Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the sum $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$ Please help me calculate the following sum
$$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}$$
| $$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}\\=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{99}-\frac{1}{100}\\=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}=0.49$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/194846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Strategies to denest nested radicals $\sqrt{a+b\sqrt{c}}$ I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?
Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as:
$$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}},
\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$
Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.
I'm a just a beginner, can anyone give me some ideas? Thank you.
| You can derive a formula for $\sqrt{a+b\sqrt{c}}$. You will have to assume that $\sqrt{a+b\sqrt{c}}$ can be rewritten as the sum of two surds (radicands). So $$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt{e}$$
Squaring both sides yields $$a+b\sqrt{c}=d+e+2\sqrt{de}$$
From that, we can see that $a=d+e$ so $e=a-d$ and $b\sqrt{c}=2\sqrt{de}\rightarrow b^{2}c=4de$.
Substituting $e$ with $a-d$ gives $b^{2}c=4d(a-d)$. So $b^{2}c=4ad-4d^{2}$. Rearranging the terms gives us $4d^{2}-4ad+b^{2}c=0$
Using the Quadratic Equation, we have $$d=\frac {a\pm\sqrt{a^{2}-b^{2}c}}{2}$$
And since $a=d+e$, $e$ is the conjugate of $d$. So $e=\frac {a-\sqrt{a^{2}-b^{2}c}}{2}$ and $d=\frac {a+\sqrt{a^{2}-b^{2}c}}{2}.\,$ Thus
$$\sqrt{a+b\sqrt{c}}\,=\, \sqrt{\frac {a+\sqrt{a^{2}-b^{2}c}}{2}} +\sqrt{\frac {a-\sqrt{a^{2}-b^{2}c}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
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"answer_id": 4
} |
Proving the trigonometric identity Please help me in proving the following idenity:
$$8\cdot \cos 40^\circ\cdot \cos 20^\circ \cdot \cos 10^\circ = \cot 10^\circ$$
| Before proving the trigonometrics relation note some trigonometrics formula which will utilize during the proof.
a) $2\cos\alpha\cdot\sin\alpha=\sin 2\alpha$
b) $\sin\alpha=\cos(90^\circ-\alpha)$
c) $\frac{\cos\alpha}{\sin\alpha}=\cot\alpha$
\begin{align}
8\cdot \cos &40^0\cdot \cos 20^0 \cdot \cos 10^0\\
& = \frac{8\cdot \cos 40^0\cdot \cos 20^0 \cdot \cos 10^0\cdot \sin 10^0}{\sin 10^0}\\
& =\frac{4\cdot \cos 40^0\cdot \cos 20^0 \cdot 2\cos 10^0\cdot \sin 10^0}{\sin 10^0}\\
& =\frac{4\cdot \cos 40^0\cdot \cos 20^0 \cdot \sin 2\cdot 10^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot 2\cos 20^0 \cdot \sin 20^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot \sin 2\cdot 20^0}{\sin 10^0}\\
&=\frac{2\cdot \cos 40^0\cdot\sin 40^0}{\sin 10^0}\\
&=\frac{\sin 2\cdot 40^0}{\sin 10^0}\\
& =\frac{\sin 80^0}{\sin 10^0}\\
& = \frac{\cos 10^0}{\sin 10^0} = \cot 10^0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/197656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Positive definiteness of a matrix Let ${x_1,x_2,...x_n}$ be positive numbers. Consider the matrix $C$ whose $(i,j)$-th entry is
$$\min\left\{\frac{x_i}{x_j},\frac{x_j}{x_i}\right\}$$
Show that $C$ is non-negative definite (or positive semidefinite, meaning $z^t C z\geq 0$ for all $z\in \mathbb{R}^n$).When is $C$ positive definite?
| We can suppose that $x_i\leq x_j$ for $i<j$ (otherwise just permute the components). As I'll show below, the determinant of your matrix is
$$\det C_n=\frac{x_2^2-x_1^2}{x_2^2}\frac{x_3^2-x_2^2}{x_3^2}\dots\frac{x_n^2-x_{n-1}^2}{x_n^2},\qquad(*)$$
which is $\geq 0$, and it is $>0$ iff all $x_i$'s are different. Notice that if you take the top left corner of $C$ with $k$ rows and columns, you get $C_k$, and $\det C_k\geq0$. Your matrix is therefore positive semidefinite (by Sylvester criterion), and it is positive definite iff all $x_i$'s are different.
Now we need to prove $(*)$. Let $D_n$ be $C_n$ with $i,k$-th element multiplied by $x_i x_j$, so that $\det D_n=\det C_n\times\prod_i x_i^2$. We want to show
$$\det D_n=x_1^2(x_2^2-x_1^2)\dots(x_n^2-x_{n-1}^2).$$
$D_n$ looks like
$$
\begin{pmatrix}
x_1^2 & x_1^2& x_1^2&x_1^2\\
x_1^2& x_2^2& x_2^2&x_2^2\\
x_1^2& x_2^2&x_3^2&x_3^2\\
x_1^2& x_2^2&x_3^2&x_4^2
\end{pmatrix}
$$
(for $n=4$ - I hope the pattern is clear), i.e.
$$
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}.
$$
($a=x_1^2,\dots,d=x_4^2$).
If we now subtract the first row from the others and then the first column from the others, we get
$$
\begin{pmatrix}
a & 0& 0&0\\
0& b-a& b-a&b-a\\
0& b-a&c-a&c-a\\
0& b-a&c-a&d-a
\end{pmatrix}.
$$
i.e.
$$
\det
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}=
a\det \begin{pmatrix}
b-a& b-a&b-a\\
b-a&c-a&c-a\\
b-a&c-a&d-a
\end{pmatrix}
$$
Repeating this identity, we get
$$\det D_n=
\det
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}=
a\det \begin{pmatrix}
b-a& b-a&b-a\\
b-a&c-a&c-a\\
b-a&c-a&d-a
\end{pmatrix}=
a(b-a)\det
\begin{pmatrix}
c-b&c-b\\
c-b&d-b
\end{pmatrix}=
a(b-a)(c-b)(d-c)
$$
as we wanted to show.
There must be a more intelligent solution, but for the moment this one should do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/197849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof the logarithmic identity $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$ Please help me proof $\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$, for $a,b,c>0$ and $a^2+b^2=c^2$. Thanks.
| Implement the formula:
1) $A^2-B^2=(A-B)(A+B)$
2) $\log_a xy=\log_a x+\log_a y$
3) $\log_a x^n=n\log_a x$
4) $\log_a a=1$
5) $\log_a b=\frac{1}{\log_b a}$
$a^2+b^2=c^2$
$a^2=c^2-b^2$/$\cdot\log_a$
$\log_a{a^2}=\log_a {(c^2-b^2)}$
$2\log_a a=\log_a{(c-b)(c+b)}$
$2=\log_a{(c+b)}+\log_a{(c-b)}$
$2=\frac{1}{\log_{c+b} a}+\frac{1}{\log_{c-b} a}$
$2=\frac{\log_{c-b} a+\log_{c+b} a}{\log_{c-b} a\cdot\log_{c+b} a}$
$\log_{b+c} a+\log_{c-b} a=2\log_{b+c} a\cdot\log_{c-b} a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2?
Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.
My attempt
I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$?
Thanks for your help.
| Let us denote your sequence by $x_n$. Then
$$
x_1=\sqrt{2},\ x_{n+1}=\sqrt{2x_n} \quad \forall\ n \ge 1.
$$
Clearly $\sqrt{2}\le x_n<2$ for every $n \ge 1$, and
$$
x_n-x_{n+1}=x_n-\sqrt{2x_n}=\frac{x_n^2-2x_n}{x_n+x_{n+1}}=\frac{x_n(x_n-2)}{x_n+x_{n+1}}<0 \quad \forall n \ge 1.
$$
Hence $(x_n)$ is increasing and bounded above. It follows that $(x_n)$ is convergent. If $l$ denotes its limit, then $\sqrt{2}\le l \le 2$ and $l=\sqrt{2l}$. Solving the equation $l=\sqrt{2l}$ we get $l\in \{0,2\}$, and since $\sqrt{2} \le l \le 2$, we deduce that $l=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/200416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 7,
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What is the coefficient of the $x^3$ term in the expansion of $(x^2+x-5)^7$ (See details)? I fail to see a simple way to answer this.
As such, this is my long winded approach:
Using the multinomial theorem,
$$(x_1 + x_2 + \cdots + x_m)^n
= \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m}
\prod_{1\le t\le m}x_{t}^{k_{t}},$$
we have the specific parameters $m=3$, $n=7$, $x_1=x^2$, $x_2=x$, and $x_3=-5$.
Via the theorem,
$$(x^2+x-5)^7=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}\prod_{1 \le t \le 3}x_{t}^{k_t}=\sum_{k_1+k_2+k_3=7} {7 \choose k_1, k_2, k_3}x^{2k_1}x^{k_2}(-5)^{k_3}.$$
The coefficient of the $x^3$ term is the summation of the multinomial coefficient multiplied by the $(-5)^{k_3}$ factor evaluated at all the solutions of the equation $2k_1+k_2=3$ where $0\le k_1\le 7$ and $0 \le k_2 \le 7$.
Those values are $(k_1,k_2)=\{(1,1),(0,3)\}.$ Given that $k_1+k_2+k_3=7$, $k_3$ are respectively $5$ and $4$.
Hence, the coefficient of the $x^3$ is
$${7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4.$$
Given that the definition of the multinomial coefficient is
$${n \choose k_1,k_2,\ldots ,k_m}=\frac{n!}{k_1!k_2!\cdots k_m!},$$
$$
\begin{align}
{7 \choose 1,1,5}(-5)^{5}+{7 \choose 0,3,4}(-5)^4&=\frac{7!(-5)^5}{1!1!5!}+\frac{7!(-5)^4}{0!3!4!}\\
&=7\cdot 6(-5)^5+\frac{7\cdot 6\cdot 5(-5)^4}{3!}\\
&=-109375
\end{align}$$
This could be atrociously wrong. Either way, I am desperate for a much simpler process. This is ridiculous to do in a timed testing environment without the formulas given.
I would like to see a very simple but also very general way of arriving at the correct answer (preferably without college methods, but I am open to any methods).
What says you, Math.SE?
| Using the binomial theorem write this as $\sum_{k=0}^7\binom{7}{k}x^{2k}(x-5)^{7-k}$. Now the non-zero coefficients of $x^3$ occurs in the terms for $k=0$ and $k=1$. For $k=0$ we want the coefficient of $x^3$ in $(x-5)^7$ which is $\binom{7}{3}(-5)^4$; for $k=1$ we want the coefficient of $x$ in $(x-5)^6$ which is $6(-5)^5$. Thus the answer is
$$\binom{7}{3}(-5)^4+7\cdot 6\cdot (-5)^5=-109375$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Chinese Remainder Theorem and linear congruences I have found the following congruences:
*
*$x \equiv 2\mod 5$
*$x \equiv 12\mod27$
*$x \equiv 2\mod4$
How can I solve for x using the Chinese Remainder Theorem? Please include justifications for the steps you take.
How can I find $x$?
| I will assume that you are partly familiar with the machinery of CRT.
We want to first find $a, b,c$ such that $a(27)(4)\equiv 1\pmod{5}$, and $b(5)(4)\equiv 1\pmod{27}$, and $c(5)(27)\equiv 1\pmod{7}$. Then a solution of our system of congruences is
$$x=2(a)(27)(4)+12(b)(5)(4)+2(c)(5)(27).$$
It may be that the $x$ we obtain is kind of large, and we may want to replace it by something congruent to it modulo $(5)(27)(4)=540$ that is, say, in the interval $[0,539]$.
Let's find suitable $a$, $b$, and $c$. In general for this work, and with large moduli, we would use the Extended Euclidean Algorithm. Let's not bother, since we are concentrating on CRT.
For $a$, work modulo $5$. Since $27$ is congruent to $2$, we want $8a\equiv 1\pmod{5}$, or equivalently $3a\equiv 1\pmod{5}$. So we can take $a=2$. Or maybe $a=-3$. Doesn't matter.
For $b$, we want $20b\equiv 1\pmod{27}$. Slightly unpleasant, but note that $(20)(4)\equiv -1\pmod{27}$, so we can take $b=-4$.
For $c$, the modulus is tiny. We want $c(5)(27)\equiv 1\pmod{4}$. But $5$ is congruent to $1$, and $27$ is congruent to $3$, so we can take $c=3$.
Remark: But we can take a bit of a shortcut. Note that we wanted $x\equiv 2\pmod{5}$ and $x\equiv 2\pmod{4}$. This is the case iff $x\equiv 2\pmod{20}$. So we want to solve the system of congruences $x\equiv 2\pmod{20}$, $x\equiv 12\pmod{27}$. This is a two congruence CRT problem. We find $a$ and $b$ such that $a(27)\equiv 1\pmod{20}$ and $b(20)\equiv 1\pmod{27}$. Then we can take
$$x=2(a)(27)+12(b)(20).$$
For $a$, it is clear that $3$ will do. For $b$, $-4$ works. This gives answer
$2(3)(27)+12(-4)(20)$. For something in the range $[0.539]$, add $(2)(540)$. We get $x=282$. So the general solution is $x\equiv 282\pmod{540}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
| I solved by another way. This is my solution. Please comment to me. Put $t=\sqrt{5x-1}-\sqrt{5-2x}.$
We have
$$t^2=3x+4-2\sqrt{(5x-1)( 5-2x)}$$
and
$$t^3=(14-x)\sqrt{5x-1}-(13x+2)\sqrt{5-2x}.$$
And then, the given equation has the form
$(t - 2)^3 = 0$, that is mean $t = 2$. With $t = 2, $ we have
\begin{equation*}
\sqrt{5x-1}-\sqrt{5-2x} = 2.
\end{equation*}
This equation has a solution $x =2.$
Thus, the given equation has the only root $x = 2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$ equidistributed in $[0, 1)$ Prove
$0, \frac{1}{2}, 0, \frac{1}{3}, \frac{2}{3}, 0, \frac{1}{4}, \frac{2}{4}, ...$
equidistributed in $[0, 1)$.
A sequence of numbers $\xi_1, \xi_2, \xi_3, ...$ in $[0, 1)$ is said to be equidistributed if for every
interval $(a, b) \subset [0, 1)$
$$\lim\limits_{N\to\infty} \frac {\bigl|\{1\le n\le N: \xi_n \in (a, b)\}\bigr|} {N}
= b-a$$
It's from Chapter 4 in the book Fourier Analysis: An Introduction
| For each integer $n>1$ let $$F_n=\left\{\frac{k}n:k=0,\dots,n-1\right\}\;;$$
for $0\le a<b<1$, $|(a,b)\cap F_n|\ge\lceil n(b-a)\rceil-1$.
The first $2$ terms of the sequence are the members of $F_2$; the next $3$ terms are the members of $F_3$, and so on. Thus, the first $\sum_{k=2}^nk=\frac12n(n+1)-1$ terms are the members of blocks $F_2$ through $F_n$ in the obvious order. For $n\ge 2$ let $T_n=\frac12n(n+1)-1$, and suppose that $T_m\le N<T_{m+1}$. Then
$$\begin{align*}
|\{n\le N:\xi_n\in(a,b)\}&\ge\sum_{k=2}^m|(a,b)\cap F_k|\\
&\ge\sum_{k=2}^m\Big(\lceil k(b-a)\rceil-1\Big)\\
&=\sum_{k=2}^m\lceil k(b-a)\rceil-m+1\\
&\ge\sum_{k=2}^mk(b-a)-m+1\\
&=(b-a)T_m-m+1\;,
\end{align*}$$
so
$$\frac{|\{n\le N:\xi_n\in(a,b)\}}N\ge(b-a)\frac{T_m}N-\frac{m-1}N\;.$$
Now $$\frac{T_m}N>\frac{T_m}{T_{m+1}}=\frac{\frac12m(m+1)-1}{\frac12(m+1)(m+2)-1}\to 1\quad\text{ as }\quad m\to\infty\;,$$ and
$$\frac{m-1}N\le\frac{m-1}{T_m}=\frac{m-1}{\frac12m(m+1)-1}\to 0\quad\text{ as }\quad m\to\infty\;,$$
so $$\lim_{N\to\infty}\frac{|\{n\le N:\xi_n\in(a,b)\}|}N\ge b-a\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the trigonometric identity Please help me prove the identity:
$$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\frac{1}{2}-\frac{1}{2}\cos2\alpha$$
| $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha=\cos^2\alpha+(\sin^4\alpha-\cos^4\alpha)=$$
$$=\cos^2\alpha+(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)=\cos^2\alpha+\sin^2\alpha-\cos^2\alpha=$$
$$=\sin^2\alpha=1/2-1/2\cos2\alpha$$ Over!
| {
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"timestamp": "2023-03-29T00:00:00",
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Multivariable limit - Two variables $ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$ How can I calculate the following limit and show that it equals $0$:
$$ \lim_{(x,y) \to (0, \pi ) } \frac{x^2 y \sin y } {\sin^2x + (\pi - y )^2 }$$
Thanks in advance
| Let's prove the limit using the definition. Fix $\varepsilon > 0$. We have:
$$
\left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \left| \frac{x^2 y \sin y}{\sin^2 x} \right| = \left| \frac{x}{\sin x} \right|^2 \cdot \left|y \sin y\right|
$$
We know that $\lim_{x \to 0}\frac{x}{\sin x} = 1$ and $\lim_{y \to \pi} y \sin y = 0$. Therefore, we can pick a neighborhood of $(0, \pi)$ so that:
$$
\left| \frac{x}{\sin x} \right|^2 < 1 + \varepsilon, \ \left|y \sin y\right| < \varepsilon
$$
Thus:
$$
\left| \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} \right| \le \varepsilon(1 + \varepsilon)
$$
Since our choice of $\varepsilon$ was arbitrary, we conclude:
$$
\lim_{(x, y) \to (0, \pi)} \frac{x^2 y \sin y}{\sin^2 x + (\pi - y)^2} = 0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate a limit by using squeeze theorem We're supposed to use the Squeeze Theorem to prove that
$$\lim_{x\to 0} {1-\cos x\over x^2} = \frac12$$
I tried this:
$$-1\le \cos x \le 1$$
$$-1\le -\cos x \le 1$$
$$0\le 1-\cos x \le 2$$
$$0\le {1-\cos x\over x^2} \le {2\over x^2}$$
Then using limits we have:
$$\lim_{x\to 0}0\le \lim_{x\to 0} {1-\cos x\over x^2} \le \lim_{x\to 0}{2\over x^2}$$
And for obvious reasons the first limit is $\Bbb {0}$, and the third limit is $\Bbb \infty$
What do I do now? Or what am I doing wrong?
Thanks in advance
| This might be an overkill, but according to the Taylor theorem, for any nonzero $x$ you can find $\xi_x$ between zero and $x$ in such a way that
$$
\cos x = 1 - \frac{x^2}{2} + \frac{1}{4!} \cos(\xi_x) \cdot x^4.
$$
Thus, shuffling those terms around, you would get
$$
\frac{1}{2} - \frac{x^2}{4!} \leq \frac{1 - \cos x}{x^2} = \frac{1}{2} - \frac{x^2}{4!} \cos(\xi_x) \leq \frac{1}{2} + \frac{x^2}{4!}, \quad x \neq 0.
$$
Obviously
$$
\lim_{x\to 0} \frac{1}{2} \pm \frac{x^2}{4!} = \frac{1}{2}
$$
and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Pell equation of the special form I have tried several times to solve the following equation and finally, I was failed to complete. Help me to find the solutions of Pell equation $y^2-2x^2 = p^m$, where $p$ is prime and $8|(p-1)$ or $8|(p+1)$.
A waiting the reply.
| Since
$$ (a^2-2b^2)(c^2-2d^2) = (ac-2bd)^2 - 2(bc-ad)^2 $$
it is sufficient to find a solution to $x^2-2y^2 = p$.
Since $p\equiv\pm 1\pmod{8}$, $2$ is a quadratic residue $\pmod{p}$, so there is a natural number $a<\frac{p}{2}$ such that $a^2\equiv 2\pmod{p}$ and
$$ (\heartsuit)\quad a^2-2 = k p,$$
with $k<\frac{p}{4}$. Moreover, every odd prime $q$ that divides $k$ is $\equiv\pm 1\pmod{8}$, because from $(\heartsuit)$ we have that $a$ is a square root of $2\pmod{q}$. We can get rid of the (possible) factor $2$ in $k$: if $k$ is even then $a$ is even too, and:
$$ (a-1)^2 - 2\left(\frac{a}{2}-1\right)^2=\frac{1}{2}\left(a^2-2\right).$$
So we have:
$$ a^2-2b^2 = Qp, $$
where $Q$ is a product of primes of the form $8n\pm1$. Let $q$ be one of them, and
$$ a_q^2-2 = rq, $$
with $r<q/4$. By the initial identity we get:
$$ (a a_q-2b)^2-2(b a_q-a)^2 = rqQp, $$
but both $(a a_q-2b)$ and $(b a_q-a)$ are divisible by $q$, so:
$$ a_*^2 - 2b_*^2 = \left(\frac{a a_q-2b}{q}\right)^2-2\left(\frac{b a_q-a}{q}\right)^2 = r(Q/q)p, $$
and we can get rid of any prime factor in Q, simply starting from the greatest. This proves that any prime of the form $8n\pm1$ is represented by the quadratic form $x^2-2y^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.
| Partial fractions are the way to go. The fraction is already reduced, and the denominator is fully factored over the reals, so your setup is
$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}\;,$$
and you must find $A,B$, and $C$ so that $$2x^2+x=A(x^2+1)+(Bx+C)(x+1)=(A+B)x^2+(B+C)x+(A+C)\;.$$
Equating coefficients of powers of $x$ yields the system
$$\left\{\begin{align*}
&A+B=2\\
&B+C=1\\
&A+C=0\;,
\end{align*}\right.$$
which is easily solved: $A=\frac12,B=\frac32$, and $C=-\frac12$. Thus,
$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac1{2(x+1)}+\frac{3x-1}{2(x^2+1)}\;,$$ and
$$\int\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac12\int\frac1{x+1}dx+\frac12\int\frac{3x-1}{x^2+1}dx\;.$$
You shouldn’t have any trouble with $\int\frac1{x+1}dx$. The other term is most easily handled by splitting it:
$$\int\frac{3x-1}{x^2+1}dx=3\int\frac{x}{x^2+1}dx-\int\frac1{x^2+1}dx\;,$$
where the first integral succumbs to a $u$-substitution, and the second is one that you should know (or at least be able to work by a trig substitution).
| {
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Solve for $x$ in this equation How do I solve for $x$ algebraically?
$$\dfrac{x^2(x^2-1)}{x+3} = 12$$
| Rearranging the equation, we get $$x^2(x^2-1) = 12(x+3)$$
$$x^4 - x^2 - 12x - 36 = 0$$
First we search for integer roots. The integer root must divide $36$. Hence, the possible integer options are $\pm 1,\pm 2, \pm 3$. Checking these $6$ options, give us $x=-2$ and $x=3$. Hence, $$x^4 - x^2 - 12x - 36 = (x+2)(x-3)(x^2+ax+b)$$
Comparing coefficients, we get $a=1$ and $b=6$. Solving the quadratic, gives the other roots as $$x^2 + x + 6 =0 \implies \left(x + \dfrac12 \right)^2 + 6 - \dfrac14 = 0 \implies x = -\dfrac12 \pm i \dfrac{\sqrt{23}}2$$
| {
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finding all pairs $(x,z)$ How to find all pairs $(x,z)$ of integers for which $2(z+1)^3$ is divisible by $xz-1$
| Here's all the pairs of $(x,z)$ I could find for positive $x,z$. On the right is $2(z+1)^3/(xz-1)$
\begin{array}{cc}
\left(1,2\right) & 54 \\
\left(1,3\right) & 64 \\
\left(1,5\right) & 108 \\
\left(1,9\right) & 250 \\
\left(1,17\right) & 729 \\
\left(2,1\right) & 16 \\
\left(2,2\right) & 18 \\
\left(2,5\right) & 48 \\
\left(2,14\right) & 250 \\
\left(3,1\right) & 8 \\
\left(3,3\right) & 16 \\
\left(3,11\right) & 108 \\
\left(3,43\right) & 1331 \\
\left(5,1\right) & 4 \\
\left(5,2\right) & 6 \\
\left(5,5\right) & 18 \\
\left(5,11\right) & 64 \\
\left(5,29\right) & 375 \\
\left(9,1\right) & 2 \\
\left(9,9\right) & 25 \\
\left(9,14\right) & 54 \\
\left(11,3\right) & 4 \\
\left(11,5\right) & 8 \\
\left(11,35\right) & 243 \\
\left(14,2\right) & 2 \\
\left(14,9\right) & 16 \\
\left(17,1\right) & 1 \\
\left(29,5\right) & 3 \\
\left(29,69\right) & 343 \\
\left(35,11\right) & 9 \\
\left(43,3\right) & 1 \\
\left(69,29\right) & 27
\end{array}
See anything?
| {
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Range of a trignonmetric function I came across this in an Engineering entrance book,
What is the range of this: $a^2 \sin^2 x + b \sin x \cos x + c \cos^2 x$
What is the method to find it? I tried the graph approach but didn't know how to proceed.
| $$a^2 \sin^2 x + b \sin x \cos x + c\cos^2 x$$
$$=\frac{a^2 2\sin^2 x + b 2\sin x \cos x + 2c\cos^2 x}2$$
$$=\frac{a^2(1-\cos2x)+b\sin2x+c(1+\cos2x))}2$$
$$=\frac{a^2+c}2+\frac12 \{\cos2x(c-a^2)+b\sin2x\}$$
Now, for $A\cos y+B\sin y=C\sin(y+\theta)$(say) where $C\ge 0$
Expanding and comparing the coefficients, we get $A=C\cos\theta, B=C\sin\theta$
squaring and adding we get, $C^2=A^2+B^2,C=\sqrt{A^2+B^2}$
As $-1\le\sin(y+\theta)\le 1, -\sqrt{A^2+B^2}\le C\sin(y+\theta)\le \sqrt{A^2+B^2}$
Here $y=2x,A=b, B=c-a^2$
| {
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Evaluate the triple integral $\int_0^2\int_0^y\int_0^{\sqrt{4-y^2}}2xdxdzdy\\$. $$
\\
\int_0^2\int_0^y\int_0^{\sqrt{4-y^2}}2xdxdzdy\\
\int_0^2\int_0^y4-y^2dzdy\\
\int_0^2(4-y^2)ydy\\
-\frac{1}{2}\left( \left.4y-\frac{1}{3}y^3\right|_0^2\right )\\
=-\frac{8}{3}
$$
I have worked over this problem several times and I cannot find the step where I went wrong.
| The step where you made mistake is in evaluating the integral $\displaystyle \int_0^2 (4-y^2)ydy$
$$\int_0^2 (4-y^2)ydy = \int_0^2 \left( 4y - y^3\right)dy = \left(2y^2 - y^4/4 \right)_{y=0}^{y=2} = 8 - 2^4/4 = 8 - 4 =4$$
| {
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Prove that $\frac{1}{4-\sec^{2}(2\pi/7)} + \frac{1}{4-\sec^{2}(4\pi/7)} + \frac{1}{4-\sec^{2}(6\pi/7)} = 1$
How can I prove the fact $$\frac{1}{4-\sec^{2}\frac{2\pi}{7}} + \frac{1}{4-\sec^{2}\frac{4\pi}{7}} + \frac{1}{4-\sec^{2}\frac{6\pi}{7}} = 1.$$
When asked somebody told me to use the ideas of Chebyshev polynomial, but I haven't learnt that in school.
I tried doing this way:
Look at $y =\cos\theta + i \sin\theta$ where $\displaystyle\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7},\cdots,2\pi\Bigr\}$
Then we have
\begin{align*}
y^{7} &=1 \\ y^{7}-1 &=0 \\ (y-1) \cdot (y^{6}+y^{5}+\cdots + 1) &= 0
\end{align*}
Now the root $y=1$ corresponds to $\theta = 2\pi$, and that $$y^{6} + y^{5}+\cdots + 1 =0$$
have roots $\cos\theta + i \sin\theta$, where $\theta \in \Bigl\{\frac{2\pi}{7},\frac{4\pi}{7} ,\cdots \Bigr\}$. Looking at $y+\frac{1}{y} $ will give me the roots as $\cos\theta$ and then i can put $z=y^{2}$ to get $\cos^{2}$ as the roots and the invert to get $\sec^{2}$, but I have some problems.
Can anyone help me out with a neat solution.
Thanks.
| Where you have left of $y^6+y^5+\cdots+y+1=0$ where $y=\cos \theta+i\sin \theta$ where $\theta=\frac{2\pi}7,\frac{4\pi}7,\frac{6\pi}7,\cdots , \frac{12\pi}7$
Let us divide both sides by $y^3,$ $y^3+\frac1{y^3}+y^2+\frac 1{y^2}+y+\frac 1 y+1=0$
or $\left(y+\frac1y\right)^3-3\left(y+\frac1y\right)+\left(y+\frac1y\right)^2-2+\left(y+\frac1y\right)+1=0$
or $\left(y+\frac1y\right)^3+\left(y+\frac1y\right)^2-3\left(y+\frac1y\right)-1=0$
Now, $\displaystyle y+\frac 1 y=2\cos \theta=z$ (say)
So, $\displaystyle z^3+z^2-3z-1=0\ \ \ \ \color{Red}{(1)},$ has the roots $\displaystyle 2\cos\frac{2\pi}7, 2\cos\frac{4\pi}7, 2\cos\frac{6\pi}7$ using $\displaystyle\cos\frac{r\pi}7=\cos\left(2\pi-\frac{r\pi}7\right)=\cos\frac{(14-r)\pi}7 $ as $\color{Red}{(1)}$ does not have repeated roots
$\displaystyle\implies z^2(1+z)=3z-1,z^2=\frac{3z-1}{z+1}$
$$\text{Now,}\displaystyle\frac 1{4-\sec^2\theta}=\frac{\cos^2\theta}{4\cos^2\theta-1}= \frac{z^2}{4z^2-4}=w(say),$$
$\displaystyle\implies z^2=\frac {4w}{4w-1}$
Comparing the values of $\displaystyle z^2, \frac {4w}{4w-1}=\frac{3z-1}{z+1}$
Replacing the $z$ with $w$ in $\color{Red}{(1)}$, we shall get a cubic equation in $w,$ whose sum of roots will give us the required identity.
| {
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Finding determinants using both reduction and cofactor expansion I know that you can find the determinant of a matrix by either row reducing so that it is upper triangular and then multiplying the diagonal entries, or by expanding by cofactors. But could I reduce the matrix halfway (not entirely reduced to the point where it is in upper triangular) and then do cofactor expansion? Would that give me the same determinant?
| Yes, provided you keep track of the changes to the determinant. Any combination of row reductions and cofactor expansions can be used. For example
$$\begin{vmatrix}5 & 2 & 3 \\ 12 & 4 & 6 \\ 3 & 4 & 7\end{vmatrix} = 2\begin{vmatrix}5 & 2 & 3 \\ 6 &2 & 3 \\ 3 & 4 & 7\end{vmatrix} = 2\begin{vmatrix}5 & 2 & 3 \\ 1 & 0 & 0 \\ 3 & 4 & 7\end{vmatrix}$$
Where we have first factored out a $2$ from row $2$ and then subtracted row $1$ from row $2$. Now we expand along row $2$ to get
$$2\begin{vmatrix}5 & 2 & 3 \\ 1 &0 & 0 \\ 3 & 4 & 7\end{vmatrix} = 2(-1)^{2+1}\begin{vmatrix} 2 & 3 \\ 4 & 7\end{vmatrix} = -2\begin{vmatrix} 2 & 3 \\ 0 & 1\end{vmatrix}$$
where in the last step we subtract twice row $1$ from row $2$. Now we simply multiply the diagonal entries to get determinant equal to $-4$.
| {
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Primitive root modulo p Let $p$ be an odd prime with a primitive root $g$. Prove that $$\prod_{x=1}^{\frac{p-1}{2}}x^2 \equiv (-1)^{\frac{p+1}{2}}\pmod{p}.$$
Remark: I intend to use the relationship $g^{\frac{p-1}{2}} \equiv -1 \pmod{p}.$
| $\prod_{1\le x\le p-1}x\equiv\prod_{1\le y\le p-1}g^y$
$=g^{1+2+\cdots +p-1}=g^{\frac{p(p-1)}2}=(g^\frac{p-1}2)^p\equiv (-1)^p=-1$
$\prod_{1\le x\le p-1}x=\prod_{1\le x\le \frac{p-1}2}x(p-x)$ as to avoid omission and repetition of terms $x\le p-x\implies x\le \frac p 2$ i.e., $x\le\frac{p-1}2$ as $p$ is odd
So, $\prod_{1\le x\le p-1}x\equiv(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2$
So, $(-1)^\frac{p-1}2\prod_{1\le x\le \frac{p-1}2}x^2\equiv-1$
Multiplying either sides by $(-1)^{p-1},$ (which is legal as $(-1)^{p-1}=1$ as $p$ is odd)
$\prod_{1\le x\le \frac{p-1}2}x^2\equiv (-1)(-1)^\frac{p-1}2=(-1)^\frac{p+1}2$
| {
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Prove that $\lim\limits_{x\to 0}\frac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\frac{1}{6}$ without De L'Hôpital Rule or Taylor Expansion? Any ideas how to show $\lim\limits_{x\to 0}\dfrac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\lim\limits_{x\to 0}\dfrac{x-\sin(x)}{x^3}=\dfrac{1}{6}$ without using the De L'Hôpital rule (or proving a special case of it?).
How can we reduce this to $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2}$?
You can suppose that we know the limit in question exists
and therefore use inequalities to bound it
| $$\lim_{x\to 0}\frac{x^2-1+\cos^2x}{x^2}=\lim_{x\to 0}\frac{x^2-(1-\cos^2x)}{x^2}=\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2}=$$
$$=\lim_{x\to 0}(1-\frac{\sin^2x}{x^2})=1-\lim_{x\to 0}\frac{\sin^2x}{x^2}=1-\lim_{x\to 0}(\frac{\sin x}{x})^2=1-1=0$$
| {
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solutions of $a^2+b^2=c^2$ I am trying to figure the following out.
If you have $a^2+b^2=c^2$ and let $x=a/c$ and $y=b/c$
how can you show that
$x=\frac{m^2-n^2}{m^2+n^2}$ and $y=\frac{2mn}{m^2+n^2}$ for some relatively prime numbers $m,n \in \mathbb{Z}$
| The substitution turns the problem into finding rational points $(x,y)$ on $$x^2 + y^2 = 1,$$ given any two rational points the line through them will have rational gradiant ($\Delta y/\Delta x$). In fact we have a sort of converse: if you have one point and a line with rational gradiant through that point, then the intersection of it with the circle will have two points (because every quadratic has two roots) both rational (because the conjugate of a rational number is rational).
If we picked $(1,0)$ and intersect the line $y = (m/n)x - m/n$ with the circle $x^2 + y^2 - 1 = 0$ we get $x^2 + ((m/n)x - m/n)^2 - 1 = 0$ which must be divisible by $x-1$ since that is $x=1$ is a root, so after performing the long division we get $x = (m^2 - n^2)/(m^2 + n^2)$, back substituting that gives $y = (-2nm)/(m^2 + n^2)$ thus $(a:b:c) = (m^2-n^2:2mn:m^2+n^2)$ gives all solutions of $a^2 + b^2 = c^2$.
| {
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Need help on dividing polynomials I have trouble understanding how to divide $x^4 + y^4$ by $f_1 = x^2 + y$ and $f_2 = x^2 y + 1$ using the ordering $ y \leq x$ and separately for $ x \leq y$.
Please help! I went to the tutoring center but none of the specialists understand this and I have never done this before in high school.
| Note: I am making an assumption here about exactly what is wanted, but I think it the likeliest possibility for a pre-calculus algebra course.
For the order $y\le x$, pretend that $y$ is a constant. Then you get this long division:
$$\require{enclose}\begin{align}
x^2-\phantom{-x^2}y\phantom{+y^4+y^2}\\
x^2+y\enclose{longdiv}{x^4+\phantom{-x^2y}+\phantom{y^4+}y^4}\\
\underline{x^4+\phantom{-}x^2y\,\phantom{+y^4+y^2}}\\
-x^2y+\phantom{y^4+}y^4\\
\underline{-x^2y-\,\phantom{y^4+}y^2}\\
y^4+y^2
\end{align}$$
Thus, $x^4+y^4=\left(x^2+y\right)\left(x^2-y\right)+\left(y^4+y^2\right)$, with quotient $x^2-y$ and remainder $y^4+y^2$.
For the order $x\le y$ you treat $x$ as if it were a constant:
$$\require{enclose}\begin{align}
y^3-\phantom-x^2y^2+x^4y-\phantom{-y}x^6\;\phantom{+x^8+x^4}\\
y+x^2\enclose{longdiv}{y^4+\phantom{-x^2y^3-x^4y^2+-x^6y}+\phantom{x^8+}x^4}\\
\underline{y^4+\phantom-x^2y^3\;\phantom{-x^4y^2+-x^6y+x^8+x^4}}\\
-x^2y^3\,\phantom{-x^4y^2+-x^6y+x^8+x^4}\\
\underline{-x^2y^3-x^4y^2\phantom{+-x^6y+x^8+x^4}}\\
x^4y^2\phantom{+-x^6y+x^8+x^4}\\
\underline{x^4y^2+\phantom-x^6y\phantom{+x^8+x^4}\;\;}\\
-x^6y+\;\phantom{x^8+}x^4\\
\underline{-x^6y-\;\phantom{x^8+}x^8}\\
x^8+x^4
\end{align}$$
That is, $y^4+x^4=\left(y+x^2\right)\left(y^3-x^2y^2+x^4y-x^6\right)+\left(x^8+x^4\right)$, with quotient $$y^3-x^2y^2+x^4y-x^6$$ and remainder $x^8+x^4$.
| {
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Limit $\lim\limits_{n\to\infty} \sqrt[n]{\frac1{\sqrt3}\left(\left(\frac{1+\sqrt3}2\right)^n-\left(\frac{1-\sqrt3}2\right)^n\right)}$ I have problems with finding: $$\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$$ I tried to do it following way:
$\displaystyle\lim_{n\to\infty} \sqrt[n]{{\frac{1}{\sqrt{3}}\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n}-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$
$\displaystyle\lim_{n\to\infty} \sqrt[n]{\frac{1}{\sqrt{3}}}\cdot\lim_{n\to\infty} \sqrt[n]{\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}=$
$1\cdot\displaystyle\lim_{n\to\infty} \sqrt[n]{\Bigg(\Bigg(\frac{1+\sqrt{3}}{2}\Bigg)^n-\Bigg(\frac{1-\sqrt{3}}{2}\Bigg)^n\Bigg)}$
Now I used formula for difference of powers:
$$$$
$\Big(\frac{1+\sqrt{3}}{2}\Big)^n-\Big(\frac{1-\sqrt{3}}{2}\Big)^n=\Big(\frac{1+\sqrt{3}}{2}-\frac{1-\sqrt{3}}{2}\Big)\cdot\Big( \Big(\frac{1+\sqrt{3}}{2}\Big)^{n-2}\Big(\frac{1+\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-3}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+\Big(\frac{1+\sqrt{3}}{2}\Big)^{n-4}\Big(\frac{1+\sqrt{3}}{2}\Big)\Big(\frac{1-\sqrt{3}}{2}\Big)+...+\Big(\frac{1-\sqrt{3}}{2}\Big)^{n-1}\Big)$
In the last parenthesis, I saw two geometric series and I tried to add them, however, it quickly appeared that there will be other geometric series and here is where I am a bit helpless (it is getting very nasty very quickly). Do you have any hints to move it in maybe another way? I would be very grateful, thanks!
| Let $p = \frac{1+\sqrt{3}}2$ and $q = \frac{1-\sqrt{3}}2$ and denote your sequence as $$a_n = \frac{p^n-q^n}{\sqrt3}.$$
We have
$$\frac{a_{n+1}}{a_n} = \frac{p^{n+1}-q^{n+1}}{p^n-q^n} = \frac{p-q\left(\frac{p}{q}\right)^n}{1-\left(\frac{p}{q}\right)^n} \xrightarrow{n\to\infty} p$$
so by this question it also follows that
$$\lim_{n\to\infty} \sqrt[n]{a_n} = p$$
because if the ratio $\left(\frac{a_{n+1}}{a_n}\right)_n$ converges, then the $n$-th root $\left(\sqrt[n]{a_n}\right)_n$ also converges and to the same limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/233441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Expressing in rationals I have this question to express it in a specific form
Express $\dfrac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$ in the form $a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42}$, for some rationals $a,b,c$ and $d$.
So the solution that I have and am trying to understand is this
So what I don't understand is that why in the world is there a 2sqrt6 here where did it come from?
| $(\sqrt{2}+\sqrt{3}+\sqrt{7})(\sqrt{2}+\sqrt{3}-\sqrt{7})$
$=(\sqrt{2}+\sqrt{3})^2-(\sqrt{7})^2$
$=(\sqrt{2})^2+(\sqrt{3})^2+2\sqrt{2}\sqrt{3}-(\sqrt{7})^2$
$=2+3+2\sqrt{6}-7$
$=2\sqrt 6-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find k given determinant? So I've got this matrix here, and need to solve for $k$
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1 & 4 \\
2 & k & 6 & 5 \\
-3& 2 & 1 & 0 \\
6 & 4 & 2 & 3 \\
\end{pmatrix}=33$$
Doing some row operations $(R3+R1) \to R3\text{ and}\; (R4-2R1)\to R4$), I end up with
$$\text{det}\;\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33$$
I expand along the first column and somehow my $k$ value is a decimal. Am I doing this correctly? I've tried making this into an upper and lower diagonal matrix and it just gets messy.
| How about continuing your work! You've already got
$$
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 4 \\
2 & k & 6 & 5 \\
0 & 4 & 0 & 4 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33.
$$
Subtract column 2 (yes, we do column reduction now!) from column 4 gives
$$
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 2 \\
2 & k & 6 & 5-k \\
0 & 4 & 0 & 0 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=33.
$$
Now, only one nonzero element remains in the third row. Expand along the third row, we only need to evaluate one $3\times3$ determinant. Hooray~~~!!!
$$
33=
\text{det}\;\begin{pmatrix}
3 & 2 & -1& 2 \\
2 & k & 6 & 5-k \\
0 & 4 & 0 & 0 \\
0 & 0 & 4 & -5\\
\end{pmatrix}=-4
\ \text{det}\;\begin{pmatrix}
3 & -1& 2 \\
2 & 6 & 5-k \\
0 & 4 & -5\\
\end{pmatrix}=-48(k-12).
$$
This gives $k=\frac{33}{-48}+12=-\frac{11}{16}+12=\frac{181}{16}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Helping turning this into sum/difference logarithm? Hey guys so i'm trying to turn this equation into it's sum/difference logarithm. However, the part that messes me up is turning the bottom of the fraction
$$
\log\left(\frac{x^2 +2x+1}{x^2 -3x +2}\right)^2\;.
$$
I think it will turn into this:
$$
4\log(x+1) -2\big(\log(x-1) +\log(x-2)\big)\;,
$$
but I don't want it to be $(x-1)^2$ times $(x-2)^2$. How do I solve it so it's $\big((x-1)(x-2)\big)^2$?
| I assume you want to simplify
$$\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right)$$
First recall the following properties of logarithm.
\begin{align}
1. & \log (a^m) = m \log (a)\\
2. & \log \left( \dfrac{a}b\right) = \log a - \log b\\
3. & \log \left( ab\right) = \log a + \log b
\end{align}
Using the first property, we get that
$$\log \left( \left( \dfrac{x^2+2x+1}{x^2 - 3x+2} \right)^2\right) = 2 \log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right)$$
Now using the second property, we get that
$$\log \left( \dfrac{x^2+2x+1}{x^2 - 3x+2}\right) = \log (x^2+2x+1) - \log (x^2-3x+2)$$
Next note that $(x^2 + 2x + 1) = (x+1)^2$ and $x^2 - 3x + 2 = (x-1)(x-2)$. Hence, we have that $$\log (x^2+2x+1) = \log ((x+1)^2) = 2 \log(x+1)$$ using property $(1)$. Similarly, we have that $$\log (x^2-3x+2) = \log ((x-2)(x-1)) = \log(x-2) + \log(x-1)$$ using property $(3)$. Putting all these together, we get what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}$ I'm trying to compute:
$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}$$
(From CMJ)
Using the duplication formula:
$$ \Gamma(x)\Gamma \left(x+\frac{1}{2} \right)=\frac{\sqrt{\pi}}{2^{2x-1}}\Gamma(2x)$$
$$ \frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}}\frac{2^{m+n-1}}{\Gamma(m+n)}=\frac{1}{\sqrt{\pi}}\frac{2^{m+n-1}}{(m+n-1)!}$$
So:
$$ \sum_{m=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}}\sum_{m=0}^{\infty}\frac{2^{m+n-1}}{(m+n-1)!}=\frac{1}{\sqrt{\pi}}\sum_{m=n-1}^{\infty} \frac{2^m}{m!}$$
$$ \frac{1}{\sqrt{\pi}}\sum_{m=n-1}^{\infty} \frac{2^m}{m!} \sim_{n\rightarrow\infty} \frac{1}{\sqrt{\pi}} \frac{2^{n-1}}{(n-1)!}$$
The series $$ \sum_{n\geq1} \frac{1}{\sqrt{\pi}} \frac{2^{n-1}}{(n-1)!}$$ is convergent so:
$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}} \sum_{n=1}^{\infty} \sum_{m=n-1}^{\infty} \frac{2^m}{m!}$$
Is there a simple way to compute this quantity?
| Rearranging the terms and putting those with $k = m+n$ together (and leaving out the one where $k=0$), we get
$$\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{\Gamma(\frac{m+n}{2})\Gamma(\frac{1+m+n}{2})} = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{2^{m+n-1}}{\Gamma(m+n)\sqrt{\pi}}$$ $$= \sum_{k=1}^{\infty} \frac{(k+1)2^{k-1}}{(k-1)!\sqrt{\pi}} = \sum_{k=1}^{\infty} \frac{(k-1)2^{k-1}}{(k-1)!\sqrt{\pi}} + 2\sum_{k=1}^{\infty} \frac{2^{k-1}}{(k-1)!\sqrt{\pi}}$$$$= \frac{2}{\sqrt{\pi}}\sum_{k=2}^{\infty} \frac{2^{k-2}}{(k-2)!} + \frac{2}{\sqrt{\pi}}e^2=\frac{4e^2}{\sqrt{\pi}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/241585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Need an explanation of a particular expression transformation Please, I need an explanation of the one transformation. I have the equation set and its solution.
$$
\begin{cases}
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 3\\\\
\frac{y}{x} + \frac{z}{y} + \frac{x}{z} = 3\\\\
\ x + y + z = 3
\end{cases}
$$
In the solution three new variables were introduced:
$$
u = \frac{x}{y}; v = \frac{y}{z}; w = \frac{z}{x}
$$
And then using this new vars equation set became this:
$$
\begin{cases}
\ u + v + w = 3\\\\
\frac{1}{u} + \frac{1}{v} + \frac{1}{w} = 3\\\\\\
\ uvw = 1
\end{cases}
$$
I can't understand how the $x + y + z = 3$ has become the $uvw = 1$. Can someone explain what have been done here?
My appreciation.
| $$
uvw = \frac{x}{y}\frac{y}{z} \frac{z}{x}=\frac{xyz}{yzx}=1
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simplifying third degree polynomials I am trying to figure out this simple problem but I have no idea how to do it.
$\dfrac {2x^3 - 5x^2 -4x -3}{2x^3 + x^2 - 18x -9}$
I know that I can keep guessing at random to find a factor in the terms to factor them out but that seems like it would be far too time intensive, especially considering that if this was on a test it would likely be the first step of the problem and I can't spend 30 minutes just setting the problem up.
| Any common divisor of $2x^3 - 5x^2 -4x -3$ and $2x^3 + x^2 - 18x -9$ should also be a common divisor of their difference.
$$(2x^3 + x^2 - 18x -9) - (2x^3 - 5x^2 -4x -3)
= 6x^2 - 14x - 6 = 2(3x^2 - 7x -3)$$
$3x^2 - 7x -3$ does not have any rational-coefficient factors and it does not divide evenly into $2x^3 - 5x^2 -4x -3$. That is enough information to conclude that
$$\gcd(2x^3 - 5x^2 -4x -3, 2x^3 + x^2 - 18x -9) = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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can we have a triangle with sides $1, x$ and $x^2$? Can we have a triangle with sides $1$, $x$ and $x^2$? And what could $x$ be?
I try to approach this question by making 3 inequalities.
$1+x>x^2$,
$1+x^2>x$,
$x^2+x>1$
and they come with different quadratic inequalities
$x^2-x-1<0$ (solution is $(1+\sqrt 5)/2 > x > (1-\sqrt 5)/2$ ) ;
$x^2-x+1>0$ (solution is $x > (-1+\sqrt 5)/2$ or $x < (-1-\sqrt 5)/2$ )
$x^2+x-1>0$ (no real solution)
Then I start to struggle with the next step.... Thank you
| Your approach is fine. $1+x > x^2$ has the solution set you have indicated, but since $x$ is the side of a triangle, we cannot have negative $x$. Thus $0 < x < \frac{1 + \sqrt{5}}{2}$. The second inequality, $1 + x^2 > x$ is satisfied by all real numbers, so no further restriction here. The solutions set of $x^2 + x > 1$ is $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$, and only positive $x$ matters. Thus, any $x$ such that:
$$ \frac{-1+\sqrt{5}}{2} < x < \frac{1+\sqrt{5}}{2} $$
will give a valid triangle with sides $1$, $x$ and $x^2$. Approximately, $.618 < x < 1.618$. (Coincidentally, the larger value is the golden ratio, and the lower value is its multiplicative inverse!)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How come $2 \times 3^k + 3^k = 3 \times 3^k$ I get something else
$$\begin{align*}
&2 \times 3^k + 3^k=\\
&2 \times 3^k \times 2=\\
&4 \times 3^k
\end{align*}$$
What does $3^k + 3^k$ give exactly?
| Remember the order of precedence of operations. Here,
First Multiply $\quad2\times 3^k\quad $ ... then add: $\quad(2\times 3^k) + 3^k\;$:
$$2\times 3^k + 3^k = (2 \times 3^k) + 3^k = (3^k + 3^k) + 3^k = 3\times 3^k$$
Note that if we just let $x = 3^k$, then $$2 \times 3^k + 3^k = (2\times x) + x = (2 \times x) + (1\times x) = (2 + 1)\times x = 3\times x.$$
Now, since we let $x = 3^k$, then $3 \times x = 3\times 3^k$.
For your last question, $3^k + 3^k = 2\times 3^k$, but note that the first operation to perform in your original question is $\quad(1st)$ "$\times$" $\quad$ then $\;\;(2nd)$ "$+$".
| {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci numbers divisible by $9$ The $n$th Fibonacci number $F_n$ is defined as follows,$$F_1=F_2=1\mbox{ and } F_{n+2}=F_{n+1}+F_{n}\mbox{ for } n\geq 1$$
I want to know how many of the first $1000$ Fibonacci numbers are divisible by $9$ ? Is it possible to classify all Fibonacci numbers which are divisible by $9$ ?
I have searched in Google and found in an article that says every $12$th Fibonacci number is divisible by $9$ but no proof is given. Are these the only such numbers ? Any idea for the soloution would be helpful.
| First, let's prove if $n$ divisible by $m$, then $F_n$ is also divisible by $F_m$.
To prove it we need formula $$\Large{F_{n+k} = F_{k+1} \cdot F_n + F_k \cdot F_{n-1}} \; (1) $$
Therefore, $F_{2n} = F_{n+1} F_n + F_n F_{n-1}$ is divisible by $F_n$, and $F_{3n} = F_{2n + n} = F_{2n+1} F_n + F_{2n} F_{n-1}$ is divisible by $F_n$, and so on.
Using induction method, let's assume $F_{k \cdot n}$ is divisible by $F_n$, therefore, $$\Large{F_{(k+1)n} = F_{kn +n} = F_{n+1} F_{kn} + F_n F_{kn-1}} \; (2) $$ is also divisible by $F_n$.
Now, we need to find all Fibonacci numbers which are divisible by 9. The first Fibonacci number which is divisible by 9 is $F_{12}=144$.
Therefore, $F_{12n}$ is divisible by 9.
Now, let's prove, that there is no other Fibonacci number which is divisible by 9.
$$ \begin{array}{|c|c|}
\hline
F_{1} \equiv 1 \; (mod \; 9) & F_{13} \equiv 8 \; (mod \; 9) \\
F_{2} \equiv 1 \; (mod \; 9) & F_{14} \equiv 8 \; (mod \; 9) \\
F_{3} \equiv 2 \; (mod \; 9) & F_{15} \equiv 7 \; (mod \; 9) \\
F_{4} \equiv 3 \; (mod \; 9) & F_{16} \equiv 6 \; (mod \; 9) \\
F_{5} \equiv 5 \; (mod \; 9) & F_{17} \equiv 4 \; (mod \; 9) \\
F_{6} \equiv 8 \; (mod \; 9) & F_{18} \equiv 1 \; (mod \; 9) \\
F_{7} \equiv 4 \; (mod \; 9) & F_{19} \equiv 5 \; (mod \; 9) \\
F_{8} \equiv 3 \; (mod \; 9) & F_{20} \equiv 6 \; (mod \; 9) \\
F_{9} \equiv 7 \; (mod \; 9) & F_{21} \equiv 2 \; (mod \; 9) \\
F_{10} \equiv 1 \; (mod \; 9) & F_{22} \equiv 8 \; (mod \; 9) \\
F_{11} \equiv 8 \; (mod \; 9) & F_{23} \equiv 1 \; (mod \; 9) \\
F_{12} \equiv 0 \; (mod \; 9) & F_{24} \equiv 0 \; (mod \; 9) \\ \hline
\end{array}$$
It will continue in the same sequence. Which means only $F_{12n}$ is divisible by 9.
Prove (1) using induction, also.
| {
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Identify and sketch the quadric surface? I'm stuck trying to figure out which type of quadric surface this equation is:
$$\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$$
I have narrowed it down to a hyperboloid, but cannot determine if it is of one or two sheets. I'm guessing it's two sheets because it has all negative signs, but I'm not sure. Thanks!
|
$$\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$$
You're correct: this is an hyperbola, and it does have two sheets. You might want to explore Hyperpoloids for more information on equations of the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1,$$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1,$$
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1.$$
$\text{Graph of }\quad\dfrac{x^2}{16} - \dfrac{y^2}{9} - \dfrac{z^2}{1} = 1$
Integer solutions: $(x, y, z): (4, 0, 0), (-4, 0, 0)$.
Solutions in $z: z = \pm \dfrac{1}{12}\sqrt{9x^2 - 16y2 -144}$.
Solutions in $y: y = \pm \dfrac34 \sqrt{x^2 - 16z^2 -16}$
| {
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Finding x in $\frac{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,1-x)}{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,x)} = \sqrt{n}$ I was trying to find a closed-form for $0<x<1$ in,
$$\frac{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,1-x)}{\,_2F_1(\frac{1}{m},\,1-\frac{1}{m},\,1,\,x)} = \sqrt{n}$$
where $\,_2F_1(a,b,c,z)$ is the hypergeometric function. There are formulas for $m = 2,3,4,6$, so I was wondering if there are for other m as well. However, one thing I observed was that, let,
$$q = \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/m)}\right)$$
Conjecture:
$$\lim_{n\to \infty}\frac{x}{q} = \text{constant}$$
namely,
$$\begin{array}{cc}
m&\lim_{n\to \infty}\frac{x}{q}\\\\
2&16\\
3&27\\
4&64\\
5&25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5}=163.95\dots\\
6&432\\
7&1152.795095384373\dots\\
8&2^8\left(1+\sqrt{2}\right)^{2\sqrt{2}}=3096.65\dots\\
\end{array}$$
and so on. This implies a good approximation to x in,
$$\frac{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,1-x)}{\,_2F_1(\frac{1}{5},\frac{4}{5},\,1,\,x)} = \sqrt{n}$$
is given by,
$$x \approx 25\sqrt{5}\left(\frac{1+\sqrt{5}}{2}\right)^\sqrt{5} \exp\left(\frac{-\,\pi\sqrt{n}}{\sin(\pi/5)}\right)$$
(One can numerically solve for x for a given n using Mathematica's FindRoot command.)
Question: Is the conjecture true? And what is the closed-form for the constant $1152.79509\dots$ when $m=7$?
EDIT:
Courtesy of Sasha’s answer below, then the closed-form for m = 7, as radicals raised to radical powers is,
$$ (14)^2 \prod_{k=1}^{3} \frac{1}{\sin(\pi k/7)^{4\cos(2\pi k/7)}} = 1152.79509\dots$$
In general,
$$\lim_{n\to\infty}\frac{x}{q} = (2m)^2 \prod_{k=1}^{\lfloor (m-1)/2 \rfloor} \frac{1}{\sin(\pi k/m)^{4\cos(2\pi k/m)}} $$
| The hypergeometric function ${}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; y\right)$ is an increasing from of $y$, starting at 1 for $y=0$ and increasing boundlessly as $y$ approaches one.
Thus, for large $n$ we should expect $x$ to be small. Taking the first to terms of the series expansion at unity:
$$
{}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; 1 - x \right) = -\frac{1}{\pi} \sin\left(\frac{\pi}{m} \right) \left( \log x + 2 \left( \gamma + \psi\left(\frac{1}{m} \right) \right) \right) - \cos\left(\frac{\pi}{m} \right) + \mathcal{O}(x)
$$
which coincides with the expansion of the ratio of hypergeometric functions.
$$
\frac{ {}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; 1 - x \right) }{ {}_2F_1\left(\frac{1}{m}, 1-\frac{1}{m} ; 1; x \right)} = -\frac{1}{\pi} \sin\left(\frac{\pi}{m} \right) \left( \log x + 2 \left( \gamma + \psi\left(\frac{1}{m} \right) \right) \right) - \cos\left(\frac{\pi}{m} \right) + \mathcal{O}(x)
$$
This gives a large $n$ approximation of the root:
$$
x_n = \exp\left( -\frac{\pi}{\sin\left(\frac{\pi}{m}\right)} \left( \sqrt{n} + \cos\left(\frac{\pi}{m}\right) \right) - 2 \left( \gamma + \psi\left(\frac{1}{m}\right) \right) \right)
$$
where $\gamma$ denotes the Euler-Mascheroni constant. From here, denoting $q_n = \exp\left( -\frac{\pi \sqrt{n} }{\sin\left(\frac{\pi}{m}\right)} \right)$
$$
\lim_{n \to \infty} \frac{x_n}{q_n} = \exp\left( -\frac{\pi}{\tan\left(\frac{\pi}{m}\right)} - 2 \left( \gamma + \psi\left(\frac{1}{m}\right) \right) \right)
$$
The expression for $m=5$ agrees numerically with your expression, implying
$$
\psi\left(\frac{1}{5}\right) = -\gamma - \frac{1}{2} \left( \pi \sqrt{1 + \frac{2}{\sqrt{5}}} + \ln\left( 25 \sqrt{5} \left(\frac{1+\sqrt{5}}{2} \right)^{\sqrt{5}} \right) \right) \tag{1}
$$
I checked this identity using Mathematica, and found it to agree to 50,000 significant decimal points:
In[136]:= N[(-(Log[25*Sqrt[5]*((1 + Sqrt[5])/2)^Sqrt[5]] +
Sqrt[(1/5)*(5 + 2*Sqrt[5])]*Pi))*(1/2) - EulerGamma,
50000] - N[PolyGamma[1/5], 50000]
Out[136]= 0``49998.97559316131
I wonder if this identity is a known one. This identity is the Gauss's digamma theorem for $k=5$ and $m=1$ (thanks Edgar):
$$
\psi\left(\frac{m}{k}\right) = -\gamma -\ln(2k)
-\frac{\pi}{2}\cot\left(\frac{m\pi}{k}\right)
+2\sum_{n=1}^{\lfloor (k-1)/2\rfloor}
\cos\left(\frac{2\pi nm}{k} \right)
\ln\left(\sin\left(\frac{n\pi}{k}\right)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}$ How may I prove that
$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}?$$
I also discussed the problem in the chat, but no solution so far. Some hints? Thanks!
| For now, here is how we can prove the second equality. Let the second sum be $S.$ Note that by symmetry we also have $$S= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2(m^2+n^2)}.$$ Now adding the two forms gives: $$2S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n^2}= \left( \sum_{m=1}^{\infty} \frac{1}{m^2} \right)\left( \sum_{n=1}^{\infty} \frac{1}{n^2}\right)= \frac{\pi^4}{36}.$$
As Fabian alludes to in the comments, it appears the first equality does not hold, since the difference between the two sums is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-n}{n^3}\frac{1}{(m^2+n^2)}>0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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why $\sum_{k=0}^{\infty}(10^{-2})^k = \frac{1}{1-10^{-2}}$ i was reading a book and suddenly saw this step: $\sum_{k=0}^{\infty}(10^{-2})^k = \frac{1}{1-10^{-2}}$
i am actually not bad at calculation and also i am okay in precalculus, but i am really stuck here, not knowing how come this equality is true. i cannot see steps done between them. please dont downvote if the question is too simple. i cannot even give you any of my steps, cos i didnot do any.
| What you have here is a geometric series. The first term, denoted by $a$, is $(10^{-2})^0=1$ while the common ratio, denoted by $r$, is $10^{-2} = 1/100$. We are interested in finding the sum of all of the terms. Let us use $S$ to denote $a + ar + ar^2 + \cdots$ There is a well-known formula for $S$. Let's define:
$$\begin{array}{ccc}
S & = & 1 + 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + \cdots \,
\end{array}$$
A nice little trick to employ if to compare $S$ and $r \times S$. We have:
$$\begin{array}{ccc}
S & = & 1 + 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + \cdots \\
10^{-2} \times S & = & 10^{-2} + 10^{-4} + 10^{-6} + 10^{-8} + 10^{-10} + \cdots
\end{array}$$
The cunning step is to consider $S - 10^{-2}S = (1-10^{-2})S$. This gives:
$$(1-10^{-2})S = 1 \iff S = \frac{1}{1-10^{-2}} \, . $$
Moreover:
$$S = \frac{1}{1-10^{-2}} = \frac{1}{\frac{100}{100}-\frac{1}{100}} = \frac{1}{\frac{99}{100}} = \frac{100}{99} = 1.\overline{01} \, .$$
| {
"language": "en",
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"source": "stackexchange",
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Find the maximum and minimum of $\sum_{i=1}^{n-1}x_ix_{i+1}$ subject to $\sum_{i=1}^nx_i^2=1$. Find the maximum and minimum of
$$
\sum_{i=1}^{n-1}x_ix_{i+1}
$$
subject to
$$
\sum_{i=1}^nx_i^2=1
$$
for all $n\in\mathbb{N}-\{1,0\}$.
| Answer: max is $\displaystyle\cos{\frac{\pi}{n+1}}$, min is $\displaystyle -\cos{\frac{\pi}{n+1}}$.
It can be used Lagrange multiplier method.
Let $\mathbf{x}=(x_1,x_2, \ldots,x_n)$.
We have function $f(\mathbf{x}) = x_1 x_2 + x_2x_3+\cdots + x_{n-1}x_n \rightarrow \max$,
and we have condition $g(\mathbf{x}) = x_1^2 + x_2^2+\cdots+x_n^2-1 = 0$.
We create other function
$$
F(\mathbf{x},\lambda) = f(\mathbf{x}) + \lambda \cdot g(\mathbf{x}).
$$
If $F(\mathbf{x},\lambda)$ has extremum somewhere, then all partial derivatives must be 0 here.
So, we have (n+1) conditions:
$
\left\{
\begin{array}{r}
x_2 + 2\lambda x_1 = 0, \\
x_1 + x_3 + 2\lambda x_2 = 0, \\
x_2 + x_4 + 2\lambda x_3 = 0, \\
\cdots \\
x_{n-2}+x_n+2\lambda x_{n-1} = 0, \\
x_{n-1}+2\lambda x_n = 0; \\
x_1^2 + x_2^2+\cdots+x_n^2-1 = 0.
\end{array}
\right.
$
We can consider $\lambda$ as parameter for first $n$ equations. Then the system of first $n$ linear equations has 3-diagonal matrix
$M_n(\lambda) = \left(
\begin{array}{cccccc}
2\lambda & 1 & 0 & \cdots & 0 & 0 \\
1 & 2\lambda & 1 & \ddots & 0 & 0 \\
0 & 1 & 2\lambda & \ddots & 0 & 0 \\
\vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\
0 & 0 & 0 & \ddots & 2\lambda & 1\\
0 & 0 & 0 & \cdots & 1 & 2\lambda \\
\end{array}
\right).
$
We see that $\det M_1(\lambda) = 2\lambda$, $\det M_2(\lambda) = 4\lambda^2-1$,
$\det M_{n+1}(\lambda) = 2\lambda \det M_n(\lambda) - \det M_{n-1}(\lambda)$.
It is recurrent formula for Chebyshev polynomials of the second kind (by definition).
So, for $\lambda_j = \cos(j\pi/(n+1))$ we have $\det M_n(\lambda_j) = \frac{\sin(j\pi)}{\sin(j\pi/(n+1))}=0$, where $j=1,\ldots,n$.
It is easy to see that for $\lambda = \lambda_j$, $j=1,\ldots,n$, we have
$\mathbf{x}_j = (x_{j1}, \ldots, x_{jn} )$, where $x_{jk} = C_j \sin \frac{(n+1-j)k\pi}{n+1} $, $k=1,\ldots,n$.
Condition $g(\mathbf{x}_j)=0$ implies $C_j = \sqrt{2/(n+1)}$.
$f(\mathbf{x}_j) = \cos \frac{(n+1-j)\pi}{n+1}$.
So,
$\max\limits_{g(\mathbf{x})=0} f(\mathbf{x}) =
\max\limits_{j=1,\ldots,n} \cos\frac{(n+1-j)\pi}{n+1} = \cos \frac{\pi}{n+1}$. (It is attained when $j = n$).
$\min\limits_{g(\mathbf{x})=0} f(\mathbf{x}) =
\min\limits_{j=1,\ldots,n} \cos\frac{(n+1-j)\pi}{n+1} = \cos \frac{n\pi}{n+1} = -\cos \frac{\pi}{n+1}$. (It is attained when $j = 1$).
$\max\limits_{g(\mathbf{x})=0} |f(\mathbf{x})| = \cos \frac{\pi}{n+1}$. (obviously).
$\min\limits_{g(\mathbf{x})=0} |f(\mathbf{x})| =
\min\limits_{j=1,\ldots,n} \left|\cos\frac{(n+1-j)\pi}{n+1} \right| = $
$
\left\{
\begin{array}{cl}
0, & \mbox{when } n \mbox{ is odd}, (\mbox{ when } j=(n+1)/2) \\
\cos \frac{n\pi}{2n+2}, & \mbox{when } n \mbox{ is even}, (\mbox{ when } j=n/2 + 1).
\end{array}
\right.
$
Examples:
$n=2$: $f_{max} = \cos{\frac{\pi}{3}}$,
$$(x_1,x_2) = \left( \sqrt{\frac{2}{3}} \sin{\frac{\pi}{3}}, \sqrt{\frac{2}{3}} \sin{\frac{2\pi}{3}}\right);$$
$n=3$: $f_{max} = \cos{\frac{\pi}{4}}$,
$$(x_1,x_2,x_3) = \left(
\sqrt{\frac{2}{4}} \sin{\frac{\pi}{4}},
\sqrt{\frac{2}{4}} \sin{\frac{2\pi}{4}},
\sqrt{\frac{2}{4}} \sin{\frac{3\pi}{4}}
\right);$$
$n=4$: $f_{max} = \cos{\frac{\pi}{5}}$,
$$(x_1,x_2,x_3,x_4) = \left(
\sqrt{\frac{2}{5}} \sin{\frac{\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{2\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{3\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{4\pi}{5}}
\right);$$
$n=5$: $f_{max} = \cos{\frac{\pi}{6}}$,
$$(x_1,x_2,x_3,x_4,x_5) = \left(
\sqrt{\frac{2}{6}} \sin{\frac{\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{2\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{3\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{4\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{5\pi}{6}}
\right);$$
$\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Show $\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$ Could you help me show that
$$\lim\limits_{n\to\infty} \dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = 1$$
| $$\dfrac{n^2+3^{2n}}{(n^3+3^n)^2} = \frac{1 + \frac{n^2}{3^{2n}}}{\left( 1 + \frac{n^3}{3^{2n}}\right)^2}$$
As exponential grows more faster than polynomial, you have $1/1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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} |
How to compute the following integral? $$\int\frac{\sqrt{1+x^2}}{x}dx$$
I tried letting $x=\tan\theta\ $ where $\frac{-\pi}{2} < \theta < \frac{\pi}{2}$ so that $dx = \sec^2\theta\,d\theta$ and after making the substitution one gets to $$\int\frac{\sec^3\theta}{\tan\theta} d\theta$$ which is equivalent to $$\int\frac{1}{\cos^2\theta\sin\theta}d\theta$$
After this, I don't know how to proceed. I tried looking for the same integral elsewhere and I found a solution that involves a method called partial fraction decomposition, I believe. But, I have not been taught that method yet and this integral appears on the section of the book that I am currently working on.
| Add and subtract $\displaystyle{\frac{1}{x}}$ to the integrand:
$$
\begin{aligned}
\int\frac{\sqrt{1+x^2}-1}{x}\,\mathrm{d}x + \int\frac{\mathrm{d}x}{x}&=\ln\left|x\right| + \int \frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2 +1}+1\right)}{x\left(\sqrt{x^2 +1}+1\right)}\,\mathrm{d}x \\
&= \ln \left|x \right| + \int\frac{x}{\sqrt{x^2 + 1}+1}\,\mathrm{d}x
\end{aligned}
$$
Now, set $u = 1 + \sqrt{x^2 +1}$ and $x\,\mathrm{d}x = \left(u-1\right)\,\mathrm{d}u$ to obtain
$$
\begin{aligned}
\ln\left|x\right| + \int \frac{u-1}{u}\,\mathrm{d}u &= \ln\left|x\right| + u - \ln\left|u\right| + C_{0}\\
&=u + \ln\left|\frac{x}{u}\right| + C_{0} \\
& = \sqrt{1 + x^2} + \ln\left|\frac{x}{1 + \sqrt{1+x^2}}\right| + \underbrace{1 + C_{0}}_{C} \\
&=\sqrt{1+x^2} + \ln\left|\frac{x}{1+\sqrt{1+x^2}}\right| + C
\end{aligned}
$$
And we are done. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding Eigenvectors, is there always a row of $0$'s?
Given the Matrix $$A = \left(\begin{matrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-1 & -3 & -3
\end{matrix}\right)$$
calculate the eigenvalues and the corresponding eigenvectors.
My attempt
$$P(\lambda) = \det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynome with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$
So there is only one eigenvalue $-1$
This involves solving $(A-(-1)I)= A+I = 0$.
Is "the last row" always $0$ when calculating the eigenvectors?
Update
I've got the final answer to be $$E_{-1}=\left(\begin{matrix}1 \\ -1 \\ 1 \end{matrix}\right)$$
| Given:
$$A = \left(\begin{matrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-1 & -3 & -3
\end{matrix}\right)$$
We arrive at:
$$P(\lambda) = det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynomial with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$
Solving for the eigenvectors, we arrive at::
$$\lambda_1 = -1, v_1 = (1, -1, 1)$$
$$\lambda_2 = -1, v_2 = (2, -1, 0)$$
$$\lambda_2 = -1, v_3 = (3,-1, 0)$$
From these eigenvalues and eigenvectors, the Jordan Normal Form can be written as:
$$A = P J P^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 1 & 0 & 0\end{bmatrix} \cdot \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ -1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
Regards
| {
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The value of $(2^n+3^n+4^n)^{1/n}$ as $n \rightarrow \infty?$ I was thinking about the following problem:
How can i find the value of $(2^n+3^n+4^n)^{1/n}$ as $n \rightarrow \infty?$
Can someone point me in the right direction? Thanks in advance for your time.
| Here you have an approach that uses the definition of the constant $e=\lim_{n\to\infty}(1+\frac 1n)^n$.
$$
(2^n+3^n+4^n)^{\frac 1n}
~ = ~~
4
\left(
1 + \frac{2^n+3^n}{4^n}
\right)^{n\frac{2^n+3^n}{4^n}\frac{4^n}{2^n+3^n}}
$$
Since $n\frac{2^n+3^n}{4^n}\to 0$ and $(1 + \frac{2^n+3^n}{4^n})^{\frac{4^n}{2^n+3^n}}\to e$ you have that the above limit is equal to $4e^0=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinations in tournament A tournament includes P total players. The game played in rounds with teams of size T. Possible number of teams is (P T).
Questions
*
*How would you calculate the number of total possible combinations of matches?
*How would you calculate the number of total possible matches per round?
Examples
Example 1: 4 players in teams of 2
4 Players
1-2
1-3
1-4
2-3
2-4
3-4
=(4 2) = 6
1-2 - 3-4
1-3 - 2-4
1-4 - 2-3
= 3
Example2: 5 players in teams of 2
Teams
1-2
1-3
1-4
1-5
2-3
2-4
2-5
3-4
3-5
4-5
=(5 2) = 10
Matches
1-2 - 3-4
1-2 - 3-5
1-2 - 4-5
1-3 - 2-4
1-3 - 2-5
1-3 - 4-5
1-4 - 2-3
1-4 - 2-5
1-4 - 3-5
1-5 - 2-3
1-5 - 2-4
1-5 - 3-4
2-3 - 4-5
2-4 - 3-5
2-5 - 3-4
= 15
| you have to chose the first and the second teams. for the first team you have $\binom{P}T$ and for the second team you have $\binom{P-T}T$ however you are counting each match twice in this case so the total number of possible matches is $$\frac{\binom{P}T\binom{P-T}T}{2}$$
The number of possible combinations of matches per round (assuming you get as many teams to play as possible in each round) is $$\prod_{n=0}^{n=2\lfloor{T/2P}\rfloor-1}\binom{P-(Tn)}T/(\lfloor{T/2P}\rfloor!(2^{\lfloor{T/2P}\rfloor})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can the sum of finitely many inverses of distinct odd integers $\geq 3$ be 1? Is there a positive number $n$ of distinct odd integers $z_1,z_2, \ldots, z_n \geq 3$ such that $\frac{1}{z_1} + \frac{1}{z_2} + \cdots + \frac{1}{z_n} = 1$?
| In 1954, it was shown by Stewart and Breusch (independently) that if $\frac {p}{q} >0$ and $q$ is odd, then it can be written as the sum of finitely many reciprocals of odd numbers.
As a specific example,
$$1=\frac {1}{3} + \frac {1}{5} + \frac {1}{7} + \frac {1}{9} + \frac {1}{15} + \frac {1}{21} + \frac {1}{27} + \frac {1}{35} + \frac {1}{63} + \frac {1}{105} + \frac {1}{135}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Simplifying the sum of terms that are "polynomial fractions" I need to simplify:
$$\left(\frac{z^2-2}{1+2z}\right)^2 \;+ \;\left(\frac{z^2-2}{1+2z}\right)$$
I get $2z^4-8z^2+8$ in the numerator when I know it should be $((z^2-2) (z^2+2 z-1))$.
| $$\left(\frac{z^2-2}{1+2z}\right)^2+\left(\frac{z^2-2}{1+2z}\right)\;\;=\;\;\frac{(z^2-2)^2}{(1+2z)^2}+\frac{(z^2-2)}{(1+2z)}$$
$$=\frac{(z^2-2)^2+(z^2-2)(1+2z)}{(1+2z)^2}$$
$$=\frac{(z^2-2)\;[(z^2-2)+(1+2z)]}{(1+2z)^2}$$
$$=\frac{(z^2-2)(z^2+2z-1)}{(1+2z)^2}$$
Note that in general, $$\dfrac{m^2}{n^2} + \dfrac mn = \dfrac {m^2 + mn}{n^2} = \dfrac{m(m+n)}{n^2}.\;$$
| {
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Eulers Approximation Inductive Proof Alright so I am having some difficulty with an inductive proof. I am attempting to prove the following:
Given that the Euler Method is described by the given recursive formula:
$y_{n} = y_{n-1} + h f(x_{n-1}, y_{n-1}), n = 1, 2, 3, ….$
Show that the Euler Approximation $y_{n}$ can be represented by the following formula:
$y_{n} = ( 1 + \frac{1}{n})^{n} , n = 1, 2, 3, ….$
Given that:
$y' = y$ , $y(0) = 1$, and $h = \frac{1}{n}$
Proposed Solution:
Base Case
(first re-write the recursive definition as $y_{n} = y_{n-1} (1 + \frac{1}{n})$ to increase clarity)
Let n = 1
$y_{0} (1 + \frac{1}{1}) = (1 + \frac{1}{1})^{1}$
$1 \cdot (1 + \frac{1}{1}) = (1 + \frac{1}{1})$
2 = 2
So the base case checks out.
Inductive Hypothesis
Now assume the following:
$y_{k – 1} (1 + \frac{1}{k}) \Rightarrow (1 + \frac{1}{k})^k \space \space \forall k \in Z^{+}$
Show that:
$y_{k} (1 + \frac{1}{1+k}) \Rightarrow (1 + \frac{1}{k+1}) ^{k+1}$
And here is my problem because if I replace $y_{k}$ with $(1 + \frac{1}{k})^k$ I get the following:
$(1 + \frac{1}{k})^k \cdot (1 + \frac{1}{k+1}) ^{k+1} \nRightarrow (1 + \frac{1}{k+1}) ^{k+1}$
Can someone spot what I’m doing wrong?
| The error is in your first line. Induction is on $k$, not on $n$, which remains fixed. The first line should be:
Base Case. (first re-write the recursive definition as $y_{k} = y_{k-1} (1 + \frac{1}{n})$ to increase clarity)
| {
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Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \ldots$
Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \dotsb$
Trial: Let
$$\begin{align}
P &= 3 \cdot 9^{\frac{1}{2}} \cdot 27^{\frac{1}{4}} \cdot 81^{\frac{1}{8}} \dotsb\\
\implies \ln P &=\ln3+\frac{1}{2} \ln 3^2+\frac{1}{2^2} \ln 3^3+\frac{1}{2^3} \ln 3^4 + \dotsb\\
&=\ln 3 \sum_{x=0}^{\infty}\frac{x+1}{2^x}
\end{align}$$
Then how I proceed. Is there any other simpler way to solve. Please help.
| I think you are correct so far. Use the facts that
$$\sum_{k=0}^{\infty} r^k = \frac{1}{1-r} $$
and
$$\sum_{k=0}^{\infty} k r^k = \frac{r}{(1-r)^2} $$
So you will have, with $r=1/2$:
$$\ln{P} = (2 + 2) \ln{3} = 4 \ln{3} \implies P = 81 $$
| {
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How to find an analytic solution to $\lim_{x \to +\infty} \frac{x+\sin(x^2)}{\sqrt{x^2+1}}$ To solve this limit:
$$\lim_{x \to +\infty} \space \frac{x+\sin(x^2)}{\sqrt{x^2+1}}$$
At the beginning I didn't know how to start. Then I thought, no matter the value that $x$ takes, $sin(x^2)$ will always be between $-1$ and $1$. So for large values of $x$, $sin(x^2)$ is insignificant. One can rewrite:
$$\lim_{x \to +\infty} \space \frac{x}{\sqrt{x^2+1}}$$
And now it's easy to find the limit:
$$\lim_{x \to +\infty} \space \frac{x}{\sqrt{x^2+1}} = \lim_{x \to +\infty} \space \frac{\sqrt{x^2}}{\sqrt{x^2+1}} = \lim_{x \to +\infty} \space \sqrt{\frac{x^2}{x^2+1}} = \lim_{x \to +\infty} \space \sqrt{\frac{x^2}{x^2}}=1$$
But I know that the justification that allowed me to find the limit this way is not an analytic justification. How can I find this limit on an analytic basis?
Thanks
| You know that $\lim \limits_{x \to +\infty} \left(\dfrac{x+\sin(x^2)}{\sqrt{x^2+1}}\right)= \lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\dfrac{\sin(x^2)}{\sqrt{x^2+1}}\right)$.
Now if both $\lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}\right)$ and $\lim \limits_{x\to +\infty}\left(\dfrac{\sin(x^2)}{\sqrt{x^2+1}}\right)$ exist, then $$ \lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=\lim \limits_{x\to +\infty}\left(\frac{x}{\sqrt{x^2+1}}\right )+ \lim \limits_{x\to +\infty}\left(\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)$$
You've proved that the first one on the RHS exists and equals $1$.
To show that the second one exists and equals $0$ it suffices to prove that $\lim \limits_{x\to +\infty}\left(\left | \dfrac{\sin(x^2)}{\sqrt{x^2+1}} \right |\right)=0$ and that's a consequence of the squeeze theorem because for all $x\in \mathbb{R},$ $$0\leq \left | \dfrac{\sin(x^2)}{\sqrt{x^2+1}} \right |\leq \left | \frac{1}{\sqrt{x^2+1}} \right |$$ and $\lim \limits_{x\to +\infty}\left(\left | \frac{1}{\sqrt{x^2+1}} \right |\right)=0$.
Therefore $$ \lim \limits_{x\to +\infty}\left(\dfrac{x}{\sqrt{x^2+1}}+\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=\lim\limits_{x\to +\infty}\left(\frac{x}{\sqrt{x^2+1}}\right )+ \lim\limits_{x\to +\infty}\left(\frac{\sin(x^2)}{\sqrt{x^2+1}}\right)=1+0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/282918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ This question is from [Number Theory George E. Andrews 1-1 #3].
Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$
This problem is driving me crazy.
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$
$(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out
This is a similar problem in the book and I tried this method but it wasn't working out
$\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$
$\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$
$\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$. $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$
\eqalign{
\sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\
&= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\
&= \dfrac{x^{k+1}-1}{x-1}.
}$$
Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.
$\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m>1$, then $n<m^n.$
| $$\frac{1-(x/y)^n}{1-x/y}=1+x/y+(x/y)^2+...+(x/y)^{n-1}$$
$$\frac{(y^n-x^n)/y^n}{(y-x)/y}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$
$$\frac{y^n-x^n}{(y-x)y^{n-1}}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$
$$y^n-x^n=(y-x)(y^{n-1}+xy^{n-2}+...+x^{n-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
proof of limit for $x \to +\infty = +\infty$ My exercise book proves the limit: $lim_{x \to +\infty} \frac{x^2 - 2}{2x + 1} = +\infty$ in this way: $\lor K > 0, \exists \bar{x} = \bar{x}_k$ : if $x > \bar{x} \Rightarrow f(x) > K$
Then, the author solved the inequality $\frac{x^2 - 2}{2x + 1} > K$
Which is valid for $x > \sqrt{K^2 + K + 2} \lor x < K - \sqrt{K^2 + K + 2} \lor x > -\frac{1}{2}$
But the author writes the solution this way:
$K - \sqrt{K^2 + K + 2} < x < -\frac{1}{2} \lor x > \sqrt{K^2 + K + 2}$
How can he say that $K - \sqrt{K^2 + K + 2} < -\frac{1}{2}$
What assumptions do he make?
Also, the exercise ends with $\bar{x} = K + \sqrt{K^2 + K + 2}$
why $K + \sqrt{K^2 + K + 2}$ and not $K - \sqrt{K^2 + K + 2}$?
Thank you.
| First we find a simpler argument. Note that $x^2-2=\frac{x^2}{2}+\frac{x^2}{2}-2\ge \frac{x^2}{2}$ if $x\ge 2$.
Note also that if $x\ge 1$ then $2x+1\le 3x$. It follows that if $x\ge 2$, then
$$\frac{x^2-2}{2x+1} \ge \frac{x^2/2}{3x}=\frac{x}{6}.$$
Thus if $x\gt \max(2, 6K)$, then $\frac{x^2-2}{2x+1}\gt K$.
The point is that we need not try to find the cheapest $\bar{x}$ such that beyond $\bar{x}$ our function is $\gt K$. All we need to do is to show that there exists such a $\bar{x}$.
Now to the actual question. Suppose that we want to solve the inequality $\frac{x^2-2}{2x+1}\lt K$. Note that in this context that is not a good thing to do.
There are various approaches. The one that seems to have been taken is to separate into cases, $x\gt -1/2$ and $x\lt -1/2$.
The case $x\lt -1/2$ is obviously irrelevant. But if we insist on solving in that case, we can rewrite the inequality as $x^2-2\gt K(2x+1)$ (we are multiplying both sides by the negative quantity $2x+1$).
In the case $x\gt -1/2$, we can rewrite our inequality as $x^2-2\lt K(2x+1)$. The roots of the equation $x^2-2Kx-K-2=0$ are $x=K\pm\sqrt{K^2+K+2}$.
You ask why $K-\sqrt{K^2+K+2} \lt -1/2$. That question does not need to be answered to solve our problem. But to answer it, note that $K^2+K+2=(K+1/2)^2 +7/4$, so $\sqrt{K^2+K+2} \gt K+1/2$. Thus $K-\sqrt{K^2+K+2}\lt K-(K+1/2)=-1/2$.
But we are interested in finding a place such that beyond it our inequality holds. That place is the larger root $K+\sqrt{K^2+K+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Matrix from eigenvalues $A=\begin{bmatrix}
a & b\\
b & c
\end{bmatrix}$
Find any $(a,b,c) \in \mathbb{C}^3$ , $a \neq 0$ , $b \neq 0$ , $c \neq 0$ , so that eigenvalues of A are $\lambda_1=\lambda_2=1$
$det(A-\lambda I)=\begin{bmatrix}
a-\lambda & b\\
b & c- \lambda
\end{bmatrix}=(a-\lambda)(c-\lambda)-b^2=0$
$\Longrightarrow$
$\lambda^2-\lambda(a+c)+ac-b^2=0$
$\lambda=\frac{a+c \pm \sqrt{a^2+4b^2-2ac+c^2}}{2}$
$\Longrightarrow$
$a^2+4b^2-2ac+c^2=0$ and $\frac{a+c}{2}=1$
(discriminant is 0 because I want a double root, and (a+c)/2=1 because I want it to be 1)
but I can't find any solution, is there something wrong?
$\Longrightarrow$
$c=2-a$
and
$4a^2-8a+4b^2+4=0$
$a= 1 \pm b \cdot i $
Whatever value I give to a, it won't work...
| Your work looks OK up to $4a^2-8a+4b^2+4=0$.
This reduces to $a^2-2a+b^2+1=0$, and you can try a few values for $a$ and $b$. I tried $a=3$, which yielded $c=-1$ and $b=2i$. The resulting matrix has the eigenvalues you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/285329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating an integral without using the Digamma function or complex analysis. Suppose that $n$ is a positive integer, and let $k$ be a positive integer less than $n$.
How can one show that $\displaystyle\int_0^1 \frac{x^{k-1} - x^{n-k-1}}{1 - x^n} \; dx= \frac{\pi}{n} \cot{\frac{\pi k}{n}}$ without using the digamma (psi) function or complex analysis? I am pretty sure that I have seen such an approach (using basic calculus) at one point, but I can not quite recall how the argument worked. Thanks!
| Begin by Taylor expanding the denominator:
$$\int_0^1 dx \; \frac{x^{k-1} - x^{n-k-1}}{1 - x^n} = \int_0^1 dx \; (x^{k-1} - x^{n-k-1}) \sum_{m=0}^{\infty} x^{m n} $$
Reverse the order of the sum and the integral:
$$ \begin{align} &= \sum_{m=0}^{\infty} \int_0^1 dx \; (x^{m n + k-1} - x^{(m+1) n-k-1}) \\ &= \sum_{m=0}^{\infty} \left ( \frac{1}{m n + k} - \frac{1}{(m+1) n - k} \right ) \\ &= \frac{1}{n} \sum_{m=0}^{\infty} \left ( \frac{1}{m + k/n} - \frac{1}{m+1 - k/n} \right ) \\&= \frac{1}{n} \sum_{m=-\infty}^{\infty} \frac{1}{m + k/n} \\ \end{align}$$
Write $k/n = z/\pi$ and rewrite the sum below:
$$ \begin{align} &= \frac{\pi}{n} \left ( \frac{1}{z} + 2 z \sum_{m=1}^{\infty} \frac{1}{z^2-\pi^2 m^2} \right ) \\ \end{align} $$
The expression in parenthesis is a well-known representation of $\cot(z)$ . This result was derived by Euler, and may be derived by considering Euler's product representation of $\sin{\pi z}/(\pi z)$ and differentiating $\log{\sin{\pi z}}$.
Using the definition of $z$, we find that
$$\int_0^1 dx \; \frac{x^{k-1} - x^{n-k-1}}{1 - x^n} = \frac{\pi}{n} \cot{\frac{\pi k}{n}} $$
EDIT
Simple proof that
$$\cot{z} = \frac{1}{z} + 2 z \sum_{m=1}^{\infty} \frac{1}{z^2-\pi^2 m^2} $$
Consider the relation
$$\cot{y} = \frac{1}{2} (\cot{\frac{y}{2}} - \tan{\frac{y}{2}}) $$
We may rewrite this as
$$\pi x \cot{\pi x} = \frac{\pi x}{2} \left (\cot{\frac{\pi x}{2}} + \cot{\frac{\pi (x \pm 1)}{2}} \right ) $$
Note that this forms a sort of recurrence. That is, we can apply the above relation to each term inside the parenthesis. We use the minus sign for the first term, and the plus sign for the second term. The result of a first pass is
$$\pi x \cot{\pi x} = \frac{\pi x}{2^2} \left (\cot{\frac{\pi x}{2^2}} + \cot{\frac{\pi (x - 1)}{2^2}} + \cot{\frac{\pi (x + 1)}{2^2}} + \cot{\frac{\pi (x + 2)}{2^2}} \right ) $$
We can repeat this as much as we like; after $r$ rounds of this, we get
$$\pi x \cot{\pi x} = \frac{\pi x}{2^r} \left (\cot{\frac{\pi x}{2^r}} + \sum_{m=1}^{2^{r-1}-1} \left [\cot{\frac{\pi (x - m)}{2^r}} + \cot{\frac{\pi (x + m)}{2^r}} \right ] + \cot{\frac{\pi (x + 2^{r-1})}{2^r}} \right ) $$
Now take the limit as $r \rightarrow \infty$. Use the facts that
$$\lim_{r \rightarrow \infty} \frac{\pi x}{2^r} \cot{\frac{\pi x}{2^r}} = 1 $$
$$\lim_{r \rightarrow \infty} \frac{1}{2^r} \cot{\frac{a}{2^r}} = \frac{1}{a} $$
$$\lim_{r \rightarrow \infty} \frac{1}{2^r} \cot{\frac{\pi(x+2^{r-1})}{2^r}} = 0$$
$\forall \, a \ne 0$. and we get the expansion
$$\pi x \cot{\pi x} = 1 + x \sum_{m=1}^{\infty} \left ( \frac{1}{x+m} + \frac{1}{x-m} \right ) $$
from which the cited result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| $\sin x-1$ is always non-positive.
$\sin x - 2$ is always negative.
$\sin x -3$ is always negative.
$\sin x + 1$ is always non-negative.
Therefore the whole expression reduces to
$$
-(\sin x-1) -(\sin x-2) -(\sin x-3) + (\sin x+1).
$$
(But if you meant $\sin(x-1)$ where you wrote $\sin x-1$, then none of the above will help you.)
You get $7-2\sin x$. Since the sine oscillates between $1$ and $-1$, the smallest this function ever gets is $7-2=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Limit problem involving cube roots (without use of L'Hôpital's rule or Taylor series) How to find this limit without the help of L'Hôpital's rule nor expansion to Taylor series?
Limit:
$$\lim_{x\to -8}\frac{ (9+ x)^{1/3}+x+7}{(15+2 x)^{1/3}+1} $$
| Hint: As
$$\sqrt[3]a+\sqrt[3]b=\frac{a-b}{\sqrt[3]a^2-\sqrt[3]ab+\sqrt[3]b^2}$$
we have
$$\lim_{x\to -8}\frac{\sqrt[3]{9+x}+x+7}{\sqrt[3]{15+2x}+1}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{\sqrt[3]{9+x}^2-\sqrt[3]{9+x}(x+7)+(x+7)^2}\frac{\sqrt[3]{15+2x}^2-\sqrt[3]{15+2x}+1}{15+2x+1^3} =
\lim_{x\to -8}\frac{9+x+(x+7)^3}{3}\frac{3}{16+2x}=\lim_{x\to -8}\frac{9+x+(x+7)^3}{16+2x}
$$
Things should be straighforward from here on.
EDIT: The question was changed to $x\to -\infty$. Again everything reduces to computing
$$\lim_{x\to -\infty}\frac{9+x+(x+7)^3}{16+2x}=\lim_{x\to -\infty}x^2\frac{\frac9{x^3}+\frac1{x^2}+(1+\frac7{x})^3}{\frac{16}x+2}
$$
which is also simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\int_{0}^{\infty }\frac{x^{a-3/2}dx}{[ x^2+( b^2-2)x+1]^a}=b^{1-2a}\frac{\Gamma(1/2)\Gamma(a-1/2)}{\Gamma(a)}$ How can one prove that
$$I\left( a,b \right)=
\int_{0}^{\infty }\frac{x^{a-\frac{3}{2}}dx}{\left[ x^2+\left( b^2-2 \right)x+1 \right]^a}=b^{1-2a}\frac{\Gamma \left( \frac{1}{2} \right)\Gamma \left( a-\frac{1}{2} \right)}{\Gamma \left( a \right)},\ $$
where $a>\frac12,\ b\in \mathbb{R}^+$?
| Observation 1:
The change of variables $x\leftrightarrow x^{-1}$ yields the identity
$$\int_0^1\frac{x^{a-\frac32}dx}{\left[x^2+(b^2-2)x+1\right]^a}
=\int_1^{\infty}\frac{x^{a-\frac12}dx}{\left[x^2+(b^2-2)x+1\right]^a},$$
which in turn implies that \begin{align}
I(a,b)=\int_0^{\infty}\frac{x^{a-\frac32}dx}{\left[x^2+(b^2-2)x+1\right]^a}
&=\int_1^{\infty}\frac{2x^{a}\cdot\frac{1}{2}\left(x^{-\frac12}+x^{-\frac32}\right)dx}{\left[x^2+(b^2-2)x+1\right]^a}\tag{1} \end{align}
Observation 2:
We have
$$x^2+(b^2-2)x+1=(x-1)^2+b^2 x=x\left[\left(x^{\frac12}-x^{-\frac12}\right)^2+b^2\right].\tag{2}$$
Observation 3:
We also have
$$d\left(x^{\frac12}-x^{-\frac12}\right)=\frac{1}{2}\left(x^{-\frac12}+x^{-\frac32}\right)dx.\tag{3}$$
Now, making in (1) the change of variables $s=x^{\frac12}-x^{-\frac12}$ and using (2) and (3), one finds that
$$I(a,b)=2\int_0^{\infty}\frac{ds}{\left(s^2+b^2\right)^a}=2b^{1-2a}\int_0^{\infty}\frac{dt}{\left(t^2+1\right)^a}=b^{1-2a}\int_0^{\infty}\frac{u^{-\frac12}du}{\left(u+1\right)^a}.$$
The last integral transforms into the standard beta function integral by the change of variables $v=\frac{u}{u+1}$. It gives
\begin{align}
I(a,b)=b^{1-2a}\int_0^1v^{-\frac12}(1-v)^{a-\frac32}dv=b^{1-2a}B\left(\frac12,a-\frac12\right)=b^{1-2a}\sqrt{\pi}\frac{\Gamma\left(a-\frac12\right)}{\Gamma(a)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
find z such that $\vec{u}$ and $\vec{v}$ are linearly independent The problem is given as, "Determine the values of z such that the vectors $\vec{u} = \pmatrix{-1\\z}$ and $\vec{u} = \pmatrix{z\\-1 + z}$ are linearly independent.
Here is my work...
$\pmatrix{-1 & z& \\ z& (-1 + z)& }$
$\pmatrix{-1 & z& \\ 0& (z^2+z-1)& }$
therefore, $z^2+z-1 \neq 0$
$z \neq \frac{1}{2}(-1-\sqrt{5})$
$z \neq \frac{1}{2}(\sqrt{5} + 1)$
Does that seem right?
| $$
z^2 + z - 1 \ne 0 \Rightarrow \\
z^2 + z +\frac{1}{4} \ne \frac{5}{4} \Rightarrow \\
\left(z+\frac{1}{2}\right)^2 \ne \frac{5}{4} \Rightarrow\\
z \ne -\frac{1}{2} + \frac{\sqrt{5}}{2}, z \ne -\frac{1}{2} - \frac{\sqrt{5}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Trigonometric Inequality. $\sin{1}+\sin{2}+\ldots+\sin{n} <2$ . How can I prove the following trigonometric inequality :
$$\sin1+\sin2 +\ldots+\sin n <2$$ with $n \in \mathbb{N}^{*}$.
The problem is that I don't know how to start this problem, I try to use some formul but nothing. I'll appreciate your support.
I try to solve this inequality without series, or information about analysis mathematics.
Thanks :)
| Overall Strategy
*
*Using Euler’s Formula $ \forall \theta \in \mathbb{R}: ~ e^{i \theta} = \cos(\theta) + i \sin(\theta) $, observe that
$$
\forall \theta \in \mathbb{R}, ~ \forall n \in \mathbb{N}: \quad \sum_{k=1}^{n} e^{ik \theta} = \sum_{k=1}^{n} \cos(k \theta) + i \sum_{k=1}^{n} \sin(k \theta).
$$
*Notice that the left-hand side of this equation is a finite geometric series.
*Hence, you can obtain a closed-form expression for the left-hand side.
*Taking the complex part of this expression and letting $ \theta = 1 $, you get a closed-form expression for your sum.
*Finally, apply basic trigonometric knowledge to show that the sum is strictly bounded above by $ 2 $.
Addendum
This addendum serves to demonstrate that the required closed-form expression for $ \displaystyle \sum_{k=1}^{n} \sin(k) $ may be derived, without much difficulty, from Euler’s Formula.
For $ \theta \notin 2 \pi \mathbb{Z} $, observe that
\begin{align}
\forall n \in \mathbb{N}: \quad
\sum_{k=1}^{n} e^{ik \theta}
&= \frac{e^{i \theta} (1 - e^{in \theta})}{1 - e^{i \theta}} \\
&= \frac{e^{i \theta} (1 - e^{in \theta})}{1 - e^{i \theta}} \cdot
\frac{e^{-i \theta/2}}{e^{-i \theta/2}} \\
&= \frac{e^{i \theta/2} (1 - e^{in \theta})}{e^{-i \theta/2} - e^{i \theta/2}} \\
&= \frac{e^{i \theta/2} - e^{i[n + (1/2)] \theta}}{e^{-i \theta/2} - e^{i \theta/2}} \\
&= \frac{\left[ \cos \left( \frac{1}{2} \theta \right) + i \sin \left( \frac{1}{2} \theta \right) \right] - \left[ \cos \left( \left( n + \frac{1}{2} \right) \theta \right) + i \sin \left( \left( n + \frac{1}{2} \right) \theta \right) \right]}{-2i \sin \left( \frac{1}{2} \theta \right)} \\
&= \left[ \frac{\sin \left( \left( n + \frac{1}{2} \right) \theta \right) - \sin \left( \frac{1}{2} \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} \right] + i \left[ \frac{\cos \left( \frac{1}{2} \theta \right) - \cos \left( \left( n + \frac{1}{2} \right) \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} \right].
\end{align}
We have thus killed two birds with one stone:
\begin{equation}
\sum_{k=1}^{n} \cos(k \theta) = \left\{
\begin{array}{ll}
\frac{\sin \left( \left( n + \frac{1}{2} \right) \theta \right) - \sin \left( \frac{1}{2} \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} & \text{if $ \theta \notin 2 \pi \mathbb{Z} $}; \\
n & \text{if $ \theta \in 2 \pi \mathbb{Z} $}.
\end{array} \right.
\end{equation}
\begin{equation}
\sum_{k=1}^{n} \sin(k \theta) = \left\{
\begin{array}{ll}
\frac{\cos \left( \frac{1}{2} \theta \right) - \cos \left( \left( n + \frac{1}{2} \right) \theta \right)}{2 \sin \left( \frac{1}{2} \theta \right)} & \text{if $ \theta \notin 2 \pi \mathbb{Z} $}; \\
0 & \text{if $ \theta \in 2 \pi \mathbb{Z} $}.
\end{array} \right.
\end{equation}
Letting $ \theta = 1 $, we obtain
$$
\sum_{k=1}^{n} \sin(k) = \frac{\cos \left( \frac{1}{2} \right) - \cos \left( n + \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$
Now, define a function $ f: \mathbb{R} \to \mathbb{R} $ by
$$
\forall x \in \mathbb{R}: \quad f(x) \stackrel{\text{def}}{=} \frac{\cos \left( \frac{1}{2} \right) - \cos \left( x + \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$
As $ \text{Range}(\cos) = [-1,1] $, it follows that
\begin{align}
\text{Range}(f) &= \left[ \frac{\cos \left( \frac{1}{2} \right) - 1}{2 \sin \left( \frac{1}{2} \right)},\frac{\cos \left( \frac{1}{2} \right) + 1}{2 \sin \left( \frac{1}{2} \right)} \right] \\
&= [-0.12767096 \ldots,1.95815868 \ldots] \\
&\subseteq [-2,2].
\end{align}
Define also a function $ g: \mathbb{R} \to \mathbb{R} $ by
$$
\forall x \in \mathbb{R}: \quad g(x) \stackrel{\text{def}}{=} \frac{\sin \left( x + \frac{1}{2} \right) - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)}.
$$
As $ \text{Range}(\sin) = [-1,1] $, it follows that
\begin{align}
\text{Range}(g) &= \left[ \frac{-1 - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)},\frac{1 - \sin \left( \frac{1}{2} \right)}{2 \sin \left( \frac{1}{2} \right)} \right] \\
&= [-1.54291482 \ldots,0.54291482 \ldots] \\
&\subseteq [-2,2].
\end{align}
Conclusion: $ \displaystyle \left| \sum_{k=1}^{n} \sin(k) \right| < 2 $ and $ \displaystyle \left| \sum_{k=1}^{n} \cos(k) \right| < 2 $ for all $ n \in \mathbb{N} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 1
} |
Induction on binomial Identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ I am having trouble proving the following identity:
$0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$
Here is what I have so far:
Proof:
Base: Let $n=0$:
LHS: $0\cdot {0\choose 0} = 0\cdot1 = 0$
RHS: $0\cdot 2^{0-2} = 0$
Step: Let $k\in \mathbb{Z} s.t k \geq 0$ and assume the identity is true for k.
Consider the LHS for $k+1$ where $k$ is even (I leave out the odd case because I think it will turn out the same?):
\begin{align}
=& 0\cdot{k+1\choose 0}+2\cdot{k+1\choose 2}+4\cdot{k+1\choose 4}+... +k\cdot{k+1\choose k}
\\=&0\cdot\left[{k\choose 0}+{k\choose -1}\right] + 2\cdot\left[{k\choose 2}+{k\choose 1}\right]+ 4\cdot\left[{k\choose 4}+{k\choose 3}\right]+\ldots+ k\cdot\left[{k\choose k}+{k\choose k-1}\right]
\\=&\left[0\cdot{k\choose 0}+2\cdot {k\choose 2}+4\cdot{k\choose 4}+\ldots+k\cdot{k\choose k}\right] + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\=& k\cdot2^{k-2} + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\
\end{align}
I know I need to end up with something like:
\begin{align}
=&k\cdot2^{k-2}+ \left[k\cdot2^{k-2} + 2^{k-1}\right]
\\=&2k\cdot 2^{k-2} + 2^{k-1}
\\=&k\cdot 2^{k-1}+2^{k-1}
\\=&(k+1)\cdot 2^{k-1}\end{align}
But, how can I get what I need from the combinations above? It may not end up exactly like that, but what is the reasoning behind this?
| A non-inductive proof now. We have:
$$
\sum_{0 \le k \le n} \binom{n}{k} z^k = (1 + z)^n
$$
So we also have:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} \binom{n}{2 k} z^{2 k} = \frac{(1 + z)^n + (1 - z)^n}{2}
$$
If you differentiate this with respect to $z$ you get:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} 2 k \binom{n}{2 k} z^{2 k - 1}
= \frac{n (1 + z)^{n - 1} - n (1 - z)^{n - 1}}{2}
$$
Then evaluate at $z = 1$ you get the requested sum:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} 2 k \binom{n}{2 k}
= \frac{n 2^{n - 1}}{2} = n 2^{n - 2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Converting $x^2 + 6y - 9 = 0$ to polar. So far I got here
\begin{align}
(r\cos\phi)^2 & + 6 r \sin\phi- 9 = 0\\
(r\cos\phi)^2 & = 9 - 6r \sin\phi
\end{align}
| $$x^2+6y=9$$
As always we choose $x=r \cos \theta$ and $y=r \sin \theta$
Thus $$(r \cos \theta)^2 +6r \sin \theta -9=0$$
With the identity $\cos^2 \theta = 1- \sin^2 \theta$ we find
$$r^2 - (r \sin \theta)^2 + 6r \sin \theta -9=0$$
Which equals $$(r \sin \theta -3)^2=r^2$$
Thus $r=r \sin \theta -3$ or $r= 3-r \sin \theta$ and subsequently we see $$r=\frac{3}{\sin \theta - 1} \lor r=\frac{3}{1+ \sin \theta} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Complex sets: factoring into circle How must $|z|=3|z-1|$ be factored so I end up with a circle, plugging in $z=x+iy$ seems to just up with square roots everywhere. Detailed steps is much appreciated, thanks!
| \begin{align*}
z \overline{z} = 9 (z - 1)(\overline{z}-1) = 9 z \overline{z} - 9 z - 9 \overline{z} + 9 \\
8 z \overline{z} - 9 z - 9 \overline{z} + 9 = 0
\end{align*}
Now a circle is supposed to look like $$(z - a)(\overline{z} - \overline{a}) = r^2= z \overline{z} - a \overline{z} - \overline{a} z + |a|^2$$
So taking our old relation and diving by 8, we get
\begin{align*}
z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{9}{8} = 0 \\
z \overline{z} - \frac{9}{8} z - \frac{9}{8} \overline{z} + \frac{81}{64} = \frac{81}{64} - \frac{9}{8} = \frac{9}{64}
\end{align*}
So with some algebra we have a circle centered at $\frac{9}{8}$ with radius $\frac{3}{8}$.
For fun: you can view the locus of solutions $S$ as the inverse image of the unit circle of the fractional linear transformation $z \mapsto \frac{z}{3z-3}$. Conjugate points are sent to conjugate points and the center of $S$'s conjugate is $\infty$. Infinity maps to $1/3$ and the conjugate about the unit circle is $3$, so the inverse of that is $8/9$. Similarly I can find what is sent to 1 and that'll give me a radius.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/296661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$ I'm tutoring for a college math class and we're doing putnam problems next week and this one stumped me:
Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$.
| Let $\sin{x}+\cos{x}=\sqrt{2}\sin{(x+\frac{\pi}{4})}=a$, then $a$ can take any value between $-\sqrt{2}$ and $\sqrt{2}$. We have $\sin{x}\cos{x}=\frac{a^2-1}{2}$. Thus
\begin{align}
\left||\sin{x}+\cos{x}+\tan{x}+\cot{x}+\sec{x}+\csc{x} \right|& =\left|\sin{x}+\cos{x}+\frac{1}{\sin{x}\cos{x}}+\frac{\sin{x}+\cos{x}}{\sin{x}\cos{x}} \right| \\
& =\left|a+\frac{2+2a}{a^2-1} \right| \\
& =\left|a+\frac{2}{a-1} \right|
\end{align}
If $a<1$, let $b=1-a>0$, then $a+\frac{2}{a-1}=1-(b+\frac{2}{b}) \leq 1-2\sqrt{2}<0$, so $\left|a+\frac{2}{a-1} \right| \geq 2\sqrt{2}-1$.
($b+\frac{2}{b} \geq 2\sqrt{2}$ by AM-GM inequality or square $\geq 0$)
Equality holds when $b=\sqrt{2}, a=1-\sqrt{2}, x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.
If $1 \leq a \leq \sqrt{2}$, then $a+\frac{2}{a-1}>\frac{2}{a-1}>2$, so $\left|a+\frac{2}{a-1} \right|>2>2\sqrt{2}-1$.
Thus the minimum value is $2\sqrt{2}-1$, achieved when $x=\sin^{-1}{(\frac{1-\sqrt{2}}{\sqrt{2}})}-\frac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Different solutions of trigonometric equations Please take a look at this trigonometric equation,
$\cos9x\cos7x = \cos5x\cos3x$
To solve this equation, we can proceed as,
$2\cos9x\cos7x = 2\cos5x\cos3x$
or, $\cos(9x+7x)+\cos(9x-7x) = \cos(5x+3x)+\cos(5x-3x)$
or, $\cos16x+\cos2x = \cos8x+\cos2x$
or, $\cos16x = \cos8x$
From the above situation we can proceed in two ways,
First Way
$\cos16x = \cos8x$
or, $\cos(2\times8x) = \cos8x$
or, $2\cos^28x -1 = \cos8x$
or, $2\cos^28x-\cos8x -1 = 0$
or, $2\cos^28x-2\cos8x+\cos8x -1 = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $2\cos8x(\cos8x-1)+(\cos8x -1) = 0$
or, $(\cos8x -1)(2\cos8x+1) = 0$
Either or both of the above factors are zero.
Taking the first one,
$(\cos8x -1) = 0$
or, $\cos8x = 1$
or, $8x = 2n\pi$, where $n$ is an integer, +ve or -ve.
or, $x = {n\pi \over4}$
Taking the second one,
$(2\cos8x+1) = 0$
or, $2\cos8x = -1$
or, $\cos8x = -\frac1 2$
or, $8x = 2n\pi \pm \frac {2\pi} 3$, where $n$ is an integer, +ve or -ve.
or, $x = \frac{n\pi} 4 \pm \frac {\pi} {12}$
Second Way
$\cos16x = \cos8x$
or, $2\sin{8x-16x\over2}\sin{8x+16x\over2} = 0$
or, $2\sin(-4x)\sin{12x} = 0$
or, $-2\sin(4x)\sin{12x} = 0$
Again, either or both of the above factors are zero.
Taking the first one,
$\sin4x = 0$
or, $4x=n\pi$
or, $x=\frac{n\pi}4$
Taking the second one,
$\sin12x = 0$
or, $12x=n\pi$
or, $x=\frac{n\pi}{12}$
Now, as you must have noticed, we are getting two different sets of solutions, $x = \left\{{n\pi \over4}, \frac{n\pi} 4 \pm \frac {\pi} {12}\right\}$ and $x = \left\{{n\pi \over4}, \frac{n\pi}{12}\right\}$. The member $x=\frac{n\pi}4$ is common to both of the sets. Moreover, all these solutions satisfy the equation under consideration.
Could anybody please tell me why is this happening. In addition to the specific answer, some general insight will be most welcome. We have got a number of similar problems in hand. So, unless we can develop some acumen, life may become difficult.
| If $x\in\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\}$, then
$12x=3n\pi$ or $12x=3n\pi+\pi$ or $12x=3n\pi-\pi$ for some $n\in\mathbb Z$,
hence $12x=m\pi$ for some $m\in\mathbb Z$ (namely $m=3n$ or $m=3n+1$ or $m=3n-1$.
On the other hand, every $m\in \mathbb Z$ can be written either as $m=3n$ or $m=3n+1$ or $m=3n-1$, depending on the remainder modulo $3$. Therefore $x\in\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\}\iff \frac{12}\pi x\in\mathbb Z$.
Moreover, if $x=\frac{n\pi}4$ with $n\in \mathbb Z$, then $x=\frac{m\pi}{12}$ with $m:=3n\in\mathbb Z$, hence $\{\frac {n\pi}4\mid n\in\mathbb Z\}\subseteq \{\frac {n\pi}{12}\mid n\in\mathbb Z\}$, which ultimately makes
$$\left\{\frac{n\pi}4,\frac{n\pi}4\pm\frac\pi{12}\mid n\in\mathbb Z\right\} = \left\{\frac{n\pi}4,\frac{n\pi}{12}\mid n\in\mathbb Z\right\}=\left\{\frac{n\pi}{12}\mid n\in\mathbb Z\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding inverse of a $3\times3$ matrix I am given a $3 \times3$ matrix and am asked to find the inverse using elementary row operations. I know how they work, but have no idea of which steps to apply first, followed by which steps.
First, the matrices:
$$\begin{pmatrix}
1 & 1 & -3\\
2 & 1 & -3\\
2 & 2 & 1
\end{pmatrix}$$
All I know thus far is that, if there is a series of operations (pre-multipliers)
$E_nE_{n-1}...E_2E_1A$ that reduces to the identity matrix, the same sequence $ E_nE_{n-1}...E_2E_1I$ reduces to the inverse of $A$, $A^{-1}$.
Any help? If not, I will use another method already because this is not working thus far.
UPDATE
Thanks to the community, I got the final answer:
$$\begin{pmatrix}
-1 & 1 & 0\\
\frac8 7 & -1 & \frac 3 7\\
\frac{-2}{7} & 0 & \frac 1 7
\end{pmatrix}$$
| I was taught to augment the matrix with the identity, then apply the row operations:
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
2 & 1 & -3&0&1&0\\
2 & 2 & 1&0&0&1
\end{pmatrix}$$
Subtract twice row 1 from row 2 and twice row 1 from row 3 (yes, this is two operations)
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
0 & -1 & 3&-2&1&0\\
0 & 0 & 7&-2&0&1
\end{pmatrix}$$
Multiply row 2 by -1 and row 3 by $\frac 17$
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
0 & 1 & -3&2&-1&0\\
0 & 0 & 1&-2/7&0&1/7
\end{pmatrix}$$
Subtract row 2 from row 1, then add three times the third to the second and you are there. The right three columns will be your inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
probabilty problem how to solve Six cards are drawn with replacement form on ordinary deck. What is the probabilty that each of four suits will be represented at least once among the six cards?
| Consider the different cases:
1 suit
$$
\begin{align*}
P(\text{Only } \clubsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \diamondsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6
\end{align*}
$$
So 4 cases.
2 suits
$$
\begin{align*}
P(\text{Only } \clubsuit \text { AND } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
&.\\&.\\&.\\
\end{align*}
$$
The idea here is you EXCLUDE the cases where only $\clubsuit$ or $\heartsuit$ are chosen. The residual probability is the probability of a combination of both $\clubsuit$ and $\heartsuit$
You will realize that there are ${4 \choose 2} = 6$ cases ie
$$
(\clubsuit,\heartsuit) (\clubsuit,\spadesuit) (\clubsuit,\diamondsuit) \\
(\heartsuit,\spadesuit) (\heartsuit,\diamondsuit) (\spadesuit,\diamondsuit)
$$
So do your multiplication accordingly.
3 suits
$$
\begin{align*}
P(\text{Only } \clubsuit \text { AND } \heartsuit { AND } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
&.\\&.\\&.\\
\end{align*}
$$
Similarly to 2 suits, the idea is to exclude the case where only 2 suits are selected, and only 1 suit is selected. There are $3$ ways that only 2 suits are selected and $3$ ways that only 1 suit can be selected.
There are 4 cases, namely
$$
(\clubsuit,\heartsuit, \spadesuit) (\heartsuit, \spadesuit, \diamondsuit) \\
(\spadesuit,\diamondsuit, \clubsuit) (\diamondsuit,\clubsuit, \heartsuit)
$$
So do the addition, and take compliment.
$$
\begin{align*}
\text{Req. Probability} &=
1\\&-
4\begin{bmatrix}\begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&-
6\begin{bmatrix}\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&-
4\begin{bmatrix}\begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 + 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How can I find the possible values that $\gcd(a+b,a^2+b^2)$ can take, if $\gcd(a,b)=1$ If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
| Since $\gcd(a,b)=1$, Bezout's Identity says we have an $x$ and $y$ so that
$$
ax+by=1\tag{1}
$$
Note that
$$
\begin{align}
2a^2&=(a^2+b^2)+(a+b)(a-b)\\
2ab&=(a+b)^2-(a^2+b^2)\\
2b^2&=(a^2+b^2)-(a+b)(a-b)\\
\end{align}\tag{2}
$$
Therefore, incorporating $(1)$ and $(2)$,
$$
\begin{align}
2
&=2(ax+by)^2\\
&=2a^2x^2+4abxy+2b^2y^2\\
&=\Big((a^2+b^2)+(a+b)(a-b)\Big)x^2\\
&+2\Big((a+b)^2-(a^2+b^2)\Big)xy\\
&+\Big((a^2+b^2)-(a+b)(a-b)\Big)y^2\\
&=\color{#00A000}{(x-y)^2}\color{#C00000}{(a^2+b^2)}
+\color{#00A000}{((x^2-y^2)(a-b)+2xy(a+b))}\color{#C00000}{(a+b)}\tag{3}
\end{align}
$$
Equation $(3)$ says that
$$
\gcd(a+b,a^2+b^2)\,|\,2\tag{4}
$$
Note that $\gcd(1+2,1^2+2^2)=1$ and $\gcd(1+3,1^2+3^2)=2$, so both $1$ and $2$ are possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 0
} |
Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum? Does the following function have a global minimum:
$$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$$
where $x \in \mathbb{N}$?
I tried using WolframAlpha, but it appears to give an inconsistent result.
| $$1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}}$$
$$=1 + \frac{1}{x} + \sqrt{\frac{2}{1 + \frac{1}{x}}}$$
$$=1 + \frac{1}{x} + \frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}+\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}$$
Applying A.M. G.M. we have,
$$1 + \frac{1}{x} + \frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}+\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}\geq3.((1 + \frac{1}{x})(\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}})^2)^{\frac{1}{3}}$$
$$3.((1 + \frac{1}{x})(\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}})^2)^{\frac{1}{3}}=3.(\frac{1}{2})^{1/3}$$
Equality holds when $\displaystyle 1 + \frac{1}{x}=\frac{1}{2}\sqrt{\frac{2}{1 + \frac{1}{x}}}$
Squaring both sides we get,
$$(1 + \frac{1}{x})^{3}=\frac{1}{2}$$
$$\Rightarrow 1 + \frac{1}{x}=\frac{1}{2^{1/3}}$$
$$\Rightarrow \frac{1}{2^{1/3}}-1=\frac{1}{x}$$
$$\Rightarrow x=\frac{2^{1/3}}{1-2^{1/3}}$$
Now check the two nearest integers to x and compare the values of the expression at those values and the min. will global minimum .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/308991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
infinitely many solutions to $\displaystyle x^n + y^n = z^{n+1}$ Let $n$ be a positive integer. Show that the equation
$\displaystyle x^n + y^n = z^{n+1}$
has infinitely many integer solutions.
| Let $x$ and $y$ be such that $x \mid y$. This means that $y = x \cdot m$ where $m$ is a positive integer.
Thus
$x^n + y^n = x^n + (x \cdot m)^n = x^n + x^n \cdot m^n = x^n \cdot [1 + m^n] = z \cdot z^n = z^{n+1}$
So we know that when $x \mid y$, we have $x^n \cdot [1 + m^n] = z^n \cdot z$
Now let $x^n = z^n$. Thus $1 + m^n = z$
Then $z^n = [1 + m^n]^n = x^n$.
Substituting $[1 + m^n]^n$ for $x^n$ we get
$$x^n + y^n = [1 + m^n]^n + (m[1 + m^n])^n = [1 + m^n]^n + m^n \cdot [1 + m^n]^n
\\ = [1 + m^n] \cdot [1 + m^n]^n = [1 + m^n]^{n+1} = x^{n+1} = z^{n+1}$$
Since there are infinitely many choices for $m$ since $n$ is a positive integer, we have our answer; a triple that gives infinitely many solutions to the equation
$x^n + y^n = z^{n+1}$
is $(1 + m^n, m \cdot (1 + m^n), 1 + m^n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show a,b,c congruences when $a^2+b^2=c^2$ i would like to know how to prove that if $a^2+b^2=c^2$ then
a) $3|a$ or $3|b$
b) $5|a$ or $5|b$ or $5|c$
c) $4|a$ or $4|b$
Also that $a^2+b^2+c^2+1$ can't be divisible by 8
Thank you very much!
| Hint: Take the equation $a^2+b^2=c^2$ modulo 3. Also, note that
$$0^2\equiv 0\bmod 3,\qquad 1^2\equiv 1\bmod 3,\qquad 2^2\equiv 1\bmod 3,$$
and since any integer is equivalent modulo 3 to either 0, 1, or 2, we see that any square number, when taken modulo 3, cannot be equivalent to 2.
Do the same for the other moduli; work out the various cases according to what values $a^2$, $b^2$, and $c^2$ might take modulo $n$, rule out the cases that are impossible.
Also, keep in mind that if a prime number $p$ divides a square number $r^2$, then it must divide $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/312864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A limit without Taylor series or l'Hôpital's rule $\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$ Computing without Taylor series or l'Hôpital's rule
$$\lim_{n\to\infty}\prod_{k=1}^{n}\cos \frac{k}{n\sqrt{n}}$$
What options would I have here? Thanks!
| The following tries to use only "basic" inequalities, avoiding L'Hopital and Taylor.
The inequality
$$\tag1 e^x\ge 1+x\qquad x\in\mathbb R$$
should be well-known and after taking logarithms immediately leads to
$$\tag2 \ln(1+x)\le x\qquad x>-1$$
and after taking reciprocals
$$\tag3 e^{-x}\le \frac1{1+x}\qquad x>-1 $$
Substituting $-\frac x{1+x}$ for $x$ in $(3)$ and taking logarithms, we have
$$\tag4 \frac x{1+x}\le \ln(1+x)\qquad x>-1.$$
For $0<x<\frac\pi2$ we have
$$\ln\cos x=\ln\sqrt{1-\sin^2x}=\frac12\ln(1-\sin^2x)$$
and by $(2)$ and $(4)$
$$\tag 5 -\frac12\tan^2x=\frac12 \frac{-\sin^2x}{1-\sin^2x}\le \ln\cos x\le -\frac12\sin^2x.$$
Since $0\le \sin x \le x\le \tan x$ for $0\le x<\frac\pi2$ and the quotient of the upper and the lower bound in $(5)$ is just $\cos^2x$ we find for $0<x\le y<\frac\pi2$ (using $\cos^2y\le \cos^2x$)
$$-\frac12 x^2\cdot\frac1{\cos^2y}\le \ln\cos x\le -\frac12 x^2\cdot \cos^2y.$$
Letting (for $n\ge1$) $y=\frac1{\sqrt n}$, $x=\frac k{n\sqrt n}$ with $1\le k\le n$, this gives
$$-\frac{k^2}{2n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le \ln\cos \frac k{n\sqrt n}\le -\frac{k^2}{2n^3}\cdot\cos^2\frac1{\sqrt n}.$$
Using $1^2+2^2+\ldots + n^2=\frac{n(n+1)(2n+1)}{6}$ we find by summation
$$ -\frac{n(n+1)(2n+1)}{12n^3}\cdot\frac{1}{\cos^2\frac1{\sqrt n}}\le\sum_{k=1}^n\ln\cos\frac k{n\sqrt n}\le-\frac{n(n+1)(2n+1)}{12n^3}\cdot\cos^2\frac1{\sqrt n}$$
and by taking the limit
$$ \sum_{k=1}^\infty\ln\cos\frac k{n\sqrt n}=-\frac16$$
and ultimately
$$ \prod_{k=1}^\infty\cos\frac k{n\sqrt n}=e^{-\frac16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
bound for $a,b,c$ in $|ax^2+bx+c| \leq 1\;\forall x\in \left[0,1\right]$
Consider $a,b,c\in\mathbb{R}$ such that $|ax^2+bx+c|\leq 1\;\forall x\in \left[0,1\right]$. Prove that $|a|\leq 8\;\;,|b| \leq 8$ and $|c| \leq 1$.
My Attempt:
Set $x = 0$ in $|ax^2+bx+c|\leq 1$ to get $|c| \leq 1$. Similarly set $x = 1$ in $|ax^2+bx+c| \leq 1$ to get $|a+b+c|\leq 1$.
From here, how can I calculate bounds for $a$ and $b$?
| Let $f(x)=ax^2+bx+c$.
Hence, $f(0)=c$, $f(1)=a+b+c$ and $f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$.
After solving of this system we obtain:
$|a|=|2f(1)-4f\left(\frac{1}{2}\right)+2f(0)|\leq2|f(1)|+4|f\left(\frac{1}{2}\right)|+2|f(0)|\leq8$,
$|b|=|-f(1)+4f\left(\frac{1}{2}\right)-3f(0)|\leq|f(1)|+4|f\left(\frac{1}{2}\right)|+3|f(0)|\leq8$,
$|c|=|f(0)|\leq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/314686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Characteristic polynomial calculation method Please help me, what to put in empty matrices
$$\begin{vmatrix}3-\lambda & -4 & -2 & -4\\
-2 & 5-\lambda & 2 & 4\\
4 & -8 & -3-\lambda & -8\\
2 & -4 & -2 & -3-\lambda
\end{vmatrix}=\begin{vmatrix}3 & -4 & -2 & -4\\
-2 & 5 & 2 & 4\\
4 & -8 & -3 & -8\\
2 & -4 & -2 & -3
\end{vmatrix}+\lambda(\begin{vmatrix}5 & 2 & 4\\
-8 & -3 & -8\\
-4 & -2 & -3
\end{vmatrix}+\begin{vmatrix}3 & -2 & -4\\
4 & -3 & -8\\
2 & -2 & -3
\end{vmatrix}+\begin{vmatrix}3 & -4 & -4\\
-2 & 5 & 4\\
2 & -4 & -3
\end{vmatrix}+\begin{vmatrix}3 & -4 & -2\\
-2 & 5 & 2\\
4 & -8 & -3
\end{vmatrix}) + \lambda^{2}(\begin{vmatrix}\\
\\
\end{vmatrix}+\begin{vmatrix}\\
\\
\end{vmatrix}+\begin{vmatrix}\\
\\
\end{vmatrix}+\begin{vmatrix}\\
\\
\end{vmatrix})+\lambda^{3}(3+5-3-3)+\lambda^{4}=
$$
Thanks in avance for help
| $\def\p{\phantom{-}}$
Upgrading comment to answer, at request of OP: $$\pmatrix{\p3&-4\cr-2&\p5\cr},\pmatrix{3&-2\cr4&-3\cr},\pmatrix{3&-4\cr2&-3\cr},\pmatrix{\p5&\p2\cr-8&-3\cr},\pmatrix{\p5&\p4\cr-4&-3\cr},\pmatrix{-3&-8\cr-2&-3\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/315686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$ Can anyone tell me what I am doing wrong? need to prove for $k\ge2$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$$$(5-\frac5k )(1+\frac{1}{(k+1)^2})= 5(1-\frac1k)(1+\frac1{(k+1)^2})$$
$$=5(1+\frac1{k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}+\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}+\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}+\frac{10}{k(k+1)^2}\le5-\frac5{k+1}$$
which doesn't look true.
| $$=5(1+\frac1{(k+1)^2}-\frac1k-\frac1{k(k+1)^2})$$
$$=5(1+\frac{k}{k(k+1)^2}-\frac{(k+1)^2}{k(k+1)^2}-\frac1{k(k+1)^2})$$
$$= 5(1-\frac{k^2+k+2}{k(k+1)^2})$$
$$=5(1-\frac{k(k+1)}{k(k+1)^2}\color{red}{-}\frac2{k(k+1)^2})$$
$$=5(1-\frac{1}{k+1}\color{red}{-}\frac2{k(k+1)^2})$$
$$= 5 - \frac5{k+1}\color{red}{-10}\frac1{k(k+1)^2}$$
Where $k \geq 2$ , so $$\frac{-10}{k(k+1)^2} < 0$$
Therefore
$$5 - \frac5{k+1}-\frac{10}{k(k+1)^2} \leq 5 - \frac5{k+1}$$
$$(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$$ Which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit of $y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$ I would appreciate any help with this problem:
If
$$y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$$
Then how do I find $y^2 - y$?
I'm not sure whether this is an arithmetic or geometric series.
| Note that $(y^2-5)^2=5-y$. It is also clear that $ y^2 \geq 5$ and $0<y$.
We have $0=y^4-10y^2+y+20=(y^2-y-4)(y^2+y-5)$.
We have $y^2+y-5>0$, so $y^2-y=4$.
P.S. Incidentally, if you want to find $y$, it is the positive root of $y^2-y-4=0$, which is $\frac{1+\sqrt{17}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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generating functions, can't seem to get the correct answers. So, I've been having some issues with generating functions and counting problems. An example problem is: $$ a_n = a_{n-1} + 9a_{n-2} - 9a_{n-3} \;\;\; (n \geq 3)$$
Where $a_0 = 0, a_1 = 1, a_2 = 2$
Let's define $g(x)$ as our generating function, in which case,
$$ g(x) = a_0 + a_1x + a_2x^2+... = 0 + x + 2x^2 + (a_2 + 9a_1 - 9a_0)x^3 + (a_3+9a_2-9a_1)x^4+...$$
We can factor this like so...
$$g(x) = x + 2x^2 + x(a_2x^2 + a_3x^3 + ...) + 9x^2(a_1x + a_2x^2+...) - 9x^3(a_0 + a_1x + a_2x^2+...)$$
Now, the infinite series, $a_2x^2 + a_3x^3+...$ can be defined as $g(x) - a_0 - a_1x$, and the series $a_1x + a_2x^2+...$ can be defined as $g(x) - a_0$, and lastly, the series $a_0 + a_1x + a_2x^2+...$ is $g(x)$
So, making substitutions:
$$g(x) = x + 2x^2 + xg(x) - x^2 + 9x^2g(x) - 9x^3g(x)$$
and finally,
$$g(x) = \frac{x^2 + x}{9x^3 - 9x^2 - x + 1}$$
I decided to run this through wolfram alpha for the partial fraction decomposition and I get:
$$g(x) = \frac{1}{3}\left (\frac{1}{1+3x}\right ) + \frac{1}{12}\left (\frac{1}{1+3x}\right ) - \frac{1}{4}\left (\frac{1}{1+x}\right )$$
Each of the terms in parenthesis can be expressed as a series, in summation notation:
$$\frac{1}{3}\sum_{k = 0}^{\infty }(-3)^kx^k + \frac{1}{12}\sum_{k = 0}^{\infty }(-3)^kx^k - \frac{1}{4}\sum_{k=0}^{\infty }(-1)^kx^k$$
And, my final answer should be:
$$a_n = \frac{1}{3}(-3)^n + \frac{1}{12}(-3)^n - \frac{1}{4}(-1)^n$$
Which, is incorrect. Could someone point out what I'm doing wrong?
| Your partial fractions aren’t right. However, there’s an easier way to handle such problems.
Let $g(x)=\sum_{n\ge 0}a_nx^n$. be the generating function. With the convention that $a_n=0$ if $n<0$, we can write the recurrence as
$$a_n = a_{n-1}+9a_{n-2}-9a_{n-3}+[n=1]+[n=2]\;,\tag{1}$$
valid for all $n\ge 0$, where the last two terms are Iverson brackets; they ensure that the initial values are correct.
Now multiply $(1)$ by $x^n$ and sum over $n$ to get
$$\begin{align*}
g(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_n(a_{n-1}+9a_{n-2}-9a_{n-3}+[n=1]+[n=2])x^n\\
&=\sum_n a_{n-1}x^n+9\sum_n a_{n-2}x^n-9\sum_n a_{n-3}x^n+x+x^2\\
&=x\sum_n a_nx^n+9x^2\sum_n a_nx^n-9x^3\sum_n a_nx^n+x+x^2\\
&=g(x)\left(x+9x^2-9x^3\right)+x+x^2\;,
\end{align*}$$
so
$$\begin{align*}g(x)&=\frac{x+x^2}{1-x-9x^2+9x^3}\\
&=\frac{x+x^2}{(1-x)(1-3x)(1+3x)}\\
&=\frac{-1}{4(1-x)}+\frac1{3(1-3x)}-\frac1{12(1+3x)}\\
&=-\frac14\sum_{n\ge 0}x^n+\frac13\sum_{n\ge 0}3^nx^n-\frac1{12}\sum_{n\ge 0}(-1)^n3^nx^n\;,
\end{align*}$$
and
$$\begin{align*}
a_n&=-\frac14+3^{n-1}-\frac14(-1)^n3^{n-1}\\
&=3^{n-1}-\frac14\left(1+(-1)^n3^{n-1}\right)\;.
\end{align*}$$
| {
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"url": "https://math.stackexchange.com/questions/318213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$.
Given:
*
*$x,y,z\geq0$
*$xy+yz+zx=1$
Prove that $\displaystyle \frac 1 {x+y}+\frac 1 {y+z}+\frac 1 {z+x}\geq \frac 5 2$.
I tried using Cauchy's inequality LHS $\geq\frac 9 {2a+2b+2c}$, but failed. Please give me some ideas. Thank you.
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that
$$\frac{\sum\limits_{cyc}(x^2+3xy)}{\prod\limits_{cyc}(x+y)}\geq\frac{5}{2}$$ or
$$\frac{9u^2+3v^2}{9uv^2-w^3}\geq\frac{5}{2}$$ or
$f(w^3)\geq0$, where $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $w^3$,
which happens in the following cases.
*
*$y=x$, $z=\frac{1-x^2}{2x}$, where $0<x\leq1$ and we get something obvious;
*$w^3=0$.
Let $z=0$. Hence, we need to prove that
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{x+y}\geq\frac{5}{2}$$ or
$$x+y+\frac{1}{x+y}\geq\frac{5}{2}$$ or
$$(x+y-2)(2(x+y)-1)\geq0,$$
which is true because $x+y\geq2\sqrt{xy}=2$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
separable equation1 Does this solve the Impartible Equation is correct?
Impartible Equation:
$(x+1)y'y=y^2-1$
solve:
$y'=y^2-1/y(x+1)$
$f(x,y)=y^2-1/y(x+1)$
$y^2-1/y/(x+1)/1$
$f(x,y)=(y^2-1)/y/(x+1)/1$
$y^2-1=M(x)$
,
$(x+1)/1=N(y)$
| The differential equation $(x+1)y'y =y^2-1$ is separable. Rewrite it as $$\frac{y}{y^2-1}y'=\frac{1}{x+1}.$$
On the left we recognize the derivative of $\frac{1}{2}\ln(|y^2-1|$. Integrating, we get
$$\frac{1}{2}\ln(|y^2-1|)=\ln(|x+1|)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternative solutions to $\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$ Here is a limit that can be computed directly by performing the integration and then taking
the limit, but the way is rather ugly. What else can we do? Might we avoid the integration?
$$\lim_{n\to\infty} \frac{1}{\sqrt{n}}\int_{ 1/{\sqrt{n}}}^{1}\frac{\ln(1+x)}{x^3}\mathrm{d}x$$
| An alternate way is to notice that expanding the integrand also gives:
$$\frac{1}{x^2}-\frac{1}{x}<\frac{\ln(1+x)}{x^3}< \frac{1}{x^2}\;\,\text{for}\;\,0<x<1$$
Hence:
$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}-\frac{1}{x}\,dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< \frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{1}{x^2}dx$$
$$1-\frac{1}{\sqrt{n}}-\frac{\ln\sqrt{n}}{\sqrt{n}}<\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}dx< 1-\frac{1}{\sqrt{n}}$$
$$\frac{1}{\sqrt{n}}\int_{1/\sqrt{n}}^1\frac{\ln(1+x)}{x^3}\to 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/319632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 5
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Find sum of the Trignomertric series Q1: The sum of the infinite series $\cot ^{-1}2 + \cot ^{-1} 8+ \cot^{-1}18+ \cot^{-1}32\cdots$
1.$\pi/3$
2.$\pi/4$
3.$\pi/2$
4.None
Q2: Value of $\lim_ {n \to \infty}[ {\cos \frac{\pi}{2^2} } {\cos
\frac{\pi}{2^3} } \ldots{\cos \frac{\pi}{2^n} }$]
*
*$\pi$
*$1/\pi$
*$2/\pi$
*$\pi/e$
| For the first one,
$$\sum_{k=1}^{m} \text{arccot}(2n^2) = \text{arccot} \left(\dfrac{m+1}m\right)$$
Your sum is
$$\sum_{k=1}^{\infty} \text{arccot}(2n^2) = \lim_{m \to \infty}\text{arccot} \left(\dfrac{m+1}m\right) = \dfrac{\pi}4$$
For the second one,
$$\prod_{k=1}^m \cos\left(\dfrac{\theta}{2^{k+1}} \right) = \dfrac{\sin\left(\dfrac{\theta}{2}\right)}{2^{m}\sin \left(\dfrac{\theta}{2^{m+1}} \right)}$$
Hence, your product is
$$\prod_{k=1}^{\infty} \cos\left(\dfrac{\pi}{2^{k+1}} \right) = \lim_{m \to \infty} \dfrac{\sin\left(\dfrac{\pi}{2}\right)}{2^{m}\sin \left(\dfrac{\pi}{2^{m+1}} \right)} = \dfrac2{\pi}$$
For the first one, recall $$\cot(A+B) = \dfrac{\cot(A) \cot(B) - 1}{\cot(A) + \cot(B)}$$
\begin{align}
\cot \left(\text{arccot}\left(\dfrac{m}{m-1}\right) + \text{arccot}\left(2m^2 \right)\right) & = \dfrac{\dfrac{m}{m-1} \cdot 2 \cdot m^2-1}{\dfrac{m}{m-1} + 2 \cdot m^2}\\
& = \dfrac{2m^3-m+1}{2m^3-2m^2+m}\\
& = \dfrac{(m+1)(2m^2-2m+1)}{m(2m^2-2m+1)}\\
& = \dfrac{m+1}m
\end{align}
Now use the above two along with induction to conclude what you want.
For second one, recall $\sin(2 \phi) = 2 \sin(\phi) \cos(\phi)$ and induction to conclude what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/326590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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factorise, $x^3-13x^2+32x+20$ factorise, $x^3-13x^2+32x+20$
Let, $f(x)=x^3-13x^2+32x+20$
$f(x)=x(x^2-13x+30)+2x+20$
$f(x)=x(x-3)(x-10)+2x+20$
$f(-1)\lt 0$, $f(0)\gt 0$, which shows there is a root between $x=-1$ and $x=0$
$f(4)\gt 0$, $f(5)\lt 0$, which shows there is a root between $x=4$ and $x=5$
$f(9)\lt 0$, $f(10)\gt 0$, which shows there is a root between $x=9$ and $x=10$
| From the elementary theory of polynomials we know that the cubic polynomial $ax^3+bx^2+cx+d$ can be factored as
$$ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3),$$
where $x_k$ ($k=1,2,3$) are the roots of the general cubic equation
$$ax^3+bx^2+cx+d=0.$$
If we use the substitution $x=t+\frac{13}{3}$, the given equation
$$
\begin{equation*}
x^{3}-13x^{2}+32x+20=0\tag{1}
\end{equation*}
$$
is transformed into the reduced cubic equation
$$
\begin{equation*}
t^{3}+pt+q=0,\qquad p=-\frac{73}{3},q=-\frac{110}{27},\tag{2}
\end{equation*}
$$
a solution of which is$^1$
$$
\begin{eqnarray*}
t_{1} &=&\left( \frac{-q+\sqrt{\Delta }}{2}\right) ^{1/3}+\left( \frac{-q-
\sqrt{\Delta }}{2}\right) ^{1/3}, \qquad \Delta &=&q^{2}+\frac{4p^{3}}{27}.\tag{3}
\end{eqnarray*}
$$
When the discriminant $\Delta <0$ the three solutions of the original cubic $(1)$ are real. By using complex numbers we can write them in the form
$$
\begin{equation*}
x_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}
\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\pi (k-1)}{3}\right) -\frac{b}{3}
,
\end{equation*}\tag{4}
$$
where $k=1,2,3$, and $b=-13$ is the coefficient of $x^{2}$ in $(1)$. Since $$\Delta =-\dfrac{57\,184}{27}<0,$$ we have:
$$
\begin{eqnarray*}
x_{1} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) \right) +\frac{13}{3} \\
&\approx &9.347\,9, \\
&& \\
x_{2} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{2\pi }{3}\right) +\frac{13}{3} \\
&\approx &-0.513\,6, \\
&& \\
x_{3} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{4\pi }{3}\right) +\frac{13}{3} \\
&\approx &4.165\,7.
\end{eqnarray*}
$$
Therefore the factorization of $(1)$ is
$$
\begin{equation*}
x^{3}-13x^{2}+32x+20=(x-x_1)(x-x_2)(x-x_3).\tag{5}
\end{equation*}
$$
--
$^1$ A deduction of $(3)$ and $(4)$ can be found in this blog post of mine, in Portuguese.
| {
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"url": "https://math.stackexchange.com/questions/329928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\cos2\theta−\sqrt{3}\sin2\theta \equiv2 \cos (2\theta+\pi/3 )\equiv−2\sin(2\theta−\pi/6)$ Going from $2\cos(2\theta+\pi/3)$ to $\cos2\theta−\sqrt{3}\sin2\theta$ is simple enough, however I'm stuck on going from $2\cos(2\theta+\pi/3)$ to $−2\sin(2\theta−\pi /6)$. How do i do this?
| First, let's recap:
$\sin{\pi\over3}=\frac{\sqrt{3}}{2}$
$\cos{\pi\over3}=\frac12$
$\sin{\pi\over6}=\frac12$
$\cos{\pi\over6}=\frac{\sqrt{3}}{2}$
Let's rock 'n roll!
Solution 1:Going from $\cos2\theta−\sqrt{3}\sin2\theta$ to $2\cos\left(2\theta+\frac\pi3\right)$.
$$
\require{cancel}
\begin{align}
\cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\
&=2\cdot\frac12\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=2\left(\frac12\left(\cos^2\theta-\sin^2\theta\right)-\frac12\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=2\left(\left(\cos^2\theta-\sin^2\theta\right)\cos\frac{\pi}{3}-2\sin\theta\cos\theta\sin\frac\pi3\right)\\
&=2\left(\cos2\theta\cos\frac\pi3-\sin2\theta\sin\frac\pi3\right)\\
&=2\cos\left(2\theta+\frac\pi3\right)
\end{align}
$$
Solution 2:Going from $\cos2\theta−\sqrt{3}\sin2\theta$ to $\sin(2\theta−\frac\pi6)$.
$$
\begin{align}
\cos2\theta−\sqrt{3}\sin2\theta&=\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\\
&=-2\cdot\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta-\sqrt{3}\cdot2\sin\theta\cos\theta\right)\\
&=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\right)\left(\sqrt{3}\cdot2\sin\theta\cos\theta\right)\right)\\
&=-2\left(\left(-\frac12\right)\left(\cos^2\theta-\sin^2\theta\right)-\left(-\frac12\sqrt{3}\right)\left(2\sin\theta\cos\theta\right)\right)\\
&=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)-\left(-\cos\frac\pi6\right)\left(\sin\theta\cos\theta\right)\right)\\
&=-2\left(-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)+\cos\frac\pi6\left(2\sin\theta\cos\theta\right)\right)\\
&=-2\left(\cos\frac\pi6\left(2\sin\theta\cos\theta\right)-\sin\frac\pi6\left(\cos^2\theta-\sin^2\theta\right)\right)\\
&=-2\left(\cos\frac\pi6\sin2\theta-\sin\frac\pi6\cos2\theta\right)\\
&=-2\left(\sin2\theta\cos\frac\pi6-\cos2\theta\sin\frac\pi6\right)\\
&=-2\sin\left(2\theta-\frac\pi6\right)
\end{align}
$$
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ Calculate $\lim_{n\to\infty} ((a+b+c)^n+(a+\epsilon b+\epsilon^2c)^n+(a+\epsilon^2b+\epsilon c)^n)$ with $a,b,c \in \Bbb R$ and $\epsilon \in \Bbb C \setminus \Bbb R, \epsilon^3=1$.
Since $a+\epsilon b + \epsilon^2c=\overline {a+\epsilon^2b+\epsilon c}$, the expression above should be real, but I don't know how to deal with the $\epsilon$ in order to bring it to a form where I can calculate its limit.
Also, for this problem, $|a+b+c|<1$ and $ab+bc+ac=0$.
| Here's a more brute force method.
The condition $ab+bc+ac=0$ says that the point $(a,b,c)$ lies on a cone with axis $a=b=c$. Making the change of variables
$$
\left(
\begin{array}{c}
a \\
b \\
c
\end{array}
\right) = \left(
\begin{array}{ccc}
\frac{1}{6} \left(3+\sqrt{3}\right) & \frac{1}{6} \left(-3+\sqrt{3}\right) & \frac{1}{\sqrt{3}} \\
\frac{1}{6} \left(-3+\sqrt{3}\right) & \frac{1}{6} \left(3+\sqrt{3}\right) & \frac{1}{\sqrt{3}} \\
-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}
\end{array}
\right)
\left(
\begin{array}{c}
u \\
v \\
w/\sqrt{2}
\end{array}
\right)
$$
(see rotation matrix from axis and angle) the conditions
$$
|a+b+c| < 1 \quad \text{and} \quad ab + bc + ac = 0
$$
become
$$
|w| < \sqrt{\frac{2}{3}} \quad \text{and} \quad u^2 + v^2 = w^2,
$$
and the main quantity in question becomes
$$
\left(\sqrt{\frac{3}{2}}\,w\right)^n + \left[\frac{\sqrt{3}}{4}-\frac{3}{4} + i \left(\frac{\sqrt{3}}{4}+\frac{3}{4}\right)\right]^n \left[ (v-iu)^n + e^{n i 4\pi/3} (u-iv)^n \right].
\tag{1}
$$
Now since $|w| < \sqrt{2/3}$ and $u^2 + v^2 = w^2$ we know that $|v-iu| = |u-iv| < \sqrt{2/3}$, so, because
$$
\left|\frac{\sqrt{3}}{4}-\frac{3}{4} + i \left(\frac{\sqrt{3}}{4}+\frac{3}{4}\right)\right| = \sqrt{\frac{3}{2}},
$$
we may conclude that the absolute value of every term in $(1)$ is $<1$, and thus that the limit as $n\to\infty$ must be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/331342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Prove that $|PF_{1}|+|PF_{2}|$ is Constant in an Elipse Given an elipse with two focus $F_{1}$ an $F_{2}$, and $A$ is an arbitrary point at the elipse. Stright line $AF_{1}$ has another intersection point $B$ with the elipse, and $AF_{2}$ has another intersection point $C$ with the elipse.
And point $P$ is the intersection of line $BF_{2}$ and $CF_{1}$, now the problem is to prove that $|PF_{1}|+|PF_{2}|$ is constant.
Thanks in advance!
| OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We draw an ellipse by holding a pen taut against the string. The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable. First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic. Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$. We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
To prove that the expression for the ellipse has the sum of the distances from the foci being constant, work backward from this sequence.
| {
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"url": "https://math.stackexchange.com/questions/332539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Calculate: $\lim\limits_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$ How do I calculate the following limit without using l'Hôpital's rule?
$$\lim_{x \to \infty}\left(\frac{x^2+2x+3}{x^2+x+1} \right)^x$$
| A slightly different take. Let $L$ be the limit in question. Then we have
$$\begin{align}\log{L} &= \lim_{x \rightarrow \infty}x \log{\left(\frac{x^2+2 x+3}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty}x \log{\left( 1+\frac{x+2}{x^2+x+1}\right)}\\ &= \lim_{x \rightarrow \infty} \frac{x(x+2)}{x^2+x+1}\end{align}$$
Therefore $\log{L}=1$ and $L=e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.
| Hint:
$a+b+c=3$
$a^2+b^2+c^2=3$
EDITED
From famous Cauchy–Schwarz inequality(http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
You get:
$(a^2+b^2+c^2)(c^2+a^2+b^2) \ge(ab+bc+ca)^2$
But the equality holds here,
$(a^2+b^2+c^2)^2-(ab+bc+ca)^2=0$
When we have Cauchy Schwarz inequality:
$(a_1^2+a_2^2+..+a_n^2)(b_1^2+b_2^2+..+b_n^2) \ge (a_1b_1+a_2b_2+..a_nb_n)^2$
Equality holds iff:
$(a_1b_2-a_2b_1)=...a_{n-1}b_{n}-a_nb_{n-1}=0$
Apply it over here you get $a^2-bc=b^2-ac=c^2-bc0 \implies a=b=c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Does there exist $a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$
Does there exist $ a,b,c\in \mathbb Q$ such that $(a+b+c)^2 + 3(a+b+c)+5=2(ab+bc+ca)$
I think the answer is no
| After simplification of the equation you will get $$(a+\frac{3}{2})^2+(b+\frac{3}{2})^2+(c+\frac{3}{2})^2=\frac{7}{4}.$$Note that LHS$\geq 3.(\frac{3}{2})^2>RHS$, hence not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
How many integers between [3,000, 8,000] have digit sum 20? Well, I've been almost losing my mind over this one.
The question is: How many integers on the interval $[3000, 8000]$ have a digit sum of 20?
Well, the first thing I did was separating to groups according to the first digit.
and then:
3- A+B+C=17
4- A+B+C=16
5- A+B+C=15
6- A+B+C=14
7- A+B+C=13
Now my main problem is: Since A, B, and C are digits, it means that they are on the interval $[0, 9]$. How do I take that restriction into account?
| Method 1: Using generating functions, you want the coefficient of $x^{20}$ in $$(x^3+x^4+ \ldots +x^7)(1+x+ \ldots +x^9)^3=\frac{x^3(1-x^5)(1-x^{10})^3}{(1-x)^4}$$
Equivalently, we drop the $x^3$ factor and look for the coefficient of $x^{17}$.
We have $(1-x)^{-4}=\sum_{i=0}^{\infty}{\binom{i+3}{3}x^i}$. Expanding $(1-x^5)(1-x^{10})^3$ and ignoring $x^n, n>17$ gives $(1-x^5)(1-3x^{10})=1-x^5-3x^{10}+3x^{15}$.
Finally the coefficient of $x^{17}$ is $$\binom{(17-0)+3}{3}-\binom{(17-5)+3}{3}-3\binom{(17-10)+3}{3}+3\binom{(17-15)+3}{3}$$
which easily evaluates to give $355$.
Method 2: As you might not be familiar with generating functions, here is a more elementary approach.
We proceed exactly as you did. If the 1st digit is $i$, then we have $A+B+C=20-i$. This gives $\binom{(20-i)+2}{2}$, but we have also counted those where $A, B, C$ are $\geq 10$. For those cases, exactly 1 of the digits is $\geq 10$. If it is $A$, we have $A=k=10, 11, \ldots , 20-i$. Then $B+C=20-i-k$, giving $21-i-k$ ways. The number of ways for $B, C$ are the same. Therefore we must subtract $3\sum\limits_{k=10}^{20-i}{(21-i-k)}=3\sum\limits_{j=1}^{11-i}{j}=3\frac{(11-i)(12-i)}{2}$.
Therefore the total number is
$$\sum_{i=3}^{7}{\left(\binom{(20-i)+2}{2}-3\frac{(11-i)(12-i)}{2}\right)}=355$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Prove that $a_n = \{\left(1+\frac{1}{n}\right)^n\}$ is bounded sequence, $ n\in\mathbb{N}$ How to prove the following:
$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$
| There was a comment that you can use $\text{AM} \ge \text{GM}$ to prove the boundedness of this.
Here is a proof. As a side effect, we also prove the convergence.
First we show that $x_n = \left(1 + \frac{1}{n}\right)^n$ is monotonically increasing. We prove this, using $\text{AM} \ge \text{GM}$.
We have that, by taking $n$ copies of $\left(1 + \frac{1}{n}\right)$ and one copy of $1$ that,
$$\frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{n}\right) + 1}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$
i.e.
$$ \frac{n+2}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$
and so
$$\left(1 + \frac{1}{n+1}\right)^{n+1} \gt \left(1 + \frac{1}{n}\right)^n$$
Now we use $\text{AM} \ge \text{GM}$ (again!).
We take $n$ copies of $1$ and one copy of $\frac{1}{2}$ to get
$$\frac{n + \frac{1}{2}}{n+1} \ge \sqrt[n+1]{\frac{1}{2}}$$
i.e.
$$ 2^{\frac{1}{n+1}} \ge \frac{2n+2}{2n+1}$$
i.e
$$ 2 \ge \left(1 + \frac{1}{2n+1}\right)^{n+1} $$
And so
$$ 4 \ge \left(1 + \frac{1}{2n+1}\right)^{2n+1} $$
Since the sequence is monotonically increasing, this bound applies to the whole sequence.
(Side effect: Since the sequence is monotonic, and bounded, it is convergent)
| {
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"url": "https://math.stackexchange.com/questions/342491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What's the ammount of solutions to the Equation $A+B+C+D=24$ with restrictions on $A$ and $B$ As the title says, I have the the following equation:
$A+B+C+D=24$
$2\leq A\leq 5$ and $4\leq B\leq 7$
And i need to find the number of possible solutions. $A,B,C,D\geq 0$, They are integers.
I'm not sure if my solution is right, What i did is:
I've made a new equation stating $A_1=A-2$ and $B_1=B-4,$ So i get:
$A_1+B_1+C+D=30.$
So my restrictions now are $0\leq B \leq 3$, Marking $A_1+B_1=S, S=[0,6],$ Means:
$C+D=30-S $ And the number of possibilities for that one is $31-S$.
So now i need to do the sum: $\sum (31-S)$ from $S=0$ to $ S=6$ and multiply it by $3^2 \;$ [Which represents the different combinations of $A$ and $B$] so that equals to: $1774.$
Did I do something wrong or it seems fine?
| You've had a good idea to set $A_1=A-2$ and $B_1=B-4$, but from there, you start having issues. We can rewrite $A=A_1+2$ and $B=B_1+4$ to give us $$A_1+B_1+C+D+6=24,$$ or $$A_1+B_1+C+D=18,\tag{1}$$ with our new constraints that $A_1,B_1,C,D$ are all nonnegative integers with $A_1\le3$ and $B_1\le 3$.
As you note in the comments, there are $16$ pairs $A_1,B_1$ that meet our constraints. However, there are not $16$ different ways to make $A_1+B_1=S$ for each integer $0\leq S\leq 6$. Consider the following table of sums: $$\begin{array}{c|cccc}& 0 & 1 & 2 & 3\\\hline\\0 & 0 & 1 & 2 & 3\\1 & 1 & 2 & 3 & 4\\2 & 2 & 3 & 4 & 5\\3 & 3 & 4 & 5 & 6\end{array}$$ In general, then, given an integer $0\le S\le 6$, there are $$\min\{S,6-S\}+1=4-|3-S|$$ ways to choose $A_1,B_1$ meeting our constraints such that $A_1+B_1=S$.
Now, given any integer $0\le S\le 6,$ there are indeed $19-S$ ways to choose nonnegative integers $C,D$ such that $S+C+D=18$. Hence, the sum you're looking for is $$\begin{align}\sum_{S=0}^6\bigl(4-|3-S|\bigr)(19-S) &= 1\cdot 19+2\cdot 18+3\cdot 17+4\cdot 16+3\cdot 15+2\cdot 14+1\cdot 13\\ &= 19+36+51+64+45+28+13\\ &= 256.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/343855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $a \sin(\alpha) - c \sin^2(\alpha) = b \cos(\alpha) - c \cos^2(\alpha)$ $a, b, c$ are given positive integers. I need $\sin(\alpha)$ or $\cos$ or anything simple with $\alpha$ from the equation:
$$a \sin(\alpha) - c \sin^2(\alpha) = b \cos(\alpha) - c \cos^2(\alpha)$$
| With this much information, you can use $$\cos\alpha=\frac{1-\tan^2\frac\alpha2}{1+\tan^2\frac\alpha2}\text{ and } \sin\alpha=\frac{2\tan\frac\alpha2}{1+\tan^2\frac\alpha2} $$ which will give you a Quartic Equation in $\tan\frac\alpha2$
Once you have solved for $\tan\frac\alpha2,$ you can easily get $\cos\alpha,\sin\alpha$ using the above formulae.
Alternatively, we can also do the following:
$$b\cos\alpha=c\cos^2\alpha-c\sin^2\alpha+a\sin\alpha=c(1-\sin^2\alpha)-c\sin^2\alpha+a\sin\alpha$$
$$b\cos\alpha=c+a\sin\alpha-2c \sin^2\alpha$$
Squaring we get $$b^2(1-\sin^2\alpha)=(c+a\sin\alpha-2c\sin^2\alpha)^2 $$
On simplification we shall get a Quartic Equation in $s=\sin\alpha$
But unfortunately, the squaring has introduced extraneous roots which need exclusion.
| {
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"url": "https://math.stackexchange.com/questions/345575",
"timestamp": "2023-03-29T00:00:00",
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach:
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
| Using the fact that $\sin \theta+\cos \theta=\sqrt{2}\cdot\sin (\theta+\frac{\pi}{4})$, this problem is reduced to find an angle $\theta\in [0, 2\pi]$ such that
$$\sin (\theta+\frac{\pi}{4})=\frac{\sqrt{2}}{2}.$$
As we know, if $\sin\alpha=\sqrt{2}/2$, it must be the case that $\alpha=\frac{\pi}{4}+2k_1\pi$ or $\alpha=\frac{3\pi}{4}+2k_2\pi$, where $k_1$ and $k_2$ can be any integer. In this problem, since $\theta\in [0,2\pi]$, we have
$$\theta+\frac{\pi}{4}\in\bigg[\frac{\pi}{4}, \frac{9\pi}{4}\bigg].$$
Thus,the possible values of $k$'s are: $k_1=0$, $k_1=1$, $k_2=0$, which correspond to three angles in $[0, 2\pi]$: $\theta_1=\frac{\pi}{4}-\frac{\pi}{4}=0$, $\theta_2=\frac{9\pi}{4}-\frac{\pi}{4}=2\pi$, and $\theta_3=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $\lim \limits_{n \to \infty}(n^5+4n^3)^{1/5}-n$
$$\lim \limits_{n \to \infty}(n^5+4n^3)^{1/5}-n=?$$
I see that $$\lim_{n \to \infty}(n^5+4n^3)^{1/5}-n=\lim_{n \to \infty}n[(1+ \frac {4}{n^2})^{1/5}-1]=\lim_{z \to 0} \frac {1}{z}[(1+ {4}{z^2})^{1/5}-1]$$,where $n=\frac {1}{z}$. Now I do not know how to proceed.
Can someone point me in the right direction? Thanks in advance for your time.
| Two methods:
*
*Clever, but requiring to be clever: conjugate quantities.
Here, the identity $(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)=a^5-b^5$ with $a=(n^5+4n^3)^{1/5}$ and $b=n$ yields $a^5-b^5=4n^3$. Furthermore, $a\sim b\sim n$ hence each term in the sum $a^4+a^3b+a^2b^2+ab^3+b^4$ is equivalent to $n^4$. Can you finish?
*
*Not clever, but sure to win: limited expansions.
Here, $(1+x)^{1/5}=1+\frac15x+o(x)$ when $x\to0$ hence $(1+4/n^2)^{1/5}=1+O(1/n^2)$. Furthermore, $(n^5+4n^3)^{1/5}=n(1+4/n^2)^{1/5}$. Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Radius of convergence in a series. Ratio test. I am having a hard time with this question.
$$\sum_{k=0}^{\infty} \frac{-(1)^k (4^k -3)x^{2k}}{k^4+3}$$
I used the ratio test and got stuck here:
$$x^2 \lim_{k\to\infty} \frac {(4^{k+1}-3)(k^4+3)}{((k+1)^4+3)(4^k-3)}$$
Im not sure how to simplify it further.
| $$\lim_{k\to\infty} \frac {(4^{k+1}-3)(k^4+3)}{((k+1)^4+3)(4^k-3)}=\lim_{k\to\infty} \frac {(4^{k+1}-3)}{(4^k-3)}\cdot \lim_{k\to\infty} \frac {(k^4+3)}{((k+1)^4+3)}$$
as long as the two limits on the right hand side exist. But,
$$\lim_{k\to\infty} \frac {(4^{k+1}-3)}{(4^k-3)}=\lim_{k\to\infty} \frac {(4-3/4^k)}{(1-3/4^k)}=4$$
and
$$\lim_{k\to\infty} \frac {(k^4+3)}{((k+1)^4+3)}=\lim_{k\to\infty} \frac {(1+3/k^4)}{((1+1/k)^4+3/k^4)}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating functions Find
$$ \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{F_{2k}F_{n-k}}{10^n}, $$
where $F_n$ - $n-th$ Fibonacci number
Any hint how to start with this?
I have thought about convolution of two Fibonacci sequences, but there is $F_{2k}$, instead of $F_{k}$.
| Consider $$\left(\sum_{k=0}^{\infty} F_k x^k \right) \cdot \left(\sum_{k=0}^{\infty} F_{2k} x^k \right) = \sum_{k=0}^{\infty} C_n x^n$$
Now $C_n = \displaystyle \sum_{k=0}^n F_{2k} F_{n-k}$. We now have that
\begin{align}
\sum_{k=0}^{\infty} F_k x^k & = \dfrac{x}{1-x-x^2}\\
\sum_{k=0}^{\infty} F_{2k} x^k & = \dfrac1{\phi-\psi} \left(\sum_{k=0}^{\infty} \phi^{2k} x^k - \sum_{k=0}^{\infty} \psi^{2k} x^k\right) = \dfrac1{\phi - \psi}\left(\dfrac1{1-\phi^2x} - \dfrac1{1-\psi^2x}\right)\\
& = \dfrac{(\phi + \psi)x}{1-(\phi^2+\psi^2)x + \phi^2 \psi^2 x^2} = \dfrac{x}{1-3x+x^2}
\end{align}
Hence,
$$\sum_{k=0}^{\infty} C_n x^n = \dfrac{x}{1-x-x^2} \cdot \dfrac{x}{1-3x+x^2}$$
Now plug in $x=\dfrac1{10}$ and get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\lim_{n \to \infty} \cos \frac {\pi}{2^2}\cos \frac {\pi}{2^3}\cos \frac {\pi}{2^4}......\cos \frac {\pi}{2^n}=\frac {2}{\pi}$ I came across the following problem that says:
prove that $$\lim_{n \to \infty} \cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}=\dfrac {2}{\pi}$$.
My Attempt: Let $$P=\lim_{n \to \infty} [\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}] \implies \log P=\lim_{n \to \infty} \sum_{r=2}^{n}\log (\cos \dfrac {\pi}{2^r})$$.
Now,I am stuck and not sure which way to go. Can someone point me in the right direction? Thanks in advance for your time.
| Let
$$ g_n = \prod_{k=2}^n \cos \frac{\pi}{2^k}$$
Then
$$
g_n \sin \frac{\pi}{2^n} = \frac{1}{2} g_{n-1} \sin \frac{\pi}{2^{n-1}}
$$
Meaning that
$$
g_n \sin \frac{\pi}{2^n} = \frac{1}{2^{n-1}} g_1 = \frac{1}{2^{n-1}}
$$
Hence
$$
g_n = \frac{2}{\pi} \frac{\frac{\pi}{2^{n}}}{\sin \frac{\pi}{2^{n}}} \longrightarrow_{n \to \infty} \frac{2}{\pi}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int \sqrt{\frac{1+x}{x}}dx$ Integrating $\int\sqrt{\frac{1+x}{x}}dx$
Let, $x=\tan^{2}\theta$
$dx=2\tan\theta \sec^{2}\theta d\theta$
Integral = $\int \frac{\sec\theta}{\tan\theta}{2\tan\theta\sec^{2}\theta}d\theta$
Integral = $\int {2\sec^{3}\theta}d\theta$
| Use integration by parts to solve for $\int \sec^3 \theta d \theta$
$$u = \sec \theta$$
$$dv = \sec^2 \theta d \theta$$
$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta$$
$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta (\sec^2 \theta - 1)$$
$$\int \sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec^3 \theta d \theta + \int \sec \theta d \theta $$
$$ 2\int \sec^3 \theta d\theta = \sec \theta \tan \theta + \int \sec \theta d \theta$$
$$ \int \sec^3 \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln\left|sec \theta + \tan \theta \right| + C$$
$$ 2\int \sec^3 \theta d \theta = \sec \theta \tan \theta + \ln\left|sec \theta + \tan \theta \right| + C$$
I used parts of my answer in this question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Solving equations with 3 unknowns I have the following $3$ equations:
$$\begin{align} A_0 + A_1 + A_2 &= 2 \\ \frac{1}{2}A_1 + A_2 &= 2 \\ \frac{1}{4}A_1 + A_2 &= \frac{8}{3} \end{align}$$
What is the process of solving this? I'm pretty sure $A_0 = A_2$. I'm just confused how to get started. Suggestions?
| Simple way: solve for $A_1$ in terms of $A_2$ using the third equation, then solve for $A_2$ using the second. You now have values for $A_1$ and $A_2$. Then, solve for $A_0$.
Alternatively, set it up as a matrix-vector equation, and solve using linear algebra:
$$\begin{pmatrix} 1 & 1 & 1 \\ 0 & \frac12 & 1 \\ 0 & \frac14 & 1 \end{pmatrix}\begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ \frac{8}{3}\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Differentiation: What did I do wrong in this problem? $y=x\sqrt{1-x^2}-4\sin^{-1}x$ Differentiate $$y=x\sqrt{1-x^2}-4\sin^{-1}x$$
Here is my work for the problem: $$(x\sqrt{1-x^2})'-(4\sin^{-1}x)'$$
$$x(\frac{1}{2}(1-x^2)^{-1/2})'+x'\sqrt{1-x^2}-4(\sin^{-1}x)'$$
$$\frac{x}{2\sqrt{1-x^2}}+\sqrt{1-x^2}-4(\frac{1}{\sqrt{1-x^2}})(x')$$
$$\frac{x}{2\sqrt{1-x^2}}+\frac{2\sqrt{1-x^2}\sqrt{1-x^2}}{2\sqrt{1-x^2}}-\frac{8}{2\sqrt{1-x^2}}$$
$$\frac{x+2(1-x^2)-8}{2\sqrt{1-x^2}}$$
$$\frac{x+2-2x^2-8}{2\sqrt{1-x^2}}$$
$$\frac{-2x^2+x-6}{2\sqrt{1-x^2}}$$
$$-\frac{2x^2-x+6}{2\sqrt{1-x^2}}$$
However the correct answer was this $$-\frac{2x^2+3}{\sqrt{1-x^2}}$$
I have redone this problem three times and I cannot figure out what I have done wrong. Did I leave out a step while differentiating?
| First term:
\begin{align}
(x \sqrt{1-x^2})' &= x'(\sqrt{1-x^2}) + x(\sqrt{1-x^2})' \\
&= \sqrt{1-x^2} + \frac{x}{2 \sqrt{1-x^2}} (1-x^2)' \\
&= \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}} \\
&= \frac{1-2x^2}{\sqrt{1-x^2}}
\end{align}
Second term:
\begin{align}
(-4 \sin^{-1}{x})' &= -\frac{4}{\sqrt{1-x^2}}
\end{align}
Summing up, $$\frac{1-2x^2}{\sqrt{1-x^2}} - \frac{4}{\sqrt{1-x^2}} = -\frac{2x^2+3}{\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For every positive integer n, the number $a^{2^n}−1$ has at least n+1 distinct prime divisors Let $a>3$ be an odd integer. Prove that for every positive integer $n$, the number $a^{(2^{n})}-1$ has at least $n+1$ distinct prime divisors.
This problem smells very strongly of induction, but maybe a more complex version than I am trying.
What I have done so far:
Let $a=2k+3$ for positive integers $k$.
Proof by induction.
Base case:
If $n=1$, then $a^{(2^n)}-1$ must have at least one prime divisor (it's greater than 1).
So now assume for $n=k$, that is, $a^{(2^k)}-1$ has at least $k+1$ distinct prime divisors.
So let $a^{(2^k)}-1=p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1} \cdot m$ for some positive integer $n$ and for distinct primes $p_1, \;p_2, \;p_3, \; \cdots , \;p_{k+1}$.
Now consider $n=k+1$. If $n=k+1$ then the given equation becomes $(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1} \cdot m+1)^2-1=(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1})(p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1}+2)$. Will this necessarily have $k+2$ distinct prime divisors, because if so the induction is complete?
| Let $a=2k+3$ for positive integers $k$.
Proof by induction:
Base Case: If $n=1$, then $a^2-1=(a+1)(a-1)=4(k+1)(k+2)$. As $gcd(k+1,k+2)=1$ and both $k+1$ and $k+2$ are greater than $1$, $a^2-1$ has at least $2$ distinct prime divisors.
Now assume for $n=q$, that is, assume that $a^{2^q}-1$ has at least $q+1$ distinct prime divisors ($q$ is a positive integer).
So let $a^{2^q}-1=p_1 \cdot p_2 \cdot p_3 \cdots p_k \cdot p_{k+1} \cdot m$ for some positive integer $m$ and for distinct primes $p_1, \; p_2, \; p_3, \; \dots \; p_k, \; p_{k+1}$.
WLOG $p_1<p_2<p_3<p_4< \; \cdots \; p_k<p_{k+1}$.
Now, as $a^{2^q}$ is an odd square it is congruent to $1 \pmod{4}$. So $p_1=2$ and $m$ is even.
Consider $n+q+1$,
$a^{2^{q+1}}-1=a^{2^q+2^q}-1=(a^{2^q}-1)(a^{2^q}+1)=(2p_2p_3 \cdots p_{k+1} \cdot m)(2p_2p_3 \cdots p_{k+1} \cdot m +2)$ (from above)
$=4(p_2p_3 \cdots p_{k+1} \cdot m)(p_2p_3 \cdots p_{k+1} \cdot m+1)$ this is clearly divisible by the $k+1$ original primes. But as $m$ is even $(p_2p_3 \cdots p_{k+1} \cdot m+1)$ is odd and thus cannot be divisible by any of the $k+1$ original primes ($gcd(a^{2^q}-1,a^{2^q}+1)=2$). It is thus a new prime or divisible by a new prime.
Hence, $a^{2^{q+1}}-1$ is divisible by at least $q+2$ primes thus completing the induction.
Therefore, by the principle of induction, for odd positive integers $a>3$, $a^{2^n}-1$ is divisible by $n+1$ distinct primes for all positive integers $n$.
Note: the little hiccup is that you needed to establish that $m$ was even.
| {
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"url": "https://math.stackexchange.com/questions/356769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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.