Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
| It is easy to see (by induction) that $x_{n+1}>x_n \ge a >0$ for all $n$.
We assume that $(x_n)$ is convergent and that $l$ is the limmit. Then $l \ge a>0$.
From $x_{n+1}= x_n + (1/x_n)$ we get $l= l+ 1/l$, which is impossible. Hence the sequence is divergent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f'(1)$ given $f(x)$ and $f(1)$ Question:
Let $$f(x)=\int\frac{x-1}{x+1}\frac{dx}{\sqrt{x^3+x^2+x}}$$ such that $f(1)=\frac{2\pi}{3}$
Then $f'(1)$ is equal to
A) $0$
B) $\cfrac \pi3$
C) $\cfrac\pi4$
D) $2$
Attempt:
\begin{align}f(x)&=\int\frac{x-1}{x+1}\frac{1}{\sqrt{x^3+x^2+x}}dx\\
f(1)&=\int0dx\\
f(1)&=0+c\\
&= \frac{2\pi}{3}\\
\implies c&= \frac{2\pi}{3}
\end{align}
I don't know what to do beyond this step. This is an indefinite integral so we cannot differentiate it. How else we bring $f'(x)$ into picture? Any hints?
| Does $f'(x)$ mean $\frac{df}{dx}$. Isnt $f'(x)$ then simply the integrated term, with the integration removed? ie, $\frac{x-1}{x+1}\times \frac{1}{\sqrt{x + x^2 + x^3}}$ ? Am I missing something?
| {
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Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
Prove that
$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
What I've tried doing :
If $\theta=\frac{\pi}{7}:$
$$
3\theta+4\theta=\pi
$$
This allowed me to prove that :
$$
\tan^2\frac{\pi}{7}+\tan^2\frac{2\pi}{7}+\tan^2\frac{3\pi}{7}=21
\\
\cot^2\frac{\pi}{7}+\cot^2\frac{2\pi}{7}+\cot^2\frac{3\pi}{7}=5
$$
Is my reasoning wrong or is this entirely the wrong way to approach this question ?
| You can show that $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ are the three roots of $64x^3-112x^2+56x-7 = 0$, as follows:
Take $\zeta_7$ a $7$'th root of unity (a root of $(x^7-1)/(x-1) = x^6+x^5+x^4+x^3+x^2+x+1 = 0$), then $2 \cos(k 2\pi/7) = \zeta_7^k + \zeta_7^{-k}$, and with $\sin^2+\cos^2=1$ you can now express $\sin^2(\pi/7)$, $\sin^2(2\pi/7)$, and $\sin^2(3\pi/7)$ in terms of $\zeta_7$. After that you can verify that they are roots of $64x^3-112x^2+56x-7 = 0$.
| {
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Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?
| The logarithm is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{k}x^{3k}$. Making sure to multiply by $x$ before we integrate, the final result is $\sum_{k=1}^\infty\frac{\left( -1\right)^{k-1}}{2^{3k+2}k\left( 3k+2\right)}$.
| {
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Evaluate the integral $\int_{0}^{\pi}\frac{\log(1+a\cos x)}{\cos x}\,dx$ Please help me to evaluate the integral $\int_{0}^{\pi}\frac{\log(1 + a\cos x)}{\cos x}\,dx$. In the question they said to use the formula $\frac{d}{dy}\int_{a}^{b}f(x,y)dx=\int_{a}^{b}\frac{\partial }{\partial y}f(x,y)dx$.
| Well, we have:
$$\frac{\text{d}}{\text{d}\text{a}}\left\{\int_0^\pi\frac{\ln\left(1+\text{a}\cos\left(x\right)\right)}{\cos\left(x\right)}\space\text{d}x\right\}=\int_0^\pi\frac{\frac{1}{\text{a}+\sec\left(x\right)}}{\cos\left(x\right)}\space\text{d}x=$$
$$\int_0^\pi\frac{1}{1+\text{a}\cos\left(x\right)}\space\text{d}x=\frac{\pi}{\sqrt{1-\text{a}^2}}\tag1$$
EDIT: substitute $\text{u}=\tan\left(\frac{x}{2}\right)$:
$$\int\frac{1}{1+\text{a}\cos\left(x\right)}\space\text{d}x=\frac{2}{1+\text{a}}\int\frac{1}{1+\text{u}^2\cdot\frac{1-\text{a}}{1+\text{a}}}\space\text{d}\text{u}\tag2$$
Now, substitute $\text{p}=\text{u}\cdot\sqrt{\frac{1-\text{a}}{1+\text{a}}}$:
$$\frac{2}{1+\text{a}}\int\frac{1}{1+\text{u}^2\cdot\frac{1-\text{a}}{1+\text{a}}}\space\text{d}\text{u}=\frac{2}{\sqrt{1+\text{a}}\cdot\sqrt{1-\text{a}}}\int\frac{1}{1+\text{p}^2}\space\text{d}\text{p}=$$
$$\frac{2}{\sqrt{1+\text{a}}\cdot\sqrt{1-\text{a}}}\cdot\arctan\left(\text{p}\right)+\text{C}=$$
$$\frac{2}{\sqrt{1+\text{a}}\cdot\sqrt{1-\text{a}}}\cdot\arctan\left(\tan\left(\frac{x}{2}\right)\cdot\sqrt{\frac{1-\text{a}}{1+\text{a}}}\right)+\text{C}\tag3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a gmat exam question.
| From the definition $A=(10^{33}-1)/3$
(if that is what you meant - a $33$-digit number, all threes)
$\bmod 19$, we have
$10^2\equiv 5$,
$10^3\equiv 12$,
$10^4\equiv 6$,
$10^5\equiv 3$,
$10^6 \equiv 11$,
$10^9\equiv 12\cdot 11 \equiv 18\equiv -1$,
$10^{18}\equiv (10^9)^2\equiv 1$,
$10^{33}\equiv 10^{15}\equiv -1\cdot 11 \equiv 8$.
Then since $10^{5}\equiv 3$,
$3^{-1}\equiv 10^{18-5}\equiv 10^{13}\equiv 10^910^4 \equiv -6\equiv 13$ and
$(10^{33}-1)/3 \equiv (8-1)\cdot 13 \equiv 91 \equiv 15 \bmod 19$
If you actually meant a $165$-digit number, all threes, then the above preparatory work can support that easily:
$10^{165} \equiv 10^{18\cdot 9}10^{3} \equiv 10^{3} \equiv 12\bmod 19$ and
$(10^{165}-1)/3 \equiv (12-1)\cdot 13 \equiv -8\cdot -6 \equiv 48 \equiv 10 \bmod 19$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Function notation with term before f(x) I saw an exam question using this notation:
$$x^2\,f(x) = x^2 + 7x + 3\,.$$
I didn't understand it, so I checked the answer and I saw that was equal to:
$$f(x) = \frac{x^2 + 7x + 3}{x^2}\,.$$
Is the former a common function notation? Where or how can I find more information about it?
Thanks in advance.
| Assuming you have copied it verbatim, it's a simple redistribution of terms. Suppose for example that you wanted to subtract 3 from both sides. You would then have $$x^2 f(x) - 3 = x^2 + 7x.$$
That was just for the sake of example. Let's go back to $$x^2 f(x) = x^2 + 7x + 3.$$ Now we want to divide both sides by $x^2$. Then we have $$\frac{x^2 f(x)}{x^2} = \frac{x^2 + 7x + 3}{x^2}$$ $$\frac{x^2}{x^2} f(x) = \frac{x^2 + 7x + 3}{x^2}.$$ As we're assuming $x \neq 0$, it follows that $$\frac{x^2}{x^2} = 1,$$ and so $$1 f(x) = \frac{x^2 + 7x + 3}{x^2}$$ $$f(x) = \frac{x^2 + 7x + 3}{x^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the orthogonal trajectories of the family of curves
Find the orthogonal trajectories of the family of curves $y^5=kx^2$.
I start by knowing that I need a differential equation satisfying all members of family, which means $k$ needs to be eliminated.
Differentiating both sides with respect to $x$:
\begin{align*}
y^5 &= kx^2\\
5y^4 \frac{dy}{dx} &= 2kx\quad\text{Replacing }k\\
5y^4 \frac{dy}{dx} &= 2 \frac{y^5}{x^2}x\\
\frac{5}{y} \frac{dy}{dx} &= \frac{2}{x}\\
\frac{dy}{dx} &= \frac{2}{5} \frac{y}{x}
\end{align*}
Now I have a differential equation, and the orthogonal trajectory should be represented by the negative reciprocal.
\begin{align*}
\frac{dy}{dx}&=-\frac{5}{2}\frac{x}{y}\\
\int ydy&=\int-\frac{5}{2}xdx\\
y&=-\frac{5}{4}x^2+C
\end{align*}
This equation looks orthogonal to me if I plot it, but all the valid answers are of a higher order (e.g., $y^2+\frac{5}{2}x^2=C$). There must be a gap in my understanding somewhere.
https://www.desmos.com/calculator/v6on063jfn
| I made a mistake when taking $\int ydy$. It should be $\frac{y^2}{2}$ not $y$.
This means that,
$$
\frac{y^2}{2}=-\frac{5}{4}x^2 + C\\
\frac{y^2}{2}+\frac{5}{4}x^2=C\\
y^2+\frac{5}{2}x^2=2C=C_2
$$
Which is the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+\tan^3(x)} dx$$
Then to evaluate $\int \frac{1}{1+\tan^3(x)}dx$, I let $\tan(x)=t$ and hence $dt = \sec^2(x) dx$, which implies $dx=\frac{1}{\sec^2(x)}dt$.
Then I don't know how to continue.
| Let $x=\frac{\pi}2-u$. Then $dx=-du$ and the integral becomes:
$$I=-\int_{\pi/2}^0 \frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du\\=\int_0^{\pi/2}\frac{\cos^3(u)}{\cos^3 (u)+\sin^3(u)}du$$
So your integral satisfies:
$$2I=\int_0^{\pi/2} 1dx=\frac{\pi}2$$
| {
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If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further?
| By the cosine rule, $\cos B = \frac{1}{2}$ iff $b^2=a^2+c^2-ac$. That's the result you need to try to obtain by rearrangement.
| {
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Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$
Let $n,x,y,z$ be positive integers such that $$(x-y)^2+(y-z)^2+(z-x)^2 = (2n+1)xyz.$$ Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$.
The original equation doesn't have $x^3,y^3,z^3$, so how can we show that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$? Expanding the given equation seems to make the question more complicated, and also how do we get the $6$?
| This is linked to a well-known identity (a standard exercise in high-school):
\begin{align}x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\&=\tfrac12(x+y+z)\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone would provide some different solutions. Thanks in advance.
| Just to provide another way.
Use polar coordinates $$x=\rho \cos(\theta) \qquad y=\rho \sin(\theta)$$ to make the equations to be $$\frac{5 \left(\rho ^2+1\right) \cos (\theta )}{\rho }=12$$ $$\frac{5 \left(\rho ^2-1\right) \sin (\theta )}{\rho }=4$$ So,$$\cos(\theta)=\frac{12 \rho }{5 \left(\rho ^2+1\right)}\qquad \sin(\theta)=\frac{4 \rho }{5 \left(\rho ^2-1\right)}$$ Now, using $$\sin^2(\theta)+\cos^2(\theta)=1 \implies \frac{16 \rho ^2}{25 \left(\rho ^2-1\right)^2}+\frac{144 \rho ^2}{25 \left(\rho
^2+1\right)^2}=1$$ Now, let $t=\rho ^2$ to have $$\frac{16 t}{25 \left(t -1\right)^2}+\frac{144 t}{25 \left(t+1\right)^2}=1$$ Reduce to same denominator to get $$25 t^4-160 t^3+206 t^2-160 t+25=0$$ where you can notice the symmetry of coefficients.
Factoring gives $$25 t^4-160 t^3+206 t^2-160 t+25=(5-t) (5 t-1) \left(5 t^2-6 t+5\right)=0$$ The quadratic does not show real roots; this makes the solutions to be $$t_1=\frac 15\qquad t_2=5$$ then $$\rho_1=\frac 1 {\sqrt 5}\qquad \rho_2=\sqrt 5$$ and now compute the corresponding $\theta$'s from $$\tan(\theta)=\frac{\sin(\theta) }{\cos(\theta) }=\frac{t+1}{3 (t-1)}=-\frac 12$$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the sequence defined by $a_{n+1} = \frac{1}{a_n} + 1$ convergent? The sequence defined by $a_1=1$ and
$$a_{n+1} = \frac{1}{a_n} + 1$$
is increasing. And I can get the limit which is equal to $$\frac{1+\sqrt{5}}{2}$$
But $a_2 = 2$ which is larger than the limit. So is the sequence convergent or divergent?
| $\begin{array}\\
a_{n+1}
&=1+ \frac{1}{a_n}\\
&=1+ \frac{1}{1+ \frac{1}{a_{n-1}}}\\
&=1+ \frac{1}{1+ \frac{1}{1+ \frac{1}{a_{n-2}}}}\\
&...\\
&=1+ \frac1{1+ \frac1{1+ \frac1{1+...}}}
\qquad n \text{ deep}\\
\end{array}
$
But that continued fraction
is
$\dfrac{F_{n+1}}{F_n}
\to \phi
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this $1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$ Let $a,b,c,x,y,z\in\mathbb{R}$ such that \begin{align}\{a,b,c,x,y,z\}&\subset[-1,1]\\
1 + 2abc &\geq a^2+b^2+c^2\\
1+2xyz&\geq x^2+y^2+z^2\end{align}
Prove that:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
| Let $A=\left(\begin{matrix}1 & a & b \\ a & 1 & c \\ b & c & 1 \end{matrix}\right)$. Due to assumptions on $a,b,c$, all principal minors of $A$ are nonnegative. Therefore $A$ is positive-semidefinite.
Analogously, $B = \left(\begin{matrix}1 & x & y \\ x & 1 & z \\ y & z & 1 \end{matrix}\right)$ is positive-semidefinite.
Let $C$ be the Hadamard product of $A$ and $B$, i.e. $$C = A \circ B = \left(\begin{matrix}1 & ax & by \\ ax & 1 & cz \\ by & cz & 1 \end{matrix}\right).$$ It follows from Schur product theorem that $C$ is positive-semidefinite as well. In particular $\det(C) \ge 0$, which is precisely the inequality we wanted to prove.
| {
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Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+y))\sin (\frac{1}{2}(x-y))}{(x+y)(x-y)}\\
&=-\dfrac{2 \sin y}{2y}\dfrac 12\\
&=-\dfrac{\sin y}{2y}\end{align}
My question: Is $-\dfrac{\sin y}{2y}$ the final answer or can it be calculated further as $-\dfrac12$?
I also try different route:
Let $p=x-y$ so $x=p+y$ and $p\rightarrow 0$ as $x \rightarrow y$. Hence,
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{p\rightarrow 0}\dfrac{\cos (p+y)-\cos y}{p^2+2py+y^2-y^2}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))\sin (\frac{1}{2}p)}{p(p+2y)}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))}{(p+2y)} \dfrac{\sin (\frac{1}{2}p)}{p}\\\\
&=-\dfrac{2\sin y}{2y} \dfrac{1}{2}\\\\
&=-\dfrac{\sin y}{2y}\end{align}
Okay, so that left me with the same result. What is the correct final answer, $-\dfrac{\sin y}{2y}$ or $-\dfrac 12$?
Thanks
| we know by definition that
$$\lim_{x\rightarrow y}\left(\frac{f(x) - f(y)}{x - y}\right) = \frac{df}{dy}$$
so the given can be expressed as
$$\lim_{x\rightarrow y}\left(\frac{\cos(x) - \cos(y)}{(x - y)(x + y)}\right)$$
that is equal to
$$\left(\frac{df}{dy}\right)\left(\lim_{x\rightarrow y}\frac{1}{x + y}\right)$$
at last becomes equal to
$$\frac{- sin(y)}{2y}$$
| {
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} |
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\beta=\frac{c}{a}$$
Therefore the roots of $9x^2-2x+7=0$ are $(\alpha+2)$ and $(\beta+2)$. So, $$\alpha+\beta+4=\frac{2}{9}\implies4-\frac{b}{a}=\frac{2}{9}\implies\frac{b}{a}=\frac{34}{9}$$ and
\begin{align*}
(\alpha+2)(\beta+2)=\frac{7}{9}\\
\Rightarrow\alpha\beta+2(\alpha+\beta)+4=\frac{7}{9}\\
\Rightarrow\frac{c}{a}-2\frac{b}{a}+4=\frac{7}{9}\\
\Rightarrow\frac{4a-2b+c}{a}=\frac{7}{9}\\
\end{align*}
So, my final equations are: $\dfrac{b}{a}=\dfrac{34}{9}$ and $\dfrac{4a-2b+c}{a}=\dfrac{7}{9}$. My mind says I'm pretty close to the solution but I can't find it out.
There is another similar question: If the roots of $px^2+qx+r=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then what will be the expression of $r$ in terms of $a$, $b$, and $c$?
| My way:
If $p$ is a root of $9x^2-2x+7=0, p-2$ will be a root of $ax^2+bx+c=0$
Now writing $p-2=q\iff p=q+2$ in $9x^2-2x+7=0,$
we get $0=9(q+2)^2-2(q+2)+7=9q^2+34q+39$
So, we need $$\dfrac a9=\dfrac b{34}=\dfrac c{39}\ \ \ \ (1)$$
So, with the given conditions, $a,b,c$ can assume any set of non-zero finite values that honor $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Question on calculating $\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$ My task is to calculate $$\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$$ using the Cauchy integral formula. My question is: Is there a simple trick or do I have to perform a partial fraction decomposition?
| Yes there is a trick you don't have to use partial fraction decomposition
Note that the degree of nomenator is higher than the degree of the denomenator.
We use partial fraction usually when the degree of the nomenator is lower than the degree of the
denomenator.
I prefer solving this using another method than using partial fraction .
I wll denote $|z-1|=\frac{3}{2}$ as $C$
The trick is to devide the circle to three parts $C_1$ and $C_2$ and $C_3$
Where $0$ is inside $C_1$ and $e^{\frac{\pi i}{4}}$ is inside $C_2$ and $e^{\frac{-\pi i}{4}}$ is inside $C_3$
\begin{align*}
\int_{C} \frac{z^7 + 1}{z^2(z^4 + 1)}dz&=\int_{C}\frac{z^7 + 1}{z^2(z^4 + 1)}dz + 0\ \\
&=\int_{C} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \ + \ \int_{\gamma} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \ + \int_{-\gamma} \frac{z^7 + 1}{z^2(z^4 + 1)}dz\\
&+\int_{\beta} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \ + \ \int_{-\beta} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \\
&=\int_{C_1} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \ + \ \int_{C_2} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \ + \ \int_{C_3} \frac{z^7 + 1}{z^2(z^4 + 1)}dz \\
\end{align*}
Now you can apply Cauchy integral formula For $C_1$
$$\int_{C_1} \frac{f}{z^2}dz = 2\pi i f(0)$$
Where $f=\frac{z^7 + 1}{(z^4 + 1)}$
And for $C_2$
$$\int_{C_2} \frac{f}{(z-e^{\frac{\pi i}{4}})}dz = 2\pi i f(e^{\frac{\pi i}{4}})$$
Wher $f=\frac{z^7 + 1}{z^2(z-e^{\frac{-\pi i}{4}})(z-e^{\frac{3\pi i}{4}})(z-e^{\frac{-3\pi i}{4}})}$
Similarly for $C_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do partial fraction decomposition. I have a to solve.$$\int\dfrac{1}{(a-2x)^2(b-x)}dx$$
But I don't even know how to start it. Please help.
| Starting with
$$
\dfrac{1}{(a-2x)(b-x)}=\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1}{b-x}\right),
$$
we get
\begin{eqnarray}
\dfrac{1}{(a-2x)^2(b-x)}&=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{(a-2x)(b-x)}\right]\\
&=&\dfrac{1}{a-2b}\left[\dfrac{-2}{(a-2x)^2}+\dfrac{1}{a-2b}\left(\dfrac{-2}{a-2x}+\dfrac{1}{b-x}\right)\right]\\
&=&\dfrac{1}{2b-a}\cdot\dfrac{2}{(2x-a)^2}+\dfrac{1}{(2b-a)^2}\cdot\dfrac{2}{2x-a}-\dfrac{1}{(2b-a)^2}\cdot\dfrac{1}{x-b}.
\end{eqnarray}
It follows that
\begin{eqnarray}
\int\dfrac{1}{(a-2x)^2(b-x)}\,dx&=&\int\left[\dfrac{1}{2b-a}\cdot\dfrac{2}{(2x-a)^2}+\dfrac{1}{(2b-a)^2}\cdot\dfrac{2}{2x-a}-\dfrac{1}{(2b-a)^2}\cdot\dfrac{1}{x-b}\right]\,dx\\
&=&\dfrac{-1}{2b-a}\cdot\dfrac{1}{2x-a}+\dfrac{1}{(2b-a)^2}\cdot\ln|2x-a|-\dfrac{1}{(2b-a)^2}\cdot\ln|x-b|+c
\end{eqnarray}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
| The Taylor series for $e^x$ is given by
$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\tag 1$$
Using $(1)$ with $x=1$, we have
$$\begin{align}
e-c_n&=\sum_{k=n+1}^\infty \frac{1}{k!}\\\\
&\le\sum_{k=n+1}^\infty \frac{1}{k!}\left(1+\color{blue}{\underbrace{\frac{1}{k-1}-\frac1k}_{\text{Positive Terms}}}\right)\\\\
&=\sum_{k=n+1}^\infty \frac{1}{k!}\left(\frac{k}{k-1}-\frac1k\right)\\\\
&=\sum_{k=n+1}^\infty\color{red}{\underbrace{\left(\frac{1}{(k-1)(k-1)!}-\frac{1}{k\,k!}\right)}_{\text{Telescoping Terms}}}\\\\
&=\frac{1}{n\,n!}
\end{align}$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$. Calculate
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$$
Question: I want to verify that my next attempt is correct, I do it too exhausted and in that state I do not trust my abilities.
My attempt: Note that $$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)=\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right).$$
In this sense, we know that
$$\begin{array}{rcl}\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}&=&{\displaystyle \sum_{k=1}^{n}\left(e^{\frac{2\pi i}{2n+1}}\right)^{k} } \\ &=& {\displaystyle\frac{e^{\frac{2\pi i}{2n+1}}-\left(e^{\frac{2\pi i}{2n+1}}\right)^{n+1}}{1-e^{\frac{2\pi i}{2n+1}}} } \\
&=& {\displaystyle\frac{ \left(e^{\frac{2\pi i}{2n+1}}-e^{\frac{2(n+1)\pi i}{2n+1}}\right)\left( 1-e^{\frac{-2\pi i}{2n+1}} \right) }{\left(1-e^{\frac{2\pi i}{2n+1}}\right)\left(1-e^{\frac{-2\pi i}{2n+1}}\right) } } \\
&=& {\displaystyle\frac{e^{\frac{2\pi }{2n+1}}-e^{\frac{2\pi (n+1) }{2n+1}} -1 +e^{\frac{2\pi n }{2n+1}} }{2-\cos\left( \frac{2\pi }{2n+1} \right)} }
\end{array}$$
Therefore, we have
$${\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right)=\displaystyle\frac{\cos\left(\frac{2\pi }{2n+1}\right)-\cos\left(\frac{2\pi (n+1) }{2n+1}\right) -1 +\cos\left(\frac{2\pi n }{2n+1}\right) }{2-\cos\left( \frac{2\pi }{2n+1} \right)} }. $$
Hence, we can conclude
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)=\lim_{n\rightarrow \infty}\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right)=\displaystyle\frac{\cos\left(0\right)-\cos\left(\pi\right) -1 +\cos\left(0\right) }{2-\cos\left( 0 \right)}=0 .$$
| There is an error is the calculation. We have
$$\sum_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12 \tag 1$$
and hence the limit is $-1/2$. To show that $(1)$ is correct, we follow the approach taken in the OP.
Then, we have
$$\begin{align}
\sum_{k=1}^n e^{i\left(\frac{2k\pi}{2n+1}\right)}&=\frac{e^{i\left(\frac{2\pi}{2n+1}\right)}-e^{i\left(\frac{2(n+1)\pi}{2n+1}\right)}}{1-e^{i\left(\frac{2\pi}{2n+1}\right)}}\\\\
&=\frac{2e^{i\left(\frac{3\pi}{2(2n+1)}\right)}\cos\left(\frac{\pi}{2(2n+1)}\right)}{-2ie^{i\left(\frac{\pi}{2n+1}\right)}\sin\left(\frac{\pi}{2n+1}\right)}\\\\
&=i\,\frac{e^{i\left(\frac{\pi}{2(2n+1)}\right)}}{2\sin\left(\frac{\pi}{2n+1}\right)}\\\\
&-\frac12 +\frac i2 \cot\left(\frac{\pi}{2n+1}\right)\tag2
\end{align}$$
Taking the real part of both sides of $(2)$ yields the equality in $(1)$. Hence we have
$$\lim_{n\to \infty}\sum_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12 $$
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} dt + B\int {1 \over t+w_2}dt \\= {w - 1\over w - w_2} \ln|t + w| + {1- w_2\over w - w_2}\ln|t + w_2| + C $$
After finding the value of $w, w_2$ final answer I got is
$${\sqrt{5} + 1\over 2\sqrt{5}}\ln|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\ln|t + 1/2 + \sqrt{5}/2| + C$$
But the given answer is :
$$\bbox[7px,Border:2px solid black]{ \frac{\ln\left(\left|t^2+t-1\right|\right)}{2}+\frac{\ln\left(\left|2t-\sqrt{5}+1\right|\right)-\ln\left(\left|2t+\sqrt{5}+1\right|\right)}{2\cdot\sqrt{5}}+C}$$
Where did I go wrong ? especially that first term of the answer is a mystery to me.
| Note that we have
$$\begin{align}
& {\sqrt{5} + 1\over 2\sqrt{5}}\log|t + 1/2 - \sqrt{5}/2| + {\sqrt{5} - 1 \over 2\sqrt{5}}\log|t + 1/2 + \sqrt{5}/2| + C\\[10pt]
&= \frac12\log\left(|t + 1/2 - \sqrt{5}/2|\,|t + 1/2 + \sqrt{5}/2|\right)\\[10pt]
&{}+\frac{1}{2\sqrt{5}}\left(\log\left(|t + 1/2 - \sqrt{5}/2|\right)-\log\left(|t + 1/2 + \sqrt{5}/2|\right)\right)+C\\\\
&=\frac12\log\left(|t^2+t+1|\right)+\frac{1}{2\sqrt{5}}\left(\log\left(|t + 1/2 - \sqrt{5}/2|\right)-\log\left(|t + 1/2 + \sqrt{5}/2|\right)\right)+C\\\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Using the Lagrange interpolation formula to find a polynomial f with $f(-1)=-6, f(0)=2, ...$ The exercise goes as follows:
Use the Lagrange interpolation formula to find a polynomial $f$ with real coefficients such that $f$ has degree $\leq3$ and $f(-1)=-6, f(0)=2, f(1)=-2, f(2)=6$.
So to start, I wrote down the formula $$f=\sum^n_{i=0}f(t_i)P_i$$
But I didn't manage to get much further than that. I understand that $P_i$ are polynomials of degree $i$, but I'm not sure how I can do anything with that information.
| I would like to propose an alternative way to solve the problem. Coefficients of Lagrange interpolation polynomial can be found if one uses a determinant form of Lagrange interpolation presented in "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation" (you may check for concrete example in "Workbook on mapping simplexes affinely"). General formula looks as follows
$$
f(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & f_0 & f_1 & \cdots & f_n \\
x^n & x_0^n & x_1^n & \cdots & x_n^n \\
x^{n-1} & x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 1 & 1 & \cdots & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
x_0^n & x_1^n & \cdots & x_n^n \\
x_0^{n-1} & x_1^{n-1} & \cdots & x_n^{n-1} \\
\cdots & \cdots & \cdots & \cdots \\
1 & 1 & \cdots & 1 \\
\end{pmatrix}
}.
$$
EXAMPLE
Let's consider your case: $f(-1) = -6$, $f(0) = 2$, and $f(2) = 6$. Previous equation should be written as
$$
f(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & -6 & 2 & 6 \\
x^2 & (-1)^2 & 0^2 & 2^2 \\
x & -1 & 0 & 2 \\
1 & 1 & 1 & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
(-1)^2 & 0^2 & 2^2 \\
-1 & 0 & 2 \\
1 & 1 & 1 \\
\end{pmatrix}
} = 2 + 6 x - 2 x^2.
$$
Now it is easy to check that
$$
\begin{aligned}
f(-1) &= 2 - 6 - 2(-1)^2 = -6,\\
f(0) &= 2 + 0 - 0 = 2,\\
f(2) &= 2 + 12 - 2\, (2)^2 = 6
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, an explicit expression for $f(x)$ for all positive rational numbers $x$.
Every number I have evaluated is of the form $ f ( x ) = \frac x { x + 1 } $ and this clearly fits the functional equations, but I can't prove that it's the only solution. Can anyone help me? I have put down the start of my workings which led me to the conjecture of $ f ( x ) = \frac x { x + 1 } $.
Plugging in $ x = 1 $ clearly gives $ f ( 1 ) = \frac 1 2 $ and $ f ( 2 ) = 2 f \big( f ( 1 ) \big) = 2 f \left( \frac 1 2 \right) $ which we can plug back into the first equation to get that $ f ( 2 ) = \frac 2 3 $. Working in this vein I have been able to show that $ f ( x ) = \frac x { x + 1 } $ for particular values of $ x $, but not in general.
The most difficult part appears to be proving it for the even integers. To prove $ x = 8 $, we have
$$ f ( 12 ) = 2 f \left( \frac 6 7 \right) = 4 f \left( \frac 3 { 10 } \right) = 4 - 4 f \left( \frac { 10 } 3 \right) \\
= 4 - 8 f \left( \frac 5 8 \right) = 8 f \left( \frac 8 5 \right) - 4 = 16 f \left( \frac 4 9 \right) - 4 \\
= 32 f \left( \frac 2 { 11 } \right) - 4 = 64 f \left( \frac 1 { 12 } \right) - 4 = 60 - 64 f ( 12 ) \text , $$
giving us $ f ( 12 ) = \frac { 12 } { 13 } $. This will probably be the main area of difficulty in the proof.
| As j___d does, I will attempt to prove that $f(\frac pq) = \frac{p}{p+q}$ by strong induction on $p+q$, starting with the base case $p+q=2$ where $f(\frac11) = \frac12$.
Now assume that $f(\frac{p}{q}) = \frac{p}{p+q}$ holds when $p+q<k$, and consider fractions $\frac pq$ with $p+q=k$.
Whenever $p<q$, we have $f(\frac{p}{q-p}) = \frac pq$, so $2f(f(\frac{p}{q-p})) = 2f(\frac pq)$, and by the second identity this implies that $f(\frac{2p}{q-p}) = 2f(\frac pq)$. Whenever $p>q$, of course, we have $f(\frac qp) = 1 - f(\frac pq)$ by the first identity.
Either way, this gives us $f(\frac{p'}{q'})$ for some different $\frac{p'}{q'}$ with $p'+q'=p+q$, in terms of $f(\frac pq)$.
Now we repeat the following process. Start with $f(\frac1{k-1}) = x$, and let $\frac pq = \frac1{k-1}$. Then repeatedly apply one of the identities $$f\left(\frac{2p}{q-p}\right) = 2f\left(\frac pq\right) \qquad \text{or} \qquad f\left(\frac qp\right) = 1 - f\left(\frac pq\right)$$ (preferentially the first) to get a different value $f(\frac{p'}{q'})$ in terms of $f(\frac{p}{q})$ and therefore in terms of $x$. Set $\frac pq = \frac{p'}{q'}$ and repeat. An example for $k=11$:
\begin{array}{cccccccccc}
f(\frac{1}{10}) & f(\frac{2}{9}) & f(\frac{3}{8}) & f(\frac{4}{7}) & f(\frac{5}{6}) & f(\frac{6}{5}) & f(\frac{7}{4}) & f(\frac{8}{3}) & f(\frac{9}{2}) & f(\frac{10}{1}) \\
x & 2x & 1-8x & 4x & 16x-1 & 2-16x & & 8x & & 32x-2
\end{array}
Because there are finitely many values, we will eventually loop back to a value we have already seen, getting a second expression for it in terms of $x$. That expression will be different from the first, because the coefficient of $x$ doubles with every step from left to right in the table above. (In this case, we'll get $f(\frac1{10}) = 3 - 32x$, so $x = 3 - 32x$.)
So we can solve for $x$ and get some value for $f(\frac{1}{k-1})$, as well as all the other values we've encountered. If there are values in the table we haven't filled in yet, we can start this process again from those values, stopping when we get two expressions for the same unknown value, or an expression for a value we've already solved for.
(By the way, if $k$ is not prime, then we will have some fractions $\frac pq$ with $p+q=k$ which can be simplified, so we already know their values. In some cases, this lets us take a shortcut from the very beginning: for example, this will happen whenever $k$ is even.)
Eventually, we can fill in the entire table. This tells us that there's a unique solution for all $f(\frac pq)$ with $p+q=k$. But we know that $f(\frac pq) = \frac{p}{p+q}$ is consistent with the functional equation, so if we got some unique solution, that must be what we got.
By induction on $p+q$, we have $f(\frac pq) = \frac{p}{p+q}$ for all $p, q \ge 1$.
| {
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"url": "https://math.stackexchange.com/questions/2227678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
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Evaluate $\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) \right]_0^{2\pi}$ where $K = \sqrt{A^2 + B^2 -1}$ and $A^2 + B^2 << 1$. I am having difficulty in evaluating the following integrand:
$$\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K})
\right]_0^{2\pi}$$
where $K = \sqrt{A^2 + B^2 -1}$ and $A^2 + B^2 << 1$.
=================================================================
To evaluate the indefinite integral $I$ $$I = \int \frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for various}\ A,B <<1.$$
Wolfram Alpha provides the following solution:-
$$I= -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) $$
where $K = \sqrt{A^2 + B^2 -1}$.
In the simple case when $A=B=0$ the definite integral over the range $0...2\pi$ should obviously be given by:-
$$I_{A=B=0}=\int_0^{2\pi}\frac{1}{1 + 0 + 0}\, \mathrm{d}\theta = 2\pi.$$
This specific result has been confirmed by the restricted ( for $A^2+B^2<1$ ) Definite Solution
$$I=\int_0^{2\pi}\frac{1}{1 + A\cos(\theta) + B\sin(\theta)}\, \mathrm{d}\theta = 2\pi.$$
developed by Dr. MV in his answer to my previous question.
However I would like to know where my evaluation of the integrand is going wrong as I will need in future to apply a similar approach to other, more complex integrands for which I dont yet have Definite Solutions.
We might (naively) approach the Wolfram Alpha solution in the following way, noting that when $A=B=0$ so $K=i$, :-
$$I_{A=B=0}= \left[ -(2/i) \tanh^{-1} ( \frac{\tan(\theta/2)}{i}) \right]_0^{\infty}$$
$$= -(2/i) \tanh^{-1} ( \frac{\tan(2\pi/2)}{i}) +(2/i) \tanh^{-1} ( \frac{\tan(0/2)}{i}) $$
$$ = -(2/i) \tanh^{-1} ( \frac{\tan(\pi)}{i}) +(2/i) \tanh^{-1} ( \frac{\tan(0)}{i}) $$
$$ = -(2/i) \tanh^{-1} ( \frac{0}{i}) +(2/i) \tanh^{-1} ( \frac{0}{i}) $$
$$ = 0.$$
but this is clearly incorrect. I presume the error comes from trying to integrate across the range $0, 2\pi$ where the $\tan$ function has singularities at $\pi/2$ and $3\pi/2$.
So let us try and break the integration into the three continuous ranges $0...\pi/2$ and $\pi/2...3\pi/2$ and $3\pi/2...2\pi$ thus:-
$$I_{A=B=0}=$$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(2\pi/2)}{i}) - -(2/i) \tanh^{-1} ( \frac{\tan(3\pi/4)}{i}) + $$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(3\pi/4)}{i}) - - (2/i) \tanh^{-1} ( \frac{\tan(\pi/4)}{i}) +$$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(\pi/4)}{i}) - - (2/i) \tanh^{-1} ( \frac{\tan(0)}{i}) $$
leading to
$$I_{A=B=0}=$$
$$ -(2/i) \tanh^{-1} ( \frac{0}{i}) +(2/i) \tanh^{-1} ( \frac{-1}{i}) + $$
$$ -(2/i) \tanh^{-1} ( \frac{-1}{i}) +(2/i) \tanh^{-1} ( \frac{1}{i}) +$$
$$ -(2/i) \tanh^{-1} ( \frac{1}{i}) +(2/i) \tanh^{-1} ( \frac{0}{i}) $$
Now
$tanh^{-1}\frac{0}{i} = 0$, $tanh^{-1}\frac{-1}{i} = \frac{i\pi}{4}$, and $tanh^{-1}\frac{1}{i} = \frac{-i\pi}{4}$
and so
$$I_{A=B=0}=$$
$$ -(2/i) 0 +(2/i) \frac{i\pi}{4} + $$
$$ -(2/i) \frac{i\pi}{4}+(2/i) \frac{-i\pi}{4} + $$
$$ -(2/i) \frac{-i\pi}{4}+(2/i) 0 $$
then
$$I_{A=B=0}= +(2/i) \frac{i\pi}{4} -(2/i) \frac{i\pi}{4}+(2/i) \frac{-i\pi}{4} -(2/i) \frac{-i\pi}{4} $$
$$I_{A=B=0}= \frac{2\pi}{4} -\frac{2\pi}{4} - \frac{2\pi}{4} + \frac{2\pi}{4} $$
$$I_{A=B=0}= 0$$.
So we still get the wrong answer. Clearly the right answer ($2 \pi$) could be obtained if all the minus signs were converted into plus signs, but I cant't find justification for doing that, as yet.
Any suggestions welcomed!
| Following comments by user mickep let us look at correcting the partitions used...
So, to avoid operating across the singularity at $\frac{\pi}{2}$, let us break the integration into the two continuous ranges $0...\pi$ and $\pi...2\pi$ thus:-
$$I_{A=B=0}=$$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(2\pi/2\right)}{i}\right) - -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(\pi/2\right)}{i}\right) + $$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(\pi/2\right)}{i}\right) - - \left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(0\right)}{i}\right) +$$
leading to
$$I_{A=B=0}=$$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{0}{i}\right) +\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{-\infty}{i}\right) -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{+\infty}{i}\right) +\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{0}{i}\right) $$
Now
$tanh^{-1}\left(\frac{0}{i}\right) = 0$, $tanh^{-1}\left(\frac{-\infty}{i}\right) = \frac{i\pi}{2} $, and $tanh^{-1}\left(\frac{\infty}{i}\right) = \frac{-i\pi}{2} $
and so
$$I_{A=B=0} = 0 +\left(\frac{2}{i}\right) \frac{i\pi}{2} -\left(\frac{2}{i}\right) \frac{-i\pi}{2} + 0 $$
$$I_{A=B=0} = \pi --\pi $$
$$I_{A=B=0} = 2\pi . $$
which is the expected answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)?$ Log integral
$$\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)\tag1$$
Making an attempt:
Following from my previous post making $u={x\over 1+x}$ it doesn't worked, so I try applying binomial series
$(1)$ becomes
$$\sum_{n=0}^{\infty}(-1)^n\cdot{n+1\over 2^n}\int_{0}^{\infty}[x^n\ln^2(x)-x^n\ln^2(1+x)]\mathrm dx=2\ln^2(2)\tag2$$
Applying IBP to these indefinite integrals
$$\int x^n\ln^2(x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(x)-x^2\cdot{2\ln x-1\over 2(n+1)}+C\tag3$$
$$\int x^n\ln^2(1+x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(1+x)-{2\over n+1}\int\color{red}{{x^{n+1}\over 1+x}\ln(1+x)\mathrm dx}\tag4$$
$$\color{red}{\int{x^{n+1}\over 1+x}\ln(1+x)}\mathrm dx={x^{n+1}\over 2}\ln^2(1+x)-{n+1\over 2}\int x^n\ln^2(1+x)\mathrm dx\tag5$$
If I put $(5)$ into $(4)$ we get zero! What I am doing here doesn't seem to be working.
How can we go about to tackle $(1)?$
| On the path of Jack D'Aurizio,
$I_1=\displaystyle \int_0^{+\infty} \dfrac{(\ln(x))^2}{(x+2)^2}dx$
In $I_1$ perform the change of variable $y=\dfrac{x}{2}$,
$\begin{align} I_1&=2\int_0^{+\infty} \dfrac{(\ln(2x))^2}{(2x+2)^2}dx\\
&=\dfrac{1}{2}\int_0^{+\infty} \dfrac{(\ln 2+\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2\int_0^{+\infty}\dfrac{1}{(1+x)^2}dx+\ln 2\int_0^{+\infty}\dfrac{\ln x}{(1+x)^2}dx+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2\left[-\dfrac{1}{1+x}\right]_0^{+\infty}+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx+\\
&\ln 2\int_1^{+\infty} \dfrac{\ln x}{(1+x)^2}dx\\
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx-\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{1}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
\end{align}$
In the latter line, in the first integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\
&=\dfrac{1}{2}(\ln 2)^2+I_2
\end{align}$
Therefore,
$\boxed{I_1-I_2=\dfrac{1}{2}(\ln 2)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2} = 0.5$
$\tan\frac{B}{2} =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?
| The following identity is very useful.
$$\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{A}{2}\tan\frac{C}{2}+\tan\frac{B}{2}\tan\frac{C}{2}=1$$
From here $C=90^{\circ}$
We can get this result by geometric way.
Let $P(0,0)$,$Q(2,1)$ and $R(3,-1)$. Draw it!
Hence $PQ=QR$, $\measuredangle Q=90^{\circ}$ and from here $$\arctan{\frac{1}{2}}+\arctan{\frac{1}{3}}=\measuredangle Q=45^{\circ},$$
which says that our $\frac{C}{2}=45^{\circ}$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Shortest universal bit string: One string to contain all others Let $s$ be a string of bits. Treat it as a cycle, with the first bit following
the last.
Say that $s$ is universal for $n$ if all the $2^n$ strings
of $n$ bits can be found in $s$ as consecutive, left-to-right bits
(with wrap-around).
Define $u(n)$ as the length of the shortest string universal for $n$.
Examples:
$u(1)=2$:
$$
\begin{matrix}
\color{red}{0} & 1 \\
0 & \color{red}{1}
\end{matrix}
$$
$u(2)=4$:
$$
\begin{matrix}
\color{red}{0} & \color{red}{0} & 1 & 1 \\
0 & \color{red}{0} & \color{red}{1} & 1 \\
\color{red}{0} & 0 & 1 & \color{red}{1} \\
0 & 0 & \color{red}{1} & \color{red}{1}
\end{matrix}
$$
$u(3)=8$:
$$
\begin{matrix}
\color{red}{0} & \color{red}{0} & \color{red}{0} & 1 & 1 & 1 & 0 & 1\\
0 & \color{red}{0} & \color{red}{0} & \color{red}{1} & 1 & 1 & 0 & 1\\
\color{red}{0} & 0 & 0 & 1 & 1 & 1 & \color{red}{0} & \color{red}{1}\\
0 & 0 & \color{red}{0} & \color{red}{1} & \color{red}{1} & 1 & 0 & 1\\
\color{red}{0} & \color{red}{0} & 0 & 1 & 1 & 1 & 0 & \color{red}{1}\\
0 & 0 & 0 & 1 & 1 & \color{red}{1} & \color{red}{0} & \color{red}{1}\\
0 & 0 & 0 & 1 & \color{red}{1} & \color{red}{1} & \color{red}{0} & 1\\
0 & 0 & 0 & \color{red}{1} & \color{red}{1} & \color{red}{1} & 0 & 1
\end{matrix}
$$
Q1. What is $u(n)$?
This may be simple, but the wrap-around seems to make it not so straightforward.
Q2. What is the generalization to strings of $k$ symbols?
Let $u(k,n)$ be the length of the shortest string on $k$ symbols
that contains all strings of length $n$.
Likely this is all known...
| What you're looking for is related to De Bruijn sequences. The formula seems to be very simple: $u(k,n) = k^n$ (and the special case $u(n) = 2^n$).
(source: Wikipedia, by Eviatar Bach)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Floor function of powers of $2$ Is there any way to know the exact value of
$$\left(\left\lfloor 2^{\frac{n}{2}} \right\rfloor\right)^2$$
for $n$ an integer $n>0$?
When $n$ is even, the solution is trivial, since we do not have to face any fractional part. On the other hand, for $n$ an odd number, it seems difficult to get the exact value.
We could try to get an aporoximation, but using $\lfloor x \rfloor = x + O(1)$ gives a pretty inaccurate result.
Any idea?
Edit: Would it be possible to get it if we knew the value of $$\left(\left\lfloor 2^{\frac{n}{2}-1} \right\rfloor\right)^2$$
? (A recursive fomula)
| Using identity:
$$\lfloor x\rfloor =x-\frac{1}{2}+\frac{\sum _{k=1}^{\infty } \frac{\sin (2 k \pi x)}{k}}{\pi }$$
Solve sum ,substitute x=2^(n/2) and raise to the power 2 we have:
$$\left\lfloor 2^{n/2}\right\rfloor ^2=\frac{\left(\left(-1+2^{1+\frac{n}{2}}\right) \pi -i \log \left(1-e^{-i
2^{1+\frac{n}{2}} \pi }\right)+i \log \left(1-e^{i 2^{1+\frac{n}{2}} \pi }\right)\right)^2}{4 \pi ^2} = 1/4\,{\frac { \left( {2}^{1+n/2}\pi-2\,{\rm arccot} \left(\cot \left(
\pi\,{2}^{n/2} \right) \right) \right) ^{2}}{{\pi}^{2}}}
$$
and:
$$\left\lfloor 2^{\frac{n}{2}-1}\right\rfloor ^2=\frac{\left(\left(-1+2^{n/2}\right) \pi -i \log \left(1-e^{-i 2^{n/2}
\pi }\right)+i \log \left(1-e^{i 2^{n/2} \pi }\right)\right)^2}{4 \pi ^2} = 1/4\,{\frac {1}{{\pi}^{2}} \left( \pi\,{2}^{n/2}-\pi-2\,\arctan
\left( {\frac {\sin \left( \pi\,{2}^{n/2} \right) }{-1+\cos \left(
\pi\,{2}^{n/2} \right) }} \right) \right) ^{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many proofs for this one, but I think this is an elegant one. I'd like you to check it out, first because I consider it to be a nice and concise and elementary, and secondly because I'd like to make sure that there are no mistakes.
Let's get started: First of all, $a_n\geq 0 \ \forall \,n \in \mathbb{N}$. We write $a_n$ as
$$\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+2^k\left(\frac{\log\left(2\right)}{2}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}.$$
It is obvious that $\left(\frac{\log\left(n\right)}{n}\right)^k \to 0$. Let $\varepsilon >0$. Then $\exists\, n_0 \in \mathbb{N}$ such that $\forall \, n \geq n_0, \left(\frac{\log\left(n\right)}{n}\right)^k < \varepsilon$.
We have
\begin{align*}
a_n&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k+ n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+2^k+\dotsb +n^k} + \frac{n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k}\\ & \qquad \qquad+ \frac{\varepsilon\left(1^k+\dotsb +\left(n_0-1\right)^k\right)+n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k} + \frac{\varepsilon\left(1^k+2^k+\dotsb +n^k\right)}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}+\varepsilon.
\end{align*}
Now, by taking the limsup and the liminf as $n \to\infty$, and since
$$\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}\to 0,$$ we have
$$0 \leq \limsup\left(a_n\right) \leq \varepsilon \quad\text{and}\quad 0\leq \liminf\left(a_n\right) \leq \varepsilon.$$ But $\varepsilon$ was arbitrarily small, so
$$\liminf\left(a_n\right)=\limsup\left(a_n\right)=0=\lim\left(a_n\right).$$
This is more of a discussion and not so much of a question :)
| Fix $k\in \mathbb N.$ Let $a_n$ be the numerator, $b_n$ the denominator. Then $b_n \to \infty.$ Time to think about Stolz-Cesaro, which suggests we consider
$$\tag 1 \frac{a_{n+1}- a_n}{b_{n+1}- b_n} = \frac{\ln^k (n+1)}{(n+1)^k} = \left (\frac{\ln (n+1)}{n+1} \right )^k.$$
Since $[\ln (n+1)]/(n+1) \to 0,$ the right side of $(1)$ $\to 0.$ By Stolz-Cesaro, the limit of interest is $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\sqrt{x^2+1}$ uniformly continuous on (0, 1)? $\sqrt{x^2+1}$ uniformly continuous on (0, 1)?
How to deal with such problems? Please help.
I know the definition of U C. but unable to handle the problem.
| Hint: Observe
\begin{align}
\left|\sqrt{x^2+1}-\sqrt{y^2+1}\right| =& \frac{|x^2-y^2|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \\
=&\ \frac{|x+y|}{\sqrt{x^2+1}+\sqrt{y^2+1}}|x-y|\\
\leq&\ \frac{|x|+|y|}{2}|x-y| \leq |x-y|
\end{align}
when $x, y \in (0, 1)$.
Edit: This is more or less the solution. However, you should try to use the definition of uniform continuity to fill in the details.
More Edit: To show $f(x)= \sqrt{x^2+1}$ is uniformly continuous on all of $\mathbb{R}$ observe
\begin{align}
\frac{|x|+|y|}{\sqrt{x^2+1}+\sqrt{y^2+1}} \leq \frac{|x|}{\sqrt{x^2+1}}+\frac{|y|}{\sqrt{y^2+1}}\leq 2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given a $2\times 2$ matrix $A$, compute $A^7$. Let $A =
\begin{pmatrix}
e^{2x} & -1 \\
0 & e^{2x}-1 \\
\end{pmatrix}
$. Compute $A^7$.
I've tried the obvious way of multiplying $A$ with $A$, then $A^2$ with $A^2$, but I arrived at a messy result in the top right member of the matrix. Is there a general form to be noticed here?
| You have $A=B+C$, where
$$
B=\begin{bmatrix} e^{2x}&0\\0&e^{2x}-1\end{bmatrix},\ \ \ C=\begin{bmatrix}0&-1\\0&0\end{bmatrix}.
$$
The key is that $C^2=0$, and that $CB^kC=0$. Also,
\begin{align}
B^kCB^m&=\begin{bmatrix} e^{2kx}&0\\0&(e^{2x}-1)^k\end{bmatrix}\begin{bmatrix}0&-1\\0&0\end{bmatrix}
\begin{bmatrix} e^{2mx}&0\\0&(e^{2x}-1)^m\end{bmatrix}
=\begin{bmatrix} 0&-(e^{2x})^k(e^{2x}-1)^m\\0&0\end{bmatrix}\\ \ \\
&=(e^{2k})^k(e^{2x}-1)^m\,C.
\end{align}
Then, after cancelling all the zero terms,
\begin{align}
(B+C)^7&=B^7+\sum_{k=0}^7B^kCB^{7-k}=B^7+\sum_{k=0}^7(e^{2x})^k(e^{2x}-1)^{7-k}C
\end{align}
So
$$
A^7=(B+C)^7=\begin{bmatrix} e^{14x}&-\displaystyle\sum_{k=0}^7(e^{2x})^k(e^{2x}-1)^{7-k}\\0&(e^{2x}-1)^7\end{bmatrix}.
$$
The sum can be simplified by noting that
$$
\sum_{k=0}^7a^kb^{7-k}=\frac{b^7-a^7}{b-a}.
$$
Thus
$$
A^7=(B+C)^7=\begin{bmatrix} e^{14x}&\displaystyle (e^{2x}-1)^{7}-e^{14x}\\0&(e^{2x}-1)^7\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove that the sequence $\left\{\frac{4n^2}{2n^3-5}\right\}$ converges to $0$. Prove that the sequence $$\left\{\frac{4n^2}{2n^3-5}\right\}$$ converges to $0$.
The first thing I would like to do is further bound the sequence so I can get rid of the $-5$ in the bottom of the fraction. How can I do this? For negatives it's tricky since you want to make it smaller, but removing a $-5$ actually makes the denominator bigger, and not smaller.
If this cannot be done, I must do the following:
$$\left\{\frac{(4/n)}{2-5/n^3}\right\}$$ and find a bound for $2-5/n^3$, separately. If I assume $n\geq 10$, then:
$$\frac{1}{n}<\frac{1}{10}\color{red}{\to} \frac{5}{n}<\frac{1}{2}
\color{red}{\to} \frac{-5}{n}>\frac{-1}{2}\color{red}{\to} 2-\frac{5}{n}>\frac{3}{2}\color{red}{\to} \frac{1}{|2-5/n^3|}<\frac{2}{3}$$
Then I have that:
$$\left|\frac{(4/n)}{2-5/n^3}\right|<\frac{2}{3}\cdot\frac{4}{n}<\frac{8}{3N}<\frac{8}{3(8/3\epsilon)}=\epsilon$$
Pick $N=\max(10,8/3\epsilon)$
This excess bounding could have been avoided if I can somehow bound the fraction even more?
| Two-stage bounding seems unavoidable. To simplify the works, I would do something like the following. If $n \geq 3$, then
$$
\frac{4n^{2}}{2n^{3}-5} < \frac{4n^{2}}{2n^{3}-n^{2}} = \frac{4}{2n-1}.
$$
Given any $\varepsilon > 0$, we have $\frac{4}{2n-1} < \varepsilon$ if in addition $n > (\frac{4}{\varepsilon+1})/2$. So taking $N := \max \{ 3, (\frac{4}{\varepsilon+1})/2 \}$ suffices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
| When you multiply by $x-2$, you have to worry about whether it's negative, in which case the direction of the inequalities would be reversed. So break things into two cases: $x>2$ and $x<2$.
If $x>2$, you can do what you did or be a little more efficient:
$$-(x-2) < 3x+1 < x-2$$
$$-x +1 < 3x < x-3$$
$$1<4x \mbox{ and } 2x<-3$$
$$\frac{1}{4}<x \mbox{ and } x<-\frac{3}{2}$$
This is the empty set because no $x$ satisfies both inequalities.
Repeat for $x<2$ but reverse the inequalities when you multiply by it,
to get
$$\frac{1}{4}>x \mbox{ and } x>-\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
| Try to write
a more general form
as the difference of two squares
which we know how to factor.
$\begin{array}\\
x^4 + a
&=(x^2+u)^2-b^2x^2\\
&=x^4+2ux^2+u^2-b^2x^2\\
&=x^4+(2u-b^2)x^2+u^2\\
\text{and}\\
x^4 + a
&=(x^2+u+bx)(x^2+u-bx)\\
\end{array}
$
so
$u^2=a$
and
$b^2 = 2u$.
Therefore
$b^4 = 4u^2
= 4a
$.
Writing $c^2$ for $a$,
this becomes
$\begin{array}\\
x^4 + c^2
&=(x^2+u)^2-b^2x^2\\
&=x^4+2ux^2+u^2-b^2x^2\\
&=x^4+(2u-b^2)x^2+u^2\\
\text{so}\\
u
&=c\\
\text{and}\\
x^4 + c^2
&=(x^2+c+bx)(x^2+c-bx)\\
\end{array}
$
so
$b^2 = 2c$.
Since $b$ and $c$
are integers,
$c = 2d^2$
and
$b^2 = 4d^2$
so $b = 2d$.
Therefore,
the expression can
only be factored
if it is of the form
$x^4+4d^4$
when the factorization is
$x^4+4d^4
=(x^2+2d^2)^2-4d^2x^2
=(x^2+2dx+2d^2)(x^2-2dx+2d^2)
$.
This is the case
$d=1$
so $b=2$
and the factorization is
$x^4+4
=(x^2+2x+2)(x^2-2x+2)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limits for the average triple integral I am trying to do the following problem:
Calculate the average of $f(x,y,z) = x^2 + y^2 + z^2$ over the set of points satisfying $|x| + |y| + |z| ≤ 1$.
So,
$\frac{1}{volume of D}\iiint Fdv$ =
$\frac{1}{volume of D}\iiint (x^2 + y^2 + z^2 )dz dy dx$
My problem is how to determine D. So, how do I find the limits of integration.
| To simplify computations, note that the integrand does not vary when the variables change sign. Because both the integrand and the region are symmetric in the three variables, the integral will be the same when restricted to each of the eight octants. Thus we may assume that $x,y,z\geq0$, and then multiply our integral by $8$.
The inequality $x+y+z\leq1$ forces $x\leq1$. So we could let $0\leq x\leq1$. Now we have
$$
y+z\leq 1-x.
$$
So we can let $$0\leq y \leq1-x,$$ and finally $z\leq1-x-y$, i.e. $$0\leq z\leq 1-x-y.$$ Thus
\begin{align}
\iiint_D(x^2+y^2+z^2)\,dV&=8\int_{0}^1\int_{0}^{1-x}\int_{0}^{1-x-y}(x^2+y^2+z^2)\,dz\,dy\,dx\\ \ \\
&=8\int_0^1\int_0^{1-x}(x^2+y^2)(1-x-y)+\frac13\,(1-x-y)^3\,dy\,dx\\ \ \\
&=8\int_0^1\frac1{12}(x-1)^2(7x^2-2x+1)+\frac1{12}(x-1)^4\,dx\\ \ \\
&=\frac8{20}=\frac25.
\end{align}
The volume of $D$ is
$$
8\,\int_0^1\int_0^{1-x} \int_0^{1-x-y} 1 \,dz\, dy\, dx=\frac86=\frac43.
$$
So the average is
$$
\frac{\frac25}{\frac43}=\frac3{10}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On Chinese remainder theorem Suppose we know $x\equiv 3 \bmod 11$ and $x\equiv 7\bmod 13$ and $0<x<143$ holds then CRT gives that $x=3\times 13[13^{-1}\bmod 11] + 7\times 11[11^{-1}\bmod 13]=39\times 6 + 77\times 6=696$ gives a solution but it is not within $0$ and $143$. So we take $696\bmod 143$ and choose $124$ as solution.
My query is say instead of $3$ we choose $3+11k$ and instead of $7$ we choose $7+13\ell$ for some $k,\ell\in\Bbb Z$ can we still recover $124$?
Instead of $696$ is there a direct way to get $124$?
We have $696=124+4(143)$. In general what is the quantity that goes in instead of $4$?
| The direct way to $124$ would be to start with $7$ and see how many $13$s to add to make it into a number that is $3\bmod 11$.
Since $13\equiv \color{red}2 \bmod 11$, we can see that we need $(3-7)/\color{red}2 = -2\equiv 9\bmod 11$ copies of $13$. $7+9\cdot 13 = 7+117 = 124$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| If you rewrite your proof as:
$3 ( x^2+y^2+z^2 ) \ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.
You would find that it is not so unintuitive after all.
Since you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives us "$3 ( x^2+y^2+z^2 ) \ge (x+y+z)^2$, and expanding out immediately tells us the solution.
In general, a technique that works for many cyclic polynomial inequalities is to try to 'smooth' the terms out. Terms that are just a power of one variable are the 'biggest', and using inequalities like "$x^2+y^2 \ge 2xy$" will 'mix' powers and thereby 'reduce' them.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now,
$$x+\dfrac {x}{2} +\dfrac {x}{3}=180°$$
$$\dfrac {6x+3x+2x}{6}=180°$$
$$\dfrac {11x}{6}=180°$$
$$x=\dfrac {180\times 6}{11}$$
So, which is the correct option?
| It looks like neither of the options is correct. $$\angle C = \frac{x}{3} = \frac{180 \times 6}{11 \times 3} = \frac{180 \times 2}{11} = \frac{360}{11} = 32 \frac{9}{11} $$
You can also calculate $A+B+C = 3C + \frac{3}{2}C + C = \frac{11}{2} C$ for all possibile answers to show that none of them gives $180^\circ$ for the sum of the angles $A+B+C$, and hence none of them is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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There exist $3$ positive numbers $x,y,z$ such that $x\cdot y\cdot z=100$ but $x^2+y^2+z^2<65$. I have to show this using a form of calculus.
I know the answer is the cube root of $100$. I did this using intuition and upper and lower bounds but I have no idea how to show it from a calculus standpoint. Maybe showing the intersection of the surfaces?
| Let $x= 2\sqrt{5}$ , $y= 2\sqrt{5}$ $z=5 $.
Then of course $xyz=100$
And $x^2+y^2+z^2=20+20+25=65$.
Now set $x=y$ so $z= \frac{100}{x^2}$ therefore $x^2+y^2+z^2=2x^2+\frac{100}{x^4}$. Now define $$f(x)=2x^2+\frac{100}{x^4}$$
By differentiating we get $$f'(x)=\frac{4x^6-4\cdot100^2}{x^5}$$
so $x=100^{1/3}$ is the minimum of $f$.
However, we have found that $f(2\sqrt{5})=65$ so $f(100^{1/3})<65$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Express the value of $s\left(m\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$ in terms of $m$. The previous question was: Find the range such that the equation $|x^2 -3x +2|=mx $ has 4 distinct real solutions: $a,b,c,d$, and that turned out to be $0<m<3-2\sqrt{2}$. The book says that the solution is $\frac{m^2+5}{2}$. I have been trying to express $a$ and $b$ as the two possible results from the quadratic equation resulting from $x^2−3x+2=mx$, and $c$ and $d$ as the ones from the equation $x^2−3x+2=-mx$, and plugging that into $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$, yet I do not see how a complicated expression can end up being the answer given by the book. Therefore I am sure I must be doing something wrong.
| The quadratic equation $x^2+px+q=0$ has two distinct roots, namely $$r_1=\frac{-p+\sqrt{p^2-4q}}{2}\qquad\text{and}\qquad r_2=\frac{-p-\sqrt{p^2-4q}}{2}\qquad\text{whenever }\,p^2-4q>0$$
Then $$r_1r_2=q\qquad\text{and}\qquad r_1^2+r_2^2=p^2-2q$$
It follows that $$\frac1{r_1^2}+\frac1{r_2^2}=\frac{r_1^2+r_2^2}{(r_1r_2)^2}=\frac{p^2-2q}{q^2}$$
So, regarding that the two solutions of the given equation are also solutions of $x^2-(m+3)x+2=0$ and the other two are solutions of $x^2+(m-3)x+2=0$, we get
$$s(m)=\frac{(-m-3)^2-2(2)}{2^2}+\frac{(m-3)^2-2(2)}{2^2}=\frac{m^2+5}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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locus of $z=w - \frac{1}{w}$ If $|w|=2$ , then set of points $z=w -\frac{1}{w}$ is equal to ?
One of my friend helped me like this:
$$|z| = \left| w - \frac{1}{w}\right| \leq |w| + \frac{1}{|w|} = 2 + 0.5 = 2.5 \\ \implies |z| \le 2.5$$
After that I am unable to proceed. Can anybody help me?
| Given $w \overline{w} = |w|^2=4\,$, it follows that $z = w - \cfrac{1}{w} = w - \cfrac{\overline{w}}{4}\,$.
Let $z=x+iy$ and $w=u+iv\,$ with $x,y,u,v \in \mathbb{R}\,$, then the above can be written as:
$$
x+iy = u+iv - \frac{1}{4}(u - iv) = \frac{3}{4}u + i\,\frac{5}{4}v \;\;\implies\;\; x = \frac{3}{4}u, \;\;y = \frac{5}{4}v
$$
The condition $|w|^2=4 \iff u^2+v^2=4$ then gives the locus of $z$ as the ellipse:
$$\left(\frac{4}{3}x\right)^2 + \left(\frac{4}{5}y\right)^2 = 4 \;\;\iff\;\; \frac{x^2}{9}+\frac{y^2}{25}=\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$ if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$
I tried as follows.
The given expression can be rewritten as
$$S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}-\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
But by symmetry $$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$
so
$$2S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}$$ and by Cauchy Scwartz inequality
$$2S \le \sqrt{4-x+4-y+4-z}\times \sqrt{3}$$ so
$$2S \le \sqrt{24}$$
so
$$S \le \sqrt{6}$$
Is this approach correct?
| I think it means that $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0$.
If $x=3$, $y=1$ and $z=0$ then $S=\frac{5}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac {x}{\sqrt {x+y}}\leq\frac{5}{4}\sqrt{x+y+z}.$$
By Cauchy-Schwarz
$$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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What am I doing wrong in computing this Taylor Series of $f(x) = \frac{1}{4+x^{2}}$ about a = 0? $$f(x) = \frac{1}{4+x^{2}}$$
What I did was notice that this is similar to the geometric series:
$$f(x) = \frac{1}{1-x} = 1 + x^{2} + x^{3} + x^{4} + ....$$
So I altered my original function from:
$$f(x) = \frac{1}{4 + x^{2}}$$
to:
$$f(x) = \frac{1}{4(1 - (-\frac{1}{4}x^{2}))}$$
And so it would now look like:
$$f(x) = \frac{1}{4(1 - \left(-\frac{1}{4}x^{2}\right))} = 1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right)^{3} + \left(\frac{-1}{4}x^{2}\right)^{4} + .... $$
now to account for the 4 out front:
$$ = 4\bigg(1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right))^{3} + \left(\frac{-1}{4}x^{2}\right))^{4} + ....\bigg) $$
Why is this wrong?
The correct answer is suppose to be:
$$= \frac{1}{4}-\frac{1}{16}x^2+\frac{1}{64}x^4-\frac{1}{256}x^6+\frac{1}{1024}x^8+\ldots $$
Thank you
| Everything is perfect, just one thing:
You have
$$ \frac{1}{4} \cdot \frac{1}{1-\left(-\frac{1}{4} x^2\right)} $$
so you need to take the series and multiply by $\frac14$, not $4$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that there is no primitive Pythagorean triple $(a,b,c)$ where one side differs from another by three I tried using all the Theorems I know about Pythagorean triples, but nothing is working out. I know I should probably go for a proof by contradiction.
| A primitive Pyth. triplet is of the form $(a,b,c)=(m^2-n^2, 2mn, m^2+n^2)$ where $m>n>0$ and $\gcd (m.n)=1 ,$ and where $m,n$ are not both odd. (BTW: If $m,n$ were both odd then $a,b,c$ would all be even.) We have $\gcd (a,b)=1.$
(1). We have $c-a=(m^2+n^2)-(m^2-n^2)=2n^2\neq 3.$
(2). We have $c-b=(m^2+n^2)-2mn=(m-n)^2\neq 3.$
(3A). If $m\equiv n \pmod 3$ then $n\not \equiv 0\pmod 3,$ else $m=(m-n)+n\equiv 0\pmod 3,$ but then $3$ divides both $m$ and $n$, contrary to $\gcd (m,n)=1.$
So if $m-n\equiv 0 \pmod 3$ then $n^2\equiv 1 \pmod 3,$ so modulo $3$ we have $a-b\equiv (m^2-n^2)-2mn=(m-n)^2-2n^2\equiv 0-2 \not \equiv 0.$
(3B). If $m-n \not \equiv 0 \pmod 3$ then $(m-n)^2\equiv 1 \pmod 3,$ so $a-b=(m-n)^2-2n^2\equiv 1-2n^2 \not \equiv 0\pmod 3.$ Because there is no solution to $1-2n^2\equiv 0 \pmod 3.$
Another way of handling cases (3A) and (3B): By contradiction: If $\{a,b\}=\{x,x+3\}$ then $x\equiv \pm 1 \pmod 3$ (Else $3$ divides both $x$ and $x+3$, that is, $3$ divides both $a$ and $b,$ contrary to $\gcd (a,b)=1$).
But then $x^2\equiv 1 \pmod 3$, so modulo $3$ we have $c^2=a^2+b^2= x^2+(x+3)^2\equiv 2x^2\equiv 2.$ But there is no solution to $c^2\equiv 2 \pmod 3.$
| {
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$
But how to find $\alpha^6+\beta^6$ ?
| Well, in general:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
So, for your example we can set:
*
*$$\alpha=x_+=\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag2$$
*$$\beta=x_-=\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag3$$
Now, for your problem:
$$\alpha^6+\beta^6=\left(\frac{-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6+\left(\frac{-\text{b}-\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\right)^6=$$
$$\frac{\left(-\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6+\left(\text{b}+\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}\right)^6}{64\cdot\text{a}^6}\tag4$$
| {
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Evaluate the following $(3-2)(5-3)(7-5)........(1995-1993)$
Evaluate the following
$(3-2)(5-3)(7-5)....(1995-1993)$
I have problem in counting the terms since the answer will be $2^n$ , $n$ is the number of terms.
So I count them as follows:
$1995=2n-1$
$1996=2n \to\ n= 998$ ; but since the first term equals $1$
The the answer will be $2^{997}$
Is my answer right?
| Since the first factor is indeed $3-2=1$, we will simply ignore it and start by counting $5-3$ as factor #1.
Factor $\#\color{red}{1}$ is $5-3$, which is $(2 \cdot \color{red}{2} + 1) - (2 \cdot \color{red}{2} - 1)$.
Factor $\#\color{red}{2}$ is $7-5$, which is $(2 \cdot \color{red}{3} + 1) - (2 \cdot \color{red}{3} - 1)$.
Factor $\#\color{red}{3}$ is $9-7$, which is $(2 \cdot \color{red}{4} + 1) - (2 \cdot \color{red}{4} - 1)$.
See the pattern?
Factor $\#\color{red}{n}$ is $(2 \cdot \color{red}{[n+1]} + 1) - (2 \cdot \color{red}{[n+1]} - 1)$.
Which $n$ gives us $2[n+1] + 1 = 1995$ and $2[n+1] - 1 = 1993$? It's the same $n$ for both so we only need to solve one of them. Either way we get $n = 996$.
So yes, there are $997$ total factors. One of them is $1$ and the other $996$ are $2$, so the answer is $1 \cdot 2^{996} = 2^{996}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solve the equation in integers $a,b$: $20a^3-b^3=1$
Solve the equation in integers $a,b$: $$20a^3-b^3=1.$$
Assume that $a \neq 0$. Then simplifying and rearranging the equation gives $$20a^3 = 2^2 \cdot 5 \cdot a^3 = b^3+1 = (b+1)(b^2-b+1).$$ Note that neither factors can be divisible by $3$, since both of them must be divisible by $3$ but $b^2-b+1 \not \equiv 0 \pmod{9}$ and $b^2-b+1 = 3$ gives a contradiction. Also since $b$ is odd, then $b+1$ is even and $b^2-b+1$ is odd. Also, $b^2-b+1 \not \equiv 0 \pmod{5}$, so $b+1 \equiv 0 \pmod{20}$.
Then since $$\gcd(b+1,b^2-b+1) = \gcd(b+1,-2b+1) = \gcd(b+1,3),$$ it follows that $(b+1)$ and $(b^2-b+1)$ are relatively prime. Thus $b+1 = 20n^3$ and $b^2-b+1 = m^3$ where $m,n$ are relatively prime. I then thought about bounding $b^2-b+1$ between two perfect cubes, but I didn't see how to do that.
Then since two solutions for $(a,b)$ are $(0,-1)$ and $(7,19)$ I wanted to show that $(b+1) \mid 20$, or equivalently that $a$ and $b+1$ are relatively prime. So suppose that for some prime $p$ we have $a \equiv 0 \pmod{p}$ and $b+1 \equiv 0 \pmod{p}$. It then follows that $b+1 \equiv 0 \pmod{p^3}$, so that $b = p^3k-1$ for some integer $k$. Then we have $$(p^3k-1)^2-(p^3k-1)+1 = p^3k(p^3k-3)+3 = m^3.$$ How can we get a contradiction from here?
| From your work, after $b+1=20n^3$ and $b^2-b+1=m^3$, multiply second by $64$
$$64b^2-64b+64=(4m)^3 \ \ \ \Longrightarrow \ \ \ (8b-4)^2+48=(4m)^3$$
All solutions to $y^2+48=x^3$ in integers are known: $(x, y)= (4,\pm4), (28,\pm148)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer is $[-2;0) $. Should there be a different approach?
| One approach:
\begin{align}
\sqrt{ x + 2} - \sqrt{ x + 3} &< \sqrt{ 2} - \sqrt{ 3} \\
\sqrt{ x + 2} + \sqrt{ 3} &< \sqrt{ x + 3} + \sqrt{ 2} \tag{1}\\
(x + 2) + 3 + 2\sqrt{ 3}\sqrt{ x + 2} &< (x + 3) + 2 +2\sqrt{ 2}\sqrt{ x + 3} \tag{2}\\
2\sqrt{ 3}\sqrt{ x + 2} &< 2\sqrt{ 2}\sqrt{ x + 3} \tag{3}\\
3(x + 2) &< 2( x + 3) \tag{4}\\
x &< 0 \tag{5}
\end{align}
where we
$(1)$ added $\sqrt{ x + 3} + \sqrt{3}$
$(2)$ squared both sides (they are positive)
$(3)$ subtracted $x+5$
$(4)$ divided by $2$ and squared (again both sides are positive)
$(5)$ subtracted $2x+6$
Now just combine it with the condition $x\geq -2$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals
$$4abc-c^3 < 4abc$$ so
$$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$ but i am unable to proceed here
| $$\sum_{cyc}\frac{abc}{a^3+b^3+abc}\leq\sum_{cyc}\frac{abc}{a^2b+ab^2+abc}=\sum_{cyc}\frac{c}{a+b+c}=1$$
because by Rearrangement $a^3+b^3=a^2\cdot a+b^2\cdot b\geq a^2b+b^2a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq y$, $A_{n \times n}$.
The determinant expression appears to be Dickson polynomial of second kind.
Let $D_n$ be the determinant of $A_n$. We can see that the appropriate recurrence relation is $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Base cases:
$$D_1=x+y=\frac{x^2-y^2}{x-y}$$
$$
D_2=(x+y)^2-xy=x^2+xy+y^2=\frac{x^3-y^3}{x-y}
$$
Suppose that $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Then we need to prove that $$D_{n+1}=(x+y)D_{n}-xyD_{n-1}$$
Which can be developed as:
$$
D_{n+1}=(x+y)((x+y)D_{n-1}-xyD_{n-2})-xyD_{n-1}=
$$
$$
=(x+y)^2D_{n-1}-xy(x+y)D_{n-2}-xyD_{n-1}=
$$
$$
=(x^2+xy+y^2)D_{n-1}-xy(x+y)D_{n-2}=
$$
$$
=\frac{x^3-y^3}{x-y}D_{n-1}-xy(x+y)D_{n-2}
$$
I tried doing this up to $D_{n-6}$ in order to get any insights into possible simplification but I'm pretty stuck.
| Your base cases are good. For the induction step, assume that
$$D_n=\frac{x^{n+1}-y^{n+1}}{x-y}\quad \text{and}\quad D_{n-1}=\frac{x^{n}-y^{n}}{x-y}.$$
Then you have that
$$D_{n+1}=(x+y)D_n-xyD_{n-1},$$so you only have to show that
$$(x+y)\frac{x^{n+1}-y^{n+1}}{x-y}-xy\frac{x^{n}-y^{n}}{x-y}=\frac{x^{n+2}-y^{n+2}}{x-y}.$$
Developping the LHS gives you
\begin{align}(x+y)\frac{x^{n+1}-y^{n+1}}{x-y}-xy\frac{x^{n}-y^{n}}{x-y} & = \frac{x^{n+2}+x^{n+1}y-xy^{n+1}-y^{n+2}-x^{n+1}y+xy^{n+1}}{x-y}\\ & = \frac{x^{n+2}-y^{n+2}}{x-y}.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
pythagorean theorem extensions are there for a given integer N solutions to the equations
$$ \sum_{n=1}^{N}x_{i} ^{2}=z^{2} $$
for integers $ x_i $ and $ z$
an easier equation given an integer number 'a' can be there solutions to the equation
$$ \sum_{n=1}^{N}x_{i} ^{2}=a^2 $$
for N=2 this is pythagorean theorem
| Just to avoid notation bloat, let's let $N=4$ and let the implied generalization take care of the rest of the question. The question can be restated as asking whether there are rational points on the $4$-sphere $\mathbb S_4:x_1^2 + x_2^2 +x_3^2+x_4^2=1$. We know that $(1,0,0,0)$ is a (trivial) rational point on
$\mathbb S_4$. From that point, we pick a rational direction,
$(\xi_1, \xi_2, \xi_3, \xi_4)$ where $\xi_1, \xi_2, \xi_3, \xi_4$ are rational numbers, and see if the line $(1,0,0,0)+t(\xi_1, \xi_2, \xi_3, \xi_4)$ intersects
$\mathbb S_4$ at another rational point.
\begin{align}
(1+t\xi_1)^2 + t^2 \xi_2^2 + t^2 \xi_3^2 + t^2 \xi_4^1 &= 1 \\
2t\xi_1 + t^2(\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2) &= 0 \\
t &= -\dfrac{2\xi_1}{\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2}
\end{align}
So, for any four rational numbers $\xi_1, \xi_2, \xi_3, \xi_4$
$$\left(
\dfrac{-\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2}
{\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2},
-\dfrac{2\xi_1 \xi_2}{\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2},
-\dfrac{2\xi_1 \xi_3}{\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2},
-\dfrac{2\xi_1 \xi_4}{\xi_1^2 + \xi_2^2 + \xi_3^2 + \xi_4^2},
\right)$$
is a rational point on $\mathbb S_4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $2^n + 4^n \leq 5^n$ I'm trying to prove by induction that $2^n + 4^n \leq 5^n$. Through some value plugging I've established that the induction must start from $n = 2$ because $2^2 + 4^2 \leq 5^2 \equiv 20 \leq 25$; for $n = 1$ it doesn't hold since $2 + 4 \geq 5$.
Now I assume that $2^k + 4^k \leq 5^k$ is true and I want to prove that implies $k+1$. Using the inductive hypothesis I multiply both sides by $4$ to get this:
$$4 \cdot 2^{k} + 4\cdot 4^{k} \leq 4 \cdot 5^{k}$$
$$2^{k+2} + 4^{k+1} \leq 4 \cdot 5^{k}$$
I will use again the induction hypothesis, this time I'll multiply both side by $5$ to get:
$$5 \cdot (2^{k} + 4^{k}) \leq 5^{k+1}$$
I can say that $2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k})$ and $4 \cdot 5^{k} \leq 5^{k+1}$ so I concatenate them:
$$2^{k+2} + 4^{k+1} \leq 5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k} \leq 5^{k+1}$$
However this doesn't feel right. I'm assuming that $5 \cdot (2^{k} + 4^{k}) \leq 4 \cdot 5^{k}$ which there's no way I can be sure about. At this point I'm stuck since the whole reasoning seems wrong.
| Assume that $2^k + 4^k \leq 5^k$ is true.
Then
\begin{align} 2^{k+1} + 4^{k+1} & \leq 2 ( 2^{k} + 4^{k}) + 2\cdot 4^k \\
& \le 2 \cdot 5^k + 2 \cdot 4^k \\
& \le 2 \cdot 5^k + 2 \cdot 5^k \\
& \le 4 \cdot5^k \\
& \le 5^{k+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Prove $k^8 \geq (k^2 - 1)^4$ for $k \geq 1$ Can anyone please help me with following inequality:
$$k^8 \geq (k^2 - 1)^4, \text{where} \ k \geq 1.$$
I tried induction but get stuck at step $k=n+1$ and can't progress anywhere.
| Since $k^8 = (k^2)^4$, we just need to show that (for $a,b\geq0$) $a\geq b \implies a^4 \geq b^4$. Indeed, if $a\geq b$, then $a-b\geq 0$, and so $a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3) \geq 0$. Thus, $a^4 \geq b^4$.
Clearly, $k^2>k^2-1$, and so $k^8 > (k^2-1)^4$ since $k^2, k^2-1 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3$ I have to prove that $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3 $$
is always true for real numbers $a, b, c>0$ with $abc=1$.
Using the AM-GM inequality I got as far as $$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq \frac{b}{\sqrt[a+1]{b}}+\frac{c}{\sqrt[b+1]{c}}+\frac{a}{\sqrt[c+1]{a}}$$
but I do not yet know how to finish my proof from there (if this is helpful at all?!).
| Hint: $\frac{1+ab}{1+a}=\frac{abc+ab}{1+a}=ab\big(\frac{1+c}{1+a}\big)$. So what is $\big(\frac{1+ab}{1+a}\big)\big(\frac{1+bc}{1+b}\big)\big(\frac{1+ca}{1+c}\big)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$
If $\cos^2\theta+\cos\theta = 1$, express $\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta-2$ as a single term in terms of $\sin\theta$.
We have \begin{align*}\sin^8\theta+\sin^6\theta+\sin^4\theta+\sin^2\theta &= \sin^6\theta(\sin^2\theta+1)+\sin^2\theta(\sin^2\theta+1)-2\\&=\sin^6\theta(2-\cos^2\theta)+\sin^2\theta(2-\cos^2\theta)-2,\end{align*} but I didn't see how to use that $\cos^2\theta+\cos\theta = 1$.
| $$\cos^2 (t)+\cos (t)=1$$
$$\implies \sin^2 (t)=\cos(t)$$
$$\implies \sin^4 (t)=\cos^2 (t) $$
$$\implies \sin^4 (t)+\sin^2 (t)=1 $$
$$\implies \sin^8 (t)+\sin^6 (t)=\sin^4 (t) $$
$$\implies f(t)=\sin^4 (t)+1-2=$$
$$=\cos^2 (t)-1=-\sin^2 (t ) $$ which is your result .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
\alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
\alpha$.
I tried:
$$\frac{\frac{\sin \alpha}{\cos \alpha}+\sin \alpha}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha+\sqrt{5}\sin \alpha \cos \alpha}{\cos \alpha}}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha}{\cos \alpha}(\sqrt{5}\cos \alpha+1)}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\tan \alpha (5\cos^2\alpha-1)}{2} = \\
???$$
What do I do next?
| Problem:
$$
f(t) = \frac{1}{2} \left(\sqrt{5} \cos t-1\right) \left(\sqrt{5} \sin t+\tan t\right)
$$
FOIL Components:
$$
%
\begin{align}
%
F
&= \frac{1}{2} \sqrt{5} \cos t
\left(\sqrt{5} \sin t\right)
= \frac{5}{2} \sin t \cos t\\[7pt]
%
O &= \frac{1}{2} \sqrt{5} \cos t \left( \tan t\right)
= \frac{\sqrt{5}}{2} \sin t \\[7pt]
%
I &= \frac{1}{2}\left(-1\right) \sqrt{5} \sin t = -\frac{\sqrt{5}}{2} \sin t \\[7pt]
%
L &= \frac{1}{2}\left(-1\right)\tan t = -\frac{1}{2}\tan t
%
\end{align}
%
$$
Combine and simplify
$$
\begin{align}
\require{cancel}
f(t) &= F + O + I + L \\
&= \frac{5}{2} \sin t \cos t +
\cancel{\frac{\sqrt{5}}{2} \sin t} -
\cancel{\frac{\sqrt{5}}{2} \sin t} -
\frac{1}{2}\tan t \\
&=
\frac{5}{2} \sin t \cos t - \frac{1}{2}\tan t
\end{align}
$$
Noting
$$
\sin t \cos t = \frac{1}{2} \sin 2t
$$
The function reduces to the desired form
$$
\boxed{
f(t) = \frac{5}{4} \sin 2t - \frac{1}{2}\tan t
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}$ is an integer Let $x$ be a positive integer. Does there always exist integers $y,z$ such that the sum
$$\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}$$
is an integer, but none of the three terms of the sum is an integer?
For instance: Take $x=1$. Then we need the sum $$\frac{y}{z+1}+\frac{yz}{2}+\frac{z}{y+1}$$ to be an integer, with $y,z$ odd and $\frac{y}{z+1},\frac{z}{y+1}$ non-integers. We can take $y=z=3$.
For $x=2$ this becomes $\frac{2y}{z+1}+\frac{yz}{3}+\frac{2z}{y+1}$, and we can take $y=z=2$.
In general, if we take $y=z$ we have the sum $$\frac{2xy}{y+1}+\frac{y^2}{x+1}.$$
| Let $y=z=2x(x+1)-1$. Then
$$
\frac{xy}{z+1}=\frac{zx}{y+1}=x-\frac{1}{2(x+1)}
$$
and
$$
\frac{yz}{x+1}=4x^2(x+1)-4x+\frac{1}{x+1}.
$$
Since $x+1\geq2$, neither of these are integers, but
$$
\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}=4x^2(x+1)-2x
$$
is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How can we tackle this integral $\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=-1?$ Something is wrong with this integral (in terms of splitting them out)
$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$
My try:
Splitting the integral
$$\int_{0}^{1}{2x^2-2x\over x^3\sqrt{1-x^2}}\mathrm dx+\int_{0}^{1}{\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2\tag2$$
Note that $I_1$ and $I_2$ diverge, so how can we tackle it as a whole?
| I will present an answer without Gamma functions and without series' expansions. Instead, a more general problem is solved, where the OP's question is a special case.
Let
$$
I(a) = \int_{0}^{1}{2a^2x^2-2ax+\ln[(1-ax)(1+ax)^3]\over x^3\sqrt{1-x^2}}\mathrm dx
$$
The answer to the OP's question is then given by $I(a = 1)$.
Then, after partial differentiation w.r.t. $a$,
$$
I'(a) = \int_{0}^{1} \frac{2a^2(2ax - 1)}{(a^2 x^2 - 1)\sqrt{1-x^2}}\mathrm dx
$$
Note that this removes the problematic factor $x^3$ in the denominator - this issue was solved in the previous solution by series' expansion of the $\log$. Hence, convergence is established and the order of integrations $x,a$ can be exchanged.
The $x$-integration gives
$$
I'(a) = a^2 \Big[ \frac{2 \arctan(\frac{x \sqrt{1 - a^2}}{\sqrt{1 - x^2}})}{\sqrt{1 - a^2}} + \frac{4\arctan(\frac{a \sqrt{1 - x^2}}{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big]_{x=0}^{1}
$$
which is
$$
I'(a) = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arctan(\frac{a }{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big] = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big]
$$
Now we integrate w.r.t. $a$:
$$
I(a) = \int I'(a) \rm{d} a = \int a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big] \rm{d} a
$$
Replacing $a = \sin (y)$ gives
$$
I(y) = \int \sin^2(y) \Big[ {\pi} - {4 y}\Big] \rm{d} y = y \sin(2y) + (\pi y)/2 - \sin^2(y) - y^2 - (\pi \sin(2y))/4 + C
$$
We need the constant $C$. Obviously $0 = I(a=0) = I(y=0)$ which gives $C=0$.
The function in question by the OP is $I(a=1) = I(y=\pi/2) = -1$.
This solves the OP'S question. $\qquad \qquad \Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proving that $a_n=\frac{n^2+2n+6}{n^3-3}\to 0$ as $n\to\infty$ $$a_n=\frac{n^2+2n+6}{n^3-3}$$
So I want to show that "$a_n\to a\iff\forall \epsilon>0,\quad\exists N\in\mathbb{N}:n\geq N\implies |a_n-a|<\epsilon$"
Then my rough working:
$|a_n-0| =\left|\frac{n^2+2n+6}{n^3-3}\right|<\epsilon$
Estimate $\frac{n^2+2n+6}{n^3-3}<\frac{b_n}{c_n}$
Require $b_n>n^2+2n+6$, so choose $b_n=2n^2>n^2+2n+6$ for $n\geq 4$
Require $c_n<n^3-3$, so choose $c_n=\frac{n^3}{2}<n^3-3$ for $n\geq 2$
Then let $N=\frac{2n^3}{n^3/2}=\frac{4}{n}$
So now the proof:
$$\text{Fix}\quad \epsilon>0.\quad\text{Pick}\quad N:N>\frac{4}{n}\quad\text{and}\quad N\geq 4$$
$$\implies \forall n\geq N,\quad |a_n-0|=\left|\frac{n^2+2n+6}{n^3-3}\right|<\frac{4}{n}\leq\frac{4}{N}\leq\epsilon$$
Is this correct? Is it completely rigorous? I've just started doing analysis on my own, so please point out if anything is wrong/not conventional so that I can get into good habits at the start. Thanks for any help.
| If i were you, i would just multiplicate numerator and denominator by $1/n^2$ both, so that: $a_n = (1 + 2/n + 6/n^2)/(n -3/n^2)$. From the defenition of infinite number, we have $2/n$, $6/n^2$ and $-3/n^2$ all equal to 0: $a_n = (1 + 0 + 0)/(n + 0 + 0) = 1/n$ which is 0 from the same defenition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
showing $ 1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0 $ How to show that the following infinite series
$$
1-\frac{3}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{3}{6}+\frac{1}{7}+\frac{1}{8}+\cdots=0?
$$
The above series is of the form $\sum_{n \ge 1} \frac{f(n)}{n}$, where $f$ is a periodic arithmetical function of period $4$, with the values $f(1)=f(3)=f(4)=1$ and $f(2)=-3$. Since $\sum_{1 \le i \le 4} f(i)=0$, it is assured that this series is convergent.
| I'm a little wary of doing this, since it's not absolutely convergent, but can you rewrite each period as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{2}{4k-2}-\frac{2}{4k}\right)
$$
which can in turn be rewritten as
$$
\left(\frac{1}{4k-3}-\frac{1}{4k-2}+\frac{1}{4k-1}-\frac{1}{4k}\right) -
\left(\frac{1}{2k-1}-\frac{1}{2k}\right)
$$
and then you have two parallel alternating harmonic series, both of which sum to $\ln 2$. Subtract one from the other and you get $0$.
But, as I say, I'm not sure that's kosher.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \frac{2x}{1-x^2} \right) $ where $|x| < \frac {1}{\sqrt {3}}$, then what is $y$? If $\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right)$ where $|x| < \dfrac {1}{\sqrt {3}}$ then find the value of $y$.
. . .
Let $\tan^{-1} y= A$
$$y=\tan A$$
$$\tan^{-1} x=B$$
$$x=\tan B$$
$$
\tan ^{-1}\left( \dfrac{2x}{1-x^2} \right) =C
$$
$$\dfrac {2x}{1-x^2}=\tan C$$.
| We know that $$tan2A= \frac{2tanA}{1-tan^2A}$$
Now, substitute $tanA=x\implies tan^{-1}x=A$
Plugging this into first equation, $tan2(tan^{-1}x)= \frac{2x}{1-x^2}$
Again taking $tan^{-1}$ on both sides we get
$$2tan^{-1}x = tan^{-1}\frac{2x}{1-x^2}$$
Plugging this into the question,
$$tan^{-1}y= tan^{-1}x + 2tan^{-1}x = 3tan^{-1}x$$
We know that $$tan3A=tan(2A+A)= \frac{tanA+tan2A}{1-tanAtan2A}$$
This can be reduced to $$tan3A=\frac{3tanA-tan^3A}{1-3tan^2A}$$
By a similiar logic as before, $$3tan^{-1}x= tan^{-1}\frac{3x-x^3}{1-3x^2}$$
Now the question reduces to
$$tan^{-1}y=tan^{-1}\frac{3x-x^2}{1-3x^2}$$
So, we conclude that $$y=\frac{3x-x^3}{1-3x^2}$$
NOTE: Since it is given that $|x|\lt\frac{1}{\sqrt3}$ we can safely apply the above formulae without adding or subtracting $\pi$,because x is well within the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ . Find the value of $\cos \tan^{-1} \sin \cot^{-1} (x)$ .
...
Let $\cot^{-1} x=z$
$$x=\cot z$$
Then,
$$\sin \cot^{-1} (x)$$
$$=\sin z$$
$$=\dfrac {1}{\csc z}$$
$$=\dfrac {1}{\sqrt {1+\cot^2 z}}$$
$$=\dfrac {1}{\sqrt {1+x^2}}$$
| Then we have $$\cos\left(\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)\right)$$
If $\tan^{-1}\dfrac1{\sqrt{1+x^2}}=y,\dfrac\pi4\le y\le\dfrac\pi2\implies\tan y=\dfrac1{\sqrt{1+x^2}}$ and $\cos y\ge0$
$\cos\left(\tan^{-1}\dfrac1{\sqrt{1+x^2}}\right)=\cos y=+\dfrac1{\sqrt{1+\tan^2y}}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$ that is
$$\sqrt{57-x}-25=\sqrt{x+40}-10 \sqrt[4]{x+40}$$ again squaring both sides we get
$$682-x-50\sqrt{57-x}=x+40+100\sqrt{x+40}-20(x+40)^{\frac{3}{4}}$$
i got messed up here any better way and just a hint please
| Firstly, I would recommend you to make the following substitution $a = \sqrt[4]{57 - x}$ and $b = \sqrt[4]{x+40}$. Thus we have $a + b = 5$ and $a^{4} + b^{4} = 57 + 40 = 97$. Therefore:
\begin{align*}
a^{4}+b^{4} & = (a^{2}+b^{2})^{2} - 2a^{2}b^{2} = [(a+b)^{2} - 2ab]^{2} - 2a^{2}b^{2} = (5^{2} - 2ab)^{2} - 2a^{2}b^{2}\\
& = (25 - 2ab)^{2} - 2a^{2}b^{2} = 625 - 100ab + 2a^{2}b^{2} = 97
\end{align*}
Henceforward we shall also agree that $\alpha = a + b$ and $\beta = ab$. According to the last relation, we obtain that $2\beta^{2} - 100\beta + 528 = 0$, whose solutions are $\beta_{1} = 6$ and $\beta_{2} = 44$. After solving the corresponding collection of systems of equations, we obtain the next solution set $S = \{-24,41\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Functions and Sequences Problem
The function $F(k)$ is defined for positive integers as $F(1) = 1$,
$F(2) = 1$, $F(3) = -1$ and $F(2k) = F(k)$, $F(2k + 1) = F(k)$ for $k \geq
2$. Then $$F(1) + F(2) + \dotsb + F(63)$$ equals
$\begin{array}{lr}
(\text{A}) & 1 \\
(\text{B}) & -1 \\
(\text{C}) & -32 \\
(\text{D}) & 32 \\
\end{array}$
My approach: $F(4)=1$, $F(5)=1$, $F(6)=-1$, $F(7)=-1$ (i.e., all values of $F$ are either $1$ or $-1$). I tried to find a pattern for which $F(x)$ repeats after a certain integer but tried till $F(30)$ and cannot find a solution. Where am I going wrong?
| For $k \ge 2$ we have:
\begin{align}
F(2k) &= F(k) \\
F(2k+1) &= F(k)
\end{align}
This means
$$
F(n) = F(\lfloor n / 2 \rfloor) \quad (*)
$$
for $n \ge 4$.
If we are using base 2 numbers this means we have a word of length $m$ with at least three bits
$$
n = (1 b_{m-1} \dotsb b_2 b_1)_2 \quad (b_i \in \{0, 1\})
$$
and $(*)$ means we can shift right the word (dropping the bit at the end) and keep the value of $F$:
$$
F((1 b_{m-1} \dotsb b_2 b_1)_2)
= F((1 b_{m-1} \dotsb b_2)_2) \quad (m \ge 3)
$$
We repeat until we have less than three bits.
In other words
\begin{align}
F(n)
&= F((1 b_{m-1})_2) \\
&=
\begin{cases}
F(3), & \text{for } b_{m-1} = 1 \\
F(2), & \text{for } b_{m-1} = 0
\end{cases}
\\
&=
\begin{cases}
-1, & \text{for } b_{m-1} = 1 \\
+1, & \text{for } b_{m-1} = 0
\end{cases}
\end{align}
for $n\ge 4$.
So
$$
\sum_{k=4}^{63} F(k) =
F(\underbrace{(100)_2}_4) + F((101)_2) + F((110)_2) + F((111)_2) + \\
F(\underbrace{(1000)_2}_8 + \dotsb + F((1111)_2) + \\
F(\underbrace{(10000)_2}_{16} + \dotsb + F((11111)_2) + \\
F(\underbrace{(100000)_2}_{32} + \dotsb + F((111111)_2) \\
$$
We see that we can group the words by length, and that we have as many words with prefix $10$ as with prefix $11$ in each group, so that their $F$ values sum to zero.
Thus the sought sum reduces to
$$
\sum_{k=1}^{63} = F(1) + F(2) + F(3) = 1 + 1 + (-1) = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2278951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Argument and modules of complex numbers Can you help me to find the argument and modules of this complex number please?
$$z=\frac{2-i}{-\sqrt {3}-2i}$$
| Argument of $z$,
\begin{align}
\arg(z)&=\arg\left(\dfrac{2-i}{-\sqrt{3}-2i}\right)\\
&=\arg(2-i)-\arg(\sqrt{3}-2i)\\
&=\left[-\tan^{-1}\left(\dfrac{1}{2}\right)\right]-\left[-\pi+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right]\\
&=\pi-\left[\tan^{-1}\left(\dfrac{1}{2}\right)+\tan^{-1}\left(\dfrac{2}{\sqrt3}\right)\right]\\
&=\pi-\tan^{-1}\left(\dfrac{\frac{1}{2}+\frac{2}{\sqrt3}}{1-\frac{1}{2}\cdot\frac{2}{\sqrt3}}\right)\\
&=\pi-\tan^{-1}\left(\dfrac{4+\sqrt3}{2\sqrt3-2}\right)
\end{align}
And modulus of $z$,
\begin{align}
|z|&=\left|\dfrac{2-i}{-\sqrt{3}-2i}\right|\\
&=\dfrac{|2-i|}{|-\sqrt{3}-2i|}\\
&=\dfrac{\sqrt{2^2+1^2}}{\sqrt{(\sqrt{3})^2+2^2}}\\
&=\dfrac{\sqrt{35}}{7}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.
| You can use (for $n \geq 2$)
$$\frac{1}{n^3} < \frac{1}{n^2(n-1)} < \frac12\frac{2(n-1) +1}{n^2 (n-1)^2} =\frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right).$$
Write your sum as
$$1 + \frac{1}{2^3} + \sum_{2015\geq n\geq 3} \frac{1}{n^3} < \frac{9}{8} + \sum_{n\geq 3} \frac{1}{2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2}\right)
= \frac{9}{8} +\frac{1}{8} = \frac{5}{4}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Bernoulli equation for pressure (with the vortex) An incompressible inviscid fluid of constant density ρ moves subject to a gravitational acceleration −gk.
The vortex caused by water flowing down a plughole is modelled by the steady velocity field
when $x^2+y^2\le a^2$
$$u=\left(\frac{Γy}{a^2},-\frac{Γx}{a^2},0 \right) $$
when $x^2+y^2\gt a^2$
$$\left(\frac{Γy}{x^2+y^2},-\frac{Γx}{x^2+y^2},0 \right) $$
Evaluate the pressure in such a flow.
need to show that when $x^2+y^2\le a^2$
a free surface at a constant pressure takes the form of
$$C − \frac{Γ^2(2a^2-x^2-y^2)}{2ga^4} $$
and when $x^2+y^2\gt a^2$
a free surface takes the form of
$$C − \frac{Γ^2}{2g(x^2+y^2)} $$
I did work out the $x^2+y^2\gt a^2$ one by using the Bernoulli's equation for irrotational flow but don't know the other one with vorticity since for steady irrotational flow $p+\frac{1}{2}\rho|\mathbf u|^2+\rho gz=constant$
But how to get the rotational part,should I also use the Bernoulli's equation? But it doesn't have the constant part then how should I approach it?
Thank you so much!
| For inviscid, rotational flow a weaker form of the Bernoulli equation $p + \frac{1}{2} \rho | \mathbf{u}|^2 + \rho g z = C(\psi)$ is valid, where $C(\psi)$ is constant along a streamline. You can work with this or solve for the pressure field directly from the governing equations for inviscid flow, which reveals the Bernoulli relationship as well.
Using cylindrical coordinates we have $x = r \cos \theta, \, y = r \sin \theta,$ and the basis vectors
$$\mathbf{e}_r = \cos \theta \, \mathbf{e}_x + \sin \theta \, \mathbf{e}_y, \\\mathbf{e}_\theta = -\sin \theta \, \mathbf{e}_x + \cos \theta \, \mathbf{e}_y $$
The velocity field is, for $r^2 = x^2 + y^2 \leqslant a^2 $,
$$u_r = 0, \,\, u_\theta = - \frac{\Gamma r}{a^2}, \,\, u_z = 0,$$
and for $x^2 + y^2 > a^2$,
$$u_r = 0, \,\, u_\theta = - \frac{\Gamma }{r}, \,\, u_z = 0$$
The components of the Euler equation for steady, inviscid flow, $\rho\mathbf{u} \cdot \nabla \mathbf{u} \,\, = - \nabla p + \rho \mathbf{g}$, reduce to
$$\frac{\partial p}{\partial r} = \frac{\rho u_\theta^2}{r} = \begin{cases}\frac{\rho \Gamma^2r}{a^4}, \,\, x^2+y^2 \leqslant a^2 \\ \frac{\rho \Gamma^2}{r^3}, \,\,\,\, x^2+y^2 > a^2\end{cases}$$
and
$$\frac{\partial p}{\partial z}= -\rho g$$
Integrating, we get for $x^2 + y^2 \leqslant a^2$,
$$p = \alpha - \rho g z + \frac{\rho \Gamma^2 r^2}{2a^4} = \alpha - \rho g z + \frac{\rho \Gamma^2 (x^2 + y^2)}{2a^4},$$
and, for $x^2 + y^2 > a^2$,
$$p = \beta - \rho g z - \frac{\rho \Gamma^2}{2r^2} = \beta - \rho g z - \frac{\rho \Gamma^2}{2(x^2 + y^2)} $$
The integration constant $\alpha$ can be evaluated in terms of $\beta \,$ by matching the inner pressure field with the outer pressure field at $x^2 + y^2 = a$, where we have
$$\alpha - \rho g z + \frac{\rho \Gamma^2 (a^2)}{2a^4} = \beta - \rho g z - \frac{\rho \Gamma^2}{2a^2} $$
Solving for $\alpha$ we get
$$\alpha = \beta - \frac{\rho \Gamma^2}{a^2}, $$
and the inner pressure field is
$$p = \beta - \frac{\rho \Gamma^2}{a^2} - \rho g z + \frac{\rho \Gamma^2 (x^2 + y^2)}{2a^4}$$
The free-surface where $p = C$ is given by
$$z = \frac{\beta-C}{\rho g} - \frac{\Gamma^2}{g a^2} + \frac{ \Gamma^2 (x^2 + y^2)}{2ga^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Recurrence relation with variable $x$ in the denominator. A recurrence is defined such that $$f_n(x)=f_1(f_{(n-1)}(x)); x\ge2$$
$f_1(x)$ is defined as
$$f_1(x)= \frac 23 -\frac 3{3x+1}$$
How can I find the values of $x$ for which the following holds true? $$f_{1001}(x)= x-3$$
I was not able to deduce any kind of pattern in the above recurrence. I did try to find a a pattern by finding $f_2$ and $f_3$ but I could not make out any generating function. How should I proceed?
| Notice that $$f_2(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}-\frac{3}{3x+1}\right)+1} = \frac{2}{3}-\frac{3x+1}{3x-2}$$
$$f_3(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}-\frac{3x+1}{3x-2}\right)+1} = \frac{2}{3}+\frac{3x-2}{3}$$
$$f_4(x) = \frac{2}{3}-\frac{3}{3\cdot\left(\frac{2}{3}+\frac{3x-2}{3}\right)+1} = \frac{2}{3}-\frac{3}{3x+1}$$
This implies $f_4(x)=f_1(x)$. So we see that this is a cycle of $3$, meaning $$x-3=f_{1001}(x)=f_{998}(x)=\ ...\ =f_2(x)= \frac{2}{3}-\frac{3x+1}{3x-2}\implies x = \frac{5}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is it true that for $n \ge5$, ${{3n} \choose {2n}} > \frac{6^n}{n}$ For $n \ge 5, \frac{9n^2 - 9n + 2}{4n^2-2n} > 2$
$9n^2 -9n +2 -8n^2+4n = n^2 -5n +2 \ge 5^2-25+2$
Here's the basis step:
$${{9} \choose {6}} = 84 > \frac{6^3}{3} = 72$$
Here's the inductive step:
$${{3n} \choose {2n}} = {{3(n-1)}\choose {2(n-1)}}\left(\frac{3n}{n}\right)\left(\frac{3n-1}{2n-1}\right)\left(\frac{3n-2}{2n}\right) > \left(\frac{6^{n-1}}{n-1}\right)\left(\frac{3n}{n}\right)\left(\frac{9n^2 -9n +2}{4n^2-2n}\right) > \left(\frac{6^{n-1}}{n}\right)(6) = \frac{6^n}{n}$$
| $$\begin{eqnarray*}\log\binom{3n}{2n} &=& \log\Gamma(3n+1)-\log\Gamma(n+1)-\log\Gamma(2n+1)\\ &=& -\log n-\log B(n,2n+1)\end{eqnarray*}$$
and
$$ B(n,2n+1) = \int_{0}^{1}\left[x(1-x)^2\right]^n\,\frac{dx}{x} $$
is a log-convex function by the Cauchy-Schwarz inequality.
You don't even need induction.
| {
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"url": "https://math.stackexchange.com/questions/2284198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the sum of the power serie $\sum_{1}^{\infty } \frac{(n^{2} + n)x^{n-1}}{2^{n-1}}$
I have found convergence interval for series: $\left | x \right | < 2$,
but I have no idea how to find the sum of the power series.
| $$\sum\limits_{n=1}^\infty{(n^2+n)x^{n-1}\over2^{n-1}} = 4{d^2\over dx^2}\sum\limits_{n=1}^\infty{x^{n+1}\over2^{n+1}} = 4{d^2\over dx^2}{{x^2\over2^2}\over1-{x\over2}} = 2{d^2\over dx^2}{x^2\over2-x} = 2{d^2\over dx^2}{x^2-4+4\over2-x} = 2{d^2\over dx^2}\left(-x-2 + {4\over2-x}\right) = {8\cdot1\cdot2\over(2-x)^3} = {16\over(2-x)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Similar Triangles and Incircle The incircle of trianlge ABC touches the sides AB, AC and BC at points P, N, and M respectively. Denote AP=AN=x, BM=BP=y and CM=CN=z. Segment UV is tangent to the incircle at point X and parallel to the side AC.
Prove $\cfrac{UV}{AC}$= $\cfrac{y}{x+y+z}$
What I have so far
$\triangle$ ABC $\sim$ $\triangle$ BUV
which has gotten $\cfrac{BV}{BC}$ = $\cfrac{BU}{AB}$ =$\cfrac{UV}{AC}$
any help would be appreciated!
| Let $a$, $b$, $c$ be the sidelengths of $BC$, $CA$, $AB$, let $h_1$ be the height from $B$ to $AC$, and let $h_2$ be the height from $B$ to $UV$. Note that $\dfrac{UV}{AC} = \dfrac{h_2}{h_1}$ because of the similar triangles you mentioned. Also, $h_2 = h_1 - 2r$ where $r$ is the inradius of triangle $ABC$. Hence $$\dfrac{UV}{AC} = \dfrac{h_1-2r}{h_1} = 1-\dfrac{2r}{h_1}.$$ Now let $K$ be the area of triangle $ABC$. We know $2K = h_1b = r(a+b+c)$, so $$\dfrac{r}{h_1} = \dfrac{2K/h_1}{2K/r} = \dfrac{b}{a+b+c}.$$ Thus $$\dfrac{UV}{AC} = 1-\dfrac{2b}{a+b+c} = \dfrac{a-b+c}{a+b+c} = \dfrac{y}{x+y+z}.\ \blacksquare$$ (Notes: The last step follows from $x = (b+c-a)/2, y=(c+a-b)/2, z=(a+b-c)/2$, which is well-known. To get the formula $2K = r(a+b+c)$, note that $K$ is the sum of the areas of triangles $IAB$, $IBC$, and $ICA$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Matlab Question: How to compute all partitions of [1,2,...,n] for a fixed number of parts? For example: [1,2,3,4] and k=3 parts should yield
[1],[2],[3,4]
[1],[3],[2,4]
[1],[4],[2,3]
[2],[3],[1,4]
[2],[4],[1,3]
[3],[4],[1,2]
as an output.
For k=2 with the same set we would get
[1],[2,3,4]
[2],[1,3,4]
[3],[1,2,4]
[4],[1,2,3]
[1,2],[3,4]
[1,3],[2,4]
[1,4],[2,3]
instead.
Can anyone give me some advice here? Thanks a lot.
| I will solve problem for $n=4$ special case. I hope that you can find the solution method for bigger values of $n$.
For $k=1$, there is only $1$ partition: $[1234]$
For $k=2$, there are two groups that $[ab],[cd]$ and $[a],[bcd]$. In the case $[ab],[cd]$: $\dbinom{4}{2} \cdot \dbinom{2}{1}$. But $[ab],[cd]$ and $[cd],[ab]$ are identical. So we have to divide by $2!$, hence $\dfrac{1}{2!}\dbinom{4}{2}=3$. Now in the case $[a],[bcd]$: $\dbinom{4}{1} \cdot \dbinom{3}{3}=4$. Sum of these $3 + 4 = 7$
For $k=3$, there are $[a],[b],[cd]$ partitions. Therefore $\dbinom{4}{1} \cdot \dbinom{3}{1} \cdot \dbinom{2}{2}$. But $[a],[b],[cd]$ and $[b],[a],[cd]$ gives identical partitions. Thus we have to divide by $2!$ and hence $\dfrac{1}{2!}\dbinom{4}{1} \cdot \dbinom{3}{1}=6$.
For $k=4$, there are $[a],[b],[c],[d]$ partitions. We get $\dbinom{4}{1} \cdot \dbinom{3}{1}\cdot \dbinom{2}{1} \cdot \dbinom{1}{1}$. But permutating of $[a],[b],[c],[d]$ are identical. Thus we have to divide by $4!$ and we find $ \dfrac{1}{4!}\dbinom{4}{1} \cdot \dbinom{3}{1}\cdot \dbinom{2}{1}=1$.
Note: Number of total partitions is $1+7+6+1=15$.
| {
"language": "en",
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solve the set of simultaneous congruences using Chinese Remainder Theorem 2x= 1 (mod 5), 3x= 9 (mod 6), 4x = 1 (mod 7), 5x = 9 (mod 11) Solve the set of simultaneous congruences using Chinese Remainder Theorem
$\begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 9 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases}$
This is what I got so far:
I simplify:
$2x \equiv 1 \pmod 5$ becomes $x \equiv 3 \pmod 5$
$3x \equiv 9 \pmod 6$ has $3$ solutions since $\gcd(3,6)=3$ they are $x = 17$, $19$, or $21 \pmod 6$.
$4x \equiv 1 \pmod 7$ becomes $x \equiv 2 \pmod 7$
$5x \equiv 9 \pmod {11}$ becomes $x \equiv 4 \pmod {11}$
By CRT, and substituting the $3$ different solutions, I got:
$x \equiv 653 \pmod{2310}$, $x \equiv 1423 \pmod{2310}$ and $x \equiv 2193 \pmod {2310}$.
But, the key answer from the book is $x \equiv 653 \pmod{770}$.
I am struggling to figure out how to get $x = 653 \pmod{770}$. I need help. Thank you.
| $\begin{cases}
2x \equiv 1 \pmod{5} & \times 3 \quad 6\equiv 1\pmod 5 & x\equiv \color{green}3 \pmod 5\\
3x \equiv 9 \pmod{6} & /3 \text{ whole equation} & x\equiv 3\equiv \color{green}1\pmod 2\\
4x \equiv 1 \pmod{7} & \times 2 \quad 8\equiv 1\pmod 7 & x\equiv \color{green}2\pmod 7\\
5x \equiv 9 \pmod{11} & \times 9 \quad 45\equiv 1\pmod {11} & x\equiv 81\equiv \color{green}4\pmod{11}\\
\end{cases}$
By Chinese remainder theorem :
$\begin{cases}
N=2.5.7.11=770\\
N_5=2.7.11=\color{red}{154} &\quad\equiv 4\equiv \color{blue}4^{-1}\pmod 5\\
N_2=5.7.11=\color{red}{385} &\quad\equiv 1\equiv \color{blue}1^{-1}\pmod 2\\
N_7=2.5.11=\color{red}{110} &\quad\equiv 5\equiv \color{blue}3^{-1}\pmod 7\\
N_{11}=2.5.7=\color{red}{70} &\quad\equiv 4\equiv \color{blue}3^{-1}\pmod {11}\\
\end{cases}$
$x\equiv (\color{red}{154}\times \color{blue}4\times \color{green}3)+(\color{red}{385}\times \color{blue}1\times \color{green}1)+(\color{red}{110}\times \color{blue}3\times \color{green}2)+ (\color{red}{70}\times \color{blue}3\times \color{green}4)\equiv 3733\pmod{770}$
$x\equiv 653\pmod{770}$
Let's have $m=\gcd(a,b,c)$ and $ax\equiv b\pmod c$
$\begin{cases}
a=m\alpha\\
b=m\beta\\
c=m\gamma\\
\end{cases}$
$\exists k\mid ax=b+kc\iff m\alpha x=m\beta+km\gamma\iff \alpha x=\beta+k\gamma$
Which is $\alpha x\equiv \beta \pmod{\gamma}$, so you can divide the whole equation by $m$.
This is the step you missed you to conclude easily.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| standard trick from contests: $5 \equiv 2 \pmod 3.$ Therefore, if $\omega^3 = 1$ but $\omega \neq 1,$ we get
$$ \omega^5 = \omega^2 $$
$$ \omega^5 + \omega + 1 = \omega^2 + \omega + 1 = 0 $$
Which means, various ways of saying this, $x^5 + x + 1$ must be divisible by
$$ (x - \omega)(x - \omega^2) = x^2 + x + 1. $$
This is what we call a "minimal polynomial" for $\omega$
The same idea would work for
$$ x^{509} + x^{73} + 1 $$
although the other factor would be worse
SEE
Prove that $n^5+n^4+1$ is composite for $n>1.$
Prime factor of $A=14^7+14^2+1$
| {
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} |
The eigenvalues and the eigenvectors of $y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0$ To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ I proceed as follows
$$\begin{align}
\bigg(\frac{d^{
4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{\lambda} \bigg) \bigg(\frac{d^{2}}{dt^{2}} - \sqrt{\lambda} \bigg) y\\
&= \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - i\lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} + \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - \lambda^{\frac{1}{4}} \bigg) y \\
&= 0
\end{align}$$
Therefore
$$\begin{align}y(t) &= A\cos(\lambda^{\frac{1}{4}}t)+B\sin(\lambda^{\frac{1}{4}}t)+ C\exp(\lambda^{\frac{1}{4}}t)+D\exp(-\lambda^{\frac{1}{4}}t)\\
y'(t) &= -\lambda^{\frac{1}{4}}A\sin(\lambda^{\frac{1}{4}}t)+ \lambda^{\frac{1}{4}}B\cos(\lambda^{\frac{1}{4}}t)+ \lambda^{\frac{1}{4}}C\exp(\lambda^{\frac{1}{4}}t)- \lambda^{\frac{1}{4}}D\exp(-\lambda^{\frac{1}{4}}t)
\end{align}$$
Let $v = \lambda^{\frac{1}{4}}$ then
$$\begin{align}y(t) &= A\cos(vt)+B\sin(vt)+ C\exp(vt)+D\exp(-vt)\\
y'(t) &= -Av\sin(vt)+ Bv\cos(vt)+ Cv\exp(vt)- Dv\exp(-vt)
\end{align}$$
Now, applying the boundary conditions
$$\begin{align}y(0) &= 0 \implies A+C+D = 0 \\
y(1) &= 0 \implies A\cos(v)+B\sin(v)+ C\exp(v)+D\exp(-v)=0\\
y'(0) &= 0 \implies B+ C- D = 0\\
y'(1) &= 0 \implies -A\sin(v)+ B\cos(v)+ C\exp(v)- D\exp(-v)=0
\end{align}$$
Actually I stuck here.
| This is my solution of the problem. Please let me know if you have any comments (Ahmed).
To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ we proceed as follows
\begin{align*}
\bigg(D^4 - \lambda \bigg) y &= \bigg( D^2 + \sqrt{\lambda} \bigg) \bigg(D^2 - \sqrt{\lambda} \bigg) y\\
&= \bigg(D + i \sqrt[4]{\lambda} \bigg) \bigg(D - i\sqrt[4]{\lambda} \bigg) \bigg(D + \sqrt[4]{\lambda} \bigg) \bigg(D - \sqrt[4]{\lambda} \bigg) y \\
&= 0
\end{align*}
Therefore
\begin{align*}
y(x) &= A\cosh(\sqrt[4]{\lambda}x)+B\sinh(\sqrt[4]{\lambda}x)+ C\cos(\sqrt[4]{\lambda}x)+D\sin(\sqrt[4]{\lambda}x)\\
y'(x) &= \sqrt[4]{\lambda}A\sinh(\sqrt[4]{\lambda}x)+ \sqrt[4]{\lambda}B\cosh(\sqrt[4]{\lambda}x)- \sqrt[4]{\lambda}C\sin(\sqrt[4]{\lambda}x)+ \sqrt[4]{\lambda}D\cos(\sqrt[4]{\lambda}x)
\end{align*}
Let $r = \sqrt[4]{\lambda}$ then
\begin{align*}
y(x) &= A\cosh(rx)+B\sinh(rx)+ C\cos(rx)+D\sin(rx)\\
y'(x) &= Ar\sinh(rx)+ Br\cosh(rx)- Cv\sin(rx)+ Dr\cos(rx)
\end{align*}
Now, applying the boundary conditions
\begin{align*}
y(0) &= 0 \implies A+C = 0 \\
y(1) &= 0 \implies A\cosh(r)+B\sinh(r)+ C\cos(r)+D\sin(r)=0\\
y'(0) &= 0 \implies B+D = 0\\
y'(1) &= 0 \implies r\left(A\sinh(r)+ B\cosh(r)- C\sin(r)+ D\cos(r)\right) =0
\end{align*}
So that $C = -A$ and $D = -B$. So that
$$A(\cosh(r) - \cos(r)) + B(\sinh(r) - \sin(r))=0$$
$$A(\sinh(r) + \sin(r)) + B(\cosh(r) - \cos(r))=0$$
Therefore
$$
\begin{bmatrix}
\cosh(r) - \cos(r) & \sinh(r) - \sin(r) \\
\sinh(r) + \sin(r) & \cosh(r) - \cos(r)
\end{bmatrix}
\begin{bmatrix}
A \\
B
\end{bmatrix} =
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
From the basic theory of system of linear algebraic equations, compatibility requires the vanishing of the determinant of the matrix, i.e.,
$$\left(\cosh(r) - \cos(r)\right)^2 - \left(\sinh^2(r)-\sin^2(r) \right)=0$$
Hence
$$\cosh(r) \cos(r) = 1$$
Using numerical methods Maple finds the positive possible $r$'s to be $$r = [4.730, 7.853, 10.996, 14.137, \cdots ] \simeq \left[\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \cdots\right] = \frac{(2n+1)\pi}{2},~n=1,2, \cdots .$$ And since $r^4 = \lambda$, the possible approximate eigenvalues for the eigenvalue problem are
$$\lambda = r^4 \simeq \frac{(2n+1)^4\pi^4}{16},~n=1,2, \cdots,$$
and the approximate eigenvectors are given by
\begin{align*}
y_n(x) \simeq &A_n \left(\cosh \frac{(2n+1)\pi}{2} x - \cos \frac{(2n+1)\pi}{2} x\right) \\
+& B_n \left(\sinh \frac{(2n+1)\pi}{2} x - \sin \frac{(2n+1)\pi}{2} x\right),~n=1,2, \cdots .
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intersection of two parabolas Given $a>0$ and $b>0$, I want to find the points of intersection of the two parabolas
\begin{align}
y&=1-ax^2 \\x&=1-by^2
\end{align}
Clearly I can just eliminate one of the variables, and I'll get a quartic equation, whose general solutions will be an enormous mess (according to Mathematica, anyway).
I also tried using this approach, but again got stuck in a quagmire of algebra.
Or, I could just use numerical methods, but that's what I'm trying to avoid.
The general problem of intersecting two conic section curves is well understood, and can only be solved by the techniques I described above (as far as I know). But my problem is not the general one, it's a very specific special case, and I'm wondering if someone can see some clever shortcut.
According to this question, the intersection points all lie on a circle, but I don't know if that helps.
| Perhaps this is an improvement.
$$
y = 1 + a x^{2}
\tag{1}
$$
$$
x = 1 + b y^{2}
\tag{2}
$$
Substitute $(2)$ into $(1)$ to obtain
$$
y = 1 + a \left(b y^2+1\right)^2
$$
and solve for $y$.
$$
y = \color{blue}{\pm} \frac{1}{2} \sqrt{-\frac{4 \sqrt[3]{2} (4 a+3)}{3 \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-\frac{\sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}{3 \sqrt[3]{2} a b^2}-\frac{2 \sqrt{6}}{a b^2 \xi }\color{red}{\pm} \frac{8}{3 b}}
\color{red}{\pm} \frac{\xi }{2 \sqrt{6}}
$$
where
$$
\alpha = a^2 b^4 (27-32 a b (8 a (b+1)+8 b+9))
$$
$$
\beta = a b^2 (27-16 a (8 a+9) b)
$$
$$
\xi = \sqrt{\frac{\frac{2^{2/3} \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}{a}+\frac{8 \sqrt[3]{2} (4 a+3) b^2}{\sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-8 b}{b^2}}
$$
There are a total of $4$ cases. The $\color{blue}{blue}$ and $\color{red}{red}$ signs are independent.
Intrigued by your comment about the intersection points, a few cases were plotted.
| {
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"source": "stackexchange",
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Manipulating a Summation Series
Summation: How does$$\begin{align*} & \frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac 12\sum\limits_{k=1}^n\frac 1k+\frac n{2n+1}\tag1\\ & =\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k\tag2\\ & =\sum\limits_{k=1}^{2n}\frac 1k-\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^n\frac 1{k+n}\tag3\end{align*}$$
Given that$$\sum\limits_{k=1}^n\frac 1{(2k)^3-2k}=\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac 12\sum\limits_{k=1}^n\frac 1k$$
I'm not sure how they got from the first step to the second. The $\tfrac 12\sum\limits_{k=1}^n\tfrac 1k$ didn't change, so that must mean that$$\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}+\frac n{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}$$But I don't see how. I tried expanding the LHS, and rearranging to get the RHS, but that didn't go far.$$\begin{align*}\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k+1}\right)+\frac n{2n+1} & =\frac 12\left(\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 1{2k+1}\right)+\frac n{2n+1}\\ & =\frac 12\sum\limits_{k=1}^n\frac {4k}{4k^2-1}+\frac n{2n+1}\\ & =2\sum\limits_{k=1}^n\frac k{(2k)^2-1}+\frac n{2n+1}\end{align*}$$I feel like I'm close, but I just can't seem to finish it. I believe that you'll need to use a summation identity.
Questions:
*
*How do you get from step $(1)$ to step $(2)$?
*How do you get from $(2)$ to $(3)$
*Is there a PDF that lists all the summation rules and identities?
| When you face problem like this, always try to decompose the summation into simple addition form. If you decompose your summation into simple series addition form these summations are settled down easily.
Step 1-2:
\begin{align*}
&\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}\\&=\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=0}^{n-1}\frac 1{2k+1}-\frac12+\frac12\cdot\frac{1}{2n+1}\\
&=\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\left(1-\frac{1}{2n+1}\right)\\
&=\sum\limits_{k=1}^n\frac 1{2k-1}-\frac{n}{2n+1}\\
\implies&\boxed{\frac 12\sum\limits_{k=1}^n\frac 1{2k-1}+\frac 12\sum\limits_{k=1}^n\frac 1{2k+1}-\frac12\sum\limits_{k=1}^n\frac1k+\frac{n}{2n+1}=\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\sum\limits_{k=1}^n\frac1k}
\end{align*}
Step 2-3:
\begin{align*}
&\sum\limits_{k=1}^n\frac 1{2k-1}\\
&=\sum\limits_{k=1}^n\frac 1{2k-1}+\sum\limits_{k=1}^n\frac 1{2k}-\sum\limits_{k=1}^n\frac 1{2k}\\
&=\sum\limits_{k=1}^{2n}\frac1k-\frac12\sum\limits_{k=1}^n\frac 1k\\
\implies&\boxed{\sum\limits_{k=1}^n\frac 1{2k-1}-\frac12\sum\limits_{k=1}^n\frac 1k=\sum\limits_{k=1}^{2n}\frac1k-\sum\limits_{k=1}^n\frac1k=\sum\limits_{k=1}^n\frac 1{k+n}}
\end{align*}
| {
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limit $\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right)$
$$\lim_{n\to\infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 2+\sqrt4}+\frac{1}{\sqrt4+\sqrt6}+\cdots+\frac{1}{\sqrt{2n} +\sqrt{2n+2}}\right)$$
To find the limit I think I can use the sandwich theorem: if $f,g$ and $h$ are real functions such that $f(x)\le g(x) \le h(x)$ for all $x$ and if $\lim_{x\to 0} f(x)=L= \lim_{x\to 0} h(x)$, then $\lim_{x\to 0} g(x)= L$.
I don't know that which series I can take for $f(x)$ and $h(x)$ to compare with $g(x)$.
| $$\lim _{ n\to \infty } \frac { 1 }{ \sqrt { n } } \left( \frac { 1 }{ \sqrt { 2 } +\sqrt { 4 } } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 6 } } +.........+\frac { 1 }{ \sqrt { 2n } +\sqrt { 2n+2 } } \right) =\\ =\lim _{ n\to \infty } \frac { 1 }{ \sqrt { n } } \left( \frac { \sqrt { 2 } -\sqrt { 4 } }{ -2 } +\frac { \sqrt { 4 } -\sqrt { 6 } }{ -2 } +.........+\frac { \sqrt { 2n } -\sqrt { 2n+2 } }{ -2 } \right) =\\ =\lim _{ n\to \infty } -\frac { 1 }{ 2\sqrt { n } } \left( \sqrt { 2 } -\sqrt { 4 } +\sqrt { 4 } -\sqrt { 6 } +.........+\sqrt { 2n } -\sqrt { 2n+2 } \right) =\\ =\lim _{ n\to \infty } -\frac { 1 }{ 2\sqrt { n } } \left( \sqrt { 2 } -\sqrt { 2n+2 } \right) =\lim _{ n\to \infty } \left( \frac { \sqrt { 2n+2 } }{ 2\sqrt { n } } -\frac { \sqrt { 2 } }{ 2\sqrt { n } } \right) =\frac { \sqrt { 2 } }{ 2 } $$
| {
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"source": "stackexchange",
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Derivative of vector consisting of euclidean distances I have $g: \Bbb R^2 \to \Bbb R^4$ given by $g(x) = (\|c_1 - x\|, \dots, \|c_4 - x\|)$, where $c_1, \dots, c_4 \in \Bbb R^2$.
I want to find $\left( \dfrac {\partial g} {\partial x_1}, \dfrac {\partial g} {\partial x_2} \right)$. Any hints?
| You are esssentially asking about the partial derivatives of the Euclidean distance function. Given that $\|c - x\| = \sqrt {(c_1 - x_1)^2 + (c_2 - x_2)^2}$, it follows that
$$\begin{eqnarray} \frac {\partial \|c - x\|} {\partial x_1} = \frac {x_1 - c_1} {\sqrt {(c_1 - x_1)^2 + (c_2 - x_2)^2}} \\
\frac {\partial \|c - x\|} {\partial x_2} = \frac {x_2 - c_2} {\sqrt {(c_1 - x_1)^2 + (c_2 - x_2)^2}} \end{eqnarray}$$
therefore
$$\frac {\partial g} {\partial x_1} = \\ \left( \frac {x_1 - c_{1,1}} {\sqrt {(c_{1,1} - x_1)^2 + (c_{1,2} - x_2)^2}}, \frac {x_1 - c_{2,1}} {\sqrt {(c_{2,1} - x_1)^2 + (c_{2,2} - x_2)^2}}, \frac {x_1 - c_{3,1}} {\sqrt {(c_{3,1} - x_1)^2 + (c_{3,2} - x_2)^2}}, \frac {x_1 - c_{4,1}} {\sqrt {(c_{4,1} - x_1)^2 + (c_{4,2} - x_2)^2}} \right)$$
and
$$\frac {\partial g} {\partial x_2} = \\ \left( \frac {x_2 - c_{1,2}} {\sqrt {(c_{1,1} - x_1)^2 + (c_{1,2} - x_2)^2}}, \frac {x_2 - c_{2,2}} {\sqrt {(c_{2,1} - x_1)^2 + (c_{2,2} - x_2)^2}}, \frac {x_2 - c_{3,2}} {\sqrt {(c_{3,1} - x_1)^2 + (c_{3,2} - x_2)^2}}, \frac {x_2 - c_{4,2}} {\sqrt {(c_{4,1} - x_1)^2 + (c_{4,2} - x_2)^2}} \right)$$
where $c_i = (c_{i1}, c_{i2})$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\frac{b}{3x-2}\right) dx=k$$
then
$$\lim_{N \to \infty} \left[\frac{2}{3}ln(x)+(a-\frac{2}{3})ln (x+3)-\frac{b}{3}ln(3x-2)\right]_3^N =k$$
By logarithm properties:
$$\lim_{N \to \infty} \left[ln\frac{(x)^\frac{2}{3}(x+3)^\left(a-\frac{2}{3}\right)}{(3x-2)\frac{b}{3}}\right]_3^N =k$$
Evaluating:
$$\lim_{N \to \infty} \left[ln\frac{(N)^\frac{2}{3}(N+3)^\left(a-\frac{2}{3}\right)}{(3N-2)^\frac{b}{3}}\right]-\left[ln\frac{(3)^\frac{2}{3}(6)^\left(a-\frac{2}{3}\right)}{(7)^\frac{b}{3}}\right] =k$$
But, I do not know what to do after this step. Any recommendation will be appreciated
| Hint. From your third line, one may write, as $x \to \infty$,
$$
\begin{align}
&\color{blue}{\frac{2}{3}}\ln(x)+\color{blue}{\left(a-\frac{2}{3}\right)}\ln (x+3)\color{blue}{-\frac{b}{3}}\ln(3x-2)
\\=&\color{blue}{\left(\frac{2}{3}+a-\frac{2}{3}-\frac{b}{3}\right)}\ln x+\left(a-\frac{2}{3}\right)\ln \left(1+\frac 3x\right)-\frac{b}{3}\cdot\ln\left(3-\frac 2x\right)
\\=&\color{blue}{\left(a-\frac{b}{3}\right)}\ln x+O(1)
\end{align}
$$ then all one needs is to annihilate the factor of $\ln x$, that is taking
$$
\color{blue}{a=\frac b3}
$$ in order to guarantee the convergence of the given integral.
| {
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Prove that $\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$ I friend of mine sent me this problem a while ago, but I still can't figure it out. (He can't figure it out either.) I figured here would be a good place to ask for help.
Prove:
$$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
| $$\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n=\frac{e^3}{4\pi}$$
$$\log\left(\prod_{n=2}^\infty \frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \log\left(\frac{1}{e^2}\left(\frac{n+1}{n-1}\right)^n\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(\log({e^{-2}}) + n\log\left(\frac{n+1}{n-1}\right)\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log\left(\frac{n+1}{n-1}\right) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(n\log(n+1) - n\log(n-1) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$\sum_{n=2}^\infty \left(2n\coth^{-1}(n) - 2\right) = \log(\frac{e^3}{4\pi})$$
$$2\sum_{n=2}^\infty \left(n\coth^{-1}(n) - 1\right) = 3 - \log(4\pi)$$
Here I get stuck however.
| {
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Sum of Squares of odd numbers I just wanted to know how we get from the first step to the next. I spent too much time on this but could not get to the next step. This seems a simple sum of squares of odd numbers but was not able to get the next step.
$$ \frac{2 \mathcal{E}_p}{M} \left(1^2 + 3^2 + 5^2 + \dots + (M-1)^2\right) $$
$$ \frac{2 \mathcal{E}_p}{M} \times \frac{M(M^2-1)}{6} $$
| By Faulhaber's formulas, $$1^2+2^2+\cdots + (M-1)^2=\frac{(M-1)M(2M-1)}{6}$$
Since $M-1$ is odd, we must have $M-2$ even. We now calculate, using the same formula,
$$2^2+4^2+\cdots+(M-2)^2=2^2(1^2+2^2+\cdots+(\frac{M-2}{2})^2)=2^2\frac{\frac{M-2}{2}\frac{M}{2}(M-1)}{6}=\frac{(M-2)M(M-1)}{6}$$
Subtract the sum of the even squares, from the sum of all the squares, to get
$$1^2+3^2+\cdots+(M-1)^2=\frac{(M-1)M(2M-1)}{6}-\frac{(M-2)M(M-1)}{6}=\frac{(M-1)M}{6}(2M-1-(M-2))=\frac{(M-1)M(M+1)}{6}=\frac{M(M^2-1)}{6}$$
| {
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Inclusion exclusion distribution problem Prompt: How many ways are there to give 20 different presents to 4 different children, so that no child gets exactly 6 presents. All presents are different.
Here's what I tried doing
Since there are 20 presents to be distributed among 4 children,
$$C_1 + C_2 + C_3 + C_4 = 20$$
By bars and stars,
$$\binom{ 20 + 4-1 }{ 4-1 } = \binom{ 23 }{ 3 } = 1771$$
$C_2 + C_3 + C_4 = 14 $ (after giving 6 presents to $C_1$)
Assuming $C_1$ gets 6 present = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_2$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_3$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
Assuming $C_4$ gets 6 presents = $\binom{ 14 + 3 - 1 }{ 3 - 1 } = \binom{ 16 }{ 2 } = 120$
No one gets exactly 6 presents = Total - Everyone gets 6 presents
= $\binom{ 23 }{ 3 } - \binom{ 4 }{ 1 }\binom{ 16 }{ 2 } = 1771 - 480 = 1291$
I am new to combinatorics and not sure if I'm on the right path any help would be appreciated.
| Your use of combinations with repetition would be appropriate if the presents are identical. However, they are different, so it matters which child receives which present.
If there were no restrictions, there would be four possible recipients for each of the twenty presents, so there would be $4^{20}$ ways to distribute the presents to the children.
From these, we must exclude those distributions in which one or more children receive exactly six presents. Since $4 \cdot 6 = 24 > 20$, at most three of the children can receive exactly six presents.
There are $\binom{4}{1}$ ways to select a child who will receive six presents and $\binom{20}{6}$ ways to select the presents that child receives. The remaining $20 - 6 = 14$ presents can be distributed among the remaining three children in $3^{14}$ ways. Hence, there are
$$\binom{4}{1}\binom{20}{6}3^{14}$$
distributions in which a child receives exactly six presents.
There are $\binom{4}{2}$ ways to select two of the four children to receive exactly six presents. There are $\binom{20}{6}$ ways to select six of the presents for the younger of the two selected children and $\binom{20 - 6}{6} = \binom{14}{6}$ ways to select six presents for the older of the two selected children. The remaining $20 - 2 \cdot 6 = 8$ presents can be distributed to the remaining two children in $2^6$ ways. Hence, the number of distributions in which two children receive exactly six presents is
$$\binom{4}{2}\binom{20}{6}\binom{14}{6}2^8$$
There are $\binom{4}{3}$ ways to select three of the four children to receive exactly six presents. There are $\binom{20}{6}$ to distribute six of the twenty presents to the youngest of the selected children, $\binom{14}{6}$ ways to distribute six of the fourteen remaining presents to the next youngest of the selected children, and $\binom{8}{6}$ ways to distribute six of the remaining eight presents to the oldest of the selected children. There is one way to distribute the remaining $20 - 3 \cdot 6 = 2$ presents to the other child. Hence, the number of distributions in which three of the children receive exactly six presents is
$$\binom{4}{3}\binom{20}{6}\binom{14}{6}\binom{8}{6}1^2$$
By the Inclusion-Exclusion Principle, the number of distributions in which no child receives exactly six of the twenty presents is
$$4^{20} - \binom{4}{1}\binom{20}{6}3^{14} + \binom{4}{2}\binom{20}{6}\binom{14}{6}2^{8} - \binom{4}{3}\binom{20}{6}\binom{14}{6}\binom{8}{6}1^2$$
| {
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every integer can be represented in the form $x^2+y^2-z^2$ Every integer can be represented in the form $x^2+y^2-z^2$ and show that $6$ actually requires all three terms.
I put
*
*$z=y+1$
*$x=n^2+3$
*$y=3n^2+4+(n^4-n)/2$
what does it mean that $6$ actually requires all three terms?
| Any integer number that is not of the form $4m+2$ can be represented as a difference of two squares, since $n=y^2-z^2=(y-z)(y+z)$ has a solution as soon as $n$ has a divisor $d$ such that $d$ and its complementary divisor $\frac{n}{d}$ have the same parity. That is the same as stating "$n$ is not twice an odd number". It follows that $n=x^2+y^2-z^2$ always has a solution with $x=0$ or $x=1$.
$6$ is a number of the form $4m+2$, hence $x\neq 0$, and it is not the sum of two squares, since $3\mid 6$ but $3^2\nmid 6$. It follows that $6=x^2+y^2-z^2$ has a lot of solutions (like $6=5^2+9^2-10^2$) but for all of them $x,y,z\neq 0$.
| {
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Egyptian fraction for $\varphi- {F(2n+2) \over F(2n+1)}$ The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$.
https://math.stackexchange.com/a/746678/134791
This may be written as the following closed form for an Egyptian fraction.
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{F(2^{k+2})}$$
where $\varphi$ is the golden ratio
$$\varphi = \frac{1+\sqrt{5}}{2}$$
and $F(n)$ are the Fibonacci numbers as described by
$$F(n)=F(n-1)+F(n-2)$$
with $F(0)=0, F(1)=1$.
The result generalizes to other samplings.
$$\varphi = \frac{F(2n+1)}{F(2n)}-\sum_{k=0}^\infty \frac{1}{F(n2^{k+2})}$$
https://math.stackexchange.com/a/2307929/134791
Is there a similar formula for $\varphi- \dfrac{F(2n+2)
}{F(2n+1)}$?
Related questions
Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions?
Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$
| From a different expansion for the expression given by @achillehui $\frac{\alpha-\beta}{\alpha^{2m}+1}$, as a geometric series, particular cases are
$$\varphi = 1+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{2k}$$
$$\varphi = \frac{3}{2}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{3·2k}$$
$$\varphi = \frac{8}{5}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{5·2k}$$
$$\varphi = \frac{21}{13}+\sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{7·2k}$$
so the general form seems to be
$$\varphi = \frac{F(2n+2)}{F(2n+1)} + \sqrt{5}\sum_{k=1}^\infty (-1)^{k+1}\left(\frac{2}{1+\sqrt{5}}\right)^{(2n+1)2k}$$
This is alternating, an increasing version would be nice.
| {
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Note that the right hand side is real, so if we take complex conjugates it remains unchanged.
$$
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{2(\sqrt{13}-3)} \\
& = \sqrt{-3+2i}+\sqrt{-3-2i} \\
\end{align}
$$
Multiply the 2 conjugate forms to obtain
$$
\begin{align}
&(\sqrt{-3-2i}+\sqrt{-3+2i})\cdot (\sqrt{-3+2i}+\sqrt{-3-2i}) \\
&= \sqrt{13}+(-3-2i)+(-3+2i)+\sqrt{13} \\
&= 2(\sqrt{13}-3) \\
&= \left(\sqrt{2(\sqrt{13}-3)}\right)^2
\end{align}
$$
Alternatively just rearrange the terms of the right-hand side rather than looking at it as a complex conjugate.
| {
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\frac{x}{1-x^2}=x+x^3+x^5+\cdots$$
$$\frac{x^2}{1-x^4}=x^2+x^6+x^{10}+\cdots$$
$$\cdots$$
Add them all and get the answer. Unfortunately, I havn't got a direct method to calculate it. Appreciate for your help!
| Adding term by term:
$$S=\frac{x(1+x^2)+x^2}{1-x^4}+\frac{x^2}{1-x^8}+...=$$
$$\frac{(x+x^2+x^3)(1+x^4)+x^4}{1-x^8}+\frac{x^8}{1-x^{16}}+...=$$
$$\frac{\sum_{n=1}^{\infty} x^n}{1-x^{2\cdot \infty}}=$$
$$\frac{x}{1-x}$$
because $0<x<1$.
| {
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Hensel Lifting Explanation
I'm trying to understand this example for my exam tomorrow, I understand everything apart from where the recurrence relation comes from, is there a set formula for this? Any help is greatly appreciated.
| Let's do this slowly. At each stage we have a number $a_n$ with both $a_n\equiv 3\pmod 5$ and with $a_n^3\equiv2\pmod{5^n}$. We try $a_{n+1}=a_n+5^nt$. Then
$$a_{n+1}^3\equiv a_n^3+3\times 5^na_n^24
=2+(a_n^3-2)+3\times 5^na_n^2t\pmod{5^{n+1}}.$$
Now
$$3a_n^2\equiv3\times 3^2\equiv 2\pmod5$$
so
$$3\times 5^na_n^2\equiv2\times 5^n\pmod{5^{n+1}}.$$
Therefore
$$a_{n+1}^3-2\equiv(a_n^3-2)+2\times 5^nt\pmod{5^{n+1}}.$$
We need $a_{n+1}^3-2\equiv0\pmod{5^{n+1}}$. So we want
$$(a_n^3-2)+2\times 5^nt\equiv0\pmod{5^{n+1}}.$$
Multiplying by $2$, this is equivalent to
$$2(a_n^3-2)-5^nt\equiv0\pmod{5^{n+1}},$$
that is
$$5^nt\equiv 2(a_n^3-2)\pmod{5^{n+1}}.$$
So
$$a_{n+1}=a_n+5^nt\equiv a_n+2(a_n^3-2)\pmod{5^{n+1}}.$$
This is the recurrence formula!
| {
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Prove that $n=\frac{5^{125}-1}{5^{25}-1}$ is a composite number
Prove that $n=\dfrac{5^{125}-1}{5^{25}-1}$ is a composite number
My attempt,
Let $x=5^{25}$, so that $5^{125}-1=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
$=(x^4+9x^2+1+6x^3+6x+2x^2-5x^3-10x^2-5x)(x-1)$
$=((x^2+3x+1)^2-5x(x+1)^2)(x-1)$
I'm stuck at this point and don't know how to continue anymore. Hope someone can provide a detailed solution. Thanks a lot.
| The polynomial $\Phi_5(x)=x^4+x^3+x^2+x+1$ fulfills an interesting identity.
We have that $4\cdot \Phi_5(x)$ is pretty close to the square of $2x^2+x+2$, and indeed:
$$ 4 \Phi_5(x) = (2x^2+x+2)^2 - 5x^2 \tag{1}$$
as well as:
$$ \Phi_5(x) = (x^2+3x+1)^2 - 5x(x+1)^2 \tag{2} $$
so if $x=5^{2k+1}$, $\Phi_5(x)$ is the difference of two large squares:
$$\begin{eqnarray*} \Phi_5(5^{2k+1}) &=& \left(5^{4k+2}+3\cdot 5^{2k+1}+1\right)^2 - \left(5^{3k+2}+5^{k+1}\right)^2\\&=&\left(5^{4k+2}+5^{3k+2}+3\cdot 5^{2k+1}+5^{k+1}+1\right)\cdot\left(5^{4k+2}-5^{3k+2}+3\cdot 5^{2k+1}-5^{k+1}+1\right) \end{eqnarray*}$$
and $\Phi_5(5^{2k+1})$ cannot be a prime number. See Aurifeuillean factorization.
| {
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Maximum rotation angle of a rectangle inside a given rectangle I want to calculate the maximum rotation angle of a rectangle which is rotating with the center on the center of a bigger rectangle.
Here is a figure for better understanding: fig.
I have tried it out already but my solution is not plausible.
Thanks in advance!
Greets, Daniel
| Both rectangle have the same center due to their symmetry. There will be four contact points on the outer rectangle if the rectangles have same aspect ratio, else only two.
Let the inner rectangle have dimensions $(2a,2b)$. The y coordinate of contact point is $B_0/2$. Radius of inner rectangle is $ \sqrt{(a^2+b^2)} $ and its x- coordinate is $ \sqrt{(a^2+b^2)- (B_0/2)^2 } $
So coordinates of contact in first quadrant are
$$ \sqrt{(a^2+b^2)- (B_0/2)^2 } ,\, B_0/2 $$
Now we can apply polar form of straight line to find $\alpha$
$$ x \cos \alpha + y \sin \alpha = b \tag1 $$
$$ \sqrt{(a^2+b^2)- (B_0/2)^2 } \cos \alpha + (B_0/2) \sin \alpha = b\tag2 $$
Let
$$ \beta= \tan ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2)- (B_0/2)^2 } }= \sin ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2) } } \tag3 $$
Divide by $ \sqrt{a^2+b^2} $
$$ \cos(\alpha- \beta) = \frac{b}{\sqrt{a^2+b^2}} \tag4 $$
$$ \alpha -\beta = \cos^{-1}\frac{b}{\sqrt{a^2+b^2}}\tag5 $$
and finally find unknown rotation required, $\theta = \alpha -\pi/2$
$$ \theta = \alpha - \pi/2 = \sin ^{-1} \dfrac{(B_0/2)}{\sqrt{(a^2+b^2) } } - \sin^{-1}\frac{b}{\sqrt{a^2+b^2}}\tag6 $$
| {
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Finding all primes $p,q$ with $p^2+q^2=9pq-13$ The last month I was trying to solve a problem of a magazine, and I found the following equation $$p^2+q^2=9pq-13,$$
Where $p$ and $q$ are primes. We need to get solutions when $p$ and $q$ are odd, because if any of them is even, the only solution that works is $(2,17)$. Any ideas will be much appreciated.
I analyzed the discriminant of the quadratic equation and we need to find solutions of $77q^2-52=k^2$, this is a variation of Pell's equations.
| This answer will find all integer answers. Don't know if there is an easy way to find prime answers.
Multiplying by $4$ we get $4p^2-36pq+4q^2=-52$ or $(2p-9q)^2-77q^2=-52$.
Now, if there is a solution to $x^2-77y^2=-52$, with $x,y>0$ then we have that $x,y$ must have the same parity, and we take.
$$\begin{align}x_1+y_1\sqrt{77}&=(x+y\sqrt{77})\left(\frac{9}{2}-\frac{1}{2}\sqrt{77}\right)\\
&=\frac{9x-77y}{2}+\frac{9y-x}{2}\sqrt{77}
\end{align}$$
If $x$ is the smallest positive integer solution, then we have that:
$$\left|\frac{9x-77y}{2}\right|\geq x$$
So either $x\geq 11y$ or $7y\geq x$. Which means that either $-52=x^2-77y^2\geq 44y^2$ which isn't possible, or $-52=x^2-77y^2\leq -28y^2,$ so the only possible value for $y$ is $y=1$, and this gives $x=5.$
So there are infinitely many solutions, all of the form $$X+Y\sqrt{77}=(5+\sqrt{77})\left(\frac{9}{2}+\frac{1}{2}\sqrt{77}\right)^k$$
Then $q=Y, p=\frac{X+9Y}{2}$.
| {
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Two dice-related games, do the outcomes have the same distribution? In this codegolf question, it is effectively asked to find the distribution for $X$, defined below:
*
*Throw $2$ dice; is the sum $7$ then $X=2$, else continue.
*Throw $4$ dice; is the sum $14$ then $X=4$, else continue.
*...
*Throw $2n$ dice; is the sum $7n$ then $X=2n$, else continue.
*...
A few answers attempt this by calculating $Y$, defined below:
*
*Throw $2$ dice; is the sum $7$ then $Y=2$, else continue.
*Keep the $2$ dice on the table, throw $2$ additional dice; is the sum $14$ then $Y=4$, else continue.
*...
*Keep the previous $2n-2$ dice on the table, throw $2$ additional dice; is the sum $7n$ then $Y=2n$, else continue.
*...
I have the feeling that $X$ and $Y$ have (slightly) different distributions, because for $Y$ you know that the first $2n-2$ dice did not sum to zero, while for $X$ that could be the case. However, I don't remember how to approach this problem...
| $X$ and $Y$ have different distributions. To see this, consider the following:
$$P[X=2] = P[Y=2] = \frac{6}{36} = \frac{1}{6}$$
For $X$ to equal 4, we first must not hit 7 with the first two dice, and then hit 14 with the four new dice. To throw these last four, we can first throw two new dice and then another two dice. If the first two dice hit 2, we need 12 with the other two; if they hit 3, we need 11 etc. As such, we get:
$$P[X=4] = (1-P[X=2]) \cdot \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4 + 5 \cdot 5 + 6 \cdot 6 + 5 \cdot 5 + 4 \cdot 4 + 3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1}{6^4} = \frac{5}{6} \cdot \frac{146}{1296} \approx 0.09388$$
For $Y$ to equal 4, we need the first two dice to not hit 7, and given this information, two new dice to complete the sum. As such, we get:
$$P[Y=4] = \frac{1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 4 + 5 \cdot 5 + 5 \cdot 5 + 4 \cdot 4 + 3 \cdot 3 + 2 \cdot 2 + 1 \cdot 1}{6^4} = \frac{110}{1296} \approx 0.08488$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Is $\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$? Look at this limit. I think, this equality is true.But I'm not sure.
$$\lim_{k\to\infty}\frac{\sum_{n=1}^{k} 2^{2\times3^{n}}}{2^{2\times3^{k}}}=1$$
For example, $k=3$, the ratio is $1.000000000014$
Is this limit mathematically correct?
| I have taken a simpler approach to solve this.
$$\frac{\sum_{n=1}^k 2^{2X3^n}}{2^{2X3^k}}$$ can be written as:
$$\frac{\sum_{n=1}^k 4^{3^n}}{4^{3^k}}$$ which further can be written as :
$$\frac{4^{3^1}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} +\frac{4^{3^3}}{4^{3^k}}......... \frac{4^{3^{k-1}}}{4^{3^k}} +\frac{4^{3^k}}{4^{3^k}}$$
writing this in reverse order:
$$\frac{4^{3^k}}{4^{3^k}} + \frac{4^{3^{k-1}}}{4^{3^k}}......... +\frac{4^{3^3}}{4^{3^k}} + \frac{4^{3^2}}{4^{3^k}} + \frac{4^{3^1}}{4^{3^k}}$$
which becomes
$$\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}}$$
Now as $k→∞, 3^k →∞ $
which implies, $4^{3^k-3^{k-1}} →∞ $
which imply, $\frac{1}{4^{3^k-3^{k-1}}} + ......... → 0 $
which leaves, $\frac{1}{1} + \frac{1}{4^{3^k-3^{k-1}}}......... +\frac{1}{4^{3^k-3^3}} + \frac{1}{4^{3^k-3^2}} + \frac{1}{4^{3^k-3^1}} → 1 $ as $k→∞ $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
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Balls in bins, probability that exactly two bins are empty If we throw randomly $n+1$ balls into $n$ bins, what is the probability that exactly two bins are empty?
I tried to do this problem but I didn't get right solution, I would be very grateful if someone would point where am I making mistake, and how should I think about that. Also, if there are any simpler approaches, I would like to see them. I have seen that there have been similar questions but I would really like to find out where my intuition is making mistake.
First case:
$4$ balls inside one bin, and in other bins exactly one ball in each. We have $\binom{n+1}{4}$ ways of choosing $4$ balls from $n+1,$ and we can choose one from $n$ bins. Other balls we can place in $(n-3)!$ ways, so that is finally $n\cdot\binom{n+1}{4}\cdot(n-3)!$ ways. This way $2$ bins stay empty each time.
Second case:
$3$ balls inside one bin, $2$ balls in some other bin, and in left bins $n-4$ balls. I can choose $3$ balls in $\binom{n+1}{3}$ ways and place them into one of $n$ bins, after that I can choose two from $n-2$ left balls in $\binom{n-2}{2}$ ways and place them in one of $(n-1)$ left bins. Considering permutations of balls that I didn't pick, that is finally
$\binom{n+1}{3}\cdot n\cdot\binom{n-2}{2}\cdot(n-4)!$ ways.
In this case also two bins will be empty.
Third case:
Two bins with two balls in them and other bins with $1$ ball in them, that will also result in $2$ empty bins. Same as described in previous cases, this would be $\binom{n+1}{2}\cdot n\cdot\binom{n-1}{2}\cdot(n-1)\cdot(n-3)! $
Total number of cases is $n^{n+1}.$
Probability is sum of these three cases over total number of cases, but I am making some mistake in counting. Thank you.
| This is a probability question, so you have a choice of considering ways to arrange indistinguishable balls or ways to arrange balls that are all different. You just have to be consistent in how you add up the probability of each arrangement.
The number of ways to arrange $n+1$ balls (all different) into the $n$ bins is $n^{n+1}.$ All these arrangements are equally likely.
The arrangements of indistinguishable balls are not all equally likely, so I will consider the balls all different.
When counting arrangements, you must consider the fact that different choices of which bins are empty will give you different arrangements.
I did not see any consideration of that in the question.
So for the first case, after choosing one of $n$ bins to put the four balls in, you must choose two of the remaining $n-1$ bins to be empty,
so instead of $n\binom{n+1}{4}(n-3)!$ arrangements
you have $n\binom{n-1}{2}\binom{n+1}{4}(n-3)!.$
Simplifying this a bit, it is $\frac{1}{48}n(n-1)(n-2)(n+1)!.$
For the second case, you are missing a factor of $n-1$ to account for the number of ways to select the bin to put two balls in,
and a factor of $\binom{n-2}{2}$ for the choice of two empty bins,
so instead of
$n\binom{n+1}{3}\binom{n-2}{2}(n-4)!$ ways
you should have $n\binom{n+1}{3}(n-1)\binom{n-2}{2}\binom{n-2}{2}(n-4)!.$
This simplifies to $\frac{1}{24}n(n-1)(n-2)(n-3)(n+1)!.$
For the third case, you actually need to put two balls into each of
three bins.
But you also need to avoid multiple counting of the same arrangement:
for example, choosing bin $1,$ putting balls $1$ and $2$ in that bin,
then choosing bin $2$ and putting balls $3$ and $4$ in it,
is the same result as choosing bin $2,$ putting $3$ and $4$ in it,
then choosing $1$ and putting $1$ and $2$ in it.
To avoid double-counting, rather than choosing one bin, then another, then another ($n(n-1)(n-2)$ ways), you can first choose the three bins that will each have two balls ($\binom{n}{3}$ ways) and then fill them left to right.
But also remember to choose two empty bins from the remaining $n-3$
before placing the remaining balls. So instead of
$\binom{n+1}{2}n\binom{n-1}{2}(n-1)(n-3)!$ you should have
$\binom{n}{3}\binom{n+1}{2}\binom{n-1}{2}\binom{n-3}{2}\binom{n-3}{2}(n-5)!.$
This simplifies to
$\frac{1}{96}n(n-1)(n-2)(n-3)(n-4)(n+1)!.$
Adding these up, we have $M$ arrangements with exactly two empty bins, where
\begin{align}
M &= \frac{1}{48}n(n-1)(n-2)(n+1)!
+ \frac{1}{24}n(n-1)(n-2)(n-3)(n+1)! \\
&\qquad + \frac{1}{96}n(n-1)(n-2)(n-3)(n-4)(n+1)! \\
&= \frac{n(n-1)(n-2)(n+1)!}{96}\left(2 + 4(n-3) + (n-3)(n-4)\right) \\
&= \frac{n(n-1)(n-2)(n+1)!}{96}\left(n^2 - 3n + 2\right) \\
&= \frac{n(n-1)^2(n-2)^2(n+1)!}{96}.
\end{align}
And then the probability is just the number of arrangements with exactly
two empty bins divided by the total number of arrangements:
$$\frac{M}{n^{n+1}} = \frac{(n-1)^2(n-2)^2(n+1)!}{96n^n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Prove or disprove that where $a, b \in Z$, the number a+b is even if and only if a-b is even.
Can some help me improve on this or check it please?
If $a + b$ is even, then there is some number $k \in Z$ such that $a + b = 2k$.
If $a - b$ is odd, then there is some number $m \in Z$ such that $a - b = 2m +1$.
$a + b = a - b + 2b = 2m + 1 + 2b = 2(m + b) +1$. This is an odd number.
Now try for $a - b$ is even.
If $a - b$ is even, then there is a number $m \in Z$ such that $a - b = 2m$.
$a + b = a - b + 2b = 2b + 2m = 2(b + m)$. This is an even number.
Therefore, $a + b$ is even if and only if $a - b$ is even.
| We have $a+b$ even iff $a+b=2k$ for some $k$ iff $a+b-2b = 2k-2b=2(k-b)$ even iff $a-b=2(k-b)$ even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove this inequality $(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,$
let $a,b>0$,and $n\ge 4$ be postive integers, such $(a+b)^{2n}=2n+\dfrac{n}{4},a^{2n}=\dfrac{n}{4}$
show that
$$(a+b)^rb^{2n-r}\le\dfrac{n}{4}+r,\forall r\in[0,2n]$$
it seem hard for $n=4$ case.show this inequality $3^{\frac{n}{4}}\cdot (3^{\frac{1}{4}}-1)^{8-n}\le 1+n$
it is enought show that
$$((a+b)^{2n})^r(b^{2n})^{2n-r}\le\left(\dfrac{n}{4}+r\right)^{2n}$$
or
$$(\dfrac{9n}{4})^{r}(b^{2n})^{2n-r}\le\left(\dfrac{n}{4}+r\right)^{2n}$$
it seem hard to prove it
| Another version ...
From $a=\sqrt[2n]{\frac{n}{4}}$, $a+b=\sqrt[2n]{2n+\frac{n}{4}}$
we have $a\geq1$ and $a+b >1$, then:
$$(a+b)^{r}b^{2n-r}=\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(b^{2n}\right)^{\frac{2n-r}{2n}} \tag{1}$$
However
$$b=\sqrt[2n]{2n+\frac{n}{4}}-\sqrt[2n]{\frac{n}{4}}<a \tag{2}$$
because
$$\sqrt[2n]{2n+\frac{n}{4}}<2\sqrt[2n]{\frac{n}{4}} \Leftrightarrow \sqrt[2n]{\frac{9n}{4}}<2\sqrt[2n]{\frac{n}{4}} \Leftrightarrow \sqrt[2n]{9}<2, \forall n\geq2$$
Continuing $(1)$ and considering $(2)$:
$$(a+b)^{r}b^{2n-r}<\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(a^{2n}\right)^{\frac{2n-r}{2n}}=\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}} \tag{3}$$
We will now look (considering $\ln{x}$ is concave function and $\frac{r}{2n}+\frac{2n-r}{2n}=1$) at:
$$\ln{\left(\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}}\right)}=\frac{r}{2n}\ln{\left(2n+\frac{n}{4}\right)}+\frac{2n-r}{2n}\ln{\left(\frac{n}{4}\right)}\leq \\
\ln{\left(\frac{r}{2n}\left(2n+\frac{n}{4}\right)+\frac{2n-r}{2n}\left(\frac{n}{4}\right)\right)}=\ln{\left(\frac{n}{4}+r\right)}$$
which, from $(3)$, leads to
$$(a+b)^{r}b^{2n-r}<\left(2n+\frac{n}{4}\right)^{\frac{r}{2n}}\left(\frac{n}{4}\right)^{\frac{2n-r}{2n}} \leq \frac{n}{4}+r $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Discuss the monotonicity of the following function without using differentiation. Can I discuss the monotonicity of the following function without using differentiation?
$$f(x) = x + \frac{9}{x}$$
Could anyone help me?
| If $f$ is strictly increasing on an interval $I$, then $f(x)> f(y)$ for $x,y\in I$ with $x>y$.
Note that $f$ is undefined when $x=0$
\begin{align}
f(x)-f(y)&=x-y+\frac{9}{x}-\frac{9}{y}\\
&=(x-y)\left(1-\frac{9}{xy}\right)\\
&=\frac{(x-y)\left(x-\frac{9}{y}\right)}{x}\\
\end{align}
For $x>0$, $\displaystyle f(x)-f(3)=\frac{(x-3)^2}{x}\ge 0$ and so $(3,f(3))$ is a local minimum.
If $x>y\ge3$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)>0$.
If $3\ge x>y>0$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)<0$.
For $x<0$, $\displaystyle f(x)-f(-3)=\frac{(x+3)^2}{x}\le 0$ and so $(-3,f(-3))$ is a local maximum.
If $0>x>y\ge-3$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)<0$.
If $-3\ge x>y$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)>0$.
$f$ is strictly increasing on $(-\infty,-3]\cup[3,\infty)$.
$f$ is strictly decreasing on $[-3,0)\cup(0,3]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the ratio of lengths in a triangle with cevians, without using mass-points In $\triangle ABC$, $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AD$. $CE$ is extended to meet $AB$ at $F$.
The question is to find the ratio of the lengths of $AF$ to $FB$.
This is trivial using mass-points geometry:
$$C=1\to B=1\to D=2\to A=2\to E=4\to F=3$$
and thus $AF:FB=1:2$.
But how would one solve this without using mass-point geometry?
Thanks!
| Let the points $A,B,C$ corresponds to vectors $\mathbf a,\mathbf b,\mathbf c$.
The vector $\vec{AD}$ is
$$\begin{align*}
\vec{AB} + \vec{BD}
&= \vec{AB} + \frac12\vec{BC}\\
&= \mathbf b - \mathbf a+\frac12(\mathbf c-\mathbf b)\\
&= -\mathbf a+\frac12\mathbf b + \frac12\mathbf c
\end{align*}$$
The vector $\vec{AE}$ is half of $\vec{AD}$,
$$\vec{AE} =-\frac12\mathbf a+\frac14\mathbf b +\frac14\mathbf c$$
The vector $\vec{CE}$ is
$$\begin{align*}\vec{CA} + \vec{AE} &=\mathbf a-\mathbf c-\frac12\mathbf a+\frac14\mathbf b + \frac14\mathbf c\\
&= \frac12\mathbf a + \frac14\mathbf b -\frac34\mathbf c
\end{align*}$$
Let $AF:FB = 1:p$. Then $\vec{AF} = \frac{1}{1+p}(\mathbf b-\mathbf a)$. Since $CEF$ is a straight line, $\vec{CF}$ has the same direction as $\vec{CE}$.
$$\begin{align*}
\vec{CF} &= k\vec{CE}\\
\vec{CA} + \vec{AF} &= k\vec{CE}\\
\mathbf a - \mathbf c + \frac1{1+p}(\mathbf b-\mathbf a) &= \frac k2\mathbf a + \frac k4b - \frac{3k}4\mathbf c\\
\frac p{1+p}\mathbf a + \frac1{1+p}\mathbf b-\mathbf c &= \frac k2\mathbf a + \frac k4b - \frac {3k}4\mathbf c\\
p &= 2\\
AF:FB &= 1:2
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to solve problem of maximum and minimum in two variables? I have function
$$f(x,y)=\frac{3(x+y)-2}{x^2+y^2}$$
and need maximum and minimum in domain
$$D=\left\{(x,y)\in\Bbb R^2 ; x^2+y^2=8\right\}\cup\left\{(x,y)\in\Bbb R^2;|x|+|y|=1\right\}$$
I thought of Lagrange multipliers but that square above is hard: it is enclosed by lines $\;y=x+1,\,y=-x+1,\,y=x-1,\,y=-x-1\;$ , so my question is:
Can I do only two constraints and form the function
$$H(x,y,\lambda,\mu)=f(x,y)-\lambda(x^2+y^2-8)-\mu(|x|+|y|-1)$$
or else I need five constraints, one for circle and each line above
$$G(x,y,\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5)=f+\lambda_1(x^2+y^2-8)-\lambda_2(x-y+1)-\lambda_3(-x+y+1)-$$
$$-\lambda_4(x-y-1)-\lambda_5(-x-y-1)\;?$$
Both ways looks very hard, because in first taking derivatives of $\;x,\,y\;$ give two possible signs, for example:
$$H'_x=f'_x-2\lambda x\,\mp\mu\;,\;\;\text{and etc.}$$
whereas in second way, with $\;G\;$, I need to solve seven nonlinear equations...
Is there perhaps some other easier way to do? I thought: do first partial equations of $\;f(x,y)\;$ alone, with Hessian and find points inside the domain, and then do substitution for circle $\;x^2+y^2=8\;$ and then for square $\;|x|+|y|=1\;$ and find, if possible, maximal points by evaluation. For example: in $\;|x|+|y|=1\;$ , it is clear $\;-1\le x,\,y\le 1\;$ , so perhaps:
$$\frac{3(x+y)-2}{x^2+y^2}\le\frac{3-2}{1}=1\;\;\;\text{ and etc...}$$
This really confuses...Thanks for any help.
| With $|x|+|y|=1$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\leq\frac{3(|x|+|y|)-2}{\frac{(|x|+|y|)^2}{2}}=2$$
The equality occurs for $x=y=\frac{1}{2}$.
With $x^2+y^2=8$ we get a smaller value.
In another hand, for $x^2+y^2=8$ we obtain:
$$\frac{3(x+y)-2}{x^2+y^2}\geq\frac{-3\sqrt{(x+y)^2}-2}{x^2+y^2}\geq$$
$$\geq\frac{-3\sqrt{2(x^2+y^2)}-2}{x^2+y^2}=-\frac{7}{4}.$$
The equality occurs for $x=y=-2$.
With $|x|+|y|=1$ we'll get for $x=y=-\frac{1}{2}$ a value $-10$.
Now, easy to show that it's a minimal value.
Indeed, we need to prove that
$$3(x+y)-2\geq-10(x^2+y^2)$$ or
$$3\left(\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2\right)+7(x^2+y^2)-\frac{7}{2}\geq0,$$
which is true because $$x^2+y^2\geq\frac{1}{2}(|x|+|y|)^2=\frac{1}{2}.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2326017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\frac{c}{a}$
$α+δ+β+δ=\frac{-2B}{A}$
⇒ $α+β+2δ =\frac{-2B}{A}$
$(α+δ)(β+δ)=\frac{C}{A}$
⇒ $αβ+αδ+βδ+δ^2=\frac{C}{A}$
⇒$\frac{c}{a}+αδ+βδ+δ^2=\frac{C}{A}$ ⇒ $αδ+βδ+δ^2=\frac{Ca-cA}{Aa}$
$(α+β)^2=\frac{4b^2}{a^2}$
⇒ $α^2+β^2+2αβ=\frac{4b^2}{a^2}$
$α^2+β^2+\frac{2c}{a}=\frac{4b^2}{a^2}$
⇒ $α^2+β^2=\frac{4b^2-2ac}{a^2}$ -(1)
$(α+β+2δ)^2 =\frac{4B^2}{A^2}$
⇒ $α^2+β^2+(2δ)^2+2(αβ+2βδ+2αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+4δ^2+2αβ+4βδ+4αδ=\frac{4B^2}{A^2}$
⇒$α^2+β^2+2αβ+4(δ^2+βδ+αδ)=\frac{4B^2}{A^2}$
⇒$α^2+β^2+\frac{2c}{a}+4(\frac{Ca-cA}{Aa})=\frac{4B^2}{A^2}$
⇒$α^2+β^2=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$ -(2)
From (1) and (2) we get,
$\frac{4b^2-2ac}{a^2}=\frac{4aB^2-2A^2 c-4Aac+4cA^2}{A^2a}$
My problem: I tried simplifying it further but could not reach the required result. A continuation of my method would be more appreciated compared to other methods.
| Here is the LHS:
$$\frac{b^2-ac}{a^2} = \left(\frac{b}{a}\right)^2-\frac {c}{a}$$
$$=\left(\frac {-(\alpha+\beta)} {2}\right)^2-\alpha\beta$$
$$=\frac {{\alpha}^2+{\beta}^2+2\alpha\beta}{4}-\frac {4\alpha\beta}{4}$$
$$=\frac{{\alpha}^2+{\beta}^2-2\alpha\beta} {4}$$
$$=\left(\frac{\alpha-\beta} {2}\right)^2$$
And here is the RHS:
$$\frac{B^2-AC}{A^2}=\left(\frac {B}{A}\right)^2-\frac{C}{A}$$
$$=\left(\frac{-(\alpha+\beta+2\delta)}{2}\right)^2-(\alpha+\delta) (\beta+\delta)$$
$$=\frac {{\alpha}^2+{\beta}^2+4{\delta}^2+2\alpha\beta+4\alpha\delta+4\beta\delta} {4}-\frac {4\alpha\beta+4\alpha\delta+4\beta\delta} {4}$$
$$=\frac {{\alpha}^2+{\beta}^2-2\alpha\beta} {4}$$
$$=\left(\frac {\alpha-\beta} {2}\right)^2$$
Clearly, LHS = RHS. QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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.