Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2$
Let $x>1$, $y>1$ and $z>1$ be such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that:
$$x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2.$$
When $x=y=z=3$, both sides are equal to $3^6$. The difficulty is how to deal with the variables appearing in the exponents, and how to use the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
| It's just AM-GM:
$$\sum_{cyc}x^{y+1}z=xyz\sum_{cyc}\frac{1}{y}x^{y}\geq xyz\prod_{cyc}\left(x^y\right)^{\frac{1}{y}}=x^2y^2z^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$ Solve the equation $$\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$$
$$\Leftrightarrow \sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}$$
$$\Leftrightarrow \sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)$$
$$\Leftrightarrow \frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}$$
$$\Leftrightarrow \frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0$$
$$\Leftrightarrow \frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0$$
$$\Leftrightarrow \left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0$$
The equation $$\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}=0$$ have a solution but it's very ugly and i can't solve this. Help me, thanks
| After squaring and rearranging we have:
$$4x^6-12x^5+9x^4+3x^3-14x^2+8x=0$$
Since there is no constant term, $x=0$ is a solution. Using the rational root theorem, $p$ and $q$ have to be coprime. Since the only factors of $4$ and $8$ are powers of $2$, then the only solutions are when either $1$ is in the numerator or $1$ is the denominator:
$$x = ±\frac{1}{8}, ±\frac{1}{4}, ±\frac{1}{2}, ±1, ±2, ±4, ±8$$
Verifying these solutions by substitution, only $x=-1$ and $x=2$ are possible. We can factorise them (for instance, by polynomial/synthetic division) to get the cubic in Dr. Sonnhard Graubner's answer.
| {
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Find $ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$
Find the limit of
$$ \lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx$$
Let $$u= 1 +\frac{x}{n} \implies du =\frac{1}{n} dx \implies n \cdot du = dx$$
at $x=0$ $u=1$ and at $x=1$ $u=1+\frac{1}{n}$ so now limit will change from $1$ to $1+\frac{1}{n}$
Back to the integral
$$ \lim\limits_{n \rightarrow \infty} \left( n \cdot \int_{1}^{1+\frac{1}{n}} u^n du \right)= \lim\limits_{n \rightarrow \infty} \left( n \cdot \left[ \frac{nu^{n+1}}{n+1} \right]_1^{1+\frac{1}{n}} \right) = \lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1} \left[ u^{n+1} \right]_1^{1+\frac{1}{n}} \right) $$
$$\implies\lim\limits_{n \rightarrow \infty} \left(\frac{n^2}{n+1} \left[ \left(1+\frac{1}{n} \right)^{n+1}-1 \right] \right)=\infty$$
Is my finding correct? Is the procedure of taking the limit before completing the integration correct?
Much appreciated
| Since
$\left(1+ \frac{x}{n}\right)^n
\to e^x
$,
$
\lim\limits_{n \rightarrow \infty} \int_{0}^{1} \left(1+ \frac{x}{n}\right)^n dx
= \int_{0}^{1} e^x dx
=e-1
$.
| {
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Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$ Question:
Prove that for all prime numbers $ \ a,b,c \ , a^{2} + b^{2} \neq c^{2}$
My attempt:
Proof by contradiction:
Assume $ \ \exists a,b,c $ prime numbers such that $ \ a^{2} + b^{2} = c^{2}$.
Then $ \ a^2 = c^2 - b^2 \implies a^2 = (c-b)(c+b) $.
I am not sure how to show a contradiction here. Could someone please tell me the easiest way to show a contradiction from here?
| For any integer $a$, $a\equiv 0,1,2(\mod 3)\Rightarrow a^2\equiv 0,1,4(\mod 3)\Rightarrow a^2\equiv 0,1(\mod 3)$
Now $a,b,c$ are prime numbers and $a^2+b^2=c^2$. First, important thing to note that $a,b,c$ are distinct.
If $a=3$, then $(c+b)(c-b)=9$. Check that $b,c$ can not prime numbers.
If $b=3$, then $(c+a)(c-a)=9$. Check that $a,c$ can not prime numbers.
If $c=3$, then $a^2+b^2\equiv 2(\mod 3)\Rightarrow c^2\equiv 2(\mod 3)$, which is impossible.
If none of $a,b,c$ is $3$, $a^2+b^2\equiv 2(\mod 3)$ and $c^2\equiv 2(\mod 3)$, which is impossible.
The conclusion is yours to decide.$\space\space\space\space\space\space\space\space\blacksquare$
| {
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Sum to infinity An infinite G.P. has a finite sum with initial term $u_1 = 2$. Find the sum to infinity of this G.P. If it is also a Fibonacci sequence
My approach-
*
*since it is a G.P., $ u_3 = u_2 * u_2/u_1$
*since it is a Fibonacci sequence, $u_3 = u_1+u_2$
*Therefore, $u_2 * u_2/u_1 = u_1+ u_2$
By substituting $u_1 = 2$ and solving, I got $u_2= 1 \pm \sqrt{5}$.
When $u_2 = 1+\sqrt{5}$ ,
*
*$r = u_2/u_1 = (1+\sqrt{5})/2 \approx 1.618$
*$| r | = | 1.618 |$ = 1.618 > 1
*therefore the sum to infinity doesn't exist.
when $u_2 = 1-\sqrt{5}$ ,
*
*$r = u_2/u_1 = (1-\sqrt{5})/2 \approx -0.618$
*$| r | = | -0.618 | = 0.618 < 1$
therefore the sum to infinity exists.
therefore, a= 2, r = (1-√5)/2
the sum to infinity $\sum_{i=1}^\infty u_i = a / ( 1-r) = -1+\sqrt{5}$.
Can anyone pls help me if this is true?
Sorry for my poor typing form.
| Yes, your reasoning looks correct. I get the same answer you do in a slightly different form:
$$\begin{align*}
\sum_{i=1}^\infty u_i &= \frac{a}{1-r}\\
& = \frac{2}{1-\frac{1-\sqrt{5}}{2}}\\
&= \frac{4}{2-(1-\sqrt{5})}\\
&= \frac{4}{1+\sqrt{5}}
\end{align*}$$
(Since $1-\sqrt{5} < 0$, every other term is negative and the overall sum is less than the first term, $a=2$. )
Note that this is answer is equal to $-1+\sqrt{5}$, because:
$$\frac{4}{1+\sqrt{5}} = \frac{4(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})} = \frac{4(1-\sqrt{5})}{1-5} = \frac{4}{-4}(1-\sqrt{5})=-1+\sqrt{5}$$
| {
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Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
| $$\int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\,dx \stackrel{x\mapsto\frac{z}{2}}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\cos z\right)^{3/2}\,dz \stackrel{z\mapsto\frac{\pi}{2}-z}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\sin z\right)^{3/2}\,dz $$
equals:
$$ \frac{1}{2}\int_{0}^{1}u^{3/2}(1-u^2)^{-1/2}\,du \stackrel{u\mapsto v^{1/2}}{=} \frac{1}{4}\int_{0}^{1}v^{1/4}(1-v)^{-1/2}\,dv = \frac{1}{4} B\left(\frac{5}{4},\frac{1}{2}\right) $$
or:
$$ \frac{\Gamma\left(\frac{5}{4}\right)\sqrt{\pi}}{4\,\Gamma\left(\frac{7}{4}\right)}=\frac{\Gamma\left(\frac{1}{4}\right)\sqrt{\pi}}{12\,\Gamma\left(\frac{3}{4}\right)}=\color{blue}{\frac{\Gamma\left(\frac{1}{4}\right)^2}{12\sqrt{2\pi}}} $$
clearly related with the lemniscate constant. In terms of the complete elliptic integral of the first kind,
$$\begin{eqnarray*}\int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\,dx=\frac{1}{3\sqrt{2}}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\frac{1}{2}\cos^2\theta}}&=&\color{blue}{\frac{\pi}{6\sqrt{2}}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{32^n}}\\&=&\color{blue}{\frac{\pi}{6\,\text{AGM}(1,\sqrt{2})}}.\end{eqnarray*}$$
The last form is very well-suited for numerical evaluation, leading to $\color{blue}{I\approx 0.43700959238202}$.
In a similar way
$$ \int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\cos(x)\,dx\stackrel{x\mapsto\arcsin(t)}{=}\int_{0}^{1/\sqrt{2}}(1-2t^2)^{3/2}\,dt \stackrel{t\mapsto\frac{u}{\sqrt{2}}}{=}\frac{1}{\sqrt{2}}\int_{0}^{1}(1-u^2)^{3/2}\,du$$
equals (through $u\mapsto\sqrt{v}$)
$$ \frac{1}{2\sqrt{2}}\int_{0}^{1}v^{-1/2}(1-v)^{3/2}=\frac{1}{2\sqrt{2}}B\left(\frac{1}{2},\frac{5}{2}\right)=\color{blue}{\frac{3\pi}{16\sqrt{2}}}.$$
| {
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find $f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})$ of $f(x)=\frac{2}{2+4^x}$ $f(x)=\frac{2}{2+4^x}$
find
$f(\frac{1}{2014})+f(\frac{2}{2014})+.....+f(\frac{2013}{2014})$
Please guide me through it, the only step I know is probably to eliminate the denominator
ps. Not a homework
| HINT:
$$f(1-x)=\dfrac2{2+4^{1-x}}=\dfrac{2\cdot4^x}{2\cdot4^x+4}=\dfrac{4^x}{4^x+2}=\dfrac{4^x+2-2}{4^x+2}=1-f(x)$$
Set $x=\dfrac r{2014}, 1\le r\le2013$ and add
| {
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How many positive integers less than $1000$ divisible by $3$ with sum of digits divisible by $7$?
How many positive integers less than $1000$ are divisible by $3$ with their sum of digits being divisible by $7$?
Well, I got Answer: $28$.
$a+b+c = 7k$ and $a,b,c$ are multiples of 3 so $a+b+c$ is a multiple of both $7$ and $3$.
Therefore $a+b+c=21$.
$(9-a)+(9-b)+(9-c)=6$
Then $^8C_2$. Can anyone explain why there is $9-a+9-b+9-c$?
If you have any other alternative method please explain?
| The digit sum has to divisible by $3$ and by $7$, hence by $21$. This implies that we have to deduct a total of $6$ points from the maximal possible $9+9+9=27$. By stars and bars these $6$ points can be distributed in ${8\choose 2}=28$ ways onto the three decimal places.
| {
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If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$
For positive real numbers $a$, $b$, and $c$ prove that: $$\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}.$$
I let $x=\frac{a}{b}$, $y=\frac{b}{c}$, and $z=\frac{c}{a}$. Then inequality becomes
$$A=\frac{x^2}{1+xz}+\frac{y^2}{1+xy}+\frac{z^2}{1+zy} \ge \frac{3}{2}$$
By Cauchy-Schwarz
$$(3+xy+yz+zx)A \ge (x+y+z)^2 \ge 3(xy+yz+zx)$$
What then?
| You can end your way.
Since $xyz=1$, by using your work and AM-GM we obtain:
$$2(x+y+z)^2=(x+y+z)^2+(x+y+z)^2\geq9+3(xy+xz+yz),$$
which you want to get.
Also, by C-S
$$\sum_{cyc}\frac{a^2}{b^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{3}{2}$$
because the last inequality it's
$$\sum_{cyc}(2a^4+a^2b^2-3a^2bc)\geq0,$$
which is true by Muirhead or we can use SOS here:
$$\sum_{cyc}(2a^4+a^2b^2-3a^2bc)=\sum_{cyc}(2a^4-2a^2b^2)+\frac{3}{2}\sum_{cyc}(2a^2b^2-2a^2bc)=$$
$$=\sum_{cyc}(a^2-b^2)^2+\frac{3}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
Done!
There is also the following solution.
By Rearrangement and Nesbitt we obtain:
$$\sum_{cyc}\frac{a^2}{b^2+bc}=\sum_{cyc}\left(\frac{a^2}{b+c}\cdot\frac{1}{b}\right)\geq\sum_{cyc}\left(\frac{a^2}{b+c}\cdot\frac{1}{a}\right)=\sum_{cyc}\frac{a}{b+c}\geq\frac{3}{2}$$
| {
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
| I'd suggest writing the proof in a more linear fashion which is easier to read. Below is how I would approach the induction step.
Let $P(n)$ be the statement $3\mid n(n^2+2)$.
Proving $P(k)\implies P(k+1)$:
$$P(k)\implies 3\mid k(k^2+2)$$
$\implies$
$$3\mid k(k^2+2)+3(k^2+k+1)$$
$\implies$
$$3\mid k^3+3k^2+5k+3$$
$\implies$
$$3\mid(k+1)(k^2+2k+3)$$
$\implies$
$$3\mid(k+1)((k+1)^2+2)$$
$\implies$
$$P(k+1)$$
| {
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Find the value of $\frac{9}{5}(a+b)$ given that $a\sqrt{a}+b\sqrt{b}=183$ and $a\sqrt{b}+b\sqrt{a}=182$
Suppose $a,b$ are positive real numbers such that $a\sqrt{a}+b\sqrt{b}=183$, $a\sqrt{b}+b\sqrt{a}=182$. Find $\frac{9}{5}(a+b)$.
It is my equation. I subtracted the second equation from the first one and found $(a-b)(\sqrt{a}-\sqrt{b})=1$.or, $(a-b)^2=(\sqrt{a}+\sqrt{b})$.
Am I going to the right path? How to escape from this. Please give me hints. Thank you.
| Let $A^2=a$ and $B^2=b$ then we have
\begin{eqnarray*}
A^3+B^3 = 183 \\
A^2B+B^2 A =182
\end{eqnarray*}
Multiply the second equation by $3$ and add the first
\begin{eqnarray*}
A^3+B^3+ 3(A^2B+B^2 A) =(A+B)^3= 729 \\
A+B =9 \\
AB = \frac{182}{9} \\
(A-B)^2=(A+B)^2-4AB= \frac{1}{9} \\
A= \frac{14}{3} \\
B= \frac{13}{3}
\end{eqnarray*}
So $ \frac{9}{5} (a+b) = \color{red}{73}$.
| {
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?
| Let $a, b, c$ be the sides of a triangle with given angles $A, B, C$.
Recall for any $3$ positive numbers $a, b, c$, it can form the sides of a non-degenerate triangle if and only if we can find $3$ positive numbers
$u,v,w$ such that
$$a = v + w,\quad b = u + w,\quad c = u + v$$
This is known as Ravi substitution for a triangle.
In terms of them, we have
$$
\sin^2\frac{A}{2}
= \frac12( 1 - \cos A) = \frac12\left(1 - \frac{b^2+c^2-a^2}{2bc}\right)
= \frac{a^2 - (b-c)^2}{4bc} = \frac{vw}{bc}
$$
By a similar argument, we have
$$\sin^2\frac{B}{2} = \frac{uw}{ac}\quad\text{ and }\quad\sin^2\frac{C}{2} = \frac{uv}{ab}$$
Multiply these 3 relations together, taking square root and apply AM $\ge$ GM to the factors in denominator, we obtain:
$$\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}
= \frac{uvw}{abc}
= \frac{uvw}{(v+w)(u+w)(u+v)}
\le \frac{uvw}{(2\sqrt{vw})(2\sqrt{uw})(2\sqrt{uv})} = \frac18$$
| {
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Asymptotic behaviour of sum I would like to evaluate the number $c$ given by
$$
c = \lim_{m\to\infty} \frac{1}{\log m}\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)}
$$
where $\tau = (1+\sqrt{5})/2$.
My attempt: my guess was this sum would be dominated by the terms for which $n$ is a Fibonacci number. I considered the sum of this sub series using the relation $F_a \tau = F_{a+1}-(-\tau)^{-a}$, the small angle approximation, and that $F_a\tau^{-a} = 1/\sqrt{5}+\mathrm{O(\tau^{-2a})}$. This yielded
$$
\sum_{n=1}^m \frac{1}{n^2 \sin^2(\pi n \tau)} \approx
\sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{1}{F_a^2 \sin^2(\pi \tau^{-a})} \approx \sum_{a=1}^{\log_\tau m \sqrt{5}} \frac{5}{\pi^2} = \frac{5}{\pi^2} \frac{\log m \sqrt{5}}{\log \tau}
$$
this would imply $c = 5/(\pi^2 \log \tau)$, however this seems about a factor $1.2$ smaller than the correct value, implying I have missed some important terms.
Is it possible to obtain an exact expression? Or at least a better a tighter lower (and upper) bound?
| $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cI{\mathcal{I}}$Amazingly, your sum can actually be computed in closed form. The answer is
$$c=\frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$
Notation: Let $\tau = \tfrac{1+\sqrt{5}}{2}$ and $\bar{\tau} = \tfrac{1-\sqrt{5}}{2}$, so $\tau+\bar{\tau}=1$ and $\tau \bar{\tau} = -1$.
Let $R$ be the ring $\ZZ[\tau]$. The ring $R$ is known to be a PID with unit group $\pm \tau^k$. Let $\cI$ be the set of nonzero ideals of $R$. For $m+n \tau \in R$, set $N(m+n \tau) = m^2+mn-n^2 = (m+n \tau) (m+n \bar{\tau})$; for an ideal $I \subseteq R$ set $N(I) = |R/I|$. The relation between these notations is that $|N(m+n \tau)| = N(\langle m+n \tau \rangle)$.
You want to evaluate
$$\lim_{K \to \infty} \frac{1}{\log K} \sum_{n=1}^K \frac{1}{n^2 \sin^2 (\pi n \tau)}.$$
We recall the identity
$$\frac{\pi^2}{\sin^2 (\pi x)} = \sum_{m=-\infty}^{\infty} \frac{1}{(m-x)^2}$$
to rewrite this as
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{n=1}^K \sum_{m=- \infty}^{\infty} \frac{1}{n^2 (m-n \tau)^2}. \quad (1)$$
All the terms are positive, so we may rearrange the sum at will; we group together terms $m-n \tau$ which generate the same ideal $I$ in $R$, giving
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{I \in \cI} \sum_{\begin{matrix} \langle m-n \tau \rangle=I \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2} \quad (2).$$
I assume it is legitimate to exchange the limit and the outer sum in (2) (should be easy, but I haven't checked). So we want to consider
$$\lim_{K \to \infty} \frac{1}{\pi^2 \log K} \sum_{\begin{matrix} \langle m-n \tau \rangle=I \\ 1 \leq n \leq K \end{matrix}} \frac{1}{n^2 (m-n \tau)^2}. \quad (3)$$
If $\gamma$ is a generator of $I$, then the list of all generators is the numbers of the form $\pm \tau^k \gamma$. All of these points lie on the hyperbolas $m^2-mn-n^2 = \pm N(I)$, with asymptotes $m = \bar{\tau} n$ and $m = \tau n$.
As $k \to - \infty$, we approach the asymptote $m=\bar{\tau} n$. Both $n$ and $m-n \tau$ grow exponentially, so the contribution from those summands is bounded and is wiped out by the $\log K$ term.
As $k \to \infty$, we approach the $m = \tau n$ asymptote. The number of terms is $\tfrac{\log K + O(1)}{\log \tau}$ and each of those terms is
$$\frac{1}{n^2 (m-n \tau)^2} = \frac{(m- \bar{\tau} n)^2}{n^2 N(I)^2} = \frac{(m/n- \bar{\tau})^2}{N(I)^2}.$$
Since we are approaching the asymptote $m/n = \tau$, the numerator approaches $(\tau - \bar{\tau})^2 = 5$. We have a sum of $\tfrac{\log K+O(1)}{\log \tau}$ terms which approach $\tfrac{5}{N(I)^2}$, so $(3)$ is
$$\frac{5}{\pi^2 (\log \tau) N(I)^2}.$$
Plugging into $(2)$,
$$c=\frac{5}{\pi^2 \log \tau} \sum_{I \in \cI} \frac{1}{N(I)^2} .$$
That last sum is your "about 1.2"; you only computed the contribution from the ideal $\langle 1 \rangle$. (To see the connection, note that $\tau^k = F_{k} \tau + F_{k-1}$.)
I expected this to be the end of the line, but it turns out this sum can actually be evaluated!
Recall that the $\zeta$ function of $R$ is defined to be
$$Z(s) := \sum_{I \in \cI} \frac{1}{N(I)^s}.$$
So we want to evaluate $Z(2)$.
We know $Z(s)$ factors as
$$Z(s) = \zeta(s) L(s)$$
where $\zeta$ is the Riemann $\zeta$ function and
$$L(s) = \sum_{n=1}^{\infty} \frac{\left( \tfrac{5}{n} \right)}{n^s}.$$
We know that $\zeta(2) = \tfrac{\pi^2}{6}$, so we are left to evaluate $L(2)$.
Using quadratic reciprocity,
$$\left( \frac{5}{n} \right) = \begin{cases} 0 & n \equiv 0 \bmod 5 \\ 1 & n \equiv \pm 1 \bmod 5 \\ -1 & n \equiv \pm 2 \bmod 5 \end{cases}$$
from which we deduce
$$\left( \frac{5}{n} \right) = \frac{2}{\sqrt{5}} \left( \cos \tfrac{2 \pi n}{5} - \cos \tfrac{4 \pi n}{5} \right).$$
So
$$L(2) = \frac{2}{\sqrt{5}} \left( \sum_{n=1}^{\infty} \frac{\cos \tfrac{2 \pi n}{5} }{n^2} - \sum_{n=1}^{\infty} \frac{\cos \tfrac{4 \pi n}{5} }{n^2} \right).$$
We now recall that, for $0 \leq x \leq 2 \pi$, we have
$$\sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{(\pi-x)^2}{4} - \frac{\pi^2}{12}.$$
Plugging in $x=2 \pi/5$ and $4 \pi/5$ we get
$$L(2) = \frac{4 \pi^2}{25 \sqrt{5}}.$$
Turning quadratic reciprocity symbols into linear combinations of trigonometric functions, and then recognizing the Fourier series that results, is a standard way to evaluate $L$-functions.
Putting it all together, we deduce
$$c=\frac{5}{\pi^2 \log \tau} \frac{\pi^2}{6} \frac{4 \pi^2}{25 \sqrt{5}} = \frac{2 \pi^2}{15 \sqrt{5} \log \tau}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine all pairs $(a,b)$ of positive integers such that $ab^2+b+7$ divides $a^2b+a+b$
Determine all pairs $(a,b)$ of positive integers such that $ab^2 + b + 7$ divides $a^2b + a +b$.
I guess the trivial solution is $a=b=7$.
Also any $a$ and $b$ which are divisible by $7$ will seem to suit.
we can also write
$b(\frac{a}{b}+1+\frac{7}{b}) \cdot x=a\cdot(ab+1+\frac{b}{a})$
From where all $a, b$ that satisfy: $\frac{b^2}{a}=7$ and $a>b$, are OK.
Is my solution full and complete? Are there any other solutions?
| First we will dealt with the cases when $b=1,2,3,4,5,6$ and then we can assume that $b \geq 7$,
For the case $b=1$ we get that $a+8 | a^2+a+1$ which means $a(a+8) +r = a^2+a+1$ and we want $r=0 \mod a+8$ so solving the above gives that $r = 1 - 7 a$ so we want that $a+8|1-7a$ which means that $-7(a+8)+r = 1-7a$ and we want $r=0 \mod a+8$ so solving the above gives that $r = 57 \mod a+8$.
Now $57=3*19$ so $3=0 \mod a+8$ or $19 = 0 \mod a+8$ or $57 = 0 \mod a+8$ solving these equations yields that $(11,1)$ and $(49,1)$ are solutions.
Doing the same for $b=2,3,4,5,6$ yields that there are no solutions.
Now we can assume $b \geq 7$, also $a \geq b \geq 7$, because if we let $a< b$ and $a b^2 +b +7 < a^2 b+a+b$ we will get contradiction arriving at $a \geq b-\frac{1}{7}$ and since $a,b$ are integers we get that $a\geq b$.
Divide the equation by $a b$ we get that $(b+\frac{1}{a}+\frac{7}{ab})x = a+\frac{1}{a} +\frac{1}{b}$ which is just $ b x +\frac{x}{a}+ \frac{7x}{ab} = a+\frac{1}{a}+\frac{1}{b}$
Now we know that $\frac{1}{a}+\frac{1}{b} \leq \frac{2}{7}<1$ so assuming that $\frac{x}{a}+ \frac{7x}{ab} <1$ we must have that $a = b x$ substituting that we get that $b=7x$ so $a = b x = 7x x = 7x^2$ for all $x \geq 1$.
Now, what if $\frac{x}{a}+ \frac{7x}{ab}\geq y \geq 1$ we get that $a= b x+y$ substituting we get that $\frac{7 x}{b (b x+y)}+\frac{x}{b x+y}+b x = \frac{1}{b x+y}+b x+\frac{1}{b}+y$ which is $\frac{7 x}{b (b x+y)}+\frac{x}{b x+y}-\frac{1}{b x+y}-\frac{1}{b}-y=0$ solving for $x$ we get that $x =\frac{-b y^2-b-y}{b^2 y-7}$ Now since $b \geq 7 $ and $y \geq 1$ the denominator is non-negative and the numerator is negative so $x$ is negative which is contradiction thus,
the solutions $(11,1),(49,1),(7x^2,7x)$ for all $x\geq 1$ are the only solutions.
| {
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"url": "https://math.stackexchange.com/questions/2407572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$? How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$?
I've been reading the Plimpton 322 news, and this fits in the gap in the Babylonian table between 0.6996 ($a=1679,b=2400$) and 0.75 ($a=3,b=4$). The Babylonians apparently tabulated a lot of triples and then looked for the best one, but is there a more direct way?
| Pythagorean triplets are characterized by being of the form
$$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$
So you could look for two integers $m>n$ such that
$$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$
or, equivalently,
$$\frac{m}{n}- \frac{n}{m} \approx 2 \cdot \frac{5}{7}$$
Now, let $x$ be unique positive solution of
$$x-x^{-1} = 2 \cdot \frac{5}{7}$$
one can easily compute that $x= \frac{5 + \sqrt{74}}{7} \approx 1.94$. The idea is to appximate $x= m/n$with a rational number. The continued fraction of $x$ is periodic and it is $[1; \overline{1, 16, 1,1,1}]$. This gives us the first convergents of $x$: $$\frac{1}{1} , \frac{2}{1} , \frac{33}{17} , \frac{35}{18} , \frac{68}{35} , \frac{103}{53} , \frac{171}{88},\dots$$ which are the best rational approximations of $x$. Taking $m= 33$ and $n= 17$ you get the triplet
$$a= 800 ; \ \ b= 1122 ; \ \ c= 1378$$
which is not a primitive triplet. Dividing it by $2$ you get the primitive triplet
$$a= 400 ; \ \ b= 561 ; \ \ c= 689$$
with
$$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{400}{561} - \frac{5}{7} \right|= \frac{5}{3927} \approx 0.0013$$
Going on like this ($m= 35 , n=18$) you get the next solution
$$a= 901 ; \ \ b= 1260 ; \ \ c= 1549$$
with
$$\left| \frac{a}{b} - \frac{5}{7} \right|= \left| \frac{901}{1260} - \frac{5}{7} \right|= \frac{1}{1260} \approx 0.0008$$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Find the locus of $w=1/z$ when $z$ lies on $y=2x+1$ For the transformation $w=\frac{1}{z}$, find the locus of $w$ when $z$ lies on the line with equation $y=2x+1$
I'm not quite sure how or where to start on this question. I know for instance that the locus of $z$ such that
$$|z-2| = |z+2-2i|$$
is the line $y=2x+1$. I'm not sure where I could go from here though
Thanks in advance
| Suppose $z=x+iy$ then as $y=2x+1$ we have $w=\dfrac{1}{x +(2x+1)i}$
$\dfrac{1}{x +(2x+1)i}=\dfrac{x-(2x+1)i}{5 x^2+4 x+1}$
If $w=a+bi$ then we have $(a,b)=\left(\dfrac{x}{5 x^2+4 x+1},\dfrac{-2 x-1}{5 x^2+4 x+1}\right)$
So
$a=\dfrac{x}{5 x^2+4 x+1};\;b=\dfrac{-2 x-1}{5 x^2+4 x+1}$
$x=\dfrac{1-4a\pm\sqrt{-4 a^2-8 a+1}}{10 a}$
plugged in $b$ gives
$b=\dfrac{1}{2} \left(-1\pm\sqrt{1-4 a (a+2)}\right)$
$2b=-1\pm\sqrt{1-4 a (a+2)};\to (2b+1)^2=1-4 a (a+2)$
$a^2 + b^2+2a-b=0$
which is the equation of a circle. When $z$ varies on the blue line, $w$ varies on the orange circle
Hope it helps
remark
There are two points which are invariant
$\left(-\dfrac{4}{5}, -\dfrac{3}{5}\right),( 0,1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurence relation $D(n) \le D(\frac{2}{3}n) + 2$ We know that
$D(1) = 0$.
Now from the inequality,
$$D(n) \le D\left(\frac{2}{3}n\right) + 2$$
it should follow that
$$D(n) \le 2\cdot \log_{\frac{3}{2}}(n).$$
*Edit: the logarithm has a base $\frac{3}{2}$
How do we get that last bound?
| (Disclaimer: the following assumes this is one of those contexts where certain hand-waving is allowed/expected, otherwise see the posted comments for why the problem is not well posed.)
Let $\,n = (3/2)^k \iff k = \log_{3/2}(n)\,$, and define $T(k)=D\big((3/2)^k\big)=D(n)\,$, then:
$$
T(k)=D \left(\left(\frac{3}{2}\right)^k \right) \;\le\; D \left(\frac{2}{3} \cdot \left(\frac{3}{2}\right)^k \right) + 2 = D \left(\left(\frac{3}{2}\right)^{k-1} \right) + 2 = T(k-1) + 2
$$
Next, by telescoping and using that $T(0)=D(1)=0\,$:
$$
T(k) \le T(k-1)+ 1 \cdot 2 \le T(k-2) + 2 \cdot 2 \le \cdots \le T(0)+k \cdot 2 = 2k
$$
Rewriting the latter in terms of $n\,$, it follows that $D(n) = T(k) \le 2 k = 2 \log_{3/2} n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Period of $f(x)=\sin{2x}-\sin{\frac{x}{2}}$ I know the periods of the two functions $f(x)=\sin{2x}$ and $g(x)=\sin{\frac{x}{2}}$, which are $\pi$ and $4\pi$ respectively. But what is their period when you subtract them?
More generally, if $f(x)=\sin{(ax+c)}$ and $g(x)=\sin{(bx+d)},$ what is the period of $f(x)-g(x)?$
| In short, period of $\sin 2x$ is $\pi$ and period of $\sin \dfrac{x}{2}$ is $4\pi$. Then answer will be $\text{lcm} (1,4)\cdot\pi=4\pi$
Now, elaborately:
$\sin 2x-\sin \dfrac{x}{2}=2\sin (x-\dfrac{x}{4})\cos(x+\dfrac{x}{4})=2\sin \dfrac{3x}{4}\cos\dfrac{5x}{4}$
If $x=4k\pi$ and $k\in \mathbb{Z}$, either $\cos\dfrac{5x}{4}$ or $\sin \dfrac{3x}{4}$ is $0$. Hence period is $4\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of series $(\frac{1}{3})^{2}+(\frac{1.2}{3.5})^{2}+(\frac{1.2.3}{3.5.7})^{2}+...$ $$\left(\frac{1}{3}\right)^{2}+\left(\frac{1\cdot2}{3\cdot5}\right)^{2}+\left(\frac{1\cdot2\cdot3}{3\cdot5\cdot7}\right)^{2}+...$$
I am not able to find a general equation and that's creating problem for me as I can't proceed further without it.
| We have to deal with:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\sum_{n\geq 1}\left(\frac{2^n n!^2}{(2n+1)!}\right)^2 = \sum_{n\geq 1}\frac{4^n}{(2n+1)^2\binom{2n}{n}^2} $$
and since $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$, the above series is clearly convergent. We may recall that
$$ \frac{\arcsin z}{\sqrt{1-z^2}}=\sum_{n\geq 0}\frac{z^{2n+1}4^n}{(2n+1)\binom{2n}{n}} \tag{1}$$
$$ \int_{0}^{\pi/2}\left(\sin\theta\right)^{2n+1}\,d\theta = \frac{4^n}{(2n+1)\binom{2n}{n}}\tag{2} $$
hence we have:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2=\tfrac{1}{9}\cdot\phantom{}_3 F_2\left(1,2,2;\tfrac{5}{2},\tfrac{5}{2};\tfrac{1}{4}\right)=-1+\int_{0}^{\pi/2}\frac{4\arcsin\frac{\sin\theta}{2}}{\sqrt{4-\sin^2\theta}}\,d\theta.\tag{3} $$
or, in a more compact form:
$$ \sum_{n\geq 1}\left(\frac{n!}{(2n+1)!!}\right)^2= -1+\int_{0}^{\pi/6}\frac{4\theta\,d\theta}{\sqrt{1-4\sin^2\theta}}.\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction?
| $$x^2-x+1=y^2\\
4x^2-4x+4=4y^2 \\
(2y)^2-(2x-1)^2=3 \\
\left( 2y-2x+1\right) \left(2y+2x-1 \right)=3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
| $$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}\equiv 5^{2n}2^n-3^n2^{2n}\pmod{19}\qquad (1)$$
since $5^12^2=20\equiv 1\pmod{19}$ and $3^22^1=18\equiv -1\pmod{19}$. Hence, $(1)$ yields the following
$$2^n(5^{2n}-3^n2^n)=2^n(25^n-6^n)=50^n-12^n\equiv 12^n-12^n\pmod{19}\equiv 0\pmod{19}$$
since $50\equiv 12\pmod{19}$.
| {
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"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am having trouble in dealing with increasing number of multiples in each step. Kindly Guide in How to proceed
| The correct power of $X$ for when $n$ is the $n^{th}$ term of the series is
$$ a_n=K \times x^{2(n-1)} $$
You can deduce the correct formula starting from that
then for $ n>0 $ :
$$ a_n = \frac { K \times x^{2(n-1)} }{ 2^{2(n-1)} (2(n-1))!} $$
For $ n > 1 $ :
$$ k = \frac {(4n-7)!}{2^{2(n-2)} (2(n-2))!}$$
Therefore,
$$ a_1 =1 $$
For $n > 1$ :
$$a_n = \frac { (4n-7)! \times x^{2(n-1)} }{ 2^{2(2n-3)} (2(n-2))! (2(n-1))!} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 4
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Solving complex equation $z^2 + (1+i) \overline{z} + 4i = 0$ Consider the following equation, where $z \in \mathbb{C}$, $i$ is the imaginary unit and $\overline{z}$ is the conjugate of $z$:
$$
z^2 + (1+i) \overline{z} + 4i = 0
$$
What is the method to deal with equations such as this?
I have tried various things: I tried substituting $z$ with $a+bi$, or $re^{i\theta}$, hoping I'd notice something. I thought I could somehow transform this into a quadratic equation, but I couldn't. Now I have no idea what to try. I'd appreciate ideas greatly.
| $$a^2-b^2+2iab+(1+i)(a-ib)+4i=0\\ a^2-b^2+a+b+(2ab+a-b+4)i=0\\ \begin{cases}a^2-b^2+a+b=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}(a+b)(a-b+1)=0\\ 2ab+a-b+4=0\end{cases}\\ \begin{cases}a=-b\\ -2a^2+2a+4=0\end{cases}\vee\begin{cases}b=a+1\\ 2a^2+2a+3=0\end{cases}$$
An these can be solved for $a,b\in\Bbb R$ accordingly.
| {
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$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.) $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$
I tried like below
$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\frac{n(n+1)(2n+1)}{6}}=\\\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\
$$
Then I can use the fraction ,but $$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}+\frac{-4}{2n+1}$$ This is ugly to turn into telescopic series .
Can you help me to find :series converge to ?
Thanks in advance .
| Trying to avoid Daniel Fischer's approach.
$$\frac2{(2n)(2n+1)(2n+2)}=\frac1{(2n)(2n+1)}-\frac1{(2n+1)(2n+2)}$$
Collect the terms by their signs:
$$S=12\sum_{n=2}^\infty\frac{(-1)^n}{n(n+1)}$$
Now apply partial fractions on this nicely and it becomes
$$S=12\sum_{n=2}^\infty\frac{(-1)^n}n+\frac{(-1)^{n+1}}{n+1}$$
You can then collect the terms again to get
$$S=6+12\sum_{n=3}^\infty\frac{(-1)^n}n$$
Upon which you may use the Maclaurin expansion of $\ln(x+1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose for the sake of contradiction }f(x)=f(y)\text{ but }x\neq y.\\ \text{Therefore } 2x^3+3x-4 = 2y^3+3y-4\\
2x^3-2y^3+3x-3y=0\\
2(x-y)(x^2+xy+y^2)+3(x-y)=0\\
(x-y)\left [2(x^2+xy+y^2)+3)\right ]=0\\
\text{which implies that either } x=y\text{ or }2(x^2+xy+y^2)+3=0 \\
\text{which contradicts to the assumption that } x\neq y \Rightarrow \Leftarrow $$
Is there anything missing in my proof or there is a better solution? I got confused at the either or part because $x=y$ my or may not be true. My logic is given above, if there is something I misunderstand please give your feedback.
$\textbf{FINAL EDIT}$
Here is the continuation of the proof that it is never true that the right factor $x^2+xy+y^2=-\frac{3}{2}$.
$$\text{We know that } x^2\geq0 \text{ and } y^2\geq0\Rightarrow x^2+y^2\geq 0 \text{ and that there are two cases for }x,y.\\
\textbf{case 1: }\text{If nonzero x and y have the same sign then we are done since that implies that }\\
x^2+xy+y^2>0\\
\textbf{case 2: } \text{If nonzero x and y have differing signs }\\ xy<0
\text{ and also we know that } (x+y)^2>0\\
\Rightarrow x^2+2xy+y^2>0 \\
\Rightarrow x^2+xy+y^2>-xy\\
\text{ Since } xy<0 \Rightarrow -xy>0 \Rightarrow x^2+xy+y^2>0 \\
\text{Therefore it is never true that } x^2+xy+y^2=-\frac{3}{2}$$
Is my proof complete now?
| In your edit, Case 1 is fine, but the "proof" in Case 2 isn't. If the assumption of Case 2 is that precisely one of $x$ and $y$ is negative, then you have to add Case 3 to you proof: the case when both $x<0$ and $y<0$. If the assumption of Case 2 is that at least one of $x$ and $y$ is negative, then it also includes the possibility that both $x<0$ and $y<0$, which makes your claim that $xy<0$ invalid. (Not to mention that $x$ and $y$ can also be equal to $0$, so you have to either include that in existing cases or consider even more cases.) In any event, the next issue is: where did the claim that $x^2+xy+y^2\ge0$ come from?
A much shorter proof can be achieved by completing the squares, for example like this:
$$x^2+xy+y^2=x^2+2\cdot x\cdot\frac{1}{2}y+\frac{1}{4}y^2-\frac{1}{4}y^2+y^2=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2\ge0.$$
Then $2(x^2+xy+y^2)+3\ge\ldots?$
| {
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Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly found the maximum.
How do I proceed?
| Note that $4xy \leq (x+y)^2$ for all $x$, $y$.
You have $$ab+bc+ca = b(a+c) + ac \leq b(a+c) + \frac{1}{4}(a+c)^2 = (a+c) \frac{a+c+4b}{4} = \frac{1}{4}(a+c)(8-(a+c)) \leq \frac{1}{4}\frac{1}{4}8^2 = 4.$$
Equality is happen when $a=c$ and $a+c = 8-(a+c)$, or $a+c=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.
| another method: asymptotics
For $x>0$, we have as $x \to \infty$:
$$
\sqrt{x^2+ax} = x\sqrt{1+\frac{a}{x}} = x\left(1+\frac{a}{2x}+o(x^{-1})\right) = x+\frac{a}{2}+o(1)
\\
\sqrt{x^2+bx}= x+\frac{b}{2}+o(1)
\\
\sqrt{x^2+ax}-\sqrt{x^2+bx}= \frac{a-b}{2}+o(1)
$$
| {
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The arithmetic sequence is 13, 21, 29, 37... Find the least number of terms required for the sum of the sequence terms to exceed 1000. All I know is that a=13 and common difference=8. I am unaware how to continue from here.
| Just to do it slightly differently.
If I have $a_0 = 13$ and $a_k = 13 + 8*k$ and I have a list of $n$ of these terms $a_0=13, a_1=13+8, a_2=13+2*8,......, a_n=13 + n*8$, what is the average value of all of them.
Well as they are all exactly $8$ apart from each other, the average value will be directly in the middle or $13 + \frac n2*8$.
So what do you get when you add up all the values? Well you nave $(n+1)$ terms. And the average term is $13+\frac n2*8$. So when you add them all up you should get $(n+1)(13 + \frac n2*8)$.
So $(n+1)(13+\frac n2*8) > 1000$
....
But one thing you should teach yourself. $1 + 2 + 3 + 4 + 5 + ...... +n=???$.
Well the average is $\frac {n+1}2$. ANd there are $n$ terms so $???? = \frac {n(n+1)}2$. You will see that formula a lot!
Alternative proof:
$1 + 2 + 3 + 4 + .......... + n = S$
$n + (n-1) + (n-2) + (n-3) + ... + 1 = S$
$(n+1) + (n-1+2) + (n-2+3) + (n-3 + 4) + ..........+(1+n) = 2S$
$(n+1) + (n+1)+ (n+1) + (n+1) + ...... + (1+n) = 2S$
$n(n+1) = 2S$
$\frac {n(n+1)}2 = S$.
.....
So we have $13 + (13 + 8) + (13+2*8) + (13+2*8) + ...... + (13+n*8) = $
$[13 + 13 + 13 + ....... + ] + [8 + 2*8 + 3*8 + ........ + n*8]=$
$(n+1)13 + 8[1 + 2 + 3 + ........ + n] =$
$(n+1)13 + 8*\frac {n(n+1)}2 = $
$(n+1)(13 + 4n) > 1000$.
| {
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Solve trigonometric equation with $\arcsin$ and $\arccos$
Solve the following equation
$$2\arccos(x)-\arcsin(2x\sqrt{1-x^2})=0.$$
I tried to solve this equation, but I don't know how can I change $\arcsin$ to $\arccos$ to solve the equation.
| Hmmm what about the following approach ( there is a fair bit of trig )
Take the cosine of both sides to get
$$ \cos\left(\underbrace{2 \cos^{-1}(x)}_{\text{=Part 1}} - \underbrace{\sin^{-1}\left(2x\sqrt{1-x^2}\right)}_{\text{=Part 2}} \right) = 1. \tag{1} \label{eqn:main} $$
Look at each part individually.
$$ \cos(2 \cos^{-1}(x)) = \cos(2\theta), $$
where $\theta \equiv \cos^{-1}(x)$ and hence $x = \cos(\theta)$. Then use the double angle formula
$$\cos(2\theta) = \cos^2 (\theta) - \sin^2(\theta) \\ = 2\cos^2(\theta) - 1. $$
Then using what we have defined above for $\theta$ and $x$ we may deduce that
$$ cos(2 \theta) = 2x^2 -1 .$$
Okay first part done. Second part is similar. We have
$$ \cos\left(\sin^{-1}\left(2x \sqrt{1 - x^2}\right) \right).$$
Let
$$ \sin^{-1}\left(t\right) \equiv \Theta, $$
where $t \equiv2x \sqrt{1 - x^2}.$ Then
$$ \sin^2 (\Theta) = t^2 \Rightarrow 1- \cos^2(\Theta) = t^2,$$
Or in other words
$$ \Theta = \cos^{-1} \left( \sqrt{1 - t^2} \right).$$
Finally, substituting this back in to the original expression
$$\cos\left(\sin^{-1}\left(2x \sqrt{1 - x^2}\right) \right) = \cos\left( \cos^{-1}(\Theta) \right).$$
So the full equation now reads
$$ 2x^2 - 1 - \sqrt{1 - t^2} = 1,$$
which, I believe after substituting in the expression for $t$ results with solution $ x = \pm 1$.
| {
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same area and perimeter
The perimeter and the area of a given triangle have the same numerical value when measured via a certain unit of measure. If the lengths of the three altitudes of the triangle are p, q, and r, what is the numerical value of $1/q + 1/p + 1/r$?
So first I set up the three sides' lengths as $a, b, c$ respectively, so that the perimeter and area both would be $a+b+c$. but the area is also equal to $ap/2 + bq/2 + cr/2 = a+b+c$, which leads to $ap + bq+cr = 2a + 2b + 2c$. However, $1/q + 1/p + 1/r = (qr + pr + pq)/(pqr)$, which doesn't seem to help much. How can I solve this?
| $$A = \frac{ap}{2}= \frac{bq}{2}= \frac{cr}{2}=a+b+c.$$
$$\frac{ap}{2}=a+b+c$$
$$p=\frac{2(a+b+c)}{a}$$
Similarly,
$$q=\frac{2(a+b+c)}{b}$$
$$r=\frac{2(a+b+c)}{c}$$
Now
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a}{2(a+b+c)}+\frac{b}{2(a+b+c)}+\frac{c}{2(a+b+c)}$$$$$$
$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{a+b+c}{2(a+b+c)}$$$$$$$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
| Here's a nice trick. For each $k$,
$$k^2=\frac{(k-1)k}2+\frac{k(k+1)}2=a_{k-1}+a_k$$
say.
Then
$$1^2-2^2+3^2-4^2+\cdots\pm n^2
=a_0+a_1-a_1-a_2+a_2+a_3-\cdots\pm a_{n-1}\mp a_n$$
and almost all of these terms cancel.
| {
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"timestamp": "2023-03-29T00:00:00",
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graph of $y = a\cos(bx+c)+d$
Attached is the graph of $y = a\cos(bx+c)+d$, where $a>0$, $b>0$, and $c>0$, and $c$ is as small as possible. Find $a + b + c + d$.
I'm having a hard time trying to solve this problem. First of all, the period seems to be $3\pi$, which means that $b=2/3$, and the amplitude, or a, is 2 and $d=1$. However, the horizontal shift of the cosine graph seems to be $3\pi/2$, which would give the value of $c=9\pi/4$. But when I add $a+b+c+d$, I get $11/3 + 9\pi/4$, which is not the correct answer. Is there a c-value smaller than the one I found? How?
| Firstly we need to find the midline of the graph, which will be the variable $d$.
Since it's maximun is $3$ and it's minimun is $-1$, the midline is:
$$y=\frac{(3)+(-1)}{2} = \fbox{d = 1}$$
We may also notice that two of the maximuns of the curve occur at: $-\frac{3\pi}{2}$ and $\frac{3\pi}{2}$, So the period of the function is $3\pi$.
Which implies:
$$\frac{2\pi}{b}=3\pi \Longrightarrow b = \frac{2}{3} $$
Now there two possible following approaches:
1) You may notice the graph to be an "upside-down" cosine graph (This will lead $c$ to be equal to $0$ which is not allowed since $c>0$).
or
2) You may notice the graph to be a shifted cosine graph
Since the nearest crest of the wave is at $-\frac{3\pi}{2}$, the original cosine function had to be moved $\frac{3\pi}{2}$ to the left (before being horizontaly scaled by $b$). So:
$$\frac{c}{b} = \frac{3\pi}{2}$$
and since $b = \frac{2}{3}$
$$c = \pi$$
Finally, since the amplitude of the graph is equal to 2
$$a=2$$
And the final equation is:
$$y=2\cos(\frac{2}{3}x+\pi)+1$$
$$a+b+c+d = 2+\frac{2}{3} + \pi+1 = \frac{11+3\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431339",
"timestamp": "2023-03-29T00:00:00",
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Verification of infinite square root equation proof
b) Show also that the infinite square root
$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\frac{1+\sqrt5}{2}$$
Now, I've devised two proofs for this statement (both of which include using substitution), and it entails assuming that the case is true, and then restating the problem.
Proof 1:
let $\displaystyle u=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$
hence $$u=\frac{1+\sqrt5}{2}$$ from here, I went on to express the equation as a quadratic as follows: $$2u-1=\sqrt5$$ $$4u^2-4u+1=5$$ $$4u^2-4u-4=0$$ and then, naturally, just substituted the new values into the quadratic formula: $$u=\frac{4\pm\sqrt{4^2-4(4)(-4)}}{8}$$ $$u=\frac{4\pm4\sqrt{5}}{8}$$ $$u=\frac{1\pm\sqrt5}{2}$$ the only problem I found here is that, evidently, the quadratic formula leaves the $\pm$ in its answer, as opposed to the desired $+$, so the second proof is as follows:
Proof 2:
let $\displaystyle u=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$
hence, $$\sqrt{1+u}=\frac{1+\sqrt5}{2}$$ and from here, I simply proceeded to solve for $u$, so $$1+u=\frac{(1+\sqrt5)^2}{4}$$ $$u=\frac{6+2\sqrt5}{4}-1$$ $$u=\frac{3+\sqrt5}{2}-\frac{2}{2}$$ $$u=\frac{1+\sqrt5}{2}$$
which (I think) completes the proof. The question is, are either of these approaches valid? Are there any better approaches that someone could list? Any responses are appreciated.
| Let's show that the sequence $(a_n)_{n=1}^\infty$ given by $a_1 = 1$, $a_{n+1} = \sqrt{a_n + 1}$ is monotonically increasing and bounded from above.
We shall prove by induction that $a_n \leq \frac{1+\sqrt{5}}{2}$.
$$a_1 = 1 < \frac{1+\sqrt{5}}{2}$$
Assume $a_n \leq \frac{1+\sqrt{5}}{2}$, for some $n\in\mathbb{N}$.
$$a_{n+1} = \sqrt{1 + a_n} \leq \sqrt{1 + \frac{1+\sqrt{5}}{2}} = \sqrt{\left(\frac{1+\sqrt{5}}{2}\right)^2} = \frac{1+\sqrt{5}}{2}$$
Now we shall prove that $(a_n)_{n=1}^\infty$ is increasing:
$$a_{n+1} \geq a_n \iff \sqrt{a_n + 1} \geq a_n \iff 1+a_n - a_n^2 \geq 0 \iff a_n \in \left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$$
The last statement holds, so we have $a_{n+1}\geq a_n$.
Thus, $(a_n)_{n=1}^\infty$ is increasing and bounded from above, hence it is convergent.
Set $u = \lim_{n\to\infty} a_n$.
Letting $n\to\infty$ in the recurrence relation $a_{n+1} = \sqrt{a_n + 1}$ gives $u = \sqrt{1 + u}$.
Thus, $u = \frac{1\pm\sqrt{5}}{2}$. Since $a_n \geq 0, \forall n \in \mathbb{N}$, we have $u \geq 0 \implies u = \frac{1+\sqrt{5}}{2}$.
| {
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Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?
| For $2n-1 = 1$:
$$1^2 = \frac{1(2*1-1)(2*1+1)}{3} = \frac{3}{3} = 1$$
Assuming the formula works for $n=2k-1$, where $k$ is some integer, for $n=2(k+1)-1$:
$$1^2+3^2+...+(2k-1)^2+(2(k+1)-1)^2 =$$
$$ \frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 \stackrel{?}{=} \frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} $$
By expanding the right hand side of the equation:
$$\frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} = \frac{4k^3+12k^2+11k+3}{3}$$
And by simplifying the left hand side of the equation:
$$\frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 = \frac{4k^3-k}{3} + (2k+1)^2 = \frac{4k^3-k}{3}+4k^2+4k+1 = \frac{4k^3+12k^2+11k+3}{3}$$
Finally:
$$ \frac{k(2k-1)(2k+1)}{3} + (2(k+1)-1)^2 = \frac{(k+1)(2(k+1)+1)(2(k+1)-1)}{3} $$
And so for any $n$, where $2n-1 \in \mathbb{N}$, this formula holds:
$$1^2 + 3^2 + ... + (2n-1)^2 = \frac{n(2n-1)(2n+1)}{3}$$
| {
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If $f(x+1) +f(x-1) =\sqrt{3}\,f(x)$ and $f(2) =2$, what is the value of $f(4)$? My Attempt
$f(2)=2$. So, $f(1) + f(3)=2\sqrt{3}$ and $f(2) + f(4)=\sqrt{3}\,f(3)$. After solving these equations I got the value of $f(3)=2\sqrt{3}$ and $f(4)=4$. But are there any other methods than this? Any suggestions are welcome.
Update:- @ProfessorVector pointed out that the above solutions are only true if $f(1)=0$. After checking I find that it is true. So, my above attempt is a failure. Is there a way to solve this question?
Update 2:- Is there a way to find the period of this function?
| I'll be more general.
$f(n+1)+f(n-1)
= cf(n)
$.
Suppose
$f(n)
= b^n
$.
Then
$b^{n+1}+b^{n-1}
= cb^n
$
or
$b^2-cb+1
= 0
$.
Then
$b
=\dfrac{c\pm\sqrt{c^2-4}}{2}
$.
If $c^2=4$,
$b = c/2 = \pm 1$,
so
$f(n)=1$
or
$(-1)^n$.
If
$c^2 > 4$,
then
$b$ has two possible values,
one with
$|b|>1$
and one with
$|b|<1$.
If
$c^2 < 4$,
then
$b$ has two possible
complex values
$b_1
=\dfrac{c+\sqrt{c^2-4}}{2}
=\dfrac{c+i\sqrt{4-c^2}}{2}
$
and
$b_2
=\dfrac{c-i\sqrt{4-c^2}}{2}
$.
Note that
$b_1b_2
=\dfrac{c+\sqrt{c^2-4}}{2}\dfrac{c-\sqrt{c^2-4}}{2}
=\dfrac{4}{4}
=1
$
(as can also be deduced
from the quadratic
specifying $b$)
and that
$|b_k|^2
=\dfrac{c^2+4-c^2}{4}
=1
$.
Since
$|b| = 1$,
$b
=e^{it}
=\cos(t)+i\sin(t)
$
where
$\cos(t)
=c/2
$.
In your case,
$c = \sqrt{3}$
so
$t
=\arccos(\sqrt{3}/2)
=\pi/6
$.
Therefore
the two possible solutions are
$f(n)
=e^{\pm ni\pi/6}
$
and any linear combination of these.
So any solution
is of the form
$ue^{in\pi/6}+ve^{-in\pi/6}
=(u+v)\cos(n\pi/6)+(u-v)\sin(n\pi/6)
$.
If the solution is real
for an $n$
such that
$\sin(n\pi/6) \ne 0$,
then
$u=v$,
so it is
$f(n)=2u\cos(n\pi/6)
$.
If
$f(2) = 2$,
then
$2=f(2)
=2u\cos(\pi/3)
=u
$
so
$u = 2$
and the solution is
$f(n)
=4\cos(n\pi/6)
$.
Putting $n=4$,
$f(4)
=4\cos(4\pi/6)
=-2
$.
| {
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"url": "https://math.stackexchange.com/questions/2433417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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determinant of the $7\times7$ matrix How to find determinant of the following $7\times7$ matrix
\begin{bmatrix}
a&1&0&0&0&0&1&\\
1&a&1&0&0&0&0&\\
0&1&a&1&0&0&0&\\
0&0&1&a&1&0&0&\\
0&0&0&1&a&1&0&\\
0&0&0&0&1&a&1&\\
1&0&0&0&0&1&a&
\end{bmatrix}
| An elementary proof. Let $J$ be the permutation matrix $(1\rightarrow 7,2\rightarrow 1,3\rightarrow 2\cdots)$; clearly $J^7=I$. Then we consider the matrix $A=aI+J+J^{-1}$. Let $\Delta_p=J^p+J^{-p}$ and $X=J+J^{-1}$.
$X^7=2I+7\Delta_5+21\Delta_3+35\Delta_1$. We remove $\Delta_5$:
$X^5=\Delta_5+5\Delta_3+10\Delta_1$ implies $X^7-7X^5=2I-14\Delta_3-35\Delta_1$.
We remove $\Delta_3$: $X^3=\Delta_3+3\Delta_1$ implies $X^7-7X^5+14X^3=2I +7\Delta_1$, that is
$f(X)=X^7-7X^5+14X^3-7X-2I=0$. Note that $f(x)=(x-2)(x^3+x^2-2x-1)^2$ (in irreducible factors).
Clearly $2$ is an eigenvalue of $X$. Then the characteristic polynomial of $X$ is $(x-2)^4(x^3+x^2-2x-1)$ or $f(x)$; in fact, it is $f(x)$ because $trace(X)=0$.
Conclusion: the required determinant is $-f(-a)=a^7-7a^5+14a^3-7a+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Are these sums of squares identities new? The identities are
\begin{align}
u^2 + \left( (2u-1)(2u^2-u+1) \right)^2 + \left( u (2u-1)(2u^2-u+1) \right)^2 &= \left( 4u^4-4u^3+5u^2-3u+1 \right)^2 \\
u^2 + \left( (2u+1)(2u^2+u+1) \right)^2 + \left( u (2u+1)(2u^2+u+1) \right)^2 &= \left( 4u^4+4u^3+5u^2+3u+1 \right)^2
\end{align}
I have not been able to find them online or in any of my books.
| The complete solution to $x^2 + y^2 + z^2 = w^2$ is known: $$x = (a^2-b^2-c^2)t,\; y = 2 a b t,\; z = 2 a c t,\; w = (a^2 + b^2 + c^2) t$$
Your first is the case
$$ \eqalign{a &= (2 u^2 - u + 1) \cr
b &= 2u-1\cr
c &= u (2u-1)\cr
t &= \frac{1}{2}\cr }$$
Similarly for the second, with some sign changes.
| {
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Find the limit to $\lim\limits_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$ My problem is evaluating the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{x^5+y^2}{x^4+|y|}$$
The answer should be 0. I tried to convert the limit into polar form, but it didn't help because I couldn't isolate the $r$ and $\theta$-variables of the expression. My "toolbox" for solving problems like these is very limited... If polar form doesn't work, then I usually have no clue on how to continue.
Edit: I think this is the solution.
$$
\lim_{(x,y)\to(0,0)}\left|\frac{x^5+y^2}{x^4+|y|}\right| = \frac{|x^5+y^2|}{|x^4+|y||}
$$
Applying the triangle inequality gives
$$
\frac{|x^5+y^2|}{|x^4+|y||} \leq \left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right|
$$
Inspecting the denominators on the RHS gives:
$$
\left|\frac{x^5}{x^4+|y|}\right| \leq |x|, \quad\left|\frac{y^2}{x^4+|y|}\right| \leq |y|
$$
So
$$
\left|\frac{x^5}{x^4+|y|}\right| + \left|\frac{y^2}{x^4+|y|}\right| \leq |x| + |y|
$$
Since $|x| + |y| \to 0$ when $x,y\to 0$, the sandwich theorem states that $|\frac{x^5+y^2}{x^4+|y|}| \to 0$. And if $\lim |f(x)|=0$ then $\lim f(x)=0$ which solves the original problem.
| A common tool is to homogenise the denominator by setting $\displaystyle u=\frac{y}{x^4}$, then discuss the limit in function of $u$.
$\displaystyle f(x,y)=\frac{x^5+y^2}{x^4+|y|}=\frac{x^5+u^2x^8}{x^4+|u|x^4}=\frac{x+u^2x^4}{1+|u|}=\frac{x+uy}{1+|u|}=x\left(\frac 1{1+|u|}\right)+y\left(\frac u{1+|u|}\right)$
Since both $\frac 1{1+|u|}$ and $\frac u{1+|u|}$ are bounded (by $1$ in absolute value)
Then $|f(x,y)|\le |x|+|y|\to 0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find $\lim\limits_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$ This problem
$$\lim_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$$
is from Silverman's "Modern Calculus and Analytical Geometry" Section 22, #16d. I've been struggling on it for a while and can't figure out what to do besides trying to multiply by the conjugate and/or substitution but it doesn't work out. What do you all think? Keep in mind you can't use L'Hopital's rule, only elementary math.
| Added for your curiosity.
For the limit itself, you aleary received good answers. The problem can also be addressed using Taylor series or the generalized binomial theorem
$$(1+x)^a=1+a x+\frac{1}{2} a(a-1) x^2+\frac{1}{6} a(a-1) (a-2) x^3+O\left(x^4\right)$$ making
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+O\left(x^4\right)$$
$$\sqrt{1-x}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}+O\left(x^4\right)$$
$$\sqrt[3]{1+x}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5 x^3}{81}+O\left(x^4\right)$$
$$\sqrt[3]{1-x}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{5 x^3}{81}+O\left(x^4\right)$$
$$ \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}=\frac{ x+\frac{1}{8}x^3+O\left(x^4\right)} {\frac{2 }{3}x+\frac{10 }{81}x^3+O\left(x^4\right) }=\frac{3}{2}-\frac{13 x^2}{144}+O\left(x^3\right)$$ which shows the limit and how it is approached.
But what is interesting is that this is a very good approximation even if $x$ is not small. For example, using $x=\frac 12$, the exact expression is $\approx 1.47469$ while the approximation leads to $\frac{851}{576}\approx 1.47743$.
Edit
Making the problem more general and using the same steps
$$\frac{(x+1)^p-(1-x)^p}{(x+1)^q-(1-x)^q}=\frac{p}{q}+\frac{p (p-q) (p+q-3)}{6 q}x^2+O\left(x^4\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Arithmetic and Geometric progression in 3 numbers Suppose
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally $$a=b=c=\frac{1}{2}.$$
Can there be any such triplet such that $a<b<c$ ?
| $2b=a+c$ and $a+b+c=\frac{3}{2}$ gives $b=\frac{1}{2}$ and $a+c=1$.
Also we have $b^4=a^2c^2$, which gives $$a^2c^2=\frac{1}{16}.$$
Thus, $$a^2(1-a)^2=\frac{1}{16},$$
which gives $a(1-a)=\frac{1}{4}$ and $a=c=\frac{1}{2}$ or
$$a(1-a)=-\frac{1}{4},$$
which gives $$(a,c)\in\left\{\left(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{2}\right),\left(\frac{1-\sqrt2}{2},\frac{1+\sqrt2}{2}\right)\right\}.$$
For $a<b<c$ we have the following unique solution:
$$(a,b,c)=\left(\frac{1-\sqrt2}{2},\frac{1}{2},\frac{1+\sqrt2}{2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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EV of probability given outcome Michael has a crush on a girl. Every night, he texts her and asks to go on a date. There is a 1/7
chance that the girl
says yes, a 2/7
chance that the girl says no, and a 4/7 chance that the girl asks Michael to text her again tomorrow, which
Michael does. Given that she said no, what is the expected number of days it took her to decide?
| I have assumed that the No the first day is assumed to be 1. If that were to be the case, then the expected no of days on we which she decided would be the following:
Probability of a decision made = Probability of a No + Probability of a Yes
$P(Decide) = \frac{2}{7}+\frac{1}{7} = \frac{3}{7}$
USing Bayes' theorem:
Probability of $P(No/Decided) = \dfrac{\frac{2}{7}}{\frac{3}{7}} = \frac{2}{3}$.
$P(Yes/Decided) = \dfrac{\frac{1}{7}}{\frac{3}{7}} = \frac{1}{3}$.
$P(Decided/No) = \dfrac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{3}} = \frac{2}{3}$
On each day it is this probability of decision that she has to make.
Thus S = Porbability of a Decision/No + Probability of not making decsion the previous day * Porbability of a Decision/No (2) + (Probability of not making decision the previous day)^2 *Porbability of a Decision/No (3) + ....
$S = \frac{2}{3}(1) + \frac{4}{7}\frac{2}{3}(2) + (\frac{4}{7})^2\frac{2}{3}(3) +\cdots$
$S = \frac{2}{3}\dfrac{1}{(1-\frac{4}{7})^2} = \frac{98}{27}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
Question:
Find the values of the constants $a$,$b$ and $c$ such that the ellipse $4x^2+y^2+ax+by+c=0$ passes through the point $(-1,2)$ and is tangent to the x-axis
What I've done so far:
Substitute $(-1,2)$ in $4x^2+y^2+ax+by+c=0$:
$$\begin{align}
&\implies 4+4-a+2b+c=0 \\
&\implies c=a-2b-8 \tag{1}
\end{align}$$
Rewriting the equation of the ellipse to get standard form:
$$\begin{align}
& 4x^2+y^2+ax+by=-c \\[6pt]
\implies& 4\left(x^2+\frac{a}{4}x+\frac{a^2}{64}\right)
+\left(y^2+by+\frac{b^2}{4}\right)=a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} \\[6pt]
\implies& \frac{4\left(x+\frac{a}{8}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{\left(y+\frac{b}{2}\right)^2}{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}}=1
\end{align}$$
Since the ellipse is tangent to the $x$-axis:
$$\begin{align}
&\quad\sqrt{a-2b-8+\frac{a^2}{16}+\frac{b^2}{4}} = \frac{b}{2} \\[6pt]
\implies&\quad a-2b-8+\frac{a^2}{16}+\frac{b^2}{4} = \frac{b^2}{4} \\
\implies&\quad 2b= a-\frac{a^2}{16}-8 \tag{2}
\end{align}$$
From $(1)$ and $(2)$:
$$\quad c= a-\left(a-\frac{a^2}{16}-8\right)- 8 \quad\implies\quad c= \frac{a^2}{16} \tag{3}$$
Therefore, the equation of the ellipse becomes
$$\frac{4\left(x+\frac{a}{8}\right)^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}+\frac{(y+\frac{b}{2})^2}{a-(a-\frac{a^2}{16}-8)-8+\frac{a^2}{16}+\frac{b^2}{4}}=1$$
That is,
$$\frac{\left(x+\frac{a}{8}\right)^2}{b^2/16}+\frac{\left(y+\frac{b}{2}\right)^2}{b^2/4}=1 \tag{4}$$
I'm stuck at this point and don't know how to find $a$ (or what else to do in general).
All help is appreciated.
| There's not one, but an infinite number of ellipses that satisfy the two conditions that you have listed. You have 3 unknowns ($a$,$b$ and $c$) to solve for, but only 2 conditions given by which you can determine them. So, no unique solution.
Note that the equation of the ellipse can be written as
$$\frac{(x+\frac{a}{8})^2}{(\frac{1}{2})^2} + \frac{(y+\frac{b}{2})^2}{1^2} = \frac{a^2}{4^2}+\frac{b^2}{2^2}+c \equiv D^2$$.
Some edits: I forgot to divide out by $D^2$ to put it in standard form.
So,
$$\frac{(x+\frac{a}{8})^2}{(\frac{D}{2})^2} + \frac{(y+\frac{b}{2})^2}{D^2} = 1$$.
This shows that the half axis along the y-direction has length D. Since the ellipse is tangent to the x-axis, the center of the ellipse $(\frac{-a}{8},\frac{-b}{2}$ must be exactly D above the x-axis. In other words, $\frac{-b}{2}=D$ or $\frac{b^2}{4}=D^2$.
$$\frac{a^2}{4^2}+\frac{b^2}{2^2}+c = \frac{b^2}{4}$$
This is one equation relating $a$,$b$ and $c$. The other is obtained by plugging $(x,y)=(-1,2)$ in to the equation of the ellipse to get the second equation
$$\frac{(x+\frac{a}{8})^2}{(\frac{1}{2})^2} + \frac{(y+\frac{b}{2})^2}{1^2} = D^2 = \frac{b^2}{4}$$.
Any combination of $a$, $b$ and $c$ that satisfy the two equations above, will give a valid ellipse that meets the two conditions you are given.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of three-term arithmetic progressions in [n]
Three numbers are chosen at random between 1 and $n$ (say $n=500$).What will be the probability of those numbers to be in arithmetic progression?
I don't know how to count the number of favorable events.
Sample space={(1,2,3),(4,5,6),(18,20,21)........}
favorable events={(2,4,6),(8,12,16),(10,20,30),(50,100,150... and many more)}
I can count the sample space but how do i count the favorable events.
Some insight could help?
| (Assuming three distinct numbers are chosen.)
A three-element arithmetic progress $a,b=a+k,c=a+2k$ with $k>0$ is entirely determined by $a,c$, and thus the number of them is equal to the number of ways of picking $a,c$ with the same parity (either both odd or both even.)
The number of odd numbers from $1$ to $n$ is $\lceil n/2\rceil$ and the number and the number of even numbers from $1$ to $n$ is $\lfloor n/2\rfloor$.
This means that the number of arithmetic progressions is:
$$\binom{\lfloor n/2\rfloor}{2}+\binom{\lceil n/2\rceil}{2}$$
When $n$ is even, this is $2\binom{n/2}{2}=\frac{n(n-2)}{4}$. When $n$ is odd, this is $\binom{(n-1)/2}{2}+\binom{(n+1)/2}2=\frac{(n-1)^2}{4}.$
Both cases can be written in one formula:
$$\left\lfloor\frac{(n-1)^2}{4}\right\rfloor$$
When $n>2$ is even, the probability is:
$$\dfrac{\frac{n(n-2)}{4}}{\frac{n(n-1)(n-2)}{6}}=\frac{3}{2(n-1)}$$
When $n>1$ is odd:
$$\frac{3(n-1)}{2n(n-2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this a correct way to prove by induction? I understand that most people use the inductive hypothesis, but I find that counterintuitive. Is the below proof correct? In particular, I am concerned with my use of $n$ in (I); is the reason people use another variable, e.g. $k$, for conceptual reasons, or does the use of $n$ create an error in my proof?
\begin{align*}
P(0) \land [P(k) \Rightarrow P(k+1)] \implies P(n) \tag{AI} \\
\end{align*}
\begin{align*}
\text{When }n = 2, \ \ 2 + 6 + 10 + . . . + (4n - 2) &= 2n^2 \tag{B} \\
2 + 6 &= 2 \cdot 2^2 \\
8 &= 8 \\
\end{align*}
$$
\begin{pmatrix}
2 + 6 + 10 + ... + (4n-2) = 2n^2 \\
\Big\Downarrow \\
2 + 6 + 10 + ... + (4n-2) + (4(n+1)-2) = 2(n+1)^2 \\
\end{pmatrix} \tag{I}
$$
\begin{align*}
\Big\Updownarrow
\end{align*}
\begin{align*}
4(n+1)-2 &= 2(n+1)^2 -2n^2 \\
4n+4-2 &= 2n^2+4n+2 - 2n^2 \\
4n+2 &= 4n+2 \\
\end{align*}
$$
\text{B} \land \text{I} \land \text{AI} \implies 2 + 6 + 10 + . . . + (4n - 2) = 2n^2 \text{ for } n > 1 \ \ \square
$$
Is this alternate solution correct? I am confident in my reasoning, but am unsure if it is a valid mathematical argument.
Pairing $1$st with $n$th term, $2$nd with $(n-1)$th term, etc., yields $\mathbf{\frac{n}{2}}$ pairs:
\begin{align*}
2 + (4n-2) \ \ + \ \ 6 + (4(n-1)-2) \ \ + \ \ 10 + (4(n-2)-2) \ \ + \ \ ... &= 2n^2 \\
2 + (4n-2) \ \ + \ \ 6 + (4n-6) \ \ + \ \ 10 + (4n-10) \ \ + \ \ ... &= 2n^2 \\
4n \ \ + \ \ 4n \ \ + \ \ 4n \ \ + \ \ ... &= 2n^2 \\
4n \cdot \mathbf{\frac{n}{2}} &= 2n^2 \\
2n^2 &= 2n^2 \\
&\ \square \\\ \\
\end{align*}
| The first proof is perfectly fine. The choice of letters is mostly/often just for aestetical/pedagogical reasons.
The second (and to some extent the first) uses the $\dots$ informal notation. This should probably be seen as a shorthand or more visual way of handling summations. Formally one should have used $\sum$-notation. The problem in the last is that you visually rearranges the terms which is not that good looking when using $\sum$ notation.
The second approach if done strictly would also need induction to be complete (unless you rely on propositions that already have been proven by induction).
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of an expression in two variables $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}$$
$x$ and $y$ are positive real numbers, what is the minimum value?
I have tried finding the minimum value of the first expression and the third expression, with the first expression being $\frac{7}{\sqrt{2}}$ while the second expression being $5$, I do not know how to move on from this (note that the answer is an integer from $0$ to $999$).
| Let $OABCD$ be a pentagon such that $OA=5\sqrt2$, $OB=y$, $OC=x$, $OD=7$, $\measuredangle AOB=45^{\circ}$, $\measuredangle BOC=45^{\circ}$ and $\measuredangle COD=45^{\circ}$.
Thus, $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}=$$
$$=DC+CB+BA\geq AD=\sqrt{(5\sqrt2)^2+7^2-2\cdot5\sqrt2\cdot7\cos135^{\circ}}=13.$$
The equality occurs, when $\{B,C\}\subset AD,$ which says that $13$ is a minimal value.
Done!
| {
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Confusing Integral Expression: $\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$ I have this integral expression to be evaluated:
$$\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}$$
I've followed these steps:
*
*Completed the square in $x$ and then substituted $u=x-\frac{1}{2}.$
*Trig substituted $u = \frac{\sqrt{3}}{2}\tan t.$ and then simplified the equation thereby giving:
$$\int{\frac{\sqrt{3}\sec^2t}{\sqrt{3}\tan t+\sqrt{3}\sec t+1}dt}$$
I know up to this, but I'm unclear what to do next.
Feel free to correct me if I'm wrong.
| Hint
Start considering the integrand as $$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{1}{x+\sqrt{x^2-x+1}}\times\frac{x-\sqrt{x^2-x+1}}{x-\sqrt{x^2-x+1}}$$
$$\frac{1}{x+\sqrt{x^2-x+1}}=\frac{x-\sqrt{x^2-x+1}}{x-1}=\frac x {x-1}-\frac{\sqrt{x^2-x+1}}{x-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the harmonic conjugates of the following rational function? List all the harmonic conjugates of the following rational function. The integration is almost impossible, Symbolab and Microsoft Math fails to arrive at the answer:
\begin{equation}
\mu(x,y) =\frac{x^2+x+y^2}{x^2+y^2}
\end{equation}
| Look at
$\mu(x, y) = \dfrac{x^2 + x + y^2}{x^2 + y^2} \tag 1$
in polar coordinates:
$\mu(x, y) = \dfrac{x^2 + x + y^2}{x^2 + y^2} = \dfrac{r^2 + r\cos \theta}{r^2} = 1 + \dfrac{\cos \theta}{r}; \tag 2$
with
$z = x + iy = r\cos \theta + ir\sin \theta =re^{i\theta} \tag 3$
we have
$z^{-1} = r^{-1}e^{-i\theta} = \dfrac{\cos \theta - i\sin \theta}{r} = \dfrac{\cos \theta}{r} - i\dfrac{\sin \theta}{r} \tag 4$
is holomorphic on $\Bbb C \setminus \{0\}$; it follows that, as long as we stay away from $r = 0$, where $\mu$ is not in any event defined, that the harmonic conjugate of $\cos \theta / r$ is $-\sin \theta / r$, up to an addidive real constant. In $x$-$y$ coordinates, we have
$-\dfrac{\sin \theta}{r} = -\dfrac{r\sin \theta}{r^2} = -\dfrac{y}{x^2 + y^2}. \tag 5$
Thus, every harmonic conjugate of $\mu(x, y)$ is of the form
$\alpha - \dfrac{y}{x^2 + y^2}, \tag 6$
$\alpha \in \Bbb R$ a constant, in agreement with the comments of Thomas Andrews.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5$
Evaluate
$$5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5.$$
The options are $5^6$, $5^7$, $5^8$, $5^9$, $5^{10}$.
I'm new to this site. I came across this question in an Olympiad foundation site. I have no idea how to solve it. Can I get the answer of this question. Thanks.
| Hint. Note that the given sum can be written as
$$5+(5-1)\cdot 5+(5-1)\cdot5^2+(5-1)\cdot5^3+(5-1)\cdot5^4+(5-1)\cdot5^5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$ Find Minimum value of Minimum value of $$S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}$$
My Try:
we have $$S=3 \times \left (\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\frac{b}{a+c}+\frac{2c}{a+b}$$
By Standard $AM$ $GM$ inequality we know that
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \ge \frac{3}{2}$$
Is it possible to proceed here?
| For any positive numbers $\alpha, \beta, \gamma$, we have
$$\frac{\alpha^2 a}{b+c} + \frac{\beta^2 b}{c+a} + \frac{\gamma^2 c}{a+b}
= (a+b+c)\left(\frac{\alpha^2}{b+c} + \frac{\beta^2}{c+a} + \frac{\gamma^2}{a+b}\right) - \left(\alpha^2+\beta^2+\gamma^2\right)$$
By Engel's form of Cauchy Schwarz, we have
$$\frac{\alpha^2}{b+c} + \frac{\beta^2}{c+a} + \frac{\gamma^2}{a+b}
\ge \frac{(\alpha+\beta+\gamma)^2}{(b+c)+(c+a)+(c+a)}
= \frac{(\alpha+\beta+\gamma)^2}{2(a+b+c)}
$$
This leads to
$$\frac{\alpha^2 a}{b+c} + \frac{\beta^2 b}{c+a} + \frac{\gamma^2 c}{a+b}
\ge \frac{(\alpha+\beta+\gamma)^2}{2} - \left(\alpha^2 + \beta^2 + \gamma^2\right)$$
and the equality holds when and only when
$$\alpha : \beta : \gamma = b+c : c+a : a+b$$
Subsitute $\alpha,\beta,\gamma$ by $\sqrt{3},\sqrt{4}$ and $\sqrt{5}$
We obtain
$$\frac{3 a}{b+c} + \frac{4 b}{c+a} + \frac{5 c}{a+b}
\ge \frac{(\sqrt{3} + \sqrt{4} + \sqrt{5})^2}{2} - 12
= \sqrt{15} + \sqrt{20} + \sqrt{12} - 6$$
and the equality is achievable when
$$a : b : c = -\sqrt{3}+\sqrt{4}+\sqrt{5} : \sqrt{3} - \sqrt{4} + \sqrt{5} : \sqrt{3}+\sqrt{4} - \sqrt{5}$$
This means the minimum value of $S$ is $\sqrt{15} + \sqrt{20} + \sqrt{12} - 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.
| More a comment than an answer:
According to Wolfy,
the indefinite integrals
involve arctan and
complex values.
But the definite integrals
come out like this:
$\int_0^∞ \dfrac{dx}{x^4 - \sqrt{3} x^2 + 1}
= \frac12 \sqrt{2 + \sqrt{3}} π≈3.0345
$
and
$\int_0^∞ \dfrac{dx}{x^4 + \sqrt{3} x^2 + 1} dx = \dfrac{π}{2 \sqrt{2 + \sqrt{3}}}
≈0.8131
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Existence of Limit for $x_{n+1} = 1+1/x_{n}$ I'm attempting to prove that if one defines a sequence with $x_{1} = 1$ and
$$x_{n+1} = 1+\frac{1}{x_{n}}$$
for all $n >1$ then the limit of $\{x_{n}\}_{n \ge 1}$ exists.
My Attempt:
Since $\{x_{n}\}_{n \ge 1}$ was defined recursively I immediately thought to try and prove that the sequence is monotone and bounded. The problem is that $\{x_{n}\}_{n \ge 1}$ is an alternating sequence. Its first few values are $1,2,3/2,5/8,\ldots$. I checked boundedness next. Since $x_{1} = 1 \le \frac{1+\sqrt{5}}{2}$ we induct on $n$ and assume $x_{n} \le \frac{1+\sqrt{5}}{2}$. For the $x_{n+1}$ case we find that $x_{n+1} \ge \frac{1+\sqrt{5}}{2}$ which I thought was a little odd... Is there another way I should be attacking this proof or can I fiddle with the sequence a bit to use MCT?
| This is a trick I have learned here a few years ago.
Let's say you have a sequence $x_n$ defined by some sort of iteration.
If the limit $\lambda$ of the sequence satisfies a quadratic equation with roots $\alpha, \beta$ and one construct an auxiliary sequence by
$$y_n = \frac{x_n - \alpha}{x_n - \beta}$$
Sometimes, the recurrence relation for $y_n$ will be so simple to the point we can deduce the convergence of $x_n$ or even an closed form expression of $x_n$.
For the problem at hand, the sequence is generated by the iteration.
$$x_{n+1} = 1 + \frac{1}{x_n}$$
The limit $\lambda$ need to satisfies a quadratic equation
$$\lambda = 1 + \frac{1}{\lambda}\quad\iff\quad \lambda^2 - \lambda - 1 = 0$$
which have roots $\alpha = \varphi, \beta = -\frac{1}{\varphi}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.
If one define $\displaystyle\;y_n = \frac{x_n - \varphi}{1 + \varphi x_n}\;$, we have:
$$\begin{align}y_{n+1}
&= \frac{x_{n+1} - \varphi}{1 + \varphi x_{n+1}}
= \frac{1 + \frac{1}{x_n} - \varphi}{1 + \varphi( 1 + \frac{1}{x_n} ) }
= \frac{(1-\varphi)x_n + 1}{(1+\varphi)x_n + \varphi}\\
&= \frac{1}{\varphi}\frac{(\varphi - \varphi^2)x_n + \varphi}{\varphi^2 x_n + \varphi}
= -\frac{1}{\varphi^2}\frac{x_n - \varphi}{1 + \varphi x_n}
= -\frac{1}{\varphi^2} y_n
\end{align}
$$
Since this is true for all $n \ge 1$, we can solve the recurrence relation and obtain a closed form expression for $y_n$:
$$y_n = \left(-\frac{1}{\varphi^2}\right)^{n-1} y_1 =
\left(-\frac{1}{\varphi^2}\right)^{n-1} \left(\frac{1-\varphi}{1+\varphi}\right)$$
Since $\left| -\frac{1}{\varphi^2} \right| < 1$, we have $y_n \to 0$ as $n \to \infty$. As a result,
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty} \frac{y_n + \varphi}{1 - \varphi y_n} = \frac{0 + \varphi}{1 - \varphi \cdot 0} = \varphi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Proof that no non-trivial integer solution exists? The equation system
$$2ab+2ac+2bc=4a+4b+4c$$
$$2ab+2ac+2bc=abc$$
In short $$2ab+2ac+2bc=4a+4b+4c=abc$$
has, according to Wolfram Alpha, only the trivial integer solution $a=b=c=0$. I would like to prove this. My approach (not sure whether it is helpful) is using the equation $$(a-2)(b-2)(c-2)=abc-2ab-2ac-2bc+4a+4b+4c-8$$ which simplifies to $$(a-2)(b-2)(c-2)=4(a+b+c-2)$$ if we have $2ab+2ac+2bc=abc$, but I do not see how can I make progress.
Motivation : The circumference and the area of a rectangular with sides $a$ and $b$, if we do not consider units has the same value than the area , when $$2a+2b=ab$$ holds. The only solutions in positive integers are $(3,6)$ $(4,4)$ , $(6,3)$, so besides the square with length $4$ only the rectangular with sides $3$ and $6$ has the property that area and length are equal in the described sense.
I wanted to extend this to a cuboid with lengths $a,b,c$ , for which the sum of the lengths of the edges ($4a+4b+4c$) is equal to the surface ($2ab+2ac+2bc$) and the volume ($abc$).
The desired proof would show that no integer cuboid with this proprty exists.
| One has $$(ab+bc+ca)^2 = a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 + c^2a^2 + 2(2ab+2bc+2ca)(\frac{1}{2}(ab+bc+ca))$$
Then $$a^2b^2 + b^2c^2 + c^2a^2 = -(ab+bc+ca)^2$$
So, one gets $ab=bc=ca=ab+bc+ca=0$,or $a=b=c=0$.
EDIT: i make it clear at the last step as @peter's comment:
One gets from $ab=bc=ca=0$ that $a=b=0$, but note that $c=a+b+c=\frac{1}{2}(ab+bc+ca)=0$.
| {
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"url": "https://math.stackexchange.com/questions/2457333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit Evaluation (Conjugate Method)–Further algebraic manipulation? $$\lim_\limits{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I have tried evaluating the above limit by multiplying either both the conjugate of numerator and the denominator with no avail in exiting the indeterminate form.
i.e. $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}$$ and conversely $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
To my suspicions of which–either numerator or denominator–conjugate to multiply by I chose $$\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}$$
This resulted in $$\frac{(\sqrt{6-x}-2)(\sqrt{3-x}-1)}{3-x-1}$$
Is it indeterminate? What is the reason for multiplying by a specific conjugate in a fraction (denominator or numerator) and the reason for the conjugate being either a) denominator b) numerator c) or both?
Am I simply practicing incorrect algebra by rationalizing the expression to:
$$\frac{(6-x)(3-x)-2\sqrt{3}+2x+2}{x-2}$$
Or am I failing to delve further and manipulate the expression out of the indeterminate form?
| The problem with $\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ is that, when you let $x=2$, you get $\dfrac 00$. So we have to assume that $x \ne 2$. This is not necessarily a problem because $\displaystyle \lim_{x \to x_0}f(x)$ does not care about what happens to $f(x)$ at $x=x_0$.
Notice below that a factor of $(2-x)$ appeared in both the numerator and the denominator and was cancelled out. That leaves us with the rational expresson $\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$ which is equal to $\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ for all all $x \ne 2$ and just so happens to be continuous and defined at $x=2$.
So the function $f(x)=\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ has a removable discontinuity at $x=2$. We can remove that discontinuity by defining
$\left. f(2)=\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right|_{x=2}=\dfrac 12$
For all $x\ne 2$ we can say
\begin{align}
\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
&= \left(\frac{\sqrt{6-x}-2}{1} \cdot
\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) \cdot
\left(\frac{1}{{\sqrt{3-x}-1}} \cdot
\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\right) \\
&= \frac{6-x-4}{\sqrt{6-x}+2} \cdot
\frac{\sqrt{3-x}+1}{3-x-1} \\
&= \frac{2-x}{\sqrt{6-x}+2} \cdot
\frac{\sqrt{3-x}+1}{2-x} \\
&= \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} \\
\end{align}
So $\displaystyle \lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}
=\lim_{x \to 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}
= \frac 24 = \frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Sylvester Equation over GF(2) I know that a Sylvester equation $$AX+XB=0$$ has nontrivial solutions exactly when there is a common eigenvalue of $A$ and $-B$. This is because if there is a common eigenvalue $k$, then there exists a column eigenvector $v$ of $A$ and a row eigenvector $w$ of $-B$. Then $vw$ is a nontrivial solution to the equation: $$A(vw)+(vw)B=(Av)w+v(wB)=kvw-kvw=0$$
But what if we are working with $GF(2)$? Suppose the characteristic polynomials of $A$ and $-B$ are both $x^2+x+1$. This polynomial is irreducible over $GF(2)$, so $A$ and $-B$ do not share an eigenvalue in $GF(2)$. But they share an eigenvalue in $\mathbb{C}$, so does this mean there is a nontrivial solution in $GF(2)$ or not? The argument above certainly doesn't work because $vw$ isn't necessarily in $GF(2)$.
| Let us assume that the characteristic polynomials $\chi_A$ and $\chi_{-B}$ have a common irreducible factor $p$. Let $C$ be the companion matrix of $p$. Then $A$ is similar to an upper triangular block matrix with $C$ in the upper left corner and $-B$ is similar to a lower triangular block matrix with $C$ in the upper left corner (other diagonal blocks can be chosen as the companion matrices of the other irreducible factors of the characteristic polynomials.) Hence we have
$$
A=S_A
\begin{pmatrix}
C & \star & \dots & \star \\
0 & \star & \ddots & \vdots \\
\vdots & \ddots & \ddots & \star \\
0 & \dots & 0 & \star
\end{pmatrix}
S_A^{-1}
$$
and
$$
-B=S_B^{-1}
\begin{pmatrix}
C & 0 & \dots & 0 \\
\star & \star & \ddots & \vdots \\
\vdots & \ddots & \ddots & 0 \\
\star & \dots & \star & \star
\end{pmatrix}
S_B
$$
Now choose
$$
X=S_A
\begin{pmatrix}
I & 0 & \dots & 0 \\
0 & 0 & \ddots & \vdots \\
\vdots & \ddots & \ddots & 0 \\
0 & \dots & 0 & 0
\end{pmatrix}
S_B
$$
where $I$ is the identity matrix of the same size as $C$.
Then
$$
AX = X(-B) = S_A
\begin{pmatrix}
C & 0 & \dots & 0 \\
0 & 0 & \ddots & \vdots \\
\vdots & \ddots & \ddots & 0 \\
0 & \dots & 0 & 0
\end{pmatrix}
S_B
$$
Note: We only need the first $d$ columns of $S_A$ and the first $d$ rows of $S_B$, where $d$ is the degree of $p.$ The columns of $S_A$ can be found by choosing one vector $v$ of the kernel of $p(A)$ and use $A^kv,\; k=0,\ldots d-1$ as the first $d$ columns of $S_A.$ (Similar approach for $S_B$, but we must pay attention to get the same $C$ and not its transpose.)
Edit
We have to find $d$ column vectors $v_0,\ldots,v_{d-1}$ such that
$$
A \begin{pmatrix} & & \\ v_0 & \dots & v_{d-1} \\ & & \end{pmatrix}
=
\begin{pmatrix} & & \\ v_0 & \dots & v_{d-1} \\ & & \end{pmatrix} C
$$
and $d$ row vectors $w_0,\ldots,w_{d-1}$ such that
$$
\begin{pmatrix} & w_{0} & \\ \;\; & \vdots & \;\; \\ & w_{d-1} & \end{pmatrix} (-B)
=
C \begin{pmatrix} & w_{0} & \\ \;\; & \vdots & \;\; \\ & w_{d-1} & \end{pmatrix}
$$
Let
$$
C=\begin{pmatrix}
0 & & 0 & -a_0 \\
1 & \ddots & & -a_1 \\
& \ddots & 0 & \vdots \\
0 & & 1 & -a_{d-1}
\end{pmatrix}
$$
where $a_0,\ldots,a_{d-1}$ are the coefficients of $p$, such that
$p(t)=t^d+a_{d-1}t^{d-1}+\ldots+a_0.$
Let $v_0$ be a vector in $\ker(p(A))$ and $v_k=A^kv_0.$ As $p(A)v_0=0,$ it can easily be seen that
$$
A \begin{pmatrix} & & \\ v_0 & \dots & v_{d-1} \\ & & \end{pmatrix}
=
\begin{pmatrix} & & & \\ v_1 & \dots & v_{d-1} & A^dv_0 \\ & & & \end{pmatrix} \\
=
\begin{pmatrix} & & & \\ v_1 & \dots & v_{d-1} & (A^d-p(A))v_0 \\ & & & \end{pmatrix}
=
\begin{pmatrix} & & \\ v_0 & \dots & v_{d-1} \\ & & \end{pmatrix} C
$$
We can also show that the $v_k$ are linearly independent. Assume there was a non-trivial linear combination $\sum_{k=0}^{d-1}\alpha_kv_k=0.$ This is equivalent with the statement that there is a polynomial $q$ of degree less than $d$ with $q(A)v_0=0.$ $q$ and $p$ must be coprime, because $p$ is irreducible. But if $q(A)x_1=0$ and $p(A)x_2=0$ for comprime $p$ and $q$ and $x_1\neq 0$ and $x_2\neq 0,$ then $x_1$ and $x_2$ are linearly independent, see here. We have a contradiction, because in our case $p(A)v_0=q(a)v_0=0.$ There cannot be a non-trivial linear combination $\sum_{k=0}^{d-1}\alpha_kv_k=0,$ hence the vectors $v_0,\ldots, v_{d-1}$ are linearly independent.
We select $w_{d-1}$ from the "left kernel" of $p(-B),$ such that $w_{d-1}\,p(-B)=0.$
Let $w_{k-1}=-w_kB + a_kw_{d-1}$ for $k=d-1,d-2,\ldots,0.$
Then $w_{-1}=w_{d-1}\,p(-B) = 0.$ Example ($d=4$):
\begin{eqnarray*}
w_2 = -w_3B+a_3w_3 &=& w_3(-B+a_3I) \\
w_1 = -w_2B+a_2w_3 &=& w_3((-B+a_3I)(-B)+a_2I) \\
w_0 = -w_1B+a_1w_3 &=& w_3(((-B+a_3I)(-B)+a_2I)(-B)+a_1I) \\
w_{-1}= -w_0B+a_0w_3 &=& w_3((((-B+a_3I)(-B)+a_2I)(-B)+a_1I)(-B)+a_0I) = w_3p(-B) = 0
\end{eqnarray*}
see Horner's method. Then we have
$$
\begin{pmatrix} & w_{0} & \\ \;\; & \vdots & \;\; \\ & w_{d-1} & \end{pmatrix} (-B)
=
\begin{pmatrix}
& -w_0B & \\
& -w_1B & \\
& \vdots & \\
& -w_{d-1}B &
\end{pmatrix} \\
=
\begin{pmatrix}
& -a_0w_{d-1} & \\
& w_0-a_1w_{d-1} & \\
& \vdots & \\
& w_{d-2}-a_{d-1}w_{d-1} &
\end{pmatrix}
=
C \begin{pmatrix} & w_{0} & \\ \;\; & \vdots & \;\; \\ & w_{d-1} & \end{pmatrix}
$$
It can also be shown that the row vectors $w_0,\ldots,w_{d-1}$ are linearly independent, in the same way as we have shown that $v_0,\ldots,v_{d-1}$ are linearly independent. This linear independence guarantees $X\neq 0$ when we set
$$
X=
\begin{pmatrix} & & \\ v_0 & \dots & v_{d-1} \\ & & \end{pmatrix}
\cdot
\begin{pmatrix} & w_{0} & \\ \;\; & \vdots & \;\; \\ & w_{d-1} & \end{pmatrix}
$$
The complete construction works for any field $F.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}$, or is that not enough?
| a,b positive real numbers.
Let $x^2:= a$, $y^2:=b,$ $x,y \gt 0.$
Consider a right triangle with lengths of legs $x,y$,
length of hypotenuse is $(x^2+y^2)^{1/2}.$
Sum of $2$ sides in a triangle is greater than the third side.
(Euclid)
$\Rightarrow:$
$(x^2+y^2)^{1/2} \lt x + y$,
or reverting to $a,b:$
$(a+b)^{1/2} \lt a^{1/2}+b^{1/2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.
| This is a standard example: it's equivalent to
$$(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9$$
which is AM/HM or Cauchy-Schwarz (provided all $a$, $b$, $c$ are non-negative).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Solve differential equation $yy'(yy'-2x)=x^2-2y^2$.
$$yy'(yy'-2x)=x^2-2y^2$$
I've tried to divide by $y^2$ and substitude $\frac{x}{y}=z$, but it led to:
$$y'^2-2zy'=z^2-2$$
$$(y'-z)^2=2(z^2-1)$$
$$|y'-z|=\sqrt{2}\sqrt{z^2-1}$$ hence
$y'=\sqrt{2}\sqrt{z^2-1}+z$ or $y'=-\sqrt{2}\sqrt{z^2-1}+z$ and I don't know any way to integrate it. Is there some hint for the very beginning maybe that I don't know?
| $$yy'(yy'-2x)=x^2-2y^2$$
divide by $y^2$:
$$(y')^2-2\frac{x}{y}y'+(2-(\frac{x}{y})^2)=0$$
$$y'=\frac{x}{y}±\sqrt {2(\frac{x}{y})^2-2}$$
substitute $y=vx$
$$v+x\frac{dv}{dx}=\frac{1}{v}±\sqrt {2(\frac{1}{v})^2-2}$$
$$\frac{dx}{x}=\frac{dv}{\frac{1}{v}-v±\sqrt {2(\frac{1}{v})^2-2}}$$
$$\frac{dx}{x}=\frac{vdv}{1-v^2±\sqrt {2-2v^2}}$$
As, $(1-v^2)>0, v=\sin t$ can be substituted
$$\frac{dx}{x}=\frac{\sin t \cos t \space dt}{\cos^2 t+\sqrt {2} \cos t}$$
$$\frac{dx}{x}=\frac{\sin t \space dt}{\cos t+\sqrt {2}}$$
Integrating on both sides:
$$ln|x|=-ln|\cos t+\sqrt {2}|+c$$
$$x(\cos t+\sqrt {2})=c$$
$$\cos^2 t=(\frac{c}{x}-\sqrt(2))^2$$
$$1-\frac{y}{x}^2=(\frac{c}{x}-\sqrt{2})^2$$
This can be rewritten as:
$$x^2-y^2=(\sqrt{2}\space x-c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2463150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| *
*$2^1+1=3$;
*$2^{2n+1}+1=3m\implies4(3m)-3=2^{2(n+1)+1}+1=3m'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}$? How to find
$$
\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1}?
$$
My try : $$(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=(x-1)$$
So we have :
$$\lim_{x \to 1} \frac{x+x^2+\sqrt{x}-3}{\sqrt[3]{x}-1} \cdot\frac{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}=\lim_{x \to 1}\frac{(x+x^2+\sqrt{x}-3)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}{x-1}$$
Now what ?
| Hint:
$$x+x^{ 2 }+\sqrt { x } -3=x-1+x^{ 2 }-1+\sqrt { x } -1=x-1+\left( x-1 \right) \left( x+1 \right) +\sqrt { x } -1=\left( x-1 \right) \left( 1+x-1 \right) +\sqrt { x } -1=x\left( x-1 \right) +\sqrt { x } -1=\\ =x\left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) +\sqrt { x } -1=\left( \sqrt { x } -1 \right) \left( x\sqrt { x } +x+1 \right) $$
| {
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How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\sqrt{x(x+3)}-6}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)}=\\\lim_{x \to 1} \frac{2(x-3+\sqrt{x(x+3)})}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)} $$
Now what ?
| First factor $x^2+x^2-2=(x-1)(x+2)$. $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}=$$
$$=\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x-1}\lim_{x\to 1}\frac{1}{x+2}=$$
$$=\left(\lim_{x\to 1}\left(\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x+3}-2}{x-1}\right)\right)\cdot \frac{1}{3}$$
Hint: multiply the numerator and denominator of $\frac{\sqrt{x}-1}{x-1}$ by $\sqrt{x}+1$ and of $\frac{\sqrt{x+3}-2}{x-1}$ by $\sqrt{x+3}+2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+z^4$ and the last equation is equivalent to the system of equations $$\begin{cases}
2x=4\lambda x^3 \\
2y=4\lambda y^3 \\
2z=4\lambda z^3
\end{cases}$$
After dividing into $2$ and multiplying to $x,y$ and $z$, respectively we get: $$\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}$$
Considerong the first equation we get two cases: $x=0$ or $1-2\lambda x^2=0$
After that I am stuck. How to rule out or consider each case?
Can anyone demonstrate it clearly?
Would be very thankful for help
| Yeah so you have $x^{2}=\frac{1}{2\lambda}$ which says $x^{4}=\frac{1}{4\lambda^2}$. And you want $x^{4}+y^{4}+z^{4}=1$. So we have $$\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=1\implies \frac{3}{4\lambda^2}=1 \implies\lambda =\frac{\sqrt{3}}{2}$$ Now you have $$x^{2}=\frac{1}{2\lambda}=\frac{1}{2\cdot \frac{\sqrt{3}}{2}} =\frac{1}{\sqrt{3}}$$ Also $y^{2}=z^{2}=\frac{1}{\sqrt{3}}$ and so the maximum value is $\frac{3}{\sqrt{3}}=\sqrt{3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.
| Thanks this answer's here : Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$
Consider
$$
f(x)=[( 1+x^{p+1})^{\frac{1}{p+1}}-(1+x^p)^{\frac1p}]
$$
Then
$$
f'(x)=x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]
$$
If $p>1$ therefore $f'(0)=0$. Therefore
$$
0=f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.
If $0<p<1$ therefore $f'(0)=-\infty$. Therefore
$$
-\infty =f'(0^+)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.
If $p=1$ therefore $f'(0)=-1 $. Therefore
$$
-1 =f'(0)=\lim_{x\to0^+}x^p[( 1+x^{p+1})^{\frac{1}{p+1}-1}-x^{p-1}(1+x^p)^{\frac1p -1}]=
\lim_{t\to\infty}\bigl([( 1+t^{p+1})^{\frac{1}{p+1}}-(1+t^p)^{\frac1p}]\bigr)
$$
with the substitution $t=1/x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help proving Jacobi Symbol property $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$ Using the properties of Legendre symbols, I need to show $\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$
Here is what I have so far
$\left(\frac{-1}{n}\right) = \left(\frac{-1}{p_1}\right)\left(\frac{-1}{p_2}\right) \cdots \left(\frac{-1}{p_k}\right)$ Where $p_1p_2\cdots p_k$ is the prime factorization of $n$. via the definition of Jacobi Symbols.
From the properties of Legendre symbols this is
$(-1)^{\frac{p_1 - 1}{2}}(-1)^{\frac{p_2 -1 }{2}} \cdots (-1)^{\frac{p_k-2}{2}} = (-1)^{\frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2}}$
If I can show that $\frac{n-1}{2} \equiv \frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2} \pmod{2}$. Then I would be there (I think).
I've been struggling with this for about an hour, and then decided to search. I found this question. However I don't understand the Lemma the answerer used in his proof.
Lemma 1
Let $a, b$ be odd integers.
Then $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Proof:
Since $a - 1$ and $b - 1$ are even,
$(a - 1)(b - 1) \equiv 0$ (mod $4$).
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Hence $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Specifically how we go from this (which I understand)
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
To this
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Which I'm sure really obvious, but I just cant see it :(
If I understood that, can I then just use it to finish off the proof by stating that
$\frac{p_1-1}{2}+\frac{p_2-1}{2}+\cdots + \frac{p_k-1}{2} \equiv \frac{n-1}{2}\pmod{2}$ via lemma above, and so it follows that
$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$
Thanks for any help or insight
| You are making hard work of this. The point is to prove that
$$\left(\frac{-1}n\right)=\chi(n)\tag1$$
where
$$\chi(n)=\begin{cases}1&\text{if }n\equiv1\pmod 4,\\
-1&\text{if }n\equiv-1\pmod 4\end{cases}$$
for odd $n$. Now $\chi(mn)=\chi(m)\chi(n)$ and
also $\left(\frac{-1}{mn}\right)=\left(\frac{-1}m\right)
\left(\frac{-1}n\right)$. So if ($1$) holds for $m$ and $n$
it holds for $mn$, and it holds for all odd primes....
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
| Your approach is not correct. You are using the expansions at $0$ of the functions $\cos(\pi x)$ and the $(x+5)^{1/3}$, whereas the limit is for $x$ that goes to $3$.
Note that by letting $x=t+3$, we have that $t\to 0$ and
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{1/3} -2}=\lim_{t\to 0} \frac {\cos (\pi t +3\pi)+1}{(t+8)^{1/3} -2}=\lim_{t\to 0} \frac {-\cos (\pi t)+1}{2\left(\left(1+\frac{t}{8}\right)^{1/3} -1\right)}.$$
Now we can use the "known" expansions at $0$:
$$\cos(\pi t)=1-\frac{(\pi t)^2}{2} +o(t^2)\quad \text{and}\quad
\left(1+\frac{t}{8}\right)^{1/3}=1+\frac{t}{24}+o(t).$$
Hence, the limit becomes
$$\lim_{t\to 0} \frac {-\cos (\pi t)+1}{2\left(\left(1+\frac{t}{8}\right)^{1/3} -1\right)}=\lim_{t\to 0} \frac {\frac{(\pi t)^2}{2} +o(t^2)}{2\left(\frac{t}{24}+o(t)\right)}=\lim_{t\to 0} \frac {\frac{\pi^2 t}{2} +o(t)}{\frac{1}{12}+o(1)}=0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why do I get two different indefinate integrals for the same integrand? (u-substitution) Using u-substitution I'm able to integrate the following two integrands:
$$\int \frac{1}{2x+3}dx = \frac{1}{2}\ln|2x+3|+C$$
$$\int \frac{1}{2x+5}dx = \frac{1}{2}\ln|2x+5|+C$$
Following this pattern I would assume:
$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|2x+4|+C$$
However when I enter this expression into wolfram alpha I get:
$$\int \frac{1}{2x+4}dx = \frac{1}{2}\ln|x+2|+C$$
I can even arrive at this result myself by factoring out the $\frac{1}{2}$ before integration like this:
$$\int \frac{1}{2}\cdot\frac{1}{x+2}dx = \frac{1}{2}\int \frac{1}{x+2}dx$$
But I don't understand how this can be the case? How can two different techniques give me different results? Is there a property of logs that I don't understand?
I would think that
$$\frac{1}{2}\ln|x+2|+C \neq \frac{1}{2}\ln|2x+4|+C$$
Or are they actually equal somehow? PLEASE help me understand what's going on here.
| Because $$\frac{1}{2}\ln|2x+4|+C=\frac{1}{2}\ln|x+2|+\frac{1}{2}\ln2+C,$$
where $\frac{1}{2}\ln2+C$ is a constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could reduce better to show there is a factor of $n+1$ left.
Examples show this is true for small n; for example
$$\binom{9}{6}=\binom{3\cdot 3}{2\cdot 3}=\frac{9\cdot8\cdot7}{3\cdot 2\cdot 1}=4(3\cdot 7)$$
$$\binom{16}{8}=\binom{4\cdot 4}{2\cdot 4}=\frac{16\cdot15\cdot...\cdot 10\cdot 9}{8\cdot 7\cdot...\cdot2\cdot 1}=5(2\cdot 3^2\cdot11\cdot 13)$$
| As counterexamples for every $n=p-1$ for any prime $p$ and considering $a=n+2=p+1$ and any $b$ with $0 \lt b \lt a$
Then $(an)! = (p^2-1)!$ is divisible by $p^{p-1}=p^n$ but not $p^{p}=p^{n+1}$, while $(bn)!(an-bn)!$ is divisible by $p^{b-1}p^{a-b-1}=p^{n}$, so ${an \choose bn}$ is not divisible by $p=n+1$
This gives for example the counterexamples
n a b C(an,bn) n+1
1 3 1 3 2
2 4 1 28 3
2 4 2 70 3
2 4 3 28 3
4 6 1 10626 5
4 6 2 735471 5
4 6 3 2704156 5
4 6 4 735471 5
4 6 5 10626 5
6 8 1 12271512 7
...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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let $ \ \ 0let $ \ \ 0<a \leq b \leq c \in \mathbb{R}$
then prove that :
$$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$
I do not know where to start please help me !
| Your inequality is true!
We need to prove that
$$\frac{a+b+c}{3}-\frac{3abc}{ab+ac+bc}\geq\frac{(c-a)^2}{6c}$$ or
$$\frac{\sum\limits_{cyc}c(a-b)^2}{ab+ac+bc}\geq\frac{(c-a)^2}{2c}$$ or
$$2c(a-b)^2+2a(b-c)^2\geq\left(\frac{ab}{c}+a-b\right)(c-a)^2.$$
Now, by C-S
$$2c(a-b)^2+2a(b-c)^2=2ac\left(\frac{(b-a)^2}{a}+\frac{(c-b)^2}{c}\right)\geq$$
$$\geq\frac{2ac(b-a+c-b)^2}{a+c}=\frac{2ac(c-a)^2}{a+c}.$$
Thus, it remains to prove that
$$2ac\geq(a+c)\left(\frac{ab}{c}+a-b\right)$$ or
$$c^2a+c^2b\geq a^2b+a^2c,$$ which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Self dual functions My book says that $f(x,y,z) = xy + yz + zx$ is self dual function but $(x+y)(y+z)(z+x)$ is not.
I understood how $xy + yz + zx$ is self dual but I think $(x+y)(y+z)(z+x)$ is also self dual because in k-map if we represent $(x+y)(y+z)(z+x)$ then it's minimized SOP is $xy + yz + zx$ and if we take dual of this, then we get original $(x+y)(y+z)(z+x)$. I think self dual is commutative.
Am I correct?
| For all boolean expression $e(x_{1},\ldots,x_{n})$, let us represent the dual expression of $e(x_{1},\ldots,x_{n})$ by $e(x_{1},\ldots,x_{n})^{D}$.
Expression $xy+yz+zx$ is self dual. In effect, in virtue of elementals properties of boolean algebra, the following equalities hold:
\begin{align*}
(x+y)(y+z)(z+x)&=x(y+z)(z+x)+y(y+z)(z+x)\\
&=(xy+xz)(z+x)+(yy+yz)(z+x)\\
&=(xy+xz)z+(xy+xz)x+(y+yz)z+(y+yz)x\\
&=xyz+xzz+xyx+xzx+yz+yzz+yx+yzx\\
&=xyz+xz+xy+xz+yz+yz+yx+yzx\\
&=xz+xy+yz+xyz\\
&=(xy+xyz)+yz+xz\\
&=xy+yz+zx
\end{align*}
Therefore:
\begin{align*}
(xy+yz+zx)^{D}&=(x+y)(y+z)(z+x)\\
&=xy+yz+zx
\end{align*}
Moreover:
\begin{align*}
((x+y)(y+z)(z+x))^{D}&=xy+yz+zx\\
&=(x+y)(y+z)(z+x)
\end{align*}
and so $(x+y)(y+z)(z+x)$ is too self dual. Does this answer your question?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$ exists Prove this limit exists:
$$\lim_{(x,y)\rightarrow (0,3)}\frac{\tan(xy)}{x}=3$$
My work:
Let $\epsilon >0$, $\delta=\cdots$
If $\sqrt{(x-0)^2+(y-3)^3}<\delta$
Then
\begin{align}
& \left|\frac{\tan(xy)}{x}-3\right|=\left|\frac{\tan(xy)-3x}{x}\right| = \left|\frac{\sin(xy)-3x\cos(xy)}{x\cos(xy)}\right| \\[10pt]
= {} & \frac{|\sin(xy)-3x\cos(xy)|}{|x\cos(xy)|} \leq \frac{|\sin(xy)|+3|x||\cos(xy)|}{|x||\cos(xy)|}\leq \frac{1+3|x|}{|x|}
\end{align}
In this step I'm stuck. Can someone help me?
| A not so good solution (which can be made better of course) is
\begin{align*}
\lim_{(x,y)\rightarrow(0,3)}\dfrac{\tan(xy)}{x}&=\lim_{(x,y)\rightarrow(0,3)}\dfrac{1}{x}\left(xy+\dfrac{1}{3}(xy)^{3}+\cdots\right)\\
&=\lim_{(x,y)\rightarrow(0,3)}\left(y+\dfrac{1}{3}x^{2}y^{3}+\cdots\right)\\
&=3.
\end{align*}
If we decide to go with $\epsilon$-argument, a painstaking way would be:
Observe that $\tan u\leq u+\dfrac{1}{3}u^{3}+\dfrac{1}{15}u^{5}$ for all real numbers $u$, just tackle with the function $\varphi(u)=u+3^{-1}u^{3}+15^{-1}u^{5}-\tan u$ and we know that $\varphi'(u)>0$ for $u\ne 0$ by the inequality follows by Mean Value Theorem.
On the other hand, by the similar fashion, we could have $\tan u\geq u+\dfrac{1}{3}u^{3}$ for $u\geq 0$.
Given $\epsilon>0$ for all $|x|^{2}+|y-3|^{2}<\epsilon^{2}$ and $2<y<4$, one see that for $x>0$
\begin{align*}
\frac{\tan(xy)}{x}-3&\leq\frac{xy+3^{-1}(xy)^{3}+5^{-1}(xy)^{5}-3x}{x}\\
&=(y-3)+\frac{1}{3}x^{2}y^{3}+\frac{1}{5}x^{4}y^{5}\\
&\leq\epsilon+\frac{64}{3}\epsilon^{2}+\frac{512}{3}\epsilon^{4}.
\end{align*}
For $x<0$, one do the trick with $\dfrac{\tan(xy)}{x}=\dfrac{\tan(-xy)}{-x}$ and still get the upper bound.
Now we move on to the lower bound. For $x>0$, we have
\begin{align*}
\frac{\tan(xy)}{x}-3\geq\frac{xy+3^{-1}(xy)^{3}-3x}{x}=(y-3)+\frac{1}{3}x^{2}y^{3}\geq-\epsilon+\frac{8}{3}x^{2}\geq-\epsilon,
\end{align*}
the rest should be easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k+1)-10$ is divisible by 18
$4^{k+1}+6(k+1)-10= 4^k \cdot 4 + 6k + 6 - 10$ $$= \color{green}{4^k +6k-10} + 3
\cdot 4^k + 6$$
The first term is divisible by 18 according to the inductive assumption and I have to find a way to manipulate the second term to be divisible by 18.
There is a similar question posted, but the answers for it are substituting another expression $18m$ instead of manipulating $4^{k+1}+6(k+1)-10$. I want to know how to solve this without substituting
| For $n=1$ we have $4^1+6(1)-10=0$, which is divisible by $18$. If there exists $k \in \mathbb{N}$ such that $4^n+6n-10=18k$, for some $n \in \mathbb{N}$, then $$4^{n+1}+6(n+1)-10=4\cdot 4^n+6n+6-10=(4\cdot 4^n+24n-40)-24n+40+6n-4=4\cdot 18k-18n+36=18(k-n+2),$$ and since $m-n+2 \in \mathbb{Z}$ you have that $4^{n+1}+6(n+1)-10$ is divisible by 18.
By induction, the result follows.
| {
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Sketch the contour plot and the graph of the function: $f(x,y) = \sqrt{36 - 9x^2 - 4y^2}$ I've done a bit:
range of function: $[0,6]$. So we need level curves for $k = 0,1,\ldots,6$
$$f(x,y) = k$$
$$\sqrt{36 - 9x^2 - 4y^2} = k$$
$$36 - 9x^2 - 4y^2 = k^2$$
$$36-k^2 = 9x^2 + 4y^2$$
Not really sure how to go further. I tried to divide by $9$ and $4$ to simplify.. but it didn't work out.
| $$36-k^2 = 9x^2+4y^2$$
$$1=\frac{x^2}{\left(\frac{\sqrt{36-k^2}}{3}\right)^2}+\frac{y^2}{\left( \frac{\sqrt{36-k^2}}{2}\right)^2}$$
Notice that $1=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is an equation of an ellipse.
| {
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Prove that $1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}$ is irrational I tried to follow this example to resolve this exercise, this
example to resolve this exercise, however if:
$1+3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=x$,then
$3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}}=(x-1)$, and
$(3^{\frac{1}{5}}+3^{\frac{2}{5}}+3^{\frac{3}{5}}+3^{\frac{4}{5}})^5=3720+2985\cdot 3^{\frac{1}{5}}+2385\cdot3^{\frac{2}{5}}+1905\cdot3^{\frac{3}{5}}+1545\cdot3^{\frac{4}{5}}$
but i cannot factorize the same way since that would result like $1545x+2175+1350\cdot 3^{\frac{1}{5}}+840\cdot3^{\frac{2}{5}}+360\cdot3^{\frac{3}{5}}=(x-1)^5$
| Observe that
$$3^{1/5} x = 3^{1/5} + 3^{2/5} + 3^{3/5} + 3^{4/5} + 3 = x + 2,$$
so by rearranging $x = 2/(3^{1/5} - 1)$.
Now if $x \in \mathbb{Q}$, we would get $3^{1/5} \in \mathbb{Q}$, say $3^{1/5} = a/b$ with $a, b \in \mathbb{Z}^+$ coprime. Then $$3b^5 = a^5,$$ forcing $3 \mid a$, whence $3^5 \mid a^5$, so $3 \mid b$ to get enough factors of $3$ on the LHS. But then $a$ and $b$ have a common factor; contradiction! So $3^{1/5} \notin \mathbb{Q}$ and likewise for $x$.
| {
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Proof by Contradiction: let a,b $\in$Z . Prove that if 3 $\nmid$ a and 3 $\nmid$ b then $3 | a^2 - b^2$ Just wondering if this works for this question, book had a different answer and I couldn't find another answer for the question.
Assume, to the contrary, that 3 | a and 3 | b, then a = 3k, and b = 3x for x,k $\in$ Z, then $a^2 - b^2$ = $(3k)^2 - (3x)^2$ which is equal to $3(3k^2 - 3x^2)$, since $3k^2 - 3x^2$ is an integer, 3 | $a^2 - b^2$ which is a contradiction.
Is this correct?
| If you don't have to use contradiction, here is a direct proof.
*
*$a = 3A+1$ and $b = 3B+1$.
Then $a^2-b^2=(a+b)(a-b)=(a+b)(3A-3B)=3(a+b)(A-B)$.
*$a = 3A+1$ and $b = 3B+2$.
Then $a^2-b^2=(a+b)(a-b)=(3A+3B+3)(a-b)=3(A+B+1)(a-b)$.
The case $a = 3A+2$ and $b = 3B+1$ follows by symmetry.
| {
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Prove that area of ellipse $Ax^2+Bxy+Cy^2=1$ is equal to $\frac{2\pi}{\sqrt{4AC-B^2}}$ Question:
If $B^2-4AC \lt 0$, the equation
$$Ax^2+Bxy+Cy^2=1$$
represents an ellipse. Prove that the area of the ellipse is
$\frac{2\pi}{\sqrt{4AC-B^2}}$.
I know that, if $2a$ and $2b$ are the major and minor axis of the ellipse respectively, its area is $\pi ab$.
I honestly have no idea how to go about this. Any help is appreciated.
| In polar coordinates, the ellipse is given by
$$r^2 (\theta)= \frac1{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}$$
Denote $\Delta = \sqrt{(A-C)^2+B^2}$ and integrate its area as follows
\begin{align}
A=\frac12 \int_0^{2\pi} r^2(\theta)\> d\theta
&= \frac12 \int_0^{2\pi} \frac{1}{A\cos^2\theta +B\cos\theta\sin\theta +C\sin^2\theta}\> d\theta\\
&= \int_0^{2\pi} \frac{1}{(A+C)+(A-C)\cos 2\theta +B\sin2\theta}\> d\theta\\
&= \int_0^{2\pi} \frac{1}{(A+C)+\Delta\cos 2\theta }\>d\theta\\
&= 4\int_0^{\pi/2} \frac{d(\tan\theta)}{(A+C+\Delta)+(A+C-\Delta)\tan^2\theta }\\
&=\frac{2\pi}{\sqrt{(A+C)^2 - \Delta^2}}
=\frac{2\pi}{\sqrt{4AC-B^2}}\\
\end{align}
| {
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Will the series $\sum\limits_{n=1}^\infty \frac{n^2-n-1}{n!}$ converge when it runs from 1 to infinity? $$\sum_{n=1}^\infty \frac{n^2-n-1}{n!}$$
I usually divide by $n$ or $n^2$ but I'm unable to do so because of the factorial term.
| The series evaluation:
\begin{align}
\sum_{n=1}^{\infty} \frac{n^{2} - n - 1}{n!} &= \sum_{n=1}^{\infty} \left(\frac{1}{(n-2)!} - \frac{1}{n!}\right) \\
&= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} - (e^{1} -1) \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} - (e^{1} - 1) = e^{1} - (e^{1} - 1) = 1.
\end{align}
Convergence by ratio test:
$$\lim_{n \to \infty} \frac{(n+1)^2 - (n+1) - 1}{(n+1)!} \cdot \frac{n!}{n^2 - n -1} = \lim_{n \to \infty} \frac{1}{n+1} \cdot \frac{1 + \frac{1}{n} - \frac{1}{n^{2}}}{1 - \frac{1}{n} - \frac{1}{n^{2}}} = 0.$$
Since the limit is less than one the series converges.
| {
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Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator? Would appreciate kindergarten language ;-D
| The OP asked for kindergarten language. At that grade level math rules are stressed. So if you are taking algebra-precalculus and this is your last math course, here are a couple of rules;
$\tag 1 \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} \;\text{ where } b \ne 0 \text{, } d \ne 0$
$\tag 2 x = \frac{x}{1}$
$\tag 3 x^m x^n = x^{m+n} \;\text{ where } x \gt 0 \text{ and both } m \text{ and } n \text{ are integers}$
$\tag 4 x \times 1 = x$
$\tag 5 \frac{x}{x} = 1 \;\text{ where } x \ne 0$
$\tag 6 x^0 = 1 \;\text{ where } x \gt 0$
$\tag 7 (-x) + (+x) = 0 \;\text{ for any } x$
So if for some reason your instructor wants you to get rid of negative exponents, and you see $3^{16} \times 7^{-6}$, you can use these seven rules:
$3^{16} \times 7^{-6} = \frac{3^{16} \times 7^{-6}}{1} \; \text{ by (2)}$
$\frac{3^{16} \times 7^{-6}}{1} = \frac{3^{16} \times 7^{-6}}{1} \times 1 \; \text{ by (4)}$
$\frac{3^{16} \times 7^{-6}}{1} \times 1 = \frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6} \; \text{ by (5)}$
$\frac{3^{16} \times 7^{-6}}{1} \times \frac{7^6}{7^6} = \frac{3^{16} \times 7^{-6} \times 7^{+6}} {1 \times 7^{+6} } \; \text{ by (1)}$
$\frac{3^{16} \times (7^{-6} \times 7^{+6})} {1 \times 7^{+6} } = \frac{3^{16} \times (7^{-6 + +6}) } {7^{+6} } \; \text{ by (3) & (4)}$
$\frac{3^{16} \times (7^{(-6) + (+6)}) } {7^{+6} } = \frac{3^{16} \times (7^{0})} {7^{+6}} = \frac{3^{16} \times 1} {7^{+6}} \; \text{ by (7) & (6)}$
$\frac{3^{16} \times 1} {7^{+6}} = \frac{3^{16}} {7^{6}}\; \text{ by (4)}$
You now have a new rule that you can use:
$\tag 8 x \times y^{-n} = \frac{x}{y^n} \;\text{ where } y \gt 0 \text{ and } n \text{ is any integer}$
| {
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Determining the limit of a function containing $(x-1)^5$ without l'hopital`s rule. Find the limit of $$\begin{equation*}
\lim_{x \rightarrow 0}
\frac{(x-1)^5 + (1 + 5x)}{x^2 + x^5}
\end{equation*}$$
Shall I use the binomial theorem?
Any hint will be appreciated!
| Divide top and bottom by $x^2$ to get $$\lim_{x \to 0} \frac{\frac{1}{x^2}(1+5x) + \frac{1}{x^2}(x-1)^5}{1+x^3}$$
This is a quotient where the denominator has limit $1$, so the limit exists iff the following limit exists, and it has the same value as this limit: $$\lim_{x \to 0} \left[\frac{1}{x^2}(1+5x) + \frac{1}{x^2}(x-1)^5\right]$$
Now this can be done using the binomial theorem; alternatively, substitute $u=x^2$ to get $$\lim_{u \to 0} \left[\frac{(1+5\sqrt{u})+(\sqrt{u}-1)^5}{u} \right]$$
which is the derivative of the function $y \mapsto (1+5 \sqrt{y}) + (\sqrt{y} - 1)^5$ at $y=0$; that derivative is easily evaluated and its limit to $0$ taken, and that limit is manifestly $\infty$.
| {
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Proof by induction that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$ Through Induction I tried to prove that:
$$\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$$
I first let $n=1$, so that on the left hand side and the right hand side you get 1.
Then I tried to prove that this also works when $$n=n+1$$
So the equation then becomes:
$$\sum_{k=1}^{n+1} (-1)^{k-1}k^2 = \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n-1}\frac{(n)(n+1)}{2}$$
As we know that $\sum_{k=1}^n (-1)^{k-1}k^2 = (-1)^{n-1}\frac{(n)(n+1)}{2}$, we can replace the first term after the $=$ with this. The equation becomes:
$$=(-1)^{n-1}\frac{(n)(n+1)}{2} + (-1)^{n-1}\frac{(n)(n+1)}{2}$$
The I subsitute $n+1$ for $n$, so that:
$$=(-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2} + (-1)^{n+1-1}\frac{(n+1)(n+1+1)}{2}$$
$$=(-1)^{n}\frac{(n+1)(n+2)}{2} + (-1)^{n}\frac{(n+1)(n+2)}{2}$$
$$=(-1)^{n}[\frac{(n+1)(n+2)}{2}+\frac{(n+1)(n+2)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+6n+4)}{2}]$$
$$=(-1)^{n}[\frac{(2n^2+2n+4n+4)}{2}]$$
$$=(-1)^{n}[\frac{((n+1)(2n+4))}{2}]$$
But how do I proceed from here?
| If
$$\sum_{k=1}^{n} (-1)^{k-1} \, k^{2} = (-1)^{n-1} \, \frac{n \, (n+1)}{2} = (-1)^{n-1} \, \binom{n+1}{2}$$
then when $n \to n+1$ it is seen that
$$\sum_{k=1}^{n+1} (-1)^{k-1}k^2 = \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n}(n+1)^{2}$$
which leads to
\begin{align}
\sum_{k=1}^{n+1} (-1)^{k-1} \, k^{2} &= \sum_{k=1}^n (-1)^{k-1}k^2 +(-1)^{n}(n+1)^{2} \\
&= (-1)^{n-1} \frac{n (n+1)}{2} + (-1)^{n}(n+1)^{2} \\
&= \frac{(-1)^{n-1}}{2} \, \left( n^{2} + n - 2 (n^{2} + 2n + 1) \right) \\
&= (-1)^{n} \, \frac{(n+1) \, (n+2)}{2} = (-1)^{n} \binom{n+2}{2}
\end{align}
| {
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Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)$.
Thus, $$z^{1/3} = (1351)^{2/3} \left(\cos \left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3}+ \frac{2k\pi}{3} \right] + i\sin\left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3} + \frac{2k\pi} {3}\right]\right)$$ for $k = 0,1,2.$
This does not seem like I am heading in the right direction. Any clues or hints would be greatly appreciated. Thank you!
| I would try to find a cubic root of $26+15\sqrt3$ of the form $a+b\sqrt3$. Since$$(a+b\sqrt3)^3=a^3+9ab^2+(3ab+3b^3)\sqrt3,$$I would try to find numbers $a$ and $b$ such that$$\left\{\begin{array}{l}a^3+9ab^2=26\\3a^2b+3b^2=15\iff(a^2+b)b=5.\end{array}\right.$$Of course, I would try first to find integer solutions. But then$$(a^2+b)b=5\implies a=2\wedge b=1.$$And it turns out that $a=2$ and $b=1$ is indeed a solution.
| {
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Show that $\lim_{n\to\infty}\left(\prod_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$ How can I show
$$\lim_{n\to\infty}\left(\prod\limits_{k=1}^n\frac{2k+1}{3k-2}\right)^{\frac{1}{n}}=\frac{2}{3}$$
with the aid of Gamma function？
| We have
$$ \left( \prod_{k=1}^{n} \frac{2k+1}{3k-2} \right)^{\frac{1}{n}}
= \frac{2}{3} \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}. $$
Therefore the claim follows from
$$ 1
\leq \left( \prod_{k=1}^{n} \left( 1 + \frac{7}{6k-4} \right) \right)^{\frac{1}{n}}
\leq \exp\left(\frac{1}{n}\sum_{k=1}^{n} \frac{7}{6k-4} \right)
\xrightarrow[n\to\infty]{}1. $$
Here we utilized the inequality $1+x \leq e^{x}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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query on finding numeric function of a generating function For the given generating function $A(z)=1/(1-x^3)$ what will be the numeric function..
I can find out for $A(z)=1/(1-x^2)$ but in the first case we get two factor in the denominator $(1-x)$ and $(x^2+x+1)$, now how do we find the numeric function for $1/(x^2+x+1)$?
| We don't need to factorize the denominator. Let's recall the geometric series expansion.
\begin{align*}
\frac{1}{1-y}=1+y+y^2+y^3+\cdots\qquad\qquad |y|<1
\end{align*}
We obtain with $y=x^3$
\begin{align*}
\frac{1}{1-x^3}=1+x^3+x^6+\cdots\qquad\qquad |x|<1
\end{align*}
and the coefficients are $1$ if the powers of $x$ are congruent $0\,\textrm{mod}(3)$ and $0$ otherwise.
| {
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If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:
$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$
The given expression can be rewritten as:
$$a+b+c-ab-ac-bc = S_1-S_2 \tag{1}$$
Then from the given condition:
$$\begin{align}
abc &= 1-(a+b+c)+ab+ac+bc-abc \\
S_3 &= 1-S_1+S_2-S_3\\
S_1-S_2 &= 1-2S_3 \tag{2}
\end{align}$$
Using $AM \ge GM$
$$\frac{1-S_3 -S_3}{3} \ge\sqrt[3]{S_3^2}$$
for extreme value I set them equal to get:
$$\begin{align}
\frac{(1-2S_3)^3}{27}-S^2_3 &= 0 \\
8S_3^3 +15S_3^2 +6S_3-1 &=0 \\
(S_3 +1)(8S_3^2+7S_3-1) &= 0\\
(S_3+1)^2\left(S_3-\frac{1}{8}\right) &= 0
\end{align}$$
From here we get for minimum value, $S_3 = \frac{1}{8}$. We ignore $S_3 = -1$ because all $a,b,c \ge 0$.
So we can say:
$$\min \{1-2S_3\} = \frac{3}{4}$$
Q. Is there error in any argument (highly probable)
Q. Please provide a better method for inequality, if possible!
NOTE :
I hope you understand my terminology, cannot put it in a better way:
$$S_i = \sum_{cyc} a_1 a_2 a_3 ... ('i'\text{ factors})$$
| $$a(1-c)+b(1-a)+c(1-b)=1-2abc\ge \dfrac{3}{4}$$
$$(abc)^2=(a(1-a))\cdot (b(1-b))\cdot(c(1-c))\le \dfrac{1}{4^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a sequence of reciprocals of square of odd natural numbers If $$a_{n}=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{(2n+1)^2}$$where $n\in N$.
Then prove that
$a_{n}<\frac{1}{4}$
| Here's a sketch of a direct proof that only requires calculus. Note that for each $k$, we have $$\frac{1}{(2k + 1)^2} \leq \int_{k-1}^k \frac{1}{(2x + 1)^2}\,dx.$$
This implies that each $a_n$ is less than $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \int_{3}^\infty \frac{1}{(2x + 1)^2}.$$
Calculating this integral directly gives the upper bound of $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{14} \approx .242 < \frac{1}{4}.$$
This method is a good technique for computing an upper bound on a sum (or a corresponding lower bound by altering the integral).
| {
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Expressing ${_3F_2}$ in terms of gamma functions I have a generalized hypergeometric funtion of the following form
$$\,_3F_2\left(\frac{1}{2},\frac{\beta }{2},\frac{1}{4} (\beta +2 p-2);\frac{\beta +1}{2},\frac{1}{4} (\beta +2 p+5);1\right)$$ where $\beta$ can take integer values $2,1,0,-1,\ldots$ and $p = 0$ or, $\frac{1}{2}$. Is it possible to express this function in terms of gamma-functions?
| Since $\beta$ is an integer, I will rename it $n$. So we have:
$$n=2,1,0,-1,-2,\dots$$
And two functions (for each value of $p$):
$$F_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-2}{4} ;\frac{n +1}{2},\frac{n+5}{4};1\right)$$
$$G_n=\,_3F_2\left(\frac{1}{2},\frac{n }{2},\frac{n-1}{4} ;\frac{n +1}{2},\frac{n+6}{4};1\right)$$
Before attempting the simplification for general $n$, I have decided to do some experiments in Mathematica. Here are the results:
$$\begin{array}{c,c,c} n & F_n & G_n \\ 2 & 1 & \frac{\sqrt{\pi } \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{5}{4}\right)}-\frac{4}{3} \\ 1 & \, _3F_2\left(-\frac{1}{4},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1\right) & 1 \\ 0 & 1 & 1 \\ -1 & \infty & \infty \\ -2 & -\frac{1}{3} & \frac{1}{4} \\ -3 & \infty & 0 \\ -4 & -\frac{3}{5} & -\frac{1}{4} \\ -5 & \infty & \infty \\ -6 & -\frac{3}{5} & \infty \\ -7 & \infty & -\frac{1}{6} \\ -8 & -\frac{65}{63} & -\frac{65}{112} \end{array}$$
For all further $n$ the functions either don't exist (which is denoted by $\infty$) or are rational numbers.
Values of $n$ for which $F_n$ diverges are all negative odd numbers (as far as experiments show):
$$n=-1,-3,-5,\dots$$
Values of $n$ for which $G_n$ diverges are:
$$n=-1,-5,-6,-9,-10,-13,-14,-17,-18,-21,-22,\dots$$
Except for $n=-1$ the other numbers seem to have the form $-(4k+1)$ and $-(4k+2)$. Where $k$ is a positive integer.
My attempts at general simplification (using Euler integral formulas) have not been successful so far. I will edit if I have more results.
Some comments.
Hypergeometric functions with negative integer parameters in the numerator are polynomials (which is why we have rational values) while negative integer parameters in the denominator lead to divergence, unless they are compensated by the numerator. By considering different cases for $n$, it's quite easy to see why the functions dirverge when they do.
$F_n$ diverges for odd negative $n$ because for $n=-2k-1$ we have:
$$F_{-2k-1}=\,_3F_2\left(\frac{1}{2},-k-\frac{1 }{2},\frac{-2k-3}{4} ;-k,\frac{-k+2}{2};1\right)$$
So the negative integer in the demoniators of the series will lead to divergence for any $k$.
On the other hand, let's consider $G_n$ for the values of $n$ where it apparently always diverges:
$$G_{-4k-1}=\,_3F_2\left(\frac{1}{2},-2k-\frac{1 }{2},-k-\frac{1}{2} ;-2k,-k+\frac{5}{4};1\right)$$
$$G_{-4k-2}=\,_3F_2\left(\frac{1}{2},-2k-1,-k-\frac{3}{4} ;-2k-\frac{1}{2},-k+1;1\right)$$
Again, the negative integers in the denominators of the series are not compensated (in the second case since $-2k-1 \neq -k+1$ unless $k=-2$, though we were assuming $k$ to be a positive integer).
| {
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How to solve a quartic equation using trigonometric power identitiy? According to this post, we can solve a cubic equation
$$t^3+pt+q=0$$
by the trigonometric identity
$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$
So I've tried to solve the quartic equation
$$t^4+pt^2+qt+r=0$$
using the identity
$$8\cos^4\theta-4\cos2\theta-\cos4\theta-3=0$$
$$8\cos^4\theta-4(2\cos^2\theta-1)-\cos3\theta\cos\theta+\sin3\theta\sin\theta-3=0$$
$$8\cos^4\theta-8\cos^2\theta-\cos3\theta\cos\theta+1+\sin3\theta\sin\theta=0$$
If we let $t:=A\cos\theta$, then we have
$$A^4\cos^4\theta+A^2p\cos^2\theta+Aq\cos\theta+r=0$$
or
$$8\cos^4\theta+\frac{8p}{A^2}\cos^2\theta+\frac{8q}{A^3}\cos\theta+\frac{8r}{A^4}=0$$
now $\frac{8p}{A^2}=-8$ implies that $A=\sqrt{-p}$ and we need to find $\theta$ in the following system
$$\begin{cases}\cos3\theta=-\frac{8q}{A^3}\\1+\sin3\theta\sin\theta=\frac{8r}{A^4}\end{cases}$$
What would you think on solving this system? Is it possible to solve this system? I couldn't find anything on the web about the solving a quartic equation by trigonometric power identity, so I don't know if this way works ...
| In general, you cannot parameterize a two-dimensional set, like
$$
\left(-\frac{8q}{A^3},\frac{8r}{A^4}\right)
$$
with a one-edimensional curve, like
$$
(\cos3\theta,1+\sin3\theta\sin\theta)
$$
| {
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Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of the numbers) is $[0,16/5]$. I ran across this problem in a book and am stuck on it. Any help would be appreciated.
| If we have
$$
\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}
$$
then all the five numbers are necessarily equal to $2$, as dictated by the AM-QM inequality.
PS. Technically, the power mean inequality is only valid for positive real numbers, but if any of the numbers were negative, then we could change its sign and increase the arithmetic mean without changing the quadratic mean, and the quadratic mean would still be larger. So if the two means are equal, even if allowing for negative numbers, we still get $a = b = c = d = e = 2$.
| {
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To find formal products in given expressions For each of the following expressions, list the list of all formal products in which exponents sum to 4.
(a)$(1+x+x^{2})^{2} (1+x)^{2}$
(b) $(1+x+x^{2}+x^{3}+x^{4})^{3}$
(c)$(1+x^{2}+x^{4})^{2} (1+x+x^{2})^{2}$
(d)$(1+x+x^{2}+x^{3}+...)$
The answers are
. (a) 7 products—$xxxx$,$x^{3}11x$, $x^{3}1x1$,$x^{3}x11$, $1x ^{3}1x$, $1x^{ 3}x1$, $xx^ {3}11$,
(b) 5 products—$1x ^{4}$, $xx^{ 3}$, $x^{ 2}x^{ 2}$, $x ^{3}x$, $x ^{4}$,
(c) 7 products
(d) 15 products—$x ^{4}11$,$ x^{ 3}x1$, $x^{ 3}1x$, $x ^{2}x^{ 2}1$, $x ^{2}xx$, $x ^{2}1x^{ 2}$,$ xx ^{3}1$, $xx ^{2}x$, $xxx ^{2}$,$x^{1}x ^{3}$, $1x^{ 4}1$, $1x ^{3}x$, $1x ^{2}x^{ 2}$,$ 1xx ^{3}$, $11x^{ 4}$
But I need the method to find formal products as I just know if $i$th polynomial factor contains $r_{i}$ different terms and there are n factors then there will be $r_1×r_2×...r_n$ formal products.
Please help
| Assuming that a "formal product" is an unsimplified term in the expanded product, then here is how one should reason about (a).
The expression in (a) is $(1 + x + x^2)(1 + x + x^2)(1 + x)(1 + x)$. A formal product is formed by picking a single term from each factor and then multiplying these terms together. For example $1 \cdot 1 \cdot 1 \cdot 1$ is the formal product formed by choosing $1$ from each factor. We would like to write down all of the formal products with exponents summing to $4$. There are $3$ choices for the first factor. Namely, $1$, $x$, or $x^2$.
*
*If we choose $1$ from the first factor then we $\textit{must}$ choose $x^2$ from the second factor and $x$ from the third and fourth factors. All other choices result in an exponent sum less than $4$. Thus, one of the formal products we seek is $1\cdot x^2\cdot x \cdot x$.
*If we choose $x$ from the first factor then we must either choose $x$ or $x^2$ from the second, otherwise we can't get a high enough exponent. Choosing $x$ from the second means we also need $x$ from the third and fourth. Choosing $x^2$ from the second means we need $1$ from the third and $x$ from the fourth, or vice versa. This gives us three more formal products: $x\cdot x \cdot x \cdot x$ and $x \cdot x^2 \cdot 1 \cdot x$ and $x \cdot x^2 \cdot x \cdot 1$.
*If we choose $x^2$ from the first factor then we are free to choose anything from the second. Choosing $1$ means that we need $x$ from the third and fourth. Choosing $x$ means we need $1$ from the third and $x$ from the fourth, or vice versa. Finally, choosing $x^2$ means we need $1$ from the third and fourth. This gives us four more formal products: $x^2\cdot 1 \cdot x \cdot x$ and $x^2 \cdot x \cdot 1 \cdot x$ and $x^2 \cdot x \cdot x \cdot 1$ and $x^2 \cdot x^2 \cdot 1 \cdot 1$.
So, there are $8$ formal products with exponents summing to $4$ in (a). The others can be done in a similar fashion.
| {
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Let $p(x)$ be a real $7$ degree polynomial with $p(\pi)=\sqrt 3$ and $\int_{-\pi}^{\pi}x^k p(x)=0$ for $0\le k\le 6$. Find $p(0)$ and $p(-\pi)$.
Q.Let $p(x)$ be a real $7$ degree polynomial with $p(\pi)=\sqrt 3$ and $\int_{-\pi}^{\pi}x^k p(x)=0$ for $0\le k\le 6$. Find $p(0)$ and $p(-\pi)$.
Let $p(x)=a_0 + a_1 x+ a_2 x^2+ a_3 x^3+...+a_7 x^7.$
Then we can approach this problem by evaluating the definite integral given in question for each $k$, therefore giving us $7$ equations in $8$ variables $a_0, a_1,...,a_7$. We get eighth equation by using given condition $p(\pi)=\sqrt 3$.
This way we have $8$ equations with $8$ unknowns. We form following system of equations,
$$
\left(\begin{matrix}
1 & 0 & \frac {{\pi}^2}3 & 0 & \frac {{\pi}^4}5 & 0 & \frac {{\pi}^6}7 & 0 \\
0 & \frac 13 & 0 & \frac {{\pi}^2}5 & 0 & \frac {{\pi}^4}7 & 0 & \frac {{\pi}^6}9 \\
\frac 13 & 0 & \frac {{\pi}^2}5 & 0 & \frac {{\pi}^4}7 & 0 & \frac {{\pi}^6}9 & 0 \\
0 & \frac 15 & 0 & \frac {{\pi}^2}7 & 0 & \frac {{\pi}^4}9 & 0 & \frac {{\pi}^6}{11} \\
\frac 15 & 0 & \frac {{\pi}^2}7 & 0 & \frac {{\pi}^4}9 & 0 & \frac {{\pi}^6}{11} & 0 \\
0 & \frac 17 & 0 & \frac {{\pi}^2}9 & 0 & \frac {{\pi}^4}{11} & 0 & \frac {{\pi}^6}{13} \\
\frac 17 & 0 & \frac {{\pi}^2}9 & 0 & \frac {{\pi}^4}{11} & 0 & \frac {{\pi}^6}{13} & 0 \\
1 & \pi & \pi^2 & \pi^3 & \pi^4 & \pi^5 & \pi^6 & \pi^7
\end{matrix}\right)
\cdot
\left(\begin{matrix}
a_0 \\
a_1 \\
a_2 \\
a_3 \\
a_4 \\
a_5 \\
a_6 \\
a_7 \\
\end{matrix}\right)=
\left(\begin{matrix}
0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
\sqrt 3 \\
\end{matrix}\right)$$
If we solve this system of equations, then we get the values of $a_i$s. This way we have identified the polynomial $p(x)$. Once we have identified the polynomial it is easy to calculate $p(0)$ and $p(-\pi)$. Note that $p(0)=a_0$.
But this approach is very much lengthy. Is there a faster way?
| One way.
Consider the vector space $V$ of polynomials of degree $\le7$. We know from linear algebra that $\dim V=8$ and $\{1,x,x^2,\ldots,x^7\}$ is a basis.
We also know that the subspace $U$ spanned by $\{1,x,\ldots,x^6\}$ has dimension seven. We further know that
$$
(f,g)=\int_{-\pi}^\pi f(x)g(x)\,dx
$$
is an inner product on $V$.
Given all this, the conditions $(p(x),x^k)=0$, $k=0,1,2\ldots,6$, define the orthogonal complement
$$U^\perp =\{f(x)\in V\mid (f,g)=0\,\forall g\in U\}$$ of $U$ in $V$. Complementary subspaces have complementary dimensions, so $\dim U^\perp=8-7=1$. Thus $U^\perp$ is spanned by a single polynomial $p(x)$. Because $p(x)\notin U$, we know that $\deg p(x)=7$.
Drums, please.
But if $f(x)\in U^\perp$ so is $f(-x)$. This is because the substitution $x\mapsto -x$ gives us
$$(f(-x),x^k)=(f(x),(-x)^k)=(-1)^k(f(x),x^k)=0$$
for all $k=0,1,2,\ldots,6$.
So $p(-x)\in U^\perp$ and therefore $p(-x)=\lambda p(x)$ for some constant $\lambda\neq0$. A comparison of the leading coefficients of $p(x)$ and $p(-x)$ then leaves $\lambda=-1$ as the only possibility.
Therefore any polynomial $p(x)\in U^\perp$ is odd. That is, $p(-x)=-p(x)$ for all $x$. Consequently $p(0)=0$ and $p(-\pi)=-p(\pi)$.
| {
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Find the minimum of the value $f=(1+\sin^2{x})(1+\sin^2{y})$ Let $\tan{x}\tan{y}=-\frac{1}{2}$,find the minumin of the value $$f=(1+\sin^2{x})(1+\sin^2{y})$$
My ugly solution:
$$f=\dfrac{1+\sin^2{x}}{\sin^2{x}+\cos^2{x}}\cdot\dfrac{1+\sin^2{y}}{\sin^2{y}+\cos^2{y}}=\dfrac{2m^2+1}{m^2+1}\cdot\dfrac{2n^2+1}{n^2+1}$$where $m=\tan{x},n=\tan{y},mn=-\frac{1}{2}$,so
$$f=2\cdot\dfrac{4m^4+4m^2+1}{4m^4+5m^2+1}=2\left(1-\dfrac{m^2}{4m^4+1+5m^2}\right)\ge 2\left(1-\dfrac{m^2}{4m^2+5m^2}\right)=\dfrac{16}{9}$$
some other simple methods?such as AM-GM and Cauchy-Schwarz inequality kill it?
| You have done the hard work.
On rearrangement
$$m^4(8-4f)+m^2(8-5f)+2-f=0$$ which is a Quadratic Equation in $m^2$ which is real
So, the discriminant must be $\ge0$
$$(8-5f)^2-4(8-4f)(2-f)=-16f+9f^2\ge0$$
$\iff f(9f-16)\ge0$
As $f>0, 9f-16\ge0\iff f\ge?$
| {
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Finding Christoffel Symbol
Given a sphere, find the christoffel symbols
$X(\theta,\phi)=(r \sin\phi \cos\theta,r \sin\phi \sin\theta, r \cos \phi)$
So
$$(g_{ij})=\begin{pmatrix} r^2\sin^2\phi& 0\\ 0& r^2 \end{pmatrix}$$
$$(g^{ij})=\frac{1}{r^4 \sin^2 \phi}\begin{pmatrix} r^2& 0\\ 0& r^2\sin^2\phi\end{pmatrix}=\begin{pmatrix} {\frac{1}{r^2 \sin^2 \phi}} & 0\\ 0& \frac{1}{r^2}\end{pmatrix}$$
$$g_{\{ij;1\}}=\begin{pmatrix} 0& 0\\ 0& 0 \end{pmatrix}$$
$$g_{\{ij;2\}}=\begin{pmatrix} 2r^2\sin \phi \cos \phi& 0\\ 0& 0 \end{pmatrix}$$
So $$\Gamma^1_{22}=\frac{1}{2}\frac{1}{r^2\sin^2 \phi}(0+0-0)+\frac{1}{2}\cdot0(0+0-0)=0$$
But the answer is $$\Gamma^1_{22}=-\sin \phi \cos \phi$$
| I think the Christoffel symbol you want is $\Gamma_{\theta\theta}^{\phi}$, which is
$$\begin{align}\Gamma_{\theta\theta}^{\phi}&=\frac{1}{2}g^{\phi k}(\partial_{\theta}g_{\theta k}+\partial_{\theta}g_{\theta k}-\partial_{k}g_{\theta\theta})\\
&=\frac{1}{2}g^{\phi\phi}(\partial_{\theta}g_{\theta\phi}+\partial_{\theta}g_{\theta\phi}-\partial_{\phi}g_{\theta\theta})\\
&=\frac{1}{2}g^{\phi\phi}(0+0-\partial_{\phi}g_{\theta\theta})\\
&=-\frac{1}{2}\cdot\frac{1}{r^2}\cdot2r^2\sin\phi\cos\phi\\
&=-\sin\phi\cos\phi
\end{align}$$
| {
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Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
$$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty
$$
which evaluates to be $\infty$.
| put $x = \tan \theta$ given integral is
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec^2 \theta}{\tan \theta \sec^2 \theta} = \frac{d(\sin \theta)}{\sin \theta} = \ln [|\sin \theta|]^{\frac{\pi}{2}}_{\frac{\pi}{4}} = 0-\ln \frac{1}{\sqrt{2}}=\ln \sqrt{2}$$
| {
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If $a,b,c$ are three complex numbers Find possible values of $\lvert a+b+c \rvert$ Given three complex numbers $a,b,c$ such that $\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1$ and $$\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1$$
Find which of the following are possible values of $\lvert a+b+c \rvert$
A)0
B)2
C)1.5
D)3
My try:
I assumed $a=e^{ix}$,$b=e^{iy}$, $c=e^{iz}$
Then we have
$$\cos (2x-y-z)+\cos (2y-x-z)+\cos (2z-x-y)=-1$$
$$\sin(2x-y-z)+\sin (2y-x-z)+\sin (2z-x-y)=0$$
Squaring and adding we get
$$3+2(\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=1$$
So
$$\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=-1$$
any clue here?
| The four complex numbers$\frac{a^2}{bc},\frac{b^2}{ac},\frac{c^2}{ab},1$ all have length $1$ and sum to zero. Added as vectors,'head to tail', in the Argand diagram they form a closed shape and must consist of two pairs of equal but opposite vectors.
Without loss of generality we therefore have $\frac{a^2}{bc}=-\frac{b^2}{ac},\frac{c^2}{ab}=-1$ and then $a^3=-b^3,ab=-c^2$. Write $c$ as $wa$, then $b=-w^2a$, where $w$ must be a 6th root of unity.
Then $|a+b+c|=|1+w-w^2|$ is $2$ if $w$ is a primitive 3rd or 6th root and is $1$ if $w$ is $1$ or $-1.$ The answer is B.
| {
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Sketching $(x^2-2)^2+(y^2-2)^2=4$ I was recently asked to sketch $(x^2-2)^2+(y^2-2)^2=2$, which did not prove to be too problematic, for establishing the range and domain of the expression gives nearly all of it away.
I then asked myself what would happen as I change the constant on the R.H.S. I recommend you try this yourself:
https://www.desmos.com/calculator/tlksmrpzxu
The constant that stood out as yielding the most intriguing result is $4$. The graph looks like this:
I then wondered how on earth I could deduce this just by analysing $(x^2-2)^2+(y^2-2)^2=4$.
One of my approaches involved spotting that the two 'orbits' are in the shape of ellipses with equations (derived partly experimentally, partly using the ellipse formula) $x^2-xy\sqrt{2}+y^2=2$ and $x^2+xy\sqrt{2}+y^2=2$. Can I derive these just from $(x^2-2)^2+(y^2-2)^2=4$?
| Well, you can go backwards, and multiply $x^2 - \sqrt2xy+y^2-2$ and $x^2 + \sqrt2xy+y^2-2$ to get the product $x^4-4x^2+y^4-4y^2 + 4$, equal to $(x^4-4x^2+4) + (y^4-4y^2+4)-4$, but I did not see the reverse path when I looked at your question.
| {
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Range of root of cubic equation having least absolute value Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equations $x^3-ax+b=0$ are all real. Let $\alpha$ is a root of this cubic equation with minimum absolute value. Then range of values of $\alpha$ is given by(choose the correct option)
$(A)$ $\frac{2b}{3a}<\alpha \leq \frac{b}{a}$
$(B)$ $\frac{b}{a}<\alpha \leq \frac{3b}{2a}$
$(C)$ $-\frac{b}{a}<\alpha \leq \frac{b}{a}$
$(D)$ $-\frac{b}{a}<\alpha \leq \frac{2b}{3a}$
This question has been asked on this site but answers use Viete's theorem which I am not familiar with.
Now as product of roots is negative, hence we will have either three roots negative or two roots positive and one negative but as sum of roots is $0$, hence we wil have two roots positive and one root negative.
Then I tried finding points of $f'(x)=0$ for $f(x)=x^3-ax+b$ to get $x=\pm \frac{\sqrt{a}}{\sqrt{3}}$ out of which one gives local maxima and other gives local minima. I drew the rough graph and I think the first positive root will be $\alpha$ but how to get desired range?
| Let $\alpha,\beta,\gamma$ be the solutions of $x^3-ax+b=0$ with $|\alpha|\le|\beta|$ and $|\alpha|\le|\gamma|$.
By Vieta's formula,
$$\alpha+\beta+\gamma=0,\quad \alpha\beta+\beta\gamma+\gamma\alpha=-a\lt 0,\quad \alpha\beta\gamma=-b\lt 0$$
Since $\alpha\beta\gamma\lt 0$, we have either that only one of them is negative or that all are negative.
It is impossible that all three are negative because of $\alpha\beta+\beta\gamma+\gamma\alpha\lt 0$.
So, we see that only one of them is negative.
Suppose here that $\alpha\lt 0\lt\beta\le\gamma$ with $|\alpha|\le|\beta|$.
Then, since $\gamma=-\alpha-\beta$, we have
$$\alpha\lt 0\lt\beta\le -\frac{\alpha}{2}\implies |\alpha|\le|\beta|\le \left|\frac{\alpha}{2}\right|$$which is impossible.
So, we may suppose that $\beta\lt 0\lt\alpha\le \gamma$ with $|\alpha|\le|\beta|$.
Since $\gamma=-\alpha-\beta$, we have
$$\beta\lt 0\lt\alpha\le -\frac{\beta}{2}\tag1$$
Now, since we have $$a=\alpha^2+\alpha\beta+\beta^2\quad\text{and}\quad b=\alpha\beta(\alpha+\beta)$$
and
$$(1)\implies \beta\lt\alpha\le-\frac{\beta}{2}\implies (\alpha-\beta)(2\alpha+\beta)\le 0\implies 2\alpha^2\le \alpha\beta+\beta^2$$
we have
$$\alpha=\alpha\times 1\gt \alpha\times \frac{(\alpha\beta+\beta^2)}{\alpha^2+(\alpha\beta+\beta^2)}=\frac{\alpha\beta(\alpha+\beta)}{\alpha^2+\alpha\beta+\beta^2}=\frac ba$$
$$\alpha=\alpha\times 1\le \alpha\times\frac{\alpha\beta+\beta^2+(2\alpha\beta+2\beta^2)}{2\alpha^2+(2\alpha\beta+2\beta^2)}=\frac{3\alpha\beta(\alpha+\beta)}{2(\alpha^2+\alpha\beta+\beta^2)}=\frac{3b}{2a}$$
$$\alpha=\alpha\times 1\gt \alpha\times\frac{(2\alpha\beta+2\beta^2)}{3\alpha^2+\alpha\beta+\beta^2+(2\alpha\beta+2\beta^2)}=\frac{2\alpha\beta(\alpha+\beta)}{3(\alpha^2+\alpha\beta+\beta^2)}=\frac{2b}{3a}$$
It follows from these that $\color{red}{(B)}$ is the only correct option.
| {
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"url": "https://math.stackexchange.com/questions/2536975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
My Try :
$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$
So we have :
$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$
now what ?
| from the first equation we have $$\log(x)=\log(2y)$$ and we get $$x=2x$$ plugging this in your second equation we have
$$2^{2y}+8^y=4$$
this can be written as $$(2^y)^2+(2^y)^3=4$$ with $t=2^y$ we have to solve
$$t^3+t^2-4=0$$
one solution is given by $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6
\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$ after the cubic formulas. This is the only real solution
| {
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"url": "https://math.stackexchange.com/questions/2539310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a primitive root of $71$. In my Number Theory Class we found that $7$ was a primitive root of 41 by first finding two integers who have order $5$ and $8$ $modulo 41$ respectively, these being $16$ and $3$. Since $16(3)=7(mod41)$ 7 is a primitive root.
I'm trying to do this for $71$ and so far have that $5$ has order $5(mod71)$.
I'm struggling to find integers whose order $modulo71$ are $2$ and $7$.
| $G=\mathbb{Z}/(71\mathbb{Z})^*$ is a cyclic group with order $70=2\cdot 5\cdot 7$. It follows that any $g\in G$ such that $g^{10}\not\equiv 1\pmod{71}$, $g^{14}\not\equiv 1\pmod{71}$ and $g^{35}\not\equiv 1\pmod{71}$ is a generator for $G$. There are $\varphi(70)=1\cdot 4\cdot 6=24$ generators, so a random non-quadratic residue is very likely to be a generator. $71\equiv -1\pmod{8}$ so $2$ is a quadratic residue. $71\equiv -1\pmod{3}$ so $3$ is a quadratic residue too. $4$ is a square, hence a quadratic residue. $71\equiv 1\pmod{5}$, hence $5$ is also a quadratic residue. $6$ too, since it is the product of two quadratic residues ($2$ and $3$). $7$ is the smallest reasonable candidate for a generator. We have
$$ 7^{35}\equiv \left(\frac{7}{71}\right)\stackrel{\text{reciprocity}}{\equiv}-\left(\frac{1}{7}\right)\equiv -1\pmod{71}, $$
$$ 7^5\equiv 51 \equiv -20\pmod{71},$$
$$ 7^{10}\equiv (-20)^2 \equiv 45\pmod{71}, $$
$$ 7^{14}\equiv (-20)^3\cdot 7^{-1} \equiv (-20)^3\cdot(-10) \equiv 54\pmod{71} $$
hence $7$ is actually a generator.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of series of fractions: $\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$
Find the sum of the series:
$$\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$$
I am not able to understand how to proceed. The numerator terms have nothing in common which might result in an AP or GP.
Please guide me with the approach.
| Observe that the first order differences of the numerators, $4,6,8$ are increasing linearly (and hopefully continue doing so - there are few known terms), so that the general form of the numerators must be quadratic, $N_n:=an^2+bn+c$.
By extrapolation, the zero-th term is $2$, and as the second order difference is $2$, the general expression must be
$$N_n=n^2+bn+2.$$
From $N_1=4$ you draw
$$N_n=n^2+n+2$$
which fits all values.
Now the general term, a polynomial over a factorial
$$\frac{N_n}{D_n}=\frac{n^2+n+2}{n!}$$
is a little embarrassing as we don't have just inverse factorials (as in the development of $e$). Anyway, we observe the simplification
$$\frac n{n!}=\frac1{(n-1)!},$$ and (more tricky), for the quadratic term,
$$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac1{(n-2)!}+\frac1{(n-1)!}.$$
Then we use the decomposition
$$\frac{N_n}{D_n}=\frac{n^2+n+2}{n!}=\frac1{(n-2)!}+\frac2{(n-1)!}+\frac2{n!},$$
to establish
$$\begin{matrix}
e=&&&\color{white}+\frac1{0!}&+\frac1{1!}&+\frac1{2!}&+\cdots\\
2e=&&\color{white}+\frac2{0!}&+\frac2{1!}&+\frac2{2!}&+\frac2{3!}&+\cdots\\
2e=&\frac2{0!}&+\frac2{1!}&+\frac2{2!}&+\frac2{3!}&+\frac2{4!}&+\cdots&&
\\\hline
5e=&\frac2{0!}&+\frac4{1!}&+\frac8{2!}&+\frac{14}{3!}&+\frac{22}{4!}&+\cdots\\
\end{matrix}$$
so that in the end
$$5e=2+S.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Irreducible components of $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.
I am trying to find the irreducible components of $V(I),$ where $I = (x^3 - y^3, x^2 - y^2 + 2x - 2y)$.
So far I've factorized as follows:
$$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2)$$
$$x^3 - y^3 = (x -y)(x - \omega y)(x - \omega^2 y)$$
From this I've got that we must have the following equal to zero:
$$x + y +2, -\omega^2 y + 2 $$
$$x + y + 2, -\omega y+ 2$$
or
$$x-y.$$
However I'm not sure if this is everything I need to do. Do I now just express them as subsets of $V = V(I)$, such that the union of them is equal to $V$?
| We have
$$x^2 - y^2 + 2x - 2y = (x-y)(x + y + 2), \quad x^3 - y^3= (x -y)(x^2+xy+y^2)$$
So, $(x,y) \in V(I)$ iff $x=y$ or $x + y + 2=0=x^2+xy+y^2$.
The last equation has two solutions: $x=-1 \pm \sqrt3i, y=-1 \mp \sqrt3i$.
Therefore, $V(I)$ has three irreducible components: a line and two isolated points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6},\frac{1}{24}, ...?$$
| hint
Let $f (x)=\frac {x^n-1}{x-1}$.
we expand $f (x+1) $ near zero as
$$f (x+1)=((x+1)^n-1)/x=$$
$$(1+nx+\frac {n (n-1)}{2!}x^2+\frac{n (n-1)(n-2)}{3!}x^3+...-1)/x $$
$$=n+\frac {n (n-1)}{2!}x+.... $$
to get the expansion arround $x=1$, replace $x $ by $x-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?
| Let $a^2+b^2=2kab$.
Hence, we need to prove that
$$(4a^2+4b^2-ab)^2\geq900\cdot\frac{a^4+b^4+a^2b^2}{60}$$ or
$$(8k-1)^2\geq15(4k^2-1)$$ or
$$(k-2)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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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.