) - 0) + W(\tau_1)(W(T) - W(t_1))$
The RHS side is then:
$\frac{1}{2}W^2(T) - \frac{1}{2}[W^2(t_1) + W^2(T) - 2W(T)W(t_1) + W^2(t_1)] + [W^2(\tau_0) + W^2(\tau_1) - 2W(\tau_1)W(t_1) + W^2(t_1)] + ...$
$= W(\tau_0)(W(t_1) - 0) + W(\tau_1)(W(T) - W(t_1)) + \bigg ((W^2(t_1) - \frac{1}{2}W^2(t_1) \bigg )$
So they are almost similar except the term $\bigg ((W^2(t_1) - \frac{1}{2}W^2(t_1) \bigg )$. And it seems there would be more of these terms left over for $m_n = j, j > 2$.
Possibly I made a mistake in my algebra, easy to misplace terms in expanding. If I did not make a mistake though could someone clarify because as it stands it seems incorrect.
| For ease of notation we replace $m_n,t_k^n,\tau_k^n$ with $m,t_k,\tau_k$ respectively. Starting on the right, we have the following:
\begin{align}
\text{RHS } & = \frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 + W(t_{k+1})W(t_k) - \frac{1}{2}W(t_k)^2 + W(\tau_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) - W(t_{k+1})W(t_k) - W(\tau_k)^2 + W(\tau_k)W(t_k) \\
\end{align}
We cancel $W(t_{k+1})W(t_k) - W(t_{k+1})W(t_k)$ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 + W(\tau_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) - W(\tau_k)^2 + W(\tau_k)W(t_k).
\end{equation}
Then we cancel $W(\tau_k)^2 - W(\tau_k)^2$ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) + W(\tau_k)W(t_k).
\end{equation}
Then $W(\tau_k)W(t_k) - 2W(\tau_k)W(t_k) = -W(\tau_k)W(t_k)$, and we have
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 - W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) .
\end{equation}
We then combine $W(t_k)^2 - \frac{1}{2}W(t_k)^2 = \frac{1}{2}W(t_k)^2 $ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 + \frac{1}{2}W(t_k)^2 - W(\tau_k)W(t_k) + W(t_{k+1})W(\tau_k) .
\end{equation}
Some slight rearranging yields the following:
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} \frac{1}{2}W(t_k)^2 -\frac{1}{2}W(t_{k+1})^2 + W(\tau_k)(W(t_{k+1}) - W(t_k)).
\end{equation}
Note that $\sum_{k=0}^{m-1} \frac{1}{2}W(t_k)^2 -\frac{1}{2}W(t_{k+1})^2$ is a telescoping sum, so we have
\begin{equation}
\frac{W(t_m)^2}{2} +(\frac{1}{2}W(t_0)^2 -\frac{1}{2}W(t_{m})^2) + \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)).
\end{equation}
As you noted, we have standard Brownian motion, so $W(t_0)^2 = 0$, and we obtain the left hand side:
\begin{equation}
\text{RHS } = \frac{1}{2}W(t_0)^2 + \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)) = \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)) = \text{LHS}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$? Consider Numbers:
\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is congruent to $10^m \bmod 41$ for some $m$?
| Since $1..1$ with $n$ ones is equal to $\frac{10^n-1}{9}$, we can write the general term of the sequence as
$$
a_n = 245 \cdot \Big(200 \cdot \frac{10^n-1}{9} + 11\Big) - 239.
$$
Looking modulo 41 we have $245 \equiv -1$, $200 \equiv -5$, $-239 \equiv 7$ and $9^{-1} \equiv -9$ (since $9 \times (-9) = -81 \equiv 1$). We can write
\begin{align*}
a_n
&\equiv (-1) \cdot 9^{-1} (-5) (10^{n}- 1) - 11 + 7 \\
&\equiv 9 \cdot (-5 \cdot 10^{n} + 5) -4 \\
&\equiv -45 \cdot 10^{n} + 45 - 4 \\
&\equiv -4 \cdot 10^n \\
&\equiv -40 \cdot 10^{n-1}\\
&\equiv 10^{n-1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?
| Great job so far! Notice that all terms of the form $\binom{n}{i}a^{n-i}$ are divisible by $a$ if and only if $i<n$. This means that the remainder is simply $(2x)^n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $ The limit can be rewritten as
$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$
Recall the inequality,
$$ \cos x < \frac{\sin x}{x} < 1$$
holds for $ x\in (-\pi/2, \pi/2) $.
This provides the new inequalities
$$ 0 < \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] < \frac{1-\cos x}{x}, \text{ with } x>0 $$
$$ \frac{1-\cos x}{x} < \frac{1}{x} \left[1- \frac{\sin x}{x} \right]< 0, \text{ with } x<0 $$
Applying the Squeeze Theorem, we get
$$ \lim_{x \rightarrow 0^{\pm}} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] = 0$$
Note:
$ \lim_{ x \rightarrow 0} \frac{1-\cos x}{x} = 0 $
| It is a fine and nice argument with trig functions.
Another way to proceed is to use asymptitics (see them as Taylor expansions at $x=0$.
Then you can write $$\lim_{x \rightarrow 0} \frac{-(\sin(x) -x)}{x^2} = \lim_{x \rightarrow 0} -(\frac{-x^3/3!}{x^2} )= \lim_{x \rightarrow 0} x/6 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculating conditional variance using two different methods
The stock prices of two companies at the end of any given year are modeled with random variables $X$ and $Y$ that follow a distribution with joint density function $f(x, y) = 2x$ for $0<x<1, x<y<x+1$ and $0$ otherwise. What is the conditional variance of $Y$ given $X=x$?
My attempt:
$$Var[Y|X=x] = E[Y^2|X=x] - (E[Y|X=x])^2$$
Now, $E[Y^2|X=x] = \int_0^2y^2\cdot f(y|X=x) \; dy$. Note that $f(y|X=x) = \dfrac{f(x,y)}{f_X(x)} = \dfrac{2x}{\int_x^{x+1}2x \; dy} = 1$. So, $E[Y^2|X=x] = \int_0^2y^2 \; dy = \dfrac{8}{3}$.
Next, $E[Y|X=x] = \int_0^2y\cdot f(y|X=x) \; dy = \int_0^2y\; dy = 2$.
So, $Var[Y|X=x] = \dfrac{8}{3} - 4 = \dfrac{-4}{3}$
which is not the correct answer. Where have I gone wrong?
Alternatively, when I recognize that $f(y|X=x) = 1 \implies W \sim Unif[0, 2]$ where $W = Y|X=x$, I can use the formula for the variance of a uniform distribution to calculate the conditional variance as follows:
$$Var[W] = \frac{(2-0)^2}{12} = \frac{1}{3}$$
which is not the correct answer either. Where have I gone wrong with this approach? Thanks!
| The main mistake is that the conditional density is actually $$f(y \mid X=x) = \begin{cases} 1 & x < y < x+1 \\ 0 & \text{otherwise}\end{cases}$$
It is very important to consider the support of the density.
In the first approach, you have
$$E[Y \mid X=x] = \int_x^{x+1} y \, dy = \frac{(x+1)^2 - x^2}{2} = x + \frac{1}{2}$$
and
$$E[Y^2 \mid X=x] = \int_x^{x+1} y^2 \, dy = \frac{(x+1)^3 - x^3}{3} = x^2 + x + \frac{1}{3}.$$
So the variance is
$$x^2 + x + \frac{1}{3} - (x+1/2)^2 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}.$$
In the second approach, you have $W \sim \text{Unif}[x, x+1]$, so the conditional variance is $\frac{((x+1)-x)^2}{12} = \frac{1}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $A=(-1,0),B=(1,0), C$ be points in $\mathbb{R}^2.$ What is the locus of points $\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$}? Let $A=(-1,0),\ B=(1,0),\ C$ be points in $\mathbb{R}^2.$ What (shape(s)) is the locus of points $L(C)=\{s\in\mathbb{R}^2:C$ lies on the angle bisector of $AsB$ } ?
Obviously the set $L$ depends on where $C$ is, which is why I've written $L$ as a function of $C.$
If $C$ is the midpoint of $AB$, or on the perpendicular bisector of $AB$, it is not clear to me that $L= \{$ the perpendicular bisector of $AB\},$ because maybe $L$ contains more points. [Edit: Karan Elangovan has answered this part of the question.]
| Let $C(a,b),s(x,y)$ where $y\not=0$.
Also, let $U(u,0)$ be the intersection point of the angle bisector with $x$-axis.
Then, we have
$$\begin{align}&sA:sB=AU:BU
\\\\&\iff (1-u)\sqrt{(x+1)^2+y^2}=(1+u)\sqrt{(x-1)^2+y^2}
\\\\&\iff u=\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}\end{align}$$
So, $C$ is on the angle bisector ($sU$) if and only if
$$\begin{align}&b(x-u)=y(a-u)
\\\\&\iff (y-b)u=ay-bx
\\\\&\iff (y-b)\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}=ay-bx
\\\\&\iff (y-b)\bigg(\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\bigg)
\\&\qquad\qquad=(ay-bx)\bigg(\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\bigg)
\\\\&\iff (y-b-ay+bx)\sqrt{(x+1)^2+y^2}=(y-b+ay-bx)\sqrt{(x-1)^2+y^2}\end{align}$$
So, we have to have
$$(y-b-ay+bx)(y-b+ay-bx)\ge 0$$
under which we have
$$(y-b-ay+bx)^2\bigg((x+1)^2+y^2\bigg)=(y-b+ay-bx)^2\bigg((x-1)^2+y^2\bigg)$$
i.e.
$$\color{red}{bx^3- a b x^2- b x+(a^2 -b^2 + 1) x y - a x^2 y + b x y^2 - a y^3+ a b y^2 - a y + a b =0}$$
Examples :
*
*If $a=b=0$, then we get a line $x=0$.
*If $b=0$ with $a\not=0$, then we get a circle$$ \bigg(x-\frac{a^2+1}{2a}\bigg)^2 + y^2 =\bigg(\frac{1-a^2}{2a}\bigg)^2$$under the condition that $-1\lt a\lt 1$ as jlammy commented.
*If $a=0$ with $b\not=0$, then we get a line $x=0$ or a circle $$x^2 +\bigg(y-\frac{b^2-1}{2b}\bigg)^2 =\bigg(\frac{b^2+1}{2b}\bigg)^2$$under the condition that $(y+bx-b)(y-bx-b)\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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What's the most elementary way to solve this trigonometric problem?
A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.
This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.
Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).
Thanks in advance.
| Here's an alternative trigonometric approach that doesn't assume knowledge of the divided triangle formulae, but at the expense of additional work algebraically.
Half the battle with trigonometry is getting started: which formulae capture the key details of your set-up? If we want to do trigonometry separately on the smaller triangles $\triangle ABC$ and $\triangle BCD$, the following points are essential:
*
*$ABD$ is a straight line, so $\angle ABC$ and $\angle CBD$ are supplementary,
*The two triangles share the length $BC = x$.
So we expect to need these angles and sides in our equations. For $\triangle ABC$ we also know $AB = 2$ and $AC = 4 - x$. A way to incorporate all three sides and $\angle ABC$ would be the law of cosines:
\begin{align}
\cos(\angle ABC ) &= \frac{2^2 + x^2 - (4-x)^2}{2 \cdot 2 \cdot x} \\
\implies -\cos(\angle CBD ) &= \frac{8x - 12}{4x} \\
\implies \cos(\angle CBD) &= \frac{3}{x} - 2 \tag{1} \\
\end{align}
where we used the fact $\angle ABC$ and $\angle CBD$ are supplementary, because we want to relate this finding to the other triangle.
Since $\triangle BCD$ is right-angled, we can stick to basic "SOHCAHTOA" rather than anything fancy, and as discussed above we certainly want to make use of $BC = x$ and $\angle CBD$. There are two choices which both give the solution, though in very different ways.
Take the sine of $\angle CBD$; combine with cosine by harmonic addition
Since $\sin (\angle CBD) = \frac{1}{x}$ then $(1)$ implies that
$$ \cos \theta - 3 \sin \theta = -2$$
This can be solved by the harmonic addition ("R") formula to deduce an acute angle of approximately 57.7°; using $\tan (\angle CBD) = \frac{1}{y}$ we have
$$y = \frac{1}{\tan (\angle CBD) } \approx 0.63$$
Note that the harmonic addition formula also produces an obtuse solution around 159.2°, for which we obtain $y \approx -2.63$, corresponding to the bead lying 0.63 vertically above $A$ instead of 0.63 vertically below $B$.
Take the cosine of $\angle CBD$; compare results and form a quadratic
From $\triangle ABC$ we had deduced $\cos(\angle CBD) = \frac{3}{x} - 2$.
From $\triangle BCD$ we have $\cos(\angle CBD) = \frac{y}{x}$.
Comparing results we have:
\begin{align}
\frac{y}{x} &= \frac{3}{x} - 2 \\
\implies y &= 3 - 2x
\end{align}
Filling that into our diagram we see $AD = y+2 = 5-2x$. We now have all the lengths of the right-angled triangle $\triangle ADC$ in terms of $x$, so we can form a quadratic using Pythagoras. Note we could instead write the sides in terms of $y$, but there would be fractional coefficients which make unnecessarily hard work. It's easier to find $x$ then use $y = 3-2x$ to convert back to $y$ at the end.
\begin{align}
(5 - 2x)^2 + 1^2 &= (4-x)^2 \\
\implies 3x^2 -12x + 10 &= 0\\
\implies 3(x-2)^2 &= 2 \\
\implies x = 2 \pm \sqrt {\frac 2 3} &= 2 \pm \frac{\sqrt 6 }{3}
\end{align}
Note that it's the lower of these two values which yields the positive root for $y$,
$$y = 3 - 2x = -1 + \frac{2\sqrt 6 }{3} \approx 0.63$$
while the negative root
$$y= -1 - \frac{2\sqrt 6 }{3} \approx -2.63$$
corresponds to the solution above $A$, as before.
Alternative: $\angle BAC = \angle DAC$ is common to $\triangle ABC$ and $\triangle ACD$
This works like the above, but has the slight advantage it doesn't require students to know $\cos (\pi - \theta) = - \cos (\theta)$. It makes use of the fact $ABD$ is a straight line by looking at how a small triangle lies inside the larger one. Using the law of cosines in $\triangle ABC$:
\begin{align}
\cos(\angle BAC) &= \frac{2^2 + (4-x)^2 - x^2}{2 \cdot 2 \cdot (4-x)} \\
\implies \cos(\angle BAC) &= \frac{20 - 8x}{4(4-x)} \\
\implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \tag{2}
\end{align}
Then in the right-angled triangle $\triangle ACD$, we have
\begin{align}
\cos (\angle BAC) &= \frac{y+2}{4-x} \tag{3} \\
\sin (\angle BAC) &= \frac{1}{4-x} \tag{4}
\end{align}
We can use $(2)$ and $(3)$ together to write $y$ in terms of $x$.
\begin{align}
\frac{5-2x}{4-x} &= \frac{y+2}{4-x} \\
\implies 5-2x &= y + 2 \\
\implies y = 3 - 2x
\end{align}
We can apply Pythagoras to $\triangle ACD$ and solve for $x$ and $y$, as above.
Alternatively, we can use $(2)$ and $(4)$ together to relate the sine and cosine:
\begin{align}
(2) \implies \cos(\angle BAC) &= \frac{5-2x}{4-x} \\
&= \frac{8-2x-3}{4-x} \\
&= 2 - 3 \cdot \frac{1}{4-x}\\
(4) \implies \cos(\angle BAC) &= 2 - 3 \sin (\angle BAC) \\
\implies \cos(\angle BAC) + 3 \sin (\angle BAC) &= 2
\end{align}
Which can then be solved e.g. by harmonic addition to produce an acute angle of 20.8° and obtuse angle of 122.3°. We can then solve for $y$ e.g. by using $\tan(\angle BAC) = \frac{1}{2+y}$.
| {
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"url": "https://math.stackexchange.com/questions/3969860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____
My approach is as follow
Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below
$z = \cos \theta + i\sin \theta ;\theta \notin \left\{ 0 \right\}$ because $z\ne1$
$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta - i\sin \theta }}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
How do we get the required answer which if $\frac{\pi}{2}$
| $$\left| {\arg\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$$ $$= \left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right) - \arg\left(2\sin \frac{\theta }{2}\right)\right|$$
$$=\left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right)\right| = \left|\frac{\pi-\theta}{2} \right| \leq \frac{\pi}{2}$$
Here, $\operatorname{cis}(x) = \cos x + i\sin x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
| The solution makes use of the Pythagorean identities.
I. $\arcsin(3/5)=\theta\implies \sin(\theta)=3/5$. Using the identity: $\cos^2(\theta)=1-\sin^2(\theta)\iff \cos(\theta)=\sqrt{1-\sin^2(\theta)}$. Hence, $\cos(\arcsin(3/5))=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{16}{25}}=4/5$.
II. $\arctan(7/24)=\theta\implies \tan(\theta)=7/24$. Using the identity $1+\tan^2(\theta)=\sec^2(\theta) \iff \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}$. Hence, $\cos(\arctan(7/24))=\frac{1}{\sqrt{1+(\frac{7}{24})^2}}=24/25$.
III. $\sin(\arcsin(3/5))=3/5$ is trivial.
IV. $\arctan(7/24)=\theta \implies \tan(\theta)=7/24$ Using the identity $1+\cot^2(\theta)=\csc^2(\theta) \iff \sin(\theta)=\sqrt{\frac{\tan^2(\theta)}{1+\tan^2(\theta)}}$. Hence, $\sin(\arctan(7/24))=\sqrt{\frac{(\frac{7}{24})^2}{1+(\frac{7}{24})^2}}=7/25$.
Combining the 4 results we have: $I*II-III*IV=4/5*24/25-3/5*7/25=3/5$ yields the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1 & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.
If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0 & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.
My questions are :
a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.
b. What is every element of factor group $G/H?$
| as for $H$ in $K$ you could notice that $H$ is of index 2 in K so it is normal.
| {
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"url": "https://math.stackexchange.com/questions/3975642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$ Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).
My Work :-
Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\frac{1}{15}$ into $x*\frac{1}{1}\pm y*\frac{1}{3}\pm z*\frac{1}{5} $ or something like that so that we can cancel some terms in the whole sum.
| Hint: $$\frac{1}{105}=\frac{1}{4}(\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7})$$ $$\frac{1}{315}=\frac{1}{4}(\frac{1}{5\cdot 7}-\frac{1}{ 7\cdot 9})$$ do you
observe a pattern........
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral residue $\int_{0}^{\infty} {\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x. $ $$
\mbox{I have the integral:}\quad
\int_{0}^{\infty}
{\frac{\cos\left(x\right)}{x^{2} + 2x + 4}}\,{\rm d}x.
$$
I tried to use the residue theorem and semicircle and calculate the integral form minus infinity to infinity, but my function is not even. Do you have any hints $?$.
| I suppose you are calculating
$$\int_{-\infty }^{\infty } \frac{\cos (z)}{z^2+2 z+4} \, dz=\frac{e^{-\sqrt{3}} \pi \cos (1)}{\sqrt{3}}$$
Which can be done by integrating $\frac{e^{i z}}{z^2+2 z+4}$ along large upper semicircular contour, using Jordan's lemma, calculating residue at $-1+\sqrt{3} i$ and taking real part. For your original question,
$$\small \int_{0}^{\infty } \frac{\cos (z)}{z^2+2 z+4} \, dz=-\frac{i \text{Ci}\left(i \sqrt{3}+1\right) \cos \left(1+i \sqrt{3}\right)}{2 \sqrt{3}}+\frac{i \text{Ci}\left(1-i \sqrt{3}\right) \cos \left(1-i \sqrt{3}\right)}{2 \sqrt{3}}-\frac{i \text{Si}\left(i \sqrt{3}+1\right) \sin \left(1+i \sqrt{3}\right)}{2 \sqrt{3}}+\frac{i \text{Si}\left(1-i \sqrt{3}\right) \sin \left(1-i \sqrt{3}\right)}{2 \sqrt{3}}-\frac{\pi \cos (1) \sinh \left(\sqrt{3}\right)}{2 \sqrt{3}}$$
Which can be done by shifting $z\to z-1$, expanding $\cos(z-1)$ into $\sin/\cos$ and factorizing $\frac{1}{z^2+3}$ into $\frac{1}{z\pm \sqrt{3}i}$ (for partial fractions), using N-L formula based on elementary formula that $\int \frac{\cos (z)}{z-a} \, dz=\cos (a) \text{Ci}(z-a)+\sin (a) \text{Si}(a-z)$ and its sine counterparts. No essential simplification seems to exist for this case, though.
| {
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"timestamp": "2023-03-29T00:00:00",
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$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$ I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$
My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.
Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .
What am i missing ,can you help me?
| The minimal polynomial of your matrix is $x^3$. To find $f(A)$ for a holomorphic function $f$ defined on some neighbourhood of $\sigma(A) = \{0\}$, you need to find a polynomial $p \in \Bbb{C}[x]$ of degree $< 3$ such that
$$p(0) = f(0), \quad p'(0) = f'(0), \quad p''(0)=f''(0)$$
and then $f(A)=p(A)$.
Trying out $p(x) = ax^2+bx+c$, we find that $p(x) = \frac12 x^2+x+1$ is the desired polynomial if $f$ is the exponential function.
Therefore
$$e^A = p(A) = \frac12A^2+A+I = \begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}.$$
| {
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How do i solve this integration question using the washer and shell method? What is the volume of a solid enclosed by $y = (x-1)^2$ and $y = 4$ revolved around $x = - 3$?
I tried the washer method and the shell method and got different answers each time and I'm really confused please help!
My set up for the washer method was:
$$ \pi \int_0^4 {(4 + \sqrt{y} + 1)^2 - 4^2 \ \ dy} $$
My set up for the shell method was:
$$ 2 \pi \int_{-1}^3 {(x + 3)(4 - \sqrt{x} + 1)^2} \ \ dx $$
| Washer Method
The larger radius comes from the right side of the parabola $y = (x - 1)^2$, while the smaller radius comes from the left side of that parabola. Rewriting that parabola in terms of $x$, we have:
$$ y = (x - 1)^2 \Rightarrow x = 1 \pm \sqrt{y} . $$
Then, $ R(y) = (1 + \sqrt{y}) - (-3) $ and $ r(y) = (1 - \sqrt{y}) - (-3) $, where in both functions, the $ - (-3) $ comes from rotating about $x = -3$. These radii are also a bit easier to see graphically.
Then we have
$$ R(y) = 4 + \sqrt{y} , $$
$$ r(y) = 4 - \sqrt{y} . $$
Then, our formula for the volume is
$$ V_w = \pi \int_0^4 {R(y)^2 - r(y)^2 \ dy} $$
$$ = \pi \int_0^4 {(4 + \sqrt{y})^2 - (4 - \sqrt{y})^2 \ dy} $$
$$ = \frac{256\pi}{3} . $$
Shell Method
You've set up the radius of your cylindrical shells correctly:
$$ r(x) = x + 3 . $$
But, the height is just the difference in the $y$-coordinates of the two curves bounding your region:
$$ h(x) = 4 - (x - 1)^2 . $$
So, our volume is then
$$ V_s = 2 \pi \int_{-1}^3 {r(x)h(x) \ dx} $$
$$ = 2 \pi \int_{-1}^3 {(x + 3)(4 - (x - 1)^2) \ dx} $$
$$ = \frac{256\pi}{3} . $$
| {
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Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $ Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
I did these steps:
*
*Basis step
$$P(10): \ \ 1024 \geq 1000 \ (True)$$
*Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.
does this demonstration work?
I know there's a similar question but the solution is different, I want to know if my demonstration is valid as well.
| You already proved the base case
Now suppose that $n^3\le 2^n$ is true and
let's prove it for $n+1$
$(n+1)^3=n^3+\left(3n^2+3n+1\right)\tag{1}<\ldots$
for $n\ge 4$ we have $3n^2+3n+1 < n^3$ indeed, adding $n^3$ to both sides we get
$n^3+3 n^2+3 n+1<n^3+n^3\to (n+1)^3<2n^3\to n+1<n\sqrt[3]{2}\to n(\sqrt[3]{2}-1)>1\to n>\frac{1}{\sqrt[3]{2}-1}\approx 3.8$
continue $(1)$ $$\ldots<n^3+n^3\le 2^n+2^n=2^{n+1}$$
| {
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Solutions in a field Suppose $F$ is a field. Suppose further that $F$ is an ordered field. Consider the following equation $x^3 = b$, where $b \in F$. Prove that the equation above has, at most, one solution.
My attempt:
Suppose we have $x = b^{1/3} \in F$. We have that
$$x^3 = (b^{1/3})^3) = b^{\frac{1}{3} \cdot 3} = b^1 = b$$
Thus, there is one solution.
Assume there exists another solution and denote this by $c^{1/3}$. Since $b^{1/3}$ is a solution, $x = b^{1/3} \implies (b^{1/3})^3 = b \implies (b^{1/3})^3 - b = 0$. In a similar fashion, since $c^{1/3}$ is another solution, $x = c^{1/3} \implies (c^{1/3})^3 = b \implies (c^{1/3})^3 - b = 0$. Together we see that
$$(b^{1/3})^3 - b = 0 = (c^{1/3})^3 - b \implies (b^{1/3})^3 =(c^{1/3})^3 \implies b^{1/3} = c^{1/3} \implies b = c$$
Therefore there is at most one solution.
Is this correct for a field, $F$? Or did I make a mistake? Any assistance is much appreciated.
| You offer no justification for your claim that $b^{1/3} = c^{1/3} \implies b = c$.
In essence, that's what the problem requires you to prove.
As an indication that your proof is not valid, note that you never used the hypothesis that $F$ is an ordered field.
Thus, let $F$ be an ordered field, and suppose $x,y\in F$ are such that $x^3=y^3$.
We want to show $x=y$.
If $x=0$, then $x^3=0$, so $y^3=0$, hence $y=0$, so $x=y$.
Similarly, if $y=0$, then $y^3=0$, so $x^3=0$, hence $x=0$, so $x=y$.
Thus suppose $x,y\ne 0$, and assume $x\ne y$.
Our goal is to derive a contradiction.
\begin{align*}
\text{Then}\;\;&
x^3=y^3
\\[4pt]
\implies\;&
x^3-y^3=0
\\[4pt]\implies\;&
(x-y)(x^2+xy+y^2)=0
\\[4pt]
\implies\;&
x^2+xy+y^2=0
\\[4pt]
\implies\;&
4x^2+4xy+4y^2=0
\\[4pt]
\implies\;&
(4x^2+4xy+y^2)+3y^2=0
\\[4pt]
\implies\;&
(2x+y)^2+3y^2=0
\\[4pt]
\end{align*}
conradiction, since $(2x+y)^2 \ge 0$ and $3y^2 > 0$.
| {
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The formulas of prostapheresis: memorization technique This question is related purely for my students of an high school and indirectly for me. The formulas below are the formulas of prostapheresis,
\begin{cases}
\sin\alpha+\sin\beta=2\,\sin \dfrac {\alpha+\beta}{2}\, \cos \dfrac {\alpha-\beta}{2} \\
\sin\alpha-\sin\beta=2\sin \dfrac {\alpha-\beta}{2} \,\cos \dfrac {\alpha+\beta}{2}\\
\cos\alpha+\cos\beta=2\cos \dfrac {\alpha+\beta}{2}\,\cos \dfrac {\alpha-\beta}{2}\\
\cos\alpha-\cos\beta=-2 \,\sin \dfrac {\alpha+\beta}{2} \,\sin \dfrac {\alpha-\beta}{2}
\end{cases}
and while I am able to find them, I am not able to find a technique to memorize them.
Is there a technique to be able to memorize them?
| You can memorize the pattern
$$f(a)+\varepsilon f(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$
where $f$, $g$ and $h$ are either $\sin$ or $\cos$, and $\varepsilon$ and $\delta$ are either $1$ or $-1$. Given $f$ and $\varepsilon$, you then need a strategy to find $g$, $h$ and $\delta$.
Step 1: Find $g$ and $h$
Specializing this for $b=a$ we get
$$f(a)+\varepsilon f(a)=2\delta g\left(a\right)h\left(0\right)$$
and for $b=-a$ we get
$$f(a)+\varepsilon f(-a)=2\delta g\left(0\right)h\left(a\right)$$
Note that if the function that is given $0$ as input is $\sin$ then the right-hand side is $0$ for all $a$, while it if is $\cos$, the right-hand side is non-zero for some $a$. Using the (a)symmetry of $f$, we can easily determine if the left-hand side is zero for all $a$, and hence what $g$ and $h$ are:
*
*If $f(a)+\varepsilon f(a)=0$ for all $a$ then $h=\sin$ and otherwise $h=\cos$;
*If $f(a)+\varepsilon f(-a)=0$ for all $a$ then $g=\sin$ and otherwise $g=\cos$.
Step 2: Find $\delta$
Taking $a=\frac{\pi}{2}$ and $b=0$ yields
$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=2\delta g\left(\frac{\pi}{4}\right)h\left(\frac{\pi}{4}\right)$$
Since both $\sin$ and $\cos$ evaluate to $\frac{\sqrt{2}}{2}$ at $\frac{\pi}{4}$, we therefore have
$$f\left(\frac{\pi}{2}\right)+\varepsilon f(0)=\delta $$
which gives us $\delta$.
Plugging other specific values for which both sides are easy to evaluate would also work as long as both $g\left(\frac{a+b}{2}\right)$ and $h\left(\frac{a-b}{2}\right)$ are non-zero (so you do not need to remember to take $a=\frac{\pi}{2}$ and $b=0$).
Example
$$\cos(a)- \cos(b)=2\delta g\left(\frac{a+b}{2}\right)h\left(\frac{a-b}{2}\right)$$
Taking $a=b$ on the left-hand side makes it $0$ so $h=\sin$. Taking $a=-b$ also makes it zero so $g=\sin$:
$$\cos(a)- \cos(b)=2\delta \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$
Specializing to $b=0$ we get
$$\cos(a)-\cos(0)=2\delta \sin\left(\frac{a}{2}\right)\sin\left(\frac{a}{2}\right)$$
so we just need to pick a value of $a$ such that $\sin\left(\frac{a}{2}\right)\not = 0$. Taking $a=\pi$ gives:
$$\cos(\pi)-\cos(0)=2\delta \sin\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)$$
i.e.
$$-2=2\delta $$
which allows to conclude $\delta=-1$ and hence:
$$\cos(a)- \cos(b)=-2 \sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$
| {
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Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.
I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:
$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.
| $$x=\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{abc}{abc}\cdot\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{\frac{a+b+c+ab+bc+ca}{abc}}{\frac{1+abc}{abc}}$$
$$=\frac{\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}{\frac{1}{abc}+1}$$
$$=\frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{bc}+\overline{ca}}{1+\overline{abc}}$$
$$\overline{\left(\frac{a+b+c+ab+bc+ca}{1+abc}\right)}=\overline{x}$$
using if $|a|=1$ then $a\overline{a}=1$, and if $x=\overline{x}$, then $x\in\mathbb{R}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Link between thetrahedral numbers and combinatorics The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1}{2}$ pairs. But you could also count the pairs differently (see here): You can match the first person with $n$ different persons, the second person can be paired with $n-1$ different persons, the third with $n-2$, $\ldots$. Hence $\binom{n+1}{2}=n+(n-1)+(n-2)+\ldots$.
Now the tetrahedral numbers $1, 4=1+3, 10=1+3+6,\ldots$ can be written as $\binom{n+2}{3}$. Is there also an combinatorical explaination as for triangular numbers? I just found this explanation but I can't see clearly why $\binom{n+2}{3}$ should be the same as $1+3+6+\ldots+\frac{n(n+1)}{2}$.
| Your last link says
The numbers you're seeing are binomial coefficients. To go from $(0,0)$ to $(3,n)$ you have to take $3+n$ steps: $3$ right and $n$ up. You have to choose which $3$ of the $3+n$ steps that are going to go to the right, and the number of ways to choose is $\binom{3+n}{3}$
Your question is why $\binom{n+3}{3} =\sum\limits_{m=0}^n \binom{m+2}{2}$
In the context of the link, this says the number of ways of going $3$ right and $n$ up is equal to the number of ways of going $2$ right and $m$ up in any pattern, followed immediately by $1$ right and then by $n-m$ up, summed over all the possible $m$. In other words all the possible ways, measured by how many steps up before the last step right.
| {
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How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$ How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.
I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(\sqrt{x}+x\right)}{1+x^2}\:dx-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:dx}_{-G}$$
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x\right)}{1+x^2}\:dx}_{\frac{\pi }{8}\ln \left(2\right)-G}-\int _0^1\frac{\ln \left(1-\sqrt{x}\right)}{1+x^2}\:dx$$
Yet I'm still stuck with those other integrals, any help is appreciated.
Managed to numerically find that:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\frac{\pi }{16}\ln \left(2\right)+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
But I still dont know how to approach it.
| Substitute $t=\sqrt x$ in $I= \displaystyle\int _0^1\frac{\ln (1+\sqrt{x})}{1+x^2}\:dx$
\begin{align}
I &
=2\int _0^1\frac{t \ln \left(1+t\right)}{1+t^4}dt
\overset{t\to\frac1t}= \int _0^\infty\frac{t \ln \left(1+t\right)}{1+t^4}dt - \int _1^\infty\overset{t^2\to t}{\frac{t \ln t}{1+t^4}}dt \\
&= -\frac12\int _0^\infty{\ln \left(1+t\right)}\>d(\cot^{-1}t^2) -\frac14\int_1^\infty\frac{\ln t}{1+t^2}dt\\
&= \frac12 \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt - \frac14 G\tag1
\end{align}
Let $J(a) = \displaystyle \int _0^\infty\frac{\cot^{-1}(at^2)}{1+t}dt$ and differentiate under the integral
\begin{align}
J’(a) &= -\int _0^\infty\frac{t^2}{(1+t)(1+a^2t^4)}dt\\
&= \frac1{1+a^2} \int _0^\infty\left(-\frac1{1+t}+\frac{a^2t^3}{1+a^2t^4}-\frac t{1+ a^2t^4}+\frac{1-a^2t^2}{1+a^2t^4}\right)dt\\
&=\frac1{1+a^2} \left(\frac12\ln a -\frac\pi{4a}-\frac{\pi(a-1)}{2\sqrt{2a}}\right)
\end{align}
Then, with $J(\infty) =0$
\begin{align}
& \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt = J(1)=- \int_1^\infty J’(a)da\\
=&-\frac12\int_1^\infty \frac{\ln a}{1+a^2}da +\frac\pi4 \int_1^\infty \frac{1}{a(1+a^2)}da
+\frac\pi2\int_1^\infty \frac{a-1}{\sqrt {2a}(1+a^2)}da\\
=& -\frac12G +\frac\pi8\ln2 + \frac\pi2\ln(1+\sqrt2)
\end{align}
Plug into $(1)$ to obtain
$$I= \frac{\pi }{16}\ln 2+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
| {
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show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$ let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$
maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks
| Hint:
It suffices to prove that
$$\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \ge \left(\frac{x^{2n+1} + y^{2n+1}}{2}\right)^{\frac{1}{2n+1}}.$$
Since this inequality is symmetric and homogeneous, assume $y = 1$ and $x \ge 1$. It suffices to prove that
$$\ln (x^{n+1} + 1) - \ln (x^n + 1) \ge \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2}.$$
Take derivative.
Update
Let $f(x) = \mathrm{LHS} - \mathrm{RHS}$. We have
$$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Using matrix algebra to solve a differential vector equation So I am solving a simple problem under the influence of gravity, with no resistance or extra forces.
Labeling the axes, let us use the downward vertical as our positive $y%-axis, with $x$ being standard.
Of course, the easiest way to actually define the position of the pendulum (or it's center of mass) is via
\begin{align*}
\text{position: }\vec{r} &= \begin{bmatrix} x\\ y\\ \end{bmatrix} = R\begin{bmatrix} \sin\theta\\\cos\theta \end{bmatrix}\\
\text{velocity: }\vec{v} &=\frac{d}{dt}(\vec{r})=\begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix}= R\begin{bmatrix}\dot{\theta}\cos\theta\\-\dot{\theta}\sin\theta\end{bmatrix}\\
\text{acceleration: }\vec{a} &=\frac{d^2}{dt^2}(\vec{r})= R\begin{bmatrix}\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta\\-\ddot{\theta}\sin\theta - \dot{\theta}^2\cos\theta\end{bmatrix}
\end{align*}
where $\theta$ is the angle formed with the vertical by the pendulum.
Normally we would write the EOM as two separate equations:
$$\begin{cases}m\ddot{x}=0\\m\ddot{y}=F_g=mg\end{cases}$$
Of course, the two equations of motion are coupled by the relationship $x^2+y^2=R^2$, where but I wanted to try something a little different.
Instead, I wanted to try solving this system through matrix algebra, so rewriting the vector equation in polar coordinates by substituting for $\ddot{x}$ and $\ddot{y}$ yields:
$$m\vec{a} =
m \begin{bmatrix} \ddot{x}\\\ddot{y}\end{bmatrix} =
mR \begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
-\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix}
= \begin{bmatrix}0\\mg\end{bmatrix}$$
Normally we would express this system like so:
$$\begin{cases}m\ddot{x} = mR(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta)=0\\
m\ddot{y} = mR(-\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta)=mg\end{cases}$$
Here is my first question: Instead of solving the two equations separately, does it make sense to factor the matrix version of the equation, then employ some matrix multiplications to both sides to solve the problem?
The way I am imagining this process is as follows (beginning from Newton's second law)
\begin{align*}
m\vec{a} &= \vec{F}_{net} = \vec{F}_g\\
%
m\begin{bmatrix}\ddot{x}\\\ddot{y}\end{bmatrix} &= \begin{bmatrix}0\\ mg\end{bmatrix}\\
%
\text{Subsitute for $x,y$} \rightarrow m\begin{bmatrix}\ddot{x}\\\ddot{y}\end{bmatrix} &= mR\begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
-\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix}\\
%
\text{Factor} \rightarrow \begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
-\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix} &=
\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}\\
%
\text{So we have: }\rightarrow \begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}
\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}
\begin{bmatrix} \cos\theta&-\sin\theta \\ \sin\theta&\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}\begin{bmatrix}0\\ -\frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix} \cos^2\theta&-\sin\theta\cos\theta\\ \sin^2\theta&\sin\theta\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}0\\ -\frac{g}{R}\sin\theta\end{bmatrix}\\
%
\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix} \cos^2\theta&-\sin\theta\cos\theta\\ \sin^2\theta&\sin\theta\cos\theta \\\end{bmatrix} \begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}0\\ -\frac{g}{R}\sin\theta\end{bmatrix}\\
%
\begin{bmatrix}\cos^2\theta+\sin^2\theta&\sin\theta\cos\theta-\sin\theta\cos\theta\end{bmatrix}\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&= -\frac{g}{R}\sin\theta\\
%
\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&= -\frac{g}{R}\sin\theta\\
\ddot{\theta}&= -\frac{g}{R}\sin\theta\\
%
\end{align*}
$$\frac{d\dot{x}}{dx}\dot{x}= -\frac{g}{R}\sin\theta$$
$$\dot{x}d\dot{x}= -\frac{g}{R}\sin\theta dx$$
Is this a valid derivation of the equation of motion for a pendulum?
Edit: Here is another question. Instead of left multiplying by $\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}$, if we left multiply by $\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}$, we get:
\begin{align}
\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}
\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}
\begin{bmatrix}0\\-\frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\sin\theta\cos\theta&\cos^2\theta\\-\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}-\frac{g}{R}\cos\theta\\0\end{bmatrix}\\
%
\begin{bmatrix}1&1\end{bmatrix}
\begin{bmatrix}\sin\theta\cos\theta&\cos^2\theta\\-\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}1&1\end{bmatrix}
\begin{bmatrix}-\frac{g}{R}\cos\theta\\0\end{bmatrix}\\
%
\begin{bmatrix}0&1\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
-\frac{g}{R}\cos\theta\\
%
\dot{\theta}^2 &= -\frac{g}{R}\cos\theta \\
%
\end{align}
If the above is valid, this should be valid too, correct?
| The problem is that you started with a wrong assumption, that the gravity is the only force. If you just have a mass suspended at some point, and you release it, it will not oscillate, but instead it will go straight down. You need to add the tension in the suspending wire.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$
I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.
For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.
| Hint:
$f(x)=[\tan^{-1}(x)][\tanh^{-1}(x)]$
$f'(x)=\dfrac{ \tan^{-1}(x)}{1-x^2}+\dfrac{ \tanh^{-1}(x)}{1+x^2} $
If $|x|<1$:
$\displaystyle f'(x)=\left(\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{2n+1} x^{2n+1} \right) \left(\sum_{n=0}^{+\infty} x^{2n} \right) + \left(\sum_{n=0}^{+\infty} \dfrac{1}{2n+1} x^{2n+1} \right) \left(\sum_{n=0}^{+\infty} (-1)^n x^{2n} \right) $
Edit to show the idea
$f′(x)=\left(x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1+x^2+x^4+x^6+x^8...)+\left(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+\dfrac{x^7}{7}+\dfrac{x^9}{9}...\right)(1−x^2+x^4−x^6+x^8...) $
By using Cauchy product:
$f′(x)=2x+ 2\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)x^5+2\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)x^9+...$
Then we integrate:
$f(x)=x^2+\left(1−\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3} +\left(1 −\dfrac{1}{3}+\dfrac{1}{5}−\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5} +... $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.
\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}
so $x > 100$
| You can make the problem more general since you have
$$S_p=\frac{\prod_{n=1}^p (2n) }{\prod_{n=1}^p (2n-1) }=\frac{2^p \Gamma (p+1) } {\frac{2^p \Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi }} }=\sqrt{\pi }\,\,\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}$$ Take logarithms, use Stirling approximation and continue with Taylor series for large $p$
$$A=\log (\Gamma (p+1))-\log \left(\Gamma \left(p+\frac{1}{2}\right)\right)$$
$$A=\frac{1}{2}\log (p)+\frac{1}{8 p}-\frac{1}{192
p^3}+O\left(\frac{1}{p^5}\right)$$
$$\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}=e^A=\sqrt{p}\left(1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)$$
$$S_p >\sqrt{p\pi}\left(1+\frac{1}{8 p}\right) $$ Using it for $p=5000$ gives $125.334547017$ while the exact value is $125.334547056$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
find the matrix exponential Let
\begin{equation*}
A =
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}\end{equation*}
\begin{equation*}
B =
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix}\end{equation*}
\begin{equation*}
C =
\begin{pmatrix}
4 & 1 & 0 & -4 & 0 & 0\\
0 & 4 & -1 & 0 & -4 & 2\\
0 & 0 & 6 & 0 & 0 & -6\\
2 & 0 & 0 & -2 & 1 & 0\\
0 & 2 & -1 & 0 & -2 & 2\\
0 & 0 & 3 & 0 & 0 & -3\\
\end{pmatrix}\end{equation*}
Suppose $\exp(z)=e^z$, for $z\in \mathbb{C}$. Find $\exp(C)$.
Hint:
\begin{equation*}
D =
\begin{pmatrix}
2 & 1 \\
1 & 1 \end{pmatrix}\end{equation*}
is invertible.
I realize that this is already asked. But I'm hoping to get a way of solving this problem in the Banach algebra prospective.
Attempt:
We know that $\mathbb{C}^n$ is a Hilbert space. and $B(\mathbb{C}^n) \simeq \mathbb{M_n}$. So we can consider each matrix as a function in $B(\mathbb{C}^n)$
And we can write $Ae_j= \sum_{k=1}^n a_k e_k$, where $(e_k)$ is the orthonormal basis for $\mathbb{C}^n$.
Also I know that the spectrum $\sigma(A)= \{\text{eigenvalues of} A \}$
Since the $\mathcal{M_n}$ is over $\mathcal{C}$, it is a matrix ring.
By the hint, I know that
\begin{equation*}
D^{-1}=
\begin{pmatrix}
1 & -1 \\
-1 & 2 \end{pmatrix}\end{equation*}
And we have $D\begin{pmatrix}
x & 0 \\
0 & y \end{pmatrix}D^{-1}=
\begin{pmatrix}
-x+2y & 2x-2y \\
-x+y & 2x-y \end{pmatrix}$.
From here, I wonder if we could perform some sort of induction from $B(\mathbb{C}^2) \simeq \mathbb{M_2}$ to $B(\mathbb{C}^6) \simeq\mathbb{M_6}$.
Any suggestions will be much appreciated.
| Rye, since we discussed this a little bit in your previous question: Note that for the case $f=\exp$ there is a standard way of defining $f(a)$ for non-normal elements of a unital Banach algebra:
$$f(a)=\sum_{k=0}^\infty\frac{a^k}{k!}$$
You can find more details about this in the first pages of Murphy's book. This can also be done for any entire function: if $g(z)=\sum_{k=0}^\infty a_kz^k$, then set $g(x)=\sum_{k=0}^\infty a_kx^k$ for $x\in A$. The series converges in the Banach algebra since it converges absolutely and we are in a Banach space. Anyway, back to your question:
We can use block-form: set
Note that $$C=\begin{pmatrix}-A+2B& 2A-2B\\ -A+B&2A-B\end{pmatrix}$$
If $P=\begin{pmatrix}B&0\\0&A\end{pmatrix}$ and $$E=\begin{pmatrix}2I_3&I_3\\I_3&I_3\end{pmatrix}$$
then $E$ is invertible, $E^{-1}=\begin{pmatrix}I_3&-I_3\\-I_3&2I_3\end{pmatrix}$ and $C=EPE^{-1}$. From this we have that $C^n=EP^nE^{-1}$ and from the definition of $\exp(\cdot)$ we will have that $\exp(C)=E\cdot\exp(P)\cdot E^{-1}$.
It is easily verified that $\exp(P)=\begin{pmatrix}\exp(B)&0\\0&\exp(A)\end{pmatrix}$, so the whole problem has been reduced to computing the exponentials of $A$ and $B$ (which are in $M_3(\mathbb{C})$, life is much easier now).
But $A^3=0$, so $\exp(A)=I_3+A+\frac{1}{2}A^2$. On the other hand, for $n\geq1$
$$B^n=\begin{pmatrix}2^n&n\cdot 2^{n-1}&0\\0&2^n&0\\0&0&3^n\end{pmatrix} $$
can be easily verified (use induction). Observe that
$$\sum_{n=0}^\infty\frac{n2^{n-1}}{n!}=\sum_{n=1}^\infty\frac{2^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{2^n}{n!}=e^2 $$
so
$$\exp(B)=\begin{pmatrix}e^2&e^2&0\\0&e^2&0\\0&0&e^3\end{pmatrix} $$
Now that we know $\exp(A)$, $\exp(B)$ and $E$, we plug everything back to
$$\exp(C)=E\exp(P)E^{-1}$$
and the result follows.
| {
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Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$ This is a math Olympiad problem.
my attempt :
By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:
$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$
By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.
| hint
$$x^2+y^2-8x=9\iff$$
$$(x-4)^2+y^2=25$$
$$=(\pm 3)^2+(\pm 4)^2$$
$$=0^2+(\pm 5)^2$$
| {
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"source": "stackexchange",
"question_score": "1",
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How to find a lower bound on a smallest solution of $\ln( \frac{x}{a} ) +\frac{1}{x}=0$. The equaiton
\begin{align}
\ln( \frac{x}{a} ) +\frac{1}{x}=0
\end{align}
has two solutions for a sufficiently large $a$ (e.g., $a>3$).
I want a lower bound on the smaller of the two solutions.
I have found an upper bound on the largest of the two solutions as follows:
\begin{align}
0= \ln( \frac{x}{a} ) +\frac{1}{x}\le 1- \frac{a}{x} +\frac{1}{x}
\end{align}
where we have used a lower boung $\ln(x) \ge 1-\frac{1}{x}$. This implies that $x_2<a-1$.
However, I have no idea how to find a lower bound on the smallest solution.
| The function$$f(x)=\log \left(\frac{x}{a}\right)+\frac{1}{x}$$ has two solutions as soon as $a >e$.
As said in comments and answers, the smallest solution is given by
$$x=-\frac{1}{W_{-1}\left(-\frac{1}{a}\right)}$$
For the secondary branch, there exist nice bounds
$$-1 - \sqrt{2u} - u < W_{-1}\left(-e^{-(u+1)}\right) < -1 - \sqrt{2u} - \tfrac{2}{3}u$$ which make for your case
$$-\log (a)- \sqrt{2(\log (a)-1)}< W_{-1}\left(-\frac{1}{a}\right)$$
$$ W_{-1}\left(-\frac{1}{a}\right)< -\frac {2\log(a)+1}3 - \sqrt{2(\log (a)-1)}$$
For $a=e^2$,the smallest solution is then bounded by
$$1-\frac{1}{\sqrt{2}} \sim 0.292893 < x <\frac{3}{7} \left(5-3 \sqrt{2}\right)\sim 0.324583$$ while the exact solution is $\sim 0.317844$.
So, a first lower bound is
$$\color{red}{x_0=\frac 1 {A+ \sqrt{2(A-1)} }}\quad \text{where}\qquad \color{red}{A=\log(a)}$$
However, this can be improved a lot since $f(x_0) >0$ and $f''(x_0)>0$. So, by Darboux theorem, the solution will be reached without any overshoot of the solution and better bounds will be given by
$$\color{red}{x_{n+1}=x_n+\frac{x_n \left(x_n\log \left(\frac{x_n}{a}\right)+1\right)}{1-x_n}}$$ For example
$$\color{red}{x_1=\frac{\left(A+2 \sqrt{2(A-1)}-1\right)-\log \left(A+
\sqrt{2(A-1)}\right)}{\left(A+ \sqrt{2(A-1)}-1\right) \left(A+ \sqrt{2(A-1)}\right)}}$$
For the working case $(a=e^2)$, the left bounds are sucessively
$(n=0,1,2)$
$$\{0.2928932188,0.3154911206,0.3178230036\}$$
| {
"language": "en",
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Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$ I was solving a question and saw a pattern. Can someone prove it, please?
We know
$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
So we see the pattern... Can anyone give a proof of why:
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
| This answer gives a proof by induction. Here instead I will give a proof using Hockey-stick identity:
$$\sum_{i=r}^n \binom ir = \binom{n+1}{r+1}$$
Note that
$$k(k+1)\cdots(k+r) = \frac {(k+r)!}{(k-1)!} = (r+1)!\binom{k+r}{r+1}$$
Hence your sum
$$=(r+1)!\sum_{k = 1}^{n} \binom {k+r}{r+1} =(r+1)!\sum_{k = {r+1}}^{n+r} \binom {k}{r+1} = (r+1)!\binom{n+r+1}{r+2} \\= \frac {(r+1)!(n+r+1)!}{(n-1)!(r+2)!} = \frac {n(n+1)\cdots(n+r+1)}{r+2}$$
| {
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"source": "stackexchange",
"question_score": "5",
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Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
| It is useful to think about a problem like that in geometric terms.
We want to have $ A^3=A $ which could be thought as some kind of rotation of the plane.
Along these lines what we want is if we rotate a vector by $\theta $ under $A$ to be equal to a rotation by $3 \theta$ corresponding to $A^3$.
A rotation by $\theta$ is given by $A_\theta=
\left (\begin{array}{cc}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{array} \right )$.
So if we choose $\theta = \pi = 3\theta ~\text{mod} 2 \pi $
We get
$$A_{\pi/2}=
\left (\begin{array}{cc}
-1 & 0\\
0 & 1
\end{array} \right ).$$
Which should satisfy $$ A_{\pi}^3 = A_{\pi} $$
We can also change the coordinates by $\pi/3$ so to eliminate the zeros.
So we define
\begin{align*}
A &= A_{\pi/3} A_{\pi} A_{-\pi/3} \\
&= \left (\begin{array}{cc}
\frac{1}{2} & -\frac{\sqrt{3}}{{2}}\\
\frac{\sqrt{3}}{{2}} & \frac{1}{2}
\end{array} \right ) \cdot \left (\begin{array}{cc}
-1 & 0\\
0 & 1
\end{array} \right ) \cdot \left (\begin{array}{cc}
\frac{1}{2} & \frac{\sqrt{3}}{{2}}\\
-\frac{\sqrt{3}}{{2}} & \frac{1}{2}
\end{array} \right )
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Domain of a function quick question $$f(x)=\arcsin\left(\frac{x-1}{2x}\right)$$
We need to find the domain of this function.
My try:
$$-1\leq \frac{x-1}{2x} \leq 1 $$
We can split this into
$$-1\leq \frac{x-1}{2x} \quad\text{and} \quad\frac{x-1}{2x} \leq 1$$
My idea is to solve for this two inequalities and then take the intersection of them both.
$$-2x\leq x-1\\
-3x\leq -1\\
x\geq\frac{1}{3}$$
$$x-1 \leq 2x\\
-1 \leq x$$
By doing the intersection I get the wrong answer which is: $x\geq\frac{1}{3}$
What have I done wrong?
| Here it is another approach for the sake of curiosity.
I would start with squaring both sides as it is done next:
\begin{align*}
\left|\frac{x-1}{2x}\right| \leq 1 & \Longleftrightarrow \left(\frac{x-1}{2x}\right)^{2} \leq 1\\\\
& \Longleftrightarrow \frac{x^{2} - 2x + 1 - 4x^{2}}{4x^{2}} \leq 0\\\\
& \Longleftrightarrow \frac{3x^{2} + 2x - 1}{4x^{2}} \geq 0
\end{align*}
We may now study the behavior of the function at the numerator.
To begin with, let us rearrange its expression:
\begin{align*}
3x^{2} + 2x - 1 & = \left(3x^{2} + 2x + \frac{1}{3}\right) - \frac{4}{3}\\\\
& = \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} - \frac{4}{3}
\end{align*}
Consequently, the solution set to problem proposed is given by
\begin{align*}
3x^{2} + 2 x - 1 \geq 0 & \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}}\right)^{2} \geq \frac{4}{3}\\\\
& \Longleftrightarrow \left(\sqrt{3}x + \frac{1}{\sqrt{3}} \geq \frac{2}{\sqrt{3}}\right)\vee\left(\sqrt{3}x + \frac{1}{\sqrt{3}} \leq -\frac{2}{\sqrt{3}}\right)\\\\
& \Longleftrightarrow \left(x \geq \frac{1}{3}\right)\vee\left(x \leq -1\right)
\end{align*}
Since both intervals does not contain the value $x = 0$, we conclude the sought domain is given by
\begin{align*}
D_{f} = (-\infty,-1]\cup\left[\frac{1}{3},+\infty\right)
\end{align*}
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\frac{DK}{DI}$ In triangle $ABC,$ where $AB = 8, AC = 7,$ and $BC = 10,$ $I$ is the incenter. If $AI$ intersects $BC$ at $K$ and the circumcircle of $\triangle ABC$ at $D,$ find $\frac{DK}{DI}.$
I first drew a diagram, but I was unsure where to go from here.
| I have doubts that you correctly compute the result, since the value $IK$ given in comments is incorrect. The correct result is:
$$\frac{DK}{DI}=\frac23.
$$
The details are given below.
Let $x,y,z$ being the distances from the vertices $A,B,C$ to the tangent points of the incircle. From the equations $x+y=c, y+z=a, z+x=b$ one obtains $x=\frac{b+c-a}2=s-a$, where $s$ is the semiperimeter. Then:
$$AI=\sqrt{x^2+r^2}=\sqrt{(s-a)^2+\frac{(s-a)(s-b)(s-c)}s}=\sqrt {\frac {bc (s-a)}{s}}=2\sqrt{\frac{14}5}.$$
The angle bisector length is:
$$AK=\sqrt{bc\left[1-\left(\frac a{b+c}\right)^2\right]}=\frac {\sqrt{4bc (s-a)s}}{b+c}=\frac{10}3\sqrt{\frac{14}5}.$$
By the power of point $K$ and angle bisector theorem:$$\color {red}{DK}=\frac{BK\cdot KC}{AK}=\frac{a^2\frac{bc}{(b+c)^2}}{AK}=\frac{a^2}{b+c} \sqrt {\frac{bc}{4 (s-a)s}}=\frac83\sqrt{\frac{14}5},$$
so that:
$$\color {red}{DI}=DK+AK-AI=4\sqrt{\frac{14}5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2 dt$$
$t= \sqrt{1-r^2}\implies r^2= t^2-1$
$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$
| First of all, since $t= \sqrt{1-r^2}$, we should have $dt/dr = \frac {-r}{\sqrt{1-r^2}}$ and $t^2 + r^2 = 1$.
Now
$$\int_0^1 \frac {r^3}{\sqrt{1-r^2}}dr = \int_{\color{red}1}^{\color{red}0} \frac {r^3}{\sqrt{1-r^2}}\frac {\sqrt{1-r^2}}{-r}dt = \int_0^1r^2dt = \int_0^1(\color{red}{1-t^2})dt = [t-\frac{t^3}3]_0^1 = \frac23$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Equilateral triangle $ABC$ with $P$ inside, $PA= x$, $PB=y$, $PC=z$ and $z^2 =x^2+y^2$. Find side length of $ABC$
$ABC$ is an equilateral triangle $ABC$ with $P$ inside it such that $PA= x$, $PB=y$, $PC=z$. If $z^2 =x^2+y^2$ , find the length of the sides of $ABC$ in terms of $x$ and $y$?
If $z^2=x^2+y^2$ then how can I find measures of angles around $P$ so that the sides can be expressed in terms of $x$ and $y$. I've tried everything I can think of.
| The relationship between $x,y,z$ and the side of the triangle, $a$ is interesting:
$$3(x^4+y^4 + z^4 + a^4) = (x^2+y^2+z^2 + a^2)^2 \tag{1}$$
Please see here, near reference $19$
If you put $z^2 = x^2 + y^2$ in $(1)$, you'll get
$$3(x^4 + y^4 + (x^2+y^2)^2 + a^4) = (2x^2 + 2y^2 + a^2)^2$$
Solving for $a$, you get
$$a = \pm \sqrt{x^2+y^2 \pm\sqrt{3}xy}$$
Consider only the positive roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Help finding Eigenvectors The matrix is \begin{equation*}
A =
\begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & -2 \\
3 & 2 & 1
\end{pmatrix}
\end{equation*}
I got the eigenvalues $\lambda_1 = 1, \lambda_2 = 1 + 2i$, and $\lambda_3 = 1-2i$. I am only concerned with the complex valued eigenvectors. For $\lambda_2$, I got the eigenvector
\begin{equation*} v_2=
\begin{pmatrix}
0 \\
i \\
1
\end{pmatrix}\end{equation*} and for $\lambda_3$, I got the eigenvector
\begin{equation*}v_3=
\begin{pmatrix}
0 \\
-i \\
1
\end{pmatrix}\end{equation*}
In the back of the book, it is saying the eigenvectors for $\lambda_2$ and $\lambda_3$ are \begin{equation*} v_2=
\begin{pmatrix}
0 \\
1 \\
-i
\end{pmatrix}\end{equation*}
and \begin{equation*}v_3=
\begin{pmatrix}
0 \\
1 \\
i
\end{pmatrix}\end{equation*}
When I checked on Wolfram Alpha, it is saying that my answers are correct. Did I do something wrong or is the back of my book wrong?
| You're correct. The eigenvalues are $\lambda_1=1,\lambda_2=1-2i,\lambda_3=1+2i$
$$v_1=\left[\begin{array}{rrr|r}
0 & 0 & 0 & 0 \\
2 & 0 & -2 & 0 \\
3 & 2 & 0 & 0 \\
\end{array}\right]\sim\left[\begin{array}{rrr|r}
1 & 0 & -1 & 0 \\
0 & 1 & \frac{3}{2} & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right]\longrightarrow\left[\begin{array}{rrr|r}
x_3 \\
-\frac{3}{2}\cdot x_3 \\
x_3 \\
\end{array}\right]=\left[\begin{array}{rrr|r}
1 \\
-\frac{3}{2} \\
1 \\
\end{array}\right]$$
$$v_2=\left[\begin{array}{rrr|r}
2i & 0 & 0 & 0 \\
2 & 2i & -2 & 0 \\
3 & 2 & 2i & 0 \\
\end{array}\right]\sim\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 & 1 & i & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right]\longrightarrow\left[\begin{array}{rrr|r}
0 \\
-i\cdot x_3 \\
x_3 \\
\end{array}\right]=\left[\begin{array}{rrr|r}
0 \\
-i \\
1 \\
\end{array}\right]$$
$$v_3=\left[\begin{array}{rrr|r}
-2i & 0 & 0 & 0 \\
2 & -2i & -2 & 0 \\
3 & 2 & -2i & 0 \\
\end{array}\right]\sim\left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 & 1 & -i & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right]\longrightarrow\left[\begin{array}{rrr|r}
0 \\
i\cdot x_3 \\
x_3 \\
\end{array}\right]=\left[\begin{array}{rrr|r}
0 \\
i \\
1 \\
\end{array}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$ Let $ABC$ be acute triangle. Prove that
$$\sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$$
My attempt:
$$\sqrt{\tan A} + \sqrt{\tan B}\geq 2\sqrt[4]{\tan A\cdot\tan B}$$
At here I think I need to prove $$2\sqrt[4]{\tan A\cdot\tan B}\ge2\sqrt{\cot\frac{A}{2}}$$
$$\Leftrightarrow \tan A\cdot\tan B\ge \cot^2\frac{A}{2}$$
And I was stuck here.
Help me... Thanks
| Remark: $\tan A,\ \tan B,\ \tan C,\ \cot\frac{A}{2},\ \cot\frac{B}{2},$ and $\cot\frac{C}{2}$ are all positive because $0<A,\ B,\ C<\frac{\pi}{2}$. This is used in this proof, for example in cancellings or taking square roots of both sides of inequalities.
The function $f\left(x\right)=\tan x$ is convex on the interval $\left(0,\ \frac{\pi}{2}\right)$, so we can use Jensen's Inequality to show that
$\begin{align}
\sum_\text{cyc}\frac{\tan A+\tan B}{2}\ge\sum_\text{cyc}\tan\frac{A+B}{2}=\sum_\text{cyc}\cot\frac{C}{2}\Rightarrow\sum_\text{cyc}\tan A\ge\sum_\text{cyc}\cot\frac{A}{2}\ \left(\#\right)
\end{align}$
It remains to prove that $\sum_\limits\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\limits\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\ \left(\#\#\right)\Leftrightarrow\\$
$\begin{align}
&\Leftrightarrow\sum_\text{cyc}\tan A\tan B\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{2\cot\frac{A}{2}}{\cot^{2}\ \frac{A}{2}-1}\cdot\frac{2\cot\frac{B}{2}}{\cot^{2}\ \frac{B}{2}-1}\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{4}{\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)}\ge0\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)\ge0
\end{align}$
which is true because $0<A,\ B<\frac{\pi}{2}\Leftrightarrow$
$\Leftrightarrow1<\cot\frac{A}{2},\ \cot\frac{A}{2}<+\infty$.
From $\left(\#\right)$ and $\left(\#\#\right)$ we have
$\begin{align}
&\sum_\text{cyc}\tan A+2\sum_\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\text{cyc}\cot\frac{A}{2}+2\sum_\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\Leftrightarrow\\
&\Leftrightarrow\left(\sum_\text{cyc}\sqrt{\tan A}\right)^{2}\ge\left(\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\right)^{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\sqrt{\tan A}\ge\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\ \cdot
\end{align}$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):
What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:
*
*$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$
*$$ x_{3/2} = x_{5}^2 $$
*the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$
*the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$
| Let $f(x)=x^3-x^2+2x-1, g(x)=x^3+x^2-1$. We have
$$-g(x)\cdot g(-x)=-(x^3+x^2-1)((-x)^3+(-x)^2-1)\\
=(x^3+x^2-1)(x^3-x^2+1) = x^6 - (x^2-1)^2\\
=x^6-x^4+2x^2-1=f(x^2)$$
Now if $g(x)=(x-r)(x-s)(x-t)$, then
$$f(x^2)= - (x-r)(x-s)(x-t) (-x-r)(-x-s)(-x-t)\\=(x^2-r^2)(x^2-s^2)(x^2-t^2)
\\\implies f(x)=(x-r^2)(x-s^2)(x-t^2)$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
x &= (-1)^{2/9} \\
x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
x &= (-1)^{4/9} \\
x &= -(-1)^{5/9} \\
x &= (-1)^{2/3} \\
x &= -(-1)^{7/9} \\
x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?
| Note that
$x^9=1=\cos 2n\pi +\sin 2n\pi=e^{i2n\pi}$
$\implies x=e^{i2n\pi/9}; n=0,1,2,...,8$ are roots.
Assume $e^{i2\pi/9}=a$. Then the powers of it coprime to $9$ viz. $a,a^2,a^4,a^5,a^7 $ and $a^8$ are generators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Words of length $10$ in alphabet $\{a,b,c\}$ such that the letter $a$ is always doubled
Compute the number of words of length $10$ in alphabet $\{a,b,c\}$ such that letter $a$ is always doubled (for example "$aabcbcbcaa$" is allowed but "$abcbcaabcc$" is forbidden).
I am looking for a quick/efficient way to resolve this problem. I thought of fixing "$aa$" in the beginning then draw a tree of the next possibilities but this tree will end up to be a whole forest. Can you help me ?
| A word of length $n$ is either a word of length $n - 1$ and a $b$ or a $c$, or a word of length $n - 2$ and $aa$. Call the number of words of length $n$ $x_n$, you see $x_0 = 1$, $x_1 = 2$, and:
$\begin{align*}
x_{n + 2}
&= 2 x_n + x_{n - 2}
\end{align*}$
Solving this using generating functions is routine. Define:
$\begin{align*}
g(z)
&= \sum_{n \ge 0} z_n z^n
\end{align*}$
Multiply your recurrence by $z^n$, sum over $n \ge 0$ and recognize the resulting sums:
$\begin{align*}
\sum_{n \ge 0} x_{n + 2} z^n
&= 2 \sum_{n \ge 0} x_{n + 1} z^n + \sum_{n \ge 0} x_n z^n \\
\frac{g(z) - x_0 - x_1 z}{z^2}
&= 2 \frac{g(z) - x_0}{z} + g(z)
\end{align*}$
Solve for $g(z)$ using the initial values, express as partial fractions:
$\begin{align*}
g(z)
&= \frac{1}{1 - 2 z - z^2} \\
&= \frac{1}{(1 - (1 - \sqrt{2}) z) (1 - (1 + \sqrt{z}) z)} \\
&= \frac{\sqrt{2} - 1}{2^{3/2} (1 + (\sqrt{2} - 1) z)}
+ \frac{\sqrt{2} + 1}{2^{3/2} (1 - (\sqrt{2} + 1) z)}
\end{align*}$
We need the coefficient of $z^n$, as this is just geometric series:
$\begin{align*}
[z^n] g(z)
&= (-1)^n \frac{\sqrt{2} - 1}{2^{3/2}} (\sqrt{2} - 1)^n
+ \frac{\sqrt{2} + 1}{2^{3/2}} (\sqrt{2} + 1)^n
\end{align*}$
Noting that $\rvert \sqrt{2} - 1 \lvert < 1$, and furthermore the respective coefficient is also less than $1$, you have finally that:
$\begin{align*}
x_n
&= \operatorname{round}
\left( \frac{\sqrt{2} + 1}{2^{3/2}} (\sqrt{2} + 1)^n \right)
\end{align*}$
(No, this isn't the best way to compute $x_n$ for large $n$, but it shows how the value grows).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why is the function differentiable in the point $(0,0)$? I am trying to figure out why my function is differentiable and therefore continuous in the point $(0,0)$ which is also a critical point and a saddle point.
Considering my function: $f(x, y) = x^3 - 3xy^2$
The partial derivative of $x$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2 $$
The partial derivative of $y$ at the position $x,y$ is:
$$ \frac{\partial f }{\partial y}(x,y) = -6xy $$
Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy \end{pmatrix} $$
How can I show that my function is point-wise differentiable in $(0,0)$?
| Besides from the solution proposed by @SonGohan, you can prove it by the definition.
The candidate to be the derivative is given by $L(x,y) = (3x^{2} - 3y^{2},-6xy)$.
Now you can prove that
\begin{align*}
\lim_{(x,y)\to(0,0)}\frac{|f(x,y) - f(0,0) - L(x,y)((x,y) - (0,0))^{T}|}{\|(x,y)\|} = 0
\end{align*}
Indeed, this is the case.
To begin with, notice the numerator can be written as
\begin{align*}
x^{3} - 3xy^{2} - (3x^{3} - 3xy^{2} - 6xy^{2}) = -2x^{3} - 6xy^{2}
\end{align*}
Thence we conclude that the proposed limit equals zero indeed.
Such claim results from the convenient application of the squeeze theorem.
To comprehend this, let us notice that
\begin{align*}
0 < x^{2} \leq x^{2} +y^{2} & \Rightarrow 0 < |x| \leq \sqrt{x^{2} + y^{2}}\\\\
& \Rightarrow 0 < \frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|x|^{3}}{\sqrt{x^{2}+y^{2}}} \leq |x|^{2}\\\\
& \Rightarrow 0 < \left|\frac{x^{3}}{\sqrt{x^{2}+y^{2}}}\right| \leq |x^{2}|
\end{align*}
Similarly, we have that
\begin{align*}
0 < y^{2} \leq x^{2} + y^{2} & \Rightarrow 0 < |y| \leq \sqrt{x^{2}+y^{2}}\\\\
& \Rightarrow 0 < \frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
& \Rightarrow 0 < \frac{|xy^{2}|}{\sqrt{x^{2}+y^{2}}} \leq |xy|\\\\
& \Rightarrow 0 < \left|\frac{xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |xy|
\end{align*}
Finally, since we also have
\begin{align*}
0 < \left|\frac{2x^{3} + 6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq \left|\frac{2x^{3}}{\sqrt{x^{2}+y^{2}}}\right| + \left|\frac{6xy^{2}}{\sqrt{x^{2}+y^{2}}}\right| \leq |2x^{2}| + |6xy|
\end{align*}
we are done.
Hopefully this helps!
| {
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"url": "https://math.stackexchange.com/questions/4037361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$
If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$
I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.
I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?
Please help me with this problem. Any more innovative methods would be appreciated.
Answer: $4$.
Edit $[7^{th}$ March, $2021]$:
I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?
| Using the denesting formula
$$\sqrt{a\pm\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have
$$f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}} \\= 2\sqrt{\frac{x+\sqrt{x^2-16(x-4)}}{2}}=2\sqrt{\frac{x+\sqrt{(8-x)^2}}{2}}$$
Now $\sqrt7<8$, therefore
$$f(\sqrt 7)=2\sqrt{\frac{\sqrt 7+8- \sqrt 7}{2}}=2\cdot 2=4.$$
However as one of the members pointed out, the square root of a nonreal complex number is not well defined. We may get around this by requiring that the "square root" have nonnegative real part, but this is not standard practice.
Update: You can just square both sides directly:
$$(f(x))^2=2x+2\sqrt{x^2-(4-x)^2} = 2x+2\sqrt{(8-x)^2}$$
So if $x\leq 8$ then $f(x)=\sqrt{2x+2(8-x)}=4.$
| {
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Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:
Option 1
The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.
That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.
Option 2
This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.
The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.
The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.
| Note that (just observe to know it, no need to expand it!)
$\mathrm{LHS}(zx + x + 1)(xy + y + 1)(yz + z + 1) - (xyz - 1)^2$ is a polynomial in $z$ of degree at most $2$.
Thus, if we can find three distinct constants $a, b, c$ such that $\mathrm{LHS} = \mathrm{RHS}$ for $z = a, b, c$ respectively,
then we have $\mathrm{LHS} = \mathrm{RHS}$ for all $x, y, z$.
We choose $z = 1, -1, 0$.
For $z = 1$, $\mathrm{LHS} = 1 - \frac{x}{2x + 1} - \frac{y}{xy + y + 1} - \frac{1}{y + 2}$
and $\mathrm{RHS} = \frac{(xy - 1)^2}{(2x + 1)(xy + y + 1)(y + 2)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, 1, -1$.
For $z = -1$, $\mathrm{LHS} = 1 - x - \frac{y}{xy + y + 1} - \frac{1}{y}$
and $\mathrm{RHS} = \frac{(xy + 1)^2}{(xy + y + 1)(-y)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 1, -1, 2$.
For $z = 0$, $\mathrm{LHS} = 1 - \frac{x}{x+1} - \frac{y}{xy + y + 1}$ and
$\mathrm{RHS} = \frac{1}{(x+1)(xy + y + 1)}$. Verify it directly (or only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, -1$).
We are done.
| {
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!
| Just to give a different approach, calculating mod $2000$ and using the fact that
$$3^{2000}=(1-10)^{1000}=1-1000\cdot10+\cdots\equiv1\mod2000$$
we have
$$\begin{align}
7^{2005}-1&=-1-(3-10)^{2005}\\
&\equiv-1-3^{2005}+2005\cdot3^{2004}\cdot10-{2005\choose2}3^{2003}\cdot10^2+{2005\choose3}3^{2002}\cdot10^3\\
&\equiv-1-3^5+5\cdot3^4\cdot10-{5\choose2}3^3\cdot100+{5\choose3}3^2\cdot1000\\
&\equiv-1-243+4050-27000+90000\\
&\equiv-244+50-1000\\
&=-1194\mod2000
\end{align}$$
and thus
$${7^{2005}-1\over6}\equiv{-1194\over6}=-199\equiv801\mod1000$$
| {
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Use the $\epsilon$-$\delta$ definition of a limit to prove $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$ $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$
You should start by writing $\frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6} + 1$ in the way $\left ( x-3 \right )g\left ( x \right )$
(a) Determine $g(x)$
(b) Could we choose $\delta = \min\lbrace 1, \frac{\varepsilon }{n} \rbrace$ for some $n$? Why?
(c) If we choose $\delta = \min\lbrace \frac{1}{4}, \frac{\varepsilon }{m} \rbrace$ , what is the smallest integer $m$ that we could use?
I managed to answer the part (a): By making the sum of the fraction with +1, and factoring the numerator, I get the following:
$\frac{\left ( x-3 \right )\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$
where
$g\left ( x \right ) = \frac{\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$
But my problem is when developing the part (b).
In the part (b) I am using the formal limit definition, but I am not managing to reach a result that allows me to find the answer.
I need help to solve this exercise, please
| This is a bit complicated question. You need to find a suitable bound for the rational function $g(x) $ for values of $x$ in a certain neighborhood of $3$. In formal terms we need two positive numbers $h, K$ such that $|g(x) |<K$ whenever $|x-3|<h$.
This clearly requires an analysis of numerator and denominator of $g$. Note that the function can not be bounded if the denominator of $g$ vanishes. Since it does not vanish at $3$ we have a guarantee by continuity of polynomials that the denominator will not vanish in some neighborhood of $3$.
Finding such a neighborhood (ie finding $h$ described above) is possible using the limit of denominator $q(x) =x^4-4x^3+x^2+x+6$ as $x\to 3$. But before we do that let's observe that $q(3)<0,q(4)>0$ and hence $q$ vanishes somewhere between $3,4$ and thus $h$ must be less than $1$.
Thus for part b) the answer is no. You can't expect a $\delta$ like $\min(1,\dots)$ to work. You should observe that $h$ described above forms the first argument of $\min$ in expression for $\delta$.
Part c) is about showing that we can take $h=1/4$. Let us now prove that $q(x)\to - 9$ as $x\to 3$. Clearly $$|q(x) +9|=|x-3||x^3-x^2-2x+5|$$ and $$|x^3-x^2-2x+5|=|x(x-2)(x+1)+5|$$ If $|x-3|<1/2$ then $5/2<x<7/2$ and hence $$|x(x-2)(x+1)+5|<(7/2)(3/2)(9/2)+5<32$$ and thus for any $\epsilon>0$ we have $$|q(x) +9|<\epsilon$$ if $|x-3|<\min(1/2,\epsilon /32)$. Using $\epsilon =8$ we can see that $$|q(x) +9|<8$$ if $|x-3|<1/4$. Thus $|q(x) |>1$ if $|x-3|<1/4$.
Next we deal with numerator $p(x) =x^3-x^2-2x-4$ of $g(x) $. Clearly if $|x-3|<1/4$ then $2.5<x<3.5$ and $$|p(x) |=|x(x-2)(x+1)-4|<28$$ Thus it follows that if $|x-3|<1/4$ then $$|g(x) |=|p(x) |/|q(x) |<28$$
It should now be clear that $$|(x-3)g(x)|<28|x-3|$$ and this is less than $\epsilon$ if $|x-3|<\min(1/4,\epsilon/28)$. This proves the desired limit in question.
It does not however make sense to find the smallest positive integer which can replace $28$ above. Finding such numbers becomes an algebraic problem related to solution of inequalities and is really not in the spirit of epsilon-delta definition of limit.
| {
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Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below
However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
| The problem is not too hard if you consider
$$f(x)=\tan (A(x))\,\tan (B(x))$$ Using logarithmic differentiation
$$\frac{f'(x)}{f(x)}=A'(x) \csc (A(x)) \sec (A(x))+B'(x) \csc (B(x)) \sec (B(x))$$
This gives $f'(0)=0$.
Repeating the process knowing that $f(0)=\frac 13$ and $f'(0)=0$ (this simplifies a lot the calculations), we have
$$f''(0)=\frac{16}{81} \left(\sqrt{3}-\pi \right) \pi <0$$ So, by the second derivative test $x=0$ corresponds to the maximum of the function.
We could also have composed Taylor series around $x=0$ to obtain
$$f(x)=\frac{1}{3}+\frac{8}{81} \left(\sqrt{3}-\pi \right) \pi x^2+\frac{8 \pi
\left(\pi \left(27+2 \sqrt{3} \pi -3 \pi ^2\right)-9 \sqrt{3}\right)
}{2187}x^4+O\left(x^6\right)$$ which a perfect approximation of the function for $-\frac 12 \leq x \leq \frac 12$ (maximum error $< 0.00012$ at the bounds).
| {
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Can someone help me finish the solution to this differential equation? Find the general solution of $(x-4y-3)dx-(x-6y-5)dy=0.$
The rest of my solution:
$x-4y=3$
$x-6y=5$
The point of intersection is $(-1, -1)$.
Let
$x=u-1$; $dx=du$
$y=v-1$; $dy=dv$
$(u-4v)du-(u-6v)dv=0$
Let $u=vz; du=vdz+zdv$
$(vz-4v)(vdz+zdv)-(vz-6v)dv=0$
$(z-4)(vdz+zdv)-(z-6)dv=0$
$(z^2-5z+6)dv-v(z-4)dz=0$
$\frac{dv}{v}-\frac{z-4}{(z^2-5z+6)}dz=0$
$\int\frac{dv}{v}-\int \frac{z-4}{(z^2-5z+6)}dz=0$
$\int\frac{dv}{v}-2\int \frac{dz}{(z-2)}+\int\frac{dz}{z-3}=0$
$\ln \lvert v\rvert -2\ln \lvert z-2\rvert +\ln \lvert z-3 \rvert + \ln \lvert c \rvert=0$
$\ln\lvert cv(z-3) \rvert =2\ln \lvert z-2 \rvert$
$\ln \lvert (z-2)^2 \rvert = \ln\lvert cv(z-3) \rvert $
I do not know how the final answer became $(x-2y-1)^2=c(x-3y-2).$
| You find :
$$\ln((z-2)^2)=\ln(cv(z-3))$$
$$(z-2)^2=cv(z-3)$$
$z=\frac{u}{v}$
$$(\frac{u}{v}-2)^2=cv(\frac{u}{v}-3)$$
$$(u-2v)^2=cv^2(u-3v)$$
$u=x+1$ and $v=y+1$
$$((x+1)-2(y+1))^2=c(y+1)^2((x+1)-3(y+1))$$
$$(x-2y-1)^2=c(y+1)^2(x-3y-2)$$
| {
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If $f'(x)=\sqrt{x+1}$ and $h(x)=f(x^2+x)$, what is $h'(x)$? I was thinking it could be plugging in $x^2+x$ to the $f'(x)$, then using the Chain Rule to solve it, but I'm not sure if it is right. Please help!
| If $f'(x) = \sqrt{x+1}$, then $f(x) = \int \sqrt{x+1} dx = \frac{2}{3}(x+1)^{\frac{3}{2}} + c$. You can solve that integral using u-substitution. Then we also know that $h(x) = f(x^2 + x)$. So then using function notation, we know that $h(x) = f(x^2 + x) = \frac{2}{3}(x^2 + x+1)^{\frac{3}{2}} + c$. We can now differentiate this using chain rule as follows:
$h(x) = \frac{2}{3}(x^2 + x+1)^{\frac{3}{2}} + c$
$h'(x) = \frac{2}{3} \cdot \frac{3}{2}(x^2 + x+1)^{\frac{1}{2}} \cdot (2x+1) = (x^2 + x+1)^{\frac{1}{2}} \cdot (2x+1) $
Notice that when we do so, the derivative of $c$ will just be $0$, so it doesn't matter that we don't know the specific value of $c$.
| {
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What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$? What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)
I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.
Is there any way by which I can determine where these functions intersect?
| Here is another answer I came up with:
Notice that $\lfloor x\rfloor$ and $3$ are integers, so $x^3$ is an integer, say $x=\sqrt[3]{y^3+z}$, where $0\leq z<(y+1)^3-y^3$, and y and z are integers. Now we have $y^3+z-y=3$, or $(y-1)y(y+1)=3-z.$ Since $z\geq0$, $(y-1)y(y+1)\leq 3$ and $y\leq1$. However, notice that if $y<-1$, the inequality $z<(y+1)^3-y^3$ fails, so $y\geq-1$, $(y-1)y(y+1)=0$, $z=3$, and $y=1$ to make the inequality true, so $x=\sqrt[3]4$.
| {
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Finding $\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$
Find $$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r^3}\right)$$
I solved the simpler versions of these:
$$\sum_{r=2}^{\infty}\ln\left(1-\frac{1}{r}\right)\;\;,\;\;\sum_{r=2}^{\infty}\ln\left(1-\frac{1}
{r^2}\right)$$
by factorizing which yields a telescoping sum. However the cube version certainly breaks this trend.
What else can be done here? Does a closed form exist?
| Hint:
$$1-\dfrac1{r^3}=\dfrac{(r-1)(r^2+r+1)}{r^3}=\dfrac{f(r-1)g(r-1)}{f^3(r)}$$
where $g(r-1)=r^2+r+1\implies g(r)=(r+1)^2+(r+1)+1=r^2+3r+3$
and $f(r)=r$
Similarly, $$1-\dfrac1{(r+1)^3}=\dfrac{r(r^2+3r+3)}{(r+1)^3}=\dfrac{f(r)g(r)}{f^3(r+1)}$$
$$\implies\prod_{r=2}^n\left(1-\dfrac1{r^3}\right)=\prod_{r=2}^ng(r)\cdot\dfrac{f(1)}{f^2(2)f^2(3)\cdots f^2(n)f^3(n+1)}=\dfrac{f(1)}{f(n+1)}\prod_{r=2}^n\dfrac{g(r)}{f^2(r)}$$
$\prod_{r=2}^nf(r)=n!$
| {
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Prove this solution is genuine I cannot prove that the solution to :$$y'=-\frac{x\sqrt{1-y^2}}{y\sqrt{1-x^2}}$$
is:
$$y=\sqrt{x^2-2C\sqrt{-x^2+1}-C^2},\:y=-\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}$$
By substituting the solution(s) back into the original equation. I have taken derivative and attempted to equate LHS with RHS but I just can't simplify enough to obviously justify the solution(s) $$\frac{x\left(C+\sqrt{-x^2+1}\right)}{\sqrt{-x^2+1}\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}}$$
| Square the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$
Taking the derivative:
$$2yy'= 2x+2C\frac{x}{\sqrt{1-x^2}}$$
So
$$yy' = x + C\frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
Assuming $x$ is not $0$, then
$$\sqrt{1-x^2} + C=\sqrt{1-y^2}$$
Squaring both sides:
$$1-x^2+2C\sqrt{1-x^2}+C^2=1-y^2$$
Then,
$$y^2=x^2-2C\sqrt{1-x^2}-C^2$$
QED
================
The solution to this ODE is by first noticing it is separable. Then you arrive at the equation:
$$-\sqrt{1-y^2}=\sqrt{1-x^2}+C$$
This gives you the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$
| {
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Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$ It's an inequality based on two found on the website MSE (see the reference):
Let $a,b,c>0$ then we have:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Lemma 1 :
$a,b,c>0$ then we have :
$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Proof of lemma 1 :
First we remark that the inequality is homogenous and we can try the substitution $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ and apply the uvw's method .
We have :
$$a^4+b^4+c^4=(9u^2-6v^2)^2-2(9v^4-6uw^3)$$
$$a^2b^2+b^2c^2+c^2a^2=9v^4-6uw^3$$
$$a^2+b^2+c^2=9u^2-6v^2$$
And :
$\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+(2.25u+0.75w)(9u^2-9v^2)(3u)}{(2.25u+0.75w)(3u3v^2-w^3)}$
it's enough to find an extreme value of our expression for the extreme value of $w^3$ wich happens for an equality case of two variables .
Since the last inequality is homogeneous, we can assume that $b=c=1$.
$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{((a+2)\frac{3}{4}+\frac{3}{4}(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$
Now it seems to be clear : we get a polynomial with a root equal to one . See the factorization by Wolfram alpha .
End of the proof of the lemma 1
Remains to show that $ a\geq b \geq c>0$:
$$P(a,b,c)\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Wich is not hard I think .
Question :
How to show it ?
Reference :
M. A. Rozenberg, “uvw–Method in Proving Inequalities”, Math. Ed., 2011, no. 3-4(59-60), 6–14
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
Stronger than Nesbitt inequality
| As username says it's not symmetric so here there is a refinement and a symmetric formula :
$a,b,c\in[0.95,1]$:
$$\sum_{cyc}\frac{a}{b+c}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{(0.75(abc)^{\frac{1}{3}}+0.75(a+b+c))(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}+\frac{9}{4}\left(abc\frac{\frac{a^2+b^2+c^2}{ab+bc+ca}-1}{a^3+b^3+c^3}\right)^2}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$ I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$?
I tried to solve it as follows:
I state $a\in\mathbb{N}$, so that $\sqrt{x^2+2x+4}<a<\sqrt{4x^2+2x+1}$
Hence $x^2+2x+4<a^2<4x^2+2x+1$
$4x^2+2x+1-x^2-2x-4=3x^2-3=3*2^{4004}-3$.
Here is where I got stuck. Could you please explain to me how to solve this question?
| With the lower limit,
$$\begin{equation}\begin{aligned}
& (x + 1)^2 \lt x^2 + 2x + 4 \lt (x + 2)^2 \\
& x + 1 \lt \sqrt{x^2 + 2x + 4} \lt x + 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This means, with $a \in \mathbb{N}$ as you stated,
$$a \ge x + 2 \tag{2}\label{eq2A}$$
With the upper limit,
$$\begin{equation}\begin{aligned}
& (2x)^2 \lt 4x^2 + 2x + 1 \lt (2x + 1)^2 \\
& 2x \lt \sqrt{4x^2 + 2x + 1} \lt 2x + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This thus means
$$a \le 2x \tag{4}\label{eq4A}$$
Combining \eqref{eq2A} and \eqref{eq4A} gives
$$x + 2 \le a \le 2x \tag{5}\label{eq5A}$$
Thus, as zwim's hints state, the number of integers in between the $2$ square root limits would be
$$2x - (x + 2) + 1 = x - 1 = 2^{2002} - 1 \tag{6}\label{eq6A}$$
| {
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$b=a^p+1$ is a perfect square. Show that $p|(b-9)$ $p$ is a prime, and $a$ is a positive integer, $b=a^p+1$ is a perfect square. Show that $p|(b-9)$
It seem very interesting problem.if let $a^p+1=x^2$,it is clear $p\neq 2$
I have prove : $x$ is odd
proof:if $x$ is even number,then $(x+1,x-1)=1$,and note $a^p=(x+1)(x-1)$,then exist $r>s\ge 1\in N^{+}$ such $x+1=r^p,x-1=s^p$
so
$$2=(x+1)-(x-1)=r^p-s^p=(r-s)(r^{p-1}+r^{p-2}s+\cdots+s^{p-1})
\ge p\ge 3$$which is contradiction
which is a pretty interesting and nice result. I wonder in which ways we may approach it.
| I am in 8th grade, so please if I make mistakes in the calculation then please tell me, I would delete the answer.
You proved that $x$ is odd. That would mean that $x^2$ is odd too.
So the equation $x^2=a^p+1$
If we subtract $1$ from both sides then $a^p$ would be even. Which implies that $a$ is even.
Since $x$ is odd, $x=2k+1$ and since $a$ is even, $a=2n$
Putting these in the equation, we get:
$(2k+1)^2=(2n)^p+1$
$\implies 4k^2+4k+1=2^p n^p+1$
$\implies k(k+1)=2^{p-2}n^p$
Since one of $k$ and $k+1$ has to be even
So assume $k$ is even, then $k+1$ is odd.
Now consider the following:
If $n^p$ is even
Then $n$ is even
$2l(2l+1)=2^{p-2}n^p$
$\implies l=2m$
$\implies k=4m$
$\implies k\geq 4$
$\implies x\geq 9$
$\implies b-9 \geq 72$-----------$(1)$
AND, If $n^p$ is odd
Then $n$ is odd, so let $n=2\alpha+1$
By the comment of John L and Kieren MacNillan, it seems that $p$ cannot be greater than $3$ so consider the case $p=3$:
$2l(2l+1)=2(2\alpha+1)^3$
$\implies 2l^2+l=8\alpha^3+6\alpha^2+6\alpha+1$
$\implies l-1=2\beta$
$\implies l\geq 3$
$\implies k\geq 6$
$\implies x\geq 13$
$\implies b-9\geq 160$---------$(2)$
Assuming that the statement in the question is true, 3 cannot divide 160. Which means that $n$ can't be even.
So $n$ is odd.
So from $(1)$, we get $b\geq 81$
$\implies (2n)^3+1\geq 81$
$\implies n\geq\sqrt[3]{10}$
Since $n$ is a positive integer,
$n\geq\lceil \sqrt[3]{10} \rceil=3$
$\implies a\geq 6$
If the statement is not correct then $(2)$ can also be the case, then from $(2)$ we get $b\geq 169$
$\implies (2n)^3+1\geq 169$
$\implies n\geq\sqrt[3]{21}$
Since $n$ is a positive integer,
$n\geq\lceil\sqrt[3]{21}\rceil=3$
$\implies a\geq 6$
So in both cases $a\geq 6$
These followings I derived in this answer might help you prove the statement:
$●b-9\geq 72$ if the statement is correct. Otherwise $b-9\geq 72$ and $b-9\geq 160$ both could be the case.
$●a=2n\geq 6$ where $n$ is an odd number.
$●$The only $p$ to consider is $p=3$ by the comment of John L and Kieren MacMillan.
| {
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$63^{63^{63}} \mod 100$ I need to find $63^{63^{63}} \bmod 100$.
This is what I've got so far:
Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$
Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$
Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$
I'm stuck here and don't know what to do next, what could I do now?
| $63^7 = (60 +3)^7= ..... + {7\choose 2}60^2\cdot 3^5 + 7\cdot 60\cdot 3^6 + 3^7 \equiv 20\cdot 3^6 +3^7\pmod{100}$.
And as $2\cdot 3^6 \equiv 2\cdot 3^2 \equiv 8 \pmod {10}$ and as $3^7 = 9^3\cdot 3 = (10-1)^3 \cdot 3 = 3(100^3 - 3\cdot 100 + 3\cdot 10 - 1)\equiv 90-3 \equiv 87 \pmod {100}$
So have $63^7\equiv 20\cdot 3^6 + 3^7 \equiv 80 + 87 \equiv 67 \pmod{100}$.
.....
Also nothing wrong with successive squaring. $7$ is a small number.
$63^2 = 3969 \equiv 69$.
So $63^3 \equiv 69\cdot 63 = 4347 \equiv 40 + 80 + 27\equiv 47\pmod {100}$.
So $63^6 \equiv 47^2 \equiv 2209\equiv 9\pmod{100}$ and so
$63^7 \equiv 9\cdot 63 \equiv 567 \equiv 67 \pmod {100}$
| {
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Prove that there is no positive rational number a such that $a^2 = 3$. I solved a similar question $a^4 = 2$ but I didn't find it hard. This one is taking a lot of my time and I still don't understand how to solve this.
Here's what I have tried so far.
edit:
\begin{align}
a^2 &= 3 \\
a &= \left(\frac mn\right) \\
a^2 &= \left(\frac mn\right)^2 = 3 \\
m^2 &= 3n^2 \\
m &= 3p \\
m^2 &= (3p)^2 \\
9p^2 &= 3n^2
\end{align}
In another similar question $a^3 = 2$, I had to divide a common number from each side to finally solve the problem. But with this problem, I don't know how to continue.
| Alright, make the straightforward assumption that $a= \frac mn$ with $m,n$ in lowest terms and they are integers. Thus, we get $(\frac mn)^2 = 3$, which soon becomes $m^2 = 3n^2$. Thus, $m^2$ is a multiple of $3$.
However, for $m^2$ to be a multiple of $3$, it needs to have an even number of factors of $3$. If it had an odd number of factors of $3$, you wouldn't get an integer for $m$ when you take the square root of $m^2$ because squares come in factors of even powers like $3^2$ or $2^2$. Thus, you must have an even number of $3$s in $m^2$, for it to be divisible by $3$. Now, when we take the square root of $m^2$ to get $m$, half of the $3$s disappear, but not all of them, so $m$ is divisible by $3$. Since $m$ is divisible by $3$, it can be expressed as $m=3p$. Thus, we get $(3p)^2 = 3n^2 \implies 3p^2 = n^2 $. Using the same logic as before, we can show that $n$ is divisible by $3$. This means that both $m,n$ are divisible by $3$, contradicting our initial assumption that $\frac mn$ is in lowest terms and shouldn't have any common factors besides $1$. Thus, there is no such $a = \frac mn$ with $m,n$ in lowest terms and integers such that $a^2=3$.
Here is a bonus easy proof. Rearrange $a^2=3$ to get $a^2-3=0$. By the rational root theorem, all the POSSIBLE rational roots of this polynomial are $a=-1,1,-3,3$. Let's see if any of these $a$ make $a^2=3$. Testing all these $a$, we see that none of these work, thus there is no RATIONAL $a$ such that $a^2=3$.
| {
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Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$
Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$
For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$
Am I correct, especially for the right direction ?
| The left direction is correct. An alternative way to prove this is using the Cauchy condensation test.
The right direction is not correct. You say: for $0 < p \le 1$ the series $\sum_n 1/n^p$ is divergent but you did not prove this!
Alternatively, you can also use the integral test or the Cauchy condensation test for this direction.
| {
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Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$
| As @Aderinsola Joshua commented, use successive eliminations
From $(1)$, you have $y=-\frac{a x+c z}{b}$.
Plug in $(2)$ and then $z=\frac{c x \left(a^2-b^2\right)}{a \left(b^2-c^2\right)}$
Plug in $(3)$ and simplify to get
$$\frac{b c (a-b) (a+b) (a-c) (a+c) \left(a^2
\left(b^2-c^2\right)^2-x^2\right)x}{a^2 \left(b^2-c^2\right)^2}=0$$
So, assuming $a\neq b \neq c$, you are left with
$$\left(a^2
\left(b^2-c^2\right)^2-x^2\right)x=0$$
| {
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Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step solution, because I am now somewhat lost. Using the substitution $t=\tan(x)$, I got to
$$\int \left(\frac{t^2}{2t^4+2t^2+1}+\frac{1}{2t^4+2t^2+1}\right)\mathrm dt$$
By expanding with 1:
$$\int \frac{1}{1+\sin^4x}\cdot \frac{\frac{1}{\cos^4x}}{\frac{1}{\cos^4x}}\mathrm dx$$
$$\int \:\frac{1}{\frac{1}{\cos^4x}\cdot \frac{\sin^4x}{\cos^4x}}\cdot \frac{1}{\cos^4x} \mathrm dx$$
$$\int \:\frac{1}{\left(\frac{1}{\cos^2x}\right)^2\cdot \tan^4x}\cdot \frac{1}{\cos^2x}\cdot \frac{1}{\cos^2x}\mathrm dx$$
And using the substitution: $t=\tan\left(x\right)$
$$\mathrm dt=\frac{1}{\cos^2x}\mathrm dx$$
$$t^2=\tan^2\left(x\right)$$
$$t^2=\frac{\sin^2x}{\cos^2x}$$
$$t^2=\frac{1-\cos^2x}{\cos^2x}$$
$$t^2=\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1$$
$$t^2+1=\frac{1}{\cos^2x}$$
Using it:
$$\int \:\frac{t^2+1}{2t^4+2t^2+1}\mathrm dt$$
I don't think I got to the expected result but I can't seem to be able to find why…
| For $n=4$, you can make it shorter since
$$A=\frac 1{1+\sin^4(x)}=\frac 1{(\sin^2(x)+i)(\sin^2(x)-i)}$$
Using partial fractions and double angle formula
$$A=\frac{i}{\cos (2 x)-(1-2 i)}-\frac{i}{\cos (2 x)-(1+2 i)}$$ Using now the tangent half-angle
$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan
^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$
| {
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Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When is that last statement true?
| Using induction,
If $10^r=1+3^sk$ where $3\nmid k$
$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$
for $s+1\le2s+1\iff s\ge0,3s\ge s+1\iff 2s\ge1$
Now for the base case $r=1=3^0, s=2$
So, by using weak induction, $$10^{3^a}-1$$ is divisible by $3^{a+2}$ for $a\ge0$
| {
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Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?
|
Let set: $\begin{cases}f(x,y)=x^2+y^2\\g(x,y)=6x^2+2xy+6y^2\end{cases}$
The curve $(\mathcal E): g(x,y)=9$ is an ellipse centered at the origin with major axis supported by $y=-x$ and minor axis supported by $y=x$.
Indeed if you calculate its equation in the $45^\circ$ rotated basis then you get it in its reduced form:
$$g(\tfrac{X+Y}{\sqrt{2}},\tfrac{X-Y}{\sqrt{2}})=7X^2+5Y^2=9$$
On the other hand, the curve $(\mathcal C): f(x,y)=k$ is just a circle centered at origin.
The linked extrema occurs when this circle is tangent to the ellipse, therefore the minimum occurs on the minor axis (i.e. when $x=y$) and the maximum occurs on the major axis (i.e. when $x=-y$).
$\begin{cases}g(x,x)=14x^2=9\\g(x,-x)=10x^2=9\end{cases}\implies \begin{cases}f(x,x)=2x^2=\frac 97&\text{min}\\f(x,-x)=2x^2=\frac 95&\text{max}\end{cases}$
Note: you can also extrapolate these values directly from the reduced equation of the ellipse.
| {
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Find a closed formula for $\sum_{n\geq 0} f(n)x^n.$ Given that $f(n) =f(n-1) + 2f(n-2)$ for $n \geq 2$, with $f(1) = 1$ and $f(2) = 3$ find a closed formula for $$\sum_{n\geq 0} f(n)x^n.$$
Here I try to do the following $ \sum_{n=0}^{\infty}f(n)x^{n} = \sum_{n=1}^{\infty}f(n-1)x^{n} + 2\sum_{n=2}^{\infty}f(n-2)x^{n}$ but when I simplify and re-index I get $$1=x+2x^2,$$ which is not correct. Can someone help me?
| Let $F(x)$ denote the generating function for $f(n)$, i.e.
$$F(x)=\sum_{n\ge1}f(n)x^n$$
Multiply both sides of the recurrence by $x^n$ and sum both sides starting from $n=3$:
$$\sum_{n\ge3}f(n)x^n=\sum_{n\ge3}f(n-1)x^n+2\sum_{n\ge3}f(n-2)x^n$$
Now manipulate this as need to get everything in terms of $F(x)$:
$$\begin{align}
\sum_{n\ge3}f(n)x^n&=\sum_{n\ge3}f(n-1)x^n+2\sum_{n\ge3}f(n-2)x^n\\[1ex]
\sum_{n\ge1}f(n)x^n-f(1)x-f(2)x^2&=\left(x\sum_{n\ge2}f(n-1)x^{n-1}-f(1)x^2\right)+2x^2\sum_{n\ge3}f(n-2)x^{n-2}\\[1ex]
F(x)-f(1)x-f(2)x^2&=x\sum_{n\ge1}f(n)x^n-f(1)x^2+2x^2\sum_{n\ge1}f(n)x^n\\[1ex]
F(x)-x-3x^2&=xF(x)-x^2+2x^2F(x)
\end{align}$$
which is easily solved for $F(x)$.
| {
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Standard normal distribution - Calculating the probability of an electron Figure $1$ shows $Bohr$'s model for the hydrogen atom. We let $(X, Y)$ denote the position of the electron in relation to the proton-core. In quantum mechanics one can never determine $(X, Y)$ exactly, but we are simplifying the model by using probability distribution. We assume $X∼N(0,1)$ and $Y∼N(0,1)$, i.e. standard normal distribution with expectation $E(X)=0$ and variance $V(X)=1$. $X$ and $Y$ are independent.
Figure 1: Bohrs model with an electron in orbit around the proton-core. The black circle has radius r = 2,15.
(A)
What is the probability that $-2,15 <= x <= 2,15$? What is the probability that the electron is inside the black square? Hint: We assume $X$and $Y$ are independent random variables.
My answer:
To determine $P(a <= x <= b)$ when $X$ is a norml random variable with parameters $\mu$ and $\sigma$, we must evaluate
\begin{align}\\
& \int_{a}^b\frac{1}{\sqrt{2\pi}\sigma}e^{\left( \frac{-(x-\mu)^2}{2\sigma ^2} \right)} dx\\
\end{align}
So we have:
\begin{align}\\
& \Rightarrow P(-2,15\le X\le 2,15)=\int_{-2,15}^{2,15}\frac{1}{\sqrt{2\pi }} * {e^{\frac{-x^2}{2}}} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^{\left( \frac{x}{2}\right) ^2} dx \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{1}{\sqrt{2\pi }} * \int_{-2,15}^{2,15} e^u du \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left[ e^u + C \right]_{-2,15}^{2,15} \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( e^{\left( \frac{2,15}{2}\right) ^2} - e^{\left( \frac{-2,15}{2}\right) ^2} \right) \\
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\ = \frac{2}{\sqrt{2\pi }} * \left( 0 \right) \\
\end{align}
I feel like I have done something wrong with the calculations but I donw know what. Hpw do I calculate the probability for if the electron is within the black square?
| You cannot integrate the normal density by substitution over a finite interval. You need tables or a computer.
Your answer should have been $\Phi(2.15)-\Phi(-2.15) \approx 0.968$
Among other errors, you have the false steps $$\int_{-2,15}^{2,15} e^{\frac{-x^2}{2}} dx \to \int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx$$ $$\int_{-2,15}^{2,15} e^{(\frac{x}{2})^2} dx \to \int_{-2,15}^{2,15} e^{u} du$$ neither of which can be justified
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question related to internal angle bisector Internal angle bisector of $\angle A$ of triangle $\Delta ABC$, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of ∆ABC then.
(a) $AD = \frac{{2bc}}{{b + c}}\cos \frac{A}{2}$
(b) $PQ = \frac{{4bc}}{{b + c}}\sin \frac{A}{2}$
(c) The triangle ∆APQ is isosceles
(d) AP is HM of b and c
My approach is as follow
This is the rough image that I have drawn
$\frac{{\cos \frac{A}{2}}}{{AD}} = \frac{{\sin {{90}^o}}}{{{b_1}}}$
$\frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{PD}}$
$PD = AD\tan \frac{A}{2}$
$\frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{PD}} \Rightarrow \frac{{\sin X}}{{{b_2}}} = \frac{{\sin C}}{{AD\tan \frac{A}{2}}}$
The official answer is a,b,c and d.
I am not able to approach from here
| You know that $\displaystyle \small \frac{BD}{DC} = \frac{c}{b}$. Add $1$ to both sides and you get,
$\displaystyle \small \frac{a}{DC} = \frac{b+c}{b}$
Now we know that $\displaystyle \small \frac{a}{\sin A} = \frac{c}{\sin C} \implies \sin C = \frac{c}{a} \sin A$
In $\triangle ADC, \displaystyle \small \frac{\sin C}{AD} = \frac{\sin (A/2)} {DC}$
$AD = \displaystyle \small \frac{2 \cdot DC \cdot c}{a} \cos {\frac{A}{2}} = \frac{2 bc}{b+c} \cos \frac{A}{2}$
Next, in $\triangle APQ$, $AD$ is angle bisector of $\angle A$ and is also perpendicular to base $PQ$, therefore $\triangle APQ$ must be isosceles and $QD = DP = \frac{PQ}{2}$.
Can you take the rest of it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction with divisibility QUESTION: Prove that $16 \mid 19^{4n+1}+17^{3n+1}-4$ for all $n \in \mathbb{N}$
This is what I have so far but I'm not sure where to go from here.
PROOF:
Let $16 \mid 19^{4n+1}+17^{3n+1}-4$ equal $S(n)$.
Base case, let $n=0$.
\begin{align*}
16 &\mid 191+171-4\\
= 16 & \mid 32\\
= 2
\end{align*}
Therefore, $S(0)$ is true.
Using induction hypothesis, suppose $19^{4k+1}+17^{3k+1}-4$ is divisible by $16$ for all $n \in \mathbb{N}$.
Claim,
$16 \mid 19^{4k+1}+17^{3k+1}-4$,
that is $19^{4k+1}+17^{3k+1}-4=16m$, whereby $m$ is a multiple of $16$.
The above equation simplifies into,
\begin{align*}
16 & \mid 19^{4n+5}+17^{3n+4}-4\\
16 & \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4
\end{align*}
photo of my working out
| To show $16 \mid 19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4$ given $16 \mid 19^{4k+1}+17^{3k+1}-4$, note that
$19^4 \cdot 19^{4k+1} + 17^3 \cdot 17^{3n+1}-4=$
$17^3\color{blue}{(19^{4k+1}+17^{3n+1}-4)} + \color{blue}{(19^4-17^3)}\cdot19^{4n+1}+
\color{blue}{(17^3\cdot4-4)},$
and all of the blue terms are divisible by $16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate $\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$? $$\sum_{n=1}^{\infty}\frac{n}{2n-1} - \frac{n+2}{2n+3}$$
I've tried combining the sum, telescoping series and even trying to make an Nth partial sum but nothing seems to budge. I'm not sure where to go...
| We have \begin{align}
&\sum _{n=1}^m \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right)=\sum _{n=1}^m \frac{n}{2 n-1}-\sum _{n=1}^m \frac{n+2}{2 (n+2)-1}\\
&=\sum_{n=1}^m \frac{n}{2 n-1}-\sum _{n=3}^{m+2} \frac{n}{2 n-1}\\
&=\frac{1}{2\cdot 1-1}+\frac{2}{2\cdot 2-1}+\sum _{n=3}^m \frac{n}{2 n-1}-\sum
_{n=3}^m \frac{n}{2 n-1}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\\
&=1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\end{align}
Taking the limit when $m \to \infty$ we have
\begin{align}
&\sum _{n=1}^\infty \left(\frac{n}{2 n-1}-\frac{n+2}{2 (n+2)-1}\right) \\&= \lim_{m\to\infty}\left( 1+\frac{2}{3}-\frac{m+1}{2 (m+1)-1}-\frac{m+2}{2 (m+2)-1}\right)\\
&= 1+\frac{2}{3}-\frac{1}{2}-\frac{1}{2} = \frac{2}{3}\end{align}
The others answer don't take into account that the "remainder" terms after "telescoping" doesn't tends to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?
$120$
| We must select the $5$ digits which must appear, although we know the $0$ must appear. So in reality we must only select $4$ digits out of $6$ possibilities.
Notice the sum of all those numbers is a multiple of $3$ ($1+2+6+7+8+9 = 33)$. Therefore the condition is that the two numbers we do not pick add up to a multiple of $3$.
There are $4$ possible ways to do this if these numbers are $1$ and $2\bmod 3$ and there is one option when both unselected numbers are multiples of $3$.
We conclude there are $5$ valid selections for the $5$ digits that appear. For each configuration there are $4!$ rearrangements.
We conclude the answer is $5\times 4! = 120$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find value of $x^3+y^3+z^3$ if $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$ If $x,y$ and $z$ are positive real numbers such that $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$, then what is the value of $x^3+y^3+z^3$?
I first thought of making them all equal because in that case the product is maximum, but obviously that was wrong and that is only valid for integers. By trial-and-error, I found that $(x,y,z)=(3,4,5)$ does satisfy the conditions, but I can't get a solid proof and method for this, other than just trial-and-error.
Thank you.
| Let's find the minimal value of $x+y+z$ when $x^3y^4z^5 = C$.
At point of tangency between $x^3y^4z^5 = C$ and $x+y+z=A$ the normal is proportional to:
$$
\left(\frac{3}{x},\frac{4}{y},\frac{5}z\right)
$$
On the other hand, it should be proportional to $(1,1,1)$. It is possible when $x/3 = y/4= z/5 = k$. From
$$
(3k)^3(4k)^4(5k^5) = (3\times4\times 5\times10)^3 \frac{4\times 5^2}{1000}k^{12} = 0.1\times 600^3
$$
we find that $k=1$. By sheer luck (actually not), this point also lies on $x+y+z=12$. Since, it's a tangency point by construction and $x^3y^4z^5$ is convex in positive octant, there is no other solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do you find the least value when an inverse trigonometric function is in the denominator?
The function:
$f(x)=\frac{4\pi^2}{3\arccos{(x^4-2x^2)}}+\frac{5\pi}{3}$
If $B=\frac{m}{\pi}$
where $m$ is the minimum value which $f(x)$ can take. Find the value of $B$.
The choices given in my book are as follows:
$\begin{array}{ll}
1.&4\\
2.&3\\
3.&5\\
4.&2\\
\end{array}$
Gee, I'm confused on exactly what should I do here to solve this problem. Can someone help me here?. I think it has to be greater than $\frac{5\pi}{3}$ because that's a vertical shift for that function. But I don't know how to get the range here.
| $x^4-2x^2=x^2(x^2-2) \ge -1$, and it takes value the value $-1$ when $x^2=1$.
As $x^2=1$, $\arccos(x^4-2x^2)$ has the largest value of $\pi$. Hence $$m=\frac{4\pi^2}{3\pi}+\frac{5\pi}{3}=\frac{9\pi}{3}=3\pi$$
Hence you are right, $$B=\frac{m}{\pi}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the form $0 \cdot \infty$.
Moreover, we obtained a similar result here, which is related to the form $\lim\limits_{n \to \infty} (a_n-3)9^n$.
How should I proceed?
| In this answer we consider the sequence $b_n=|3-a_n|6^n$, where $a_{n+1}=\sqrt{6+a_n}$ with initiial condition $a_0\geq0$. We show that $b_n$ is
*
*bounded and monotone increasing if $0\leq a_0<3$ (see \eqref{three}).
*bounded and monotone decreasing when $a_0>3$ (see \eqref{twop}).
As a consequence,
*$b_n(a_0)$ (dependence of initial condition) is convergent for all $a_0\geq0$:
$$b_\infty(a_0)=|3-a_0|\prod^\infty_{n=1}\Big(1-\frac{|3-a_n|}{6}\Big)^{-1}$$
*When $0\leq a_0<3$, $(a_n-3)6^n=-b_n$ converges to a negative number; when $a_0>3$, $(a_n-3)6^n=b_n$ converges to a positive number.
I don't think the limit $b_\infty(a_0)=\lim_nb_n(a_0)$ can be expressed in terms of elementary functions. A numerical estimate (with double precision arithmetic) gives $b_\infty(0)\approx3.3657$.
The map $f:[0,\infty)\rightarrow[0,\infty)$ defined by
$$ f(x)=\sqrt{6+x}$$
is a contraction: First notice that $f(x)\geq\sqrt{6}$ for all $x\in[0,\infty$, and
\begin{align}
f(x)-f(y)=\frac{y-x}{\sqrt{6+x}+\sqrt{6+y}}\tag{1}\label{one}
\end{align}
Then,
$$|f(x)-f(y)|\leq\frac{1}{2\sqrt{6}}|x-y|
$$
and so, for any $a_0\geq0$, $a_{n+1}=f(a_n)$ converges to a fixed point $x=f(x)$ with $x\geq0$, which in this case is $x=3$.
Further analysis of the map $f$ shows that
*
*If $3<a_0$, then $3<a_n<a_{n-1}$ for all $n\geq1$
*If $0\leq a_0<3$, then $a_{n-1}<a_n<3$ for all $n\geq1$.
We now study the speed of convergence of the sequence $a_n$. Let $d_n=|3-a_n|=|f(3)-f(a_{n-1})|$. It follows from \eqref{one} that
$$d_n=\frac{d_{n-1}}{3+a_n}$$
If $0\leq a_0<3$, then
\begin{align}
\frac{1}{6}d_{n-1}<d_n=\frac{d_{n-1}}{3+a_n}=\frac{d_{n-1}}{6-d_n}\tag{2}\label{two}
\end{align}
As $a_n\xrightarrow{n\rightarrow\infty}3$, we have that
$\frac{d_{n+1}}{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$;
consequently $\sqrt[n]{d_n}\xrightarrow{n\rightarrow\infty}\frac{1}{6}$.
This alone however, does not provide convergence of $b_n=(6\sqrt[n]{d_n})^n$.
Now we prove that the sequence $b_n$ indeed converges. From \eqref{two} we have that
\begin{align}
b_{n-1}=6^{n-1}d_{n-1}< 6^nd_n=b_n=\frac{6^{n-1}d_{n-1}}{1-\tfrac{d_n}{6}}=\frac{b_{n-1}}{1-\tfrac{d_n}{6}}\tag{3}\label{three}
\end{align}
and
\begin{align}
\frac{d_{n-1}}{6}<d_n\leq\frac{d_{n-1}}{3+a_0}\leq \frac{d_0}{(3+a_0)^{n-1}}\tag{4}\label{four}
\end{align}
Putting this together, we obtain
\begin{align}
b_n=b_0\prod^n_{k=1}\frac{b_k}{b_{k-1}}=b_0\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)^{-1}\tag{5}\label{five}
\end{align}
From \eqref{four} we have that $\sum^\infty_{k=1}d_k<\infty$; from
$$\prod^n_{k=1}\Big(1-\tfrac{d_k}{6}\Big)\leq \exp\Big(-\frac{1}{6}\sum^n_{k=1}d_k\Big)$$
we conclude that $b_n$ converges, and that
\begin{align}
b_n\xrightarrow{n\rightarrow\infty} b_0\prod^\infty_{k=1}\big(1-\tfrac{d_k}{6}\big)^{-1}\tag{6}\label{6}
\end{align}
Remarks:
*
*An interesting observation is that when $a_0>3$, convergence of $b_n$ is much easier to check, for in this case, the inequalities in \eqref{two} are reverse, that is
\begin{align}
\frac{d_{n-1}}{6+d_n}=\frac{d_{n-1}}{3+a_n}=d_n<\frac{1}{6}d_{n-1}\tag{2'}\label{twop}
\end{align}
This implies $ b_n<b_{n-1}$ and convergence follows immediately. This can be explained in part to the fact that $f'(x)=\big(2\sqrt{6+x}\big)^{-1}$ is a decreasing function, and so $f'(y)<f'(x)$ whenever $x<3<y$. So convergence tens to be faster to the right of $3$.
*Notice that the limit of the sequence $b_n$ depends on the initial condition $a_0$. For $a_0=0$, simple numerical implementation gives $\lim_nb_n\approx 3.3657$; for $a_0=6$, $\lim_nb_n\approx 2.7426$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 1
} |
Computing or approximating $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ I'm trying to compute $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ when $N$ is large. Is there an asymptotic formula that tells me $\lim_{n \to \infty} \frac{1}{f(N)} \sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ for some function $f(N)$?
| By Stirling's formula
$$
\log \binom{N}{n} \!= n\log \left( {\frac{N}{n}} \right) - (N - n)\log \left( {1 \!-\! \frac{n}{N}} \right) - \frac{1}{2}\log \left( { n\left( {1 \!-\! \frac{n}{N}} \right)} \right) + \mathcal{O}(1),
$$
provided $1\leq n \leq N-1$.
Now
$$
\sum\limits_{n = 1}^{N-1} {\log \left( {n\left( {1 - \frac{n}{N}} \right)} \right)\log \left( {1 + \frac{1}{n}} \right)} \le \sum\limits_{n = 1}^{N-1} {\log \left( { \frac{N}{4}} \right)\frac{1}{n}} = \mathcal{O}(\log ^2 N)
$$
and
$$
\sum\limits_{n = 1}^{N-1} {\mathcal{O}(1)\log \left( {1 + \frac{1}{n}} \right)} \le \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}} = \mathcal{O}(\log N).
$$
Also, using $\log \left( {1 + \frac{1}{n}} \right) = \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)$,
\begin{align*}
\sum\limits_{n = 1}^{N-1} {n\log \left( {\frac{N}{n}} \right)\log \left( {1 + \frac{1}{n}} \right)} & = \sum\limits_{n = 1}^{N-1} {\log \left( {\frac{N}{n}} \right)} + \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}\log \left( {\frac{N}{n}} \right)} \\ & = N + \mathcal{O}(\log ^2 N)
\end{align*}
and
\begin{align*}
& \sum\limits_{n = 1}^{N-1} {(N - n)\log \left( {1 - \frac{n}{N}} \right)\log \left( {1 + \frac{1}{n}} \right)} \\ & = N\frac{1}{N}\sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{N}{n}\log \left( {1 - \frac{n}{N}} \right)} + \mathcal{O}(1) \sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{{N }}{{n^2 }}\log \left( {1 - \frac{n}{N}} \right)} \\ & = N\frac{1}{N}\sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{N}{n}\log \left( {1 - \frac{n}{N}} \right)} + \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}} \\ & = N\int_0^1 {\frac{{(1 - x)\log (1 - x)}}{x}dx} + \mathcal{O}(1)+\mathcal{O}(\log N) = \left( {1 - \frac{{\pi ^2 }}{6}} \right)N + \mathcal{O}(\log N).
\end{align*}
Thus, in summary,
$$
\sum\limits_{n = 1}^{N-1} {\log \binom{N}{n}\log \left( {1 + \frac{1}{n}} \right)} = \frac{{\pi ^2 }}{6}N + \mathcal{O} (\log^2 N).
$$
With a bit more work, you can improve the error term, for example, into
$$
-\frac{1}{2} \log^2 N +\mathcal{O} (\log N).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4099336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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