Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Demonstration of sum of powers of $2$ Theorem : For every natural number $p$:
$$\sum^p_{i=0} 2^i = 2^{p+1}-1$$
I trieed to demonstrate the theorem using induction
Demonstration :
$1)$ If we have $p=0$ then we get $2^0=2^{0+1}-1$ that is always true.
$2)$ Supposing that the first statement is true then we get that
$$\sum^{p+1}_{i=0} 2^i = 2^{p+2}-1$$
Now we know that:
$$\sum^{p+1}_{i=0} 2^i - \sum^p_{i=0} 2^i = 2^{p+1}$$
and if that is true, must be also true that:
$$2^{p+2}-1-(2^{p+1}-1) = 2^{p+1}$$
$$2^{p+2}-1-(2^{p+1}-1)=2^{p+2}-2^{p+1}=2^{p+1}(2^1-2^0)=2^{p+1}$$
QED
Is this a valid demonstration?
| Inductive hypothesis: $\quad\displaystyle\sum_{i=0}^p 2^i=2^{p+1}-1$ for some $p\ge 0$.
We must deduce that $\quad\displaystyle\sum_{i=0}^{p+1} 2^i=2^{p+2}-1$.
Decompose the sum: \begin{align*}\sum_{i=0}^{p+1} 2^i=2^{p+2}-1 &= \sum_{i=0}^p 2^i+2^{p+1}\\
&=2^{p+1}-1 +2^{p+1}\qquad\scriptstyle(\text{inductive hypothesis)}\\
&=2\cdot2^{p+1}-1=2^{p+2}-1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Find $E(|X-Y|^a)$ where $X$ and $Y$ are independent uniform on $(0,1)$
Let $X,Y$ be independent $Uniform(0,1)$ random variables. Find $E(|X-Y|^a)$ where $a>0$.
My working:
Define $W=1$ if $X>Y$ and $W=0$ if $X<Y$. We seek $E(|X-Y|^a)=E[E(|X-Y|^a|W)]=E(|X-Y|^a|W=1)P(W=1)+E(|X-Y|^a|W=0)P(W=0)=E((X-Y)^a)P(X>Y)+E((Y-X)^a)P(X<Y)$.
Now $P(X<Y)=\dfrac{1}{2}$ and $P(X>Y)=\dfrac{1}{2}$ by symmetry.
$E((X-Y)^a)=\int_{0}^{1}\int_{0}^{x}(x-y)^af_{X,Y}(x,y)dydx=\int_{0}^{1}\int_{0}^{x}(x-y)^adydx=\dfrac{1}{(a+1)(a+2)}$.
Similarly $E((Y-X)^a)=\dfrac{1}{(a+1)(a+2)}$ by symmetry when $Y>X$.
Required expectation=$\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}+\dfrac{1}{(a+1)(a+2)}\dfrac{1}{2}=\dfrac{1}{(a+1)(a+2)}$
But correct answer is $\dfrac{2}{(a+1)(a+2)}$.
Where am I going wrong?
| You should have something like:
$$
E(|X-Y|^a)=\iint_{x>y}(x-y)^a\;dxdy+\iint_{y>x}(y-x)^a\;dxdy=2\iint_{x>y} (x-y)^a\;dxdy
$$
Now:
$$
\iint_{x>y}(x-y)^a\;dxdy=\int_{y=0}^1\int_{x=y}^1(x-y)^a\;dxdy=\frac{1}{(a+1)(a+2)}
$$
So:
$$
E(|X-Y|^a)=2\iint_{x>y}(x-y)^a\;dxdy=\frac{2}{(a+1)(a+2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How does $(x+3)^2 - 2^2$ become $(x+1)(x+5)$? I don't understand how $(x+3)^2 - 2^2$ can be transformed to equal $(x+1)(x+5)$. A short demonstration and/or reference to math rules would be very kind.
| Just multiply each part out:
$(x + 3)^{2}$
$= (x + 3)(x + 3)$
$= x^{2} + 3x + 3x + 9$
$= x^{2} + 6x + 9$
Also, $2^{2} = 4$.
So, putting the two above together, we get:
$(x + 3)^{2} - 2^{2}$
$= x^{2} + 6x + 9 - 4$
$= x^{2} + 6x + 5$
And, if you understand how to factor, we get $x^{2} + 6x + 5 = (x + 5)(x + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that for any integer, $n^2 + 5$ is not divisible by $4$. So I got that there is two cases: odd or even.
If odd then say $n^2$ is $(2k+1)^2 = 4k^2 + 4k + 1.$
then $4k^2 + 4k + 1 + 5$ would need to be divisible by 4 and I don't know where to go from there.
| $n$ is even or odd.
$n$ is even: $n = 2m \Rightarrow n^2 + 5 = 4m^2 + 5 = 4(m^2 + 1) + 1$, which isn't divisible by $4$.
$n$ is odd: $n = 2m+1 \Rightarrow n^2 + 5 = 4m^2 + 4m + 1 + 5 = 4(m^2 + m + 1) + 2$, which isn't divisible by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 3
} |
Infinite Product computation How can we compute the infinite product:
Do we need Gamma function?
Edited: I forgot to add 1/e factor so the product converges.
The product becomes
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Note that
$$
\prod_{k=2}^{\infty}\bracks{\frac{1}{\expo{}}\pars{1 + \frac{1}{k^{2} - 1}}
^{k^{2}\ - 1}}
=\exp\pars{\color{#66f}{%
\sum_{k=2}^{\infty}
\braces{\bracks{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}} - 1}}}
$$
We'll study the above $\ds{\color{#66f}{blue}}$ expression:
\begin{align}
&\color{#66f}{\lim_{N \to \infty}\
\sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}}- 1}}
=\lim_{N \to \infty}\ \sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ \overbrace{%
2\int_{0}^{1}\frac{x\,\dd x}{k^{2} - x^{2}}}
^{\dsc{\ln\pars{1 + \frac{1}{k^{2} - 1}}}}\ -\ 1}
\\[5mm]&=\lim_{N \to \infty}\ \sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}
2\int_{0}^{1}\frac{x\,\dd x}{k^{2} - x^{2}}\ -\
\overbrace{2\int_{0}^{1}x\,\dd x}^{\dsc{1}}}
\\[5mm]&=2\lim_{N \to \infty}\ \sum_{k=2}^{N}\
\int_{0}^{1}\frac{\pars{k^{2}x - x}-\pars{k^{2}x - x^{3}} }{k^{2} - x^{2}}\,\dd x
\\[5mm]&=2\int_{0}^{1}x\pars{x^{2} - 1}
\sum_{k=0}^{\infty}\frac{1}{\pars{k + 2 + x}\pars{k + 2 - x}}\,\dd x
\\[5mm]&=\int_{0}^{1}
\pars{x^{2} - 1}\bracks{\Psi\pars{2 + x} - \Psi\pars{2 - x}}\,\dd x\tag{1}
\end{align}
where $\ds{\Psi}$ is the Digamma Function. With the Digamma Recurrence Formula:
\begin{align}
\Psi\pars{2 + x} - \Psi\pars{2 - x}
&=\bracks{\Psi\pars{1 + x} + \frac{1}{1 + x}}
-\bracks{\Psi\pars{1 - x} + \frac{1}{1 - x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \Psi\pars{1 + x}
-\bracks{\Psi\pars{-x} + \frac{1}{-x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \frac{1}{x}
+ \bracks{\Psi\pars{1 + x} - \Psi\pars{-x}}
\\[5mm]&=\frac{1}{1 + x} - \frac{1}{1 - x} + \frac{1}{x} - \pi\cot\pars{\pi x}
\tag{2}
\\[5mm]&=\totald{}{x}\bracks{\ln\pars{1 + x} + \ln\pars{1 - x} + \ln\pars{x}\
-\ln\pars{\sin\pars{\pi x}}}
\end{align}
where we used, in expression $\pars{2}$, the Euler Reflection Formula. Expression $\pars{1}$ is reduced to:
\begin{align}
&\color{#66f}{\lim_{N \to \infty}\
\sum_{k=2}^{N}\bracks{\pars{k^{2} - 1}\ln\pars{1 + \frac{1}{k^{2} - 1}}- 1}}
\\[5mm]&=-\ln\pars{\pi}
-2\ \underbrace{%
\int_{0}^{1}\bracks{\ln\pars{1 + x} + \ln\pars{1 - x} + \ln\pars{x}\
-\ln\pars{\sin\pars{\pi x}}}x\,\dd x}
_{\dsc{\frac{2\ln\pars{2} - 3}{4}}}
\end{align}
Finally,
$$
\color{#66f}{\large\prod_{k=2}^{\infty}\bracks{\frac{1}{\expo{}}\pars{1 + \frac{1}{k^{2} - 1}}^{k^{2}\ - 1}}}
=
\color{#66f}{\large\frac{\expo{3/2}}{2\pi}}\approx{\tt 0.7133}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to calculate $\int e^{\frac{x^2}{2}} \cdot x^{3} dx$? $\int e^{\frac{x^2}{2}} \cdot x^{3} dx=?$
I tried to do the substitution $du = x^3dx$, so that $u=\frac{x^4}{4}$. Then,
$\int e^{\frac{x^2}{2}} \cdot x^{3} dx= \int e^{\sqrt{u}} du$
However, I would have to do a second substitution ($v = \sqrt{u}$) which would give a product again.
How should I proceed?
| Let $t=\frac{x^2}{2}, dt=xdx$ to get
$\displaystyle \int e^{\frac{x^2}{2}}x^3\;dx=\int e^t(2t)dt=2\int te^t\;dt=2\left[te^t-e^t\right]+C=2\left[\frac{x^2}{2}e^{\frac{x^2}{2}}-e^{\frac{x^2}{2}}\right]+C$
$\;\;\;\;=\displaystyle e^{\frac{x^2}{2}}\left[x^2-2\right]+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve an Hyperbolic triangle when all is given except angle C and side c) Another Hyperbolic triangle problem (all given except angle $\angle C$, and side $c$)
I thought that after asking How to solve an hyperbolic Angle Side Angle triangle? I could solve all hyperbolic triangles, but I am still stumped with SSA and AAS from both you can via the hyperbolic law of sinus calculate one extra value to get to AASS but then I got stuck again therefore:
If from an hyperbolic triangle $ \triangle ABC$ the angles $\angle A$, $\angle B$ and sides $a$ and $b$ are given. (so a combination of ASS or AAS )
How can I calculate the angle $\angle C$ or side $c$?
I thought with the two variations of the Hyperbolic law of cosines ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )
or the hyperblic law of sinus ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )
It must be easy, but neiter law can solve this problem.
Or am I overlooking something (obious) again?
| Somehow the following formula's work:
$$ \tanh\left(\frac{c}{2}\right) = \tanh\left(\frac{1}{2} (a-b)\right) \frac{\sin{\left(\frac{1}{2}(A+B)\right)}}{\sin{\left(\frac{1}{2}(A-B)\right)}} $$
and
$$ \tan\left(\frac{C}{2}\right) = \frac{1}{\tan\left(\frac{1}{2} (A-B)\right)} \frac{\sinh\left(\frac{1}{2}(a-b)\right)}{\sinh\left(\frac{1}{2}(a+b)\right)} $$
They are hyperbolic alternatives to the spherical formulas mentioned at https://en.wikipedia.org/wiki/Solution_of_triangles#A_side.2C_one_adjacent_angle_and_the_opposite_angle_given
(maybe they can be simplified)
I don't know exactly why they work (and i won't accept this answer till i know)
user MvG advised me to investigate
http://en.wikipedia.org/wiki/Spherical_trigonometry#Napier.27s_analogies
http://en.wikipedia.org/wiki/Tangent_half-angle_formula
and
http://en.wikipedia.org/wiki/Hyperbolic_function#Hyperbolic_functions_for_complex_numbers
so maybe after that I can explain them :)
proof based on the proof for spherical geometry by I Hothunter, section 52
(http://www.gutenberg.org/ebooks/19770 )
But then rewritten for hyperbolic geometry
The hyperbolic sinus rule ( https://en.wikipedia.org/wiki/Law_of_sines#Hyperbolic_case )
$$ \frac{\sin A}{sinh(a)} = \frac{\sin B}{sinh(b)}= \frac{\sin C}{sinh(c)} $$
from it follows that
$ \frac{\sin A + \sin B}{sinh(a) + sinh(b)} = \frac{\sin C}{sinh(c)} $
and
$ \frac{\sin A - \sin B}{sinh(a) - sinh(b)} = \frac{\sin C}{sinh(c)} $
In the following $ \frac{\sin C}{sinh(c)} $ or any of the other equivalents is $ SQ $
The hyperbolic cosinus rule ( https://en.wikipedia.org/wiki/Hyperbolic_law_of_cosines )
has two forms
(CR1) $ cosh(a) = cosh (b) cosh(c) - sinh(b) sinh(c) \cos A $
and
(CR2) $ \cos A = - \cos B \cos C + \sin B \sin C cosh(a) $
CR2 rewritten
(1) $ \sin B \sin C cosh(a) = \cos A + \cos B \cos C $
and for $ \angle B $
(2) $ \sin A \sin C cosh(b) = \cos B + \cos A \cos C $
1 and 2 involving SQ
(3) $ SQ \sin C cosh(a) sinh(b) = \cos A + \cos B \cos C $
(4) $ SQ \sin C cosh(b) sinh(a) = \cos B + \cos A \cos C $
adding 3 and 4 together
(5) $ SQ \sin C ( cosh(a) sinh(b) + cosh(b) sinh(a) ) = (\cos B + \cos A ) * (1 + \cos C )$
(6) $ sinh(x + y) = sinh (x) cosh (y) + cosh (x) sinh (y) $
therefore
(7) $ SQ \sin C sinh(a+b) = (\cos B + \cos A ) (1 + \cos C )$
(8) $ \frac {\sin X}{1 + \cos X } = \tan (\frac {X}{2}) $
therefore
(9) $ SQ \tan (\frac {C}{2}) sinh(a+b) = (\cos B + \cos A ) $
(10) $ SQ = (\frac {\sin A + \sin B }{ sinh(a) + sinh(b)} $
therefore
(11) $ (\sin A + \sin B ) \tan (\frac {C}{2}) sinh(a+b) = (\cos B + \cos A ) (sinh(a) + sinh(b))$
(12) $ \tan(\frac {1}{2} (A + B) ) = \frac {\sin A + \sin B}{\cos B + \cos A} $
(13) $ \tan(\frac {1}{2} (A + B) ) \tan (\frac {C}{2}) sinh(a+b) = (sinh(a) + sinh(b)) $
(14) $ \tan (\frac {C}{2}) = \frac {1}{\tan(\frac {1}{2} (A + B) ) } \frac {sinh(a) + sinh(b)}{sinh(a+b)} $
That should do :)
| {
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"url": "https://math.stackexchange.com/questions/1131152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In a triangle, prove that $a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$ I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$
where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS:
$$
\begin {align}
a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\
&=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\
&=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\
&=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc]
\end{align}$$
where $$s=\frac{a+b+c}{2}$$
I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$
But I'm stuck there. Any help please?
| From the cosine rule:
$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
So
$$\begin{align}a\cos(A)+b\cos(B)+c\cos(C)&=a\frac{b^2+c^2-a^2}{2bc}+b\frac{a^2+c^2-b^2}{2ac}+c\frac{a^2+b^2-c^2}{2ab}\\
&=a^2\frac{b^2+c^2-a^2}{2abc}+b^2\frac{a^2+c^2-b^2}{2abc}+c^2\frac{a^2+b^2-c^2}{2abc}\\
&=\frac{a^2b^2+a^2c^2-a^4+a^2b^2+b^2c^2-b^4+a^2c^2+b^2c^2-c^4}{2abc}\\
&=\frac{-(a-b-c)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\
&=\frac{(b+c-a)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\
&=\frac{(2s-2a)(2s-2c)(2s-2b)(2s)}{2abc}\\
&=\frac{8(s-a)(s-c)(s-b)(s)}{abc}\\
&=\frac{8\Delta^2}{abc}\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrating $ \int \frac{1}{\sqrt{x^2+4}}\,dx $ using Trigonometric Substitution I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answer which is very close to being correct. However, I somehow end up dividing by two where I should not. I'm hoping another set of eyes can quickly set me right so I can stop frustrating myself reworking the problem to the same apparently wrong answer repeatedly! Thanks in advance!
The problem asks to solve:
$$
\int \frac{1}{\sqrt{x^2+4}}\,dx
$$
The answer is given as:
$$
ln\lvert x + \sqrt{x^2+4} \rvert + C
$$
Somehow I keep getting:
$$
ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C
$$
Here's my work:
$$
\int \frac{1}{\sqrt{x^2+4}}\,dx = \int \frac{1}{\sqrt{4(\frac{1}{4}x^2+1)}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}x^2+1}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{(\frac{1}{2}x)^2+1}}\,dx
$$
At this point I substitute as follows:
$$
\frac{1}{2}x = \tan\theta
$$
$$
x = 2\tan\theta
$$
$$
dx = 2\sec^2\theta
$$
So I continue on with:
$$
\frac{1}{2}\int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta+1}}\,d\theta = \int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,d\theta = \int\frac{\sec^2\theta}{\sec\theta}\,d\theta = \int\sec\theta\,d\theta = ln\lvert\sec\theta + \tan\theta\rvert + C
$$
Finally, to get the answer in terms of x I essentially draw a right triangle and use the fact that $\tan\theta = \frac{x}{2}$. The side opposite $\theta$ I take to be x, the side adjacent $\theta$ is 2, and the hypotenuse is $\sqrt{x^2+4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ and $\tan\theta = \frac{x}{2}$.
So, substituting these values back in, as mentioned, I end up with:
$$
ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C
$$
Can anyone help me see where I'm going wrong or failing to understand something?
| Your answer is also correct. Remember that antiderivates are only determined up to an additive constant.
$$\ln\left|{\sqrt{x^2+4}\over 2}+{x\over 2}\right|+C=\ln|\sqrt{x^2+4}+x|-\ln2+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the trace of all elements in $F_8$ I got the following exercise where you have to calc the trace of all elements in ${F_8}$ which is constructed as ${F_2}[x]$/(${x^3+x+1}$)${F_2}[x]$.
Up to now I did those steps:
1) Find all elements in ${F_8}$ which are in my opinion: $0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1$
2) Then I found those traces for the elements:
$Tr(0)=0\\
Tr(1)=1\\
Tr(x)=x+x^2+x^4\\
Tr(x+1)=Tr(x)+Tr(1)=x^4+x^2+x+1\\
Tr(x^2)=x^8+x^4+x^2 \text{(must this be reduced!?)}\\
Tr(x^2+1)=Tr(x^2)+Tr(1)=x^8+x^4+x^2+1\\
Tr(x^2+x)=Tr(x^2)+Tr(x)=x^8+x\\
Tr(x^2+x+1)=Tr(x^2)+Tr(x)+Tr(1)=x^8+x+1$
Is this procedure correct? Thank you!
| You can take advantage of the linearity of trace. To make the notation less cumbersome I denote the coset $x+\langle x^3+x+1\rangle$ by $\alpha$. In that case the elements of the field are
$$
\Bbb{F}_8=\{a_0+a_1\alpha+a_2\alpha^2\mid a_0,a_1,a_2\in\Bbb{F}_2\}.
$$
The minimal polynomial of $\alpha$ is $m(x)=x^3+x+1$. Its other zeros are the conjugates $\alpha^2$ and $\alpha^4=\alpha\cdot\alpha^3=\alpha(\alpha+1)$. Therefore
$$
\begin{aligned}
m(x)&=(x-\alpha)(x-\alpha^2)(x-\alpha^4)=x^3-(\alpha+\alpha^2+\alpha^4)x^2+\cdots\\
&=x^3-tr(\alpha)x^2+\cdots
\end{aligned}
$$
Therefore we can conclude that $tr(\alpha)=0$. This also follows from the earlier calculation (that also appears in Timbuc's answer) giving us that $\alpha^4=\alpha+\alpha^2$.
Because $m(x)$ is also a minimal polynomial of $\alpha^2$, we see that $tr(\alpha^2)=tr(\alpha)$, so also $tr(\alpha^2)=0$. Clearly
$$
tr(1)=1+1^2+1^4=3=1.
$$
Linearity thus gives us that
$$
tr(a_0+a_1\alpha+a_2\alpha^2)=a_0\, tr(1)+a_1\, tr(\alpha)+a_2\, tr(\alpha^2)=a_0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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If $\log_{12}54=a$ then $ \log_{6}12=?$ I am given $$\log_{12}54=a$$
So what will be value of $ \log_{6}12?$
I used base changing theorem and wrote expression as
$$\frac{\log_{6}54}{ \log_{6}12} =a$$
And then $$ \frac{1+\log_{6}9}{ a} = \log_{6}12$$ now what to do?
| Observe first that
$$\begin{align}a&=\log_{12}(54)=\log_{12}(2)+3\log_{12}(3)\\&=\frac{1}{\log_2(12)}+\frac{3}{\log_3(12)}=\frac{1}{2+\log_2(3)}+\frac{3}{1+\frac{2}{\log_2(3)}}\end{align}$$
From this, we can compute $\log_2(3)$ in terms of $a$. We just need to solve for $x$ in the equation $$a=\frac{1}{2+x}+\frac{3}{1+\frac{2}{x}}$$
Then we can use this value to compute:
$$\log_6(12)=2\log_6(2)+\log_6(3)=\frac{2}{\log_2(6)}+\frac{1}{\log_3(6)}=\frac{2}{1+\log_2(3)}+\frac{1}{1+\frac{1}{\log_2(3)}}$$
The equation above is $$a=\frac{1+3x}{x+2}$$
from where $$x=\frac{2a-1}{3-a}$$
Therefore
$$\begin{align}\log_{6}(12)&=\frac{2}{1+\frac{2a-1}{3-a}}+\frac{1}{1+\frac{1}{\frac{2a-1}{3-a}}}\\&=\frac{5}{2+a}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Derivative of an inverse Let $f(x)=2x^3+7x−1$, and let $g(x)$ be the inverse of $f(x)$. Then find $g′(191/4)$. I know only one way of doing this. Solving the cubic equation for x and then differentiating it. This is too too long. How to solve it more easily?
| In order to not confuse the variables of the inverse function with the variables of the function itself, let :
$y=f(x)=2x^3+7x−1=X$
$Y=g(X)=x$
One little picture says more than a long speech! : See below
$\frac{dg}{dX}=\frac{1}{\frac{df}{dx}}$
For $X=\frac{191}{4}$ we have $y=\frac{191}{4}=2x^3+7x-1$ which real root is $x=\frac{5}{2}$
$\frac{1}{\frac{df}{dx}}=\frac{1}{6x^2+7}=\frac{1}{6\left(\frac{5}{2}\right)^2+7}=\frac{2}{89}$
$\frac{dY}{dX}=\frac{1}{\frac{df}{dx}}=\frac{2}{89}$
where $X=y=\frac{191}{4}$ and $Y=x=\frac{5}{2}$
One see on the graph on right, which represents the inverse function, that the point considered is at abscissa $\frac{191}{4}$ as specified in the wording of the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&=
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong.
What am I doing wrong here?
| Here are the steps
$$\lim\limits _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}
$$
$$= \lim\limits _{n\to \infty}\sqrt{\frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2+1}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+\sqrt{n^2\left(1+\frac{1}{n^2}\right)}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{n+|n|\sqrt{1+\frac{1}{n^2}}}{3n+1}} $$
$$= \sqrt{ \lim\limits _{n\to \infty} \frac{1+\sqrt{1+\frac{1}{n^2}}}{3+\frac{1}{n}}} $$
$$= \sqrt{ \frac{1+\sqrt{1+0}}{3+0}} $$
$$= \sqrt{ \frac{2}{3}} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a, b, c,x$ be elements of a group $G$. In each of the following, solve for $x$ in terms of $a, b, c$. Just checking my work.
$axb =c.$
$(axb)a^{-1}b^{-1} = ca^{-1}b^{-1}$
$x = ca^{-1}b^{-1}$
$x^2b = xa^{-1}c$
$x^{-1}(x^2b) = (xa^{-1}c)x^{-1}$
$xb = a^{-1}c$
$x = a^{-1}b^{-1}c$
$x^2a = bxc^{-1}$ and $acx = xac.$
$(x^2a)x^{-1} = (bxc^{-1})x^{-1}$
$xa = bc^{-1}$
$(xa)c = (bc^{-1})c$
$acx = b$
$x = a^{-1}bc^{-1}$
$ax^2 = b$ and $x^3 = e.$
$(ax^2)x = bx$
$ae = bx$
$x = ab^{-1}$
$x^2 = a^2$ and $x^5 = e$
$(x^2)^2 = (a^2)^2$
$(x^4)x = a^4x$
$e = a^4x$
$(a^4)^{-1} = x$
$(xax)^3 = bx$ and $x^2a = (xa)^{-1}$
$((xax)^3)^{-2} = (bx)^{-2}$
$xax = (bx)^{-2}$
$((xa)^{-1})^2 = ((bx)^{-2})^2$
$xa = 0$
$xaa^{-1} = 0a^{-1}$
$x = 0$
| You’re forgetting that the group operation need not be commutative: the order of the operands matters. In the first problem, for example, $(axb)a^{-1}b^{-1}$ does not simplify to $x$; in fact, it doesn’t simplify at all without further knowledge of the group in question. You need to multiply $axb$ on the left by $a^{-1}$ and on the right by $b^{-1}$:
$$a^{-1}(axb)b^{-1}=(a^{-1}a)x(bb^{-1})=x\;,$$
so $x=a^{-1}cb^{-1}$.
You’ve made similar errors in the second and third problems. The fourth is correct, as is the fifth, though its answer could better be expressed as $x=a^{-4}$. The last is wrong: $(y^3)^{-2}=y^{-6}$, not $y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration using Algebraic Manipulation Trying to solve
$$\int^1_2\frac{x-4}{x^2}dx$$
All I have so far is re-writing it into two integrals which yields the following:
$$\int^1_2\frac{x}{x^2} \, dx - \int^1_2\frac{-4}{x^2} \,dx$$
Then $\frac{x}{x^2}$ can be re-written as $\frac{1}{x}$ which yields:
$$\int^1_2\frac{1}{x} \,dx - \int^1_2\frac{-4}{x^2} \, dx$$
I know that $\displaystyle \int\frac{1}{x} \, dx =\ln(x)$ but I'm not sure where to go from here. Could someone perhaps help?
| For one, you can rewrite $$\begin{align}\int^1_2\frac{x-4}{x^2}dx = \int^2_1\frac{4-x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{x}{x^2}dx \\ =4\int^2_1\frac{1}{x^2}dx - \int^2_1\frac{1}{x}dx \end{align}$$ and you are correct that $\int \frac{1}{x}dx = \ln|x|+C$. As for the other term to be integrated, consider this. For any constants $a,n,C \in \Bbb{R}$ we know by the power rule that $$\frac{d}{dx}ax^n+C = anx^{n-1} \quad \implies \quad \int anx^{n-1}dx = ax^n+C$$ Can you pattern match your integral of $\int\frac{4}{x^2}dx = \int 4x^{-2}dx$ to the general form of $\int anx^{n-1}dx$ for some $n$ and $a$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Given $f(x)=3x+2$ for $x<0$ and $f(x)= 3x +3$ for $x\ge0$, what is $f(t^2 + 1)$? I need to find $f(t^2 + 1)$ given $$f(x) = \begin{cases} 3x+2 & x \lt 0\\ 3x+3 & x\ge 0\end{cases}$$
The textbook says the answer is $3t^2 + 6$, but how do you know which piecewise to plug $t^2 + 1$ into?
| $$t^2 + 1 \geq 1 \quad \forall t \in \mathbb R$$
(The square of any number is greater than or equal to $0$. So $t^2 + 1\geq 1$.)
So we use $f(x) = 3x + 3$. Then we have $$f(t^2 + 1) = 3(t^2+1) + 3 = 3t^2 + 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
| $\cot\dfrac{11\pi}{18}=\cot\left(\pi-\dfrac{7\pi}{18}\right)=-\cot\dfrac{7\pi}{18}$
Now $\cot\dfrac{2\pi}9-\cot\dfrac{7\pi}{18}=\dfrac{\sin\left(\dfrac{7\pi}{18}-\dfrac{2\pi}9\right)}{\sin\dfrac{2\pi}9\cdot\sin\dfrac{7\pi}{18}}=\dfrac1{2\sin\dfrac{2\pi}9\cos\dfrac\pi9}$
( as $\sin\dfrac{7\pi}{18}=\cos\left(\dfrac\pi2-\dfrac{7\pi}{18}\right)=\cos\dfrac\pi9$)
which needs to be $=\sin\dfrac\pi{18}\cot\dfrac{2\pi}9$
which will be true if $8\sin\dfrac\pi{18}\cos\dfrac\pi9\cos\dfrac{2\pi}9=1$
$\iff8\cos\dfrac\pi9\cos\dfrac{2\pi}9\cos\dfrac{4\pi}9=1$
as $\dfrac{4\pi}9+\dfrac\pi{18}=\dfrac\pi2$
Now use: Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle?
| {
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"url": "https://math.stackexchange.com/questions/1146390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$ is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did
| HINT:
If $f(n)=2^{4n+2}+3^{n+2}$
$f(m+1)-3f(m)=2^{4(m+1)+2}+3^{(m+1)+2}-3[2^{4m+2}+3^{m+2}]=2^{4m+2}(2^4-3)$ which is divisible by $13$
So, $13|f(m+1)\iff13|f(m)$
Alternatively try with $f(m+1)-2^4f(m)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Rationalize the denominator: $(\frac{3}{2x^2}) ^{1/4}$ My answer is $\frac{(6x^2)^{1/4}}{ 2x^2}$
However the book's answer is $\frac{(24x^2)^{1/4}}{2x}$
Where did the book get that from?
Here's my work:
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}}$$
$$= \frac{\sqrt[4]{6x^2}}{2x^2}$$
Since the product of two roots with the same index is the root of the product:
$$\sqrt[N]{a} \cdot \sqrt[N]{b} = \sqrt[N]{ab}$$
| The simplest way is to multiply numerator and denominator by $(2x^2)^{\color{red}3/4}$, just because $\frac34+\frac14=1$:
\begin{align*}
\biggl(\frac3{2x^2}\biggr)^{1/4}&=\frac{3^{1/4}}{(2x^2)^{1/4}}=\frac{3^{1/4}(2x^2)^{3/4}}{2x^2}=\frac{(3\cdot8x^6)^{1/4}}{2x^2}\\
&=\frac{(24x^2)^{1/4}\cdot x}{2x^2}=\frac{\Bigl(\cfrac32x^2\Bigr)^{1/4}}x.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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derivative of $\sec^2(x/12)$ Alright, so the derivative of $\sec^2(x/12)$ is $\frac{1}{6} \tan\left(\frac{x}{12}\right) \sec^2\left(\frac{x}{12}\right)$
But if you use chain rule, you get:
$$2 \sec\left(\frac{x}{12}\right) \left(\sec\left(\frac{x}{12}\right)\right)'$$
$$2 \sec\left(\frac{x}{12}\right) \sec\left(\frac{x}{12}\right) \tan\left(\frac{x}{12}\right)$$
So why do they multiply by the derivative of $\frac{x}{12}$ as well?
I don't get it, what formula do they use then?
| You forgot to apply the chain rule again for the second equation on your post. You have $\sec\left(\frac{x}{12}\right)$, which is a composite. Applying the chain rule, we have
\begin{align}\frac{d}{dx} \left[sec^2\left(\frac{x}{12}\right)\right] &= 2* \sec\left(\frac{x}{12}\right) * \frac{d}{dx} \left[\sec\left(\frac{x}{12}\right)\right] \\
&= 2* \sec\left(\frac{x}{12}\right) * \sec\left(\frac{x}{12}\right) * \tan\left(\frac{x}{12}\right) * \frac{d}{dx}\left(\frac{x}{12}\right) \\
&= 2* \sec^2\left(\frac{x}{12}\right) * \tan\left(\frac{x}{12}\right) * \frac{1}{12} \\
&= \frac{1}{6} * \sec^2\left(\frac{x}{12}\right) * \tan\left(\frac{x}{12}\right).
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also be proved by AM–GM inequality." But I can't reason out the answer to this question.
| By C-S
$$\left(\sum_{cyc}\sqrt{a^2+(b-1)^2}\right)^2=\sum_{cyc}\left(a^2+(b-1)^2+2\sqrt{(a^2+(b-1)^2)(b^2+(c-1)^2}\right)\geq$$
$$\geq\sum_{cyc}(a^2+b^2-2b+1+2(ab+(b-1)(c-1)))=$$
$$=\sum_{cyc}(2a^2-2a+1+2ab+2ab-4a+2)=$$
$$=\sum_{cyc}(2a^2+4ab-6a+3)=2(a+b+c)^2-6(a+b+c)+9.$$
Thus, it remains to prove that
$$2(a+b+c)^2-6(a+b+c)+9\geq\frac{9}{2}$$ or
$$4(a+b+c)^2-12(a+b+c)+9\geq0$$ or
$$(2(a+b+c)-3)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Closed form for $\int_0^\infty\arctan\Bigl(\frac{2\pi}{x-\ln\,x+\ln(\frac\pi2)}\Bigr)\frac{dx}{x+1}$ I'm trying to find a closed form for this integral:
$$I=\int_0^\infty\arctan\left(\frac{2\pi}{x-\ln\,x+\ln\left(\frac\pi2\right)}\right)\frac{dx}{x+1}$$
Its approximate numeric value is
$$I\approx3.3805825284453469793953592216276992165696856825906055108192183...$$
Any help is appreciated. Thanks!
| Computing a Related Contour Integral:
Define
$$f(z)=\frac{i}{2}\frac{z-1}{1+az}\left(\frac{1}{z-\ln{z}+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{z-\ln{z}+2\pi i+\ln\left(\frac{\pi}{2}\right)}\right)$$
and let $\gamma$ denote a keyhole contour deformed around $[0,\infty]$. Restricting the argument between $0$ and $2\pi$, it is not hard to see that $f(z)$ has poles at $z=-\dfrac{1}{a}$, $z=-W_{-1}\left(-\dfrac{\pi}{2}\right)=\dfrac{\pi i}{2}$, and $z=-W_0\left(-\dfrac{\pi}{2}\right)=-\dfrac{\pi i}{2}$. The residues at these poles are
\begin{align}
\operatorname*{Res}_{z=\frac{\pi i}{2}}f(z)
&=\frac{i}{2}\frac{\frac{\pi i}{2}-1}{\frac{\pi i}{2}a+1}\frac{1}{1-\frac{2}{\pi i}}\\
\operatorname*{Res}_{z=-\frac{\pi i}{2}}f(z)
&=\frac{i}{2}\frac{\frac{\pi i}{2}+1}{\frac{\pi i}{2}a-1}\frac{1}{1+\frac{2}{\pi i}}\\
\operatorname*{Res}_{z=-\frac{1}{a}}f(z)
&=-\frac{i}{2}u'(a)\left(\frac{1}{u(a)+\ln\left(\frac{\pi}{2}\right)-\pi i}+\frac{1}{u(a)+\ln\left(\frac{\pi}{2}\right)+\pi i}\right)\\
\end{align}
where $u(a)=\ln{a}-\dfrac{1}{a}$. By the residue theorem,
\begin{align}
\oint_{\gamma}f(z)\ dz
&=2\pi i\sum_{z_k\in\left\{-a^{-1}, \pm\pi i/2\right\}}\operatorname*{Res}_{z=z_k}f(z)\\
&=\pi\left(\frac{2\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)}{\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}u'(a)-\frac{2\pi^2a}{\pi^2a^2+4}\right)
\end{align}
Parameterisation of the Contour Integral:
We take the argument of $z$ to be $0$ above the branch cut, and $2\pi$ below the branch cut. Also, the contribution from the big arc is clearly $2\pi i\times\dfrac{i}{2}\times\dfrac{1}{a}\times (1+1)=-\dfrac{2\pi}{a}$. Taking all of these points into consideration, we eventually arrive at
\begin{align}
\oint_\gamma f(z)\ dz+\frac{2\pi}{a}
&=\small\frac{i}{2}\int^\infty_0\frac{x-1}{1+ax}\left(-\frac{1}{x-\ln|x|-2\pi i+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{x-\ln|x|+2\pi i+\ln\left(\frac{\pi}{2}\right)+\pi^2}\right)\ dx\\
&=2\pi\int^\infty_0\frac{x-1}{\left(x-\ln{x}+\ln\left(\frac{\pi}
{2}\right)\right)^2+4\pi^2}\frac{dx}{1+ax}\\
\end{align}
Obtaining the Closed Form:
Integrating with respect to $a$, we obtain
\begin{align}
\small\int^\infty_0\frac{2\pi\left(1-\frac{1}{x}\right)\ln(1+ax)}{\left(x-\ln{x}+\ln\left(\frac{\pi}{2}\right)\right)^2+4\pi^2}\ dx
&\small=\ \pi\int\left(\frac{2\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)}{\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}u'(a)-\frac{2\pi^2a}{\pi^2a^2+4}+\frac{2}{a}\right)\ da\\
&=\small\pi\left(\ln\left(\left(u(a)+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2\right)-\ln\left(\pi^2a^2+4\right)+\ln{a^2}\right)+\text{const.}\\
&=\small\pi\ln\left(\frac{\left(\ln{a}-\frac{1}{a}+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}{\pi^2+\frac{4}{a^2}}\right)+\text{const.}
\end{align}
Letting $a\to 0$, we find that the constant term is $\pi\ln{4}$. Plugging in $a=1$ and integrating by parts, we finally arrive at the closed form.
\begin{align}
\int^\infty_0\frac{2\pi\left(1-\frac{1}{x}\right)\ln(1+x)}{\left(x-\ln{x}+\ln\left(\frac{\pi}{2}\right)\right)^2+4\pi^2}\ dx
&=\int^\infty_0\arctan\left(\frac{2\pi}{x-\ln{x}+\ln\left(\frac{\pi}{2}\right)}\right)\frac{dx}{1+x}\\
&=\left.\pi\ln\left(\frac{\left(\ln{a}-\frac{1}{a}+\ln\left(\frac{\pi}{2}\right)\right)^2+\pi^2}{\frac{\pi^2}{4}+\frac{1}{a^2}}\right)\right|_{a=1}\\
&=\color{red}{\pi\ln\left(\frac{\ln^2\left(\frac{\pi}{2}\right)-2\ln\left(\frac{\pi}{2}\right)+1+\pi^2}{\frac{\pi^2}{4}+1}\right)}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1150822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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If $n$ is an odd integer, does $x+y$ divide $x^n + y^n$? I believe that the answer is yes.
Here's my thinking:
*
*$x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$
*$-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$
*So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c})$ with $c$ increasing by $1$ at each step and
*Since $n$ is odd, eventually, we get: $(-1)^c(x^{c}y^{c+1} + x^{c}y^{c+1}) = (x+y)(-1)^c(x^{c}y^{c})$
Am I correct? Am I making any mistakes in my thinking?
If this is correct, is the standard formula:
$$x^n + y^n = (x+y)(x^{n-1} -x^{n-2}y -xy^{n-2} + \dots + (-1)^{\frac{n-1}{2}}x^{\frac{n-1}{2}}y^{\frac{n-1}{2}} + y^{n-1})$$
Thanks,
-Larry
Edit: Fixed the equation based on comment.
| Your reasoning is fine.
A more general result is that for any positive integer $n$, and integers $u,v$, $u^n-v^n$ is divisible by $u-v$. This theorem is easier to prove by induction since you are proving for all $n$.
Your theorem then follows when $n$ is odd by setting $u=x,v=-y$ so $u-v=x+y$ and $u^n-v^n=x^n+y^n$.
| {
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"question_score": "2",
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} |
Find the domain of this function in the x,y,z graph Find the domain of this function in set notation:
$f(x,y)=\dfrac{\sqrt{x^2+y^2+4}}v$
$v=y-x$
Could I say that $x^2+y^2+4 \geq 0$ and $y-x \geq 0$ and based off those statements put that into set notation? and get:
$D= \{(x,y)|x^2+y^2+4 \neq 0$ and $y-x \neq 0\}$
| Since $v \neq 0$ by definition, then $y-x \neq 0$ ---> $y \neq x$. Therefore, the line y= x is excluded from the domain of f. Also, since $\sqrt{x^2+y^2+4}\geq 0$ and $y \neq x$, $\sqrt{x^2+y^2+4}$. So the domain is $ D = \{(x,y)| \sqrt{x^2+y^2+4}\geq 0$ where $y\neq x$} Surb gives an excellent sketch of the domain above in his/her reply.
| {
"language": "en",
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"source": "stackexchange",
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} |
Solve the quadratic congruences, or show them to be unsolvable.
Solve the following congruences, or show them to be unsolvable:
$(a) 3x^2 - 5x +7 \equiv 0 \mod 13$
Since $9 \cdot 3 \equiv 1 \mod 13$, $9 \cdot 5 \equiv 6 \mod 13$ and $9 \cdot 7 \equiv 11 \mod 13$ we can multiply the above congruence by $9$ to obtain $x^2 - 6x + 11 \equiv 0 \mod 13$. Completing the square, we obtain $x^2 - 6x + 9 = (x-3)^2 \equiv -2 \mod 13$. Since the congruence $x^2 \equiv -2 \mod 13$ has no solutions, it follows that $(x-3)^2 \equiv -2 \mod 13$ has no solutions. $\textbf{Is this correct?}$
$(b) 5x^2 - 6x + 2 \equiv 0 \mod 13$
Since $8 \cdot 5 \equiv 1 \mod 13, 8 \cdot 6 \equiv 9 \mod 13,$ and $16 \equiv 3 \mod 13$, we can multiply the congruence $(b)$ by $8$ to obtain $x^2 - 9x +3 \equiv 0 \mod 13$. Completing the square, we obtain $(x - \frac{9}{2})^2 \equiv \frac{69}{4} \mod 13$. $\textbf{How do I proceed from here?}$
| About b). Yes multiplication by 8 is correct. The polynomial you get is the same as $ x^2 +4 x + 3 \equiv 0 \pmod {13} $. Clearly you get $(x+1)(x+3) \equiv 0 \pmod {13} $. So $ x=10$ or $ x=12 \pmod {13} $.
| {
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Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \frac{1}{4} & \geq \frac{7}{12}\\
\frac{5}{6} & \geq \frac{7}{12}
\end{align*}
therefore base case is true.
Let's assume it is a true statement for $p(n+1)$ for all $n$ greater than or equal to $2.$
\begin{align*}
\frac{1}{(n+1)+1} + \frac{1}{(n+1)+2} + \frac{1}{2(n+1)} \geq \frac{7}{12}\\
\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{2n+2} \geq \frac{7}{12}
\end{align*}
then I wrote that if $n$ is greater than or equal to $2$ is a true statement then the inequality is true. This is a proof by induction and I'm not sure if I concluded properly
| what would be p(n+1)? think about where that falls in your solution. p(n+1) is 1/(2n+2)
| {
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How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? How can I prove that
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$
I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
| $$
\frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1}
$$
So
$$
\begin{align}\sum_n^\infty \frac{1}{n(n+1)} &= \left( \frac{1}{1}-\frac{1}{2} \right)
+ \left( \frac{1}{2}-\frac{1}{3} \right) + \cdots\\ &= \frac{1}{1} + \left( -\frac{1}{2}+\frac{1}{2} \right) + \left( -\frac{1}{3}+\frac{1}{3} \right) + \cdots \\
&=1+ 0 + 0 + 0 + \cdots = 1
\end{align}
$$
| {
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Composition of two rotations of the same angle $\alpha$ fixing points $a,b \in{S^2}$ Let $g,h$ be rotations of the same angle $\alpha$ around fixed points $a,b \in{S^2}$. Show $gh$ fixes $c \in{S^2}$ on the great circle that forms the perpendicular bisector of the segment $ab$, such that $ac$ and $bc$ make an angle $\alpha/2$ with $ab$.
All I know is that two rotations of the same angle conjugate, but othe than that i don't know a lot. $gh$ is the composition of two rotations and hence is a rotation, so by definition must fix a point (technically antipodal points), but i can't see how to show it lies n this great circle. It makes intuitive sense that the point is in some way an 'average' of points $a$ and $b$, but i cannot see how one would show this.
Any pointers in the right direction are greatly appreciated.
| Let
$$
\mathbf{K_a}=
\left[\begin{array}{ccc}
0& -a_3 & a_2 \\
a_3 & 0 & -a_1 \\
-a_2 & a_1 & 0
\end{array}\right]
$$
where $a = [a_1, a_2, a_3]$
Then
$$
\mathbf{R_a} = \mathbf{I} + (\sin\alpha) \mathbf{K_a} + (1-\cos\alpha)\mathbf{K_a}^2
$$
is the rotation by the vector $a$. (This is called Rodrigues' Formula.)
You can, with this, write out the formula for $K_b$ as well, and multiply $K_b K_a$ by $a+b$ (which is just twice the average) to see that it fixes the correct vector. It helps to remember that $K_x x = 0$ for any vector $x$. Let's just do it. I'm going to let $c$ and $s$ denote the cosine and sine of $\alpha$.
\begin{align}
\mathbf{R_b}\mathbf{R_a}(\mathbf{a}+\mathbf{b})
&= (\mathbf{I} + (\sin\alpha) \mathbf{K_b} + (1-\cos\alpha)\mathbf{K_b}^2)(\mathbf{I} + (\sin\alpha) \mathbf{K_a} + (1-\cos\alpha)\mathbf{K_a}^2)(\mathbf{a}+\mathbf{b}
)\\
&= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)(\mathbf{I}(\mathbf{a}+\mathbf{b}
) + s \mathbf{K_a}(\mathbf{a}+\mathbf{b}
) + (1-c)\mathbf{K_a}^2 (\mathbf{a}+\mathbf{b}
)\\
&= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\mathbf{a}+\mathbf{b}
) + s \mathbf{K_a}\mathbf{b}
+ (1-c)\mathbf{K_a}^2 \mathbf{b}
)
\end{align}
Now we use the fact that $K_a \mathbf{u}
= \mathbf{a} \times \mathbf{u}
$ for any $\mathbf{u}$, and similarly for $K_b$ to get
$$
\newcommand{\a}{\mathbf{a}}
\newcommand{\b}{\mathbf{b}}
\newcommand{\u}{\mathbf{u}}
\newcommand{\v}{\mathbf{v}}
$$
\begin{align}
\mathbf{R_b}\mathbf{R_a}(\a+\b)
&=(\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\a+\b) + s \mathbf{K_a}\b + (1-c)\mathbf{K_a}^2 \b)\\
&= (\mathbf{I} + s\mathbf{K_b} + (1-c)\mathbf{K_b}^2)((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b))\\
&= \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right] + s \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right] + (1-c) \b \times \left( \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)(\a \times (\a \times \b)) \right]\right )
\end{align}
Ugh. Pretty ugly. Let's let $\u$ denote $\a \times (\a \times \b)$, and $\v$ denote $\b \times (\b \times \a)$. Then use the fact that $\mathbf{w \times w} = 0$ for any vector $\mathbf{w}$, repeatedly, and we get
\begin{align}
\mathbf{R_b}\mathbf{R_a}(\a+\b)
&= \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + s \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b\ ) + s \a \times \b + (1-c)\u \right]\right ) \\
&= \left[ (\a+\b) + s \a \times \b + (1-c)\u \right] + \left[ (s \b \times (\a+\b) + s^2 (\b \times \a) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ ((\a+\b) + s \a \times \b + (1-c)\u \right]\right ) \\
&= \left[\a+\b + s \a \times \b + (1-c)\u \right] + \left[ -s \a \times \b - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right]\right )
\end{align}
And at this point, we have our first cancellation -- the second and fourth terms -- so we cancel them and further expand cross products in the latter terms:
\begin{align}
\mathbf{R_b}\mathbf{R_a}(\a+\b)
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left( \b \times \left[ (\a+\b) + s \a \times \b + (1-c)\u \right]\right ) \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) + s \b \times (\a \times \b) + (1-c)\b \times\u \right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \b \times (\b \times \a) + (1-c)\b \times\u \right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 (\a \times \b) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ + s^2 (\b \times \a) \times \b + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \b \times (\b \times \a) + (1-c)s \b \times \u \right] + (1-c) \b \times \left[ (\b \times\a) - s \v + (1-c)\b \times\u \right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \v + (1-c)s \b \times \u \right] + (1-c) \left[ \b \times(\b \times\a) - s \b \times\v + (1-c)\b \times (\b \times\u )\right] \\
&= \left[\a+\b + (1-c)\u \right] + \left[ - s^2 \v + (1-c)s \b \times \u \right] + (1-c) \left[ \v - s \b \times\v + (1-c)\b \times (\b \times\u )\right] \\
\end{align}
Now it's a matter of figuring out simpler expressions for $\b \times \u$ and things like that. I suggest that you define $\a^\perp$ to be $\a \times (\a \times \b)$, which is a vector perpendicular to $\a$ in the plane spanned by $\a$ and $\b$, and define $\b^\perp$ similarly. You can then use these to simplify things like $\b \times \u$.
Processing each equation I write is taking longer and longer, and I can't bring myself to write out any more...but since this is your problem, I'll bet that you can do so.
This isn't a very elegant solution, I admit...but it'll get you there.
| {
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If the sum of $n$ terms of an A.P. is $2n+3n^2$, find the $r^{th}$ term. If the sum of $n$ terms of an Arithmetic Progression is $2n+3n^2$, find the $r^{th}$ term.
Note: This question is from the book Higher Algebra by H.S. Hall & S.R. Knight and its answer is $6r-1$. (As listed at the end of the book).
What I've already done:
$$
S_n = 2n+3n^2\\
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}\\2n+3n^2=\frac{n}{2}\left \{ 2a+(n-1)d \right \}
$$
| We know that the sum of the first $n$ terms is $S_n = 2n + 3n^2$, so the sum of the first $(n - 1)$ terms is $S_{n-1} = 2(n - 1) + 3(n - 1)^2$. But if the $n$th term is $t_n$, then recall that $S_{n-1} + t_n = S_n$. Hence, it follows that:
\begin{align*}
t_n
&= S_n - S_{n-1} \\
&= (2n + 3n^2) - (2(n - 1) + 3(n - 1)^2) \\
&= (2n + 3n^2) - (2n - 2 + 3(n^2 - 2n + 1)) \\
&= (2n + 3n^2) - (2n - 2 + 3n^2 - 6n + 3) \\
&= (2n + 3n^2) - (3n^2 - 4n + 1) \\
&= 6n - 1 \\
\end{align*}
as desired.
Remark: Notice that the formula for $S_n$ gives us a lot of information. Indeed, we didn't need to know that the sequence was an arithmetic progression, and we didn't need to memorize any fancy formulas.
| {
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Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| For the equation.
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$
The easiest way is to factor the square.
$$p^2=ks$$
Then decisions can be recorded.
$$x=p$$
$$y=s-p$$
$$z=p-k$$
For the more General case formula there. Number of solution for $xy +yz + zx = N$
| {
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Technique to calculate rank of this matrix. 1) How can we calculate the rank of this matrix for $n > 3$ ?
\begin{matrix}
1^2 & 2^2 & 3^2 & \cdots & n^2 \\
2^2& 3^2 & 4^2 &\cdots & (n+1)^2 \\
3^2& 4^2 & 5^2 &\cdots & (n+2)^2 \\
\vdots&\vdots &\vdots &\ddots & \vdots\\
n^2& (n+1)^2 & (n+2)^2 &\cdots &(2n-1)^2
\end{matrix}
2) What are some of the tricks one must learn to calculate rank of a matrix ? Can anyone summarize the important ones ?
| We claim that the determinant of matrix $M$ given by $M_{i,j}=(a_i+b_j)^m$ where $i,j\in\mathbb{N}_n$ is $0$ if $n>m+1$. We have
$$
\begin{aligned}
\det(M)
&=\begin{vmatrix}
(a_1+b_1)^m & (a_1+b_2)^m & \ldots & (a_1+b_n)^m\\
(a_2+b_1)^m & (a_2+b_2)^m & \ldots & (a_2+b_n)^m\\
\ldots & \ldots & \ldots & \ldots \\
(a_n+b_1)^m & (a_n+b_2)^m & \ldots & (a_n+b_n)^m\\
\end{vmatrix}\\
&=\begin{vmatrix}
\sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_1^{m-k_1} & \sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_2^{m-k_1} & \ldots & \sum\limits_{k_1=0}^m\binom{m}{k_1} a_1^{k_1}b_n^{m-k_1}\\
\sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_1^{m-k_2} & \sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_2^{m-k_2} & \ldots & \sum\limits_{k_2=0}^m\binom{m}{k_2} a_2^{k_2}b_n^{m-k_2}\\
\ldots & \ldots & \ldots & \ldots\\
\sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_1^{m-k_n} & \sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_2^{m-k_n} & \ldots & \sum\limits_{k_n=0}^m\binom{m}{k_n} a_n^{k_n}b_n^{m-k_n}\\
\end{vmatrix}\\
&=\sum\limits_{k_1=0}^m\sum\limits_{k_2=0}^m\ldots\sum\limits_{k_n=0}^m\binom{m}{k_1}\binom{m}{k_2}\ldots\binom{m}{k_n}
\begin{vmatrix}
a_1^{k_1} b_1^{m-k_1} & a_1^{k_1} b_2^{m-k_1} & \ldots & a_1^{k_1} b_n^{m-k_1} \\
a_2^{k_2} b_1^{m-k_2} & a_2^{k_2} b_2^{m-k_2} & \ldots & a_2^{k_2} b_n^{m-k_2} \\
\ldots & \ldots & \ldots & \ldots \\
a_n^{k_n} b_1^{m-k_n} & a_n^{k_n} b_2^{m-k_n} & \ldots & a_n^{k_n} b_n^{m-k_n} \\
\end{vmatrix}\\
&=\sum\limits_{k_1=0}^m\sum\limits_{k_2=0}^m\ldots\sum\limits_{k_n=0}^m\binom{m}{k_1}\binom{m}{k_2}\ldots\binom{m}{k_n}a_1^{k_1}a_2^{k_2}\ldots a_n^{k_n} b_1^m b_2^m\ldots b_n^m
\begin{vmatrix}
b_1^{-k_1} & b_2^{-k_1} & \ldots & b_n^{-k_1} \\
b_1^{-k_2} & b_2^{-k_2} & \ldots & b_n^{-k_2} \\
\ldots & \ldots & \ldots & \ldots \\
b_1^{-k_n} & b_2^{-k_n} & \ldots & b_n^{-k_n} \\
\end{vmatrix}\\
\end{aligned}
$$
Denote
$$
M_{k_1,k_2,\ldots,k_n}=
\begin{vmatrix}
b_1^{-k_1} & b_2^{-k_1} & \ldots & b_n^{-k_1} \\
b_1^{-k_2} & b_2^{-k_2} & \ldots & b_n^{-k_2} \\
\ldots & \ldots & \ldots & \ldots \\
b_1^{-k_n} & b_2^{-k_n} & \ldots & b_n^{-k_n} \\
\end{vmatrix}
$$
Note that if $k_i=k_j$ for some $i,j\in\mathbb{N}_n$, then the determinant $M_{k_1,k_2,\ldots,k_n}$ have to equal columns and therefore equals zero. Now note that if $n>m+1$ there alway exist $i,j\in\mathbb{N}_n$ such that $k_i=k_j$, so $M_{k_1,k_2,\ldots,k_n}=0$ for all tuples $(k_1,k_2,\ldots,k_n)\in \{0,\ldots,m\}^n$ and as the consequence $\det(M)=0$.
Now we return to the original problem. For $n=1,2,3$ we have the following matrices
$$
A=(1)\qquad
A=\begin{pmatrix}1&4\\4&1\end{pmatrix}\qquad
A=\begin{pmatrix}1&4&9\\4&9&16\\6&16&25 \end{pmatrix}
$$
respectively. Their ranks are $1$, $2$ and $3$ respectively. If $n>3$ by previous claim all minors of $A$ of size greater than $3$ are zero, hence $\operatorname{rank}(A)\leq 3$. Since $n>3$, then $A$ contains minor
$$
\begin{pmatrix}1&4&9\\4&9&16\\6&16&25 \end{pmatrix}
$$
whose rank equal to $3$, so $\operatorname{rank}(A)=3$ for $n>3$. Finally
$$
\operatorname{rank}(A)=
\begin{cases}
1 & \quad\mbox{if}\quad n=1\\
2 & \quad\mbox{if}\quad n=2\\
3 & \quad\mbox{if}\quad n>2\\
\end{cases}
$$
| {
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Maximum of a trigonometric expression Maximize $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$ I broke it in 2 parts but then realized they don't have their maxima at the same points. So I am stuck. This is what I did. I wrote it as $$3\cos x+4\sin x+2\sin x(24\cos x+7\sin x)$$ Now we can maximize the first two terms and the terms inside the bracket.
| Note that if $ \tan (x) = \frac {3}{4} $, then $ \tan (2x) = \frac {24}{7} $
With $ 48 \sin x \cos x = 24 \sin (2x) $ and $ 14 \sin^2 x = 7 - 7 \cos (2x) $
And consider the Pythagoras Triplets $ (3,4,5), (7,24,25) $
We have
$ f(x) = ( 4 \sin x + 3 \cos x ) + (24 \sin (2x) - 7 \cos (2x) ) + 7 $
Express them in terms of $ R \sin (\theta \pm \alpha ) $, link for reference
$ \text{max} (f(x)) = 5 + 25 + 7 = 37 $
| {
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AIME 2013 Solutions (divisiblity) Problem 2
Find the number of five-digit positive integers, $n$, that satisfy the following conditions:
(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$
Obviously, if $n$ satisfies divisiblity by five and takes the form,
$$n = abcda$$ then $a=5$, deductively,
$$n = 5bcd5$$ And:
$$10 + b + c + d \equiv 0 \pmod{5} \implies b + c + d \equiv -10 \equiv 0 \pmod{5}$$
What do I do next? HINTS are appreciated!
| Well you need to find the number of solution to the following equations:
$$b + c + d = 0$$
$$b + c + d = 5$$
$$b + c + d = 10$$
$$b + c + d = 15$$
$$b + c + d = 20$$
$$b + c + d = 25$$
I would recommend you using generating funtions. So since $0\ge b,c,d \ge 9$, consider the following generating function:
$$(1 + x + x^2 + ... + x^9)^3 $$
$$=\left(\frac{1-x^{10}}{1-x}\right)^3 = \left(\binom{2}{0}x^0 + \binom{3}{1}x^1 + \binom{4}{2}x^2 +...\right)(1 - 3x^{10} + 3x^{20} -x^{30})$$
To find the number of solution to let's say $b+c+d=k$, check the coefficient in front of $x^k$.So for $k=0$, there's only one solution, for $k=5$, there are $\binom{7}{5} = 21$ solutions. But pleas be careful for the others equation. For example for $k=10$, there are:
$$\binom{12}{10} - 3\binom{2}{0} = 66-3 = 63 \text{ solutions}$$
For $k=25$, you'll have:
$$\binom{27}{25} - 3\binom{17}{15} + 3\binom{7}{5} = 351 - 3\cdot136 + 3\cdot21=6 \text{ solutions}$$
Now calculate for $k=10,15,20$ and add all the numbers and you have the final solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Sum of the roots of an equation.
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.
I used Vieta's to get the wrong answer.
If $$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3$$
And $p, b, c$ are the roots then: (here let $a = a_3$ the high degree coefficient)
$$f(x) = a(x - p)(x - b)(x - c) = ax^3 - apx^2 - acx^2 + acpx - abx^2 + abpx + abcx - abcp$$
$$ = ax^2 + x^2\overbrace{(-ap - ac - ab)}^{a_2} + x\overbrace{(acp + abp + abc)}^{a_1} - abcp$$
$$f(x) = 8x^3 - 3x^2 - 3x - 1$$
We are after $a_2$
$$-3 = -(ap + ac + ab) \implies p + c + b = 3/a = \frac{3}{8}$$
Which is wrong because, the answer is supposed to be an integer, and the answer is: $$A=98$$
| We have:
$$ p(x)=8x^3-3x^2-3x-1 = 9x^3-(x+1)^3 $$
hence $p(x)=0$ iff
$$ \left(1+\frac{1}{x}\right)^3 = 9 $$
and the only real root of the above equation is given by
$$ x = \frac{1+\sqrt[3]{81}+\sqrt[3]{9}}{8} $$
leading to the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(x) = x^3$ is continuous at $ x=2.$ I need to give an $\epsilon, \delta$ proof that $x^3$ is continuous at $x=2$. I got that $|x^3 -8| = |x-2| |x^2 +2x +4|$, and I need to assume $|x-2| < \delta$. How can I choose $\delta$ in terms of $\epsilon$ so that $|x-2| |x^2 +2x +4| < \epsilon$ ?
We learned the "adding in zero" trick, but that gets me $\mid x^2 +2x+4 \mid = \mid x^2 - 4x + 6x + 4 \mid \leq \mid (x-2)^2\mid + \mid 6x \mid$. Is that the right step to take?
We haven't learned about bounding the function yet. In our in-class example, we learned how to make the terms cancel nicely to constants using the "adding in zero". Is that not possible here?
| Let $\epsilon > 0$ and $\delta < \min{1, \epsilon/19}$. Let $x$ be within $\delta$ of $2$. Observe that
$$
|x^3 - 8 | = |x-2||x^2+2x+4| = |x-2| |(x-2)^2 + 6x| = |x-2| |(x-2)^2 + 6(x-2) + 12|.
$$
Applying the triangle inequality to the final absolute value, we find that
$$
|x^3 - 8 | \leq |x-2|^3 + 6 |x-2|^2 + 12 |x-2| < \delta^3 + 6 \delta ^2 + 12 \delta.
$$
Since $\delta < 1$, $\delta^n < \delta$ for all $n$. Thus
$$
|x^3 - 8 | < \delta^3 + 6 \delta ^2 + 12 \delta < 19 \delta \leq \epsilon.
$$
So the function is continuous at $2$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
3 Gents, 3 Ladies, and one round table The question is how many ways the aforesaid ladies and gentlemen be seated around the round table, such that two and only two ladies sit together.
The answer says 72.
My understanding:
(#ways to arrange 3 gentlemen at the table) // (3-1)!
*(#ways to select 2 out of 3 ladies) // 3P2
*(#ways to arrange the pair of ladies among themselves) // 2
= 2*6*2 = 24
So where am i going wrong ?
EDIT :
(#ways to arrange 3 gentlemen at the table) // (3-1)!
*(#ways to select 2 out of 3 ladies) // 3P2
*(#ways to select 2 out of 3 available positions in between the men) // 3P2
= 2*6*6 = 72
Am i understanding correctly ?
| There are $\binom{3}{2}$ to select the two women who sit together and $2!$ ways to arrange them within that block. That leaves four open seats, two of which are adjacent to the two women who sit together, so there are $2$ ways to sit the third woman in a seat that is not adjacent to the other two women and $3!$ ways to arrange the men in the remaining seats. Thus, there are
$$\binom{3}{2} \cdot 2! \cdot 2 \cdot 3! = 3 \cdot 2 \cdot 2 \cdot 6 = 72$$
ways to seat three men and three women at a round table if exactly two of the women sit together.
Seating the men first is more difficult. You have to choose which two men are to sit three seats apart (to accommodate the pair of women who sit together), arrange those two men in those seats, then choose which of the two remaining seats will be filled by the third man. This can be done in $\binom{3}{2} \cdot 2! \cdot 2$ ways. You can then choose two of the three women to sit together, arrange them in two ways, and seat the third woman in the remaining seat, which can be done in $\binom{3}{2} \cdot 2! \cdot 1$ ways. Thus, the number of possible arrangements is
$$\binom{3}{2} \cdot 2! \cdot 2 \cdot \binom{3}{2} \cdot 2! \cdot 1 = 3 \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 1 = 72$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Finding the Laurent series of a given function How do I find the Laurent series expansion for the function
$\frac{z}{(z+1)(z+2)}$
which converges for $1 < |z| < 2$ and diverges elsewhere.
I have done the partial fractions for $f(z) = \frac{2}{z+2} - \frac{1}{z+1}$.
| As you found already, $f(z) = \frac{2}{2+z} - \frac{1}{1+z}$
Let's look at each fraction separately:
$-\frac{1}{1+z}$ is analytic in $|z| \gt 1$ and since $|\frac{1}{z}| \lt 1$ we have:
$$
-\frac{1}{1+z} = -\frac{1}{z}\cdot\frac{1}{1-(-\frac{1}{z})} = -\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}-\dots) = -\frac{1}{z}+\frac{1}{z^2}-\frac{1}{z^3}+\dots
$$
$\frac{2}{2+z}$ is analytic in $|z|\lt2$, we have $|\frac{z}{2}| \lt 1$ and we can represent it with it's Taylor series:
$$
\frac{2}{2+z} = \frac{1}{1-(-\frac{z}{2})} = 1 -\frac{z}{2}+\frac{z^2}{4}-\dots
$$
Therefore the Laurent series:
$$
\dots -\frac{1}{z^3}+\frac{1}{z^2}-\frac{1}{z}+1-\frac{z}{2}+\frac{z^2}{4}-\dots
$$
Converges in $1 \lt |z| \lt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The number of ways of writing an integer as a sum of two squares Given an integer $m=pq$, where $p,q$ are both primes such that $p\equiv 1 \pmod{4}, q\equiv 1 \pmod{4}$.
It is known that $p$ can be written as a sum of two squares (of positive integers) in a unique way, and the same for $q$. Prove that $m$ can be written as a sum of two squares (of positive integers) in exactly two distinct ways.
Attempt
Notice that for positive integers $u,v,A,B$, we have
\begin{align*}(u^2+v^2)(A^2+B^2)&=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2\end{align*}
Therefore if $p=a^2+b^2,q=c^2+d^2$, then
\begin{align*}m&=(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(bc-ad)^2=(bc+ad)^2+(ac-bd)^2\end{align*}
meaning that $m$ can be written as a sum of two squares (of positive integers) in at least two distinct ways.
How do I prove that there doesn't exist a third way of writing $m$ as a sum of two squares (of positive integers)?
| @Elaqqad
This is my solution based on your idea and notations. Please point out wherever there is a mistake.
There exist coprime integers $x,y,z,t$ such that
\begin{align*}
&a-c=xy\\&
a+c=zt\\&
d-b=xz\\&
d+b=yt
\end{align*}
Therefore we have that
\begin{align*}
&a=\frac{1}{2}(xy+zt)\\&
c=\frac{1}{2}(zt-xy)\\&
d=\frac{1}{2}(xz+yt)\\&
b=\frac{1}{2}(yt-xz)
\end{align*}
Consequently, we obtain that
\begin{align*}
&a^2+c^2=\frac{1}{2}(x^2y^2+z^2t^2)\\&
b^2+d^2=\frac{1}{2}(x^2z^2+y^2t^2)
\end{align*}
which imply that
\begin{align*}
&2pq=(a^2+b^2)+(c^2+d^2)=(a^2+c^2)+(b^2+d^2)=\frac{1}{2}(x^2+t^2)(y^2+z^2)
\end{align*}
So we have that
\begin{align*}pq=\frac{1}{4}(x^2+t^2)(y^2+z^2)\end{align*}
meaning that there are three possible cases:
\begin{align*}
&\{p,q\}=\{\frac{1}{2}(x^2+t^2),\frac{1}{2}(y^2+z^2)\}\\&
\{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}\\&
\{p,q\}=\{(x^2+t^2),\frac{1}{4}(y^2+z^2)\}
\end{align*}
Case 1.$\{p,q\}=\{\frac{1}{2}(x^2+t^2),\frac{1}{2}(y^2+z^2)\}$:
If a prime $p_0$ satisfies that $p_0\equiv 1 \pmod 4$, then $p_0=m^2+n^2$ for some positive integers $m,n$ with different parity.
Using $(u^2+v^2)(A^2+B^2)=(uA+vB)^2+(vA-uB)^2=(vA+uB)^2+(uA-vB)^2$,
we obtain that $2p_0=(1^2+1^2)(m^2+n^2)=(m+n)^2+(m-n)^2$. The right-hand side is the unique way of writing $2p_0$ as a sum of two squares.
Up to permutation, we may assume that $p=\frac{1}{2}(x^2+t^2),q=\frac{1}{2}(y^2+z^2)$.
If $p=u^2+v^2,q=A^2+B^2$, where $u>v,A>B$,and $u>A)$, then we obtain that
$t>x,z>y$, so $t=u+v,x=u-v,z=A+B,y=A-B$.
\begin{align*}
&a=\frac{1}{2}(xy+zt)=uA+vB\\&
c=\frac{1}{2}(zt-xy)=vA+uB\\&
d=\frac{1}{2}(xz+yt)=uA-vB\\&
b=\frac{1}{2}(yt-xz)=vA-uB
\end{align*}
Case 2.$\{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}$:
If a prime $q_0$ satisfies that $q_0\equiv 1 \pmod 4$, then $q_0=m^2+n^2$ for some positive integers $m,n$ with different parity.
Likewise, we obtain that $4q_0=(0^2+2^2)(m^2+n^2)=(2m)^2+(2n)^2$. The right-hand side is the unique way of writing $4q_0$ as a sum of two squares.
Using the same notations, we have that $t=2u,x=2v,z=A,y=B$, then
\begin{align*}
&a=\frac{1}{2}(xy+zt)=vB+uA\\&
c=\frac{1}{2}(zt-xy)=uA-vB\\&
d=\frac{1}{2}(xz+yt)=vA+uB\\&
b=\frac{1}{2}(yt-xz)=uB-vA
\end{align*}
Case 3.$\{p,q\}=\{\frac{1}{4}(x^2+t^2),(y^2+z^2)\}$:
This is analogous to case 2.
Consequently, we know that $\{(a,b),(c,d)\}=\{(uA+vB,vA-uB),(vA+uB,uA-vB)\}$, so the product of two distinct primes can be written as a sum of two squares in exactly two ways. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Minimise $x+y$ under the condition $2x+32y \leq 9xy$ where $x,y > 0$ Let $x$ and $y$ be positive real numbers such that $2x+32y\leq 9xy$. Then find the smallest possible value of $x+y$. I tried using AM-GM but it is difficult to find the case of equality. Is Cauchy-Schwarz also helpful? Thanks.
| here is how i approached. i will look at the boundary of $$2x + 32 y = 9xy $$ which is given by the hyperbola $$ y = \frac{2x}{9x - 32}, \frac{dy}{dx} = -\frac{64}{(9x-32)^2} $$ this hyperbola has slope of $-1$ at $$ -\frac{64}{(9x-32)^2} = -1 \to \left(x = \frac{40}9,y = \frac{10}9, x + y = \frac{50} 9\right) \text{ and} \left(x = \frac{24}9,y = -\frac{6}9, x + y = 2\right) $$ the second solution must be rejected because of the condition that $x > 0$ and $y > 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
A cubic factor of a nine degree polynomial The polynomial $x^3−3x^2+4x−1$ is a factor of $x^9+px^6+qx^3+r$. Find $p+q+r$. How do I start with the problem?
| As David wrote, the division must be done very carefully. To avoid it, we could use a real brute force approach writing $$\Delta=(x^9+px^6+qx^3+r)-(x^3−3x^2+4x−1)\sum_{i=0}^6 a_ix^i=0$$ This should give $$\Delta=(a_0+r)+(a_1-4 a_0) x+(3 a_0-4 a_1+a_2) x^2+ (-a_0+3 a_1-4
a_2+a_3+q)x^3+(-a_1+3 a_2-4 a_3+a_4) x^4+(-a_2+3 a_3-4 a_4+a_5) x^5+
(-a_3+3 a_4-4 a_5+a_6+p)x^6+(-a_4+3 a_5-4 a_6) x^7+(3 a_6-a_5) x^8+(1-a_6)
x^9=0$$ Now, we cancel the coefficient of each power of $x$ starting from the highest. This will give $a_6=1,a_5=3,a_4=5,a_3=p+4,a_2=3p-5,a_1=5p-26,a_0=-r$. After this quite laborious task, what is left is $$\Delta= (4 p+q+r-54)x^3+(-17 p-3 r+99)x^2 + (5 p+4 r-26)x=0$$ and again we cancel the coefficients to get three linear equations for the three unknowns $p,q,r$ and the solutions as already given in the previous answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to show $\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$ How to show
$$\sum_{n=0}^{\infty }\frac{1}{(24n+5)(24n+19)}=\frac{\pi }{336}(\sqrt{6}-\sqrt{3}-\sqrt{2}+2)$$
| hint: $\displaystyle \sum_{n=0}^\infty\dfrac{1}{(24n+5)(24n+19)} = \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\dfrac{1}{24n+5} - \dfrac{1}{24n+19}\right)= \dfrac{1}{14}\displaystyle \sum_{n=0}^\infty\left(\displaystyle \int_{0}^1\left(x^{24n+4} - x^{24n+18}\right)dx\right)= \dfrac{1}{14}\displaystyle \int_{0}^1\displaystyle \left(\sum_{n=0}^\infty\left(x^{24n+4} - x^{24n+18}\right)\right)dx$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Algebra of trigonometry and some limits I am singlehandedly trying to get through a book on divergent series, however one sequence is really strange and I cant seem to understand what the authors use to come to these conclusions!
Imagine the following is our sum. ($0 < x < 2{\pi}$)
As the authors attempt to determine the mean of the partial sums, they go on to do the following:
Is there any way to understand any of the following steps, including the determination of the limit as $n$ approaches positive infinity?
Thanks a lot for your help!
| let us pick a small $n$ and try what they are doing. i will pick $n = 10.$ we have
$$S_1 = \frac{\sin(3x/2) - \sin (x/2)}{2\sin x/2}, S_2 = \frac{\sin(5x/2) - \sin (x/2)}{2\sin x/2}, \cdots, S_{10}=\frac{\sin(21x/2) - \sin (x/2)}{2\sin x/2} $$
now we can compute $$\begin{align} S &= 2\sin^2(x/2)\left(S_1 + S_2 + S_3 + \cdots + S_{10}\right)\\ &= 2\sin(x/2)\sin(3x/2) + 2\sin(x/2)\sin(5x/2) + \cdots + 2\sin(x/2)\sin(21x/2) - 20\sin^2 (x/2)\\
& = (\cos(3x/2-x/2) -\cos(3x/2 + x/2)) + (\cos(5x/2-x/2) -\cos(5x/2 + x/2)) + \cdots + (\cos(21x/2-x/2) -\cos(21x/2 + x/2)) - 20\sin^2 (x/2)\\
&=\cos x-\cos(11x) - 20\sin^2 (x/2)\end{align}$$
so we have $$ \frac{S}{20\sin^2(x/2)} = \frac{S_1 + S_2 + S_3 + \cdots + S_{10}}{10} = \frac{\cos x - \cos(11x)}{20\sin(x/2)} -\frac 12$$
therefore we can see that $$ \frac{S_1 + S_2 + S_3 + \cdots + S_{n}}{n} = \frac{\cos x - \cos((n+1)x)}{2n\sin(x/2)} -\frac 12 \to -\frac 12 \text{ as } n \to \infty.$$
we have used the fact that $$\big|\cos x - \cos((n+1)x)\big| \le 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $f$ is differentiable at $x=4$ $f$ is continuous. $f$ is defined on $(0,1) \cup (1, \infty)$.
$$f(x) = \begin{cases} \frac{4\sqrt x}{(\ln x)^2} & \text{if } x\in (0,1) \cup (1,e^4),\\
\frac{e^2}{64} (ln x) (8-lnx) & \text{ if } x\in[e^4, \infty)
\end{cases}$$
Need to show that $f$ is differentiable at $x = e^4$
I tried differentiating and subbing in $e^4$ into $x$
$f'-(x) = 1/(4e^2)$
$f'+(x) = 0$
| i will try this. make a change of variable $$x = e^4(1+h).$$ we can now write
$$\begin{align} \ln x &= \ln \left(e^4(1+h) \right) = 4 + \ln(1+h) = 4 + h + \cdots\\
(\ln x)^{-2} &=(4+h \cdots)^{-2} = 4^{-2} - 24^{-3}h+ \cdots=\frac 1{16}-\frac h{32}+\cdots\\
\sqrt x &= e^2(1 + \frac 12 h + \cdots)= e^2 + \frac{e^2h}2+\cdots\\
\frac{e^2}{64}(\ln x)(8-\ln x) &= \frac{e^2}{64}(4 + h + \cdots)(4 - h + \cdots) = \frac{e^2}{4} + \cdots \tag 1
\end{align}$$
collecting all these into $$\frac{4\sqrt x}{(\ln x)^2} = 4(e^2 +\frac{e^2h}2+\cdots )(\frac 1{16}-\frac h{32}+\cdots) = 4(\frac{e^2}{16} + \cdots)=\frac{e^2}{4} + \cdots \tag 2$$
you can see from $(1)$ and $(2)$ that $f$ is differentiable at $x = 4$ and its derivative is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$ for $a,b,c>0$ If $a$, $b$, and $c$ are positive real numbers, show that
$$
\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \geq \sqrt{b^2 + bc + c^2}.
$$
What values of $a$, $b$, and $c$ ( or relation among them) make both sides equal ( makes it an equality)?
| Consider the points $A\equiv (0,0),B \equiv (a,0), C,D$ on the plane such that $AC=b$ and $\angle CAB=60^{\circ}$ and $AD=c$ and $\angle DAB=60^{\circ}$. Then $BC=\sqrt{a^2-ab+b^2}$ and $BD=\sqrt{a^2-ac+c^2}$. Now we also have $CD=\sqrt{b^2+bc+c^2}$. Consider the triangle $BCD$ and apply triangle inequality to get $BC+BD\geq CD$ which is the given inequality. For equality we need $B\in CD$. Finding the exact condition should be easy now.
| {
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"url": "https://math.stackexchange.com/questions/1187798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve trigonometric equations with sin and cos I have two equations for two unknowns $u_k$ and $v_k$:
$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0$
$\tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$
where $u_1$, $v_1$, $u_2$ and $v_2$ are parameters.
I managed to find a general solution for $v_k$ by expanding $\cos{(v_i - v_k)} = \cos{v_i} \cos{v_k} + \sin{v_i} \sin{v_k}$ and then dividing to obtain a general solution for $v_k$:
$\tan{v_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}}$
However, I cannot find a general algebraic solution for $u_k$ not containing $v_k$. I have only obtained the following identities
$\tan{u_k} = \frac{\sin{(v_2 - v_1)}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}} \cos{v_k}$
$\tan{u_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\sin{(v_2 - v_1)}} \sin{v_k}$
but have been unable to get rid of the $v_k$ from either of them as I always start with both a sine and a cosine of it. Squaring the equations did not help, too.
Is there a way to get $u_k$ as a function of just $u_1$, $v_1$, $u_2$ and $v_2$?
| $$\tan{u_1} \tan{u_k} + \cos{(v_1 - v_k)} = 0\\
\tan{u_2} \tan{u_k} + \cos{(v_2 - v_k)} = 0$$
You reached:
$$\tan{u_k} = \frac{\sin{(v_2 - v_1)}}{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}} \cos{v_k},\tan{u_k} = \frac{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}{\sin{(v_2 - v_1)}} \sin{v_k}$$
Try calculating: $\sin^2v_k+\cos^2v_k=1$:
$$\left(\frac{\tan u_k\sin(v_2-v_1)}{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}\right)^2+
\left(\frac{\tan u_k(\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1})}{\sin{(v_2 - v_1)}}\right)^2=1$$
So:
$$u_k=\arctan\left(\left(\frac{\sin(v_2-v_1)}{\tan{u_2} \cos{v_1} - \tan{u_1} \cos{v_2}}\right)^2+
\left(\frac{\tan{u_1} \sin{v_2} - \tan{u_2} \sin{v_1}}{\sin{(v_2 - v_1)}}\right)^2\right)^{-1/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differentiate. $y = (1 − \sec x) / \tan x$ I have no idea what I have done wrong. Please criticise.
Differentiate $y = \frac{1 - \sec x}{\tan x}$.
\begin{align*}
\frac{d}{dx} \left[ \frac{1 - \sec x}{\tan x} \right]
&= \frac{\tan x \frac{d}{dx} [ 1 - \sec x ] - (1 - \sec x) \frac{d}{dx} \tan x }{(\tan x)^2} \\
&= \frac{\tan x (- \sec x \tan x) - [(1 - \sec x) (-\csc^2 x)]}{(\tan x)^2} \\
&= \frac{-\tan^2 x \sec x + \csc^2 x - \sec x \csc^2 x}{(\tan x)^2}
\end{align*}
| Hint:
$$
\dfrac{d}{dx} \tan x= \sec^2 x
$$
Here you was wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$? I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
(6, 6, 5, 4),
(5, 6, 6, 4),
(6, 4, 5, 6),
(6, 5, 4, 6),
(4, 6, 6, 5),
(4, 5, 6, 6),
(4, 6, 5, 6),
(6, 4, 6, 5),
(6, 6, 4, 5),
(5, 4, 6, 6),
(5, 6, 4, 6)\}$$
But how would I calculate this?
I know we talk about permutation as order matters also I know that repetition is allowed (at least for $x = 6$).
| For a list of $n$ distinct elements the number of permutations if $n!$. If you have $4$ distinct elements, the number of permutations is $24$.
However in this case we have the number $6$ appearing twice. For every ordering of the four numbers $6,6',5,4$ there is another ordering switching the $6$ and the $6'$, so you need to divide $24$ by $2$ which gives $12$ as you've got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\sqrt{2}\cos\theta = 1$ and $\sqrt{2}\sin\theta = 1$, so the angle $\theta$ is $\frac{\pi}{4}$.
However, if we do $k=0$, then we get that one of the roots is $1$, which is obviously not true since $1^4 \neq 1 + i$. The professor says that the solutions are: $k=1: \cos(\frac{9\pi}{16}) + i\sin(\frac{9\pi}{16})$, $k=2: \cos(\frac{17\pi}{16}) + i\sin(\frac{17\pi}{16})$, and $k=3: \cos(\frac{25\pi}{16}) + i\sin(\frac{25\pi}{16})$. I plugged these into WolfRamAlpha and rose them to the $4$th power, but none of them return the form $1+i$.
What is incorrect about these steps?
| From the classic for $a+bi$ can be seen in polar form as
\begin{align}
a + b i = \sqrt{a^{2} + b^{2}} \, e^{i \tan^{-1}(b/a)}
\end{align}
for which
\begin{align}
1 + i = \sqrt{2} \, e^{i \tan^{-1}(1)} = \sqrt{2} \, e^{\pi i/4} = 2^{1/2} e^{\pi i/4 + 2n \pi i}
\end{align}
for $n$ being integer. Now
\begin{align}
(1+i)^{1/4} = 2^{1/8} e^{\pi i/16 + n\pi i/2} = 2^{1/8} \left[ \cos\left(\frac{(8n+1)\pi }{16} \right) + \sin\left(\frac{(8n+1)\pi }{16} \right) \right]
\end{align}
where $n = 0,1,2,3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Problem with definite integral $\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$ $$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better.
$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
$$u=1-2\sin x$$
$$du=-2\cos x dx$$
$$=-\frac{1}{2}\int_{1}^{0}\sqrt{u}du$$
$$=\frac{1}{2}\int_{0}^{1}\sqrt{u}du$$
$$=\left | \frac{u^{\frac{3}{2}}}{3} \right |_{0}^{1}$$
$$=\frac{1}{3}$$
My question is how to solve it by using $u=\cos x$. Can anyone show the solution for it? Thanks a lot!
| Consider the integral
\begin{align}
\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx
\end{align}
with the substitution $u = \cos(x)$. Making the desired substitution the integral becomes
\begin{align}
I = \int_{\sqrt{3}/2}^{1} u \sqrt{ \frac{1 - 2 \sqrt{1-u^{2}}}{1 - u^{2} } } du
\end{align}
Now make the substitution $t = 1 - u^{2}$ to obtain the integral
\begin{align}
I = \int_{0}^{1/2} \sqrt{1 - 2t} \, dt.
\end{align}
Making one last change of $y = 1-2t$ which leads to
\begin{align}
I = \frac{1}{2} \int_{0}^{1} \sqrt{y} \, dy = \frac{1}{3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: integral $$\frac{1 + t^2}{2(t + 2)(t + 3)}$$ or $$\frac{1 + t^2}{12 + 2t^2 + 10t}.$$
Could someone give me a hint?
Many thanks.
| Let $t = x - \tan^{-1}\frac15$ to integrate
\begin{align}
& \int\frac{1}{5\cos x+\sin x+7}~dx
= \int \frac1{\sqrt{26}\cos t +7}dt\\
=& \int \frac{2d(\tan\frac t2)}{(7+\sqrt{26})+(7-\sqrt{26})\tan^2\frac t2}\\
=& \frac2{\sqrt{23}}\tan^{-1} \left( \sqrt{\frac{ 7-\sqrt{26}}{7+\sqrt{26}}}\tan\frac t2 \right)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.
| Setting $u=\sqrt{x-1}$ gives you
$$ \int \frac{2u \, du}{(1+u^2)^2u}= 2\int \frac{du}{(1+u^2)^2} \tag{1}$$
There are two ways to do this: the first is to change variables again to $u=\tan{\theta}$, so $du=\sec^2{\theta} \, d\theta$:
$$ \begin{align*}
2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} &= 2\int \frac{\sec^2{\theta} \, d\theta}{(1+\tan^2{\theta})^2} \\
&= \int 2\cos^2{\theta} \, d\theta \\
&= \int (1+\cos{2\theta}) d\theta, \end{align*}$$
using the double-angle formula $\cos{2\theta}=2\cos^2{\theta}-1$.
$$ \int (1+\cos{2\theta}) = \theta + \frac{1}{2}\sin{2\theta}, $$
and substituting back, $\theta = \arctan{\sqrt{x-1}}$, and in particular,
$$ \int \frac{dx}{x^2\sqrt{x-1}} = \frac{1}{2} \sin{(2\arctan{\sqrt{x-1}})} + \arctan{\sqrt{x-1}}. $$
The first term can be simplified using the formula $\sin{(2 \arctan{t})} = \frac{2t}{1+t^2}$ to give the answer.
Alternatively:
Having got to (1), we can write $1=(1+u^2)-u^2$ integrate by parts:
$$ 2\int \frac{(1+u^2)-u^2}{(1+u^2)^2}\,du = \int \frac{2du}{1+u^2} - \int u\frac{2u}{(1+u^2)^2} \, du \\
= \int \frac{2du}{1+u^2} + \frac{u}{1+u^2} - \int \frac{du}{1+u^2} \, du, $$
and then proceed to substitute $u=\tan{\theta}$ in the first one to get the $\arctan$ term. This way, we avoided having to use lots of double angle formulae.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding x in matrix algebra The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.
$$
\left(
\begin{matrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
x & x & 3x & 1\\
0 & x & x & 0\\
\end{matrix}
\right)
$$
I got my answer to be $-2x^3 + 2x^3 = 0$, indicating that the matrix is not invertible because the determinant is a zero, regardless of the value of $x$.
However, I'm not certain whether I'm right after doing all the calculations. Could you help me verify?
| Recall that performing row replacement operations does not affect the determinant. Hence, if we replace row $3$ with (row $3$ + $(-x) \cdot $ row $1$), then replace row $3$ with (row $3$ - row$2$) and replace row $4$ with (row $4$ - row $2$), we obtain:
$$
\begin{vmatrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
x & x & 3x & 1\\
0 & x & x & 0\\
\end{vmatrix}
=
\begin{vmatrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
0 & x & x & 1-x\\
0 & x & x & 0\\
\end{vmatrix}
=
\begin{vmatrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
0 & 0 & 0 & 1-2x\\
0 & 0 & 0 & -x\\
\end{vmatrix}
$$
but since this last matrix is upper triangular, its determinant is the product of the diagonal entries. But one of the diagonal entries is a zero, so $\det A = 0$, as you computed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is this a theorem or a conjecture? A few months ago, I read The Irrationals by Julian Havil. I remember reading that if you order the rationals between 0 and 1 in this way: $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\cdots$ and then construct intervals, the first one being $[\frac{1}{3},\frac{1}{2}]$ and the following ones being constructed from the next two (not necessarily consecutive) terms in the sequence that are included in the last constructed interval, you get smaller and smaller intervals converging to a single point: $\sqrt2 - 1$.
My question is whether this result was arrived at numerically or is there a formal proof? [and if the latter, what is the proof or where can I find it?] The wording in the book did not make this very clear, if I recall correctly.
| First few intervals : $(1/3,1/2),(2/5,3/7),(7/17,5/12),(12/29,17/41),...$
The general construction is related to the Farey sequence.
Define freshman's sum $\frac{a}{b}\oplus \frac{c}{d}=\frac{a+c}{b+d}$.
This sum satisfies the median property : $\frac{a}{b}<\frac{c}{d}\Rightarrow \frac{a}{b}<\frac{a}{b}\oplus \frac{c}{d}<\frac{c}{d}$.
Then, the intervals are constructed as the following rule :
Start with $\frac{1}{3}<\frac{1}{2}$. Then add the (freshman's sum) $\frac{1}{3}\oplus \frac{1}{2}=\frac{2}{5}$, which is the next entry of the sequence between $\frac{1}{2}$ and $\frac{1}{3}$.
Now the modified sequence of appearing fractions is $\frac{1}{3}<\frac{2}{5}<\frac{1}{2}$. As we know, the next entry appearing is $\frac{2}{5}\oplus\frac{1}{2}=\frac{3}{7}$.(the sum of leftmost two has greater denominator than this sum, so this sum should appear first.)
So the modified sequence in this step is $\frac{1}{3}<\frac{2}{5}<\frac{3}{7}<\frac{1}{2}$.
Note that the second interval $(\frac{2}{5},\frac{3}{7})$ is constructed from this sequence in this step.
In the next step, the fraction added is $\frac{2}{5}\oplus\frac{3}{7}=\frac{5}{12}$(Note that we should find a number between the second interval $(\frac{2}{5},\frac{3}{7})$), and the modified seq is $\frac{1}{3}<\frac{2}{5}<\frac{5}{12}<\frac{3}{7}<\frac{1}{2}$.
Following step add $\frac{2}{5}\oplus\frac{5}{12}=\frac{7}{17}$(sum with less-denominator one) to the modified seq and we get the third interval $(\frac{7}{17},\frac{5}{12})$.
Now I think you'll be able to develop all following step with ease; in summary, choosing the terms in the sequence corresponds to the freshman's sum in Farey sequence, and the intervals constructed is the center two terms in the modified sequence in each even step.
Because the intervals are determined in each even step, I'll describe the patterns of the modified sequence in even steps from now on.
As you can check, the order relation between the center two terms reverses as 2 steps go along, thus our algorithm is a period 4 calculation, i.e, starting with $\frac{1}{3}<\frac{1}{2}$, we proceed the following 4 sum in one period:
Start at the centermost two fractions, say $A,B$.
$\cdots < A<B<\cdots \Rightarrow \cdots < A < A\oplus B < B < \cdots$
$ \Rightarrow \cdots < A < A\oplus B <(A\oplus B)\oplus B < B < \cdots$
$\Rightarrow \cdots < A < A\oplus B <(A\oplus B )\oplus \left\{(A\oplus B)\oplus B\right\}<(A\oplus B)\oplus B < B < \cdots$
$\Rightarrow \cdots < A < A\oplus B<(A\oplus B)\oplus [(A\oplus B )\oplus \left\{(A\oplus B)\oplus B\right\}] <(A\oplus B )\oplus \left\{(A\oplus B)\oplus B\right\}<(A\oplus B)\oplus B < B < \cdots$
Then retake the centermost two terms in the final sequence and iterate the above algorithm.
Hence, the $(2n+1)$th interval is $\left( (A\oplus B)\oplus [(A\oplus B )\oplus \left\{(A\oplus B)\oplus B\right\}] ,(A\oplus B )\oplus \left\{(A\oplus B)\oplus B\right\}\right)$, where $(A,B)$ is the $(2n-1)$th interval.
Let the two endpoints of the $(2n-1)$th inverval $\frac{a_n}{c_n} < \frac{b_n}{d_n}$.
This sequence $a_n,b_n,c_n,d_n$ satisfies the following recurrence formula:
$a_{n+1}=3a_n+4b_n, b_{n+1}=2a_n+3b_n, c_{n+1}=3c_n+4d_n, d_{n+1}=2c_n+3d_n$
($a_1=b_1=1,c_1=3,d_1=2$)
$\therefore \begin{pmatrix}
a_n&c_n \\
b_n & d_n
\end{pmatrix}=
\begin{pmatrix}
3&4 \\
2& 3
\end{pmatrix}^{n-1}\begin{pmatrix}
a_1 &c_1 \\
b_1 & d_1
\end{pmatrix}=
\begin{pmatrix}
3&4 \\
2& 3
\end{pmatrix}^{n-1}\begin{pmatrix}
1 &3 \\
1 & 2
\end{pmatrix}$
$\therefore a_n=\frac{1+\sqrt{2}}{2}\xi_1^{n-1}-\frac{\sqrt{2}-1}{2}\xi_2^{n-1},b_n=\frac{2+\sqrt{2}}{4}\xi_1^{n-1}+\frac{2-\sqrt{2}}{4}\xi_2^{n-1},c_n=\frac{\xi_1^n+\xi_2^n}{2},d_n=\frac{4+3\sqrt{2}}{4}\xi_1^{n-1}+\frac{4-3\sqrt{2}}{4}\xi_2^{n-1}$
($\xi_1=3+2\sqrt{2},\xi_2=3-2\sqrt{2}$)
$\therefore \lim_{n\to \infty}\frac{a_n}{c_n}=\lim_{n\to \infty}\frac{b_n}{d_n}=\sqrt{2}-1$
P.S. The fractions in some interval $(a,b)$ indeed lies between $a$ and $b$ in a Farey sequence(at least one of order the denominator of this fraction).
And, if $\frac{p}{q}$ has neighbors $a,b$ in some Farey sequence, then $\frac{p}{q}=a\oplus b$.
Because we find the numbers first appearing in a squence having lexicographic order in each step, it is obviouse that if $\frac{p}{q}$ is the first-appearing fraction between $a$ and $b$, then $\frac{p}{q}$ has neighbors $a,b$ in $q$th Farey seqence.
To find properties I mentioned, this wikipedia page would be helpful.
| {
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"url": "https://math.stackexchange.com/questions/1198564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Proving divergence of a sequence by proving the sequence is increasing. We define a sequence recursively by $$a_{n+1}=\frac{1}{4}({a_n}^2+a_n+2)~~~~~~~(a_1=3)$$
By showing $a_n$ is increasing prove that $a_n$ does not converge.
Not sure how to do this one. I tried showing that $a_{n+1}-a_n>0$ but couldn't get it to work also not sure how this would be used to imply it doesn't converge. (Although I think you can just state it is unbounded so eventually it will be bigger than any limit you could imagine)
Thanks
| $\begin{array}\\
a_{n+1}
&=\frac{1}{4}({a_n}^2+a_n+2)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+8)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+1+7)\\
&=\frac{1}{16}((2{a_n}+1)^2+7)\\
\end{array}
$
Therefore
$2a_{n+1}+1
=\frac{1}{8}((2{a_n}+1)^2+7)+1
=\frac{1}{8}((2{a_n}+1)^2+15)
$.
Letting
$b_n
=2a_n+1
$,
$b_{n+1}
=\frac{1}{8}(b_n^2+15)
$.
Since $b_1
=2a_1+1
=7
$,
$b_2
=\frac{49+15}{8}
=8
$,
$b_3
=\frac{64+15}{8}
=\frac{79}{8}
> 9
$.
If $b_n
\ge 8+k
$
where $k \ge 1$
(which is true for
$n=3$),
then
$b_{n+1}
=\frac{1}{8}(b_n(8+k)+15)
=\frac{1}{8}(8 b_n+kb_n+15)
=b_n+\frac{1}{8}(kb_n+15)
=b_n+\frac{1}{8}(k(8+k)+15)
$
so
$b_{n+1}-b_n
=\frac{1}{8}(k(8+k)+15)
= k+\frac{1}{8}(k^2+15)
\ge k+2
$.
Therefore
$b_n$
is unbounded
and strictly increasing
and so is
$a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Converting a rotated ellipse in parametric form to cartesian form I have a rotated ellipse in parametric form:
$$\begin{pmatrix}y \\ z\end{pmatrix} = \begin{pmatrix}a\cos t + b\sin t \\ c\cos t + d\sin t\end{pmatrix} \tag{1} $$ or,
$$(y,z) = (a\cos t + b\sin t , c\cos t + d\sin t) \tag{2} $$
By using $$\cos^2 t + \sin^2 t = 1 $$
I can rewrite into:
$$ \frac{(d^2 + c^2)y^2 + (-2bd-2ac)yz + (a^2+b^2)z^2}{(ad-bc)^2} = 1 \tag{3} $$
I need to compare it with the standard form of a rotated ellipse (the input format in a program I am writing):
$$\left(\frac{\cos\theta(y-h) + \sin\theta (z-k)}{r_1}\right)^2 + \left(\frac{\sin\theta(y-h) - \cos\theta (z-k)}{r_2}\right)^2 = 1 \tag{4} $$
To solve for $r_1, r_2, \theta $ (namely the semi-major, minor axis and angle of rotation).
However I realized that this will involve 3 non-linear equations. Although it is solvable, I was wondering if there is a simpler way to find the values?
| You can also find the eigenvalues of the matrix
$\begin{pmatrix}\frac{c^2+d^2}{(ad-bc)^2}&\frac{-ac-bd}{(ad-bc)^2}\\\frac{-ac-bd}{(ad-bc)^2}&\frac{a^2+b^2}{(ad-bc)^2}\end{pmatrix}$
in order to diagonalise it.
The eigenvalues are $\lambda=\frac{s^2\pm\sqrt{(s^2-2\det{})(s^2+2\det{})}}{2(\det{})^2}$, where $\det{} =ad-bc$ and $s^2=a^2+b^2+c^2+d^2$
giving your rotated ellipse the form $\lambda_1(y')^2+\lambda_2(z')^2=1$, which gives you the $r_1$ and $r_2$. The eigenvector matrix will give the $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$ How is $x^\frac{9}{4} \cdot x^\frac{1}{2} = x^\frac{11}{4}$?
I don't really understand it.
| $$x^{9/4}=(x^{1/4})^9.$$
$$x^{1/2}=(x^{1/4})^2.$$ Letting $y=x^{1/4}$, we get:
$$x^{9/4}\cdot x^{1/2} = y^9\cdot y^2=y^{11} = (x^{1/4})^{11}=x^{11/4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Parametrization of an intersection Hello I'm trying to parametrize the intersection of the sphere of centre $(0,0,0)$ and radius $R$ and the cylinder of centre $(R/2,0,0)$ and radius $R/2$. (the intersection is the surface)
Any hints of how to proceed? My guess is to switch into cylinder coordinates but I can't figure out what value I have to put for my $z$ coordinate. I don't understand the concept of parametrization very well, any comment is welcome.
| Given your equations:
$${x^2} + {y^2} + {z^2} = {R^2}$$
$${\left( {x - \frac{R}{2}} \right)^2} + {y^2} = {\left( {\frac{R}{2}} \right)^2}$$
Indroducing:
$$\begin{gathered}
x = \frac{R}{2}(\cos (\varphi ) + 1) \hfill \\
y = \frac{R}{2}\sin (\varphi ) \hfill \\
\end{gathered} $$
the second equation is satisfied.
From first equation now you get:
$$\begin{gathered}
{\left( {\frac{R}{2}(\cos (\varphi ) + 1)} \right)^2} + {\left( {\frac{R}{2}\sin (\varphi )} \right)^2} + {z^2} = {R^2} \hfill \\
{\left( {\frac{R}{2}} \right)^2}({\cos ^2}(\varphi ) + 2cos(\varphi ) + 1) + {\left( {\frac{R}{2}} \right)^2}{\sin ^2}(\varphi ) + {z^2} = {R^2} \hfill \\
{\left( {\frac{R}{2}} \right)^2}{\sin ^2}(\varphi ) + {\left( {\frac{R}{2}} \right)^2}{\cos ^2}(\varphi ) + 2{\left( {\frac{R}{2}} \right)^2}cos(\varphi ) + {\left( {\frac{R}{2}} \right)^2} + {z^2} = {R^2} \hfill \\
2{\left( {\frac{R}{2}} \right)^2}(cos(\varphi ) + 1) + {z^2} = {R^2} \hfill \\
{z^2} = {R^2} - 2{\left( {\frac{R}{2}} \right)^2}(cos(\varphi ) + 1) \hfill \\
\hfill \\
\hfill \\
\end{gathered}$$
With $0 \leqslant \varphi < 2\pi $
$$\begin{gathered}
x = \frac{R}{2}(\cos (\varphi ) + 1) \hfill \\
y = \frac{R}{2}\sin (\varphi ) \hfill \\
z = \pm \frac{{R\sqrt {1-\cos (\varphi )} }}{{\sqrt 2 }} \hfill \\
\end{gathered}$$
is given the parametrization, that describes two surfaces bounded by magenta coloured curve. We used cylindric parametrization.
For $R=1$ picture shows upper half.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Help evaluating $ \int \sqrt{{x}^{2} + 3} \; dx $ Can you help me evaluating the following indefinite integral?
$$
\int \sqrt{{x}^{2} + 3} \; dx
$$
Please, don't give a full solution, just some hint on which method to use...
** UPDATE **
Thank you very much to everybody for the useful comments and suggestions. I'm sorry for the delay with my reply, unfortunately do some mathematics as an hobby and often don't have time to work at it.
I tried to take Lucian suggestion on board and use trigonometric substitution as follows.
$$ x = \sqrt{3} \tan\theta $$
and
$$ \int \sqrt{{x}^{2} + 3} \; dx = \int \sqrt{{3\tan}^{2}\theta + 3} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{{3\sec}^{2}\theta} \; \sqrt{3}\sec^{2}\theta \; d\theta = \int \sqrt{3}\sec\theta \; \sqrt{3}\sec^{2}\theta \; d\theta = 3 \int \sec^{3}\theta \; d\theta $$
which (according to common integral tables) is equal to
$$ 3 \left[ \frac{1}{2} \sec\theta \tan\theta + \frac{1}{2} \ln\left| \sec\theta + \tan\theta \right| + C \right] $$
Problem arises when I try to substitute back the variable from $ \theta $ to $ x $ because I know $ \tan\theta = \frac{x}{\sqrt{3}} $ but I don't know how to substitute back $ \sec\theta $, so basically I stopped here:
$$
\frac{3}{2}\sec\theta\;\frac{x}{\sqrt{3}} + \frac{3}{2}\ln\left| \sec\theta + \frac{x}{\sqrt{3}} \right| + 3C
$$
It looks close to the final answer but still not there...any suggestions?
| Substitution $\sqrt{x^2+3}=u-x$ will work
$\sqrt{x^2+3}=u-x\\ x^2 +3 = u^2 - 2ux + x^2\\3=u^2-2ux \\ 2ux = u^2-3 \\ x = \frac{u^2-3}{2u} \\ u - x = u - \frac{u^2-3}{2u}\\u-x=\frac{u^2+3}{2u} \\
dx = \frac{2u\cdot2u-2(u^2-3)}{4u^2}du \\ dx = \frac{2u^2+6}{4u^2}du \\ dx = \frac{u^2+3}{2u^2}du \\ \int{\frac{u^2+3}{2u^2}\cdot\frac{u^2+3}{2u}du} \\ \frac{1}{4}\int{\frac{u^4+6u^2+9}{u^3}du}\\
\frac{1}{4}\left(\frac{1}{2}u^2-\frac{9}{2}\frac{1}{u^2}\right)+\frac{3}{2}\ln{|u|}+C\\\frac{1}{2}(\frac{u^4-9}{4u^2})+\frac{3}{2}\ln{|u|}+C\\
\frac{1}{2}\frac{u^2-3}{2u} \cdot \frac{u^2+3}{2u}+\frac{3}{2}\ln{|u|}+C \\
\frac{1}{2}x\sqrt{x^2+3} + \frac{3}{2}\ln{|x+\sqrt{x^2+3}|}+C
$
$x=\sqrt{3}\tan{\theta}\\
\frac{x}{\sqrt{3}} = \tan{\theta}
$
Recall identity $1+\tan^2{\theta}=\sec^2(\theta)$
$\frac{x^2}{3}=\tan^2{\theta}\\
1+\frac{x^2}{3} = 1+ \tan^2{\theta} \\
1+\frac{x^2}{3} = \sec^2{\theta}\\
\sqrt{\frac{3+x^2}{3}} = \sec{\theta}
$
but you must be sure that secant is nonnegative
In my opinion subsitution which i showed is better for this integral because after this substitution we only need to integrate power function
| {
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"url": "https://math.stackexchange.com/questions/1211555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$
Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?
| Hint: It is related to the Fibonacci numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$ \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\} \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$ If $\displaystyle 0 < \theta < \frac{\pi}{2}$ and $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin \left\{\theta+\left(m-1\right)\cdot \frac{\pi}{4}\right\}\cdot \sin \left\{\theta+m\cdot \frac{\pi}{4}\right\}} = 4\sqrt{2}$. Then value of $\theta=.$
$\bf{My\; Try::}$ Here the question is given in the form of $\displaystyle \frac{1}{\sin A \cdot \sin B}$. Then we can write it as
$$\displaystyle \Rightarrow \frac{1}{\sin A \cdot \sin B} = \frac{1}{\sin (B-A)}\cdot \left[\frac{\sin (B-A)}{\sin A\cdot \sin B}\right] $$
$$\displaystyle \Rightarrow \frac{1}{\sin(B-A)}\cdot \left[\frac{\cos A\cdot \sin B-\sin A\cdot \cos B}{\sin A \cdot \sin B}\right]$$
$$\displaystyle = \frac{1}{\sin (B-A)}\cdot \left[\cot A-\cot B\right]$$
Now Let $\displaystyle \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}=A$ and $\displaystyle \left\{\theta+m\cdot \frac{\pi}{4}\right\}=B.$
So We Get $\displaystyle \sum_{m=1}^{6}\frac{1}{\sin A \cdot \sin B}=\sum_{m=1}^{6}\sqrt{2}\left[\cot A-\cot B\right] = 4\sqrt{2}.$
So we get $$\displaystyle \sum_{m=1}^{6}\left[\cot \left\{\theta+(m-1)\cdot \frac{\pi}{4}\right\}-\cot \left\{\theta+m\cdot \frac{\pi}{4}\right\}\right]=4.$$
Now after expanding we get $$\displaystyle \cot \theta - \cot \left(\theta+\frac{6\pi}{4}\right) = 4.\Rightarrow \cot \theta - \cot \left(\theta+\frac{2\pi}{3}\right)=4$$
Again after simplifying we get $$\displaystyle \cot \theta - \left(\frac{\cot \theta-\sqrt{3}}{1-\sqrt{3}\cot \theta}\right)=4\Rightarrow \frac{\sqrt{3}-\sqrt{3}\cot^2 \theta}{1-\sqrt{3}\cot \theta} = 4.$$
So We get $$\displaystyle \sqrt{3}-\sqrt{3}\cot^2 \theta = 4-4\sqrt{3}\cot \theta.\Rightarrow \sqrt{3}\cot^2 \theta - 4\sqrt{3}\cot \theta+(4-\sqrt{3})=0$$.
Now Let $\cot \theta = y\;,$ Then equation is $\sqrt{3}y^2-4\sqrt{3}y+(4-\sqrt{3}) = 0.$
We Get $$\displaystyle y = \cot \theta = \frac{2\pm \sqrt{15-4\sqrt{3}}}{3}$$
But I did not understand how can i get value of $\theta.$
Help me, Thanks
| HINT:
Can you try tis way?
$$\cot\theta-\cot\left(\frac{2\pi}3+\theta\right)=\frac{\sin\dfrac{2\pi}3}{\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}=\frac{\sqrt3}{2\cos\theta\cdot\cos\left(\frac{2\pi}3+\theta\right)}$$
$$=\frac{\sqrt3}{\cos\dfrac{2\pi}3+\cos\left(\dfrac{2\pi}3+2\theta\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
| Useful inequality chain
Given $a \ge 0$ and $b > -\frac{1}{2}a$:
$a+b-2\frac{b^2}{a} \le a+b-\frac{b^2}{a+b} = \frac{a^2+2ab}{a+b} \le \sqrt{a^2+2ab} \le a+b-\frac{1}{2}\frac{b^2}{a+b} \le a+b$
Solution
Given $x \ge 1$:
$\sqrt{x}+\frac{1}{2}-\frac{1}{4\sqrt{x}+2} \le \sqrt{x+\sqrt{x}} \le \sqrt{x}+\frac{1}{2}$.
$\sqrt{x}-\frac{1}{2\sqrt{x}}-\frac{1}{4x\sqrt{x}} \le \sqrt{x-1} \le \sqrt{x}-\frac{1}{2\sqrt{x}}$.
Therefore $\sqrt{x+\sqrt{x}}-\sqrt{x-1} \to \frac{1}{2}$ as $x \to \infty$.
Notes
This kind of inequalities are useful when we do not want to use asymptotic expansion but otherwise as shown by abel expansion is the most widely applicable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| Use this trigonometric identity:
$$
\frac{3\sin x+2\cos x}{2\sin x+3\cos x} = \overbrace{\frac{12}{13} + \frac5{13}\tan\left( x - \varphi \right)}^{\text{So integrate this function.}} \text{ where }\varphi = \arctan\frac 2 3.
$$
NOTE: I initially mislaid the denominator in $5/13$ and just had $5$ as the coefficient. @robjohn pointed out the error in comments below.
Proof: The graph looks like a tangent function with period $\pi$ except that the inflection point is higher than the $x$-axis and the asymptotes are not at $\pm\pi/2$. The asymptotes occur where the denominator is $0$, so at those points $2\sin x+3\cos x=0$, so $\tan x = -3/2$. Hence the inflection points are at
$$
\frac\pi2 -\arctan\frac 3 2 = \arctan\frac 2 3.
$$
We have
$$
\frac{3\sin x+2\cos x}{2\sin x+3\cos x} = \frac{3\tan x + 2}{2\tan x + 3}
$$
and when $\tan x = \dfrac 2 3$ this simplifies to $\dfrac{12}{13}$. So we want
$$
\frac {12}{13} + c\tan(x-\varphi) = \frac{12}{13} + c\frac{\tan x - \frac 2 3}{1+ \frac 2 3 \tan x} = \frac {12}{13} + c\frac{3\sin x - 2\cos x}{3\cos x + 2\sin x}.
$$
We need to find the value of $c$ for which we get the right function, and a bit of algebra tells us $c=5/13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
What will be $E(X)$ and $Var(X)$? Given a discrete random variable
$$P(X = k) = \frac{1}{2^k}, \ \ \ k=1,2,...$$
How to calculate the expected value and Variance of $X$?
| Let $E(x)=\sum_{k=1}^\infty kx^k$ and $D(x)=\sum_k x^k$. Then
$$
E(x)+D(x)=\sum_k(k+1)x^k=\frac{d}{dx}\left(\sum_{k=1}^\infty x^{k+1}\right)=\frac{d}{dx}\frac{x^2}{1-x}=\frac{x(2-x)}{(1-x)^2}\cdot
$$
In particular,
$$
E(0.5)=\frac{0.5(2-0.5)}{(1-0.5)^2}-D(0.5)=3-1=2.
$$
This takes care of the expectation. Similarly, let $F(x)=\sum_{k=1}^\infty k^2x^k$. Then:
$$
F(x)+3E(x)+2D(x)=\sum_k(k+1)(k+2)x^k=\frac{d^2}{dx^2}\left(\sum_{k=1}^\infty x^{k+2}\right)=\frac{d^2}{dx^2}\frac{x^3}{1-x}=\frac{2x(-3+3x-x^2)}{(1-x)^3}.
$$
And so
$$
F(0.5)=\frac{2(0.5)(-3+1.5-0.5^2)}{(1-0.5)^3}-3E(0.5)-2D(0.5)=14-6-2=6.
$$
The variance is thus $6-2^2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Mean of the difference between uniform random variables. I have two uniform random variables $B$ and $C$ distributed between $(2,3)$ and $(0,1)$ respectively. I need to find the mean of $\sqrt{B^2-4C}$. Could I plug in the means for $B$ and $C$ and then solve or is it more complicated than that?
The original question is here: Difference between two real roots with uniformly distributed coefficents
| Use change of variables:
$$X = \sqrt{B^2-4C}
$$
$$Y=C$$
$$B = \sqrt{X^2+4Y}$$
$$C=Y$$
Jacobian:
$$
J = det\left( \begin{array}{ccc}
\frac{\partial B}{\partial X} &\frac{\partial C}{\partial X} \\
\frac{\partial B}{\partial Y} & \frac{\partial C}{\partial Y} \\
\end{array} \right) = det \left( \begin{array}{ccc}
\frac{X}{\sqrt{X^2+4Y}} & 0 \\
\frac{2Y}{\sqrt{X^2+4Y}} & 1 \\
\end{array} \right) = \frac{X}{\sqrt{X^2+4Y}}
$$
Join distribution of (X,Y) is given by:
$$f_{X,Y}(X,Y) = |J|\ f_{B,C}(B,C) = |J|\ f_B(B) f_C(C) $$ where $f_{B,C}(B,C) = f_B(B) f_C(C)=1$ because $B,C$ are independent.
Thus, $$f_{X,Y}(X,Y) = \frac{X}{\sqrt{X^2+4Y}}$$ where $ 0 \leq X \leq 3$ and $0 \leq Y \leq 1$
We need to find $E[X] = \int_0^3 xf_X(x) dx$
$f_X(x) = \int_0^1 f_{X,Y}(x,y)dy = \int_0^1 \frac{x}{\sqrt{x^2+4y}}dy = \frac{x.\sqrt{x^2+4y}}{2}|_0^1 = \frac{x}{2}(\sqrt{x^2+4}-x)$
Further:
$E[X] = \frac{1}{2}\int_0^{3} x^2(\sqrt{x^2+4}-x)dx$ which probably further requires $x=2\ tan\theta$ substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why can $5^{2n+3} + 3^{n+3} \cdot 2^{n} \quad \forall n \in \mathbb{N}$ never be a prime number? It seems to be true for the first thousand $n$ but I really can't think of a way to prove this statement.
Any kind of help will be appreciated!
| Oops...I think I found a proof.
We can notice that for every $n$ our sum seems to be divisible by $19$.
So we have to prove that
$$5^{2n+3} + 3^{n+3} \cdot 2^{n} \equiv 0 \pmod{19}\quad \forall n \in \mathbb{N}$$
We can prove this using induction.
Base-case:
\begin{align*}
&19|152&
\end{align*}
Inductive step:
\begin{align*}
0 &\equiv 5^{2(n+1)+3} + 3^{n+4} \cdot 2^{n+1} &\pmod{19} \\
0 &\equiv 25\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\
0 &\equiv 6\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\
0 &\equiv 6\cdot ( 5^{2n+3} + 3^{n+3} \cdot 2^{n} )\\
0 &\overset{\text{IH}}{\equiv} 6 \cdot 0 \equiv 0 \\&&\square
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
| If we consider the equation of a certain type.
$$(n-4)^2+(n-3)^2+(n-2)^2+(n-1)^2+n^2+(n+1)^2+(n+2)^2+(n+3)^2+$$
$$+(n+4)^2=(k-3)^2+(k-2)^2+(k-1)^2+k^2+$$
$$+(k+1)^2+(k+2)^2+(k+3)^2+(k+4)^2$$
The solutions can be written simply using the sequence. The next element which is obtained from the previous one.
$$p_2=3p_1+4s_1$$
$$s_2=2p_1+3s_1$$
If we use the first element of the sequence. $( p ; s) - (2 ; 1)$
Then the formula will look like.
$$n=2p^2\pm6ps+4s^2$$
$$k=2p^2\pm6ps+5s^2$$
If we use the first element of the sequence. $(p ; s ) - ( 3 ; 2 )$
Then the formula will look like.
$$n=2(2p^2\pm6ps+4s^2)$$
$$k=2(2p^2\pm6ps+5s^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Prove this form $2^{n+2}-7a^2_{n}$ always is square numbers. Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=-1$,and
$$a_{n}+a_{n-1}+2a_{n-2}=0,n\ge 3$$
show that
$2^{n+2}-7a^2_{n}$ is square numbers.
such as
$$2^3-7=1,n=1$$
$$2^4-7=9=3^2,n=2$$
$a_{3}=-1$
$$2^{5}-7=25=5^2,n=3$$
$a_{4}=3$
$$2^6-7\cdot 9=1^2,n=4$$
| We have always $2^{n+2}-7a^2_{n}=b_n^2$, where $b_1=-1$, $b_2=-3$ and $b_n+b_{n-1}+2b_{n-2}=0$ (induction).
| {
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"url": "https://math.stackexchange.com/questions/1223767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find all primitive triples (a,b,c)? (Pythagorean Triples) how to find all primitive triples when one value of $(a,b,c)$ is given? For example in this case $a = 45$.
What is the procedure to find the primitive triples ?
Conditions for primitive triples are:
*
*$a^2 + b^2 = c^2$
*$(a,b) = 1$
*$a$ and $c$ are odd and $b$ is even.
| We can find Pythagorean triples for a given side $A$ by solving Euclid's formula for $n$ and then testing a finite number of $m$-values to see which generates an integer for $n$.
$$A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil$$
The lower limit ensures $m^2>A$ and the upper limit ensures $m-n\ge 1$.
In this case, we have $n=\sqrt{m^2-45}\quad \text{where}\quad \lceil\sqrt{46}\space\rceil=7\le m \le \lceil 22.5\rceil=23$
Texting, we find $3$ values of $m$ in this range that yield integers for $n$.
$\sqrt{7^2-45}=4,\quad F(7,4)=(33,56,65)\qquad GCD(33,56,65)=1$
$\sqrt{9^2-45}=6, \quad F(9,6)=(45,108,117)\qquad GCD(45,108,117)=9$
$\sqrt{23^2-45}=22, \quad F(23,22)=(45,1012,1013)\qquad GCD(45,1012,1013)=1$
We can see that $1$ of the triangles found not primitive. No other values of $m$ yield an integer so no other primitive triples exist where side $A=45$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find $11^{644} \mod 645$ Can someone just explain to me the basic process of what is going on here? I understand everything until we start adding 1's then after that it all goes to hell. I just need some guidance. The Problem with the solution is attached.
Thanks in advance..
Find $11^{644} \mod 645$
| The rather intimidating notation used in the textbook is effectively just finding the binary representation of 644.
$644 = 512+128+4 = 1010000100_{2}$
It does this by extracting factors of two and noting when the result is odd, corresponding to successive digits (from the right) of the binary number above.
so putting this knowledge into the require exponent,
$\begin{align} \\
x^{644} &= x^{512}\cdot x^{128} \cdot x^4 \\
&= x^{4\times 32 \times 4}\cdot x^{32\times 4} \cdot x^4\\
&= ((x^4\cdot x)^{32}\cdot x)^4\\
\end{align}$
Once all the powers can be represented as powers of two, the bulk of the exponentiation can be undertaken by squaring, and of course taking the numbers modulo 645 can be done at each step. Your book example works by calculating the three values for $11^n \bmod 645$ on the top row, but also this process may be undertaken effectively using the decomposition on the third row to accumulate the final value:
Starting with $x$, we square twice, multiply by $x$ again, square $5$ times, multiply by $x$ again, and square twice more.
This is equivalent to stepping through the binary representation above and squaring at each step, then also multiplying by $x$ if the digit is a $1$.
Your sample problem solution looks like,
$$\begin{array}{c|c|c|c}
\text{step} & \text{power} & 11^n \bmod 645 & \text{accumulate} \\ \hline
0 & 1 & 11 & \\
1 & 2 & 121 & \\
2 & 4 & \fbox{451} & \to 451 \\
3 & 8 & 226 & \\
4 & 16 & 121 & \\
5 & 32 & 451 & \\
6 & 64 & 226 & \\
7 & 128 & \fbox{121} & \to 391 \\
8 & 256 & 451 & \\
9 & 512 & \fbox{226} & \to 1 \\
\end{array}$$
and the binary-digit-driven version looks like:
$$\begin{array}{c|c|c|c}
\text{step} & \text{binary digit} & \text{action} & \text{value} \\ \hline
0 & 1 & & 11 \\
1 & 0 & \text{square} & 121 \\
2 & 1 & \text{sq+mult} & 446 \\
3 & 0 & \text{square} & 256 \\
4 & 0 & \text{square} & 391 \\
5 & 0 & \text{square} & 16 \\
6 & 0 & \text{square} & 256 \\
7 & 1 & \text{sq+mult} & 431 \\
8 & 0 & \text{square} & 1 \\
9 & 0 & \text{square} & 1 \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving by strong induction for a sequence of integers, $2^n$ divides term $n$ Provided the following sequence of integers $t_1, t_2, t_3$,... is defined as:
$t_1 =4, t_2 =8$ and $t_n= $ $ 6t_n$$_-$$_1$ - $4t_n$$_-$$_2$ for all integers $n \geq 3$
How do we prove that $2^n$ divides $t_n$ for all integers $n \geq 1$ by strong induction?
So far, I am having problems with the induction step as I am still trying to grasp the concept of strong induction, but this is what I came up with so far:
Base step:
$t_3 = 32$, which is divisible by 8; and $t_4 = 160$, which is divisible by 16.
$\therefore$ Statement is true when $n=3$ and $n=4$.
Induction step:
Assuming $k=n$ and the statement is true for $3,4,...k$. Prove that $2^k$ divides $t_k$ for $k+1$:
$t_3 = ((2^3 - 2)(2^3) - (2^2)(2^2)) = 2^6 - 2^4 - 2^4 $; dividing this by $2^3$ gives $2^3 - 2^2$ which would be an integer result.
$t_4 = ((2^3 - 2)(2^5) - (2^2)(2^3)) = 2^8 - 2^6 - 2^5 $; dividing this by $2^4$ gives $2^4 - 2^2 - 2$ which would also be an integer result.
$\therefore$ By the logic above, the statement also holds for $k-1$, $k$ as well as $k+1$, thus as required to prove by strong induction.
I believe this is a fundamentally flawed way of trying to prove by strong induction, but this is the only way I can possibly think of. Can someone show a correct method of solving this?
| Let
$u_n = t_n/2^n$,
or
$t_n = 2^n u_n$.
Since
$t_n
=6 t_{n-1}-4t_{n-2}
$,
$2^nu_n
=6 \cdot 2^{n-1}u_{n-1}-4\cdot 2^{n-2}u_{n-2}
=3 \cdot 2^{n}u_{n-1}- 2^{n}u_{n-2}
$,
or
$u_n
=3 u_{n-1}- u_{n-2}
$.
Since $u_1$ and $u_2$
are integers
(
$u_1 = t_1/2 = 2$
and
$u_2 = t_2/4 = 2$
),
all following
$u_n$ are integers.
Even more,
since $u_1$ and $u_2$
are even integers,
so are all following
$u_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lagrange's multiplier not working
Given the function $f(x,y):=xy+x-y$. Let $D:=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq25\wedge x \geq 0\}$. Find the absolute maximum and minimum of $f$ on $D$.
My working is as follows:
$\begin{array}
& f_x(x,y)=y+1=0 & \qquad \qquad f_y(x,y)=x-1=0 \\
\Rightarrow y=-1 & \qquad \qquad \Rightarrow x=1
\end{array}$
$D(x,y)=\begin{vmatrix}
f_{xx}(x,y) & f_{xy}(x,y) \\
f_{xy}(x,y) & f_{yy}(x,y)
\end{vmatrix} = \begin{vmatrix}
0 & 1 \\
1 & 0
\end{vmatrix} = -1$
$D(x,y) = D(1,-1) < 0 \Rightarrow (1,-1)$ is a saddle point.
Also, just for interest, $f(1,-1)=1$
To find the maximum and minimum of $f$ subject to $x^2+y^2=25$ I will use a Lagrange multiplier.
$\nabla f(x,y) = \lambda \nabla g(x,y)$ where $g(x,y)=x^2+y^2-25$
$\langle y+1, x-1 \rangle = \lambda \langle 2x, 2y \rangle$
$\left\{\begin{array}{llll}
y+1=2\lambda x & \Rightarrow & y=2\lambda x -1 & (1) \\
x-1=2\lambda y & \Rightarrow & x=2\lambda y +1 & (2) \\
x^2+y^2=25 & & & (3)
\end{array}\right.$
Putting (1) into (2) and (2) into (1) gives
$$x=\frac{1}{1+2\lambda} \qquad \text{and} \qquad y=-\frac{1}{1+2\lambda}\tag{4}$$ Where $\lambda \neq \pm \frac12$
Putting (4) into (3) gives
$$\lambda = \frac{-5\pm \sqrt2}{10} \approx -0.64 \quad \text{or} \quad -0.36$$
Subsequently,
$$x \approx \pm 3.54 \quad \text{and} \quad y \approx \mp 3.54$$
Note that $x=-y$. So,
$$f(3.54,-3.54) \approx -5.43 \quad \text{and} \quad f(-3.54,3.54) \approx -19.57$$
By this calculation, (-3.54, 3.54, -19.57) would be a point of absolute minimum on the circle $x^2+y^2=25$. But $x\geq 0$.
Hmm. Let me try evaluating $f(0,5)$. $$f(0,5)=-5 \nless f(3.54,-3.54) \approx -5.43$$ Nope. What should I do now to find the minimum in a procedurally correct way?
I also cannot find the absolute maximum. $(1,-1,1)$ is not the absolute maximum because I have found that $(3.54, 3.54, 12.5)$ exists on $D$. Why did my calculation using Lagrange's multiplier not give me this point?
| Let us focus our attention on the boundary of $D$. From your system we deduce
$$
(4\lambda^2-1)y=1-2\lambda,
$$
and then $x=-y$. This gives us a single admissible point $P_0$ whose coordinates are $x=\frac{5}{\sqrt{2}}$ and $y=-\frac{5}{\sqrt{2}}$. At this point, $f(P_0)=-\frac{25}{2}+5 \sqrt{2} \approx -5.42893$.
Now we must remember that the case $4\lambda^2-1=0$ must be treated separately. If $\lambda=1/2$, then the systems becomes
$$
\begin{array}{ll}
y+1=x \\
x-1 = y \\
x^2+y^2=25,
\end{array}
$$
which gives $x=4$ (the root $x=-3$ cannot be accepted), $y=3$, $f(4,3)=13$. Similarly, for $\lambda=-1/2$,
$$
\begin{array}{ll}
y+1=-x \\
x-1 = -y \\
x^2+y^2=25,
\end{array}
$$
and the same solution.
Finally, on the vertical segment $\{(0,y)\mid -5 \leq y \leq 5\}$ we have $g(x,y)=x=0$ as a constraint and therefore the system
$$
\begin{array}{ll}
y+1=\lambda \\
x-1 = 0 \\
x=0
\end{array}
$$
and no solutions. However we must also consider the two singular points $(0,-5)$ and $(0,5)$, where
$$
f(0,-5)=5, \quad f(0,5)=-5.
$$
To summarize, the point $(5/\sqrt{2},-5/\sqrt{2})$ is the global minimum, the point $(4,3)$ is the global maximum of $f$. Here is the graph of $x \mapsto f(x,\sqrt{25-x^2})$:
Remark. I do not believe that the methods of Lagrange multipliers is the best one to apply. It is much better to parametrize the boundary by polar coordinates $x=25 \cos \phi$, $y=25\sin \phi$, $-\pi/2 \leq \phi \leq \pi/2$, and to look at $f$ on the vertical segment. Alternatively, use the parametrization $y=\sqrt{25-x^2}$ for $0 \leq x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Question about a solution of a system of three non linear equations in three unknowns Let $a$, $b$ and $c$ be positive real numbers such that $$ a + \frac{1}{b} = 3$$ $$b + \frac{1}{c} = 4$$ $$ c + \frac{1}{a} = \frac{9}{11} $$ then $$ a \times b \times c =?$$
I tried doing this problem but I was unsuccessful. Tried a lot but couldn't get the answer! The answer is a numerical value...
|
Let $a$, $b$ and $c$ be positive real numbers such that
\begin{align}
a + \frac{1}{b} &= 3 \tag{1}\label{1}
\\
b + \frac{1}{c} &= 4 \tag{2}\label{2}
\\
c + \frac{1}{a} &= \frac{9}{11} \tag{3}\label{3}
\end{align}
\eqref{1}$\times$\eqref{2}$\times$\eqref{3}
$-$\eqref{1}$-$\eqref{2}$-$\eqref{3} gives:
\begin{align}
abc+\frac{1}{abc}&=2,
\end{align}
hence $abc=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\frac{t^2}{t+1}$ to the form $\frac{1}{t+1} +t -1$ One of my attempts would look like below.
$\frac{t^2}{t+1}$ = $\frac{t \times t+1-1}{t+1}$ = 1+ $\frac{t-1}{t+1}$ = $\frac{t-1+1-1}{t+1} + 1$ = $\frac{-2}{t+1} +2$
Also, I put into t an arbitrary number in both equations and have gotten the same answer. The first number was 18785 to which both contained the answer $\frac{352876225}{18786}$.
| Several responses have you make the following factorization:
$$\frac{t^2}{t+1} = \frac{(t+1)(t-1) + 1}{t+1}$$
But in general, you can take a slightly more iterative approach:
$$\frac{t^2}{t+1} = \frac{t^2 + t - t}{t+1} = \frac{t\cdot (t+1)}{t+1} - \frac{t}{t+1} = t - \big(\frac{t+1-1}{t+1}\big) = t - 1 + \frac1{t+1}$$
You just need to consider which terms are required to turn the largest term in the numerator into a multiple of the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rewrite also as: $p^2+pq\equiv 3 \pmod5$.
Now I analyzed five possible cases: $p^2\equiv 0$, $p^2\equiv 1$, $p^2\equiv 2$, $p^2\equiv 3$ and $p^2\equiv 4$.
*
*We can not that if $p^2\equiv 0$ also $pq\equiv 0$ therefore the equation is impossible
*$p^2\equiv 2$ is impossible because $2$ is a non-quadratic residue $\pmod5$
*also $p^2\equiv 3$ is impossible because $3$ is a non-quadratic residue $\pmod5$.
*If $p^2\equiv 1 \pmod5$, $p\equiv \pm 1$, therefore $p\equiv -1\equiv 4 \pmod5$ and $p2\equiv 1 \pmod5$. These values are not acceptable because if $p=9$ ($9\equiv 4 \pmod5$) $275\cdot 9>2008$, if $p=11$ ($11\equiv 1\pmod5$) $275\cdot 9>2008$ therefore two values aren't acceptable.
*If $p^2\equiv 4 \pmod5$, $p\equiv \pm2$ and if $p\equiv -2\equiv 3 \pmod5$ there not exist solutions for the same reason of previous example. If $p\equiv 2 \pmod5$ possible values of $p$ are $2$ or $7$ because $275p<2008$. Replacing $p=2$ we obtain a valor not integer of $q$ while replacing $p=7$ we obtain $q=2$ therefore this is the only solution of equation.
Is correct my process or are there other solutions?
Thanks:)
| Because there is alot of solutions, let me present you mine: the equation you stated is equivalent to $$(p+10)(p+q+265)=4658=2\cdot 17\cdot 137$$
and because $p+10>2$ then $p+10\geq 10$ but also $p+q+265\geq 2\cdot 137$ hence $p+2=17$ and $p+q+265=2\cdot 137$
In negative solutions (without (negative) prime assumptions) there is exactly $16$ solutions, because every factor can take any possible value of the divisors of $4658$ and there are exactly $16$ possible divisors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly from the definition of limit. One of my homework questions is:
Prove $\lim\limits_{n \to \infty} \frac{2n^2+2}{3n^3+1}=0$ directly
from the definition of limit.
In trying to follow: Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1} = 0$ from the definition , i have so far:
Proof:
It must be shown that for any $\epsilon>0$, there exists an integer $N$ such that $\left|\frac{2n^2+2}{3n^3+1}-0\right| <\epsilon$ whenever $n>N$.
$\left|\frac{2n^2+2}{3n^3+1}-0\right|=\frac{2n^2+2}{3n^3+1}>\frac{2n^2}{3n^3}=\frac{2}{3n}$.
Because in the problem im given, the left side of the inequality that im trying to simplify i guess, is instead greater than the right side, im stuck.
And i dont understand the how or why behind the algebraic manipulations that the original author and one of the answers performs, in the linked question:
$$\left|\frac{n}{n^2 + 1}\right| < \epsilon \text{ whenever }x \gt M.$$
$$ n \lt \epsilon(n^2 + 1) $$
$$n \lt \epsilon n^2 + \epsilon$$
Truly do feel clueless at this point!
| Let $\epsilon>0$ be given and let $N>\frac{4}{3\epsilon}$. If $n>N$, then $n>\frac{4}{3\epsilon}$ or $$\epsilon>\left|\frac{4}{3n}\right|$$
Thus :
$$\epsilon>\left|\frac{4n^2}{3n^3}\right|$$
$$\epsilon>\left|\frac{2n^2+2n^2}{3n^3}\right|\geq\left|\frac{2n^2+2}{3n^3}\right|>\left|\frac{2n^2+2}{3n^3+1}\right|=\left|\frac{2n^2+2}{3n^3+1}-0\right|$$
Do these manipulations make sense?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof by Induction $3^n > n^3$ I am trying to prove the following, however I'm stuck at the Induction hypothesis
Prove by induction that, for all integers $n$, if $n\geq 5$, then $3^n>n^3$
What I have Done:
Base Case:
$n = 5$
$3^5 > 5^3$
$243 > 125$ so TRUE
Assume True for $n = k$
$3^k > k^3$ must be TRUE
Should be TRUE for $n = k+1$
$3^k + 3^{k+1} > k^3 + 3^{k+1}$
Im a little stuck here. Any help would be appreciated!
| Supposing that $3^k\gt k^3$, you need to prove $3^{k+1}\gt (k+1)^3$.
For inductive step : $$\begin{align}3^{k+1}&\gt 3k^3\\&=k^3+k^3+k^3\\&\gt k^3+3k^2+3^2k\\&=k^3+3k^2+3k+6k\\&\gt k^3+3k^2+3k+1\\&=(k+1)^3.\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Working out $\tan x$ using sin and cos expansion
Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$
Find the series expansions of the $\tan x$ function up to the $x^5$ term.
So it is:
$$
\frac {x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...} {1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...}
$$
However many times I retry, trying long division results :
$$
\require{enclose}
\begin{array}{rll}
Q(x) \\[-3pt]
1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ... \enclose{longdiv}{x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...}\kern-.2ex \\[-3pt]
\underline{x-\dfrac{x^3} {2!} + \dfrac{x^5} {4!}\phantom{0000}} \\[-3pt]
\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}+ \dfrac{x^5} {5!}-\dfrac{x^5} {4!} \phantom{00}&& \\[-3pt]
\underline{\phantom{0}\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}- \dfrac{x^5} {2!2!}+ \dfrac{x^5} {3!2!}} && \\[-3pt]
\dfrac{x^5} {5!} - \dfrac{x^5} {4!} + \dfrac{x^5} {2!2!}-\dfrac{x^5} {3!2!}&& \\[-3pt]
\end{array}
$$
Sorry for the horrible format. but this is the best i can do..
$$
Q(x) = x + (\frac {x^3}{2!} - \frac {x^3}{3!}) + (\dfrac {x^5}{5!} - \dfrac {x^5}{4!} + \dfrac {x^5}{2!2!} - \dfrac {x^5}{3!2!}) + ...
$$
Which results the coefficients of the $x^5$ term $\dfrac 1 {20}$.
This is clearly a wrong value.
I can't seem to get $\dfrac 2 {15}$ instead.
Where did I get wrong?
Many thanks in advance.
| The following way may be easier.
Since $\tan x$ is an odd function, it can be represented as $\tan x=ax+bx^3+cx^5+\cdots.$
Since $\sin x=\cos x\tan x$, one has
$$x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\left(ax+bx^3+cx^5+\cdots\right).$$
Comparing the coefficients gives you
$$1=a$$
$$-\frac{1}{3!}=b-\frac{a}{2!}$$
$$\frac{1}{5!}=c-\frac{b}{2!}+\frac{a}{4!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238965",
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"source": "stackexchange",
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Which is larger, $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$? The clue given by the text is to "use the fact that $\sqrt{x}$ is increasing."
I was able to get the correct answer here by squaring both expressions. But I don't think I made use of the text-prided clue, so am wondering if perhaps I'm missing the "deeper" lesson.
What do you think the author is hinting at?
Thanks!
| Since $x\mapsto\sqrt{x}$ is an increasing function we have the following
\begin{align*}
2\sqrt{15}&>2\sqrt{12}\\
8+2\sqrt{15}&>8+2\sqrt{12}\\
3+2\sqrt{15}+5&>2+2\sqrt{12}+6\\
(\sqrt{3}+\sqrt{5})^2&>(\sqrt{2}+\sqrt{6})^2
\end{align*}
Therefore, $\sqrt{3}+\sqrt{5}>\sqrt{2}+\sqrt{6}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integral by using substitution (How to proceed?) Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.
My attempt,
$x=a\sin\theta$
$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$
The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$
How to proceed then?
| Using your substition:
$$\int\frac1{x^2\sqrt{a^2-x^2}}\;\rightarrow\;\int\frac{a\cos\theta\;d\theta}{a^2\sin^2\theta\cdot a\cos\theta}=-\frac1{a^2}\int\frac{d\theta}{-\sin^2\theta}=-\frac1{a^2}\cot\theta+C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating the value of $\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$ I need help finding the value of $$\cos \left(\frac{1}{2} \arccos \frac{3}{5}\right)$$
My try:
let $\theta = \frac{1}{2} \arccos \frac{3}{5}$, then $2\theta = \arccos \frac{3}{5}$ and $\theta \in [0,\frac{\pi}{2}]$
Now let $A = 2\theta$, and $A \in [0,\frac{\pi}{2}]$ So $\cos A = \frac{3}{5} $ and $\sin A = \frac{4}{5} $. I noticed that I got stuck and wasn't getting closer to the answer.
Can someone please give me a HINT AND NOT THE ANSWER? Thanks.
| Hint with a trigonometric identity:
$$\cos x=2\cos^2\frac x2-1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How can I solve this integral with complex number? $n$ here is a complex number such that $n=n_r+in_i$
How can I solve this integral?
$$\int_{0}^{\infty}\frac{x^4}{|x^2-n^2|^2} d x=? $$
| Let's change the notation to a more friendly one: $\;n:= a+bi\;$ , so:
$$x^2-n^2=x^2-(a^2-b^2+(2abi))=(x^2-a^2+b^2)-2abi\implies$$
$$|x^2-n^2|^2=(x^2-a^2+b^2)^2+4a^2b^2$$
Thus your integral is
$$\int_0^\infty\frac{x^4}{(x^2-a^2+b^2)^2+4a^2b^2}dx\stackrel{M=\max\{a,b\}}\ge\frac14\int_M^\infty\frac{x^4}{x^4+M^4}dx=$$
$$=\frac14\int_M^\infty\left(1-\frac{x^4}{x^4+M^4}\right)dx$$
and this last integral diverges.
| {
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Given $n \in \Bbb Z$, determine $\gcd(3n^2 + 7n + 4, n + 2)$. I factored $3n^2+7n+4$ to $(3n+4)(n+1)$ and because there isn't a common factor of those and $n+2$ I said that the gcd is $1$, but is there any othere way to go about it that would come up with a gcd that isn't $1$?
| We have
$$3n^2+7n+4 = (n+2)(3n+1) + 2$$
Hence, if $d$ is a common divisor of $3n^2+7n+4$ and $n+2$, then $d$ must divide $2$. Hence, $\gcd(3n^2+7n+4,n+2) = 1 \text{ or }2$. $3n^2+7n+4$ is always even. Further, $n+2$ has the same parity as $n$.
Hence, if $n$ is odd, $$\gcd(3n^2+7n+4,n+2) = 1$$
while if $n$ is even $$\gcd(3n^2+7n+4,n+2) = 2$$
| {
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Find the limit of $\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$ Find the limit of $$\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$$
I'm not sure how to evaluate this limit. Any hints or solutions are greatly appreciated. I turned this into $\int_o^1{1\over {1+x^3}}dx$ using algebra. Is this correct or is there a better way?
| Your integral is correct, since the terms of your sum can be rewritten as $$\frac{n^2}{n^3+i^3} = \frac{1}{n} \cdot \frac{1}{1+(i/n)^3}$$, which means that your sum is a Riemann sum for the corresponding integral.
If you are having a hard time getting started on evaluating the integral, you can factor $x^3+1 = (x+1)(x^2-x+1)$, and then focus on rewriting your integrand in the following form (using the method of partial fractions)
$$
\frac{1}{x^3+1} = \frac{A}{x+1} + \frac{B(2x-1)}{x^2-x+1} +\frac{C}{x^2-x+1}
$$
(also note that $x^2-x+1 = \left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$.)
| {
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Intriguing Indefinite Integral: $\int ( \frac{x^2-3x+1/3 }{x^3-x+1})^2 \mathrm{d}x$
Evaluate
$$\int \left( \frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 \mathrm{d}x$$
I tried using partial fractions but the denominator doesn't factor out nicely. I also substituted $x=\dfrac{1}{t}$ to get
$$\frac{-1}{9} \int \left(\frac{t^2-9t+3}{t^3-t^2+1}\right)^2 \, \mathrm{d}t $$
But I don't know how to solve this either.
Please Help.
Thanks in advance.
| Partial fractions also works, this may be a bit more work than trying to find a clever ad hoc solution, but the advantage is that it is a straightforward method.
If $\alpha$ is a root of the denominator, then putting $x = \alpha + t$ and performing a so-called Laurent expansion around $t=0$ (which is analogous to a Taylor expansion, except that negative power terms may appear in the expansion) yields:
$$\frac{x^2-3x+\frac{1}{3}}{x^3 - x + 1} = \frac{\alpha^2 - 3\alpha + \frac{1}{3}}{3\alpha^2 -1}\frac{1}{t} +\left[\frac{2\alpha-3}{3\alpha^2-1} -\frac{3\alpha\left(\alpha^2-3\alpha+\frac{1}{3}\right)}{\left(3\alpha^2-1\right)^2}\right]+\mathcal{O}(t)$$
Squaring both sides yields:
$$\left(\frac{x^2-3x+\frac{1}{3}}{x^3 - x + 1}\right)^2 = \left( \frac{\alpha^2 - 3\alpha + \frac{1}{3}}{3\alpha^2 -1}\right)^2\frac{1}{t^2}+\mathcal{O}(1)$$
The $\frac{1}{t}$ term cancels out due to $\alpha$ satisfying the equation $\alpha^3 - \alpha +1 = 0$.
We thus have the partial fraction expansion:
$$\left(\frac{x^2-3x+\frac{1}{3}}{x^3 - x + 1}\right)^2 = \sum_{\alpha}\left( \frac{\alpha^2 - 3\alpha + \frac{1}{3}}{3\alpha^2 -1}\right)^2\frac{1}{(x-\alpha)^2}$$
(Proof: the difference between the l.h.s. and the r.h.s. is a rational function but which by construction has no singularities, so it must be a polynomial. But because the terms all tend to zero for $x\to\infty$, that polynomial must be identical to zero.)
This is obviously trivial to integrate, but how do we sum over the roots $\alpha$? Consider the function:
$$f(\alpha) = \frac{1}{\alpha^3 -\alpha +1}\frac{\left(\alpha^2 - 3\alpha + \frac{1}{3}\right)^2}{3\alpha^2 -1}\frac{1}{\alpha-x}$$
obtained by multiplying the summand of the answer by the logarithmic derivative of the denominator inside the square of the integrand. Then consider doing a partial fraction expansion using the Laurent expansion method as we did above. Since the function tends to zero for $\alpha \to \infty$, faster than $1/\alpha$, the sum of the coefficients of the terms with exponent -1 will sum to zero. By construction, the desired summation is part of this summation.
The answer is thus given by minus the sum of the coefficients the Laurent expansion terms with exponent -1 of $f(\alpha)$ around $\alpha = x$ and $\alpha = \pm \frac{1}{\sqrt{3}}$. Such coefficients are referred to as "residues" in the theory of complex functions.
The residue at $\alpha = x$ is obviously:
$$\frac{1}{x^3 -x +1}\frac{\left(x^2 - 3x + \frac{1}{3}\right)^2}{3x^2 -1}$$
The residue at $\alpha = \pm \frac{1}{\sqrt{3}}$ can be written as:
$$\frac{x+\frac{3}{2}}{1-3 x^2} \mp\frac{\frac{\sqrt{3}}{3}+\frac{3}{2}\sqrt{3}x}{1-3 x^2}$$
The integral, which is minus the sum of these residues, can thus be written as:
$$\frac{x^4+9x^3-\frac{35}{3}x^2+x+\frac{26}{9}}{\left(x^3-x+1\right)\left(3 x^2-1\right)}$$
| {
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Compute the values of two infinite products whose factors are the same I have the following question:
How to prove that
$(1-\frac{1}{2})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{6})\cdot (1+\frac{1}{7})\cdot (1-\frac{1}{8})\cdot\ ...\ = \frac{1}{2}\ $, but
$(1-\frac{1}{2})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{6})\cdot (1-\frac{1}{8})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{10})\cdot (1-\frac{1}{12})\cdot (1+\frac{1}{7})\cdot\ ...\ = \frac{1}{2 \sqrt{2}}\ ?$
Thanks for the help!
| The first is straightforward as already answered by @MarcoCantarini.
For the second
$$P = (1-\tfrac{1}{2})(1-\tfrac{1}{4}) (1+\tfrac{1}{3})(1-\tfrac{1}{6})(1-\tfrac{1}{8}) (1+\tfrac{1}{5}) (1-\tfrac{1}{10}) (1-\tfrac{1}{12}) (1+\tfrac{1}{7}) \cdots
$$
here's an alternate approach. Add a factor $(1+1/1)$ to get
$$
\begin{align}
2P &= (1+\tfrac{1}{1})(1-\tfrac{1}{2})(1-\tfrac{1}{4}) (1+\tfrac{1}{3})(1-\tfrac{1}{6})(1-\tfrac{1}{8}) (1+\tfrac{1}{5}) (1-\tfrac{1}{10}) (1-\tfrac{1}{12}) \cdots \\
&= \prod_{k=0}^{\infty} \left(\frac{2k+2}{2k+1}\right)\left(\frac{4k+1}{4k+2}\right)\left(\frac{4k+3}{4k+4}\right) \\
& = \prod_{k=0}^{\infty} \frac{(4k+1)(4k+3)}{(4k+2)^2} = \frac{1\cdot 3}{2 \cdot 2}\cdot \frac{5 \cdot 7}{6\cdot 6}\cdot \frac{9\cdot 11}{10\cdot 10}\cdots \\
& = \prod_{k=0}^{\infty} \left(1-\frac{1}{(4k+2)^2}\right)
\end{align}
$$
which looks something like the Wallis product for $\pi$, and can be resolved the same way using the infinite product for sine
$$
\frac{\sin(x)}{x} = \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2\pi^2}\right)
$$
Taking $x=\pi/2$ and $x=\pi/4$ we get
$$
\frac{2}{\pi} = \prod_{k=1}^\infty \left(1-\frac{1}{(2k)^2}\right) \\
\frac{4}{\sqrt{2}\pi} = \prod_{k=1}^\infty\left(1-\frac{1}{(4k)^2}\right) \\
2P \cdot \frac{4}{\sqrt{2}\pi} = \frac{2}{\pi}\\
P = \frac{1}{2\sqrt{2}}
$$
| {
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Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \implies 4 a^2+4 b^2+4 c^2 \geq (a-b)^2+(b-c)^2+(a-c)^2,$$
then the geometric mean inequality gives
$$ (a-b)^2+(b-c)^2+(a-c)^2 \geq 3[(a-b)^2(b-c)^2(a-c)^2]^{1/3} .$$
Combining the two we get
$$ a^2+b^2+c^2 \geq \frac{3}{4}[(a-b)^2(b-c)^2(a-c)^2]^{1/3},$$
which is not strong enough.
Can anyone prove this for $2^{1/3}$ in place of $3/4$?
| Let $x=a-b$ and $y=b-c$. Then $a-c=x+y$. Thus, we are left with
$$
(x+y+c)^2+(y+c)^2+c^2\ge(2x^2y^2(x+y)^2)^{1/3}\tag{1}
$$
To find the minimum of the left-hand side of $(1)$, we set the derivative with respect to $c$ to $0$:
$$
2(x+y+c)+2(y+c)+2c=0\tag{2}
$$
$(2)$ says that $c=-\frac{x+2y}3$. Plugging this into the left-hand side of $(1)$ yields
$$
\begin{align}
(x+y+c)^2+(y+c)^2+c^2
&=\left(\frac{2x+y}3\right)^2+\left(\frac{\vphantom{2}y-x}3\right)^2+\left(\frac{x+2y}3\right)^2\\[6pt]
&=\frac23\left(x^2+xy+y^2\right)\tag{3}
\end{align}
$$
Thus, inequality $(1)$ is equivalent to the two variable inequality
$$
\frac4{27}\left(x^2+xy+y^2\right)^3\ge x^2y^2(x+y)^2\tag{4}
$$
$(4)$ is obvious if $y=0$, so let $z=\frac xy$ and divide $(4)$ by $y^6$. We then get that $(1)$ is equivalent to the one variable inequality
$$
\frac4{27}\left(z^2+z+1\right)^3\ge z^2(z+1)^2\tag{5}
$$
and
$$
\frac4{27}\left(z^2+z+1\right)^3-z^2(z+1)^2=\frac1{27}(2z+1)^2(z-1)^2(z+2)^2\ge0\tag{6}
$$
Thus, $(1)$ is true; and using $(2)$ and $(6)$, we can determine that equality holds precisely when
$$
\overbrace{a+b+c=0}^{\large c=-\frac{x+2y}3}\qquad\text{and}\qquad\frac{a-b}{b-c}\in\left\{-\frac12,1,-2\right\}\tag{7}
$$
which is when one of $a$, $b$, or $c$ is $0$ and the sum of the other two is $0$.
| {
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Proving $\frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n}$ for $a,b>0, n\in\mathbb{N}$ by induction prove using induction:
$$ \frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n} $$
$$a,b \gt 0 , n \in N$$
my attempt:
*
*base $n=1$:
$$ \frac {2}{(a+b)} \le \frac {1}{a} + \frac {1}{b}$$
$$2ab \le b(a+b) + a(a+b)$$
$$2ab \le ab + b^2 + a^2 + ab$$
$$0 <= a^2 + b^2$$
sum of $2$ not negatives, done.
*assume it's right for $n=k \in N$.......
$$ \frac {2k}{(a+b)^k} \le \frac {1}{a^k} + \frac {1}{b^k}$$
*prove for $n=k+1$......
$$ \frac {2(k+1)}{(a+b)^{k+1}} \le \frac {1}{a^{k+1}} + \frac {1}{b^{k+1}}$$
i've tried various things... and get stuck everytime.
this is the attempt mimicking what the lecturer/teaching aid do usually with such problems (problems without parameters tho)...
$$ \frac {2(k+1)}{(a+b)^{k+1}} = $$
$$ \frac {1}{(a+b)} \times \left( \frac {2k}{(a+b)^k} + \frac {2}{(a+b)^k} \right) = $$
$$\left(M + \frac {2}{(a+b)^k} \right) \times \frac {1}{(a+b)}$$
go back to the assumption...
$$M \le \frac {1}{a^k} + \frac {1}{b^k}$$
$$M + \frac {2}{(a+b)^k} \le \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k}$$
$$\left( M + \frac {2}{(a+b)^k} \right) \times \frac {1}{a+b} \le \left( \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k} \right) \times \frac {1}{a+b}$$
now we need to prove...
$$\left( \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k} \right) \times\frac {1}{a+b} \le \frac {1}{a^{k+1}} + \frac {1}{b^{k+1}}$$
no matter how i try messing around with this... i reach a dead end.
please help.
| Direct Proof
First, I would re-write the inequality so that you want to show $2n\leq \frac{(a+b)^{n}}{a^n}+\frac{(a+b)^{n}}{b^n}$. Then, we will use two facts which I will prove before finishing.
Fact 1: Since $a,b>0$ it follows that $(a+b)^{n}=\sum_{i=0}^{n}\binom{n}{i}a^{i}b^{i}\geq \binom{n}{1}ab^{n-1}+\binom{n}{n-1}a^{n-1}b=n(ab^{n-1}+ba^{n-1})$. (Here I've used the Binomial theorem and just left off some positive terms, so the sum was made smaller. I could put strict inequality if I like.)
Fact 2: Since $a,b>0$ it follows that $\left(\frac{a}{b}+\frac{b}{a}\right)\geq 2$. This can be seen as follows:
\begin{align*}
0 &\leq (b-a)^{2} = b^2 - 2ab + a^2 \\
\Rightarrow 2ab &\leq a^{2}+b^{2} \\
\Rightarrow 2 &\leq \frac{a}{b} + \frac{b}{a}
\end{align*}
Now, putting these two facts to use:
\begin{align*}
\frac{(a+b)^{n}}{a^{n}} + \frac{(a+b)^{n}}{b^{n}} &\geq \frac{n(ab^{n-1}+ba^{n-1})}{a^{n}} + \frac{n(ab^{n-1}+ba^{n-1})}{b^{n}} \\
&\geq n\left(\frac{ba^{n-1}}{a^{n}}+\frac{ab^{n-1}}{b^{n}}\right) \\
&=n\left(\frac{b}{a}+\frac{a}{b}\right) \\
&\geq 2n.
\end{align*}
Inductive Proof
Base Case: (sure)
The key fact that we will need is that $(a+b)^{n}\geq a^{n}$ and $(a+b)^{n}\geq b^{n}$. This is only true when $a,b>0$.
Inductive Step: We know that $2n\leq (a+b)^{n}\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}\right)$ and we want to show that $2n+2=2(n+1)\leq (a+b)^{n+1}\left(\frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}\right)$.
The first thing (as you rightly did) is that we need to make our inductive hypothesis useful by pulling out something that looks like it.
\begin{align*}
(a+b)^{n+1}\left(\frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}\right) &=(a+b)^{n}\left(\frac{a+b}{a^{n+1}}+\frac{a+b}{b^{n+1}}\right) \\
&=(a+b)^{n}\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}\right) + (a+b)^{n}\left(\frac{b}{a^{n+1}}+\frac{a}{b^{n+1}}\right) \\
&\geq 2n + (a^{n}+b^{n})\left(\frac{b}{a^{n+1}}+\frac{a}{b^{n+1}}\right) \\
&\geq 2n + \frac{a^{n}b}{a^{n+1}} + \frac{ab^{n}}{b^{n+1}} \text{ (selective distribution)} \\
&= 2n + \frac{b}{a}+\frac{a}{b} \\
&\geq 2n + 2 \\
&=2(n+1).
\end{align*}
Here, we again used the fact that $\frac{a}{b}+\frac{b}{a}\geq 2$ that I referenced in the other solution. If it wasn't clear, the selective distribution means that I dropped the middle terms of my "FOIL"ing... i.e. I multiplied only the first two terms together and the last two terms together; the result is smaller because I dropped the middle terms that were positive.
| {
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$4$-element subsets of the set $\{1,2,3,\ldots,10\}$ that do not contain any pair of consecutive numbers Find the number of subsets of the set $\{1,2,3,\ldots,10\}$ that contain $4$ elements and do not include any pair of consecutive numbers.
For example $\{1,2,5,7\}$ is not an example of such a subset since $1,2$ are consecutive. I think that we can set the $4$ cases to solve it, case one exactly $2$ numbers of consecutive elements, case two exactly $3$ numbers of consecutive elements, case three exactly $3$ numbers of consecutive elements and case four exactly $4$ numbers of consecutive elements, and we sum $4$ cases and minus them from all choices.
| Let's apply your strategy of using the Inclusion-Exclusion Principle.
There are $\binom{10}{4}$ four-element subsets of a ten-element set. From these, we must exclude those subsets containing at least two consecutive numbers.
A string of at least two consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the nine numbers less than $10$. Once we select a pair of consecutive numbers beginning with one of these nine numbers, we must select two additional elements of the ten-element set from the eight elements that are not in the pair of consecutive numbers, which we can do in $\binom{8}{2}$ ways. Thus, it appears we have $9 \cdot \binom{8}{2}$ sets containing at least two consecutive numbers. However, we have overcounted since we have counted sets containing two disjoint pairs twice. There are $\binom{8}{2}$ such sets since we must exclude from the $\binom{9}{2}$ ways of selecting two of the nine numbers less than $10$ as starting points for the two pairs of consecutive integers the $\binom{8}{1}$ ways of selecting consecutive starting points that would prevent them from being disjoint. Note that it is a consequence of Pascal's Identity that $\binom{9}{2} - \binom{8}{1} = \binom{8}{2}$. Hence, there are
$$9 \cdot \binom{8}{2} - \left[\binom{9}{2} - \binom{8}{1}\right] = 9 \cdot \binom{8}{2} - \binom{8}{2} = 8 \cdot \binom{8}{2}$$
four-element subsets containing at least two consecutive numbers.
A string of at least three consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the eight numbers less than $9$. Once we select a triple of consecutive numbers, we can select the remaining element in the subset from one of the seven elements in the set that are not in the triple, giving
$$8 \cdot \binom{7}{1}$$
four-element subsets that contain at least three consecutive numbers.
A string of four consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the seven numbers less than $8$.
Thus, by the Inclusion-Exclusion Principle, the number of four-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ that do not contain consecutive numbers is
$$\binom{10}{4} - 8 \cdot \binom{8}{2} + 8 \cdot \binom{7}{1} - 7 = 210 - 224 + 56 - 7 = 35$$
Addendum: A more efficient approach is to arrange six blue and four green balls in a row so that no two of the green balls are consecutive, then number the balls from left to right. The numbers on the green balls are the desired subset of four numbers, no two of which are consecutive, selected from the set $[10] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Line up six blue balls in a row. This creates seven spaces, five between successive blue balls and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$
To ensure that no two of the green balls are consecutive, we select four of these seven spaces in which to place a green ball, which can be done in
$$\binom{7}{4} = 35$$
ways. We now number the balls from left to right. The numbers on the green balls are the desired subset of four numbers selected from the set $[10]$, no two of which are consecutive.
| {
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40th derivative of a function I would like to have some verification to see if my answer is correct. The given function is $f(x)=ln(1+x^2)$ and I need the 40th derivative at $x=0$. Here is my work: Using series one can manipulate $\frac{1}{1+x}=1-x+x^2-x^3+x^4...=SUMx^n(-1)^n$ into $\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8...=SUMx^{2n}(-1)^n$. Then $\frac{2x}{1+x}=2x-2x^3+2x^5-2x^7+2x^9...=2SUMx^{2n+1}(-1)^n$. Integrating gives $ln(1+x^2)=...=2SUM\frac{x^{2n+2}(-1)^n}{2n+2}$ where a $2$ cancels to arrive at a $n+1$ in that denominator. Now for the 40th derivative, $2n+2=40$ gives $n=19$ and thus I believe the answer is $\frac{-40!}{20}$ Do you concur? If not could you correct me? Thanks.
| Taking the $40$th derivative at 0 will yield you the coefficient for the $x^{40}$ term multiplied by $40!$
The general series for natural log is:
$$
ln(x+1) = \frac {x^1}1 - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4...
$$
replacing $x^2$ makes:
$$
ln(x^2+1) = \frac {x^2}1 - \frac {x^4}2 + \frac {x^6}3 - \frac {x^8}4...
$$
From this it is apparent that the 20th term of the first series shares its coefficient with the 40th term of the second series.
The coefficient equals $-\frac 1{20}$. Multiplying $40!$ to cancel the implied division yields: $-\frac {40!}{20}$
From this I confirm your answer to be correct; however, your proof and solution is incredibly messy. There is a known series for natural log. Using a fractional series and integrating was a waste of time and effort.
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How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
| Define $f(\alpha)=\int\dfrac{dx}{\alpha+x^2}=\frac{1}{\sqrt\alpha}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)$
You can recover the original integral by putting $\alpha=1$ later,
Now, differentiating both sides.
$f'(\alpha)=\int\dfrac{-dx}{(\alpha+x^2)^2}=\frac{-1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{-x}{2(\alpha)^{3/2}}$
$\int\dfrac{dx}{(\alpha+x^2)^2}=\frac{1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{x}{2(\alpha)^{3/2}}$
Put $\alpha=1$
$$\int\dfrac{dx}{(1+x^2)^2}=\frac{1}{2}\tan^{-1}x+\frac{1}{1+x^2}.\frac{x}{2}$$ and of course the remaining $C$
And there you have your integral found by differentiating only.^^
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, with $\mathbf{a+b+c=0}$. The angle between any two of these vectors is $120^\circ$. Let $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ be unit vectors, such that $\mathbf{a}+\mathbf{b}+\mathbf{c} = \mathbf{0}$. Show that the angle between any two of these vectors is $120^\circ$.
Hi,
I have been having some trouble with this problem.
I have tried to assign variables to the vectors, and creating various equations with them. However, I can't figure out how to use the $\mathbf{a+b+c=0}$ information.
Thanks!
| Geometrically, $\mathbf a + \mathbf b + \mathbf c = \mathbf 0$ means we can form a triangle $\triangle XYZ$ whose oriented sides $\overrightarrow{XY}, \overrightarrow{YZ}, \overrightarrow{ZX}$ are parallel copies of $\mathbf a$, $\mathbf b$, $\mathbf c$.
Since $\mathbf a$, $\mathbf b$, $\mathbf c$ are unit vectors, $\triangle XYZ$ is equilateral, so $\angle X = \angle Y = \angle Z = 60^\circ$.
But $\angle X$ is the angle between $\overrightarrow{XY}$ and $\overrightarrow{XZ}$, or between $\mathbf a$ and $-\mathbf c$. If the angle between $\mathbf a$ and $-\mathbf c$ is $60^\circ$, then the angle between $\mathbf a$ and $\mathbf c$ is its supplementary angle, $120^\circ$.
Similarly, we conclude from $\angle Y = \angle Z = 60^\circ$ that the angle between $\mathbf a$ and $\mathbf b$, and the angle between $\mathbf b$ and $\mathbf c$, are both $120^\circ$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help understanding how to factor completely $x^3-x^2-x+1$ I need someone to help explain the steps to completely factor the problem $x^3-x^2-x+1$.
Here is what I have done so far:
$x^3-x^2-x+1$
to
$x^3-x^2+-1(x+1)$
Since there is a negative I changed $-x+1$ to $-1(x+1)$
to
$x^2(x-1)+-1(x+1)$
to
$(x^2-1)(x-1)(x+1)$.
Is this correct? If not please explain what I did wrong and what I should do. If it is then thank you for saying so, if not, thank you for helping me to understand.
| $$x^3 - x^2 - x + 1 \to x^2(x - 1) + (-1)(x-1)$$
$$\to (x-1)(x^2 - 1) \to (x-1)^2(x+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be? I tried solving this question by
$1.$ $-1$ and $4$ will not be in domain because denominator can not be zero .
$2.$ Either both denominator and numerator will be positive or negative so that whole term in root becomes positive.
But I am not able to solve the upper part , can it be done by taking $\log$ ?
| The zeroes of the denominator are $\frac{4 \pm \sqrt{32}}{2} = 2(1 \pm \sqrt{2})$.
Therefore the denominator is negative iff $2(1 - \sqrt{2}) < x < 2(1 + \sqrt{2})$.
The upper part is positive if $x<0$, zero if $x=0$, and negative is $x>0$
So they ae both positive if $x < 2(1 - \sqrt{2})$.
They are both negative if $0 \leq x < 2(1 + \sqrt{2})$.
The answer is therefore $(\leftarrow, 2(1 - \sqrt{2})) \cup [0, 2(1 + \sqrt{2}))$
| {
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"timestamp": "2023-03-29T00:00:00",
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Question regarding proving by induction I am struggling with a math problem I have been assigned. The problem is as follows:
Let $X_1 = -3$ and $X_2 = 0$. Given that for every natural number $n \geq 2, X_{n+1} = 7X_n - 10X_{n-1}$, prove by induction that for every $n$ belonging to $\mathbb{N}$, $X_n = 2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}$
Right now all I have proven so far is the two base cases, I am not sure how to take this proof any further. Any assistance you can offer is greatly appreciated, thank you.
| Assume true for $n = k$, Which means
${X_k} = 2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}$
${X_{k + 1}} = 7{X_k} - 10{X_{k - 1}}$
Required to prove for $n = k + 1$ That is required to prove
${X_{k + 1}} = 2 \cdot {5^k} - 5 \cdot {2^k} = 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) - 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}})$
or
$2 \cdot {5^k} - 5 \cdot {2^k} - 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) + 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}}) = 0$
Now
$2 \cdot {5^k} - 5 \cdot {2^k} - 7(2 \cdot {5^{k - 1}} - 5 \cdot {2^{k - 1}}) + 10(2 \cdot {5^{k - 2}} - 5 \cdot {2^{k - 2)}}) =$
$ = 2 \cdot {5^k} - 14 \cdot {5^{k - 1}} + 20 \cdot {5^{k - 2}} - 5 \cdot {2^k} + 35 \cdot {2^{k - 1}} - 50 \cdot {2^{k - 2}}$
$ = 2 \cdot {5^{k - 2}}({5^2} - 7 \times 5 + 10) + 5 \cdot {2^{k - 2}}( - {2^2} + 7 \times 2 - 10)$
$ = 2 \cdot {5^{k - 2}}(0) + 5 \cdot {2^{k - 2}}(0)$
$ = 0$
| {
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Describe the set of odd primes such that $\left(\frac{-5}{p}\right) = 1$ (Legendre Symbol) Okay, so $\left(\frac{-5}{p}\right) = 1$.
I am assuming that I can start this by saying
$\left(\frac{-5}{p}\right) = \left(\frac{5}{p}\right) \times \left(\frac{-1}{p}\right)$.
There are well defined rules for $\left(\frac{-1}{p}\right)$.
Such that if I want $\left(\frac{-1}{p}\right) = 1$, then I must have $p \equiv 1$ mod $4$.
Since $5$ is congruent to $1$ mod $4$, I can say $\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right)$.
So now I must find $p$ such that $\left(\frac{p}{5}\right) = 1$. By quadratic reciprocity I can find that $\left(\frac{p}{5}\right) = 1$ when $p \equiv \{1,4\}$ mod $5$.
By CRT I get $p \equiv \{1,9\}$ mod $20$.
I must also consider the case when both $\left(\frac{-1}{p}\right) = -1$ and $\left(\frac{p}{5}\right) = -1$.
$\left(\frac{-1}{p}\right) = -1$ when $p \equiv 3$ mod $4$.
$\left(\frac{p}{5}\right) = -1$ when $p \equiv \{2,3\}$ mod $5$.
So by CRT I get $p \equiv \{3,7\}$ mod $20$.
So I am assuming it would be safe to say that
$\left(\frac{-5}{p}\right) = 1$ when $p \equiv \{1,3,7,9\}$ mod $20$.
Is there anything wrong with this? Or is there a more obvious and faster way to accomplish this?
| $\left(\frac{-5}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{5}{p}\right)=(-1)^{\frac{p-1}{2}}\left(\frac{5}{p}\right)\stackrel{\text{QR}}=(-1)^{\frac{p-1}{2}}\left(\frac{p}{5}\right)$
$(-1)^{\frac{p-1}{2}}=\begin{cases}1, &p\equiv 1\pmod{\!4}\\-1, &p\equiv -1\pmod{\!4}\end{cases}$
$0^2\equiv \color{#0bc}{0},\ 1^2\equiv \color{#0bc}{1},\ 2^2\equiv\color{#0bc}{4},\ 3^2\equiv (-2)^2\equiv 2^2,\ 4^2\equiv 1^2\pmod{\!5}.$
$\left(\frac{p}{5}\right)=\begin{cases}1, &p\equiv \{\color{#0bc}{1},\color{#0bc}{4}\}\pmod{\!5}\\-1, &p\equiv \{2,3\}\pmod{\!5}\\ 0, &p\equiv \color{#0bc}{0}\pmod{\!5}\end{cases}$
Notice that above didn't use QR but used basic checking, like André said in a comment.
$(-1)^{\frac{p-1}{2}}\left(\frac{p}{5}\right)=\left\{\begin{array}{}1,\ \ \ \ \left\{\left\{\begin{array}{}p\equiv 1\pmod{\!4},\ &p\equiv \{1,4\}\pmod{\!5}\\p\equiv -1\pmod{\!4},\ &p\equiv \{2,3\}\pmod{\!5}\end{array}\right\}\right\}\\-1,\ \left\{\left\{\begin{array}{}p\equiv 1\pmod{\!4},\ &p\equiv \{2,3\}\pmod{\!5}\\p\equiv -1\pmod{\!4},\ &p\equiv \{1,4\}\pmod{\!5}\end{array}\right\}\right\}\\0, \ \ \ \ \ \ \ p\equiv 0\pmod{\!5}\end{array}\right\}$
$\stackrel{\text{CRT}}=\left\{\begin{array}{}1,&p\equiv \{1,3,7,9\}\pmod{\!20}\\-1,&p\equiv \{11,13,17,19\}\pmod{\!20}\\ 0,&p\equiv 0\pmod{\!5}\end{array}\right\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\int_{-1}^{1} (x^{4/3} + 4x^{1/3}) dx$? I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?
| You are asking "What is wrong with it?"
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
Check the derivative
$$(\frac{3}{7}\frac{7}{3} x^{4/3} + \color{red}{\frac{16}{4}\frac{4}{3}}x^{1/3}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271847",
"timestamp": "2023-03-29T00:00:00",
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$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$, & excluding $(x+y)$ the sum of the digits $(3+5+5+6+4+2+8+9+6) =48$.Thus excluding option d) the ans can be any of a),b) or c).
Please help me to find the exact answer.
Thank you
| Testing divisibility by 9 gives
$$
48 + x + y \equiv 0 \pmod {9}.
$$
Testing divisibility by 11 gives
$$
12 - x + y \equiv 0 \pmod {11}
$$
So $x - y \equiv 1 \pmod {11}$ and $x + y \equiv 6 \pmod{9}$.
Since $x, y$ are both digits, $x = 1 + y$, and $x + y$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $\sum \limits_{k=1}^{n} x_n=1$ then $\sum \limits_{k=1}^{n} x_n^2 \geq {1 \over n}$ Prove that if $\sum \limits_{k=1}^{n} x_n=1$ then $\sum \limits_{k=1}^{n} x_n^2 \geq {1 \over n}$ where $\{x_k\}_1^n$ are real numbers which are not all the same.
I tried to prove it by induction. The base was trivial. Now, suppose for every numbers $\{x_k\}_1^n$ that satisfy $\sum \limits_{k=1}^{n} x_n=1$ the following inequality is true: $\sum \limits_{k=1}^{n} x_n^2 \geq {1 \over n}$.
Let $\{x_k\}_1^{n+1}$ be real numbers that satisfy $\sum \limits_{k=1}^{n+1} x_k$, and let us denote $y:=x_n +x_{n+1}$, so according to the induction hypothesis $\sum \limits_{k=1}^{n-1} x_k^2 +y^2\geq {1 \over n}$.
$$\sum \limits_{k=1}^{n+1} x_k^2= \sum \limits_{k=1}^{n-1} x_k^2+x_n+x_{n+1}+y^2-y^2\geq {1 \over n}+x_n^2+x_{n+1}^2-y^2=\frac{1}{n}-2x_n x_{n+1}$$
And I'm stuck. I thought that there must be an element in $\{x_k\}_1^{n+1}$ that is less or equal to $1 \over {n+1}$ (since $\sum \limits_{k=1}^{n+1} x_k$ and the elements in the sum are not all the same), but it didn't help much.
| This is false. For instance, $x_1 = 1$ and rest of $x_i$'s zero violates the inequality. Your inequality should be the other way around, which follows from Cauchy Schwarz inequality.
By Cauchy Schwarz, we have
$$\left \vert\vec{a} \cdot \vec{b} \right \vert \leq \left \Vert \vec{a}\right \Vert \left \Vert \vec{b}\right \Vert$$
where $\vec{a},\vec{b} \in \mathbb{R}^n$.
Take $\vec{a} = \begin{bmatrix}x_1 & x_2 & \cdots & x_n\end{bmatrix}^T$, and $\vec{b} = \begin{bmatrix}1 & 1 & \cdots & 1\end{bmatrix}^T$. We then have
$$\left \vert \vec{a} \cdot \vec{b} \right \vert = x_1 + x_2 + \cdots + x_n = 1$$
whereas $\left \Vert \vec{a}\right \Vert = \sqrt{x_1^2+x_2^2 + \cdots + x_n^2}$ and $\left \Vert \vec{b}\right \Vert = \sqrt{1+1+\cdots+1} = \sqrt{n}$. This gives us that
$$\sqrt{x_1^2+x_2^2 + \cdots + x_n^2} \geq \dfrac1{\sqrt{n}} \implies x_1^2+x_2^2 + \cdots + x_n^2 \geq \dfrac1n$$
Here is a proof by induction. Clearly, base case is trivial. Assume that for some $n$, we have that
$$x_1+x_2+\cdots+x_n = 1 \implies x_1^2+x_2^2 + \cdots + x_n^2 \geq \dfrac1n$$
Now consider
$$x_1+x_2+\cdots+x_n+x_{n+1} = 1$$
Clearly for $n>1$, there exists one $x_i$, which is not $1$, say let it be $x_{n+1}$. We then have
$$x_1+x_2+\cdots+x_n = 1-x_{n+1} \implies \dfrac{x_1}{1-x_{n+1}} + \dfrac{x_2}{1-x_{n+1}} + \cdots + \dfrac{x_n}{1-x_{n+1}} = 1$$
which from induction hypothesis gives us
$$\dfrac{x_1^2}{(1-x_{n+1})^2} + \dfrac{x_2^2}{(1-x_{n+1})^2} + \cdots + \dfrac{x_n^2}{(1-x_{n+1})^2} \geq \dfrac1n$$
This gives us that
$$x_1^2 + x_2^2 + \cdots +x_n^2 \geq \dfrac{(1-x_{n+1})^2}n$$
which gives us that
\begin{align}
x_1^2 + x_2^2 + \cdots +x_n^2 + x_{n+1}^2 & \geq \dfrac{(1-x_{n+1})^2}n + x_{n+1}^2 = \dfrac{1-2x_{n+1}+(n+1)x_{n+1}^2}n\\
& = \dfrac1{n(n+1)} \left((n+1)^2x_{n+1}^2 - 2(n+1)x_{n+1} + 1 + n\right)\\
& = \dfrac{((n+1)x_{n+1}-1)^2 + n}{n(n+1)} = \dfrac{((n+1)x_{n+1}-1)^2}{n(n+1)} + \dfrac1{n+1}\\
\geq \dfrac1{n+1}
\end{align}
which completes our induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof without using induction How to prove that
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
without using induction.
If we don't know the right side of this expression, how to get right expression. I tried with partial sums and binomial formula but can't get it.
So the problem is:
$$1^2+2^2+...+n^2=?$$
Thanks for replies.
| Assume we know $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$. Compute the following cubes
$$\begin{align}
1^3&=1\\
(1+1)^3&=1^3+3\cdot 1^2+3\cdot 1+1^3\\
\vdots&=\vdots\\
n^3&=(n-1)^3+3(n-1)^2+3(n-1)+1^3\\
(n+1)^3&=n^3+3n^2+3n+1^3\end {align}$$
Add these equations together and cancel the cubes you have on both sides you get
$$\begin{align}(n+1)^3&=1+3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+n\\
&=(n+1)\frac{3n+2}{2}+3\sum_{k=1}^nk^2\end{align}$$
This yields
$$\sum_{k=1}^nk^2=\frac{n+1}{3}\left(n^2+2n+1-\frac{3n+2}{2}\right)=\frac{(n+1)(2n^2+n)}{6}$$
Factoring $n$ we get the result expected
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding roots of 5 th degree taylor expansion of $e^x$ I need to find the roots [actually their count and sign] of :
$$
{\mathcal T}(e^x,5)=\frac{x^5}{5!}+\frac{x^4}{4!}+\frac{x^3}{3!}+\frac{x^2}{2!}+x+1$$
*
*It[Root] is clearly negative.
*It should have odd number of roots.
*Either One or Three or Five.
*It is easy to find the derivatives, but only the third derivative, a quadratic can be easily analyzed.
Any help?
Basics Only.
| A similar way to @Daniel Fischer's answer (if not quite as clean) to show that the derivative is positive:
$$\begin{aligned}
\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1 &= \frac{1}{24}\big(x^4+4x^3+12x^2+24x+24\big)\\
&= \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+23)\big)\\
&> \frac{1}{24}\big((x^4+4x^3+6x^2+4x+1) + (6x^2+20x+\frac{50}{3})\big)\\
&= \frac{1}{24}\big((x+1)^4 + 6(x+\frac53)^2\big)\\
&> 0.
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.