Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\[6pt]
& \frac{14x^2-64x-9x^2+21x+30}{(x+1)(x-4)} < 0 \\[6pt]
& \frac{5x^2-43x+30}{(x+1)(x-4)}<0
\end{align}
using quadratic formula, roots of $5x^2-43x+30$ comes $7.83$(approx.) and $0.763$(approx.)
so rewriting the equation as
$$\frac{(x-7.83)(x-0.763)}{(x+1)(x-4)}<0$$
I then did the plotting of zeroes and poles on number line for finding the values for $x$ but I donot get 2 integral values (which is the answer). Can anyone tell where I did wrong?
| Inequality is equivalent with $$\frac{(x-1)(x-6)}{(x+1)(x-4)}<0.$$ Function $f(x)=\frac{(x-1)(x-6)}{(x+1)(x-4)}$ has sign $+$ on intervals $(-\infty,-1)$, $(1,4)$, $(6,\infty)$ and sign $-$ on intervals $(-1,1)$ and $(4,6)$.
Therefore, only integer solutions that satisfy inequality are $0$ and $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using substitution while using taylor expansion I am trying to prove to myself why can substitution be used while using taylor expansion. for example: the taylor expansion of $e^{(x-3)^2}$ around $a=3$ and order of $6$ can be done by first evaluating $e^x$ and than plugging in $(x-3)^2$ getting $P_{6}(x)=1+(x-3)^2+\frac{(x-3)^4}{4}+\frac{(x-3)^6}{6}$
We know that $f(x)=P_{n}(x)+R_{n}(x)$ when $lim_{x\to a} \frac{R_n{x}}{(x-a)^n}=0$
So we first use taylor expansion around $a=0$ and of order $k=3$
So we in general we have $f(x)=P_{k}(x)+R_{k}(x)$ now we plug in $(x-3)^2$ and get $f(x)=P_{k*m}(x)+R_{k*m}(x)$ where $m=2$
Because we overall got a polynomial in the form of $f(x)=P_{n}(x)+R_{n}(x)$ the used substitution can be made? is that a valid proof?
| This problem is related to series composition and what you did is almost perfectly correct.
Make a change of variable $(x-3)^2=y$ and use the fact that, around $y=0$ $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)\tag 1$$ Now, you just need to replace $y$ by $(x-3)^2$ to get $$e^{(x-3)^2}=1+(x-3)^2+\frac{1}{2} (x-3)^4+\frac{1}{6} (x-3)^6++O\left((x-3)^{8}\right)$$
Take care : there is a typo in your formula ($\frac 14$ should be $\frac 12$).
The problem would have been different say for $e^{\sin(x)}$. You could start using $(1)$ and write $$e^{\sin(x)}=1+\sin(x)+\frac{\sin^2(x)}{2}+\frac{\sin^3(x)}{6}+O\left(\sin^4(x)\right)$$ and use $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)\tag 2$$ Replacing you would get $$e^{\sin(x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+O\left(x^5\right)$$ The other way, would have been to write $$e^{\sin(x)}=e^{x-\frac{x^3}{6}+O\left(x^5\right)}$$ and use $y=x-\frac{x^3}{6}+O\left(x^5\right)$ in $(1)$. For sure, the result would be the same.
| {
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"timestamp": "2023-03-29T00:00:00",
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About the integral $\int _{\ln 2} ^{\ln 3} \frac { e ^x} {e ^{4x}-1} \,d x$ How do I compute $$\int \limits _{\ln 2} ^{\ln 3} \frac {e ^x} {e ^{4x}-1} \, d x $$
? Thank you so much for your answer.
| $$\int_{\ln(2)}^{\ln(3)}\frac{e^x}{e^{4x}-1}\space\text{d}x=$$
Substitute $u=e^x$ and $\text{d}u=e^x\space\text{d}x$.
This gives a new lower bound $u=e^{\ln(2)}=2$ and upper bound $u=e^{\ln(3)}=3$:
$$\int_{2}^{3}\frac{1}{u^4-1}\space\text{d}u=$$
Use partial fractions:
$$\int_{2}^{3}\left[\frac{1}{4(u-1)}-\frac{1}{4(u+1)}-\frac{1}{2(u^2+1)}\right]\space\text{d}u=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{2}^{3}\frac{1}{u+1}\space\text{d}u-\frac{1}{2}\int_{2}^{3}\frac{1}{u^2+1}\space\text{d}u=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{2}^{3}\frac{1}{u+1}\space\text{d}u-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
For the integral $\int\frac{1}{u+1}\space\text{d}u$:
Substitute $s=u+1$ and $\text{d}s=\text{d}u$.
This gives a new lower bound $s=2+1=3$ and upper bound $s=3+1=4$:
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\int_{3}^{4}\frac{1}{s}\space\text{d}s-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\int_{2}^{3}\frac{1}{u-1}\space\text{d}u-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
Substitute $p=u-1$ and $\text{d}p=\text{d}u$.
This gives a new lower bound $p=2-1=1$ and upper bound $p=3-1=2$:
$$\frac{1}{4}\int_{1}^{2}\frac{1}{p}\space\text{d}p-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\left[\ln|p|\right]_1^2-\frac{1}{4}\left[\ln|s|\right]_3^4-\frac{1}{2}\left[\arctan(u)\right]_2^3=$$
$$\frac{1}{4}\left(\ln|2|-\ln|1|\right)-\frac{1}{4}\left(\ln|4|-\ln|3|\right)-\frac{1}{2}\left(\arctan(3)-\arctan(2)\right)=$$
$$\frac{1}{4}\left(\ln|2|-0\right)-\frac{1}{4}\left(\ln|4|-\ln|3|\right)-\frac{1}{2}\left(\arctan(3)-\arctan(2)\right)=$$
$$\frac{\ln(2)-\ln(4)+\ln(3)}{4}-\frac{\arctan(3)-\arctan(2)}{2}\approx0.03041774972495909$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Anti-associativity of product of sum of squares $\newcommand{\P}{\mathbb{P}}$$\newcommand{\Z}{\mathbb{Z}}$ Let $\P$ be the set of prime numbers congruent to $1 \pmod 4$. I know that for every $p \in \P$ there's a unique couple $(a^2,b^2)\in \Z^2$ such that $p= a^2+b^2$ and $a < b \in \Z$. Moreover, if we take $p,q \in \P$ with $p=a^2+b^2$, $q=c^2+b^2$, we'll have $pq=(ac+bd)^2+(ad-bc)^2$. If we take $p,q,r \in \P$ different from each other, we have $2$ different associations: we can write $pq$ as a sum of squares and then, multiplying by $r$, we obtain a sum of squares; but we can also start with $qr$ and then multiply by $p$ (written as sums of squares) and possibly get different squares. For example,
\begin{align}
5 \cdot 13 \cdot 17 &= [(1^2+2^2)(2^2+3^2)](1^2+4^2) = (1^2+8^2)(1^2+4^2) = 4^2+ 33^2 \\
&=(1^2+2^2)[(2^2+3^2)(1^2+4^2)] = (1^2+2^2)(5^2+14^2) = 24^2+ 23^2
\end{align}
My question is, if we start with $n$ different primes of $\P$, is it always true that every different association of those primes gives different squares?
| I address a differently phrased question, but it should provide the answer to yours.
Suppose the positive integer $z$ is a product of primes $p_j$ all congruent to $1$ modulo $4$. Write $z = a^ 2 + b^2$, with $a, b$ positive, $a < b$. Then $$\tag{dec}z = (a + i b) (a - i b)$$ is the product of two conjugate Gaussian integers.
Now if $p_j = a_j^2 + b_j^2 = (a_j + i b_j)(a_j - i b_j)$ is the decomposition of $p_j$ as the product of two Gaussian primes, you obtain a decomposition (dec) by choosing, for each $j$, one of $a_j \pm i b_j$.
For instance $5 = 1^2 + 2^2$ and $13 = 2^2 + 3^2$, so we get
$$
5 \cdot 13 = (1 + 2 i) (2 + 3 i) \cdot (1 - 2 i) (2 - 3 i),
= (-4 + 7 i) (-4 - 7 i)
$$
which yields $65 = 4^2 + 7^2$, or
$$
5 \cdot 13 = (1 + 2 i) (2 - 3 i) \cdot (1 - 2 i) (2 + 3 i) = (8 - i) (8 + i),
$$
which yields $65 = 8^2 + 1^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to show that any group of order four is either cyclic or isomorphic to V I know the question has already been asked. But I have trouble with the answer.
Having a non-cyclic group $\,G=\{1,a,b,c\}\,$, how can I show that $ab=c$?
In my attempt, I assume that $ab=1$, and then $c^2=1$. And I see no problem with that. What am I missing?
UPD: is it possible to do without Lagrange's theorem?
| We surely have $a^2\ne a$. So $a^2=1$, $a^2=b$ or $a^2=c$. If $a^2\ne1$, it's not restrictive to assume $a^2=b$. So we can start building the Cayley table:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & & \\
b & b & & & \\
c & c & & &
\end{array}
$$
We can see that $ab=c$ (otherwise $ab=1$ and we'd remain with $ac=c$, a contradiction); thus $ac=1$. Similarly, $ba=c$ and $ca=1$.
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & c & 1 \\
b & b & c & & \\
c & c & 1 & &
\end{array}
$$
Now we are forced to $b^2=1$ and $bc=a$; similarly, $cb=a$; finally, $c^2=b$:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & b & c & 1 \\
b & b & c & 1 & a \\
c & c & 1 & a & b
\end{array}
$$
Thus we see that $a^0=1$, $a^2=b$, $a^3=c$; hence the group is cyclic.
In order not to have a cyclic group we therefore need $a^2=1$:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & & \\
b & b & & & \\
c & c & & &
\end{array}
$$
We then must have $ab=c$ and $ac=b$; similarly, $ba=c$ and $ca=b$:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & & \\
c & c & b & &
\end{array}
$$
If $b^2=a$, we'd be in the same situation as before, because then $b^3=c$ and the group would be cyclic. So, for a non cyclic group we must have $b^2=1$ and $bc=a$. Now we can complete the Cayley table:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & 1 & a \\
c & c & b & a & 1
\end{array}
$$
which shows the group is the Klein group.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
What is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$, if $\sin(\alpha+\beta) = 1$ and $\sin(\alpha-\beta) = \frac 12$ If $\sin(\alpha+\beta) = 1$, $\sin(\alpha-\beta) = \frac 12$ , where $\alpha$, $\beta$ belongs to $[0,\frac \pi2],$ then what is $\tan(\alpha+2\beta)\tan(2\alpha+\beta)$?
What should be my approach to the question?
| Since
\begin{align}
0\leq \alpha \leq \frac{\pi}{2}, ~~~~~~~~~~~~~~0\leq \beta \leq \frac{\pi}{2},
\end{align}
we have
\begin{align}
0\leq \alpha+\beta \leq \pi,~~~~~~~~~~~~-\frac{\pi}{2}\leq \alpha-\beta \leq \frac{\pi}{2}.
\end{align}
Thus, from the equations given, we find that
\begin{align}
\sin(\alpha+\beta)&=1\Rightarrow \alpha+\beta=\frac{\pi}{2},\\
\sin(\alpha-\beta)&=1/2\Rightarrow \alpha-\beta=\frac{\pi}{6}.
\end{align}
Solving, we get $\alpha=\frac{\pi}{3}$ and $\beta=\frac{\pi}{6}$. Thus,
\begin{align}
\tan(2\alpha+\beta)\tan(\alpha+2\beta)=\left(-\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)=1.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find one-sided limits *algebraically*? Find $$\lim_{x\to\ -0.5^-}\sqrt{\frac{x+2}{x+1}}$$
Sorry, I have no idea where to start. I know how to find regular limits algebraically, but not one-sided.
Thanks
| *
*$$\lim_{x \to -0.5} f(x) = \lim_{x \to -0.5^{-1}} f(x)$$
if $f$ is continuous at $x=-0.5$
Is $$\sqrt{\frac{x+2}{x+1}}$$ continuous at $x=-0.5$?
What we need to check is if $$\lim_{x \to -0.5} \sqrt{x} = \lim_{x \to -0.5^{-1}} \sqrt{x}$$
and if $$\lim_{x \to 0-.5} \frac{x+2}{x+1} = \lim_{x \to -0.5^{-1}} \frac{x+2}{x+1}$$
It turns out both are true because
*
*$\sqrt{x}$ is continuous at all points $x > 0$ and $\frac{x+2}{x+1}|_{x = -0.5} > 0$.
*$\frac{x+2}{x+1}$ is continuous at all points $x \ne -1$
So this what we have:
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \sqrt{\lim_{x \to 0.5^{-1}}\frac{x+2}{x+1}} \ \text{by continuity of} \ \frac{x+2}{x+1}$$
$$\sqrt{\lim_{x \to 0.5}\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
$$\lim_{x \to 0.5^{-1}}\sqrt{\frac{x+2}{x+1}} = \lim_{x \to 0.5}\sqrt{\frac{x+2}{x+1}} \ \text{by continuity of} \ \sqrt{x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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solving trigonometry equation $20\cot\theta + 15\cot\theta\operatorname{cosec}\theta - 4\operatorname{cosec}\theta = 3(1 + \cot^2 \theta) ?$ how do you solve for $0 < \theta < 360$:
$$20\cot\theta + 15\cot\theta\operatorname{cosec}\theta - 4\operatorname{cosec}\theta = 3(1 + \cot^2 \theta) ?$$
I tried turning the $(1 + \cot^2 \theta)$ part into $\operatorname{cosec}^2$ then dividing through or changing every term into $\sin$ and $\cos$ but I always end up with a dodgy equation that I can't factorise.
| Let try to convert the equation to $\sin$ and $\cos$.
$$20\frac{\cos \theta}{\sin\theta} + 15 \frac{\cos \theta}{\sin\theta}\frac{1}{\sin\theta} - 4\frac{1}{\sin\theta} = 3\frac{1}{\sin^2\theta}$$
$$20\cos\theta\sin\theta + 15\cos\theta-4\sin\theta = 3$$
$$(4\sin\theta+3)(5\cos\theta-1) = 0$$
Now, you can solve the equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Calculate limit for, $\lim\limits_{x\to 0}\frac{1-cos(x^6)}{x^{12}}$, but in there have suprize. Let's think about this function, $\quad \to f(x)=\dfrac{x^2-1}{x-1}$,
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=0/0$ ,
First Solution :
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}=\lim\limits_{x\to 1}\dfrac{x+1}{1}$
$=\lim\limits_{x\to 1}\dfrac{x+1}{1}=\dfrac{1+1}{1}=2$
And Second Solution with l'hopital:
$\lim\limits_{x\to 1}\dfrac{x^2-1}{x-1}\overbrace{\longrightarrow}^{l'hopital}\lim\limits_{x\to 1}\dfrac{2x}{1}=2$
But...
Let's we take this function ,$f(x)=\dfrac{1-cos(x^6)}{x^{12}}$
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0/0$,
For this funciton couldn't simplification,I have just , graph of function and l'hôpital,
1-L'hôpital;
$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=\lim\limits_{x\to 0}\dfrac{6.x^5.sin(x^6)}{12.x^{11}}=\underbrace{\lim\limits_{x\to 0}\dfrac{sin(x^6)}{x^6}}_1.\dfrac{1}{2}=\dfrac{1}{2}$
2-Graph of function;
Graph telling me ,$\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=0$
l'hôpital $\lim\limits_{x\to 0}\dfrac{1-cos(x^6)}{x^{12}}=1/2$
What we do now?
And link of graph , if you can't see clearly,https://www.desmos.com/calculator/jfz4kslm4w
| Setting $x^6=2y$ and using $\cos2y=1-2\sin^2y$ to find $$\lim_{y\to0^+}\dfrac{1-\cos2y}{(2y)^2}=\dfrac24\left(\lim_{y\to0^+}\dfrac{\sin y}y\right)^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$ I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.
Thanks!
| $$\begin{align}
\tan 70-\tan 20-2\tan 40 &= \left( \tan { 70-\tan { 40 } } \right) -\left( \tan { 40+\tan { 20 } } \right) \\
&=\frac { \sin { 70 } }{ \cos { 70 } } -\frac { \sin { 40 } }{ \cos { 40 } } -\left( \frac { \sin { 40 } }{ \cos { 40 } } +\frac { \sin { 20 } }{ \cos { 20 } } \right) \\
&=\frac { \sin { 30 } }{ \cos { 70\cos { 40 } } } -\frac { \sin { 60 } }{ \cos { 20\cos { 40 } } } \\
&=\frac { 1 }{ 2\cos { 40 } } \left( \frac { 1 }{ \cos { 70 } } -\frac { \sqrt { 3 } }{ \cos { 20 } } \right) \\
&=\frac { 1 }{ 2\cos { 40 } } \left( \frac { \cos { 20-\sqrt { 3 } \cos { 70 } } }{ \cos { 70\cos { 20 } } } \right) \\
&=\frac { 1 }{ 2\cos { 40 } } \left( \frac { \cos { 20-\sqrt { 3 } \sin { 20 } } }{ \sin { 20 } \cos { 20 } } \right) \\
&=\frac { 1 }{ \cos { 40 } } \left( \frac { \frac { 1 }{ 2 } \cos { 20 } -\frac { \sqrt { 3 } }{ 2 } \sin { 20 } }{ \sin { 20 } \cos { 20 } } \right) \\
&=\frac { 2 }{ \cos { 40 } } \frac { \cos { 60\cos { 20 } -\sin { 60\sin { 20 } } } }{ 2\sin { 20 } \cos { 20 } } \\
&=\frac { 2 }{ \cos { 40 } } \frac { \cos { 80 } }{ \sin { 40 } } \\
&= 4\frac { \cos { 80 } }{ 2\sin { 40 } \cos { 40 } } \\
&= 4\frac { \cos { 80 } }{ \sin { 80 } } =4\cot { 80 } =4\tan { 10 }
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$
| It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^{12}]\frac{1}{(1-x^2)^5}
&=[x^{12}]\sum_{k=0}^\infty \binom{-5}{k}(-x^2)^{k}\tag{1}\\
&=[x^{12}]\sum_{k=0}^\infty \binom{k+4}{4}x^{2k}\tag{2}\\
&=\binom{10}{4}\tag{3}\\
&=210
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion of $\frac{1}{(1-x^2)^5}$
*In (2) we use the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k
\end{align*}
*In (3) we select the coefficient of the series with $k=6$ accordingly
Note: Since the function is even, the coefficient of $x^{17}$ is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$
Prove that $$\sum_{k=0}^n \binom{3n-k}{2n}=\binom{3n+1}{n}$$
I've tried multiple things that didn't work.
Maybe this would help
$$\sum_{k=0}^n \binom{3n-k}{2n}=\sum_{k=0}^n \binom{3n-(n-k)}{2n}=\sum_{k=0}^n \binom{2n+k}{2n}$$
| Your identity is a special case of the more general identity $$S(m,n) = \sum_{k=0}^n \binom{m+k}{m} = \binom{m+n+1}{m+1},$$ which you can prove by induction on $n$: note $S(m,0) = 1 = \binom{m+1}{m+1}$. Then observe $$\begin{align*} S(m,n+1) &= S(m,n) + \binom{m+n+1}{m} \\
&= \binom{m+n+1}{m+1} + \binom{m+n+1}{m} \\
&= \binom{m+n+2}{m+1} = \binom{m + (n+1) + 1}{m+1},
\end{align*}$$
hence by the induction hypothesis, the claim is proven. Then choose $m = 2n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
So,
$$A=0$$
$$2A+B=0$$
$$A+B=1$$
but that does not make sense, because if $A=0$, then $2A+B$ can't be zero, could you please tell me what's the problem?
| The fraction becomes
$$
\frac{A}{(x+1)} + \frac{B}{(x+1)^2}=
\frac{A(x+1)+B}{(x+1)^2}=
\frac{Ax+(A+B)}{(x+1)^2}
$$
so $A=0$ and $B=1$. Indeed
$$
\frac{1}{(x+1)^2}
$$
is already in “partial fractions” form. And
$$
\int\frac{1}{(x+1)^2}\,dx=-\frac{1}{x+1}+C
$$
It would be different if you started with
$$
\frac{x}{(x+1)^2}
$$
because then the decomposition would give $A=1$ and $B=-1$, so
$$
\int\frac{x}{(x+1)^2}\,dx=
\int\left(\frac{1}{x+1}-\frac{1}{(x+1)^2}\right)\,dx=
\log|x+1|+\frac{1}{x+1}+C
$$
Where's the problem with your computations? You have done
$$
\frac{A}{(x+1)} + \frac{B}{(x+1)^2}=
\frac{A(x+1)^2+B(x+1)}{(x+1)^3}
$$
and then equalled this with $\frac{1}{(x+1)^2}$ without taking into account the different denominators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve:
$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$
I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?
| $$\int \frac { x+1 }{ \sqrt { 1-x^{ 2 } } } dx=-\left( \int \frac { -x }{ \sqrt { 1-x^{ 2 } } } dx-\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } } \right) =\\ =-\frac { 1 }{ 2 } \int \frac { d\left( 1-{ x }^{ 2 } \right) }{ \sqrt { 1-x^{ 2 } } } +\int { \frac { dx }{ \sqrt { 1-x^{ 2 } } } =-\sqrt { 1-x^{ 2 } } +\arcsin { x } +C } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 3
} |
Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do next ?
Please also tell me alternative way to do this .
| Hint: if we multiply by $\sin^{-6}\left(x\right)
$ we get $$I=\int\frac{\sin^{4}\left(x\right)}{\sin^{4}\left(x\right)+\cos^{4}\left(x\right)}dx=\int\frac{\csc^{2}\left(x\right)}{\csc^{2}\left(x\right)+\cot^{4}\left(x\right)\csc^{2}\left(x\right)}dx
$$ $$=\int\frac{\csc^{2}\left(x\right)}{\cot^{6}\left(x\right)+\cot^{4}\left(x\right)+\cot^{2}\left(x\right)+1}dx
$$ and now taking $\cot\left(x\right)=u,\,-dx \csc^{2}\left(x\right)=du
$ we have $$I=-\int\frac{1}{u^{6}+u^{4}+u^{2}+1}du=-\int\frac{1}{\left(u^{2}+1\right)\left(u^{4}+1\right)}du$$ $$=\int\frac{1}{\left(u^{2}+1\right)\left(u^{2}+\sqrt{2}x+1\right)\left(u^{2}-\sqrt{2}x+1\right)}du$$ can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$
Show that for all $z \in \overline{D}(0;1)$, $(3-e)|z| \leq |e^z - 1|\leq |z|(e-1)$
I think I'm supposed to use the following chain of inequalities
$$|e^z -1|\leq e^{|z|}-1 \leq |z|e^{|z|}$$
But every time I try to solve it I get stuck because some term doesn't vanish or I get something similar but no the same I'm asked to prove. Any help or hint will be appreciate and thanks in advance!
| Start with the series expansion
$$e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$
and use the triangle and reverse triangle inequalities to get the result.
By the triangle inequality and the condition $\lvert z\rvert \le 1$, we have\begin{align}\lvert e^z - 1\rvert &\le \lvert z \rvert + \frac{\lvert z\rvert^2}{2!} + \frac{\lvert z\rvert^3}{3!} + \cdots\\ &\le \lvert z\rvert + \frac{\lvert z\rvert}{2!} + \frac{\lvert z\rvert}{3!} + \cdots = \lvert z\rvert\left(1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\right) = \lvert z\rvert(e - 1)\end{align}
On the other hand, by the reverse triangle inequality,\begin{align}\lvert e^z - 1\rvert &\ge \lvert z\rvert - \left\lvert\frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\right\rvert \ge \lvert z\rvert - \frac{\lvert z\rvert^2}{2!} - \frac{\lvert z\rvert^3}{3!} -\cdots \\
&\ge \lvert z\rvert - \frac{\lvert z\rvert}{2!} - \frac{\lvert z\rvert}{3!} - \cdots = \lvert z\rvert\left[1 - \left(\frac{1}{2!} + \frac{1}{3!} + \cdots\right)\right] = \lvert z\rvert(1 - (e - 2)) = \lvert z\rvert(3 - e)\end{align}Hence, $(3 - e)\lvert z\rvert \le \lvert e^z - 1\rvert \le (e - 1)\lvert z\rvert$ on $\overline{D}(0;1)$.
| {
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"url": "https://math.stackexchange.com/questions/1865618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $, then $x$ equals $-4$ For the equation
$$5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $$
$x$ is equal to $-4$, but I'm not sure why.
I've taken the right side of the equation ${ \left( \frac { 1 }{ 5 } \right) }^{ \frac { 1 }{ x } }$ and converted it to $5^{-(1/x)}$, but haven't been able to perform the mathematical gymnastics to prove this.
| Since $125 = 5^3$, we have $$5 \cdot 5^{\frac{3}{x}} = \frac{1}{5^{\frac{1}{x}}} \Rightarrow 5^{1 + \frac{3}{x}} = 5^{-\frac{1}{x}}$$
Equating powers gives $1 + \frac{3}{x} = -\frac{1}{x} \Rightarrow x = -4$.
Some explanations:
We have $a^b \cdot a^c = a^{b+c}$, this can be reasoned as such: $$a^b \cdot a^c = \underbrace{a \cdot a \cdots a}_{b \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{c \, \text{times}} = \underbrace{a \cdot a \cdots a}_{(b+c) \, \text{times}} = a^{b+c}$$ This gives $5 \cdot 5^{3/x} = 5^{1 + 3/x}$ as promised.
Next, we have $(a^b)^c = (a^c)^b = a^{bc}$, this should make sense if you think of it being $a^b$ multiplied together $c$ times, giving a total of $a$ being multiplied together $bc$ times, since $a^b$ is $a$ multiplied together $b$ times. Symbolically $$(a^b)^c = \underbrace{a^b \cdot a^b \cdots a^b}_{c \, \text{times}} = a^{\overbrace{b + \cdots + b}^{c \, \text{times}}} = a^{bc}$$
This is what lets us write $125^{1/x} = (5^3)^{1/x} = 5^{3/x}$.
I do need to emphasise that my 'explanations' are simply that, visual and handy intuitive explanations, they do not constitute proper proof (as we are implicitly assuming $a,b,c$ are (positive) integers, whilst the stated rules hold for real numbers.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.
| A tricky way to verify the monotonicity is by applying AM-GM inequality as follows:
\begin{align*}
a_n & = \left(1 + \frac{1}{n}\right)^n \\
& = 1\times \left(1 + \frac{1}{n}\right) \times \cdots \times \left(1 + \frac{1}{n}\right)\\
& \leq \left[\frac{1 + n\left(1 + \frac{1}{n}\right)}{n + 1}\right]^{n + 1} \\
& = \left(1 + \frac{1}{n + 1}\right)^{n + 1} = a_{n + 1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$
$\nabla f_x$ $=$ $2xy^{2}z^{2}$,
$\nabla f_y$ $=$ $2yx^{2}z^{2}$,
$\nabla f_z$ $=$ $2zx^{2}y^{2}$
$\nabla g_x$ $=$ $2x$,
$\nabla g_y$ $=$ $2y$,
$\nabla g_z$ $=$ $2z$
setting the sets equal to each other and multiplying each to get the equality where $\nabla f_{x..z} = 2x^2y^2z^2$ gives me(namely multiplying $x$ on $\nabla f_x = \lambda g_x$...and so on) :
$\lambda2x^2 = \lambda2y^2=\lambda2z^2 \to x=y=z$ after division and taking the roots. So:
$x=y=z=a \to 3a^2=1 \to a= \pm \frac{1}{\sqrt{3}}$
Thus a maximum at $f\Big(\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}\Big) = \frac{1}{27}$.
However looking at wolrfram alpha, it also gives a minimum at $f\Big(-\frac{1}{2},-\frac{\sqrt{3}}{2},0\Big) = 0$. I would like to know how there are more solutions from $\lambda 2x^2=\lambda 2y^2=\lambda 2z^2$ than just x=y=z
Using the corrections from posters lets say I let $x=0$ now I have $y^2+z^2=1$ Then if I wanted one scenario just to show there is a minimum I could say $y^2=z^2$ and then say $a^2+a^2=1 \to 2a^2=1 \to a^2=\frac{1}{2} \to a=\pm\frac{1}{\sqrt{2}}$ therefore have a minimum at $f\Big(0,\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)$ Is this correct. I feel the only correct solution thought is to use set notation and proof to show there are infinitely many cases. But does my above asnwer for one of those cases hold?
| You didn't consider the case $x = y = \lambda = 0, z = 1$ which satisfies the nabla equations, and gives the minimum value for $f$ of $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{\pi}{3\sqrt{3}}$$
Usually, this can be proven by differentiating $\sum_{n=1}^\infty \frac{x^{2n}}{n^2 \binom{2n}{n}}=2(\arcsin{\frac{x}{2}})^2$, but I have an alternative proof.
Using the result of:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{y+n+1}}=\frac{1}{y \binom{y+n}{n}} \tag1$$
, which can be easily proved.
I can substitute $y=n$ to obtain
$$\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\frac{1}{n \binom{2n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{2n+1}}=\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}$$
Therefore,
\begin{align}
\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}} & = \int_0^\infty \frac{xdx}{(x+1)(x^2+x+1)} \\
& = \lim_{L\to \infty} \frac{1}{2}\ln(x^2+x+1)-\ln(x+1)+\frac{\tan^{-1}(\frac{2x+1}{\sqrt{3}})}{\sqrt{3}}\large{|_0^L} \\
&= \frac{\pi}{3\sqrt{3}}
\end{align}
Now for $k=3$, I tried to substitute $y=2n$ into $(1)$:
$$\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\frac{1}{2n \binom{3n}{n}}$$
$$\sum_{n=1}^\infty\int_0^\infty \frac{x^ndx}{(x+1)^{3n+1}}=\sum_{n=1}^\infty\frac{1}{2n \binom{3n}{n}}$$
So we can have
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\int_0^\infty\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}$$
However, by partial fraction $$\frac{2xdx}{(x+1)(x^3+3x^2+2x+1)}=-\frac{2}{1+x}+\frac{2x^2+4x+2}{x^3+3x^2+2x+1}$$
The left part does not seem to converge.
Feeling frustrated, Wolfram Alpha plays its part. It spits out these results:
$$\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}=\frac{1}{3}{}_3F_2\left(\left.\begin{array}{c} 1,1,\frac{3}{2}\\ \frac{4}{3}, \frac{5}{3} \end{array}\right| \frac{4}{27}\right)$$
$$\sum_{n=1}^\infty\frac{1}{n \binom{4n}{n}}=\frac{1}{4}{}_4F_3\left(\left.\begin{array}{c} 1,1,\frac{4}{3},\frac{5}{3}\\ \frac{5}{4}, \frac{6}{4}, \frac{7}{4} \end{array}\right| \frac{27}{256}\right)$$
However, I am not very familiar with hypergeometric function.
The pattern suggests that $$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{n}}=\frac{1}{2}{}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{\pi}{3\sqrt{3}}$$
Thus, $${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)=\frac{2\pi}{3\sqrt{3}}$$
Arriving these results, I have the following questions:
How can ${}_2F_1\left(\left.\begin{array}{c} 1,1\\ \frac{3}{2} \end{array}\right| \frac{1}{4}\right)$ be expressed into this simple elementary form?
How can we arrive to the result for $\sum_{n=1}^\infty\frac{1}{n \binom{3n}{n}}$ given by Wolfram Alpha?
Ultimately, can we evaluate $\sum_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ for all integers k $\ge$ $2$?
| For the most general case,
$\sum\limits_{n=1}^\infty\dfrac{1}{n\binom{kn}{n}}$
$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+1)}{n\Gamma(kn+1)}$
$=\sum\limits_{n=1}^\infty\dfrac{\Gamma(n)\Gamma((k-1)n+1)}{\Gamma(kn+1)}$
$=\sum\limits_{n=0}^\infty\dfrac{\Gamma(n+1)\Gamma((k-1)n+k)}{\Gamma(kn+k+1)}$
$=~_3\Psi_1\left[\begin{matrix}(1,1)~~(1,1)~~(k,k-1)\\(k+1,k)\end{matrix};1\right]$ (according to http://en.wikipedia.org/wiki/Fox%E2%80%93Wright_function)
| {
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"url": "https://math.stackexchange.com/questions/1870690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 1
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Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
| Hint $\,\ \sqrt a + \sqrt b\, = \sqrt{(\sqrt a + \sqrt b)^2} = \sqrt{a+b +2{\sqrt{ab}}}.\ $ Here $\ a+b=12,\ \color{}{\sqrt{ab}} = 4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Chemical reaction modeled by a differential equation I am badly stuck on the following question. Thus, I am asking for some help :)
Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as
$$A + B \xrightarrow{k} X$$
If $2 \, \rm{g}$ of $A$ and $1 \, \rm{g}$ of $B$ are required to produce $3g$ of compound $X$, then the amount of compound $x$ at time $t$ satisfies the differential equation
$$ \frac{dx}{dt} = k \left(a - \frac{2}{3}x\right)\left(b - \frac{1}{3}x\right) $$
where $a$ and $b$ are the amounts of $A$ and $B$ at time $0$ (respectively), and initially none of compound $X$ is present (so $x(0) = 0$). Time is in units of minutes, and $k$ is the reaction rate, per minute per gram.
Use separation of variables (and integration by partial fractions) to show that solution can be expressed in the form
$$ \ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct $$
where the constant c depends on k,a,b. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.
So, first of all, showing this solution can be expressed as second equation,
by using separation of variables ( and integration by partial fractions), i got
$$ \ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c $$
how can this can be expressed in the form of the second equation ? am I doing something wrong ?, I dont get how I should get $$ \frac{b}{a} $$ inside the ln... and on the RHS, how kt + c becomes just ct ?
and by the second euqation,
"""
where the constant $c$ depends on $k$, $a$ and $b$. Now suppose that a = 15g, b = 20g and after 10min, 15g of compound X has been formed, find the amount of X after 20 mins.
"""
how do we find the amount of $X$ after $20$ minutes?
Thank you !
| Without checking if the two forms are both correct :
$$\begin{cases}
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = ct \qquad [1]\\
\ln \left(\frac{a - \frac{2}{3}x}{b - \frac{1}{3}x}\right) * \frac{3}{a-2b} = kt + c \qquad [2]
\end{cases}$$
The relationship between $a,b,c,k$ can be derived :
$\begin{cases}
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) =\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)}\right) +\ln\left(\frac{b}{a} \right) =ct \\
\ln \left(\frac{a - \frac{2}{3}x(t)}{b - \frac{1}{3}x(t)} \cdot \frac{b}{a}\right) = \frac{a-2b}{3}(kt+c)
\end{cases} \quad\to\quad ct-\ln\left(\frac{b}{a} \right)=\frac{a-2b}{3}(kt+c)$
$$c=\frac{a-2b}{3}k=-\ln\left(\frac{b}{a} \right)$$
If this relationship is true, Eqs. $[1]$ and $[2]$ are equivalent.
$\begin{cases}
a=\frac{3c}{k}\frac{e^c}{e^c-1}\\
b=\frac{3c}{k}\frac{1}{e^c-1}\\
\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to express $\cos(20^\circ)$ with radicals of rational numbers? In showing that the trisection of an angle with ruler and compass is not possible in general one shows that $\cos(20^\circ)$ cannot be constructed (thus the angle $60^\circ$ cannot be trisected) by determining its minimal polynomial, which is $8x^3-6x-1$.
Solving $8x^3-6x-1=0$ yields a solution $x_1=\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}+\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$. Expressing $\sqrt[3]{\frac{1}{16}+\frac{\sqrt{3}}{16}i}$ and $\sqrt[3]{\frac{1}{16}-\frac{\sqrt{3}}{16}i}$ in polar form yields $x_1=\cos(20^\circ)$.
Is it possible to express $\cos(20^\circ)$ with radicals without complex numbers?
| $$
\cos (20^\circ) = \cos (\pi/9) = -\frac12 (-1)^{8/9} \left(1+(-1)^{2/9}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the real and imaginary parts of an equation
Find the real and imaginary parts of $\frac{1}{3z+2}$
So I have expanded it out to get $\frac{1}{3x+3iy+2}$
Thus giving $Re(\frac{1}{3z+2})=\frac{1}{3x+2}$ and $Im(\frac{1}{3z+2})=\frac{1}{3y}$
However in my answer book it says: $Re(\frac{1}{3z+2})=\frac{3x+2}{(3x+2)^2+9y^2}$ and $Im(\frac{1}{3z+2})=\frac{-3y}{(3x+2)^2+9y^2}$
Is the book incorrect/outdated or if not could someone explain how to gain these answers, thanks
| you find it as the form $z=a+ib$ so
$$z=\frac { 1 }{ 3x+3iy+2 } =\frac { 3x+2-3iy }{ \left( 3x+2+3iy \right) \left( 3x+2-3iy \right) } =\frac { 3x+2 }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } +i\frac { -3y }{ { \left( 3x+2 \right) }^{ 2 }+9{ y }^{ 2 } } \\ $$
| {
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$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).
Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.
The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).
Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.
| Other way
$$x^2\left(x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}\right)=0$$
$x=0$ is not a solution of this equation , therefore
$$x^2-4x+6-\frac{4}{x}+\frac{1}{x^2}=0$$
we have
$$\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+6=0$$
set $t=x+\frac{1}{x}$, thus
$$t^2-4t+4=(t-2)^2=0$$
Hence
$$t=x+\frac{1}{x}=2$$
i.e. $x=1$
| {
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Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My attempt:
Consider the line $y=mx+c$ as it passes through $(0,4)$ then $c=4$ --> Condition 1
Plug in $y=mx+c$
$$ 4x^2-(mx+c)^2=36$$
$$4x^2-36=m^2x^2+2mcx+c^2$$
$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$
For tangency $b^2-4ac=0$
Hence
$$ 4m^2c^2-4(c^2+36)(m^2-4)=0$$
$$ 16c^2-144m^2+576=0$$
But $c=4$ due to the line condition hence
$$ m = \pm \frac{2\sqrt{13}}{3}$$
So the lines are
$y= \frac{2\sqrt{13}}{3}x + 4$ and $y= -\frac{2\sqrt{13}}{3}x + 4$
How to find the coordinates for the points on the hyperbola for this to occur tho?
| This is more easily solved using pole-polar relationships, I think. The chord of contact the the two tangents is the polar of their intersection point, namely $$4(0\cdot x)-(4\cdot y)=36,$$ i.e., the horizontal line $y=-9$. Substitute this value into the equation of the hyperbola and solve for $x$.
| {
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Inequality Challenge with $a+b+c=3$ I tried to prove this inequality, but I failed. It seemed that if one uses Cauchy-Schwarz inequality on the right hand side, it may expand RHS's value larger than LHS's. Problem is presented as follows:
If $a,b,c\ge0$ and $a+b+c=3$, prove that $$2(ab+bc+ca)-3abc\ge a\cdot \sqrt{\frac{b^2+c^2}{2}}+b\cdot\sqrt{\frac{c^2+a^2}{2}}+c\cdot\sqrt{\frac{a^2+b^2}{2}}$$
| After using $\sqrt{\frac{a^2+b^2}{2}}\leq\frac{3a^2+2ab+3b^2}{4(a+b)}$,
which is just $(a-b)^4\geq0$, you'll get something obvious.
Indeed, it remains to prove that
$$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\sum\limits_{cyc}\frac{c(3a^2+2ab+3b^2)}{4(a+b)}$$ or
$$2(ab+ac+bc)-\frac{9abc}{a+b+c}\geq\frac{3}{2}(ab+ac+bc)-abc\sum\limits_{cyc}\frac{1}{a+b}$$ or
$$\frac{1}{2}(ab+ac+bc)-\frac{9abc}{2(a+b+c)}-\frac{9abc}{2(a+b+c)}+abc\sum\limits_{cyc}\frac{1}{a+b}\geq0$$ or
$$\sum\limits_{cyc}c(a-b)^2+abc\left(\sum\limits_{cyc}(a+b)\sum\limits_{cyc}\frac{1}{a+b}-9\right)\geq0$$
Done!
| {
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A Problem with the Generating Function of Fibonacci. So basically I want to find the closed form of $G_n = \sum_{k = 1}^n \binom{n+k - 1}{2k-1}$.
After checking for $n = 1,2,3,4$ the values are $1, 3, 8, 21$ respectively. I claim that it is $F_{2n}$ where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$. My idea was to prove this using generating functions.
We have that $\binom{n+k-1}{2k-1}$ is the $n$th coefficient of $X^k \sum_{i \geq 0} \binom{i + 2k - 1}{i} X^i = \frac{X^k}{(1-X)^{2k}}$ so the $$\sum_{n \geq 0} G_n X^n = \sum_{k = 1}^n \frac{X^k}{(1-X)^{2k}}$$
Now we want to find what $F_0 + F_2X + F_4X^2 + ...$ is and we know that $F_0 + F_1X + ... = \frac{X}{1-X-X^2}$.
So
\begin{align*}
F_0 + F_2X^2 + F_4X^4 + ... & = \frac{1}{2} \left ( \frac{X}{1-X-X^2} + \frac{-X}{1 + X - X^2} \right ) \\
& = \frac{X^2}{(1-X^2)^2 - X^2}
\end{align*}
which means $$F_0 + F_2X + F_4X^2 + ... = \frac{X}{(1-X)^2 - X}.$$
So if suffices to show
\begin{align*}
\frac{X}{(1-X)^2 - X} & = \sum_{k = 1}^n \frac{X^k}{(1-X)^{2k}} \\
\frac{1}{(1-X)^2 - X} & = \sum_{k = 1}^n \frac{X^{k-1}}{(1-X)^{2k}} \\
\frac{(1-X)^{2n}}{(1-X)^2 - X} & = ((1-X)^2)^{n-1} + ((1-X)^2)^{n-2}X + ... + X^{n-1} \\
\frac{(1-X)^{2n}}{(1-X)^2 - X} & = \frac{(1-X)^{2n} - X^n}{(1-X)^2 - X}
\end{align*}
which is not true. I don't see where I went wrong with this.
| Hint: There is just a small calculation error when transforming $\binom{n+k-1}{2k-1}$.
We obtain
\begin{align*}
\sum_{n=0}^\infty G_nx^n&=\sum_{n=0}^\infty\sum_{k=1}^n\binom{n+k-1}{2k-1}x^n\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\binom{n+k-1}{2k-1}x^n\tag{1}\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty\binom{n+2k-1}{2k-1}x^{n+k}\tag{2}\\
&=\sum_{k=1}^\infty x^k\sum_{n=0}^\infty\binom{-2k}{n}(-x)^n\tag{3}\\
&=\sum_{k=1}^\infty \frac{x^k}{(1-x)^{2k}}\tag{4}\\
&=\frac{\frac{x}{(1-x)^2}}{1-\frac{x}{(1-x)^2}}\tag{5}\\
&=\frac{x}{1-3x+x^2}\\
&=x+3x^2+8x^3+21x^4+55x^5+144x^6+\cdots=\sum_{n=1}^\infty F_{2n}x^n
\end{align*}
and the claim follows according to OPs expectation.
Commment:
*
*In (1) we exchange the summation order.
*In (2) we shift the index of the inner series by $k$.
*In (3) we use the binomial identity $\binom{p+q-1}{q}=\binom{-p}{q}(-1)^q$.
*In (4) we use the binomial series expansion.
*In (5) we apply the geometric series expansion.
| {
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} |
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:
$$\begin{split}
\tan\theta + \tan 2\theta+\tan 3\theta
&= (\tan\theta + \tan 2\theta)
+ \dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta} \\
&= \dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}
{1-\tan\theta\tan2\theta}
\end{split}$$
And, RHS yields
$$\begin{split}
\tan\theta\tan2\theta\tan3\theta
&= (\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta}
\end{split}$$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$$
\begin{split}
2-\tan\theta\tan2\theta &= \tan\theta\tan2\theta\\
\tan\theta\tan2\theta &= 1,
\end{split}$$
which can be further reduced as:
$$\tan^2\theta=\frac{1}{3}\implies\tan\theta=\pm\frac{1}{\sqrt3}$$
Now, we can yield the general solution of this equation:
$\theta=n\pi\pm\dfrac{\pi}{6},n\in Z$. But, setting $\theta=\dfrac{\pi}{6}$ in the original equation is giving one term $\tan\dfrac{\pi}{2}$, which is not defined.
What is the problem in this computation?
| Clearly, $\tan\theta\tan2\theta\tan3\theta$ is undefined if $\theta\equiv\dfrac\pi2,\dfrac\pi4,\dfrac\pi6\pmod\pi$
Otherwise,
$$\tan3\theta=-\dfrac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=-\tan(\theta+2\theta)=\tan(-3\theta)$$
$$\implies3\theta=n\pi-3\theta\iff\theta=\dfrac{n\pi}6$$ where $n$ is any integer
But $3\nmid n$ else $\tan\theta\tan2\theta\tan3\theta$ will be undefined
| {
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What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos(x)-1+\frac{x^2}{2}}{x^4})$$
then using the properties of logarithms I came to the conclusion that
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+\ln(x^4)$$
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x)$$
So then I took the limit as the $\ln(y)$ approaches 0.
$$\lim_\limits {x \to 0} (\ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x))$$
Here I used L'Hopital's Rule and got
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x}{\cos(x)-1+x^2}+\frac{4}{x}$$
then got a common denominator
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x+4 \cdot (\cos(x)-1+x^2)}{x\cdot (\cos(x)-1+x^2)}$$
I clearly made a mistake somewhere because the denominator is 0. The answer by the way is $\frac{1}{24}$. I have no idea how to arrive at that conclusion.
| Here is another take with minimum number of applications of L'Hospital's Rule. We have
\begin{align}
L &= \lim_{x \to 0}\dfrac{\cos x - 1 + \dfrac{x^{2}}{2}}{x^{4}}\notag\\
&= \lim_{t \to 0}\dfrac{\cos 2t - 1 + 2t^{2}}{16t^{4}}\text{ (putting }x = 2t)\notag\\
&= \frac{1}{16}\lim_{t \to 0}\dfrac{2t^{2} - 2\sin^{2}t}{t^{4}}\notag\\
&= \frac{1}{8}\lim_{t \to 0}\dfrac{t^{2} - \sin^{2}t}{t^{4}}\notag\\
&= \frac{1}{8}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\cdot\frac{t + \sin t}{t}\notag\\
&= \frac{1}{8}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\cdot\left(1 + \frac{\sin t}{t}\right)\notag\\
&= \frac{1}{4}\lim_{t \to 0}\dfrac{t - \sin t}{t^{3}}\notag\\
&= \frac{1}{4}\lim_{t \to 0}\dfrac{1 - \cos t}{3t^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{12}\lim_{t \to 0}\dfrac{1 - \cos^{2} t}{t^{2}(1 + \cos t)}\notag\\
&= \frac{1}{24}\lim_{t \to 0}\dfrac{\sin^{2} t}{t^{2}}\notag\\
&= \frac{1}{24}\notag
\end{align}
If you have studied Taylor series expansions then that is the best technique to solve such problems (see the answer by Claude Leibovici for details).
| {
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"source": "stackexchange",
"question_score": "1",
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Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathcal o \left(1/x+\sqrt{1/x^3}\right)}-1)=...$$
Main question: Now wouldn't it have to be $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ instead? So I could write $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ or $\mathcal O(1/x+\sqrt{1/x^3})$?
Are there other ways to calculate the limit?
| An overkill is to exploit that for any sufficiently large $x>0$ we have that
$$ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \frac{1}{2}+\sqrt{x+\frac{1}{4}}$$
since both terms are a positive solution of $y^2=x+y$. On the other hand
$$ \sqrt{x+\sqrt{x}} \geq \frac{1}{2}+\sqrt{x-\frac{1}{4}} $$
is a trivial inequality by just squaring both sides, hence
$$ \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}-\frac{1}{2} $$
is bounded between $\sqrt{x-\frac{1}{4}}-\sqrt{x}$ and $\sqrt{x+\frac{1}{4}}-\sqrt{x}$, and the limit is $\color{red}{\large\frac{1}{2}}$ by squeezing.
| {
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Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0$ has a solution modulo $p$ I need to solve this question:
Find the primes $p$ such that the equation: $x^{2} + 6x + 15 = 0 $ has a solution modulo $ p $.
My approach was: I checked for $p = 2$ and there is no solution.
Now if $p \neq 2 $ so the equation has a solution $\iff 6^{2} -4 \times 15 = -24$ is a square modulo $p$.
Now $-24 $ is a square modulo $p \iff \left(\frac{-24}{p}\right) = 1$ (Legendre symbol) $ \iff \left(\frac{-1}{p}\right) \left(\frac{2}{p}\right) \left(\frac{3}{p}\right) = 1.$
And now there are many cases to check and I'm not sure how to do it...
For example - case 1:
$ \left(\frac{-1}{p}\right) =1 $ and $ \left(\frac{2}{p}\right) =1 $ and $ \left(\frac{3}{p}\right) = 1$ :
$ \left(\frac{-1}{p}\right) =1 \iff p \equiv 1 (4) $ ,
$ \left(\frac{2}{p}\right) =1 \iff p \equiv 1, -1 (8) $ ,
$ \left(\frac{3}{p}\right) = 1 \iff p \equiv 1, -1 (12) $
so how do I combine these results for case 1? and after that, do I really need now to check all the other cases - that 2 of the Legendre symbols are $(-1)$ and one is $1$ ?
| $x^2+6x+15 \equiv 0\pmod{p}\Rightarrow(x+3)^2+6\equiv0\pmod{p}\Rightarrow (x+3)^2\equiv-6\pmod{p}$.
In other words, $y^2\equiv -6\pmod{p}$ or to rephrase the argument, $\left(\frac{-6}{p}\right)=1$
| {
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How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
| It is easy to check that if $a,b$ are two positive real numbers, then $\frac{a^a b^b}{(a+b)^{a+b}}$ is the maximum value of the function $f(x)=x^a(1-x)^b$ over the interval $[0,1]$, attained at $x=\frac{a}{a+b}$. By Euler's beta function we have:
$$ \int_{0}^{1}f(x)\,dx = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}=\frac{1}{a+b+1}\cdot\frac{a! b!}{(a+b)!}\tag{1}$$
Over the interval $\left(0,\frac{a}{a+b}\right)$ the function $(1-x)^b$ is bounded by one, as well as over the interval $\left(\frac{a}{a+b},1\right)$ the function $x^a$ is bounded by one. That leads to:
$$ \int_{0}^{1}f(x)\,dx \leq \int_{0}^{\frac{a}{a+b}}x^a\,dx + \int_{\frac{a}{a+b}}^{1}(1-x)^b\,dx \\=\frac{1}{a+1}\left(\frac{a}{a+b}\right)^{a+1}+\frac{1}{b+1}\left(\frac{b}{a+b}\right)^{b+1}\tag{2} $$
On the other hand, over the interval $\left(0,\frac{a}{a+b}\right)$ the function $(1-x)^b$ is lower-bounded by $\left(\frac{b}{a+b}\right)^b$, as well as over the interval $\left(\frac{a}{a+b},1\right)$ the function $x^a$ is lower-bounded by $\left(\frac{a}{a+b}\right)^a$. That leads to:
$$ \int_{0}^{1}f(x)\,dx \geq \frac{1}{a+1}\left(\frac{a}{a+b}\right)^{a+1}\left(\frac{b}{a+b}\right)^b+\frac{1}{b+1}\left(\frac{b}{a+b}\right)^{b+1}\left(\frac{a}{a+b}\right)^a\tag{3}$$
that is equivalent to:
$$ \int_{0}^{1}f(x)\,dx \geq \frac{a^a b^b}{(a+b)^{a+b}}\left(\frac{\frac{a}{a+1}+\frac{b}{b+1}}{a+b}\right)\tag{4}$$
so:
$$ \frac{a!b!}{(a+b)!}\geq \frac{a^a b^b}{(a+b)^{a+b}}\left(\frac{\frac{a}{a+1}+\frac{b}{b+1}}{a+b}\right)(a+b+1)\tag{5} $$
and:
$$ \frac{a^a b^b}{a! b!}\leq \frac{(a+b)^{a+b}}{(a+b)!}\cdot\frac{a+b}{(a+b+1)\left(\frac{a}{a+1}+\frac{b}{b+1}\right)}\tag{6} $$
is a stronger inequality than the one we wanted to prove, since $(a+b+1)\left(\frac{a}{a+1}+\frac{b}{b+1}\right)\geq (a+b)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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asymptotics of the Gamma function and remainder I have found the following asymptotic formula in a book:
$$\lim_{\vert y\vert\rightarrow\infty}\vert \Gamma(x+iy) \vert e^{\frac{\pi}{2}\vert y\vert}\vert y \vert^{\frac{1}{2}-x}= \sqrt{2\pi}. $$
I would like to know if there are even sharper estimates, in particular is it true that:
$$\vert \Gamma(x+iy) \vert = \sqrt{2\pi}\cdot e^{-\frac{\pi}{2}\vert y\vert}\vert y \vert^{-\frac{1}{2}+x}\left( 1+O\left( \frac{1}{\vert y \vert}\right) \right) , \quad \vert y \vert\rightarrow \infty$$
Unfortunately I could not find any reference. Do you know some literature where I can find this result stated?
Best wishes
| There is an exact solution for $x=\dfrac 12$ as shown in this answer :
$$\tag{1}\left|\Gamma\left(\frac 12+iy\right)\right|=\sqrt{\frac{\pi}{\cosh\left(\pi y\right)}}$$
More general asymptotic results were obtained here like your :
$$\tag{2}\left|\Gamma\left(x+iy\right)\right|=\sqrt{2\pi}\,e^{-\pi|y|/2}\,|y|^{x-1/2}(1+r(x,y))$$
with $|r(x,y)|\to 0$ uniformly for $x<K$ as $|y|\to\infty$.
and the more precise (also for $|y|\to\infty$) :
\begin{align}
|\Gamma(x+iy)|&\sim \sqrt{2\pi}\,\exp\left[\frac{\left(x-\frac 12\right)\;\ln\bigl(x^2+y^2\bigr)}2-y\;\arg(x+iy)-x+\sum\limits_{m=1}^\infty \frac{B_{2m}\;\Re\;{z^{1-2m}}}{2m\,(2m-1)}\right]\\
(3)\qquad&\sim \sqrt{2\pi}\,\bigl(x^2+y^2\bigr)^{\frac x2-\frac 14}\exp\left[-x-y\,\arg(x+iy)+\frac 1{12}\frac x{x^2+y^2}-\frac 1{360}\frac{x^3-3xy^2}{(x^2+y^2)^3}+\cdots\right]\\
\end{align}
Addition
We may find more about the error term $r(x,y)$ by dividing the more precise expansion $(3)$ by $(2)$ ex-purged of $(1+r(x,y))$ of course!
For $x> 0\,$ fixed and using the expansion $\;\displaystyle\arg(x+iy)=\arctan\frac yx\sim\frac{\pi}2-\frac xy+\frac{x^3}{3\,y^3}+O\left(\frac 1{y^5}\right)$ we get as $\,y\to +\infty$ :
\begin{align}
1+r(x,y)&\sim \frac{\bigl(x^2+y^2\bigr)^{\large{\frac x2-\frac 14}}}{|y|^{x-1/2}}\exp\left(-y\,\arg(x+iy)-(-\pi|y|/2)-x+\frac 1{12}\frac x{x^2+y^2}+O\left(\frac 1{y^4}\right)\right)\\
&\sim \left(1+\frac{x^3-\frac {x^2}2}{2\,y^2}+O\left(\frac 1{y^4}\right)\right)e^{\Large{-y\frac{\pi}2+x-\frac{x^3}{3\,y^2}+\frac{\pi}2|y|-x+\frac x{12\,y^2}+O\left(\frac 1{y^4}\right)}}\\
&\sim \left(1+\frac{x^3-\frac {x^2}2}{2\,y^2}+O\left(\frac 1{y^4}\right)\right)\left(1-\frac{x^3}{3\,y^2}+\frac x{12\,y^2}+O\left(\frac 1{y^4}\right)\right)\\
&\sim 1+\frac{\frac x{12}-\frac{x^3}3+\frac{x^3}2-\frac{x^2}4}{y^2}+O\left(\frac 1{y^4}\right)\\
\end{align}
and conclude (with numerical confirmation) that :
$$r(x,y)\sim \dfrac{2\,x^3-3\,x^2+x}{12\,y^2}+O\left(\frac 1{y^4}\right)$$
(note that for $x\in\left\{0,\frac 12,1\right\}$ the $\dfrac 1{y^2}$ term disappears and that this formula is valid for any $x\in\mathbb{R}$)
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 =
4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$
But this is not possible . Help?
EDIT $k\equiv 43 \pmod{64}$ thus,
$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$.
| $x\equiv 32 (\mod 81) \ and \ x\equiv 59 (\mod 64) \Rightarrow x=81a+32 \ and \ x=64b+59 \\ $
then $81a+32=64b+59 \\
81a-64b=27 \ ...I \\$
with the euclidean algorithm we get a solution
$ (a,b)=(-405,-513) \\
81(-405)-64(-513)=27 \ ...II \\ $
subtracting II from I we get $ 81(a+405)-64(b+513)=0 \Leftrightarrow 81(a+405)=64(b+513) \ ...III \\ $
$64\mid a+405\Rightarrow \boxed{a=64k-405} \\ $
plugging $a$ into III gives $\boxed{b=81k-513} \\$
plugging $a$ into $x=81a+32$ or $b$ into $x=64b+59\\$
$\boxed{x=5184k-32773} \ \ k\in \mathbb{Z}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:
1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.
Now I want you to help me with these problems.Thanks.
| I hope this would be useful for you.
So your trying to evaluate $\displaystyle\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}$. Consider first $\displaystyle\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}$, after developing the product properly. Now you have to make sense of $$\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}$$
, so to reach a specific value, note that $\sum_{n=1}^{k}\frac{n}{2^{n}}=\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}$, therefore
\begin{eqnarray*}
\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}&=&\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}
{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}\\
&=&\lim_{k\to\infty}-\frac{k}{2^{k}}+\frac{1-\frac{1}{2^{k+1}}}{1-\frac{1}{2}}\\
&=&2
\end{eqnarray*}
Finally $$\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{2}=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many k-digit numbers are both divisible by 3 and include the digit 3? At first I thought I could simply count how many (k-1)-digit numbers are divisible by 3 and multiply by k, accounting for the different possible placements of the final 3. But It seems that method includes duplicates.
Any direction, such as Inclusion-Exclusion or something cleverer would be appreciated. I'd like to be able to apply it to 9 instead of 3 as well. (A general method for any digit d would actually solve the original problem eliciting this question.)
| We get from first principles the generating function
$$f(z) = (z+z^2+A+z+z^2+1+z+z^2+1) \\ \times
\prod_{q=1}^{k-1} (1+z+z^2+A+z+z^2+1+z+z^2+1).$$
This is
$$f(z) = (A + 2 + 3z + 3z^2)
\prod_{q=1}^{k-1} (A + 3 + 3z + 3z^2).$$
We obtain for the desired count
$$\frac{1}{3}
\sum_{q=0}^2
\left. \left(f(z) - \left. f(z)\right|_{A=0}\right)
\right|_{A=1, z=\exp(2\pi i q/3)}.$$
Here we subtract the values for $A=0$ to remove contributions where
the digit three does not occur.
The value for $q=0$ is
$$(A + 8)
\prod_{q=1}^{k-1} (A + 9).$$
Subtracting the value for $A=0$ and setting $A=1$ yields
$$9\times 10^{k-1} - 8 \times 9^{k-1}.$$
For $q=1$ and $q=2$ we get
$$(A - 1) \prod_{q=1}^{k-1} A.$$ This is zero when we set $A=0$ and
also when $A=1$ and hence makes no contribution.
This yields for the end result
$$\frac{1}{3} (9\times 10^{k-1} - 8 \times 9^{k-1})
= 3\times 10^{k-1} - 24 \times 9^{k-2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}$$
But I know that is wrong because it looks nothing like the graph of the real part of 1/z^2 on wolfram alpha found here: http://www.wolframalpha.com/input/?i=1%2F(x%2Bi*y)%5E2
Then I thought I could must multiply $1/z^2$ by $z/z$ to get $\frac{x}{z^3}$ and $\frac{iy}{z^3}$ however graphing these again shows that they are not the real and complex parts of $\frac{1}{z^2}$.
| Your first proposition should be corrected by noting that when you multiply numerator and denominator by the conjugate of $x^2+2ixy-y^2$, the denominator becomes $(x+iy)^2(x-iy)^2=(x^2+y^2)^2$ and not $x^2+y^2$. Moreover, there's no $i$ in the imaginary part.
| {
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Analytic Geometry Question: Computing the equation of a circle given 2 points and center I want to derive the equation of a circle passing through the intersection of the two circles. My method gave me a nonsensical solution. I would love if someone could tell me how my method went wrong, although since my solution is kinda long (but it's mostly algebra) I would welcome an alternative solution.
$$
x^2+y^2=100\\
(x-11)^2+(y-4)^2=9
$$
and the centered at $(22,8)$.
The way I went about solving this was first starting with the family of curves $F(x,y,h,k)=0$ which go through the point of intersection
$$
h(x^2+y^2-100)+k(x-11)^2+(y-4)^2-9)=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+121k+16k-9k-100h=0\\
\Rightarrow (h+k)(x^2+y^2)-22kx-8ky+128k-100h=0\\
\Rightarrow x^2+y^2-\frac{22k}{h+k}x-\frac{8k}{h+k}y+\frac{128k}{h+k}-\frac{100h}{h+k}=0\\
\Rightarrow(x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2
+\frac{128k+100h}{h+k}=\frac{121k^2+16k^2}{(h+k)^2}=\frac{137k^2}{(h+k)^2}\\
\Rightarrow (x-\frac{11k}{h+k})^2+(y-\frac{4k}{h+k})^2=\frac{137k^2}{(h+k)^2}-\frac{128k+100h}{h+k}
$$
Then imposing the restriction that this circle is centered at $(22,8)$ we have the system
$$
\frac{11k}{h+k}=22\\
\frac{4k}{h+k}=8\Rightarrow 2h=-k
$$
Which yields a potential solution of $(h,k)=(1,-2)$ and a radius of $392$.
edit: I fixed an algebraic mistake at the end, would still be curious to hear alternative solutions!
| Less elegant than Mick's answer.
Considering $$x^2+y^2=100 \tag 1$$ $$(x-11)^2+(y-4)^2=9 \tag 2$$ Subtract $(1)$ from $(2)$ to get $$-22 x-8 y+228=0\implies y=\frac{1}{4} (114-11 x)\tag 3$$ Replace in $(1)$ to get, after simplification $$137 x^2-2508 x+11396=0\tag 4$$ One of the solutions corresponds to $$x=\frac{2}{137} \left(627-16 \sqrt{11}\right)\implies y=\frac{8}{137} \left(57+11 \sqrt{11}\right)$$ So, $$(x-22)^2+(y-8)^2=r^2$$ Replacing by the above values of $x,y$ leads to $r^2=192$.
I then suppose that I made some mistakes somewhere. However, a plot of the three circles seems to confirm this answer.
| {
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Generalizing Variant Proof of Basel Problem Recently I have been thinking a lot about variations of the Basel Problem, and methods to solve them. Here I found the following solution to the Basel Problem by Alfredo Z. (I include the entire answer due to its brevity)
Define the following series for $ x > 0 $
$$\frac{\sin x}{x} = 1 -
\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\quad.$$
Now substitute $ x = \sqrt{y}\ $ to arrive at
$$\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 1 -
\frac{y}{3!}+\frac{y^2}{5!}-\frac{y^3}{7!}+\cdots\quad.$$
if we find the roots of $\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 0 $ we
find that
$ y = n^2\pi^2\ $ for $ n \neq 0 $ and $ n $ in the integers
With all of this in mind, recall that for a polynomial
$ P(x) = a_{n}x^n + a_{n-1}x^{n-1} +\cdots+a_{1}x + a_{0} $ with roots
$ r_{1}, r_{2}, \cdots , r_{n} $
$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} =
-\frac{a_{1}}{a_{0}}$$
Treating the above series for $ \frac{\sin \sqrt{y}\ }{\sqrt{y}\ } $
as polynomial we see that
$$\frac{1}{1^2\pi^2} + \frac{1}{2^2\pi^2} + \frac{1}{3^2\pi^2} +
\cdots = -\frac{-\frac{1}{3!}}{1}$$
then multiplying both sides by $ \pi^2 $ gives the desired series.
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots =
\frac{\pi^2}{6}$$
This solution fascinated me. Although it takes a few things for granted (such as that the Fundamental Theorem of Algebra applies to infinite polynomials) I nevertheless thought the proof was one of the most beautiful I've seen. In attempting to generalize this I stumbled across the following function and its associated power series:
$$\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{4k}}{(4k)!}(-1)^k$$
By a simple substitution we find that
$$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right)\cosh\left(\frac{x^{1/4}}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{k}}{(4k)!}(-1)^k$$
Which is a polynomial in $x$; noting that $\cosh$ is nowhere zero (on the real line) we solve for the roots as such:
$$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right) = 0 \implies \frac{x^{1/4}}{\sqrt{2}} = n\pi + \frac{\pi}{2} \implies x = \frac{\pi^4(2n+1)^4}{4}$$
Using the argument for Viete's Formula in the solution above, we find that
$$\frac{4}{\pi^4}\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = -\frac{-\frac{1}{4!}}{1}$$
Upon manipulation we find that
$$\color{red}{\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = \frac{\pi^4}{96}}$$
From here we can bring this into a form more like that of the Basel Problem by noting that
$$\sum_{k=0}^\infty \frac{1}{(2k+2)^4} = \frac{1}{16}\sum_{k=0}^\infty \frac{1}{(k+1)^4} = \frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}$$
Noting that this computes the even terms of $\sum_{k=1}^\infty k^{-4}$ we find that the odd terms must equal $\frac{15}{16}\sum_{k=1}^\infty k^{-4}$ which is our series calculated above
Applying this, we find that, as desired,
$$\color{red}{\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4}{90}}$$
However, I have been unable to generalize this approach further. Intuition tells me that the functions desired would have power series similar to those used, and would be a combination of trigonometric functions. Nevertheless, I have only been able to solve this for the case above (which required me to switch the power series from $\sin(x)$ to $\cos(x)$ from Alfredo's proof, as neglecting $\cosh(x)$ was desirable).
My question is thus: Can this proof format be applied to solve series of the form $\sum_{k=1}^\infty k^{-n}$ for any $n$ other than $2$ and $4$?
Note: I apologize for the length of this post, but I felt that a full presentation of the proof might assist in generalizing it. If anyone has any suggestions concerning this please let me know!
Note 2: In case anyone is troubled by this seemingly naive application of Viete's Formula to infinite polynomials, know that this is perfectly valid, and is known as the Root Linear Coefficient Theorem.
| To start, the "factoring" step is known as the Weierstrass factorization theorem, which asserts that you can express some functions as products of their factors.
From here, take note that you had
$$\frac{\sin(x)}x=\dots\left(1+\frac{x^2}{2^2\pi^2}\right)\left(1+\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\dots$$
which is basically what you noted.
From here, I multiply similar terms to get
$$f(x):=\frac{\sin(x)}x=\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\dots$$
and define this as $f(x)$.
using $(1-ue^{\frac{2\pi i}n})(1-ue^{\frac{4\pi i}n})\dots(1-ue^{2\pi i})=1-u^n$, we can then get
$$f(xe^{\frac{2\pi i}n})f(xe^{\frac{4\pi i}n})\dots f(xe^{2\pi i})=\left(1-\frac{x^{2n}}{1^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{2^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{3^{2n}\pi^{2n}}\right)\dots$$
And since this is not generalizable to $n\notin\mathbb N$, we can only solve for
$$\sum_{k=1}^\infty\frac1{k^{2n}}$$
By using this method. As for the odd values, no closed form yet exists.
| {
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$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ if $x^2+y^2=1$. Show that $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$. It is given that $x^2+y^2=1$. $x,y$ are positive real numbers.
[From a Regional Mathematical Olympiad, 2013 in India]
| $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{1+x^2+1+y^2}{(1+x^2)(1+y^2)}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{3}{2+x^2y^2}+\frac{1}{1+xy}\ge\frac{3}{1+(\frac{x+y}{2})^2} \rightarrow \frac{3+3xy+2+2x^2y^2}{2+2xy+x^2y^2+x^3y^3}\ge \frac{12}{4+x^2+y^2+2xy} \rightarrow \frac{5+3xy+2x^2y^2}{2+2xy+x^2y^2+x^3y^3}\ge \frac{12}{5+2xy}.$$
Consider $xy=z$ then we have:
$$\frac{5+3z+2z^2}{2+2z+z^2+z^3}\ge \frac{12}{5+2z}. $$ We know that $$x^2+y^2 \ge 2 (\sqrt{x^2y^2})\rightarrow -0.5\le z \le 0.5 \rightarrow (5+2z)\ge 0, (z+1)\ge 0, (1+2z) \ge 0$$. So we have:
$$\frac{5+3z+2z^2}{2+2z+z^2+z^3} - \frac{12}{5+2z}\ge 0 \rightarrow \frac{25+25z+16z^2+4z^3-24-24z-12z^2-12z^3}{(2+2z+z^2+z^3)(5+2z)} \rightarrow \frac{1+z+4z^2-8z^3}{(z+1)(2+z^2)(5+2z)}\ge 0 \rightarrow 1+z+4z^2-8z^3 \ge 0 \rightarrow (1+z) + 4z^2(1-2z) \ge 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A quadratic equation Find all values of a for which the quadratic $$\cos^2x - (a^2 + a + 5) |\cos x| + (a^3 + 3a^2 + 2a + 6) = 0$$ has real solution(s)
A. $a=-3$, B. $a=-2$, C. $a=-1$, D. $a=0$
I have solved this question by putting the values of the given options :
Let $\cos x = t$
then it will look like $$t^2 - (a^2 + a + 5) |t| + (a^3 + 3a^2 + 2a + 6)$$
Comparing this quadratic with the standard quadratic and then finding out the discriminant for all four values of $a$ we get $D>0$ for all four values while the answer is given only $A$ and $B$
Kindly help.
| Doing the same as Robert Z in his answer, considering $$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=0$$ the roots of the quadratic are $$t_1=3+a\qquad , \qquad t_2=2+a^2$$ since, after simplification $$\Delta=\left(a^2-a-1\right)^2$$ Obviousle $t_2$ must be discarded and $t_1$ would be acceptable if $3+a \leq 1$. So, if $a$ is suppose to be an integer, only $a=-2$ and $a=-3$ are acceptable.
| {
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Mathematical Proof - Cube Roots of Perfect Cubes Using Vedic Mathematics
Please refer this site. A method is provided for finding cube
roots of perfect cubes.
As per the method explained, suppose we are finding cube root of
$157464$
First we write as $157,\quad 464$
Last digit of $464$ is $4.$ Hence RHS=$4$
$157-5^3 \ge 0$ ($5$ is the maximum). So LHS$=5$
Hence, $\sqrt[3]{157464}=54$
How we can prove this mathematically? Please give directions on how to start towards writing a proof for this. Thanks.
| Lemma. Then function $x\mapsto x^3:\mathbb Z\to\mathbb Z$ induces a bijection $\mathbb Z/10\mathbb Z\to \mathbb Z/10\mathbb Z$.
Proof. It's enough to show that $x^3\equiv y^3\pmod{10}$ implies $x\equiv y\pmod{10}$.
If $x^3\equiv y^3\pmod{10}$, then $x\equiv x^3\equiv y^3\equiv y\pmod{2}
$ and $xy^2\equiv x^5y^2\equiv (xy)^2x^3\equiv (xy)^2y^3\equiv x^2y^5\equiv x^2y\pmod{5}$, hence $xy(x-y)\equiv 0\pmod{5}$.
Assume $x-y\not\equiv 0\pmod{5}$ so that (wlog) $x\equiv 0\pmod{5}$.
From $0\equiv x^3-y^3=(x-y)(x^2+xy+y^2)\pmod{5}$ we get $0\equiv x^2+xy+y^2\equiv y^2\pmod{5}$ from which $y\equiv 0\equiv x\pmod{5}$ which contradicts $x\not\equiv y\pmod 5$.
Thus $x\equiv y\pmod 5$, hence $x\equiv y\pmod{10}$ which concludes the proof.
Proposition
Let $n\in\mathbb N$.
There exists unique $q,r\in\mathbb N$ with $r<10^3$ such that $n=10^3q+r$.
There exists unique $d,u\in\mathbb N$ with $u<10$ such that $r\equiv u^3\pmod{10}$ and $d^3\leq q<(d+1)^3$.
Proposition. Let $n=10^3q+r$ and $m=10d+u$ with $r<10^3$ and $u<10$.
If $n=m^3$ then $r\equiv u^3\pmod{10}$ and $d^3\leq q<(d+1)^3$.
Proof. Clearly, we have $n\equiv r\equiv u^3\pmod{10}$.
If $q<d^3$, then $q\leq d^3-1$ hence $10^3d^3\leq n=10^3q+r\leq 10^3d^3+r-10^3$ from which $10^3\leq r$ - a contradiction which proves $d^3\leq q$.
If $q\geq (d+1)^3$, then $(10d+u)^3\geq (10d+10)^3$ from which $u\geq 10$ - a contradiction which proves $q<(d+1)^3$.
| {
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} |
Using a derivative to find fourier coefficients The previous question was:
Find the fourier coefficients of $f(x)=x^2+1$
To which I found $a_0=\frac{\pi^2}{3}+1$ and $a_n=\frac{2}{n^2}(-1)^n$ (I am unsure on $b_n$ I get $b_n=(\frac{-2}{n\pi}-\frac{2\pi}{n}+\frac{4}{n^3\pi})(-1)^n$ but this is more method than correcting that)
I want to complete the next part:
By considering the derivative of $f$, write down the fourier series of $g(x)=x $ without explicitly calculating the coefficients
So as $\frac{1}{2}f'(x)=g(x)$ I get (again ignoring $b_n$ for now):
\begin{align}
g(x) & =\frac{1}{2}\int_{-\pi}^\pi \left(\frac{\pi^2}{3} + 1 + \sum_{n=1}^\infty \frac{2}{n^2} (-1)^n\cos(nx)\,dx\right) \\[10pt]
& =\frac{1}{2} \left[\frac{\pi^2x}{3} + x + \sum_{n=1}^\infty \frac{2}{n^3}(-1)^n\cos(nx)\right]_{-\pi}^\pi =\frac{2\pi^3}{6} + \frac{\pi}{2} + \sum_{n=1}^\infty \frac{1}{n^3}
\end{align}
Is this correct?
| Firstly: $f(x)$ is an even function, so you expect all $b_n = 0$.
Now for $g(x)$. You are right to note that $g(x) = \frac{1}{2} f'(x)$, but why are you integrating? Futhermore: the last equation: LHS is a function of $x$ and RHS is just a constant.
To solve the problem write $f(x)$ as a sum of cosines (so in in it's Fourier form), take the derivative and then divide by $2$ to get $g(x)$. From there you can get the coefficients without much hustle.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometry proof based on if $\cos A+$.... If $$\cos A+\cos B+\cos C=\sin A.+\sin B+\sin C=0$$ then prove that: $$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$
My Attempt;
Here,
$$\cos A+\cos B+\cos C=0$$
$$\cos A+\cos B=-\cos C$$
squaring on both sides,
$$\cos^2A+2\cos A.\cos B+\cos^2B=\cos^2C$$
Again,
$$\sin A+\sin B+\sin C=0$$
$$\sin A+\sin B=-\sin C$$
squaring on both sides
$$\sin^2A+2\sin A.\sin B+\sin^2B=\sin^2C$$
Now,
$$L.H.S=\cos 3A+\cos 3B+\cos 3B$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=4(\cos^3A+\cos^3B+\cos^3C)-3(\cos A+\cos B+\cos C)$$
$$=4(\cos^3A+\cos^3B+\cos^3C)-3\times 0$$
$$=4(\cos^3A+\cos^3B+\cos^3C$$
Now, please help me to continue from.here.
NOTE: PLEASE DO NOT USE COMPLEX NUMBERS IN THE SOLUTION
| Maybe you could try a complex approach (which is simpler ;-))
Let $z,u,v$ be three complex numbers:
$z=\cos(A)+i\sin(A)$, $u=\cos(B)+i\sin(B)$, $v=\cos(C)+i\sin(C)$.
then by hypothesis $z+u+w=0$. Hence
$$0=(z+u+v)^3=z^3+u^3+v^3+3uv(u+v)+3uz(u+z)+3zv(z+v)+6zuv\\=
z^3+u^3+v^3+3uv(-z)+3uz(-v)+3zv(-u)+6zuv=z^3+u^3+v^3-3zuv.$$
Therefore $z^3+u^3+v^3=3zuv$
which implies, after separating real and imaginary parts,
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C)$$
and
$$\sin 3A+\sin 3B+\sin 3C=3\sin(A+B+C).$$
So you get two (real) trigonometric relations at one stroke!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show that formally one can firstly fix $0<c \le 3$, introduce a variable $0\le\alpha\le1$ and express $a$ and $b$ through $\alpha$
$$a= \alpha(3-c)$$ $$b=(1-\alpha)(3-c)$$
Then we minimize $(a+2)^3+(b+2)^3+(c+2)^3$ w.r.t $\alpha$ treating $c$ as a parameter. We get $\alpha=\frac{1}{2}(3-c)$. Further we maximize $(a+2)^3+(b+2)^3+(c+2)^3$ one more time w.r.t. $c$.
That is quite tedious. I am wondering whether their is a more elegant way. Probably using idea of norms. Basically we need to show that $\lVert(a,b,c)+(2,2,2)\rVert_3 \ge (81)^{1/3}$ while $\lVert(a,b,c)\rVert_1=3$ and $a,b,c>0$.
| Because $(a+2)^3+(b+2)^3+(c+2)^3\geq\frac{1}{9}(a+2+b+2+c+2)^3=81$.
| {
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"url": "https://math.stackexchange.com/questions/1898007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is there a simple function that generates the series; $1,1,2,1,1,2,1,1,2...$ or $-1,-1,1,-1,-1,1...$ I'm thinking about this question in the sense that we often have a term $(-1)^n$ for an integer $n$, so that we get a sequence $1,-1,1,-1...$ but I'm trying to find an expression that only gives every 3rd term as positive, thus it would read;
$-1,-1,1,-1,-1,1,-1,-1...$
Alternatively a sequence yielding $1,1,2,1,1,2,1,1,2...$ could also work, as $n$ could just be substituted by it in $(-1)^n$
| Let $\omega \neq 1$ be a cubic root of unity. We have that $1 + \omega + \omega^2 = 0$, i.e. $\omega + \omega^2 = -1$. Also, $w^{-1} = \omega^2$ and $\omega^3 = 1$.
Put $a_n = \omega^n + \omega^{-n}$. We get the sequences
\begin{eqnarray*}
a_0 &=& \omega^{0} + \omega^{0} = 2 \\
a_1 &=& \omega^{1} + \omega^{-1} = \omega^{1} + \omega^{2} = -1 \\
a_2 &=& \omega^{2} + \omega^{-2} = \omega^{2} + \omega^{1} = -1 \\
\end{eqnarray*}
and then the sequence repeats.
If you put $b_n = \frac{2 a_n -1}{3}$, then you get $1, -1, -1, 1, -1, -1, \cdots$.
| {
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"url": "https://math.stackexchange.com/questions/1899416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b = 0$$
from which i got $a = -610, b = 987$
but from wolfram alpha the remainder of $-610x^{17}+987x^{16}+1 \over x^2-x-1$ is $x-1$
Somebody please show me where i went wrong ?
Thanks.
| A slightly different (but longer) method:
Let roots of $x^2-x-1$ be $c$ and $d$
Using Vieta's formulas, we get
$$c+d=1\tag{1}$$
$$cd=-1\tag{2}$$
Using Factor theorem, we get
$$ac^{17}+bc^{16}=-1\tag{3}$$
$$ad^{17}+bd^{16}=-1\tag{4}$$
Multiplying $(3)$ with $d^{16}$ and using $(2)$, we get
$$ac+b=-d^{16}\tag{5}$$
Multiplying $(3)$ with $c^{16}$ and using $(2)$, we get
$$ad+b=-c^{16}\tag{6}$$
Subtracting $(6)$ from $(5)$, we get
$$a(c-d)=c^{16}-d^{16} \Rightarrow a=\frac{c^{16}-d^{16}}{c-d}=(c+d)(c^2+d^2)(c^4+d^4)(c^8+d^8)$$
Likewise, eliminating $a$, we get $$-b=\frac{p^{17}-q^{17}}{p-q}=p^{16}+p^{15}q+ \cdots + q^{16}$$
Using simple algebraic manipulations we get $a=987$ and correspondingly $b=-1597$, which are left as an exercise to reader.
Hint:
Using $(1)$ and $(2)$, we give the finishing touch to this problem.
Answer to the exercise left to readers:
$c+d=1$
$c^2+d^2=(c+d)^2-2cd=3$
$c^4+d^4=(c^2+d^2)^2-2c^2d^2=7$
$c^8+d^8=(c^4+d^4)^2-2c^4d^4=47$
So, $\color{red}{a=1 \cdot 3 \cdot 7 \cdot 47=987}$
$-b=(c^{16}+d^{16})-(c^{14}+d^{14})+ \cdots -(c^2+d^2)+1$
Let $k_{2n}=c^{2n}+d^{2n}$
Also, $ k_{2n+4}=3k_{2n+2}-k_{2n}$
We have
$$\begin{array}{c c c} \\
k_0 & 2 \\
k_2 & 3\\
k_4 & 7\\
k_6 & 18\\
k_8 & 47\\
k_{10} & 123\\
k_{12} & 372\\
k_{14} & 843\\
k_{16} & 2207\\
\end{array}$$
Thus, $\color{blue}{b=-(2207-843+322-123+47-18+7-3+2)=-1597}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan 10^{\circ}}{\tan 20^{\circ}}}{\tan 10^{\circ}+\frac{1}{\tan 20^{\circ}}}$$
assuming $y=10$ and using $\tan 2y=\frac{2 \tan y}{1-\tan ^2 y}$ we get
LHS as
$$\frac{3+\frac{\tan y}{\tan 2y}}{\tan y+\frac{1}{\tan 2y}}=\frac{\tan y(7-\tan ^2 y)}{1+\tan ^2 y}$$
how to proceed further?
| If $\cot3A=\cot3x,3x=n180^\circ+3A$ where $n$ is any integer
$x=n60^\circ+A$ where $n\equiv0,1,2\pmod3$
$$\cot3A=\cot3x=\dfrac1{\tan3x}=\dfrac{1-3\tan^2x}{3\tan x-\tan^3x}=\dfrac{\cot^3x-3\cot x}{3\cot^2x-1}$$
$$\iff\cot^3x-3\cot3A\cot^2x-3\cot x+\cot3A=0\ \ \ \ (1)$$
If $\cot x_1,\cot x_2,\cot x_3$ are roots of $(1),$
$$\cot x_1\cot x_2+\cot x_2\cot x_3+\cot x_3\cot x_1=-\dfrac31$$
$$\implies\cot A\cot(A+60^\circ)+\cot(A+60^\circ)\cot(A+120^\circ)+\cot(A+120^\circ)\cot A=-3$$
Here $A=20^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900836",
"timestamp": "2023-03-29T00:00:00",
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Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?
| To find $\sum_{k=1}^{n}(k^4)$ you may follow this process.
Consider the identity, $(x+1)^5-x^5=5x^4+10x^3+10x^2+5x+1$.
Putting $x=1,2,3,...,(n-1),n$ successively, we get,
$2^5\space-\space 1^5=5\cdot1^4+10 \cdot1^3+10\cdot1^2+5\cdot1+1$
$3^5\space-\space 2^5=5\cdot2^4+10 \cdot2^3+10\cdot2^2+5\cdot2+1$
$\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\cdot$
$\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\space\space\space\cdot\space\space\space\space\space\space\space\cdot$
$n^5\space-\space (n-1)^5=5\cdot(n-1)^4+10 \cdot(n-1)^3+10\cdot(n-1)^2+5\cdot(n-1)+1$
$(n+1)^5\space-\space n^5=5\cdot n^4+10 \cdot n^3+10\cdot n^2+5\cdot n+1$
Adding column wise we get,
$(n+1)^5-1^5=5(1^4+2^4+\cdot\cdot\cdot+n^4)+10(1^3+2^3+\cdot\cdot\cdot+n^3)+10(1^2+2^2+\cdot\cdot\cdot+n^2)+5(1+2+\cdot\cdot\cdot+n)+(1+1+\cdot\cdot\cdot+n)$
$\implies n^5+5n^4+10n^3+10n^2+5n=5\sum_{k=1}^{n}(k^4)+10\sum_{k=1}^{n}(k^3)+10\sum_{k=1}^{n}(k^2)+5\sum_{k=1}^{n}(k)+n$
Knowing the sum of the series' $\sum n^3$ , $\sum n^2$ and $\sum n$, you can solve for $\sum n^4$ from the above equation to get the result:
$\sum n^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{30}$
| {
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"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.
Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.
OBS. $O_2$ is the zero matrix
Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$
This is all I could get.
UPDATE
The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.
| Let's rewrite the - correct - identity you obtained as follows:
\begin{align*}& A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2\\
\iff& t(A-B)=A^2 -B^2- (\det(A) - \det(B))I_2\end{align*}
Squaring both sides of your original identity gives
$$(A^2 -B^2)^2=t^2\underbrace{(A-B)^2}_{=0} + 2(\det(A) - \det(B))t(A-B) + (\det(A) - \det(B))^2I_2.$$
Now substitute $t(A-B)$ given by the second equation:
\begin{align*}(A^2 -B^2)^2& =2(\det(A) - \det(B))[(A^2 -B^2)- (\det(A) - \det(B))I_2] + (\det(A) - \det(B))^2I_2\\
& = 2(\det(A) - \det(B))(A^2 -B^2)- (\det(A) - \det(B))^2I_2
\end{align*}
and lastly apply Cayley Hamilton to $A^2-B^2$ and compare the coefficients.
Note that no cancellation takes place, as, using that $tr(A)=tr(B)=t$, one can prove that
$$2(\det(A) - \det(B))=-(tr(A^2+B^2))$$
Indeed this is again a consequence of Cayley Hamilton, taking its trace:
$$A^2-tr(A)A+\det(A)I_2=0\Rightarrow tr(A^2)-tr(A)^2+2\det(A)=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $9|2^{2n+1}+2^{4n+1}-4$
Show that $9|2^{2n+1}+2^{4n+1}-4$ ( n is a positive integer):
1-Using induction
2-Don't use induction
I have noticed that $2^m \equiv 2^{m+6} \pmod 9 $ but couldn't use it to solve the problem!
| It's true for all non-negative integers $n$.
Without induction:
$9\mid 2^{2n+1}+2^{4n+1}-4$ is equivalent to $$9\mid 2\cdot 2^{2n+1}+\left(2^{2n+1}\right)^2-8$$
$$=\left(2^{2n+1}+1\right)^2-9$$
I.e. $9=3^2\mid \left(2^{2n+1}+1\right)^2$, i.e. $3\mid 2^{2n+1}+1$, which is true,
because $2^{2n+1}\equiv (-1)^{2n+1}\equiv -1\pmod{3}$, because $2n+1$ is odd for all $n\in\mathbb Z$, $n\ge 0$.
With induction:
Clearly $9\mid 2^{2\cdot 0+1}+2^{4\cdot 0+1}-4=0$.
If $9\mid 2^{2n+1}+2^{4n+1}-4$, then
$$9\mid 2^{2(n+1)+1}+2^{4(n+1)+1}-4$$
$$=\left(2^{2n+1}+2^{4n+1}-4\right)+3\left(2^{2n+1}+5\cdot 2^{4n+1}\right)$$
because $$2^{2n+1}+5\cdot 2^{4n+1}\pmod{3}$$
$$\equiv (-1)^{2n+1}-(-1)^{4n+1}\pmod{3}$$
$$\equiv -1-(-1)\equiv 0\pmod{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A really interesting logarithm integral I need to prove that
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x =
\frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$
My approach
We know that
$$\int^1_0 \frac{\log x}{x^2-2\cos(\theta)x+1}\mathrm{d}x =- \frac{\mathrm{cl}_2(\theta)}{\sin(\theta)}$$
Let $\theta=\pi/3$
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x =- \frac{2}{\sqrt{3}}\mathrm{cl}_2(\pi/3)$$
By definition
\begin{align}
\mathrm{cl}_2(\pi/3) &= -\int^{\pi/3}_0 \log(2\sin \frac{t}{2})\mathrm{d}t\\
& = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\sin(\pi x))\,\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\pi)-\log\Gamma(x)-\log\Gamma(1-x)\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+\log\Gamma(1-x)\mathrm{d}x\\
& = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+2\pi\int^{1}_{5/6}\log\Gamma(x)\mathrm{d}x
\end{align}
Now use the loggamma integral
$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$
Implies the two integrals
$$\int_{0}^{1/6} \log \Gamma(x) \, \mathrm dx = \frac{1}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,1/6)$$
$$\int_{5/6}^{1} \log \Gamma(x) \, \mathrm dx =\frac{1}{2}\log(2\pi)-( \frac{5}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,5/6))$$
Hence we have
$$\mathrm{cl}_2(\pi/3) = -\frac{\pi}{3}\log(2\pi)+2\pi (\frac{1}{6}\log(2\pi)+\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6)) \\=2\pi (\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6)) $$
This implies that
$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x = \frac{-4\pi}{\sqrt{3}}\left(\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6) \right) = \frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$
Maybe using
$$ \zeta(s,p/q)=2\Gamma(1-s)(2\pi q)^{s-1}\sum_{n=1}^q \sin \left[\frac{\pi s}{2}+ \frac{2 \pi n p}{q}\right] \zeta(1-s,n/q) $$
Question
*
*My approach seems to be so indirect I am really interested in seeing more directed approaches that don't use loggamma integral.
*I am not sure about the last step (relating the derivative of the Hurwitz zeta function to the trigamma.
| Hint. One may write
$$
\int^1_0 \frac{\log x}{x^2-x+1}dx=\int^1_0 \frac{(1+x)\log x}{1+x^3}dx=\left.\partial_s\int^1_0 \frac{(1+x)\:x^s}{1+x^3}dx\right|_{s=0}
$$ then after a change of variable one may use
$$
\int^1_0 \frac{t^s}{1+t}dt=\frac12\psi\left(\frac{s}2+1\right)-\frac12\psi\left(\frac{s}2+\frac12\right), \quad s>-1.
$$
| {
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Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| As $P(0)=1\cdot3\cdot5\cdot7\cdot9,$ and $9\nmid P(10),7\nmid P(8)$
The highest divisors of $P(n)$ must divide $3\cdot5$
$P(n)\equiv(n+1)n^2(n+2)^2\pmod3$
Clearly, $3\mid n(n+1)(n+2)$
$P(n)\equiv(n-2)(n-1)n(n+1)(n+2)\pmod5$
Clearly, $5\mid(n-2)(n-1)n(n+1)(n+2)$
| {
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If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the same expression can be written as $$a^2 + (a^2 + 1)(a+1)^2$$ which would imply that there is no real root. Would really appreciate some help, thanks.
| Since the expression is symmetric in $a$ and $b$, it is not restrictive to assume $b=a+1$, so
$$
a^2+b^2+a^2b^2=
a^2+(a+1)^2+a^2(a+1)^2=
a^4+2a^3+3a^2+2a+1
$$
If you don't see an easy factorization, note that for $a=0$ the statement is clear; for $a\ne0$ we can write
$$
a^4+2a^3+3a^2+2a+1
=
a^2\left(a^2+\frac{1}{a^2}+2a+\frac{2}{a}+3\right)
$$
Now remember that
$$
a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2
$$
so the expression becomes
$$
a^2\left(\left(a+\frac{1}{a}\right)^2+2\left(a+\frac{1}{a}\right)+1\right)=
a^2\left(\left(a+\frac{1}{a}\right)+1\right)^2=(a^2+a+1)^2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sum^{6}_{i=1} a_{i}=\frac{15}{2}$ and $ \sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4} \implies \prod_{i=1}^{6} a_{i} \leq \frac{5}{2}$ Let $a_{i}$, $1 \leq i \leq 6,$ be real numbers such that
$\displaystyle\hspace{1.2 in}\sum^{6}_{i=1} a_{i}=\frac{15}{2}\;\;$ and $\;\;\displaystyle\sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4}$.
Prove that $\hspace{.15 in}\displaystyle\prod_{i=1}^{6} a_{i} \leq \frac{5}{2} $.
I was thinking if I consider the first summation and extended it, it going be pretty long which $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}= \frac{15}{2}$ and the second one like $a^{2}_{1}+a^{2}_{2}+a^{2}_{3}+a^{2}_{4}+a^{2}_{5}+a^{2}_{6}= \frac{45}{4}$. But I do not think this is the shortest of doing that, I am wondering if someone would be able to give me a hint so I can think better than this. Thank you
| Here is a Calculus of Variations approach. Perhaps not terribly elegant, but it works.
$$
\sum_{j=1}^6a_j=\frac{15}2\implies\sum_{j=1}^6\delta a_j=0\tag{1}
$$
$$
\sum_{j=1}^6a_j^2=\frac{45}4\implies\sum_{j=1}^6a_j\delta a_j=0\tag{2}
$$
To maximize $\prod\limits_{j=1}^6a_j$, we want
$$
\sum_{j=1}^6\frac{\delta a_j}{a_j}=0\tag{3}
$$
for all variations under the conditions $(1)$ and $(2)$. Linearity implies that there are constants $b$ and $c$ so that
$$
\frac1{a_j}=b+ca_j\tag{4}
$$
Multiplying $(4)$ by $a_k$ and shuffling yields
$$
ca_j^2+ba_j-1=0\tag{5}
$$
Equation $(5)$ implies that there are only two possible values for $a_j$, $h$ and $k$. All of the $a_j$ cannot be equal since then $(1)$ implies $a_j=\frac54$, which does not satisfy $(2)$.
Thus, we have $3$ cases:
$$
h+5k=\frac{15}2\quad\text{and}\quad h^2+5k^2=\frac{45}4\implies(h,k)\in\left\{\left(0,\frac32\right),\left(\frac52,1\right)\right\}\tag{6}
$$
This gives products of $0$ and $\frac52$.
$$
2h+4k=\frac{15}2\quad\text{and}\quad2h^2+4k^2=\frac{45}4\implies(h,k)=\left(\frac{5\pm\sqrt{10}}4,\frac{10\mp\sqrt{10}}8\right)\tag{7}
$$
This gives products of $\frac{125(247\pm14\sqrt{10})}{16384}$.
$$
3h+3k=\frac{15}2\quad\text{and}\quad3h^2+3k^2=\frac{45}4\implies(h,k)=\left(\frac{5+\sqrt5}4,\frac{5-\sqrt5}4\right)\tag{8}
$$
This gives a product of $\frac{125}{64}$.
The greatest of these products is $\frac52$ given by the second solution in $(6)$: $\left\{\frac52,1,1,1,1,1\right\}$.
Therefore,
$$
\prod_{j=1}^6a_j\le\frac52\tag{9}
$$
| {
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Show product of $(r^5-r+2)$ over the 5 roots of $x^5+x^3+1=0$ is $1$ Let $(a,b,c,d,e)$ be the five roots of $x^5+x^3+1=0$, and let $g(x) = x^5-x+2$.
Prove
$$
g(a)g(b)g(c)g(d)g(e) = 1
$$
preferably without pages of messy algebra.
| We'll do some computations. As noted in G. Sassatelli's comment, for roots of our polynomial,
$$x^5-x+2 - (x^5+x^3+1)=-x^3-x+1.$$
Use polynomial division to get
$$x^2+1 + x^2(x^3+x-1)=x^5+x^3+1=0$$
and
$$-(x^3+x-1)=(1+x^2)/x^2,$$
since none of the roots are zero.
Realize that $abcde=-1$, so our product is equivalent to
$$h(a)\cdots h(e)$$
with $h(x)=(1+x^2)$.
Use polynomial division again:
$$(1+x^2)x^3+1=x^5+x^3+1=0$$
$$(1+x^2)=-1/x^3$$
So our product is equivalent to
$$(-1)^5/(a^3b^3c^3d^3e^3)=1.$$
Please tell me if there is a sign error.
I believe there is a sort of euclidean algorithm behind all this here that will allow you to compute $g(a)\cdots g(e)$ for any $g$.
| {
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How do I prove thcd It makes me think of means... The RHS Is like a geometric mean but the "divided by 4" annoyes me. The LHS is probably an arithmetic mean... Can they be combined?
| By Cauchy-Schwarz,
$$\sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}}
=\sqrt{a^2 + b^2 + c^2 + d^2}\cdot \sqrt{4\cdot\frac{1}{4^2}}\geq \frac{a+b+c+d}{4}\\\geq \sqrt[3]{\frac{abc + abd + acd + bcd}{4}}$$
where in the last step we use
Inequality. $\frac{1}{16}(a+b+c+d)^3 \geq abc+bcd+cda+dab$
| {
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Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
| Hint: the left hand side equals $$(1+n)^3=(1+n)^2+n(1+n)^2$$
and $$\sum_{k=0}^nk=\frac{n(n+1)}{2}$$
| {
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Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.
| Hint: The given inequality is equivalent to
$$\sum_{cyc} \frac1{1-a} - \frac32\sum_{cyc}a^2 \ge 4 $$
which follows from noting that for $x \in (0, 1)$,
$$f(x) = \frac1{1-x}-\frac32x^2-\frac43 -\frac54\left(x-\frac13\right) = \frac{(1+2x)(1-3x)^2}{12(1-x)}\geqslant 0$$
| {
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Showing $\frac{z}{1+z}+\frac{2z^2}{1+z^2}+...+\frac{2^{k}z^{2^k}}{1+z^{2k}}+...=\frac{z}{1-z}$
Prove that for $\left\lvert z \right\rvert<1$, $\dfrac{z}{1+z}+\dfrac{2z^2}{1+z^2}+...+\dfrac{2^{k}z^{2^k}}{1+z^{2k}}+...=\dfrac{z}{1-z}$. Also, justify any change in the order of summation.
This is a exercise from my textbook, but I have no idea. I have shown that: $$\dfrac{z}{1+z}+\dfrac{2z^2}{1+z^2}+...+\dfrac{2^{k}z^{2^k}}{1+z^{2k}}-\dfrac{z}{1-z}= \dfrac{2^{k+1}}{1-\dfrac{1}{z^{2^{k+1}}}}$$
but I do not know how to show $2^{k+1}$
is growing 'slow enough' in order to make the whole fraction tending zero.
The hint from the book is: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+...+2^k$
Could you please give me some more hint? Thank you.
| Note that for each $k \ge 1$,
$$\frac{2^k z^{2^k}}{1+z^{2^k}} = \frac{2^k z^{2^k}(1 - z^{2^k})}{(1 + z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}(1+z^{2^k} - 2z^{2^k})}{(1+z^{2^{k}})(1-z^{2^k})} = \frac{2^kz^{2^k}}{1-z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}. $$
Further, since $\lvert z \rvert < 1$,
$$\lim_{k\to \infty} \frac{2^k z^{2^k}}{1 - z^{2^k}} = 0$$
With these two facts, you'll be able to prove the result.
| {
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Solution of $\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$
Find the solution of $$\frac{xdx - ydy}{xdy - ydx} = \sqrt{\frac{1 - y^{2} + x^{2}}{x^{2} - y^{2}}}$$
I was able to bring it down to $$\frac{d(x^2-y^2)}{\sqrt{1+x^2-y^2}}=2\left(\frac{x.d(y/x)}{\sqrt{1-(y/x)^2}}\right)$$
Any help would be greatly appreciated. Thanks in advance!
$$$$
Edit: An answer exists here, but I'm trying to solve the question $without$ using trigonometric or hyperbolic substitution (I was told it could be done without both, and that the solution was quite neat).
$$$$
EDIT: Also the question that is mentioned as a possible duplicate is $\textbf{different}$ from mine.
| We can write the equation as follows:
$$ \frac{xdx-ydy}{x^2-y^2}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}\frac{xdy-ydx}{x^2-y^2}$$
It's
$$ \frac{xdx-ydy}{x^2-y^2}=\frac{d\sqrt{x^2-y^2}}{\sqrt{x^2-y^2}}$$
and
$$\frac{xdy-ydx}{x^2-y^2}=\frac{1}{2}d\ln\frac{x+y}{x-y}$$
$\,\displaystyle \sqrt{x^2-y^2}\,$ and $\,\displaystyle \ln\frac{x+y}{x-y}\,$ used as variables gives the wished result.
| {
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Trigonometric Integral - Quotient Rule: $\int \frac{x\cos(x)}{(x+\cos(x))^2} dx$? $$\int \frac{x\cos(x)}{(x+\cos(x))^2} dx$$
I suspect that the integral can be solved by thinking of it as the derivative of a quotient with a denominator of $x + \cos(x)$. How else could this integral be evaluated?
| We have $$I=\int\frac{x\cos\left(x\right)}{\left(x+\cos\left(x\right)\right)^{2}}dx=\int\frac{1-\sin\left(x\right)}{\left(x+\cos\left(x\right)\right)^{2}}\frac{x\cos\left(x\right)}{1-\sin\left(x\right)}dx
$$ $$\stackrel{IBP}{=}-\frac{1}{x+\cos\left(x\right)}\frac{x\cos\left(x\right)}{1-\sin\left(x\right)}+\int\frac{1}{1-\sin\left(x\right)}dx
$$ and the last integral is easy to evaluate using the tangent half substitution $$\int\frac{1}{1-\sin\left(x\right)}dx\stackrel{u=\tan\left(\frac{x}{2}\right)}{=}2\int\frac{1}{\left(u-1\right)^{2}}=-\frac{2}{\tan\left(\frac{x}{2}\right)-1}
$$ so $$ \begin{align}I=
& -\frac{x\cos\left(x\right)}{\left(x+\cos\left(x\right)\right)\left(1-\sin\left(x\right)\right)}-\frac{2}{\tan\left(\frac{x}{2}\right)-1}+C
\\ =
& -\frac{x\cos\left(x\right)}{\left(x+\cos\left(x\right)\right)\left(1-\sin\left(x\right)\right)}-\frac{2\left(\cos\left(x\right)+1\right)}{\sin\left(x\right)-\cos\left(x\right)-1}+C
\\=
& \color{red}{\frac{1+\sin\left(x\right)}{x+\cos\left(x\right)}+C'}
\end{align}$$ as wanted.
| {
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How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy?
For small values of $x$, how good is the approximation $cos(x)\approx 1$? How small must $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy?
My teacher told me, it would be easier to do the second part of the question first, so I need to find out how small $x$ be to have $\frac{1}{2} \cdot 10^{-8}$ accuracy
So as far as I understand, I need to solve for $x$ in the following equation
$|cos(x)-1|<\frac{1}{2} \cdot 10^{-8} \Rightarrow \\
|x| < \ arccos(1-\frac{1}{2} \cdot 10^{-8}) \Rightarrow \\
x \in \ (-arccos(1-\frac{1}{2} \cdot 10^{-8}),arccos(1-\frac{1}{2} \cdot 10^{-8}))$
but that just seems like a messy answer. Any ideas?
| If you haven't learned about Taylor series.
Will you accept that $\sin x \approx x$ when $x$ is small?
and $\sin x < x$ for all $x.$
You can do this geometrically, plot your unit circle and see that for small $x$, the distance traveled about the $x$ axis is very nearly equal to the length of the curve.
$\cos x = \sqrt{1-sin^2 x}$
$\cos x \approx (1-x^2)^{\frac 12}$
In fact, $\cos x$ is slightly greater than (1-x^2)^{\frac 12}$ when $x$ is small, but we will stick with this approximation.
Now do a binomial expansion on $(1-x^2)^{\frac 12}$
$(1-x^2)^{\frac 12} 1 - \frac 12 x^2$
$cos x \approx 1 - \frac 12 x^2 $
Your error is:
$\frac 12 x^2 < \frac 12 10^{-8}$
If $|x| < 10^{-4}, |\cos x - 1| < \frac 12 10^{-8}$
| {
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How to prove $\sqrt{5}-\sqrt{3}$ is bigger than $\sqrt{15}-\sqrt{13}$ Although I can determine using a calculator that $\sqrt{5}-\sqrt{3}$ is larger than $\sqrt{15}-\sqrt{13}$, how would I go about proving that? My teacher gave us a hint which was to use the difference of two squares identity $(a^2-b^2) = (a-b)\cdot(a+b)$, but I don't see how to proceed.
Thanks in advance!
| Note that $$\sqrt { 5 } -\sqrt { 3 } =\frac { \left( \sqrt { 5 } -\sqrt { 3 } \right) \left( \sqrt { 5 } +\sqrt { 3 } \right) }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } ,\\ \quad \sqrt { 15 } -\sqrt { 13 } =\frac { \left( \sqrt { 15 } -\sqrt { 13 } \right) \left( \sqrt { 15 } +\sqrt { 13 } \right) }{ \sqrt { 15 } +\sqrt { 13 } } =\frac { 2 }{ \sqrt { 15 } +\sqrt { 13 } } \\ $$
$$\frac { 2 }{ \sqrt { 5 } +\sqrt { 3 } } >\quad \frac { 2 }{ \sqrt { 15 } +\sqrt { 13 } } $$
| {
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What is the fewest number of squares required to cover a $11\times13\text{ cm}$ rectangle without overlap? I need help figuring out this math puzzle: I have a $11\times13\text{ cm}$ rectangle and I need help figuring out the least number of squares I need to cover the rectangle without overlap. I'm told the answer should be at most 5. If you can, provide a picture to help me understand.
| I can prove there is no 5-square solution.
The partitions of $11\times 13 = 143$ into sums of five squares can be enumerated:
$$ \matrix{1^2 &+ 1^2 &+ 2^2 &+ 4^2 &+ 11^2\cr
1^2 &+ 1^2 &+ 4^2 &+ 5^2 &+ 10^2\cr
1^2 &+ 2^2 &+ 5^2 &+ 7^2 &+ 8^2\cr
1^2 &+ 3^2 &+ 4^2 &+ 6^2 &+ 9^2\cr
2^2 &+ 4^2 &+ 5^2 &+ 7^2 &+ 7^2\cr
2^2 &+ 5^2 &+ 5^2 &+ 5^2 &+ 8^2\cr
3^2 &+ 3^2 &+ 3^2 &+ 4^2 &+ 10^2\cr
3^2 &+ 3^2 &+ 5^2 &+ 6^2 &+ 8^2\cr
}$$
All but one of these can be eliminated out of hand, by looking at the two largest squares: $a \times a$ and $b \times b$ squares can't fit in a rectangle without overlapping unless the rectangle has one dimension at least $a+b$. The remaining possiblility is $2^2 + 5^2 + 5^2 + 5^2 + 8^2$, but it's easy to see that an $8 \times 8$ square and three $5 \times 5$ squares won't fit in the rectangle.
EDIT: There's no tiling of the $11 \times 13$ rectangle with $5$ squares even if you don't require integer sides. It's best to work up to $5$ tiles one at a time.
With one tile ($a \times a$) you can only tile an $a \times a$ rectangle.
With two tiles, both must be $a \times a$, and you get an $a \times 2a$ rectangle. Henceforth, I'll leave out the $a$, and assume the greatest
common divisor of edge lengths is $1$, so call this $1 \times 2$.
With three tiles, at least one must be on an edge of your rectangle, and the rest of the rectangle is a two-tile rectangle. There are two cases, depending on how that two-tile rectangle is oriented:
With four tiles, you could put another square on one side of a three-tile rectangle, or you could have four equal squares, each taking one corner of a $2 \times 2$ square. There are five possibilities.
With five tiles, you could put one square on one side of a four-tile rectangle, obtaining a $4 \times 7$, $7 \times 3$, $5 \times 1$, $4 \times 5$,
$4 \times 2$, $8 \times 5$, $3 \times 8$, $7 \times 2$ or $5 \times 7$ rectangle.
Or if no square takes a whole side of the rectangle, you must have one square in each of the four corners of the rectangle and one square not on a corner.
If so, it's not hard to see that this non-corner square must be on an edge, let's say the right edge.
On the left edge, the two squares may be the same size (resulting in a $6 \times 5$ rectangle) or different sizes. If they are different sizes, the smaller one must be the same size as its neighbour to the right, resulting
in a $7
\times 6$ rectangle.
(I hope) that's all the possibilities, not counting rotations and reflections.
None of the possibilities has an $11$ to $13$ ratio.
| {
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What is a good upper bound $n^n(n-1)^{n-1}\ldots2^21^1$? Given an integer $n \ge 1$, I'd like to have a not-very-loose upper bound for the integer $$u(n) := \Pi_{k=1}^n k^k = n^n(n-1)^{(n-1)}\ldots2^21^1.$$
It's easy see that, $u(n) \le n^{n(n+1)/2}$, but this is not very interesting.
Update
We have $u(n) \le e^{\left(\frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)\right)}$, and we can't really do much better!
Indeed, using Euler-Maclaurin, we have
$ \log(u(n)) = \int_2^nx\log x dx = \frac{1}{4}n^2(2\log(n) - 1) - 2\log(2) + \frac{1}{4} + \text{error terms}$, which is comparable to the bound $\log(u(n)) \le \frac{1}{2}n(n+1)\log\left(\frac{2n + 1}{3}\right)$ in the accepted answer (see below). In particular, we can conclude that accepted answer's bound is tight!
| Using Jensen's inequality:
Letting $A= \sum_{k=1}^n k = \frac{n (n+1)}{2}$ and $B= \sum_{k=1}^n k^2 = \frac{n (n+1)(2n+1)}{6}$
We have
$$
\begin{align}
\log(u(n))
&=\sum_{k=1}^n k \log(k) \\
&= A \sum_{k=1}^n \frac{k}{A} \log(k) \tag{1}\\
&\le A \log \left( \sum_{k=1}^n \frac{k}{A} k \right) =
A \log \left( \frac{B}{A} \right) \tag{2}
\end{align}
$$
Hence
$$\log(u(n)) \le \frac{n(n+1)}{2} \log\left(\frac{2 n+1}{3}\right) \tag{3}$$
The bound seems to be quite tight:
Update: as noted by comments and OP, the bound $(3)$ agrees with the true order of growth; this can be checked by applying the trapezoidal rule to the integral:
$$ -\frac{1}{4}=\int_{0}^{1} x \log(x) dx \approx \frac{1}{n+1} \sum_{k=1}^n \frac{k}{n} \log\left(\frac{k}{n}\right) $$
which gives
$$ \log(u(n)) \approx\frac{ n(n+1)}{2}\left( \log n -\frac{1}{2} \right) \tag{4}$$
If one is interested in an approximation (instead of a bound), $(4)$ might be preferable.
Better asymptotics here (from comments).
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How to solve the following first order ODE I want to solve the following ODE: $$\left(x^2+x y(x)\right) y'(x)+y(x)^2+3 x y(x)=0.$$ However, I didn't find this fit any types in the textbook. For example, I expressed it as $$M(x,y)dx+N(x,y)dy=0$$ and checked whether $M_y=N_x$ which doesn't hold though. Does anyone know how to solve this?
| $$\left(x^2+x y\right) y'+y^2+3 x y=0$$
Make substitution $y=xz$, where $z$ is function of $x$.
$$
\left(x^2+x^2z\right)\left(z+xz^{'}\right)+x^2z^2+3x^2z=0
$$
Factor out $x^2$ and solve separable ODE
$$
\left(1+z\right)\left(z+xz^{'}\right)+z^2+3z=0\to\\
x\frac{dz}{dx}=-z\frac{z+3}{z+1}-z=-z\frac{2z+4}{z+1}\to\\
\frac{z+1}{2z(z+2)}
dz=-\frac{1}{x}dx\to\frac{1}{4}(\ln(z)+\ln(z+2))=-\ln(x)+C^*\to\\
\ln(z(z+2))=\ln(1/x^4)+\ln(C)\to\\
z(z+2)=\frac{C}{x^4}
$$
Which after inverse substitution leads to
$$
\frac{y}{x}\left(\frac{y}{x}+2\right)=\frac{C}{x^4}\to \color{red}{y(y+2x)=\frac{C}{x^2}}
$$
Check solution
$$
(x+y)y^{'}=(-1)\left(\frac{C}{x^3}+y\right)\\
\left(x^2+xy\right)y'+y^2+3xy=0\to x\left(x+y\right)y'+y(y+2x)+xy=0\to\\
-x\left(\frac{C}{x^3}+y\right)+\frac{C}{x^2}+xy\equiv 0
$$
Note: factoring out $x^2$, mentioned above, gives us additional particular solution, which is not covered by general formula. $\color{red}{y=0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| The quadratic we get is $x^4-2ax^2-x+a^2-a$. We can factor it by trying to guess the coefficients (this will depend heavily on the equation and how good-written the problem is): Let's assume all coefficients will be integers. Let $p(x)=x^2-2ax^2+a^2-a$. There are two cases:
If we could factor $p(x)$ as a product of a polynomial of degree $3$ and one of degree $1$, we would be finding a root of $p(x)$. Roots of polynomials must divide the constant term, which is $a^2-a=a(a-1)$. The only "obvious" divisors of it are $a$ and $a-1$, which are not roots of $p(x)$, so this is not a case of interest.
Let's try to factor $p(x)$ as a product of two quadratics. The coefficient of $x^4$ is $1$, so the coefficients of $x^2$ in the quadratics must be either both $1$ or both $-1$. Up to sign, assume they are $1$:
$$x^4-2ax^2-x+a^2-a=(x^2+b x+c)(x^2+d x+e).$$
Now, when we do the product on the right-hand side, we get a term of the form $(b+d)x^3$, so we must have $b=-d$, i.e., we are looking for a decomposition of the form
$$x^4-2ax^2-x+a^2-a=(x^2+bx+c)(x^2-bx+e)$$
The constant term on the right-hand side is $ce$, which should be equal to $a^2-a=a(a-1)$. Again, since we are assuming the decomposition is "simple", we need to is decompose $a(a-1)$ as a product of two terms. There are a few options:
$$a(a-1)=1\cdot(a(a-1))=(-a)(1-a)$$
Then we try $c$ to be one of the terms, $d$ to be the other one, expand the right-hand side with this option, which will give us a polynomial of degree 4, with the coefficients multiplying $x^2$ and $x$ depending on $b$. By equaling them to the respective coefficients on the left-hand side, we get a linear equation on $b$ and a quadratic equation on $b$, with coefficients depending on $a$. For the first two options, the solutions ($b$ in terms of $a$) of the linear equation is not a solution of the quadratic, so these don't work.
So let's try the third option:
\begin{align*}
x^4-2ax^2-x+a^2-a&=(x^2+bx-a)(x^2-bx+1-a)\\
&=x^4+(b-b)x^3+x^2-ax^2-b^2x^2-ax^2+bx-abx+abx+a^2-a\\
&=x^4+(1-b^2-2a)x^2+bx+a^2-a,
\end{align*}
and $b=-1$ works. All of this means that
$$x^4-2ax^2-x+a^2-a=(x^2-x-a)(x^2+x+1-a)$$
The solutions of these are
$$x=\frac{1\pm\sqrt{4a+1}}{2}\qquad\text{ and }x=\frac{-1\pm\sqrt{4a-3}}{2}$$
Since $x$ is a square root, it is non-negative. Since $a\geq 1$, the signs before the square roots should be $+$. Now we need to see if these are really solutions. You can try a few values of $a$, (for example, $a=6$ gives $x=3$ and the equation does not hold), just to have some ideas.
On one hand, $x=\sqrt{a-\sqrt{a+x}}<\sqrt{a}$, but $\frac{1+\sqrt{4a+1}}{2}\geq\sqrt{a}$, so this solution does not work.
(Interestingly, however, you can check that $x=\frac{1+\sqrt{4a+1}}{2}$ is a solution for $\sqrt{a+x}=x$ and for $\sqrt{a+\sqrt{a+x}}=\sqrt{a+x}=x$)
The only remaining possibility is $x=\frac{-1+\sqrt{4a-3}}{2}$. It is easy to check, in this case, that $x^2=a-(x+1)$, and since the left-hand side is non-negative the right-hand side is also non-negative, and since $x\geq 0$ it follows that
$$x=\sqrt{a-(x+1)}$$
Now note, similarly, that $(x+1)^2=a+x$, and therefore, again because both $x+1$ and $a+x$ are non-negative, we have $(x+1)=\sqrt{a+x}$, from which we conclude that $x=\sqrt{a-\sqrt{a+x}}$, as we wanted, and therefore $x=\frac{-1+\sqrt{4a-3}}{2}$ is the only solution of the initial equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Picking certain number of balls without replacement and finding its probability distribution Suppose I am to choose three balls without replacement from a bag containing $5$ white and $4$ red balls. What will be the probability distribution of the red balls drawn ?.
According to my book, probability function will be
$$
{3\choose x}\left(\,{4 \over 9}\,\right)^{x}\left(\,{5 \over 9}\,\right)^{3 - x}
$$
What I didn't understand is why my book is taking probability of choosing red ball to be $4/9$ and the probability of choosing a white ball to be $5/9$. I think the above probabilities are of choosing the red and the white balls in the first trial. In other trials the probability of the above two events will change as we are drawing balls without replacement.
| Simplest Approach
Without replacement, the simplest method is to compute how many ways there are to pick $k$ from the $4$ red balls and $3-k$ from the $5$ white balls. Then divide that by the total number of ways to pick $3$ from the $9$ balls in total:
$$
\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}
$$
Pick by Pick Approach
Without replacement, there are $\binom{3}{k}$ orders in which to draw $k$ red balls and $3-k$ white balls. Suppose we draw all the red balls first, the probability of that draw would be
$$
\underbrace{\overbrace{\frac{4}{9}\cdot\frac{3}{8}\cdots}}^{\text{drawing $k$ red balls}}_{k\text{ terms}}
\quad\underbrace{\overbrace{\frac{5}{9-k}\cdot\frac{4}{8-k}\cdots}}^{\text{drawing $3-k$ white balls}}_{3-k\text{ terms}}
=\frac{\frac{4!}{(4-k)!}\frac{5!}{(5-(3-k))!}}{\frac{9!}{(9-3)!}}
=\frac{\binom{4}{k}k!\binom{5}{3-k}(3-k)!}{\binom{9}{3}3!}
=\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}\binom{3}{k}}
$$
But no matter which order we drew the balls, the probability would be the same, the numerators would just change order. Thus, we just multiply the probability of each order by the number of orders:
$$
\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}\binom{3}{k}}\binom{3}{k}
=\frac{\binom{4}{k}\binom{5}{3-k}}{\binom{9}{3}}
$$
Values
$k=0:\dfrac{\binom{4}{0}\binom{5}{3}}{\binom{9}{3}}=\dfrac{10}{84}\doteq0.11904762$
$k=1:\dfrac{\binom{4}{1}\binom{5}{2}}{\binom{9}{3}}=\dfrac{40}{84}\doteq0.47619048$
$k=2:\dfrac{\binom{4}{2}\binom{5}{1}}{\binom{9}{3}}=\dfrac{30}{84}\doteq0.35714286$
$k=3:\dfrac{\binom{4}{3}\binom{5}{0}}{\binom{9}{3}}=\dfrac{4}{84}\doteq0.04761905$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| $$ax^3+8x^2+bx+6=(ax+c) (x^2-2x-3)=ax^3+(c-2a)x^2+(-2c-3a)x-3c$$
$$\begin{align}6&=-3c\Rightarrow c=-2\\8&=c-2a\Rightarrow 2a=c-8=-10\Rightarrow \color{red}{a=-5}\\b&=-2c-3a=4+15=\color{red}{19}\end{align}$$
Using this approach, you will also find out the third root is given by $ax+c=0\Rightarrow x=-\frac{2}{5}$
Also $x^2-2x-3x=(x-3)(x+1)$ gives $x=3$ and $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 7
} |
CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the story:
Consider a sequence of independent binomial trials, each one producing the result success or failure, with probabilities $p$, and $1-p$, respectively.
Let $x$ be the total number of trials which must be carried out in order to attain exactly $r$ successes.
Knowing that the probability mass function for this Negative Binomial Distribution is as follows,
$P(x=X)=\binom{X-1}{r-1}p^r(1-p)^{X-r}$, (for $X \geq r$),
I was trying to prove the following about the corresponding Cumulative Distribution Function.
$P(x \leq X)=\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1-\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}$
I started out with the following:
$\sum_{x=r}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$,
which can be recast as below. (Relation I)
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}+\sum_{x=X+1}^{\infty}\binom{x-1}{r-1}p^r(1-p)^{x-r}=1$
In addition, from binomial theorem, we have:
$\left ( p+(1-p) \right )^X=\sum_{x=0}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Which can be restated in Relation II as below.
$\sum_{x=0}^{x=r-1}\binom{X}{x}p^x(1-p)^{X-x}+\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}=1$
Comparing the relations I and II with the expression for the CDF, the proof boils down to verification of the following:
$\sum_{x=r}^{x=X}\binom{x-1}{r-1}p^r(1-p)^{x-r}=\sum_{x=r}^{x=X}\binom{X}{x}p^x(1-p)^{X-x}$
Any idea how to continue form this point onward?
Thanks.
| We may prove it this way too. (Symbols are a bit different, apologies for that).
We want to show that
$$
\sum_{j=r}^n \binom{j-1}{r-1} (1-p)^{j-r}
= \sum_{j=r}^{n} \binom{n}{j} p^{j-r} (1-p)^{n-j}
$$
For $n=r$, LHS = 1 = RHS.
Assume the result is true for $n = r + m$, we prove it is true for $n = r + m + 1$.
LHS =
$$ \sum_{j=r}^{r+m+1} \binom{j-1}{r-1} (1-p)^{j-r}
= \sum_{j=r}^{r+m} \binom{j-1}{r-1} (1-p)^{j-r} + \binom{r+m}{r-1} (1-p)^{m+1}\\
=^{(Assumption)} \sum_{j=r}^{r+m} \binom{r+m}{j} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} \\
=(*) \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} - p\sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m-j} \\
= \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m+1-j} + \sum_{j=r}^{r+m} \binom{r+m+1}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} + p^{m+1} - p^{m+1}\\
= \sum_{j=r}^{r+m+1} \binom{r+m+1}{j} p^{j-r} (1-p)^{r+m+1-j} + A.
$$
Here
$$A = (*) \sum_{j=r}^{r+m} \Bigg\{ \binom{r+m}{j} + \binom{r+m}{j-1} \Bigg\} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m-j} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1}\\
= \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j=r}^{r+m} \binom{r+m}{j-1} p^{j-r} (1-p)^{r+m+1-j} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1}\\
= \sum_{j=r}^{r+m} \binom{r+m}{j} p^{1+j-r} (1-p)^{r+m-j} - \sum_{j'=r-1}^{r+m-1} \binom{r+m}{j'} p^{j'-r+1} (1-p)^{r+m-j'} + \binom{r+m}{r-1} (1-p)^{m+1} - p^{m+1} = 0.
$$
The equalities with $(*)$ make use of [Pascal rule]https://en.wikipedia.org/wiki/Pascal%27s_rule#:~:text=In%20mathematics%2C%20Pascal's%20rule%20(or,natural%20numbers%20n%20and%20k%2C&text=is%20a%20binomial%20coefficient%3B%20one,(1%20%2B%20x)n.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Deriving a recurrence relationship for derivatives of $\frac{\arctan(x)}{x}$ I was trying to derive a recurrence relantionship for computing the $k$-th derivative of the function (hoping no error during copy)
$$
f(x) = \frac{\arctan(x)}{x}
$$
Using maple I've seen the following derivatives
$$
\begin{array}{l}
f^{(0)}(x) = \frac{\arctan(x)}{x} \\
f^{(1)}(x) = \frac{1}{(x^2 + 1)x} - \frac{f(x)}{x} \\
f^{(2)}(x) = \frac{-2}{(x^2+1)^2} + \frac{-2}{(x^2 + 1)x^2} + \frac{2f(x)}{x^2} \\
f^{(3)}(x) = \frac{8}{(x^2+1)^3x^{-1}} + \frac{4}{(x^2+1)^2x} + \frac{6}{(x^2+1)x^3} - \frac{6f(x)}{x^3} \\
f^{(4)}(x) = \frac{-48}{(x^2+1)^4x^{-2}} + \frac{-8}{(x^2+1)^3} + \frac{-16}{(x^2+1)^2 x^2} + \frac{-24}{(x^2+1)x^4} + \frac{24 f(x)}{x^4} \\
f^{(5)}(x) = \frac{384}{(x^2+1)^5 x^{-3}} + \frac{-48}{(x^2+1)^4x^{-1}} + \frac{64}{(x^2+1)^3x} + \frac{80}{(x^2+1)^2x^3} + \frac{120}{(x^2+1)x^5} - \frac{120f(x)}{x^5}
\end{array}
$$
This lead me to the following expression, as summation, for the $k$-th derivative
$$
f^{(k)}(x) = \sum_{j=1}^k \frac{c_{j,k}}{(x^2+1)^jx^{k-2j+2}} + (-1)^k \frac{k! f(x)}{x^k}
$$
Firstly do you agree with me that the coefficients $c_{j,k}$ fully describes the $k$-th derivative of $f$? If not can you point out the mistake I made?
The next step I would writing $f^{(k+1)}$ in terms of the coefficients $c_{j,k}$ from there I should be able to derive a recurrence relationship for the coefficients (like a triangular table that would allow me to derive the coefficients $c_{j,k}$ for given $k$ and $j = 1 \ldots k$).
Do you think there's a smarter way to do achieve the same result? maybe easier? the computations involved here are quite messy.
Update : here is my attempt
$$
f^{(k+1)}(x) = \sum_{j=1}^k c_{j,k}\left[ \frac{-2jx}{(x^2+1)^{j+1}x^{k-2j+2}} + \frac{k-2j+2}{(x^2+1)^jx^{k-2j+3}}\right] + \frac{(-1)^k {k!}}{(x^2+1)x^{k+1}} + (-1)^{k+1} {(k+1)!} \frac{f(x)}{x^{k+1}}
$$
I got stuck now...
| $n\in\mathbb{N}$
It's $$\frac{d^n}{dx^n }\arctan x=\frac{d^n}{dx^n}(xf(x))=n\frac{d^{n-1}}{dx^{n-1} }f(x)+ x\frac{d^n}{dx^n }f(x)$$ and for $x>0$ (and proofed by induction) one gets $$\frac{d^n}{dx^n }\arctan x=(-1)^{n-1}(n-1)!\frac{\sin(n \arctan\frac{1}{x})}{\sqrt{1+x^2}^n}$$
It follows that the recursion for the derivatives of $\displaystyle f(x):=\frac{\arctan x}{x}$ is
$$\frac{d^n}{dx^n }f(x)=(-1)^{n-1}(n-1)!\frac{\sin(n \arctan\frac{1}{x})}{x\sqrt{1+x^2}^n}-\frac{n}{x} \frac{d^{n-1}}{dx^{n-1}}f(x)$$
Note:
Instead of $\arctan\frac{1}{x}$ I should write $\,$arccot(x)$\,$ (so that we can also use $x<0$) but it's not declared here.
Hint:
A proof can be found e.g. in http://www.math.nthu.edu.tw/~amen/2010/090408-2.pdf with $\arcsin\frac{1}{\sqrt{1+x^2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
$$
z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \cdots = \frac{3-2i \pm \sqrt{-8i-15}}{2}
$$
I am guessing I cannot go further than this.
Other idea: Being $z = ai+b$,
$$
(ai+b)^2 + (2i-3)(ai+b) + 5 - i = 0
$$
Which will give me a system of two equations and two variables
$$
\begin{cases}
-a^2-2a+b^2-3b = -5\\
2ab+2b-3a = 1
\end{cases}
$$
which seems almost impossible to solve.
What am I missing? Is there a simpler solution to this that I am not thinking about?
| Here's the general method:
You have to find the roots of $\Delta=-15-8i$. Let $(x+iy)^2=\Delta$. This means
$$x^2-y^2+2ixy=-15-8i\tag{1}$$
This means $x^2-y^2=-15$, $\;xy=-4$. As it would be too complicated to eliminate one of the unknowns to determine the other, we'll obtain another equation from the moduli.
Observe that $\lvert x+iy\rvert^2=\lvert -15+8i\rvert=\sqrt{289}=17$, whence the linear system in $x^2$ and $y ^2$:
$$\begin{cases}x^2-y^2=-15\\x^2+y^2=17\end{cases}\iff\begin{cases}x^2=1\\y^2=16\end{cases}\iff\begin{cases}x=\pm1\\y=\pm4\end{cases}$$
There seems to be $4$ square roots. However, note the equation $xy=-4$ implies $x$ and $y$ have opposite signs. So there are really only two square roots: $\;\pm(1-4i)$, and the roots of the initial equation are
$$z=\frac{-2i+3\pm(1-4i)}2=\begin{cases}1+i,\\2-3i.\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of positive integer solutions to the equation: $ (x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77. $ Q: Find the number of positive integer solutions to the equation:
$$\begin{align}
(x_1+x_2+x_3)(y_1+y_2+y_3+y_4)=77.
\end{align}$$
The solution given is $\begin{pmatrix} 6\\2 \end{pmatrix}$ $\begin{pmatrix} 10\\3 \end{pmatrix}$$+$ $\begin{pmatrix} 10\\2 \end{pmatrix}$ $\begin{pmatrix} 6\\3 \end{pmatrix}$.
I am clueless as to how can I go about solving this, given that the method of solving this question should centre around the concepts of combinations, distribution problems and multisets, in particular the formulas for $r$ $\mathbb identical $ objects, $n$ $distinct$ boxes:
*
*Each box can hold at most 1 object:$$\begin{align} C_r^n = \begin{pmatrix} n\\r \end{pmatrix} \end{align}$$
*Each box can hold any number of obects: $$\begin{align} H_r^n = \begin{pmatrix} {r+n-1}\\r \end{pmatrix} \end{align}$$
*Each box must hold at least one object - Put 1 object in each box first: r-n objects left. Apply (2), then number of ways: $$\begin{align} \begin{pmatrix} {r-n+n-1}\\{r-n} \end{pmatrix} = \begin{pmatrix} {r-1}\\{n-1} \end{pmatrix}\end{align}$$
Some help will be deeply appreciated.
| Hint:
You probably have specific examples of how to find the number of integer solutions to either $x_1+x_2+x_3=k$ or $y_1+y_2+y_3+y_4=k$, for any constant $k$. (If you don't: think about trying to distribute those $k$ $1$'s into boxes, such that each box gets at least one.)
To reduce there: note that $77=7\cdot11$, which are both prime. So, if $A\cdot B=77$, you must have one of:
*
*$A=1$, $B=77$
*$A=7$, $B=11$
*$A=11$, $B=7$
*$A=77$, $B=1$
Only two of these are possible with all $x_i$ and $y_j$ positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I go from $33/64 - 2x + x^2 \longrightarrow \bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$?
How do I go from $\frac{33}{64} - 2x + x^2 \longrightarrow\bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$ and then to $\big(x - 1-\frac{\sqrt{33}}{8}\big)\big(x - 1 + \frac{\sqrt{31}}{8}\big)?$
For the first, I know that it has something with the quadratic equation to do, but what is that? And how do I go from the second to the third?
| $$
\begin{align}
&\frac{33}{64} - 2x + x^2 \\
&= x^2 - 2x + \frac{33}{64} \\
&= x^2 - 2x + 1 - 1 + \frac{33}{64} \\
&= (x-1)^2 - 31/64 \\
&= (x-1)^2 - 31/64 \\
&= (x-1)^2 - 31/64 \\
&= \left( (x-1) - \sqrt{31}/8 \right)\cdot\left( (x-1) + \sqrt{31}/8 \right) \\
&= \color{red}{\left( x-1 - \sqrt{31}/8 \right)\cdot\left( x-1 + \sqrt{31}/8 \right)} \\
&= \left( x-\frac {2 + \sqrt{31}/4}2 \right)\cdot\left( x-\frac{2 - \sqrt{31}/4}{2} \right) \\
&= \left( x-\frac {2 + \sqrt{31/16}}2 \right)\cdot\left( x-\frac{2 - \sqrt{31/16}}{2} \right) \\
& = \left(x - \frac{2 + \sqrt{4 \color{red}-33/16}}{2}\right)\cdot\left(x - \frac{2 - \sqrt{4 \color{red}-33/16}}{2}\right)
\end{align}
$$
Please note the minus inside square roots!
| {
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} |
Solutions to ceiling equation system
Prove that there does not exist an $x$ with $1000 \leq x \leq 1990$ that can be expressed in the forms $$\dfrac{10000}{x} = \left \lceil\dfrac{10000}{x} \right \rceil -\dfrac{1}{m} \quad \text{and} \quad \dfrac{10000}{x-1} = \left \lceil\dfrac{10000}{x-1} \right \rceil-\dfrac{1}{k}$$ where $m$ and $k$ are positive integers and $\left \lceil\dfrac{10000}{x} \right \rceil = \left \lceil\dfrac{10000}{x-1} \right \rceil$.
I wasn't sure how to go about solving this question. I found an example that worked but instead of $10000$ it was $1000$. We have $$\dfrac{1000}{144} = 7-\dfrac{1}{18} \quad \text{and} \quad \dfrac{1000}{143} = 7-\dfrac{1}{143}.$$ How do we go about finding solutions here?
| Denote $\lceil\frac{10000}{x}\rceil=A$ and $\lceil \frac{10000}{x-1}\rceil=B$. We have $1000\le x\le 1990$.
A tedious calculation gives five values to be discarded because $A\ne B$ and five sets of candidates to be considered: $$\begin{cases}x=1000 \Rightarrow A=10\text{ and } B=11\\x=1012\Rightarrow A=9\text{ and } B=10\\x=1250\Rightarrow A=8\text{ and } B=9\\x=1429\Rightarrow A=7\text{ and } B=8\\x=1667\Rightarrow A=6\text{ and } B=7\end{cases}$$
$$\begin{cases}1001\le x\le 1111\Rightarrow A=B=10\\1013\le x\le 1249\Rightarrow A=B=9\\1251\le x\le 1428\Rightarrow A=B=8\\1430\le x\le 1666\Rightarrow A=B=7\\1668\le x\le 1990\Rightarrow A=B=6\end{cases}$$
Denoting $a=\lceil\frac{10000}{x}\rceil$ we need $$\frac Nx=a-\frac1m\\\frac{N}{x-1}=a-\frac 1k$$ which implies $$1=\frac{N(k-m)}{(am-1)(ak-1)}\qquad (*)$$
Among the five possible values for $a$, there is not solution for $a=10,8,6$ (almost visible because the corresponding denominators in $(*)$ become odd) Hence we get two necessary conditions: $$1=\frac{10000(k-m)}{(9m-1)(ak-1)}\qquad (**)\\1=\frac{10000(k-m)}{(7m-1)(7k-1)}\qquad (***)$$
The equations $(**)$ and $(***)$ have the solutions $(m,k)=(139,-1111),(-1111,-111)$ and $(x,y)=(268,-857),(3393,-217),(199,7768)$ respectively.
These necessary solutions of $(*)$, calculation shows are not compatible for the problem.
Thus there is not solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$
Q: How can we prove the above equality?
The integrand has an easily computable anti-derivative but I don't know if it's useful in this context.
| $$\newcommand{\PV}{\operatorname{PV}}
\begin{align}
R(x,y)
&=\frac12\,\PV\!\!\int_0^\infty\frac{\mathrm{d}z}{(z+y)\sqrt{z+x}}\tag{1}\\[6pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\frac{2\,\mathrm{d}z}{z^2-1}\tag{2}\\[3pt]
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}}^\infty\left(\frac1{z-1}-\frac1{z+1}\right)\mathrm{d}z\tag{3}\\
&=\frac1{2\sqrt{x-y}}\,\PV\!\!\int_{\sqrt{\frac{x}{x-y}}\,-1}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{4}\\
&=\frac1{2\sqrt{x-y}}\int_{\left|\sqrt{\frac{x}{x-y}}\,-1\right|}^{\sqrt{\frac{x}{x-y}}\,+1}\frac{\mathrm{d}z}z\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac1{2\sqrt{x-y}}\log\left|\frac{\sqrt{x}+\sqrt{x-y}}{\sqrt{x}-\sqrt{x-y}}\right|}\tag{6}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\frac1{2\sqrt{u-v}}\log\left|\frac{\sqrt{u-v}+\sqrt{u}}{\sqrt{u-v}-\sqrt{u}}\right|\tag{7}\\[6pt]
&=\sqrt{\frac{u-v}{u}}\,R(u,v)\tag{8}\\[9pt]
&=\sqrt{\frac{x}{x-y}}\,R(x-y,-y)\tag{9}
\end{align}
$$
Explanation:
$(2)$: substitute $z\mapsto(x-y)z^2-x$
$(3)$: partial fractions
$(4)$: substitute $z\mapsto z+1$ and $z\mapsto z-1$ and subtract integrals
$(5)$: Principal Value is easy with an odd function
$(6)$: evaluate integral
$(7)$: $u=x-y$ and $v=-y$
$(8)$: apply $(6)$ to $(7)$
$(9)$: undo the substitution in $(7)$
Substituting $y\mapsto-y$ in $(9)$ yields
$$
R(x,-y)=\sqrt{\frac{x}{x+y}}\,R(x+y,y)\tag{10}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
| An interesting way may be the following: if we define
$$ a_n = \int_{0}^{\pi/2}\sin^n(x)\,dx \tag{1}$$
we clearly have that $\{a_n\}_{n\geq 1}$ is a decreasing sequence. On the other hand, integration by parts gives:
$$ a_{2n} = \frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad a_{2n+1}=\frac{1}{2n+1}\cdot\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}\tag{2} $$
and we want to give an upper bound to
$$\frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{(2n)!!^2} = \binom{2n}{n}\frac{1}{4^n}\tag{3} $$
so we may just exploit $a_{2n+1}>a_{2n+2}$, leading to:
$$\frac{1}{2n+1}\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}>\frac{\pi}{2}\binom{2n+2}{n+1}\frac{1}{4^{n+1}}=\frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n}\cdot\frac{2n+1}{2n+2} \tag{4}$$
or to:
$$\left(\binom{2n}{n}\frac{1}{4^n}\right)^2< \color{red}{\frac{2}{\pi}\cdot\frac{2n+2}{(2n+1)^2}}\tag{5}$$
that is much stronger than the wanted inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sove it?
Meanwhile,I remember a similar integral:
$$\large{\int_0^\infty \frac{1}{(x^2+1)(x^a+1)}\ dx=\frac{\pi}{4}}$$
And I want to know the following one:
$${\color{red}{\large{\int_0^\infty \frac{\ln x}{(x^2+1)(x^a+1)}\ dx = \huge{?}}}}$$
Using the Mathematica I got the follow result.
Could you suggest some ideas how to prove this? Any hints will be appreciated.
| Preliminary Result:
$\displaystyle \frac{2}{(x^2+1)(x^3+1)} \equiv \frac{x+1}{x^2+1}-\frac{x^2+x-1}{x^3+1}$
Proof: Obvious.
Consider the parametrised integral:
$\displaystyle f(\alpha) = \int \frac{x^\alpha \, \text{d}x}{(x^2+1)(x^3+1)} = \frac{1}{2} \int \frac{x^{\alpha+1}+x^\alpha}{x^2+1}-\frac{x^{\alpha+2}+x^{\alpha+1}-x^\alpha}{x^3+1} \, \text{d}x$
For the first part, substitute $t = x^2$, and $t = x^3$ for the second part
$\displaystyle f(\alpha) = \frac{1}{4} \int_0^\infty \frac{t^{\frac{\alpha}{2}}+t^\frac{\alpha-1}{2}}{1+t} \text{d}t +\frac{1}{6}\int_0^\infty \frac{t^{\frac{\alpha}{3}}+t^{\frac{\alpha-1}{3}}-t^{\frac{\alpha-2}{3}}}{1+t} \text{d}t$
Use the Integral Representations of the Beta Function to obtain:
$\displaystyle f(\alpha) = \frac{\text{B}(1+\frac{\alpha}{2},-\frac{\alpha}{2})+\text{B}(\frac{1+\alpha}{2},\frac{1-\alpha}{2})}{4} - \
\frac{\text{B}(1+\frac{\alpha}{3},-\frac{\alpha}{3})+\text{B}(\frac{2+\alpha}{3},\frac{1-\alpha}{3})-\text{B}(\frac{1+\alpha}{3},\frac{2-\alpha}{3})}{6}$
$\displaystyle \frac{f(\alpha)}{\pi} = \frac{\csc{\left(\pi+\frac{\pi \alpha}{2} \right)}+\csc{\left(\frac{\pi}{2}+\frac{\pi \alpha}{2}\right)}}{4}-\frac{\csc{\left( \pi + \frac{\pi \alpha}{3} \right)}+\csc{\left(\frac{2\pi}{3} +\frac{\pi \alpha}{3}\right)}-\csc{\left( \frac{\pi}{3} + \frac{\pi \alpha}{3}\right)}}{6}$
$\displaystyle \frac{12f(\alpha)}{\pi} = 3\sec{\frac{\pi \alpha}{2}} -3\csc{\frac{\pi \alpha}{2}}+2\csc{\frac{\pi \alpha}{3}}+2\csc{\left( \frac{2\pi}{3}+\frac{\pi \alpha}{3}\right)}-2\csc{\left(\frac{\pi}{3}+\frac{\pi \alpha}{3}\right)}$
Differentiate both sides with respect to $\alpha$ and evaluate at $0$ to obtain the integral you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inductive proof for $\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$ I want to prove the following identity using induction (not double counting method). Although it is a specific version of Vandermonde's identity and its inductive proof is presented here, but I need a direct inductive proof on this, not the general form.
$$\binom{2n}{n}=\sum\limits_{k=0}^n\binom{n}{k}^2$$
I have tried to simplify $\binom{2n+2}{n+1}$ using Pascal's theorem, but did not get any result. Any help?
| Suppose
$$
\sum_{k=0}^n\binom{n}{k}^2=\binom{2n}{n}\tag{1}
$$
then
$$
\begin{align}
\sum_{k=0}^{n+1}\binom{n+1}{k}^2
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]^2\tag{2a}\\
&=\sum_{k=0}^{n+1}\left[\binom{n}{k}^2+\binom{n}{k-1}^2+2\binom{n}{k}\binom{n}{k-1}\right]\tag{2b}\\
&=\binom{2n}{n}\left(1+1+\frac{2n}{n+1}\right)\tag{2c}\\
&=\binom{2n+2}{n+1}\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: Pascal's Triangle identity
$\text{(2b)}$: algebra
$\text{(2c)}$: apply $(1)$ and $(3)$
$\text{(2d)}$: $\binom{2n+2}{n+1}=\frac{4n+2}{n+1}\binom{2n}{n}$
Lemma:
$$
\sum_{k=1}^n\binom{n}{k}\binom{n}{k-1}=\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2\tag{3}
$$
Proof:
Since $\binom{n}{k-1}=\frac{k}{n-k+1}\binom{n}{k}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{n-k+1}\binom{n}{k}$. Therefore,
$$
\frac{n-k+1}{n+1}\left[\binom{n}{k}+\binom{n}{k-1}\right]\binom{n}{k-1}=\binom{n}{k}\binom{n}{k-1}\tag{3a}
$$
Since $\binom{n}{k}=\frac{n-k+1}{k}\binom{n}{k-1}$, we have $\binom{n}{k}+\binom{n}{k-1}=\frac{n+1}{k}\binom{n}{k-1}$. Therefore,
$$
\frac{k}{n+1}\left[\binom{n}{k-1}+\binom{n}{k}\right]\binom{n}{k}=\binom{n}{k-1}\binom{n}{k}\tag{3b}
$$
Adding $(3a)$ and $(3b)$ and cancelling yields
$$
\frac{n-k+1}{n+1}\binom{n}{k-1}^2+\frac{k}{n+1}\binom{n}{k}^2=\binom{n}{k-1}\binom{n}{k}\tag{3c}
$$
Summing $(3c)$ over $k$, and substituting $k\mapsto k+1$ in the leftmost sum, gives
$$
\frac{n}{n+1}\sum_{k=0}^n\binom{n}{k}^2=\sum_{k=1}^n\binom{n}{k-1}\binom{n}{k}\tag{3d}
$$
QED
| {
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Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
| The following standard formula is well known $$\lim_{x \to 1}\frac{x^{n} - 1}{x - 1} = n = \lim_{t \to 0}\frac{(1 + t)^{n} - 1}{t}\tag{1}$$ and it appears that we can go very easily to the next step if $n$ is a positive integer and derive the formula $$\lim_{x \to 1}\frac{x^{n} - 1 - n(x - 1)}{(x - 1)^{2}} = \lim_{t \to 0}\frac{(1 + t)^{n} - 1 - nt}{t^{2}} = \frac{n(n - 1)}{2}\tag{2}$$ The simplest approach to prove $(2)$ is to use Binomial theorem. Hence we have $$x^{n} - 1 = n(x - 1) + \frac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})$$ and therefore
\begin{align}
\frac{n}{1 - x^{n}} &= \dfrac{n}{n(1 - x) - \dfrac{n(n - 1)}{2}(x - 1)^{2} + o((x - 1)^{2})}\notag\\
&= \frac{1}{1 - x}\left(1 - \frac{n - 1}{2}(1 - x) + o((1 - x))\right)^{-1}\notag\\
&= \frac{1}{1 - x}\left(1 + \frac{n - 1}{2}(1 - x) + o(1 - x)\right)\notag\\
&= \frac{1}{1 - x} + \frac{n - 1}{2} + o(1)\tag{3}
\end{align}
It follows that $$\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}} = \frac{n - m}{2} + o(1)$$ and hence $$\lim_{x \to 1}\left(\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\right) = \frac{n - m}{2}$$ and putting $n = 23, m = 11$ we get the desired limit as $6$.
The gymnastics of series division to reach $(3)$ can be avoided in another manner by using $(2)$ directly. We have
\begin{align}
L &= \lim_{x \to 1}\frac{n}{1 - x^{n}} - \frac{m}{1 - x^{m}}\notag\\
&= \lim_{x \to 1}\frac{n(1 - x^{m}) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\
&= \lim_{x \to 1}\frac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{(1 - x^{n})(1 - x^{m})}\notag\\
&= \lim_{x \to 1}\dfrac{n(1 - x^{m}) - mn(1 - x) + mn(1 - x) - m(1 - x^{n})}{\dfrac{(1 - x^{n})(1 - x^{m})}{(1 - x)^{2}}\cdot(1 - x)^{2}}\notag\\
&= \frac{1}{mn}\lim_{x \to 1}n\frac{1 - x^{m} - m(1 - x)}{(1 - x)^{2}} - m\frac{1 - x^{n} - n(1 - x)}{(1 - x)^{2}}\notag\\
&= \frac{1}{mn}\left(\frac{nm(1 - m)}{2} - \frac{mn(1 - n)}{2}\right)\text{ (using (2))}\notag\\
&= \frac{n - m}{2}\notag
\end{align}
| {
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A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers
A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers.
My Working
Let original number $n$ be $\ a_1+10a_2+10^2a_3+...+10^{k-1}a_k$.
The new number $n'$ is given by $\ a_2+10a_3+10^2a_4+...+10^{k-2}a_k$.
$n'$|$n$
$\frac{n-a_1}{10} |\ n$
$n=m \frac{n-a_1}{10} $
$a_1+10a_2+10^2a_3+...+10^{k-1}a_k=m \frac{(a_1+10a_2+10^2a_3+...+10^{k-1}a_k)-a_1}{10} $
$a_1+10a_2+10^2a_3+...+10^{k-1}a_k= \frac{m(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= \frac{m(10a_2+10^2a_3+...+10^{k-1}a_k)-10(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= \frac{(m-10)(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= (m-10)(a_2+10a_3+...+10^{k-2}a_k) $
For
$a_1\ |\ n'$, or $n'=ka_1$, the above equation is true*.
Original number is given by
$n=10n'+a_1$
$n=10ka_1+a_1$
$n=(10k+1)a_1$
Problem
If we put values into this equation we don't get numbers that satisfy the condition. For example if $a_1=1$ and $k=1$, The number comes out to be 201 which does not satisfy the condition. I am not able to figure out where I went wrong.
*My equation might not give all solutions but the ones it gives should satisfy the condition.
| Assume that $n$ has at least two digits, then $n=10a+b$ where $a\geq 1$ and $b\in \{0,1,2,3,4,5,6,7,8,9\}$.
By assumption $a$ divides $n$, therefore $a$ divides also $b=n-10a$.
Hence if $b\not=0$ then $a\leq b\leq 9$, which implies that $n$ has only two digits.
It follows that the number has this property iff it ends with a $0$ or it is in this finite list:
$$1,2,3,4,5,6,7,8,9,11, 12, 13, 14, 15, 16, 17, 18, 19,\\ 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99.$$
| {
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Obtaining the bound without using only the trivial inequality
If $x,y,z$ are 3 real numbers such that $$x+y+z = 4$$ and $$x^2 + y^2 + z^2 = 6$$ show that each of $x,y,z$ lie in the closed interval $[\frac{2}{3},2]$
I obtained the upper bound, but I am unable to obtain the lower bound.
My approach
$x^2 + y^2 + (4-(x+y))^2 = 6$
$\therefore 2x^2 + 2y^2 + 2xy - 8x - 8y = -10 \implies (2-y-x)^2 - 4 -xy = -5 \implies xy -1 \geq 0$
$$\therefore \color {blue}{xy \geq 1}$$
similarly we get $yz \geq 1$ and $zx \geq 1$
Now suppose that $z >2 \:\:$ Then $x + y < 2$ and $x^2 + y^2 <2 \implies (x+y)^2 - x^2 - y^2 = \color{red}{2xy < 2}$
Contradiction. A similar argument works for $x$ and $y$. Therefore we have $x,y,z \leq 2$
If possible, I'd like a similar argument for the lower bound, i.e, using only the trivial inequality. Thank you.
| $y+z=4-x$ and $(4-x)^2-2yz=6-x^2$, which gives $yz=x^2-4x+5$.
Thus, $(4-x)^2-4(x^2-4x+5)\geq0$, which is $(x-2)(3x-2)\leq0$ and we are done!
| {
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Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$. Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$.
Since $p \mid m^2 + 1$, we have $m^2 \cong -1( \mod p)$ and hence $\left ( \frac{-1}{p} \right)$ (Legendre Symbol) is $1$.
Again, $p \mid n^2 + 2$, we have $n^2 \cong -2( \mod p)$ and hence $\left ( \frac{-2}{p} \right)$ (Legendre Symbol) is $1$.
I am stuck here...How to proceed with the problem...Help Needed.
Thank You.
| Let $p$ divide some numbers of the form $m^2 + 1$ and $n^2 + 2$. Then both $-1$ and $-2$ are quadratic residues modulo $p$. This means that $p \equiv 1 \pmod 4$ and that $p \equiv 1,7 \pmod 8$. This means that $p$ is of the form $8s + 1$.
Now from the Euler's Criterion we have that an coprime number $a$ of $p$ is a biqaudratic residue iff $a^{\frac{p-1}{4}} \equiv 1 \pmod p$. For $-1$ we have that:
$$(-1)^{\frac{p-1}{4}} \equiv (-1)^{2s} \equiv 1 \pmod p$$
As $-1$ is a biquadratic residue modulo $p$ we have that there exists some $k$ s.t. $k^4 + 1$.
Then you should be able to easy revert back the proof for the other side.
| {
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Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{\frac{x^2-1}{x^2}}=\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{\frac{1}{x^2}}{1-\frac{1}{x^2}}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2}\cdot\frac{x^2}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}$$
As you can see I'm going in circles. Can anyone give me a hint on how to start on this problem?
| Hint: $x^2 - 1 = (x - 1)(x + 1)$.
| {
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By Induction, prove $1^n+2^n+3^n+...+n^n < (n+1)^n$ Here is the question:
Given that $$x^n + n( x^{n-1} ) \leq (x + 1) ^n \tag{A}$$
Prove: $$1^n + 2^n +3^n +...+n^n < (n+1)^n. \tag{B}$$
Here is my thinking.
Replace $n$ with $x$ in A can get $n^n + n(n^{n-1})$ which is $2n^n$
In inductive step, $1^{n+1} + 2^{n+1} + .... + (n+1)^{n+1}$
I separate it into $(1^n +2^n+...+n^n) + (2^n + 2(3^n)+3(4^n) +.....)$
But then I don't know how to do next.
| A rough sketch of the induction step for n:
Assume $1^n + \dots + n^n < (n+1)^n$. Our goal is to show that $1^{n+1} + \dots + (n+1)^{n+1} < (n+2)^{n+1}$. So,
\begin{alignat}{2}
(n+2)^{n+1} &\geq (n+1)^{n+1} + (n+1)(n+1)^n &\qquad \mbox{(by A)} \\
&> (n+1)^{n+1} + (n+1)(1^n + \dots + n^n) &\\
&\geq (n+1)^{n+1} + 1^{n+1} + \dots + n^{n+1} &
\end{alignat}
| {
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Confused with changes of bases. I have a problem to solve, and it makes me realise that I have a bad understanding of basis change and the linear applications that go along. I have a linear application who's matrix is expressed like this:
$A=\begin{bmatrix}
\frac{4a-b}{3}&\frac{2a-2b}{3} \\
\frac{-2a+2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$
I suppose in order to get the basis of this linear appplication I should compute the values of $Av_{1}$ and $Av_{2}$ where $v_{1}=\binom{1}{0}$ and $v_{2}=\binom{0}{1}$. Am I correct so far?
Now, where my lack of understand lacks is here: how to I determine the matrix $A^{'}$ that is the associated matrix in the base $x_{1} \:x_{2}$? Should I just compute $Ax_{1}$ and $Ax_{2}$?
EDIT 1:
I made a small mistake in the matrix:
it's not$$A=\begin{bmatrix}
\frac{4a-b}{3}&\frac{2a-2b}{3} \\
\frac{-2a+2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$$ but
$$A=\begin{bmatrix}
\frac{4a-b}{3}&\frac{-2a+2b}{3} \\
\frac{2a-2b}{3}& \frac{-a+4b}{3}
\end{bmatrix}
$$
I calculated the determinant of this matrix, and found that it is equal to:
$3ab$.
I know that to find the inverse of the matrix A, I can use the adjudant matrix. So normally I should be able to compute this:
$$A^{-1}=\frac{1}{3ab}*adj(A)$$
For $adj(A)$ I did this:
*
*Find the cofactor matrix:
$Com(A)$:
$$Com(A)=\begin{bmatrix}
\frac{-a+4b}{3}&\frac{2a-2b}{3} \\
\frac{2b-2a}{3}& \frac{4a-b}{3}
\end{bmatrix}
$$
The adjugate matrix will be equal to the transpose of the $Com(A)$. So we have:
$$adj(A)=\begin{bmatrix}
\frac{-a+4b}{3}&\frac{2b-2a}{3} \\
\frac{2a-2b}{3}& \frac{4a-b}{3}
\end{bmatrix}
$$
*
*Find the expression of the inverse:
I have now:
$$A^{-1}=\frac{1}{3ab}*\begin{bmatrix}
\frac{-a+4b}{3}&\frac{2b-2a}{3} \\
\frac{2a-2b}{3}& \frac{4a-b}{3}
\end{bmatrix}$$
Or in a simplified way:
$$A^{-1}=\frac{1}{9ab}*\begin{bmatrix}
-a+4b&2b-2a\\
2a-2b& 4a-b
\end{bmatrix}$$
So now all I have to do is to chose the two vectors I want $v_{1},v_{2}$ and calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$? And this will give me the basis of the linear transformation in the basis ($v_{1},v_{2}$), right?
EDIT 2:
I made a mistake. I should not calculate $A^{-1}v_{1}$ and $A^{-1}v_{2}$ but solve $v_{1}=A^{-1}\begin{bmatrix}
\frac{4a-b}{3}\\
\frac{2a-2b}{3}
\end{bmatrix}$ and $v_{2}=A^{-1}\begin{bmatrix}
\frac{-2a-2b}{3}\\
\frac{-a+4b}{3}
\end{bmatrix}$
| HINT.-The determinant of $A$ is equal to $ab$ so if $ab\ne 0$ then the matrix is invertible. If you want to know the base $\{b_1,b_2\}$ where $b_1=\begin{pmatrix} x_1\\y_1\end{pmatrix}$ and $b_2=\begin{pmatrix} x_2\\y_2\end{pmatrix}$
you have to solve
$$A\begin{pmatrix} x_1\\y_1\end{pmatrix}=\begin{pmatrix} \frac{4a-b}{3}\\\frac{-2a+2b}{3}\end{pmatrix}$$
$$A\begin{pmatrix} x_2\\y_2\end{pmatrix}=\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$
For this you multiply both equations for the inverse $A^{-1}$ of $A$ (you know that if $ab\ne 0$ then $A$ is invertible). Finally you have $$\begin{pmatrix} x_1\\y_1\end{pmatrix}=A^{-1}\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$
$$\begin{pmatrix} x_2\\y_2\end{pmatrix}=A^{-1}\begin{pmatrix} \frac{2a-2b}{3}\\\frac{-a+4b}{3}\end{pmatrix}$$
Do you know calculate the inverse matrix of A?
| {
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Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
| Here is an answer based upon a transformation of generating series.
We show
\begin{align*}
\sum_{j=0}^k\binom{k-j}{j}(-1)^j
=\frac{(-1)^{\lfloor k/3\rfloor}+(-1)^{\lfloor (k+1)/3\rfloor}}{2}
\qquad\qquad k\geq 0\tag{1}
\end{align*}
where $\lfloor x \rfloor$ denotes the floor function. We set as upper limit of the sum $j=k$ and use $\binom{p}{q}=0$ if $q>p$. We will also use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series.
Note, the sum at the LHS of (1) is of the form
\begin{align*}
\sum_{j=0}^k\binom{k-j}{j}a_j
\end{align*}
We can find in Riordan Array Proofs of Identities in Gould's Book by R. Sprugnoli in section 1.4 (A) a useful transformation formula:
Let $A(z)=\sum_{j=0}^\infty a_jz^j$ be a series, then the following holds
\begin{align*}
\frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)
=\sum_{k=0}^\infty\left(\sum_{j=0}^{k}\binom{k-j}{j}a_j\right)z^k
\end{align*}
So, we have the following relationship
\begin{align*}
[z^k]A(z)=a_k\qquad\longleftrightarrow\qquad
[z^k]\frac{1}{1-z}A\left(\frac{z^2}{1-z}\right)=\sum_{j=0}^{k}\binom{k-j}{j}a_j
\tag{2}\end{align*}
We obtain from (1) with $a_j=(-1)^j$ the generating function $A(z)$
\begin{align*}
A(z)=\sum_{j=0}^\infty(-z)^j=\frac{1}{1+z}
\end{align*}
and conclude according to (2)
\begin{align*}
\sum_{j=0}^k\binom{k-j}{j}(-1)^j&=[z^k]\frac{1}{1-z}\cdot\frac{1}{1+\frac{z^2}{1-z}}\tag{3}\\
&=[z^k]\frac{1}{1-z+z^2}\tag{4}\\
&=[z^k]\left(1+z-z^3-z^4+z^6+z^7-z^9-z^{10}+\cdots\right)\tag{5}\\
&=\frac{(-1)^{\lfloor k/3\rfloor}+(-1)^{\lfloor (k+1)/3\rfloor}}{2}\tag{6}
\end{align*}
and the claim follows.
Comment:
*
*In (3) we apply the transformation formula (2).
*In (4) we do some simplifications.
*In (5) we expand the series with the help of Wolfram Alpha.
*In (6) we select the coefficient of $z^k$.
Note: This binomial identity can also be found as (1.75) in R. Sprugnoli's paper.
| {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 6
} |
Describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$. I need to describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$.
After testing out some $n \in \mathbb{N}$, I came to the conclusion that $3^{n} - 2^{n}$ is divisible by $5$ iff $n$ is even - i.e., if $n$ is of the form $2k$ for $k \in \mathbb{N}$.
Here is my attempt so far:
$(\implies)$:
To show that if $n$ is even, then $3^{n}-2^{n}$ is divisible by $5$, we prove that $\forall k \in N$, $3^{2k}-2^{2k} \equiv \,0 \mod 5$ by induction on $k$:
*
*Basis Step: For $k = 1$, $3^{2(1)}-2^{2(1)}=3^{2}-2^{2}=9-4=5 \equiv\, 0 \mod 5 $.
*Suppose true for $\mathbf{k=m}$: $3^{2m}-2^{2m}\equiv \, 0 \mod 5 \, \implies \, 3^{2m}-2^{2m}=5l$, $l \in \mathbb{Z}$.
*Show true for $\mathbf{k = m+1}$: Consider $3^{2(m+1)}-2^{2(m+1)}\\ = 3^{2m+2}-2^{2m+2}\\ = 3^{2m}\cdot 3^{2}- 2^{2m}\cdot 2^{2}\\ = 9\cdot 3^{2m}-4 \cdot 2^{2m} \\= 5 \cdot 3^{2m} + 4\cdot 3^{2m} - 4 \cdot 2^{2m} \\ = 5(3^{2m}) + 4(3^{2m}-2^{2m}) \\ = 5(3^{2m})+4(5l)\, \text{(by the induction hypothesis)} \\ = 5(3^{2m} + 4l) \, \text{which is divisible by}\, 5.$
So, by induction, then the statement $3^{2k}-2^{2k} \equiv \, 0 \mod 5$ holds $\forall k \geq 1 \, \implies \, $ the statement $3^{n}-2^{n} \equiv \, 0 \mod 5$ holds $\forall n = 2k$, $k \geq 1$.
$(\Longleftarrow)$:
To show that $3^{n}-2^{n}$ divisible by $5$ $\implies$ $n$ even, we will show that $n$ odd $\implies$ $3^{n}-2^{n}$ is not divisible by $5$. Now, suppose the proposition is false. I.e., assume $\exists n \in \mathbb{N}$ for which $n$ is odd, but $3^{n}-2^{n}$ is divisible by $5$.
Since $n$ is odd, it is of the form $2k+1$, $k \in \mathbb{N}$.
So, we have that $3^{2k+1}-2^{2k+1} \equiv \, 0 \mod 5 \, \implies \, 3^{2k+1} - 2^{2k+1} = 5l$, $l \in \mathbb{N}$.
So, $\displaystyle \frac{3^{2k+1}}{5} - \frac{2^{2k+1}}{5} = l$.
At this point, I got stuck. I think my $(\implies)$ direction is fine, but I definitely need help on my $(\Longleftarrow)$ direction.
Could somebody please help me complete this proof?
Thank you.
| I think this is a little easier to see using modular arithmetic. Note that $5$ divides $3^n - 2^n$ means $3^n \equiv 2^n \pmod{5}$. Since $3 \cdot 2 = 6 \equiv 1 \pmod{5}$, then $3 = 2^{-1}$ in $\mathbb{Z}/5\mathbb{Z}$. Then
\begin{align*}
2^n \equiv 3^n \equiv (2^{-1})^n = 2^{-n} \pmod{5}
\end{align*}
and multiplying through by $2^n$ yields $2^{2n} \equiv 1 \pmod{5}$. This means that the order of $2$ in the unit group $(\mathbb{Z}/5\mathbb{Z})^\times$ divides $2n$. Since $2$ has order $4$ in $(\mathbb{Z}/5\mathbb{Z})^\times$, then $4$ divides $2n$, so $2$ divides $n$, i.e., $n$ is even. All the implications above are actually two-sided (iff), so we find that $5$ divides $3^n - 2^n$ iff $n$ is even.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find a Sum for $\left(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\right)^2$ How can one find a formula for $\big(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\big)^2$ in the form
$$
\Big(1+\cos(x)+\cos(2x)+\cos(3x)+...+\cos((n-1)x)\Big)^2=\sum _{i=0}^N A_i \cos (B_i \,x),
$$
where $N$ depends only on $n,$ and the coefficients $A_i$ and $B_i$ depend only on $i$ and $n,$ but none of these depend on $x.$
| We'll show that
\begin{align}\boxed{\text{ }
\\\quad\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2= \frac{n+1}{2}+\sum_{k=1}^{n-1} \Big(n-\frac{k-1}{2} \Big)\cos(kx)+\sum_{k=n}^{2n-2} \Big(n-\frac{k+1}{2} \Big)\cos(kx).\quad\\}
\end{align}
Note that this is in the form requested since the constant term is just the $\cos (0x)$ term.
Here's how to prove this:
\begin{align}
\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2&=\sum_{j=0}^{n-1}\sum_{k=0}^{n-1}\cos(jx)\cos(kx)
\\&=\sum_{j=0}^{n-1}\sum_{k=0}^{n-1}\frac{\cos\big((j+k)x\big)+\cos\big((j-k)x\big)}{2}
\\&=\frac12\Big(\sum_{j=0}^{n-1}\sum_{k=0}^{n-1} \cos\big((j+k)x\big)+\sum_{j=0}^{n-1}\sum_{k=0}^{n-1}\cos\big((j-k)x\big)\Big)
\\&=\frac12\Big(\sum_{m=0}^{2n-2}\big(\text{# of ways of writing }m\text{ as a sum }j+k\text{ with }0\le j\le n-1\text{ and }0\le k\le n-1\big) \cos(mx)
\\&\quad+\sum_{m=-(n-1)}^{n-1}\big(\text{# of ways of writing }m\text{ as }j-k\text{ with }0\le j\le n-1\text{ and }0\le k\le n-1\big) \cos(mx)\Big).
\end{align}
If $\boxed{0\le m \le n-1},$ there are $\boxed{m+1\text{ ways of writing }m\text{ as }j+k},$ namely $0+m, 1+(m-1), 2+(m-2), \dots, m+0.$
If $\boxed{n \le m \le 2n-2},$ then $0 \le m-n \le n-2,$ and there are $\boxed{2n-m-1\text{ ways to write }m\text{ as }j+k},$ namely $(m-n+1)+(n-1), (m-n+2)+(n-2), \dots, (n-1)+(m-n+1).$
Turning to the differences now, if $\boxed{0\le m \le n-1},$ then there are $\boxed{n- m \text{ ways of writing }m\text{ as }j-k},$ namely $m-0, (m+1)-1, (m+2)-2, \dots, (n-1)-(n-m-1).$
Finally, if $\boxed{-(n-1)\le m \lt 0},$ then there are $\boxed{n+m \text{ ways of writing }m\text{ as }j-k},$ namely $0-\lvert m \rvert, 1-(1+\lvert m \rvert), 2-(2+\lvert m \rvert), \dots, (n+m-1)-(n-1).$
It follows that the quantity that we're computing, which is half of $$\sum_{m=0}^{2n-2}\big(\text{# of ways of writing }m\text{ as a sum }j+k\text{ with }0\le j\le n-1\text{ and }0\le k\le n-1\big) \cos(mx)
\\\quad\quad+\sum_{m=-(n-1)}^{n-1}\big(\text{# of ways of writing }m\text{ as }j-k\text{ with }0\le j\le n-1\text{ and }0\le k\le n-1\big) \cos(mx),$$ is equal to half of
\begin{align}
&\sum_{m=0}^{n-1} (m+1) \cos(mx) + \sum_{m=n}^{2n-2}(2n-m-1) \cos(mx)
+ \sum_{m=0}^{n-1}(n-m)\cos(mx) + \sum_{m=-(n-1)}^{-1} (n+m)\cos(mx)
\\&= \sum_{m=0}^{n-1} (n+1) \cos(mx) + \sum_{m=n}^{2n-2}(2n-m-1) \cos(mx)
+ \sum_{m=-(n-1)}^{-1} (n+m)\cos(mx).
\end{align}
So we find that
\begin{align}{\quad\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2= \sum_{m=-(n-1)}^{2n-2} \frac12 B_m \cos(mx)\quad}
\end{align}
where
$${\quad B_m=\begin{cases}
m+n, &\text{ if }-(n-1)\le m\lt 0,\quad
\\n+1, &\text{ if }0 \le m \lt n,\quad
\\2n-m-1,&\text{ if }n \le m \le 2n-2.\quad
\end{cases}}$$
Finally, we can use the identity $\cos(-mx)=\cos(mx)$ to combine terms, and get
\begin{align}\boxed{\quad\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2= \sum_{m=0}^{2n-2} A_m \cos(mx)\quad}
\end{align}
where
$$\boxed{\quad A_m=\begin{cases}
\frac{n+1}{2}, &\text{ if }m=0,\quad
\\n-\frac{m-1}{2}, &\text{ if }1 \le m \lt n,\quad
\\n-\frac{m+1}{2},&\text{ if }n \le m \le 2n-2.\quad
\end{cases}}$$
And you can write this as
\begin{align}\boxed{\text{ }
\\\quad\Big(\sum_{k=0}^{n-1} \cos(kx)\Big)^2= \frac{n+1}{2}+\sum_{k=1}^{n-1} \Big(n-\frac{k-1}{2} \Big)\cos(kx)+\sum_{k=n}^{2n-2} \Big(n-\frac{k+1}{2} \Big)\cos(kx).\quad\\}
\end{align}
| {
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Prove $4p-3$ is a square knowing that $n\mid p-1$ and $p\mid n^3-1$, $p$ prime I really need some help at this problem:
Let $p$ be a prime number and $n$ a natural number, $n\ge2$ such that $n \mid p-1$ and $p \mid n^3-1$. Prove that $4p-3$ is a square.
So $p \mid (n-1)(n^2+n+1)$
What if $p \mid n-1$?
Treating the cases wasn't too efficient.
I was thinking about Fermat's theorem but it didn't helped really much.
A hint would be really apreciated.
Thanks!
| Note $p \mid n-1$ is impossible because $n \le p-1$, so we have $p \mid n^2+n+1$.
Since $n \mid p-1$, we can write $p = an+1$ for some integer $a \ge 1$. Since $p \mid n^2+n+1$, we can write $$n^2 + n + 1 = bp = b(an+1)$$ for some integer $b \ge 1$.
Reducing modulo $n$ gives $1 \equiv b \pmod{n}$, so write $b = rn+1$ for some integer $r \ge 0$. Putting this in the above equation gives
$$n^2 + n + 1 = (rn+1)(an+1).$$
If $r \ge 1$, then $(rn+1)(an+1) \ge (n+1)^2 > n^2+n+1$, which is a contradiction. So $r = 0$, $b = 1$, and $n^2 + n + 1 = p$, so we get $4p-3 = (2n+1)^2$.
| {
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} |
Prove this trigonometry equation: $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
Prove that $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
I've tried writing $\sin 40^\circ$ as $\sin(40^\circ+10^\circ)$, then wrote $\sin(50^\circ+10^\circ)$ as $\sin 40^\circ \cos 10^\circ + \cos 40^\circ \sin 10^\circ$, but I don't know what to do next.
| You can use the general formula
$$ \sin\alpha \sin\beta = \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}.$$
There are also the similar formulas
$$ \cos\alpha \cos\beta = \frac{\cos(\alpha+\beta) + \cos(\alpha-\beta)}{2}$$
and
$$ \sin\alpha \cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}.$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Right Triangles and a Circle In a test sometime ago, I had to deal with the following question:
Let $\triangle ABC$ be a right angled triangle. Let $\triangle DEF$ be a right triangle inscirbed in the incircle of $\triangle ABC$. Find the smallest possible ratio of $\frac{[ABC]}{[DEF]}$ where $[ABC]$ denotes the area of $\triangle ABC$.
What I did was to break the problem into 2 parts:
$1$. Find the largest right triangle with a circumcircle of a fixed radius.
$2$. Find the smallest right triangle with a incircle with a fixed radius.
For the first part, I was able to prove that the isosceles right triangle had the largest area using similar triangles but I got stuck on the second part during the test and was unable to complete the question. I have a suspicion that it is also an isosceles right triangle but I could not prove it.
I have looked up some info on the incircle and tried the following method:
Let $P_{ABC}$ denote the perimeter of $\triangle ABC$ and let $\angle BAC = 90^{\circ}$.
WLOG Let the inradius $r$ of $\triangle ABC$ be 1
Since $[ABC] = \frac{1}{2} \cdot P_{ABC} \cdot r$ and $[ABC] = \frac{1}{2} \cdot AB \cdot AC$, we have:
$$
\begin{align*}
P_{ABC} & = AB \cdot AC\\
(AB + AC + BC) & = AB \cdot AC\\
\end{align*}
$$
This is where I got stuck. Is it possible to further develop this method or is there a simpler method all together?
| If a right triangle $DEF$ is inscribed in a circle with radius $r$, its maximum area is given by $r^2$, since the longest side of such a triangle is also a diameter of the circumcircle of $DEF$. It follows that $\frac{[ABC]}{[DEF]}$ is lower bounded by $\frac{[ABC]}{r^2}$ where $r$ is the inradius of $ABC$. On the other hand, if the lengths of the legs of $ABC$ are $x$ and $y$, we have $[ABC]=\frac{xy}{2}$ and $r=\frac{2[ABC]}{x+y+\sqrt{x^2+y^2}}=\frac{xy}{x+y+\sqrt{x^2+y^2}}$, so:
$$ \frac{[ABC]}{[DEF]}\geq \frac{\left(x+y+\sqrt{x^2+y^2}\right)^2}{2xy}\geq\frac{\left(2\sqrt{xy}+\sqrt{2}\sqrt{xy}\right)^2}{2xy}=\color{red}{3+2\sqrt{2}} $$
by the QM-AM-GM inequality:
$$ \sqrt{\frac{x^2+y^2}{2}}\geq \frac{x+y}{2}\geq \sqrt{xy}. $$
Equality is achieved at $x=y$ as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability that the lot contains defective articles. I tried much in the problem but I didn't get my answer correct.
The question is---
A lot contains 20 articles.the probability that the lot contains exactly 2 defective articles is 0.4 and that the lot contains exactly 3 defective articles is 0.6.articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found.then the probability that the testing procedure ends at the twelfth testing is------
My attempt -----
let E1 be the event that the lot contains exactly 2 defective articles and E2 be the event that the lot contains exactly 3 defective articles.
I noticed that $P(E1)+P(E2)=1 $and also it is easy to see E1 and E2 ate mutually exclusive. So that implies that the lot contains exactly 3 or 2 defective articles.
Hence my approach was $$\frac{\binom{12}{2} \times 0.4 +\binom{12}{4} \times 0.6}{\binom{20}{12}}$$
But I am not getting the answer.please help me in this regard.
Thanks.
| If the lot contains two defective articles and the testing procedure ends at the $12^{th} $ testing, then the first defective article must be chosen among the first $11$ and the second must be the $12^{th} $. The probability that this occurs with the first defective article taken at the first test is
$$ \frac{2}{20} \cdot \frac{18}{19} \cdot \frac{17}{18} \cdot \frac{16}{17}.... \cdot \frac{9}{10} \cdot \frac{1}{9}$$
where the first and last fraction express the probability of picking the two defective articles. It is not difficult to show that the same probability (with different fractions but identical resulting factors in the overall product of the numerators and denominators) is obtained is we consider that the first defective article is chosen as the $2^{nd} $, $3^{rd} $... $11^{th} $. So the probability is
$$ 11 \cdot \frac{2}{20} \cdot \frac{18}{19} \cdot \frac{17}{18} \cdot \frac{16}{17}.... \cdot \frac{9}{10} \cdot \frac{1}{9}$$
$$=11 \cdot 2 \cdot \frac {18!}{8!} \cdot \frac {8!}{20!} $$
$$=\frac {11}{190} $$
By similar considerations, we get that, if the lot contains three defective articles and the testing procedure ends at the $12^{th} $ testing, the probability that this occurs with the first two defective articles chosen in the first two choices is
$$ \frac{3}{20} \cdot \frac {2}{19} \cdot \frac{17}{18} \cdot \frac{16}{17}.... \cdot \frac{9}{10} \cdot \frac {1}{9} $$
where the first two fractions and the last one express the probability of picking the three detective articles. Again, it is not difficult to show that the same probability, with different fractions but identical resulting factors in the product of the numerators and denominators, is obtained if we consider that the first two defective articles are chosen in any of the possible $\binom {11}{2} $ different combinations. So the probability is
$$ \binom {11}{2} \cdot \frac{3}{20} \cdot \frac{2}{19} \cdot \frac{17}{18} \cdot \frac{16}{17}.... \cdot \frac{9}{10} \cdot \frac{1}{9}$$
$$=55 \cdot 6 \cdot \frac {17!}{8!} \cdot \frac {8!}{20!} $$
$$=\frac {11}{228} $$
This, the final probability is
$$0.4 \cdot \frac {11}{190} + 0.6 \cdot \frac {11}{228}$$
$$ =\frac {99}{1900} \approx 5.2 \%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Second partial derivative with $f(x,y)=x^3+5x^2y+y^3$ I have this problem:
I found $F_x=3x^2+10yx$ and $F_y=5x^2+3y^2$, then $D_uf=\frac{9}{5}x^2+6yx+4x^2$
$F_{x2}= \frac{58}{5}x + 6y, F_{y2}=6x+{24}{5}y$
$D_uf_2=\frac{174}{25}x+\frac{18}{5}y+\frac{32}{5}x+\frac{96}{25}y=>at(2,1)=>534/25$
The answer however is $\frac{774}{25}$
| If $f(x,y)= x^3 + 5x^2y + y^3 $ and $u=\langle\frac35, \frac45\rangle $
$$\nabla f(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle$$
$$D_uf(x,y)=\langle 3 x^2 + 10xy, 5x^2 +3y^2\rangle \cdot \langle\frac{3}{5},\frac{4}{5}\rangle =\frac{1}{5}(29x^2 + 30xy +12y^2) $$
$$D^2_uf(x,y)=\frac{3}{5}\underbrace{\frac{1}{5}(58x+30y)}_{\frac{\partial D_uf(x,y)}{\partial x}}+\frac{4}{5}\underbrace{\frac{1}{5}(30x + 24y)}_{\frac{\partial D_uf(x,y)}{\partial y}} = \frac{1}{25}(174x + 90y + 120x + 96y) = \frac{294x+186y}{25}$$
which gives
$$D^2_uf(2,1)=\frac{774}{25}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A trigonometric equation - Finding the value of theta I have to find the value of $\theta$ in this equation "$\cos2\theta \sin\theta = 1$"
The problem is that I have to solve using algebra (making equations...etc) like this one, for example: \begin{align} 4\sin^2\theta - 1 = 0\\ \sin^2\theta = 0.25\\ \sin\theta = +0.5, -0.5\\ \theta = 30, 150, 210, 330\end{align}
But for "$\cos2\theta \sin\theta = 1$", I figured, according to the fact that $\cos\alpha$ can be only between -1 and 1. And the same for $\sin\alpha$, that the only two expected solutions are when $\cos2\theta$ and $\sin\theta$ are both equal to 1 or -1 so that the product is equal to 1 as in the main equation.
So $\theta = 270$. \begin{align} \cos(540)\sin(270) = 1\\ \cos(180)\sin(270) = 1\\ -1 * -1 = 1 \end{align}
I'm trying to find another way to solve it like the way I solved the other example $4\sin^2\theta - 1 = 0$..
| Hint:
Use the identity-$\cos2\theta=\cos^2\theta-\sin^2\theta$
So,$$\cos2\theta\sin\theta=1$$
$$\implies(\cos^2\theta-\sin^2\theta)\sin\theta=1$$
Now,replace $\cos^2\theta$ by $1-\sin^2\theta$
So,$$(1-2\sin^2\theta)\sin\theta=1$$
$$\implies\sin\theta-2\sin^3\theta-1=0$$
Now,let,$\sin\theta=x$.So,this reduces to a cubic equation.Now,solve for $\sin\theta$ and just plug the value in the equation.
Hope this helps!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Difficulty in the evaluation of double integral of a continuous function. Evaluate the double integral
$$\iint_D \frac{xy}{\sqrt{1-y^2}} dxdy,$$
where $D$ is the region of the first quadrant bounded by standard circle with unit radius $1$: that is $x^2+y^2=1$.
I have difficulty with its evaluation.
First of all I want an answer to verify. I tried this using polar coordinate. I have also tried by taking $x$ changing from $0$ to $1$ and $y$ changing from $0$ to $\sqrt{1-y^2}$.
The answer comes to be 1/6.
Need help. As I am not getting the same answer always when I try to use new method.
| The fact that $D$ is a disc tells you that you should use polar coordinates. Therefore, let $x = r \cos t$ and $y = r \sin t$ with $r \in [0,1]$ and $t \in [0, \frac \pi 2]$, in order to transform your integral into
$$\iint \limits _{[0,1] \times [0, \frac \pi 2]} \frac {r \cos t r \sin t } {\sqrt {1 - r^2 \sin^2 t}} r \Bbb d r \Bbb d t .$$
Notice that the derivative with respect to $t$ of the expression under the square root is $-r^2 2 \sin t \cos t$, which is almost what you already have in the numerator (save for a factor of $-2$). Threrefore, the computation can be continued as
$$\int \limits _0 ^1 -r \left( \int \limits _0 ^{\frac \pi 2} \frac {\frac {\partial} {\partial t} (1 - r^2 \sin^2 t)} {\sqrt {1 - r^2 \sin^2 t}} \Bbb d t \right) \Bbb d r = \int \limits _0 ^1 -r \left( \sqrt {1 - r^2 \sin^2 t} \ \Bigg| _0 ^{\frac \pi 2} \right) \Bbb d r = \int \limits _0 ^1 -r \left( \sqrt {1 - r^2} - 1\right) \Bbb d r = \\
\int \limits _0 ^1 r \left( 1 - \sqrt {1 - r^2} \right) \Bbb d r = \frac {r^2} 2 \Bigg| _0 ^1 + \frac 1 2 \int \limits _0 ^1 (1-r^2)' \sqrt {1-r^2} \ \Bbb d r = \frac 1 2 + \frac 1 2 \frac 2 3 \sqrt{1 - r^2} \Big| _0 ^1 = \frac 1 2 - \frac 1 3 = \frac 1 6.$$
The fact that the result should be $\frac 1 6$ can be also seen by using Fubini's thorem: your integral is
$$\int \limits _0 ^1 x \left( \int \limits _0 ^{\sqrt {1 - x^2}} \frac y {\sqrt {1-y^2}} \ \Bbb d y \right) \ \Bbb d x = \int \limits _0 ^1 x \left( - \sqrt{1 - y^2} \Bigg| _0 ^{\sqrt{1-x^2}} \right) = \int \limits _0 ^1 x ( -x + 1) \ \Bbb d x = - \frac 1 3 + \frac 1 2 = \frac 1 6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$ $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x}{x^2}$
without D l hospital rule and series expansion.
i have solved it series expansion of $\cos x$ but want be able to go further without series expansion
| $b_n:=1-\cos a_{1}x \cdot \cos a_{2}x\cdot \cos a_{3}x\cdot \cdot \cdot \cdot \cdot \cos a_{n}x$
$b_n-1=(b_{n-1}-1)\cdot \cos a_n x$ => $b_n=b_{n-1}\cdot \cos a_n x+1-\cos a_n x$
$\lim_{x\to 0}\frac{b_n}{x^2}=\lim_{x\to 0}(\frac{b_{n-1}}{x^2\cos a_n x}+\frac{1-\cos a_n x}{x^2\cos a_n x})=\lim_{x\to 0}\frac{b_{n-1}}{x^2}+\frac{a_n^2}{2}$
=> $\lim_{x\to 0}\frac{b_n}{x^2}=\frac{1}{2}\sum\limits_{k=1}^n a_k^2$
I have assumed here that $\lim_{x\to 0}\frac{1-\cos ax}{x^2}=\frac{a^2}{2}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Real values of $k$ for which $x^2+(k+1)x+k^2$ has one root double the other
For what real values of $k$ does $x^2+(k+1)x+k^2$ have one root double the other?
For a start, I found the range of $k$ which endows this equation with real roots:
$$-\frac13\le k\le1$$
| Suppose the equation has one root double the other. Then it can be written as
$$(x-a)(x-2a)=x^2-3ax+2a^2$$
Comparing coefficients between this expression and $x^2+(k+1)x+k^2$ we have
$$k+1=-3a\qquad2a^2=k^2$$
From the first equation we have $k=-3a-1$; substituting this into the second equation yields
$$2a^2=(-3a-1)^2=9a^2+6a+1$$
$$7a^2+6a+1=0$$
$$a=\frac{-6\pm\sqrt{6^2-4\cdot7\cdot1}}{2\cdot7}=\frac{-3\pm\sqrt2}7$$
From here we can recover the possible values of $k$:
$$k=-3\left(\frac{-3\pm\sqrt2}7\right)-1=\frac{2\pm3\sqrt2}7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.