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Find $\lim_{x\to 0} \cos \big(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \big) $ Find the limit $$\lim_{x\to 0} \cos \bigg(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \bigg)$$
I dont even know where to get started... Some hints and solutions would be appreciated! Thanks in advance!
P.S typed this on an iphone, sorry for any mistakes will edit soon.
EDIT
Here are my current workings.
$$\lim_{x\to 0} \cos \bigg(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \bigg)=\lim_{x\to 0} \cos \bigg(\pi x^2 (\frac {\cos (6x)} {\sin (\frac {x} {2}) \sin 6x} \bigg)$$
| Note that
$$\pi x^2 \csc (\frac {x} {2}) \cot (6x)=\frac{\pi}{3}\cdot\frac{\frac{x}{2}}{\sin(\frac{x}{2})}\cdot\frac{6x}{\sin(6x)}\cos(6x).$$
Since $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\cos(x)=1$, we have
$$\lim_{x\to 0}x^2 \csc (\frac {x} {2}) \cot (6x)=\frac{1}{3}.$$
Since $\cos$ is a continuous function, we have
$$ \lim_{x\to 0}\cos \Big(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \Big)
=\cos \Big(\lim_{x\to 0}\pi x^2 \csc (\frac {x} {2}) \cot (6x) \Big)=\cos\frac{\pi}{3}=\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this composition of polynomial correct? Suppose $p(x)=x^2+5x+3$ and $q(x)=3x^3-3x+7$. Write the expression $(q \circ p)(x)$ as a sum of terms, each which is a constant times the power of $x$.
Here is my work for the problem:
$(q\circ p)(x)=3(x^2+5x+3)^3-3(x^2+5x+3)+7$
$(q\circ p)(x)=3(x^6+125x^3+27)-3x^2-15x-9+7$
$(q\circ p)(x)=3x^6+375x^3+81-3x^2-15x-2$
$(q\circ p)(x)=3x^6+375x^3-3x^2-15x+79$
Did I do something wrong while working with this problem? The final answer I got wasn't in the multiple choice answers. I went over this problem 3 times and cannot find what I am doing wrong.
| As Jasper noted you did wrong here: $$...=3(x^2+5x+3)^3...$$ $$...=3(x^6+125x^3+27)-...$$ In fact $$(a+b+c)^3=a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^3$$ we have $q(p(x))=3x^6+45x^5+252x^4+645x^3+753x^2+390x+79$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many triangles I saw this question today, it asks how many triangles are in this picture.
I don't know how to solve this (without counting directly), though I guess it has something to do with some recurrence.
How can count the number of all triangles in the picture ?
| Say that instead of four triangles along each edge we have $n$. First count the triangles that point up. This is easy to do if you count them by top vertex. Each vertex in the picture is the top of one triangle for every horizontal grid line below it. Thus, the topmost vertex, which has $n$ horizontal gridlines below it, is the top vertex of $n$ triangles; each of the two vertices in the next row down is the top vertex of $n-1$ triangles; and so on. This gives us a total of
$$\begin{align*}
\sum_{k=1}^nk(n+1-k)&=\frac12n(n+1)^2-\sum_{k=1}^nk^2\\
&=\frac12n(n+1)^2-\frac16n(n+1)(2n+1)\\
&=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\
&=\frac16n(n+1)(n+2)\\
&=\binom{n+2}3
\end{align*}$$
upward-pointing triangles.
The downward-pointing triangles can be counted by their by their bottom vertices, but it’s a bit messier. First, each vertex not on the left or right edge of the figure is the bottom vertex of a triangle of height $1$, and there are $$\sum_{k=1}^{n-1}=\binom{n}2$$ of them. Each vertex that is not on the left or right edge or on the slant grid lines adjacent to those edges is the bottom vertex of a triangle of height $2$, and there are
$$\sum_{k=1}^{n-3}k=\binom{n-2}2$$ of them. In general each vertex that is not on the left or right edge or on one of the $h-1$ slant grid lines nearest each of those edges is the bottom vertex of a triangle of height $h$, and there are
$$\sum_{k=1}^{n+1-2h}k=\binom{n+2-2h}2$$ of them.
Algebra beyond this point corrected.
The total number of downward-pointing triangles is therefore
$$\begin{align*}
\sum_{h\ge 1}\binom{n+2-2h}2&=\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-2k}2\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)(n-2k-1)\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}\left(n^2-4kn+4k^2-n+2k\right)\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+2\sum_{k=0}^{\lfloor n/2\rfloor-1}k^2-(2n-1)\sum_{k=0}^{\lfloor n/2\rfloor-1}k\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+\frac13\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\left(2\left\lfloor\frac{n}2\right\rfloor-1\right)\\
&\qquad\qquad-\frac12(2n-1)\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\;.
\end{align*}$$
Set $\displaystyle m=\left\lfloor\frac{n}2\right\rfloor$, and this becomes
$$\begin{align*}
&m\binom{n}2+\frac13m(m-1)(2m-1)-\frac12(2n-1)m(m-1)\\
&\qquad\qquad=m\binom{n}2+m(m-1)\left(\frac{2m-1}3-n+\frac12\right)\;.
\end{align*}$$
This simplifies to $$\frac1{24}n(n+2)(2n-1)$$ for even $n$ and to
$$\frac1{24}\left(n^2-1\right)(2n+3)$$ for odd $n$.
The final figure, then is
$$\binom{n+2}3+\begin{cases}
\frac1{24}n(n+2)(2n-1),&\text{if }n\text{ is even}\\\\
\frac1{24}\left(n^2-1\right)(2n+3),&\text{if }n\text{ is odd}\;.
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/203873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove or disprove: For all positive integers $ n$ , $\sqrt[3]{n}+\sqrt[3]{n+1}$ are irrational numbers. Prove or disprove: For all positive integers $ n$ ,
$\sqrt[3]{n}+\sqrt[3]{n+1}$ are irrational numbers.
| Note that $(x+y)^{3} = x^{3} +y^{3} + 3xy(x+y).$ If $n$ or $n+1$ is the cube of an integer, the result is clear by the uniqueness of prime factorization, so we assume that neither is the cube of an integer. Then neither is $n(n+1)$. Set $x = n^{\frac{1}{3}}$ and $y = (n+1)^{\frac{1}{3}}.$ If $x+y$ is rational, then so are $(x+y)^{3}$ and $3xy(x+y)$. Hence $3xy$ is rational. Thus $27n(n+1)$ must be the cube of an integer, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/204262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How does $\zeta(1 - s)$ become $(-1/s + \cdots)$? Why is
$$\zeta(1 - s) = -\frac{1}{s} + \cdots$$
for small negative values of $s$?
A detailed explanation would be appreciated.
| For $s < 0$
$$ \begin{eqnarray}
\zeta(1-s) &=& \sum_{n=1}^\infty n^{s-1} = \int_1^\infty x^{s-1} \mathrm{d}x + \sum_{n=1}^\infty \left( n^{s-1} - \int_{n}^{n+1} x^{s-1} \mathrm{d}x \right) \\ &=& -\frac{1}{s} + \sum_{n=1}^\infty \left( n^{s-1} - \frac{(n+1)^s - n^s}{s} \right)
\end{eqnarray}
$$
Now, observe that the following limit is finite
$$
\lim_{s \to 0^-} \zeta(1-s) + \frac{1}{s} = \sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \gamma \tag{1}
$$
where $\gamma$ is the Euler-Mascheroni constant. The sum on the right-hand-side of $(1)$ converges , since for large $n$
$$
\frac{1}{n} - \log\left( 1+\frac{1}{n}\right) = \frac{1}{2 n^2} + \mathcal{o}\left(\frac{1}{n^2}\right)
$$
Alternatively:
$$
\sum_{n=1}^\infty \left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \sum_{n=1}^m\left( \frac{1}{n} - \log\frac{n+1}{n} \right) = \lim_{m \to \infty} \left( \sum_{n=1}^m \frac{1}{n} - \log(m+1) \right) = \gamma
$$
Hence, this establishes that, for small negative $s$:
$$
\zeta(1-s) = -\frac{1}{s} + \gamma + \mathcal{o}(1)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find the oblique asymptote of root of a function? In a test example I'm solving, the question asks to find the oblique asymptote of the following function:
$f(x) = \sqrt{4x^2+x+6}$
$x$ at $+\infty$
We have only learned how to do so with rational functions. Is there any general way of finding the oblique asymptote that works with any kind of function? Perhaps using limits?
| By completing the square we get $f(x) = \sqrt{ (2x + 1/4)^2 + 95/16}$.
This means (after squaring both sides and taking $(2x + 1/4)^2$ to the left hand side and factoring) that $$( f(x) - (2x + 1/4) ) ( f(x) + (2x + 1/4) ) = 95/16$$ and hence $$f(x) - (2x + 1/4) = \frac{95/16}{f(x) + (2x + 1/4)}.$$
But $f(x) + (2x + 1/4) \rightarrow \infty$ as $x \rightarrow \infty$. This implies that
$$f(x) - (2x + 1/4) \rightarrow 0$$ as $x \rightarrow \infty$.
Remark: Likewise $$f(x) + (2x + 1/4) = \frac{95/16}{f(x) - (2x + 1/4) }$$ and hence as $x \rightarrow -\infty, f(x) - (2x + 1/4) \rightarrow \infty$ and hence
$$\frac {95/16}{f(x) - (2x + 1/4)} \rightarrow 0.$$ This implies that $f(x) + (2x + 1/4) \rightarrow 0$ and thus $y = -(2x + 1/4)$ is another asymptote.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/205587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Double module in a inequality Can somebody explain to me (or link me a site which does) how to solve this?
$$
||x+1| -1| \geq 3
$$
I have no idea how to work out this double absolute value sign.
| Here is a solution with squaring technique
$$
\begin{align}
||x+1|-1|\geq 3&\Longleftrightarrow (|x+1|-1)^2\geq 3^2\quad\text{ since both sides of inequality are non negative}\\
&\Longleftrightarrow (|x+1|-1)^2-3^2\geq 0\\
&\Longleftrightarrow ((|x+1|-1-3)((|x+1|-1)+3)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\
&\Longleftrightarrow (|x+1|-4)(|x+1|+2)\geq 0\\
&\Longleftrightarrow |x+1|-4\geq 0\quad\text{ since }\quad\quad|x+1|+2>0\quad\text{for all }x\\
&\Longleftrightarrow |x+1|\geq 4\\
&\Longleftrightarrow |x+1|^2\geq 4^2\quad\text{ since both sides of inequality are non negative}\\
&\Longleftrightarrow (x+1)^2\geq 4^2\quad\text{ since }\quad |a|^2=a^2\\
&\Longleftrightarrow (x+1)^2 - 4^2\geq 0\\
&\Longleftrightarrow ((x+1)-4)((x+1)+4)\geq 0\quad\text{ since}\quad a^2-b^2=(a-b)(a+b)\\
&\Longleftrightarrow (x-3)(x+5)\geq 0\\
&\Longleftrightarrow x\in(-\infty,-5]\cup[3,+\infty)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/208297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Using generating functions find the sum $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ I am quite new to generating functions concept and I am really finding it difficult to know how to approach problems like this. I need to find the sum of $1^3 + 2^3 + 3^3 +\dotsb+ n^3$ using generating functions. How do I proceed about it?
| You can do this more prettily with exponential generating functions. Note that
$$e^{kx} = \sum_n \frac{k^n x^n}{n!}$$
so
$$1+e^x+e^{2x} + e^{3x} + \cdots + e^{kx} = \sum_n \frac{(1+2^n+3^n + \cdots + k^n) x^n}{n!}.$$
The left hand side is
$$(e^{kx}-1) \cdot \frac{1}{e^x-1} = \left( kx + \frac{k^2 x^2}{2} + \frac{k^3 x^3}{6} + \cdots \right) \left( \frac{1}{x} - \frac{1}{2} + \frac{x}{12} - \frac{x^3}{720} + \cdots \right)$$
where the second factor can be expressed in terms of Bernoulli numbers.
Now compare coefficients of $x^3$ on both sides.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/209268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 0
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Common tangent to two circles
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$
and
$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.
| The parametric equation of $(x-a)^2+(y-b)^2=r^2$ is $(x=a+r\cos C,y=b+r\sin C)$
Let $$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=m$$ be a tangent at $(a+r\cos C,b+r\sin C)$, then the distance of line from the center is equal to the radius.
$$r=\frac{\mid m(a-a-r\cos C)-b+b+r\sin C\mid}{\sqrt{m^2+1}}=\frac{r\mid\sin C-m\cos C\mid}{\sqrt{m^2+1}}$$
$m^2+1=(\sin C-m\cos C)^2$
$\implies (m\sin C+\cos C)^2=0\implies m=-\frac{\cos C}{\sin C}$
So, the equation of the tangent becomes
$$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=-\frac{\cos C}{\sin C}$$
$x\cos C+y\sin C-a\cos C-b\sin C -r=0$ (this can also be reached using calculus)
For $(x-2)^2+y^2=9,a=2,b=0,r=3$
So, equation of the tangent will be $x\cos A+y\sin A-2\cos A -3=0$
For $(x-5)^2+(y-4)^2=4,a=5,b=4,r=2$
So, equation of the tangent will be $x\cos B+y\sin B-5\cos B-4\sin B -2=0$
For common tangent, these two lines must be same,
So, $$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}=\frac{2\cos A+3}{5\cos B+4\sin B+2}$$
$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}\implies \sin(A-B)=0$
$\implies A=B$ or $A=\pi+B$
(1)If $A=B,1=\frac{2\cos B+3}{5\cos B+4\sin B+2}\implies 4\sin B+3\cos B=1$
(2)If $A=\pi+B,\cos A=\cos(\pi+B)=-\cos B, -1=\frac{-2\cos B+3}{5\cos B+4\sin B+2}$
For (1), $4\sin B+3\cos B=1$
Putting $4=R\sin D,3=R\cos D\implies R=5,D=cos^{-1}\frac 3 5,$
$\cos(B-D)=\frac 1 5, B-\cos^{-1}\frac 3 5 =2n\pi \pm \cos^{-1}\frac 1 5$ where $n$ is any integer.
$\cos B=\frac{3 \pm 8\sqrt{6}}{25},\sin B$ can be calculated uniquely using (1).
So, there will be two tangent in this case.
For(2) $ 3\cos B+4\sin B=-5$
Applying the same approach like in (1),
$\cos(B-\cos^{-1}\frac 3 5)=-1=\cos \pi, B=\cos^{-1}\frac 3 5+\pi$
$\cos B=\cos(\cos^{-1}\frac 3 5+\pi)=-\cos(\cos^{-1}\frac 3 5)=-\frac 3 5$
$\sin B$ becomes $-\frac 4 5$
So, the tangent becomes, $(x-5)(-\frac 3 5)+(y-4)(-\frac 4 5) -2=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Circle Least Squares Fit So my question is this:
Find the equation of the circle that gives the best least squares circle fit to the points $(-1,-2), (0,2.4), (1.1,-4),$ and $(2.4,-1.6).$
So far I have this general equation:
$2xc_1+2yc_2+(r^2-c_1^2-c_2^2)=x^2+y^2$ where $r^2-c_1^2-c_2^2 = c_3$
So then I think I create matrix:
$\begin{pmatrix} 2x_1 & 2y_1 & 1 \\ . & . & . \\ 2x_n & 2y_n & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} x_1^2+y_1^2 \\ ... \\ x_n^2+y_n^2\end{pmatrix}$
and after replacing $x_1 = -1$ and $y_1 = -2$ until $x_4$ and $y_4$, I have this matrix:
$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} - \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$
Which i guess I try to solve by setting equal to 0, then moving the last matrix to the other side and make it look like this:
$\begin{pmatrix} -2 & -4 & 1 \\ 0 & 4.8 & 1 \\ 2.2 & -8 & 1 \\ 4.8 & -3.2 & 1 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \\ c_3\end{pmatrix} = \begin{pmatrix} 5 \\ 5.76 \\ 17.21 \\ 8.32\end{pmatrix}$
And finally solve for $c_1$, $c_2$, $c_3$. But this system has no solution as shown here. Am I missing something huge here or what am I doing wrong?
| It is true that the given equation $Ac = B$ has no solution. That is why we are trying to get an approximate answer (least-squares fit).
An explanation on Khan Academy
will give you an understanding of the result after approximating it.
If we cannot find the solution to $Ac = B$, we find solution to $A^TAc=A^TB$.
This equation is the connection missing.
| {
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"timestamp": "2023-03-29T00:00:00",
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Frequency of a trigonometric function - Where is my mistake? I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$
The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$
This is what i did:
Finding $\sin(x)^4$ frequency (I'll call it F1):
$$\cos(2x)=1-\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$
Finding $\cos(2x)^2$ frequency:
$$\cos(4x)=2\cos^2(2x)-1$$
$$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$
$$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$
Finding $\sin(x)^2$ frequency:
$$\cos(2x)=1-2\sin^2(x)$$
$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$
$$f_2=\frac{2\pi}{2}=\pi$$
$$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_1=\frac{\pi}{2}\times2=\pi$$
Finding $cos(x)^4$ frequency (I'll call it F2):
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
$$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$
Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1):
$$f_3=\frac{2\pi}{2}=\pi$$
$$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$
$$F_2=\frac{\pi}{2}\times2=\pi$$
Finding $y$'s frequency:
$$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$
$$F_y=\pi\times1=\pi$$
| You have proved that $\pi$ is a period, but you have not shown that it is the smallest period.
I would tackle the problem in more or less the same way that you did, using double angle identities, but the algebra can be simplified. Note that
$$1=(\cos^2 x+\sin^2 x)^2=\cos^4 x+\sin^4 x +2\cos^2 x\sin^2 x.$$
It follows that our function is equal to
$$1-2 \cos^2 x\sin^2 x.$$
The only interesting part is $2\cos^2 x\sin^2 x$, which is $\frac{1}{2}\sin^2 2x$.
It is clear that this has period $\dfrac{\pi}{2}$.
If you wish, you can use the trigonometric identity $\cos 2u=1-2\sin^2 u$ to express our function in terms of $\cos 4x$. We get $\frac{1}{2}\sin^2 2x=\frac{1}{4}(1-\cos 4x)$, and therefore
$$\cos^4 x+\sin^4 x=\frac{3}{4}+\frac{1}{4}\cos 4x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this equation:$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5$ Please, help me to solve this equation:
$$\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}=5$$
I think if we continue to square, it would have to face a high-order equation.
I guess $x = 30$, but do not know how to solve.
| Observe that $\pm \sqrt{[17,33] + [-7,7]} = \pm [\sqrt{10}, \sqrt{40}] \subset [-7;7]$
This implies that for $x \in [17,33]$, all the quantities $f_1(x), \ldots, f_8(x) = \sqrt{x \pm \sqrt {x \pm \sqrt {x \pm \sqrt {x - 5}}}}$ are well-defined.
Furthermore, for $x=17$, $\sqrt{17 + [-7,7]} = [\sqrt{10}, \sqrt{24}] < 5$, so $f_i(17)<5$
and for $x=33$, $\sqrt{33 + [-7,7]} = [\sqrt{26}, \sqrt{40}] > 5$, so $f_i(33)>5$ .
Thus each of the $8$ equations $f_i(x)=5$ has at least one root in $[17,33]$.
But, a solution of any of these equations is a root of the polynomial $(((5^2-x)^2-x)^2-x)^2-x+5 = 0$, which is of degree $8$, so can have at most $8$ roots.
Reciprocally, if $x$ is a root of this polynomial, then it is a solution of exactly one $f_i(x) = 5$, because the signs of $((5^2-x)^2-x)^2-x, (5^2-x)^2-x, 5^2-x$ determine the signs in front of the square roots from the innermost one to the outermost one.
Therefore, since each of the $8$ equations have at least a solution, they give at least $8$ distinct roots of the polynomial, so there isn't any other root, and thus there is no other solutions to the equations $f_i(x)=5$.
This means that the only solution to $\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}} = 5$ has to be the solution to $\sqrt{x-5} = 5$, which is easily solvable and gives $x = 30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 4
} |
How to show that $\left(\frac{y}{y-1}\right)^y = \left( 1 + \frac 1{y-1}\right)^y $ I need to show
\[\left(\frac{y}{y-1}\right)^y = \left( 1 + \frac 1{y-1}\right)^y \]
I cannot understand how they made this step! Can someone explain how this works?
| We have
\[
\frac y{y-1} = \frac{y-1+1}{y-1} = \frac{y-1}{y-1} + \frac 1{y-1} = 1 + \frac 1{y-1}. \]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/216944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Two interesting and basic integrals. Show: $$\int_0^\pi \sqrt{1 + \sin(x)}\,dx.\ = 4$$
and $$\int_0^\pi \frac{xdx}{1 + \cos^2(x)} = \frac{\pi^2}{2\sqrt{2}}$$
| For the first one, you can use the identity $\sin(x)=-\cos(\frac{\pi}{2}+x)$.
Then by half angle formula
$$1+\sin(x)=1- \cos(\frac{\pi}{2}+x)=2 \sin^2(\frac{\pi}{4}+\frac{x}{2})$$
Thus
$$\int_0^\pi \sqrt{1 + \sin(x)}\,dx.\ = \int_0^\pi \sqrt{2}\sin(\frac{\pi}{4}+\frac{x}{2})\,dx=-2\sqrt{2}\cos(\frac{\pi}{4}+\frac{x}{2})|_0^\pi=4$$
Second one
Let $I=\int_0^\pi \frac{xdx}{1 + \cos^2(x)}$.
Let $u= \pi-x$. Then
$$I= \int_0^\pi \frac{(\pi-u)}{1 + \cos^2(\pi-u)}du=\int_0^\pi \frac{\pi}{1 + \cos^2(u)}du -I $$
Thus
$$2I=\pi \int_0^\pi \frac{1}{1+\cos^2(u)} du=2 \pi \int_0^\frac{\pi}{2} \frac{1}{1+\cos^2(u)} du$$
by the symmetry about $u=\frac{\pi}{2}$.
Thus
$$I= \pi \int_0^\frac{\pi}{2} \frac{1}{1+\cos^2(u)} du= \pi \int_0^\frac{\pi}{2} \frac{\sec^2(u)}{1+\sec^2(u)} du=\pi \int_0^\frac{\pi}{2} \frac{\sec^2(u)}{2+\tan^2(u)} du$$
This integral becomes a trivial improper integral after the obvious substitution $v =\tan(u)$.
P.S. $I= \pi \int_0^\frac{\pi}{2} \frac{1}{1+\cos^2(u)} du$ can also be calculated with the standard substitution $t=\tan(\frac{u}{2})$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to eliminate these extra solutions? (finding the gcd of two expressions)
Prove that for any integer $n$, $\gcd (3n^2+5n+7, n^2+1)=1$ or $41$.
The following answer is convoluted because I've intentionally created excess solutions. However, I can't figure out how to eliminate them! Anyone?
Let $$d=\gcd (3n^2+5n+7, n^2+1).$$
Then $$d|[(3n^2+5n+7)-3(n^2+1)]$$
$$d |(5n+4)$$
And
$$d | [5(3n^2+5n+7)-3n(5n+4)]$$
$$d |(13n+35)$$
And
$$d |[5(13n+35)-13(5n+4)]$$
$$d |123$$
Therefore, $d= 1$ or $3$ or $41$ or $123$.
| From your last step, we get that $d = 1,3,41,123$.
Recall that $$n^2 \equiv 0,1 \pmod{3} \text{ (Why?)}$$ Hence, $3$ (or) $123$ does not divide $n^2+1$.
EDIT
Note that any $n$ is either $0 \pmod{3}$ or $\pm1 \pmod{3}$.
Hence, $n^2 \equiv 0,1 \pmod{3}$. (Recall that if $x \equiv y \pmod{a}$, then $x^k \equiv y^k \pmod{a}$.)
Hence, $n^2 + 1 \equiv 1,2 \pmod{3}$. This means that $3$ does not divide $n^2+1$. Hence, $3$ cannot divide any divisor of $n^2+1$. This enables us to rule out $d=3$ and $d=123$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/218915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How did my teacher gets this result? Algebra/Limits $$\begin{align*}&\lim_{x\to0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt[3]{1+x}-\sqrt[3]{1-x}}=\\
&\lim_{x\to0}\left(\frac{(1+x)-(1-x)}{(1+x)-(1-x)}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{(1+x)(1-x)}+\sqrt[3]{(1-x)^2}}{\sqrt{1+x}+\sqrt{1-x}}\right)
\end{align*}$$
My teacher was using this to calculate the value of the first limit as seen above.
I am not sure how he pull this huge rabbit out of the hat XD.
Also I am not sure why is it useful...
| To get from the first line to the second line, your teacher is multiplying by
$$
\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \cdot \frac{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}{\sqrt[3]{(1+x)^2} + \sqrt[3]{(1+x)(1-x)} + \sqrt[3]{(1-x)^2}}.
$$
Why did he make this choice? He is using the identities
$$
(\sqrt{a} - \sqrt{b}) \cdot (\sqrt{a} + \sqrt{b}) = a - b
$$
and
$$
(\sqrt[3]{a} - \sqrt[3]{b}) \cdot (\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}) = a - b.
$$
As with many limit problems, the reason for these manipulations is to try to obtain an algebraic cancellation to deal with the fact that both the numerator and denominator are approaching zero.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A couple of GRE questions Look for help with the following GRE questions
Question 1. If $C$ is the circle in the complex plane whose equation $|z|=\pi$, oriented counterclockwise, find the value of the integral $\oint_C(\cos z-z\cos\frac{1}{z})dz$.
Question 2. How to show the sequence $\{x_n\}_{n=1}^\infty$ definted by $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$, $x_1\ne 0$ converge.
Question 3. Let $L$ be the curve whose equation in the polar coordinates $r$ and $\theta$ is $r^2=4\cos 2\theta$. Fine the largest value of $y$ such that the point with rectangular coordinates $(x,y)$ is on $L$.
Question 4. Consider the set $S$ of all real-valued functions defined on $[a, b]$, $a<b$. Is it true that if the inverse of $f$ is a constant function, then $f$ is a constant function?
| For the first, $\oint_C \cos z dz = 0$ since $\cos$ is an entire function. Thus, only the term $z\cos\frac{1}{z}$ is relevant. That term has a single pole at $z=0$, and by the residue theorem you therefore have $$
\oint_C z\cos\frac{1}{z} dz = i2\pi R
$$
where $R$ is the residue (i.e. the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$) at $0$. Now, $\cos z = \frac{1}{2}(e^{iz} + e^{-iz})$, hence the coefficient of $z^2$ in the taylor expansion of $\cos z$ is $\frac{1}{2}(\frac{-1}{2} + \frac{-1}{2})$ = $-\frac{1}{2}$. Which is the same as the coefficient of $z^{-2}$ in the laurent expansion of $\cos\frac{1}{z}$, and hence the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$, i.e. the residue. Together, you get that $$
\oint_C \cos z - z\cos\frac{1}{z} dz = 0 - i2\pi\frac{-1}{2} = i\pi
$$
For the second, first find a fixed point, i.e. solve $x_{n+1} = x_n$. You get $$
\frac{1}{2}\left(x_n + \frac{2}{x_n}\right) = x_n \iff
\frac{1}{x_n} = \frac{x_n}{2} \iff
x_n = \sqrt{2}
$$
Then show that $\sqrt{2} < x_{n+1} < x_n$ if $x_n > \sqrt{2}$, which works similarly $$
\begin{eqnarray}
\sqrt{2} < \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) < x_n &\iff&
\sqrt{2} - \frac{x_n}{2} < \frac{1}{x_n} < \frac{x_n}{2} \iff \\
2\sqrt{2}x_n - x_n^2 < 2 < x_n^2 &\iff&
2 - (x_n - \sqrt{2})^2 < 2 < x_n^2
\end{eqnarray}
$$
The last inequality shows not only that the sequence decrases monotonically if started at any value larger than $\sqrt{2}$, it also shows that regardless of where it it started, it always immediatly assumes a value larger than $\sqrt{2}$. Thus, the sequence converges regardless of its initial value (Strictly speaking, the above verified that only for positive $x_1$, but things work identically for negative $x_1$)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What are the steps to simplify the following modulus expression? I have no clue how to do this exactly. Is there a systematic way of doing this or you just have to do it by trial and error?
$n^2 \equiv 9 \pmod {72}$
to $n \equiv a \pmod b$?
| Note that $n^2\equiv 9 \pmod {72}$ is equivalent to
$$
(n-3)(n+3)\equiv 0 \pmod {72}
$$
So it is sufficient to solve the simultaneous congruences
$$
(n-3)(n+3)=0\pmod 8
$$
and
$$
(n-3)(n+3)\equiv 0 \pmod 9
$$
The second congruence is true iff $n$ is divisible by $3$, because then both $n-3$ and $n+3$ are divisible by $3$, so their product is divisible by $9$. If $n$ is not divisible by $3$, then neither $n-3$ nor $n+3$ is divisible by $3$, so their product isn't. Thus, $n\equiv 0 \mod 3$.
In the first congruence, we similarly obtain that $n$ must be odd. We claim that this is also sufficient. Indeed, since clearly $n-3$ and $n+3$ are both even, assume that they are both $2 \pmod 4$. This is clearly impossible because they have a difference of $6$. Thus, one of $n-3$ and $n+3$ must be divisible by $4$, and the other is even, so their product is $0 \pmod 8$. Then $n\equiv 1 \pmod 2$.
Then, the set of solutions is just all $n$ that satisfies $n\equiv 1\pmod 2$ and $n \equiv 0 \pmod 3$, which is equivalent to $n \equiv 3 \mod 6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the degree of a field extension I have to determine the degree of $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$
over $\mathbb{Q}$ and show that $\sqrt{2}+\sqrt{3}$ is a primitive
element ? Could someone please give me any hints on how to do that
?
| Clearly $[\mathbb Q(\sqrt 2):\mathbb Q]\le 2$ becasue of the polynomial $X^2-2$ and $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt 2)]\le 2$ because of the poylnomial $X^2-3$.
In fact, $\sqrt 2\notin \mathbb Q$ implies $[\mathbb Q(\sqrt 2):\mathbb Q]=2$.
We also have $\sqrt 3\notin \mathbb Q(\sqrt 2)$ because $(a+b\sqrt 2)^2 = 3$ implies $(a^2+2b^2) + 2ab\sqrt 2 = 3$, hence $2ab = 0$ and $a^2+2b^2=3$; thus either $a=0$ and $b^2=\frac 32$, or $b=0$ and $a^2=3$. But both $\sqrt{\frac32}$ and $\sqrt 3$ are irrational.
Therefore $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt 2)]=2$ and finally
$$[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=4.$$
For the second part , note that $\mathbb Q(\sqrt 2+\sqrt 3)$ contains $(\sqrt 2+\sqrt 3)^2=2+2\sqrt 6+3$, hence also $\sqrt 6$ and $\sqrt6(\sqrt 2+\sqrt 3)=2\sqrt 3+3\sqrt 2$, and finally both $3(\sqrt2+\sqrt 3)-(2\sqrt 3+3\sqrt 2)=\sqrt 2$ and $(2\sqrt 3+3\sqrt 2)-2(\sqrt2+\sqrt 3)=\sqrt 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Prove that $\lim\limits_{x\to 0}\frac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\frac{1}{6}$ without De L'Hôpital Rule or Taylor Expansion? Any ideas how to show $\lim\limits_{x\to 0}\dfrac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\lim\limits_{x\to 0}\dfrac{x-\sin(x)}{x^3}=\dfrac{1}{6}$ without using the De L'Hôpital rule (or proving a special case of it?).
How can we reduce this to $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2}$?
You can suppose that we know the limit in question exists
and therefore use inequalities to bound it
| Squeeze Theorem.
First, there is some factoring and cancellation:
$$
\begin{align}
\frac{x^2-1+\cos^2x}{x^4+x^3\sin x} &=\frac{x^2-(1-\cos^2{x})}{x^3(x+\sin x)}\\
&=\frac{x^2-\sin^2(x)}{x^3(x+\sin x)}\\
&=\frac{(x-\sin x )(x+\sin x)}{x^3(x+\sin x)}\\
&=\frac{x-\sin x}{x^3}\\
\end{align}
$$
The function $f$ with $f(x)=\frac{1}{6}x^3$ and $g$ with $g(x)=x-\sin x$ satisfy the inequality $g(x)\leq f(x)$ for all $x\in(0,\epsilon)$. This is because at $x=0$, the two functions have the same value, derivative, second derivative, third derivative, and fourth derivative, but $g^{(5)}(0)<f^{(5)}(0)$.
And if $h(x)=\frac{1}{6}x^3 - x^5$, then $h(x)\leq g(x)$ in $(0,\epsilon)$ for basically the same reason, but now $h^{(5)}(0)<g^{(5)}(0)$.
So for all $x\in (0,\epsilon)$,
$$h(x)\leq g(x)\leq f(x)$$
$$\implies\frac{h(x)}{x^3}\leq \frac{g(x)}{x^3}\leq \frac{f(x)}{x^3}$$
$$\implies\frac{1}{6}-x^2\leq \frac{x-\sin x}{x^3}\leq \frac{1}{6}$$
Applying the Squeeze Theorem as $x\to0^+$ gives that $$\lim_{x\to0^+}\frac{x-\sin x}{x^3}=\frac{1}{6}$$ You can either alter this argument to work on both sides at once, or make a separate similar argument for the other side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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General term of this interesting sequence A sequence of positive integer is defined as follows
*
*The first term is $1$.
*The next two terms are the next two even numbers $2$, $4$.
*The next three terms are the next three odd numbers $5$, $7$, $9$.
*The next $n$ terms are the next $n$ even numbers if $n$ is even or the next $n$ odd numbers if $n$ is odd.
What is the general term $a_n?$
Please, proofs of all these formulas would be nice
| The general term is
$$\tag1a_n=2n-\left\lceil \sqrt{2 n}-\frac12\right\rceil.$$
Why?
We have $a_{n+1}=a_{n}+2$ unless there is an integer $\ge\frac12 +\sqrt{2n}$ and $<\frac12+\sqrt{2(n+1)}$.
We need $a_{n+1}=a_n+1$ iff $n$ is one of the numbers $1, 3, 6, 10, \ldots$, i.e. a number of the form $k \choose 2$. This is accounted for by the ceiling/sqrt term.
Proof:
That $a_{n}=a_{n-1}+1$ iff $n-1=\frac{k(k+1)}2$ because of the well-known summation $1+2+3+\cdots+k=\frac{k(k+1)}2$ should be clear.
For which $\nu$ does there exist a $k$ such that $\nu-1=\frac{k(k+1)}2$? This is a quadratic in $k$ with solutions
$$\tag2k_{1,2}=\frac{-1\pm\sqrt{1+8(\nu-1)}}2=\frac{-1\pm\sqrt{8\nu-7}}2.$$
Thus the $\nu$ with $a_{\nu}=a_{\nu-1}+1$ are characterized by the fact that $8\nu-7$ is a square (of an odd number).
In total we have $a_n = 2n-m_n$ where $m_n$ is the number of $\nu\le n$ for which $(2)$ has a positive integer solution $k$. But since there is one $\nu$ (namely $\frac{k(k+1)}2$) for each $k=1, 2, \ldots$, we find that
$m_n=\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor$.
This proves a different formula, but it can be simplified to $(1)$ by observing that $\sqrt {2n}-\frac12$ is never an integer (that would make $2n$ the square of an odd integer). Therefore $\lfloor\sqrt{2n}+\frac12\rfloor$ may be used instead of $\lceil\sqrt{2n}-\frac12\rceil$ in $(1)$ (or just round $\sqrt{2n}$ to nearest integer). We have
$$\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor\le\left\lfloor \sqrt{2 n}-\frac12\right\rfloor $$
and inequality can only hold if there is a natural number $r$ such that
$$\frac{-1+\sqrt{8n-7}}2<s<\sqrt{2 n}-\frac12$$
i.e.
$$\sqrt{8n-7}<2s+1<2\sqrt{2 n}$$
$$8n-7<(2s+1)^2<8 n.$$
The latter is impossible because odd squares are $\equiv 1\pmod 8$.
Therefore, rounding $\left\lfloor\frac{-1+\sqrt{8n-7}}2\right\rfloor$ may be replaced with e.g. $\left\lfloor \sqrt{2 n}-\frac12\right\rfloor$ or (with adjustment of the complete formula by a constant) with $\left\lceil \sqrt{2 n}-\frac12\right\rceil$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac 1{x+1}<\ln\left(\frac{x+1}x\right)<0.5\left(\frac 1x+\frac 1{x+1}\right)$ I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this.
$$\frac 1{x+1}<\ln\left(\frac{x+1}x\right)<0.5\left(\frac 1x+\frac 1{x+1}\right)$$
| For LHS:use this well konw inequality
$$\dfrac{x}{x+1}\le\ln{(1+x)},x>-1\Longrightarrow \ln{\left(\dfrac{x+1}{x}\right)}=\ln{\left(1+\dfrac{1}{x}\right)}\ge\dfrac{\dfrac{1}{x}}{1+\dfrac{1}{x}}=\dfrac{1}{x+1}$$
For RHS: we only prove
$$2\ln{(1+x)}<x+\dfrac{x}{x+1}=\dfrac{x^2+2x}{x+1}$$
let
$$g(x)=2\ln{(1+x)}-\dfrac{x^2+2x}{x+1}\Longrightarrow g'(x)=\dfrac{2}{x+1}-1-\dfrac{1}{(x+1)^2}=\dfrac{-x^2}{(x+1)^2}\le 0$$
so
$$g(x)\le g(0)=0$$
so
$$\ln{(1+x)}\le\dfrac{1}{2}(x+\dfrac{x}{x+1})$$
so
$$\ln{\left(\dfrac{x+1}{x}\right)}\le 0.5\left(\dfrac{1}{x}+\dfrac{1}{x+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Power Series: Ratio test, but $x$ cancels out?!? The question is:
Determine the interval of convergence of the power series
$$\sum_{n=1}^\infty\left(\frac{2n+1}{n^2+1}\right)(2x+1)^{12}$$
My attempt at an answer:
$$u_n=\frac{(2n+1)}{(n^2+1)}(2x+1)^{12}$$
Applying the ratio test:
$$\begin{align}
\require{enclose}
\frac{|u_{n+1}|}{|u_n|}&=\left|\frac{(2n+2)(2x+1)^{12}}{((n+1)^2+1)}.\frac{(n^2+1)}{(2n+1)(2x+1)^{12}}\right|\\
&=\left|\frac{(2n+2)\enclose{horizontalstrike}{(2x+1)^{12}}}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)\enclose{horizontalstrike}{(2x+1)^{12}}}\right|\\
&=\left|\frac{(2n+2)}{(n^2+2n+1)}.\frac{(n^2+1)}{(2n+1)}\right|\\
\end{align}$$
But now I just got rid of all the $x$ components which is obviously wrong!?!.
$_{help!!!}$
| If the series really is $$\sum_{n=1}^\infty\left(\frac{2n+1}{n^2+1}\right)(2x+1)^{12}\;,$$ the loss of $x$ is obviously right: this is simply
$$(2x+1)^{12}\sum_{n=1}^\infty\frac{2n+1}{n^2+1}\;,$$
which is either $a(2x+1)^{12}$ for $$a=\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$$ if that series converges, or undefined if it does not. And in fact it’s undefined, since
$$\sum_{n=1}^\infty\frac{2n+1}{n^2+1}$$ is readily seen to diverge, e.g., by limit comparison with the harmonic series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\lim_{n \to +\infty} \sqrt{n}\int_0^\pi{\cos(\frac{t}{2})^n}dt>0$ I want to prove
$$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}dt>0.$$
First, I consider $$\lim_{n \to +\infty}\sqrt{n}\int_0^\pi{\cos\left(\frac{t}{2}\right)^n}\sin\left(\frac{t}{2}\right)dt,$$ which is smaller than what I want, but the second integral leads to $$\lim_{n \to +\infty}\frac{\sqrt{n}}{2(n+1)}.$$ So it does not work.
| By a change of variables $t\mapsto 2x$, you want to show $$\mathop {\lim }\limits_{n \to + \infty } 2\sqrt n \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx > 0.$$
Now, you need to evaluate $$I\left( n \right) = \int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx$$
For $n\geq 2$ we can reduce this integral with integration by parts
$$\eqalign{
& {\cos ^{n - 1}}x = u \cr
& \cos xdx = dv \cr} $$
then $$\eqalign{
& - \left( {n - 1} \right){\cos ^{n - 2}}x\sin xdx = du \cr
& \sin x = v \cr} $$
thus $$\int_0^{\frac{\pi }{2}} {{{\cos }^n}x} dx = \int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx} $$ for the other term vanishes. But $$\displaylines{
\int_0^{\frac{\pi }{2}} {\left( {n - 1} \right){{\cos }^{n - 2}}x{{\sin }^2}xdx} = \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x\left( {1 - {{\cos }^2}x} \right)dx} \cr
= \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}xdx} - \left( {n - 1} \right)\int_0^{\frac{\pi }{2}} {{{\cos }^{n - 2}}x{{\cos }^2}xdx} \cr
= \left( {n - 1} \right)I\left( {n - 2} \right) - \left( {n - 1} \right)I\left( n \right) \cr} $$
which means $$I\left( n \right) = \left( {\frac{{n - 1}}{n}} \right)I\left( {n - 2} \right)$$
If $n=2m$ is even, or if $n=2m+1$ is odd we have
$$\displaylines{
I\left( n \right) = \frac{{2m - 1}}{{2m}}\frac{{2m - 3}}{{2m - 2}} \cdots \frac{3}{4}\frac{1}{2}I\left( 0 \right) \cr
I\left( n \right) = \frac{{2m}}{{2m + 1}}\frac{{2m - 2}}{{2m - 1}} \cdots \frac{4}{5}\frac{2}{3}I\left( 1 \right) \cr} $$
for we reduce the number by $2$ each time, so if we start with an odd number, we'll be left with $1$.
So finally since
$$\displaylines{
I\left( 0 \right) = \frac{\pi }{2} \cr
I\left( 1 \right) = 1 \cr} $$
we get for $n$ even $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}\frac{\pi }{2}$$ and for $n$ odd $$I\left( n \right) = \frac{{\left( {n - 1} \right)!!}}{{n!!}}$$
Thus we have two sequences to consider
$$\displaylines{
{a_{2m}} = \sqrt {2m} \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}\frac{\pi }{2} \cr
{a_{2m + 1}} = \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr} $$
But Wallis' approximation says $$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{2n} = \frac{\pi }{2}$$
so that
$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} = \sqrt \pi $$
All things considered $$\displaylines{
\mathop {\lim }\limits_{m \to \infty } {a_{2m + 1}} = \mathop {\lim }\limits_{m \to \infty } \sqrt {2m + 1} \frac{{\left( {2m} \right)!!}}{{\left( {2m + 1} \right)!!}} \cr
= \mathop {\lim }\limits_{m \to \infty } \frac{{\sqrt {2m + 1} }}{{2m + 1}}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr
= \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt {2m + 1} }}\frac{{\sqrt m }}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} \cr
= \mathop {\lim }\limits_{m \to \infty } \sqrt {\frac{m}{{2m + 1}}} \mathop {\lim }\limits_{m \to \infty } \frac{1}{{\sqrt m }}\frac{{\left( {2m} \right)!!}}{{\left( {2m - 1} \right)!!}} = \frac{1}{{\sqrt 2 }}\sqrt \pi = \sqrt\frac\pi2 \cr} $$
$$\mathop {\lim }\limits_{m \to \infty } {a_{2m}} = \frac{\pi }{{\sqrt 2 }}\mathop {\lim }\limits_{m \to \infty } \sqrt m \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}} = \frac{\pi }{{\sqrt 2 }}\frac1{\sqrt\pi} = \sqrt\frac\pi2 $$
Hence, the limit exists and is positive, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/231526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Encode the message $[1,1,0,1,1,0,1]$ in BCH code based on the field $\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1}$ So here's what I understand so far:
$\mathbb F = \frac{\mathbb Z_{2}[x]}{x^4+x+1} = GF(16)$
The code is written as $[x^{14},x^{13},x^{12},x^{11},x^{10},x^{9},x^{8}$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
So we write the message $[1,1,0,1,1,0,1] = x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ and we get the first part of the code:
$[1,1,0,1,1,0,1$ $|$ $x^{7},x^{6},x^{5},x^{4},x^{3},x^{2},x^{1},x^{0}]$
We divide $x^{14}+x^{13}+x^{11}+x^{10}+x^{8}$ by $(x^8+x^7+x^6+x^4+1)$ and we get $(x^6+x^4+x^2+x)$ with remainder $(x^7+x^5+x^4+x^2+x)$
Taking the coefficients of the remainder we get the encoded BCH code
$[1,1,0,1,1,0,1$ $|$ $1,0,1,1,0,1,1,0]$
My question is: Why do we divide by $(x^8+x^7+x^6+x^4+1)$? Where did that come from?:
| Yes, the question is somewhat incomplete (as pointed by Dilip Sarwate and Jyrki Lahtonen).
This is a 2-error correcting (primitive narrow-sense) BCH code. The generator polynomial is the smallest binary polynomial that has $\alpha, \alpha^2, \alpha^3, \alpha^4$ as its roots. This is precisely the LCM of the minimal polynomials of those four roots. Also, the minimal polynomial of $\alpha, \alpha^2, \alpha^4$ are the same polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Fibonacci conjecture: $(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} $. So this is the question I have
The Fibonacci sequence is a recurrence system given by $$F_1 = 1, \ F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \qquad (n = 1, 2, 3, \ldots).$$ This question concerns the following conjecture: $$(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} \qquad (n = 1, 2, 3, \ldots).$$ (a) Confirm that the conjecture is true when $n = 6$.
(b) Prove that the conjecture is true for all integers $n \geq 1$.
But when I go through the solution I don't understand one line (separated and marked).
(b) Let $n$ be any integer such that $n \geq 1$.
$$\begin{align}
\left( F_{n+5} \right)^2 - \left( F_n \right)^2
&= \left( F_{n+5} - F_n \right) \left( F_{n+5} + F_n \right) \\
&= \left( F_{n+4} + F_{n+3} - F_{n+2} + F_{n+1} \right) \left( F_{n+4} + F_{n+3} + F_{n+2} - F_{n+1} \right) \\
\\
&= \left( \color{red}{2} F_{n+3} + F_{n+1} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+1} \right) \tag{*} \\
\\
&= \left( 2 F_{n+3} - F_{n+3} - F_{n+2} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+3} + F_{n+2} \right) \\
&= \left( 3 F_{n+3} - F_{n+2} \right) \left( F_{n+3} + 3 F_{n+2} \right) \\
&= 3 \left( \left( F_{n+3} \right)^2 - \left( F_{n+2} \right)^2 \right) + 8 F_{n+2} F_{n+3}
\end{align}$$
Hence, the conjecture is true for all integers $n \geq 1$
In the marked line, where does the red $\color{red}{2}$ come from?
| He substituted $F_{n+3}+F_{n+2}$ for $F_{n+4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/234877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality with constraint I've been trying to prove the following inequality without success.
For $a,b,c \in \mathbb{R}$ such that $abc=1$, prove that:
$$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1} + \frac{1}{c^2+c+1} \geq 1$$
| Multiply both sides of the inequality by $(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)$ to get
$$
\begin{align*}
&(b^2 + b + 1)(c^2 + c + 1) + (a^2 + a + 1)(c^2 + c + 1) + (a^2 + a + 1)(b^2 + b + 1)\\
\geq &(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1).
\end{align*}
$$
Suffer through the algebra (or enlist Maple's help) to simplify this to
$$
a^2 + b^2 + c^2 - ab - ac - bc \geq 0.
$$
Note that during the simplification, we use the constraint $abc = 1$ several times.
Now use F'OlaYinka's observation: The inequality above is true if and only if
$$
2(a^2 + b^2 + c^2 - ab - ac - bc) \geq 0.
$$
The lefthand side of this expression can be factored as
$$
(a-b)^2 + (a-c)^2 + (b-c)^2.
$$
Since this is the sum of three positive terms, it is surely greater than or equal to zero, which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/235950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Visualization of Singular Value decomposition of a Symmetric Matrix The Singular Value Decomposition of a matrix A satisfies
$\mathbf A = \mathbf U \mathbf \Sigma \mathbf V^\top$
The visualization of it would look like
But when $\mathbf A$ is symmetric we can do:
$\begin{align*}
\mathbf A\mathbf A^\top&=(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf U\mathbf \Sigma\mathbf V^\top)^\top\\
\mathbf A\mathbf A^\top&=(\mathbf U\mathbf \Sigma\mathbf V^\top)(\mathbf V\mathbf \Sigma\mathbf U^\top)
\end{align*}$
and since $\mathbf V$ is an orthogonal matrix ($\mathbf V^\top \mathbf V=\mathbf I$), so we have:
$\mathbf A\mathbf A^\top=\mathbf U\mathbf \Sigma^2 \mathbf U^\top$
I have two questions:
*
*Is the above statement correct? when Matrix $\mathbf A$ is symmetric and we compute SVD we would get $\mathbf U\mathbf \Sigma^2 \mathbf U^\top$
*How would the decomposition looks like in a symmetric matrix? As we are getting the eigenvectors and squared eigenvalues in matrices $\mathbf U $ and $\mathbf \Sigma$
| Singular value decomposition
Start with a matrix with $m$ rows, $n$ columns, and rank $\rho$,
$$
\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}
$$
which has the singular value decomposition
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
where the color denotes $\color{blue}{range}$ spaces and $\color{red}{null}$ spaces. The dimensions of the domain matrices are
$$
%
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\in \mathbb{C}^{m\times \rho}, \quad
%
\color{red}{\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\in \mathbb{C}^{m \times m - \rho}, \quad
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\in \mathbb{C}^{n\times \rho}, \quad
%
\color{red}{\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\in \mathbb{C}^{n\times n - \rho}.
$$
The domain matrices are unitary:
$$
\begin{align}
\mathbf{U}\mathbf{U}^{*} &= \mathbf{U}^{*}\mathbf{U} = \mathbf{I}_{m} \\
\mathbf{V}\mathbf{V}^{*} &= \mathbf{V}^{*}\mathbf{V} = \mathbf{I}_{n}
\end{align}
$$
The dimensions of the singular value matrices are
$$
%
\Sigma \in \mathbb{R}^{m\times n}, \quad
%
\mathbf{S}
\in \mathbb{R}^{\rho\times \rho}.
$$
The hermitian conjugate is constructed according to
$$
\mathbf{A}^{*} =
\mathbf{V} \, \Sigma^{\mathrm{T}} \, \mathbf{U}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
where $\Sigma^{\mathrm{T}}\in \mathbb{R}^{n\times m}$.
The Moore-Penrose pseudoinverse is constructed according to
$$
\mathbf{A}^{\dagger} =
\mathbf{V} \, \Sigma^{\dagger} \, \mathbf{U}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
where $\Sigma^{\dagger}\in \mathbb{R}^{n\times m}$.
The product matrix rules you stated always hold:
$$
\begin{align}
%
\mathbf{A} \mathbf{A}^{*} &=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)^{*}
%
=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{V} \, \mathbf{\Sigma}^{\mathrm{T}} \, \mathbf{V}^{*}
\right) \\
%
\mathbf{A}^{*} \mathbf{A} &=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)^{*}
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
=
%
\left(
\mathbf{V} \, \mathbf{\Sigma}^{\mathrm{T}} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right) \\
%
\end{align}
$$
Examples follow.
Square, full rank $m = n = \rho$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \mathbf{S} \right]
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &= \color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &= \color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*}
%
\end{align}
$$
Tall, full column rank $n = \rho$, $m \ge n$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{S} \\ \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &= \color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &=
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
\end{align}
$$
Wide, full row rank $m = \rho$, $n \ge m$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &=
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &=
%
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} \,
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} \\
%
\end{align}
$$
For the hermitian matrix,
$$
\begin{align}
\mathbf{A} &= \mathbf{A}^{*} \\
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} &=
\mathbf{V} \, \Sigma \, \mathbf{U}^{*}
\end{align}
$$
because in this case $\Sigma = \Sigma^{\mathrm{T}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/243811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Best way to discuss a system of linear equations that depends on two parameters. I know there is no specific way, but what is, in your opinion, the best way to discuss a system of linear equations like:
$$
\left\{
\begin{array}{l}
ax+by+z = 1\\
x+aby+z=b\\
x+by+bz=1
\end{array}
\right.
$$
$$(A|B) = \left(\begin{array}{ccc|c}
a & b & 1 & 1\\
1 & ab & 1 & b\\
1 & b & b & 1
\end{array}
\right)
$$
If $A$ is the coefficient matrix, $\det A = b(a-1)(-2+b+a b)$, so it is a little bit hard to discuss the system using determinants, right?
| If $b=0$ the system leads to also $a=2$ and then the solution is $x=1,z=1$ with $y$ arbitrary.
If $a=1$ it leads to also $b=1$ and in this case the system becomes the single relation $x+y+z=1$.
Now if we assume that $b$ is not zero (since the case $b=0$ already dealt with), but that $-2+b+ab=0$, then we have $a=(2-b)/b$ [OK since $b$ nonzero]. In this case the system is inconsistent unless it happens that $b^2+b-2=0$, i.e. $b=1,-2$. The case $b=1$ leads again to simply $x+y+z=1$, while the case $b=-2$ has also $a=-2$ and the solution may be written $x=-z$, $y=-1/2-1/2z$, and $z$ arbitrary.
In the remaining case where the determinant is nonzero, Cramer's rule can be used. It leads to
$$x = \frac{ (a-b)(b-1)}{(a-1)(ab+b-2)},$$
$$y=\frac{(ab+a-2)(b-1)}{(a-1)(ab+b-2)b},$$
$$z=\frac{a-b}{ab+b-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What are $2222^{5555}+5555^{2222} \pmod 7$ and $9^{2n+1}+8^{n+2} \pmod{73}$? Tell me hint for solve :
1) $ 2222^{5555}+5555^{2222} \equiv \mathord? \pmod 7$
2) $ 9^{2n+1}+8^{n+2} \equiv \mathord ?\pmod{73}$
thank you.
| For problem 1) You can show by elementary division that $$2222 \equiv 3 \mod 7$$
Now $$3^6 \equiv 1 \mod 7$$
And therefore (all mod 7)
$$2222^{5555} \equiv 3^{5555} \equiv 3^{6\times 925+5} \equiv 3^5 \mod 7 $$
Now since inverse of 3 is 5 in mod 7, the above becomes $3^6.3^{-1}\equiv 5 \mod 7$
Similarly you can show $5555 \equiv 4 \mod 7$. Now $4^3 \equiv 1\mod7$. So
$$5555^{2222} \equiv 4^{2222} \equiv 4^{3\times 740+2} \equiv 4^2 \equiv 2 \mod 7$$
Therefore the sum will be $0 \mod 7$. This means that it is divisible by 7.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to calcualate how many unique set of 6 can i have in a given set. Hello my question is quite simple i would think but i just cant seem to find an answer. I have a set of $\{1,2,3,4,5,6,7,8,9,10\}$ and i would like to calculate how many unique given sets of $6$ can i get from this set.
In other words for the number $1$
i would end up with
$[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]$
I would move down the line with the number $2$ to compare to unique sets of $6$
note:
when moving to two I would no longer do this
$[2,1,3,4,5,6]$ because it repeats my first case above.
its there a formula to figure this sort of thing?
Thanks in advance.
when I work this out on paper i end up with 15 sets
here is how
for 1
[1,2,3,4,5,6]
[1,3,4,5,6,7]
[1,4,5,6,7,8]
[1,5,6,7,8,9]
[1,6,7,8,9,10]
for 2
[2,3,4,5,6,7]
[2,4,5,6,7,8]
[2,5,6,7,8,9]
[2,6,7,8,9,10]
for 3
[3,4,5,6,7,8]
[3,4,6,7,8,9]
[3,5,6,7,8,9,10]
for 4 [4,5,6,7,8,9]
[4,6,7,8,9,10]
for 5 [5,6,7,8,9,10]
after that i cant make any more groups of $6$ thus i end up with $15$ sets.
| Binomial coefficients count the number of distinct subsets of $k$ elements from a set containing $n$ elements. The notation for this is $\binom{n}{k}$ which is equal to $\frac{n!}{k!(n-k)!}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How come $2 \times 3^k + 3^k = 3 \times 3^k$ I get something else
$$\begin{align*}
&2 \times 3^k + 3^k=\\
&2 \times 3^k \times 2=\\
&4 \times 3^k
\end{align*}$$
What does $3^k + 3^k$ give exactly?
| $2\times 3^k+3^k=(2+1)\times3^k=3\times3^k$
| {
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} |
How to integrate $(x^{1/2} + 5 x^{1/3})^{-1}$? $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13} $$
Let $$u = 5+x^\frac16 $$
I have change x into u, and find $$\int_5^6 \frac{7}{6u(u-5)^3} $$
Then, i don't know how to integrate it.
| Putting $x=(t-5)^6, dx=6(t-5)^5$ and $x=0\implies t=5,x=1\implies t=6$
We get $$ \int_0^1 \frac{dx}{x^\frac12+5x^\frac13}=\int_5^6 \frac{6(t-5)^3dt}{t} =6\int_5^6 (t^2-15t+75-125\frac1t)dt=6(\frac{t^3}3-15\frac{t^2}2+75t-125\log t)\mid_5^6$$
$$=2(6^3-5^3)-45(6^2-5^2)+450(6-5)-750\log\frac65$$
| {
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"source": "stackexchange",
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} |
Solve${{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$ Please help me to evaluate the integral:
$\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$
Thanks.
| Notice that:
$$x^8 = (x^6 + 1)x^2 - x^2$$
So that your integrand takes the form:
$$\int\frac{x^2}{1+x^6}dx - \int\frac{x^2}{(1+x^6)^2}dx$$
Now substitute $u = x^3, du = 3 x^2 dx$:
$$\int\frac{1}{3(1+u^2)}du - \int\frac{1}{3(1+u^2)^2}du$$
The first integral is a multiple of $\arctan(u)$, and the second can be solved dividing again into two parts:
$$\int\frac{1}{3(1+u^2)^2}du = \int\frac{1}{3(1+u^2)}du - \int\frac{u^2}{3(1+u^2)^2}du$$
I think you get the idea..
| {
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No natural numbers sucht that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$. Prove that there are not exist the natural numbers, $x,y,z$ such that $x+3y+5z=2012$ and $x^2+y^2+3z^2=2013$.
tell me please if my proof is ok ?
$x^2+y^2=3\cdot671-3z^2=3(671-z^2)=3k$ but if we want we can find out $z$. $671 \geq z^2$ so $z=\overline{0\ldots25}.$
But we know that a perfect square has the following form: $4k$ or $4k+1$. For example $x^2=4p+1$ and $y^2=4q$ so $x^2+y^2=4(p+q)+1\neq3k$ for $k\neq 3$ and $p+q \neq 2$.
I cannot find anothers numbers such that $4(p+q)+1=3k$
Is OK?
thanks :)
| You are fine to $x^2+y^2=3(671-z^2)=3k$, but I don't know what you mean by $z=\overline{0\ldots25}$. You are correct that perfect squares are $4k$ or $4k+1$ (but be careful about reusing $k$. You might confuse yourself thinking they are the same). This means $x^2+y^2 \equiv 0,1,2 \pmod 4$ but that doesn't say anything about modulo $3$. In fact, $x^2+y^2$ can be anything $\pmod 3$ If you want to combine modulos $3$ and $4$ you can work modulo $12$.
| {
"language": "en",
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"source": "stackexchange",
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Second order partial differential equation Second order partial differential equation:
$$(x-y)^2/4 u_{xx} + (x-y) \sin(x^2+y^2) u_{xy} + \cos(x^2+y^2) u_{yy} +\dots=0$$ is
a. Elliptic in $\{(x,y): x≠y, x^2+y^2<\frac{\pi}{6}\}$
b. Hyperbolic in $\{(x,y): x≠y, \frac{\pi}{4}<x^2+y^2<\frac{3\pi}{4}\}$
c. Elliptic in $\{(x,y): x≠y, \frac{\pi}{4}<x^2+y^2<\frac{3\pi}{4}\}$
d. Hyperbolic in $\{(x,y): x≠y, x^2+y^2<\frac{\pi}{4}\}$
I am stuck on this problem . Can anyone help me please......
| compute the expression $S^2-4RT$ . here $S=(x-y)\sin(x^2+y^2)$ , $R=(x-y)^2/4$ , $T=\cos^2(x^2+y^2)$. after calculating we have
$S^2-4RT=(x-y)^2\cos 2(x^2+y^2)$
so the option ($3$) is correct
| {
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how to compute the matrix polynomial Let $A$ be the matrix $$A = \begin{pmatrix} 1 & \sqrt{2}\\ -\sqrt{2} & -1\\ \end{pmatrix}$$
Compute the matrix $B = 3A -2A^2 - A^3 -5A^4 + A^6$.
Could any one give me any hint for this one? I have calculated the eigenvalues they are $(1+\sqrt{2}i),(1-\sqrt{2}i)$
| We have the characteristic polynomial $ch_A(x)=\begin{vmatrix}1-x&\sqrt{2}\\-\sqrt{2}&1-x\end{vmatrix}=(x-1)^2+2=x^2-2x+3$.
Then by Cayley-Hamilton theorem we know that $A^2-2A+3=0$.
So if you divide your polynomial $p(x)$ by $(x^2-2x+3)$ and you get $p(x)=q(x)(x^2-2x+3)+r(x)$, you only need to calculate $r(A)$, i.e., to plug the matrix $A$ into the remainder.
| {
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Do there exist any odd prime powers that can be represented as $n^4+4^n$? Well, I wrote up a solution on it, but according to the place I found the problem, it isn't quite correct. Ah, I'm simply hoping someone will point out where I got wrong.
Now, let, $n^4+4^n = p^k$, where $p$ is an odd prime and $k$ is a positive integer.
Further, $p^k \equiv 1 \pmod 2$. Therefore, $n^4 + 4^n \equiv n^4 \equiv 1 \pmod 2$, and so $n \equiv 1 \pmod 2$. $n$ must be odd.
Okay, now let, $n = 2m +1$. Substituting it in $n^4+4^n$ and using the Sophie Germain inequality, we have,
$$n^4+4\cdot4^{2m} = n^4 + 4(2^m)^4 = (n^2 + 2^n+2^{m+1}\,n)(n^2 + 2^n-2^{m+1}\,n) = p^k$$
Now, as $p^k$ can only be factorized into smaller powers of $p$, let $n^2 + 2^n+2^{m+1}\,n = p^i$, and let $n^2 + 2^n-2^{m+1}\,n = p^j$ where $i+j= k$, obviously, and $i>j$.
Now consider this:
$$\begin{align}
p^i - p^j & \equiv 0\\
2\cdot2^{m+1}\,n = 2^{m+2}\,n &\equiv 0 \pmod p
\end{align}$$
But, as $p$ is odd, $\gcd(p, 2) = 1$, so $n \equiv 0 \pmod p$.
Now look at this:
$$\begin{align}
p^i + p^j &\equiv 0 \\
2(n^2 + 2^n) &\equiv 0 \\
n^2 + 2^n &\equiv 0 \pmod p
\end{align}$$
But we just established that $n \equiv 0 \pmod p$, so $2^n \equiv 0 \pmod p$.
Therefore, let $2^n = jp$ for some integer $j$.
Now, $2^n$ is its own prime factorization, which is unique according to the Fundamental Theorem of Arithmetic and does not include $p$.
Therefore, the above statement is an impossibility!
There exist no such $p$ and $n$, and no odd prime powers can be written as $n^4+4^n$.
Ah, well, that's it.
Sorry for the tediousness of it. I've still no clue how to use $\LaTeX$.
Thank you everybody,
Cheers.
| There is the small counterexample
$$5^1=1^4+4^1$$
| {
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incircle and circumcircle Given a triangle $\Delta ABC$ with circumcenter $O$, and incenter $I$.
If $∠ABC = 45°$, $OI = d$, and $c − b = d√2$,
Find the value of $\sin A$.
I've thought of using the fact that $d^2 = R(R-2r)$ where $R$ is the circumradius and $r$ is the inradius. But I didn't know where to apply it.
| Seat your belt, it will be great...
You can use your $d^2 = R(R-2r)$, which gives you
$(c-b)^2 = 2d^2 = 2R(R-2r) = 2R^2-4rR$
Then, you remark that the area of the triangle can be obtained with $r$ or with $R$ :
$\displaystyle\text{Area} = \frac{(a+b+c)r}{2}$ and $\displaystyle\text{Area} = \frac{ab\sin{C}}{2} = \frac{abc}{4R}$
So you have $\displaystyle\frac{(a+b+c)r}{2} = \frac{abc}{4R}$, thus $\displaystyle2rR=\frac{abc}{a+b+c}$
Putting it in the first equality gives :
$\displaystyle c^2+b^2-2bc = 2R^2-2\frac{abc}{a+b+c}$
But $\displaystyle \frac{b}{\sin{B}} = 2R$ and $\displaystyle\sin{B} = \frac{\sqrt{2}}{2}$, so $b=\sqrt{2}R$ and
$\displaystyle c^2+b^2-2bc = b^2-2\frac{abc}{a+b+c}$
$\displaystyle\Leftrightarrow c^2 - 2bc = -2\frac{abc}{a+b+c}$
$\displaystyle\Leftrightarrow c - 2b = -2\frac{ab}{a+b+c}$
$\displaystyle\Leftrightarrow (c - 2b)(a+b+c) = -2ab$
$\displaystyle\Leftrightarrow ac+bc+c^2-2ab-2b^2-2bc = -2ab$
$\displaystyle\Leftrightarrow ac+c^2-2b^2-bc = 0$
Now, we can use $a = 2R\sin A$, $b = 2R\sin B$, and $c = 2R\sin C$ to
$\displaystyle \sin A \sin C+\sin^2 C-2\sin^2 B-\sin B \sin C = 0$
$\displaystyle\Leftrightarrow \sin A \sin C+\sin^2 C-1-\frac{\sqrt{2}}{2} \sin C = 0$
We also know that
$\sin C = \sin(180°-45°-A) = \sin 135° \cos A - \sin A \cos 135° = \frac{\sqrt{2}}{2}(\cos A + \sin A)$
Thus,
$\displaystyle \sin A \frac{\sqrt{2}}{2}(\cos A + \sin A)+\frac{1}{2}(\cos A + \sin A)^2-1- \frac{1}{2}(\cos A + \sin A) = 0$
$\displaystyle\Leftrightarrow \sqrt{2} \sin A \cos A + \sqrt{2}\sin^2 A+\cos^2 A + \sin^2 A + 2\sin A \cos A-2- \cos A - \sin A = 0$
$\displaystyle\Leftrightarrow (2+\sqrt{2}) \sin A \cos A + \sqrt{2}\sin^2 A -1 - \cos A - \sin A = 0$
To only have $\sin$, we can isolate $\cos A$ and square terms...
$\sqrt{2}\sin^2 A - \sin A - 1 = \cos A (1 - (2+\sqrt{2}) \sin A)$
$\Rightarrow (\sqrt{2}\sin^2A - \sin A - 1)^2 = \cos^2 A (1 - (2+\sqrt{2}) \sin A))^2$
Let $x = \sin A$ :
$(\sqrt{2}x^2-x-1)^2 = (1-x^2)(1 - (2+\sqrt{2}) x)^2$
$\Leftrightarrow 2x^4+x^2+1-2\sqrt{2}x^3-2\sqrt{2}x^2+2x = (1-x^2)(1 - (4+2\sqrt{2})x + (6+4\sqrt{2})x^2)$
$\Leftrightarrow 2x^4-2\sqrt{2}x^3+(1-2\sqrt{2})x^2+2x+1 = -(6+4\sqrt{2}) x^4 + (4+2\sqrt{2})x^3+(5+4\sqrt{2})x^2 - (4+2\sqrt{2})x + 1$
$\Leftrightarrow (8+4\sqrt{2})x^4-(4+4\sqrt{2})x^3-(4+6\sqrt{2})x^2+(6+2\sqrt{2})x = 0$
A solution is $x = 0$ which is impossible so it gives us :
$(4+2\sqrt{2})x^3-(2+2\sqrt{2})x^2-(2+3\sqrt{2})x+(3+\sqrt{2}) = 0$
Aaaand... I don't want to find the solution, especially since I've probably make a mistake somewhere :D. There surely is something shorter but perhaps It could help you to find a nice solution!
| {
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Showing that: $(\frac{a}{b+c})^2+(\frac{b}{a+c})^2+(\frac{c}{a+b})^2+\frac{10abc}{(a+b)(b+c)(c+a)}\ge 2$ Let a;b;c>0. Prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{a+c}\right)^2+\left(\frac{c}{a+b}\right)^2+\frac{10abc}{(a+b)(b+c)(c+a)}\geq 2$$
I think
$$\frac{2a}{b+c}=x;\frac{2b}{c+a}=y;\frac{2c}{a+b}=z$$
We have: $xy+yz+zx+xyz=4$
$$(\frac{x}{2})^2+(\frac{y}{2})^2+(\frac{z}{2})^2+\frac{10xyz}{8} \ge 2$$
$\Leftrightarrow x^2+y^2+z^2+5xyz \ge 8$
$\Leftrightarrow x^2+y^2+z^2-5(xy+yz+zx) +12 \ge 0$
deadlock
Can you help me? Thank you very much
| I collected a solution:
Need to prove: $x^2+y^2+z^2+5xyz \ge 8$
Put: $x+y+z=p;xy+yz+zx=q;xyz=r$
We have: $q+r=4$
Need to prove inequality is equivalent to:
$p^2-2q+5r \ge 8 \Leftrightarrow p^2-7q+12 \ge 0$
*)If: $p \le 4$. Applying Schur's inequality; we have:
$r \ge \dfrac{p(4q-p^2)}{9} \Rightarrow 4 \ge q+\dfrac{p(4q-p^2)}{9} \Leftrightarrow q \le \dfrac{p^3+36}{4p+9}$
WE will prove: $p^2-\dfrac{7(p^3+36)}{4p+9}+12 \ge 0 \Leftrightarrow (p-3)(p^2-16) \ge 0$ TRUE
Because: $4 \ge p \ge \sqrt{3q} \ge 3$
*) If: $p \ge 4 $. We have:
$p^2-2q+5r \ge p^2-2q \ge \dfrac{p^2}{8} \ge 5$
Equality occurs if and only if: $(x;y;z) \in [(1;1;1);(2;2;0);(0;2;2);(2;0;2)]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Range of a function $f(x)$
As x varies over all real numbers, the range of the function $$f(x)=\frac{x^2-3x+4}{x^2+3x+4}$$ is (1) $[\frac{1}{7},7]$, (2) $[-\frac{1}{7},7]$, (3) $[-7,7]$ (4) $(-\infty,\frac{1}{7})\bigcup(7,\infty)$.
Trial:$$\begin{align} \frac{x^2-3x+4}{x^2+3x+4} &=\frac{(x-\frac{3}{2})^2+\frac{7}{4}}{(x+\frac{3}{2})^2+\frac{7}{4}} >0 \end{align}$$ So,(1) will be the right answer. But how I show that $\frac{1}{7} \leq f(x) \leq 7$. Please help.
Note: Originally it was stated that
$$f(x)=\frac{x^2-3x+1}{x^2+3x+1},$$
and this is what the initial answers were based on.
| Let $\dfrac{x^2-3x+1}{x^2+3x+1} = y$. We then have $$(y-1)x^2 + 3(y+1)x + (y-1) = 0$$
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$9(y+1)^2 - 4 (y-1)^2 \geq 0 \implies 5y^2 +26y+5 \geq 0 \implies (5y+1)(y+5) \geq 0$$
This gives us $y \geq -\dfrac15$ and $y \leq -5$. Hence, the right answer is
$$\left(-\infty,-5\right] \cup \left[ -\dfrac15, \infty \right)$$
EDIT
If the constant term is $4$ i.e. if $\dfrac{x^2-3x+4}{x^2+3x+4} = y$. We then have $$(y-1)x^2 + 3(y+1)x + 4(y-1) = 0$$
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$9(y+1)^2 - 16 (y-1)^2 \geq 0 \implies -7y^2 +50y-7 \geq 0 \implies 7y^2 - 50y +7 \leq 0 \implies (7y-1)(y-7) \leq 0$$
This gives us $y \in \left[\dfrac17,7 \right]$. Hence, the right answer is
$$\left[\dfrac17,7\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of $(x_n)$ defined by $x_{n+1}=c_nu(x_n)$ Find $\displaystyle \lim_{n\rightarrow \infty }x_n$ :
$$\left\{\begin{matrix}x_1=a>0\\ \\ x_{n+1}=\frac{2x_n\cdot \cos\left(\frac{\pi}{2^n+1}\right)}{x_n+1}\end{matrix}\right.$$
I have tried that :
Let $a_n=\dfrac{1}{x_n}$ . So :
$$a_{n+1}=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}.a_n+\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
So I tried to have geometric series:
By let :
$$a_{n+1}+f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}(a_n+f(n))$$
So we must find one $f(n)$:
$$\frac{f(n)}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}-f(n+1)=\frac{1}{2\cos\left(\frac{\pi}{2^{n+1}}\right)}$$
As $$f(n)-f(n+1)\cdot 2\cos\frac{\pi}{2^{n+1}}=1$$
Can you give me the way to find one $f(n)$ sastisfied that ; or anyone has nice way to solve this problem
|
$$\lim\limits_{n\to\infty}x_n=1$$
To show this, note that: first, $x_n\gt0$ for every $n\geqslant1$; second, $x_n\leqslant x^0_n$ for every $n\geqslant1$, where $x^0_1=a$ and
$$
x^0_{n+1}=\frac{2x^0_n}{x^0_n+1},
$$
for every $n\geqslant1$; third, $x^0_n\to1$ when $n\to\infty$. Thus, $\limsup\limits_{n\to\infty}x_n\leqslant1$.
On the other hand, $\cos\left(\frac\pi{2^n+1}\right)\to1$ when $n\to\infty$. For every positive $b\lt1$, choose some positive integer $n(b)$ such that $\cos\left(\frac\pi{2^n+1}\right)\geqslant b$ for every $n\geqslant n(b)$ and define $(x^b_n)_{n\geqslant n(b)}$ by $x^b_{n(b)}=x_{n(b)}$ and
$$
x^b_{n+1}=\frac{2bx^b_n}{x^b_n+1},
$$
for every $n\geqslant n(b)$. Then, $x_n\geqslant x^b_n$ for every $n\geqslant n(b)$ and, provided $b\gt\frac12$, $x^b_n\to x^b=2b-1$ when $n\to\infty$.
Hence $\liminf\limits_{n\to\infty}x_n\geqslant x^b$. Since $x^b\to1$ when $b\to1$, this proves the result.
The same method shows that every sequence $(x_n)_{n\geqslant1}$ defined by $x_1=a\gt0$ and
$$
x_{n+1}=\frac{c_nx_n}{x_n+1},
$$
for every $n\geqslant1$, with $c_n\gt0$ for every $n\geqslant1$ and $\lim\limits_{n\to\infty}c_n=c$ with $c\geqslant1$, converges to $c-1$. If $0\leqslant c\leqslant1$, $\lim\limits_{n\to\infty}x_n=0$.
| {
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$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$
Find $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$
I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
| The limit is zero, because $|x| \sin{\frac{1}{3 \sqrt{x}}} \rightarrow 0$ as $x \rightarrow 0$. (The denominator is nonzero in this limit.).
| {
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Direct proof that $(5/p)=1$ if $p\equiv 1\pmod{5}$. I'm trying to show without use of quadratic reciprocity that $(5/p)=1$ if $p\equiv 1\pmod{5}$. If $p\equiv 1\pmod{5}$, then there exists some $x\in U(\mathbb{Z}/p\mathbb{Z})$ with order $5$.
I note that modulo $p$, $(x+x^4)^2+(x+x^4)-1=0$, since
$$
(x+x^4)^2+(x+x^4)-1=x^2+2x^5+x^8+x+x^4-1=x^4+x^3+x^2+x+1=0.
$$
Investigating this relation was the hint in my book, (Ireland and Rosen, question 15 on page 63), but I don't see how it helps. I squared $(x+x^4)^2=1-(x+x^4)$, and got
$$
(x+x^4)^4=2-3(x+x^4)
$$
and
$$
(x+x^4)^8=13-21(x+x^4)
$$
so I think I'm going off the track.
| Multiply your relation by $4$. We get
$$4(x+x^4)^2+4(x+x^4)-4\equiv 0\pmod{p}.\tag{$1$}$$
Complete the square: $(1)$ is equivalent to $\left(2(x+x^4)+1\right)^2\equiv 5\pmod{p}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\epsilon$-$\delta$ proof that $f(x) = x \sin(1/x)$, $x \ne 0$, is continuous I'm doing an exercise that asks me to prove that $f$ is continuous using a $\epsilon$-$\delta$ proof. I have that
$$
f(x) = \begin{cases}
x\cdot \sin \frac1x,&x\neq 0
\\
0,&x = 0
\end{cases}
$$
I've already managed to show this property for $x=0$. How can I show it for $x \ne 0$, also using a $\epsilon$-$\delta$ proof?
Thank you very much.
| Although the answer given by @Nameless is complete in itself, a slight change in the choice of added terms in the triangle inequality helps a lot. A modified version of his answer is as follows.
Consider
\begin{align*}
\big|f(x)-f(a)\big|&=\left|x\sin\frac{1}{x}-a\sin\frac{1}{a}\right|\\
&=\left|x\sin\frac{1}{x}-x\sin\frac{1}{a}+x\sin\frac{1}{a}-a\sin\frac{1}{a}\right|\\
&=\left|x\left(\sin\frac{1}{x}-\sin\frac{1}{a}\right)+\sin\frac{1}{a}(x-a)\right|\\
&\leq|x|\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|+\left|\sin\frac{1}{a}\right|\big|x-a\big|
\end{align*}
By the trigonometric identity $\sin\alpha-\sin\beta=2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$, we have
\begin{align*}
\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|&=\left|2\sin\frac{\frac{1}{x}-\frac{1}{a}}{2}\cos\frac{\frac{1}{x}-\frac{1}{a}}{2}\right|=2\left|\sin\frac{a-x}{2ax}\cos\frac{x+a}{2ax}\right|\\
&\leq2\left|\sin\frac{a-x}{2ax}\right|\leq2\left|\frac{x-a}{2ax}\right|=\frac{|x-a|}{|a||x|}
\end{align*}
Therefore,
\begin{align*}
\big|f(x)-f(a)\big|&=|x|\left|\sin\frac{1}{x}-\sin\frac{1}{a}\right|+\left|\sin\frac{1}{a}\right|\big|x-a\big|\\
&\leq\frac{|x-a|}{|a|}+\frac{|x-a|}{|a|}=\frac{2|x-a|}{|a|}
\end{align*}
Thus we see that for any $\varepsilon>0$, we can find $\delta=\frac{|a|\varepsilon}{2}$ such that
$$\big|f(x)-f(a)\big|<\varepsilon\qquad\text{whenever}\qquad|x-a|<\delta$$
showing that $f$ is continuous at all points $a\neq0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$
where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums.
Can someone provide a nice proof that
$$A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2?$$
I worked for a while on this today but was unsuccessful. Summation by parts, swapping the order of summation, and approximating $H_k$ by $\log k$ were my best ideas, but I could not get any of them to work. (Perhaps someone else can?) I would like a nice proof in order to complete my answer here.
Bonus points for proving $A(1,2) = \frac{5}{8} \zeta(3)$ and $A(2,1) = \zeta(3) - \frac{1}{2}\zeta(2) \log 2$, as those are the other two alternating Euler sums needed to complete my answer.
Added: I'm going to change the accepted answer to robjohn's $A(1,1)$ calculation as a proxy for the three answers he gave here. Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying. I am pleased to see that it can be used to do the $A(1,1)$, $A(1,2)$, and $A(2,1)$ derivations.
| Interestingly, $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}H_{n}^{-}}{n} = \frac{\zeta(2)}{2} {\color{red}{+}} \frac{\log^{2} (2)}{2}$$ where $H_{n}^{-}$ are the alternating harmonic numbers defined as $$H_{n}^{-} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} .$$
One way to show this is to notice that $$ \begin{align} \log (2) - H_{n}^{-} &= \sum_{k=n+1}^{\infty} \frac{(-1)^{k-1}}{k} \\ &= (-1)^{n}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n} \\ &= (-1)^{n} \sum_{k=1}^{\infty} (-1)^{k-1} \int_{0}^{1} x^{k+n-1} \ dx \\ &= (-1)^{n} \int_{0}^{1} x^{n}\sum_{k=1}^{\infty}(-1)^{k-1} x^{k-1} \ dx \\ &= (-1)^{n} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx . \end{align}$$
Thus an integral representation of the alternating harmonic numbers is $$ H_{n}^{-} = \log (2) + (-1)^{n-1} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx .$$
The integral on the right can be evaluated in terms of the digamma function, and you'll get a closed-form expression for the alternating harmonic numbers.
But getting back to evaluating that sum,
$$ \begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}H_{n}^{-}}{n} &= \log(2) \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} + \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{n}}{1+x} \ dx \\ &= \log^{2} (2) + \int_{0}^{1} \frac{1}{1+x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \ dx \\ &= \log^{2} (2) - \int_{0}^{1} \frac{\log (1-x)}{1+x} \ dx \\ &=\log^{2} 2 - \int_{1/2}^{1} \frac{\log \big(1-(2t-1) \big)}{2t} \ 2 \ dt \\ &= \log^{2}(2) - \int_{1/2}^{1} \frac{\log \big(2(1-t) \big)}{t} \ dt \\ &= \log^{2}(2) - \int_{1/2}^{1} \frac{\log 2}{t} \ dt - \int_{1/2}^{1} \frac{\log (1-t)}{t} \ dt \\ &= \log^{2}(2) - \log^{2}(2) + \text{Li}_{2}(1) - \text{Li}_{2} \left( \frac{1}{2}\right) \\ &= \zeta(2) - \frac{\zeta(2)}{2} + \frac{\log^{2} (2)}{2} \\ &= \frac{\zeta (2)}{2} + \frac{\log^{2} (2)}{2} . \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "139",
"answer_count": 11,
"answer_id": 8
} |
Decomposition of polynomial into irreducible polynomials I'm preparing to my algebra exam. And I have problem and I have no idea how to solve it.
Given polynomial
$$x^4+4x^3+4x^2+1.$$
The task is find expansion of the polynomial as a product of irreducible polynomials in $\mathbb{R}$.
I will be happy if you show me the way how to solve such problems
| It suffices to show that $f(x)=x^4+4x^3+4x^2+1$ has no linear factors over $\mathbb{Z}_3$: $f(0)=f(1)=1$ and $f(2)=2$, so $f(x)$ has no linear factors. Then $f(x)$ must factor to two quadratic polynomials: $$f(x)=(ax^2+bx+c)(ux^2+vx+w)$$ We then have that $au=1$. Multiplying the first polynomial by $u$ and the second by $a$, we may assume that $a=u=1$. Equating the coefficients of the powers of $x$, we have
$$
\begin{eqnarray*}
4&=&v+b\\
4&=&w+c+bv\\
0&=&bw+cv\\
1&=&cw
\end{eqnarray*}
$$
Some algebra shows that $$\begin{eqnarray*}
b&=& 2+\sqrt{2 \left(1+\sqrt{2}\right)}\\c&=& 1+\sqrt{2}+\sqrt{2 \left(1+\sqrt{2}\right)} \\ v&=& 2-\sqrt{2 \left(1+\sqrt{2}\right)}\\ w&=& 1+\sqrt{2}-\sqrt{2 \left(1+\sqrt{2}\right)}
\end{eqnarray*}
$$
So, letting $\alpha=1+\sqrt{2}$, we see that $x^4+4x^3+4x^2+1$ factors to $$\left(x^2+\left(2+\sqrt{2\alpha}\right)x+\alpha+\sqrt{2\alpha}\right)\left(x^2+\left(2-\sqrt{2\alpha}\right)x+\alpha-\sqrt{2\alpha}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/276362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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} |
Calculate $\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $ How can I calculate the following integral :
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx $$
I try to write the integral like :
$$\int\frac{1+x-x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{x}{(x+1)\sqrt{1+x^2}}=\int\frac{1}{\sqrt{1+x^2}}-\int\frac{(\sqrt{x^2+1})'}{(x+1)}$$
but still nothing .
thanks :)
| Substitute $x=\frac{1-t}{1+t}$
$$\int\frac{1}{(x+1)\sqrt{1+x^2}} dx =-\int\frac1{\sqrt{2(1+t^2)}}dt=-\frac1{\sqrt2}\sinh^{-1}t
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/279526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Pick out the true statements. Let $a, b$ and $c$ be the lengths of the sides of an arbitrary triangle. Define
$$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}.$$
Pick out the true statements.
(a) $1/2 ≤ x ≤ 2$.
(b) $1/2 ≤ x ≤ 1$.
(c) $1/2 < x ≤ 1$.
How can I able to solve this problem
| Triangle inequality $$\Rightarrow\left | a-b \right |<c \Rightarrow \sum_{a,b,c}(a-b)^{2}<\sum_{a,b,c}c^{2}\Rightarrow 2\sum_{a,b,c}a^{2}-2\sum_{a,b,c}ab<\sum_{a,b,c}a^{2}\Rightarrow \frac{1}{2}<\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}=x$$ $x\in \mathbb{R}\Rightarrow x^{2}\geq 0$
$$\Rightarrow(a-b)^2\geq 0\Rightarrow \sum_{a,b,c}(a-b)^2\geq 0\Rightarrow 2\sum_{a,b,c}a^2-2\sum_{a,b,c}ab\geq 0\Rightarrow x=\frac{\sum\limits_{a,b,c}ab}{\sum\limits_{a,b,c}a^{2}}\leq 1$$
Answer:
The true statements are:
(a) $1/2 ≤ x ≤ 2$
(b) $1/2 ≤ x ≤ 1$
(c) $1/2 < x ≤ 1$
(c) $1/2 < x ≤ 1$ gives the tightest bounds, but since the other two intervals are supersets of this interval, (a) and (b) are also satisfied. To further clarify:
$\forall\Delta ABC$, with sides a,b,c
$x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Rightarrow 1/2 < x ≤ 1\Rightarrow
1/2 ≤ x ≤ 1\Rightarrow 1/2 ≤ x ≤ 2$
Also, $\forall x\in (\frac{1}{2},1],\exists \Delta ABC$ with sides $ a,b,c \ni x =\frac{ab + bc + ca}{a^2 + b^2 + c^2}$ since condition (c) gives the tightest bound.
However, $\exists x\in[\frac{1}{2},1],\forall\Delta ABC$ with sides $a,b,c ;\frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x$ and hence $\exists x\in[\frac{1}{2},2],\forall\Delta ABC$ with sides $a,b,c ; \frac{ab + bc + ca}{a^2 + b^2 + c^2}\not=x.$ Thus the implication is one sided for conditions (a) and (b).
$\therefore x=\frac{ab + bc + ca}{a^2 + b^2 + c^2}\Leftrightarrow1/2 < x ≤ 1 $but $1/2 ≤ x ≤ 1\vee1/2 ≤ x ≤ 2\not\Rightarrow \frac{ab + bc + ca}{a^2 + b^2 + c^2}=x $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Integral of $\frac{x^3-2}{x^3-x}$ As the title says, I need some help with this function:
$$ \int\frac{x^3-2}{x^3-x}$$
I tried it with different versions/forms of the function to get it with substitution or partiel integration:
$$ \int\frac{x^3-2}{x^3-x} = \frac{x^3-2}{x(x-1)(x+1)} = \frac{x^3}{x^3-x} - \frac{2}{x^3-x}$$
Do you have some advice?
| try this
$$\frac{x^3-2}{x^3-x} = 1 + \frac{x-2}{x^3-x}$$
Now use partial fraction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $x,y,z$ if $3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7$. How can I solve the following system with $x,y,z$ real numbers:
$$x^2+y^2+z^2=6$$
$$3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7.$$
I observe that $x=y=z=\sqrt{2}$ and I feel it must apply the inequality $AM \geq GM$ but I don't know how to do.
Thanks :)
| Applying $AM\geq GM$ to your second equation you get
$$3^6\geq ({3^{x^4+y^2+y^4+z^2+z^4+x^2}})^{1/3}=(3^{x^4+y^4+z^4})^{1/3}\cdot(3^{x^2+y^2+z^2})^{1/3}=(3^{x^4+y^4+z^4})^{1/3}\cdot 3^2$$
and therefore $$3^4\geq 3^{\frac{x^4+y^4+z^4}{3}}$$
Now we know that $\sqrt{\frac{x^4+y^4+z^4}{3}}\geq \frac{x^2+y^2+z^2}{3}=2$ where the equality holds if and only if $x^2=y^2=z^2$. This implies that $$3^4\geq 3^{\frac{x^4+y^4+z^4}{3}}\geq 3^4.$$
Then we must have $\sqrt{\frac{x^4+y^4+z^4}{3}}=\frac{x^2+y^2+z^2}{3}$ and therefore $x^2=y^2=z^2=2$. Then this is your unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/280820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving an inequality for positive numbers $a, b, c$
Let be $a,b,c$ positive numbers such that $a+b+c=3$. Prove that
$$\frac{b+c+bc}{a^2+b^3+c^4}+\frac{c+a+ca}{b^2+c^3+a^4}+\frac{a+b+ab}{c^2+a^3+b^4} \le 3$$
| By Holder
$$\sum_{cyc}\frac{a+b+ab}{c^2+a^3+b^4}=\sum_{cyc}\frac{9(c^2+a+1)(a+b+ab)}{(c^2+a+1)(c^2+a^3+b^4)(1+1+1)(1+1+1)}\leq$$
$$\leq\frac{9\sum\limits_{cyc}(c^2+a+1)(a+b+ab)}{(a+b+c)^4}=\frac{1}{9}\sum_{cyc}(a^2b+a^2c+a^2bc+a^2+ab+a^2b+2+ab)=$$
$$=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+a^2bc+a^2+2ab+2)=\frac{1}{9}\sum_{cyc}(2a^2b+a^2c+abc+5)=$$
$$=\frac{2(a^2b+b^2c+c^2a+abc)+(a^2c+b^2a+c^2b+abc)+15}{9}\leq\frac{2\cdot4+4+15}{9}=3.$$
We used the following lemma.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$a^2b+b^2c+c^2a+abc\leq4$$
A proof of the lemma.
Let $\{a,b,c\}=\{x,y,z\}$ such that $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integral via partial fractions I have to find the integral of this rational function:
$$\frac{x^5}{(x-1)^2(x^2-1)}$$
Since the power in the numerator is bigger than the power in the denominator, I have to divide the first one by the second one. So I divide it and I have $(x+2)$ and the quotient $(4x^3-2x^2-3x-2)$. Now what I do is write:
$$x^5 = x+2 + \frac{4x^3-2x^2-3x-2}{(x-1)^2 (x^2-1)}$$
Now, when I find the integral of the fraction, I have something like $$x^5 = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{x+1}.$$
When I try to find $A ,B,C$ and $D$, I can only find $D=1/8$ and not $A, B$ or $C$. How do I find $A$, $B$, and $C$?
| Assuming you have done your algebra correctly, you have arrived at $${4x^3-2x^2-3x-2\over(x-1)^3(x+1)}={A\over x-1}+{B\over(x-1)^2}+{C\over(x-1)^3}+{D\over x+1}$$ and you want to find $A,B,C,D$. Clear fractions to get $$4x^3-2x^2-3x-2=A(x-1)^2(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^3$$ Now you have some options. You can multiply out everything on the right side, gather like powers of $x$, and get $4$ equations for your $4$ unknowns by comparing the coefficients of $x^3$, of $x^2$, od $x$, and of $1$ on both sides of the equation.
Alternatively, if you stick in $x=1$ you immediately get the values of $C$; if you stick in $x=-1$ you immediately get the value of $D$. Then you can stick in two more values of $x$, say, $x=0$ and $x=2$, to get two equations relating $A$ and $B$; or you can differentiate and then stick in $x=1$ to get $B$, and look at the coefficient of $x^3$ to get $A$; and so on.
If you are going to use this website regularly, you ought to learn something about formatting mathematics here, and also something about English grammar and punctuation.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| This will have derivatives. Substitute $a=\sqrt{x}, b=\sqrt{y}$ and $c=\sqrt{z}$. Then $x+y+z=3$. Consider the function $f(x)=\ln(x\sqrt{x}+\sqrt{x}+1)$. It is concave.
Hence Jensen yields: $f(x)+f(y)+f(z)<=3f(1)=3\ln{3}$
This is equivalent to what is asked.
| {
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"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
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"answer_id": 2
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Inequality on the side lengths of a triangle: $\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1$. This problem is taken from the Kosovo Mathematical Olympiad for Grade-$ 10 $ students.
Let $ a $, $ b $ and $ c $ be the lengths of the edges of a given triangle.
How can one prove the following inequality?
$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1.
$$
| You can write the question as
$|\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}|<1$. This means you would like to show that
$-1<\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}<1$. Since $a,b,c$ are the sides of a triangle, you have that $a,b,c>0$. This means you want to show $-abc<a^2c+b^2a+c^2b-b^2c-c^2a-a^2b<abc$. We write the inside term more compactly, we are trying to show $-abc<ac(a-c)+ab(b-a)+bc(c-b)<abc.$ Because the expression we are trying to show is cyclic, without loss of generality, we assume $a\geq b\geq c >0$. Since they are the sides of a triangle, you have that $a-b<c$, $b-c<a$ and $a-c<b$ (sum of two sides exceeds the third one). Therefore, this gives you $ac(a-c)<abc$. Also $a\geq b\geq c>0$ give you $ab(b-a)\leq 0$ and $bc(c-b)\leq 0$. Adding the three inequalities gives $ac(a-c)+ab(b-a)+bc(c-b)<abc$.
Now since $b-c<a$ you have that $bc(b-c)<abc$. Call this (1). Since $b+c > a$ and since $b-c\geq 0$, it follows that $(b+c)(b-c)\geq a(b-c)$, which means $b^2-c^2-ab+ac\geq 0$ implying $b(a-b)+c(c-a)\leq 0$. This means that $ab(a-b)+ac(c-a)\leq 0$. Call this (2).
Adding (1) and (2) side-by-side gives $ab(a-b)+ac(c-a)+bc(b-c)< abc$ which is the same as $-abc<ab(b-a)+ac(a-c)+bc(c-b)$, which is what we wanted to show.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determinant of a block matrix with $\mathrm{Id}$ and $0$ in the diagonal How to compute the determinant $\det A$ depending on $B$ and $C$, where
$$ A = \left(\begin{matrix}\mathrm{Id} & B \\ C & 0 \end{matrix} \right), $$
a) when $C$ is square,
b) $C$ has more rows than columns
| Hints:
*
*note that $$\begin{pmatrix} \text{Id} & B \\
C & 0 \end{pmatrix} = \begin{pmatrix} \text{Id} & 0 \\
C & \text{Id} \end{pmatrix} \cdot\begin{pmatrix} \text{Id} & B \\
0 & - C\cdot B\end{pmatrix}.$$
*furthermore, in general $$\det\begin{pmatrix}A& 0\\ C& D\end{pmatrix} = \det\begin{pmatrix}A& B\\ 0& D\end{pmatrix} = \det(A) \det(D)$$ and $\det (A\cdot B) = \det (A) \det(B)$ for square matrices $A$ and $B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/286074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $a_{n} < 2$ if $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}$ Let $(a_{n})_{n \geq1}$ be a real sequence such that $a_{1}=a_{2}=1$ and $\displaystyle a_{n+2}=a_{n+1}+\frac{a_{n}}{3^n}, n\geq 1$.
Prove that $a_{n} < 2, \forall n \geq 1.$
I write $$\sum a_{k+2}-a_{k+1}=\sum \frac{a_{k}}{3}$$
and I obtained :
$$3a_{n+2}=a_{1}+\ldots+a_{n}+3$$
or
$$a_{n}=\frac{a_{1}+\ldots+a_{n-2}+3}{3} < 2$$
And what remains to prove it is :
$$a_{1}+\ldots +a_{n-2} < 3,$$ but from this point I don't know how I have to do.
I need a proof without derivatives.
Thanks :)
| Claim:
$$a_n < 2-\frac{1}{3^n} \,.$$
$P(1), P(2)$ are easy to check.
Inductive step:
$$a_{n+1}= a_{n+1}+\frac{a_n}{3^n} \leq 2-\frac{1}{3^{n+1}}- \frac{2-\frac{1}{3^n}}{3^n}=2-\frac{1}{3^{n+1}}- \frac{2}{3^n}+\frac{1}{9^n} $$
If we can prove taht
$$2-\frac{1}{3^{n+1}}- \frac{2}{3^n}+\frac{1}{9^n} < 2-\frac{1}{3^{n+2}}$$ we are done.
But this is equivalent to
$$\frac{1}{3^{n+2}}+\frac{1}{9^n} <\frac{1}{3^{n+1}}+ \frac{2}{3^n}$$
which is obvious.
P.S. This is a pretty standard but not well known technique. If $a_n$ is increasing, then $a_n \leq C$ cannot be proven directly by induction, but one might be able to find a decreasing $b_n \geq 0$, and then prove by induction the stronger claim
$$a_n < C-b_n \,.$$
The standard well known example of this phenomena is
$$1+\frac{1}{2^2}+..+\frac{1}{n^2} <2 $$
vs
$$1+\frac{1}{2^2}+..+\frac{1}{n^2} <2 -\frac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove this product: $\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$ How to prove this product?
$$\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$$
| HINT: It never hurts to gather some data by doing some actual computation:
$$\begin{array}{c|l}
n&\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}\\ \hline
2&\frac73\\
3&\frac{\color{red}7}3\cdot\frac{13}{\color{red}7}\\
4&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{21}{\color{blue}{13}}\\
5&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{\color{green}{21}}{\color{blue}{13}}\cdot\frac{31}{\color{green}{21}}
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/287880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ Prove $|\frac{1}{n} - \frac{1}{x}| \le \frac{1}{n^2}$ for all $n \le x \le n+1$, using the mean value theorem applied to $f(x) = \frac{1}{x}$
Immediately, I can recognize some components of the mean value theorem. $\frac{1}{n^2}$ likely comes from the slope of $f$ at $n$, and the $|\frac{1}{n} - \frac{1}{x}|$ expression likely comes from the secant expression $\frac{\frac{1}{n} - \frac{1}{x}}{x}$. But I cannot figure out how they fit together in the end.
| If $x = n$, then the left hand side is $0$ and the result holds. So let's assume that $n < x \leq n+1$. Then applying the mean-value theorem to the function $f(t) \colon= 1/t$ for $t \in [n, x]$, we find that there is a real number $c \in (n,x)$ such that
$$ f(x)-f(n) = f\prime(c) \cdot (x-n) $$
or
$$\frac{1}{x}-\frac{1}{n} = - \frac{1}{c^2} \cdot (x-n). $$
So
$$ |\frac{1}{n}-\frac{1}{x}| = \frac{1}{n}-\frac{1}{x} = \frac{1}{c^2} \cdot (x-n) \leq \frac{1}{c^2} < \frac{1}{n^2}. $$
We of course assume that $n$ is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ I have to find $\displaystyle\lim_{x \rightarrow 0} \frac{e^{\sin x} - e^x}{\sin^3 2x}$ using Taylor polynomials.
Here's what I've done so far:
*
*$e^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$
*$\sin x = x - \frac{1}{6} x^3 + o(x^4)$
*$e^{\sin x} = 1 + x - \frac{1}{6} x^3 + o(x^4)$
*$\sin 2x = 2x + o(x^2)$
*$\sin^3 2x = 8x^3 + o(x^6)$
Therefore I can rewrite my limit as: $\displaystyle\lim_{x \rightarrow 0} \frac{-\frac{1}{3} x^3 - \frac{1}{2} x^2 + o(x^3)}{8x^3 + o(x^6)}$
In this form, the limit appers to be $-\frac{1}{24}$, but the correct result is $-\frac{1}{48}$.
Could you tell me what I'm doing wrong?
| $\mathbb{e}^x = 1 + x + \frac{1}{2} x^2 + \frac{1}{6} x^3 + o(x^3)$ and $\sin x=x-\frac{1}{6}x^3+o(x^3)$
$\Rightarrow \mathbb{e}^{\sin x}=1+x-\frac{1}{6}x^3+\frac{1}{2}x^2+\frac{1}{6}x^3+o(x^3)=1+x+\frac{1}{2}x^2+o(x^3)$
$\Rightarrow \mathbb{e}^{\sin x}-\mathbb{e}^x=-\frac{1}{6}x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/288893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum of $\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$ I would like to know the minimum value of $$\left| \sin x- 1\right| + \left|\sin x- 2\right| + \left| \sin x -3\right| + \left| \sin x+1\right|$$
for $x \in \mathbb{R}$.
| For every $x \in \mathbb{R}$ we have
$$
-1 \le \sin x\le 1 \quad \forall\ x \in \mathbb{R},
$$
and therefore:
\begin{eqnarray}
f(x)&:=&|\sin x-1|+|\sin x-2|+|\sin x-3|+|\sin x+1|\\
&=&(1-\sin x)+(2-\sin x)+(3-\sin x)+(\sin x+1)\\
&=&5+2(1-\sin x).
\end{eqnarray}
The minimum is therefore equal to $5=f(\pi/2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/289185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral Anyone remember the method for this? I think this should been done on the site
$$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$$
| I think @kiwi ment
$$
\fbox {$I$} = \int_0^\infty \frac {\ln x}{x^2+a^2} dx = \left | u = \frac {a^2}x \Longrightarrow\left\{\begin{array}{c}
\ln x = 2 \ln a - \ln u \\
dx = -\frac {a^2du}{u^2}
\end{array}\right\} \right | = -\int_\infty^0 \frac{2\ln a - \ln u}{\frac {a^4}{u^2}+a^2}\frac {a^2 du}{u^2} = \\
2\int_0^\infty \frac{\ln a}{u^2+a^2}du-\int_0^\infty \frac{\ln u}{u^2+a^2}du = \fbox{$2\int_0^\infty \frac{\ln a}{u^2+a^2}du - I$}
$$
From last part it's clear that
$$
I = \int_0^\infty \frac{\ln a}{u^2+a^2}du
$$
This integral can be easily found
$$
I = \ln a\int_0^\infty \frac {du}{u^2+a^2} = \frac {\ln a}a \ \left.\mbox{atan}\ \frac ua \right|_0^\infty = \frac {\pi \ln a}{2a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
Help solving summation series of a recursive function Yesterday in class, we were analyzing the Karatsuba multiplication algorithm and how it applies to recurrence equations. Time ran short, and I feel I missed how to solve the final summation.
First, we defined the recurrence equation as
$$T(n) = 3T \left(\frac{n}{2}\right) + 4n$$
and applied a recurrence tree such like
$$T(n) = 3T \left(\frac{n}{2}\right) + 4n \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^0$$
$$T\left(\frac{n}{2}\right) = 3T \left (\frac{n}{4} \right) + 4\left(\frac{n}{2}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^1$$
$$T\left(\frac{n}{4}\right) = 3T \left (\frac{n}{8} \right) + 4\left(\frac{n}{4}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^2$$
$$T\left(\frac{n}{8}\right) = 3T \left (\frac{n}{16} \right) + 4\left(\frac{n}{8}\right) \Rightarrow 4n \cdot \left(\frac{3}{2}\right)^3$$
Because the denominiator increases in a logarithmic fashion, we defined the summation as
$$\sum_{x=0}^{log_2n} 4n \cdot \left(\frac{3}{2}\right)^x$$
Time was running short, so several steps were skipped, and the final solution was given as
$$9\cdot 3^{log_2n} = 9n^{log_23} = 9n^{1.58} = O(n^{1.58})$$
based on the properties
$$a^{lg\, b} = b^{lg\, a}\: \text{and}\: log_2 3 \approx 1.58$$
I've tried applying the summation formula
$$\sum_{x=0}^{n}r^x = \frac{r^{n+1}-1}{r-1}$$
with this result, and end up with
$$\sum_{x=0}^{n}r^x = \frac{r^{n+1}-1}{r-1} = \sum_{x=0}^{log_2 n} 4n \cdot \left(\frac{3}{2}\right)^x $$
$$= 4\left(n\cdot \frac{\frac{3}{2}^{lg_2n+1}-1}{\frac{3}{2}-1}\right) = 4\left(n \cdot \frac{\frac{3}{2}^{log_2n+1}-1}{\frac{1}{2}}\right)
= 2\left(n \cdot \frac{3}{2}^{log_2n+1}+1\right) $$
$$=2n \cdot 3^{log_2n+1} + 2$$
which is very different than the solution given. Where did I go wrong?
| This recurrence has the nice property that we can compute explicit values for $T(n)$ the same way as was done here, for example.
Let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$ be the binary digit representation of $n.$ It is not difficult to see that with $T(0)=0$ we have $$ T(n) = 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor} 3^j \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j} = 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left( \frac{3}{2} \right)^j \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Now for an upper bound consider $n$ consisting only of one digits, giving
$$ T(n) \le 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left( \frac{3}{2} \right)^j \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^k =
4 \sum_{j=0}^{\lfloor \log_2 n \rfloor} \left( \frac{3}{2} \right)^j
\left( 2^{\lfloor \log_2 n \rfloor + 1} - 2^j\right) $$
which is $$ 4 \left( 2^{\lfloor \log_2 n \rfloor + 1} \frac{(3/2)^{\lfloor \log_2 n \rfloor + 1}-1}{3/2-1} - \sum_{j=0}^{\lfloor \log_2 n \rfloor}3^j \right) =
4 \left(2 \left( 3^{\lfloor \log_2 n \rfloor + 1} - 2^{\lfloor \log_2 n \rfloor + 1}\right)
- \frac{3^{\lfloor \log_2 n \rfloor + 1}-1}{3-1} \right)
= 2\times 3^{\lfloor \log_2 n \rfloor + 2} - 2^{\lfloor \log_2 n \rfloor + 4} + 2.$$
For a lower bound, take all digits zero except the leading one, getting
$$ T(n) \ge 4 \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left( \frac{3}{2} \right)^j 2^{\lfloor \log_2 n \rfloor} =
2^{\lfloor \log_2 n \rfloor + 2} \frac{(3/2)^{\lfloor \log_2 n \rfloor + 1}-1}{3/2-1} =
4\times 3^{\lfloor \log_2 n \rfloor + 1} - 2^{\lfloor \log_2 n \rfloor + 3} .$$
The lower bound and the upper bound taken together show that
$$ T(n) \in \Theta\left(3^{\lfloor \log_2 n \rfloor}\right) =
\Theta\left(2^{\log_2 3 \lfloor \log_2 n \rfloor} \right) =
\Theta\left(n^{\log_2 3}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Comparing $\large 3^{3^{3^3}}$, googol, googolplex How to show that $\large 3^{3^{3^3}}$ is larger than a googol ($\large 10^{100}$) but smaller than googoplex ($\large 10^{10^{100}}$).
Thanks much in advance!!!
| $$3^{3^{3^3}} > 3^{300} > 10^{100}$$ since $$3^{3^3} > 3^7 = 3 \cdot (3^3)^2 > 3 \cdot 10^2 = 300$$ since $$3^3 > 7$$
$$3^{3^{3^3}} < 10^{3^{3^3}} < 10^{10^{100}}$$ since $$3^{3^3} < 3^{100} < 10^{100}$$ since $$3^3 < 100$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/291729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Uniqueness of solution for the functional equation $f(x)=\frac{x}{2}f\left(\frac{x}{2}\right)+h(x)$ Let $h \in \mathcal{C}:=C([-a,a])$, where $a>0$. Prove that there exists a unique function $f \in \mathcal{C}$ such that
$$
f(x)=\frac{x}{2}f\Big(\frac{x}{2}\Big)+h(x)\quad \forall x \in [-a,a].
$$
| If $f$ is a solution then for all $x \in [-a,a]$ we get by applying the functional equation $n$ times
\begin{align*}
f(x) &= \frac{x}{2} f\left(\frac{x}{2}\right) + h(x)\\
&= \frac{x^2}{2^{1+2}} f\left(\frac{x}{4}\right) + \frac{x}{2} h\left(\frac{x}{2}\right) \\
&= \frac{x^2}{2^{1+2+3}} f\left(\frac{x}{8}\right) + \frac{x^2}{2^{1+2}} h\left(\frac{x}{4}\right) + \frac{x}{2} h\left(\frac{x}{2}\right) + h(x)\\
&= \ldots\\
&= \frac{x^n}{2^{n(n-1)/2}} f\left(\frac{x}{2^n}\right) + \sum_{k=0}^{n-1} \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right)
\end{align*}
The first summand tends to zero for $n \to \infty$, as do the summands inside the second sum for $k \to \infty$. So we would like to define
$$f(x) = \sum_{k=0}^{\infty} \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right).$$
This $f$ satisfies the functional equation, we just have to show that the series converges and defines a continuous function on $[-a,a]$. If we can show that the series converges uniformly on $[-a,a]$ then the our limit function $f$ will be continuous as a uniform limit of continuous functions.
For large $k$, say $k \geq k_0$, we have $|h(x/2^k)| \leq |h(0)| + \varepsilon$, by continuity of $h$, so
$$\sum_{k \geq k_0} \left| \frac{x^k}{2^{k(k-1)/2}} h\left(\frac{x}{2^k}\right) \right|
\leq \left(|h(0)| + \varepsilon\right) \sum_{k \geq k_0} \frac{a^k}{2^{k(k-1)/2}}$$
and the right hand side converges independently of $x$.
For the uniqueness, let $f$ and $g$ be two solutions to the functional equation. Then $f-g$ satisfies the functional equation for $h \equiv 0$. Therefore we can assume $h \equiv 0$ and show that $f \equiv 0$ is the only solution. Applying the functional equation $n$ times as above shows
$$f(x) = \frac{x^n}{2^{n(n-1)/2}} f\left(\frac{x}{2^n}\right)$$
for all $n$. But $f(x/2^n)$ is bounded by continuity of $f$, so this converges to $0$ for $n \to \infty$. Thus $f \equiv 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/292481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Combinatorics. Please explain me how to do it. In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively? This isn't a homework question. Please explain me the steps. Thank you!
| Let $a,b,c$ be the amounts that $A,B$ and $C$ give resp. Then we want to count the number of distinct solutions to $a + b + c = 10$ under the condition that $a,b,c$ are positive integers and $a \le 6, b \le 7, c \le 8$.
One way to compute this is to compute the coefficient of $x^{10}$ in the expression $(1+x+\ldots +x^6)(1+x+\ldots +x^7)(1+x+\ldots +x^8)$ ($a$ is the exponent we choose in the first term, $b$ in the second etc., so the fact that we go up to $x^6$ in the first term expresses the $a \le 6$ and so on.)
We can write these terms as $\frac{1-x^7}{1-x}$, $\frac{1-x^8}{1-x}$ and $\frac{1-x^9}{1-x}$, respectively, so this product equals
$$(1-x^7)(1-x^8)(1-x^9)(1-x)^{-3}$$ and then we can use the general Newton formula to compute the coefficient of $x^{10}$ (we expand $(1-x)^{-3}$ using that, and then count the (not too many) ways the first terms give rise to a power of $x$ that is $\le 10$).
Expanded: the general binomial implies
$$(1-x)^{-3} = \sum_{n=0}^{\infty} \binom{k+2}{2} x^k$$
e.g. see here, and now note that we can form $x^{10}$ by picking 1's in the first 3 terms and the coefficient of $x^{10}$ in this expansion, so $\binom{12}{2}$, and also by picking $-x^7$,1,1 and $\binom{5}{3}$ (term for $x^3$), $1,-x^8,1$ and the $x^2$ term and finally $1,1,-x^9$ and the term for $x^1$ in the infinite expansion.
So we get $$\binom{12}{2} - \binom{5}{3} - \binom{4}{2} - \binom{3}{1} = 47$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/293610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry with multiple angle and exact value of $\tan\pi/5$ By considering the equation $\tan5\theta=0$, show that the exact value of $\tan\pi/5$ is $\sqrt{5-2\sqrt{5}}$.
Do I need to evaluate the multiple angle for $\tan5\theta=0$?
| Using Euler's Formula and Binomial Theorem, we get
$$
\begin{align}
\cos(5\theta)+i\sin(5\theta)
&=\left(\cos(\theta)+i\sin(\theta)\right)^5\\
&=\cos^5(\theta)+5i\cos^4(\theta)\sin(\theta)-10\cos^3(\theta)\sin^2(\theta)\\
&-10i\cos^2(\theta)\sin^3(\theta)+5\cos(\theta)\sin^4(\theta)+i\sin^5(\theta)\tag{1}
\end{align}
$$
Taking the ratio of the real and imaginary parts of $(1)$, we get
$$
\tan(5\theta)=\frac{5\tan(\theta)-10\tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta)+5\tan^4(\theta)}\tag{2}
$$
Thus, if $\tan(5\theta)=0$, but $\tan(\theta)\ne0$, $(2)$ says that
$$
5-10\tan^2(\theta)+\tan^4(\theta)=0\tag{3}
$$
The Quadratic Formula yields
$$
\tan^2(\theta)=5\pm2\sqrt{5}\tag{4}
$$
Therefore,
$$
\tan(\theta)=\pm\sqrt{5\pm2\sqrt{5}}\tag{5}
$$
Matching up the least positive values of $\theta$ for which $\tan(5\theta)=0$ yields
$$
\begin{align}
\tan\left(\frac\pi5\right)&=+\sqrt{5-2\sqrt{5}}\\
\tan\left(\frac{2\pi}5\right)&=+\sqrt{5+2\sqrt{5}}\\
\tan\left(\frac{3\pi}5\right)&=-\sqrt{5+2\sqrt{5}}\\
\tan\left(\frac{4\pi}5\right)&=-\sqrt{5-2\sqrt{5}}\\
\end{align}\tag{6}
$$
Note that these values support this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/294884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Induction on binomial Identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ I am having trouble proving the following identity:
$0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$
Here is what I have so far:
Proof:
Base: Let $n=0$:
LHS: $0\cdot {0\choose 0} = 0\cdot1 = 0$
RHS: $0\cdot 2^{0-2} = 0$
Step: Let $k\in \mathbb{Z} s.t k \geq 0$ and assume the identity is true for k.
Consider the LHS for $k+1$ where $k$ is even (I leave out the odd case because I think it will turn out the same?):
\begin{align}
=& 0\cdot{k+1\choose 0}+2\cdot{k+1\choose 2}+4\cdot{k+1\choose 4}+... +k\cdot{k+1\choose k}
\\=&0\cdot\left[{k\choose 0}+{k\choose -1}\right] + 2\cdot\left[{k\choose 2}+{k\choose 1}\right]+ 4\cdot\left[{k\choose 4}+{k\choose 3}\right]+\ldots+ k\cdot\left[{k\choose k}+{k\choose k-1}\right]
\\=&\left[0\cdot{k\choose 0}+2\cdot {k\choose 2}+4\cdot{k\choose 4}+\ldots+k\cdot{k\choose k}\right] + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\=& k\cdot2^{k-2} + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\
\end{align}
I know I need to end up with something like:
\begin{align}
=&k\cdot2^{k-2}+ \left[k\cdot2^{k-2} + 2^{k-1}\right]
\\=&2k\cdot 2^{k-2} + 2^{k-1}
\\=&k\cdot 2^{k-1}+2^{k-1}
\\=&(k+1)\cdot 2^{k-1}\end{align}
But, how can I get what I need from the combinations above? It may not end up exactly like that, but what is the reasoning behind this?
| Add the first two to get the third:
$$
\begin{align}
\sum_{k=0}^nk\binom{n}{k}&=\sum_{k=0}^nn\binom{n-1}{k-1}&=n(1+1)^{n-1}&=n2^{n-1}\\
\sum_{k=0}^n(-1)^kk\binom{n}{k}&=\sum_{k=0}^n(-1)^kn\binom{n-1}{k-1}&=n(1-1)^{n-1}&=0\\
\sum_{k=0}^{\lfloor n/2\rfloor}2k\binom{n}{2k}&&&=n2^{n-1}\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/295019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Determinant of block matrix Here is a determinant of a $(k+m) \times (k+m)$ Block matrix.
\begin{align}
D=\begin{vmatrix}
a_{11} &a_{12} & \cdots & a_{1k} &0 &\cdots &0 \\
a_{21}& a_{22}& \cdots & a_{2k} & 0 &\cdots &0 \\
\vdots& \vdots & & \vdots & \vdots & &\vdots\\
a_{k1} & a_{k2} & \cdots & a_{kk} & 0 &\cdots & 0\\
c_{11}& c_{12} & \cdots& c_{1k} & b_{11} & \cdots & b_{1m}\\
\vdots& \vdots & & \vdots & b_{21}&\cdots & b_{2m}\\
c_{m1}& c_{m2} & \cdots & c_{mk} & b_{m1}& \cdots & b_{mm}
\end{vmatrix}
\end{align}
If I have got a determinant $$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Then $D_1$ is equal to $(-1)^{k \times m}$$D$.
I know that the existence of the factor -1 is due to the interchange of 2 row, but i have a question on that $k \times m$.In my book,it said that i have to do $k\times m$ times row operations to transform $D_1$ into $D$.However,i thought only k times is needed for $D_1$ transform into $D$.
If i have done 1 times row operation for $D_1$
$$D_1=
\begin{vmatrix}
0 &\cdots &0&a_{11} &a_{12} & \cdots & a_{1k} \\
0 &\cdots &0 &a_{21}& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
0 &\cdots & 0&a_{k1} & a_{k2} & \cdots & a_{kk} \\
b_{11} & \cdots & b_{1m}& c_{11}& c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
b_{m1}& \cdots & b_{mm}&c_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
=\begin{vmatrix}
a_{11} &\cdots &0&0 &a_{12} & \cdots & a_{1k} \\
a_{21} &\cdots &0 &0& a_{22}& \cdots & a_{2k} \\
\vdots& & \vdots & \vdots & \vdots & &\vdots\\
a_{k1} &\cdots & 0& 0& a_{k2} & \cdots & a_{kk} \\
c_{11} & \cdots & b_{1m}&b_{11} & c_{12} & \cdots& c_{1k}\\
\vdots& & \vdots & \vdots & \vdots& & \vdots\\
c_{m1}& \cdots & b_{mm}&b_{m1}& c_{m2} & \cdots & c_{mk}
\end{vmatrix}
$$
Correct me if i have made any mistakes
| The discrepancy is that you have to interchange two adjacent rows. From this it is quite clear you have to perform $k\cdot m$ such operations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve for $ x$, $-\frac{1}{2}x^2 + 2x + 5 = 0$ I'm having trouble solving this equation for $x$:
$$-\frac{1}{2}x^2 + 2x + 5 = 0$$
What's the steps to take to solve it?
Thanks.
| $$x^2-4x-10=0$$
$$x^2-2\cdot x\cdot2+2^2-2^2-10=0$$
$$x^2-2\cdot x\cdot2+2^2=4+10$$
$$(x-2)^2=14$$
$$x-2=\pm \sqrt{14}$$
$$x=2\pm \sqrt{14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/297578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What are the three cube roots of $1-i$? I am trying to find the three cube roots of $1-i$ and I am unsure of how to proceed. My math teacher recommended I use de Moivre's formula, however I do not know how to set it up. How do you suggest I continue?
de Moivre's formula
$$(\cos x + i\sin x)^n = \cos(nx) + i\sin(nx)$$
| Another way without using DeMoivre's formula that works for cube roots is set
$x=1$ and $y=-i$ and then find any real roots of the following equation
$$\frac{(-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3)}{-64} = 0$$
This expands to
$$a^9 - \frac{3}{4}a^6 - \frac{21}{32}a^3 - \frac{1}{64} = 0$$
and has three real roots $a_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16}); $$a_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})$; $a_3=\frac{1}{2}(-\sqrt[6]{16})$.
Next solve the equations
$$a_1^3+3a_1b_1^2 = 1$$
$$a_2^3+3a_2b_2^2 = 1$$
$$a_3^3+3a_3b_3^2 = 1$$
(where $1$ is the value set as $x$) for $b_1$, $b_2$ and $b_3$, or $b_1=\pm\frac{1}{4}(\sqrt[6]{432}-\sqrt[6]{16})i$; $b_2=\pm\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i; $$b_3=\pm\frac{1}{2}(\sqrt[6]{16})i$ . So the three cube roots of $$\sqrt[3]{1-i}$$ are
$$a_1+b_1=\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})i$$
$$a_2+b_2=\frac{1}{4}(-\sqrt[6]{432}+\sqrt[6]{16})+\frac{1}{4}(\sqrt[6]{432}+\sqrt[6]{16})i$$
$$a_3+b_3=\frac{1}{2}(-\sqrt[6]{16})+\frac{1}{2}(-\sqrt[6]{16})i$$
Combinations of $a_1$, $a_2$ and $a_3$ and $b_1$, $b_2$ and $b_3$ will also give the three roots of $\sqrt[3]{1+i}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Finding limit of a function ($0$ appears in denominator) This is the function:
$$f(x)= \frac{x^3 + 3x^2+2x}{x^2-2x-8} $$
So I tried to find this limit:
$$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8}=\left[\frac{-8-12-4}{4+4-8}=\frac{-24}{0}\right]= \lim_{x\to -2}\frac{x(x-1)(x-2)}{(x-4)(x+2)}$$
This isn't a case where numerator and denominator, after factorization, have some elements which can reduce a "problematic" element... I need to get rid of $(x+2)$ which makes the fraction explode but I don't know how.
| You have factor wrong your function, just do it in the following way
$$\lim_{x\to -2}\frac{x^3 + 3x^2+2x}{x^2-2x-8} = \lim_{x \to -2} \frac{x(x+1)(x+2)}{(x-4)(x+2)} = \lim_{x \to -2} \frac{x(x+1)}{(x-4)} = \frac{-2(-2+1)}{-2-4} = -\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction Proof: $\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $ Prove by Mathematical Induction . . .
$$\sum_{i=1}^{n+1} i \cdot 2^i = n \cdot 2^{n+2}+2 $$
for all $n \geq 0$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$1\cdot 2^1 = 0\cdot 2^{0+2}+2$
$2 = 2$
b. Inductive Hypothesis
$$\sum_{i=1}^{k+1} i \cdot 2^i = k \cdot 2^{k+2} +2 $$
for $k \geq 0$
Prove k+1 is true.
$$\sum_{i=1}^{k+2} i \cdot 2^i = (k+1)\cdot 2^{k+3}+2 $$
$\big[RHS\big]$
$k\cdot 2^{k+3}+2^{k+3}+2$
$\big[LHS\big]$
$$\sum_{i=1}^{k+2} {i \cdot 2^{i}} $$
$= \underbrace{\sum_{i=1}^{k+1} i \cdot 2^i} + (k+2)\cdot 2^{k+2}$ (Explicit last step)
$= \underbrace{k\cdot 2^{k+2}+2}+(k+2)\cdot 2^{k+2}$ (Inductive Hypothesis Substitution)
$= k\cdot 2^{k+2}+2+k\cdot 2^{k+2}+2^{k+3}$
$= 2k\cdot 2^{k+2} + 2^{k+3} + 2$
My [LHS] has one too many $2k\cdot 2^{k+2}$ or did it just do it completely wrong?
| Attention: you replaced $\sum_{i=1}^{k+1}2^i=2+2\cdot 2^2+3\cdot 2^3+\ldots+(k+1)2^{k+1}$ by $(k+1)2^{k+1}$, and this is wrong for $k\geq 2$.
The formula holds for $n=0$ as you observed.
Now assume it holds for some $n\geq 0$, i.e.
$$
\sum_{i=1}^{n+1}i2^i=n2^{n+2}+2.
$$
Then
$$
\sum_{i=1}^{n+2}i2^i=\sum_{i=1}^{n+1}i2^i+(n+2)2^{n+2}
$$
(so, using the induction hypothesis)
$$
=n2^{n+2}+2+(n+2)2^{n+2}
$$
(so, simplifying)
$$
=2n2^{n+2}+2^{n+3}+2=n2^{n+3}+2^{n+3}+2=(n+1)2^{n+3}+2.
$$
So, by induction, the formula holds for all $n\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the exact value of $ \cos(36^\circ) $? The problem reads as follows:
Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of
$$\cos(36^\circ)$$
it says that you may find useful the following identities:
$$\cos^2 t+\sin^2 t = 1,\\
\sin 2t = 2\sin t\cos t,\\
\sin 3t = 3\sin t - 4\sin^3 t.
$$
Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?
Thanks in advance :)
| Here is a way using roots of unity.
We have $\cos 36 = \frac{\omega + \omega^{-1}}{2}$ where $\omega = \text{exp} \left ( \frac{2 i\pi}{10} \right )$.
We have $-w$ is a primitive $5^{th}$ root of unity, since $ -\exp(\frac{2i\pi}{10}) = \exp(i\pi)\exp(\frac{2i\pi}{10}) = \exp(\frac{12i\pi}{10}) = \exp(\frac{2ki\pi}{5})$, so it follows $\omega^4 - \omega^3 + \omega^2 - \omega + 1 = 0$.
Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}) = x^2\left ( \left (x + \frac{1}{x} \right )^2 - \left (x + \frac{1}{x} \right ) -1 \right )$.
Thus $0 = \omega^2 \cdot ((2 \cos 36)^2 - (2 \cos 36) - 1)$.
Therefore $4 \cos^3 36 - 2 \cos 36 - 1 = 0$.
Using the quadratic formula we then arrive at $$\cos 36 = \frac{1 + \sqrt{5}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Hint:
$(a_1 + a_2 + \dots + a_n) \mod 8 = [(a_1 \mod 8) + (a_2 \mod 8) + \dots + (a_n \mod 8)] \mod 8$
How may terms are there in the sequence? Take the $n^{th}$ term to be $2n - 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/304041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Max. distance of Normal to ellipse from origin How Can I calculate Maximum Distance of Center of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ from the Normal.
My Try :: Let $P(a\cos \theta,b\sin \theta)$ be any point on the ellipse. Then equation of Normal at that point is
$ax\sec \theta-by\csc \theta = a^2-b^2$. Then How can I find Max. distance of Center of the ellipse from the Normal
| So, the distance of the normal from the origin $(0,0)$ is $$\left| \frac{a^2-b^2}{\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}} \right|$$
So, we need to minimize $(a\sec\theta)^2+(-b\csc\theta)^2=a^2\sec^2\theta+b^2\csc^2\theta=f(\theta)$(say)
So, $\frac{df}{d\theta}=a^22\sec\theta\sec\theta\tan\theta+b^22\csc\theta(-\csc\theta\cot\theta)=2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}$
For the extreme value of $f(\theta),\frac{df}{d\theta}=0$
$\implies 2a^2\frac{\sin\theta}{\cos^3\theta}-2b^2\frac{\cos\theta}{\sin^3\theta}=0$ or $\tan^4\theta=\frac{b^2}{a^2}$
Assuming $a>0,b>0$, $\tan^2\theta=\frac ba$
Now, $\frac{d^2f}{d\theta^2}=2a^2\left(\frac1{\cos^2\theta}+\frac{3\sin^2\theta}{\cos^4\theta}\right)+2b^2\left(\frac1{\sin^2\theta}+\frac{3\cos^2\theta}{\sin^2\theta}\right)>0$ for real $\theta$
So, $f(\theta)$ will attain the minimum value at $\tan^2\theta=\frac ba$
So, $f(\theta)_\text{min}=a^2\sec^2\theta+b^2\csc^2\theta_{\text{at }\tan^2\theta=\frac ba}=a^2\left(1+\frac ba\right)+b^2\left(1+\frac ab\right)=(a+b)^2$
So, the minimum value of $\sqrt{(a\sec\theta)^2+(-b\csc\theta)^2}$ is $a+b$
If $\tan\theta=\sqrt \frac ba, \frac{\sin\theta}{\sqrt b}=\frac{\cos\theta}{\sqrt a}=\pm\frac1{b+a}$
If $\sin\theta=\frac{\sqrt b}{a+b}\implies \csc\theta=\frac{a+b}{\sqrt b},\cos\theta=\frac{\sqrt a}{a+b}\implies \sec\theta=\frac{a+b}{\sqrt a}$
There will be another set $(\csc\theta=-\frac{a+b}{\sqrt b},\sec\theta=-\frac{a+b}{\sqrt a})$
There will be two more set of values of $(\csc\theta,\sec\theta)$ for $\tan\theta=-\sqrt\frac ba$
So, we shall have four normals having the maximum distance from the origin.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/305534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -6]$ which is not my answer.
| The quadratic form$(a+4)x^2-2ax+2a-6=\begin{bmatrix} x & 1\end{bmatrix}\begin{bmatrix}a+4& -a \\ -a & 2a-6\end{bmatrix}\begin{bmatrix}x\\ 1\end{bmatrix}$
So if you want the quadratic form to be negative for all $x$, you need a negative definite $\begin{bmatrix}a+4& -a \\ -a & 2a-6\end{bmatrix}$. By Slvester's criterion, this implies $a^2+2a-24<0=(a+6)(a-4)<0$ and $a<-4$. Together these conditions mean $a\in(-\infty,-6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer Find all integers such that $\frac{n^3-3}{n^2-7}$ is an integer.
I have no idea how to approach these types of proofs.
But I tried a few things, did not get me anywhere.
$n^3 -3 = an^2-7a$ then $n^3-an^2 = 3-7a$, and hence $n^2(n-a) = 3-7a$
And then I have no where to go...
Any help is appreciated thanks.
| Note that
$$
\frac{n^3-3}{n^2-7} = n + \frac{7n-3}{n^2-7}.
$$
So it suffices to determine when $n^2-7$ divides $7n-3$. It is not hard to show that $|n^2-7| > |7n-3|$ when $|n|>8$; so for those $n$, it is impossible for $n^2-7$ to divide $7n-3$. That leaves only the cases $n=-8,-7,\dots,8$ to check, revealing the two solutions $n=-3$ and $n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
What is closed-form expression for $F(n)$ when $F(n)=F(n-1)+F(n-2)$ and $F(0)=a$,$F(1)=b$ and $a,b>0$? What is closed-form expression for $F(n)$ when $F(n)=F(n-1)+F(n-2)$ and $F(0)=a$, $F(1)=b$ and $a,b>0$ ? It seems to be simple generalization of Fibonacci sequence but I can't find closed form for it nor by myself neither using google.
| Note that
$$
\binom{F(n)}{F(n+1)}
=
\begin{bmatrix}
0 & 1\\[6pt]
1 & 1
\end{bmatrix}
\cdot
\binom{F(n-1)}{F(n)}
=
\begin{bmatrix}
0 & 1\\[6pt]
1 & 1
\end{bmatrix}^n
\binom{a}{b}
$$
so the problem essentially reduces to diagonalize the matrix $M=\begin{bmatrix}0&1\\1&1\end{bmatrix}$.
Denoting $\phi_1=\frac{1-\sqrt 5}{2}$ and $\phi_2=\frac{1+\sqrt 5}{2}$ (the roots of $x^2-x-1=0$) you have
$$
\begin{bmatrix}
0 & 1\\
1 & 1
\end{bmatrix}
=
T
\begin{bmatrix}
\phi_1 & 0\\
0 & \phi_2
\end{bmatrix}
T^{-1}
$$
where,
$$
T
=
\begin{bmatrix}
-\phi_2 & -\phi_1\\
1 & 1
\end{bmatrix}
\quad\text{and}\quad
T^{-1}
=
\frac{1}{\sqrt 5}
\begin{bmatrix}
-1 & -\phi_1\\
1 & \phi_2
\end{bmatrix}
$$
so that
$$
\binom{F(n)}{F(n+1)}
=
T
\begin{bmatrix}
\phi_1^n & 0\\[6pt]
0 & \phi_2^n
\end{bmatrix}
T^{-1}
\binom{a}{b}
$$
The first row tells you that
$$
F(n)
=
\phi_1\phi_2
\frac
{
(\phi_1^{n-1}-\phi_2^{n-1})a
+
(\phi_1^n-\phi_2^n)b
}
{
\sqrt 5
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
how to prove $\displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = .... $ How to prove $$ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{\sin^2 \theta}{\sin^2 \left( \frac{k\pi }{2n+1} \right ) } \right ) $$
So far, I manage to prove $ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{2n}\left(1 - \frac{\sin \theta}{\sin \left( \frac{k\pi }{2n+1} \right ) } \right ) $ though I am not sure I am aright.
| Put $z = e^{i\theta}.$ The LHS becomes
$$ \frac{z^{2n+1} - 1/z^{2n+1}}{z-1/z} =
\frac{z^{4n+2} - 1}{z^{2n+2} - z^{2n}} =
\frac{1}{z^{2n}} \frac{z^{4n+2}-1}{z^2-1}.$$
Let $\zeta_k = e^{\frac{2 \pi i k}{4n+2}} = e^{\frac{\pi i k}{2n+1}} $ be the $k$th root of unity.
The RHS is
$$ (2n+1) \prod_{k=1}^n
\left( 1 - \frac{\sin^2 \theta}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) =
(2n+1) \prod_{k=1}^n \left(1 - \frac{(z-1/z)^2}{(\zeta_k-1/\zeta_k)^2} \right)\\ =
(2n+1) \frac{1}{z^{2n}} \prod_{k=1}^n
\left(z^2- \frac{(z^2-1)^2}{(\zeta_k-1/\zeta_k)^2} \right) =
(2n+1) \frac{1}{z^{2n}} \prod_{k=1}^n
\left(z^2 - \zeta_k^2 \frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)$$
So we have an equality between two polynomials that we need to show:
$$ \frac{z^{4n+2}-1}{z^2-1} = (2n+1) \prod_{k=1}^n
\left(z^2 - \zeta_k^2\frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)$$
But these two vanish at the same set of points, namely the roots $\pm\zeta_k$ of unity and their multiplicative inverses $\pm 1/\zeta_k$, where $1\le k\le n$ and are of the same degree ($4n$), so they are scalar multiples of each other. We just need to determine the scalar.
To do this, note that $\theta = 0$ was not in fact a singularity of the original LHS since in a neighborhood of zero, we have
$$ \frac{\sin (2n+1)\theta}{\sin\theta} \sim 2n+1.$$ The same goes for $\theta = \pi.$
Therefore $z=1$ and $z=-1$ are not singularities of the LHS in $z$ either and we are justified in writing
$$ \frac{z^{4n+2}-1}{z^2-1} = z^{4n} + z^{4n-2} + \ldots + z^4 + z^2 + 1.$$
Now the LHS is equal to $2n+1$ at $z=1$, and the product is
$$ \left. \prod_{k=1}^n
\left(z^2 - \zeta_k^2\frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)\right|_{z=1} = 1.$$
Therefore the value of the scalar is $2n+1$ and we are done.
| {
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"url": "https://math.stackexchange.com/questions/310213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Ordered pairs $(x,y)$ that satisfy the equation $x^2+y^2 = 2013$ (1) The no. of Integer ordered pair $(x,y)$ that satisfy the equation $x^2+y^2 = 2013$
(2) The no. of Integer ordered pair $(x,y)$ that satisfy the equation $x^2+y^2 = 2014$
My solution for fisrt::
Here R.H.S is a odd quantity means L.H.S must be odd which is possibly only when one square
quantity is even and other is odd
So Let $x=2r$ and $y=2s+1$ and equation is $4(r^2+s^2+s)+1=2013$
Now How can I calculate after that
Thanks
| Note that $2013=3\cdot11\cdot61, 2014=2\cdot19\cdot53$.
$x^2+y^2=2013 \Rightarrow 3 \mid x^2+y^2 \Rightarrow 3\mid x, y \Rightarrow 9 \mid x^2+y^2=2013$, a contradiction.
$x^2+y^2=2014 \Rightarrow 19 \mid x^2+y^2 \Rightarrow 19\mid x, y \Rightarrow 19^2 \mid x^2+y^2=2014$, a contradiction.
Thus there are no integer solutions.
Here we have used the fact that $-1$ is not a quadratic residue $\pmod{p}$ if $p \equiv 3 \pmod{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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help for proving an equation by induction For this equation:
$$-1^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)$$
how can I prove this by induction?
When I set $n = 1$ for the base case I got:
$$-1^3 + (-3)^3 + (-5)^3 + \ldots + (-3)^3 = -28$$
but am having trouble with the following inductive steps
| You’d have had an even simpler base case had you started with $n=0$: $(-1)^3=(-1)^2(-1)$.
For the induction step your induction hypothesis should be that the result is true for some particular $n$, i.e., that
$$(-1)^3+(-3)^3+(-5)^3+\ldots+(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)\;,\tag{1}$$
and you’ll try to prove the corresponding statement about $n+1$. The first step is to figure out what that statement is:
$$\begin{align*}
(-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2(n+1)-1\big)^3\\
&=\big(-(n+1)-1\big)^2\big(-2(n+1)^2-4(n+1)-1\big)
\end{align*}$$
or, after a bit of algebraic simplification,
$$\begin{align*}
(-1)^3+(-3)^3+\ldots+(-2n-1)^3+\big(-2n-3\big)^3&=(-n-2)^2(-2n^2-8n-7)\\
&=-(n+2)^2(2n^2+8n+7)\;.
\end{align*}\tag{2}$$
The lefthand side of $(2)$ can be split into two pieces as
$$\Big((-1)^3+(-3)^3+\ldots+(-2n-1)^3\Big)+\big(-2n-3\big)^3\;,$$
and the induction hypothesis $(1)$ tells us that the first piece is $$(-2n-1)^3=(-n-1)^2(-2n^2-4n-1)=-(n+1)^2(2n^2+4n+1)\;.$$
Thus,
$$\begin{align*}
(-1)^3+(-3)^3&+\ldots+(-2n-1)^3+\big(-2n-3\big)^3\\
&=-(n+1)^2(2n^2+4n+1)+(-2n-3)^3\;,
\end{align*}$$
and all that you have to do now in order to prove $(2)$ (and thereby complete the induction step) is show that
$$-(n+1)^2(2n^2+4n+1)+(-2n-3)^3=-(n+2)^2(2n^2+8n+7)\;.$$
This is just algebra, and I’ll leave it to you.
| {
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Prove that ${\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}}$. Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$
I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.
| Hint: take logarithms of both sides and approximate for sufficiently large $n$.
$$\log{\sqrt{n}^{\sqrt{n+1}}} = \frac{1}{2} \sqrt{n+1} \log{n}$$
For sufficiently large $n$, this is approximately
$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)-\frac{1}{4} \sqrt{\frac{1}{n}} \log
\left(\frac{1}{n}\right)+\frac{1}{16} \left(\frac{1}{n}\right)^{3/2} \log
\left(\frac{1}{n}\right)+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
$$\log{\sqrt{n+1}^{\sqrt{n}}} = \frac{1}{2} \sqrt{n} \log{(n+1)}$$
For sufficiently large $n$, this is approximately
$$-\frac{1}{2}\sqrt{n} \log \left(\frac{1}{n}\right)+\frac{1}{2}\sqrt{\frac{1}{n}}-\frac{1}{4}\left(\frac{1}{n}\right)^{3/2}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
The latter expression is greater than the former when $n > e^2$. Because the log is monotonic, then
$$\sqrt{n}^{\sqrt{n+1}} > \sqrt{n+1}^{\sqrt{n}}$$
when $n > e^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
how to find $\int_{0}^{1}h_n(x)dx?$ I would appreciate if somebody could help me with the following problem:
$$h(x)=\begin{cases}
2x&\left(0\leq x\leq \frac{1}{3}\right)\\
\frac{1}{2}x+\frac{1}{2} &\left(\frac{1}{3}<x\leq 1\right)
\end{cases}$$
let $h_2(x)=h(h(x)), h_3(x)=h_2(h(x)),\cdots, h_{n+1}(x)=h_n(h(x))$
find $\int_{0}^{1}h_n(x)dx ?$
| Playing around with the patterns of this recursion, I get the following:
$$h_n(x) = \begin{cases}\\ 2^n x, & 0 < x < \frac{1}{3 \cdot 2^{n-1}}\\2^{n-2 m} x + \left (1-\frac{1}{2^m}\right ),&\frac{1}{3 \cdot 2^{n-m}}<x<\frac{1}{3 \cdot 2^{n-m-1}} \\ \frac{1}{2^n} x +\left (1-\frac{1}{2^n}\right ),&\frac{1}{3}<x<1 \end{cases}$$
where $m \in \{1,2,\ldots,n-1\}$. You can verify this is true by induction, using $h_{n+1}(x) = h(h_n(x))$. (I leave this to the reader and OP.)
Once you have this definition set, doing the integral is a matter of careful bookkeeping:
$$\begin{align}\int_0^1 dx \: h_n(x) = \int_0^{1/(3 \cdot 2^{n-1})} dx \: 2^n x + \sum_{m=1}^{n-1} \int_{1/(3 \cdot 2^{n-m})}^{1/(3 \cdot 2^{n-m-1})} dx \left [(2^{n-2 m} x + \left (1-\frac{1}{2^m}\right )\right ]\\ + \int_{1/3}^1 dx \: \left [\frac{1}{2^n} x +\left (1-\frac{1}{2^n}\right )\right ] \\\end{align}$$
I again leave the algebra to the reader and OP; the result is
$$\int_0^1 dx \: h_n(x) = 1 - \frac{n+3}{6 \cdot 2^n} $$
You can verify that this result agrees with the $n=1$ case.
ADDENDUM
Here is a plot of the first few cases for $h_n$:
| {
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Evaluate $\frac{1}{2\pi}\int _{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos (\varphi)} \, \mathrm{d}\varphi$ I couldn't evaluate this integral. Could you please help me?
$$\frac{1}{2\pi}\int _{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos (\varphi)} \, \mathrm{d}\varphi,\text{ where $k$ is a constant}$$
| Following up on Yimin's suggestion:
$$
\int_{-\pi}^\pi \frac{e^{-i\varphi}}{1-k\cos\varphi} \, d\varphi = \int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi - i\int_{-\pi}^\pi \frac{\sin\varphi}{1-k\cos\varphi} \, d\varphi.
$$
Since the second integral is that of an odd function over an interval symmetric about $0$, it is $0$, and we only need to work on the first integral.
The Weierstrass substitution is
\begin{align}
x & = \tan\frac\varphi2 \tag{1}\\[8pt]
\frac{1-x^2}{1+x^2} & = \cos\varphi \tag{2}\\[8pt]
\frac{2\,dx}{1+x^2} & = d\varphi\tag{3}
\end{align}
How to get $(2)$ and $(3)$ from $(1)$ using trigonometric identities could be discussed here (I wonder if anyone's ever posted that as a question?). There is also the identity $\dfrac{2x}{1+x^2}=\sin\varphi$, but we won't need that here.
We get
$$
\int_{-\pi}^\pi \frac{\cos\varphi}{1-k\cos\varphi} \, d\varphi = \int_{-\infty}^{\infty}\frac{\frac{1-x^2}{1+x^2}}{1-k\frac{1-x^2}{1+x^2}} \cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \frac{1-x^2}{(1+x^2)-k(1-x^2)}\cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \frac{1-x^2}{(1-k)+(1+k)x^2}\cdot\frac{2\,dx}{1+x^2}
$$
$$
= \int_{-\infty}^\infty \left( \frac{Ax+B}{(1-k)+(1+k)x^2} + \frac{Cx+D}{1+x^2} \right) \, dx
$$
etc. The antiderivative should involve logarithms and/or arctangents.
| {
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"url": "https://math.stackexchange.com/questions/325683",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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What would the space under the image of this function look like? The function is as follows: $$\phi(x,y,z) = \left(\frac{yz}{x^2+y^2+z^2},\frac{zx}{x^2+y^2+z^2}\right)$$
And when $z=0$,the function becomes $ \; \phi(x,y,0)=\frac{xy}{x^2+y^2}$. Some properties of this function are that for any point $ \; p = (x,y,z) \in \mathbb{R}^3/(0) \; and \; \forall \lambda \in \mathbb{R} \neq 0$; if $\; \lambda p=(\lambda x,\lambda y,\lambda z)$ then
$$\begin{align}
\phi(\lambda p) & =\left(\frac{\lambda y \lambda z}{ \lambda^2 x^2 + \lambda^2 y^2 + \lambda z^2},
\frac{\lambda z \lambda z}{ \lambda^2 x^2 + \lambda^2 y^2 + \lambda z^2}\right)\\
& = \left(\frac{\lambda^2(yz)}{\lambda^2(x^2+y^2+z^2)},\frac{\lambda^2(zx)}{\lambda^2(x^2+y^2+z^2)}\right) \\
& = \phi(p)\end{align}$$ and so of course $\phi(-x,-y,-z)=\phi(x,y,z)$. I'm tempted to say this surface is going to look an awful lot like the projective plane and share certain features of it (like mapping 2 antipodal points into point under the image of $\phi$) but I can't exactly visualize it.
| I claim that the image equals the closed ball $B$ with radius $1/2$ and center $(0,0)$.
For $\mathrm{im}(\phi) \subseteq B$, observe that
$$\left(\frac{yz}{x^2+y^2+z^2}\right)^2 + \left(\frac{zx}{x^2+y^2+z^2}\right)^2=\frac{z^2 (y^2+x^2)}{(x^2+y^2+z^2)^2} \leq \frac{1}{4}$$
since for all $a,b,c \geq 0$ we have $(a+b+c)^2-4c(a+b)=(a+b-c)^2 \geq 0$.
I leave the other inclusion as an exercise.
| {
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"timestamp": "2023-03-29T00:00:00",
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trigonometric equation involving two variables Show that the equation $\sec^2\theta = \frac{4xy}{(x+y)^2}$ is only possible when $x =y$.
Can we write it this way : $$\cos^2\theta = \frac{(x+y)^2}{4xy}$$ and $$-1 \leq \cos\theta \leq 1 \Longrightarrow -1 \leq \frac{(x+y)^2}{4xy} \leq 1 $$
or $-4xy \leq (x+y)^2 \leq 4xy $ Please guide...
| $$\sec^2\theta = \frac{4xy}{(x+y)^2}$$
$$\implies \tan^2\theta=\sec^2\theta-1$$
$$=\frac{4xy}{(x+y)^2}-1=\frac{4xy-(x+y)^2}{(x+y)^2}=-\frac{(x-y)^2}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
As $\theta$ is real, $\tan^2\theta\ge0$
$\implies -\left(\frac{x-y}{x+y}\right)^2\ge0$
$\implies \left(\frac{x-y}{x+y}\right)^2\le0$
But the square of any real number is $\ge0,$
as $x,y$ are real $\implies \left(\frac{x-y}{x+y}\right)^2\ge0$
So, $\left(\frac{x-y}{x+y}\right)^2=0\implies x-y=0$ for finite $x,y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/329051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
Let s be the side of the equilateral triangle
Using cosine formula for the triangle
$$\frac{13^2 + 5^2 - s^2}{2.5.13} = \cos\alpha\tag1$$
$$\frac{12^2 + 5^2 - s^2}{2.5.12} = \cos\beta\tag2$$
$$\frac{12^2 + 13^2 - s^2}{2.13.12} = \cos\gamma\tag3$$
Now we have $\gamma + \alpha + \beta = \pi$
$$\cos\gamma = \cos(\pi - \alpha - \beta)$$
Knowing $\cos(\pi - \theta) = \cos(\theta)$
$$\Rightarrow \cos\gamma = \cos(\alpha + \beta)$$
$$\Rightarrow \cos\gamma = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$
$$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = \sin^2\alpha\sin^2\beta$$
$$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = (1 - \cos^2\alpha)(1 - \cos^2\beta)$$
$$\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma - 2.\cos\alpha.\cos\beta.\cos\gamma -1 = 0\tag4$$
Applying $(1)$, $(2)$, $(3)$ to $(4)$ and simplifying we have
$$s^4 - 338.s^2 + 17761 = 0$$
using Heron's formula we have
$$\Rightarrow s^2 = \pm 60 . \sqrt3 + 169$$
$$\Rightarrow s = \pm\sqrt{\pm 60 . \sqrt3 + 169}$$
as Area of a triangle is real and positive
$$s = \sqrt{60 . \sqrt3 + 169}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this identity $\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;$? How to prove this identity? $$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}\;$$
I found the above interesting identity in the book $\bf \pi$ Unleashed.
Does anyone knows how to prove it?
Thanks.
| There is a simple way to compute the sum. Note
$$ \sum_{k=-\infty}^\infty\frac{\sin^2k}{k^2}=1+2\sum_{k=1}^\infty \frac{\sin^2k}{k^2}=1+\sum_{k=1}^\infty\frac{1-\cos(2k)}{k^2}=1+\frac{\pi^2}{6}-\sum_{k=1}^\infty\frac{\cos(2k)}{k^2}.$$
Letting $x=e^{2\theta i}$ in $\sum_{k=1}^\infty\frac{x^k}{k}=-\ln(1-x)$ gives us
$$ \sum_{k=1}^\infty\frac{1}{k}(\cos(2k\theta)+i\sin(2k\theta))=-\ln(1-\cos(2\theta)-i\sin(2\theta)). $$
So
$$ \sum_{k=1}^\infty\frac{1}{k}\sin(2k\theta)=-\Im\ln(1-\cos(2\theta)-i\sin(2\theta))=-\arctan(-\cot \theta)=\frac{\pi}{2}-\theta. $$
Integrating this derives
$$ -\sum_{k=1}^\infty\frac{1}{2k^2}\cos(2k\theta)=\frac{\pi}{2}\theta-\frac{1}{2}\theta^2+C. $$
Letting $\theta=0$, we have $C=-\frac{\pi^2}{12}$. Thus
$$ \sum_{k=1}^\infty\frac{1}{k^2}\cos(2k\theta)=-\pi \theta+\theta^2+\frac{\pi^2}{6}. $$
Letting $\theta=1$, we have
$$ \sum_{k=1}^\infty\frac{\cos(2k)}{k^2}=-\pi+1+\frac{\pi^2}{6} $$
and hence
$$ \sum_{k=-\infty}^\infty\frac{\sin^2k}{k^2}=1+\frac{\pi^2}{6}-\sum_{k=1}^\infty\frac{\cos(2k)}{k^2}=1+\frac{\pi^2}{6}-(-\pi+1+\frac{\pi^2}{6})=\pi.$$
It is easy to use the same trick to generalize this result to $\sum_{k=-\infty}^\infty\frac{\sin^2(ak)}{k^2}$. I omit the detail.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area between 3 curves I need to calculate an area between those 3 curves.
I have no idea what area should I calculate (because of y=-x-1)
Regards.
| It is very simple(this is school example?).
Simply compute $\int_a^0 [(x-1)^3 - 3\cdot x - 5] {d}x$ where a is intesect of $(x-1)^3$ with $3\cdot x - 5$(if in Wolfram|Alpha enter "$[solve (x-1)^3 = 3\cdot x - 5][1]$" you get $x = -1$ and $x = 2$, so $a = -1$): $\int_{-1}^0 [(x-1)^3 - 3\cdot x + 5] {d}x = \left.{{x}\over{4}}\cdot [(x - 4)\cdot x^2 + 16]\right|_{-1}^0 = {{11}\over{4}}$. If now in Wolfram|Alpha enter "solve 3\cdot x-5 = -x - 1" you get $x = 1$, so second part is ${4}\cdot \int_{0}^1 [- x + 1] {d}x = \left.-{2\cdot x}\cdot (x - 2)\right|_0^1 = 2$.
Total is ${{11}\over{4}} + 2 = {{15}\over{4}}$
| {
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Help with inequality please Once again I have come across an olympiad-type problem which probably requires some sort of insight even though it looks simple. The question is as follows:
Let $a$, $b$ and $c$ be positive real numbers. Prove that:
$(a+b)(b+c)(c+a)$ $\geqslant$ $8(a+b-c)(b+c-a)(c+a-b)$
I have tried to multiply out the LHS but unfortunately it didn't get me much...
I found that if one of $a$, $b$ or $c$ is greater than or equal to the sum of the other two, then the inequality is trivially true, since LHS is positive while RHS isn't.
Would there be a quick and easy formula or known inequality that I could use to make this problem simpler? Or is this just a 'bash-and-solve' type question?
Any help, comments or edits are greatly appreciated! Thanks! :)
This question appeared in the South African Mathematics Olympiad in 2008.
| Given $\quad(a+b)(b+c)(c+a) \ge 8(a+b-c)(b+c-a)(c+a-b)\quad$ show $x\ge y$
$$x=(a+b)(b+c)(c+a) \quad=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2$$
$$y=8(a+b-c)(b+c-a)(c+a-b) \\ =
-8 a^3 + 8 a^2 b + 8 a^2 c + 8 a b^2 - 16 a b c + 8 a c^2 - 8 b^3 + 8 b^2 c + 8 b c^2 - 8 c^3$$
We negate the RHS and subtract it from both sides by addition so $x-y \ge 0$
$$x-y\quad =\quad a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\
+8 a^3 - 8 a^2 b - 8 a^2 c - 8 a b^2 + 16 a b c - 8 a c^2 + 8 b^3 - 8 b^2 c - 8 b c^2 + 8 c^3\\
=8a^3-7a^2b-7a^2c-7ab^2+18abc-7ac^2+8b^3-7b^2c-7bc^2+8c^3\quad \ge\quad 0\\
\implies 8(a^3+b^3+c^3)\quad\ge\quad 7(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)$$
If $\quad a=b=c=1\quad$ then $\quad 24\ge 42\quad $which is a contradiction.
If $\quad a=b=c=2\quad$ then $\quad 192 \ge 336\quad $ which is a contradiction.
If $\quad a=b=c=3\quad$ then $\quad 648 \ge 1134\quad $ which is a contradiction.
This math could be wrong but it appears that the original statement is reversed.
| {
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How to find the $f\left(x_i\right)$ for midpoint Riemann sum
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\mbox{Time} & \mbox{9am} & \mbox{10:30 am} & \mbox{Noon} & \mbox{1:30 pm} & \mbox{3 pm} & \mbox{4:30 pm} & \mbox{6 pm} & \mbox{7:30 pm} & \mbox{9 pm} \\
\hline
t & 0 & 1.5 & 3 & 4.5 & 6 & 7.5 & 9 & 10.5 & 12 \\
\hline
P\left(t\right) & 200 & 728 & 1193 & 1329 & 1583 & 1291 & 804 & 256 & 0 \\
\hline
\end{array}
$
Use a midpoint Riemann sum with four intervals of equal size to estimate the total number of people seeking care during the 12-hour period.
I know the formula for the midpoint Riemann sum is $ \int_a^b f\left(x\right) \; \mathrm{d}x \approx \frac{b-a}{n} \left[ f\left(x_1\right) + f\left(x_2\right) + \ldots + f\left(x_n\right) \right]$
I know that $b = 12$ and $a = 0$ and $n = 4$, but what is my $f\left(x_1\right), \; f\left(x_2\right), \; f\left(x_3\right), \; f\left(x_4\right)$
| For $n$ points from $a$ to $b$,
$x_i = a + \frac{b-a}{n}(i-\frac1{2})$.
Note that $x_1 = a + \frac1{2}\frac{b-a}{n}$
is just to the right of $a$,
and $x_n = a+(n-\frac1{2})\frac{b-a}{n}
= b - \frac1{2}\frac{b-a}{n}$
is just to the left of $b$.
| {
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Find all solutions of $7x^2 \equiv 1 \pmod {17}$ Find all solutions of $7x^2 \equiv 1 \pmod {17}$
I found out all the primitive root of $U_{17}$ to be : $\{3,5,6,7,10,12,14\}$.
To continue with the computation, I think i need to use the theorem which is $x^2 \equiv1\pmod{n}$. But Im not sure how to connect it with the question?
Thanks!!
| $$7x^2 \equiv 1 \pmod{17} \implies x^2 \equiv 7^{-1} \pmod{17} \implies x^2 \equiv 5 \pmod{17}$$
When $(x,17) = 1$ from little-Fermat, we have
$$x^{16} \equiv 1 \pmod{17}$$
But we also need $x^2 \equiv 5 \pmod{17}$. This means $$x^{16} = {x^2}^8 \equiv 5^8 \pmod{17} \equiv (25)^4 \pmod{17} \equiv 8^4 \pmod{17} \equiv (64)^2 \pmod{17}$$
$$(64)^2 \pmod{17} \equiv 13^2 \pmod{17} \equiv 169 \pmod{17} \equiv -1 \pmod{17}$$
contradicting little-Fermat.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$
Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way too much time to test if each of them is primitive root or not...
But Im not quite familiar with the method of Quadratic Residue.
We have $x^8\equiv5\pmod3$ and $x^8\equiv 5 \pmod {11}$
so.... follows the rule of $x^2\equiv q \pmod n$,
we have $(x^4)^2 \equiv 5 \pmod 3$ and $(x^4)^2 \equiv 5 \pmod {11}$
and then how do I continue with it??
| Hint $\rm\ mod\ 3\!:\ x\not\equiv 0\:\Rightarrow\:x\equiv \pm 1\:\Rightarrow\:x^2\equiv 1\:\Rightarrow\:x^{2n}\!\equiv 1\not\equiv 5$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing series expansion How would I show $$\frac{\sinh z}{z^2}=\frac{1}{z}+\sum_0^\infty \frac{z^{2n+1}}{(2n+3)!}$$
I have $\sinh z=\sum_0^\infty\frac{z^{2n+1}}{(2n+1)!}$ and if I multiply by $\frac{1}{z^2}$ then $\frac{1}{z^2}\sinh z=\sum_0^\infty \frac{1}{z^2}\frac{z^{2n+1}}{(2n+1)!}$ but I how would I would factor out $\frac{1}{z}$?
| Write out a few terms to see what is going on.
$$\begin{align}
\sinh z &= \frac{z}{1!}+\frac{z^3}{3!} + \frac{z^5}{5!}+\dots
\\
\frac{\sinh z}{z^2} &= \frac{1}{1! z}+\frac{z}{3!} + \frac{z^3}{5!}+\dots
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that :
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ .
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $
What to do next?
| First note that $\dfrac{A}2, \dfrac{B}2, \dfrac{C}2 \in (0, \pi/2)$. Now consider the function $f(x) = \tan^2(x)$. For $x \in (0,\pi/2)$, $f(x)$ is convex. Hence, we have
$$\dfrac{f(x_1) + f(x_2) + f(x_3)}3 \geq f \left(\dfrac{x_1 + x_2 + x_3}3 \right)$$
Take $x_1 = \dfrac{A}2$, $x_2 = \dfrac{B}2$ and $x_3 = \dfrac{C}2$. We have $x_1 + x_2 + x_3 = \dfrac{A+B+C}2 = \dfrac{\pi}2$.
Hence, we get
$$\dfrac{f(A/2) + f(B/2) + f(C/2)}3 \geq f \left(\dfrac{\pi}6 \right)$$
This gives us
$$\dfrac{\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2)}3 \geq \tan^2(\pi/6) = \dfrac13$$
Hence,
$$\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \geq 1$$
If $f(x)$ is convex, a $n$-gon with vertices on the curve will always lie on the upper side of the curve (In fact, this is the definition of convexity. A quick way to check if a function is convex is to check if its second derivative (if it exists) is non-negative.) and hence the center of mass of the $n$-gon will also lie on the upper side of the curve. In our case, $n=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit (if it exists): $\lim_{(x,y,x)\rightarrow(0,0,0)}\frac{xyz}{x^2+y^2+z^2}$ Find the limit (if it even exists). If not, prove it doesn't exist.
$$\lim_{(x,y,x)\rightarrow(0,0,0)}\frac{xyz}{x^2+y^2+z^2}$$
| If you use the facts
$$ |x|=\sqrt{x^2}\leq \sqrt{x^2+y^2+z^2}, \quad |y|=\sqrt{y^2}\leq \sqrt{x^2+y^2+z^2},\quad |z|=\sqrt{z^2}\leq \sqrt{x^2+y^2+z^2}, $$
then the problem falls apart as
$$ \Bigg|\frac{xyz}{x^2+y^2+z^2}\Bigg|= \frac{|xyz|}{x^2+y^2+z^2}= \frac{|x||y||z|}{x^2+y^2+z^2} $$
$$ \leq \frac{(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2}=\sqrt{x^2+y^2+z^2}< \epsilon=\delta. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/339400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\sum_{n=1}^{\infty}\frac{1}{n^3+3 n^2+2 n}$ Summing this series from $0$ to $\infty$, the result is $\frac{1}{4}$. I tried a lot, but I could not get this result. I think it´s wrong.
Can anybody help me?
| Hints - Partial Fractions:
$$n^3+3n^2+2n=n(n+1)(n+2)\Longrightarrow$$
$$\frac{1}{n^3+3n^2+2n}=\frac{1}{n(n+1)(n+2)}=\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}\Longrightarrow$$
$$\sum_{n=1}^\infty\frac{1}{n^3+3n^2+2n}=\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{3}+\frac{1}{8}\right)=\ldots$$
Further hint: find the cancelation pattern in the above telescopic-ish series...
| {
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"timestamp": "2023-03-29T00:00:00",
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Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity
Find the sum of the following series:
$ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity
The answer is given as $5e$ but I got it as $5e+1$
$T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e + 3(e-1) +(e-2)$
$=5e-1$
Can you teell me which is correct and how?
| The answer is $5e-1$. There might be some confusion with the initial values of the sum.
By letting $T_n = \frac{1}{(n-2)!} + \frac{3}{(n-1)!} + \frac{1}{n!}$, as you did:
$S = 2^2 + T_2 + T_3 + \dots = 4 + (\frac{1}{0!} + \frac{3}{1!} + \frac{1}{2!}) + (\frac{1}{1!} + \frac{3}{2!} + \frac{1}{3!}) + (\frac{1}{2!} + \frac{3}{3!} + \frac{1}{4!}) + \dots$
Then we can take the $\frac{1}{0!}$, $\frac{3}{1!}$, and $\frac{1}{1!}$ out of the brackets, giving us $S=4 + \frac{1}{0!} + \frac{3}{1!} + \frac{1}{1!} + 5(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots) = 4+1+3+1+5(e-2) = 5e-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$? Let $r,s,t$ be the roots of the equation $ x^3 - 6x^2 + 5x + 1$. What is the value of $(2-r)(2-s)(2-t)$?
The question is mentioned in my math olympiad. Please explain how to solve the problem. I have factorised the equation to $$-x^2+1, x-6x, -5x +1.$$
I am only in year 6.
| If $c$ is a root of some polynomial, then $x - c$ is a a factor of that polynomial. Furthermore, if the polynomial has degree $n$, then it has at most $n$ roots. So $$x^3 - 6x^2 + 5x + 1 = (x - r)(x - s)(x - t)$$ Thus,
\begin{align*}
(2 - r)(2 - s)(2 - t) &= 2^3 -6(2^2) + 5(2) + 1 \\
&= 8 - 6(4) + 10 + 1 \\
&= 8 - 24 + 10 + 1 \\
&= -5
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.