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$\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove : $\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$ $\triangle ABC;AB=c;AC=b;BC=a$ such that $a\geq b\geq c$. Prove :
$$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}$$
I have tried that :
$a\geq b\geq c\Rightarrow \frac{a^{2}-b^{2}}{c}\geq 0;\frac{b^{2}-c^{2}}{a}\geq 0;\frac{3a^{2}}{b}\geq \frac{3ac}{b}$
$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2+2a^2}{b}\geq \frac{2ab-2bc+3ca}{b}\Leftrightarrow \frac{c^{2}-a^{2}}{b}+\frac{3a^{2}}{b}\geq 2(a-c)+\frac{3ac}{b}\Leftrightarrow \frac{(c-a)(c+a)}{b}\geq 2(a-c)\Leftrightarrow c+a\geq -2b$
!!??
| $abc*(LHS-RHS)=-bc^3+ac^3+2abc^2-3a^2c^2+b^3c-2a^2bc+2a^3c-ab^3+a^3b=ab(a^2-b^2)+bc(b^2-c^2)+ac(a-c)(2a-c) \ge 0$
it is trivial when$a=b=c$, the = is hold.$\implies LHS \ge RHS$
| {
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"url": "https://math.stackexchange.com/questions/597664",
"timestamp": "2023-03-29T00:00:00",
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Solution to a trigonometric system Find the solutions of the system
$$\sin a-\frac{\sqrt{3}}{3}\sin b=0$$
$$\frac{\tan 2a-2\tan a}{\tan 2b}\cdot\frac{\tan 2b-2\tan b}{\tan 2a} =1$$
How to work with them ? Thanks
| Let's take a look at the second equation :
$\frac{\tan 2a-2\tan a}{\tan 2b}\cdot\frac{\tan 2b-2\tan b}{\tan 2a} =1$
$\frac{\tan 2a-2\tan a}{\tan 2a}\cdot\frac{\tan 2b-2\tan b}{\tan 2b} =1$
Let's take a look at :
$1-\frac{2\tan b}{\tan 2b} \implies 1-\frac{2\frac{\sin b}{\cos b}}{\frac{2\sin b \cos b}{cos^2 b - \sin^2 b}}=\tan^2 b$
$\tan^2(a) \tan^2(b) = 1$
Then
$$3\sin^2 a =\sin^2 b$$
$$\tan^2(a) \tan^2(b) = 1$$
I think it's easier to see that
$\sin^2 a = \frac{1}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$ I have to determine the following:
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)$
$\lim\limits_{n \rightarrow \infty}(\sqrt{x^8+4}-x^4)=\lim\limits_{x \rightarrow \infty}(\sqrt{x^8(1+\frac{4}{x^8})}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4\sqrt{1+\frac{4}{x^8}}-x^4 = \lim\limits_{x \rightarrow \infty}(x^4(\sqrt{1+\frac{4}{x^8}}-1)= \infty$
Could somebody please check, if my solution is correct?
| This is indeterminate because $x^4\rightarrow \infty$, but $\sqrt{1+\frac{4}{x^8}}-1\rightarrow 0$. You can multiply by the conjugate
$$
\left(\sqrt{x^8+4}-x^4\right)\left(\frac{\sqrt{x^8+4}+x^4}{\sqrt{x^8+4}+x^4}\right)=\frac{x^8+4-x^8}{\sqrt{x^8+4}+x^4}=\frac{4}{\sqrt{x^8+4}+x^4}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove this is an equivalence relation. Define a relation of $\mathbb{Q} -$ {$0$} as follows:
$x$ ~ $y$ $\Leftrightarrow$ $\dfrac {x} {y} = 2^k $ for some $k \in \mathbb{Z}$
Prove this is an equivalence relation.
ATTEMPT:
Reflexive: For any $x\in\mathbb Z$, $\dfrac {x} {x}=1$ and $2^0 = 1$.
Symmetric: For any $x,y\in\mathbb Z$, if $\dfrac {x} {y}= 2^k$ for some $k\in\mathbb Z$, then $\dfrac {y} {x}= 2^k$ where $k\in\mathbb Z$.
Transitive: For any $x,y,z\in\mathbb Z$, if $\dfrac {x} {y}= 2^k$ and $\dfrac {x} {z}= 2^l$ for some $k,l\in Z$, then $\dfrac {x} {z}= 2^\frac{k} {l} - ^y$ where $\frac{k} {l}-y\in\mathbb Z$.
| As we're working with the rationals of form $\frac{x}{y}$ here, we'll need the restriction that $x,y \neq 0$, else we'd get $\frac{x}{0}$, etc. I will work with the assumption that $x,y \neq 0$ in all problems.
Reflexivity:
$x \sim x \Rightarrow \frac{x}x = 2^k$. Obviously, $\frac{x}x = 1 \forall x \in \mathbb{Z}$, so let $k = 0 \in \mathbb{Z}$.
Symmetry:
WTS if $x \sim y$ then $y \sim x$, i.e. $\frac{x}{y} = 2^k$ implies $\frac{y}{x} = 2^l$ for some $k,l \in \mathbb{Z}$. Note that $\frac{y}{x} = \frac{1}{x/y}$. So take $\frac{x}{y} = 2^k$. Then $\frac{1}{x/y} = \frac{1}{2^k}$. Note that $\frac{1}{2^k} = 2^{-k}$. Let $-k = l$, where $l \in \mathbb{Z}$ because of additive inverses. Then we have $\frac{y}{x} = 2^l$.
In this discussion, it is important to note that our fractions $\frac{1}{2^k}, \frac{1}{x/y}$, etc. are never equal to $0$ (because of nonzero numerators, and we have $\frac{x}{y}$ defined as nonzero, and $2^k \neq 0 \forall k$.), so they are in $\mathbb{Q} \setminus \{0\}$.
Transitivity:
Note that your approach is confused: either you made a typo, or you were trying to show that if $x \sim y$ and $x \sim z$, then you can draw some conclusion about $x\sim z$.
WTS: For any $x,y,z \neq 0 \in \mathbb{Z}$, if $x \sim y$ and $y \sim z$, then $x \sim z$. So WTS if $\frac{x}{y} = 2^k$ and $\frac{y}{z} = 2^l$, then $\frac{x}{z} = 2^m$ for some $k,l,m \in \mathbb{Z}$.
Take $\frac{y}{z} = 2^l$. Then $y = 2^l \cdot z$. Now take $\frac{x}{y} = 2^k$. By substitution, we get $\frac{x}{2^l \cdot z} = 2^k$. Now note $\frac{x}{2^l \cdot z} = \frac{x}{z} \cdot \frac{1}{2^l}$. So $\frac{x}{z} \cdot \frac{1}{2^l} = 2^k$. So $\frac{x}{z} = 2^l \cdot 2^k = 2^{l+k}$.
I.e. if $\frac{x}{y} = 2^k$ and $\frac{y}{z} = 2^l$, then $\frac{x}{z} = 2^{k+l}$. Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this $a^3+b^3+c^3-20(a+3)(b+3)(c+3)=2013$ equation integer solution if $a,b,c\in Z$,and $a\le b\le c$
and such
$$\begin{cases}
a+b+c=-3\\
a^3+b^3+c^3-20(a+3)(b+3)(c+3)=2013
\end{cases}$$
Find the value $3a+b+2c=?$
my try
$$a+b+c=-3\Longrightarrow (a+1)+(b+1)+(c+1)=0$$
so
$$(a+1)^3+(b+1)^3+(c+1)^3=3(a+1)(b+1)(c+1)$$
This problem is from a few month ago comption.
then I can't,Thank you
| Let's forget about $a\le b \le c$ for the moment.
$$a+b+c=-3$$
$$\implies (a+b+c)^3 =a^3+b^3+c^3+3(a+b)(b+c)(a+c)=-27$$
Let's substitute $a^3+b^3+c^3=x$, $(a+b)(b+c)(a+c)=y$.Then, we have that
$$x+20y=2013$$
$$x+3y=-27$$
$$\implies y=120$$
Let's sub $a+b=m,b+c=n,a+c=p$.Now we have
$$m+n+p=-6,$$
$$mnp=120$$
Since by an easy argument either all $m,n,p$ are even or only one of them is, and since $mnp=120$, they are all even, or one is a multiple of $8$. So let's sub again! $m=2r,n=2s,p=2t$.So
$$r+s+t=-3$$
$$rst=15$$
So we have $(r,s,t)=(-5,-1,3)$ and permutations, so a solution is $(a,b,c)=(-1,-9,7)$.
The other case
$$8r+s+t=-6$$
$$rst=15$$
So we have $(r,s,t)=(-1,-3,5)$ and $(a,b,c)=(-8,0,5)$.
Remark: I was running out of letters.
| {
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Finding mod of X^2+1 = 0 to have exactly 4 solutions Find a natural number $m$ that is product of 3 prime numbers, and that the equation $x^2+1 \equiv 0 \text { (mod m)}$ has exactly 4 solutions mod m.
| We sketch, mostly without proof, some standard theory. Let $m$ have prime power factorization
$$m=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}.$$
The congruence $x^2\equiv -1\pmod{m}$ is solvable if and only if the congruence $x^2\equiv -1\pmod{p_i^{a_i}}$ is solvable for all $i$.
Moreover, let $e_i$ be the number of solutions of the congruence $x^2\equiv -1\pmod{p_i^{a_i}}$. Then the number of solutions of $x^2\equiv -1\pmod{m}$ is $e_1e_2\cdots e_k$.
Briefly, this is because given solutions $c_i$ of $x^2\equiv -1\pmod{p_i^{a_i}}$, we can use the Chinese Remainder Theorem to find a solution $x$ of the system of congruences $x\equiv c_i \pmod{p_i^{a_i}}$, and this will satisfy $x^2\equiv -1\pmod{p_i^{a_i}}$.
So far we have not used any properties of $x^2+1$ other than it is a polynomial.
But now we need to examine the solvability of $x^2 \equiv -1\pmod{p^a}$.
The prime $2$ is special. The congruence $x^2\equiv -1\pmod{2}$ has one solution, while for $a\gt 1$, the congruence $x^2\equiv -1 \pmod{2^a}$ has no solutions.
We turn to the odd primes $p$. It is a standard theorem that the congruence $x^2\equiv -1 \pmod{p^a}$ has no solutions if $p$ is of the form $4q+3$, and has exactly two solutions if $p$ is of the form $4q+1$.
This is enough information to solve any problem of the same general type as the problem in the OP.
To be specific, if we have $3$ distinct primes $p_1,p_2,p_3$, and $m=p_1p_2p_3$, then the congruence $x^2\equiv -1\pmod{m}$ has $e_1e_2e_3$ solutions, where $e_i$ is the number of solutions of $x^2\equiv -1\pmod{p_i}$. Since $e_i=1$ if $p_i=2$, and $e_i=0$ or $2$ otherwise, if $e_1e_2e_3=4$ we must have one of the $p_i$ (say $p_1$) equal to $2$, and the other two must be of shape $4q+1$.
The first few possibilities are $m=(2)(5)(13)$, $m=(2)(5)(17)$, $(2)(5)(29)$, $(2)(5)(37)$, $(2)(5)(41)$, and $(20(13)(17)$.
For example, let us look at $m=(2)(5)(17)$. Solutions of $x^2\equiv -1$ modulo the primes $2, 5,17$ are respectively $1$, $\pm 2$, $\pm 4$. So we want to find $x$ which is odd and is congruent to $\pm 2\pmod{5}$, and to $\pm 4\pmod{17}$.
That gives us $4$ systems of congruences to solve, using the machinery of the CRT, or by inspection. Let us solve for example $x$ odd, congruent to $2$ mod $5$, and to $-4$ modulo $17$. A quick search yields $x=47$.
This shows that $-47$ solves the problem for $x$ odd, congruent to $-2$ modulo $5$, and to $4$ modulo $17$.
Only two more to go, really only one, since we can switch signs.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\lim_{x\to 0}\dfrac{\sqrt[m]{\cos x}-\sqrt[n]{\cos x}}{x^2}$ For two positive integers $m$ and $n$, compute
$$\lim_{x\to 0}\dfrac{\sqrt[m]{\cos x}-\sqrt[n]{\cos x}}{x^2}$$
Without loss of generality I consider $m>n$ and multiply the numerator with its conjugate. But what next? Cannot proceed further! Help please!
| $\displaystyle \begin{aligned}L &= \lim_{x \to 0}\frac{\sqrt[m]{\cos x} - \sqrt[n]{\cos x}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)} - \sqrt[n]{1 - 2\sin^{2}(x/2)}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)}}{x^{2}} - \frac{\sqrt[n]{1 - 2\sin^{2}(x/2)}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\sqrt[m]{1 - 2\sin^{2}(x/2)} - 1}{x^{2}} - \frac{\sqrt[n]{1 - 2\sin^{2}(x/2)} - 1}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{x^{2}} - \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{-2\sin^{2}(x/2)}\cdot\frac{-2\sin^{2}(x/2)}{x^{2}}\\
&\,\,\,\,\,\,\,\,\,\,\,- \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{-2\sin^{2}(x/2)}\cdot\frac{-2\sin^{2}(x/2)}{x^{2}}\\
&= \lim_{x \to 0}\frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/m} - 1^{1/m}}{(1 - 2\sin^{2}(x/2)) - 1}\cdot\frac{-2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&\,\,\,\,\,\,\,\,\,\,\,- \frac{\left(1 - 2\sin^{2}(x/2)\right)^{1/n} - 1^{1/n}}{(1 -2\sin^{2}(x/2)) - 1}\cdot\frac{-2\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x^{2}}\\
&= \frac{1}{m}\cdot 1^{(1 - m)/m}\cdot (-2)\cdot\frac{1}{4} - \frac{1}{n}\cdot 1^{(1 - n)/n}\cdot (-2)\cdot\frac{1}{4}\\
&= \frac{1}{2n} - \frac{1}{2m}\end{aligned}$
We have used standard limits $$\lim_{y \to a}\frac{y^{n} - a^{n}}{y - a} = na^{n - 1}$$ where $y = 1 - 2\sin^{2}(x/2), a = 1$ and $n$ is $1/m$ at one place and $1/n$ at another place. Also we make use of the $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form for $\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}$ I want to express $$\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}$$ in closed form. What comes to mind is the formula $$\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}$$ and also $$\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.$$ But neither of these gives the term $(z+n)^2+a^2$ that we want. Perhaps we can adjust somehow?
| You can use the residue theorem, based on the following formula (not going to prove here):
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_k \operatorname*{Res}_{\zeta=\zeta_k} [\pi \cot{(\pi \zeta)} f(\zeta)]$$
where $\zeta_k$ are the non-integer poles of $f$. Here,
$$f(\zeta) = \frac1{(z+\zeta)^2+a^2}$$
so that the poles $\zeta_{\pm} = -z\pm i a$. The sum is therefore
$$\frac{\pi \cot{\pi (z-i a)}}{i 2 a} - \frac{\pi \cot{\pi (z+i a)}}{i 2 a}= \frac{\pi}{a} \Im{[\cot{\pi(z-i a)}]}$$
Now,
$$\cot{\pi(z-i a)} = \frac{\cos{\pi z} \cosh{\pi a} + i \sin{\pi z} \sinh{\pi a}}{\sin{\pi z} \cosh{\pi a} - i \cos{\pi z} \sinh{\pi a}} $$
Therefore, then sum is
$$\sum_{n=-\infty}^{\infty} \frac1{(z+n)^2+a^2} = \frac{\pi}{2 a} \frac{\sinh{2 \pi a}}{\sin^2{\pi z} \cosh^2{\pi a} + \cos^2{\pi z} \sinh^2{\pi a}}$$
EDIT
This may be simplified even further to
$$\sum_{n=-\infty}^{\infty} \frac1{(z+n)^2+a^2} = \frac{\pi}{2 a} \frac{\sinh{2 \pi a}}{\sin^2{\pi z}+\sinh^2{\pi a}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that no $n,m, 0Prove or disprove the claim that there are integers $n,m, 0<n<m$ such that $m^2 +mn+n^2$
is a perfect square.
| For instance, let $m=5$ and $n=3$.
Remark: Here is a beginning to a general analysis. We want $x^2+xy+y^2$ to be a perfect square. It is easier to multiply by $4$, and solve $4(x^2+xy+y^2)=z^2$.
Complete the square. So we want $(2x+y)^2+3y^2=z^2$. More generally look for solutions of $u^2+3v^2=z^2$. To produce solutions, we take a reasonable $v^2$, and express $3v^2$ as a difference of squares.
Added: In a comment OP asked whether there are infinitely many (relatively prime) pairs $(m,n)$.
Let $n$ be odd. Note that
$$\left(\frac{3n^2-1}{2}\right)^2 + 3n^2=\left(\frac{3n^2+1}{2}\right)^2.$$
So if $2m+n=\frac{3n^2-1}{2}$, then the pair $(m,n)$ should do the job. That gives $m=\frac{3n^2-2n-1}{4}$. (It is easy to check that $3n^2-2n-1$ is divisible by $4$).
This by no means produces all pairs $(m,n)$ with the desired property.
| {
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"url": "https://math.stackexchange.com/questions/608355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you please explain me the solution?
| Well...
The range of $\sin^2$ is $[0,1]$. That is the first step.
Now, $\frac{x^2+y^2+1}{2x} \leq 1$ implies $\frac{x^2+y^2+1-2x}{2x} \leq 0$. Now, rearranging, $\frac{x^2-2x+1+y^2}{2x} = \frac{(x-1)^2+y^2}{2x} \leq 0$, which is true if and only if the numerator is zero or $x < 0$. Now $x>0$, since otherwise $\sin^2(x) < 0$ and this is outside the acceptable range. So, $(x-1)^2+y^2 = 0 \iff x=1,y=0$. Hence, $x=1$ as desired.
| {
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Find Minimum value of this expression: $P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc$ Let $a,b,c>0$ and satisfying $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le\frac{a+b+c}{2}$. Find Min of this expression?
$P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc$
From the condition, I have :$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}$
So, I get: $\frac{a+b+c}{2}\ge\frac{9}{a+b+c+3}\Leftrightarrow (a+b+c-3)(a+b+c+6)\ge 0\Leftrightarrow a+b+c\ge 3$
And then, I can't find the way to mutate P with one variable... Please guide me. Thanks
| first, we prove when $a+b+c=3$ ,$P$ will get min when $a=b=c=1$
Let $3u=a+b+c=3,3v^2=ab+bc+ac,w^3=abc, \to u=1 \ge v \ge w, w^3\ge3v^2-2-2\sqrt{(1-v^2)^3}$
$(1-v^2)\le 1 \implies \sqrt{(1-v^2)^3} \le (1-v^2) \implies w^3\ge3v^2-2-2(1-v^2) =5v^2-4$,
$P=54-42v^2+7w^3 \ge 26-7v^2\ge 19$ when $v=1=u \implies a=b=c=1, P_{min}=19 $
if $a'+b'+c'>3,a'+b'+c'=k(a+b+c) \implies k>1,a'=ka,b'=kb,c'=kc$
$P'=2(a'^3+b'^3+c'^3)+4(a'b'+b'c'+c'a')+a'b'c'=k^32(a^3+b^3+c^3)+k^24(ab+bc+ca)+k^3abc>P=2(a^3+b^3+c^3)+4(ab+bc+ca)+abc \ge 19$
QED.
| {
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Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} \right) > \dfrac{\sqrt{2}}{2}$
To find bounds: $\sin(x) = \dfrac{\sqrt{2}}{2},\ x = \dfrac{\pi}{4},\ \dfrac{3\pi}{4}$
Resolving to $x + \dfrac{\pi}{3} \implies \dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$
But wolfram gives a way different result, where's my mistake ?
| We need $$2\sin\left(x-60^\circ\right)-\sqrt2>0$$
But as $\sin45^\circ=\frac1{\sqrt2},$ it essentially implies and is implied by $$2\sin\left(x-60^\circ\right)-2\sin45^\circ>0$$
using Prosthaphaeresis Formulas,
$$\sin\left(x-60^\circ\right)-\sin45^\circ=2\sin\dfrac{x-105^\circ}2\cos\frac{x-15^\circ}2\ \ \ \ (1)$$
Now, $$\sin\frac{x-105^\circ}2>0\iff n360^\circ<\frac{x-105^\circ}2<n360^\circ+180^\circ$$
$$\iff n720^\circ+105^\circ<x<n720^\circ+(540+105)^\circ$$
Setting $n=0, 105^\circ<x<(540+105)^\circ$
and setting $n=-1,$
$(-720+105)^\circ<x<(-720+540+105)^\circ\iff -615^\circ<x<-75^\circ$
As we have $-180^\circ<x<180^\circ,$
$\displaystyle\sin\frac{x-105^\circ}2>0\iff-180^\circ<x<-75^\circ$ or $105^\circ<x<180^\circ\ \ \ \ (2)$
Again, $$\cos\frac{x-15^\circ}2>0\iff m360^\circ-90^\circ<\frac{x-15^\circ}2<m360^\circ+90^\circ$$
$$m720^\circ-165^\circ<x<m720^\circ+195^\circ$$
Set $n=0$ as $-180^\circ<x<180^\circ,$
$\displaystyle \cos\frac{x-15^\circ}2>0\iff -165^\circ<x<180^\circ \ \ \ \ (3)$
So, $(1)$ will be $>0$
if $105^\circ<x<180^\circ$(where both the multipliers in $(2)<(3)$ are positive )
or if $-180^\circ<x<-165^\circ$ (where both the multipliers are negative )
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility Mathematical Induction Help Could you please help me with this question prove that $\displaystyle5^n + 2\cdot(11)^n$ is a multiple of $3$.
Thanks
| Hint:
$$\begin{align*}5^{k+1}+2\cdot11^{k+1}&=5\cdot5^k+2\cdot11\cdot11^k\\
&=5\cdot5^k+2\cdot(\color{red}5+\color{blue}6)\cdot11^k\\
&=5\cdot5^k+\color{red}{2\cdot5\cdot11^k}+\color{blue}{2\cdot6\cdot11^k}\\
&=5(5^k+\color{red}{2\cdot11^k})+\color{blue}{2\cdot6\cdot11^k}\\
&=5(5^k+\color{red}{2\cdot11^k})+\color{blue}{3\cdot2\cdot2\cdot11^k}\\\end{align*}$$
| {
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Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}$. Are these the only 10 solutions?
First, none of $x$, $y$ or $z$ can be $1$ ($x$, $y$ and $z$ are positive integers)
If I let $x=2$, then finding all solutions to $1/y+1/z = 1/2$ leads to $(4,4), (3,6)$ and $(6,3)$ which gives me $(x,y,z)$ as $(2,4,4), (2,3,6), (2,6,3)$ but this also means $(4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)$ are all valid triples for this equation.
If I let $x=3$, the only different values of $y$ and $z$ are $(3,3)$
How do I prove these are the only ten solutions? (without using any programming)
Known result: If we denote $d(n^2)$ as the number of divisors of $n^2$, then the number of solutions of ${\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }$ = $d(n^2)$ (For positive $x$, $y$)
For ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$
$z = \frac{xy}{y(x-1)-x}$ where $xy \neq 0$
What happens after that?
Question is: how do we sho there are the only ten solutions? I'm not asking for a solution.
Assuming $x \le y \le z$
$1 \le y \le \frac{xy}{y(x-1)-x}$
$\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} $
Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later)
Liked @user44197 answer.
| HINT : You may suppose that $1\le x\le y\le z.$ This will make it easier to find the solutions.
| {
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Dealing with Generating Functions accurately I'm currently working through Iven Niven's "Mathematics of Choice." In the chapter on Generating Functions, the exercises include problems like:
How many solutions in non-negative integers does the equation $2x+3y+7z+9r=20$ have?
This of course ends up being the same as finding the coefficient of $x^{20}$ in $(1+x^2+x^4+x^6+\cdots+x^{20})(1+x^3+x^6+\cdots+x^{18})(1+x^7+x^{14})(1+x^9+x^{18})$. Maybe I'm just being a big weenie, but dealing with polynomials this large ends up being really tedious and error-prone.
I've realized that you can save the most involved multiplication for last, since that requires the least detail, and I'm of course not bothering to keep track of any powers greater than the one I'm interested in.
Any tips for keeping these giant polynomials manageable?
| Frankly, I always resort to a computer algebra system at this point, but here is how I think Niven probably expected the problem to be done.
There are some labor-saving tricks. First, it usually pays to start with the sparser polynomials (those with fewer non-zero terms). Second, you only have to work up to the highest degree needed for your answer-- $x^{20}$ in this case; higher powers can be discarded. Finally, it may not be necessary to perform the final multiplication. In detail, for this problem, let
$P_2 = 1 + x^2 + x^4 + x^6 + x^8 + x^{10} + x^{12} + x^{14} + x^{16} + x^{18} + x^{20}\\
P_3 = 1 + x^3 + x^6 + x^9 + x^{12} + x^{15} + x^{18} \\
P_7 = 1 + x^7 + x^{14} \\
P_9 = 1 + x^9 + x^{18} \\$
Then, discarding any terms higher than $x^{20}$, we have
$P_7 P_9 = 1 + x^7 + x^9+ x^{14} + x^{16} + x^{18} + \dots$
next
$P_3 P_7 P_9 = 1 + x^3 + x^6 + x^7 + 2x^9 + x^{10} + 2x^{12} + x^{13} + x^{14} + 2x^{15} + 2x^{16} +x^{17} + 3x^{18} + 2x^{19} + x^{20} + \dots$
Now it is not necessary to compute $P_2 P_3 P_7 P_9$, because we are only interested in the coefficient of $x^{20}$ in the final product. Since $P_2$ has only even powers of $x$, we can just read off the coefficients of $1, x^2, x^4, \dots , x^{20}$ in $P_3 P_7 P_9$ (the even-powered terms) and add them up:
$1 + 1 + 1 + 2 + 1 + 2 + 3 + 1 = 12$
So the final answer to the number of solutions is 12.
| {
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Properties of Inequalities I am unable to understand the following statements. If someone can explain with the help of numerical examples it would be great:
*
*If sides of an inequality are both positive or both negative, taking the reciprocal reverses the inequality
*If $0<x<1$ and $m$ and $n$ are integers with $m>n$, then $x^m<x^n<x$
*If $0<x<1$, then $\sqrt{x}>x$
*If $0<x<1$, then $\dfrac{1}{x}>x$ and $\dfrac{1}{x}>1$
| Below are some numerical examples of each of these inequalities.
*
*You know that $2<4$. Now, $0.5=\dfrac{1}{2} > \dfrac{1}{4}=0.25$. Similarly, $-2>-4$. now, $-0.5=\dfrac{-1}{2}<\dfrac{-1}{4}=0.25$.
*Consider $a=4$ and $x=0.5$. Everything meets the stipulations. Then $xa=0.5(4)=2<4=a$.
*Suppose $x=\dfrac{1}{2}$ and $m=3$ and $n=2$. Then we have that $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$. Furthermore, $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$. We know that $\dfrac{1}{8}<\dfrac{1}{4}$.
*Suppose that $x=\dfrac{1}{4}$. Then we have that $\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$. Furthermore, we know that $\dfrac{1}{2}>\dfrac{1}{4}$.
*Suppose that $x=\dfrac{1}{2}$. Then we know that $\dfrac{1}{\left(\frac{1}{2}\right)}=2$. Furthermore, $2>\dfrac{1}{2}$. Also, $\dfrac{2}>1$.
| {
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How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$
show that
$$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$
I know this Weitzenböck's_inequality
$$a^2+b^2+c^2\ge 4\sqrt{3}S$$
But my inequality is stronger than this Weitzenbock's inequality.
my try:
let the semiperimeter, inradius, and circumradius be $s,r,R$ respectively
$$a+b+c=2s,ab+bc+ac=s^2+4Rr+r^2,S=rs$$
$$\Longleftrightarrow (s^2+4Rr+r^2)4s^2\ge 12\sqrt{3}\cdot rs[4s^2-2(s^2+4Rr+r^2)]$$
$$\Longleftrightarrow s^3+4Rrs+r^2s\ge 6\sqrt{3}rs^2-24\sqrt{3}Rr^2-6\sqrt{3}r^3$$
and use this Gerretsen inequality:
$$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$
and Euler inequality
$$R\ge 2r$$
But seems is not usefull,
Thank you
| It follows immediately from known
$$2(ab+ac+bc)-a^2-b^2-c^2\geq4\sqrt3S$$
because it remains to prove that
$$\frac{(ab+ac+bc)(a+b+c)^2}{3(a^2+b^2+c^2)}\geq\sum_{cyc}(2ab-a^2).$$
Let $a^2+b^2+c^2=k(ab+ac+bc)$.
Hence, $k\geq1$ and we need to prove that
$$\frac{k+2}{3k}\geq2-k$$ or
$$(k-1)(3k-2)\geq0,$$
which is obvious.
Done!
| {
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proving the inequality $\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$ If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$
and also show that equality hold if $a=b=c$.
$\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$
Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$
So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)>0$
$$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$
So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$
But I did not understand how can I prove above inequality
help Required
Thanks
| It helps to write down what you want to prove:
$$(b+c-a)(c+a-b)(a+b-c) \le abc$$
You viewed the left-hand expression as a geometric mean, but replacing it by a arithmetic mean does not work because the resulting inequality is an AM-GM inequality in the wrong direction, so you need to try something else.
There are many ways to proceed, but a good routine first step is a substitution that replaces $a, b, c$ who are linked by triangle inequalities with $2x=b+c-a$, $2y=c+a-b$, $2z=a+b-c$ who are simply positive numbers.
The desired inequality becomes:
$$8xyz \le (y+z)(x+z)(x+y).$$
But this is simply a product of three AM-GM inequalities:
$$\sqrt{yz}\sqrt{xz}\sqrt{xy}\le \frac{y+z}2\frac{x+z}2\frac{x+y}2.$$
So everything is proved.
Alternatively, if you do not see it, you can simply expand the product and apply one AM-GM inequality to the 8 terms.
| {
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if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$.
$\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \mathbb{R}$.Then $\displaystyle kx^2+2kx+3k=x^2+ax+b$
$\Rightarrow (k-1)x^2+(2k-a)x+(3k-b) = 0$
Now we will form $2$ cases::
$\bf{\bullet}$ If $(k-1)=0\Rightarrow k=1$, Then equation is $(2-a)x+(3-b)=0$
$\bf{\bullet}$ If $(k-1)\neq 0\Rightarrow k\neq 1$ means either $k>1$ or $k<1$
How can i solve after that
Help Required
Thanks
| We need $$(2k-a)^2-4(k-1)(3k-b)=-8k^2+(b+3-4a)k+a^2-4b\ge0$$
$$\iff 8k^2-(b+3-4a)k-(a^2-4b)\le0$$
Now, we know $\displaystyle (x-a)(x-b)\le0\iff a\le x\le b$ where $a\le b$
So, here $$x^2-(-5+4)x+(-5)4=0\iff x^2+x-20=0\ \ \ \ (1)$$ will be same as $$8k^2-(b+3-4a)k-(a^2-4b)=0\ \ \ \ (2)$$
Now use Find $k, m$ so solutions to $(iz+k)^2=-2+2\sqrt3i$ are the same as those to $z^2-2iz+m=0$ to find $a,b$
| {
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$\alpha,\beta,\gamma$ are roots of cubic equation $x^3+4x-1=0$ If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+4x-1=0$ and $\displaystyle \frac{1}{\alpha+1},\frac{1}{\beta+1},\frac{1}{\gamma+1}$ are the roots of the equation
$\displaystyle 6x^3-7x^2+3x-1=0$. Then value of $\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2}=$
$\bf{My\; Try}::$ Given $x=\alpha,\beta,\gamma$ are the roots of the equation $x^3+4x-1=0$, Then
$x^3+4x-1 = (x-\alpha)\cdot(x-\beta)\cdot (x-\gamma)$
put $x=-1$, we get $(1+\alpha)(1+\beta)(1+\gamma) = 6$
Now $\displaystyle \frac{(\beta+1)(\gamma+1)}{\alpha^2}+\frac{(\gamma+1)(\alpha+1)}{\beta^2}+\frac{(\alpha+1)(\beta+1)}{\gamma^2} = 6\left\{\frac{1}{(\alpha+1)\alpha^2}+\frac{1}{(\beta+1)\beta^2}+\frac{1}{(\gamma+1)\gamma^2}\right\}$
Now how can I solve after that,
Help required
thanks
| From the second equation, $$\frac6{(\alpha+1)^3}-\frac7{(\alpha+1)^2}+\frac3{\alpha+1}-1=0$$
$$\implies 6-7(\alpha+1)+3(\alpha+1)^2-(\alpha+1)^3=0\iff \alpha^3+4\alpha-1=0$$
So, both equations are actually equivalent
$\displaystyle y=\frac{(\beta+1)(\gamma+1)}{\alpha^2}=\frac{\beta\gamma+\beta+\gamma+1}{\alpha^2}=\frac{\alpha\beta\gamma+\alpha(\beta+\gamma)+\alpha}{\alpha^3}=\frac{1-\alpha^2+\alpha}{1-4\alpha} $ as $\alpha^3=1-4\alpha$
$\displaystyle\implies \alpha^2-\alpha(1+4y)+y-1=0\ \ \ \ (1)$
$\displaystyle y=\frac6{\alpha^2(\alpha+1)}=\frac6{(1-4\alpha)+\alpha^2}$
$\displaystyle\implies \alpha^2y-4y\alpha+y-6=0\ \ \ \ (2) $
Solve $(1),(2)$ for $\alpha^2,\alpha$
and use $\alpha^2=(\alpha)^2$ to eliminate $\alpha$ and form a cubic equation in $y$ whose roots are $\displaystyle y_1,y_2,y_3$(say)
Now use Vieta's formulas to find the required $\displaystyle y_1+y_2+y_3$
We can also find $\displaystyle y_1y_2+y_2y_3+y_3y_1$ from here if need be
| {
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If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$ Problem :
If real number x and y satisfy $(x+5)^2 +(y-12)^2=14^2$ then find the minimum value of $x^2 +y^2$
Please suggest how to proceed on this question... I got this problem from [1]: http://www.mathstudy.in/
| $$(x+5)^2 +(y-12)^2=14^2$$
$$x^2+y^2-27=24y-10x$$
Now, by Cauchy Schwarz
$$(24y-10x)^2 \leq (24^2+10^2)(x^2+y^2)$$
with equality if and only if $\frac{y}{24}=\frac{x}{-10}$.
Thus
$$(x^2+y^2-27)^2 \leq (24^2+10^2)(x^2+y^2)$$
Let $t := x^2+y^2$. Then
$$(t-27)^2 \leq 676t$$
$$t^2-54t+729 \leq 676 t$$
$$t^2-730t+729 \leq 0$$
or
$$(t-1)(t-729) \leq 0$$
Hence
$$1 \leq t \leq 729$$
This proves that
$$1 \leq x^2+y^2 \leq 729 $$
Note that to get $x^2+y^2=1$ or $x^2+y^2=729$ we must have
$$ \frac{y}{24}=\frac{x}{-10} \,.$$
| {
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How to prove this equation (given two inequalities)? How to prove that:
$$\frac{au+bv}{a+b} < y$$
given that:
$$u<y, v < y$$
Here, a, b are positive integers and u, v and y are real numbers between 0 and 1 (inclusive).
| Note that
$\dfrac{au + bv}{a + b} = \dfrac{a}{a + b} u + \dfrac{b}{a + b} v \tag{1}$
and
$\dfrac{a}{a + b} + \dfrac{b}{a + b} = 1; \tag{2}$
now if $u = v$, (1) and (2) show that
$\dfrac{au + bv}{a + b} = v < y \tag{3}$
by hypothesis; otherwise we can assume $u < v$ in which case (1) and (2) yield
$\dfrac{au + bv}{a + b} = \dfrac{a}{a + b} u + \dfrac{b}{a + b} v < \dfrac{a}{a + b} v + \dfrac{b}{a + b} v = v < y. \tag{4}$
QED.
Points worth of note: the restriction $0 < u, v, y < 1$ is not necessary for the result, as long as $u, v < y$; also the expression (1) shows that $(au + bv) / (a + b)$ is in fact a weighted average of $u, v$ with coefficients $0 < a/a + b, b / a + b < 1$; the assumptions on $a, b$ easily show this to be the case. Indeed, taking $u < v$, $(au + bv) / (a + b)$ is precisely the real number which is $b / a+ b$ of the way between $u$ and $v$, going in the positive direction, since
$u + \dfrac{b}{a + b}(v - u) = (1 - \dfrac{b}{a + b}) u + \dfrac{b}{a+ b} v = \dfrac{au + bv}{a + b}, \tag{5}$
which provides another way of seeing $(au + bv) / (a + b) < y$.
Hope this helps. Happy New Year to One and All,
and as always,
Fiat Lux!!!
| {
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Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
| Note that the expansion of $(\sqrt{x - 5} - 1)^2 = (\sqrt{x - 5})^2 - {\bf 2}\sqrt {x - 5} + 1$.
But that correction doesn't change the fact that indeed, no solution exists.
| {
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Solving an irrational equation Solve for $x$ in:
$$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=\sqrt{5}$$
I used the property of proportions ($a=\sqrt{3+x}$, $b=\sqrt{3-x})$:
$$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac{a}{b}$$
I'm not sure if that's correct.
Or maybe the notations $a^3=3+x$, $b^3=3-x$ ? I don't know how to continue. Thank you.
| $$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot 1=\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\cdot\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}+\sqrt{3-x}} =\sqrt{5}$$
$$\frac{(\sqrt{3+x}+\sqrt{3-x})^2}{\sqrt{3+x}^2-\sqrt{3-x}^2} =\sqrt{5}$$
$$\frac{3+x+3-x+2\sqrt{9-x^2}}{{3+x}-3+x} =\sqrt{5}$$
$$\frac{2(3+\sqrt{9-x^2})}{2x} =\sqrt{5}$$
$$\frac{3+\sqrt{9-x^2}}{x} =\frac{\sqrt{5}}{1}$$
$$3+\sqrt{9-x^2}=x\sqrt{5}$$
$$\sqrt{9-x^2}=x\sqrt{5}-3/^2$$
$$9-x^2=5x^2-6x\sqrt 5+9 $$
$$6x^2-6x\sqrt 5=0$$
$$6x(x-\sqrt 5)=0$$
$$x_1=0, x_2=\sqrt 5$$
But $x=0$ is not a rots of the given equation, becouese: $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}\neq\sqrt 5$
| {
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"url": "https://math.stackexchange.com/questions/628314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three Variables Inequality $a$,$b$,$c$ are non-negative numbers such that $(a+b)(b+c)(c+a)=1$.
show that $\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\geq\sqrt[3]{2}\sqrt{ab+bc+ca}$
I can't guess where the constant $\sqrt[3]{2}$ come from.
| let $ x=b+c,y=a+c,z=a+b,\implies xyz=1,a = min${$a,b,c$}$,a=\dfrac{y+z-x}{2} \ge 0 \implies x \le y+z$
$1=xyz \le x\left(\dfrac{y+z}{2}\right)^2 \le \dfrac{(y+z)^3}{4} \implies y+z \ge \sqrt[3]{4} \implies 2a+b+c \ge \sqrt[3]{4} $
RHS$=\sqrt{\sqrt[3]{4}(ab+bc+ca)} \le \sqrt{(2a+b+c)(ab+bc+ca)}=\sqrt{bc(b+c)+(2a^2b+2a^2c+4abc+ab^2+ac^2)} $
$\sqrt{ab(a+b)}+\sqrt{ca(c+a)}\ge \sqrt{2(ab(a+b)+ca(c+a))} \iff \sqrt{2(ab(a+b)+ca(c+a))} +\sqrt{bc(b+c)} \ge \sqrt{bc(b+c)+(2a^2b+2a^2c+4abc+ab^2+ac^2)}\iff ab^2+ac^2 +2\sqrt{2(ab(a+b)+ca(c+a))bc(b+c)} \ge 4abc$
$ab^2+ac^2 \ge 2abc \iff \sqrt{2(ab(a+b)+ca(c+a))bc(b+c)} \ge abc \iff abc(2b^3+2ac^3+2ab^2+2ac^2+3abc) \ge 0$
all above "=" hold when $a=0,b=c$ must be hold both or either. so the inequality will be hold "=" when $a=0$ and $b=c$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n})$ , $n \ge 1$ convergent? $$x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n}) ,\quad n \ge 1$$
My work is:
First i analyzed the convergence of this sequence in 2 parts:
a) $\lim \sqrt{n+1}− \sqrt{n} = \lim \frac{(\sqrt{n+1}− \sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{n+1-n}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{1}{ \infty} = 0$ when $n \rightarrow \infty$.
But now how can I conclude that $\lim_{n \rightarrow \infty}x_n = 0$
| A basic trick you should learn is that:
$$\sqrt{n+1}- \sqrt{n} = \frac{(\sqrt{n+1}- \sqrt{n}) \cdot (\sqrt{n+1}+ \sqrt{n})}{1 \cdot (\sqrt{n+1}+ \sqrt{n})} = \frac{1}{\sqrt{n+1} + \sqrt{n}} $$
Using that yields:
$$x_n=\frac{\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac{\sqrt n}{\sqrt n\left(\sqrt{1+\frac{1}{n}}+1\right)} = \frac{1}{\sqrt{1+\frac{1}{n}} + 1}$$
Hence:
$$\lim_{x\to\infty} x_n = \frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)
n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...).
For each n element N with n>=n0 we have:
\begin{align}
0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\
&=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\
&=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon
\end{align}
(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)
From this we get
$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$
, thus the proof is complete.
Thanks for the clarifications!
| $\sqrt{n^4+n^2+20n+7}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+20n+7}}\leq \dfrac{1}{\sqrt{n^4}}$
$\sqrt{n^4+n^2+1}\geq \sqrt{n^4}\Rightarrow \dfrac{1}{\sqrt{n^4+n^2+1}}\leq \dfrac{1}{\sqrt{n^4}}$
your $n$ is choosen such that $n\geq 3$ so : $6\leq 2n$
That choosen $n_1>11$ is not random and it is not choosen before solving that out...
After solving everything i got $\dfrac{11}{n}$ but then I want that $\dfrac{11}{n}$ to be so small so if i choose $n>11$ then I am done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$ Title says all. If $f$ is an analytic function on the real line, and $f\left(\dfrac{1}{x}\right)+f(x+1)=x$, what, if any, is a possible solution for $f(x)$?
Additionally, what are any solutions for $f\left(\dfrac{1}{x}\right)-f(x+1)=x$?
| $$f(x)+f\left(\frac{x+1}{x}\right)=\frac1x\\
f\left(\frac{x+1}{x}\right)+f\left(\frac{2x+1}{x+1}\right)-\frac1\phi=\frac{x}{x+1}-\frac1\phi\\
f\left(\frac{2x+1}{x+1}\right)+f\left(\frac{3x+2}{2x+1}\right)-\frac1\phi=\frac{x+1}{2x+1}-\frac1\phi$$
If $f$ is continuous at $\phi$, then
$$f(x)=(\frac1x-\frac1{2\phi})-(\frac{x}{x+1}-\frac1{\phi})+(\frac{x+1}{2x+1}-\frac1{\phi})-...\\
=(\frac1x-\frac1{2\phi})-(1-\frac1{\phi}-\frac1{x+1})+(\frac12+\frac1{2(2x+1)}-\frac1{\phi})-...\\
=C+\frac1x+\frac1{x+1}+\frac1{2(2x+1)}+\frac1{3(3x+2)}+\frac1{5(5x+3)}+...
$$
for $C-\frac1{2\phi}-\frac1{1\times2}-\frac1{3\times5}-\frac1{8\times13}-...$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\,\,\frac{x^2}{cy+bz}=\frac{y^2}{az+cx}=\frac{z^2}{bx+ay}=1,$ then show that .... I am stuck on the following problem that one of my friends gave me:
If $\,\,\frac{x^2}{cy+bz}=\frac{y^2}{az+cx}=\frac{z^2}{bx+ay}=1,$ then show that $$\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}=1$$. I did a problem which was similar to this one but I could not tackle this particular one. Can someone help?
| It's wrong. Put $x=y=1, z=-1$. Then from the first condition we get the system $$ -a+c=1, \\b+a=1,\\ -b+c=1. $$ It follows $ a= \frac 12,b= \frac 12,c= \frac 32. $
Then
$$
{\frac {a}{a+x}}+{\frac {b}{b+y}}+{\frac {c}{c+z}}={\frac {a}{a+1}}+{\frac {b}{b+1}}+{\frac {c}{c-1}}=\frac{11}{3}\neq 1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $12x\equiv9\pmod{15}$ Question:
Solve $12x\equiv9\pmod{15}$
My try:
$\gcd(12,15)=3$ so it has at least $3$ solutions.
Now
$15=12\times1+3\\
3=15-12\times 1\\
3=15+2\times(-1)\\
\implies9=15\times3+12\times(-3)\\
\implies12\times(-3)\equiv9\pmod{15}$
So $x\equiv-3\pmod{15}$
Am I correct?
| Lemma: Let $gcd(a,n)=d$ and suppose that $d|b$. Then the linear congruence
$ax \equiv b $ (mod $n$)
has exactly $d$ solutions modulo $n$. These are given by
$t,t+\dfrac{n}{d},t+\dfrac{2n}{d},\cdots,t+\dfrac{(d-1)n}{d}$
where $t$ is the solution, unique modulo $n/d$, of the linear congruence
$\dfrac{a}{d}x\equiv \dfrac{b}{d}$ (mod $\dfrac{n}{d}$).
By Lemma, we just need to consider the linear congruence $4x\equiv 3$ (mod $5$), easy to see the unique solution is $2$. Hence the all solutions of $12x\equiv 9$ (mod $15$) is $x\equiv 2,7,12$(mod $15$).
| {
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Functional Equation $ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) $ I have the following functional equation:
$$ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) $$
Under the precondition that $ f ( n ) = \omega ( 1 ) $, monotonicity and an initial value $ f ( 1 ) = \Theta ( 1 ) $, one can show (by induction) that $ \Omega \big( ( \log n ) ^ k \big) = f ( n ) = O ( n ^ \alpha ) $ for arbitrary large $ k $ and small $ \alpha $ (see below).
It seems to me that these bounds are very tight so the question is:
Is there a (simple) function that satisfies these conditions?
Proof sketch:
$$ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) \le 2 \left( \frac n { f ( n ) } \right) ^ \alpha = \left( \frac 2 { f ( n ) ^ \alpha } \right) n ^ \alpha \le n ^ \alpha $$
for sufficiently large $ n $.
$$ f ( n ) = 2 f\left( \frac n { f ( n ) } \right) \ge 2 \left( \log \frac n { f ( n ) } \right) ^ k \ge 2 \left( \log \frac n { n ^ \alpha } \right) ^ k = 2 ( 1 - \alpha ) ^ k ( \log n ) ^ k \ge ( \log n ) ^ k $$
for sufficiently large $ n $ and small $ \alpha $.
| First of all, note that for a nonnegative constant $ c $, the function $ f $ defined by
$$ f ( n ) = 2 ^ { - \frac 1 2 + \sqrt { 2 \log _ 2 n + c } } $$
satisfies the functional equation
$$ f ( n ) = 2 f \left( \frac n { f ( n ) } \right) \text . $$
Also, $ f $ is strictly positive and strictly increasing, and as $ n \to + \infty $, $ f ( n ) \in o ( n ^ \alpha ) \cap \omega \big( ( \log _ 2 n ) ^ k \big) $ for any $ k $ and any positive $ \alpha $.
But how can one find such solutions? For more simplicity in the upcoming calculations, first define the function $ g $ with $ g ( n ) = \log _ 2 f ( 2 ^ n ) $, so that we have $ f ( n ) = 2 ^ { g ( \log _ 2 n ) } $. Then, the functional equation is simplified to
$$ g ( n ) = g \big( n - g ( n ) \big) + 1 \text . $$
Assuming we have $ g ( n ) = m $ for some fixed $ n $, using induction we get
$$ g \left( n - m k + \frac { k ( k - 1 ) } 2 \right) = m - k $$
for any nonnegative integer $ k $. Assuming that $ f $, and thus $ g $, is well-behaved enough, we can consider the above equation for all nonnegative real values of $ k $. By letting $ y = m - k $ and $ x = n - m k + \frac { k ( k - 1 ) } 2 $, we can see that
$$ m - y = k = m + \frac 1 2 \pm \sqrt { \left( m + \frac 1 2 \right) ^ 2 - 2 ( n - x ) } \text , $$
which for $ y > 0 $ gives
$$ g ( x ) = y = - \frac 1 2 + \sqrt { 2 x + c } \text , $$
where $ c = \left( m + \frac 1 2 \right) ^ 2 - 2 n $. This holds for any $ x \le n $, as we had the previous equations for any $ k \ge 0 $. Thus, we can see that $ \left( g ( x ) + \frac 1 2 \right) ^ 2 - 2 x $ is in fact constant on the whole domain (the constant must be nonnegative if we want the square root above to be well-defined for all positive values of $ x $). Therefore, the above formula for $ g ( x ) $ holds for any desired input, and writing it in terms of $ f $ gives the claimed result for all $ n \ge 1 $.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing $\lim_{n\to \infty}{\frac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}}$ Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.
I.) In the first step I studied monotony:
$a_{n+1}-a_{n}=\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)}{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)}\cdot\cfrac{-2}{4n+7}<0$, $\{a_n\}$ is decreasing.
II.) In the second step I studied boundary.
$$1>a_{1}=\cfrac{5}{7}>a_{2}=\cfrac{45}{77}>\dots>a_{n}>0$$
III.) In the last step I know that $\{a_n\}$ converges to $a\in\mathbb R$.
$$a_{n+1}=a_{n}\cdot\cfrac{4n+1}{4n+3}$$
Taking the limit as $n\to\infty$:
$$a=a$$
No conclusion.
But if I apply Cesaro-Stolz?
IV.) Let $\{x_n\}_{n\ge1}^{\infty}=\{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)\}$ and $\{y_n\}_{n\ge1}^{\infty}=\{7\cdot11\cdot15\cdot\dots.\cdot(4n+3)\}$. Then
$$\lim_{n\to \infty}{\cfrac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}}=\lim_{n\to \infty}{\cfrac{5\cdot9\cdot13\cdot\dots.\cdot(4n+1)^2}{7\cdot11\cdot15\cdot\dots.\cdot(4n+5)}=?}}$$
If you have a simple solution, I would appreciate it. Thank you!
| Using the relation $x\Gamma(x)=\Gamma(x+1)$, we get
$$
\frac{\Gamma\left(\frac54\right)}{\Gamma\left(\frac74\right)}\frac{\frac54\frac94\frac{13}4\cdots\frac{4n+1}4}{\frac74\frac{11}4\frac{15}4\cdots\frac{4n+3}4}
=\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}
$$
Therefore,
$$
\frac{5\cdot9\cdot13\cdots(4n+1)}{7\cdot11\cdot15\cdots(4n+3)}
=\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\frac{\Gamma\left(\frac74\right)}{\Gamma\left(\frac54\right)}
$$
Using Gautschi's Inequality, we get
$$
\left(\frac{4n+7}4\right)^{-1/2}\le\frac{\Gamma\left(\frac{4n+5}4\right)}{\Gamma\left(\frac{4n+7}4\right)}\le\left(\frac{4n+3}4\right)^{-1/2}
$$
By the Squeeze Theorem, we get that the limit is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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(Highschool Pre-calculus) Solving quadratic via completing the square I'm trying to solve the following equation by completing the square:
$x^2 - 6x = 16$
The correct answer is -6,1. This is my attempt:
$x^2 - 6x = 16$
$(x - 3)^2 = 16$
$(x - 3)^2 = 25$
$\sqrt(x -3)^2 = \sqrt(25)$
$x - 3 = \pm5$
$x =\pm5 - 3$
$x = -8,2$
I did everything according to what I know,but my answer was obviously wrong. Any help is appreciated. Thanks
| no ,no,what do you need is following
we have
$x^2-6*x=16$
$x^2-6*x+9=16+9$
$(x-3)^2=25$
now $x-3=5$ or $x=8$
and $x-3=-5$ or $x=-2$
there is no mistake,why is answer $-6$?
$(-6)^2-6*(-6)=36+36=72$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $n,k\in\mathbb N$, solve $2^8+2^{11}+2^n=k^2$. If $n,k\in\mathbb N$, solve $$2^8+2^{11}+2^n=k^2$$
It's hard for me to find an idea. Some help would be great. Thanks.
| First assume $n\ge 8$ and write $n=m+8$, we get:
$$(2^4)^2(1+2^3+2^m)=k^2$$
This is possible only if $1+2^3+2^m=9+2^m$ is a perfect square. If it is, then we can write it as $(3+p)^2$ (since it is obviously greater than $3$), and thus:
$$9+6p+p^2=9+2^m$$
$$\Leftrightarrow p(6+p)=2^m$$
In particular, $p=2^r$ and $6+p=2^{m-r}$ for some $1\le r\le m$ (notice that $r=0$ is impossible). From this we get $6+p=6+2^r=2(3+2^{r-1})=2^{m-r}$, and the only solution to this equation is $r=1$, $m=4$. Thus the only solution for $n\ge8$ is $n=12$, which gives us:
$$2^8+2^{11}+2^{12}=80^2$$
Now for $0\le n<8$, we have:
$$2^n(2^{8-n}+2^{11-n}+1)=k^2$$
Since the number in the brackets cannot be a multiple of $2$, we observe that $n$ must be even. This leaves us with the condition that $2^{8-n}+2^{11-n}+1$ is a square. That number is exactly:
$$2^{8-n}3^2+1$$
For $n=6$ we have $2^23^2+1=37$ which is not a square.
For $n=4$ we have $2^43^2+1=145$ which is not a square.
For $n=2$ we have $2^63^2+1=577$ which is again not a square.
Finally for $n=0$ we have $2^83^2+1=1153$ which is not a square either.
Thus the only solution is $n=12$.
| {
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How do I use the Euclidean Algorithm in the ring $\mathbb Z[\sqrt{3}]$? I was asked to find the GCD of $7+\sqrt{3}$ and $6-2\sqrt{3}$ in the ring $\mathbb Z[\sqrt{3}]$, but have no idea where to start. Any tips would be appreciated!
| I didn't use the Euclidean algorithm but here's my approach:
$6-2\sqrt{3}=2(3-\sqrt{3})=2\sqrt{3}(\sqrt{3}-1)$, $ 2=-(1-\sqrt{3})(1+\sqrt{3})$
Therefore: $$6-2\sqrt{3}=(1-\sqrt{3})(1+\sqrt{3})\sqrt{3}(1-\sqrt{3})$$
On the other hand: $$7+\sqrt{3}=7+3\sqrt{3}-2\sqrt{3}=(1+\sqrt{3})(1+2\sqrt{3})+(1-\sqrt{3})(1+\sqrt{3})\sqrt{3}=(1+\sqrt{3})(-2+3\sqrt{3})$$
You can check that the elemnts in both factorizations are irreducible and since our integral domain is a principal ideal domain, the elements are prime as well. So the GCD is $1+\sqrt{3}$. Can anyone confirm my result?
| {
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Is the Series : $\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + ..........+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $ convergent?
Is the Series :
$$\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$ convergent?
Attempt:
$$\sum_{n=0}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta $$
$$= \sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) (1- \cos( 2n\theta)) \frac 1 2$$
$$=\frac 1 2 \left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)\right]- \frac 1 2\left[\sum_{n=1}^{\infty} \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \cos(2n\theta)\right] $$
The left part by sandwich theorem has limiting value to $0$
The right part by Dirichlets theorem is convergent as $ \left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right)$ is a positive, monotonically decreasing sequence and the sequence of partial sum of $\sum_{n=1}^{\infty} \cos 2n\theta$ is bounded.
Hene, the given series is convergent. Am i correct?
| if $\theta=k\pi$, the series is convergent.
If $\theta$ is not a multiple of $\pi$,
$$\left( \frac 1 {(n+1)^2} + \ldots+\frac 1 {(n+n)^2} \right) \sin^2 n\theta \geq \frac n{(n+n)^2} \sin^2 n\theta =\frac {\sin^2 n\theta}{4n} =\frac1{8n}- \frac{\cos 2n\theta}{8n}$$
$\sum \frac{\cos 2n\theta}{8n} $ is convergent, hence, the series is divergent
| {
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"timestamp": "2023-03-29T00:00:00",
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An identity involving the Pochhammer symbol I need help proving the following identity:
$$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$
Here,
$$(a)_n = a(a + 1)(a + 2) \cdots (a + n - 1), \quad n > 1, \quad (a)_0 = 1,$$
is the Pochhammer symbol. I do not really know how one converts expressions involving factorials to products of the Pochhammer symbols. Is there a general procedure? Any help would be appreciated.
| By using the formula
\begin{align}
(a)_{kn} = k^{kn} \prod_{r=0}^{n-1} \left( \frac{a+r}{k} \right)_{n}
\end{align}
it is evident that the desired quantity,
\begin{align}
(1728)^{n} \left( \frac{1}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{5}{6} \right)_{n},
\end{align}
can be seen as
\begin{align}
2^{6n} 3^{3n} \left( \frac{1}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{5}{6} \right)_{n} = \frac{ 6^{6n} \left( \frac{1}{6} \right)_{n} \left( \frac{2}{6} \right)_{n} \left( \frac{3}{6} \right)_{n} \left( \frac{4}{6} \right)_{n} \left( \frac{5}{6} \right)_{n} }{ 3^{3n} \left( \frac{1}{3} \right)_{n} \left( \frac{2}{3} \right)_{n} } = \frac{(1)_{6n}}{(1)_{3n}} = \frac{(6n)!}{(3n)!}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\sum x_k=\frac12$, then $\prod\frac{1-x_k}{1+x_k}\geq\frac13$ The question is this
The sum of positive numbers $x_1,x_2,x_3,\dotsc,x_n$ is $\frac{1}{2}$. Prove that $$\frac{1-x_1}{1+x_1}\cdot\frac{1-x_2}{1+x_2}\cdot\frac{1-x_3}{1+x_3}\cdots\frac{1-x_n}{1+x_n}\geq\frac{1}{3}.$$
My process was something like this:
Using Weirstrass' Inequality in the numerators, we get $$\prod_{i=1}^{n}(1-x_i)>1-\sum_{i=1}^{n}(x_i)=1-\frac12=\frac12.$$ So now we need to prove that $$\prod_{i=1}^{n}(1+x_i)<\frac32.$$
But then, again using Weirstrass' Inequality, we get $$\prod_{i=1}^{n}(1+x_i)>1+\sum_{i=1}^{n}(x_i)=\frac32.$$
So now I'm stuck. Please help me solve this problem.
Note: This question if from Mathematical Circles (Russian Experience) so I don't think that this question is supposed to be wrong
| We will prove it using induction over $n$.
*
*It is seen that for $n=1$ it follows that $x_{1} = \frac{1}{2}$. Then
$$
\frac{1-x_{1}}{1+x_{1}} = \frac{1}{3}.
$$
*Assume it holds for $n=k$, then for $n = k+1$, we have $\sum_{i=1}^{k+1} x_{i} = \frac{1}{2}$. Define the element $y_{i} = x_{i}$ for $i=1, \ldots,n-1$ and $y_{n} = x_{n}+x_{n+1}$. Then still $\sum_{i=1}^{k} y_{i} = \frac{1}{2}$ and by assumption we have:
$$
\frac{1-y_{1}}{1+y_{1}} \cdot \frac{1-y_{2}}{1+y_{2}} \cdots \frac{1-y_{k}}{1+y_{k}} \geq \frac{1}{3}.
$$
It is seen that
$$
\frac{1-x_{k}}{1+x_{k}} \cdot \frac{1-x_{k+1}}{1+x_{k+1}} = \frac{1 - x_{k} - x_{k+1} + x_{k} x_{k+1}}{1+ x_{k} + x_{k+1} + x_{k} x_{k+1}} > \frac{1 - x_{k} -x_{k+1}}{1+x_{k}+x_{k+1}} = \frac{1-y_{k}}{1+y_{k}}.
$$
Combining, we have:
$$
\frac{1-x_{1}}{1+x_{1}} \frac{1-x_{2}}{1+x_{2}} \cdots \frac{1-x_{k}}{1+x_{k}} \cdot \frac{1-x_{k+1}}{1+x_{k+1}} \geq \frac{1-y_{1}}{1+y_{1}} \cdot \frac{1-y_{2}}{1+y_{2}} \cdots \frac{1-y_{k}}{1+y_{k}} \geq \frac{1}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Equation of a circle given one point and two lines Find the equation of the circle that pass through $(2,3)$ and are tangent to both the lines $3x - 4y = -1$ and $4x + 3y = 7$.
| Using the angle bisector equation, we obtain that the center of the required circle has to lie on the line $\frac{3x-4y+1}{5} = \pm \frac{4x+3y-7}{5}$. Draw a figure and you can see that we need to consider the equation with minus on RHS. Hence, the required angle bisector $L : 7x-y-6=0$.Or, the center is of the form $(x, 7x-6)$.
We know that the center has to be equidistant from the point $(2,3)$ and both the lines. This condition gives $(x-2)^2 + (7x-6-3)^2 = \frac{(3x - 4(7x- 6) + 1)^2}{25}$ which on simplification gives $x = 2 \text{ or } 6/5$. You can easily verify from the diagram that the required point is $(2,8)$. And the radius $r = 5$.
| {
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"url": "https://math.stackexchange.com/questions/660434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.
| For what it's worth, good ol' induction works, too -- $ \ n = 1 \ $ gives $ \ \frac{1}{3} \ $ and the sum for the $ \ (n+1) \ $ case gives
$$ \frac{n}{2n+1} \ + \ \frac{1}{(2n+1) \ (2n+3)} \ = \ \frac{n \ (2n+3) \ + \ 1}{(2n+1) \ (2n+3)} \ = \ \frac{ 2n^2 + 3n + 1}{(2n+1) \ (2n+3)} $$
$$= \ \frac{ (2n + 1) \ (n + 1)}{(2n+1) \ (2n+3)} \ = \ \frac{ n + 1}{ 2n+3} \ = \ \frac{(n + 1)}{2(n+1) + 1} \ \ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Question about the congruence classes of a square I've noticed that for any prime $p$ where $p = 2x+1$, a square $w^2$ can only take on $x$ possible congruence classes modulo $p$.
I've also noticed that:
$x^2 \equiv (x+1)^2 \pmod p$
$(x-1)^2 \equiv (x+2)^2 \pmod p$
$\vdots$
$1 \equiv (2x)^2 \pmod p$
For example, for $7$, we have:
$3^2 \equiv 4^2 \pmod 7$
$2^2 \equiv 5^2 \pmod 7$
$1 \equiv 6^2 \pmod 7$
Is this always true for $p > 7$? If so, why is it true? If it is not always true, can you give an example where it is not true?
Thanks,
-Larry
| Based on the comments, given, I will attempt to answer my own question.
First, the reason that the following is true:
$x^2 \equiv (x+1)^2 \pmod p$
$(x-1)^2 \equiv (x+2)^2 \pmod p$
$\vdots$
$1 \equiv (2x)^2 \pmod p$
is because since $p$ is odd:
$x^2 \equiv (-x)^2 \equiv (p-x)^2 \equiv (2x+1 -x)^2 = (x+1)^2 \pmod p$
$(x-1)^2 \equiv (-x+1)^2 \equiv (p -x+1)^2 \equiv (2x+1-x+1)^2 \equiv (x+2)^2 \pmod p$
$\vdots$
$1^2 \equiv (-1^2) \equiv (p-1)^2 \equiv (2x+1-1)^2 \equiv (2x)^2 \pmod p$
There are $x$ possible congruence classes because there are $2x$ congruence classes that not congruent to $0 \pmod p$ and since for each $c^2 \equiv (p-c)^2$, it follows that $w^2$ can only take on $x$ possible values.
If we include $p^2$, it follows that there are $x+1$ distinct congruence classes. Since each congruence class is called a quadratic residue, it follows that there are $x+1$ quadratric residues.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/663065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square. Prove that $2xy\mid x^2+y^2-x$ implies $x$ is a perfect square.
My work:
$2xy\mid x^2+y^2-x \implies x^2+y^2-x=2xy\cdot k$
So,$x^2+y^2+2xy-x=(x+y)^2-x=2xy \cdot (k+1)$
And,$x^2+y^2-2xy-x=(x-y)^2-x=2xy \cdot (k-1)$
I found that for $x,y$ both odd, no solution exists. For $x$ even, and $y$ odd,no solution exists. Solution exists only for $x$ odd, $y$ even and $x$ even and $y$ even solution exists. Cannot do anything more. Please help!
| By hypothesis $x\mid y^2\,$ and $\, \color{#c00}x = x^2\!+\!y^2 \!-\! 2kxy \color{#c00}{\equiv (x\!-\!ky)^2} \pmod{\!y^2}\,$ so below applies.
Theorem $\quad\, x\mid y^2\,$ and $\ \color{#c00}{x\equiv z^2}\pmod{\!y^2}\,\Rightarrow\, {\pm}x\, =\, (z,y)^2,\ $ if $\,x,y,z\in\Bbb Z$.
Proof $\ {\pm}x = (\color{#c00}x,y^2) = (\color{#c00}{z^2},y^2) = \color{#0a0}{(z,y)^2}\, $ by GCD mod $\rm\color{#c00}{reduce}$ & Freshman's $\rm\color{#0a0}{ Dream}$. $\ \small\bf QED$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/663283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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} |
Hyperboloid with different sheets? I just took a quiz and have a couple basic questions because I am too anxious to wait a week for my score.
My questions are:
1) *I put hyperboloid of one sheet for this
$$
\frac{(x-1)^{2}}{2^{2} } - \frac{(y)^{2}}{2^{2}} + \frac{(z)^{2}}{2^{2}} = 1
$$
2) *I put hyperboloid of one sheet again, but feeling like it's two sheets.
$$
\frac{-(x)^{2}}{2^{2} } + \frac{(y)^{2}}{2^{2}} + \frac{(z)^{2}}{2^{2}} + 10 = 0
$$
| The general equation for a hyperboloid of one sheet is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$$
For $1)$, the equation is precisely this form (with the roles of $y$ and $z$ reversed so that the axis of symmetry is the $y$ axis).
The general equation for a hyperboloid of two sheets is
$$\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$$
Your equation is
$$\frac{-x^2}{2^2} + \frac{y^2}{2^2} + \frac{z^2}{2^2} + 10= 0 \implies \frac{x^2}{10*2^2} - \frac{y^2}{10*2^2} - \frac{z^2}{10*2^2} = 1$$
shiwch is indeed an equation of a hyperboloid of two sheets (comparing it with the general equation).
| {
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"timestamp": "2023-03-29T00:00:00",
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All points at which the surfaces $x^2+y^2+z^2-1=0$ and $x^2+y^2-z^2-2y=0$ are intersect orthogonally $f: x^2+y^2+z^2-1=0$
$g: x^2+y^2-z^2-2y=0$
I set these two surfaces equal to each other to solve for the intersection, getting
$y=(1-z^2)/2$...then attempted to insert this value of $y$ in terms of $z$ back into the original equations, took the gradient of each, and set the dot product of the two gradients equal to $0$. My result was that $4x^2+16z^2(z-1)(z-2)=0$ which doesn't seem to have a great solution that exists on my two surfaces. Further, I realize that the intersection I found at the beginning doesn't make much sense as it is a parabola in the $yz$ plane and f is the unit sphere. I also tried equating both surfaces to where they equal $z^2$ and got the intersection
$x^2+(y-1/2)^2=5/8$...but am unsure of how to proceed in terms of orthogonality. Where did I go wrong? thanks in advance.
| If we set $f(x,y,z)=x^2+y^2+z^2-1=0$ and $g(x,y,z)=x^2+y^2-z^2-2y=0$ then $$\nabla f=(2x,2y,2z),~~\nabla g=(2x,2y-2,-2z)$$ If we want to find that points so they are lying on the a curve in which $(\nabla f\cdot\nabla g) (x,y,z)=0$. This gives us $$4(x^2+y^2-z^2)-4y=0$$ But $x^2+y^2-z^2=2y$ on the intersection so $$4(x^2+y^2-z^2)-4y=0\to y=0$$ Now we should think of the possible solution(s) of the following system:
$$ \left\{
\begin{array}{ll}
x^2+z^2=1 \\
x^2-z^2=0
\end{array}
\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorise: $a^4-b^4+c^4-d^4-2(a^2c^2-b^2d^2)+4ac(b^2+d^2)-4bd(a^2+c^2)$ Factorise: $a^4-b^4+c^4-d^4-2(a^2c^2-b^2d^2)+4ac(b^2+d^2)-4bd(a^2+c^2)$.
My working: $(a^4-2a^2c^2+c^4)-(b^4-2b^2d^2+d^4)+4ac(b^2+d^2)-4bd(a^2+c^2)$
$=(a^2-c^2)^2-(b^2-d^2)^2+4ac(b^2+d^2)-4bd(a^2+c^2)$
$=(a+c)^2(a-c)^2-(b+d)^2(b-d)^2+4ac(b^2+d^2)-4bd(a^2+c^2)$
Solution: $(a+c+b+d)(a+c-b-d)(a^2-2ac+c^2+b^2-2bd+d^2)$
My question is: it is not hard to work backward from the solution and see how it works, but how could I proceed from my stage of working if we were not told about the solution? And how many ways are there to factorise this expression?
Many thanks!
| Last question first: If you take integers coefficients and factor all the way down to irreducibles (no further factoring possible), then up to ordering, there is only one factorization.
Factoring this one by hand is all about recognizing differences of squares. The hints are "$a^4+c^4$" with "$a^2+c^2$" and "$a^2c^2$" and likewise for $b$ with $d$. The general method is "you don't want to know". If there's interest, I can describe an entirely automatable method that no human should ever undertake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I have is divisible by $3$ not $12$...
| Write $a = 2t+1$. Then $a^2+(a+2)^2+(a+4)^2+1=12 (t^2+3 t+3)$.
It may be simpler to write $b=a+2$. Then $a^2+(a+2)^2+(a+4)^2+1=(b-2)^2+b^2+(b+2)^2+1=3b^2+9$. Now $a$ odd implies $b$ odd, and so write $b=2u+1$. Then $3b^2+9=12 (u^2+u+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\
-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0.
\end{cases}
\end{align}
I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.
| Take the sine of both sides, and use the angle addition formula, then further simplify it by using the fact that $\cos\arcsin t=\sqrt{\cos^2\arcsin t}=\sqrt{1-\sin^2\arcsin t}=\sqrt{1-t^2}$. Then apply the $\arcsin$ function to both sides, and you're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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evaluation of $\int\frac{1}{\sin^3 x-\cos^3 x}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx$
$\bf{My\; Try::}$ Given $\displaystyle \int\frac{1}{\sin^3 x-\cos^3 x}dx = \int\frac{1}{(\sin x-\cos x)\cdot (\sin^2 x-\sin x\cos x+\cos^2 x)}dx$
$\displaystyle = 2\int\frac{(\sin x-\cos x)}{(\sin x-\cos x)^2\cdot (2-\sin 2x)}dx = 2\int \frac{(\sin x-\cos x)}{(1-\sin 2x)\cdot (2-\sin 2x)}dx$
Let $(\sin x+\cos x) = t\;,$ Then $(\cos x -\sin x)dx = dt\Rightarrow (\sin x -\cos x)dx = dt$
and $(1+\sin 2x)=t^2\Rightarrow \sin 2x = (t^2-1)$
So Integral Convert into $\displaystyle 2\int\frac{1}{(2-t^2)\cdot (3-t^2)}dt = 2\int\frac{1}{(t^2-2)\cdot (t^2-3)}dt$
My Question is , is there is any better method other then that
Help me
Thanks.
| I had just learnt an useful technique to deal with $1+\sin x \cos x$ and want to share with you now.
$$
\begin{array}{l}
\displaystyle \because \frac{3}{\sin ^{3} x-\cos ^{3} x}=\frac{2}{\sin x-\cos x}+\frac{\sin x-\cos x}{1+\sin x \cos x}\\ \displaystyle
\therefore \int \frac{d x}{\sin ^{3} x-\cos ^{3} x}\\ \displaystyle =\frac{2}{3} \int \frac{d x}{\sin x-\cos x}+\frac{1}{3} \int \frac{\sin x-\cos x}{1+\sin x \cos x} d x\\ \displaystyle
=-\frac{2}{3 \sqrt{2}} \int \frac{d x}{\cos \left(x+\frac{\pi}{4}\right)}-\frac{2}{3} \int \frac{d(\sin x+\cos x)}{1+(\sin x+\cos x)^{2}}\\
\displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\sec \left(x+\frac{\pi}{4}\right)+\tan \left(x+\frac{\pi}{4}\right)\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C\\
\displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\frac{1+\frac{1}{\sqrt{2}}(\sin x+\cos x)}{\frac{1}{\sqrt{2}}(\cos x-\sin x)}\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C\\
\displaystyle =-\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}+\sin x+\cos x}{\cos x-\sin x}\right|-\frac{2}{3} \tan ^{-1}(\sin x+\cos x)+C
\end{array}
$$
By the way, replacing $x$ by $-x$ yields its partner integral
$$
\int \frac{d x}{\sin ^{3} x+\cos ^{3} x}=-\frac{\sqrt{2}}{3} \ln \left|\frac{\sqrt{2}-\sin x+\cos x}{\cos x+\sin x}\right|+\frac{2}{3} \tan ^{-1}(\sin x-\cos x)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
| Taking the absolute value of both sides: $|z^3| = |i|$, gives $|z| = 1$. So, $z = \cos (\theta) + i \sin (\theta)$ for some real $\theta$.
Using De Moivre's formula gives $z^3 = \cos(3\theta) + i \sin(3\theta)$. Given that $z^3 = i = 0 + 1i$, this means that $\cos(3\theta) = 0$ and $\sin(3\theta) = 1$. Solving this system gives $3\theta = \frac{\pi}{2} + 2\pi n$, or $\theta = \frac{\pi}{6} + \frac{2 \pi n}{3}$, for any $n \in \mathbb{Z}$.
Plugging in a few values for $n$ gives:
*
*$n = 0$ → $\theta = \frac{\pi}{6}$ → $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i$
*$n = 1$ → $\theta = \frac{5\pi}{6}$ → $z = \frac{-\sqrt{3}}{2} + \frac{1}{2} i$
*$n = 2$ → $\theta = \frac{3\pi}{2}$ → $z = -i$
And we can stop there because this is a polynomial equation of degree 3, and the Fundamental Theorem of Algebra guarantees that it has at most 3 distinct roots. The solution set is thus $z \in \{ \frac{\sqrt{3}}{2} + \frac{1}{2} i, \frac{-\sqrt{3}}{2} + \frac{1}{2} i, -i \}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)
$$
I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick. That doesn't seem to work. I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :(
Thanks.
| So:
$$
\lim_{x\to-\infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}) = \lim_{x\to-\infty}(\frac{x^2 + 2x - x^2 + 2x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}) = \lim_{x\to-\infty}(\frac{-4}{\sqrt{1 + \frac{2}{x}} + \sqrt{1 - \frac{2}{x}}}) = -2
$$
| {
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"answer_id": 2
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Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we must find a pattern for $k$ that grants this. We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$. Solving for $k$, we have $k=4+7m$.
Substituting this into our value for $x$, we get:
\begin{align*}
x & = \dfrac{15(4+7m) + 3}{7}.\\
&= \dfrac{63}{7} + \frac{105m}{7}.\\
&= 9+15m.
\end{align*}
So, $x = 9+15m, m\in \mathbb{Z}.$
So, is this what I was looking for? I'm not exactly sure what is meant by incongruent solutions.
| $$\left ( 7,15 \right )=1 \Rightarrow$$ $$ x \equiv 3(7)^{^ \phi(15)-1 } (mod 15)$$$$x \equiv 3(7)^{^7}\equiv 3.13 \equiv 9 (mod15)$$, which is the unique solution by Euler-Fermat theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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nonlinear diophantine equation $x^2+y^2=z^2$ how to solve a diophantine equation $x^2+y^2=z^2$ for integers $x,y,z$
i strongly believe there is a geometric solution ,since this is a pythagoras theorem form
or a circle with radius $z$
$x^2+y^2=z^2$
$(\frac{x}{z})^2+(\frac{y}{z})^2=1\implies x=y=\pm z$ or $0$
so we consider a line passing through points $P_1(- z,0)$ and $P(x,y)$ both on the circle
$m=\frac{y}{x+z}$
$x^2+m^2(x+z)^2=z^2$
$(m^2+1)x^2+2xzm^2+(m^2-1)z^2=0$
$((m^2+1)x+(m^2-1)z)(x+z)=0$
$\frac{x}{z}=-\frac{m^2-1}{m^2+1}$ or $-1$
let $m=\frac{a}{b}\implies \frac{x}{z}=\frac{b^2-a^2}{b^2+a^2}$
$\frac{y}{z}=\frac{2a^2}{b^2+a^2}$
how to get explicit $z,x,y$
| Generally, if a conic curve with rational coefficients has one rational point $\rm\:P\:$ then it has infinitely many, since any rational line through $\rm\:P\:$ will intersect the curve in another point, necessarily rational, since if one root of a rational quadratic is rational then so is the other. Therefore, by sweeping lines of varying rational slopes through $\rm\:P\:$ we obtain infinitely many rational points on the conic. Projecting these points onto a line leads to a rational parametrization of the conic. For a very nice exposition see Chapter $1$ of Silverman and Tate: Rational Points on Elliptic Curves. There are also many online expositions, e.g. search on "rational parametrization Pythagorean triple".
For Pythagorean Triples there is even more beautiful geometrical structure. One may generate Ternary Tree of Pythagorean Triples using arithmetical reflections.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show
$$(a+1)(b+1)(c+1)<4 $$
I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $.
I am looking for a non-calculus solution (no Lagrange multipliers).
Do you know how to do it?
| $(1+a)(1+b)(1+c) = 2 + a+b+c + abc < 2 + 2(a+b) + ab(a+b) = 2 + (a+b)(2 + ab)$ (because $c < a+b)$
$ab + bc + ac = ab + (a+b)c < ab + (a+b)^2$ for the same reason, that is $ab > 1 - (a+b)^2$, and $2 + ab > 3 - (a+b)^2$
Combining, we have $(1+a)(1+b)(1+c) < 2 + 3(a+b) - (a+b)^3$
The latter achieves the maximum 4 at $a+b = 1$ (we are not interested in negative $a+b$), QED
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Show $1+\cosθ+\cos(2θ)+\cdots+\cos(nθ)=\frac{1}{2}+\frac{\sin[(n+1/2)θ]}{2\sin(θ/2)}$ Show
$$1+\cosθ+\cos(2θ)+\cdots+\cos(nθ)=\frac12+\frac{\sin\left(\left(n+\frac12\right)θ\right)}{2\sin\left(\frac\theta2\right)}$$
I want to use De Moivre's formula and $$1+z+z^2+\cdots+z^n=\frac{z^{n+1}-1}{z-1}.$$ I set $z=x+yi$, but couldn't get it.
| Here's a very simple solution that doesn't use complex numbers, just some basic trigonometric identities. Recall that $$2 \cos \alpha \sin \beta = \sin(\alpha + \beta) - \sin(\alpha - \beta).$$ With the choice $\alpha = k \theta$, $\beta = \theta/2$, we then have $$2 \cos k\theta \sin \frac{\theta}{2} = \sin\bigl((k + {\textstyle \frac{1}{2}})\theta\bigr) - \sin\bigl((k - {\textstyle \frac{1}{2}})\theta\bigr).$$ Summing of both sides over $k = 0, 1, \ldots, n$ and observing that the RHS telescopes, $$\sum_{k=0}^n 2 \cos k\theta \sin \frac{\theta}{2} = \sin\bigl((n + {\textstyle \frac{1}{2}})\theta\bigr) - \sin\bigl(-{\textstyle \frac{1}{2}}\theta\bigr),$$ from which it immediately follows that $$\sum_{k=0}^n \cos k\theta = \frac{1}{2}\left( 1 + \frac{\sin((k+\frac{1}{2})\theta)}{\sin\frac{\theta}{2}}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Finding system with infinitely many solutions The question asks to find equation for which the system has infinitely many solutions.
The system is:
\begin{cases}
-cx + 3y + 2z = 8\\
x + z = 2\\
3x + 3y + az = b
\end{cases}
How should I approach questions like this?
I tried taking it to row reduced echelon form but it got kind of messy.
The answer is supposed to be:
$$a - c -5 = 0$$ and $$b- 2c +2 = 0$$
| Use Cramer's rule. If the determinant:
$$\begin{vmatrix}
-c && 3 && 2 \\
1 && 0 && 1 \\
3 && 3 && a
\end{vmatrix} = 0$$
then the system has either no solution or infinite amount of solutions. If all three determinants:
$$\begin{vmatrix}
8 && 3 && 2 \\
2 && 0 && 1 \\
b && 3 && a
\end{vmatrix} = \begin{vmatrix}
-c && 8 && 2 \\
1 && 2 && 1 \\
3 && b && a
\end{vmatrix} = \begin{vmatrix}
-c && 8 && 2 \\
1 && 2 && 1 \\
3 && b && a
\end{vmatrix} = 0$$
Then the system has infinite amount of solution. Can you solve it on your own now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Calculating $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\dots}}} $ If $ x= \sqrt{4+\sqrt{4-\sqrt{ 4+\sqrt{4-\sqrt{ 4+\sqrt{4-\dots}}}}}} $
then find value of 2x-1
I tried the usual strategy of squaring and substituting the rest of series by x again but could not solve.
| We have
$$-\left((x^2-4)^2-4\right)=x\iff x^4-8x^2+x+12= (x^2+x-4)(x^2-x-3) =0$$
solve this equation and notice that $x\ge2$ we find that the acceptable answer is
$$x=\frac12\sqrt{13}+\frac12$$
hence
$$2x-1=\sqrt{13}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.
I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?
| Note that by Little Fermat,
$$
(x^3)^3\equiv x^{2\phi(5)+1}\equiv x\pmod{5}
$$
Thus, if $x^3\equiv y^3\pmod{5}$, by cubing both sides, we must have $x\equiv y\pmod{5}$. Therefore,
$$
5\mid x^3-y^3\implies5\mid x-y
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\sum_{x=1}^n (3x^2+x+1) = n^3+2n^2+3n$? I wanna check if the following equation involving a sum is true or false? How do I solve this? Please help me.
$$ \sum_{x=1}^n (3x^2+x+1) = n^3+2n^2+3n$$
for all $n \in \{0,1,2,3, \dots\}$.
| $(x+1)^3=x^3+3x^2+3x+1\iff3x^2+3x+1$ is the difference of two consecutive cubes, so the sum will be telescopic, since $3x^2+x+1=(3x^2+3x+1)-2x=\Big[(x+1)^3-x^3\Big]-2x$. $$\sum_{x=1}^n(3x^2+x+1)=\sum_{x=1}^n\Big[(x+1)^3-x^3\Big]-2\sum_{x=1}^nx=\Big[(n+1)^3-1^3\Big]-2\bigg[\frac{n(n+1)}2\bigg]=$$ $$=\Big[(n^3+3n^2+3n+1)-1\Big]-(n^2+n)=n^3+2n^2+2n.$$ See Faulhaber's formula for more information.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Solving $\arcsin(1-x)-2\arcsin(x)=\pi/2$ \begin{eqnarray*}
\arcsin(1-x)-2\arcsin(x) & = & \frac{\pi}{2}\\
1-x & = & \sin\left(\frac{\pi}{2}+2\arcsin(x)\right)\\
& = & \cos\left(2\arcsin(x)\right)\\
& = & 1-2\left(\sin\left(\arcsin(x)\right)\right)^{2}\\
& = & 1-2x^{2}\\
x & = & 2x^{2}\\
x\left(x-\frac{1}{2}\right) & = & 0
\end{eqnarray*}
So $x=0$ or $x=\frac{1}{2}$
But puttig $x=\frac{1}{2}$ in the original expression gives $-\frac {\pi} 4 \ne \frac \pi 2$
So, why do we get $x=-1/2$ as an answer?
| Beside the good answers you already received, you can also consider the problem from an algebraic point of view. Let $$f(x)=\arcsin(1-x)-2\arcsin(x)-\frac{\pi }{2}$$ Its derivative $$f'(x)=-\frac{2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{(2-x) x}}$$ is always negative (with infinite branches at $x=0$ and $x=1$). For $x=0$, $f(0)=0$ and since the function decreases, you cannot have any root beside $x=0$ (remember that $f(x)$ is only defined for [$0<x<1$]).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\int_1^\infty\frac{K(x)^2}x dx=\frac{i\,\pi^3}8$? How can I prove the following identity?
$$\int_1^\infty\frac{K(x)^2}x dx\stackrel{\color{#B0B0B0}?}=\frac{i\,\pi^3}8,\tag1$$
where $K(x)$ is the complete elliptic integral of the 1ˢᵗ kind:
$$K(x)={_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)\cdot\frac\pi2.\tag2$$
| When $0<x<1$, we have the identity $K(x^{-1})=x(K(x)-iK(\sqrt{1-x^2}))$.
Therefore
$$\int^{\infty}_{1}K^2(x)\frac{dx}{x}=\int^{1}_{0}K^2(x^{-1})\frac{dx}{x}\\
=\int^{1}_{0}\left(x(K(x)-iK(\sqrt{1-x^2}))\right)^2\frac{dx}{x}\\
=\int^{1}_{0}x\left(K(x)-iK(\sqrt{1-x^2})\right)^2dx\\
=\int^1_0(xK^2(x)-xK^2(\sqrt{1-x^2})-2ixK(x)K(\sqrt{1-x^2}))dx.$$
We also have $\int^1_0xK^2(\sqrt{1-x^2})dx=\int^1_0yK^2(y)dy$ using the substitution $x\mapsto\sqrt{1-y^2}$. Thus the problem is reduced to proving
$$ \int^1_0xK(x)K(\sqrt{1-x^2})dx\stackrel?=\frac{\pi^3}{16}.$$
We first prove that $\int^1_0x^nK(\sqrt{1-x})dx=\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}$.
$$\int^1_0x^nK(\sqrt{1-x})dx=\frac{\pi}{2}\sum_{m=0}^{\infty}\frac{(2m)!^2}{2^{4m}(m!)^4}\int^1_0x^n(1-x)^mdx\\
=\frac{\pi n!}{2}\sum_{m=0}^{\infty}\frac{(2m)!^2}{2^{4m}(m!)^3(m+n+1)!}\\
=\frac{\pi n!}{2\Gamma(n+2)}{}_2F_1(\frac12,\frac12;n+2;1)\\
=\frac{\pi n!}{2\Gamma(n+2)}\frac{\Gamma(n+2)\Gamma(n+1)}{\Gamma(n+3/2)^2}\\
=\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}.$$
Thus we have $$\int^1_0xK(x)K(\sqrt{1-x^2})dx=\frac12\int^1_0K(\sqrt{x})K(\sqrt{1-x})dx\\
=\frac12\int^1_0(\frac\pi2\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}x^n)K(\sqrt{1-x})dx\\
=\frac\pi4\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}\int^1_0x^nK(\sqrt{1-x})dx\\
=\frac\pi4\sum^{\infty}_{n=0}\frac{(2n)!^2}{2^{4n}(n!)^4}\frac{2^{4n+1}(n!)^4}{(2n+1)!^2}\\
=\frac\pi2\sum^{\infty}_{n=0}\frac{1}{(2n+1)^2}=\frac{\pi}{2}\frac{\pi^2}{8}=\frac{\pi^3}{16}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Interpretation of a homogeneous transformation matrix of the plane I have the transformation matrix $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}$. This $3\times 3$ matrix is a homogeneous transformation matrix in $2-D$ space.
My book says that this matrix translates the line $y=x+1$ to $y=x$. I don't see how. Let us take the point $(a,b)$. After the translation by $-1$ along the $y$-axis, the point should become $(a,b-1)$. Now let us determine $\begin{pmatrix} 1&0&0\\0&1&0\\0&-1&1\end{pmatrix}\begin{pmatrix} a\\b\\1\end{pmatrix}$. We get $\begin{pmatrix} a\\b\\1-b\end{pmatrix}$. After transforming this resultant matrix to 2-D form, we get $\begin{pmatrix} \frac{a}{1-b}&\frac{b}{1-b}\end{pmatrix}$. How is this equal to $(a,b-1)$?
| Work backwards. The translation of $y = x - 1$ to $y = x$:
$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ -1 \end{bmatrix}$$
Affine to linear:
$$\begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}$$
So it seems the book is representing the vectors as row vectors, $\begin{bmatrix} x & y \end{bmatrix}$, so your matrix is transposed.
| {
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Find rectangular equation of a cardioid Given the equation in polar form $$r = 1 - \sin\theta,$$ find the rectangular equation.
So far, I found:
$$x^2 + y^2 = 1 - 2\sin\theta + \sin^2\theta\quad x = \cos\theta - \sin\theta\cos\theta\quad y = \sin\theta - \sin^2\theta$$
Where should I go from here?
| First, note that $y = r\sin{\theta} \Rightarrow \sin{\theta} = \frac{y}{r}$. This gives us:
$$
r = 1 - \frac{y}{r}
$$
Now, multiplying by $r$ on each side, we get:
$$
r^2 = r - y
$$
Noting that $r^2 = x^2 + y^2$, so this is the same as
$$
x^2 + y^2 = \sqrt{x^2+y^2} - y
$$
You can do what you wish from there to simplify it: I'm not sure what form you want it in.
| {
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Proof: $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ iff R is commutative We want to show that for some ring $R$, the equality $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ if and only if $R$ is commutative.
Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someone could take a look.
Forward: $a^2 -b^2 = (a-b)(a+b) \forall a,b \in R$ implies $R$ is commutative
Let $x = (a-b)$. Then \begin{align}
x(a+b) &= xa+xb\\
&= (a-b)a + (a-b)b\\
&= a^2 -ba + ab - b^2\end{align}
Then we note that $a^2 - ba + ab - b^2 = a^2 - b^2$ iff $-ba + ab = 0$ if and only if $ab=ba$ iff $R$ is commutative.
Backwards: $R$ is commutative implies $a^2 - b^2 = (a-b)(a+b) \forall a,b \in R$.
Let $x = (a+b)$. Then $(a-b)x = ax - bx = a(a+b) - b(a+b) = a^2 + ab - ba - b^2$. $R$ is commutative, so $ab-ba = 0$, so $a^2 + ab - ba - b^2 = a^2 - b^2$.
| I think there are not many word necessary.
$$a^2 -b^2 = a^2+ab-ab-b^2=a^2+ab-ba-b^2=a(a+b)-b(a+b)=(a-b)(a+b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to simplify this easy expression $3\cdot(4x)^2 \cdot 4$
How do I simplify that expression? I know it's easy but I suck at math so a detailed explanation would be much appreciated.
| We are given $3(4x)^2 \cdot 4 $, and so here (because we have multiple operations) we must follow the order of operations, starting with parentheses. The parentheses in this case can't be simplified further, and so now we move on to exponents: $$ 3\cdot (4c)^2 \cdot 4 = 3 \cdot (4c)(4c) \cdot 4 \ .$$
Here we can take advantage of the fact that multiplication is associative and communicative by re-arranging the expression: $$\begin{align} 3 \cdot (4c)(4c) \cdot 4 &=3 \cdot 4 \cdot c \cdot 4 \cdot c \cdot 4\\ &= 3 \cdot 4 \cdot 4 \cdot 4 \cdot c \cdot c\end {align}$$
and because we know that $3 \cdot 4 \cdot 4 \cdot 4 = 192$ and that $c \cdot c = c^2$ we can say $$3\cdot (4c)^2 \cdot 4 = 192c^2$$
| {
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Differential equation and substitution. If the substitution square root of $x = \sin y$ is made in the $\int^0_{0.5}\dfrac{\sqrt{x}}{\sqrt{1-x}}dx$, what is the resulting integral?
| I will do $0$ to $\frac{1}{2}$, i.e., $\displaystyle \int_0^{0.5}\frac{\sqrt{x}}{\sqrt{1-x}}dx$.
First, as $g(x)=\frac{\sqrt{x}}{\sqrt{1-x}}$ is continuous in $(0,\frac{1}{2})$, it follows that this integral is not unfit.
Now, replace $1-x =u^2$, then $x=1-u^2$ and $dx=-2udu$.
Calculating the indefinite integral
$$\int \frac{\sqrt{x}}{\sqrt{1-x}}dx=-2\int \frac{\sqrt{1-u^2}\cdot u}{u}du =-2\int \sqrt{1-u^2} du =(*).$$
Now, do
$$\sin t=\sqrt{1-u^2} \text{ and } \cos t=u,$$
then $du=-\sin t dt$ and so
$$(*)=-2\int \sin t(-\sin t)dt=2\int \sin^2t dt=2\int \frac{1-\cos 2t}{2}dt= $$
$$=\int dt -\int \cos 2t dt=t-\frac{1}{2}\sin 2t+c=t-\sin t\cos t +c= $$
$$=\arccos u-u\sqrt{1-u^2}+c=\arccos\sqrt{1-x}-\sqrt{1-x}\cdot\sqrt{x}+c. $$
Lastly,
$$\int_0^{\frac{1}{2}}g(x)dx= \left.(\arccos\sqrt{1-x}-\sqrt{1-x}\cdot\sqrt{x})\right|_0^{\frac{1}{2}}= $$
$$=\arccos \frac{\sqrt{2}}{2}-\sqrt{\frac{1}{2}}\cdot \sqrt{\frac{1}{2}}-\left( \arccos 1-0\right) =\frac{\pi-2}{4}.$$
| {
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How to prove that $d \sin(x)/dx = \cos(x)$ without circular logic such as L'Hôpital's rule? How do I prove that the derivative of $\sin$ is $\cos$ without resorting to L'Hôpital's rule (circular logic)?
This part is easy:
$$
\begin{align*}
\sin'(x) &= \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x} \\
\sin'(x) &= \lim_{\Delta x \to 0} \frac{\cos(x)\sin(\Delta x) + \sin(x) \cos(\Delta x) - \sin(x)}{\Delta x} \\
\sin'(x) &= \lim_{\Delta x \to 0} \left(\cos(x)\frac{\sin(\Delta x)}{\Delta x} + \sin(x)\frac{\cos(\Delta x) - 1}{\Delta x}\right) \\
\end{align*}
$$
but where do I go from here?
| Starting off where you finished and credits to this pdf:
$$\sin'(x) = \lim_{\Delta x \to 0} \left(\cos x \frac{\sin \Delta x}{\Delta x} + \sin x \frac{\cos \Delta x - 1}{\Delta x} \right)$$ $$= \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x}$$
Part I: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$.
Assuming that $OA = 1$ (i.e. it's a unit circle), we have $\sin \theta = BC$. We then note that
[Area of triangle $AOD$] > [Area of sector $AOC$] > [Area of triangle $AOC$]
which means
$$\frac{1}{2} \frac{\sin \theta}{\cos \theta} > \frac{1}{2} \theta > \frac{1}{2} \sin \theta$$
From this we get $$\cos \theta < \frac{\sin \theta}{\theta} < 1$$
and since $\displaystyle \lim_{\theta \to 0} \cos \theta = 1$, we have that $\displaystyle \lim_{\theta \to 0} = 1$.
Part II: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = 0$.
$$\lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \lim_{\Delta x \to 0} \frac{(1 - \cos \Delta x) (1 + \cos \Delta x)}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{1 - \cos^2 \Delta x}{(1 + \cos \Delta x)\Delta x}$$
$$ = \lim_{\Delta x \to 0} \frac{\sin^2 \Delta x}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} \frac{\sin \Delta x}{1 + \cos \Delta x}$$
We have proven in part I that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$, and we note that $\displaystyle \frac{\sin \Delta x}{1 + \cos \Delta x} = \frac{0}{1+1} = 0$.
We finally put all this together to get that $$\boxed{\displaystyle \sin'(x) = \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \cos x \cdot 1 - \sin x \cdot 0 = \cos x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/707469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove $ \left |\sin(x) - x + \frac{x^3}{3!} \right | < \frac{4}{15}$ Prove $ \left |\sin(x) - x + \dfrac{x^3}{3!} \right | < \dfrac{4}{15}$ $\forall x \in [-2,2]$
By Maclaurin's formula and Lagrange's remainder we have $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{\sin(\xi)}{5!}x^5$ for some $0<\xi<2$
subbing this in we get $\left|\dfrac{\sin(\xi)}{5!}x^5 \right| \leq \left |\dfrac{x^5}{5!} \right| \leq \dfrac{2^5}{5!} = \dfrac{4}{15}$, but the question has $<$ rather than $\leq$ - where have I done wrong?
edit: thinking the $\cos(\xi)$ should be there rather than $\sin(\xi)$
| Use the exact form of the Taylor formula:
$$
\sin x - x + \frac{x^3}3 = \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt
\\
\left| \sin x - x + \frac{x^3}3 \right| =
\left| \int_0^x \frac{(x-t)^4}{4!}\cos(t) dt \right|
\le\int_0^x \left| \frac{(x-t)^4}{4!}\cos(t) \right|dt \\
\le\int_0^x \left| \frac{(x-t)^4}{4!} \right|dt
= \int_0^x \frac{t^4}{4!} dt = \frac{2^5}{5!}
$$
Now if there is equality anywhere, every inequality becomes an equality, but considering the first implies that $x=0$, and the last implies that $x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the exact coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? What is the coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$?
I figure you need to pick $x^3$ 4 times so $C(10,4)$...but what happens with the other numbers/variables???
Can someone explain to me how this is done properly?
Thanks.
EDIT:
$(x + y)^n = C(n,k) \cdot x^{n-k} \cdot y^k$
EX: Find the term for $x^5$ in $(5-2x)^8$
Answer: $C(8,5) \cdot (-2)^5 \cdot 5^3$
How can I use this info to solve a polynomial based question such as the featured?
Answer:
To sum up all information provided by everyone (Thanks!!!):
$(C(10,4) * 2^6) + (C(10,2) * 2^8) + (C(10,1) * C(9,2) + 2^7) = 71040 $
| You can reach $x^{12}$ in the product from $x^6x^6\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{2}$ ways with a coefficient of $2^8$.
Or you can reach it with $x^6x^3x^3\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{1}\binom{9}{2}$ ways with a coefficient of $2^7$.
Or you can reach it with $x^3x^3x^3x^3\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2$, which can be done in $\binom{10}{4}$ ways with a coefficient of $2^6$.
$$\begin{align}2^8\binom{10}{2}+2^7\binom{10}{1}\binom{9}{2}+2^6\binom{10}{4}&=64(4\cdot45+2\cdot10\cdot36+210)\\
&=64(1110)\\
&=71040
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Formula for $\int_0^\infty \frac{\log(1+x^2)}{\sqrt{(a^2+x^2)(b^2+x^2)}}dx$ Is it possible to express the following integral in terms of known special functions?
$$I(a,b)=\int_0^\infty \frac{\log(1+x^2)}{\sqrt{(a^2+x^2)(b^2+x^2)}}dx$$
*
*I have managed to solve the special case when $b=1$ and $a>1$. $$I(a,1)=\frac{\pi}{2a}K\left(\frac{1}{a} \right)+\frac{\log(a^2-1)}{2a}K\left( \sqrt{1-\frac{1}{a^2}}\right) $$
where $K(k)$ is the complete elliptic integral of the first kind. (Please refer to this answer for it's derivation) Unfortunately, I was not able to solve any other case.
*Is it possible to express $I(a,b)$ in terms of complete elliptic integrals?
Update
Using the substitution $x\mapsto1/x$, it can be shown that
$$I(a,b)=\frac{1}{ab}I\left(\frac{1}{a},\frac{1}{b}\right)+\frac{\log(ab)}{b}K\left( \frac{\sqrt{b^2-a^2}}{b}\right) \quad b>a$$
| Assume $a > b$, define $m$ and $s$ such that $\frac{b^2}{a^2} = 1 - m = \frac{1-3s}{2}$. Substitute $x^2$ by $\frac{b^2}{p-\frac{1-s}{2}}$, we can rewrite $I(a,b)$ as
$$I(a,b) = -\frac{1}{a}\int_\infty^{\frac{1-s}{2}}
\frac{\log\left[\frac{p-\left(\frac{1-s}{2}-b^2\right)}{p-\left(\frac{1-s}{2}\right)}\right]}{
\sqrt{4p^3 - g_2 p - g_3}} dz
\quad\text{ where }\quad
\begin{cases}g_2 = 1+3s^2\\g_3 = s(s^2-1)\end{cases}$$
Let $\wp(z), \zeta(z), \sigma(z)$ be the Weierstrass elliptic, zeta and sigma functions associated with the ODE:
$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3 =
4 \left(\wp(z) - \frac{1-s}{2}\right)\left(\wp(z) + \frac{1+s}{2}\right)\left(\wp(z) - s\right)$$
The double poles of $\wp(z)$ lies on a rectangular lattice with half period
$$\omega = K(\sqrt{m}),\quad \omega' = iK'(\sqrt{m}) = iK(\sqrt{1-m})$$
where $K(k)$ is the complete elliptic integral of the first kind. Furthermore, $\wp(\omega) = \frac{1-s}{2}$ and $\wp(\omega') = -\frac{1+s}{2}$.
Choose $\rho$ such that $\wp(\omega \pm \rho ) = \frac{1-s}{2} - b^2$.
When $b$ increases from $0$ to $\infty$, $\wp(\omega\pm\rho)$ decreases from $\wp(\omega) = \frac{1-s}{2}$ to $-\infty$. For concreteness, we will choose $\rho$ such that as $b$
varies from $0$ to $\infty$. $\omega + \rho$ will move along the polygonal path joining
$\omega, \omega+\omega', \omega'$ and $0$. i.e. for small $b$, $\rho$ is purely imaginary.
In terms of all these, we can simplify $I(a,b)$ as
$$\begin{align}
I(a,b)
&= \frac{1}{a}\int_0^\omega\log\left[
\frac{\wp(z)-\wp(\omega+\rho)}{\wp(z)-\wp(\omega)}
\right]dz
= \frac{1}{2a}\int_{-\omega}^\omega\log\left[
\frac{\wp(z)-\wp(\omega+\rho)}{\wp(z)-\wp(\omega)}\right]dz\\
&= \frac{1}{2a}\int_{-\omega}^\omega\log\left[C
\frac{\sigma(z+\omega+\rho)\sigma(z+\omega-\rho)}{\sigma(z+\omega)^2}
\right]
\end{align}
$$
where $\displaystyle\;C = \frac{\sigma(\omega)^2}{\sigma(\omega+\rho)\sigma(\omega-\rho)}$.
Using the same tricks as this answer, we can evaluate the $z$ dependent part and get
$$\begin{align}
I(a,b)
&= \frac{1}{2a}\left(
2\omega\log C
+ \varphi_{+}(\omega+\rho) + \varphi_{-}(\omega-\rho)-\varphi_{+}(\omega)-\varphi_{-}(\omega)\right)\\
&= \frac{1}{2a}\left(
2\omega\log C
+ \zeta(\omega)((\omega+\rho)^2 + (\omega-\rho)^2 - 2\omega^2) -\pi i((\omega + \rho) - (\omega - \rho))\right)\\
&= \frac{1}{a}\left( \omega \log C + \zeta(\omega) \rho^2 - \pi i \rho\right)
\end{align}$$
Using various identities for the elliptic functions:
$$\begin{align}
\wp(z) - \wp(u) &= -\frac{\sigma(z+u)\sigma(z-u)}{\sigma(z)^2\sigma(u)^2}\\
\wp(\omega) + \wp(\rho) + \wp(\omega+\rho) &= \frac14\left(\frac{\wp'(\omega)-\wp'(\omega+\rho)}{\wp(\omega)-\wp(\omega+\rho)}\right)^2
\end{align}
$$
We find
$$C = -\frac{1}{\sigma(\rho)^2(\wp(\omega)-\wp(\rho))}
\quad\text{ and }\quad
\wp(\omega) - \wp(\rho) = \frac{1}{a^2}$$
This finally gives us
$$I(a,b) = \frac{1}{a}\left\{ \omega\log\left[ -\frac{a^2}{\sigma(\rho)^2}\right] + \zeta(\omega) \rho^2 - \pi i\rho \right\}
\quad\text{ with }\quad \rho = \wp^{-1}\left(\frac{a^2+b^2-3}{3a^2}\right)
$$
As a double check, let us study two special cases $b = 1$ and $a = 1$.
Case I: $b = 1$.
When $b = 1$, $\wp(\rho) = s = \wp(\omega'\pm\omega) \implies \rho = \omega'-\omega$.
Let $\eta = \zeta(\omega)$ and $\eta' = \zeta(\omega')$. The addition formula for sigma function
$$\wp'(z+\rho) = -\frac{\sigma(2z+2\rho)}{\sigma(z+\rho)^4}
= e^{2(\eta'-\eta)(2z + \omega'-\omega)}\frac{\sigma(2z)}{\sigma(z+\rho)^4}$$
implies
$$\sigma(\rho)^4 = \frac{2}{\wp''(\rho)}e^{2(\eta'-\eta)(\omega'-\omega)}
= -\frac{a^4}{b^2(a^2-b^2)}e^{2(\eta'-\eta)(\omega'-\omega)}
$$
Up to a sign factor in the intermediate imaginary pieces, this allow us to fix $I(a,b)$ to the form
$$\frac{1}{a}\left\{\omega\left(\log(b\sqrt{a^2-b^2}) \stackrel{?}{\pm} \frac{\pi i}{2}\right)
-\omega ( (\eta'-\eta)(\omega'-\omega) )
+ \eta(\omega'-\omega)^2 - \pi i(\omega'-\omega)
\right\}$$
Using the identity $\eta\omega'-\omega\eta' = \frac{\pi i}{2}$, the mess after
the logarithm can be simplified:
$$\begin{align}
& -\omega ( (\eta'-\eta)(\omega'-\omega) ) + \eta(\omega'-\omega)^2 - \pi i(\omega'-\omega)\\
=& (\omega'-\omega)( -\omega(\eta'-\eta) + \eta(\omega'-\omega) - \pi i)\\
=& -\frac{\pi i}{2}(\omega' - \omega)
\end{align}
$$
We know $I(a,b)$ is a real number, the imaginary part $\frac{\pi i}{2}\omega$
in above expression will get cancelled by the imaginary piece associated with the logarithm.
As a result, we find when $b = 1$,
$$\begin{align}
I(a,1)
&= \frac{1}{a}\left\{\omega\log(b\sqrt{a^2-b^2}) - \frac{\pi i}{2}\omega'\right\}\\
&= \frac{1}{a}\left\{\log(b\sqrt{a^2-b^2})K\left(\sqrt{1-\frac{b^2}{a^2}}\right) + \frac{\pi}{2}K\left(\frac{b}{a}\right)\right\}
\end{align}$$
reproducing what is known.
Case II: $a = 1$.
When $a = 1$, $\wp(\rho) = -\frac{1+s}{2} = \wp(\pm\omega') \implies \rho = \omega'$. Once again, by the addition formula for sigma function, we have
$$\sigma(\rho)^4 = \frac{2}{\wp''(\rho)}e^{2\eta'\omega'} = \frac{a^2}{a^2-b^2}e^{2\eta'\omega'}$$
This allow us to conclude $I(a,b)$ has the form
$$\frac{1}{a}\left\{\omega\log(a\sqrt{a^2-b^2})
- \omega\eta'\omega' + \eta\omega'^2 - \pi i\omega'\right\}
= \frac{1}{a}\left\{\omega\log(a\sqrt{a^2-b^2}) -\frac{\pi i}{2}\omega'\right\}
$$
and hence $I(1,b)$ can be casted to a form very similar to what we have for $b = 1$.
$$I(1,b) =
\frac{1}{a}\left\{\log(a\sqrt{a^2-b^2})K\left(\sqrt{1-\frac{b^2}{a^2}}\right) + \frac{\pi}{2}K\left(\frac{b}{a}\right)\right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
Optimization problem for several variables calculus For $a,b,c > 0$, let
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$
How can I find the axes parallel box of maximal volume inscribed in this ellipsoid ?
| Suppose that the corner of the box in the first octant is $(x,y,z)$. Then the volume is $8xyz$. We want to maximize $x^2y^2z^2$ subject to the condition $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$.
By the Arithmetic Mean Geometric Mean Inequality, we have
$$\frac{1}{3}=\frac{1}{3}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) \ge \sqrt[3]\frac{x^2y^2z^2}{a^2b^2c^2},$$
with equality precisely if $\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}$.
Since the sum of these terms is $1$, each is $\frac{1}{3}$, and therefore for the maximum we have $x=\frac{a}{\sqrt{3}}$, $y=\frac{b}{\sqrt{3}}$, $z=\frac{c}{\sqrt{3}}$.
Remark: Alternately, we can use tools from the calculus. Lagrange multipliers work smoothly.
In addition to the constraint equation, we get by partial differentiation that for a maximum we need to have
$$8yz=\lambda\frac{2x}{a^2}, 8zz=\lambda\frac{2y}{b^2},\quad 8xy=\lambda\frac{2z}{c^2}.$$
From the first two equations, we get
$$\frac{8yz}{8zx}=\frac{(2x)(b^2)}{(2y)(a^2)},$$
or equivalently $\frac{x^2}{a^2}=\frac{y^2}{b^2}$. Similarly, $\frac{y^2}{b^2}=\frac{z^2}{c^2}$, and we are essentially finished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to Prove This Trigonometry Identity? I have to prove that:
$$\tan^2\theta \sin^2\theta = \tan^2\theta - \sin^2 \theta$$
Here is what I have tried
$$\tan^2\theta \sin^2\theta$$
$$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$
$$=\frac{\sin^4\theta}{\cos^2\theta}$$
Not much of an attempt, but now I am stuck. What should I do next? Thanks in advance for your answers ;)
| $$\tan^2\theta \sin^2\theta$$
$$=\left(\frac{\sin^2\theta}{\cos^2\theta}\right)\left(\sin^2\theta\right)$$
$$=\frac{\sin^4\theta}{\cos^2\theta}$$
$$=\frac{\sin^2\theta\sin^2\theta}{\cos^2\theta}$$
$$=\frac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}$$
$$=\frac{\sin^2\theta-\sin^2\theta\cos^2\theta}{\cos^2\theta}$$
$$=\frac{\sin^2\theta}{\cos^2\theta}-\frac{\sin^2\theta\cos^2\theta}{\cos^2\theta}$$
$$=\tan^2\theta-\sin^2\theta$$
$$\displaystyle \boxed{\therefore \tan^2\theta \sin^2\theta=\tan^2\theta - \sin^2\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/711580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What values of a is the set of vectors linearly dependent? The question is is "determine conditions on the scalars so that the set of vectors is linearly dependent".
$$ v_1 = \begin{bmatrix} 1 \\ 2 \\ 1\\ \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}
$$
When I reduce the matrix I get
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & a-2 & 0 \\ 0 & 0 & b - \frac{4}{(a-2)} \end{bmatrix}$$
If the matrix is linearly independent then shouldn't $a-2 = 0$ and $b - \frac{4}{(a-2)} = 0$? So, I said the solution is when $a-2 \neq 0 $ and $b - \frac{4}{(a-2)} \neq 0$. The textbooks says the answer is when $ b(a-2) = 4 $. I understand how they got to $ b(a-2) = 4 $ but why is it equals instead of not equals?
| let be $\{v_1,v_2,v_3\} \subseteq \Bbb{R}^3$, with
$$ v_1 = \begin{bmatrix} 1 \\ 2 \\ 1\\ \end{bmatrix}, v_2 = \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix}, v_3 = \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}
$$
$\{v_1,v_2,v_3\}$ is linearly dependent iff $\{v_1,v_2,v_3\}$ is not independent, therefore if $$\forall \alpha_1,\alpha_2,\alpha_3 \in \Bbb{R}(\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3 \cdot v_3=0_{\Bbb{R}^3} \to \alpha_1=\alpha_2=\alpha_3=0)$$ is false. I consider $$\alpha_1 \cdot v_1 + \alpha_2 \cdot v_2 + \alpha_3 \cdot v_3=\alpha_1 \cdot \begin{bmatrix} 1 \\ 2 \\ 1\\ \end{bmatrix} + \alpha_2 \cdot \begin{bmatrix} 1 \\ a \\ 3 \\ \end{bmatrix} + \alpha_3 \cdot \begin{bmatrix} 0 \\ 2 \\ b \\ \end{bmatrix}=$$$$=\begin{bmatrix} \alpha_1 \\ 2\alpha_1 \\ \alpha_1\\ \end{bmatrix} + \begin{bmatrix} \alpha_2 \\ a\alpha_2 \\ 3\alpha_2 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 2\alpha_3 \\ b\alpha_3 \\ \end{bmatrix}=\begin{bmatrix} \alpha_1+\alpha_2 \\ 2\alpha_1+a\alpha_2+2\alpha_3 \\ \alpha_1+3\alpha_2+b\alpha_3 \\ \end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$$ I consider the linear system $\Sigma:=\left\{\begin{matrix}
\alpha_1+\alpha_2=0\\
2\alpha_1+a\alpha_2+2\alpha_3=0\\
\alpha_1+3\alpha_2+b\alpha_3=0
\end{matrix}\right.$ it is homogeneous system and $$\mathbf{rnk}(\Sigma)\neq3 \leftrightarrow Sol(\Sigma)\neq\{(0,0,0)\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Gram-Schmidt of 2 by 2 matrix Given $A=\begin{bmatrix}
2 &-1 \\
-1 & 2
\end{bmatrix}$, I take the first column of $A$, and divide it by its norm to find $q_{1}=\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix}$
Now if I denote the second column of $A$ as $b$, then $b-(q_{1}^{T}b)q_{1}$ divided by $||b-(q_{1}^{T}b)q_{1}||$ will give me $q_{2}$, correct?
$\begin{bmatrix}
-1 \\
2
\end{bmatrix}-(\begin{bmatrix}\frac{2}{\sqrt{5}} \frac{-1}{\sqrt{5}} \end{bmatrix}\begin{bmatrix}
-1 \\
2
\end{bmatrix})\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix}$
$=\begin{bmatrix}
-1 \\
2
\end{bmatrix}-(\frac{-4}{\sqrt{5}}\begin{bmatrix}\frac{2}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}}\end{bmatrix})$
$=\begin{bmatrix}
-1 \\
2
\end{bmatrix}-\begin{bmatrix}
-8/5 \\
1/5
\end{bmatrix}$
$=\begin{bmatrix}
13/5 \\
9/5
\end{bmatrix}$
And its norm is $\sqrt{10}$, so $q_{2}=\begin{bmatrix}
13/(5\sqrt{10}) \\
9/(5\sqrt{10})
\end{bmatrix}$
However this is not the right answer in the book. Where did I go wrong?
| The error was: $\displaystyle \left(\frac{-4}{\sqrt{5}}\right)\left(\frac{-1}{\sqrt{5}}\right)\neq \frac{1}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of numbers not divisible by $2,3,5,7$ or $11$ between multiples of $2310$ Looking at partitions of the natural number line of the form $P=[a,b)$, I noted that
*
*if $a$ and $b$ are multiples of $6$, there exist at least $2$ numbers in the partition which are not divisible by $2$ or $3$
*if $a$ and $b$ are multiples of $30$, there exist $8$ numbers in the partition which are not divisible by $2, 3$ or $5$
*if $a$ and $b$ are multiples of $210$, there exist $54$ numbers in the partition which are not divisible by $2,3,5$ or $7$.
This leads me to guess that if $a$ and $b$ are multiples of $2310$, there exist $592$ numbers in the partition which are not divisible by $2,3,5,7$ or $11$. Is this true?
(I arrived at $592$ because it is equal to $54 \times 11 - 2$ and $2310$ because it is equal to $210 \times 11$.)
| Use chinese remainder representation with basis $[2,3,5,7,11]$.
Given a CRR basis $[a,b,c,d,e]$ with $M = abcde = 2\times 3 \times 5 \times 7 \times 11 = 2310$, each valid tuple represents exactly 1 number from $0$ to $M - 1 \pmod M$. So you want the number of tuples that don't contain a zero, or $(a - 1)(b - 1)(c - 1)(d - 1)(e - 1)$ or $1 \times 2 \times 4 \times 6 \times 10 = 480$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Another Verify the identity: $\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$ Another Verify the identity that I can't get:
$$\sec^2 \frac{x}{2} = \frac{2}{1+\cos x}$$
$$ = \frac{1 + \left(\frac{1}{\cos x}\right)}{2}$$
$$ = \frac{\cos x + 1}{2 \cos x}$$
| Again, both ways are possible :
Going backwards (which is simpler, coincidentally)
$$\frac{2}{1 + \cos x} = \frac{2}{1 + (2\cos^2\frac{x}{2} - 1)}\\
= \frac{2}{2\cos^2\frac{x}{2}}\\
= \sec^2\frac{x}{2}$$
Going forward, start by rewriting as:
$$\sec^2\frac{x}{2} = \frac{1}{\cos^2\frac{x}{2}}$$
By the double angle formula for $\cos$, $\cos 2\theta = 2\cos^2\theta - 1$ we have (by letting $\theta = \frac{x}{2}$):
$$\cos x = 2\cos^2\frac{x}{2} - 1$$
Rearrange to get
$$\cos^2\frac{x}{2} = \frac{1 + \cos x}{2}$$
Simply substitute this back, to deduce
$$\sec^2\frac{x}{2} = \frac{1}{\frac{1 + \cos x}{2}} = \frac{2}{1 + \cos x}$$
It's probably worth mentioning that both approaches are equivalent, differing only by which direction you choose to prove the identity.
| {
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Linear independence and dependence $u = (1, 1, 0)$, $v = (1, 4, 1)$, $w = (r^2, 1, r^2)$, $b = (3+2r, 5+12r, 2r)$
a) For which values of $r$, is the set $(u,v,w)$ linearly independent? --> I got $r = \pm 0.5$ via ERO. Could someone just show me the matrix, so I know if I was on the right track?
b) For which values of $r$ is the vector $b$ a linear combination of $u,v,w$, and for which of these values of $r$ can be written as a linear combination in more than one way? --> I got $r = 1$ or $-2$ but how to work out this part "or which of these values of $r$ can be written as a linear combination in more than one way?"
Thanks folks
| Let the matrix A be the $3\times3$ matrix whose columns are the vectors $u, v\text{ and } w$. Form the augmented matrix, $[A | b]$ .
$$\left[ \begin{array}{ccc|c}1&1&r^2&3+2r\\1&4&1&5+12r\\0&1&r^2&2r\end{array} \right]$$
Do elementary row operations to get the left side of the augmented matrix into row echelon form.
$$\left[ \begin{array}{ccc|c}1&1&r^2 &3+2r\\1&4&1&5+12r\\0&1&r^2&2r
\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\1&4&1&5+12r\\0&1&r^2&2r\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\0&4&1&2+12r\\0&1&r^2&2r\end{array} \right] \Rightarrow \left[ \begin{array}{ccc|c}1&0&0&3\\0&4&1&2+12r\\0&0&r^2-\frac{1}{4}&-r-\frac{1}{2}\end{array} \right]$$
The last row is the key to the whole question.
For part a) if $r^2-\frac{1}{4}=0$, the the rows of $A$ are linearly dependent and so too are the columns of the original matrix, $A$. Thus, If $r = \pm \frac{1}{2}$, the the vectors $u, v, w$ are linearly dependent. This is easily verified since in this case, $v = 4w$. The vectors are independent for any other value of $r$. Namely the vectors $u, v\text{ and } w$ are independent for any $r$ such that $r \ne \pm \frac{1}{2}$.
For part b) we again see that if $r \ne \pm \frac{1}{2}$, the matrix $A$ is full rank and the system $Ax = b$ has a unique solution for any $b$. If, however, $r = \pm \frac{1}{2}$ the system has either infinitely many solutions or no solutions. Notice that when we do back substitution the last row says $(r^2-\frac{1}{4})x_3 = -(r+\frac{1}{2})$. If we let $r = \frac{1}{2}$, then we get $0\,x_3 = -1$ and the system of equations is inconsistent and there is no solution. If however, we let with $r = -\frac{1}{2}$, we get $0\,x_3 = 0$ and there are infinitely many solutions.
| {
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Prove that $\frac{x \log(x)}{x^2-1} \leq \frac{1}{2}$ for positive $x$, $x \neq 1$. I'd like to prove
$$\frac{x \,\log(x)}{x^2-1} \leq \frac{1}{2} $$
for positive $x$, $x \neq 1$.
I showed that the limit of the function $f(x) = \frac{x \,\text{log}(x)}{x^2-1}$ is zero as $x$ tends to infinity. But not sure what to do next.
| By the standard inequality $e^x>1+x$, we have $x-1>\log(1+x-1)=\log(x)$ thus
$$\frac{x\log(x)}{x^2-1}<\frac{x(x-1)}{x^2-1}=\frac{x}{x+1}\le \frac{1}{2} \text{ for } x< 1.$$
For $x>1$, note that the derivative is
$$\frac{(\log(x)+1)(x^2-1)-2x^2\log(x)}{(x^2-1)^2}=\frac{x^2-1-(x^2+1)\log(x)}{(x^2-1)^2}$$
which is negative since the numerator is $0$ at $1$, clearly negative for large $x$, and can only be $0$ when $\log(x)=\frac{x^2-1}{x^2+1}$, but the derivative of $\log(x)$ is $1/x$ while the derivative of the $\frac{x^2-1}{x^2+1}$ is $4/(x^2+1)^2<1/x$ for $x>1$ so this can never be the case. Finally, since the limit as $x\to 1$ is
$$\lim\limits_{x\to 1}\frac{x\log(x)}{x^2-1}=\lim\limits_{x\to 1}\frac{\log(x)+1}{2x}=\frac{1}{2}$$
by L'Hopital's rule, the desired inequality follows.
| {
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Integrate $\int\frac{5x-7}{x^2-3x+2}$ I want to integrate $\int\frac{5x-7}{x^2-3x+2}$ but my result differs from the one on Wolframalpha http://www.wolframalpha.com/input/?i=integrate+%285x-7%29%2F%28x%5E2-3x%2B2%29
I did the following steps:
$$\frac{5x-7}{(x-2)(x-1)} = \frac{A}{x-2}+\frac{B}{x-1}$$
$$5x-7 = A(x-1)+B(x-2)$$
$$5 = A + B$$
$$-7 = -A-2B$$
$$A=5-B\\
-7= -(5-B)-2B\\
-2 = -B\\
2 = B$$
Therefore $A = 3$ and $B=2$
$$\int\frac{3}{x-2}+\int\frac{2}{x-1}=3\ln(x-2)+2\ln(x-1) + C$$
While on Wolframalpha it is $3\ln(2-x)+2\ln(1-x)$ Where did I do the error?
| Neither is quite right. We have $\int \frac{1}{u}\,du=\ln(|u|)+C$.
| {
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Proof that 6 divides $a \in \mathbb{Z}, a(a^2 - 7)$ I am trying to prove a question from my tutorial sheet, is this an acceptable proof?
Six cases exist:
$$a,k \in \mathbb{Z}, a(a^2 - 7) = 6k
\\\text{Proof:}\\
a = 0 \mod 6 \longrightarrow a^2 = 0 \mod 6 \longrightarrow a^2 - 1\mod 6 = 5 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 1 \mod 6 \longrightarrow a^2 = 1 \mod 6 \longrightarrow a^2 - 1\mod 6 = 0 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 2 \mod 6 \longrightarrow a^2 = 4 \mod 6 \longrightarrow a^2 - 1\mod 6 = 3 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 3 \mod 6 \longrightarrow a^2 = 3 \mod 6 \longrightarrow a^2 - 1\mod 6 = 2 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 4 \mod 6 \longrightarrow a^2 = 4 \mod 6 \longrightarrow a^2 - 1\mod 6 = 3 \longrightarrow a(a^2 - 7) = 0\mod 6\\
a = 5 \mod 6 \longrightarrow a^2 = 1 \mod 6 \longrightarrow a^2 - 1\mod 6 = 0 \longrightarrow a(a^2 - 7) = 0\mod 6\\
$$
Therefore, since for all $a$, $a(a^2 - 7) = 0 \mod 6$, $6$ divides $a(a^2 - 7)$.
Is there an easier proof, or a better way to lay this out via $LaTeX$?
| HINT:
$$a(a^2-7)= a(a^2-1)-6a=(a-1)a(a+1)-6a$$
Now the first term being the product of $3$ consecutive integers is divisible by $3!$
| {
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Why $\cos^3 x - 2 \cos (x) \sin^2(x) = {1\over4}(\cos(x) + 3\cos(3x))$? Wolfram Alpha says so, but step-by-step shown skips that step, and I couldn't find the relation that was used.
| Start by the left hand side. You can calculate it known that:
*
*$\cos (x+2x)=\cos x\cos 2x-\sin x\sin 2x,$
*$\cos 2x=2\cos^2 x -1,$
*$\sin 2x=2\sin x\cos x,$
*$\cos^2 x+\sin^2 x=1.$
$$\frac{1}{4}\left(\cos x+3\cos 3x\right)=\frac{1}{4}\left(\cos x+3\left(4\cos^3 x-3\cos x\right)\right)=3\cos^3 x-2\cos x=\cos^3 x-2\cos x\left(1-\cos^2 x\right)$$
| {
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Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after equating the remainders.
$p(x)= kx^3 + 4x^2 + 3x - 4/(x – 3)$: remainder= $(30k+32)$
$q(x)= x^3 - 4x + k/(x – 3)$: remainder= $(k+15)$
So,
$30k+32=k+15$
=> $k=(-17/29)$
Any help would be appreciated. :)
| Your first remainder is wrong since $$\frac{k x^3+4 x^2+3 x-4}{x-3}=k x^2+(3 k+4) x+3 (3 k+5)+\frac{27 k+41}{x-3}$$ I am sure that you can take from here.
| {
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Distribution of different objects into different boxes We want to put n different objects into n different boxes. In how many ways can we do this if we want that exactly two boxes remain empty?
| The number of ways to choose the two boxes that are to remain empty is a combination:
$$\binom{n}{2} = \frac{n(n-1)}{2}$$
As for the remaining boxes, there can either be $1$ box with $3$ objects and $n-3$ boxes with $1$ object each, or $2$ boxes with $2$ objects each and $n-4$ boxes with $1$ object each.
Case 1: One box has three objects.
The number of ways to choose this special box is $n-2$. The number of ways to choose the three objects that go in this one box is $$\binom{n}{3}$$. Then, the number of ways to permute the remaining $n-3$ objects in the $n-3$ boxes is simply $(n-3)!$. Thus, the total number of ways to place the objects in this case is $$\binom{n}{2} \cdot (n-2) \cdot \binom{n}{3} \cdot (n-3)!$$
Case 2: Two boxes have two objects.
The number of ways to choose the two special boxes is $$\binom{n-2}{2}$$ Once these boxes are chosen, the number of ways to choose the objects that go in the first of these is $$\binom{n}{2}$$ After choosing these, the number of ways to choose the objects that go in the second box with two things in it, since there are $n-2$ objects left to choose from, is $$\binom{n-2}{2}$$ Finally, the number of ways to permute the remaining $n-4$ objects in the remaining $n-4$ boxes is $(n-4)!$. This gives the number of placements in this case as $$ \binom{n-2}{2} \cdot \binom{n}{2} \cdot \binom{n-2}{2} \cdot (n-4)!$$
The final answer is the sum of the answers in cases 1 and 2, which is
$$\boxed{\binom{n}{2} \cdot (n-2) \cdot \binom{n}{3} \cdot (n-3)! + \binom{n-2}{2} \cdot \binom{n}{2} \cdot \binom{n-2}{2} \cdot (n-4)!}$$
Edit: I noticed that the final answer, as I wrote it, could be simplified quite nicely. First, we can expand the binomial coefficients and combine $n-2$ and $(n-3)!$:
$$\frac{n(n-1)}{2} \cdot (n-2)! \cdot \frac{n(n-1)(n-2)}{6} + \frac{(n-2)(n-3)}{2} \cdot \frac{n(n-1)}{2} \cdot \frac{(n-2)(n-3)}{2} \cdot (n-4)!$$
Then, we find that both terms have $n!$ in them:
$$\frac{n!}{2}\binom{n}{3} + \frac{n!}{4}\binom{n-2}{2}$$
$$\frac{n!}{4}\cdot\frac{n(n-2)(n-3)}{3} + \frac{n!}{4}\cdot\frac{(n-2)(n-3)}{2}$$
This ends up as the relatively concise expression
$$\frac{n!(n-2)(n-3)}{4}\left(\frac{n}{3}+\frac{1}{2}\right)$$
| {
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If $\sum x_k= \sqrt{n}$, then $\sum\frac{x_k^2}{(x_k^2+1)^2}\leq\frac{n^2}{(n+1)^2}$ A new question has emerged after this one was successfully answered by r9m: If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$. I thought of this generalization. Does it hold?
$$\dfrac{x_1^2}{(x_1^2+1)^2}+\dfrac{x_2^2}{(x_2^2+1)^2}+\cdot\cdot\cdot+\dfrac{x_n^2}{(x_n^2+1)^2}\le \dfrac{n^2}{(n+1)^2}$$ with $$x_1+x_2+\cdot\cdot\cdot+x_n= \sqrt{n}$$ $$x_1,x_2,\cdot\cdot\cdot,x_n \ge0$$ $$ n \in \mathbb{N}$$
| I'm afraid the inequality is wrong. Note that RHS $< 1$. However, if we take $x:=1$, then $x^2/(x^2+1)^2=1/4$. Thus, we choose a large $n$, and let as much $x_i$ as possible be $1$, then the inequality fails. But I believe that there exists a bound of $n$ to let the inequality hold.
| {
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Integrate with square root in square $\int \left(1 + \sqrt{\frac{x-1}{x+1}}\right)^2 dx$
How would you attack this? I've tried variable substitution with $t = \sqrt{\frac{x-1}{x+1}}$
| OK
$$
t = \sqrt{\frac{x-1}{x+1}} \\
x = \frac{1+t^2}{1-t^2} \\
dx = \frac{4t}{(1-t^2)^2}\;dt = \frac{4t}{(1-t)^2(1+t)^2}\;dt \\
\int\left(1+\sqrt{\frac{x-1}{x+1}}\right)^2 dx =
\int (1+t)^2\frac{4t}{(1-t)^2(1+t)^2}\;dt
= \int \frac{4t}{(1-t)^2}\;dt \\
\qquad = \int\left(\frac{4}{t-1} + \frac{4}{(t-1)^2}\right) dt
= 4\log(t-1) - \frac{4}{t-1} +C
$$
and substitute back to get the answer in terms of $x$.
| {
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Finding the matrix exponantial I have to calculate the following:
$e^{xA} , A=
\begin{pmatrix}1 & 1 & 1 \\
1&1&1\\
1&1&1\\
\end{pmatrix}$
I use the following rule:
$e^{xA} = \sum \frac{x^k}{k!}A^k$
Now I'm looking for a matrix such that:
$A=S D S^{-1}$.
So I calculated the eigenvalues and eigenvectors of A, these are:
$\lambda _0=3 , \lambda_{1,2} = 0 $
$\vec{y_0} = \begin{pmatrix}
1\\
1\\
1
\end{pmatrix} $,$\vec{y_1} = \begin{pmatrix}
1\\
-1\\
0
\end{pmatrix} $,$\vec{y_2} = \begin{pmatrix}
1\\
0\\
-1
\end{pmatrix}$
So I have the following 3 matrices:
$
S = \begin{pmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{pmatrix}
, D = \begin{pmatrix}
3&0&0\\
0&0&1\\
0&0&0
\end{pmatrix} , S^{-1}=
\begin{pmatrix}
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & \frac{1}{3} & -\frac{2}{3} \\
\frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \\
\end{pmatrix}
$
But when I calculate $SDS^{-1}$ I don't get A.
Can someone help me where I have gone wrong. I thought I understood it up until this point, but there must be something wrong with my reasoning because I have checked every thing with mathematica.
I also don't know what to do when I have found the correct S and D.
| Hint: for $n\gt0$, $A^n=3^{n-1}A$ This should be enough to compute $e^{xA}$.
You have the wrong Jordan Normal Form for $A$. Since $A$ is Hermitian, it is diagonalizable:
$$
\begin{bmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{bmatrix}
\begin{bmatrix}
3&0&0\\
0&0&0\\
0&0&0
\end{bmatrix}
\begin{bmatrix}
1&1&1\\
1&-1&0\\
1&0&-1
\end{bmatrix}^{-1}
=
\begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1
\end{bmatrix}
$$
| {
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Recurrence problem for $a_5$ Assume that the sequence $\{a_0,a_1,a_2,\ldots\}$ satisfies the recurrence $a_{n+1} = a_n + 2a_{n−1}$. We know that $a_0 = 4$ and $a_2 = 13$. What is $a_5$?
| From the recurrence you get $a_1=5$. Write the recurrence as:
$$
a_{n + 2} = a_{n + 1} + 2 a_n \qquad a_0 = 4, a_1 = 5
$$
Define the generating function:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Multiply the recurrence by $z^n$ and sum over $n \ge 0$. Notice that:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2} \\
\end{align}
to get:
$$
\frac{A(z) - 4 - 5 z}{z^2} = \frac{A(z) - 4}{z} + 2 A(z)
$$
Solving this equation for $A(z)$, and splitting into partial fractions:
$$
A(z) = \frac{4 + z}{1 - z - 2 z^2} = \frac{1}{1 + z} + \frac{3}{1 - 2 z}
$$
This are just geometric series:
$$
a_n = (-1)^n + 3 \cdot 2^n
$$
So the value sought is $a_5 = (-1)^5 + 3 \cdot 2^5 = 95$
| {
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Express this polar equation in cartesian form Having trouble converting this polar equation into Cartesian form:
$r = 2 + \sin(\theta)$
This is how far I get:
$(r = 2 + \sin(\theta))\cdot r$
$r^2 = 2r + r\sin(\theta)$
$x^2 + y^2 = 2r + y$, since $r^2 = x^2 + y^2$ and $y = r\sin(\theta)$
Where do I go from here or where did I go wrong? Thanks
| If $r=2+\sin \theta$, as $r^2=x^2+y^2$ and $y=r\sin \theta$, then multiplying the equation by $r$ we obtain
$$ r^2=2r+r\sin \theta ,$$
and then
$$x^2+y^2=2\sqrt{x^2+y^2}+y, $$
which is a cardioid.
Hope this helps.
| {
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Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$
Why is $\lim\limits_{n\to\infty}(1+\frac{1}{n^2})^n = 1$?
Could someone elaborate on this? I know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.
| You can do this without knowing anything about $e$. Using Bernoulli's Inequality we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \geq 1 - \frac{1}{n^{3}}$$ and clearly we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}$$ so that $$1 - \frac{1}{n^{3}} \leq \left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}\tag{1}$$ and via Squeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{4}}\right)^{n} = 1\tag{2}$$ Using similar technique we get the inequality $$1 - \frac{1}{n} \leq \left(1 - \frac{1}{n^{2}}\right)^{n} \leq 1 - \frac{1}{n^{2}}$$ and via Sqeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{2}}\right)^{n} = 1\tag{3}$$ and dividing $(2)$ by $(3)$ we get $$\lim_{n \to \infty}\left(1 + \frac{1}{n^{2}}\right)^{n} = 1\tag{4}$$
| {
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Proof of the image of a set $\mathbb R$ be the set of real numbers.
I have a set A = {$x$ ∈ $\mathbb R$ | $0< x$} . A function g : A --> $\mathbb R$ is defined by
$f(x) = x^2
How do i prove whether the Image(A) = $\mathbb R$
| Let $r$ be any real number. Consider $g(x) = r$. So: $\dfrac{2x-1}{2x - 2x^2} = r$. So: $2x - 1 = 2rx - 2rx^2$, and $2rx^2 + 2(1 - r)x - 1 = 0$. So: $ D = b'^2 - ac = (1 - r)^2 + 2r = r^2 + 1 > 0$ implying the equation always has 2 distinct real zeroes. Let's look at one of the zeroes $x = \dfrac{r - 1 + \sqrt{r^2 + 1}}{2r}$. There are three cases to consider:
Case 1: $r < 0$. Then $r^2 - 2r + 1 > r^2 + 1$. So $(r - 1 - \sqrt{r^2 + 1})(r - 1 + \sqrt{r^2 + 1}) > 0$. This implies $r - 1 + \sqrt{r^2 + 1} < 0$. Hence: $\dfrac{r - 1 + \sqrt{r^2 + 1}}{2r} > 0$. Thus $x > 0$. Now $x - 1 = \dfrac{r - 1 + \sqrt{r^2 + 1}}{2r} - 1 = \dfrac{-1 -r + \sqrt{r^2 + 1}}{2r} < 0$ because $\sqrt{r^2 + 1} > 1 + r$ when $r < 0$ and this inequality is easily verified. So $x < 1$, and $0 < x < 1$. Thus if $r < 0$ we can always choose an $x$ such that $0 < x < 1$ and $g(x) = r$.
Case 2: $r > 0$. Observe that for this case $\sqrt{r^2 + 1} > 1 - r$. So $x > 0$, and $-1 - r + \sqrt{r^2 + 1} < 0$ because $\sqrt{r^2 + 1} < 1 + r$ when $r > 0$. So $x < 1$. Thus for this case we also have $0 < x < 1$ such that $g(x) = r$.
Case 3: $r = 0$. Simply choose $x = \frac{1}{2}$, and we readily check that $g(\frac{1}{2}) = 0$.
So all three cases imply that Image($A$) = $\mathbb R$ as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding an asymptotic expansion for $\sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}}$ It is well known that an asymptotic expansion of the n-th harmonic number is $$H_{n}= \sum_{k=1}^{n} \frac{1}{k} \sim \ln(n) + \gamma + \frac{1}{2n} -\frac{1}{12n^{2}} + O(n^{-4}).$$
How could we find an asymptotic expansion for the sum $ \displaystyle \sum_{k=0}^n \frac{1}{1+\frac{k}{n}}$ to a similar order?
| Let $ \displaystyle f(x) = \frac{1}{1+\frac{x}{n}}$.
Using the Euler-Maclaurin summation formula, we get
$$ \begin{align} \sum_{k=0}^{n} \frac{1}{1+\frac{k}{n}} &\sim \int_{0}^{n} \frac{1}{1+\frac{x}{n}} \, dx + \frac{f(n)+f(0)}{2} + \sum_{m=1}^{\infty} \frac{B_{2m}}{(2m)!} \left(f^{(2m-1)}(n) -f^{(2m-1)}(0) \right) \\ &= n \ln \left( 1+\frac{x}{n} \right) \Bigg|^{n}_{0} + \frac{\frac{1}{2}+1}{2} + \frac{1}{6} \left(\frac{1}{2!} \right) \left(-\frac{1}{n} \frac{1}{(1+\frac{x}{n})^{2}} \right) \Bigg|^{n}_{0} \\ &- \frac{1}{30} \frac{1}{4!} \left( - \frac{6}{n^{3}} \frac{1}{(1+ \frac{x}{n})^{4}}\right) \Bigg|^{n}_{0} + \mathcal{O}(n^{-5}) \\ &= n \ln (2) + \frac{3}{4} + \frac{1}{16n} - \frac{1}{128 n^{3}} + \mathcal{O}(n^{-5}) \end{align}$$
For $n=20$, the above approximation is correct to $8$ digits after the decimal point.
| {
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Find the last two digits of the number $9^{9^9}$
Find the last two digits of the number $9^{9^9}$ .
[Hint: $9^9 \equiv 9 \pmod {10} $; hence, $9^{9^9}$ = $9^9+10k$ ;now use the fact that $9^9 \equiv 89 \pmod {100}$]
| Our observations start like this;
$9=-1 ($mod $10) \Rightarrow 9^9=(-1)^9 ($mod $10)$ or $9^9=-1 $(mod $10)$
We have $-1=9 ($mod $10)$, thus $9^9=9 ($mod $10)$ ...i
This statement directly implies that upon dividing $9^9$ by $10$ we get a remainder $9$.Thus $9^9=9+10k$
Also,$9^9=89 ($mod $100) \Rightarrow 9*9^9=9*89 ($mod $100)$
or $9^{10}=801($mod $100)$. Also, $801=1 ($mod $100)$ thus making
$9^{10} = 1 \pmod{100}$ ....ii
$(9^{10})^k=1^k ($mod $100)$ from ii
$9^9*(9^{10})^k=9^9 ($mod $100)$ or, $9^{(9+10k)}=9^9 ($mod $100)$. Since,$9^9=89 ($mod $100)$, we have $9^{(9+10k)}=89 ($mod $100)$
This statement implies that upon division of $9^{9^9}$ by $100$ we get a remainder of $89$. Thus last two digits are $89$.
| {
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Show that $y/x$ tends to a finite limit as $x \to + \infty$ and determine this limit.
Let $y=f(x)$ be that solution of the differential equation $$y' = \frac{2y^2+x}{3y^2+5}$$ which satisfies the initial condition $f(0)=0$. (Do not attempt to solve this differential equation.)
(a) The diffeerential equation shows that $f'(0)=0$ . Discuss whether $f$ has a relative maximum or minimum or neither at $0$.
(b)Notice that $f'(x) \geq 0 $ for each $ x \geq0$ and that $f'(x) \geq \frac{2}{3}$ for each $x \geq \frac{10}{3}$.Exhibit two positive numbers $a$ and $b$ such that $f(x) > ax-b$ for each $x\geq \frac{10}{3}$.
(c) Show that $x/y^2 \to 0$ as $x \to + \infty$. Give full details of your reasoning.
(d) Show that $y/x$ tends to finite limit as $x \to + \infty$ and determine this limit.
a) From the equation I have $$y''= \frac{(2yy'+1)(3y^2+5)-(2y^2+x)(3yy')}{(3y^2+5)^2}=\frac{3y^2+5-x3yy'}{(3y^2+5)^2}$$ so $y''>0$ when $x=0$ ,$f$ has a relative minimum at 0.
b) Let $a= \frac{2}{3},b=\frac{20}{9}$,we know $f(x)>0=ax-b$ when $x=\frac{10}{3}$.
Let $g(x)=f(x)-ax+b$ then $g'(x)=f'(x)-a \geq 0$ for $x \geq \frac{10}{3}$ . so $g(x)\geq g(\frac{10}{3})>0$ for $x \geq \frac{10}{3}$ . It means $f(x) > ax-b$ for each $x\geq \frac{10}{3}$.
c) For those $x\geq 10/3$ I have $0< \frac{x}{y^2} <\frac{x}{(\frac{2}{3}x-\frac{20}{9})^2}$ as $x \to + \infty$.By the sandwiches theorem $\frac{x}{y^2} \to 0$ as $x \to + \infty$.
d) Notice $y > ax -b$ ,so $y \to + \infty$ as $ x \to + \infty$ the limit $\frac{y}{x}$ has the form $\frac{+ \infty}{+ \infty}$.It seems like L'Hopital's rule stuff.But the book only gives application and proof of LHR in the form $\frac{0}{0}$.So I write $y/x=\frac{1/x}{1/y}$ in the form $0/0$, use L'Hopital's rule I get $$\frac{y^2}{x^2} \frac{2+\frac{x}{y^2}}{3+\frac{5}{y^2}}$$ So I'm stuck.
Any help is appreciated.
| Even though the book gives L'Hospital rule only for the case when $f(x)$ and $g(x)$ tend to $0$, the rule holds also for $\frac{\infty}{\infty}$ case, too.
Since $\lim_{x \to +\infty} \frac{y'}{x'} = \lim_{x \to +\infty} \frac{2y^2}{3y^2 + 5} + \lim_{x \to +\infty} \frac{x}{3y^2 + 5} = \frac{2}{3} + 0$, the answer to the part d) is $\frac{2}{3}$. The last equality is coming from composition theorem $3.5$, and the part c).
| {
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Convert from base $10$ to base $5$ I am having a problem converting $727$(base $10$) to base $5$. What is the algorithm to do it?
I am getting the same number when doing so: $7\times 10^2 + 2\times10^1+7\times10^0 = 727$, nothing changes.
Help me figure it out!
| You have to divide each number by 5 :
$$\begin{aligned}
727 &= 5 \cdot 145 &+ \color{red}{2}\\
145 &= 5 \cdot 29 &+ \color{green}{0}\\
29 &= 5 \cdot 5 &+ \color{blue}{4}\\
5 &= 5 \cdot 1 &+ \color{magenta}{0}\\
1 &= 5 \cdot 0 &+ \color{brown}{1}
\end{aligned}
$$
Hence the answer is $\color{brown}{1}\color{magenta}{0}\color{blue}{4}\color{green}{0}\color{red}{2}_5 = 1 \cdot 5^4 + 0 \cdot 5^3 + 4 \cdot 5^2 + 0 \cdot 5^1 + 2 \cdot 5^0 = 727$
| {
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"timestamp": "2023-03-29T00:00:00",
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Remainder of a high fibonacci number I found a question in my assessment book:
What is the remainder when the 1995th number of the
fibonacci sequence is divided by 8?
How to solve?
| *
*$a_{ 0} \equiv 0 \bmod 8$
*$a_{ 1} \equiv 1 \bmod 8$
*$a_{ 2} \equiv 1 \bmod 8$
*$a_{ 3} \equiv 2 \bmod 8$
*$a_{ 4} \equiv 3 \bmod 8$
*$a_{ 5} \equiv 5 \bmod 8$
*$a_{ 6} \equiv 0 \bmod 8$
*$a_{ 7} \equiv 5 \bmod 8$
*$a_{ 8} \equiv 5 \bmod 8$
*$a_{ 9} \equiv 2 \bmod 8$
*$a_{10} \equiv 7 \bmod 8$
*$a_{11} \equiv 1 \bmod 8$
You can prove by induction that $a_{i} \equiv a_{i-12} \bmod 8$, hence $a_{i} \equiv a_{i \bmod 12} \bmod 8$
Therefore $a_{1995} \equiv a_{3} \bmod 8 \equiv 2 \bmod 8$
So the remainder of the $1995th$ ($0$-based count) Fibonacci number divided by $8$ is $2$.
| {
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Show that $\sin 10^\circ$ is irrational So, this is the problem I am working on.
Show that $\sin 10^\circ$ is irrational.
The solution to the problem is $$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ .$$
Let $$x = 2\sin 10^\circ.$$
Then we have, $$x^3 - 3x + 1 = 0.$$
And, we have to work on this to find out the roots. But, what I don't understand is that why do I have to subtract $4\sin^3 10^\circ$ from $3\sin 10^\circ$. And, how did they come up with $x^3 - 3x+1 = 0?$ I am confused. Can someone please explain this in details and is there any other way we can do this problem?
| identity: $\sin(3a)=3\sin(a)-4\sin^3(a)$ By using this identity,
$$1/2 = \sin 30^\circ = 3 \sin 10^\circ - 4\sin^3 10^\circ$$
$$1=2\sin 30^\circ = 6 \sin 10^\circ - 8\sin^3 10^\circ$$ Then if you set $x=2\sin(10)$ you will get $$x^3 - 3x+1 = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Multiplying Adjacent Matrices? My teacher hasn't explained it too well, so i'm looking for an explanation:
$$A =
\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}$$
$$A^2 =
\begin{pmatrix}
2 & 1 & 1\\
1 & 2 & 1\\
1 & 1 & 2
\end{pmatrix}$$
I would have thought the first number in $A^2$ would be $0$, as $0 \cdot 0 = 0$, and am struggling to see where the $2$ comes from.
| The first entry in $A^2$ comes from the first entry in $AA$, and that entry is $0\cdot 0 + 1\cdot 1 + 1 \cdot 1 = 2$.
Likewise, the entry in the first row, second column of $A^2$ is $0\cdot 1 + 1\cdot 0 + 1 \cdot 1 = 1$.
Perhaps you'll want to revisit matrix multiplication.
In particular, Wikipedia provides this example for arbitrary $3\times 3$ matricies:
Square matrices
If
$$\mathbf{A} = \begin{pmatrix} a & b & c \\ p & q & r \\ u & v & w \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} \alpha & \beta & \gamma \\ \lambda & \mu & \nu \\ \rho & \sigma & \tau \\ \end{pmatrix}\,,$$
their matrix products are:
$$\mathbf{AB} = \begin{pmatrix} a & b & c \\ p & q & r \\ u & v & w \end{pmatrix} \begin{pmatrix} \alpha & \beta & \gamma \\ \lambda & \mu & \nu \\ \rho & \sigma & \tau \\ \end{pmatrix} =\begin{pmatrix} a\alpha + b\lambda + c\rho & a\beta + b\mu + c\sigma & a\gamma + b\nu + c\tau \\ p\alpha + q\lambda + r\rho & p\beta + q\mu + r\sigma & p\gamma + q\nu + r\tau \\ u\alpha + v\lambda + w\rho & u\beta + v\mu + w\sigma & u\gamma + v\nu + w\tau \end{pmatrix}\,,$$
and
$$\mathbf{BA} = \begin{pmatrix} \alpha & \beta & \gamma \\ \lambda & \mu & \nu \\ \rho & \sigma & \tau \\ \end{pmatrix} \begin{pmatrix} a & b & c \\ p & q & r \\ u & v & w \end{pmatrix} =\begin{pmatrix} \alpha a + \beta p + \gamma u & \alpha b + \beta q + \gamma v & \alpha c + \beta r + \gamma w \\ \lambda a + \mu p + \nu u & \lambda b + \mu q + \nu v & \lambda c + \mu r + \nu w \\ \rho a + \sigma p + \tau u & \rho b + \sigma q + \tau v & \rho c + \sigma r + \tau w \end{pmatrix}\,.$$
| {
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Find the integral of $\frac{x^5+x^2+4x+\sin(x)}{64+x^6} dx$ from $-2$ to $2$ Find $$\int\limits_{-2}^{2}\frac{x^5+x^2+4x+\sin(x)}{64+x^6}dx$$
I understand this questions is trying to make the point about the integrals of symmetric functions.
So I separated all of these to get $\frac{x^5}{64+x^6}+\frac{x^2}{64+x^6}+\frac{4x}{64+x^6}+\frac{\sin(x)}{64+x^6}$
thinking it would be better to look at them individually.
So the first part, $\frac{x^5}{64+x^6}$ is easy to integrate.
But I'm not sure how to go about the rest.
I know that the integral of an odd function is zero because the limits of integration are symmetrical about the origin. But I'm not sure how to directly use this in the question, considering only the terms in the numerator are odd.
Please help!
| $$\int^2_{-2}\dfrac{x^5+x^2+4 x+\sin(x)}{64+x^6} dx=\int^2_{-2}\dfrac{x^5}{64+x^6} dx+\int^2_{-2}\dfrac{x^2}{64+x^6} dx+\\ \int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
\left.\dfrac{1}{6}\ln|64+x^6|\right|^2_{-2}+\int^2_{-2}\dfrac{x^2}{1+\left(\frac{x^3}{8}\right)^2} dx+\int^2_{-2}\dfrac{4 x}{64+x^6} dx+\int^2_{-2}\dfrac{\sin(x)}{64+x^6} dx=\\
0+\left.\dfrac{1}{24}\arctan\left(\frac{x^3}{8}\right)\right|^2_{-2}+0+0\text{ (by properties of odd functions)}=\boxed{\dfrac{\pi}{48}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Complex Numbers and Hyperbolic Functions How would you evaluate: $\mathfrak{R}\left[(1+i)\sin\left(\dfrac{(2+i)\pi}{4}\right)\right]$?
I know that $\cos x = \dfrac{e^{ix}+e^{-ix}}{2}$ and $\sin x = \dfrac{e^{ix}-e^{-ix}}{2i}$.
I have also tried expanding the $\sin\left(\dfrac{(2+i)\pi}{4}\right)$ part to $\sin\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{i\pi}{4}\right)+\cos\left(\dfrac{\pi}{2}\right)\sin\left(\dfrac{i\pi}{4}\right)$ using the addition formula for $\sin$ angle.
| We know that
$$
\sin\left(\dfrac{(2+i)\pi}{4}\right)=\sin\left(\dfrac{\pi}{2}+\frac{i\pi}{4}\right)=\cos\left(\frac{i\pi}{4}\right),
$$
then
$$
\begin{align}
(1+i)\sin\left(\dfrac{(2+i)\pi}{4}\right)&= (1+i)\cos\left(\frac{i\pi}{4}\right)\\
&=(1+i)\left(\dfrac{e^{i\frac{i\pi}{4}}+e^{-i\frac{i\pi}{4}}}{2}\right)\\
&=(1+i)\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right)\\
&=\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right)+i\left(\dfrac{e^{-\frac{\pi}{4}}+e^{\frac{\pi}{4}}}{2}\right).
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5 \begin{pmatrix}n+1\\2\end{pmatrix}+1$ $\sum\limits_{k=0}^n\ (3k^2+2k+1) = n^3 + 5
\begin{pmatrix}n+1\\2\end{pmatrix}+1$
How would you go on proving this equation? Doesn't have to be induction..
| Recall the following identities:
$$\sum_{k=0}^n 1 = n+1$$
$$\sum_{k=0}^n k = \dfrac{n(n+1)}2$$
$$\sum_{k=0}^n k^2 = \dfrac{n(n+1)(2n+1)}6$$
And
$$\sum_{k=0}^n (3k^2+2k+1) = 3 \sum_{k=0}^n k^2 + 2\sum_{k=0}^n k + \sum_{k=0}^n 1$$
Now plug in and obtain what you want.
| {
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Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from solve this question...
| Let $a = 2^x, b = 2^y$. Use some algebraic manipulation to arrive at $(a - 1)^2 + (b - 1)^2 = 2$
This is a circle of radius $\sqrt{2}$ centered at $(1, 1)$, and we want to find the minimum and maximum values of $a + b$.
So $a + b = M \rightarrow b = -a + M$, which is the equation of a line with slope $-1$. The maximum and minimum will be when we find the maximum and minimum y-intercept of this line such that it intersects at least one point on the circle with $a, b > 0$
It's tangent to the circle at the top right when $a = b = 2$ which gives a value of $M = 4$. The smallest y-intercept occurs at $m = 2$, however, we do not include $2$ because it makes one of the coordinates zero. This establishes $(2, 4]$ as the range.
| {
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Subsets and Splits