Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
| Symmetry is always a powerful and nice instrument
$$
\eqalign{
& 0 = \cos (3x) + \cos x = \cr
& = \cos (2x + x) + \cos (2x - x) = \cr
& = \cos (2x)\cos x - \sin (2x)\sin x + \cos (2x)\cos x + \sin (2x)\sin x = \cr
& = 2\cos (2x)\cos x \cr}
$$
and then
$$
\eqalign{
& \cos (2x) = 0\quad \vee \quad \cos x = 0 \cr
& 2x = {\pi \over 2} + k\pi \quad \vee \quad x = {\pi \over 2} + j\pi \cr
& x = {\pi \over 4} + k{\pi \over 2}\quad \vee \quad x = {\pi \over 2} + j\pi \cr
& x = {\pi \over 4} + 2k{\pi \over 4}\quad \vee \quad x = 2{\pi \over 4} + 4j{\pi \over 4} \cr
& x = \left( {2k + 1} \right){\pi \over 4}\quad \vee \quad \left( {2j + 1} \right){\pi \over 2}\quad \left| {\,k,j \in Z} \right. \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Pseudo determinant of product of two square matrices Let $A$ and $B$ be square symmetric matrices. $A$ is singular and $B$ is non-singular. Is there a way to decompose:
$Det(AB)$ in terms of $Det(A)$ and $det(B)$.
$Det(.)$ refers to the pseudo determinant and $det(.)$ refers to the usual determinant of a square non-singular matrix.
| No. For example, consider
\begin{align*}
A_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \\
A_2 &= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \\
B &= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.
\end{align*}
Then, $\operatorname{Det} A_1 = \operatorname{Det} A_2 = 1$, but $\operatorname{Det}(A_1 B) = 1$ and $\operatorname{Det} (A_2 B) = 2$. That is, $\operatorname{Det}(AB)$ is not purely a function of $\operatorname{Det} A$ and $\det B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3732573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $pFor what value of $m$ the is sum
$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$
My approach
$$\sum_i^{m} {10 \choose i}{20 \choose m - i} = {10 \choose 0}{20 \choose m} + {10 \choose 1}{20 \choose m - 1} + \dots + {10 \choose m}{20 \choose 0}$$
$$(1 +x)^{20} = {20 \choose 0} + {20 \choose 1}x + \dots + {20 \choose m-1}x^{m-1} + {20 \choose m}x^{m} + \dots + {20 \choose 20}x^{20}$$
$$(1 +x)^{10} = {10 \choose 0} + {10 \choose 1}x + \dots + {10 \choose 10}x^{10}$$
Later what to do??
Any other method or hint will be greatly welcomed.
| Vandermonde's Identity says that
$$
\sum_{i=0}^m\binom{10}{i}\binom{20}{m-i}=\binom{30}{m}\tag1
$$
The central binomial coefficient is the greatest. Therefore, the maximum of $(1)$ is when $m=15$; that is, $\binom{30}{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx \tag 1$$
must be split thus:
$$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\color{red}{\int A\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +B \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; dx}$$
or it can be splitted in a different way.
Using the added angle formula (for numerator and denominator of the $(1)$) $$a\sin x+b\cos x=\lambda \sin (x+\phi)$$ if $\lambda=\sqrt{a^2+b^2}$ and $\tan \phi=b/a \ $ or $$a\sin x+b\cos x=\lambda \cos (x+\varphi)$$ with $\tan \varphi=-a/b \ $ is it possible to obtain the same result?
| Answer for @Koro: Yes with some calculus I obtain:
$$\int \left(\cos \left(a+b\right)-\tan \left(t\right)\sin \left(a+b\right)\right)dt=t\cos \left(a+b\right)+\sin \left(a+b\right)\ln \left|\cos \left(t\right)\right|+k, \quad k\in\mathbb R$$
Hence,
$$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }dx=$$
$$=\sqrt{\frac{5}{34}}\left[(x-b)\cos \left(a+b\right)+\sin \left(a+b\right)\ln \left|\cos \left(x-b\right)\right|\right]+k=$$
But $a=\arctan (1/2)$ and $b=\arctan (3/5)$. But I will have many calculus having,
$$\cos(\arctan (1/2)+\arctan (3/5))=\dotsb$$
$$\sin(\arctan (1/2)+\arctan (3/5))=\dotsb$$
$$(x-\arctan (3/5))=\dotsb, \quad \cos(x-\arctan (3/5))=\dotsb$$
I think that it takes a lot of time for this way.....to obtain the solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to simplify $\cos(\frac{1}{3}\cos^{-1}(x))$ and $\sin(\frac{1}{3}\cos^{-1}(x))$? How to simplify this trigonometric expression
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right)
$$
I want to write this equation as
$$
\cos \left(\frac{1}{3} \cos ^{-1}(b)\right) = \cos(\cos^{-1}(B)) = B
$$
Where $B$ is written in terms of $b$.
I couldn't find any identity that would let me do it.
This term appears in the solution of the depressed cubic equation as given in wikipedia.
In the original equation, the term is like
$$
\cos\left(\frac{1}{3}\cos^{-1}(b) - \frac{2\pi n}{3}\right) \\
= \cos\left(\frac{1}{3}\cos^{-1}(b)\right)\cos\left(\frac{2\pi n}{3}\right)+\sin\left(\frac{1}{3}\cos^{-1}(b)\right)\sin\left(\frac{2\pi n}{3}\right)
$$
Here the only term that are bothering me are only those $\cos\left(\frac13\cos^{-1}(b)\right)$ and $\sin\left(\frac13\cos^{-1}(b)\right)$
| Use trig. identity $$4\cos^3x-3\cos x=\cos 3x$$
substitute $x=\frac13\cos^{-1}b$
$$4\cos^3\left(\frac13\cos^{-1}b\right)-3\cos\left(\frac13\cos^{-1}b\right)=b$$
$$4B^3-3B-b=0$$
Solve above cubic equation for $B$ i.e. $\cos\left(\frac13\cos^{-1}b\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739741",
"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$ by change of order of integration. I was trying to prove the following by changing the order of integration:
$$\int_0^1dx\int_x^{1/x}\frac{y^2dy}{(x+y)^2\sqrt{1+y^2}}=\sqrt{2}-\frac{1}{2}$$
Splitting the region $R=\big\{0\leq x\leq1, x\leq y\leq \frac{1}{x}\big\}$ of integration into $R_1=\big\{0\leq x\leq y, 0\leq y\leq1\big\}$ and $R_2=\Big\{1\leq y\leq\infty, 0\leq x\leq\frac{1}{y}\Big\}$ and integrating by changing the order of integration on these two regions, I get the integral to equal $\frac{\sqrt{2}}{2}$.
While using the regions as done in the answer of To prove $\int_0^1dx\int_x^{1/x}\frac{ydy}{(1+xy)^2(1+y^2)}=\frac{\pi-1}{4}$, I get $\frac{\sqrt{2}}{2}+\frac{1}{2}$.
Any suggestion/help would be appreciated. Thanks!
| Without seeing your computation, it is not possible to pinpoint your error. Let $$f(x,y) = \frac{y^2}{(x+y)^2\sqrt{1+y^2}} .$$
Then $$\int f(x,y) \, dx = -\frac{y^2}{(x+y)\sqrt{1+y^2}} + C$$ and in particular, for $g > 0$,
$$F(g,y) = \int_{x=0}^g f(x,y) \, dx = \frac{y^2}{\sqrt{1+y^2}}\left(\frac{1}{y} - \frac{1}{g+y}\right).$$ therefore, the given integral equals
$$\begin{align}
I &= \int_{x=0}^1 \int_{y=x}^{1/x} f(x,y) \, dy \, dx \\
&\int_{y=0}^1 \int_{x=0}^y f(x,y) \, dx \, dy + \int_{y=1}^\infty \int_{x=0}^{1/y} f(x,y) \, dx \, dy \\
&= \int_{y=0}^1 F(y,y) \, dy + \int_{y=1}^\infty F(1/y,y) \, dy \\
&= \int_{y=0}^1 \frac{y}{2\sqrt{1+y^2}} \, dy + \int_{y=1}^\infty \frac{y}{(1+y^2)^{3/2}} \, dy.
\end{align}$$
The common substitution $$u = 1 + y^2, \quad du = 2y \, dy,$$ easily yields
$$\begin{align}
I &= \frac{1}{4}\int_{u=1}^2 u^{-1/2} \, du + \frac{1}{2} \int_{u=2}^\infty u^{-3/2} \, du \\
&= \frac{1}{4}\left[2u^{1/2} \right]_{u=1}^2 + \frac{1}{2} \left[-2u^{-1/2}\right]_{u=2}^\infty \\
&= \frac{1}{4}\left(2 \sqrt{2} - 2\right) + \frac{1}{2} \left(0 + \frac{2}{\sqrt{2}}\right) \\
&= \frac{\sqrt{2} - 1}{2} + \frac{1}{\sqrt{2}} \\
&= \sqrt{2} - \frac{1}{2}
\end{align}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1})$.
I am stuck at this point and I cannot prove that $\log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}) > 0$.
I know by the definition $\log (1+\frac{1}{n}) =\int_1^{1+\frac{1}{n}}\frac{1}{x}dx$.
But I can only reach to the point where $ \frac{1}{n+1} < \log (1+\frac{1}{n}) < \frac{1}{n}$.
Thanks for your help in advance.
| When in doubt, plot the function. As you can see, the function is strictly decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740793",
"timestamp": "2023-03-29T00:00:00",
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How prove this $\sum_{i=n+2}^{+\infty}\frac{1}{i^2}>\frac{2n+5}{2(n+2)^2}$ let $n$ be postive integer,show that
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\dfrac{2n+5}{2(n+2)^2}\tag{1}$$
I know
$$\sum_{i=n+2}^{+\infty}\dfrac{1}{i^2}>\int_{n+2}^{+\infty}\dfrac{1}{x^2}dx=\dfrac{1}{n+2}$$
But $$\dfrac{1}{n+2}-\dfrac{2n+5}{2(n+2)^2}=-\dfrac{1}{2(n+2)^2}<0$$
then this integral method can't solve (1),so How to prove it?Thanks
| By using $\frac{1}{q^2} = \int_0^\infty \mathrm{e}^{-x q} x \mathrm{d} x$ for $q > 0$, we have (the summation and the integral are interchangeable by Fubini/Tonelli theorems)
\begin{align}
\sum_{i=n+2}^\infty \frac{1}{i^2} &= \sum_{i=n+2}^\infty \int_0^\infty \mathrm{e}^{-x i} x \mathrm{d} x\\
&= \int_0^\infty \sum_{i=n+2}^\infty \mathrm{e}^{-x i} x \mathrm{d} x\\
&= \int_0^\infty \mathrm{e}^{-x(n+2)} \frac{x}{ 1 - \mathrm{e}^{-x}} \mathrm{d} x\\
&> \int_0^\infty \mathrm{e}^{-x(n+2)} \left(1 + \frac{x}{2}\right) \mathrm{d} x\\
&= \frac{2n+5}{2(n+2)^2}
\end{align}
where we have used
$$\sum_{i=n+2}^\infty \mathrm{e}^{-x i} = \mathrm{e}^{-x(n+2)} \frac{1}{ 1 - \mathrm{e}^{-x}}$$
and (see the remark later)
$$\frac{x}{ 1 - \mathrm{e}^{-x}} > 1 + \frac{x}{2}, \ \forall x > 0. \tag{1}$$
Remark 1: To prove (1), it suffices to prove that $\mathrm{e}^{-x} > \frac{2-x}{2+x}$ or $-x > \ln \frac{2-x}{2+x}$ for $x \in (0, 2)$.
Let $f(x) = -x - \ln \frac{2-x}{2+x}$. We have $f'(x) = \frac{x^2}{(2-x)(2+x)} > 0$ for $x \in (0, 2)$. Also, $f(0) = 0$. Thus, we have $f(x) > 0$ for $x\in (0, 2)$.
Remark 2: $1 + \frac{x}{2}$ is first two terms of the Taylor expansion of $\frac{x}{ 1 - \mathrm{e}^{-x}}$ around $x = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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how can I integrate $\int \dfrac{2x+3}{x^3-9x}dx$? How can I integrate this $$\int \dfrac{2x+3}{x^3-9x}dx?$$
My work
$$\dfrac{2x+3}{x^3-9x}=\dfrac{2x+3}{x(x+3)(x-3)}$$
$$=\dfrac{A}{x-3}+\dfrac{B}{x}=\dfrac{C}{x+3}$$
after solving, I got $A=1/2$, $B=-1/3$, $C=-1/6$
partial fractions decompositions:
$$\dfrac{2x+3}{x^3-9x}=\dfrac{1/2}{x-3}+\dfrac{-1/3}{x}=\dfrac{-1/6}{x+3}$$
$$\int \left(\dfrac{1}{2(x-3)}- \dfrac{1}{3x}- \dfrac{1}{6(x+3)}\right)\ dx$$
$$=\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C$$
I am not sure if my answer is correct. my question is can I use substitution to solve above integration? if yes please help me solve it by substitution. thanks
| $$\left(\dfrac{1}{2}\ln (x-3)- \dfrac{1}{3}\ln (x)- \dfrac{1}{6}\ln (x+3)+C\right)'=\frac1{2(x-3)}-\frac1{3x}-\frac1{6(x+3)}
\\=\frac{18(x^2+3x)-12(x^2-9)-6(x^2-3x)}{12x(x^2-9)}
\\=\frac{72x+108}{36(x^3-9x)}.$$
It seems that you are right.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problematic solution related to finding the function range
The Question: Find the range of the following function
$$y=\dfrac{x}{2}+\dfrac{8}{x}$$
Solution $-1.$ (the solution given to me)
By Cauchy inequality,
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$ $$\dfrac{x}{2}+\dfrac{8}{x}≤-2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=-4$$ where $x<0$ which implies $y \in(-\infty, -4] ∪ [4, +\infty).$
But, as far as I know, we don't define inequality of arithmetic and geometric means for negative numbers.For this reason, I strange this mathematical way.
$$\dfrac{x}{2}+\dfrac{8}{x}≤-2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=-4$$ where $x<0.$
Okay, If our equation were equal to a
$$y=\dfrac{x-2}{2}+\dfrac{8}{2-x}$$ or $$ y=\dfrac{x}{2}+\dfrac{8}{|x|}$$ then we can not apply,
$$y=\dfrac{x-2}{2}+\dfrac{8}{2-x} \leq -2\sqrt{ \dfrac{x-2}{2}× \dfrac{8}{2-x}} \in {\emptyset}.$$
$$ y=\dfrac{x}{2}+\dfrac{8}{|x|} ≤-2\sqrt{ \dfrac{x}{2} × \dfrac{8}{|x|}} \in {\emptyset}.$$ where $x<0$.
What I mean,
For $x<0$ ,the arithmetic meaning of $\dfrac{\dfrac{x}{2}+\dfrac{8}{x}}2$ doesn't exist and the geometric meaning of $\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}$ doesn't exist.
So, for $x<0$ to write the $\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}$, I think, it doesn't make sense.
I would go on like this.
$$y=\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$ where $x>0$
Then, for $x<0$ we have both $\dfrac{x}{2}$ and $\dfrac{8}{x}$ are negative. In this sense we can write
$$\dfrac{x}{2}+\dfrac{8}{x}≥2\sqrt{ \dfrac{x}{2}× \dfrac{8}{x}}=4$$
$$-\left(\dfrac{x}{2}+\dfrac{8}{x}\right)\leq-4$$
$$- \dfrac{x}{2}+\left(-\dfrac{8}{x}\right) \leq-4$$ where $x>0$.
It seems more sense to me.
So we get, $y \in(-\infty, -4] ∪ [4, +\infty)$
I don't know how right I am.
My solution:
$$\begin{align} y=\dfrac{x}{2}+\dfrac{8}{x} \Longrightarrow 2yx=x^2+16 \Longrightarrow x^2-2yx+16=0 \Longrightarrow \Delta=y^2-16 \geq0 \Longrightarrow y \in(-\infty, -4] ∪ [4, +\infty). \end{align}$$
*
*Question $-1$ :Do you find the solution $-1$ perfect?
*Question$-2$ :Is my own solution correct?
Remark.
If our function were as follows, we could easily apply the arithmetic-geometric inequality.
$$y=\dfrac{x^2}{2}+\dfrac{8}{x^2}$$
$$y=\dfrac{x^2}{2}+\dfrac{8}{x^2}≥2\sqrt{ \dfrac{x^2}{2}× \dfrac{8}{x^2}}=4$$
$$y \in [4, +\infty)$$
| y' = 1/2 - 8/x$^2$
Solving y' = 0, x$^2$ = 16.
Thus for positive x,
the minimum of y is y(4) = 4
and the range of y is [4,$\infty$).
Simular for negative x.
| {
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"timestamp": "2023-03-29T00:00:00",
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Line that passes through the Centroid of the ABC triangle and the sectioned areas. (ratio of the triangle areas: $2≤r≤2.25$) The line through Centroid $G$ divides the triangle $ABC$ into two figures. I need help with formal proof that the ratio between the areas of triangle ABC and the one that was sectioned is between $2$ and $2.25$
$r∩ΔABC=\left\{ D,E \right\}$
distance between point and line
$dCr=dAr+dBr$
$\begin{array}{} \text{similar triangles:} & ΔAEJ∼ΔCEI & ΔBDK∼ΔCDI \end{array}$
$\begin{array}{} \frac{dAr}{dCr}=\frac{AE}{CE} & \frac{dAr+dCr}{dCr}=\frac{AE+CE}{CE}=\frac{b}{CE} & CE=\frac{b·dCr}{dAr+dCr} \end{array}$
$\begin{array}{} \frac{dBr}{dCr}=\frac{BD}{CD} & \frac{dBr+dCr}{dCr}=\frac{BD+CD}{CD}=\frac{a}{CD} & CD=\frac{a·dCr}{dBr+dCr} \end{array}$
$ratioC=\frac{Δ(ABC)}{Δ(CDE)}$
$\begin{array}{} Δ(ABC)=\frac{1}{2}·a·b·sin(\hat{C}) & Δ(CDE)=\frac{1}{2}·CE·CD·sin(\hat{C}) \end{array}$
$\begin{array}{} rC=\frac{\left( dAr+dCr \right)\left( dBr+dCr \right) }{dCr^2} & rC=\frac{\left( 2·dAr+dBr \right)\left( dAr+2·dBr \right) }{\left( dAr+dBr \right)^2 } \end{array}$
$\begin{array}{} if & dBr=0 & rC=\frac{2·dAr·dAr}{dAr^2}=2\\ if & dAr=dBr & rC=\frac{3·dAr·3·dAr}{\left( 2·dAr \right)^2 }=\frac{9}{4}=2.25\end{array}$
The Euler line is a particular case of the line $r$ passing through the centroid
$rA=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 1 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 1\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(A)+m·y(A)+n)^{2}}|$
$rB=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 1 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 1 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(B)+m·y(B)+n)^{2}}|$
$rC=|\frac{\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 0 \\ x(B) & y(B) & 1 & 1\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|·\left| \begin{array}{} -m & l & 0 & 0 \\ x(A) & y(A) & 1 & 1 \\ x(B) & y(B) & 1 & 0\\x(C) & y(C) & 1 & 0 \\ \end{array} \right|}{(l·x(C)+m·y(C)+n)^{2}}|$
$\begin{array}{} \text{Euler line:} & l·x+m·y+n=0 \end{array}$
The ratio $r$ is between $2$ and $2.25$ as shown by “Geogebra” ($r = 2$ when the Euler line passes through one of the vertices of the triangle ABC and $r = 2.25$ when the Euler line is parallel to one side of the triangle $ABC$. The parallelism condition is given by:
$\begin{array}{} \text{Euler line parallel to the side a} & \text{Euler line parallel to the side b} & \text{Euler line parallel to the side c} \\ \frac{b^2+c^2}{a^2}+\left( \frac{b^2-c^2}{a^2} \right)^2=2 & \frac{a^2+c^2}{b^2}+\left( \frac{a^2-c^2}{b^2} \right)^2=2 & \frac{a^2+b^2}{c^2}+\left( \frac{a^2-b^2}{c^2} \right)^2=2 \\ \end{array}$
If we consider the sectioned areas (triangle and quadrilateral) the ratio $r'$ will be defined as: $r'=\frac{1}{r-1}$ $(0.8\le r'\le1)$
| If you are so inclined to fill out the details, then you can solve it by a simple Menelaus.
Let $ED$ intersect $AB$ at $F.$ If the two are parallel, then the ratio is precisely $\dfrac 94$, so we can assume they must intersect.
Then, draw the median $CM$, passing through $G$, and write the Menelaus' theorem for the triangles $\triangle CMB$ and $\triangle CMA.$ After manipulating the ratios obtained from them, you will find that:
$$\dfrac{CB}{CD}+\dfrac{CA}{CE} = 3.$$
Call these two ratios $x$ and $y$, then by your assumption, $x,y\in[1,2]$ and now we know that $x+y = 3.$ Now AM-GM immediately yields that $xy\leq\dfrac 94.$
For the other direction:
$$xy-2 = x(3-x) - 2 = (x-1)(2-x)\geq 0.$$
Finally, notice that the ratio of area which are you trying to bound is precisely:
$$\dfrac{\triangle CAB}{\triangle CED} = \dfrac{CA\cdot CB}{CE\cdot CD} = xy$$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1$ Prove The following inequality $(ax+by)^2 \le ax^2+by^2$ for $a+b=1, 0 \le a,b \le 1$
I tried expanding the equation and substituting $b=1-a$
\begin{equation}
(ax+by)^2=a^2x^2+2abxy+b^2y^2=a^2x^2+2axy-2a^2xy+b^2y^2
\end{equation}
The middle member $2axy-2a^2xy$ is negative only for $a>1$, so I'm not sure how to proceed from here.
In the hints I was given it was also said that it can be done by proving that the quadratic form $q(x,y)=ax^2+bx^2-(ax+by)^2$ is always positive. I tried finding the eigenvalues but ended up with huge equation I couldn't make sense of.
| We can use $\frac{x^2}{p}+\frac{y^2}{q}\geq\frac{(x+y)^2}{p+q}$,
$$ax^2+by^2=\frac{(ax)^2}{a}+\frac{(by)^2}{b}\geq\frac{(ax+by)^2}{a+b}$$
$$\Rightarrow ax^2+by^2\geq{(ax+by)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3749567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions of $2a+2b-ab\gt 0$ Let $a\in\mathbb{N}_{\ge 3}$ and $b\in\mathbb{N}_{\ge 3}$. What are the solutions of the Diophantine inequality
$$2a+2b-ab\gt 0?$$
By guessing, I found 5 solutions:
$$\text{1)}\, a=3,\, b=3$$
$$\text{2)}\, a=3,\, b=4$$
$$\text{3)}\, a=4,\, b=3$$
$$\text{4)}\, a=5,\, b=3$$
$$\text{5)}\, a=3,\, b=5.$$
Are these all the solutions? How could I find all the solutions rigorously?
| My thinking is: in general $2a +2b$ is only two $a$s and two $b$s but $ab$ is $b$ number of $a$s which can be much more than $2$ if $b$ is large. Likewise $ab$ is $a$ number of $b$s and even if we used two "of those $a$s" to account for $2a$ we still have more than $2b$ is $a$ is large enough. So if $a,b$ are "large enough" then $ab$ will "outweigh" $2a + 2b$. ... So now I need to find the limits of $a$ and $b$ before the "outweighing" occurs.
In detail:
$2a + 2b -ab > 0$
$2a + 2b > ab$
$2b > a(b-2)$
$0 > a(b-2) - 2b$
$0 > a(b-2) - 2(b-2) -4$
$4 > (a-2)(b-2)$
So we may have the following positive integer possibilities (as $a-2 \ge 3-2=1;b-2\ge 3-2=1$ and bothe $a-2,b-2$ are integers):
$(b-2) = 1$ and $a-2 = 1,2,3$
$(b-2) = 2$ and $a-2 = 1$
$(b-2) = 3$ and $a-2 =1$
ANd that's that.
$(a,b) = \{(3,3),(4,3),(5,3), (3,4),(3,5)\}$.
==== Alternatively ====
$2a + 2b - ab > 0$
$2a-ab > -2b$ and $2b -ab > -2a$
$a(2-b) > -2b$ and $b(2-a) > -2a$
$a(b-2) < 2b$ and $b(2-a) < 2b$ and as $a,b \ge 3$
$3\le a < \frac {2b}{b-2}= \frac {2b-4}{b-2} + \frac 4{b-2} = 2+\frac 4{b-2}$ and $3 \le b < 2+\frac {a-2}$.
So $\frac 4{b-2} > 1$ and $\frac 4{a-2}>2$ so $b-2,a-2 < 4$ and $a,b < 6$.
And if $b = 3,4,5$ we have
$3 \le a < 2 +\frac 4{3,2,1} = 3\frac 13, 4, 6$.
So $a = 3$ if $b > 3$ (and vice versa)
So the five solutions you found are the only ones that exist.
| {
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"url": "https://math.stackexchange.com/questions/3749995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-2=0\implies x=2$$
as $x>0$. Now for the proof:
I was attempting some different infinite expansions/square roots when trying to solve another question of mine (Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ ) and I came across this:
$$x+\frac{1}{x}=\sqrt{(x+\frac{1}{x})^2}=\sqrt{2+x^2+\frac{1}{x^2}}=\sqrt{2+\sqrt{(x^2+\frac{1}{x^2}}})^2=\sqrt{2+\sqrt{2+x^4+\frac{1}{x^4}}}=\sqrt{2+\sqrt{2+\sqrt{(x^4+\frac{1}{x^4})^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^8+\frac{1}{x^8}}}}=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=2$$
if you keep on applying this and using the result found at the start of the question. So we have that for any real number $x$ that
$$x+\frac{1}{x}=2\implies x^2-2x+1=0\implies (x-1)^2=0$$
so we finally have:
$$x=1$$
Where have I gone wrong, for surely this cannot be correct?
| As you go along the last square root has $x^{2n}+\frac{1}{x^{2n}}$ which diverges, so it can't be ignored as $n\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Remainder when $\prod_{n=1}^{100}(1- n^{2} +n^{4})$ is divided by $101$
What is the remainder when the expression
$$\prod_{n=1}^{100}(1- n^{2} +n^{4})$$
is divided by $101$?
If $\zeta=\dfrac{-1+\sqrt{-3}}{2}$, then
$$1-n^2+n^4=(1-n+n^2)(1+n+n^2)=(-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n).$$
We then have
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv \prod_{n=1}^{100}\big((-\zeta-n)(-\bar{\zeta}-n)(\zeta-n)(\bar{\zeta}-n)\big)\pmod{101}\,.$$
Since
$$\prod_{n=1}^{100}(x-n)\equiv x^{100}-1\pmod{101},$$
we obtain
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv\big((-\zeta)^{100}-1\big)\big((-\bar\zeta)^{100}-1\big)\big(\zeta^{100}-1\big)\big(\bar{\zeta}^{100}-1\big)\pmod{101}\,.$$
Since $\zeta^3=1$ and $\bar{\zeta}^3=1$, we get
$$(-\zeta)^{100}=\zeta^{100}=\zeta\text{ and }(-\bar\zeta)^{100}=\bar\zeta^{100}=\bar\zeta\,.$$
Therefore,
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (\zeta-1)^2(\bar{\zeta}-1)^2=\big((1-\zeta)(1-\bar{\zeta})\big)^2\pmod{101}\,.$$
As
$$(x-\zeta)(x-\bar{\zeta})=x^2+x+1\,,$$
we get
$$\prod_{n=1}^{100}(1-n^2+n^4)\equiv (1^2+1+1)^2=9\pmod{101}\,.$$
Are there other solutions? How do we solve this problem without resorting to complex numbers?
| Modulo $101$, the set of values $0^3, 1^3,\dots,100^3$ is a permutation of $0,1,2,\dots,100.$ This is because $101$ is prime and $3$ is not a divisor of $100.$
But $$n^4-n^2+1=\frac{n^6+1}{n^2+1}$$
Now, if $n=10,91$ then $n^2+1$ is divisible by $101.$ The other terms are a permutation, so:
$$\begin{align}\prod_{n=1}^{100} (n^4-n^2+1)&=(10^4-10^2+1)(91^4-91+1)\prod_{n\neq 10,91}\frac{n^6+1}{n^2+1}\\
&\equiv (10^4-10^2+1)((-10)^4-(-10)^2+1)\pmod{101}\\
&\equiv 3\cdot 3=9\pmod{101}
\end{align}$$
This works more generally if $p\equiv 5\pmod {12}:$
$$\prod_{n=1}^{p-1}\left(n^4-n^2+1\right)\equiv 9\pmod p$$
If $p\equiv 11\pmod{12},$ the remainder is $1.$
I think when $p\equiv 1\pmod{12},$ the remainder is $0.$
Not sure about $p\equiv 7\pmod{12}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry (For what values of $(\cos A-1/5)(\cos B-1/5)(\cos C-1/5) \leq k$ hold in all triangles $ABC$? For what values of $k$ does
$$
\bigl(\cos A-\frac{1}{5}\bigr)\bigl(\cos B-\frac{1}{5}\bigr)\bigl(\cos C-\frac{1}{5}\bigr ) \leqslant k\:
$$
hold in all triangles $ABC$?
| \begin{align}
(\cos A-\tfrac15)(\cos B-\tfrac15)(\cos C-\tfrac15)
\le k
\tag{1}\label{1}
.
\end{align}
Given that for $\triangle ABC$ we know its semiperimeter $\rho$,
inradius $r$ and circumradius $R$,
let $u=\rho/R$, $v=r/R$. It is known that
for a valid triangle (including degenerate cases)
\begin{align}
v&\in[0,\tfrac12]
,\\
u&\in[u_{\min}(v),u_{\max}(v)]
\tag{2}\label{2}
,
\end{align}
where the boundary expressions
\begin{align}
u_{\min}(v)&=
\sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}}
\tag{3}\label{3}
,\\
u_{\max}(v)&=
\sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}}
\tag{4}\label{4}
\end{align}
correspond to two kinds of isosceles triangles with given $v$.
Consider the left-hand side of expression \eqref{1}
in terms of $u,v$;
\begin{align}
(\cos A-\tfrac15)
&(\cos B-\tfrac15)(\cos C-\tfrac15)
=(\cos A\cos B\cos C)
\tag{5}\label{5}
\\
&-\tfrac15(\cos A\cos B+\cos B\cos C+\cos C\cos A)
\\
&+
\tfrac1{25}(\cos A+\cos B+\cos C)
-\tfrac1{125}
\tag{6}\label{6}
,\\
&=
(\tfrac14\,(u^2-(v+2)^2))
-\tfrac15(\tfrac14\,(u^2+v^2)-1)
\\
&\phantom{=}
+\tfrac1{25}(v+1)
-\tfrac1{125}
\tag{7}\label{7}
,\\
&=
\tfrac15\,u^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125}
\tag{8}\label{8}
,
\end{align}
which leads to
\begin{align}
\max_{v\in[0,\frac12]}&
(\tfrac15\,u(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\tag{9}\label{9}
\\
&=\max_{v\in[0,\frac12]}
(\tfrac15\,u_{\max}(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\tag{10}\label{10}
\\
&=(\tfrac15\,u_{\max}(v)^2-\tfrac3{10}\,v^2-\tfrac{24}{25}\,v-\tfrac{96}{125})
\Big|_{v=0}
\tag{11}\label{11}
\\
&=\tfrac{4}{125}
.
\end{align}
Similarly, a minimal value of the left-hand side of \eqref{1}
can be found to be $-\tfrac{96}{125}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve Complex Equation $z^3 = 4\bar{z}$ I'm trying to solve for all z values where $z^3 = 4\bar{z}$.
I tried using $z^3 = |z|(\cos(3\theta)+i\sin(3\theta)$ and that $|z| = \sqrt{x^2+y^2}$ so:
$$z^3 = \sqrt{x^2+y^2}(\cos(3\theta)+\sin(3\theta))$$ and $$4\bar z = 4x-4iy = 4r\cos(\theta)-i4r\sin(\theta)$$
but I have no idea where to go from there.
| Just for fun, let's do this by expanding $(x+iy)^3=4(x-iy)$ with $x,y\in\mathbb{R}$ and separate the real and imaginary parts. We wind up with
$$x(x^2-(3y^2+4))=0\quad\text{and}\quad y(y^2-(3x^2+4))=0$$
If $x=0$ then $y(y^2-4)=0$, so $y=0,\pm2$, hence $z=0$, $2i$ and $-2i$ are solutions.
If $x\not=0$, then we must have $x^2=3y^2+4$, which implies $y(y^2-(9y^2+12+4))=-8y(y^2+2)=0$. The only real solution is $y=0$, which leads to $x^2=3\cdot0^2+4=4$, or $x=\pm2$. So $z=2$ and $-2$ are also solutions.
In all we have $z=0,2,2i,-2$, and $-2i$ as solutions of $z^3=4\overline{z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdot 5^3 \cdot 2^4\,.$$ Let $p = 3^a5^b7^c2^d$ and $r = 3^e 5^f 7^g 2^s$. Notice, that given $p$ and $r$, $h$ is determined, so we can count $p$ and $r$. Multiplying $p$ and $r$ we get $$pr = 3^{(a+e)} 5^{(b+f)} 7^{(c+g)} 2^{(d+s)}\,,$$ and so $a+e = 0,1,2$.
For the first case we have $0+1 = 1$ possibility, similarly $2$ and $3$ for the other cases, so the total number is $6$. For $b+f$ we have $b + f = 0,1,2,3$ giving $10$ options. Similarly for $c + g$ we have $3$ choices and for $d + h$ we have $$1+2+3+4+5 = 15$$ choices. Multiplying these together, we get $$15\cdot 3\cdot 6\cdot 10 = 60\cdot 45 = 2700\,,$$ which is not equal to the given answer of $473$.
Edit: sorry for the weird variables. I think I've fixed everything, if not, please do point it out
| $126000 = 2^4\cdot 3^2\cdot 5^3\cdot 7^1$
For two numbers to have an LCM of $126000$, either $a$ or $b$ has to have $2^4$ in its prime factorization, and similarly with $3^2, 5^3, $ and $7$.
We need either $a$ and/or $b$ to have $2^4$, so we assign $2^4$ to $a$ first. Then, $b$ can have a factor from $2^0$ to $2^4$, for a total of $5$ total possibilities. We do the same for if $b$ as $2^4$, so $5\cdot 2 = 10$. However, we double-counted the case where both $a$ and $b$ have $2^4$ in it, so we subtract one to get $9$.
Similarly for $3^2$, set $a$ to have $3^2$ in its factorization. There are three choices for $b$, then swap $a$ and $b$ to get $3\cdot 2$, subtract one at the end to get $5$.
Same for $5^3$, we get $7$ and for $7^1$, we get $3$. $9\cdot 5\cdot 7\cdot 3 = 945$. However, we have a special case of $(126000, 126000)$ because this can be ordered in just one way. Therefore, we add one to $945$ to get $946$, then divide by $2$ because the pairs are unordered.
Hence, the answer is $$\fbox{473}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3760396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
| Note
$$LHS^2= \left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} \right)^2\\
=2+2\sqrt{(1 -x)^2}= {4-2x}=RHS^2
$$
So, the solution is just the domain of the equation, which is jointly determined by
$$2x-x^2\ge 0,\>\>\>1-\sqrt{2x-x^2}\ge 0,\>\>\>4-2x\ge 0$$
which leads to $0\le x\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or something like this and then find some kind of linear recurrence, but I'm not sure.
| The hint given by @ThomasAndrews indicates a purely algebraic approach.
We obtain
\begin{align*}
\color{blue}{\sum_{j=0}^n\frac{2^{j+1}}{j+1}\binom{n}{j}}
&=\frac{1}{n+1}\sum_{j=0}^n2^{j+1}\binom{n+1}{j+1}\tag{1}\\
&=\frac{1}{n+1}\sum_{j=1}^{n+1}2^j\binom{n+1}{j}\tag{2}\\
&\,\,\color{blue}{=\frac{1}{n+1}\left(3^{n+1}-1\right)}\tag{3}
\end{align*}
Comment:
*
*In (1) we use the binomial identity $\frac{n+1}{j+1}\binom{n}{j}=\binom{n+1}{j+1}$.
*In (2) we shift the index by one and start with $j=1$.
*In (3) we apply the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplifying trigonometry function by substitution I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {{\tan }^2}{1^{\circ}}} \right)\left( {1 + {{\tan }^2}{2^{\circ}}} \right)}}$
My approach is as follow
$T = \left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
$\Rightarrow \tan {58^{\circ}}\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right) = \sqrt 3 - \tan {2^{\circ}}$
$\Rightarrow \tan {59^{\circ}}\left( {1 + \sqrt 3 \tan {2^0}} \right) = \sqrt 3 - \tan {1}$
$\tan {60^{\circ}} = \frac{{\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)}}{{\left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)}} \Rightarrow \left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)$
$\tan {60^{\circ}} = \frac{{\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)}} \Rightarrow \left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)$
$T = \frac{{\sqrt 3 - \tan {1^{\circ}}}}{{\tan {{59}^{\circ}}}} \times \frac{{\sqrt 3 - \tan {2^{\circ}}}}{{\tan {{58}^{\circ}}}} \times \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right) \times \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$
After this step I am confused
| $$
\tan 59^{\circ} = \frac{\sqrt{3}-\tan 1^{\circ}}{1+\sqrt{3} \tan 1^{\circ}}
$$
$$
\tan 59^{\circ}+\tan 1^{\circ} = \frac{\sqrt{3}(1+\tan^2 1^{\circ})}{1+\sqrt{3} \tan 1^{\circ}},
$$so we have
$$
(1+\sqrt{3} \tan 1^{\circ})(\tan 59^{\circ}+\tan 1^{\circ})=\sqrt{3}(1+\tan^2 1^{\circ})
$$A similar argument works for $\tan 2^{\circ}$ and $\tan 58^{\circ}$. So the answer is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i do next?
| You can prove by induction that $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$
$=20\cdot50^{n}+18\cdot12^{n}$ is divisible by $19$.
It's obviously true for $n=0$, since $20+18=38=2\cdot19$.
Now assume $19$ divides $20\cdot50^{n}+18\cdot12^{n}$.
Then $20\cdot50^{n+1}+18\cdot12^{n+1}=50\cdot20\cdot50^{n+1}+12\cdot18\cdot12^{n+1}$
$=12\cdot(20\cdot50^n+18\cdot12^{n})+38\cdot20\cdot50^n$ is divisible by $19$.
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $|a + b| = |a| + |b| \iff a\overline{b} \ge 0$ I'm reading a complex analysis book. In this book, the author establishes the following statement
If $a,b \in \mathbb{C}$, then $|a + b| = |a| + |b| \iff \left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right)$
This statement didn't seem intuitive to me, so I decided to try to prove it. I denoted the complex numbers $a$ and $b$ as $a = \alpha + i \beta$ and $b = \gamma + i \delta$. Using this, I get that $a\overline{b} = (\alpha \gamma + \beta\delta) + i(\beta \gamma - \alpha\delta)$, which tells us that
$$
\left(a\overline{b}\in \mathbb{R}\right) \wedge \left(a\overline{b}\ge 0\right) \iff (\alpha \gamma + \beta \delta \ge 0) \ \ \wedge \ \ (\alpha\delta= \beta \gamma )
$$
From here, I do the following
\begin{align}
&2(\alpha\delta)^2 = 2(\alpha\delta)^2 \iff 2(\alpha\delta)(\alpha\delta) = (\alpha\delta)^2 + (\alpha\delta)^2 \iff 2\alpha\delta\beta \gamma = (\alpha\delta)^2 + (\beta \gamma)^2 \notag \\
\iff& (\alpha\gamma)^2 + 2\alpha\gamma\beta \delta + (\beta \delta)^2 =(\alpha\gamma)^2 + (\alpha\delta)^2 + (\beta \gamma)^2 + (\beta \delta)^2 \iff (\alpha\gamma + \beta \delta)^2 = \left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right) \notag \\
\iff& 2(\alpha\gamma + \beta \delta) = 2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)} \qquad \ \text{(here using the hypothesis that $\alpha \gamma + \beta \delta \ge 0$)} \notag \\
\iff & \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\gamma + \beta \delta) = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 +2\sqrt{\left(\alpha^2 + \beta^2\right)\left(\gamma^2 + \delta^2\right)}\notag\\
\iff& \left(\alpha^2 +2\alpha\gamma + \gamma^2 \right)+ \left(\beta^2 +2 \beta \delta+ \delta^2\right) = \left(\sqrt{\alpha^2 + \beta^2}\right)^2 +2\sqrt{\alpha^2 + \beta^2}\sqrt{\gamma^2 + \delta^2} + \left(\sqrt{\gamma^2 + \delta^2}\right)^2\notag\\
\iff& (\alpha + \gamma)^2 + (\beta + \delta)^2 = \left(\sqrt{\alpha^2 + \beta^2} +\sqrt{\gamma^2 + \delta^2}\right)^2 \iff |a+ b|^2 = \left(|a| + |b|\right)^2 \iff |a+ b| = |a| + |b|
\end{align}
where in the last equivalence I used the fact that $|z|\ge 0, \ \forall z \in \mathbb{C}$.
Is my proof correct? And also, does anyone know a different (possibly shorter) method of proving the above statement? Any and all help would be greatly appreciated. Thank you!
| HINT:
Note that
$$|a+b|^2=|a|^2+|b|^2+2\text{Re}(a\bar b)$$
while $(|a|+|b|)^2=|a|^2+|b|^2+2|a||b|$
| {
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Use Cauchy formula to solve $\int _0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c} $ given $\sqrt{(a^2+b^2)}=1Use Cauchy formula to solve $$\int _0^{2\pi} \frac{dt}{a \cos t+ b \sin t +c} $$ Given $\sqrt{(a^2+b^2)}=1<c$.
I tried a variable substitution, but nothing elegant. Can anyone solve it using Cauchy formula?
Edit: there are simpler solutions, but the challenge is to solve it with complex analysis.
| It may still be desirable to integrate as follows,
\begin{align}
& \int_0^{2\pi} \frac{dt}{a\cos t+ b\sin t +c}\\
=& \int_0^{2\pi} \frac{dt}{\sqrt{a^2+b^2}\cos (t-\theta) +c}
= \int_0^{\pi} \frac{2dt}{\sqrt{a^2+b^2}\cos t +c}\\
= & \int_0^{\pi} \frac{2dt}{2\sqrt{a^2+b^2}\cos^2\frac t2+c-\sqrt{a^2+b^2}}\\
= & \int_0^{\pi} \frac{2\sec^2\frac t2 dt}{(c+ \sqrt{a^2+b^2} )+( c- \sqrt{a^2+b^2} )\tan^2\frac t2}\\
=& \frac{4}{\sqrt{c^2-a^2-
b^2} }\tan^{-1} \left( \sqrt{\frac{c-\sqrt{a^2+b^2}}{c+\sqrt{a^2+b^2}} }\tan\frac t2 \right)_0^{\pi}\\
=& \frac{2\pi}{\sqrt{c^2-a^2-b^2} }
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that satisfy the above conditions. I was also trying to find whether there are additional requirements on $c$ and $d$ so that positive integer solutions for $a$ and $b$ can even exist.
I found that this equation simplifies to $abc - ad - bd = 0$ so that $abc = d(a+b)$.
Also, since the equation is equivalent to $a+b = ab(\frac{c}{d})$, this means $a$ and $b$ are the roots of the quadratic $dx^2-abcx+abd = 0$ since their product is $ab$ and their sum is $a+b = ab(\frac{c}{d})$.
However, after I analyzed the quadratic I just ended up with $a = a$ and $b = b$.
Any ideas on how to solve this further?
Again, I need to find all the conditions on the positive integers $c$ and $d$ (where $\gcd(c,d) = 1$) such that positive integer solutions for $a, b$ can exist. And then also find the positive integer solutions for $a$ and $b$ given that those conditions are satisfied.
| $\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$
Above has solution:
$a=(3k-2)$
$b=(k-2)(3k-2)$
$c=(k-1)(3k-2)$
$d=(k-2)(3k-2)^2$
For k=5 we get:
$(a,b,c,d)=(13,39,52,507)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show $C(t)=(2\cos^3(t),\sin(2t),2\sin(t))$ is on the intersection of $x^2+y^2+z^2=4$ and $x^2+y^2=2x$ Given a curve defined as: $$C(t)=(2\cos^2(t),\sin(2t),2\sin(t))$$ for $0\le t\le2\pi$
Show that the curve is on the intersection of the sphere $x^2+y^2+z^2=4$ and $x^2+y^2=2x$.
The intersection of the two equation satisfies:
$$x=2-\frac{z^2}{2}$$
I think the curve is on the the intersection of the two curves if :
$$2\cos^3(t)=2-\frac{4\sin^2(t)}{2}$$
$$\cos^3(t)=\cos^2(t)$$
For all $0\le t\le2\pi$.
But solving the equation implies this is not true in general.
So what is the main strategy and how should I continue?
| Check if it belongs to the first surface
Replacing the curve in the first equation we get
$$x^2 + y^2 + z^2 =
4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) + 4\sin^2(t)$$
which is
$$4\cos^4(t )+ \sin^2(2t) + 4\sin^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)] + 4\sin^2(t) = 4\cos^2(t) + 4\sin^2(t) = 4$$
So $C(t)$ is on the first surface.
Check if it belongs to the second surface
Replacing the curve in the second equation we get
$$x^2 + y^2 = 4\cos^4(t )+ \sin^2(2t) =4\cos^4(t )+ 4\sin^2(t)\cos^2(t) =4\cos^2(t)[\cos^2(t) + \sin^2(t)]=4\cos^2(t)=2x$$
So $C(t)$ is on the second surface.
| {
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"url": "https://math.stackexchange.com/questions/3770695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
| Hint
You have
$$f(x)=(x^2-1)(ax+b)+(2x-5)$$
and
$$f(x)=(x^2-4)(cx+d)+(-3x+4)$$
From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus
$$\color{red}{c+d=\frac{4}{3}}$$
From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus
$$\color{red}{2a+b=\frac{-1}{3}}$$
You also get $$f(0)=-b-5=-4d+4 \implies\color{red}{4d-b=9}$$
NOTE You can make a simple observation about the coefficients of $x^3$ and $x^2$ in both the expressions for $f(x)$ to conclude that $a=c$ and $b=d$. With that the first two equations can suffice.
| {
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"source": "stackexchange",
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Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here
| First of all we need $-1\le x\le1$ for real $\arcsin(x)$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
the equality will be valid
if $$x^2+\dfrac12\le1\iff x^2\le1-\dfrac12\iff?\le x\le?$$
Or if $x^2+\dfrac12>1$ and $(\sqrt2x<0\iff x<0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \implies DE =\frac{1}{5}EC$$
| HInt:$$AB = 6,BC = 8,AC = 10\\AC^2=BC^2+AB^2\\10^2=6^2+8^2$$ SO $$\triangle ABC$$ Is right angle, $\widehat{B}=90$
IF you draw a line $OF$ then $BGOF$ is a rectangle so,$ BF =FG$
then for $OG=OF=OD$ they are the radius of the circle .
after that
$GA,HA$ are tangent to the circle, so
$$OH=OG=R\\H=G=90\\OA=OA \\\to \triangle AOG=\triangle AOH \Longrightarrow AG=AH$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}$$
Using $$\frac{n!}{2^n}=\frac{4!}{2^4}\frac{5}{2}\frac{6}{2}\cdots \cdots >1\Longrightarrow \frac{n!}{2^n}>1$$
So $$\frac{2^n}{n!}\leq 1\Longrightarrow \frac{2^n\cdot n^5}{n!}<n^5$$
How do i solve it Help me please, Thanks
| All solutions given are very good and effective, as a further alternative by root test we obtain
$$\sqrt[n]{\frac{2^n\cdot n^5}{n!}}=\frac{2\cdot \sqrt[n]{n^5}}{\sqrt[n]{n!}}\to 0$$
indeed by standard limits
*
*$\sqrt[n]{n^5}\to 1$
*$\sqrt[n]{n!}\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write the factored form of the polynomial with the above information. I get:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Whereas the provided solution is:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$
Here's my working:
Write out in factored form:
$f(x) = a\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
I know that $f(0)=18$ so:
$$18 = a\big(-\frac{1}{2}\big)^2(2)(-6)$$
$$18 = a\big(\frac{1}{4}\big)(2)(-6)$$
$$18 = -3a$$
$$a = -6$$
Thus my answer:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Where did I go wrong and how can I arrive at:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ ?
| Both forms are equivalent:
$$
-6(x-\frac{1}{2})^2(x+2)(x-6)\\
=-\frac{3}{2}\cdot 2^2(x-\frac{1}{2})^2(x+2)(x-6)\\
=-\frac{3}{2}(2(x-\frac{1}{2}))^2(x+2)(x-6)\\
=-\frac{3}{2}(2x-1)^2(x+2)(x-6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Multiplication group for $\mathbb Z_n$ modulo $n$ By definition:
Let $\mathbb Z^+_n = \{[0],[1],[2],\ldots,[n−1]\}$
$\mathbb Z^+_4 = \{[0],[1],[2],[3]\},$
but
how $\mathbb Z^*_{12}$ is $\{[1],[5],[7],[11]\}$ ?
how $\mathbb Z^*_{7}$ is $\{[1],[2],[3],[4],[5],[6]\}$ ?
| $\require{cancel}$
\begin{align}
\frac 1 {12} & = \frac 1 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 2 {12} & = \frac {\cancel2\times1}{\cancel2\times6} = \frac 1 6 \\[6pt]
\frac 3 {12} & = \frac {\cancel3\times 1}{\cancel3\times 4} = \frac 1 4 \\[6pt]
\frac 4 {12} & = \frac{\cancel4\times1}{\cancel4\times3} = \frac 13 \\[6pt]
\frac 5 {12} & = \frac 5 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 6 {12} & = \frac{\cancel6\times1}{\cancel6\times2} = \frac 1 2 \\[6pt]
\frac 7 {12} & = \frac 7 {12} & \text{No cancellation occurs here.} \\[6pt]
\frac 8 {12} & = \frac{\cancel4\times2}{\cancel4\times3} = \frac 2 3 \\[6pt]
\frac 9 {12} & = \frac{\cancel3\times3}{\cancel3\times4} = \frac 3 4 \\[6pt]
\frac{10}{12} & = \frac{\cancel2\times5}{\cancel2\times6} = \frac 5 6 \\[6pt]
\frac{11}{12} & = \frac{11}{12} & \text{No cancellation occurs here.} \\ {}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773712",
"timestamp": "2023-03-29T00:00:00",
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For which values of $\eta\in\mathbf{N}$, $\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$ is an integer ($\varphi\in\mathbf{N}_0$)? Recently, I have found this problem:
Determine for which values of $\eta\in\mathbf{N}$ and $\varphi\in\mathbf{N}_0$, the quantity:
$$\gamma=\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$$
is a positive integr ($\gamma\in\mathbf{N}$).
To solve this problem, I began studying the case where $\gamma=0$. Here $\varphi$ can be every number and $\eta=0$.
After that, I passed to the case where $\varphi=1$. So, I have:
$$\gamma=\frac{10^2\cdot\eta}{\eta-1}$$
The tecnique I used, is trying to transform the numerator into the denominator, so:
$$\gamma=\frac{10^2\cdot\eta-10^2+10^2}{\eta-1}=\frac{10^2\cdot(\eta-1)+10^2}{\eta-1}=10^2+\frac{10^2}{\eta-1}$$
Now, $10^2$ is an integer, so also $\frac{10^2}{\eta-1}$ must be an integer. In conclusion if $\varphi=1$ then the solutions are: $\eta=\{2,3,5,6,11,21,26,51,101\}$.
Then, I studied the case when $\varphi\mid 10^2$. Here, I can use the tecnique shown above. In fact:
$$\gamma=\frac{10^2\cdot\eta-\frac{10^2}{\varphi}+\frac{10^2}{\varphi}}{\varphi\cdot\eta-1}=\frac{\frac{10^2\cdot\varphi\cdot\eta-10^2\cdot\varphi}{\varphi}+\frac{10^2}{\varphi}}{\varphi\cdot\eta-1}=\frac{10^2}{\varphi}+\frac{10^2}{\varphi\cdot(\varphi\cdot\eta-1)}$$
In the hypotesis written abvove $\frac{10^2}{\varphi}$ is an integer, so $\frac{10^2}{\varphi\cdot(\varphi\cdot\eta-1)}$ must be an integer for a fixed $\varphi$.
The case where $\varphi\nmid10^2$, I think, is more complicated.
So, can we build a general method to find $\eta$ when $\varphi$ varies over the positive integers?
| Observe that $\varphi\cdot\eta -1$ is prime with $\eta$ (use Euclidean Division). So $\gamma$ is an integer iff $\varphi\cdot\eta -1$ divides $10^2$, do you know how to proceed?
| {
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How to find the f-related vector fields So lets say you have a vector field $X = (2x+y) \partial x + x \partial y$ and a diffeomorphism $f(x,y) = (x-2y,2x+y)$, then I need to know how to find the f-related vector fields of X, lets say $\overline{X}$. That is, those that meet:
$\overline{X} \circ f = \partial f \circ X \Rightarrow \overline{X} = \partial f \circ X \circ f^{-1}$
Since $f^{-1} = \left(\frac{x+2y}{5},\frac{y-2x}{5}\right)$
$X \circ f^{-1} = \left[2(x \circ f^{-1})+(y \circ f^{-1})\right] \partial (x \circ f^{-1}) + \left[x \circ f^{-1}\right] \partial (y \circ f^{-1}) = \left[2(\frac{x+2y}{5})+(\frac{y-2x}{5})\right] \partial (\frac{x+2y}{5}) + \left[\frac{x+2y}{5}\right] \partial (\frac{y-2x}{5}) = y (\frac{1}{5}\partial x +\frac{2}{5}\partial y) + \left[\frac{x+2y}{5}\right] (-\frac{2}{5}\partial x +\frac{1}{5}\partial y) = \frac{y-2x}{25}\partial x + \frac{12y+x}{25}\partial y$
And since $\partial f = (\partial x -2 \partial y , 2 \partial x +\partial y)$
$\overline{X} = \partial f \circ X \circ f^{-1} = (\partial x -2 \partial y , 2 \partial x +\partial y) \circ \left(\frac{y-2x}{25}\partial x + \frac{12y+x}{25}\partial y\right)$
And the above composition would give a two-component object, which differs from what I was trying to get, a vector field.
Please let me know how to approach this.
Edit1: So I've seen that I had two main mistakes in what I presented above.
*
*The first mistake is that is not $\partial f$ but $d f$, which corresponds to the jacobian matrix.
*The other mistake was related to $(x \circ f^{-1})$ and $(y \circ f^{-1})$
So the correct way to solve the problem would be:
$\overline{X} \circ f = d f \circ X \Rightarrow \overline{X} = d f \circ X \circ f^{-1}$
Since $f^{-1} = \left(\frac{x+2y}{5},\frac{y-2x}{5}\right)$
$X \circ f^{-1} = \left[2(x \circ f^{-1})+(y \circ f^{-1})\right] \partial x + \left[x \circ f^{-1}\right] \partial y = \left[2(\frac{x+2y}{5})+(\frac{y-2x}{5})\right] \partial x + \left[\frac{x+2y}{5}\right] \partial y = y \partial x + \frac{x+2y}{5}\partial y $
And the above result can be seen as the column vector $\begin{pmatrix}y \\ \frac{x+2y}{5} \end{pmatrix}$
And since $d f =\begin{pmatrix}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y}\\\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{pmatrix} = \begin{pmatrix}1 & -2\\2 & 1\end{pmatrix}$
$\overline{X} = d f \circ X \circ f^{-1} = \begin{pmatrix}1 & -2\\2 & 1\end{pmatrix}\begin{pmatrix}y \\ \frac{x+2y}{5} \end{pmatrix} = \begin{pmatrix}\frac{y-2x}{5} \\ \frac{x+12y}{5} \end{pmatrix} = \frac{y-2x}{5}\partial x + \frac{x+12y}{5} \partial y$
| I'll use the notation of An Introduction to Manifolds (second edition) written by Loring W.Tu (in particular, chapter one), so that $$X = (2x+y) \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$
A vector field $\overline{X}$ is $f$-related to $X$ if and only if $f_{*}X=\overline{X}$. More precisely, $f_{*, ~p}(X_p) = \overline{X}_{f(p)}$ for every point $p$ in the domain of $X$. In our case, $\overline{X}_q=f_{*, ~p}(X_{p})$ where $p=f^{-1}(q)$.
Note that \begin{align} \ f_* = \begin{bmatrix} 1 & -2 \\ 2 & 1\end{bmatrix} \end{align}
in a sense that $$f_{*} \left( \frac{\partial}{\partial x}\bigg|_{p} \right) = \frac{\partial}{\partial x}\bigg|_{q} + 2\frac{\partial}{\partial y}\bigg|_{q} $$
$$f_{*} \left( \frac{\partial}{\partial y}\bigg|_{p} \right) = -2\frac{\partial}{\partial x}\bigg|_{q} +\frac{\partial}{\partial y}\bigg|_{q} $$
Write $$\overline{X}=f(x,y)\frac{\partial}{\partial x}+g(x,y)\frac{\partial}{\partial y} $$
Put $q=(a,b)$. Then $$\overline{X}_{q}=f(a,b)\frac{\partial}{\partial x}\bigg|_{q} +g(a,b)\frac{\partial}{\partial y} \bigg|_{q}$$
On the other hand, $p=f^{-1}(q)=\frac{1}{5}(a+2b, b-2a )$.
Thus \begin{align} X_p &= \left( \frac{ 2a+4b+b-2a}{5} \right)
\frac{\partial}{\partial x}\bigg|_{p} + \left( \frac { a+2b }{5} \right)\frac{\partial}{\partial y}\bigg|_{p}
\\ &=b \frac{\partial}{\partial x}\bigg|_{p} + \left( \frac { a+2b }{5} \right)\frac{\partial}{\partial y}\bigg|_{p}\end{align}
Finally, \begin{align} f_{*, ~p}(X_p) &=b \left(\frac{\partial}{\partial x}\bigg|_{q} + 2\frac{\partial}{\partial y}\bigg|_{q} \right)+ \left( \frac { a+2b }{5} \right) \left( -2\frac{\partial}{\partial x}\bigg|_{q} +\frac{\partial}{\partial y}\bigg|_{q}\right) \\ &=
\left( \frac { -2a+b }{5} \right) \frac{\partial}{\partial x}\bigg|_{q} + \left( \frac { a+12b }{5} \right) \frac{\partial}{\partial y}\bigg|_{q} \end{align}
Therefore, $f(a,b) = \frac{1}{5}(-2a+b) $ and $g(a, b)= \frac{1}{5}(a+12b)$.
In short, $$\overline{X}= \left( \frac { -2x+y }{5} \right)\frac{\partial}{\partial x} +\left( \frac { x+12y }{5} \right)\frac{\partial}{\partial y}$$
| {
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Prove by induction $1 + z + z^2 + .... +z^n = \frac{z^{n+1}-1}{z-1}$ for $z \neq$ 1 Prove $$1 + z + z^2 \ldots +z^n = \frac{z^{n+1}-1}{z-1} \quad \text{ where } z\ne 1 $$
So I started to prove this by induction.
What have I done so far:
Let $n= 1$
$$1+ z + z^2 \ldots+z^n = \frac{z^{n+1}-1}{z-1} \\
1+z \stackrel {?}{=}\frac{z^{2}-1}{z-1} \\~\\
1+z \stackrel {?}{=} \frac{(z+1)(z-1)}{z-1} \\
1+ z \stackrel{?}{=} z+1 \quad \checkmark$$
But I think this is wrong? And I don't know what to do next
and how to do it? But for (a) and (b) use
$e^{i\theta}$ and go from there. Or is this proof an assumption?
Actual question:
Prove that if $z \ne 1$, then
$$
1+z +z^{2}+\cdots+z^{n}=\frac{z^{n+1}-1}{z-1}
$$
Use this result and De Moivre's formula to establish the
following identities.
(a) $1+ \cos \theta+\cos 2 \theta+\cdots+\cos n\theta=\dfrac{1}{2}+\dfrac{\sin\left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)}$
(b) $\sin \theta+ \sin 2 \theta+\cdots+\sin n
\theta=\dfrac{\sin (n \theta / 2) \sin ((n+1) \theta / 2)}{sin (\theta / 2)}$,
where $0<\theta<2 \pi$.
| This is provable using standard algebra; however, if you wish to do this by induction:
For ${n=1}$, we get ${1 + z = \frac{z^2-1}{z-1}=z+1}$, so it works. Now assume
$${1 + z + z^2 + ... + z^k = \frac{z^{k+1}-1}{z-1}}$$
This would imply that
$${1 + z + z^2 + ... + z^{k} + z^{k+1} = \frac{z^{k+1}-1}{z-1} + z^{k+1}}$$
Now we simplify the right hand side
$${=\frac{z^{k+1}-1}{z-1} + \frac{z^{k+1}(z-1)}{z-1} = \frac{z^{k+2}-1}{z-1}}$$
Altogether that's
$${1 + z + z^2 + ... + z^k + z^{k+1} = \frac{z^{k+2}-1}{z-1}}$$
which is the correct formula. And so we have shown that if it works for ${n=k}$, that would imply it works for ${n=k+1}$. We have shown it works for ${n=1}$, and so it works for ${n=2}$ then ${n=3}$... so by the Principle of Mathematical Induction,
$${1 + z + z^2 + z^3 + ... + z^{n} = \frac{z^{n+1}-1}{z-1}}$$
as required.
Algebra proof
Thought I'd add the Algebra proof because why not.
Define
$${S_n = 1 + z + z^2 + ... + z^n}$$
Then you see this is
$${= 1 + z(1 + z + z^2 + ... + z^{n-1})}$$
But
$${1 + z + z^2 + ... + z^{n-1} = (1 + z + z^2 + ... + z^n)-z^n = S_n-z^n}$$
So we get overall
$${S_n = 1 + z(S_n - z^n)}$$
We can now rearrange for ${S_n}$;
$${\Rightarrow S_n(1-z) = 1-z^{n+1}}$$
And so
$${S_n = \frac{1-z^{n+1}}{1-z}}$$
This looks different from your form, but it's the same since we just multiplied by ${\frac{-1}{-1}}$:
$${\Rightarrow S_n = \frac{1-z^{n+1}}{1-z}\times \frac{-1}{-1} = \frac{z^{n+1}-1}{z-1}}$$
And so
$${1 + z + z^2 + ... + z^n = \frac{z^{n+1}-1}{z-1}}$$
Part (a) and (b)
Indeed for part (a) and (b) you want to use the fact that
$${(e^{i\theta})^{n} = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)}$$
After applying this formula with ${z=e^{i\theta}}$, you want to equate both real and imaginary parts on the left and right hand sides of the formula. Can you take it from here?
| {
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Graphing $ \lim _{n \rightarrow \infty} \frac{{|x+1 \mid}^{n}+x^{2}}{|x|+x^{2n}} $ Graph: $$
\lim _{n \rightarrow \infty} \frac{{|x+1 \mid}^{n}+x^{2}}{|x|+x^{2n}}
$$
Here I tried first to resolve the limit, dividing by ${x}^{2n}$ in both numerator and plugging in $\infty$ I obtain the limit to be zero for all values of $x$. But when it tried to graph it using a graphing calculator I obtained the following graph (I put n=10000),
Can anyone please point out my mistake?
| Case by case:
If $x>\varphi = \frac{1+\sqrt{5}}{2}$, then $x+1<x^2$ and we can divide by $x^{2n}$ to se that:
$$
\lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \frac{(\frac{x+1}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{0+0}{0+1}=0
$$
If $0<x<\varphi$, then $x+1>x^2$ , thus
$$
\lim_n \frac{(x+1)^n + x^2}{x+(x^2)^n}=\lim_n \frac{(\frac{x+1}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{\infty+0}{0+1}=\infty
$$
If $-1<x<0$, then $x^2<1$ and $0<x+1<1$, and the limit turns out to be:
$$
\lim_n \frac{(x+1)^n + x^2}{-x+(x^2)^n}=\frac{x^2}{-x}=-x
$$
If $-2<x<-1$, then $|x+1|<1$ and $x^2>1$, resulting in
$$
\lim_n \frac{|x+1|^n + x^2}{-x+(x^2)^n}=\frac{0+x^2}{-x+\infty}=0
$$
And finally, for $x<-2$ we can divide by $x^2$,
and taking into account that $\frac{|x+1|}{x^2}<1$ we get that
$$
\lim_n \frac{|x+1|^n + x^2}{-x+(x^2)^n}=\lim_n \frac{(\frac{-(x+1)}{x^2})^n + x^{2-2n}}{x^{1-2n}+1}=\frac{0+0}{0+1} = 0
$$
Maybe we could have considered the last two together, but I think that separating them helps conceptually.
| {
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
| Another way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $v^2=1$ and we need to prove that:
$$a^2b^2c^2+\sum_{cyc}(a^2b^2+a^2)\geq7$$ or $f(w^3)\geq0,$ where $$f(w^3)=w^6+(9v^4-6uw^3)v^2+(9u^2-6v^2)v^4-7v^6.$$
But, $$f'(w^3)=2w^3-6uv^2w^3<0,$$ which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which by $uvw$ happens for equality case of two variables.
Since our inequality is symmetric, we can assume that $b=a$, which gives $c=\frac{3-a^2}{2a}.$
Can you end it now?
For $a^2=x$ I got $$(x-1)^2(x^2+2x+9)\geq0.$$
| {
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Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.
| You have $x^{1/4}+(x+1)^{1/4}=(2x+1)^{1/4}$
Raise both sides to the power $4$ and you have:
$$x+(x+1)+4x^{3/4}(x+1)^{1/4}+6x^{1/4}(x+1)^{1/4}+4x^{1/4}(x+1)^{3/4}=2x+1$$
$$x^{1/4}(x+1)^{1/4}\Big(2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}\Big)=0$$
From here we have $x=0$ or $x=-1$ as solution.
Now if $x\neq 0$ and $x\neq -1$, then $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=0$
Observe that $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=2(x^{1/2}+2x^{1/4}(x+1)^{1/4}+(x+1)^{1/2})-x^{1/4}(x+1)^{1/4}=2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2-x^{1/4}(x+1)^{1/4}$
$$2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2=x^{1/4}(x+1)^{1/4}$$
$$2(2x+1)^{1/2}=x^{1/4}(x+1)^{1/4}$$
Raising both the sides to the power of $4$ gives:
$$16(2x+1)^2=x(x+1)$$
This in fact will give us an quadratic equation. I hope you can solve from here.
| {
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Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit numbers.
Five digit numbers:
Let $x \in \{2, 3, 4, 5\}$. If the first position in a five digit number is not $x$, then there are $5$ possibilities for this position as there are four values for $x$ and $0$ is inadmissable. The rest of the four positions will have the various permutations of four values of $x$. There are $5 \times 4!$ such numbers. If the non-$x$ value is in the second position, then there are $4$ ways to choose an $x$-value for the first position, $6$ integers for the second position and $3!$ permutations for the rest of the positions. There are $4\times 6 \times 3!$ such numbers. If the non-$x$ value is in the third position, then there are $\binom 42$ ways to choose two $x$-values, $2!$ ways to permute them and $6$ integers for the third position meaning there are $6 \times 2 \times 6$ such numbers. When the non-$x$ value is in the fourth position, there are $4 \times 3! \times 6$ such numbers. Finally, if non-$x$ is in the fifth position, there are $4! \times 6$ such numbers.
Four digit numbers:
We just need to permute the number $2345$. There are $4!$ such permutations.
Thus the number of numbers with the given restrictions is $5\times 4! + 4\times 3!\times 6 + 2\times 6 \times 6 + 4 \times 3! \times 6 + 6 \times 4! + 4! = 648$.
What did I forget to take into account? Thanks.
| Four digit numbers are accounted for by the $4!=\underline{24}$ ways to arrange the four digits $2,3,4,5$.
Five digit numbers not containing $0$ are accounted for by $5\times5!=\underline{600}$ (arranging the four digits $2,3,4,5$ then one from $\{1,6,7,8,9\}$).
Five digit numbers containing $0$ are accounted for by the $5!-4!=\underline{96}$ ways to arrange the five digits $0,2,3,4,5$ while fixing $0$ to the second to fifth digit.
In total, we have $24+600+96=\boxed{720}$ as desired.
| {
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Find value of $AB$
Let
$$A=\dfrac{1}{\sin20^\circ\cos40^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ\cos40^\circ}$$
and
$$B=\cot40^\circ+\cot230^\circ-\tan185^\circ(\tan230^\circ-\cot50^\circ)$$
Find $AB$.
My attempt :
\begin{align*}
A&=\dfrac{1}{\cos40^\circ}(\dfrac{1}{\sin20^\circ}+\dfrac{1}{\sqrt{3}\cos20^\circ})\\
&=\dfrac{2(\sqrt{3}\cos20^\circ+\sin20^\circ)}{\sqrt{3}\cos^2 40^\circ}\\
B&=\tan50^\circ(1-\tan5^\circ)+\tan40^\circ(1+\tan5^\circ)
\end{align*}
I tried to factor $A$ and $B$ because I think it will cancel each other out.
| \begin{align*}
A &= \frac{4(\sin 60^{\circ} \cos 20^{\circ} + \cos 60^{\circ} \sin 20^{\circ})}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} = \frac{4\sin 80^{\circ}}{\sqrt{3} \sin 40^{\circ}\cos 40^{\circ}} \\&= \frac{8\sin 80^{\circ}}{\sqrt{3}\sin 80^{\circ}} = \frac{8\sqrt{3}}{3}\\
B &= \cot 40^{\circ} + \cot 230^{\circ} - \tan 185^{\circ}(\tan 230^{\circ} - \cot 50^{\circ}) \\ &=\cot 40^{\circ} + \tan 40^{\circ} - \tan 5^{\circ}(\tan 50^{\circ} - \cot 50^{\circ}) \\ &= \cot 40^{\circ} + \tan 40^{\circ} - \tan 5^{\circ}(\cot 40^{\circ} - \tan 40^{\circ})\\
\therefore AB \tan 40^{\circ} &=\frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - \tan 5^{\circ} ( 1 - \tan^{2} 40^{\circ}) \right ) \\&= \frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - (\frac{1- \tan 40^{\circ}}{1 + \tan 40^{\circ}}) ( 1 - \tan^{2} 40^{\circ}) \right ) \\ &= \frac{8\sqrt{3}}{3} \left ( 1 +\tan^{2} 40^{\circ} - ( 1 - \tan 40^{\circ})^{2} \right ) \\ &= \frac{16\sqrt{3}}{3} \tan 40^{\circ}
\end{align*}
| {
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Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore
\begin{align*}
(\gamma - \beta)^3 &= \alpha^3 = -\alpha - 1 \\
\gamma^3 - 3\gamma^2\beta+3\gamma\beta^2-\beta^3 &= -\alpha - 1
\end{align*}
and using $\beta^2+\beta-3=0$, i get $\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1$.
I think the minimal polynomial of $\alpha+\beta$ has degree $6$, then i was trying raise to the $2^{nd}$ power the last equation, but i don't get nothing to obtain the minimal polynomial of $\alpha + \beta$.
| Let $\alpha=\alpha_1$ be one of the roots, $\alpha_i,i=1,2,3,$ of $x^3+x+1$ and let $\beta=\beta_1$ be one of the roots, $\beta_j,j=1,2,$ for $x^2+x-3$. Consider the polynomial $$f(x)=\prod_{i,j}(x-(\alpha_i+\beta_j)).$$ Then by the relation between coefficients of a polynomial and its roots, one sees that $f(x)\in {\mathbb Q}[x]$ and $f(\alpha+\beta)=0$. To find $f(x)$, it just takes some algebraic manipulations: $$f(x)=\prod_j\prod_i((x-\beta_j)-\alpha_i)$$
$$=\prod_j((x-\beta_j)^3+(x-\beta_j)+1)$$
$$=\prod_j((x^3-3\beta_jx^2+3\beta_j^2x-\beta_j^3)+(x-\beta_j)+1)$$
$$=\prod_j((x^3+10x+4)-\beta_j(3x^2+3x+5)),\quad (1)$$ where one uses the relations $\beta_j^2=3-\beta_j,\beta_j^3=-3+4\beta_j$ derived from $\beta_j^2+\beta_j-3=0$. Note that $$\beta_1\beta_2=-3,\beta_1+\beta_2=-1.\quad (2)$$ Now expanding (1) and using (2), one sees that $$f(x)=(x^3+10x+4)^2-(\beta_1+\beta_2)(x^3+10x+4)(3x^2+3x+5)+\beta_1\beta_2(3x^2+3x+5)^2$$
$$=x^6+3x^5-4x^4-11x^3+25x^2+52x-39.$$
To show that $f(x)$ is the minimal polynomial of $\alpha+\beta$, reduce its coefficients mod $2$ to get $$\overline{f(x)}=x^6+x^5+x^3+x^2+1,$$ which is irreducible over ${\mathbb F}_2$, since it is not divisible by $x,x+1,x^2+x+1,x^3+x^2+1$ and $x^3+x+1$. It follows that $f(x)$ irreducible over ${\mathbb Q}$. QED
| {
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Limit of ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }}$ This question comes from Hubbard's textbook "Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach" (5th edition, exercise 1.5.16b)
I wrote a $\epsilon -\delta$ proof to show ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }} = 0$, but wanted to ask to check if what I wrote was correct:
Suppose we have $\epsilon > 0$.
*
*First, it can be shown (with L'Hoptial for example) that $\lim_{x\to 0} |x| \ln(x^k) = 0$ for even positive integers $k$. Thus there exists $\delta_{1;k} > 0$ such that $|x| < \delta_{1;k}$ implies $\left| |x| \ln(x^k) \right| = \left| k |x| \ln(x) \right| < \frac{k\epsilon}{6}$. Therefore: $|x| \ln(x) > - \frac{\epsilon}{6}$.
*Secondly, clearly we have $\lim_{x \to 0} {|x| \ln(x^2 + c)} = \lim_{y \to 0} {|y| \ln(y^4 + c)} = 0$ for any $c > 0$. So there exists $\delta_{2; c}, \delta_{3;c} > 0$ such that:
*
*$|x| < \delta_{2;c} \implies \left| |x| \ln(x^2 + c) \right| < \frac{\epsilon}{2}$.
*$|y| < \delta_{3;c} \implies \left| |y| \ln(y^4 + c) \right| < \frac{\epsilon}{2}$.
Let $\delta = \min(\epsilon, \delta_{1;2}, \delta_{1;4}, \delta_{2;\epsilon}, \delta_{3;\epsilon})$. So if $\sqrt{x^2 + y^2} < \delta$, note that this implies $|x| < \delta$ and $|y| < \delta$. Then:
$$\begin{align} (|x| + |y|) \ln(x^2 + y^4) &= |x| \ln(x^2 + y^4) + |y| \ln(x^2 + y^4) \\ &\leq |x| \ln(x^2 + \epsilon^4) + |y| \ln(\epsilon^2 + y^4) \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}$$
As $x^2 + y^4 < \epsilon^2 + y^4$ (as $|x| < \delta \leq \epsilon$) and $\ln(x)$ is a strictly increasing implies $\ln(x^2 + y^4) < \ln(\epsilon^2 + y^4)$. Since $|y| \geq 0$ then $|y| \ln(x^2 + y^4) \leq |y| \ln(\epsilon^2 + y^4)$.
Moreover:
$$\begin{align} (|x| + |y|) \ln(x^2 + y^4) &= |x| \ln(x^2 + y^4) + |y| \ln(x^2 + y^4)
\\ &\geq |x| \ln(x^2) + |y| \ln(y^4) \\ &= 2|x|\ln(x) + 4|y| \ln(y) \\ &> -\frac{2\epsilon}{6} - \frac{4\epsilon}{6} = -\epsilon \end{align}$$
Putting these together we have $\left| (|x| + |y|) \ln(x^2 + y^4) \right| < \epsilon$.
Therefore ${ \lim_{(x,y)\to(0,0)} {(\left| x \right| + \left| y \right|) \ln{(x^2 + y^4)} }} = 0$.
| Given that the function is even with respect to $x$ and with respect to $y,$ it is enough to consider $x,y\geq0,$ and $(x,y)\neq(0,0).$ I would write
$$
\lim_{(x,y)\to(0,0)}\frac{x+y}{\sqrt[4]{x^2+y^4}}\cdot\sqrt[4]{x^2+y^4}\log(x^2+y^4).
$$
Now, it should be rather easy to see that
$$
\lim_{(x,y)\to(0,0)}\sqrt[4]{x^2+y^4}\log(x^2+y^4)=0.
$$
In fact, set $z=g(x,y)=x^2+y^4,$ prove that $g(x,y)\to0$ (use the continuity), then
$$
\lim_{z\to0}\sqrt[4]{z}\log(z)=0,
$$
while the other factor is bounded in a neighbourhood of $(0,0).$ We can equivalently consider its fourth power
$$
\frac{(x+y)^4}{x^2+y^4},\qquad x,y\geq0,\ (x,y)\neq(0,0),
$$
and use polar coordinates
$$
\frac{\rho^2(\cos\theta+\sin\theta)^4}{\cos^2\theta+\rho^2\sin^4\theta},\qquad\rho>0,\ 0\leq\theta\leq\pi/2
$$
It is easy to see that, whenever $0<\rho<1/\sqrt{2}$, the denominator satisfy
$$
\cos^2\theta+\rho^2\sin^4\theta\geq\rho^2
$$
so that
$$
0<\frac{(x+y)^4}{x^2+y^4}=\frac{\rho^2(\cos\theta+\sin\theta)^4}{\cos^2\theta+\rho^2\sin^4\theta}\leq(\cos\theta+\sin\theta)^4\leq4
$$
and
$$
0<\frac{x+y}{\sqrt[4]{x^2+y^4}}\leq\sqrt{2},\qquad x,y\geq0,\ 0<x^2+y^2<1/2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting limit: $\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}$ I would like to prove that
$$\lim_\limits{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3\sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}=0$$
but I am stuck.
I tried to solve it by using Euler-Maclaurin formula, but I could not to.
Euler-Maclaurin formula applied to the function $f(x)=\sin x \sin\sqrt{x}\;\;$ is the following:
$$\sum_{h=1}^n\sin h\sin\sqrt{h}=\int_\limits{0}^n\left[\sin x\sin\sqrt{x}+\left(x-\lfloor x\rfloor\right)\left(\cos x\sin\sqrt{x}+\frac{\sin x\cos\sqrt{x}}{2\sqrt{x}}\right)\right] \, dx$$
but I could not manage to prove that
$$\frac{1}{n}\int_\limits{0}^n\left(x-\lfloor x\rfloor\right)\left(\cos x \sin\sqrt{x} \right) \, dx\rightarrow 0 \text{ as } n\to\infty.$$
Moreover I tried to write the limit as a limit of a Riemann sum, but I did not manage to.
Furthermore I tried to prove the following inequality:
$$\left|\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\cdots+\sin n \sin\sqrt{n} \right|\le\sqrt[4]{n^3}\\\text{for all }\;n\in\mathbb{N},$$
but it was not successful.
I managed to prove that
$$\lim_{n\to\infty}\frac{\sin 1+\sin 2 +\sin 3+\ldots+\sin n}{n}=0$$
and
$$\lim_{n\to\infty}\frac{\sin\sqrt{1}+\sin\sqrt{2}+\sin\sqrt{3}+\cdots+\sin\sqrt{n}}{n}=0.$$
Is it possible to use these last two limits in order to prove that
$$\lim_{n\to\infty}\frac{\sin 1\sin\sqrt{1}+\sin 2\sin\sqrt{2}+\sin 3 \sin\sqrt{3}+\cdots+\sin n\sin\sqrt{n}}{n}=0\text{ ?}$$
I tried to use Cauchy-Schwartz inequality, but I got $$\lim_{n\to\infty}\frac{\sin^21+\sin^22+\cdots+\sin^2n}{n}$$ and $$\lim_{n\to\infty}\frac{\sin^2\sqrt{1}+\sin^2\sqrt{2}+\cdots+\sin^2\sqrt{n}}{n}$$ and these last two limits are not zero in fact there are both $\frac{1}{2}$.
| Let $S_n$ be given by
$$S_n=\sum_{k=1}^n \sin(k)\sin\sqrt{k}\tag1$$
Applying summation by parts to the sum in $(1)$ reveals
$$S_n=\sin(\sqrt {n+1})\sum_{k=1}^{n}\sin(k)-\sum_{k=1}^n \left(\sum_{\ell=1}^k \sin(\ell)\right)\left(\sin(\sqrt {k+1})-\sin(\sqrt{k})\right)\tag 2$$
ESTIMATES:
The sum $\sum_{\ell=1}^k \sin(\ell)$ can be evaluated in closed form which provides the estimate
$$\begin{align}
\left|\sum_{\ell=1}^n \sin(\ell)\right|&=\left|\csc(1/2)\sin(n/2)\sin((n+1)/2)\right|\\\\
\le \csc(1/2)\tag3
\end{align}$$
Moreover, from the Prosthaphaeresis identities, we have the estimate
$$\begin{align}
\left|\sin(\sqrt {k+1})-\sin(\sqrt{k}\right|&=\left|\frac12\cos\left(\frac{\sqrt{k+1}+\sqrt{k}}{2}\right)\sin\left(\frac{\sqrt{k+1}-\sqrt{k}}{2}\right)\right|\\\\
&=\left|2\cos\left(\frac{\sqrt{k+1}+\sqrt{k}}{2}\right)\sin\left(\frac{1}{2(\sqrt{k+1}+\sqrt{k})}\right)\right|\\\\
&\le \frac{1}{\sqrt{k}}\tag4
\end{align}$$
Using the estimates in $(3)$ and $(4)$ in $(2)$, we find that
$$\begin{align}
|S_n|&\le \csc(1/2)\left(1+\sum_{k=1}^n\frac1{\sqrt k}\right)\\\\
&\le \csc(1/2)(1+2\sqrt n)\tag5
\end{align}$$
Finally, using the estimate in $(5)$ we have
$$\left|\frac{S_n}{n}\right|\le \frac{\csc(1/2)(1+2\sqrt n)}{n}$$
whence application of the squeeze theorem recovers the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{\sum_{k=1}^n \sin(k)\sin(\sqrt k)}{n}=0}$$
NOTE: We have tacitly found that $$\limsup_{n\to \infty}\frac{S_n}{\sqrt n}\le 2\csc(1/2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$
Compute:
$$\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$$
Answer: $\frac{\pi^2}{6\sqrt 3}$
My Attempt:
The obvious substitution: $\arcsin \sqrt x=t$. This transforms my integral (say $I$) to:
$$I=\int_0^{\frac{\pi}{2}} \frac{t\cdot 2\sin t\cos t}{\sin^4-\sin^2+1}dt$$
Then the substitution $t\rightarrow \left(\frac{\pi}{2}+0\right)-t$. This leads to the denominator remaining the same.
Then, I added the "versions" of $I$ and on some simplifications, obtained:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\underbrace{\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}}_{f(t)}dt$$
Since $f(x)=f\left(\frac{\pi}{2}-x\right)$ we say:
$$I=\frac{\pi}{2}\int_0^{\frac{\pi}{4}}\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}dt$$
Now the substitution $2t=u$ transforms it to:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\frac{\sin u}{1-\frac{\sin^2 u}{4}}du$$
Or,
$$I=\pi\int_0^{\frac{\pi}{2}}\frac{\sin u}{4-\sin^2 u}du$$
Then I tried to decompose it into partial fractions and integrate them individually, by converting them in terms of $\tan \frac u2$ but the answer I'm getting doesn't match the given one.
Thanks in advance!
| We could complete this integral with one integration-by-parts:
$$\int_0^1 \frac{\arcsin\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}dx = \frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\arcsin\sqrt{x}\Biggr|_0^1 - \frac{1}{\sqrt{3}}\int_0^1\frac{\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{x}\sqrt{1-x}}dx$$
$$= \frac{\pi^2}{6\sqrt{3}}-\frac{1}{\sqrt{3}}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{\arctan\left(\frac{2x}{\sqrt{3}}\right)}{\sqrt{\frac{1}{2}+x}\sqrt{\frac{1}{2}-x}}dx = \frac{\pi^2}{6\sqrt{3}}$$
where the second integral vanishes by odd symmetry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Further, the expression $[x_1^2+x_2^2+x_3^2+x_4^2]$ can be written as $[(x_1+x_3)^2+(x_2+x_4)^2-2(x_1x_3+x_2x_4)].$ I'm having trouble eliminating $(x_1x_3+x_2x_4)$ from this expression. Failing to make any sense out of this, I manipulated the existing expressions to deduce $$x_1x_2+x_1x_4+x_2x_3+x_3x_4=\frac{a^2-b^2}{4}$$and $$(x_1^2+x_3^2)-(x_2^2+x_4^2)+2(x_1x_3-x_2x_4)=a\cdot b$$Beyond this, I cannot make sense of the expressions anymore. I have no idea how to proceed with simplifying the expressions further, and would appreciate hints in the same direction.
| By your work and by C-S
$$x_1^2+x_2^2+x_3^2+x_4^2\geq\frac{1}{2}\left(\frac{a+b}{2}\right)^2+\frac{1}{2}\left(\frac{a-b}{2}\right)^2=\frac{a^2+b^2}{4}.$$
The equality occurs for $x_1=x_3=\frac{a+b}{4}$ and $x_2=x_4=\frac{a-b}{4},$ which says that we got a minimal value.
We used the following C-S:
$$x^2+y^2=\frac{1}{2}(1^2+1^2)(x^2+y^2)\geq\frac{1}{2}(x+y)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
| We can rewrite,
$$-8^{\frac13}=2.(-1)^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi + \pi}{3}\right)$$
And on substituting,
$k=0,1 $ and $2$ we get our desired answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate integral $\int^{+\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx$? I've posted a similar integral earlier, in which the Goodwin-Staton Integral is involved, making the integral unsolvable. Now I make a little modification to make it solvable and give my answer below.
| $$\begin{array}\ \int^{+\infty}_0 \frac{e^{-x^2}}{(x^2+\frac{1}{2})^2} dx &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx - \int^{+\infty}_0 \frac{x^2e^{-x^2}}{(x^2+\frac{1}{2})^2} dx\bigg]\\ &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx + \frac{1}{2}\int^{+\infty}_0xe^{-x^2}d(\frac{1}{x^2+\frac{1}{2}})\bigg]\\ &= 2\bigg[\int^{+\infty}_0 \frac{e^{-x^2}}{x^2+\frac{1}{2}} dx - \frac{1}{2}\int^{+\infty}_0\frac{1-2x^2}{x^2+\frac{1}{2}}e^{-x^2}dx\bigg]\\ &= 2\int^{+\infty}_0e^{-x^2}dx\\ &= \sqrt{\pi} \end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
find a value of $a$ so the equation below has some solutions. I need find a value of $a$ so the equation below has some solutions.
$\begin{pmatrix}
3 & 1 &-5&2\\
-1 & 0 &2&-1\\
2&1&-3&1\\
3&-2&4&-1
\end{pmatrix}
\left(
\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{array}
\right)=
\left(
\begin{array}{c}
2a-3 \\
-a+1 \\
a-2 \\
-2a+5 \\
\end{array}
\right)$
I have tried that find the value by setting$$\operatorname{rank}(A)=\operatorname{rank}(\tilde A)$$
where $\tilde A$ is an enlarged coefficient matrix of $A$.
Then I have found $\operatorname{rank}(A)=3$ by row manipulation.
Similarly, I did the row manipulation and ended up with
$$\begin{pmatrix}
1 & 0 &0&1&-a-3\\
0 & 1 &0&0&1\\
0&0&1&-1&-a-1\\
0&0&0&0&-2
\end{pmatrix}$$
Doesn't this mean that we cannot have solution whatever the value is as we end up with $0=2$ at the bottom line?
Can anyone help me?
| You would be right, but you made a mistake in your computations. Last row should be all zeroes. Actually, the RREF is
$$
\begin{bmatrix}
1&0&0&0&-a/6+1/3\\
0&1&0&-1/2&-5a/12-2/3\\
0&0&1&-1/2&-7a/12+2/3\\
0&0&0&0&0
\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum of the function $F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}$ I am stuck with finding the maximum of the function
\begin{align}
F(x,y,z):=\frac{x(1-x)y(1-y)z(1-z)}{1-(1-xy)z}
\end{align}
on the compact interval $[0,1]\times[0,1]\times[0,1].$
I think the Lagrange multiplier may work, but I have no idea to set a constraint function. Is there any hint I can follow or another method ? Any suggestions I will be grateful.
| Taking partial derivatives with respect to $x,y,z$ we have
$$\begin{cases}
\dfrac{\partial F}{\partial x} = -\dfrac{(y - 1) y (z - 1) z (x^2 y z - 2 x (z - 1) + z - 1)}{(z (x y - 1) + 1)^2}\\
\dfrac{\partial F}{\partial y} =-\dfrac{(x - 1) x (z - 1) z (x y^2 z - 2 y z + 2 y + z - 1)}{(x y z - z + 1)^2}\\
\dfrac{\partial F}{\partial z} =-\dfrac{(x - 1) x (y - 1) y (x y z^2 - z^2 + 2 z - 1)}{(x y z - z + 1)^2}
\end{cases}$$
Equating them to $0$ and dropping unnecessary factors which clearly don't lead to a maximum, we get
$$\begin{cases}
x^2 y z - 2 x (z - 1) + z - 1=0\\
x y^2 z - 2 y z + 2 y + z - 1=0\\
x y z^2 - z^2 + 2 z - 1 =0
\end{cases}$$
Then the desired solution is, according to WA
$$\begin{cases}
x = \sqrt{2} - 1\\
y = \sqrt{2} - 1\\
z = \frac{1}{\sqrt{2}}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
which does not match with the answer on Desmos.
| Hint:
*
*If $A>B>0$ then $A^2>B^2$. Apply this to $\sqrt{9-x^2}>x^2+1$.
*As $x^2+1$ is always positive then $9-x^2>0$.
Both must be satisfied. So you need $\cap$ to intersect the solution sets from both cases.
The hint is exapanded.
\begin{gather}
9-x^2>x^4+2x^2+1\\
x^4+3x^2-8<0\\
(x^2-\frac{-3+\sqrt{41}}{2})(x^2-\frac{-3-\sqrt{41}}{2})<0
\end{gather}
that must be intersected with
\begin{gather}
9-x^2>0\\
x^2<9
\end{gather}
The solution is $x^2< \frac{-3+\sqrt{41}}{2}$ or $$-\sqrt{\frac{-3+\sqrt{41}}{2}}<x< \sqrt{\frac{-3+\sqrt{41}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $S=2016^2 + 2015^2 +2014^2 -2013^2 -2012^2 -2011^2 .....+6^2 +5^2 +4^2-3^2-2^2-1^2$ Simplifying the expression as
$$2016^2+2015^2+2014^2+2013^2+2012^2+2011^2...6^2+5^2+4^2+3^2+2^2+1^2-2(2013^2+2012^2+2011^2 +2007^2+2006^2+ 2005^2....3^2+2^2+1^2)$$
Now the left part can be evaluated by using $\sum = \frac{(n)(n+1)(2n+1)}{6}$, but I am not sure about the right part. How should I solve that one?
| Note that $2016=6\cdot 336.$ Then rewrite the whole sum as:
$$S = \sum_{k=1}^{336}\left((6k)^2+(6k-1)^2+(6k-2)^2-(6k-3)^2-(6k-4)^2-(6k-5)^2\right) = $$
$$ = \sum_{k=1}^{336}(108k-45)=...$$
Or, if you insist on your own method, then the thing in the bracket you are struggling with is simply:
$$T = \sum_{k=1}^{335}\left((6k+3)^2 + (6k+2)^2+(6k+1)^2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$$x \to \frac{x-1}{x+1}$$
After considerable algebra, a cubic would be obtained, and then the sum can easily be found out
My question: This method is touted around a lot, but I never around how it actually works, especially in this case. If $\frac{x-1}{x+1} =\alpha$, then the transformed equation would be $\alpha^3 -2007 \alpha +2002$, and the sum of all $\alpha$ should still be $2007$. Can I get an explanation for this method?
| The equation is expanded to $$\frac{3abc+(ab+ac+bc)-(a+b+c)-3}{abc+(ab+ac+bc)+(a+b+c)+1}.$$
By Vieta's, $$abc = -2002$$ $$ab+ac+bc = -2007$$ $$a+b+c = 0$$
Therefore, the expression is $$\frac{3(-2002) + (-2007) - 0 - 3}{-2002 + (-2007) + 0 + 1}.$$
This simplifies to $$\boxed{+2.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove $\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d}$? My question:
How do I prove the following property of ratio?
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d} \tag{1}$$
$a,b,c,d \in \mathbb{R} \backslash \{ 0 \}$
I want to use the result in argument below.
Use the sine rule to establish the following identities for triangles:
$$\frac{a+b}{c}=\frac{\sin(A) +\sin(B)}{\sin(C)}$$ and \begin{equation}\frac{a-b}{c}=\frac{\sin(A) -\sin(B)}{\sin(C)}\end{equation}
To prove both these identities the equations rearrange to the following
$$\frac{\sin(C)}{c}=\frac{\sin(A)+\sin(B)}{a+b}=\frac{\sin(A)-\sin(B)}{a-b} $$
From here, $(1)$ would complete the argument.
Thanks in advance
| To prove your requested expression (note there are limitations as Gerry Myerson's question comment explains, e.g., $b \neq 0$, $d \neq 0$, $b + d \neq 0$, $b - d \neq 0$, etc.), cross-multiply and use other manipulations to get
$$\begin{equation}\begin{aligned}
\frac{a}{b} & = \frac{c}{d} \\
ad & = bc \\
ad + cd & = bc + cd \\
d(a + c) & = c(b + d) \\
\frac{a + c}{b + d} & = \frac{c}{d}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You can similarly prove that $\frac{c}{d} = \frac{a - c}{b - d}$. Note, however, you can prove the $2$ requested identities more simply as shown below. The Law of sines gives
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \tag{2}\label{eq2A}$$
Cross-multiplying the left & right parts gives
$$\begin{equation}\begin{aligned}
a\sin(C) & = c\sin(A) \\
\frac{a}{c} & = \frac{\sin(A)}{\sin(C)}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Similarly, using the middle & right parts in \eqref{eq2A} gives
$$\frac{b}{c} = \frac{\sin(B)}{\sin(C)} \tag{4}\label{eq4A}$$
Adding \eqref{eq3A} and \eqref{eq4A} gives
$$\frac{a+b}{c}=\frac{\sin(A) +\sin(B)}{\sin(C)} \tag{5}\label{eq5A}$$
while \eqref{eq3A} minus \eqref{eq4A} gives
$$\frac{a-b}{c}=\frac{\sin(A) -\sin(B)}{\sin(C)} \tag{6}\label{eq6A}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given four edge-lengths of a quadrangle $a,b,c,d$ so that $a\leq b\leq c\leq d$. Prove that $a^{2}+b^{2}+c^{2}+d^{2}<2\left(ab+ac+ad+bc+bd+cd\right).$
Given four edge-lengths of a quadrangle $a, b, c, d$ so that $a\leq b\leq c\leq d$. Prove that
$$a^{2}+ b^{2}+ c^{2}+ d^{2}< 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right )$$
My solution
$$a+ b+ c> d$$
$$\begin{align}\Rightarrow \left ( 2a+ 2b+ 2c \right )d+ \left ( 2a+ 2b \right )c+ 2ab & > 2\left ( a+ b+ c \right )d+ ac+ bc\\
& > 2d^{2}+ a^{2}+ b^{2}\\
& \geq c^{2}+ d^{2}+ a^{2}+ b^{2}
\end{align}$$
How about you ?
| \begin{align}\Rightarrow 2\left ( ab+ ac+ ad+ bc+ bd+ cd \right ) & = 2\left ( a+ b+ c \right )d+ ab+bc+bc+ca+ca+ab\\
& \gt 2d^{2}+ \left (c+ a \right)b + \left (a+ b \right)c + \left (b+ c \right)a\\
& \gt 2d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\
& \gt c^2 + d^{2} + b^{2} + \left (a + b \right)c + a^{2} \\
& \gt a^2 + b^{2} + c^{2} + d^{2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the infinite sum $\sum_{n\geq0} \frac{2^n}{1+x^{2^n}}$ for $|x|>1$ I tried finding a pattern when expanding the sum to get
$$\frac{x^{n-1}+2x^{n-2}+\cdots+(n-1)x+n}{x^n+x^{n-1}+\cdots+x+1}$$
where nothing seems to cancel out, and the solution to the sum seems to be $\frac{1}{x-1}$, but how would one get to a simplified answer as such?
| You are trying to solve $$\sum_{n = 0}^{\infty} \frac{2^n}{1+x^{2^n}}$$
Each term can be be split as $\frac{2^n}{1+x^{2^n}} = \frac{2^n}{-1+x^{2^n}} - \frac{2^{n+1}}{-1+x^{2^{n+1}}}$. Let $a_n = \frac{2^n}{-1+x^{2^n}}$. The sum is then $$\sum_{n = 0}^{\infty} \left( a_n - a_{n+1}\right)$$
which telescopes to $a_0$, or equivalently $$\frac{2^0}{x^{2^0}-1} = \frac{1}{x-1}$$
| {
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Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'm not sure.
| $$f(x)=x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$
$$f(x)=\frac{x^6+x^5+8 x^4+3 x^3+12 x^2+2 x+4}{x \left(x^2+1\right) \left(x^2+2\right)}$$
$$f'(x)=\frac{ x^{10}+x^8-2 x^6-8 x^4-12 x^2-8}{\big[x \left(x^2+1\right) \left(x^2+2\right)\big]^2}$$
$$x^{10}+x^8-2 x^6-8 x^4-12 x^2-8=\left(x^3-2 x^2+2 x-2\right) \left(x^3+2 x^2+2 x+2\right) \left(x^4+x^2+2\right)$$
*
*The last factor does not show any real root.
*The first cubic shows only one real root $x_1$
*The second cubic shows only one real root $x_2$ which in fact $x_2=-x_1$
So, let us solve $x_1$ using the hyperbolic method
$$x_1=\frac{2}{3} \left(1+\sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2
\sqrt{2}}\right)\right)\right)$$
Rrplacing, we have the simple $f(x_1)=5$ and $f(x_2)=-3$.
| {
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Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\omega^n$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{1+2+3…2n}$
$\Rightarrow \prod_{n=1 \to 2n}(-2\omega^n) =2^{2n}\omega^{n(2n+1)}$
If $n$ is $3k$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+1$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{3m}=2^{2n}$
If $n$ is $3k+2$ type: $2^{2n}\omega^{n(2n+1)}=2^{2n}\omega^{(3k+2)(6k+5)} =2^{2n}\omega^{(3m+1)}=2^{2n}\omega^{3m}\omega=2^{2n}\omega$
What have I done wrong here?
| Since $\omega$ is the third root of unity, and also since $2^{2n}-1\equiv 0 \mod{3}$
$$
\begin{align}
1+\omega^{2^{i}}+\omega^{2^{i+1}}&=0\\
\\
1-\omega^{2^{i}}-\omega^{2^{i+1}}&=2\omega^{2^{i}}\\
\\
\prod_{i=0}^{2n-1}{\left(1-\omega^{2^{i}}+\omega^{2^{i+1}}\right)}&=\prod_{i=0}^{2n-1}{\left(2\omega^{2^{i}}\right)}\\
&=2^{2n}\omega^{2^{2n}-1}\\
&=2^{2n}
\end{align}
$$
| {
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Values of the Diophantine equation in an interval Set $a$ and $b$ so that for each $k$, with $ a < k \leq a + b $,
there exist $ x $ and $ y $ with $ x \geq 2 $ and $ y \geq 1 $ such that $ k = 6xy + x -y $
Is it possible to proof that, set $a$, there is a value of $b$ for which this cannot happen?
For example by proving that $ b> c $
with $c$ maximum number of $k$ possible values that can be obtained for $ a < k \leq a + b $
Edit
I checked for $a <10000$ that $b < \sqrt{a}$
in fact for
$$a<k \leq a+\sqrt{a}$$
the number of possible values for $k=6xy+x-y$ are lower than $c$ and $c<\sqrt{a}$
$c = \sum \limits_{\substack{x=2 \\ 6x-1 \text { prime}}}^{\left\lfloor{\frac{a+\lfloor\sqrt{a}\rfloor+1}{7}}\right\rfloor} {\left\lfloor \frac{\displaystyle a+\lfloor\sqrt{a}\rfloor-x}{ \displaystyle 6x-1}\right\rfloor } - \sum \limits_{\substack{x=2 \\ 6x-1 \text { prime}}}^{\left\lfloor{\frac{a+1}{7}}\right\rfloor} { \left\lfloor \frac{\displaystyle a-x}{ \displaystyle 6x-1}\right\rfloor}+ \sum \limits_{\substack{z=1 \\ 6z+1 \text { prime}}}^{\left\lfloor{\frac{a+\lfloor\sqrt{a}\rfloor-6}{35}}\right\rfloor} { \left\lfloor \frac{\displaystyle a+\lfloor\sqrt{a}\rfloor-(5z+1)}{ \displaystyle 6(5z+1)-1}\right\rfloor } -\sum \limits_{\substack{z=1 \\ 6z+1 \text { prime}}}^{\left\lfloor{\frac{a-6}{35}}\right\rfloor} {\left\lfloor \frac{\displaystyle a-(5z+1)}{ \displaystyle 6(5z+1)-1}\right\rfloor }$
Is it possible to find a value of the upper limit of $c$ to verify it is always less than $\sqrt{a}$?
| $k = 6xy + x - y = (6x - 1)y + x$
imagining this number as if $x$ was fixed, and $y$ runs through the positive integers. With this logic we see that incrementing/decrementing $y$ by 1, the next value of $k$ is $6x - 1$ away.
Now if $1 < k - a < 6x - 1$, then it means $a$ is strictly smaller than $k$ and decrementing $y$ by 1 makes $k < a$, so that can not happen, the actual value of $k$ is the smallest to have $k > a$. On the other hand, $k-a$ always exists $\forall x, y$, as $x \geq 2 \Rightarrow 6x - 1 \geq 11$, $\forall x, y$, so we can have such values of $a$ $\forall x, y$.
For such values of $a$ there exist $b < k - a > 1 \Rightarrow a + b < k$, $\forall x, y$.
| {
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Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$ Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$
$\textbf{My Attempt}$
$$3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$$
$$\Rightarrow \log(3^x4^y)=\log(12^z)\Rightarrow \log \big(\frac{3^x4^y}{12^z}\big)=0\Rightarrow \frac{3^x4^y}{12^z}=1 \Rightarrow 3^x4^y=12^z$$
$$\Rightarrow 3^x4^y=3^x+4^y$$
How do you find what $z$ equals from this? Any help would be appreciated.
| It's wrong.
Try $x=z=1$ and $y=\log_23.$
| {
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Proving $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin\frac{ab\pi}{2}$ for $a,b\in\mathbb{N}$? I was playing with the graphs of $f:\,\mathbb{R}\to\mathbb{R},\, f(x)=\sin axt\pi$ and $g:\,\mathbb{R}\to\mathbb{R},\, g(x)=\sin xt\pi\sin at\pi$.
What I conjectured is the following:
If $t=\frac{m}{2}$ where $m$ is $1$ above a multiple of $4$ (or if $m$ is even), then $f=g$ for all $a,x\in\mathbb{N}$.
I'm the most interested in the simplest non-trivial case, which is $m=1$:
$$\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}\quad a,b\in\mathbb{N}.$$
If this is true, how could it be rigorously proved?
I tried rewriting $\sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}$ as
$$\frac{1}{4}\left(e^{\frac{\pi i}{2}(a-b)}+e^{-\frac{\pi i}{2}(a-b)}-e^{\frac{\pi i}{2}(a+b)}-e^{-\frac{\pi i}{2}(a+b)}\right)$$
and rewriting $\sin\frac{ab\pi}{2}$ as
$$\frac{1}{2i}\left(e^{\frac{ab\pi i}{2}}-e^{-\frac{ab\pi i}{2}}\right),$$
but that doesn't seem to help when I want to couple $a$ and $b$ so that there is a product of them in both expressions.
| We have that
*
*$a\lor b$ even
$$\implies \sin\frac{a\pi}{2}\,\sin\frac{b\pi}{2}=\sin \frac{ab\pi}{2}=\sin (k\pi)=0$$
*
*$a\land b$ odd
$$\implies \sin\frac{(2A+1)\pi}{2}\,\sin\frac{(2B+1)\pi}{2}=\sin \frac{(2A+1)(2B+1)\pi}{2}=$$
$$\sin \frac{(4AB+2A+2B+1)\pi}{2}=\sin \left(\frac \pi 2+k\pi\right)=\pm1$$
| {
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Counting and Symmetry The digits $1,2,3,4$ and $5$ can be arranged to form many different $5$-digit positive integers with five distinct digits. In how many such integers is the digit $1$ to the left of the digit $2$? (The digits 1 and 2 do not have to be next to each other.)
Case 1: 2 is in the second spot
We get $3!=6$ ways.
Case 2: 2 is in the third spot
We get $4\cdot2=8$ ways.
Case 3: 2 is in the 4th spot
We get $3\cdot4=12$ ways.
Case 4: 2 is in the last spot.
We get $4!=24$ ways.
Thus, the probability should be $\frac{6+8+12+24}{5!}=\frac5{12}.$
Is this correct?
| A different way is $\binom{5}{2}$ ways to pick 2 slots for 1 and 2, where 1 is always the left one, and $3!$ ways to arrange the rest of the numbers in their own 3 slots. So total number of cases is
$$
\binom{5}{2} \cdot 3! = 10 \cdot 6 = 60.
$$
This also changes the probability:
$$
\frac{60}{5!} = \frac{60}{120} = \frac12 = 50 \%.
$$
Now that you see a tedious way to find it's $50\%$, note there is an obvious symmetry. In the set of all $5!$ permutations, the number of cases where $1$ precedes $2$ is by symmetry the same as number of cases where $2$ precedes $1$, so the result must be $1/2$.
| {
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Prove $\sum_{m=i}^{n}2^{n-m}\binom{m}{i}=\binom{n+1}{i+1}+\ldots+\binom{n+1}{n+1}=\sum_{m=i}^{n}\binom{n+1}{m+1}$ without induction I would like to prove the following: $$\sum_{m=i}^{n}2^{n-m}\binom{m}{i} = \binom{n+1}{i+1}+\binom{n+1}{i+2}+\ldots+\binom{n+1}{n+1} = \sum_{m=i}^{n}\binom{n+1}{m+1}$$
which is quite easy to prove by induction. I'm looking for an algebraic or combinatoric approach. This shouldn't be very complicated from a combinatoric point of view, since the right hand side is the number of different subsets of the set $\{1, 2, \ldots, n+1\}$ with at least $i+1$ elements. Also on the left hand side $2$ appears, which arises when one counts the total number of subsets. But I struggle to find a method to count this number to make it look like the left hand side; what do you suggest?
| Here is an algebraic proof. We get for the RHS
$$\sum_{m=q}^n {n+1\choose m+1} =
\sum_{m=q}^n {n+1\choose n-m}
= [z^n] \sum_{m=q}^n z^m (1+z)^{n+1}.$$
Here the coefficient extractor enforces the range and we get
$$[z^n] (1+z)^{n+1} \sum_{m\ge q} z^m
= [z^n] (1+z)^{n+1} \frac{z^q}{1-z}.$$
This is
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{z^q}{1-z}.$$
Letting $z/(1+z) = w$ we get $z=w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$
so this becomes
$$\mathrm{Res}_{w=0} \frac{1}{w^{n+1}}
\frac{w^q}{(1-w)^q} \frac{1}{1-w/(1-w)}
\frac{1}{(1-w)^2}
\\ = \mathrm{Res}_{w=0} \frac{1}{w^{n-q+1}}
\frac{1}{(1-w)^{q+1}} \frac{1}{1-2w}.$$
We get at last
$$[w^{n-q}] \frac{1}{(1-w)^{q+1}} \frac{1}{1-2w}
= \sum_{m=0}^{n-q} [w^m] \frac{1}{(1-w)^{q+1}}
[w^{n-q-m}] \frac{1}{1-2w}
\\ = \sum_{m=q}^{n} [w^{m-q}] \frac{1}{(1-w)^{q+1}}
[w^{n-m}] \frac{1}{1-2w}
\\ = \sum_{m=q}^{n} {m-q+q\choose q} 2^{n-m}
= \sum_{m=q}^n 2^{n-m} {m\choose q}.$$
This is the claim.
| {
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prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a^3-a^2b-a^2c+abc)=\sum_{cyc}(a^3-abc-a^2b-a^2c+2abc)=$$
$$=\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-c\right)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq$$
$$\geq\frac{1}{2}((a-c)^2(a+c-b)+(b-c)^2(b+c-a))\geq$$
$$\geq\frac{1}{2}((b-c)^2(a-b)+(b-c)^2(b-a))=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.
| Another way: note that
$$
x^8-x^5+x^2-x+1=\left(x^4-\frac{1}{2}x\right)^2+\frac{1}{2}x^2+\left(\frac{1}{2}x-1\right)^2>0.
$$
Equality cannot occur since $x^4-\frac{1}{2}x$, $x$ and $\frac{1}{2}x-1$ cannot vanish together.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of generators for $\Bbb Z_9\times \Bbb Z_{18}$. Is $\{(4,3),(3,5)\}$ a system of generators for $\Bbb Z_9\times \Bbb Z_{18}$?
I tried to generate the elements using the straightforward method $(4,3), (8,6), (3, 9)$ ... but it takes too long.
Can I use a faster method?
| Let $v_1=(4,3)$ and $v_2=(3,5)$. Then
$$
\begin{pmatrix}v_1\\v_2\end{pmatrix}
=
\begin{pmatrix}4&3\\3&5\end{pmatrix}
\begin{pmatrix}e_1\\e_2\end{pmatrix}
$$
where $e_1=(1,0)$ and $e_2=(0,1)$.
The matrix has determinant $11$, which is invertible in the ring $\Bbb Z_9\times \Bbb Z_{18}$. Therefore, $v_1$ and $v_2$ generate the same additive group as $e_1$ and $e_2$.
Explicitly,
$$
\begin{pmatrix}e_1\\e_2\end{pmatrix}
=
\begin{pmatrix}7&3\\3&2\end{pmatrix}
\begin{pmatrix}v_1\\v_2\end{pmatrix}
$$
| {
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combinatorial identity another solution? I would appreciate if somebody could help me with the following problem:
I need to show that,
$$\sum^{33}_{k=0}\binom{100}{3k}=\sum^{49}_{k=0}4^k$$
My proof is the following:
$$(1+x)^{100} = a_0 + a_1x + a_2x^2 + \cdots + a_{100}x^{100}$$
Let
$A = a_0 + a_3 + a_6 + \cdots + a_{99}, B = a_1 + a_2 + a_4 + a_5 + a_7 + a_8 + \cdots + a_{98} + a_{100}$$
then
$$A+B = 2^{100}$$
$$x^3=1 \to x=1,w,w^2$$
$x$ put $w$, $w^2$ and sum
$$\begin{align*}
(1+w)^{100} + (1+w^2 )^{100}
&=w^{200} + w^{100}\\
&=w^{2} + w\\
&=-1\\
&= 2A + a_1(w+w^2) + a_2(w^2 + w^4) + a_4(w^4 + w^8) + \cdots + a_{100}(w^{100} + w^{200}) \\
&= 2A - B
\end{align*}$$
therfore
$$A+B = 2^{100}, 2A - B=-1$$
$$A = \frac{2^{100} - 1}{3} = \frac{4^{50} - 1}{3} =\sum^{49}_{k=0}4^k $$
But I want to know if exists another proof for this problem.
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\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
My answer is somehow similar to the one by
$\ds{\tt @Joshua\ P.\ Swanson}$. However, there are some differences.
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{33}{100 \choose 3k}} =
\sum_{k = 0}^{99}{100 \choose k}{1 + \expo{2k\pi\ic/3} + \expo{-2k\pi\ic/3} \over 3}
\\[5mm] = &\
{1 \over 3}\sum_{k = 0}^{99}{100 \choose k} +
{2 \over 3}\,\Re\sum_{k = 0}^{99}{100 \choose k}
\pars{\expo{2\pi\ic/3}}^{k}
\\[5mm] = &\
{1 \over 3}\pars{2^{100} - 1} +
{2 \over 3}\,\Re\bracks{\pars{1 + \expo{2\pi\ic/3}}^{100} -
\expo{200\pi\ic/3}}
\\[5mm] = &\
{2^{100} \over 3} +
{2 \over 3}\,\Re\bracks{\pars{{1 \over 2} +
{\root{3} \over 2}\,\ic}^{100}}
\\[5mm] = &\
{2^{100} \over 3} +
{2 \over 3}\,\Re\bracks{\pars{\expo{\ic\pi/3}}^{100}} =
{2^{100} \over 3} +
{2 \over 3}\cos\pars{100\pi \over 3}
\\[5mm] = &\
{2^{100} \over 3} - {1 \over 3} =
{4^{50} - 1 \over 4 - 1} = \bbx{\sum_{k = 0}^{49}4^{k}} \\ &
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Book: Actuarial Mathematics for Life Contingent Risks Exercise 2.4 Exercise $2.4$ Let $$S_0(x) =\exp\left({-\left( Ax+ \frac{Bx^2}2 +\frac{C(D^x-1)}{\ln D}\right)}\right)$$
where $A, B, C , D>0$
$(b)$ Derive a formula for $S_x(t)$
I know the equation for $S_x(t)= \dfrac{S_0(x+t)}{S_0(x)}$
Right now I am at: $\dfrac{\exp\left({-\left( Ax+ At+\frac{B(x^2+2xt+t^2)}2 +\frac{C(D^{x+t}-1)}{\ln D}\right)}\right)}{\exp\left({-\left( Ax+ \frac{Bx^2}2 +\frac{C(D^x-1)}{\ln D}\right)}\right)}$
Do the $2$ exp's combine to just be a single exp? I am unsure of how to cancel things out. Essentially the basic algebra is tripping me up. Please Help!
| It is better to let $$f(x) = Ax + \frac{Bx^2}{2} + \frac{C(D^x - 1)}{\log D}$$ for fixed constants $A, B, C, D$, then observe $$S_x(t) = \frac{S_0(x+t)}{S_0(x)} = \frac{\exp(-f(x+t))}{\exp(-f(x))} = \exp(-f(x+t)+f(x)).$$ Then $$\begin{align}
f(x) - f(x+t)
&= \left(Ax - A(x+t)\right) + \left(\frac{Bx^2}{2} - \frac{B(x+t)^2}{2}\right) + \left(\frac{C(D^x - 1)}{\log D} - \frac{C(D^{x+t} - 1)}{\log D}\right) \\
&= -At - \frac{B}{2}\left((x^2 + 2xt + t^2) - x^2\right) - \frac{C}{\log D} \left(D^x D^t - D^x \right) \\
&= -\left( At + \frac{B}{2}t(2x + t) + \frac{C D^x}{\log D}(D^t - 1)\right).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3819986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the answer = value of $k$ is $11$.
| Hint:
It simply means that $x+11$ divides the numerator, i.e. $-11$ is a root of the numerator, the other root being $4/3$. Use Vieta's relations to determine $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the following limit and prove its value by definition. $\lim _{n\to \infty} (n^2 + 3n + 7)/(2n^2 −5n − 4)$. Find the following limit and prove its value by definition. $\lim_{n\to \infty} \frac{n^2 + 3n + 7}{2n^2 −5n − 4}$. I am supposed to use$$ \left|\frac{n^2 + 3n + 7}{2n^2 −5n − 4} - \frac{1}{2}\right| < ε $$then combine the fraction of the limit and the result and eventually get rid of the absolute value bars with some algebra and end up with one $n <$ some $ε$ but since neither the top nor bottom of the limit factor I do not know how I can manipulate the limit so my final inequality can be a single $n$ and some $ε$.
| Since
$$\frac{n^2 + 3n +7}{2n^2-5n+4} - \frac{1}{2} = \frac{2n^2 + 6n +14 - 2n^2+5n-4}{4n^2-10n-8} = \frac{11n-10}{4n^2-10n-8} $$
Then you could simply bound the right hand side,
$$|\frac{11n-10}{4n^2-10n-8} | \leq |\frac{11n}{4n^2-10n-8}| \leq |\frac{11n}{4n^2-11n}| = |\frac{11}{4n-11}| \leq |\frac{11}{4n-n}| \leq \frac{11}{3n}$$
Where we used that $n$ has to bigger than $8$ and $11$, thus the above inequality holds for $n \geq 11$. We want, given $\varepsilon >0$, find an $n \in \mathbb N$ such
$$|\frac{11n-10}{4n^2-10n-8} |<\varepsilon,$$
but we can bound the upper term, in the following way:
$$|\frac{11n-10}{4n^2-10n-8} |< \frac{11}{3n}<\varepsilon$$
for $n \geq 11$. So we only have to find the $n$ that makes
$$\frac{11}{3n}<\varepsilon$$
true. But the previous inequality is equivalent to
$$ \frac{11}{3 \varepsilon} < n. $$ This means the inequality is true of $n$'s bigger than $\frac{11}{3 \varepsilon}$. We can take, for example, $n = \lceil \frac{11}{3 \varepsilon} \rceil + 1$. But don't forget a condition is also that $n \geq 11$! So finally,
$$n_0 = \max\left\{11, \left \lceil \frac{11}{3 \varepsilon} \right \rceil + 1 \right\} $$ is an $n$ that makes $|\frac{11n-10}{4n^2-10n-8} |<\varepsilon$ true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Divisibility of higher power polynomials I am stuck with this problem suggested to me by a friend . The problem reads,
The power of $x^2+xy+y^2$ by which the polynomial $(x+y)^7-x^7-y^7$ is divisible are ?
I tried it by writing $x^7+y^7$ expansion and simplifying the polynomials in factors but that doesn't help as the resulting expression is ugly . Someone kindly suggest a method to solve this problem.
| By the Binomial theorem we obtain:
$$(x+y)^7-x^7-y^7=7xy(x^5+3x^4y+5x^3y^2+5x^2y^3+3xy^4+y^5)=$$
$$=7xy(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4+3xy(x^2-xy+y^2)+5x^2y^2)=$$
$$=7xy(x+y)(x^4+2x^3y+3x^2y^2+2xy^3+y^4)=7xy(x+y)(x^2+xy+y^2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximizing $y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$ for $x\in[-\frac{5\pi}{12}, -\frac{\pi}{3}]$
Let $x \in [-\frac{5\pi}{12}, -\frac{\pi}{3}]$. Find the maximum value of $$y=\tan(x+\frac{2\pi}{3})-\tan(x+\frac{\pi}{6})+\cos(x+\frac{\pi}{6})$$
The above question is from the China Mathematical Competition from 2003, held for students in Shaanxi.
The solution that the testmakers made is provided below:
The solution (outline):
Let $z=-x-\frac{\pi}{6}$, then $\tan(x+\frac{2\pi}{3})=\cot(z)$ and then $y=\cot(z)+\tan(z)+\cos(z)=\frac{2}{\sin(2z)}+\cos(z)$. Then, both are monotonic decreasing, then $z=\frac{\pi}{6}$, and thus the minimum is $\frac{11}{6}\sqrt{3}$.
I just want to know any other methods to solve this question, especially because I find trigonometric inequalities and maxima and minima interesting.
| The given proof is very clever and efficent.
As an alternative we have that
$$y=\frac{2}{\sin(2z)}+\cos(z)=\frac{1}{\sin(z)\cos(z)}+\cos(z)$$
which is equivalent to
$$y(t)=\frac{1}{t\sqrt{1-t^2}}+t \quad t=\left(\cos \left(\frac \pi 4\right),\cos \left(\frac \pi 6\right)\right)=\left(\frac{\sqrt 2}2,\frac{\sqrt 3}2\right)$$
with
$$\frac d{dt}\left(t\sqrt{1-t^2}\right)=\frac{1-2t^2}{\sqrt{1-t^2}}\le 0$$
$$0<t\sqrt{1-t^2}<1 \iff 0<t^2-t^4<1 \iff t^4-t^2+1 >0$$
therefore $y(t)$ is increasing and the maximum is reached for $z=\frac \pi 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\ln(\sqrt{1-x}+\sqrt{1+x})$ Integrate $$\ln(\sqrt{1-x}+\sqrt{1+x})$$
My attempts:
i) Multiplying and dividing its conjugate, we get
$$I=\int \ln(-2x)dx-\int\ln(\sqrt{1-x}-\sqrt{1+x})dx$$
The second term is equally challenging as the original one, so its of no use.
ii) $\sqrt{1-x}+\sqrt{1+x}=t \implies\frac{1}{2}(\frac{1}{\sqrt{1+x}}-\frac{1}{\sqrt{1-x}})dx=dt$
Though $\sqrt{1-x}-\sqrt{1+x}$ can be written in terms of $t$, nothing can be done to $\sqrt{1-x^2}$
iii) Let $x=\cos(2\theta) \implies dx=-2\sin(2\theta)d\theta$
$$I=\ln(\sqrt2(|\sin(\theta)|+|\cos(\theta)|))\cdot-2\sin(2\theta)d\theta$$
$$I=\ln(2(2\sin(\theta)\cos(\theta)+1))\cdot-\sin(2\theta)d\theta$$
$$I=\ln(2(\sin(2\theta)+1))\cdot-\sin(2\theta)d\theta$$
$y=2(1+\sin(2\theta)) \implies dy=4\cos(2\theta)d\theta$
$$I=-\ln(y)\cdot (\frac{y}{2}-1)\cdot \frac{1}{4\sqrt{y-\frac{y^2}{4}}} dy$$
Which doesnt seem promising.
iv) Using IBP(as suggested by @AnginaSeng),
$$I=x\ln(\sqrt{1-x}+\sqrt{1+x})-\int x \frac{1}{\sqrt{1-x}+\sqrt{1+x}}\cdot \frac{1}{2}\cdot \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x^2}}$$
$$I=x\ln(\sqrt{1-x}+\sqrt{1+x})-\frac{1}{2}\int \frac{x}{\sqrt{1-x^2}} \frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}$$
A new method, or some hints for existing ones is appreciated
| I think IBP approach you did (as @AnginaSeng said) is on the right track. We have
$$
\frac{x}{\sqrt{1-x^{2}}}\frac{\sqrt{1-x} - \sqrt{1+x}}{\sqrt{1-x} +\sqrt{1+x}} = \frac{x}{\sqrt{1-x^{2}}} \frac{(\sqrt{1-x} - \sqrt{1+x})^{2}}{-2x} =- \frac{2 - 2\sqrt{1-x^{2}}}{2\sqrt{1-x^{2}}} = 1 - \frac{1}{\sqrt{1-x^{2}}}
$$
and I bet you that you can deal with this integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we prove that $\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$? I would like to verify the claim:
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(CBA)$$
I tried verifying through an example:
Given the following $3$ different matrices:
\begin{align}
A & = \begin{pmatrix}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12 \\
13 & 14 & 15 & 16
\end{pmatrix}\\[2ex]
B & = \begin{pmatrix}
4 & 3 & 2 & 1 \\
3 & 6 & 4 & 2 \\
2 & 4 & 6 & 3 \\
1 & 2 & 3 & 4
\end{pmatrix}\\[2ex]
C & = \begin{pmatrix}
4 & 3 & 2 & 1 \\
5 & 6 & 7 & 8 \\
8 & 7 & 6 & 5 \\
4 & 3 & 2 & 1
\end{pmatrix}
\end{align}
I calculated $\operatorname{Tr}(ABC) = 7930$ and
$\operatorname{Tr}(CBA) = 7510$.
Is there any thing wrong in my calculation, or does this prove that $\operatorname{Tr}(ABC) \neq \operatorname{Tr}(CBA)$?
Many thanks!
| As suggested in the comments, the statement $\text{Tr}(ABC)=\text{Tr}(CBA)$ is not true in general, so you should not be concerned that you found a counterexample. Actually, there is a simpler counterexample. Take
$$ A=
\begin{bmatrix}
5&2\\
0&0
\end{bmatrix},\:\:\:
B=
\begin{bmatrix}
2&7\\
5&3
\end{bmatrix},\:\:\:
C=
\begin{bmatrix}
1&1\\
1&1
\end{bmatrix}.
$$
Then $\text{Tr}(ABC)=49$ while $\text{Tr}(CBA)=61$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$ is non negative Prove that the sequence
$$s_k = \frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$ is non-negative.
I would appreciate an elementary proof. I tried using series / integral comparison without success.
This question is a follow up of this one. There is an answer to that question that uses the integral representation of Hurwitz zeta function, that I'd like to avoid... if possible!
| $$\Delta_k=\frac{1}{(2k+1)^4} - 2 \sum_{n=k+1}^\infty \left(\frac{1}{(2n)^4} - \frac{1}{(2n+1)^4}\right)$$
$$\sum_{n=k+1}^\infty \frac{1}{(2n)^4}=\frac{1}{96} \psi ^{(3)}(k+1)\qquad \qquad\sum_{n=k+1}^\infty \frac{1}{(2n+1)^4}=\frac{1}{96} \psi ^{(3)}\left(k+\frac{3}{2}\right)$$ Using asymptotics
$$\Delta_k=\frac 1 {16k^5} \left(1-\frac{5}{2 k}+\frac{25}{8 k^2}+O\left(\frac{1}{k^3}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Concrete Mathematics: Clarifying expressing sum in terms of $H_n$ leading to equation 2.14 In Concrete Mathematics (Knuth, Patashnik, and Graham) we have an intermediate solution to the quick-sort recurrence of
$$
C_n = 2(n + 1)\sum^n_{k=1}\frac{1}{k + 1}
$$
Given the sum part of that resembles the Harmonic summation of $\sum^n_{k=1}\frac{1}{k}$ we can transform the sum part of the intermediate form above to a Harmonic sum thereby providing a "closed form" solution.
The book does this in the following steps
$$
\begin{align}
\sum^n_{k=1}\frac{1}{k + 1} &= \sum_{1 \leq k - 1 \leq n}\frac{1}{k} \\
&= \sum_{2 \leq k \leq n+1}\frac{1}{k} \\
&= \left( \sum_{1\leq k \leq n}\frac{1}{k} \right) - \frac{1}{1} + \frac{1}{n+1} = H_n - \frac{n}{n+1}
\end{align}
$$
I think the $\frac{1}{1} + \frac{1}{n+1}$ terms are to "trim" the resultant the sum so that we can use $H_n$ in its natural form with clean boundary conditions. So $-\frac{1}{1}$ removes the "start" and $\frac{1}{n+1}$ adds on what the $n+1$ boundary condition would have done. Is this what is happening?
And, if you'll indulge me, add on a follow-up question. The eventual, closed form solution is then presented as follows (2.14 in book)
$$
C_n = 2(n + 1)H_n - 2n
$$
Where do they get $2n$ from? I would have thought the eventual closed form would then be $C_n = 2(n+1)H_n - \frac{n}{n+1}$ Obviously 2.14 is correct though; maybe I'm being insufficiently creative in my algebraic manipulations (or need more coffee).
| $$\sum_{k=1}^n \frac{1}{k+1} = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n+1}.$$ So if we compare this to $$H_n = \sum_{k=1}^n \frac{1}{k} = 1 + \frac{1}{2} + \cdots + \frac{1}{n},$$ we can see that in the first sum we are missing the first term $1$, and we have an extra last term, $\frac{1}{n+1}$. Therefore, $$\sum_{k=1}^n \frac{1}{k+1} = H_n - 1 + \frac{1}{n+1}.$$ Then $$\begin{align}
C_n &= 2(n+1) \sum_{k=1}^n \frac{1}{k+1} \\ &= 2(n+1) \left( H_n - 1 + \frac{1}{n+1} \right) \\ &= 2(n+1)H_n - 2(n+1) + 2 \\ &= 2(n+1)H_n - 2n. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Compute the limit $\lim\limits_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin tx} $ I have been working on this limit for days, but I am not getting it. The question is
Compute the limit $$\lim_{t \to + \infty} \int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (tx)}$$
Note that the integral is well defined and convergent for every $t >0$. Indeed the integrand function is a positive function for every $t >0$ since
$$e^x + \sin tx > e^x-1 > x>0$$
And as $x \to + \infty$ the integrand function behaves like $e^{-x}$.
WHAT I TRIED:
I consider $t=2n \pi$ a multiple of $2 \pi$, and see what happens:
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} = \sum_{k=0}^\infty \int_{k /n}^{(k+1) /n} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)}$$
Making the change of variables $u = 2n \pi x$ I get
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{u/2n \pi}+ \sin (u)} &\ge
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(2k+2) \pi/2n \pi}+ \sin (u)} \\&=
\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)}\end{align} where I write the lower bound with the minimum of the function at $u=(2k+2) \pi$. Now I use the fact that the integrand function does is integrated over a period of $2 \pi$, and using the result for $C>1$
$$\int_0^{2 \pi} \frac{ \mathrm d u}{C+ \sin (u)} = \frac{2 \pi}{\sqrt{C^2-1}}$$
I get the estimate
\begin{align}\sum_{k=0}^\infty \frac{1}{2n \pi} \int_{2k \pi}^{(2k+2) \pi} \frac{ \mathrm d u}{e^{(k+1)/n}+ \sin (u)} &= \sum_{k=0}^\infty \frac{1}{2n \pi} \frac{2 \pi}{\sqrt{e^{2(k+1)/n} -1 }} \\&= \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}\end{align}
Summing all up, I got that
$$\int_0^{+ \infty} \frac{ \mathrm d x}{e^x+ \sin (2n \pi x)} \ge \frac{1}{n} \sum_{k=0}^\infty \frac{1}{\sqrt{e^{2(k+1)/n} -1 }}$$
As $n \to \infty$ the series converges to the Riemann integral
$$\int_0^{+ \infty} \frac{\mathrm d y}{\sqrt{e^{2y}-1}} = \frac{\pi}{2}$$
Hence the limit should be a number larger than $\pi/2$, or $+ \infty$.
Using WA I got for large values of $t$ that the integral is between $1$ and $2$, thus $\pi/2$ could be the actual limit.
| For each $t > 0$, we have
\begin{align}
\int_0^\infty \frac{1}{\mathrm{e}^x + \sin t x}\mathrm{d} x
&= \int_0^\infty \sum_{k=0}^\infty (-1)^k \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k \int_0^\infty \mathrm{e}^{-(k+1)x} (\sin t x)^k\mathrm{d} x\\
&= \sum_{k=0}^\infty (-1)^k\frac{1}{k+1}
\int_0^\infty \mathrm{e}^{-y} (\sin \tfrac{t y}{k+1})^k\mathrm{d} y.
\end{align}
(Hint: Use Fubini's theorem for the interchange of integration and summation.)
Denote $I_k = \int_0^\infty \mathrm{e}^{-y} (\sin \tfrac{t y}{k+1})^k\mathrm{d} y, \ k=0, 1, 2, \cdots$. Using integration by parts, we have
$$I_k = \frac{(k-1)kt^2}{(k+1)^2 + k^2t^2}I_{k-2}, \ k=2, 3, 4, \cdots.$$
Also, $I_0 = 1$ and $I_1 = \frac{2t}{t^2+4}$.
Thus we have, for $k = 0, 1, 2, \cdots$,
$$\lim_{t\to\infty} I_{2k+1} = 0,$$
and
$$\lim_{t\to\infty} I_{2k} = \frac{2k-1}{2k}\cdot \frac{2k-3}{2k-2}\cdots \frac{1}{2}I_0 = \frac{(2k)!}{4^k (k!)^2}.$$
By Tannery's theorem (https://en.wikipedia.org/wiki/Tannery%27s_theorem),
we have
$$\lim_{t\to \infty} \int_0^\infty \frac{1}{\mathrm{e}^x + \sin t x}\mathrm{d} x
= \sum_{k=0}^\infty \frac{1}{2k+1} \frac{(2k)!}{4^k (k!)^2}
= \frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Convergence of the series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ The series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
(A) is uniformly convergent for all x
(B) is convergent for all x but the convergence is not uniform
(C) is convergent only for $|x|\le \frac{1}{2}$ but the convergence is not uniform
(D) is uniformly convergent on $[-\frac{1}{2},\frac{1}{2}]$
My approach is if we take $ f(x)=\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
then $$ \int f(x)dx=\log((1+x^2)(1+x^4)(1+x^8)...)=\log\left(\frac{1-x^{2^n}}{1-x^2}\right)$$
Hereafter, I am stuck.
| The answer is D). For $|x| \leq \frac1 2 $ we have $|\frac {n x^{2n-1}} {1+x^{2n}}| \leq n|x|^{2n-1} $ and $ \sum n |\frac 1 2 |^{n}$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many 5-digit sequences of $\{1, 2, 3, 4\}$ are there in which $1$, $2$, $3$ and $4$ all appear? I'm trying to work out how many possible 5 digit sequences there are using the numbers $\{1, 2, 3, 4\}$ in which each number appears at least once.
I thought it might be $4\times 3\times 2\times 1\times 4$ but it's not one of the possible answers.
| Here's an alternative approach via inclusion-exclusion, where the four properties to be avoided are that $j\in\{1,2,3,4\}$ does not appear:
\begin{align}
\sum_{k=0}^4 (-1)^k \binom{4}{k} (4-k)^5
&= \binom{4}{0} 4^5 - \binom{4}{1} 3^5 + \binom{4}{2} 2^5 - \binom{4}{3} 1^5 + \binom{4}{4} 0^5 \\
&= 1024 - 972 + 192 - 4 + 0 \\
&= 240
\end{align}
More generally, the number of surjections from an $n$-set onto an $m$-set is
$$\sum_{k=0}^m (-1)^k \binom{m}{k} (m-k)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
| Here is a verification approach by factorization:
We want to show
$$x^2+3x+\frac1x \ge \frac{15}4$$
We know that $x > 0$,
Hence it is equivalent to showing:
$$4x^3 + 12x^2 + 4 \ge 15x$$
$$4x^3+12x^2-15x+4 \ge 0$$
Observe that $x=-4$ is a root.
$$(x+4)(4x^2-4x+1) \ge 0$$
$$(x+4)(2x-1)^2 \ge 0$$
Since $x>0$, clearly, $x+4>0$ and $(2x-1)^2 \ge 0$, hence the product must be nonnegative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
Assume the statement is true for some integer $k\geq 2$.
$$f(k)=5^{2k+1}-3^{2k+1}-2=30t$$
For $n=k+2$,
\begin{align}
f(k+2)&=(5^{2(k+2)+1})-(3^{2(k+2)+1})-2\\
&=(5^4)\cdot (5^{(2k+1)})-(3^4)\cdot (3^{(2k+1)})-2
\\&=30(80t)+160+544\cdot (5^{2k+1})
\end{align}
I'm stuck here. Any help would be appreciated!
| Suppose $f(k)=5^{2k+1}-3^{2k+1}-2=30t$
for $n=k+2$,
\begin{align}f(k+2)&=5^{2(k+2)+1}-3^{2(k+2)+1}-2\\
&=(5^4)\cdot (5^{2k+1})-(3^4)\cdot (3^{2k+1})-2\\
&=30(80t)+160+544\cdot (5^{2k+1}) \\
&= 30(80t) + 150 + 10 + 540(5^{2k+1}) + 4(5^{2k+1})\end{align}
It suffices to prove that $10+4(5^{2k+1})$ is divisible by $30$.
\begin{align}
10 + 4(5^{2k+1}) &= 10 + 20(5^{2k})
\end{align}
It is clearly divisible by $10$, we just have to show that it is divisible by $3$.
$$10+20(5^{2k}) \equiv 1-2^{2k}\equiv 1-4^k = 1-1^k \equiv 0 \pmod{3}$$
Hence is it divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Starting with $1,9,9,3$ the sequence continues as the last digit of the sum of preceding 4 terms. Will subsequence $7, 3, 6, 7$ ever appear in it? Formally, $a_n = (a_{n-1} + a_{n-2} + a_{n-3} + a_{n-4})$ mod $10$.
Is there a mathematical way of knowing whether $7, 3, 6, 7$ will ever occur in the sequence?
One idea I was delving upon was to prove the sequence is analogous to a decimal expansion of an irrational number, that way any arbitrary four numbers would appear.
| If $X_n = \pmatrix{a_n\cr a_{n+1}\cr a_{n+2}\cr a_{n+3}\cr}$, we have $X_{n+1} = M X_n \mod 10$ where $M = \pmatrix{0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 1 & 1 & 1 & 1\cr}$. That matrix has characteristic polynomial $P(\lambda) = \lambda^4 - \lambda^3 - \lambda^2 - \lambda - 1$, which is irreducible both mod $2$ and mod $5$. The period of
$M$ mod $p$ must be a divisor of $p^4-1$. We find the period mod $2$ is $5$ and the period mod $5$ is $312$. Therefore the period mod $10$ is $5 \cdot 312 = 1560$.
With $X_0 = \pmatrix{1\cr 9\cr 9\cr 3\cr}$ I find $X_{1552} = \pmatrix{7\cr 3\cr 6\cr 7\cr}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\frac{\pi}{3} - x\right) \\
&= \frac{1}{\sin x} - \frac{1}{\sin\left(\frac{\pi}{3} + x\right)} + \frac{1}{\sin\left(\frac{\pi}{3} -x\right)} \\
&= \frac{\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x }{\sin x \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right)} \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} + x\right) \sin\left(\frac{\pi}{3} - x\right) - \sin\left(\frac{\pi}{3} - x\right) \sin x\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(\sin x - \sin\left(\frac{\pi}{3} + x\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) - \sin\left(\frac{\pi}{3} - x\right) \left(2\sin\frac{-\pi}{6}\cos\left(x + \frac{\pi}{6}\right)\right)\right) \\
&=\frac{4}{\sin 3x}\left(\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\right) \\
\end{align}
How should I proceed? Or did I make some mistakes somewhere? Thanks in advance.
| Using the identities
$$\sin A \sin B = \frac12 (\cos (A-B) - \cos (A+B))$$
$$\sin A \cos B = \frac12 (\sin (A+B) + \sin (A-B))$$
we have
\begin{align}
&\phantom{=}\sin x \sin\left(\frac{\pi}{3} + x\right) + \sin\left(\frac{\pi}{3} - x\right) \cos\left(x + \frac{\pi}{6}\right)\\&=\frac12\left(\cos \left(-\frac\pi3\right)-\cos\left(2x+\frac\pi3\right)+\sin\frac\pi2+\sin\left(\frac\pi6-2x\right)\right)\\
&=\frac 12\left(\frac12+1-\cos\left(2x+\frac\pi3\right)+\cos\left(\frac\pi2 - \left(\frac\pi6-2x\right)\right)\right)\\
&=\frac34+\frac 12\left(-\cos\left(2x+\frac\pi3\right)+\cos\left(2x+\frac\pi3\right)\right)\\
&=\frac34
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the magic number as following Let $s$ and $t$ be distinct positive integers with $s+t$ and $s-t$ are a square numbers.
A pair $(s,t)$ called magic if there is exist positive integer $u$, such that
$12s^2 + t^2 = 4t^2u^3$. Does it exist a magic number?
I try that $s+t = m^2$ and $s-t = n^2$ for some positive integer $m, n$, such that
$2t = (m-n)(m+n)$. LHS is even, so RHS must be even. There are 2 cases, when both $m$ and $n$ are odd, and, when both $m$ and $n$ are even.
And then, what next? I stuck at here. Any idea?
| Too long for a comment. As Ross Molikan points out, the problem boils down to solve $3k^2+1=4u^3$.
We work in the ring $R=\mathbb{Z}[j]$, where $j=e^{2i\pi/3}$. The ring $R$ is a PID (it is even euclidean), hence a UFD.
Set $z=1+k\sqrt{-3}=k+1+2k j$. The equation may be rewritten $zz^*=4u^3$, where $*$ denotes complex conjugation (which induces an automorphism of $R$).
Since $2$ is known to be irreducible in $R$, $2$ divides $z$ or $z^*$ in $R$, but then $2$ divides $z$ in both cases (apply complex conjugation). Since $z=(k+1)+2kj$, this implies that $k+1$ is even, and that $k$ is odd. We then have $z=2y$ with y=$\frac{k+1}{2}+kj$, with $k$ odd. In particular, $2\nmid y$ in $R$.
Now the equation is equivalent to $yy^*=u^3$.
We claim that $y$ and $y^*$ are coprime in $R$. Indeed , if $t\in R$ is a common divisor of $y$ and $y^*$, it divide $y+y^*=\frac{z+z^*}{2}=1$, and so $t$ is a unit.
Since $y, y^*$ are coprime and $yy^*$ is a cube, $y=\alpha w^3$, where $\alpha$ is a unit of $R$ and $w\in R$. Notice now that the units of $R$ are $\pm 1,\pm j,\pm j^2$
Assume first that $\alpha=\pm 1.$ Changing signs (since $-1$ ) , one may assume that $\alpha=1$.
Hence $y=w^3$, so $z=2w^3$. We now use the fact that an element $w$ of $R$ may be written under the form $w=\frac{a+b\sqrt{-3}}{2}$, where $a,b$ have same parity.
We then get $z=2w^3=\dfrac{a^3-9 ab^2+(3a^2b-3b^3)\sqrt{-3}}{4}=1+k\sqrt{-3}$.
In particular, $4=a(a^2-9b^2)$. Note that if $a$ and $b$ are even, then $a^2-9b^2$ must be divisible by $4$, and then $a (a^2-9b^2)$ is divisible by $8$, contradiction. Hence $a$ and $b$ are odd, so $a=\pm 1$. If $a=1$, then $3=-9b^2\leq 0$, contradiction. Hence $a=-1$, so $9b^2=5$, another contradiction.
It remains to examine the case $\alpha=\pm j, \pm j^2$. Since $-1$ is a cube, one may assume that $\alpha=j$ or $j^2$. If $\alpha=j^2$, conjugating yields that $z^*=2j (w^*)^3$.
So replacing $k$ by $-k$, one may assume that $z=2jw^3$. This seems to be the difficult case. Still thinking about it...Maybe somebody will be able to continue further.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit of multiple absolute values Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.
I broke up each absolute value into parts:
$|1-x^2| = \begin{cases}
1-x^2 & -1 \leq x\leq 1 \\
-(1-x^2) & x>1,x<-1
\end{cases}
$ ,
$|x+1| = \begin{cases}
x+1 & x\geq -1 \\
-(x+1) & x<-1
\end{cases}
$,
$|x^2+x| = \begin{cases}
x^2+x & x\geq0,x\leq -1 \\
-(x+1) & -1<x<0
\end{cases}
$
Thus, when $x\rightarrow -1$, the function will approach the positive value, because by my definitions of the absolute values,each has the positive value at $x=-1$. So then you can take
$\lim_{x\rightarrow -1}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1}\frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(x^2+x)}=\lim_{x\rightarrow -1}\frac{(4x-2)(x+1)}{(x+1)(-x+2)}=-2$.
But looking at the graph, the limit is -6. So I must have messed up in my absolute value declarations, most likely in the third one.
So, how would I correctly declare and solve this limit?
| $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$
$\lim_{x\rightarrow -1^-}f(x)=\lim_{x\rightarrow -1^-}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1^-} \frac{x^2+2x-3(x^2-1)+1}{2(-(x+1))-(x^2+x)}=\lim_{x\rightarrow -1^-}f(x)\frac{-2x^2+2x+4}{-x^2-3x-2}=\lim_{x\rightarrow -1^-} \frac{-2(x+1)(x-2)}{-(x+1)(x+2)}=\lim_{x\rightarrow -1^-} \frac{-2(x-2)}{-(x+2)}=-6$.
$\lim_{x\rightarrow -1^+}f(x)=\lim_{x\rightarrow -1^+}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1^+} \frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(-(x^2+x))}=\lim_{x\rightarrow -1^+}f(x)\frac{4x^2+2x-2}{x^2+3x+2}=\lim_{x\rightarrow -1^+} \frac{2(2x-1)(x+1)}{(x+1)(x+2)}=\lim_{x\rightarrow -1^+} \frac{2(2x-1)}{(x+2)}=-6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim_{x\to \infty} \frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}$ $$\lim_{x\to \infty} \frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{(2\tan^{-1}(x^2)-\pi)x^2}$$
$$=\lim_{x\to \infty} \frac{1}{x^2(2\tan^{-1} x^2 -\pi)}$$
How do I proceed?
| From here
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}$$
we have that by standard limit
$$\frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \to 1$$
and by $\arctan x = \frac \pi 2 -\arctan \frac1x$
$$\frac{1}{\left(2\tan^{-1}(x^2)-\pi\right)x^2}=-\frac12\frac{\frac1{x^2}}{\tan^{-1}\left(\frac1{x^2}\right)} \to -\frac12$$
To proceed by l'Hospital by $\frac1{x^2}=t \to 0$ we have
$$\lim_{x\to \infty }\frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}=\lim_{t\to 0 }\frac{e^{t}-1}{2\tan^{-1} \left(\frac1t\right)-\pi}=\lim_{t\to 0 }\frac{e^{t}}{-\frac2{t^2+1}}=-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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If $\frac1x+\frac1y+\frac1z=0, xyz \neq 0$, $\sqrt[3]{\tiny\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} =?$ If $\frac1x+\frac1y+\frac1z=0,$ and $xyz \neq 0$, What is $$K =\sqrt[3]{\frac{x^9+y^9+z^9-3xyz(x^6+y^6+z^6)+6x^3y^3z^3}{x^6+y^6+z^6-3x^2y^2z^2}} ?$$
Source: Lumbreras Editors
I found this way:
$ (xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$
$ \Rightarrow (xy)^2+(yz)^2+(zx)^2 =-2xyz(x+y+z) $
For Gauss:
$ (xy)^3+(yz)^3+(zx)^3=3x^2y^2z^2$
Whit:
$(x+y+z)(\underbrace{xy+yz+zx}_{0})=(x+y)(y+z)(z+x)+xyz$
$ ⇒(x+y)(y+z)(z+x) =-xyz$
And
$ (x+y+z)^2 =x^2+y^2+z^2+2(\underbrace{xy+yz+zx}_{0})$
$ ⇒ (x+y+z)^2 =x^2+y^2+z^2$
$ ⇒ (x+y+z)^6 = x^6+y^6+z^6+3(x^2+y^2+z^2)\underbrace{x^2y^2+y^2z^2+z^2x^2}_{-2xyz(x+y+z)})\\
-3x^2y^2z^2$
$ ⇒ (x+y+z)^3[(x+y+z)^3+6xyz] = x^6+y^6+z^6-3x^2y^2z^2$
\intertext{Trinomio al cubo:}
$(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(\underbrace{xy+yz+zx}_{0})-3xyz $
$ ⇒ (x+y+z)^3=x^3+y^3+z^3-3xyz $
Then:
$ (x+y+z)^6=x^6+y^6+z^6+6x^2y^2z^2+3[x^3(y^6+z^6)+y^3(x^6+z^6)+z^3(x^6+y^6)]$
So far I got, I didn't know what else to do
| Use the identity:
$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$$
Then the denominator, $A$, inside the square root is:
\begin{equation}A = (x^9+y^9+z^9-3x^3y^3z^3) -3xyz(x^6+y^6+z^6 - 3x^2y^2z^2)\end{equation}
The first summand is then further decomposed as:
\begin{align}
x^9+y^9+z^9-3x^3y^3z^3 = (x^3+y^3+z^3)(x^6+y^6+z^6 - x^3y^3-y^3z^3-z^3x^3)
\end{align}
However, the same identity gives us:
$$\sum\dfrac{1}{x^3} = \dfrac{3}{xyz}\iff \sum x^3y^3 = 3x^2y^2z^2.$$
Therefore, the cube of your fraction is:
$$\dfrac{\sum x^9 - 3x^3y^3z^3}{\sum x^6 - 3x^3y^2z^2} - 3xyz = x^3+y^3+z^3 - 3xyz.$$
Now, let $x+y+z=p$ and $xyz = r$ with $xy+yz+zx = 0.$ Then,
$$x^3+y^3+z^3 - 3xyz = (x+y+z)((x+y+z)^2-3xy-3yz-3zx) = p^3.$$
This means that your fraction is equal to:
$$p = x+y+z.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| The inequality
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
become equality when $a-b=b-c=c-a,$ or $a=b=c,$ but for this value then
$$(a-b)(b-c)(c-a)=0 \ne -8.$$
This is my solution, we have
$$P^2 = \frac{4(a^2+b^2+c^2-ab-bc-ca)^3-(a+b-2c)^2(b+c-2a)^2(c+a-2b)^2}{27} \quad (1)$$
$$\leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3.$$
Therefore
$$P^2 \leqslant \frac{4}{27}(a^2+b^2+c^2-ab-bc-ca)^3 = 32,$$
or
$$-4\sqrt 2 \leqslant (a-b)(b-c)(c-a) \leqslant 4\sqrt 2.$$
So $P_{\min} = -4\sqrt 2,$ equality occur when $a=1,\;b=1+2\sqrt 2,\;c=1+\sqrt 2.$
Note. How to find constant $\frac{4}{27}?$
For $(a-b)(b-c)(c-a) \ne 0,$ setting $x=a-b,\,y=b-c,$ then
$$F = \frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2+c^2-ab-bc-ca)^3} = \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}.$$
From $x^2+xy+y^2 \geqslant \frac{3}{4}(x+y)^2$ and the AM-GM inequality, we have
$$F \leqslant \frac{64}{27} \cdot \frac{x^2y^2}{(x+y)^4} \leqslant \frac{4}{27}.$$
From this proof we get
$$\frac{4}{27}- \frac{x^2y^2(x+y)^2}{(x^2+xy+y^2)^3}=\frac{(x-y)^2(2y+x)^2(2x+y)^2}{(x^2+xy+y^2)^3}.$$
It's equivalent to the identity $(1).$
| {
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Simplification of Combinatorial Expression I have been solving a problem and have been stuck on it for hours now.
I can't recall any combinatorial approach/formula for the below expression :
$$\sum_{i=a+1}^n {i-1 \choose a}{n-i \choose k-a}$$
Any sort of help would be appreciated. I know it is a norm to share my approach as well to show that I have put an effort but honestly I don't have any clue on this particular question.
Thanks!
| What follows is admittedly not the simplest proof. We post here by way
of enrichment and to showcase four techniques, coefficient extractors,
the Iverson bracket, residues and the Leibniz rule.
We seek to show that
$$\sum_{q=a+1}^n {q-1\choose a} {n-q\choose k-a}
= {n\choose k+1}$$
where $k\ge a$ for the binomial coefficient to be defined,
and $n\ge a+1$ or alternatively
$$\sum_{q=0}^{n-a-1} {q+a\choose a} {n-a-1-q\choose k-a}
= {n\choose k+1}.$$
The LHS is
$$[z^{k-a}] (1+z)^{n-a-1}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q} [[q\le n-a-1]]
\\ = [z^{k-a}] (1+z)^{n-a-1}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q}
[w^{n-a-1}] \frac{w^q}{1-w}
\\ = [z^{k-a}] (1+z)^{n-a-1} [w^{n-a-1}] \frac{1}{1-w}
\sum_{q\ge 0} {q+a\choose a} (1+z)^{-q} w^q
\\ = [z^{k-a}] (1+z)^{n-a-1} [w^{n-a-1}] \frac{1}{1-w}
\frac{1}{(1-w/(1+z))^{a+1}}
\\ = [z^{k-a}] (1+z)^{n} [w^{n-a-1}] \frac{1}{1-w}
\frac{1}{(1+z-w)^{a+1}}.$$
This is
$$[z^{k-a}] (1+z)^n (-1)^a
\mathrm{Res}_{w=0} \frac{1}{w^{n-a}}
\frac{1}{w-1} \frac{1}{(w-(1+z))^{a+1}}.$$
Now the residue at infinity for $w$ is zero by inspection, residues
sum to zero and the residue at $w=1$ yields
$$[z^{k-a}] (1+z)^n (-1)^a \frac{1}{(-1)^{a+1} z^{a+1}}
= - {n\choose k+1}.$$
This is the claim if we can show that the contribution from the pole at
$w=1+z$ is zero. We get (Leibniz rule)
$$\frac{1}{a!} \left(\frac{1}{w^{n-a}} \frac{1}{w-1}\right)^{(a)}
= \frac{1}{a!}
\sum_{q=0}^a {a\choose q}
\frac{(-1)^q (n-1-a+q)!}{(n-1-a)! \times w^{n-a+q}}
\frac{(-1)^{a-q} (a-q)!}{(w-1)^{a+1-q}}
\\ = (-1)^a \sum_{q=0}^a {n-1-a+q\choose q}
\frac{1}{w^{n-a+q}} \frac{1}{(w-1)^{a+1-q}}.$$
We thus obtain for the contribution
$$[z^{k-a}] (1+z)^n \sum_{q=0}^a {n-1-a+q\choose q}
\frac{1}{(1+z)^{n-a+q}} \frac{1}{z^{a+1-q}}
\\ = \sum_{q=0}^a {n-1-a+q\choose q}
[z^{k+1-q}] (1+z)^{a-q} = 0$$
because $a\ge q$ and $k+1\gt a.$ This concludes the argument.
| {
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If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971) If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$
I was trying to do the question above using homogenization. I was trying to do it as follows:
$\sqrt{xy+yz+xz}=\sqrt{26}$
Hence $x+y+z>\frac{35}{4*\sqrt{26}}*\sqrt{xy+yz+xz}$
So it is enough to prove that $16*26(x^2+y^2+z^2+2xy+2yz+2xz)>35*35*(xy+yz+xz)$. Which is true since $x^2+y^2+z^2+2xy+2yz+2xz\ge 3(xy+yz+xz)$ Could you please share other simple ways to solve the question, as I am intrigued by its simplicity.
| We can use AM-GM inequality to get
$$26 = xy + yz + xz \leq \frac{x^2+y^2}{2} + \frac{y^2+z^2}{2} + \frac{x^2+z^2}{2} = x^2+y^2+z^2$$
Then add the two inequalities
$$x^2+y^2+z^2+2xy+2yz+2xz = (x+y+z)^2 > 78$$
$$\implies x+y+z > \sqrt{78} > \frac{35}{4}$$
Oddly enough, this only relies on $x+y+z$ being positive, and we never used the more restrictive range.
| {
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If AM=a, NC=b, BL=? Source: "Problemas selectos" by Lumbreras Editors.
The bisectors harmonically divide the segment AC: $→ \dfrac{NC}{MN}=\dfrac{AM+MN+NC}{AM}$
By metric relation in triangle ABC: $ → (AM + MN + NC)(MN + NC) = (BL + LC)^2$
It is what I managed to relate, I did not advance further.
| Let $O$ be the center of the circle, $r$ be the radius, $h=MT$ and $x=MN$. We have
$$2r=AN=AM+MN=a+x$$
$$r^2=MT^2+OM^2=h^2+(ON-MN)^2=h^2+(r-x)^2$$
$$2rx=h^2+x^2$$
$$(r+b)^2=(ON+NC)^2=OC^2=OT^2+TC^2=r^2+MT^2+MC^2=r^2+h^2+(MN+NC)^2=r^2+h^2+(x+b)^2$$
$$2br=h^2+x^2+2bx=2rx-x^2+x^2+2bx=2xr+2bx$$
$$2r(b-x)=2bx$$
$$(a+x)(b-x)=2bx$$
$$ab-bx-ax-x^2=0$$
$$x^2+x(a+b)-ab=0$$
$$x=\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}-a-b\right)$$
Let $\xi=\angle BTL=\angle MTN$ and $\alpha=\angle BAC=\angle MBC=\angle TBL$. Then
$$(a+x+b)\cos^2\alpha=AC\cos^2\alpha=AM=a$$
$$\tan\xi=\frac{x}{h}$$
$$BM^2=AM\cdot MC=a(x+b)$$
By sine law in triangle $\triangle BLT$:
$$\frac{BL}{\sin\xi}=\frac{BT}{\sin(180-\alpha-\xi)}$$
$$BL=(BM+MT)\frac{\sin\xi}{\sin(\alpha+\xi)}=\frac{\sqrt{ax+ab}+\sqrt{ax}}{\frac{\sin\alpha}{\tan\xi}+\cos\alpha}=\frac{\sqrt{ax+ab}+\sqrt{ax}}{\sqrt{\frac{x+b}{a+x+b}}\cdot\frac{\sqrt{ax}}{x}+\sqrt{\frac{a}{a+x+b}}}=\sqrt{x(a+b+x)}=\sqrt{\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}-a-b\right)\cdot\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}+a+b\right)}=\sqrt{\frac{1}{4}\left(a^2+6ab+b^2-(a+b)^2\right)}=\sqrt{ab}$$
| {
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Is $(-2)^{\frac{7}{5}}=((-2)^7)^{\frac{1}{5}}$ or not? In Wolfram Alpha this statement is false. But how? Because $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}$. Is there any condition. Please tell me.
| We have that $(a)^{\frac{b}{c}}=(a^b)^{\frac{1}{c}}=(a^\frac1c)^{b}$ holds for any $a\ge 0$.
For the case $(-a)^{\frac{b}{c}}$ with $b$ and $c$ coprime we need to consider the following cases:
*
*$b$ odd and $c$ even
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\sqrt[c] {(-1)^b}\cdot a^{\frac{b}{c}} $$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left(\sqrt[c] {-1}\right)^b\cdot a^{\frac{b}{c}} $$
*
*$b$ even and $c$ odd
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\left[1\cdot(a^b)\right]^\frac1c=a^{\frac{b}{c}} $$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left(\sqrt[c] {-1}\right)^b\cdot(a)^{\frac{b}{c}} =(-1)^b \cdot (a)^{\frac{b}{c}}=a^{\frac{b}{c}}$$
*
*$b$ odd and $c$ odd
$$\left[(-a)^b\right]^\frac1c=\left[(-1)^b\cdot(a^b)\right]^\frac1c=\left[(-1)\cdot(a^b)\right]^\frac1c=\sqrt[c] {-1}\cdot(a)^{\frac{b}{c}}=-\left(a^{\frac{b}{c}}\right)$$
$$\left[(-a)^\frac1c\right]^b=\left[(-1)^\frac1c \cdot a^\frac1c\right]^b=\left[(-1) \cdot a^\frac1c\right]^b=(-1)^b\cdot(a)^{\frac{b}{c}} =-\left(a^{\frac{b}{c}}\right)$$
Therefore for the given example
$$(-2)^{\frac{7}{5}}=-\left(2^{\frac{7}{5}}\right)$$
| {
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Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim_{(x,y)\to(2,5)} xy = 10$.
Question:
Use the $\varepsilon - \delta$ definition of the limit to verify that
$\lim\limits_{(x,y)\to(2,5)}xy = 10$.
Hint given:
$xy − 10 = (x − 2)(y − 5) + 5(x − 2) + 2(y − 5).$
My solution
$\forall \space\varepsilon\gt0\space\space\exists\space\delta(\varepsilon)\gt0:|xy-10|\lt\varepsilon$ where $(x,y)\in D$, whenever $0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$
Note:
$(x-2)^2 \le (x-2)^2+(y-5)^2$ $\implies |x-2|\le\sqrt{(x-2)^2+(y-5)^2}$, similarly, $|y-5|\le\sqrt{(x-2)^2+(y-5)^2}$
So if $(x,y)\in D\space$ and $\space0\lt\sqrt{(x-2)^2+(y-5)^2}\lt\delta$. We choose $ \delta :=\frac{-7+\sqrt{49+4\varepsilon}}2$.
We obtain...
$\begin{aligned}|xy-10|&=|(x − 2)(y − 5) + 5(x − 2) + 2(y − 5)|\\&\le|x − 2||y − 5|+ 5|x − 2| + 2|y − 5|\\&\le \left((x-2)^2 +(y-5)^2\right) +5\sqrt{(x-2)^2+(y-5)^2}+2\sqrt{(x-2)^2+(y-5)^2}\\&=\left(\sqrt{(x-2)^2+(y-5)^2}\right)^2+7\sqrt{(x-2)^2+(y-5)^2}\\&\lt \delta^2+7\delta\\&=\frac{49\pm\sqrt{49-4\varepsilon}+49+4\varepsilon}4+\frac{-98\pm14\sqrt{49+4\varepsilon}}4\\&=\frac{4\varepsilon}4\\&=\varepsilon\end{aligned}$
Comments
This is a question on my Analysis 2 module. Would be great if someone could check my solution. Also, I realise you have to use a minimum for $\delta$, but unsure on how to unpack it when writing the solution, would be great if anyone can make my understanding of this clearer.
| Note that since $\epsilon >0$, $\sqrt{ 49 + 4\epsilon } > \sqrt{49} = 7$. So your $\delta$ is negative.
To remedy this, you can choose
$$ \delta = \frac{-7+\sqrt {49+4\epsilon}}{4},$$
which is positive and $\delta^2 + 7\delta = \epsilon$.
Another method: note that you obtained
$$ |xy-10| \le \delta^2 + 7\delta$$
and you need that $\delta ^2 + 7\delta <\epsilon$. For example, if you choose $\delta <1$, then
$$ \delta ^2 + 7\delta < \delta + 7\delta.$$
Then if you choose $ \epsilon = \min\{ 1, \delta/8\}$, you have
$$|xy-10| \le \delta^2 + 7\delta<\delta + 7\delta < \epsilon. $$
| {
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Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z}$ such that $|mn|\ne 1$ and $y=x+m$ and $z=x-n$? Another way would be assuming at least 2 of them are non-consecutive?
I would like to see general ideas on how to proceed.
| Use the well-known identity $$x^3+y^3+z^3-3xyz = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right)$$ to have, from your given equation (since the left hand side equals $3(x+y+z)$, $$\begin{aligned} 3(x+y+z) & = \dfrac12(x+y+z)\left((x-y)^2+(y-z)^2+(z-x)^2\right) \\ \implies 6 &= (x-y)^2+(y-z)^2+(z-x)^2 \ (\because x,y,z>0 \implies x+y+z\ne 0) \end{aligned}$$
also, $x,y,z$ being positive integers, the squares in the last line $(x-y)^2, (y-z)^2, (z-x)^2$ are perfect square integers, and the only way three perfect squares add up to $6$ is $$6=1+1+4$$
so, two of the differences $|x-y|$ and $|y-z|$ (without loss of generality) must be $1$, naturally the $|x-z|$ must be $2$.
| {
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Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is ..... Let $f(x) = x^2+ax+b$ for $a,b \in \mathbb{R}$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is
(A) real
(B) imaginary
(C) real and distinct
(D) equal roots
My attempts:
\begin{align}
f(1) &= 1+a+b \\
f(2) &= 4+2a+b \\
f(3) &= 9+3a+b \\
f(1)+f(2)+f(3)
&= 1+a+b+4+2a+b+9+3a+b \\
0 &= 14+6a+3b
\end{align}
now how can we take it further about the nature of the roots , whether the roots of $f(x)=0$ is imaginary or real , please help, thanks...
| Brute force,
\begin{align}
\Delta &= a^2-4b \\
&=a ^2-4\left( -\frac{14+6a}{3} \right) \\
&= a^2+8a \color{red}{+16} +\frac{56}{3} \color{red}{-16} \\
&= (a+4)^2+\frac{8}{3} \\
&> 0
\end{align}
| {
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If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.
From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.
| It's true even for any reals $a$, $b$ and $c$ such that $a+b+c=3.$
Indeed, by C-S $$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}\right)}=$$
$$=\sqrt{\sum_{cyc}\left(2a^2+ab+2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3}{4}b^2\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3}{4}c^2\right)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(2a^2+ab+2\left(\left(a+\frac{b}{2}\right)\left(a+\frac{c}{2}\right)+\frac{3}{4}bc\right)\right)}=$$
$$=\sqrt{\sum_{cyc}(4a^2+5ab)}\geq\sqrt{3(a+b+c)^2}=\sqrt3|a+b+c|=3\sqrt3>\sqrt3.$$
For positive variables we have:
$$\sum_{cyc}\sqrt{a^2+ab+b^2}\geq a+b+c=3>\sqrt3.$$
| {
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Have I done my homogenization correctly for this question and if so how do I finish it off? We have $a,b,c>0$ with $a^2+b^2+c^2=1$ prove that:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>1$
As soon as I saw this question, I immediately thought of using homogenization in the following manner:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>\frac{1}{\sqrt{a^2+b^2+c^2}}$
Then I tried using Andreescu:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}\ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{(1+bc)(1+ac)(1+ab)}$
Which didn't work out.
Could you please explain to me if I have done my homogenization correctly? If I have done it correctly, could you please show me how to finish it off and if I haven't how to think of doing the correct homogenization?
| Observe that $$\frac{a}{1+bc}+\frac{b}{1+ca}+\frac{c}{1+ab} = \sum_{cic} \frac{a^2}{a+abc} \geqslant \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}$$ Using $a^2+b^2+c^2=1$, we obtain $$ \sum_{cyc} \frac{a^2}{a+a \cdot \frac{b^2+c^2}{2}}= \sum_{cyc} \frac{a^2}{a+a \cdot \frac{1-a^2}{2}} = \sum_{cyc} \frac{a^2}{1-\frac{(a+2)(a-1)^2}{2}}$$ This final sum is clearly smaller than $a^2+b^2+c^2=1$.
| {
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Triple integration with $x^2+y^2-z^2$ I'm having this problem.
$$\iiint_{K}(x^2+y^2-z^2)dxdydz,\quad K: x^2+y^2+z^2\leq 1$$
My solution.
$$x=\rho \sin\phi \cos\Theta,\ y=\rho \sin\phi \sin\Theta, z=\rho \cos \phi\\
dxdydz=\rho^2 \sin\phi d\rho d\phi d\theta\\
0\leq \rho \leq 1\\
0 \leq \phi \leq \pi\\
0 \leq \theta \leq 2\pi\\
$$
$$\int _0^{2\pi }\int _0^{\pi }\int _0^1\left(\left(\rho \sin\left(\phi\right)\cos\left(\theta\right)\right)^2+\left(\rho \sin\left(\phi\right)\sin\left(\theta\right)\right)^2+\left(\rho \cos\left(\phi\right)\right)^2\right)d\rho d\phi d\theta$$
$$\int _0^1\left(\rho\sin \left(\phi\right)\cos \left(\theta\right)\right)^2+\left(\rho\sin \left(\phi\right)\sin \left(\theta\right)\right)^2-\left(\rho \cos \left(\phi\right)\right)^2d\rho\\ =-\cos ^2\left(\phi\right)\frac{1}{3}+\cos ^2\left(\theta\right)\sin ^2\left(\phi\right)\frac{1}{3}+\sin ^2\left(\phi\right)\sin ^2\left(\theta\right)\frac{1}{3}\\
\int _0^{\pi }\left(-\cos ^2\left(\phi\right)\frac{1}{3}+\cos ^2\left(\theta\right)\sin ^2\left(\phi\right)\frac{1}{3}+\sin ^2\left(\phi\right)\sin ^2\left(\theta\right)\frac{1}{3}\right)d\phi\\
=-\frac{\pi }{6}+\frac{\pi }{6}\cos ^2\left(\theta\right)+\frac{\pi }{6}\sin ^2\left(\theta\right)=\frac{\pi }{6}\left(\cos ^2\left(\theta\right)+\sin ^2\left(\theta\right)-1\right)=\frac{\pi }{6}\left(-1+1\right)=0\\
\int _0^{2\pi }0d\theta=0
$$
It's the minus sign in $-z^2$ that creates multiplication with $0$, and therefore the answer is $0$. Is this correct?
When I plot this in Geogebra I get a sphere inside a hyperboloid.
Thanks for all the help!
| The integrals over $y^2-z^2$ cancel due to symmetry. It remains to integrate over $x^2$ over the volume which gives $\frac{4\pi}{15}$
| {
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System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} \\ \alpha^2r(1+r+2r^2+r^3+r^4)=\frac{35}{128} \\ \alpha^3 r^3(1+r+r^2+r^3)=\frac{15}{512} \\ \alpha^4 r^6=\frac{1}{1024}\end{gather}
But this looks like a rather complex system and I can't see any obvious way solve this for $\alpha$ and $r$.
How can this system be solved?
EDIT
I can see that all of the denominators are powers of $2$, but I can't see how that will help me here.
| Since you see powers of $2$, you can use a factorization:
$$3072x^4-2880x^3+840x^2-90x+3=3(1024x^4-960x^3+280x^2-30x+1)=$$
$$=3(1024x^4-64x^3-896x^3+56x^2+224x^2-14x-16x+1)=$$
$$=3(16x-1)(64x^3-56x^2+14x-1)=$$
$$=3(16x-1)(64x^3-8x^2-48x^2+6x+8x-1)=$$
$$=3(16x-1)(8x-1)(8x^2-6x+1)=2(16x-1)(8x-1)(4x-1)(2x-1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.
If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.
So, we cannot prove as follows:
$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
| It all depends on how much you "want to prove".
The function $f(x)= \sqrt{1-x^2}$ has domain $x\in [-1,1]$. The derivative of $f$, if well-defined, can only be defined on $(-1,1)$. If possible, you can extend the definition to $x=\pm 1$ by taking the left- and right-limits.
The derivative of $f$ is
$$f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{\sqrt{1-(x+h)^2} - \sqrt{1-x^2}}{h} = \lim_{h\to 0} \frac{\sqrt{1-(x+h)^2} - \sqrt{1-x^2}}{h}\frac{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}= \lim_{h\to 0} \frac{1-(x+h)^2 -1+x^2}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}}=\lim_{h\to 0} \frac{-2x-h}{\sqrt{1-(x+h)^2} + \sqrt{1-x^2}} = -\frac{x}{\sqrt{1-x^2}}.$$
As you can see the function $f$ is not defined on $x=-1$ or $x=1$, hence $f$ is not differentiable at those points. There is not much proof other than differentiating $f$ and looking at its definition domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Variance of $X_n$ if we have convergence in probability for a random variable with $|X_n| >1$ Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$. Assume $X_n \overset{p}{\to} X$. True or false: $Var(X_n)\to 0$. If false, give a counterexample. If true, give an argument.
Convergence of $\operatorname{Var}(X_n)$ if $|X_n| \le 1$ seems relevant. How much does the $X_n \to 0$ change what I have to say? Would appreciate a start in the right direction thanks!
__
Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$. Assume $X_n \overset{p}{\to} X$. True or false: $Var(X_n)\to 0$. If false, give a counterexample. If true, give an argument. \
\indent The statement is false. Here we can offer a counter example.\
\
\indent Let $X \sim U[0,1]$. Since $X_n \overset{p}{\to} X$ we know $X_n \overset{d}{\to} X$. This implies that $Var(X_n) = Var(X)$. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can then follow the steps outlined here and say:
\begin{equation*}
\begin{aligned}
& \operatorname{Var}(X) = \int_{-\infty}^\infty x^2 f_X(x) \, dx - \left( E(X) \right)^2 \\
& = \int_{-\infty}^0 0 f_X(x) \, dx + \int_0^1 x^2 f_X(x) \, dx + \int_0^\infty 0 f_X(x) \, dx - \left( \frac{0+1}{2} \right)^2 \\
& = \int_0^1 x^2 (1) \, dx - \frac{1}{4} \\
& = \left( \frac{x^3}{3} \right) \bigg\rvert_0^1 \\
& = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\
& = \frac{1}{12}
\end{aligned}
\end{equation*}\
| I added an edit that posted my answer after following some advice given! Thanks to Teresa!
Let $X \sim U[0,1]$. Since $X_n \overset{p}{\to} X$ we know $X_n \overset{d}{\to} X$. This implies that $X_n \sim U[0,1]$ $Var(X_n) = Var(X)$ and we will show this $\neq 0$ for at least the following case. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can say:
\begin{equation*}
\begin{aligned}
& Var(X) = \int_{-\infty}^{\infty} x^2 f_X(x) dx - \left( E(X) \right)^2 \\
& = \int_{-\infty}^{0} 0 f_X(x) dx + \int_{0}^{1} x^2 f_X(x) dx + \int_{0}^{\infty} 0 f_X(x) dx - \left( \frac{0+1}{2} \right)^2 \\
& = \int_{0}^{1} x^2 (1) dx - \frac{1}{4} \\
& = \left( \frac{x^3}{3} \right) \bigg\rvert_{0}^{1} \\
& = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\
& = \frac{1}{12}
\end{aligned}
\end{equation*}\
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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