Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the remainder when $7^{7^{7}}$ is divided by 1000 I need help with this problem please
Find the remainder when $7^{7^{7}}$ is divided by $1000$
My try follow
$1000=8×125$ , now
$7 \equiv -1 \;\bmod\; (8)$
$\to$ $7^{7^{7}} \equiv -1 \;\bmod\; (8)$ and
$7^{100}\equiv 1 \;\bmod\; (125)$
Any help to complete this solution?
| As $7^4=2400+1$ and $7\equiv-1\pmod4\implies 7^7\equiv-1\equiv3\pmod4$
We can write $7^7=4m+3$ where integer $m\ge0$
$$7^{4m+3}=7^3(1+2400)^m$$
Now $(1+2400)^m\equiv1+\binom m12400\pmod{1000}$
$7^8=(1+2400)^2\equiv1\pmod{40}\implies7^7\equiv7^{-1}\equiv23$
We can choose $7^7=40n+23=4(10n+5)+3$
$\implies m\equiv5\pmod{10}, m=10r+5$(say)
$1+2400m\equiv1+2400(10r+5)\equiv1\pmod{1000}$
| {
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"url": "https://math.stackexchange.com/questions/2353314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is?
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
$\frac{a+b} {c}$ is how much?
Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
| Write the relation as an equation in $c$
$$c^2 - (a + b) c + a^2 + b^2 - a b = 0$$
The discriminant is $\Delta = (a+b)^2-4(a^2 + b^2 - a b)=-3 (a - b)^2$
It is given that $c\in\mathbb{R}$ thus $\Delta$ must not be negative and the only way is to set $a=b$ and $c=\dfrac{a+b}{2}$
I like this approach not because it is mine, but because is understandable by the average student.
It's not tricky, I mean :)
| {
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"timestamp": "2023-03-29T00:00:00",
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Telescoping(?) an infinite series Find the value of the sum $\displaystyle \sum\limits_{n=1}^{\infty} \frac{(7n+32) \cdot3^n}{n(n+2) \cdot 4^n}.$
Using partial fraction decomposition, I found the above expression is equivalent to $\displaystyle \sum\limits_{n=1}^{\infty} \frac{25}{n} \cdot \left(\frac{3}{4}\right)^n - \sum\limits_{n=1}^{\infty} \frac{18}{n+2} \cdot \left(\frac{3}{4}\right)^n,$ where I got couldn't find a closed form of either expression because of the $\left(\dfrac{3}{4}\right)^n.$
Similarly, trying to telescope one of the the sums with $\displaystyle \sum\limits_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+2}\right) \cdot \left(\frac{3}{4}\right)^n$ fails for the same reason. How can I further simplify the above expression? Thanks.
| Consider:
$$\frac{7 \, n +32}{n(n+2)} = \frac{7(n+2) + 9}{n(n+2)} = \frac{16}{n} - \frac{9}{n+2}$$
from which
\begin{align}
\sum_{n=1}^{\infty} \frac{7 \, n +32}{n(n+2)} \, t^{n} &= - 16 \, \ln(1-t) - \frac{9}{t^2} \, \left( \sum_{n=1}^{\infty} \frac{t^{n}}{n} - t - \frac{t^{2}}{2} \right) \\
&= \left(\frac{9}{t^{2}} - 16 \right) \, \ln(1-t) + \frac{9}{t} + \frac{9}{2}.
\end{align}
When $t = 3/4$ then this becomes
\begin{align}
\sum_{n=1}^{\infty} \frac{7 \, n +32}{n(n+2)} \, \left( \frac{3}{4}\right)^{n} = \frac{33}{2}.
\end{align}
It becomes evident that the only two values of $t$ for which the logarithmic term has a zero coefficient are $t = \pm (3/4)$. In this view it is developed:
$$\sum_{n=1}^{\infty} (-1)^{n-1} \, \left(\frac{7 \, n +32}{n(n+2)}\right) \, \left( \frac{3}{4}\right)^{n} = \frac{15}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}.$ Additional data added
If $\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x}=k$, find $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$
I raised $\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x}$ to the second power in order to express $k$:
\begin{align}&\color{white}=1-\cos^2x+2\sqrt{1-\cos^2x}\sqrt{1+\sin^2x}+1+\sin^2x\\&=1-\cos^2x+\sqrt{1+\sin^2x}(2\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x})\\&=1-\cos^2x+\sqrt{1+\sin^2x}2k.\end{align}
Since it doesn't give the answer, I also raised the second expression equalling $k$ to the second power. That didn't work out. The variants are:
A)$1.5k$ B)$2k$ C)$\frac{2}{k}$ D)$-k$ E)$-\frac{1}{k}$
I have solved it in the way Fred did (having learnt from him) but as is obvious there is NOT such an answer in the variants above.
The correct answer is E)$-\frac{1}{k}$ (according to the answer tables in the book).
How to get to that correct answer E)???
| $k(\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x})=(\sqrt{1-\cos^2x}-\sqrt{1+\sin^2x})(\sqrt{1-\cos^2x}+\sqrt{1+\sin^2x})=1-\cos^2x-(1+\sin^2x)=-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $\tan(\alpha)+\cot(\alpha)=p,$ denote $\tan^3(\alpha)+\cot^3(\alpha)$ in terms of $p$ I wrote the second expression according to the identity:
\begin{align}
& \tan^3(\alpha) + \cot^3(\alpha) \\[10pt]
= {} & \bigl(\tan(\alpha) + \cot(\alpha)\bigr) \bigl( \tan^2(\alpha) - \tan(\alpha) \cot(\alpha) + \cot^2(\alpha)\bigr) \\[10pt]
= {} & p\bigl(\tan^2(\alpha) + \cot^2(\alpha) - 1\bigr).
\end{align}
Then tried to use the identity $a^2+b^2=(a-b)^2+2ab$ but that was to no vail. How to solve it?
| Let $u=\tan(\alpha)$.
\begin{align*}
\text{Solu}&\text{tion #1:}\\[6pt]
&u+\frac{1}{u}=p\\[4pt]
\implies\;&\left(u+\frac{1}{u}\right)^2=p^2\\[4pt]
\implies\;&u^2+\frac{1}{u^2}=p^2-2\\[4pt]
\implies\;&u^2 - u\left(\frac{1}{u}\right) + \frac{1}{u^2} = p^2-3\\[4pt]
\implies\;&\left(u+\frac{1}{u}\right)\left(u^2 - u\left(\frac{1}{u}\right) + \frac{1}{u^2}\right) = p(p^2-3)\\[4pt]
\implies\;&u^3+\left(\frac{1}{u}\right)^3 = p(p^2-3)\\[12pt]
\text{Solu}&\text{tion #2:}\\[6pt]
&u+\frac{1}{u}=p\\[4pt]
\implies\;&\left(u+\frac{1}{u}\right)^3=p^3\\[4pt]
\implies\;&u^3+3u^2\left(\frac{1}{u}\right)+3u\left(\frac{1}{u}\right)^2+\left(\frac{1}{u}\right)^3=p^3\\[4pt]
\implies\;&u^3+3\left(u+\frac{1}{u}\right)+\left(\frac{1}{u}\right)^3=p^3\\[4pt]
\implies\;&u^3+\left(\frac{1}{u}\right)^3=p^3-3p\\[4pt]
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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finding relationship between roots of quadratic and equal area above and under x-axis
i am trying to find the relationship between the x value (last column) and the root of the function f(x) which will result in a the exact same area under and above the x-axis. i have tried finding a factor between each 'k' in the last column but it may be to complex to prove?
| It seems like some brute-force tactic may be the best here. In that case we want to find a number $k>n$ such that:
$$\begin{align*}&\left|\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx\right|=\left|\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx\right|\Rightarrow\\
\Rightarrow&-\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx=\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx\Rightarrow\\
\Rightarrow&\int_{m+\sqrt{n}}^k\left((x-m)^2-n\right)dx+\int_{m-\sqrt{n}}^{m+\sqrt{n}}\left((x-m)^2-n\right)dx=0\Rightarrow\\
\Rightarrow&\int_{m-\sqrt{n}}^k\left((x-m)^2-n\right)dx=0\overset{y=x-m}{\underset{dy=dx}{\Rightarrow}}\\
\Rightarrow&\int_{-\sqrt{n}}^{k-m}\left(y^2-n\right)dy=0\Rightarrow\\
\Rightarrow&\left.\frac{y^3}{3}-ny\right|_{-\sqrt{n}}^{k-m}=0\Rightarrow\\
\Rightarrow&\frac{(k-m)^3}{3}-n(k-m)-\frac{-\sqrt{n}}{3}-n\left(-\sqrt{n}\right)=0\overset{s=k-m}{\underset{a(n)=\frac{\sqrt{n}}{3}+n\sqrt{n}}{\Rightarrow}}\\
\Rightarrow&\frac{s^3}{3}-ns+a(n)=0\Rightarrow\\
\Rightarrow&s^3-3ns=-3a(n)
\end{align*}$$
Now, using Tartaglia's formula - or his poem about it! - we get that:
$$\begin{align*}s=&\sqrt[3]{\sqrt{\left(\frac{-3a(n)}{2}\right)^2+\left(\frac{-3n}{3}\right)^3}+\frac{-a(n)}{2}}-\sqrt[3]{\sqrt{\left(\frac{-3a(n)}{2}\right)^2+\left(\frac{-3n}{3}\right)^3}-\frac{-a(n)}{2}}=\\
=&\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}-\frac{a(n)}{2}}-\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}+\frac{a(n)}{2}}
\end{align*}$$
So
$$k=m+\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}-\frac{a(n)}{2}}-\sqrt[3]{\sqrt{\frac{9a^2(n)}{4}-n^3}+\frac{a(n)}{2}}$$
where
$$a(n)=\frac{\sqrt{n}}{3}+n\sqrt{n}$$
Note: What really affects $k$ is the value of $n$ since, $m$ just "transports" our entire problem right or left.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| Your second and third approaches are workable, but they lack the part proving that the fraction otherwise is between $-1$ and $1$.
For the second you use the sign of the derivate between and beyond the stationary points, especially beyond. You then can see that for $x>1$ the function is decreasing and for $x<-1$ too (and for $-1\le x\le 1$ it's increasing.
For the discriminant you consider for a given candidate value of $y$ the equation $2x/(x^2+1) = y$ which becomes
$$yx^2 - 2x + y = 0$$
which has the discriminant $(-2)^2 - 4y^2 = 4(1-y^2)$ and the equation only have solutions if $1-y^2=(1-y)(1+y)\ge 0$ which means that the factors are both positive or both negative (but they can't be both negative). So $1-y \ge 0$ and $1+y\ge 0$ that is $-1\le y\le 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $x + \sqrt{x + \sqrt{11+x}}=11$ algebraically? How can I solve the following equation algebraically?
$x + \sqrt{x + \sqrt{11+x}}=11$
| $x-11=-\sqrt{x+\sqrt{11+x}}$. Square both sides to get $(x-11)^2=x+\sqrt{11+x}$.
Then $(x-11)^2-x=\sqrt{11+x}$ and you square again
$((x-11)^2-x)^2=11+x$. Unfortunately you get a 4th order equation. Try this quartic equation
| {
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Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.
My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
| Remember that $\displaystyle \cos(\alpha) - \cos(\beta) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\beta-\alpha}{2}\right)$. Hence we obtain
\begin{align*}
\cos(\theta) - \sin(\theta) = \cos(\theta) - \cos\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}-\theta\right) = \sqrt{2}\sin\left(\frac{\pi}{4}-\theta\right)
\end{align*}
Finally, we get:
\begin{align*}
\lim_{\theta\rightarrow\pi/4}\frac{\cos(\theta)-\sin(\theta)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{\theta\rightarrow\pi/4}\frac{\sqrt{2}\displaystyle\sin\left(\frac{\pi}{4}-\theta\right)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{x\rightarrow 0}\frac{-\sqrt{2}\sin(x)}{x} = -\sqrt{2}
\end{align*}
Where it has been used the well known result
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the simplest way to factor the following polynomial $x^4-8x^2+x+12$?
What is the simplest way to factor the following polynomial $$x^4-8x^2+x+12$$ ?
Note : I already knew the factorization which is
$$(x^2-x-3)(x^2+x-4)$$
I need the way to get that
Thank you for your help.
| We can write $$x^4-8x^2+x+12=(x^2+ax+b)(x^2+cx+d)$$
Comparing the coefficients of $x^3,$ $$0=a+c\iff c=-a$$
Comparing the coefficients of $x,$ $$1=ad+bc=a(d-b)$$
$$\implies a=d-b=\pm1$$
Comparing the constants, $$bd=12$$
If $d-b=1, d=4,b=3$ or $d=-3.b=-4$
What if $d-b=-1?$
| {
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Perimeter and area of hyperbolic octagon How does one find the area and perimeter of the hyperbolic octagon with interior angles $\frac{\pi}2$? I'm completely stuck.
I have subdivided the octagon into eight hyperbolic triangles with two interior angles $\frac{\pi}4$. How can I now find the third angle?
| Imagine we have your octagon we divide it in 8 equal sectors, (all meeting at the centre of the octagon) then every sector has a top angle at the centre of $ \frac {2 \pi} {8} = \frac {\pi} {4} $ and base angles are the internal angle / 2 $ = \frac {\pi / 2} {2} = \frac {\pi} {4} $
We have 8 of these sectortriangles.
(Only while writing this realised it were all equilateral triangles, but for being more general and following the reasoning in the comments, we just overlook this)
For trigonomic reasons we cut all sectortriangles in half from the top angle to the middle of the base
so we now have 16 triangles each having angles of
*
*half internal angle $ (= \frac {\pi} {4} )$
*half top angle of sectortriangles $ (= \frac {\pi} {4 * 2} = \frac {\pi} {8} ) $
*right angle $ (=\frac {\pi} {2} )$
The area of this triangle is
$ (\pi- \frac {\pi} {4} - \frac {\pi} {8} - \frac { \pi} {2} = \frac {\pi} {8} $
Following wikipedia.https://en.wikipedia.org/wiki/Hyperbolic_triangle#Relations_between_angles
$ \cos A = \cosh a \sin B $
$ \sin A = \frac{\cos B}{\cosh b} $
$ \tan A = \frac{\cot B}{\cosh c} $
Or rewritten
$ \cosh a = \frac{ \cos A} { \sin B} $
$ \cosh b = \frac{ \cos B} { \sin A } $
$ \cosh c = \frac{ \cot B} { \tan A} $
filled in
$ \cosh a = \frac{\cos \frac{\pi} {4}} {\sin \frac{\pi} {8}} $
$ \cosh b = \frac{\cos \frac {\pi} {8}} {\sin \frac{\pi} {4} } $
$ \cosh c = \frac{\cot \frac {\pi} {8}} {\tan \frac {\pi} {4}} $
caculated (not exact values)
$ \cosh a = 1.848 \to a = 1.224 $
$ \cosh b = 1.307 \to b = 0.764 $
$ \cosh c = 2.414 \to c = 1.529 $
Now we have all there is to say about a regular octagon with internal right angles:
the area is $ 16 * \frac {\pi} {8} = 2{\pi}$
the perimater is $ 16 * 0.764 = 12.229 $
the radius of the inscribed circle is $ 1.224$
the radius of the circumscribed circle is $ 1.529 $
Thats all :)
| {
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"timestamp": "2023-03-29T00:00:00",
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Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ Question
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$
Definition
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation}
\begin{aligned}
P_n(x) = f(a)
+ \frac{f'(a)}{1!}(x - a)
+ \frac{f''(a)}{2!}(x - a)^2
& + \ldots
+ \frac{f^{(n)}(a)}{n!}(x - a)^n \\
\end{aligned}
\end{equation}
And $R_n (x) $ is (\emph{where $\xi$ is between $a$ and $x$ })
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Working
I'm not sure how to go about this, would I say that this is a first order
approximation as
\begin{equation}
\begin{aligned}
P(x) & = 1 - \frac{1}{2}x \\
P'(x) &= - \frac{1}{2} \\
P''(x) &= 0
\end{aligned}
\end{equation}
Then the remainder term would be
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)}
\end{aligned}
\end{equation}
Where $n + 1 = 2$. For $f(x) = \sqrt{1 + x}$ this would be
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 + \xi)^{-3/2}}{(3)!}
\end{aligned}
\end{equation}
And $\xi $ is between $-0.01$ and $0.01$
This would give the maximum error as
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 \pm 0.01 )^{-3/2}}{(3)!} \approx -0.0423
\end{aligned}
\end{equation}
The error is greatest when $\xi = -0.01$.
| You can just expand $f(x) \approx 1+\frac 12x-\frac 18x^2+\ldots$ as a Taylor series. The first two terms of the replacement are correct, so the error for small $x$ is dominated by the $-\frac 18x^2$ term. The error will then be negative and no less than $-\frac 18(0.01)^2=-.0000125$
| {
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"timestamp": "2023-03-29T00:00:00",
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An integral problem: I am working on a proof in lecture notes, but there are two steps that I couldn't understand
The first one: $$\int_1^n \frac{n-\lfloor x \rfloor}{x} dx=\int_1^n \frac{n+\frac{1}{2}+(\{x\}-\frac{1}{2})-x }{x} dx$$
Edit: Knowing that $\{x\}=x-\lfloor x\rfloor\quad$ , I am done with this part.
The second:
$$\int_1^n \frac{\{x\}-\frac{1}{2}}{x} dx=\int_1^\infty \frac{\{x\}-\frac{1}{2}}{x}dx+o(1),$$ since $$\int_1^\infty \frac{\{x\}-\frac{1}{2}}{x}dx \qquad (*) $$can be written as an alternating sum which is convergent so that its tail will converge to $0$.
I couldn't understand how they get from one equation to another and also the last part that says $(*)$ can be written as an alternating sum which is convergent so that its tail will converge to $0$
Actually in the second part, I don't know what is the meaning/use of $\{x\}$
| That
\begin{equation*}
\int_1^n \frac{\{x\} - \frac{1}{2}}{x} \,dx = \int_1^\infty \frac{\{x\} - \frac{1}{2}}{x} \, dx + o(1)
\end{equation*}
will follow from
\begin{equation*}
\int_n^\infty \frac{\{x\} - \frac{1}{2}}{x} \, dx = o(1).
\end{equation*}
This is equivalent to
\begin{equation*}
\lim_{n \to \infty} \int_n^\infty \frac{\{x\} - \frac{1}{2}}{x} \, dx = 0
\end{equation*}
which will follow from the fact that
\begin{equation*}
\int_1^\infty \frac{\{x\} - \frac{1}{2}}{x} \, dx
\end{equation*}
is convergent.
The idea is that we should partition the integral into an alternating sum. Recall that an alternating sum $\sum_n (-1)^n a_n$ is convergent if and only if $a_n \to 0$ as $n \to \infty$. After some thinking one realises that the partition to use is $1, 1 + \frac{1}{2}, 2, 2 + \frac{1}{2}, \ldots$. Indeed, for a natural $k$,
\begin{equation*}
\int_k^{k + \frac{1}{2}} \frac{\{x\} - \frac{1}{2}}{x} \, dx
= \int_k^{k + \frac{1}{2}} \frac{x - k - \frac{1}{2}}{x} \, dx < 0
\end{equation*}
and
\begin{equation*}
\int_{k + \frac{1}{2}}^{k + 1} \frac{\{x\} - \frac{1}{2}}{x} \, dx
= \int_{k + \frac{1}{2}}^{k + 1} \frac{x - k - \frac{1}{2}}{x} \, dx > 0.
\end{equation*}
What is left is to see that the terms of the sum converges to 0. I am lazy and will only do the positive ones. Well,
\begin{align*}
\int_{k + \frac{1}{2}}^{k + 1} \frac{x - k - \frac{1}{2}}{x} \, dx
&= \int_{k + \frac{1}{2}}^{k + 1} 1 - (k + \frac{1}{2}) \frac{1}{x} \, dx \\
&= \frac{1}{2} - (k + \frac{1}{2}) (\log(k + 1) - \log(k + \frac{1}{2})
\end{align*}
which converges to 0 as $k \to \infty$. This feels a bit cumbersome but it involved alternative sums so it was probably what was intended.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2371294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Cosecants of half angles
The tangent of an angle is $2.4$. Find the cosecant of half the angle.
My tries:
As $\tan A=2.4=\dfrac{12}{5}\implies \sin A=\pm\dfrac{12}{13}=\dfrac{1}{\csc A}$
Also:
$\sin \frac{A}{2}+\cos \frac{A}{2}=\pm\sqrt{1+\sin A}=\pm\dfrac{5}{{\sqrt{13}}}$ , considering $\sin A=\dfrac{12}{13}$
$\sin \frac{A}{2}-\cos \frac{A}{2}=\pm\sqrt{1-\sin A}=\pm\dfrac{1}{{\sqrt{13}}}$, considering $\sin A=\dfrac{13}{13}$
Adding them gives: $\sin \frac{A}{2}=\pm\dfrac{3}{\sqrt{13}}=\dfrac{1}{\csc{\frac A2}}$
I did same by considering $\sin A=-\dfrac{12}{13}$ then also it gave the same result.
But answer provided by the author is $\dfrac{\pm\sqrt{13}}{2}$ and $\dfrac{\pm\sqrt{13}}{3}$.
So what did I miss? please help.
| You have
$$
\tan^2A=\frac{144}{25}
$$
so
$$
\frac{1}{\cos^2A}=1+\tan^2A=\frac{169}{25}
$$
and therefore
$$
\cos A=\pm\frac{5}{13}
$$
Hence
$$
\sin\frac{A}{2}=\pm\sqrt{\frac{1-\cos A}{2}}
\qquad
\csc\frac{A}{2}=\pm\sqrt{\frac{2}{1-\cos A}}
$$
Plug in the values.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Relationship between the Picard-Fuchs differential equation and the hypergeometric differential equation Consider the Picard-Fuchs differential equation:
$$\frac{d^2 \Omega}{dJ^2} + \frac{1}{J} \frac{d\Omega}{dJ} + \frac{31J - 4}{144J^2 (1 - J)^2} \Omega = 0.$$
The author of this article claims that the solutions to this equation are those to the hypergeometric differential equation
$$J(1 - J) \frac{d^2 \Omega}{dJ^2} + [c - (a + b + 1)J] \frac{d\Omega}{dJ} - ab\Omega = 0$$
with $a = 11/12$, $b = 11/12$, $c = 4/3$, multiplied by $J^{1/6} (1 - J)^{3/4}$.
My question is: how does one know that? How can one look the Picard-Fuchs differential equation above and know that its solutions are related to the solutions of the hypergeometric differential equation? Is it because the Picard-Fuchs differential equation is related to Riemann's differential equation?
| Let $\Omega=J^p(1-J)^qX$ ,
Then $\dfrac{d\Omega}{dJ}=J^p(1-J)^q\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)X$
$\dfrac{d^2\Omega}{dJ^2}=J^p(1-J)^q\dfrac{d^2X}{dJ^2}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X=J^p(1-J)^q\dfrac{d^2X}{dJ^2}+2J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X$
$\therefore J^p(1-J)^q\dfrac{d^2X}{dJ^2}+2J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X+\dfrac{1}{J}\left(J^p(1-J)^q\dfrac{dX}{dJ}+J^p(1-J)^q\left(\dfrac{p}{J}-\dfrac{q}{1-J}\right)X\right)+\dfrac{31J-4}{144J^2(1-J)^2}J^p(1-J)^qX=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{p(p-1)}{J^2}-\dfrac{2pq}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X+\dfrac{1}{J}\dfrac{dX}{dJ}+\left(\dfrac{p}{J^2}-\dfrac{q}{J(1-J)}\right)X-\left(\dfrac{1}{36J^2}-\dfrac{23}{144J}-\dfrac{23}{144(1-J)}-\dfrac{3}{16(1-J)^2}\right)X=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p+1}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{p^2}{J^2}-\dfrac{(2p+1)q}{J(1-J)}+\dfrac{q(q-1)}{(1-J)^2}\right)X-\left(\dfrac{1}{36J^2}-\dfrac{23}{144J(1-J)}-\dfrac{3}{16(1-J)^2}\right)X=0$
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{2p+1}{J}-\dfrac{2q}{1-J}\right)\dfrac{dX}{dJ}+\left(\dfrac{36p^2-1}{36J^2}-\dfrac{144(2p+1)q-23}{144J(1-J)}+\dfrac{16q(q-1)+3}{16(1-J)^2}\right)X=0$
Choose $p=\dfrac{1}{6}$ and $q=\dfrac{3}{4}$ , the ODE becomes
$\dfrac{d^2X}{dJ^2}+\left(\dfrac{4}{3J}-\dfrac{3}{2(1-J)}\right)\dfrac{dX}{dJ}-\dfrac{121}{144J(1-J)}X=0$
$J(1-J)\dfrac{d^2X}{dJ^2}+\left(\dfrac{4}{3}-\dfrac{17}{6}J\right)\dfrac{dX}{dJ}-\dfrac{121}{144}X=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the **Local Truncation error** in the form of $ \ o(h^k) \ $ of the multi-step method Find the Local Truncation error in the form of $ \ o(h^k) \ $ of the multi-step method $ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $.
Also determine it is convergent or not.
Answer:
The scheme is $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i + 6hf(t_{i+2}, u_{i+2}) $.
We know that $ f(t_i,u_i )=\frac{u_{i+1}-u_i}{h} ,\ by \ Taylor \ series \ $
Then , $ \ 2u_{i+3} = −3u_{i+2} + 6u_{i+1} − u_i +6 u_{i+3}-6u_{i+2} $,
or,$ \ 4u_{i+3}-9u_{i+2}+6u_{i+1}-u_i=0 $
or, $ 4(u_{i+3}-u_{i+2})-5(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)=0 $
Dividing by $ h \ $ ,
$ 4 .\frac{u_{i+3}-u_{i+2}}{h}-5. \frac{u_{i+2}-u_{i+1}}{h}+\frac{u_{i+1}-u_i}{h}=0 $ ,
or, $ 4 u'''+o(h^4)-5 u'' +o(h^2)+u'+o(h)=0, $
So the Local Truncation error is $ \ (h^4) \ $.
Am I right ? Is there any help ?
| I'll assume this is approximating $y'(t) = f(t,y)$. The method is
$$
2y_{n+3} =-3 y_{n+2} + 6 y_{n+1} - y_n + 6 h f (t_{n+2}, y_{n+2})
$$
To find the (local) error, taylor expand each term:
\begin{align*}
2 y_{n+3} &= 2\left( y_n + 3h y_n ' + \frac{(3h)^2}{2} y_n '' + \frac{(3h)^3}{6} y_n ''' + \frac{(3h)^4}{24} +O(h^5)\right) \\
-3 y_{n+2} &= -3 \left( y_n + 2h y_n ' + \frac{(2h)^2}{2} y_n '' + \frac{(2h)^3}{6} y_n ''' + \frac{(2h)^4}{24} + O(h^5) \right) \\
6 y_{n+1} &= 6 \left( y_n + h y_n ' + \frac{h^2}{2} y_n '' + \frac{h^3}{6} y_n ''' + \frac{h^4}{24} + O(h^5) \right) \\
-y_n &= -y_n \\
6 h f(t_{n+2}, y_{n+2}) &= 6h \left( y_n' + 2h y_n '' + \frac{(2h)^2}{2} y_n ''' + \frac{(2h)^3}{6} y_n '''' + O(h^4) \right)
\end{align*}
and matching the coefficients of the scheme reveals truncation error $O(h^4)$.
To check if it converges, we look to the root condition (i.e. apply to the test equation $y'=f(t,y)=0$, and see if the scheme is zero-stable.
$$
2 y_{n+3} = -3 y_{n+2} + 6 y_{n+1} - y_n \implies 2 r^3 = -3 r^2 + 6r -1
$$
One of the roots is larger than 1 in absolute value, so by Dahlquist Equivalence Theorem, the scheme does not converge (since it doesn't satisfy the root condition)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $P(8- \frac12X_1+X_2<0)$ for $(X_1,X_2)$ i.i.d. uniform on $[-8;8]$
Random variables $X_{1}$ and $X_{2}$ are stochastically independent, uniformly distributed in a range $[-8;8]$. Let $g(x_{1},x_{2}) = 8- \frac{1}{2}x_{1}+x_{2}$. What is the probability that $g(X_1,X_2)<0$ ?
My solution looks like this.
Firstly I limit $x_{2}$:
$x_{2}< \frac{1}{2}x_{1}-8$
Main equation that I use to calculate the probability:
$P = \int_{x_{1}=- \infty }^{x_{1}= \infty} \int_{x_{2}=- \infty }^{x_{2}= \frac{1}{2}x_{1}-8 } f_{x_{2}}(x_{2})dx_{2} f_{x_{1}}(x_{1})dx_{1}$
Where
$\int_{x_{2}=- \infty }^{x_{2}= \frac{1}{2}x_{1}-8 } f_{x_{2}}(x_{2})dx_{2}=F_{x_{2}}( \frac{1}{2}x_{1}-8 )= \frac{\frac{1}{2}x_{1}-8-(-8)}{8-(-8)}= \frac{1}{32}x_{1}$
So
$P = \int_{x_{1}=- \infty }^{x_{1}= \infty}\frac{1}{32}x_{1}f_{x_{1}}(x_{1})dx_{1}
=\frac{1}{32} E(X_{1})=\frac{1}{32} * \frac{-8+8}{2}=0$
I know that the result should be $\frac{1}{16}$. I would be thankful for pointing out where I do the mistake.
| The lower limit and the upper limit of $x_2$ are $-8$ and $\frac12x_1-8$ respectively. That means that $\frac12x_1-8\geq -8\Rightarrow \frac12x_1 \geq 0\Rightarrow x_1\geq 0$ Therefore the calculation is
$$P(8- \frac{1}{2}x_{1}+x_{2}<0)=\int_{0}^8 \frac{1}{16} \frac{x_1}{32} \, dx_1 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many solutions does the equation $a + b + c + d + e = 21$ have in the nonnegative integers if $a \leq 3$, $0 < b < 4$, and $c \geq 15$? How many solutions are there to the equation in the nonnegative integers:
$$a+b+c+d+e = 21$$
Conditions:
a) $ a \le 10$
I understood the solution which is total number of possibilities - the number of possibilities with $a \gt 10$ which is $\binom{5+21-1}{21} - \binom{5+10-1}{10}$.
b) $a \le 3$, $0 \lt b \lt 4$, $c \ge 15$.
How do I solve this part? Thanks in advance.
| Put $b=1+b'$, $c=15+c'$. Then we have to count the nonnegative solutions of
$$a+b'+c'+d+e=5\tag{1}$$
satisfying $a\leq3$, $b'\leq2$. Denote by $n_r$ the number of admissible pairs $(a,b')$ with $a+b'=r$. By stars and bars the total number $N$ of admissible solutions of $(1)$ is then given by
$$N=\sum_{r=0}^5n_r\cdot{7-r\choose 2}=1\cdot21+2\cdot15+3\cdot10+3\cdot6+2\cdot3+1\cdot1=106\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$
If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then
$$S\gt 1,$$
but how?
I understood that there are $451$ pair of terms. So clubbed two terms together.
$\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$
But I am not able to solve it further. Such a tricky question for me.
| There are $901$ terms in $S$
Consider the first $100.$ The smallest term of this set is $\frac 1{199}$
The sum of the first $100$ terms is greater than $\frac 12$
Now consider the next $200$ terms.
the smallest term in this set is $\frac {1}{399}$
and the sum of that subset is greater that $\frac 12$
and you are done. And there are still $601$ you haven't even considered yet.
The harmonic series:
$S = \sum_\limits{n=1}^\infty \frac 1n = 1 + \frac 12 + \frac 13 + \frac 14 \cdots\\
\frac 13 + \frac 14 > \frac 1{4} + \frac 1{4} = \frac 12\\
\frac 15 + \frac 16 + \frac 17 + \frac 18 > \frac 18 + \frac 18 + \frac 18 + \frac 18 = \frac 12\\
$
And we can break infinite series into (infintely many) finite sub-series, and each sub-series sums to something that is greater than $\frac 12$
$S > 1+\frac 12 + \frac 12 + \frac 12 \cdots$ and as we add more terms $S$ marches off to infinity (albeit slowly)
For you problem I have employed a very similar tactic. I have found subsets of your series, each of which sum to something greater than $\frac 12$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Evaluate $\lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$ Evaluate
$$ \lim_{n \to \infty} \frac{1}{1^2+n^2}+\frac{2}{2^2+n^2}+\frac{3}{3^2+n^2}+\cdots+\frac{n}{n^2+n^2}$$
I used definite integral as a limit of a sum as:
$$S= \lim_{ n \to \infty}\frac{1}{n} \sum_{r=1}^{n} \frac{\left(\frac{r}{n}\right)}{1+\left(\frac{r}{n}\right)^2}$$
So
$$S=\int_{0}^{1}\frac{ x \:dx}{1+x^2}=\frac{1}{2} \log 2$$
Is there any other approach?
| Using a double limit with uniform convergence,
$$\begin{align}\lim_{n \to \infty} \sum_{k=1}^n \frac{k}{n^2 + k } &= \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k / n^2} \\&=\lim_{m \to \infty}\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^n \frac{k/n}{1 + k /(nm)} \\ &= \lim_{m \to \infty} \int_0^1 \frac{x}{1 + x/m} \, dx \\ &= \int_0^1 x \, dx \\ &= \frac{1}{2}\end{align}$$
Note that if $a_{nm} \to b_n$ uniformly, then
$$\lim_{n\to \infty}a_{nn} = \lim_{n\to \infty} \lim_{m \to \infty}a_{nm} = \lim_{m\to \infty} \lim_{n \to \infty}a_{nm}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Polynomial divisibility exercise How do I prove that $(X+1)^{6n+1}-X^{6n+1}-1$ is divisible by $(X^2+X+1)^2$ without derivative functions? (I am 10th grade).
| Let us denote $q_n(x)=(x+1)^{6n+1}-x^{6n+1}-1$. We may notice that
$$ q_{n+1}(x)-(x+1)^6 q_n(x) = \left[(x+1)^6-x^6\right]x^{6n+1}+\left[(x+1)^6-1\right] $$
where both $(x+1)^6-x^6$ and $(x+1)^6-1$ are multiples of $x^2+x+1$:
$$ q_{n+1}(x)-(x+1)^6 q_n(x) = (x^2+x+1)\left[(1 + 2 x)(1 + 3 x + 3 x^2)x^{6n+1}+x (2 + x) (3 + 3 x + x^2)\right] $$
In order to prove the claim by induction, it follows that it is enough to show
$$ (x^2+x+1)\mid \left[(1+2x)x^{6n+1}+x^3(2+x)\right] $$
for any $n\geq 0$. This can be done by induction, again. Let us denote $r_n(x)=(1+2x)x^{6n+1}+x^3(2+x)$. The last claim holds for $n=0$ and
$$ r_{n+1}(x)-x^6 r_n(x) = x^3 \left(2+x-2 x^6-x^7\right) = x^3(2+x)(1-x^6) $$
so we are done, since $x^2+x+1$ is a divisor of $1-x^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Hints to prove $\tan^2{\theta}=\tan{A}\tan{B}$, given $\frac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\frac{\sin{2A}}{\sin{2B}}}$ I need some hints on solving this trigonometry problem.
Problem
If $\dfrac{\sin{(\theta + A)}}{\sin{(\theta + B)}} = \sqrt{\dfrac{\sin{2A}}{\sin{2B}}}$, then prove that $\tan^2{\theta}=\tan{A}\tan{B}$.
I tried to expand the left hand side of the equation, but no clue what to do next. I also tried to use $\sin{2\alpha} = \dfrac{2\tan{\alpha}}{1 + \tan^2{\alpha}}$ for the right hand side of the equation with no result.
I appreciate for any help. Thank you.
| $$\dfrac{\sin(\theta+A)}{\sin(\theta+B)}=\cdots=\dfrac{\tan\theta\cos A+\sin A}{\tan\theta\cos B+\sin B}$$ Dividing numerator & the denominator by $\cos\theta$
$$\implies\dfrac{(\tan\theta\cos A+\sin A)^2}{(\tan\theta\cos B+\sin B)^2}=\dfrac{2\sin A\cos A}{2\sin B\cos B}$$
$$\iff\dfrac{(\tan\theta+\tan A)^2}{(\tan\theta+\tan B)^2}=\dfrac{\tan A}{\tan B}$$
Dividing numerator of both sides by $\cos^2A$
and the denominator of both sides by $\cos^2B$
Now simplify assuming $\tan A\ne\tan B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\binom{n}{0} \binom{2n}{n}-\binom{n}{1} \binom{2n-2}{n}+\binom{n}{2} \binom{2n-4}{n}+\cdots$ Find $$\binom{n}{0} \binom{2n}{n}-\binom{n}{1} \binom{2n-2}{n}+\binom{n}{2} \binom{2n-4}{n}+\cdots$$
I have taken $r$th term and modified as follows:
$$T_r =(-1)^r \binom{n}{r} \binom{2n-2r}{n}=(-1)^r \frac{n!}{(n-r)!r!} \times \frac{ (2n-2r)!}{n! (n-2r)!}=(-1)^r \frac{(2n-2r)!}{(n-r)! r!(n-2r)!}$$
Can we continue from here?
| A successful application of Euler´s Finite Difference Theorem
Given $f(x) = \sum_{j=0}^ra_jx^j$, Euler´s Finite Fifference Theorem statest that:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k) = (-1)^n \Delta_1^nf(x)\big|_{x=0}=\left\{
\begin{array}{ll}
0, & 0 \leq r<n\\
(-1)^na_nn!,& r = n
\end{array}
\right.
$$
We have
\begin{align*}
\tag1\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{an-ak}{n} &= \sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{a(n-k)}{n}\\
\tag2&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{(n-k)}\binom{a(n-(n-k))}{n}\\
\tag3&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{n-k}\binom{ak}{n}\\
\tag4&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{k}\binom{ak}{n}\\
\tag5&=(-1)^{n}\sum_{k=0}^{n}(-1)^{-k}\binom{n}{k}\binom{ak}{n}\\
\tag6&=(-1)^{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{ak}{n}\\
\end{align*}
Let $f(k) = \binom{ak}{n}$ where $a$ is any nonzero complex number. The definition of the binomial coefficient implies that $f(k)$ is a polynomial of degree $n$ in $k$, i.e. $\binom{ak}{n} = \sum_{j=0}^na_jk^j$.
The coefficient of $k^n$ is $\frac{a^n}{n!}$ sience
\begin{align*}
f(k)=\binom{ak}{n} = \tfrac{(ak)(ak-1)(ak-2)\cdots(ak-n+1)}{n!}=\sum_{j=0}^na_jk^j
\end{align*}
By definition
\begin{align*}
\tag7(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{ak}{n}
&=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k)\\
\tag8&=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sum_{j=0}^na_jk^j\\
\tag9&=(-1)^n\underbrace{\sum_{j=0}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j}_{(-1)^n \Delta_1^nf(x)\big|_{x=0}}=\left\{\begin{array}{ll}
0, & 0\leq j <n\\
a_n n!,& j = n
\end{array}\right.
\end{align*}
Because here the coefficient of $k^n$ is $\frac{a^n}{n!}$ and $j=n$, implies that
\begin{align*}
\tag{10} (-1)^n\sum_{j=n}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j&=(-1)^na_n
\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^n\\
\tag{11} &=(-1)^n\big[a_n(-1)^nn!\big]\\
\tag{12} &= \frac{a^n}{n!}n! \\
\tag{13} &= a^n\\
&&\Box
\end{align*}
In this way we have for $a=2$:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n-2k}{n} = 2^n$$
Note In $(10)$: When we study Stirling numbers of the second kind, $S(n, k)$, we will discover that
$\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j= (-1)^nn!S(n, k),\quad j\leq n$
Bibliographic references:
Gould, H. W. (Oktober 2015). Combinatorial Identities for Stirling Numbers, 68, 69, 70.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the limit when $x = \pi / 6$ Let $a_0 = 1$ and $a_n = a_{n-1}(\cos \frac{x}{2^n})$ .Find the limit $\{a_n\}$ when $x = \pi / 6$ . (i.e. find the $\lim_{n \to \infty} a_n$)
My try : I solved it but the solution was time-consuming .
$\frac{a_n}{a_{n-1}} = \cos \frac{x}{2^n} \to \frac{a_n}{a_0} = \frac{a_n}{a_{n-1}} \times \frac{a_{n-1}}{a_{n-2}} \times \dots \times\frac{a_1}{a_0} = \cos {\frac{x}{2^n}} \times \cos {\frac{x}{2^{n-1}}} \times \dots \times \cos {\frac{x}{2}} \to (2^n \sin{\frac{x}{2^n}})(a_n) = \sin x \to a_n = \frac{\sin x}{2^n\sin{\frac{x}{2^n}}} = \frac{\frac{x}{2^n}}{\sin{\frac{x}{2^n}}} \times \frac{\sin x }{x} \to \lim_{n \to \infty } a_n = 1 \times \frac{\sin{\frac{\pi}{6}}}{\frac{\pi}{6}} = \frac{3}{\pi} $
Note : I did some calculations by hand and didn't write them .
| You already know that $$a_n = \cos \left({\frac{x}{2^n}}\right) \times \cos \left({\frac{x}{2^{n-1}}}\right) \times \dots \times \cos \left({\frac{x}{2}}\right) = \prod_{i=1}^n \cos \frac{\theta}{2^i}
$$
Therefore $\displaystyle \lim_{n\to\infty} a_n = \prod_{i=1}^{\infty} \cos \dfrac{\theta}{2^i}$, if the product exists.
If you have only 1 minute to solve this, it is maybe expected that you know Viète's infinite product:
$$\prod_{n=1}^\infty \cos \frac{\theta}{2^n}
= \frac{\sin \theta}{\theta}$$
Using this, we immediately get that the answer is $\dfrac{\sin (\pi/6)}{\pi/6} = \dfrac3{\pi}$.
So this is basically your solution, but with the addition that the second, perhaps more difficult step, is actually already a known identity. If you know the identity, you can do it in a minute.
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
|
This is my solution after alot of tries
Note:The special limit which used in the solution can be proved without niether Lospital rule nor series
| {
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"question_score": "3",
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"answer_id": 5
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Small powers of small numbers: Is there really a pattern or am I just used to looking for them? This probably is just a coincidence, but I've always found it interesting and I wanted to put some feelers out there to see if maybe there really is something to it after all.
There are these collections of powers, and in one case a sum of powers in two different ways, that have similar digits to each other.
$$3^7 = 2187$$
$$6^4 = 1296$$
$$12^3 = 1728$$
$$13^3 = 2197$$
$$1^3 + 12^3 = 9^3 + 10^3 = 1729$$
$$3^6 = 729$$
(I suppose the last two are inevitable in any radix.)
We also have
$$2^{10} = 1024$$
$$7^4 = 2401$$
and
$$2^8 = 256$$
$$5^4 = 625$$
$$24^2 = 576$$
I can see the following class is inevitable in any radix > 4.
$$12^2 = 144$$
$$21^2 = 441$$
$$13^2 = 169$$
$$31^2 = 961$$
Is there an overarching mechanism at play here, or am I just finding statistically insignificant patterns?
| This is really only a tiny piece of an answer, but...
This is a very interesting question. My first reaction is to think that it can't be a coincidence, but I'm not certain.
Now, a statistical explanation seems more or less plausible. You're mostly looking at numbers up to $13$ raised to numbers up to $10$, so there are about $12 \cdot 9 = 108$ "interesting" powers. Meanwhile, there are $286$ ways of choosing $4$ digits, with duplicates allowed but disregarding order. So it shouldn't be too uncommon for powers to be anagrams.
Still, this "family" of prime powers seems pretty compelling:
$5^2 = 25 = 24 + 1$
$7^4 = 2401 = 24 \cdot 10^2 + 1$
$2^{10} = 1024 = 24 + 10^3$
We can examine the relationships between these equations a little more clearly by pairing them up, multiplying by $10^2$ as needed, and subtracting them to cancel the $24$s:
$2^2 \cdot 5^4 - 7^4 = 10^2 - 1$
$2^{10} - 5^2 = 10^3 - 1$
$2^{12} \cdot 5^2 - 7^4 = 10^5 - 1$
By coincidence or otherwise, all of the powers on the left-hand side are even, so we can apply the difference-of-squares formula. We can also factor the right-hand sides, of course. These equations factor as:
$(2 \cdot 5^2 - 7^2) (2 \cdot 5^2 + 7^2) = (10 - 1) (10 + 1)$
$(2^5 - 5) (2^5 + 5) = (10 - 1) (10^2 + 10 + 1)$
$(2^6 \cdot 5 - 7^2) (2^6 \cdot 5 + 7^2) = (10 - 1) (10^4 + 10^3 + 10^2 + 10 + 1)$
Then, if we replace each factor with its own prime factorization, we end up with:
$(1) (3^2 \cdot 11) = (3^2) (11)$
$(3^3) (37) = (3^2) (3 \cdot 37)$
$(271) (3^2 \cdot 41) = (3^2) (41 \cdot 271)$
But this definitely doesn't seem like a compelling explanation of anything. We started with some pretty interesting equations, like $2^{10} - 5^2 = 10^3 - 1$, and "explained" them by progressively factoring them, but there's no explanation of why the prime factors on the left ended up matching the prime factors on the right.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Any integer is of the form $n=a_1^2+a_2^2+a_3^2-a_4^2-a_5^2$ I've struggled with this conjecture, which probably can be proved:
Any natural number $n$ can be written as $n=a_1^2+a_2^2+a_3^2-a_4^2-a_5^2$ for some
$a_1,a_2,a_3,a_4,a_5\in\mathbb Z^+$.
I guess there are some polynomials
$a_k=p_k(n)$ that solves the problem by substitution. But which?
I'm almost sure that ME can prove the conjecture.
| $\color{Green}{\text{Lemma}}$:
Let $1 < m$ be an odd number;
then the equation $m=A^2-B^2$
has a solution in natural numbers.
Proof:
Let $A:=\dfrac{m+1}{2}$ and
let $B:=\dfrac{m-1}{2}$.
*
*If $n$ is odd, then let
$a_5=2, \ a_1=1, \ a_2=1$.
So the equation changes to $n+2=a_3^2-a_4^2$;
and then
let $a_3=\dfrac{n+3}{2}$ and
let $a_4=\dfrac{n+1}{2}$.
*If $4 \leq n$ is even, then let
$a_5=1, \ a_1=1, \ a_2=1$.
So the equation changes to $n-1=a_3^2-a_4^2$;
and then
let $a_3=\dfrac{n}{2}$ and
let $a_4=\dfrac{n-2}{2}$.
*If $n=2$ , then let
$a_1=1, \ a_2=1, \ a_3=5; \ a_4=4, \ a_5=3$.
| {
"language": "en",
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"question_score": "2",
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How many solutions does the equation $n_1 + n_2 + n_3 + n_4 + n_5 = 20$ have in the positive integers if $n_1 < n_2 < n_3 < n_4 < n_5$? Let $n_1 < n_2 < n_3 < n_4 < n_5$ be positive integers such that $n_1 + n_2 + n_3 + n_4 + n_5 = 20$. Then the number of such distinct arrangements $(n_1, n_2, n_3, n_4, n_5)$ is......
I have no idea how to proceed. Manually, I have done it
$$1+2+3+4+10$$
$$1+2+3+5+9$$
$$1+2+3+6+8$$
$$1+2+4+5+8$$
$$1+2+4+6+7$$
$$1+3+4+5+7$$
$$2+3+4+5+6$$
But is there any way I can do it by Permutation and Combination method?
| A variation based upon generating functions. We introduce positive integers $a,b,c,d$ and put
\begin{align*}
n_2&=n_1+a\\
n_3&=n_2+b=n_1+a+b\\
n_4&=n_3+c=n_1+a+b+c\\
n_5&=n_4+d=n_1+a+b+c+d
\end{align*}
The equation $n_1+n_2+n_3+n_4+n_5=20$ transforms to
\begin{align*}
5n_1+4a+3b+2c+d=20\tag{1}
\end{align*}
with $n_1,a,b,c,d>0$.
In order to find the number of solutions of (1) we consider the generating function $A(x)$
\begin{align*}
A(x)&=\frac{x^5}{1-x^5}\cdot\frac{x^4}{1-x^4}\cdot\frac{x^3}{1-x^3}\cdot\frac{x^2}{1-x^2}\cdot\frac{x}{1-x}\\
&=x^{15}+x^{16}+2x^{17}+3x^{18}+5x^{19}+\color{blue}{7}x^{20}+10x^{21}+\cdots
\end{align*}
and obtain with some help of Wolfram Alpha the solution
\begin{align*}
[x^{20}]A(x)\color{blue}{=7}
\end{align*}
Add-on: Some details
We first transform the equation with restrictions by introducing positive integers $a,b,c,d$ in an equivalent equation with more convenient restrictions
\begin{align*}
&n_1 + n_2 + n_3 + n_4 + n_5 = 20\qquad&\qquad&5n_1+4a+3b+2c+d=20\\
&0<n_1<n_2<n_3<n_4<n_5\qquad&\qquad&0<n_1,0<a,0<b,0<c,0<d
\end{align*}
We now consider admissible $5$-tuples $(n_1,a,b,c,d)$. Increasing $n_1$ by $1$ adds $5$ to the equation. Similarly, increasing $a$ by $1$ adds $4$ to the equation. We encode these increments via exponents of generating functions:
*
*$n_1$: Increment by $5$ gives
\begin{align*}
x^5+x^{10}+x^{15}+\cdots=x^5(1+x^5+x^{10}+\cdots)=\frac{x^5}{1-x^5}
\end{align*}
*$a$: Increment by $4$ gives
\begin{align*}
x^4+x^8+x^3+\cdots=x^4(1+x^4+x^8+\cdots)=\frac{x^4}{1-x^4}
\end{align*}
and similarly for $b,c$ and $d$. Observe that each of $n_1,a,b,c,d$ is positive, i.e. has at least value $1$. This is respected by smallest values $x^5,x^4,x^3,x^2$ and $x^1$.
The number of admissible solutions is therefore
\begin{align*}
[x^{20}]&\frac{x^5}{1-x^5}\cdot\frac{x^4}{1-x^4}\cdot\frac{x^3}{1-x^3}\cdot\frac{x^2}{1-x^2}\cdot\frac{x}{1-x}\\
&=[x^{20}]\frac{x^{15}}{(1-x^5)(1-x^4)(1-x^3)(1-x^2)(1-x)}\\
&=[x^{5}]\frac{1}{(1-x^5)(1-x^4)(1-x^3)(1-x^2)(1-x)}\tag{2}\\
&=[x^{5}](1+x^5)(1+x^4)(1+x^3)(1+x^2+x^4)(1+x+x^2+x^3+x^4+x^5)\tag{3}\\
&=\cdots\tag{4}\\
&\color{blue}{=7}
\end{align*}
Comment:
*
*In (2) we use the coefficient of operator rule: $[x^{p}]x^qA(x)=[x^{p-q}]A(x)$.
*In (3) we expand the geometric series restricted to powers less or equal to $x^5$ since other terms do not contribute to $[x^5]$.
*In (4) we expand further and can omit terms with powers greater than $5$.
Hint: Instructive examples can be found in H.S. Wilf's book generatingfunctionology.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
$\forall n \in \mathbb{N}_{>5}\implies\exists (a,b)\in \mathbb Z^+:a^2+b^2\notin\mathbb P\;\wedge\;n=a+b$ Conjecture:
$\forall n \in \mathbb{N}_{>5} \implies\exists (a,b)\in \mathbb Z^+:a^2+b^2\notin\mathbb P\;\wedge\;n=a+b$
Tested $\forall n\leq 100,000$. Small exceptions: {1,2,3,5}.
I would like to see proofs or counterexamples.
The conjecture seems to be related to:
Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$
Any odd number is of form $a+b$ where $a^2+b^2$ is prime
| If $n=5m$ with $m>1$ then $n=4m+m$ and $(4m)^2+m^2=17m^2$ is composite because $m^2>1.$
If $n\equiv 4 \pmod 5$ then $(n-1)^2+1^2>(n-1)^2\geq 3^2>5$ and $(n-1)^2+1^2 \equiv 0\pmod 5.$
If $5<n\equiv 1\pmod 5$ then $(n-2)^2+2^2>(n-2)^2\geq 4^2>5$ and $(n-2)^2+2^2\equiv 0\pmod 5.$
If $5<n\equiv 3\pmod 5$ then $(n-2)^2+2^2>(n-2)^2\geq 6^2>5$ and $(n-2)^2+2^2\equiv 0 \pmod 5.$
If $5<n\equiv 2 \pmod 5$ then $(n-4)^2+4^2>(n-4)^2\geq 3^2>5$ and $(n-4)^2+4^2\equiv 0\pmod 5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Range of the integral of a trigonometric function Given $$g(x) = \int_0^{x} \sqrt{1+\sin t} dt -\sqrt{1+\sin x}$$
I would like to show that $g(x) \geq 2\sqrt{2} -10,\,\forall x\in \left(0,\infty\right)$.
Any help is appreciated.
| $\int_{0}^{x}\sqrt{1+\sin t}\,dt$ is clearly a non-negative, differentiable and increasing function on $\mathbb{R}^+$.
The minimum of $g(x)$ occurs at a zero of $g'(x)$, and since $g'(x)=\sqrt{1+\sin x}-\frac{\cos x}{2\sqrt{1+\sin x}}$, the minimum value of $g(x)$ is attained at some point such that $1+\sin x-\frac{1}{2}\cos x=0$, or
$$ \frac{2}{\sqrt{5}}\sin x-\frac{1}{\sqrt{5}}\cos x = \sin\left(x-\arcsin\frac{1}{\sqrt{5}}\right)=-\frac{2}{\sqrt{5}} $$
so the first interesting point is $x=\pi+\arcsin\frac{1}{\sqrt{5}}+\arcsin\frac{2}{\sqrt{5}}=\pi+\arctan\frac{1}{2}+\arctan 2=\frac{3\pi}{2}$, at which $\sqrt{1+\sin(x)}$ equals zero, $g(x)$ is positive and $g'(x)$ has a jump discontinuity. Not a minimum for sure. The second interesting point is
$x=2\pi+\arcsin\frac{1}{\sqrt{5}}-\arcsin\frac{2}{\sqrt{5}}=\frac{3\pi}{2}+2\arctan\frac{1}{2}$ or $x=2\pi-\arctan\frac{3}{4}$, at which $\sqrt{1+\sin x}=\sqrt{\frac{2}{5}}$. We have $\int_{0}^{2\pi}\sqrt{1+\sin t}\,dt=4\sqrt{2}$ and
$$ \int_{0}^{\arctan\frac{3}{4}}\sqrt{1+\sin t}\,dt = \int_{0}^{2\arctan\frac{1}{3}}\sqrt{1+\sin t}\,dt = 2\int_{0}^{\arctan\frac{1}{3}}\sqrt{1+\sin(2z)}\,dz $$
equals:
$$ 2\int_{0}^{\frac{1}{3}}\sqrt{1+\frac{2t}{1+t^2}}\,dt=2\int_{0}^{1/3}\frac{1+t}{\sqrt{1+t^2}}\,dt =2-2\sqrt{\frac{2}{5}}$$
and the given function is positive at $x=2\pi-2\arctan\frac{1}{3}$, too. The other interesting points provide larger values, hence the minimum of the given function is attained at $x=0$ and trivially equals $-1$:
$$ \forall x\geq 0,\qquad g(x)=\int_{0}^{x}\sqrt{1+\sin t}\,dt-\sqrt{1+\sin x}\geq -1.$$
Additionally, $g(x)$ roughly behaves like $-1+\frac{2\sqrt{2}}{\pi}x$ on $\mathbb{R}^+$:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
When is $\sqrt[3]{a+\sqrt b}+\sqrt[3]{a-\sqrt b}$ an integer? I saw a Youtube video in which it was shown that
$$(7+50^{1/2})^{1/3}+(7-50^{1/2})^{1/3}=2$$
Since there are multiple values we can choose for the $3$rd root of a number, it would also make more sense to declare the value of this expression to be one of $2, 1 + \sqrt{-6},$ or $1 - \sqrt{-6}$
We may examine this more generally. If we declare $x$ such that
$$x=(a+b^{1/2})^{1/3}+(a-b^{1/2})^{1/3}$$
$$\text{(supposing } a \text{ and } b \text{ to be integers here)}$$
one can show that
$$x^3+3(b-a^2)^{1/3}x-2a=0$$
Which indeed has $3$ roots.
We now ask
For what integer values of $a$ and $b$ is this polynomial solved by an integer?
I attempted this by assuming that $n$ is a root of the polynomial. We then have
$$x^3+3(b-a^2)^{1/3}x-2a$$
$$||$$
$$(x-n)(x^2+cx+d)$$
$$||$$
$$x^3+(c-n)x^2+(d-nc)x-nd$$
Since $(c-n)x^2=0$ we conclude that $c=n$ and we have
$$x^3+3(b-a^2)^{1/3}x-2a=x^3+(d-c^2)x-cd$$
And - to continue our chain of conclusions - we conclude that
$$3(b-a^2)^{1/3}=d-c^2 \quad\text{and}\quad 2a=cd$$
At this point I tried creating a single equation and got
$$108b=4d^3+15c^2d^2+12c^4d-c^6$$
This is as far as I went.
| Use $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ identity. Things will cancel out.
| {
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} |
if quadratic $x^2-(a+b+c)x+ab+bc+ca=0$ has complex roots, Then prove that $\sqrt{a}$,$\sqrt{b}$ and $\sqrt{c}$ are sides of triangle
The following quadratic equation $$x^2-(a+b+c)x+ab+bc+ca=0$$ has complex roots.
Prove that $\sqrt{a}$,$\sqrt{b}$ and $\sqrt{c}$ are sides-lengths of triangle, where $a,b,c \in \mathbb{R^+}$.
Since the quadratic has complex roots, we have Discriminant negative. So
$$(a+b+c)^2 \lt 4ab+4bc+4ca$$ $\implies$
$$a^2+b^2+c^2 \lt 2(ab+bc+ca)$$ $\implies$
$$c^2-2c(a+b)+(a-b)^2 \lt 0$$ $\implies$
$$(a+c-b)^2 \lt 4ac$$ $\implies$
we get
$$a+c-b \lt 2 \sqrt{ac}$$
any clue further?
| From the given we obtain:
$$\sum_{cyc}(2ab-a^2)>0$$ or
$$(\sqrt{a}+\sqrt{b}+\sqrt{c})\prod_{cyc}(\sqrt{a}+\sqrt{b}-\sqrt{c})>0$$
because
\begin{align}
\sum_{cyc}(2ab-a^2)&=4ab-(a+b-c)^2\\
&=\left(2\sqrt{ab}\right)^2-(a+b-c)^2\\
&=(2\sqrt{ab}-a-b+c)(2\sqrt{ab}+a+b-c)\\
&=(c-(\sqrt{a}-\sqrt{b})^2)((\sqrt{a}+\sqrt{b})^2-c)\\
&=(\sqrt{c}-\sqrt{a}+\sqrt{b})(\sqrt{c}+\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+\sqrt{c}).
\end{align}
Let $a\geq b\geq c$.
Hence, $\sqrt{b}+\sqrt{c}-\sqrt{a}>0$ and rest multipliers are also positives.
Done!
| {
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A $3\times 3$ matrix to the power of $n$. I can't find a formula for :
$$
A =\begin{pmatrix}
1 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 3 \\
\end{pmatrix}^n
$$
I tried to separate $A = I + J$ with $J$ nilpotent but I didn't success.
Can you give me a hint?
Thanks.
| You ask for a hint...
$A^2 =
\begin{pmatrix}
1 & 3 & 1 \\
0 & 4 & 5 \\
0 & 0 & 9 \end{pmatrix}$
$A^3 =
\begin{pmatrix}
1 & 7 & 6 \\
0 & 8 & 19 \\
0 & 0 & 27 \end{pmatrix}$
$A^4 =
\begin{pmatrix}
1 & 15 & 25 \\
0 & 16 & 65 \\
0 & 0 & 81 \end{pmatrix}$
$A^5 =
\begin{pmatrix}
1 & 31 & 90 \\
0 & 32 & 211 \\
0 & 0 & 243 \end{pmatrix}$
Diagonal elements are easy, other elements are fairly obvious...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the closed form of the integral$\int_{0}^{\infty}f(x)\frac{\sin^nx}{x^m}dx$ Where f(x) is an even function,and also is a periodic functions,the period is $\pi$,and $n,m\in\Bbb N$,and n+m is odd number.
When n+m is even, I already have a solution.But when n+m is odd, I don't know how to solve it.I'd appreciate it if someone could help me with it.
When n+m is even,my answer is as follows:
If f (x) is an even function, and the period is $\pi$,we have:
$$\int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx=\int_{0}^\frac{\pi}{2}f(x)g_m(x)\sin^nxdx \qquad (1)$$
$n,m\in\Bbb N$,
Where the n+m is an even,and $g_m(x)$ in (1) is as follows:
$$g_m(x)=\begin{cases}\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\csc x\right),& \text{for n is odd and}\\[2ex]
\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\cot x\right),& \text{ for n is even .}
\end{cases}$$
——————————————————————————————————————————————————
Proof:
\begin{align}
\int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin ^nx}{x^m}\right)dx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin^n x}{x^m}\right)dx\\
&=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx\\
&=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin^n x}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty(-1)^{nk+n+m}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin^n x}{(x-k\pi)^m}\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^m}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^m}+\frac{(-1)^{n+m}}{(x-k\pi)^m}\right]\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_m(x)dx
\end{align}
When n+m is an even,and we know by the Fourier series
\begin{align}
\csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\
\end{align}
and
\begin{align}
\cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)
\end{align}
Take the m-1 order derivative,thus we obtain $g_m(x)$.
——————————————————————————————————————————————————
Example:
\begin{align}
(1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\
&=\frac{\pi}{4}\\
\end{align}
\begin{align}
(2.)
\int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\
\end{align}
\begin{align}
(3.)
\int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\
\end{align}
| There is an excellent result related with this integral, enjoy!
$$I=\int_{0}^{\infty }x^{p}\ \left ( \frac{\sin(x)}{x} \right )^ndx\ \ \ \ \ \ , n=1,2,3,\dots , \ \ \ 0\geq p\geq -1,\\
\\
\\
I=\frac{\pi }{2(2i)^{n}\Gamma (n-p)}\sum_{m=0}^{n }(-1)^{n-m}\frac{n!}{m!(n-m)!}\left | n-2m \right |^{n-p-1}\left ( \frac{1}{\sin(\frac{n-p+1}{2})\pi }-\frac{\operatorname{sgn}(n-2m)}{\sin(\frac{n-p}{2})\pi }i \right ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proof writing for polynomials Let the polynomial $f(x)=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n$ have integral coefficients. If there exists four distinct integers $a,b,c$ and $d$ such that $f(a)=f(b)=f(c)=f(d)=5$ show that there is no integer $k$ such that $f(k)=8$
I have tried to prove it but I somehow feel that my proof is incorrect please point out the errors and suggest some other way to do the question.
This how I proceeded
$\frac{x^{n+1}-1}{x-1}=x^n+a_1 x^{n-1}+a_2 x^{n-2}+.....a_{n-1}x+a_n $
$\frac{a^{n+1}-1}{a-1}=5$
hence $\frac{a^{n+1}-1}{5}=a-1$
Since $a$ is an integer $a-1$ will also be an integer hence $5|a^{n+1}-1$
Using fermat's little theorem we know that $a^4-1\equiv_5 0$
Since $f(a)=5$ we get $n+1=4$
Let k be an integer such that $f(k)=8$
$\frac{a^4-1}{8}=a-1$ Since $a-1$ is an integer $8|a^4-1$
We will consider two cases
*
*when $a$ is an even integer
*when $ a$ is an odd integer
If $a$ is even then $a^4-1$ will be odd and it not be divisible by $8$
If $a$ is odd then it will be relatively prime to $8$ hence $8$ will be Carmichael number
$a^7-1\equiv0(mod8)$ which proves that $a^4-1$ is not divisible by 8.
$\therefore$ there is no integer k such that $f(k)=8$
| Hint: show that if $f(a) = f(b) = f(c) = f(d) = 5$, then
$$f(x) = 5 + (x-a)(x-b)(x-c)(x-d) g(x)$$
for some polynomial $g$ with integer coefficients. If $f(e) = 8$, then
$$(e-a)(e-b)(e-c)(e-d) g(e) = 3,$$
but a product of four distinct integers can't be a divisor of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$(\frac{x +1}{x^{2/3}-x^{1/3}+1} -\frac{x+1}{x-x^{1/2}})^{10}$ which term doesn't contain ${x}$? As I know $T_{r+{1}}$=$C(n,r)$
I can't able to apply the formula for the Term which not contain $x$,which is applicable for this (${x +\frac{1}{x}})^n$
So, please help me to solve this problem . Any help will be appreciated
| Assuming the expression is
$$\left(\frac{x +1}{x^{2/3}-x^{1/3}+1} -\frac{x-1}{x-x^{1/2}}\right)^{10}$$
The first term can be simplified setting $x^{1/3}=z\to x=z^3$
$\dfrac{x +1}{x^{2/3}-x^{1/3}+1}=\dfrac{z^3+1}{z^2-z+1}=z+1=x^{1/3}+1$
The second term simplifies in a similar way $x^{1/2}=z\to x=z^2$
$\dfrac{x-1}{x-x^{1/2}}=\dfrac{z^2-1}{z^2-z}=1+\dfrac{1}{x^{1/2}}$
The original expression simplifies to
$$\left(x^{1/3}+1-1-\dfrac{1}{x^{1/2}}\right)^{10}=\left(\sqrt[3]{x}-\frac{1}{\sqrt{x}}\right)^{10}$$
Thus the terms non containing $x$ are those who simplify the cube and square roots, that is the $+a^6b^4$ term whose coefficient is $+\dbinom{10}{6}=210$
Hope this is useful
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show $5n^3+7n^5 \equiv 0$ mod 12 for all integers $n$. I've been stuck on this question for quite a while now. I could obvious show this by setting $n$ to all numbers from 0-11, however this is not very efficient.
Note $5n^3+7n^5 = n^3(5+7n^2)$. We could use the Chinese remainder theorem, we show $n^3(5+7n^2) \equiv 0 $ mod 3 and mod 4, but this doesn't seem to be any easier. What would be the best way to solve this problem?
| The only quadratic residue mod $3$ is $1$.
The only quadratic residue mod $4$ is also $1$.
Therefore:
\begin{align*}
n^3 &\equiv n^5 \pmod{3} &\qquad n^3 &\equiv n^5 \pmod{4}
\end{align*}
and so
$$ n^3 \equiv n^5 \pmod{12} $$
Applying this to the original equation:
$$ 5n^3 + 7n^5 \equiv 12n^3 \equiv 0 \pmod{12} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2411862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies
\begin{cases}
A+B=0
\\[2ex]
A-B=\frac{1}{i}=-i
\end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
| You found
$$\arctan x=\dfrac{i}{2}\Big(\ln(x+i)-\ln(x-i)-i\pi\Big)=\dfrac{i}{2}\ln\dfrac{x+i}{x-i}+\dfrac{\pi}{2}=-\dfrac{i}{2}\ln\dfrac{x-i}{x+i}+\dfrac{\pi}{2}$$
let $x=\dfrac1z$ so
$$\arctan\dfrac1z=-\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}+\dfrac{\pi}{2}$$
and then
$$\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}=\dfrac{\pi}{2}-\arctan\dfrac1z=\arctan z$$
as you want!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int {\sqrt{x^4+1}}dx$ Evaluate $$\int \sqrt{x^4+1}dx$$
My Try: I used parts as follows:
$$I=x{\sqrt{x^4+1}}-4 \int \frac{x^4 dx}{\sqrt{x^4+1}}$$ $\implies$
$$I=\frac{x}{\sqrt{x^4+1}}-4I+4 \int \frac{dx}{\sqrt{x^4+1}}$$ $\implies$
$$5I=\frac{x}{\sqrt{x^4+1}}+4J$$ where
$$J=\int \frac{dx}{\sqrt{x^4+1}}$$
any clue here?
| When $|x|\leq1$ ,
$\int\sqrt{x^4+1}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{4n+1}}{4^n(n!)^2(1-2n)(4n+1)}+C$
When $|x|\geq1$ ,
$\int\sqrt{x^4+1}~dx$
$=\int x^2\sqrt{1+\dfrac{1}{x^4}}~dx$
$=\int x^2\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{4n}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2-4n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3-4n}}{4^n(n!)^2(1-2n)(3-4n)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(2n-1)(4n-3)x^{4n-3}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Inequality : $\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z}\geq1 $
Prove that if $a, b, c, x, y, z >0$ and $n$ is positive integer such that
$(a^n+b^n+c^n)^{n+1}=x^n+y^n+z^n$ , then $$\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z}\geq1 $$
My attempt :
By Holder Inequality,
$$(a^{n+1}+b^{n+1}+c^{n+1})^n(1+1+1) \geq (a^n+b^n+c^n)^{n+1}$$
$$(a^{n+1}+b^{n+1}+c^{n+1})^n \geq \frac{x^n+y^n+z^n}{3}$$
I don't know how to proceed.
| By Holder we have $(\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z})^n(x^n+y^n+z^n)\geq(a^n+b^n+c^n)^{n+1}$.
Thus we have $(\frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z})^n\geq\frac{(a^n+b^n+c^n)^{n+1}}{x^n+y^n+z^n}=1 \rightarrow \frac{a^{n+1}}{x}+\frac{b^{n+1}}{y}+\frac{c^{n+1}}{z}\geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit of a sequence : $x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)}$ Can someone help me with this problem?
Finding the limit $\lim_{n \to \infty}\ x_n$ where
$$x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)},\quad n\in\mathbb{N}.$$
I don't have a clue how to do this.
| Use $$\frac{1}{n(n+1)(n+2)}=\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Generating functions for the solutions to $3x_1 + x_2 = n$ I need to find the amount of solutions with non negative integers to the equation $x_1 + x_2 + x_3 = n$
with $x_2 = 2x_1$.
I transformed it to $y_1 + y_2 = n$ when $y_1=3x_1, y_2 = x_3$
(because $x_1 + x_2 + x_3 = n$ here is the same as $x_1 + 2x_1 + x_3 = n$, which is $3x_1 + x_3 = n$).
Using generating functions $$[x^n] = \frac {1}{(1-x^3)(1-x)}$$ which is by expanding $1-x^3$:$$[x^n] = \frac {1}{(1-x)^2(1+x+x^2)}$$
The problem here is that that the expansion of $(1+x+x^2)$ involves complex numbers so I'm not sure how to proceed from here.
| You can proceed in two ways. First, you can use the formula
$$ \frac{1}{1 - x} \sum_{n} a_n x^n = \sum_n \left( \sum_{k = 0}^n a_k \right) x^n. $$
That is, multiplication by $(1 - x)^{-1}$ gives you the partial sums of a series. The partial sums of
$$ \frac{1}{1 - x^3} = 1 + x^3 + x^6 + x^9 + \cdots $$
are given by
$$ (1 + x + x^2) + 2(x^3 + x^4 + x^5) + 3(x^6 + x^7 + x^8) + 4\cdots. $$
Which can be described as
$$ [x^n] \frac{1}{(1 - x)^2(1 + x + x^2)} = \left\lfloor \frac{n}{3} \right\rfloor + 1. $$
The second way, which is what you are trying to do, uses the fundamental property of rational generating functions (c.f. Stanley EC1 Theorem 4.1.1) which implies that
$$ [x^n] \frac{1}{(1 - x)^2(1 + x + x^2)} = (A + Bn)1^n + C\left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)^n + D \left( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3} \right)^n $$
Which if you look at the first 4 coefficients give you enough information to determine (here it's helpful to use a computer rather than do this by hand)
$$ [x^n] \frac{1}{(1 - x)^2(1 + x + x^2)} = \frac{2}{3} + \frac{1}{3}n + \frac{1}{3} \cos\left(\frac{2 n \pi}{3}\right) + \frac{1}{3 \sqrt 3} \sin\left(\frac{2 n \pi}{3}\right). $$
Then, using the periodicity of sine and cosine, we see that when $n \equiv 0 \pmod 3$ we have
$$ \frac{2}{3} + \frac{1}{3}n + \frac{1}{3} = \frac{n}{3} + 1. $$
When $n \equiv 1 \pmod 3$ we have
$$ \frac{2}{3} + \frac{1}{3}n - \frac{1}{6} + \frac{1}{6} = \frac{n - 1}{3} + 1. $$
And finally, when $n \equiv 2 \pmod 3$ we have
$$ \frac{2}{3} + \frac{1}{3}n - \frac{1}{6} - \frac{1}{6} = \frac{n - 2}{3} + 1. $$
This three cases can be described together as
$$ \left\lfloor \frac{n}{3} \right\rfloor + 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$
What's
$$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
What have I tried?
$$(n+2)^4=n^4+8n^3+24n^2+32n+16$$
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
Remainder:
$$4n^3+18n^2+28n+15$$
mod:
$$2n^2-1\pmod{4}$$
I can compute $\sum x^2$ but I don't know what to do with $$\sum_{n=1}^{2017}\left(2n^2-1\mod{4}\right)$$
| What a mess.
First of I must point out that the is meaningless garbage.
$a \mod b + c \mod d$ is an abuse of notation. The statement $a \mod b$ is not a number. It is a class of equivalent numbers. It may but add moduloly to other classes in the same modulo residue system but not to classes in others.
However I will take this (under objection) as an abuse of notation for the remainder function where, for example $28 \mod 5 = 3$ and not $-2$ or $7$ as we are taking one value between $0$ and $5$. This is wrong and you should use a different notation for the remainder function that returns a number.
But I'll use your abuse.
======
But note: $(k+1)^4 \mod k^4 \equiv 4k^3 + 6k^2 + 4k + 1 \mod k^4$
So $(n+2)^4 \mod (n+1)^4 \equiv 4(n+1)^3 + 6(n+1)^2 + 4(n+1) + 1 \mod (n+1)^4$
And modulo $4$ we get:
$[ (n+2)^4 \mod (n+1)^4] \mod 4 \equiv [2(n+1)^2 + 1]\mod 4 \equiv 2n^2 + 4n + 3\equiv 2n^2 + 3 \mod 4$.
If $n \equiv 0,1,2,3 \mod 4$ we have $2n^2 + 3 \equiv 3,1,3,1 \mod 4$.
So $\sum\limits_{i= 2k+1}^{2k+2} ([ (n+2)^4 \mod (n+1)^4] \mod 4) = 1+ 3 = 4$.
So $\sum\limits_{i= 1}^{2017} ([ (n+2)^4 \mod (n+1)^4] \mod 4)=\sum\limits_{i=1}^{2017}[3\text{ if i is odd}|1\text{ if i is even}]= \frac {2016}2*4 + 3 = 4035$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fibonacci numbers prove by induction that $F(n)>2n$ for $ n>7$ So I am having trouble proving that $F(n)>2n$ for all $n>7$.
for base case I took n=8, so obviously $f(8)=21> 2*8= 16$
Assuming$ f(n)>2n$
$f(n+1)>2(n+1)$
$f(n)+f(n-1)>2n+2$
$f(n)> 2n$ according to the induction hypothesis and since $n>7$ ,$f(n-1)$ should be at least $f(7)$ and therefore more than 2, however I'm not sure if this is the correct way to prove this by induction on n...
| Another way :
It is easier to prove :
$$\begin{array}{l}
{F_n} = \frac{{{\varphi ^n} - {\psi ^n}}}{{\varphi - \psi }} = \frac{{{\varphi ^n} - {\psi ^n}}}{{\sqrt 5 }} > \frac{{{\varphi ^n}}}{{\sqrt 5 }} = \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}\\
\varphi = \frac{{1 + \sqrt 5 }}{2} \approx 1.61803{\mkern 1mu} 39887 \cdots
\end{array}$$
so prove that $$\forall n>7 :F_n>\frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}>2n$$ so
$$p(8) \space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^8}}}{{\sqrt 5 }}>2\times 8 \\20.57>16 \checkmark $$
$$p(n) :\space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n}}}{{\sqrt 5 }}>2n\\p(n+1): \space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^{n+1}}}}{{\sqrt 5 }}>2(n+1)$$ multiply $p(n) $ by $\frac{{1 + \sqrt 5 }}{2}$
$$\space \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^{n+1}}}}{{\sqrt 5 }}>2n(\frac{{1 + \sqrt 5 }}{2})>2n(1.5)=\underbrace{3n >2n+2}_{\forall n>7} \checkmark$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\lfloor\frac{(n+1)^2}2\rfloor$ equal $\lfloor\frac{n^2}2\rfloor + n$? By expanding the left, I found that $\lfloor\frac{(n+1)^2}2\rfloor$ = $\lfloor\frac{n^2}2 + n + \frac{1}2\rfloor$. I am not sure how to relate $\lfloor n + \frac{1}2 \rfloor$ to n so as to show that the right is true/false. Any ideas?
| We can start looking at what would happen if you actually have some values for $n$ for the left hand side when you expand it: $$\lfloor \frac{n^2}{2} + n + \frac{1}{2} \rfloor$$
We can clearly see that IF $n$ is an even number, then $\frac{n^2}{2} + n$ is an even number (won't show the proofs for even and odd number properties here); in that case the $\frac{1}{2}$ at the end won't do anything, the floor function will essentially delete it.
This means we're just left with $\lfloor \frac{n^2}{2} + n \rfloor$. Which is the same as $\lfloor\frac{n^2}{2}\rfloor + n$, because a floor function on an even number does nothing. We can just put floors all over the place but we'll leave it like that just to make it the same as in the question.
Now let's try and see what happens IF $n$ is an odd number; then the fraction $\frac{n^2}{2}$ will always have a rest of $\frac{1}{2}$ (i.e. an odd number divided by two will always have that rest). If we were to put the $+\frac{1}{2}$ fraction outside of the floor, it would just become a $+1$ (because this fraction will always bump $\frac{n^2}{2}$ up to the next integer), similarly the floor function will do nothing to the final term $n$ (because it is an integer and floor functions do nothing to those). This means we can rewrite the expanded equation as $\lfloor\frac{n^2}{2} \rfloor + n + 1$ (if $n$ is odd). As before, on this final equation, we can just put extra floor functions on there however we want to, but as before let's leave it like that for simplicity.
Now, let's compile the information:
\begin{cases}
\lfloor \frac{(n+1)^2}{2} \rfloor = \lfloor\ \frac{n^2}{2} \rfloor + n& \text{ if } n = even\\
\lfloor \frac{(n+1)^2}{2} \rfloor = \lfloor\frac{n^2}{2} \rfloor + n + 1 & \text{ if } n= odd
\end{cases}
$$$$
So only when $n = even$ are they the same, else it's always one off.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$, then $\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $.
If $$\cos(z-x) + \cos(y-z) + \cos(x-y) = -\frac{3}{2}$$ then how can I show that the sum of cosines of each angle ($x$, $y$, $z$) and sines of each angle sum up to zero? i.e. $$\sin x + \sin y + \sin z = 0 = \cos x + \cos y + \cos z $$
I tried:
• Expanded using $\cos(A-B) = \cos A\cos B+\sin A\sin B $, but it did nothing.
After spending one hour to this problem, I thought that there must be a shorter and ideal way.
| Let $z_1 = \cos x +i \sin x$ etc.
Then $2 \cos (x-y) = \dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}$ etc.
We are given that $\dfrac{z_1}{z_2}+\dfrac{z_2}{z_1}+\dfrac{z_2}{z_3}+\dfrac{z_3}{z_2}+\dfrac{z_3}{z_1}+\dfrac{z_1}{z_3} = -3$
or $\dfrac{z_2+z_3}{z_1}+\dfrac{z_3+z_1}{z_2}+\dfrac{z_1+z_2}{z_3} = -3$
$\Rightarrow \displaystyle \sum_{cyc} \frac{z_2+z_3}{z_1}+1 =0 \Rightarrow (z_1+z_2+z_3)\left(\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right) = 0$
Thus $z_1+z_2+z_3 = 0$ or $\displaystyle \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=0$
from which the required result follows
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of $a_1=\sqrt{3}$, $a_2=\sqrt{3\sqrt{3}}$, $a_3=\sqrt{3\sqrt{3\sqrt{3}}}$... Find the limit of:
$a_1=\sqrt{3}$
$a_2=\sqrt{3\sqrt{3}}$
$a_3=\sqrt{3\sqrt{3\sqrt{3}}}$
...
By using induction, I found that the limit is 3 but it seems pretty strange to me, I thought it would go to a further number.
I just want to know if I'm right (proof-verification).
| The sequence is defined recursively as
$x_0=\sqrt 3;\;x_n=\sqrt{3x_{n-1}}$
It's not difficult to prove that $x_n=3^{1-\frac{1}{2^{n+1}}}$
Indeed if $x=1$ we have $x_1=\sqrt{3\sqrt{3}}=\sqrt[4]{3^3}=3^{\frac34}=3^{1-\frac14}$
If the formula is valid for $n$ then $x_n=3^{1-\frac{1}{2^{n+1}}}$ we show that it is valid for $x_{n+1}=\sqrt{3x_n}=\sqrt{3\cdot 3^{1-\frac{1}{2^{n+1}}}}=\sqrt{3^{2-\frac{1}{2^{n+1}}}}=\left(3^{2-\frac{1}{2^{n+1}}}\right)^{\frac12}=3^{1-\frac{1}{2^{n+2}}}$
Therefore we proved by induction that $x^n=3^{1-\frac{1}{2^{n+1}}}$ and as $n\to\infty$ we have $x\to 3$
More quickly,assuming that the sequence converges,
$$x=\sqrt{3\sqrt{3\sqrt{3}}}\ldots$$
$$x^2=3\sqrt{3\sqrt{3}}\ldots$$
$$x^2=3x$$
$$x=3$$
hope it helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim\limits_{x \to 1} \frac{x+2}{x^2+1} = \frac{3}{2}$ I am required to prove the following limit using the epsilon-delta definition:
$$\lim_{x\to 1}\frac{x+2}{x^2+1}=\frac{3}{2}$$
So, ($\forall\epsilon>0)(\exists\delta>0)[0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert<\epsilon]$
Below is my working on getting the $\delta$ that I need:
$\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert\le\vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}=\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}$ (Triangle Inequality)
But I know that $x^2+1\ge1$ for all values of $x$, so I can say that $\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}$
So for every $\epsilon$, I choose $\delta=\epsilon-\frac{9}{2}$, then:$$0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert \le \vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}\le\vert x+2\vert+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}=\epsilon$$
Am I doing this right? Is there something wrong to assume that $x^2+1\ge1$ for all values of $x$?
| Hint:)
Notice that you can't never conclude from
$$|f(x)-\ell|\leq|f(x)|+|\ell|<\epsilon~~~~,~~~~~\ell\neq0$$
obtained by triangle inequality, a limit yield.
| {
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How to find Ackermann(3,n)? { y+1 if x=0;
A(x,y)= { A(x-1,1) if y=0;
{ A(x-1,A(x,y-1)) otherwise
So I was trying to prove A(3,y) = $2^{(y+3)} -3 $
Though I was able to figure out
A(1,y) = y + 2
A(2,y) = 2y + 3
But for A(3,y) = A(2,A(3,y-1))
= 2A(3,y-1) + 3
= 2(A(2,A(3,y-2))) + 3
= 2(2(A(3,y-2) + 3) + 3
= 2(2(A(3,y-2)) + 6 + 3
= 2(2(2....(y times)..(A(3,0) + 3))))) + 3.2^y + 3.2^(y-1) ... + 3
= 2^y(A(3,0) + 3 ) + 3(1-2^y)/(1-2)
and as A(3,0) = 5
$$=2^y.8 + 3(1-2^y)/(1-2)$$
$$=2^{(y+3)} - 3(1-2^y)$$
However it seems like I am doing something wrong as A(3,y) = $2^{(y+3)} -3 $and not $2^{(y+3)} - 3(1-2^y) $
| There is a mistake in your decomposition. The result looks like this:
$$2(2(...(2\cdot A(3,0) + 3)+3)+3)...+3$$
So, in fact you have $5\cdot2^y$ and then there is a geometric sequence with overall sum $3(2^y-1)$, add it together and you get $8\cdot 2^y-3$.
| {
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Show that the prime must be $13$ if $p\mid n^2+3$ and $p\mid (n+1)^2+3$ Let $p$ be a prime for which it exists a $n\in \mathbb{Z}$ with $p\mid n^2+3$ and $p\mid (n+1)^2+3$. I want to show that it must hold that $p=13$ and that there are infinitely many integers $n$ so that the above relations hold.
$$$$
We have that $p\mid n^2+3$ and $p\mid (n+1)^2+3 \Rightarrow p\mid n^2+2n+1+3$, so we get $p\mid (n^2+2n+1+3)-(n^2+3) \Rightarrow p\mid 2n+1$, right?
How could we continue?
| Since $p \mid 2n+1$ we have $p\mid(2n+1)^2=4n^2+4n+1$.
We also have $p\mid2(n^2+3)+2((n+1)^2+3)=4n^2+4n+14$. Thus we have $p\mid13$ and $p=13$.
Also, any $n=13k+6, k\in \mathbb{Z}$ satisfies the condition.
| {
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"url": "https://math.stackexchange.com/questions/2429174",
"timestamp": "2023-03-29T00:00:00",
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What is a basis for $V_A$? PROBLEM
If $A \in \mathbb{F}^{3 \times 3}$ is given by
$$A := \left[\begin{array}{ccc} 1 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 1 \\ \end{array}\right],$$
find a basis for
$$V_A = \{B \in \mathbb{F}^{3 \times 3} | AB = BA\}.$$
MY TRY
I write out $B$ explicitly as
$$B := \left[\begin{array}{ccc} a & b & c \\
d & e & f \\
g & h & i \\ \end{array}\right],$$
then $$AB = BA$$ is equivalent to
$$\left[\begin{array}{ccc} 1 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 1 \\ \end{array}\right]\left[\begin{array}{ccc} a & b & c \\
d & e & f \\
g & h & i \\ \end{array}\right] = \left[\begin{array}{ccc} a & b & c \\
d & e & f \\
g & h & i \\ \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 1 \\ \end{array}\right]$$
which in turn is equivalent to the simultaneous system of equations
$$\left\{\begin{array}{ccc} a=a & 3b=b & c=c \\
d=3d & 3e=3e & f=3f \\
g=g & 3h=h & i=i \\ \end{array}\right\},$$
which then gives the general form of $B$, as follows:
$$B = \left[\begin{array}{ccc} a & 0 & c \\
0 & 3e & 0 \\
g & 0 & i \\ \end{array}\right].$$
Note that we can write $B$ as the following linear combination:
$$B = \left[\begin{array}{ccc} a & 0 & c \\
0 & 3e & 0 \\
g & 0 & i \\ \end{array}\right] = \left[\begin{array}{ccc} a & 0 & 0 \\
0 & 3e & 0 \\
0 & 0 & i \\ \end{array}\right] + \left[\begin{array}{ccc} 0 & 0 & c \\
0 & 0 & 0 \\
0 & 0 & 0 \\ \end{array}\right] +
\left[\begin{array}{ccc} 0 & 0 & 0 \\
0 & 0 & 0 \\
g & 0 & 0 \\ \end{array}\right].$$
This implies that the set
$$S = \Bigg\{\left[\begin{array}{ccc} a & 0 & 0 \\
0 & 3e & 0 \\
0 & 0 & i \\ \end{array}\right], \left[\begin{array}{ccc} 0 & 0 & c \\
0 & 0 & 0 \\
0 & 0 & 0 \\ \end{array}\right],
\left[\begin{array}{ccc} 0 & 0 & 0 \\
0 & 0 & 0 \\
g & 0 & 0 \\ \end{array}\right]\Bigg\}$$
spans $V_A$. In addition, $S$ is clearly linearly independent. Therefore, $S$ is a basis for $V_A$.
QUERY
Is my solution correct? If it is wrong, where is the error and how can it be mended?
| There's a problem near the end. An example of a basis will be$$S=\left\{\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}\right\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the likelihood of two line segments crossing? Consider a square space.
Randomly select 4 points.
Randomly connect two sets of two to each other with line segments.
What is the chance of the line segments intersecting?
(I will maybe try and solve this with a little Monte-Carlo simulation but would be very interested in an analytic solution.)
| This is my approach:
Consider a line $y=mx+c$ that divides the unit square into two trapeziums. Let us call the left trapezoid $a$, and the right trapezoid $b$.
By solving the equations $(y=mx+c)=1$ and $(y=mx+c)=0$, or $mx+c=1$ and $mx+c=0$, the $x$-values of the intersections are $x = \frac{-c+1}{m}$, and $\frac{-c}{m}$. Since the $y$-values are $0$ and $1$, the intersection coordinates are $(\frac{-c}{m}, 0)$ and $(\frac{-c+1}{m}, 1)$.
Using this information, the area of trapezium $a$ is $1 * \frac{(-c/m+1)+(-c/m)}{2}$, and of $b$ is $1 - (1 * \frac{(-c/m+1)+(-c/m)}{2})$ (since $a+b=1$). Simplifying gives the area of $a$ as $-\frac{c}{m}+ \frac{1}{2}$, and the area of $b$ as $\frac{c}{m}+\frac{1}{2}$.
For the line connecting two points to not cross the green line, the points cannot be in $a$ and $b$. Since using geometric probability, the probability of this is $ab-ba$, we can subtract this condition from $1$ to get $1 - 2(-\frac{c}{m}+ \frac{1}{2})(\frac{c}{m}+\frac{1}{2})$, and after simplifying we have the probability as $1 - 2(-\frac{c^2}{m^2}+\frac{1}{4})$ or $$\frac{2c^2}{m^2}+\frac{1}{2},$$
where $c$ is the $y$-intercept of the line (when extended), and $m$ is its slope.
Note: This example doesn't work when $c=0$, because regardless of what $m$ is chosen, it would equal $0+\frac{1}{2}=\frac{1}{2}$, and when the line crosses a point not between $0$ and $1$ when $x=0$ or $x=1$.
| {
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"url": "https://math.stackexchange.com/questions/2430323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove this inequality $\sum_{cyc}\sqrt{\frac{a}{3b^2+1}}\ge \frac{3}{2}$
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove that: $$\sqrt{\frac{a}{3b^2+1}}+\sqrt{\frac{b}{3c^2+1}}+\sqrt{\frac{c}{3a^2+1}}\ge \frac{3}{2}$$
I tried the AM-GM, Holder, C-S inequalities but was unsuccessful.
| Holder helps!
$$\left(\sum_{cyc}\sqrt{\frac{a}{3b^2+1}}\right)^2\sum_{cyc}a^2(3b^2+1)(a+b+3c)^3\geq$$
$$\geq\left(\sum_{cyc}a(a+b+3c)\right)^3=\left(\sum_{cyc}(a^2+4ab)\right)^3.$$
Thus, it remains to prove that
$$4\left(\sum_{cyc}(a^2+4ab)\right)^3\geq9\sum_{cyc}a^2(3b^2+1)(a+b+3c)^3,$$
which after homogenization and full expanding gives
$$\sum_{cyc}(a^6+36a^5b+18a^5c+24a^4b^2+12a^4c^2+16a^3b^3+$$
$$+342a^4bc+348a^3b^2c+354a^3c^2b-1151a^2b^2c^2)\geq0,$$
which is obvious by AM-GM.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$ Determine: $\sum_\limits{k=1}^{100}\left \lfloor \sqrt{k}+\frac{1}{k} \right \rfloor$.
It's well-known that $x-1<\lfloor x \rfloor \leq x, \forall x \in \mathbb{R}$, but this didn't help me at all.
I computed it and the sum equals 627, but I found no useful properties.
| Note that for $k\ge5$,
$$
\sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}}\gt\frac1k\tag{1}
$$
so that
$$
\sqrt{k}\lt\sqrt{k}+\frac1k\lt\sqrt{k+1}\tag{2}
$$
Taking the floor of $(2)$ says that
$$
\left\lfloor\sqrt{k}\right\rfloor\le\left\lfloor\sqrt{k}+\frac1k\right\rfloor\le\left\lfloor\sqrt{k+1}\right\rfloor\tag{3}
$$
The only time that $\left\lfloor\sqrt{k}\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$ is when $k+1$ is a perfect square, which, in light of $(2)$ means that $\left\lfloor\sqrt{k}+\frac1k\right\rfloor\lt\left\lfloor\sqrt{k+1}\right\rfloor$. Thus, we get that for $k\ge5$
$$
\left\lfloor\sqrt{k}+\frac1k\right\rfloor=\left\lfloor\sqrt{k}\right\rfloor\tag{4}
$$
We can also verify $(4)$ for the case $k=4$.
Thus,
$$
\begin{align}
\sum_{k=1}^{100}\left\lfloor\sqrt{k}+\frac1k\right\rfloor
&=\overbrace{2+1+2+10}^{k=1,2,3,100}+\sum_{k=4}^{99}\left\lfloor\sqrt{k}\right\rfloor\\
&=15+\sum_{j=2}^9j\left((j+1)^2-j^2\right)\quad\text{floor times number of terms with that floor}\\
&=15+\sum_{j=2}^9\left(2j^2+j\right)\\
&=15+\sum_{j=2}^9\left(4\binom{j}{2}+3\binom{j}{1}\right)\\
&=15+\left[4\binom{j+1}{3}+3\binom{j+1}{2}\right]_1^9\\[3pt]
&=15+4\binom{10}{3}+3\binom{10}{2}-4\binom{2}{3}-3\binom{2}{2}\\[12pt]
&=15+480+135-0-3\\[18pt]
&=627
\end{align}
$$
| {
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Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$
Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$.
From the equation, I know that $3x^2$ must be odd and therefore equal $2k + 1$ for some integer $k$.
But I am unsure what to do after that.
I have also worked out that $k = 2(y^2 + 3)$, but I don't know if that helps at all.
My instructor noted that I should look at whether $x^2$ is even or odd, but I am at a loss.
| $3x^2=\underbrace{13+4y^2}_\text{odd}$
We know that x is odd lets' pose $x=2k+1$
\begin{align}3(2k+1)^2=13+4y^2\\
3(4k^2+4k+1)=13+4y^2\\
3(4k^2+4k)=10+4y^2\\
12(k^2+k)=10+4y^2\\
\underbrace{6(k^2+k)}_\text{even}=\underbrace{5+2y^2}_\text{odd}
\end{align}
but $5+2y^2$ is odd not even which contradict the factor 6 on the left side of the equation
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the roots of the quadratic equation
Show that the roots of the quadratic equation $$(b-c)x^2+2(c-a)x+(a-b)=0$$
are always reals.
My Attempt:
$$(b-c)x^2+2(c-a)x+(a-b)=0$$
Comparing above equation with $Ax^2+Bx+C=0$
$$A=b-c$$
$$B=2(c-a)$$
$$C=(a-b)$$
Now,
$$B^2-4AC=[2(c-a)]^2-4(b-c)(a-b)$$
$$=4(c-a)^2 - 4(ab-b^2-ac+bc)$$
$$=4c^2-8ac+4a^2-4ab+4b^2+4ac-4bc$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
How do I proceed further?
| The first part of your solution is true.
We can end it by the following way.
$b-c\neq0$ by given and we need to prove that
$$(c-a)^2-(b-c)(a-b)\geq0$$ or
$$a^2+b^2+c^2-ab-ac-bc\geq0$$ or
$$\sum_{cyc}(a^2-ab)\geq0$$ or
$$\sum_{cyc}(2a^2-2ab)\geq0$$ or
$$\sum_{cyc}(a^2-2ab+b^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2\geq0,$$
which is obvious.
| {
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Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.
My Attempt:
Equation of the line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ is:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
| Prove that the perpendicular bisector from origin to the line joining the points $(c\cos\alpha,c\sin\alpha)$ and $(c\cos\beta,c\sin\beta)$ passes through the center as well.
| {
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"answer_id": 5
} |
Solve $z\cdot |z|-2z-i+1=0$ I have an equation with complex numbers that I can't solve
$$z\cdot |z|-2z-i+1=0$$
My problem is that $z$ appears not only as an unknown quantity but also its modulus.
| Let $z=a+bi$, where $a$ and $b$ are reals.
Thus, $$(a+bi)\sqrt{a^2+b^2}-2(a+bi)-i+1=0,$$
which gives
$$a\sqrt{a^2+b^2}=2a-1$$ and
$$b\sqrt{a^2+b^2}=2b+1$$ or
$$\sqrt{a^2+b^2}=\frac{2a-1}{a}$$ and
$$\sqrt{a^2+b^2}=\frac{2b+1}{b},$$
which gives
$$\frac{2a-1}{a}=\frac{2b+1}{b}$$ or
$$a=-b.$$
Thus, $\sqrt2a|a|=2a-1$ and the rest is smooth.
| {
"language": "en",
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"source": "stackexchange",
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Solving differential equation by separation of variables In the following question I am trying to solve the following ODE by the method of separation of variables. From there I want to use the initial condition given to estimate $y(8)$
The given equation is,
$$\frac{dy}{dx}=\frac{y(2-y)}{x+3}$$
Separating the variables gives:
$$\int \frac{1}{y(2-y)}dy=\int \frac{1}{x+3}dx$$
Using partial fractions for the left side gives,
$$\frac {1}{y(2-y)}=\frac{A}{y}+\frac{B}{2-y}$$
$$=-\frac{\frac{1}{2}}{y}+ \frac{\frac{1}{2}}{2-y}$$
$$y=-\frac{1}{2y}+ \frac{1}{2(y-2)}$$
So therefore,
$$\int \frac{1}{y(2-y)}dy=\int \frac{1}{x+3}dx=$$
$$-\frac{1}{2}lny-\frac{1}{2}ln(y-2)=ln(x+3)+c$$
$$ln(y^2-2y)=\frac{e^c}{(x+3)^2}$$
$$y^2-2y=\frac{e^c}{(x+3)^2}$$
$$y^2-2y-\frac{e^c}{(x+3)^2}=0$$
Applying the quadratic formula,
$$y=1+\sqrt{1+\frac{e^c}{(x+3)^2}}$$
$$y=1-\sqrt{1+\frac{e^c}{(x+3)^2}}$$
So my thinking is that $y$ has to be $>2$ in order for $ln(y-2)$ to be defined
Using the initial condition $y(0)=6$ gives
$$6=1+\sqrt{1+\frac{e^c}{(0+3)^2}}$$
$$c=ln216$$
and hench $y(8)=1+\sqrt {\frac{337}{121}}$
So I am just wondering if I did this correctly, and the quadratic formula gave two answers so I'm wondering if I chose the correct one.
| $$\frac {1}{y(2-y)}=\frac{A}{y}+\frac{B}{2-y}=\frac {2A+(B-A)y}{y(2-y)}$$
means $2A=1$ and $A=B$. Then
$$\dfrac12\left(\frac{1}{y}+\frac{1}{2-y}\right)dy=\dfrac{1}{x+3}dx$$
after integration
$$\dfrac12\ln\frac{y}{2-y}=\ln(x+3)+\ln C$$
or
$$\frac{y}{2-y}=C(x+3)^2$$
with $y(0)=6$ you find $\dfrac{y}{2-y}=-\dfrac16(x+3)^2$ and finally $x=8$ gives us $\color{blue}{y(8)=\dfrac{242}{115}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summation of Products $\sum\limits_{n=1}^\infty \frac 1{11}\cdot \frac {10}{20}\cdot \frac {19}{29}\cdots \frac{9n+1}{9n+11}$ This is an extension of a nice question posted recently on MSE.
One of the solutions posted was very interesting and this is an attempt to extend it to a general case.
Evaluate
$$\sum_{n=1}^\infty a_n,\qquad a_n=\frac 1{11}\cdot \frac {10}{20}\cdot \frac {19}{29}\cdot \frac{28}{38}\cdots \frac{9n+1}{9n+11}$$
$$a_n = \prod_{k=0}^n\frac{9k+1}{9k+11}$$
| In general, consider:
$$\sum_{n=1}^\infty \frac{\Gamma(n+a)}{\Gamma(n+a+k)} = \frac{1}{k-1} \sum_{n=1}^\infty [(n+a+k-1) - (n+a)] \cdot \frac{\Gamma(n+a)}{\Gamma(n+a+k)} = \\ \frac{1}{k-1} \sum_{n=1}^\infty \left( \frac{\Gamma(n+a)}{\Gamma(n+a+k-1)} - \frac{\Gamma(n+a+1)}{\Gamma(n+a+k)} \right).$$
If $k > 1$, then this is a telescoping series for which the term $\frac{\Gamma(n+a)}{\Gamma(n+a+k-1)} \to 0$ as $n \to \infty$. Therefore, the sum is equal to $\frac{1}{k-1} \cdot \frac{\Gamma(a+1)}{\Gamma(a+k)} = \frac{\Gamma(a+1)}{(k-1) \Gamma(a+k)}$.
Now, the originally desired sum is equal to
$$\sum_{n=1}^\infty \frac{\Gamma(11/9)}{\Gamma(1/9)} \cdot \frac{\Gamma(n+10/9)}{\Gamma(n+20/9)} = \frac{\Gamma(11/9)}{\Gamma(1/9)} \cdot \frac{\Gamma(19/9)}{(10/9-1) \Gamma(20/9)} = \frac{10}{11}.$$
Note: Expanding the argument a bit, this is equivalent to observing that
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty [(9n+11)a_n - (9n+10)a_n] = \sum_{n=1}^\infty [(9n+11)a_n - (9n+20)a_{n+1}]$$
which again telescopes, so the sum is equal to $20 a_1 = \frac{10}{11}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$ converges
Prove $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$ converges, where the sums are for all primes $p$.
I've found in this link that $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$
converges to $K$, where $K<1$.
I know that $\sum_p \frac{1}{p^2}\le \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6},$
because see Basel problem and specific values of Riemann zeta function.
Therefore also $$\sum_p \frac{1}{p^k}\le \sum_p \frac{1}{p^2}\le\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
for all $k\ge 3$, $k\in\mathbb Z.$
| Since $2$ is the smallest prime, we can see by comparing terms that
$$\frac{1}{2}\sum_p \frac{1}{p^2}>\sum_p \frac{1}{p^3}$$
$$\frac{1}{2}\sum_p \frac{1}{p^3}>\sum_p \frac{1}{p^4}$$
and so on.
So it follows that
$$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots < \frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{2}\sum_p \frac{1}{p^3}+\cdots <$$
$$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{4}\sum_p \frac{1}{p^2}+\cdots =\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\sum_p\frac{1}{p^2}=$$
$$\sum_p \frac{1}{p^2}<\sum_n \frac{1}{n^2} =\frac{\pi^2}{6}$$
Since the sum is increasing, the presence of an upper bound means it converges.
| {
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"url": "https://math.stackexchange.com/questions/2442234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to simplify this fraction using algebraic logic? Well, I need to simplify this fraction using algebraic logic
$$\frac{(2^4+2^2+1)(4^4+4^2+1)(6^4+6^2+1)(8^4+8^2+1)(10^4+10^2+1)}{(3^4+3^2+1)(5^4+5^2+1)(7^4+7^2+1)(9^4+9^2+1)(11^4+11^2+1)}$$
For resolve this, I think I will use this concept.
If you have:
$$\frac{(2^2)(4^2)(6^2)(8^2)(10^2)}{(3^2)(5^2)(7^2)(9^2)(11^2)}$$
Simplifying, you must get:
$$\frac{3840^2}{10395^2}$$
Because all are terms with same exponent, like:
$$(2^2)(2^2) = 4^2 $$
Aplying this, I get:
$$\frac{3840^4+3840^2+1}{10395^4+10395^2+1}$$
But, its not the same, I resolve the first fraction and I get: 0.0225539...
The second one result in: 0.0186220...
This is embarrassing, probably i´m complety lost, but I need help
| After using $$x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$$ and
$$x^2-x+1=(x-1)^2+(x-1)+1$$ we obtain
$$\frac{2^2-2+1}{11^2+11+1}=\frac{3}{133}.$$
| {
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"url": "https://math.stackexchange.com/questions/2442616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.
However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants.
In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.
Any help would be appreciated, thank you all!
|
I think, from the second expression you will get easy a result :
$$
\lim_{x\to0} { {\sin^2({x \over 2})} - {{x^2} \over 4} \over
{ e^{x^2}+ e^{-{x^2}}-2} } = \lim_{y\to0} { {\sin^2({\sqrt y \over 2})} - {y \over 4} \over
{ e^y+ e^{-y}-2} }=?
$$
I do $y=x^2$. Clearly numerator is 0 and denominator is $0$.
I will correct here. I expand function $\sin^2{y \over 2}={y \over 4} - {{y^2} \over 48}+ ...$. So, at numerator we get ${{-{y^2}}\over 48}+ ...$. At denominator a Taylor expansion get after resting 2 : $y^2 + ...$
We do a division :
$$
\lim_{y\to0} { {\sin^2({\sqrt y \over 2})} - {y \over 4} \over
{ e^y+ e^{-y}-2} }= \lim_{y\to0} {{{ -{y^2}\over 48}} \over {y^2}} = {-1 \over 48}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the values of $A$, $B$ and $C$. There is $3$ unknowns so we need three equations but I can only get two equations.
| Using the Generalized Euclidean Algorithm, we get
$$
(x+2)\left(x^2+x+1\right)-(x+1)(x+1)^2=1\tag1
$$
Therefore, we have
$$
\begin{align}
(x+2)\left(x^2+x+1\right)&\equiv1&&\pmod{(x+1)^2}\\
(x+2)\left(x^2+x+1\right)&\equiv0&&\pmod{x^2+x+1}
\end{align}\tag2
$$
and
$$
\begin{align}
-(x+1)(x+1)^2&\equiv0&&\pmod{(x+1)^2}\\
-(x+1)(x+1)^2&\equiv1&&\pmod{x^2+x+1}
\end{align}\tag3
$$
Thus, to get
$$
\begin{align}
f(x)&\equiv x-1&&\pmod{(x+1)^2}\\
f(x)&\equiv x+5&&\pmod{x^2+x+1}
\end{align}\tag4
$$
we can use $x-1$ times $(2)$ plus $x+5$ times $(3)$ to get $\bmod{(x+1)^2\left(x^2+2x+1\right)}$
$$
\begin{align}
f(x)
&\equiv(x-1)(x+2)\left(x^2+x+1\right)-(x+5)(x+1)(x+1)^2\\
&=-6x^3-18x^2-17x-7\\
&\equiv-6x^2-5x-1\pmod{(x+1)\left(x^2+2x+1\right)}\tag5
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\lim_{(x, y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$ D.N.E Show that $$\lim_{(x,y)\to(0,0)}\frac{x^4y^4}{(x^2+y^4)^3}$$ D.N.E
In order to do this I should consider two paths and show that they do not reach the same limit.
Consider when $x=0$.
Then we have $\lim_{y\to 0}\dfrac{y^4}{y^{12}}=\dfrac{1}{y^8}=\infty$
Similarly if we choose $y=0$, then we have
Then we have $\lim_{x\to 0}\dfrac{x^4}{x^{6}}=\dfrac{1}{x^2}=\infty$
Since both of these limits approach an asympote, can I even use them to show that the limit does not exist? Or must they approach a finite number?
| Write the function as $$\frac{x^4 y^4}{(x^2 + y^4)^3} = \frac{x^4 y^4 / x^6}{(1 + (y^4/x^2))^3} = \frac{y^4/x^2}{(1+(y^4/x^2))^3}.$$ Now we can see that if $(x,y) \to (0,0)$ in such a way that the ratio $y^4/x^2$ does not tend to $0$, the resulting limit will not be zero. This furnishes the motivation behind the choice of path $x = y^2$. In fact, we can see that for any real constant $c$, if we choose the parametrization $(x,y) = (t^2, ct)$, then $y^4/x^2 = c^4$ is independent of the parameter $t$, and $$\lim_{(x,y) \to (0,0)} \frac{x^4 y^4}{(x^2 + y^4)^3} = \lim_{t \to 0} \frac{c^4}{(1+c^4)^3} = \frac{c^4}{(1+c^4)^3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Linear homogenous recurrence relations. I'm having trouble solving linear homogenous recurrence relations. I've searched for guides and seen a few video's on how to solve them. I'm confused to how it's done, some suggest matrices, solving them like two linear equations etc...
*
*Given the difference equation $x_{n+2}-2x_{n+1}-2x_n=0$ with $x_0 = 1$ and $x_1=2$
2.Show that the general solution is $x_n=C(1-\sqrt3)^n+D(1+\sqrt3)^n $ And that the initial values $x_0=1$ and $x_1=1-\sqrt3$ decides that the final solution is $x_n=(1-\sqrt3)^n$
I know that we can write $x_{n+2} -2x_{n+1}-2x_n =0 $ can be written as $r^2-2r-2=0$ and with completing the square we get $r_1r_2=1\pm\sqrt3$ thus, $x^h_n=C(1-\sqrt3)^n+D(1+\sqrt3)^n$
But how do I show that the initial values decides that the final solution is $(1-\sqrt3)^n$
ps:
*
*$x_0=C+D=1$
*$x_1=C-C\sqrt3+D+D\sqrt3$ =$1-\sqrt3$
| If $x_0=1,\;x_1=2$ then from the general solution $x_n=C(1-\sqrt3)^n+D(1+\sqrt3)^n$
put $n=0$ and get $C+D=1\to D=1-C$
put $n=1$ and get $C(1-\sqrt3)+D(1+\sqrt3)=2$
substitute and get $C(1-\sqrt3)+(1-C)(1+\sqrt3)=2$
$C-C\sqrt{3}+1+\sqrt{3}-C-C\sqrt{3}=2$
$-2C\sqrt{3}=-\sqrt{3}+1$
$C=\dfrac{\sqrt{3}-1}{2\sqrt 3}=\dfrac{3-\sqrt{3}}{6}$
$D=1-C=1-\dfrac{3-\sqrt{3}}{6}=\dfrac{3+\sqrt 3}{6}$
The solution is then
$x_n=\dfrac{3-\sqrt{3}}{6}(1-\sqrt3)^n+\dfrac{3+\sqrt 3}{6}(1+\sqrt3)^n$
$x_n=\dfrac{1}{6} \left[\left(\sqrt{3}+3\right) \left(1+\sqrt{3}\right)^n-\left(\sqrt{3}-3\right) \left(1-\sqrt{3}\right)^n\right]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Did I do something wrong when rationalizing this denominator by conjugation? Now, I know the basic steps of conjugating a number; it's basically a difference of squares.
So to solve both #'s 1 & 2, I did the following:
1.) $\frac {7{\sqrt 2}}{\sqrt{6} + 8 }\cdot\frac {\sqrt {6} - 8}{\sqrt {6} - 8}$ = $\frac {- 56{\sqrt 12}}{6 - 64 }$ = $\frac {- 56{\sqrt3\cdot4}}{- 58 }$ = $\frac {- 112{\sqrt 3}}{- 58 }$
2.) $\frac {{\sqrt 3} + {\sqrt 13}}{\sqrt{3} - \sqrt {13} }\cdot\frac {{\sqrt 3} + {\sqrt 13}}{\sqrt{3} + \sqrt {13} }$ = $\frac {3 + 13}{3 - 13 }$ = $-\frac {16}{10}$
and got $\frac {- 112{\sqrt 3}}{- 58 }$ for #1
and $-\frac {16}{10}$ for #2
The answers in the textbook are:
1.) $\frac {- 7{\sqrt 3} + 28\sqrt {2}}{29 }$
2.) $\frac {- 8 -\sqrt {39}}{5 }$
Possibly the product from the first step is wrong, but I don't know what mistake I made.
| You failed to write down all the terms while expanding the numerator after conjugation. For (1):
$$\frac{7\sqrt2}{\sqrt6+8}\cdot\frac{\sqrt6-8}{\sqrt6-8}=\frac{7\sqrt{12}-56\sqrt2}{6-64}=\frac{14\sqrt3-56\sqrt2}{-58}=\frac{28\sqrt2-7\sqrt3}{29}$$
For (2):
$$\frac{\sqrt3+\sqrt{13}}{\sqrt3-\sqrt{13}}\cdot\frac{\sqrt3+\sqrt{13}}{\sqrt3+\sqrt{13}}=\frac{3+2\sqrt{39}+13}{3-13}=\frac{16+2\sqrt{39}}{-10}=\frac{-8-\sqrt{39}}5$$
| {
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Proof verification: $\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$ Is either of the following methods correct?
Prove
$$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
First Method
Preliminary Analysis:
We know that $a = 2$, $L= \frac{1}{2}$, and $f(x)= \frac{1}{x}$. By the precise definition of limit we have the following:
$$
0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\\
0<|x-2|<\delta \implies \left|\frac{1}{x}-\frac{1}{2}\right|<\varepsilon \implies \left|\frac{2}{2x}-\frac{x}{2x}\right| <\varepsilon \implies \left|\frac{2-x}{2x}\right| <\varepsilon \\
\implies \left|\frac{-(-2+x)}{2x}\right| <\varepsilon \implies \left|\frac{x-2}{2x}\right| <\varepsilon \implies \frac{\left|x-2\right|}{\left|2x\right|} <\varepsilon \implies \left|x-2\right|<\varepsilon \left|2x\right|
$$
let $\delta = \varepsilon\left|2x\right|$ but we need to simplify it further because delta should be in terms of $\varepsilon$ only.
Assume $|x-a| < 1$
$$ |x-2| < 1 \implies -1 <x -2<1 \implies -1+2<x<1+2 \implies 1<x<3$$
Then we have to simplify $|2x|$ as well, which ends up being
$$ 2<2x<6 \implies -6<2x<6 \implies |2x| <6$$
Now consider the inequality we discovered, specifically, $\left|x-2\right|<\varepsilon \left|2x\right|$ this inequality is also valid for $\left|x-2\right|<\varepsilon\cdot6$.
Let $\delta = \min{\{1, \varepsilon\cdot6}\}$
Proof:
Given $\varepsilon > 0$ let $\delta = \min{\{1, \varepsilon\cdot6}\}$ if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$
We also have $|x - 2| < \varepsilon \cdot6$.
$$
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$
By the precise definition of limit $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Second Method
Proof by Definition/Property:
By the direct substitution property if $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then
$$\lim_{x \to a} f(x) = f(a)$$
Then by the direct substituon property of limit: $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Are either of the above methods correct? (I am putting the question here as well in case someone misses it)
| The first is incorrect.
$$|x-2|<1\implies 1 <x <3\implies $$
$$\frac {1}{6}<\frac {1}{2x}<\frac {1}{2} $$
thus
$$|\frac {1}{x}-\frac {1}{2}|<\frac {|x-2|}{2} $$
if $|x-2|<1$ and $|x-2|<2\epsilon $ then $$|\frac {1}{x}-\frac {1}{2}|<\epsilon $$
we take $$\delta=\min (1,2\epsilon) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Understanding the proof for the divergence of the harmonic series I'm trying to understand the following proof for the divergence of the harmonic series:
The parts which I can't seem to understand are:
$$1. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{r} > \sum_{i=0}^{n-1}
\sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}}$$
$$2. \sum_{i=0}^{n-1} \sum_{r=2^i+1}^{2^{i+1}} \frac{1}{2^{i+1}} = \sum_{i=0}^{n-1} (2^{i+1} - 2^i)\frac{1}{2^{i+1}}$$
I think understanding 2. follows from understanding 1. but I'm not entirely sure. Anyway for 1. I couldn't understand if the series on the RHS of the equality was $(\frac{1}{2^3})+(\frac{1}{2^3} + \frac{1}{2^4})+(\frac{1}{2^5} + \frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8})+...$ or $(\frac{1}{2^1})+(\frac{1}{2^2} + \frac{1}{2^3})+(\frac{1}{2^4} + \frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7})+...$ . And for part I simply could not understand 2. (probably because I didn't understand 1.). So I'm wondering what the series on the RHS of 1. actually is and how it equals $\sum_{i=0}^{n-1} (2^{i+1} - 2^i)\frac{1}{2^{i+1}}$ .
| Look at how the terms are grouped early on:
$$ \sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \cdots $$
Note that
$$ 1 = \sum_{r=0}^{0} 2^{0}
\qquad\text{and}\qquad
\frac{1}{2} = \sum_{r=1}^{1} 2^{-1} = \sum_{r=2^0}^{2^1-1} 2^{-1}
$$
In the next group, note that $3 < 4 = 2^2$, from which it follows that $\frac{1}{3} > 2^{-2}$. The second term is $\frac{1}{4}$, which is exactly $2^{-2}$, and so we have
$$ \frac{1}{3} + \frac{1}{4} > 2^{-2} + 2^{-2} = \sum_{r=2}^{3} 2^{-2} = \sum_{r=2^1}^{2^2-1} 2^{-2}. $$
In the next group, note that
$$ \frac{1}{5} > \frac{1}{6} > \frac{1}{7} > \frac{1}{8} = 2^{-3}. $$
That is, every element of this group is larger than $2^{-3}$. From this, it follows that
$$ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > 2^{-3} + 2^{-3} + 2^{-3} + 2^{-3} = \sum_{r=4}^{7} 2^{-3} = \sum_{r=2^2}^{2^3-1} 2^{-3}. $$
In other words, we have
$$ \sum_{k=1}^{\infty} \frac{1}{k}
\ge \sum_{j=1}^{\infty} \left( \sum_{r=2^{j-1}}^{2^{j}-1} 2^{-j} \right), \tag{1}$$
Where the inner sum is the generalized version of the sums above which represent the groups of terms. But the general term of the inner sums are constant with respect to the index of summation (i.e. the summands don't depend on $r$), and so we just have to count the number of terms. Therefore
$$ \sum_{r=2^{j-1}}^{2^{j}-1} 2^{-j}
= \big((2^{j}-1) - (2^{j-1})\big) 2^{-j}
= 2^{j-j} - 2^{-j} - 2^{j-1-j}
= 1 - 2^{-j} - 2^{-1}
= \frac{1}{2} - 2^{-j}
\ge \frac{1}{2}. $$
Plugging this back in at (1), we get
$$ \sum_{k=1}^{\infty} \frac{1}{k}
\ge \sum_{j=1}^{\infty} \left( \sum_{r=2^{j-1}}^{2^{j}-1} 2^{-j} \right)
\ge \sum_{j=1}^{\infty} \frac{1}{2}
= \infty.$$
Of course, I really should write all of that out in terms of partial sums, but I think that actually hides some of the intuition. The point is that for each $j$, we associate terms into groups that contain $2^{j}$ terms, where each term in the group is larger than $2^{-j-1}$. Thus each group sums to more than $\frac{1}{2}$. There are infinitely many such terms, so the series diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the last $3$ digits of $3^{352}$? Find the last $3$ digits of $3^{352}$ ?
Apart from Carmichael Function, any other way of solving it ?
| Note that $3^4=81$, so $$\begin{aligned}3^{352} &\equiv (80+1)^{88} \pmod{1000} \\ &\equiv 1+\binom{88}{1} 80 + \binom{88}{2} 6400\pmod{1000}
\\&\equiv 1+040+200 \pmod{1000}
\end{aligned}
$$
Because last two digits of $88\times 8$ is $04$, while last digit of $44\times 87\times 64$ is $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $(\tan(x) - 2 y + 5) dx + (\sin(2 x) + (4 - y) \cos^2(x)) dy = 0$ on $(-\pi/2, \pi/2)$ Solve $(\tan(x) - 2 y + 5) dx + (\sin(2 x) + (4 - y) \cos^2(x)) dy = 0$ on $(-\pi/2, \pi/2)$
I got stuck here
I'm not looking for a full solution, just a way of solving.
| $(\tan x-2y+5)~dx+(\sin2x+(4-y)\cos^2x)~dy=0$
$(\tan x+5-2y)~dx=((y-4)\cos^2x-2\sin x\cos x)~dy$
$(y-4-2\tan x)\dfrac{dy}{dx}=\sec^2x\tan x+5\sec^2x-2y\sec^2x$
This belongs to an Abel equation of the second kind.
Let $u=y-4-2\tan x$ ,
Then $y=u+2\tan x+4$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+2\sec^2x$
$\therefore u\left(\dfrac{du}{dx}+2\sec^2x\right)=\sec^2x\tan x+5\sec^2x-2(u+2\tan x+4)\sec^2x$
$u\dfrac{du}{dx}+2u\sec^2x=-2u\sec^2x-3(\tan x+1)\sec^2x$
$u\dfrac{du}{dx}=-4u\sec^2x-3(\tan x+1)\sec^2x$
Let $t=\tan x$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dt}\dfrac{dt}{dx}=\sec^2x\dfrac{du}{dt}$
$\therefore u\sec^2x\dfrac{du}{dt}=-4u\sec^2x-3(\tan x+1)\sec^2x$
$u\dfrac{du}{dt}=-4u-3(\tan x+1)$
$u\dfrac{du}{dt}=-4u-3(t+1)$
$\dfrac{du}{dt}=-4-\dfrac{3(t+1)}{u}$
Luckily this becomes a first-order homogeneous ODE.
Let $v=\dfrac{u}{t+1}$ ,
Then $u=(t+1)v$
$\dfrac{du}{dt}=(t+1)\dfrac{dv}{dt}+v$
$\therefore(t+1)\dfrac{dv}{dt}+v=-4-\dfrac{3}{v}$
$(t+1)\dfrac{dv}{dt}=-v-4-\dfrac{3}{v}$
$(t+1)\dfrac{dv}{dt}=-\dfrac{v^2+4v+3}{v}$
$\dfrac{v}{(v+3)(v+1)}~dv=-\dfrac{dt}{t+1}$
$\int\left(\dfrac{3}{2(v+3)}-\dfrac{1}{2(v+1)}\right)~dv=-\int\dfrac{dt}{t+1}$
$\dfrac{3\ln(v+3)-\ln(v+1)}{2}=-\ln(t+1)+c$
$(v+3)^3(t+1)^2=C(v+1)$
$\left(\dfrac{u}{t+1}+3\right)^3(t+1)^2=C\left(\dfrac{u}{t+1}+1\right)$
$(u+3t+3)^3(t+1)^2=C(u+t+1)$
$(y+\tan x-1)^3(\tan x+1)^2=C(y-\tan x-3)$
| {
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Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$
My steps
A= $\frac{a}{b+c}$,B= $\frac{b}{c+a}$ &C=$\frac{c}{a+b}$
A.M$\ge$H.M
$\frac{A+B+C}{3} \ge \frac{3ABC}{AB+BC+AC}$
I am struck after this step
${A+B+C} \ge \frac{9}{\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}}$
| Hint:
\begin{eqnarray*}
\sum_{cyc} (a+b)(a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
2(a(a+b)(a+c)+b(b+c)(b+a)+c(c+a)(c+b))-3(a+b)(a+c)(b+c) \geq 0
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $a,b$ such that each of the equations $x^2-ax+b=0$ and $x^2-bx+a=0$ has distinct positive integral roots.
Find all positive integers $a,b$ such that each of the equations $$x^2-ax+b=0$$ and $$x^2-bx+a=0$$ has distinct positive integral roots.
Suppose $\alpha,\beta$ are roots of the first equation and $\gamma,\delta$ are of the second, then we have these two sets of equations
$$a=\alpha+\beta$$ $$b=\alpha\beta$$
and
$$b=\gamma+\delta$$ $$a=\gamma\delta$$
How to proceed?
| A long answer, I'm afraid, but I think it's correct. Please let me know if anything is unclear.
Observe first that $a/b = 1/\beta + 1/\alpha < 2$ and similarly $b/a < 2$. That is, $a < 2b$ and $b < 2a$. But now, assuming wlog $\alpha > \beta$, the previous inequality says:
$$\alpha \beta < 2\alpha + 2 \beta < 4 \alpha,$$
meaning $\beta < 4$. Of course, there's a similar statement for $(\gamma, \delta)$, but let's work with $(\alpha,\beta)$, remembering that whatever we find for $(a,b)$ can be swapped to yield another solution $(b,a)$. (Which we would have found if we had used ($\gamma, \delta$).)
If $\beta = 3$, then $b = 3 \alpha < 2a = 2 \alpha + 2 \beta = 2 \alpha + 6$, i.e. $\alpha < 6$, forcing $\alpha = 4$ or $5$. Neither of these choices work, by a routine check.
If $\beta = 2$, then $4 - 2a + b = 0$ by the first equation, i.e. $b = 2a - 4$. But then
$$b^2 - 4a = 4a^2 - 20a + 16 = 4(a^2 - 5a + 4)$$
needs to be a perfect square, meaning $a^2 - 5a + 4 = c^2$ for some $c$, so $a = (10 \pm \sqrt{9+(2c)^2})/2$ by the quadratic formula. But the expression under the square root here has to be odd, so $a$ cannot be an integer – contradiction.
Thus $\beta = 1$, and we arrive at $b = a - 1$. Now, the roots of $x^2 - (a-1)x + a = 0$ are
$$x = \frac{a-1 \pm \sqrt{(a-1)^2 - 4a}}{2};$$
thus we require the discriminant $(a-1)^2 - 4a = a^2 - 6a + 1$ to be a perfect square, say $a^2 - 6a + 1 = c^2$ for some $c$, so $a = (6 \pm \sqrt{32 + (2c)^2})/2$ by the quadratic formula again.
It will have to be plus, to ensure $a > 0$, and $32 + (2c)^2$ now has to be a square, say $32 + (2c)^2 = d^2$. But then $32 = (d-2c)(d+2c)$, forcing $d - 2c = 1, 2,$ or $4$, since it must be less than $\sqrt{32} = 4 \sqrt2$. But $d$ is even, so actually $d - 2c = 2$ or $4$. Suppose $d = 2c + 2$, so
$$32+ (2c)^2 = (2c+2)^2.$$
Then $32 = 8c + 4$, which is impossible. Hence $d = 2c + 4$ and we get $32 + (2c)^2 = (2c + 4)^2$ or $32 = 16c + 16$, i.e. $c = 1$, $a = 6, b = 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine sinus and cosinus : $2\sin x + 3\cos x = 3$ $$2\sin x + 3\cos x = 3$$
Seemed easy at first but I have no idea how to determine them. I tried replacing the 3 with $3(\sin^2 x + \cos x^2x)$ But it doesnt work. I also tried switching sides from almost everything and no luck. Please help!
| Hint:
You can rewrite the equation as
\begin{align}
2\sin x +3\cos x=3&\iff 2\sin x=3(1-\cos x)\iff 4\sin\dfrac x2\cos\dfrac x2=6\sin^2\dfrac x2\\[1ex]
&\iff\begin{cases}
\sin \dfrac x2=0\\\;\text{ or}\\[-2ex]2\cos\dfrac x2=3\sin \dfrac x2
\end{cases}\iff\begin{cases}
\sin \dfrac x2=0\\\;\text{ or}\\[-2ex]\tan\dfrac x2=\dfrac23
\end{cases}\\&\iff\begin{cases}
\dfrac x2\equiv 0 \pmod\pi\\\;\text{ or}\\[-2ex]\dfrac x2\equiv\arctan\dfrac23\pmod{\pi}
\end{cases}\iff\begin{cases}x\equiv 0 \pmod{2\pi}\\\;\text{ or}\\x\equiv 2\arctan\dfrac23\pmod{2\pi}\end{cases}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How would I solve the following problem involving elementary matrices?
Let $A=\begin{pmatrix}0&5\\ 7&4\end{pmatrix}$.
1) Write $A$ as a product of $4$ elementary matrices.
2) Write $A^{-1}$ as a product of $4$ elementary matrices.
My work. I have managed to find $A^{-1}$, which came out to be this:
\begin{pmatrix}-\frac{4}{35}&\frac{1}{7}\\ \frac{1}{5}&0\end{pmatrix}
However, I am struggling to figure out how I would split each of these matrices into $4$ elementary matrices. Any help?
| For $a\not=0$ we have that
$$\begin{pmatrix} a & b \\ c & 0 \end{pmatrix}=
\begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & -bc/a \end{pmatrix}
\begin{pmatrix} 1 & b/a \\ 0 & 1 \end{pmatrix}$$
then $A^{−1}$ is a product of 4 elementary matrices with $a=-4/35$, $b=1/7$, and $c=1/5$.
Note that elementary matrices generate a group of invertible matrices. Hence if $A^{-1}=E_1E_2E_3E_4$ where $E_i$ for $i=1,2,3,4$ are some elementary matrices then $A=E_4^{-1}E_3^{-1}E_2^{-1}E_1^{-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction $\left(1 - \frac{1}{n^{2}}\right)^n \left(1 + \frac{1}{n}\right) < 1$ I'm trying to prove that for every integer $n > 0\quad (P)$ :
$$\left(1 - \frac{1}{n^{2}}\right)^n \left(1 + \frac{1}{n}\right) < 1$$
This is what I did so far, using induction :
$$\text{for }(n = 1) : (1 - 1/1^2)^1 (1 + 1/1) = 0 < 1$$
Let's suppose that $(P)$ is true for all integers from $1$ to a certain $n$
and let's prove that $P$ is true for $n+1$ also :
I simplified the formula and now it looks like this :
$$(P) : \left(\frac{n-1}{n}\right)^n < \left(\frac{n}{n+1}\right)^{(n+1)}$$
So if we define $Q(n) = (\frac{n-1}{n})^n$
We have $Q(n) < Q(n+1)$
and we need to prove that $Q(n+1) < Q(n+2)$
That's where my head stopped. Can anyone help me please ?
| Consider the function
\begin{eqnarray*}
y= \frac{x^{x+1}}{(1+x)^{x+1}} \\
\frac{1}{y} \frac{dy}{dx} =\frac{1}{x} +\ln x - \ln (1+x) \color{yellow}{>0}
\end{eqnarray*}
Now for $x>0$ we have $e^{\frac{1}{x}} > 1+ \frac{1}{x}$ & so
\begin{eqnarray*}
x+1 < x e^{\frac{1}{x}} \\
\ln (x+1) < \frac{1}{x}+ \ln(x).
\end{eqnarray*}
In other words, our function $y$ is monotonically increasing & so
\begin{eqnarray*}
\frac{(n-1)^n}{n^n} < \frac{n^{n+1}}{(1+n)^{1+n}}
\end{eqnarray*}
This can be rearranged to give your result.
| {
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Simplest way to solve radical equation $\sqrt{3x+1}-\sqrt{x+4}=1$ I have the following equation:
$$\sqrt{3x+1}-\sqrt{x+4}=1$$
I can get the answer $x=5$ through tedious and long algebraic manipulation with quite a few extraneous solutions. It's not elegant. Is there a simple, straightforward way to solve this equation?
| If we introduce the variables
\begin{align}
y^2&=&3x+1\tag1\\
z^2&=&x+4\tag2\\
y-z&=&1\tag3\end{align}
we get an extraneous solution, but it is easy to dismiss and this makes the calculations a bit neater. Subtract three times $(2)$ from $(1)$
\begin{align}
y^2-3z^2=-11\\\
\end{align}
using $(3)$
\begin{align}
(z+1)^2-3z^2+11=-2z^2+2z+12=0\\
\end{align}
$$\Leftrightarrow$$
\begin{align}
z^2-z-6=0\\
\end{align}
There are two solutions to this $z = -2$ and $z = 3$. We dismiss the first since $z$ represents a square root and we don't want this to be negative. The second leads to $x=5$, $y=4$ and $z=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution of the differential equation $2x^3dy + (1 - y^2)(x^2y^2 + y^2 - 1)dx = 0$.
Find the solution of the differential equation
$$2x^3dy + (1 - y^2)(x^2y^2 + y^2 - 1)dx = 0$$
My attempt: After arranging the above equation as $$\frac{dy}{dx}=\frac{(1-y^2)(x^2y^2+y^2-1)}{-2x^3}$$ I am not getting any standard method to get to the next step. Any suggestion will be helpful.
| Hint:
$2x^3~dy+(1-y^2)(x^2y^2+y^2-1)dx=0$
$2x^3~dy=(y^2-1)(x^2y^2+y^2-1)dx$
$(x^2y^2+y^2-1)\dfrac{dx}{dy}=\dfrac{2x^3}{y^2-1}$
Let $u=\dfrac{1}{x^2}$ ,
Then $\dfrac{du}{dy}=-\dfrac{2}{x^3}\dfrac{dx}{dy}$
$\therefore-\dfrac{(x^2y^2+y^2-1)x^3}{2}\dfrac{du}{dy}=\dfrac{2x^3}{y^2-1}$
$\left(\dfrac{1}{x^2}+\dfrac{y^2}{y^2-1}\right)\dfrac{du}{dy}=-\dfrac{4}{(y^2-1)^2x^2}$
$\left(u+\dfrac{y^2}{y^2-1}\right)\dfrac{du}{dy}=-\dfrac{4u}{(y^2-1)^2}$
This belongs to an Abel equation of the second kind.
Let $v=u+\dfrac{y^2}{y^2-1}$ ,
Then $u=v-\dfrac{y^2}{y^2-1}$
$\dfrac{du}{dy}=\dfrac{dv}{dy}+\dfrac{2y}{(y^2-1)^2}$
$\therefore v\left(\dfrac{dv}{dy}+\dfrac{2y}{(y^2-1)^2}\right)=-\dfrac{4}{(y^2-1)^2}\left(v-\dfrac{y^2}{y^2-1}\right)$
$v\dfrac{dv}{dy}+\dfrac{2yv}{(y^2-1)^2}=-\dfrac{4v}{(y^2-1)^2}+\dfrac{4y^2}{(y^2-1)^3}$
$v\dfrac{dv}{dy}=-\dfrac{2(y+2)v}{(y^2-1)^2}+\dfrac{4y^2}{(y^2-1)^3}$
| {
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Prove $\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$ I need some help showing that these are equivalent. I made a couple attempts to get this right but so far the following work is as far as I've gotten.
Here is the question in its entirety:
Let n be a natural number. Give a combinatorial proof of the following:
$\binom{2n+2}{n+1} = \binom{2n}{n+1} + 2\binom{2n}{n} + \binom{2n}{n-1}$
My first impression was that I could use the $\binom{n}{k} = \binom{n}{n-k}$ identity to make the term "$\binom{2n}{n-1}$" equal "$\binom{2n}{n+1}$" so I could simplify the equation a bit. I now have:
$\binom{2n+2}{n+1} = 2\binom{2n}{n+1} + 2\binom{2n}{n} = 2[\binom{2n}{n+1} + \binom{2n}{n}]$
Afterwards i figured I could use pascals identity $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ to get rid of yet more terms resulting in:
$\binom{2n +2}{n+1} = 2\binom{2n+1}{n+1}$
This is where I realize I made a pretty big mistake and got stuck, I'm fairly sure I made a maths error somewhere but I am unsure where after my third attempt at this.
| \begin{eqnarray*}
\binom{2n+2}{n+1}= \color{red}{\binom{2n+1}{n+1}}+\color{blue}{\binom{2n+1}{n}}=\color{red}{\binom{2n}{n+1}+\binom{2n}{n}}+\color{blue}{\binom{2n}{n}+\underbrace{\binom{2n}{n-1}}_{\binom{2n}{n-1}=\binom{2n}{n+1}}}=2 \left(\binom{2n}{n+1}+\binom{2n}{n}\right)
\end{eqnarray*}
| {
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Trying to solve $2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$ without seeing the obvious solution
Determine all real values of $x$ such that $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$
Taken from Waterloo 2012: Link
What I tried:
$$2\log_{5x+9}(x+3)+\log_{x+3}\left((x+3)(5x+9)\right)=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)=3$$
Then I was stuck and I graphed it:
It was clearly $0$:
$$2\log_{9}(3)+\log_{3}(9)=3$$
But how could I do it without seeing the "zero"? Is there a way?
| Hint: $$\log_{5x+9}(x^2+6x+9)+\log_{x+3}(5x^2+24x+27)=4$$
$$\log_{5x+9}(x+3)^2+\log_{x+3}\left[(5x+9)(x+3)\right]=4$$
$$2\log_{5x+9}(x+3)+\log_{x+3}(5x+9)+\log_{x+3}(x+3)=4$$
$$2\frac{1}{\log_{x+3}(5x+9)}+\log_{x+3}(5x+9)+1=4$$
Now, substitute $u=\log_{x+3}(5x+9)$ and solve the quadratic equation for $u$.
Note that we need to assume $x+3>0$ and $5x+9>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$ Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$
Are Vieta's formulas appropriate?
| Rearrange the equation to $z(2z^2-1)=z^4+z^2-7$ square this & substitute $y=z^2+1$
\begin{eqnarray*}
(y-1)(2y-3)^2=((y-1)^2+(y-1)-7)^2 \\
y^4+\cdots +58 =0.
\end{eqnarray*}
Note that the roots of this polynomial will be $\alpha^2+1,\beta^2+1,\gamma^2+1,\delta^2+1$ and we have not calculated all the terms explicitly (only the ones required to obtain the product of these roots.)
So $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\color{blue}{58}$.
| {
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Find the degree of $\mathbb{Q}(i,\sqrt[4]{3}, \sqrt[6]{3})$ over $\mathbb{Q}$
Find the degree of $\mathbb{Q}(i,\sqrt[4]{3}, \sqrt[6]{3})$ over
$\mathbb{Q}$
Since $\mathbb{Q}(i)$ is contained in either $\mathbb{Q}$ of the roots of $x^4-3$ or $x^6-3$, we can just take the degree of $\mathbb{Q}(\sqrt[4]{3}, \sqrt[6]{3})$ over $\mathbb{Q}$, right? Then, by the multiplicative formula of the degrees, it's gonna be $4\cdot 6 = 24$. Am I right?
| After using
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc);$$
$$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ and
$$(a-bi)(a+bi)=a^2+b^2$$ we'll get a polynomial from $\mathbb Q[x]$ with degree $3\cdot4\cdot2=24$.
| {
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Find the volume of $\frac{x^2}{36}+\frac{y^2}{16}+\frac{z^2}{25}=1$ using linear transformation Is there anyone here that could help me calculating the volume of the ellipsoid $$\frac{x^2}{36}+\frac{y^2}{16}+\frac{z^2}{25}=1$$ by turning it into an circunference of coordinates $uvw$ using a linear transformation?
I have calculated this volume by integration and got $V=160\pi$.
| Consider the transformation $(u,v,w) = f(x,y,z) = (\frac{x}{6}, \frac{y}{4}, \frac{z}{5})$
The given ellipsoid is transformed into $u^2+v^2+w^2=1$ which has volume $\frac{4\pi}{3}$. Now, the transformation also changes the volume from $V$ to $\frac{V}{6.4.5} = \frac{V}{120}$
So $\frac{V}{120} = \frac{4\pi}{3}$ so $V = 160\pi$
| {
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Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation
| You can also use rearrangement inequality, which is true for all real numbers. The inequality is symmetric in terms of $x,y,z$. Let WLOG $x\ge y\ge z$. Notice that it's only given that $xyz>0$, $x,y,z\in\mathbb R$, not necessarily $x,y,z>0$. There are two cases:
$1)$ If $z>0$, then $y>0$, $x>0$.
$2)$ If $z<0$, then $y<0$, $x>0$.
In both cases $(x,y,z)$ and $\left(\frac{1}{yz},\frac{1}{xz},\frac{1}{xy}\right)$ are sorted in the same order.
$(x,y,z)$ and $\left(\frac{1}{xy},\frac{1}{yz},\frac{1}{zx}\right)$ is another way to sort them, but we don't care if they're sorted in the same or opposite or another order.
$$x\cdot\frac{1}{yz}+y\cdot\frac{1}{xz}+z\cdot \frac{1}{xy}\ge $$
$$\ge x\cdot\frac{1}{xy}+y\cdot\frac{1}{yz}+z\cdot \frac{1}{zx}=$$
$$=\frac{1}{y}+\frac{1}{z}+\frac{1}{x}$$
Edit: as other answers have noticed, multiplying both sides by $xyz>0$ gives $x^2+y^2+z^2\ge xy+yz+zx$ and there has been a proof shown. Another proof is using rearrangement inequality. $(x,y,z)$ and $(x,y,z)$ are sorted in the same order. $(x,y,z)$ and $(y,z,x)$ is another way to sort them.
| {
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Basic Algebra Help \begin{align}
x^2-2x+1 &= x^2-2x+1\\
(1-x)^2 &= (x-1)^2\\
\sqrt{(1-x)^2} &= \sqrt{(x-1)^2}\\
1-x &= x-1\\
1-x &= -1\cdot(1-x)\\
1 &= -1
\end{align}
I definitely did something wrong.
| $a^2 = b^2$ does not mean $a=b$. For example $1^2 = (-1)^2$ but that doesn't mean $1 = -1$. Instead, all you can conclude is that $|a| = |b|$, which means that $a=b$ or $a=-b$.
| {
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Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| We know that for all n = odd positive integers,
a+b|a^n+b^n
Therefore,
3 i.e. 2+1|2^n+1^n for n is odd positive integer
| {
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Sum of sixth power of roots of $x^3-x-1=0$ Question:
Find the sum of sixth power of roots of the equation $x^3-x-1=0$
My First approach:
Let $S_i$ denote the sum of $i^{th}$ power of roots of the given equation.
Now, multiplying given equation by $x^3$ , putting each of the three roots and adding the three formed equations
we get, $S_6=S_4+S_3$
repeating same above procedure to obtain,
$S_4=S_2+S_1$ , $S_3=S_1+1$
hence, $S_6=S_2+2S_1+1=2+0+1=3$
My Second Approach:
Let $a,b,c$ be the roots of $f(x)=x^3-x-1$
Clearly, $f^/(x)/f(x)=1/(x-a)+1/(x-b)
+1/(x-c)$
=$\sum(1/x+a/x^2+a^2/x^3+a^3/x^4+\cdots)=3/x+S_1/x^2+S_2/x^3+\cdots$
hence we get, $S_6=5$
$\rule{200px}{0.5px}$
I am getting wrong answer through First Approach, Please point out my mistake or post a better solution.
Thank You
| Let $a$, $b$ and $c$ be roots of the equation, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Id est, $u=0$, $v^2=-\frac{1}{3}$ and $w^3=1$.
Thus, $$a^6+b^6+c^6=(a^3+b^3+c^3)^2-2(a^3b^3+a^3c^3+b^3c^3)=$$
$$=(27u^3-27uv^2+3w^3)^2-2(27v^6-27uv^2w^3+3w^6)=3^2-2((-1)^3+3)=5.$$
Done!
| {
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How to solve the trigonometric equation $\sin x + \cos x=\sin 2x + \cos 2x$?
Question: Solve the trigonometric equation: $\sin x + \cos x=\sin 2x + \cos 2x$.
My attempt:
$\sin x + \cos x=\sin 2x + \cos 2x$
$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - \sin^2 x$
$\implies \sin x + \cos x=2\sin x \cos x + \cos^2 x - (1-\cos^2 x)$
$\implies \sin x + \cos x=2\sin x \cos x + 2\cos^2 x - 1$
$\implies \sin x - 2\sin x \cos x + \cos x - 2\cos^2 x= - 1$
$\implies \sin x(1-2\cos x)+\cos x(1-2\cos x)=-1$
$\implies (1-2\cos x)(\sin x+\cos x)=-1$
$\implies (1-2\cos x)=-1$ or $(\sin x +\cos x)=-1$
$\implies \cos x=1$ or $\sin^2 x +\cos^2 x + 2\sin x\cos x=1$
$\implies x=2n\pi$ or $\sin 2x=0$
$\implies x=2n\pi$ or $2x=n\pi$
$\therefore x=2n\pi$ or $x=\frac{n\pi}{2}$
But the answers given in my book are $x=2n\pi$ and $x=\frac{(4n+1)\pi}{6}$. Where have I gone wrong? Please help.
| To expand on @gribouillis 's comment, the error in your argument is this step:
$(1-2\cos x)(\sin x+\cos x)=-1$
$\implies (1-2\cos x)=-1$ or $(\sin x +\cos x)=-1$
This is an incorrect implication.
$ab=c$ only implies $a=c$ or $b=c$ when $c=0$.
For $c=-1$ as in this case, you could have $a=1,b=-1$ or $a=5,b=-0.2$ or $a=-1000,b=0.001$ or an infinite number of other combinations.
| {
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Evaluating: $\lim_{x \to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}$ I was asked to evaluate the following limit:
$$\lim_{x \to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}$$
I really have no idea how to solve it!
| Using Taylor of oder 4, near $x= 0$ see here we have
$$\sin x\sim x-\frac{x^3}{6}+o(x^4)~~~and ~~~~\tan x\sim x +\frac{x^3}{3} +o(x^4)$$
but $\tan x\sim x+\frac{x^3}{3}$ stay near 0 as x is near 0 therefore,
$$\sin(\tan x) \sim \sin(x+\frac{x^3}{3}) \sim x+\frac{x^3}{3} -\frac{1}{6}(x+\frac{x^3}{3})^3 = x+ \frac{x^3}{6} +o(x^4)$$
similarly, $\sin x\sim x-\frac{x^3}{6}$ stay near 0 as x is near 0 therfore,
$$\tan(\sin x) \sim x-\frac{x^3}{6} +\frac{1}{3}(x-\frac{x^3}{6})^3 \sim x+\frac{x^3}{6} + o(x^4)$$
also
$$x-\sin x \sim \frac{1}{6}x^3 +o(x^4)$$
we conclude that
$$\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}\sim \frac{x+\frac{x^3}{6}-x-\frac{x^3}{6} +o(x^4)}{\frac{x^3}{6} +o(x^4) } = \frac{0+o(x)}{\frac{1}{6} +o(x) }$$
that is $$\color{red}{\lim_{x\to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x} = 0}.$$
OR By L'hopital rule
we have
$$\lim_{x\to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x} \\ =
\lim_{x\to 0}\frac{(1+\tan^2x)\cos(\tan x)-\cos x(1+\tan^2(\sin x))}{1-\cos x} \sim \frac{0+o(x^2)}{\frac{x^2}{2} +o(x^2)} = 0 $$
by just taking Taylor of oder 2 to get the same answer.
| {
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How to prove that $\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}={1\over 4}?$ How do we show that $(1)$
$$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}={1\over 4}?\tag1$$
$$\tan^{-1}(x)=\sum_{n=0}^{\infty}{(-1)^nx^{2n+1}\over 2n+1}$$
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\int_{0}^{\infty}x^{2n+1}\cdot{\mathrm dx\over (1+x)^3}\tag2$$
$$(1+x)^{-3}=\sum_{k=0}^{\infty}{2+k\choose k}x^k$$
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\sum_{k=0}^{\infty}{2+k\choose k}\int_{0}^{\infty}x^{2n+k+1}\mathrm dx\tag3$$
| $$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}$$
$$\int_{0}^{\infty}\arctan(x)\cdot{\mathrm dx\over (1+x)^3}$$
Apply Integration By Parts $$u = \arctan(x),\,v\prime = \frac{1}{(1 + 3)^3} $$
$$\arctan(x) \left(-\frac{1}{2(1 + 3)^2}\right) -
\int_{0}^{\infty}\frac{1}{x^2+1}\cdot\left(-\frac{1}{2(1+x)^2}\right)dx$$
$$-\frac{\arctan(x)}{2(1 + x)^2}- \underbrace{\int_{0}^{\infty}-\frac{1}{2(x+1)(x^2 +1)}dx}_I$$
$$I = -\frac{1}{2}\int_{0}^{\infty}-\frac{1}{(x+1)(x^2 +1)}dx$$
Create the partial fraction
$$I = -\frac{1}{2}\int_{0}^{\infty}\left(-\frac{x}{2(x^2 + 1)} + \frac{1}{2(x + 1)}+ \frac{1}{2(x + 1)^2}\right)dx$$
$-\int_{0}^{\infty}\frac{x}{2(x^2 + 1)} dx = -\frac{1}{4}\ln(x^2 + 1)$
$\int_{0}^{\infty}\frac{1}{2(x + 1)}dx= \frac{1}{2}\ln(x + 1)$
$\int_{0}^{\infty}\frac{1}{2(x + 1)^2}dx= -\frac{1}{2(x + 1)}$
$$I = -\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right)$$
Also
$$-\frac{\arctan(x)}{2(1 + x)^2}- \left[ -\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right)\right]$$
$$-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) + C$$
$$\lim_{x \to 0}\left(-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) \right) = -\frac{1}{4}$$
$$\lim_{x \to \infty}\left(-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) \right) = 0$$
$$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3} = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$$
| {
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Unclear math contest solution I'm having difficulty understanding this solution to a problem in an old math contest. I understand everything up to the word "Hence". Can anybody explain this to me?
Question
Answer
| Alternatively: From Vieta's theorem:
$$a+b+c=0, ab+bc+ac=-3, abc=-1.$$
We can find:
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=0^2-2(-3)=6.$$
From the equation:
$$\begin{cases} a^3=3a-1 \\ b^3=3b-1 \\ c^3=3c-1 \end{cases} \Rightarrow a^3+b^3+c^3=3(a+b+c)-3=-3.$$
Multiply the two:
$$(a^2+b^2+c^2)(a^3+b^3+c^3)=-18 \Rightarrow$$
$$a^5+b^5+c^5+(abc)^2(ab+bc+ac)=-18 \Rightarrow$$
$$a^5+b^5+c^5=-18-(-1)^2(-3)=-15.$$
| {
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If $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0} g(x) = +\infty$ prove that $\lim\limits_{x \to x_0} \frac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$
Consider the functions $f, g : \mathbb{R} \rightarrow \mathbb{R}$ for which holds that $\lim\limits_{x \to x_0} f(x) = \lim\limits_{x \to x_0} g(x) = +\infty$ for some $x_0 \in \mathbb{R}$. Prove that $\lim\limits_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=0$
Is the following proof correct? For some $x \to x_0$ holds that $f(x)>0$ and $g(x)>0$
So it is true that $2f(x)g(x)\geq 0 \\ \Rightarrow (f(x)+g(x))^2 \geq f^2(x)+g^2(x) \\ \Rightarrow \dfrac{1}{f^2(x)+g^2(x)} \geq \dfrac{1}{(f(x)+g(x))^2} \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{f(x)+g(x)}{(f(x)+g(x))^2}=\dfrac{1}{f(x)+g(x)}$
But $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}>0$ hence $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}=\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg|$
Thus $\Bigg|\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}\Bigg| \geq \dfrac{1}{f(x)+g(x)} \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \dfrac{1}{f(x)+g(x)} \text{OR} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq \lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq -\lim_{x \to x_0}\dfrac{1}{f(x)+g(x)} \\ \Rightarrow \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq 0 \text{OR} \lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \leq 0 \\ \Rightarrow \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} = 0$
| I don't see how you conclude $\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} = 0$ from $\lim_{x \to x_0}\dfrac{f(x)+g(x)}{f^2(x)+g^2(x)} \geq 0$.
In any case, the proof can be simplified. Write
$$
\lim_{x \to x_0} \dfrac{f(x)+g(x)}{f^2(x)+g^2(x)}= \lim_{x \to x_0} \dfrac{f(x)}{f^2(x)+g^2(x)} + \dfrac{g(x)}{f^2(x)+g^2(x)}
$$
Since $f^2(x), g^2(x) \geq 0$, we have
$$
\lim_{x \to x_0} \dfrac{f(x)}{f^2(x)+g^2(x)} + \dfrac{g(x)}{f^2(x)+g^2(x)} \leq \lim_{x \to x_0} \dfrac{f(x)}{f^2(x)} + \dfrac{g(x)}{g^2(x)} = \lim_{x \to x_0} \dfrac{1}{f(x)} + \dfrac{1}{g(x)} =0
$$
| {
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How to compute the following limit$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$ I am trying to find
$$ \lim\limits_{x\to \infty }[( 1+x^{p+1})^{\frac1{p+1}}-(1+x^p)^{\frac1p}],$$
where $p>0$. I have tried to factor out as
$$(1+x^{p+1})^{\frac1{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} =x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}},$$
but still was not able to make progress. Any other approach to this is welcome.
| Let try to use a general method to solve this kind of limits by
looking at the first order Taylor expansion of your expression:
\begin{aligned}\left( 1+x^{p+1}\right)^{\frac{1}{p+1}}- \left( 1+x^{p}\right)^{\frac{1}{p}} &=
x\left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- x\left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}} \\
&= x\left[ \left(1+\frac{1}{x^{p+1}}\right)^{\frac{1}{p+1}}- \left(1+\frac{1}{x^{p}}\right)^{\frac{1}{p}}\right] \\
&= x\left[1+\frac{1}{x^{p+1}{p+1}}+o(x^{-p-1}) - 1 -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&= x\left[\frac{1}{x^{p+1}{p+1}} -\frac{1}{x^{p}{p}}+o(x^{-p})\right]\\
&=\frac{1}{x^{p}{p+1}} -\frac{1}{x^{p-1}{p}}+o(x^{-p+1})
\end{aligned}
So that the whole expression tends to 0 when $x\rightarrow \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof about Fermat number without induction I want to show that $7 \mid (F_{2k + 1} + 2)$ where $k \in \mathbb{N_0}$ and $F_n := 2^{2^n} + 1$. I was able to proof this using induction, but was wondering if there is a more direct approach?
Here is a simple sketch of proof (only induction step):
$$
F_{2(k+1)+1} +2= 2^{2^{2k +3}} +3 = 2^{2^{2k + 3}} + 7q - 2^{2^{2k + 1}}
$$
for some $q \in \mathbb{Z}$.
$$
= 16^{2^{2k + 1}} + 7q - 2^{2^{2k + 1}} = 7q + 2^{2^{2k + 1}} (8^{2^{2k + 1}} -1) \\
= 7q + 2^{2^{2k + 1}} ((7 + 1)^{2^{2k + 1}} -1) \\
= 7q + 2^{2^{2k + 1}} ((\sum_{i = 0}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i}) -1) \\
= 7q + 2^{2^{2k + 1}} ((\sum_{i = 1}^{2^{2k + 1}}7^i1^{2^{2k + 1}-i}))
$$
This is clearly divisible by $7$.
| $$F_{2k+1}+2=7+2^{2^{2k+1}}-2^2=7+4\left(2^{4^k-1}-1\right)\left(2^{4^k-1}+1\right)$$ and $2^{4^k-1}-1$ is divided by $7$ because $4^k-1$ is divided by $3$ and $2^3-1=7$.
| {
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"url": "https://math.stackexchange.com/questions/2484480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the ODE $y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$ Question:
Solve the ODE given below:
$y(1+\sqrt{x^2 y^4+1})dx+2xdy=0$
My try:
The equation is not separable because a function of $x$ is added to a function of $y$.
($y+y\sqrt{x^2y^4+1}$)
Also, it is not linear with respect to $x$ or $y$, because it has the term$\sqrt{x^2y^4+1}$.
On the other hand, it's not a complete ODE because $\frac{d}{dy}(y(1+\sqrt{x^2 y^4+1})) \neq \frac{d}{dx}(2x)$
I also tried homogenous ODE's. But this ODE is not homogenous. It doesn't seem to be a Clero DE either.
Any idea?
Thanks in advance.
| Well, we have:
$$\text{y}\cdot\left\{1+\sqrt{1+x^2\cdot\text{y}^4}\right\}\space\text{d}x+2x\space\text{d}\text{y}=0\tag1$$
Let $\text{y}\left(x\right)=\frac{\text{p}\left(x\right)^\frac{1}{4}}{\sqrt{x}}$, which gives:
$$\frac{x\cdot\text{p}\space'\left(x\right)+2\cdot\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}{2\sqrt{x}\cdot\text{p}\left(x\right)^\frac{3}{4}}=0\tag2$$
So, for $\text{p}\space'\left(x\right)$:
$$\text{p}\space'\left(x\right)=-\frac{2\cdot\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}{x}\tag3$$
Divide both sides by $\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}$:
$$\frac{\text{p}\space'\left(x\right)}{\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}=-\frac{2}{3}\tag4$$
Integrate both sides with respect to $x$:
$$\int\frac{\text{p}\space'\left(x\right)}{\text{p}\left(x\right)\cdot\sqrt{1+\text{p}\left(x\right)}}\space\text{d}x=\int-\frac{2}{3}\space\text{d}x\tag5$$
Which gives:
$$\ln\left|\frac{1-\sqrt{1+\text{p}\left(x\right)}}{1+\sqrt{1+\text{p}\left(x\right)}}\right|=\text{C}-2\ln\left|x\right|\tag6$$
Taking $\exp$ of both sides, gives:
$$\left|\frac{1-\sqrt{1+\text{p}\left(x\right)}}{1+\sqrt{1+\text{p}\left(x\right)}}\right|=\frac{\text{C}}{\left|x\right|^2}\tag7$$
Set $\text{y}\left(x\right)=\frac{\text{p}\left(x\right)^\frac{1}{4}}{\sqrt{x}}$ back:
$$\left|\frac{1-\sqrt{1+x^2\cdot\text{y}\left(x\right)^4}}{1+\sqrt{1+x^2\cdot\text{y}\left(x\right)^4}}\right|=\frac{\text{C}}{\left|x\right|^2}\tag8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root I am not getting on how to approach this problem. Clearly, this polynomial can have atleast 2 real roots. And using Descartes's rule of signs, it can have a maximum of 4 positive real roots. But after that, how should I proceed ?
Any help would be highly appreciated...
| Note that $x^8-x^7$ and $x^2 - x$ look very similar. We can use this to factor some of the terms:
$$
x^8 - x^7 + x^2 - x + 15\\
= x^7(x-1) + x(x-1) + 15\\
= (x^7+x)(x-1) + 15\\
= x(x^6+1)(x-1) + 15
$$
If the entire expression is to be zero, then at the very least, we must have $x(x^6+1)(x-1)< 0$, which happens for $x\in (0, 1)$. But for $x\in (0, 1)$, we have
$$
|x|<1\\
|x^6+1|<2\\
|x-1|<1
$$
which implies $|x(x^6-1)(x-1)|<2$, which in turn implies that $x(x^6+1)(x-1) + 15$ is always larger than $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Number of solutions of $3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$ Find Number of solutions of
$$3\sin^2 x+\cos^2 x+\sqrt{3} \sin x+\cos x+1=\sqrt{3}\sin x\cos x$$ in $\left[0 \:\: 10\pi\right]$
My Try:
The given equation is
$$2+2\sin^2 x+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)$$ $\implies$
$$3+2\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\sin (2x)+\cos (2x)$$
Any idea here?
| We can rewrite the equation as $\left(\sqrt 3 \sin x - \cos x \right)^2+ \left(1+ \sqrt 3 \sin x \right) \left(1+\cos x\right) = 0$
Let $a=1+ \sqrt 3 \sin x$ and $b=1+\cos x$
Then the equation is $(a-b)^2+ab = 0 \Rightarrow a^2-ab+b^2 =0$
But $a^2 -ab+b^2 = 0 \iff a=0,b=0 $ which is not possible.
Hence the equation has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
| By definition of derivative of a function at $x=3$ we have,
$$\text{since $\cos3\pi =-1\implies$ }\lim_{x\to 3} \frac {\cos (\pi x)-(-1)}{x-3} =\lim_{x\to 3} -\pi\sin (\pi x) =0$$
and
$$\lim_{x\to 3} \frac {(x+5)^{\frac {1}{3}} -2}{x-3} =\lim_{x\to 3} \frac {1}{3}(x+5)^{\frac {-2}{3}} =\frac {1}{3} 8^{-2/3} =\frac {1}{12}.$$
Therefore, $$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}=\lim_{x\to 3} \frac {\cos (\pi x)+1}{\color{red}{x-3}}\frac {\color{red}{x-3}}{(x+5)^{\frac {1}{3}} -2}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove that for any $a,b\in \mathbb{N}$, $\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+...}}$. Prove that for any natural numbers $a$ and $b$, $$\sqrt{a^2+b}=a+\frac{b}{2a+\frac{b}{2a+...}}.$$
This question is taken from Spivak's Calculus Chapter 22 on Infinite sequences, Problem 22-7(c). Part (b) proves the continued fraction expansion for $\sqrt{2}$. I cannot seem to generalize.
EDIT: Some answers are assuming that the limit exists, however proving that the limit exists is the part I really need a solution for.
| You get the continued fraction as limit of $x_{n+1}=a+\frac{b}{a+x_n}$, $x_0=a$, if that limit exists. As rational iterations are not that easy to analyze, linearize them by setting $p_0=a$, $q_0=1$, and separate numerator and denominator in
$$
\frac{p_{n+1}}{q_{n+1}}=a+\frac{b}{a+\frac{p_n}{q_n}}=a+\frac{bq_n}{aq_n+p_n}
=\frac{ap_n+(a^2+b)q_n}{aq_n+p_n}
$$
to get the linear iteration
$$
\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix}
=
\begin{bmatrix}a&a^2+b\\1&a\end{bmatrix}
\cdot
\begin{bmatrix}p_n\\q_n\end{bmatrix}
$$
with eigenvalues $ a\pm\sqrt{a^2+b}$ and eigenvectors $[\sqrt{a^2+b},\pm 1]^T$.
The resulting sequence of partial fractions is thus
$$
x_n=\sqrt{a^2+b}\cdot\frac{(a+\sqrt{a^2+b})^{n+1}+(a-\sqrt{a^2+b})^{n+1}}{(a+\sqrt{a^2+b})^{n+1}-(a-\sqrt{a^2+b})^{n+1}}
$$
and as $|a+\sqrt{a^2+b}|>|a-\sqrt{a^2+b}|$, this always converges to $\sqrt{a^2+b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the remainder of the division Find the remainder when $f(x)=x^{2017}-1$ is divided by $(x^2+1)(x^2+x+1)$
I have expanded the function and factorized it. The remainder I had found was $-x^3-2x^2-3$
I am not sure if I was right ! So can anyone tell me the other way to solve it ..?
| The hint:
$$f(x)=x^{2017}-x+x-1$$ and
$$x^{2016}-1=(x^4)^{504}-1$$ is divisible by $x^4-1=(x^2-1)(x^2+1).$
Also, $$x^{2016}-1=(x^3)^{672}-1$$ is divisible by $x^3-1=(x-1)(x^2+x+1),$
which gives that there is polynomial $g$ for which
$$f(x)=(x^2+1)(x^2+x+1)g(x)+x-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
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