Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $\frac{n}{3} \le \sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2} \le \sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1}$.
Given $0 < y < 1 < n \in \mathbb Z^+$ and $n$ positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $\displaystyle \sum_{i = 1}^nx_i = n$.
Prove that $$\large \frac{n}{3} \le \sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2} \le \sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1}$$
We have that
$$\sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2} - \frac{n}{3}$$
$$\frac{y}{3} \cdot \sum_{i = 1}^n\frac{(x_i - 1)(2n - x_iy - 2y)}{x_i^2y^2 + x_iy^2 + y^2 - 3nx_iy - 3ny + 3n^2}$$
and
$$\sum_{i = 1}^n\frac{1}{x_i^2 + x_i + 1} - \sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}$$
$$ = n \cdot \sum_{i = 1}^n\frac{(x_i - 1)(nx_i + 2n - 2x_iy - y)}{(x_i^2 + x_i + 1)(x_i^2y^2 + x_iy^2 + y^2 - 3nx_iy - 3ny + 3n^2)}$$
So that's no help.
Besides, $$\sum_{i = 1}^n\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}$$ is a concave function.
So Jensen's inequality could help here, but I don't know how.
| The left inequality.
We need to prove that
$$\sum_{i=1}^n\left(\frac{(n - y)^2}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}-\frac{1}{3}\right)\geq0$$ or
$$\sum_{i=1}^n\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\geq0$$ or $$\sum_{i=1}^n\left(\frac{(x_i-1)(3n-2y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}-\frac{x_i-1}{n-y}\right)\geq0$$ or
$$\sum_{i=1}^n\frac{(x_i-1)^2(2n-y-yx_i)}{y^2(x_i^2 + x_i + 1) - 3ny(x_i + 1) + 3n^2}\geq0,$$ which is obvious.
For $n\geq3$ we can prove a right inequality by the same way.
Also, easy to check that the right inequality is true for $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Irrational equation high school I've been trying to solve this one without success... can anybody help me?
The result should be $x=\frac{17}{16}$ and it's correct, I've already checked.
This is the equation:
$$\frac{1}{\sqrt {x+2} - \sqrt x}+\frac{2}{\sqrt {x+4} + \sqrt x}=2$$
Thanks in advance!
| Rationalize the denominators.
$$\frac{1}{\sqrt{x+2} - \sqrt{x}} = \frac{\sqrt{x+2} + \sqrt{x}}{2},$$
and
$$\frac{2}{\sqrt{x+4} + \sqrt{x}} = \frac{2\sqrt{x+4} - 2\sqrt{x}}{4},$$
and so we find
$$\sqrt{x+2} + \sqrt{x+4} = 4.$$
Thus,
$$\sqrt{x+4} = 4 - \sqrt{x+2},$$
and so we find that
$$x+4 = x + 18 - 8\sqrt{x+2},$$
$$8\sqrt{x+2} = 14,$$
$$x+2 = \frac{196}{64},$$
and so
$$x = \frac{68}{64} = \frac{17}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$ Evaluate:
$$ \lim_{x\to0}{\frac{1}{x^2}-\cot^2(x)}$$
My approach :
$$\lim_{x\to0}{\frac{1}{x^2}-\frac{\cos^2(x)}{\sin^2(x)}}$$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} $$
Using $$\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $$
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} $$
$$ \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
$$ \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} $$
Applying L Hopital,
$$\lim_{x\to0}\frac{2\sin(x)\cos(x)}{2x}=1$$
But the actual answer is $\frac{2}{3}$. What am I doing wrong here?
| The following steps are not allowed
$$ \lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^2\sin^2(x)} \color{red}{=\lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4} }= \lim_{x\to0}\frac{\sin^2(x)}{x^2}\cdot\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} =\\ \color{red}{= \lim_{x\to0}\frac{1}{x^2}-\frac{\cos^2(x)}{x^2} }$$
since $\lim_{x\to0}\frac{\sin^2(x)}{x^2}=1 $ doesn't mean that $\frac{\sin^2(x)}{x^2}=1$.
We need to solve by l'Hopital or by Taylor's expansion starting from the first expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Convergence of product of rational functions How can I prove that the sequence $a_{n} = \frac{4}{5} * \frac{104}{105} * ... * \frac{50n^{2} - 50n + 4}{50n^{2} - 50n + 5}$ converges or not (where $a_{1} = \frac{4}{5}$, $a_{2} = \frac{4}{5} * \frac{104}{105}$, and $a_{3}= \frac{4}{5} * \frac{104}{105} * \frac{304}{305}$)?
| Hint: Your sequence is
$$a_N=\prod_{n=0}^N\frac{50n^2-50n+4}{50n^2-50n+5}=\prod_{n=0}^N\left(1-\frac1{50n^2-50n+5}\right).$$
Each additional term in the product causes it to decrease in value since the $a_n$ are positive and the term is less than $1$. Now can you think of a way to use monotone convergence theorem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Direct telescopic proof for sum of $1^2+2^2+...+n^2$ I was teaching $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\\\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$$ for the first I did direct telescopic proof like below
$$\sum_{k=1}^{n}k=\sum_{k=1}^{n}k(\frac{k+1}{2}-\frac{k-1}{2})=\sum_{k=1}^{n}(\frac{k(k+1)}{2}-\frac{k(k-1}{2}))=\\\sum_{k=1}^{n}(f(k)-f(k-1))=\frac{n(n+1)}{2}-0$$
for the second I did a classic proof $$\begin{align}
\sum_{k=1}^n k^2
& = \frac{1}{3}(n+1)^3 - \frac{1}{2}n(n+1) - \frac{1}{3}(n+1) \\
& = \frac{1}{6}(n+1) \left[ 2(n+1)^2 - 3n - 2\right] \\
& = \frac{1}{6}(n+1)(2n^2 +n) \\
& = \frac{1}{6}n(n+1)(2n+1)
\end{align}$$ some of the student asked for direct telescopic proof for the case...
I can't find this kind of proof. can anybody help me to find, or write this kind proving. \
I tried to rewrite $1=\frac{k+1}{2}-\frac{k-1}{2}$and I have
$$\sum_{k=1}^{n}k^2(\frac{k+1}{2}-\frac{k-1}{2})=$$ I can't go further more.
I promised to my students to try to find a direct proof Idea. Thanks for any help.
| Noting that
$$ (n+1)^2-n^2=2n+1, (n+1)^3-n^3=3n^2+3n+1 $$
one has
$$ n=\frac12\bigg[(n+1)^2-n^2\bigg]-\frac12,n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13. $$
So
$$ n^2=\frac13\bigg[(n+1)^3-n^3\bigg]-n-\frac13=\frac13\bigg[(n+1)^3-n^3\bigg]-\frac12\bigg[(n+1)^2-n^2\bigg]+\frac16 $$
and hence
\begin{eqnarray}
\sum_{k=1}^nk^2&=&\sum_{k=1}^n\left[\frac13\bigg[(k+1)^3-k^3\bigg]-\frac12\bigg[(k+1)^2-k^2\bigg]+\frac16\right]\\
&=&\frac13\bigg[(n+1)^3-1\bigg]-\frac12\bigg[(n+1)^2-1\bigg]+\frac n6\\
&=&\frac16n(n+1)(2n+1)
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the the least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real The least integral value of a for which all the roots of the equation $x^4-4x^3 -8x^2 +a=0$ are real.
let $f(x) =x^4-4x^3 -8x^2 +a=0$
$ f'(x) = 4x^3 -12x^2 -16x $
Put $f'(x) = 0 $ we get x =0, -1, 4. How to proceed further, I am not getting any idea on this. Please guide thanks.
| Let $f(x)=x^4-4x^3-8x^2+a$, $$f'(x)=4x^3-12x^2-16x=0 \Rightarrow x=4,-1,0$$
Next $f''(x)=12x(x-2)-16.$ so $f(x)$ has local minima at at $x=4,-1$ and loca max at $x=0.$ Hence $f_{min}=a-128, a-3$, $f_{max}=a$. For four real rootss: $f_{min}<0$ and
$f_{max}>0.$ Thus, for four real roots $$a \in( 0,3).$$
Both the integers $a=1,2$ are the solutions.
See the fig. for a=1, below
$f(x)$ for $a=1$">
See the fig. for $a=2$, below
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
I have troubles with proving the induction step: $\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$ Let's assume that the unequality is right if $n = k$, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k-1)}{2k} ≤ \frac{1}{\sqrt{2k+1}}$
Let's prove that the unequality is right for $n = k + 1$ as well, that is:
$\frac{1}{2} * \frac{3}{4} * \frac{5}{6} * ... * \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}}$
I've spent a comparable amount of time trying to prove this point but looks like I can't grasp what to do.
| Note that the induction hypotesis reduces to show that
$$ \frac{1}{\sqrt{2k+1}} \frac{(2k+1)}{2k+2} ≤ \frac{1}{\sqrt{2k+3}} \implies\sqrt{(2k+1)(2k+3)}\le 2k+2$$
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
| You can use this result: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
Since $T_n$ is of degree $4$ then its sum will be of degree $5$.
So let $S(n)=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5$.
Then you can proceed by solving the system obtained calculating the first $6$ terms $S(1)$ to $S(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Surface tangent plane Determine a plane that is tangent to the surface $x ^{2}$ + $3y ^ 2$ + $2z ^ 2$ = $\frac{11}{6}$ and parallel to the plane x + y + z = 10
$$\nabla f (x, y, z) = (2x, 6y, 4z)$$
Quotation vector $X + Y + Z = 10 \implies (1,1,1)$
$$(2x, 6y, 4z) = K (1,1,1)$$
for $K = 1$
*
*$2x = 1 x = \frac{1}{2}$
*$6y = 1 y = \frac{1}{6}$
*$4z = 1 z = \frac{1}{4}$
Putting the plane equation together: $(1,1,1) [(x - 1/2) + (y-1/6) + (z-1/4)]$
My plan equation gave: $X + Y + Z = \frac{-11}{12}$
But the answer of the book is $X + Y + Z = \frac{-11}{6}$or X + Y + Z = $\frac{11}{6}$
| I think there is an error for $x$ and $y$, I obtain $K=\pm 2$ and
*
*$x=\pm1$
*$y=\pm \frac13$
*$z=\pm \frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Perfect numbers (prime number theory) Calculate x and y, so that $2 ^ x. 3 ^ y$ be a perfect number.
I tried to produce using this formula of the sum of the divisors of a number: Let $a ^ x, b ^ y,$ and $c ^ z$ prime factors of a number: $\frac{ a^{x+1} -1}{ a-1}. \frac{b^{y-1} -1}{b-1}. \frac{c^{c +1} -1}{c-1}$
It replaces $ 2 ^ x = a$ and $ 3^y = b $, but it can't produce much else: $ a.b $ = $ \frac {(2a-1) (3b-1)} {2} $
| An even number is perfect if and only if it is of the form $2^{n-1}(2^n-1)$ for some $n$ with $2^n-1$ being prime. And $2^n-1=3^m$ is prime only for $n=2$ and $m=1$. So we obtain $2^1\cdot 3^1=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$ Convergence of $\sum_{n=1}^{\infty} \log (n\sin\frac{1}{n})$
First, we have
$$ n \sin \frac{1}{n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}} < 1 $$
for all positive integers $n$. (Since $\frac{\sin x}{x} < 1 \forall x > 0$.) Therefore,
$$ \sum_{n=1}^{\infty}\left| \log \left( n \sin \frac{1}{n} \right)\right| = \sum_{n=1}^{\infty}-\log \left( n \sin \frac{1}{n} \right) = \sum_{n=1}^{\infty} \log \left( \frac{1}{n \sin \frac{1}{n}} \right). $$
\begin{align*} \sin \frac{1}{n} &= \frac{1}{n} - \frac{1}{6n^3} + \cdots \\[9pt] \implies n \sin \frac{1}{n} &= 1 - \frac{1}{6n^2} + \cdots \\[9pt] &\geq 1 - \frac{1}{6n^2} \\[9pt] &\geq 1 - \frac{1}{n^2}. \end{align*}
Could someone please guide me on how to proceed from here?
Thanks a lot for your help.
| We have that
$$\sin \frac{1}{n} = \frac{1}{n} - \frac{1}{6n^3} + O\left(\frac1{n^5}\right)$$
$$n\sin \frac{1}{n} = 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)$$
$$\log \left(n \sin \frac {1}{n}\right)=\log \left( 1 - \frac{1}{6n^2} + O\left(\frac1{n^4}\right)\right)=-\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
therefore the given series converges by limit comparison test with $\sum \frac1{n^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3377929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$.
Then i calculated the area with
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next.
| If $AB=x$, then $AC=x-2,BC=x+1$.
Now use the cosine formula: $$x^2=(x-2)^2+(x+1)^2-(x-2)(x+1)$$ Expanding we get $x=7$. So the longest side is 8 and the height $AC\sin60^o=5\sqrt3/2$. Hence the area is $10\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove the inequality is true Here is a question that I need to prove
Prove that for $a, b \geq 0$
$$a^8+b^8\geq a^3b^5+a^5b^3$$
So far I have managed to simplify to
$$(a^3-b^3)(a^5-b^5)\geq 0$$
| If $a\geq b$ then $a^3\geq b^3$ and $a^5\geq b^5$ so $(a^3-b^3)(a^5-b^5)\geq 0$. Otherwise $a^3-b^3$ is negative, and so is $a^5-b^5$, so their product is positive, and the inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Calculating $\iiint_E\sqrt{3x^2+3z^2}\,dV$
Evaluate the following integral :$$\iiint_E\sqrt{3x^2+3z^2}\,dV$$where $E$ is the solid bounded by $y=2x^2+2z^2$ and the plane $y=8$.
I put $x=r\cos\theta$ and $z=r\sin\theta$ so, $y=2r^2\implies8\leq y\leq2r^2$. Finding limits of $r$ I got $0\leq r\leq2\csc\theta$ and for $\theta$ I got $\tan^{-1}4\leq\theta\leq\pi-\tan^{-1}4$.
$$\begin{align}\iiint_E\sqrt{3x^2+3z^2}\,dV &=\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}\int_8^{2r^2}\sqrt{3}\cdot r\ dy \ r\ dr\ d\theta\\&=2\sqrt3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\int_0^{2\csc\theta}r^2(r^2-4)dr\ d\theta\\&=2\sqrt3\ 8^3\int_{\tan^{-1}4}^{\pi-\tan^{-1}4}\csc^3\theta\bigg(\dfrac{8^2\csc^2\theta}{5}-\dfrac{4}{3}\bigg)d\theta\end{align}$$
Now, this is ugly to get through, please help, is there any efficient way to carry out this problem?
| Observe that the solid is symmetric with respect to the y-axis. So, it is efficient to integrate using cylindrical coordinates along the y-direction. Rewrite the original integral as
$$\iiint_E\sqrt{3x^2+3z^2}\,dV =\int_0^8 dy \int_0^{2\pi}d\theta\int_0^{\sqrt{y/2}} \sqrt{3}r\>rdr$$
$$=2\pi\sqrt 3 \int_0^8 dy \frac13 \left(\frac y2 \right)^{3/2}= \frac{256\sqrt 3}{15}\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Numeric sequence Beginning with $2$ and $7$, the sequence $2, 7, 1, 4,7 ,4, 2 , 8,$... is constructed by multiplying successive pairs of its members and adjoining the result as next one or two members of the sequence, depending on whether the product is a one- or a two-digit number. Prove that the digit 6 appears an infinite of times in the sequence.
I kept writing the sequence a few more terms and saw that $6$ will appear a few times, I also thought about which pairs of numbers will generate one, but not as if I assume it will happen forever
| (Too long to be a comment)
To slightly demystify how to approach this problem, one idea is to find a never-ending sequence, which then hopefully repeats. There is nothing obvious to do, so let's start naively:
Starting with 2, 7:
2, 7, ...
1, 4, ...
4, ...
At this point, we cannot continue.
So, let's start with 2, 7, 1:
2, 7, 1, ...
1, 4, 7, ...
4, 2, 8, ...
8, 1, 6, ...
8, 6, ...
4, 8, ...
3, 2, ...
6, ...
At this point, we cannot continue.
So, let's start with 2, 7, 1, 4:
2, 7, 1, 4, ...
1, 4, 7, 4, ...
4, 2, 8, 2, 8, ...
8, 1, 6, 1, 6, 1, 6, ...
8, 6, 6, 6, ...
4, 8, 3, 6, 3, 6, ...
3, 2, 2, 4, 1, 8, 1, 8, 1, 8, ...
6, 4, 8, 4, 8, 8, 8, 8, 8, ...
2, 4, 3, 2, 3, 2, 3, 2, 6, 4, 6, 4, 6, 4, 6, 4, ...
8, 1, 2, 6, 6, 6, 6, 6, 1, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, ...
8, 2, 1, 2, 3, 6, 3, 6, 3, 6, 3, 6, 6, 2, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, ...
From here, it follows that once we get that sequence of 8, 8, 8, we get the loop as pointed out by @79037662
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help solve recurrence : f(x) = 2f(x-1) + x While trying to analyze the time complexity of the heapify algorithm ,
I came up with the following recurrence for it's time complexity in terms of heap height:
T(h) = 2T(h-1) + h
Could you help me approach this recurrence?
| You can write it as a matrix equation:
$$\begin{pmatrix}1\\ T_h\end{pmatrix}=\begin{pmatrix}1&0\\ h&2\end{pmatrix}{1\choose T_{h-1}}=\begin{pmatrix}1&0\\ h&2\end{pmatrix}\begin{pmatrix}1&0\\ h-1&2\end{pmatrix}{1\choose T_{h-2}}=\\
\begin{pmatrix}1&0\\ (1+2)h-2&2^2\end{pmatrix}\begin{pmatrix}1&0\\ h-2&2\end{pmatrix}{1\choose T_{h-3}}=\\
\begin{pmatrix}1&0\\ (1+2+2^2)h-(2+2\cdot 2^2)&2^3\end{pmatrix}\begin{pmatrix}1&0\\ h-3&2\end{pmatrix}{1\choose T_{h-4}}=\cdots=\\
\begin{pmatrix}1&0\\ (1+2+\cdots+2^{h-1})h-(2+2\cdot 2^2+\cdots+(h-1)\cdot 2^{h-1}&2^h\end{pmatrix}{1\choose T_0}=\\
\begin{pmatrix}1&0\\ (2^h-1)h-(2^hh-2^{h+1}+2)&2^h\end{pmatrix}{1\choose T_0}=\\
\begin{pmatrix}1&0\\ 2^{h+1}-h-2&2^h\end{pmatrix}{1\choose T_0}=\\{1\choose 2^{h+1}-h-2+2^hT_0}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But I don't see a pattern in the left side.
| Intuitively, all those big numbers at the right are under so many square root signs they don't matter much. I would think it "obvious" that $3$ is larger. I made a spreadsheet (love copy down) and found the value to be about $2.0903$. For comparison, I changed all the numbers after $9$ to $0$ and the value only changed in the sixth decimal place.
Proving it without computer calculation is another matter. One approach is just to unpack the square roots one by one. We want to compare $9$ with ${2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}}$, which means comparing $7$ with $ \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}}$, which means comparing $49$ with $3 + \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46$ with $ \sqrt{4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}}$, which means comparing $46^2$ with $4 + \sqrt{5 + \sqrt{… + \sqrt{100}}}$ and now we can afford to take all the square root signs except the first off the right and still have the left larger- $46^2 \gt 4+\sqrt{5+6+7+8+\ldots 100}=4+\sqrt{5044}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Find the coefficient of $ x^6 $in the given expansion. Find the coefficient of $x^6$ in $\left[(1+x)(1+x^2)^2(1+x^3)^3 \cdots (1+x^n)^n\right]$.
Expansion
$$\left[\left(1+\binom11x \right)\left(1+\binom21x + \binom22x^2 \right)\left(1 +\binom31x+\binom32x^2+\binom33x^3 \right)\cdots \left(1 + \binom n1 x + \binom n2x^2 + \cdots +\binom n nx^n \right) \right]$$
I expanded it as shown above but couldn't proceed further. Any Help would be appreciated.
| Taken mod $x^7$ (meaning we drop anything with a power of $x$ greater than $6$), we have
$$(1+x)(1+x^2)^2(1+x^3)^3\cdots\equiv(1+x)(1+2x^2+x^4)(1+3x^3+3x^6)(1+4x^4)(1+5x^5)(1+6x^6)\\
\equiv(1+x+5x^5+5x^6)(1+2x^2+5x^4+8x^6)(1+3x^3+9x^6)\\
$$
where in the second step we've paired $(1+x)$ with $1+5x^5$, $(1+2x^2+x^4)$ with $(1+4x^4)$, and $(1+3x^3+3x^6)$ with $(1+x^6)$. The coefficient of $x^6$ in the resulting expansion is
$$5+8+9+1\cdot2\cdot3=28$$
(where the $1\cdot2\cdot3$ comes from the product $x\cdot2x^2\cdot3x^3$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sigma notation breakdown What does the sum below evaluate to?
$$\sum_{i=1}^n 4^i*i $$
I know the rule for a sum of a geometric series:
$$\sum_{i=1}^n a\cdot r^{i-1}=\frac{a(r^n-1)}{r-1} $$
But I cant seeem to apply to this formula because its to the power of i
and there is also the *i factor
| The formula $3S= 1+ n4^{n+1} - \sum_{i=0}^{n}4^i$ (where $S=\sum_{i=1}^n 4^i\cdot i$) from @Random-15's answer simplifies to
$$S=\frac{(3n-1)4^{n+1}+4}9$$ which does produce correct answers $4, 36$ and $228$ for $n=1,2,3,$ respectively.
Now, knowing the formula let me try to prove it by induction.
When $n=1$ we have $\sum_{i=1}^1 4^i\cdot i = 4^1\cdot1=4$ and
$\frac{(3\cdot1-1)4^{1+1}+4}9=\frac{(2)4^2+4}9=\frac{36}9=4$, so the formula
$S=\frac{(3n-1)4^{n+1}+4}9$ is correct when $n=1$.
Assume formula is correct for some $n\ge1$. Prove it for $n+1$. We have
$$\sum_{i=1}^{n+1} 4^i\cdot i=\sum_{i=1}^n 4^i\cdot i +4^{n+1}\cdot(n+1)=$$ (by induction hypothesis)
$$=\frac{(3n-1)4^{n+1}+4}9+4^{n+1}\cdot(n+1)=
\frac{(3n-1+9n+9)4^{n+1}+4}9=$$
$$=\frac{(12n+12-4)4^{n+1}+4}9=\frac{4(3n+3-1)4^{n+1}+4}9=$$
$$=\frac{\bigl(3(n+1)-1\bigr)4^{n+2}+4}9$$ which completes the proof.
Note. Wolfram alpha also gives a similar formula:
$\sum_{i=1}^n 4^i\cdot i = \frac49 (3\cdot 4^n\cdot n - 4^n + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
Prove that $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational
My attempt:-
Suppose $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is rational, then for some $x\in\mathbb{Q}$ we have $$\sqrt{6}-\sqrt{2}-\sqrt{3}=x$$
Rewriting this equation as $$\sqrt{6}-x=\sqrt{2}+\sqrt{3}$$
and now squaring this we get $$ 6-2x\sqrt{6}+x^2=5+2\sqrt{6}$$. This implies that $$\sqrt{6}=\frac{x^2-1}{2+2x}$$ but this is absurd as RHS of the above equation is rational but we know that $\sqrt6$ is irrational. Therefore , $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is irrational. Does this look good? Have I written it properly? Is there any other proof besides this..like one using geometry? Thank you.
| $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is a root of $x^4 - 22 x^2 - 48 x - 23$.
By the rational root theorem, $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is either irrational or an integer.
But
$$
1.4 < \sqrt 2 < 1.5
\\
1.7 < \sqrt 3 < 1.8
\\
2.4 < \sqrt 6 < 2.5
\\
$$
imply
$$
-0.9 < \sqrt{6}-\sqrt{2}-\sqrt{3} <-0.6
$$
and so $\sqrt{6}-\sqrt{2}-\sqrt{3}$ is not an integer. Therefore, it is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| Since the argument and output are linear, we can try to extract $2x-3$ from $4x-2$ through arithmetic means.
\begin{align}
f(\color{red}{2x-3})&=4x-2\\
&=4x\color{blue}{-6+6}-2\\
&=2(\color{red}{2x-3})+4
\end{align}
Therefore, $f(x)=2x+4$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Inequality involving Harmonic mean Let $a, b \ge 0 $ and $x,y > 1$ Show $\frac{1}{1/x+1/y}(a+b) \le \max(ax,by)$
If $a = b = 0$ then this is clear so assume not both are zero.
This seems to be related to the harmonic mean but I am not quite getting it.
$\frac{1}{1/x+1/y}(a+b) = \frac{xy}{x+y}(a+b) = \frac{1}{2}H(x,y)(a+b)$
Where $H(x,y) = \frac{2xy}{x+y}$ : the harmonic mean of $x,y$
And, $\max(ax,by) = \frac{1}{2}[ax+by+\mid ax - by \mid ] = A(ax,by) + \frac{1}{2}\mid ax - by \mid$
Where $A(ax,by) = \frac{ax+by}{2}$ : the arithmetic mean of $ax,by$
This is where I get stuck.
EDIT: I am closer but not quite:
$\displaystyle \frac{a+b}{x+y} = \frac{a}{x+y}+\frac{b}{x+y} \le \frac{a}{y} + \frac{b}{x} \implies xy\frac{a+b}{x+y} \le xy(\frac{a}{y} + \frac{b}{x} ) = ax + by \le 2\max(ax,by) $
| In fact, the conditions for $x,y,a,b$ can be loosened:
For all $x,y > 0$ and $a,b \in \mathbb{R}$, we have
$$\frac{1}{1/x+1/y}(a+b)=\dfrac{y\cdot ax+x\cdot by}{x+y} \le \dfrac{y\cdot \max(ax,by)+x\cdot \max(ax,by)}{x+y} =\max(ax,by).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finite summation $\sum _{k=0}^{n-1} \csc ^{2 m}\left(\frac{\pi (2 k+1)}{2 n}+y\right)$ How to prove
$$\sum _{k=0}^{n-1} \csc ^4\left(\frac{\pi (2 k+1)}{2 n}+y\right)=(n \sec (n y))^4-\frac{2}{3} \left(n^2-1\right) (n \sec (n y))^2$$
Moreover, is there a closed-form for higher order summations? Any help will be appreciated.
| The function
$$
f(z)=\frac{m/z}{(z/a)^m-1}
$$
has residue $1$ where $z^m=a^m$ and residue $-m$ at $0$. For this question, we want $m=2n$ and $a^{2n}=-e^{i2ny}$. Furthermore,
$$
|z|=1\implies\csc^2(\arg(z))=-\frac{4z^2}{(z-1)^2(z+1)^2}
$$
Thus,
$$\newcommand{\Res}{\operatorname*{Res}}\newcommand{\sech}{\operatorname{sech}}
\begin{align}
\sum_{k=0}^{n-1}\csc^2\left(\frac{\pi(2k+1)}{2n}+y\right)
&=\sum_{k=0}^{n-1}\csc^2\left(\frac{k\pi}n+y+\frac\pi{2n}\right)\\
&=\frac12\oint_{|z|\to\infty}\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)\mathrm{d}z\\
&-\frac12\Res_{z=1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)\right)\\
&-\frac12\Res_{z=-1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)\right)\\
&=0-\left(-\frac{n^2}{2\cos^2(ny)}\right)-\left(-\frac{n^2}{2\cos^2(ny)}\right)\\[9pt]
&=n^2\sec^2(ny)
\end{align}
$$
Furthermore,
$$
\begin{align}
\sum_{k=0}^{n-1}\csc^4\left(\frac{\pi(2k+1)}{2n}+y\right)
&=\sum_{k=0}^{n-1}\csc^4\left(\frac{k\pi}n+y+\frac\pi{2n}\right)\\
&=\frac12\oint_{|z|\to\infty}\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^2\mathrm{d}z\\
&-\frac12\Res_{z=1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^2\right)\\
&-\frac12\Res_{z=-1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^2\right)\\
&=n^4\sec^4(ny)-\frac{2n^2\left(n^2-1\right)}3\sec^2(ny)
\end{align}
$$
Using the same method,
$$
\begin{align}
\sum_{k=0}^{n-1}\csc^6\left(\frac{\pi(2k+1)}{2n}+y\right)
&=\sum_{k=0}^{n-1}\csc^6\left(\frac{k\pi}n+y+\frac\pi{2n}\right)\\
&=\frac12\oint_{|z|\to\infty}\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^3\mathrm{d}z\\
&-\frac12\Res_{z=1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^3\right)\\
&-\frac12\Res_{z=-1}\left(\left(-\frac{2n/z}{e^{-i2ny}z^{2n}+1}\right)\left(-\frac{4z^2}{(z-1)^2(z+1)^2}\right)^3\right)\\
&=\frac{2n^2\left(n^2-1\right)\left(n^2-4\right)}{15}\sec^2(ny)-n^4\left(n^2-1\right)\sec^4(ny)+n^6\sec^6(ny)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}$$\frac{1}{(n+1)n^{1/p}}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{n}{n^{1/p}}=\frac{1}{n^{1/p}}-\frac{n}{(n+1)n^{1/p}},$$
Can we go on frome here?
| I find the proper conclusion should be
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)n^{1/p}}\color{red}\leq p,$$
which is because, if $p=1$,then
$$\sum_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}=\sum_{n=1}^{\infty}\frac{1}{(n+1){n}}=1=p.$$
Here is my proof I just figured out. First, we have
\begin{align*}
a_n:&=\frac{1}{(n+1)n^{1/p}}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{n}{n^{1/p}}=\left[\left(\frac{1}{n^{1/p}}\right)^p-\left(\frac{1}{(n+1)^{1/p}}\right)^p\right]\frac{n}{n^{1/p}}.
\end{align*}
Apply Lagrange's MVT to $f(x)=x^p$ over the interal $[1/n^{1/p},1/(n+1)^{1/p}]$. We obtain
\begin{align*}
\exists \xi \in (1/(n+1)^{1/p},1/n^{1/p})~~s.t.~~&\left(\frac{1}{n^{1/p}}\right)^p-\left(\frac{1}{(n+1)^{1/p}}\right)^p\\=&p\cdot\xi^{p-1}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right)\\
<&p\cdot\left(\frac{1}{n^{1/p}}\right)^{p-1}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right)\\
=&p\cdot\frac{n^{1/p}}{n}\cdot\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right).
\end{align*}
Thus
$$a_n<p\left(\frac{1}{n^{1/p}}-\frac{1}{(n+1)^{1/p}}\right).$$
Therefore
\begin{align*}
\sum_{k=1}^na_k&<\sum_{k=1}^np\left(\frac{1}{k^{1/p}}-\frac{1}{(k+1)^{1/p}}\right)\\
&=p\sum_{k=1}^n\left(\frac{1}{k^{1/p}}-\frac{1}{(k+1)^{1/p}}\right)\\
&=p\left(1-\frac{1}{(n+1)^{1/p}}\right)\\
&<p.
\end{align*}
Let $n \to \infty$. It follows that
$$\sum_{k=1}^{\infty}a_k\leq p,$$
which is desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $x\equiv 1\pmod2$, $x\equiv 2\pmod3$, $x\equiv 3\pmod4$, $x\equiv 4\pmod5$, $x\equiv 5\pmod6$ and $x\equiv 0\pmod7$ $$\begin{align*}
x&\equiv 1\pmod2\\
x&\equiv 2\pmod3\\
x&\equiv 3\pmod4\\
x& \equiv 4\pmod5\\
x&\equiv 5\pmod6\\
x&\equiv 0\pmod7\\
\end{align*}$$
So the solution says we can eliminate $x\equiv 5(\bmod6)$ because the first two cases cover it, but I don't really know how it does. How do we solve it in cases like this where the moduli are not mutually relatively prime.
| Using the reduced equations , we get
$$ \color{ red} {x = 3k+2 \quad \quad {\text(1)}}$$
Substituting it into the next gives us :
$$\color {#F0A}{3k+2\equiv 3\mod 4 \implies k \equiv 3 \mod 4 \implies k = 4j +3 \quad \quad \text { (2)}}$$
From $(1)$ and $(2)$ we get :
$$\color{ blue}{x = 12j + 11}$$
Substituting it into the next gives us :
$$\color {#50F}{12j +11 \equiv 4 \mod 5 \implies j\equiv 4 \mod 5 \implies j = 5l +4 \quad \quad \text { (3)}}$$
From $(1)$ and $(3)$ we get :
$$\color{ green}{x = 60l +59 \quad \quad \text{ (4)}}$$
And finally from the last equation , we get :
$$\color {orange}{60l +59 \equiv 0 \mod 7 \implies l \equiv1 \mod 7 \implies l = 7y+1 \quad \quad \text{ (5)} }$$
And we get :
$$\boxed{\color{navy}{x = 60 (7y+1) +59 \implies x = 420y + 119}}$$
Hence all values of form $420y + 119$ is the solution starting from $119,539...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A term that is not bounded by 1. $f_{n}(x)= \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ I think when $0 < x <1$ this expression is bounded by 1, am I correct? if so how can I prove it?
My trials:
At $x = 1,$ the following fraction $\{ \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ becomes $\{\frac{n^3 }{ 1 + n^4}\}$ which is clearly less than 1 as $n^3 < 1 + n^4$ for all $n \in \mathbb{N}.$\
At $x = 0,$ the following fraction $\{ \frac{n^3 x^{3/2}}{ 1 + n^4 x^2}\}$ becomes $0$ which is less than 1.\
At $0 < x < 1,$ we have $x < 1,$ then $x^{3/2} < 1.$, so $n^3 x^{3/2} < n^3 < n^4 + 1.$ OTOH, $x^2 < 1$, then $1 + n^4 x^2 < 1 + n^4$ and so $\frac {1}{1 + n^4 x^2} > \frac{1}{ 1 + n^4}$, So it seems like in this case this term is not bounded by 1.
{A counterexample:}
if $n^{3}x^{3/2} \leq C (1+n^{4}x^{2})$ for all $n$ and for all $x$ then we can put $x =\frac 1 {n^{2}}$ to get $1 \leq 2C$ which is true. I am confused?
| If you differentiate $f_n$ for $n \geq 1$ you have that the function attains its maximum at $x=\frac{\sqrt{3}}{n^2}$ on $[0,1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$
How to evaluate the following challenging sum:
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+i)?$$
where $H_{n}$ is the harmonic number, $\operatorname{Li}_n$ is the polylogarithm function and $G$ is Catalan constant.
This problem was proposed by a friend and here is my approach:
Using the identity
$$\int_0^1\frac{x^n}{1+x}\ dx=H_{n/2}-H_n+\ln2, \quad x\mapsto x^2$$
$$2\int_0^1\frac{x^{2n+1}}{1+x^2}\ dx=H_{n/2}-H_n+\ln2$$
Multiply both sides by $\frac{(-1)^n}{(2n+1)^2}$ to get
$$\sum_{n=0}^\infty (-1)^n\frac{H_{n/2}-H_n+\ln2}{(2n+1)^2}=2\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}\ dx\\=2\int_0^1\frac{1}{1+x^2}\Im\sum_{n=0}^\infty\frac{(i)^nx^n}{n^2}\ dx=2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
Rearranging the terms to get
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\ln2G+2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
where
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}=-\int_0^1\ln x\sum_{n=0}^\infty(-1)^n x^{2n}H_n\ dx=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx$$
which is calculated here:
$$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$
And the question here is how to evaluate the dilogarithm integral or a different way to compute $S$? Thank you.
| From here we got
$$X=2\sum_{n=0}^\infty\frac{H_{n/2}-H_n+\ln2}{(2n+1)^3}=\frac{\pi^2}{4}G-\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx$$
Using the identity in (4)
$$\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\sin(x(2n+1))=\frac{\pi/2-x}{2\cos x}$$
multiply both sides by $x^2$ then integrate from $x=0$ to $\pi/2$ we get
\begin{align}
I&=\int_0^{\pi/2}\frac{(\pi/2-x)x^2}{2\cos x}\ dx=\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\int_0^{\pi/2} x^2 \sin(x(2n+1))\ dx\\
&=\sum_{n=0}^\infty(H_{n/2}-H_n+\ln2)\left(\frac{\pi(-1)^n}{(2n+1)^2}-\frac{2}{(2n+1)^3}\right)\\
&=\pi\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^2}-\pi\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^2}+\pi\ln2G-X\\
&=\pi\sum_{n=0}^\infty\frac{H_{n/2}}{(2n+1)^2}-\pi\sum_{n=0}^\infty\frac{H_{n}}{(2n+1)^2}+\pi\ln2G-\frac{\pi^2}{4}G+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx
\end{align}
But
\begin{align}
I&=\int_0^{\pi/2}\frac{(\pi/2-x)x^2}{2\cos x}\ dx=\int_0^{\pi/2}\frac{x(\pi/2-x)^2}{2\sin x}\ dx\\
&=\frac{\pi^2}{4}\underbrace{\int_0^{\pi/2}\frac{x}{2\sin x}\ dx}_{G}-\pi\underbrace{\int_0^{\pi/2}\frac{x^2}{2\sin x}\ dx}_{\pi G-\frac74\zeta(3)}+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx\\
&=\frac74\pi\zeta(3)-\frac34\pi^2G+\int_0^{\pi/2}\frac{x^3}{2\sin x}\ dx
\end{align}
Rearrange and divide by $\pi$ to get
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}+\frac74\zeta(3)-\frac{\pi}{2}G-\ln2G$$
Substitute the result of $\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}$ calculated in the question body, we get the desired closed form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\int \frac{dx}{1-\sin x+\cos x}$ Solve :$$ \int \frac{dx}{1-\sin x+\cos x} $$
I tried:
$$\int \frac{\text{dx}}{(1+\cos x)-\sin x} \times \frac{(1+\cos x)+\sin x}{(1+\cos x)+ \sin x} dx=\int \frac{1+\cos x+\sin x}{(1+\cos x)^2-(\sin x)^2}$$ $$=\int \frac{1+\cos x+\sin x}{1-\sin x^2+2 \cos x+\cos x^2}=\int \frac{1+\cos x+\sin x}{2 \cos x(\cos x+1)}$$ $$=\int (\frac{1}{2 \cos x}+ \frac{\sin x}{2 \cos x(\cos x+1)})dx=\frac{1}{2} \ln(\sec x+\tan x)+ \int \frac{\sin x}{2\cos x(\cos x+1)}dx$$
$ \cos x=u$ , $du=-\sin xdx$
$$\int \frac{-du}{2u(u+1)}= \frac{-1}{2} \int (\frac{1}{u}-\frac{1}{u+1})= \frac{-1}{2}(\ln(\cos x)-\ln(\cos x+1))$$
final answer : $$\frac{1}{2} \ln(\sec x+\tan x)-\frac{1}{2} \ln(\frac{\cos x}{1+\cos x})+ c$$
First: Is my answer right?
Second: Is there another approach or easier approach to solve this integral?
| My hint for your question:
$$\int \frac{dx}{1-\sin x+\cos x}$$
If I apply the $t-$substitution $t=\tan(\frac{x}{2})$ you will have $$\int \frac{dx}{1-\sin x+\cos x}=\int \frac{1}{-u+1}du$$
Using another $y-$substitution for last integral of the type $p=-y+1$ you will have:
$$\int \frac{1}{-u+1}du=\int -\frac{1}{p}dp$$
Hence remebering all substitutions $p=-u+1,\,t=\tan (\frac{x}{2} )$ you will obtain
$$\int \frac{dx}{1-\sin x+\cos x}=-\ln \left|-\tan \left(\frac{x}{2}\right)+1\right|+k, \quad k\in\mathbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Do the coefficients of these polynomials alternate in sign? Define polynomials $p_i(x)$ by the recurrence
\begin{align}
p_0(x)&=0 \\
p_1(x)&=1 \\
p_{2i}(x)&=p_{2i-1}(x)-p_{2i-2}(x) \\
p_{2i+1}(x)&=xp_{2i}(x)-p_{2i-1}(x) \\
\end{align}
The first few are given by
\begin{align}
p_0(x)&=0\\
p_1(x)&=1\\
p_2(x)&=1\\
p_3(x)&=x-1\\
p_4(x)&=x-2\\
p_5(x)&=x^2-3x+1\\
p_6(x)&=x^2-4x+3\\
p_7(x)&=x^3-5x^2+6x-1\\
p_8(x)&=x^3-6x^2+10x-4
\end{align}
It is reasonable to ask whether the coefficients of these polynomials alternate in sign. Any thoughts?
| I reckon that
$$p_{2n-1}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+1-k}k x^{n-1-k}$$
and
$$p_{2n}(x)=\sum_{k=0}^{n-1}(-1)^k \binom{n+2-k}k x^{n-1-k}$$
and that these can be proved by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}$
Does anyone know how to solve this?
| Since $y=x\exp\frac{\pm2\pi i}{3}$,$$\frac{x^n+y^n}{(x+y)^n}=\frac{1+\exp\frac{\pm2\pi i n}{3}}{(1+\exp\frac{\pm2\pi i}{3})^n}=\frac{2\exp\frac{\pm\pi i n}{3}\cos\frac{\pi n}{3}}{(2\exp\frac{\pm\pi i}{3}\cos\frac{\pi}{3})^n}=2\cos\frac{\pi n}{3}.$$In the case $n=2013$, this simplifies to $-2$ because $n/3$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
Determine the residuals of $f(z)=\frac{\log(1+z)}{\cos z-1}$ Determine the residuals of $f$
$f$ is given by
$$
f(z)=\frac{\log(1+z)}{\cos z-1}, \quad z \in \mathbb{C}-(-\infty, -1]
$$
where $\log z$ is the principal logarithm in the cutplane $\mathbb{C}_\pi:=\mathbb{C}-\{z \in \mathbb{R} \mid z \leq 0\}$.
Determining the poles:
Since
$$
\cos z-1=0 \Leftrightarrow \cos z=1 \Leftrightarrow z=2\pi k,
\quad k \in \mathbb{Z}
$$
the set of poles of $f$ is given by
$$
P=\{2\pi k \mid k \in \mathbb{N} \cup \{0\}\}
$$
For $z=0$ we can write $f$ as
$$
f(z)=\frac{\log(1+z)}{\cos z-1}
=
\frac{z-\frac{z^2}{2}+\frac{z^3}{3}-\cdots}{-\frac{z^2}{2!}+\frac{z^4}{4!}- \cdots}
=
\frac{z \left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{z^2 \left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
=
\frac{\left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{z \left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
$$
and therefore,
$$
(z-0)f(z)=
\frac{\left(1-\frac{z}{2}+\frac{z^2}{3}- \cdots \right)}
{\left(\frac{1}{2!}+\frac{z^2}{4!}- \cdots \right)}
$$
The limit of this as $z \rightarrow 0$ is $2$, so the residue in $z=0$ is 2.
But how do I determine the other residuals?
Here's one of my attemps
Since cosine is periodic we can write $f$ as
$$
f(z)=\frac{\log(1+z)}{\cos z-1}=\frac{\log(1+z)}{\cos(z-2\pi k)-1}
=
\frac{\left(z-\frac{z^2}{2}+\frac{z^3}{3}- \cdots \right)}
{(z-2\pi k)^2\left(\frac{1}{2!}+\frac{(z-2\pi k)^2}{4!}- \cdots \right)}
$$
and therefore,
$$
(z-2\pi k)^2f(z)=
\frac{\left(z-\frac{z^2}{2}+\frac{z^3}{3}- \cdots \right)}
{\left(\frac{1}{2!}+\frac{(z-2\pi k)^2}{4!}- \cdots \right)}
$$
However, taking the derivative of this is not particularly nice...
Any suggestions?
| I'll do it when $k=1$. The general case is similar.
You have\begin{align}\frac{\log(1+z)}{\cos(z)-1}&=\frac{\log(1+2\pi+z-2\pi)}{\cos(z-2\pi)-1}\\&=\frac{\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots}{-\frac{(z-2\pi)^2}{2!}+\frac{(z-2\pi)^4}{4!}-\cdots}\\&=\frac1{(z-2\pi)^2}\times\frac{\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots}{-\frac12+\frac{(z-2\pi)^2}{4!}-\cdots}\end{align}Clearly, this quotient can be written as$$\frac{a_{-2}}{(z-2\pi)^2}+\frac{a_{-1}}{z-2\pi}+a_0+a_1(z-2\pi)+\cdots$$and the residue is $a_{-1}$. But\begin{multline}\log(1+2\pi)+\frac{z-2\pi}{1+2\pi}-\frac{(z-2\pi)^2}{2(1+2\pi)^2}+\cdots=\\=\left(-\frac12+\frac{(z-2\pi)^2}{4!}-\cdots\right)\times\left(a_{-2}+a_{-1}(z-2\pi)+a_0(z-2\pi)^2+\cdots\right)=\\=-\frac{a_{-2}}2-\frac{a_{-1}}2(z-2\pi)+\cdots\end{multline}and therefore $-\frac{a_{-1}}2=\frac1{1+2\pi}$. So$$\operatorname{res}_{z=2\pi}\frac{\log(1+z)}{\cos(z)-1}=a_{-1}=-\frac2{1+2\pi}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ is bounded and has monotonicity?
How to show, that the number-sequence $a_n:=\frac{n^2}{n!}$ with $(a_n)_{n\in\mathbb{N}}$ is bounded and has monotonicity?
Boundedness: I observed, that $a_1<a_2>a_3>a_4>a_5>\dots$ That's why I assume, that there is a bound $2=K\geq \mid a_n \mid$ with $K\in\mathbb{N}$. How do I show, that $a_n$ is actually bounded at $K=2$?
Monotonicity: For $n\geq2$, we show
$\begin{align}
a_n&\geq a_{n+1}\\
\frac{n^2}{n!}&\geq \frac{(n+1)^2}{(n+1)!}\\
n^2(n+1)!&\geq (n+1)^2\cdot n!\\
(n+1)!&\geq n!
\end{align}$
For $n\leq 2$ we show
$\begin{align}
a_n&\leq a_{n+1}\\
\frac{n^2}{n!}&\leq \frac{(n+1)^2}{(n+1)!}\\
n^2(n+1)!&\leq (n+1)^2\cdot n!\\
n^2&\leq (n+1)^2
\end{align}$
but that's not really a proof. I don't know how to express formally, that $n!$ increases faster, than $n^2$ for $n\geq 2$ and the other way around for $n\leq 2$.
(Because of the fact, that, $n!$ increases faster, than $n^2$ for $n\geq 2$ and the other way around for $n\leq 2$ it's only important to look at $(n+1)!\geq n!$ or $n^2\leq (n+1)^2$ in the first place.)
| Your proof is correct and for $n\leq 2$ you can simply say that $a_1=1<a_2=2$.
With a little effort we can show more.
For $n\geq 2$: $0<a_{n+1}<a_n$ because
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n+1)!}\cdot \frac{n!}{n^2}
=\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}\leq \frac{1}{2}+\frac{1}{4}=\frac{3}{4}<1.$$
Since $a_1=1<a_2=2$, it follows that $(a_n)_{n\geq 1}\in (0,2]$ and the maximum value is just $a_2=2$.
By the above estimate we have also that for $n\geq 2$
$$0<a_{n+1}\leq \frac{3}{4}a_n\leq \left(\frac{3}{4}\right)^{2}a_{n-1}\leq \dots
\leq \left(\frac{3}{4}\right)^{n-1}a_{2}\to 0$$
and therefore $a_n=\frac{n^2}{n!}\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}$ Is this solution correct? Just want to double check whether all my operations are legal. The result seems to be correct but I want to make sure I didn't make any mistakes.
$\require{cancel}$
$$\lim_{x \to 0} \frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{\sin x}{e^x - \cos x}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2}{e^x - \cos x}+\frac{x}{e^x - \cos x}\cdot\cancelto{1}{\frac{\sin x}{x}}=$$
$$=\lim_{x \to 0} \frac{x^{11}-3x^2+x}{e^x - 1 + 1 - \cos x}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^{11} - 3x^2 + x} + \frac {e^x-1}{x^{11}-3x^2+x})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1-\cos x}{x^2\cancelto{-\infty}{(x^9-3-\frac {1}{x})}} + \frac {e^x-1}{x\cancelto{1}{(x^{10}-3x+1)}})^{-1}=$$
$$ =\lim_{x \to 0} (\cancelto{\frac{1}{2}}{\frac{1-\cos x}{x^2}}\cdot\cancelto{0}{\frac{1}{-\infty}} + \cancelto{1}{\frac {e^x-1}{x}})^{-1}=$$
$$=\lim_{x \to 0} (\frac {1}{2} \cdot0+ 1)^{-1}=$$
$$=(0+1)^{-1} = 1$$
Is this correct?
| The final result is correct, but in my opinion your approach is too complicated and you should avoid the "inverse of the inverse" part. More simply, by using the same standard limits as $x\to 0$,
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{\overbrace{x^{10}-3x}^{\to 0}+\overbrace{\frac{\sin x}{x}}^{\to 1}}{\underbrace{\frac{e^x-1}{x}}_{\to 1} + \underbrace{\frac{1- \cos x}{x^2}}_{\to 1/2}\cdot\underbrace{x}_{\to 0}}\to 1.$$
If you are comfortable with the little-o notation, there is a shorter way: as $x\to 0$,
$$\frac{x^{11}-3x^2+\sin x}{e^x - \cos x}=\frac{x^{11}-3x^2+(x+o(x))}{(1+x+o(x)) - (1-o(x))}=\frac{x+o(x)}{x+o(x)}\to 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
| Beware: overkill incoming. We may use the AM-GM inequality for producing tight bounds for $\sqrt[n]{n}$. In particular, by noticing that
$$ n = 1\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n}{n-1} $$
we have
$$ \sqrt[n]{n}=\text{GM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)<\text{AM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right) $$
and
$$ \sqrt[n]{n} < 1+ \frac{H_{n-1}}{n},\qquad \left(1+\frac{1}{\sqrt{n}}\right)^2=1+\frac{2\sqrt{n}+1}{n}, $$
so it is enough to show that $H_{n-1}\leq 2\sqrt{n}+1$, at least from some point on. On the other hand the Cauchy-Schwarz inequality ensures
$$ H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq\sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\frac{1}{k^2}} < \sqrt{n\zeta(2)} $$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D-4E+F)x^2+(4A-8+4B+4C+4D-8D+4E-4F)x+(-8A+4C+8D+4F)=1$$
So:
$$A+B+C+D+E=0$$
$$2A+4B+C-2D-4E+F=0$$
$$A+B+C-D+E=2$$
$$-8A+4C+8D+4F=1.$$
But how to find $A$, $B$, $C$, $D$, $E$, $F$?
I also tried substitution $$x=2\sec t$,$
but It caused some difficulty.
| Perhaps the neatest approach is to square $\frac{1}{x^2-4}=\frac14\left(\frac{1}{x-2}-\frac{1}{x+2}\right)$ to give$$\frac{1}{(x^2-4)^2}=\frac{1}{16}\left(\frac{1}{(x-2)^2}+\frac{1}{(x+2)^2}-\frac{2}{x^2-4}\right)=\frac{1}{32}\left(\frac{2}{(x-2)^2}+\frac{2}{(x+2)^2}+\frac{1}{x+2}-\frac{1}{x-2}\right).$$Now you can integrate:$$\int\frac{dx}{(x^2-4)^2}=\frac{1}{32}\left(-\frac{2}{x-2}-\frac{2}{x+2}+\ln\left|\frac{x+2}{x-2}\right|\right)+C,$$where $C$ is locally constant and can change at each of $x=\pm2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Number Theory (Polynomials) Find The Remainder? A polynomial $f(x) = x^{50} + 3x^{49} + 3x + 12$ when divided by $x - a$, it leaves remainder $3$ & when its quotient is further divided by $x - b$ it leaves remainder $5$, also when $f(x)$ is divided by $x^2 - ( b + a)x + ab$ the remainder is $x + 6$. Find $a$?
| $P=x^{50}+3x^{49}+3x+12=x^{49}(x+3)+3(x+3)+3=k(x-a)+3$
So the quotient k is:
$k=\frac{(x^{49}+3)(x+3)}{x-a}$
⇒$\frac{(x^{49}+3)(x+3)}{x-a}=k_1(x-b)+5$
⇒$(x^{49}+3)(x+3)=k_1(x-a)(x-b)+5(x-a)$
⇒$p=(x^{49}+3)(x+3)+3=k_1(x-a)(x-b)+5(x-a)+3$
Therefore:
$5(x-a)+3=x+6$ ⇒ $a=\frac{4x-3}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3415131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the smallest multiple of $3^{1000}$, which has only 1’s in the digits?(Like 1, 11, 111 etc) I think the answer is $3^{1000}$ ones,
and I want to prove it with induction.
$3^0 | 1$ and $1$ is the smallest.
$ 3^1 | 111 $; it’s the smallest too.
Let’s say it's true for $ 3^k | 111...111 $ ($3^k$ ones). How can I prove to $3^{k+1} $ ?
| If $m$ is coprime to $3$ then the g.c.d. of $10^m-1$ and $10^3-1$ is $10-1=9$. In particular, $9$ but not $27$ is a factor of $10^m-1$.
If $n=3^km$ where $m$ is coprime to $3$ then $$\frac {10^{3^km}-1}{10^{3^{k-1}m}-1}=1+10^{3^{k-1}m}+10^{3^{k-1}(2m)}\equiv3\pmod9$$
Repeating this process we obtain
$10^{3^km}-1=\frac {10^{3^km}-1}{10^{3^{k-1}m}-1}\frac {10^{3^{k-1}m}-1}{10^{3^{k-2}m}-1}..\frac {10^{3m}-1}{10^{m}-1}(10^{m}-1)$
$$\equiv3^{k+2}\pmod {3^{k+3}}$$
Hence the least $n$ such that $10^n-1$ is a multiple of $3^{1002}$ is obtained when $k=1000$ and $m=1$ i.e. $n=3^{1000}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction.
Please, explain how to remove the radical from the denominator. Thanks, for your help.
| Call the fraction by $r$.
$$r:=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}\frac{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
$$=\frac{8\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}{2-\sqrt{\frac{5+\sqrt{5}}{2}}}=\frac{8\left(\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\right)}{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\frac{\left(2+\sqrt{\frac{5+\sqrt{5}}{2}}\right)}{2+\sqrt{\frac{5+\sqrt{5}}{2}}}$$
Call $u:=8\left(\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\right)\left(2+\sqrt{\frac{5+\sqrt{5}}{2}}\right)$. Then,
$$r=\frac{u}{4-\frac{5+\sqrt{5}}{2}}=\left(\frac{1}{2}\right)\frac{u}{8-5-\sqrt{5}}=\left(\frac{1}{2}\right)\frac{u}{3-\sqrt{5}}=\left(\frac{1}{2}\right)\frac{u}{3-\sqrt{5}}\frac{3+\sqrt{5}}{3+\sqrt{5}}.$$
Can you procced from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Given a defined region, write down the integral and then evaluate
Let the region be defined by $x \geq 0$, $x^2 + y^2 \leq 2$, and $x^2 + y^2 \geq 1$. Write down the integral over the region in each of the two possible orders of $f(x,y) = x^2$. Evaluate both integrals.
I was wondering before I try to evaluate the integral if I set up the bounds correctly for the two integrals
\begin{align}
\int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}dx \int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2dy
\end{align}
The other order would be
\begin{align}
\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}dy \int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}x^2dx
\end{align}
edit: Updated with picture of the bounds
| None of your integrals has correct bounds. For the first one, since the integral w.r.t. $y$ is inside, $x$ should not depend on $y$ so the bounds of the outer integrals should have no $y$ at all. Similarly, for the second one, you have the integral w.r.t. $x$ inside; therefore the outer integral should have no $x$ appearing in the bounds. I recommend that you really draw the picture to see what the proper bounds in each case are.
The correct bounds for the first integral would be
$$\int_0^{1}dx\ \left(\int_{-\sqrt{2-x^2}}^{-\sqrt{1-x^2}}x^2 dy+\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2 dy\right) +\int_1^{\sqrt2}dx\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}x^2dy.$$
Similarly for the second integral, you should have
$$\int_{-\sqrt{2}}^{-1}dy\ \int_{0}^{\sqrt{2-y^2}}x^2dx+\int_{-1}^{1}dy\ \int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}x^2dx+\int_1^{\sqrt{2}}dy\ \int_0^{\sqrt{2-y^2}}x^2dx.$$
However, it is much easier to evaluate this integral in polar coordinates. If $I$ is the required integral, then by symmetry, $$I=\frac12\iint_{A} x^2 da$$ where $A$ is the annulus $\big\{(x,y)\ :\ 1\le x^2+y^2\le 2\big\}$, and $da$ is the area element. By symmetry again, $$I=\frac12\iint_{A}y^2 da.$$ So $$I=\frac14\iint_A(x^2+y^2)da=\frac14\iint_A r^2da=\frac14\int_{0}^{2\pi} \ \int_1^{\sqrt{2}} r^2\cdot r dr\ d\theta.$$ Hence $I=\frac{\pi}{2}\left(\frac{(\sqrt{2})^4-1^4}{4}\right)=\frac{3\pi}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that following sequence is bounded I have to prove that $x_n=\frac{1-n}{\sqrt{2+n^2}}$ sequence is bounded. I've solved some easier ones, but these seems to be kinda different. I have to find actual bounds of the sequence.
| Lower bound:
$$ \frac{1-n}{\sqrt{2+n^2}} = \frac{1}{\sqrt{2+n^2}} - \frac{n}{\sqrt{2+n^2}} \geq \frac{1}{\sqrt{2+n^2}} - \frac{n}{\sqrt{n^2}} = \frac{1}{\sqrt{2+n^2}} - 1 > -1$$
Upper bound:
Since $\sqrt{2+n^2} > 0$ we get
$$\frac{1-n}{\sqrt{2+n^2}} \leq \frac{1}{\sqrt{2+n^2}} < \frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a^2b^2+a^2c^2+b^2c^2\leq 3$ Let $a, b, c\geq 0$ s.t.$$(a^2-a+1)(b^2-b+1)(c^2-c+1)=1$$
Show that $ a^2b^2+a^2c^2+b^2c^2\leq 3$.
My idea:
I denote $a+b+c=x$, $ab+bc+ac=y$ and $abc=z$.
Then I have $$x^2+y^2+z^2-xy-xz-yz+2z-y-x=0$$
I have to show that $$y^2-2xz\leq 3$$
I tried to prove it with the sign of trinom, but it doesn't work.
| Remark: I just realize that my Fact 1 is the same at Atticus's solution (he gave an elegant solution).
My Fact 1 is derived from the optimization theory (the method of Lagrange Multipliers, penalty functions, etc.).
We want to minimize $f(a,b,c) = 3 - (a^2b^2 + b^2c^2 + c^2a^2)$ subject to $g(a,b,c) = (a^2-a+1)(b^2-b+1)(c^2-c+1)-1 = 0$.
Let $L(a,b,c) = f(a,b,c) + \lambda g(a,b,c)$ be the Lagrangian. At the optimum $(a, b, c) = (1,1,1)$,
from $\frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = \frac{\partial L}{\partial a} = 0$, we get $\lambda = 4$.
I use Mathematica Resolve command to see that $L(a, b, c)\ge 0$ is true surprisingly!
Actually, I first tried penalty functions. Find $\mu > 0$ such that $F(a,b,c) = f(a,b,c) + \mu [g(a,b,c)]^2 \ge 0$.
But I found it is complicated.
An alternative solution:
I found the following fact (the proof is given later):
Fact 1: For any real numbers $a, b, c$, it holds that
$$a^2b^2 + b^2c^2 + c^2a^2 \le 3 + 4\cdot [(a^2-a+1)(b^2-b+1)(c^2-c+1)-1].$$
From Fact 1, the desired result follows.
$\phantom{2}$
Proof of Fact 1: $\mathrm{RHS} - \mathrm{LHS}$ can be expressed as SOS (Sum of Squares). Indeed,
we have
$$\mathrm{RHS} - \mathrm{LHS} = \frac{1}{6} z^T Q z$$
where $z = [1, a, b, c, ab, ca, bc, abc]^T$ and ($Q$ is positive semidefinite)
\begin{align}
Q = \left(\begin{array}{rrrrrrrr} 18 & -12 & -12 & -12 & 7 & 7 & 7 & -3\\ -12 & 24 & 5 & 5 & -12 & -12 & -3 & 5\\ -12 & 5 & 24 & 5 & -12 & -3 & -12 & 5\\ -12 & 5 & 5 & 24 & -3 & -12 & -12 & 5\\ 7 & -12 & -12 & -3 & 18 & 7 & 7 & -12\\ 7 & -12 & -3 & -12 & 7 & 18 & 7 & -12\\ 7 & -3 & -12 & -12 & 7 & 7 & 18 & -12\\ -3 & 5 & 5 & 5 & -12 & -12 & -12 & 24 \end{array}\right).
\end{align}
The desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
show this inequality $\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)$ let $x,y,z>0$, and such
$xyz=1$,show that
$$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)^3+12\ge 13(x^3+y^3+z^3)\tag{1}$$
I have konwn use C-S we have
$$\left(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\right)\ge\dfrac{(x+y+z)^2}{x+y+z}=x+y+z$$
this relsut can't sove this inequality $(1)$,so How to prove it?and if use $pqr$ methods,$x+y+z=p,xy+yz+xz=q,xyz=r=1$,but
$$\sum \dfrac{x^2}{y}=\dfrac{\sum x^3z}{xyz}=\sum x^3z$$
| The hint.
By the $uvw$'s technique prove that:
$$\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}.$$
After this by $uvw$ prove that:
$$\frac{5(x^2+y^2+z^2)-2(xy+xz+yz)}{x+y+z}\geq\sqrt[3]{13(x^3+y^3+z^3)-12xyz}.$$
A proof of the second inequality.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $xyz=w^3$ and $u^2=tv^2$.
Thus, $t\geq1$ and we need to prove that
$$\frac{5(9u^2-6v^2)-6v^2}{3u}\geq\sqrt[3]{13(27u^3-27uv^2+3w^3)-12w^3}$$ or
$$(5u^2-4)^3\geq u^3(13u^3-13uv^2+w^3).$$
Now, since $uw^3\leq v^4,$ it's enough to prove that
$$(5t-4)^3\geq t(13t^2-13t+1)$$ or
$$(t-1)(112t^2-175t+64)\geq0$$ or
$$t-1+7(t-1)^2(16t-9)\geq0,$$ which is obvious for $t\geq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The sum $\sum^{n}_{k=0} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$ I was trying to evaluate:
$$\sum_{k=0}^{n} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$$
And in particular, when $x=1$.
I already have $2$ proofs for the $x=1$ case, (it is $2$) by counting subsets of a $2n-1$ size set and by splitting the term into a telescoping sum. I was looking for a solution using only generating functions.
| (I started an answer some hours ago, but had to stop, well, it is time to finish and submit.) Let $y=1/x$, (if $x$ is not zero,) and let us compute closer for some integer $n>0$
$$
\begin{aligned}
\frac{\binom nk}{\binom {2n-1}k}
&=\frac{n!}{k!(n-k)!}\cdot\frac{k!(2n-k-1)!}{(2n-1)!}
\\
&=\frac{n!(n-1)!}{(2n-1)!}\cdot\frac{(2n-k-1)!}{(n-k)!(n-1)!}
\\
&=\binom{2n-1}{n}^{-1}\cdot\binom{2n-k-1}{n-1}\ .
\\ %[3mm]
&\qquad\text{ This implies for the sum to be calculated:}
\\
f(x)
&:=
\sum_{0\le k\le n}
\frac{\binom nk}{\binom {2n-1}{k}}x^k
=
\binom{2n-1}{n}^{-1}
\sum_{0\le k\le n}
\binom{2n-k-1}{n-1} x^k
\\
&=
x^n
\binom{2n-1}{n}^{-1}
\sum_{0\le k\le n}
\binom{n-1+(n-k)}{n-1} y^{n-k}\qquad\text{ (Recall $y=1/x$)}
\\
&\qquad\text{ and let us use in the summation the variable $j=n-k$}
\\
&=
x^n
\binom{2n-1}{n}^{-1}
\sum_{n\ge j\ge 0}
\binom{n-1+j}{n-1} y^j\ .
\end{aligned}
$$
Now it is time to set $x=y=1$, and recall the formula:
$$
\binom KK +
\binom {K+1}K +
\binom {K+2}K +
\dots +
\binom NK
=
\binom{N+1}{K+1}\ .\qquad(*)
$$
This gives
$$
f(1) =
\binom{2n-1}{n}^{-1}
\sum_{n\ge j\ge 0}
\binom{n-1+j}{n-1}
=
\binom{2n-1}{n}^{-1}
\binom{(2n-1)+1}{(n-1)+1}
=
\binom{2n-1}{n}^{-1}
\binom{2n}{n}
=2\ .
$$
(Because of $\binom{2n}n=\frac{2n}n\binom{2n-1}{n-1}=2\binom{2n-1}{n}$.)
For a more general formula of $f(x)$ containing the variable $x$, we would need a corresponding relation $(*)$ containing $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$$x\in[-1,0]$$
Finding the range
$$f(x)=\sin^{-1}\left(\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left([-1,0]+\frac{1}{2}\right)^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{-1}{2},\frac{1}{2}\right]^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[0,\frac{1}{4}\right]+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{3}{4},1\right]}\right)$$
$$f(x)=\sin^{-1}\left({\left[\frac{\sqrt{3}}{2},1\right]}\right)$$
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
but actual answer is $\left[\dfrac{\pi}{3},\dfrac{\pi}{2}\right]$
| The domain of $$f(x)=\sin^{-1}(\sqrt{x^2+x+1})$$ is where you get $$0\le x^2+x+1\le 1$$ which is the interval $[-1,0]$
Note that $f(-1/2)=\pi/3$ is the minimum value and $f(-1)=f(0)= \pi/2$ is the maximum.
Thus the range is the interval $[\pi/3, \pi/2]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that $\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$ Show that $\sum_{i=1}^{k/2} {{k}\choose i}2^i < 3^{k-1}$ for $k\geq6$ and even. I tried using $(1+2)^k=\sum_{i=0}^k{{k}\choose{i}}2^i$, the first half has the same binomials as the second and larger powers of two. Also tried induction, but the step is proving that $\dfrac{\sum_{i=0}^{k/2+1}{k+2\choose i}2^i}{\sum_{i=0}^{k/2}{k\choose i}2^i}<9$.
| Here's an inductive approach. Let $a_n = \sum_{i \leq n} \binom{2n}{i} 2^i$ (including $i = 0$ for convenience). Then since $\binom{2n+2}{i} = \binom{2n}{i-2} + 2\binom{2n}{i-1} + \binom{2n}{i}$, we have
\begin{align*}
a_{n+1}
&= \sum_{i \leq n+1} \binom{2n+2}{i} 2^i \\
&= \sum_{i \leq n+1} \left(\binom{2n}{i-2} 2^i + \binom{2n}{i-1} 2^{i+1} + \binom{2n}{i} 2^i\right) \\
&= \sum_{i \leq n-1} \binom{2n}{i} 2^{i+2} + \sum_{i \leq n} \binom{2n}{i} 2^{i+2} + \sum_{i \leq n+1}\binom{2n}{i} 2^i \\
&= \left(4a_n - \binom{2n}{n}2^{n+2}\right) + 4a_n + \left(a_n + \binom{2n}{n+1} 2^{n+1}\right) \\
&< 9a_n
\end{align*}
where the last inequality holds because $\binom{2n}{n} > \binom{2n}{n+1}$. Then since $a_3 < 3^5$, by induction we have $a_n < 3^{2n-1}$ for all $n \geq 3$ (though there are better asymptotic bounds, e.g. $a_n \leq 2^n \sum_{i \leq n} \binom{2n}{i} \leq 8^n$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3426814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Comparing ranks of block matrices This seems simple, but I don't really know how to prove it, so I'd be thankful for any help.
For the following block matrix:
$A$, $B$, and $C$ are all $n\times n$ matrices. Entries of $A$, $B$, $C$ are arbitrary, $0$ is a zero matrix.
$$\operatorname{rk}\left(\left[\matrix{A&0\\0&B}\right]\right)\leqslant \operatorname{rk}\left(\left[\matrix{A&0\\C&B}\right]\right).$$
| Let $A = \begin{bmatrix} \alpha_1 & \cdots & \alpha_n\end{bmatrix}$, $B = \begin{bmatrix} \beta_1 & \cdots & \beta_n\end{bmatrix}$, $C = \begin{bmatrix} \gamma_1 & \cdots & \gamma_n\end{bmatrix}$, where $\alpha_i, \beta_i, \gamma_i$ are column vectors of $A, B, C$ respectively.
Consider any linearly independent column vectors $\{\delta_{i_1}, \ldots, \delta_{i_p}, \xi_{j_1}, \ldots, \xi_{j_q}\}$ of the matrix $\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$, where $1 \leq i_1 < \cdots < i_p \leq n, 1 \leq j_1 < \cdots < j_q \leq n$. Note $p$ or $q$ can be $0$, for which case $\delta$ or $\xi$ is absent from the group. Here, $\delta_i$ are columns taken from $\begin{bmatrix} A \\ 0 \end{bmatrix}$, while $\xi_j$ are columns taken from $\begin{bmatrix} 0 \\ B \end{bmatrix}$.
Since $\{\delta_{i_1}, \ldots, \delta_{i_p}, \xi_{j_1}, \ldots, \xi_{j_q}\}$ are linearly independent, so are $\{\delta_{i_1}, \ldots, \delta_{i_p}\}$ and
$\{\xi_{j_1}, \ldots, \xi_{j_q}\}$, which respectively implies that
\begin{align}
& a_1\alpha_{i_1} + \cdots + a_p\alpha_{i_p} = 0 \implies a_1 = \cdots = a_p = 0 \tag{1} \\
& b_1\beta_{j_1} + \cdots + b_q\beta_{j_q} = 0 \implies b_1 = \cdots = b_q = 0. \tag{2}
\end{align}
Now we can show that the corresponding columns taken from $\begin{bmatrix} A & 0 \\
C & B \end{bmatrix}$ are linearly independent. In fact, if
\begin{align}
a_1\begin{bmatrix}\alpha_{i_1} \\ \gamma_{i_1} \end{bmatrix} + \cdots +
a_p\begin{bmatrix}\alpha_{i_p} \\ \gamma_{i_p} \end{bmatrix} +
b_1\begin{bmatrix}0 \\ \beta_{j_1} \end{bmatrix} + \cdots +
b_q\begin{bmatrix}0 \\ \beta_{j_q} \end{bmatrix} =
\begin{bmatrix}0 \\ 0 \end{bmatrix},
\end{align}
then
\begin{align}
& a_1\alpha_{i_1} + \cdots + a_p\alpha_{i_p} = 0 \tag{3} \\
& a_1\gamma_{i_1} + \cdots + a_p\gamma_{i_p} + b_1\beta_{j_1} + \cdots + b_q\beta_{j_q} = 0 \tag{4}
\end{align}
$(3)$ and $(1)$ imply $a_1 = \cdots = a_p = 0$. Substitute $a_1 = \cdots = a_p = 0$ into $(4)$, we get $b_1\beta_{j_1} + \cdots + b_q\beta_{j_q} = 0$, which further implies $b_1 = \cdots = b_q = 0$, by $(2)$. That shows $a_1 = \cdots = a_p = b_1 = \cdots = b_q = 0$, i.e.,
$$\begin{bmatrix}\alpha_{i_1} \\ \gamma_{i_1} \end{bmatrix}, \ldots,
\begin{bmatrix}\alpha_{i_p} \\ \gamma_{i_p} \end{bmatrix},
\begin{bmatrix}0 \\ \beta_{j_1} \end{bmatrix}, \ldots
, \begin{bmatrix}0 \\ \beta_{j_q} \end{bmatrix}$$
are linearly independent.
In summary, we showed that, for any group of linearly independent column vectors from $\begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$, its corresponding group of column vectors from $\begin{bmatrix} A & 0 \\ C & B \end{bmatrix}$ is also linearly independent. Since the rank of a matrix is defined as the cardinality of the maximal linearly independent group, the proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to approach this? If $2m^3 - 8m^2 + 8n^3 - 12n^2 -10 \equiv 0 \mod 10$, then $8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 0 \mod 10$ Assuming
$$2m^3 - 8m^2 + 8n^3 - 12n^2 -10 \equiv 0 \mod 10$$
Prove
$$8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 0 \mod 10$$
I tried the following:
$$8m^3 - 12m^2 + 2n^3 - 8n^2 - 10 \equiv 6m^3 - 4m^2 - 6n^3 + 4n^2 \mod 10$$
I'm not sure where to go from here or even if this is correct.
| Let $$A=2m^3 - 8m^2 + 8n^3 - 12n^2 -10 $$ and $$B= 8m^3 - 12m^2 + 2n^3 - 8n^2 - 10$$
Note that $$A+B = 10 m^3-20m^2+10n^3-20n^2-20 \equiv 0 , \mod (10)$$
So if one of $A$ or $B$ is a multiple of $10$ so is the other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving a summation by induction How can I prove $$\sum_{i=2}^{n} \frac{1}{i^2-i}\lt1$$ by induction on $n$?
So far:
If $m$ is a natural number such that $m\ge2$, let $P(m)$ be the statement:
$$\sum_{i=2}^{m} \frac{1}{i^2-i}\lt1$$
We will prove $P(m)$ by induction on $m$.
Base Case:
P(2) is the statement:
$$\sum_{i=2}^{2} \frac{1}{2^2-2}=\frac{1}{2}\lt1$$
So $P(1)$ is true.
Inductive Step:
Let $k$ be a natural number such that $k\ge2$. Assume $P(k)$ for some arbitrary $k\ge2$.
$\sum_{i=2}^{n+1} \frac{1}{i^2-i}=\sum_{i=2}^{n} \frac{1}{i^2-i}+\frac{1}{(n+1)^2-(n+1)}\lt 1+\frac{1}{(n+1)^2-(n+1)}$ (by Ind. Hyp.).
I do not know where to go from here.
| Well, since $\frac 1{(n+1)^2 -n} > 0$ and knowing $\sum\limits_{k=2}^n \frac 1{k^2 - k} = M < 1$, that isn't enough to prove $M + \frac 1{(n+1)^2 -n}< 1$.
We have to prove something a little stronger that $\sum\limits_{k=2}^n \frac 1{k^2 - k} $ is not just $< 1$ but less then $1-v_n$ for some $v_n > \frac 1{(n+1)^2 - n}$.
Let's figure what some of the differences are.
$\frac 1{2^2-2} = \frac 12 = 1-\frac 12$ so $v_1 = \frac 12$.
$\frac 1{2^2- 2} + \frac 1{3^2-3}=\frac 12 + \frac 16 = \frac 23$ and so $v_1 = \frac 13$.
$\sum\limits_{k=2}^4\frac 1{k^2-k}=\frac 23+\frac 1{12}=\frac 9{12}=\frac 34$ and $v_n = \frac 14$.
Can that possible be it?
Can it be true that $\sum\limits_{k=2}^n\frac 1{k^2-k}= 1-\frac 1n < 1$?
Well...... let's see:
We've done three base cases..
Induction step: $\sum\limits_{k=2}^n\frac 1{k^2-k}= 1-\frac 1n$ then
$\sum\limits_{k=2}^{n+1}\frac 1{k^2-k}=(\sum\limits_{k=2}^n\frac 1{k^2-k}) + \frac 1{(n+1)^2 - n}=$
$ 1-\frac 1n + \frac 1{(n+1)^2 - (n+1)} =$
$\frac {n-1}n + \frac 1{(n+1)((n+1)-1)} = \frac {n-1}n + \frac 1{n(n+1)}=$
$\frac {(n-1)(n+1)}{n(n+1)} + \frac 1{n(n+1)}=\frac {n^2-1}{n(n+1)} + \frac 1{n(n+1)}$
$\frac {n^2 -1+1}{n(n+1)} = \frac {n^2}{n(n+1)}=$
$\frac n{n+1} = 1 - \frac 1{n+1}$.
Excellent! It works.
......
I should note; A La Jose Carlos Santos excellent answer, that $\frac 1{n^2 - n} = \frac 1{n-1} - \frac 1n$. That makes
$\sum_{k=2}^n \frac 1{n^2 - n} = \sum_{k=2}^n(\frac 1{n-1} - \frac 1n)=(1 - \frac 12) + (\frac 12 -\frac 13) + (\frac 13 - \frac 14) + ......(\frac 1{n-1} - \frac 1n) = 1 - \frac 1n$.
Hence JC Santos's excellent answer.
It's a clever manipulation the first time you see $\frac 1{n^2 - n} = \frac {n- (n-1)}{n(n-1)} = \frac 1{n-1} - \frac 1n$ but one should get used to it. It's more common than one would think.
It's essential in proving that $\lim_{n\to \infty}(1 + \frac 1n)^n:= e$ actually converges and exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of function averages Please let me know if you know an answer to this problem. May be you could provide a reference to some publication on this topic?
Let $f(x)$ be a real-valued strictly convex function on $[0, 1]$. For any integer $k$ between $0$ and some positive integer $n$ let $x_k =k/n$. Consider the average value function $g(n)= \sum_{k=0}^{n}f(x_k)/(n+1) $. Is it true that $g(n)$ is decreasing?
Thank you!
| $g(n-1) \geq g(n) \Leftrightarrow (n+1) \sum f( \frac{i}{n-1}) \leq n \sum f( \frac{i}{n})$.
This follows by summing up the following inequlities:
$\begin{array} { l l l l l l l }
& & n \times f(0) & = & n \times f(0) &=& n \times f(0), \\
1 \times f(\frac{0}{n-1}) &+& (n-1) \times f( \frac{1}{n-1} ) & \geq & n \times f( \frac{\frac{0}{n-1} + \frac{n-1}{n-1}}{n}) &=& n f(\frac{1}{n}), \\
2 \times f( \frac{1}{n-1}) &+& (n-2) \times f( \frac{2}{n-1}) &\geq& n \times f ( \frac{\frac{ 2}{n-1} + \frac{2n-4}{n-1}}{n}) &=& n f(\frac{2}{n}), \\
\vdots \\
i \times f( \frac{i-1}{n-1}) & + & (n-i) \times f( \frac{i}{n-1}) & \geq & n \times f( \frac{\frac{ i^2 - i}{n-1} + \frac{ni - i^2}{n-1} } { n } ) & = & n f(\frac{i}{n}), \\
(n-1) \times f( \frac{n-2}{n-1} ) &+& 1 \times f( \frac{n-1}{n-1}) &\geq& n \times f(\frac{ \frac{ n^2-3n+2}{n-1} + \frac{n-1}{n-1} }{n}) &=& n f(\frac{n-1}{n}) \\
n \times f(1) & & &=& n \times f(1)&=& n \times f(1) \\
\end{array}$
Note: Strict convexity gives strict inequalities, so the function is strictly decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all triples of non-negative real numbers $(a,b,c)$ Find all triples of non-negative real numbers $(a,b,c)$ such that:
$a^2+ab=c$
$b^2+bc=a$
$c^2+ca=b$.
This question was problem number of 3 in the RMO(India) Olympiad in 2019 held on $10^{th}$ November. link .
My attempt-
Assume $a \geq b\geq c$,
$\therefore a^2 \geq b^2$ and $ab\geq bc$.
Adding these two, $a^2+ab\geq b^2+bc$
implies $c \geq a$.
Which can be possible only if $a=c$, which also implies $a=b=c$.
Substituting in the base equations,
$a^2+a^2=a$
$\therefore (a,b,c)=(0,0,0)$ or $(0.5,0.5,0.5)$
I want to know if my method is correct because it seems a lot different than the one provided in the solutions.
| Your method is not full because the system is cyclic and not symmetric.
You can not assume that $a\geq b\geq c$.
We can solve our system by the following way.
We have: if $a=0$ so $c=0$ and from here $b=0$, which gives a solution $(0,0,0)$.
Thus, it's enough to solve our system for $abc\neq0$ and sice
$$a(a+b)=c,$$
$$b(b+c)=a$$ and $$c(c+a)=b,$$ we obtain:
$$abc\prod_{cyc}(a+b)=abc$$ or
$$\prod_{cyc}(a+b)=1.$$
In another hand, from the starting system we obtain:
$$(a-b)(a+b)=(c-a)(b+1),$$
$$(b-c)(b+c)=(a-b)(c+1)$$ and
$$(c-a)(c+a)=(b-c)(a+1).$$
Thus, $a=b$ gives $c=a$ and we got as you got: $a=b=c,$ which gives also a solution $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right).$
Now, let $(a-b)(b-c)(c-a)\neq0.$
Thus, from the last system we obtain:
$$\prod_{cyc}(a-b)\prod_{cyc}(a+b)=\prod_{cyc}(a-b)\prod_{cyc}(a+1)$$ or
$$1=\prod_{cyc}(a+1),$$ which is impossible, which gives the answer:
$$\left\{(0,0,0),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given $n$ distinct items, distribute them to $m$ distinct bins such that no bin is empty One solution can be solved by using inclusion-exclusion principle:
Let $A$ be the group that all items exist and to compute $A'$ then reduce it from the universe.
Then I thought about another way to solve it but I think it's not accurate -
let $A$ be the group that all bins have at least $1$ item.
The way I computed $A$:
Let's choose $m$ items from the $n$ items which is $\binom{n}{m}$ and distribute one item to each bin such that none of the bins is empty, there are $m!$ ways to do that.
Then I have $n-m$ items that can go to any bin, so it's $m^{n-m}$.
Total - $\binom{n}{m} \cdot m! \cdot m^{n-m}$
Any idea why it's not correct or what's wrong with this way of approaching the problem.
| Let's look at the case $n = 5$ and $m = 3$. We will solve the problem in two ways, then examine why your approach does not work.
Method 1: We use the Inclusion-Exclusion Principle.
There are three choices for each of the five objects, so the objects can be distributed in $3^5$ ways. From these, we must subtract those distributions in which one or more bins is left empty.
There are three ways to select a bin that will be left empty and $2^5$ ways to distribute the five objects to the remaining bins.
However, if we subtract $3 \cdot 2^5$ from the total, we will have subtracted the three cases in which two bins are left empty twice, once for each way we could designate one of the empty bins as the bins to be left empty. We only want to subtract those cases once, so we must add them back.
Thus, the number of ways of distributing five distinct objects to three distinct bins so that no bin is left empty is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 150$$
Method 2: This method is practical in the case $n = 5$, $m = 3$ since the difference between $n$ and $m$ is small. However, it will be useful in illustrating why your method did not work.
We can express $5$ as a sum of three positive integers in two ways:
\begin{align*}
5 & = 3 + 1 + 1\\
& = 2 + 2 + 1
\end{align*}
Three items are placed in one bin and one object each is placed in the other bins: Choose which of the three bins will receive three items. Choose three of the five objects to place in that bin. Arrange the remaining two items in the remaining two bins. There are
$$\binom{3}{1}\binom{5}{3}2! = 60$$
such distributions.
One item is placed in one bin and each of the other bins receives two objects: Choose which of the three bins receives a single object. Choose which of the five objects is placed in that bin. Choose which two of the remaining four objects to be placed in the leftmost open bin. The remaining two objects must be placed in the remaining open bin. There are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2} = 90$$
such distributions.
Total: Since these cases are mutually exclusive and exhaustive, there are a total of
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 60 + 90 = 150$$
ways to distribute five distinct objects to three distinct bins so that no bin is left empty.
What is wrong with your approach?
Suppose we wish to distribute the numbers $1, 2, 3, 4, 5$ to bins $A$, $B$, and $C$.
If we place the numbers $1$, $2$, and $3$ in bin $A$, $4$ in bin $B$, and $5$ in bin $C$, you count that case three times, once for each of the three numbers you could designate as the object that goes in bin $A$.
$$
\begin{array}{c c c c}
\text{bin $A$} & \text{bin $B$} & \text{bin $C$} & \text{additional objects in bin $A$}\\ \hline
1 & 4 & 5 & 2, 3\\
2 & 4 & 5 & 1, 3\\
3 & 4 & 5 & 1, 2
\end{array}
$$
If we place the numbers $1$ and $2$ in bin $A$, $3$ and $4$ in bin $B$, and $5$ in bin $C$, you count that case four times, twice for each way you could designate one of the two numbers in bin $A$ as the object that goes in bin $A$ and twice for each way you could designate one of the two numbers in bin $B$ as the object that goes in bin $B$.
$$
\begin{array}{c c c c}
\text{bin $A$} & \text{bin $B$} & \text{bin $C$} & \text{additional object in bin $A$} & \text{additional object in bin $B$}\\ \hline
1 & 3 & 5 & 2 & 4\\
1 & 4 & 5 & 2 & 3\\
2 & 3 & 5 & 1 & 4\\
2 & 4 & 5 & 1 & 3
\end{array}
$$
Notice that your approach yields the answer
$$\color{red}{\binom{5}{3}3! \cdot 3^2 = 10 \cdot 6 \cdot 9 = 540}$$
and that
$$\color{red}{\binom{3}{1}}\binom{3}{1}\binom{5}{3}2! + \color{red}{\binom{2}{1}\binom{2}{1}}\binom{3}{1}\binom{5}{1}\binom{4}{2} = \color{red}{3} \cdot 60 + \color{red}{4} \cdot 90 = \color{red}{180} + \color{red}{360} = \color{red}{540}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality with paramater Find all values of $a$ such that the inequality
$$\frac{a-\left(\log_3x+2\sqrt6\log_x3-5\right)}{\left(3\cos\sqrt{x-9}-4\right)-a}\le0$$
has no solution.
My work:
I used $\frac bc< 0$ if only if $bc< 0$ and considered case $a=0, b\not=0$ but I don't know how to get answer.
| First of all, we have to have
$$x\ge 9\tag1$$
Under $(1)$, we have, by AM-GM inequality,
$$\begin{align}&\left(\log_3x+2\sqrt6\log_x3-5\right)-\left(3\cos\sqrt{x-9}-4\right)\\\\&=\log_3x+\frac{2\sqrt 6}{\log_3x}-3\cos\sqrt{x-9}-1
\\\\&\ge 2\sqrt{2\sqrt 6}-3-1\gt 0\end{align}$$
from which we see that
$$\left(\log_3x+2\sqrt6\log_x3-5\right)\gt\left(3\cos\sqrt{x-9}-4\right)\tag2$$
holds for all $x\ge 9$.
Considering $(2)$, we see that $$\frac{a-\left(\log_3x+2\sqrt6\log_x3-5\right)}{\left(3\cos\sqrt{x-9}-4\right)-a}\le0$$
is equivalent to
$$a\ge \left(\log_3x+2\sqrt6\log_x3-5\right)\quad\text{or}\quad a\lt \left(3\cos\sqrt{x-9}-4\right)$$
Note here that the former has at least one real solution if and only if $a\ge 2\sqrt{2\sqrt 6}-5$, and that the latter has at least one real solution if and only if $a\lt -1$.
Hence, the answer is
$$\color{red}{-1\le a\lt 2\sqrt{2\sqrt 6}-5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Systems of equations over reals
Solve over reals
$$2x^2+xy+x-3y+3=0$$
$$3xy+y^2-2x+6y-6=0$$
Adding or subtracting these equations doesn't help, and completing the square also doesn't work. I'm considering substituting $x+y$ and $xy$, but it also doesn't show much promise. Thanks!
| $$2x^2+xy+x-3y+3=0\tag{1}$$
$$3xy+y^2-2x+6y-6=0\tag{2}$$
Take 2$\times$(1)+(2) to get,
$$4x^2+5xy+y^2=0 \implies (4x+y)(x+y)=0$$
which yields $y=-x$ and $y=-4x$. Plug them into (1)
$$x^2+4x+3=0\implies x=-1,\>x=-3$$
$$2x^2-13x-3=0\implies x=\frac{13\pm\sqrt{193}}{4}$$
Then, evaluated the corresponding $y$'s to obtain the follow solutions,
$$(-1,1),\>\>\>(-3,3),\>\>\>
\left(\frac{13+\sqrt{193}}{4},-13-\sqrt{193}\right),
\left(\frac{13-\sqrt{193}}{4},-13+\sqrt{193}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Strategy for expanding the function $e^{e^x-1}$ into Maclaurin series I am trying to find a way to expand the function $e^{e^x-1}$ into Maclaurin series but I don't know how to. I don't want to use the cumbersome Maclaurin formula of taking successive higher derivative.
Do you know how to nicely expand this function into Taylor series. I know that this is the generating function of Bell numbers.
Wolfram gives the expansion as:
$1-x+x^2+\dfrac{5x^3}{6}+\dfrac{5x^4}{8}...$
| The exponent is
\begin{eqnarray*}
e^x-1 = x+\frac{x^2}{2}+ \frac{x^3}{6}+\cdots.
\end{eqnarray*}
Now just plug this in the power series for the exponential function
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = 1&+& \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right) \\
&+& \frac{1}{2} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^2 \\
&+& \frac{1}{6} \left(x+ \frac{x^2}{2}+ \frac{x^3}{6}+\cdots \right)^3 \\
&+& \cdots \\
\end{eqnarray*}
Expand to whatever order you need ...
\begin{eqnarray*}
\operatorname{exp}(e^x-1) = \color{red}{1}+\color{red}{1}x+ \frac{\color{red}{2}x^2}{2}+ \frac{\color{red}{5}x^3}{3!} +\cdots
\end{eqnarray*}
But of course ... as Ethan Bolker says in the comments the Bell numbers can be calculated much more easily using
\begin{eqnarray*}
B_{n+1} = \sum_{k=0}^{n} \binom{n}{k} B_k.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Divisibility Induction $17|18^{(5n+1)}+13^{(4n+1)}+3$ Prove that by induction
$17|18^{(5n+1)}+13^{(4n+1)}+3$ for all n∈N.
So far I'm stuck on the proof for n=k+1:
When n=k+1:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$= (18^{5k+1}+13^{4k+1}+3)+[(18^5-1)18^{5k+1}+(13^4-1)13^{4k+1}]$$
From the assumption n=k, I can prove the first part is divisible by 17, but unsure of how to prove for the second part.
Any help is appreciated.
Edit:word
| For completeness, here is another way to achieve the same result:
$$RHS=18^{5k+6}+13^{4k+5}+3$$
$$=18^5 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+3$$
$$=13^4 \cdot 18^{5k+1}+ 13^4 \cdot 13^{4k+1}+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+ (18^5-13^4)\cdot 18^{5k+1}+3$$
$$=13^4 (18^{5k+1}+ 13^{4k+1})+3\cdot13^4+ (18^5-13^4)\cdot 18^{5k+1}+3-3 \cdot13^{4}$$
$$=13^4 (18^{5k+1}+ 13^{4k+1}+3)+ (18^5-13^4)\cdot 18^{5k+1}+3-3 \cdot13^{4}$$
The first term is divisible by $17$ by the induction hypothesis. The second term is divisible by $17$ because $17$ divides $18^5-13^4 = 1861007$, and the same goes for $3 - 3 \cdot 13^4 = -85680$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$ I'm trying to solve the following problem:
Find the coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$.
Can I use the multinomial theorem to solve it? I'm unsure how to start..
Thanks!
| Definition:
$[x^n]f(x)~$ means the coefficient of $~x^n~$ of the Taylor expansion of $~f(x)~$ at $~x=0~$ .
Calculation:
$\displaystyle [x^{10}]x^3(x^2-3x^3-1)^6 = [x^{7}](x^2-3x^3-1)^6 = [x^{7}]\sum\limits_{k=0}^6 {\binom 6 k}x^{2k}(3x-1)^k = $
$\displaystyle =\sum\limits_{k=0}^6 {\binom 6 k}[x^{7-2k}](3x-1)^k =\sum\limits_{~~~~~k=0\\0\leq 7-2k\leq k}^6 {\binom 6 k}[x^{7-2k}](3x-1)^k$
$\displaystyle =\sum\limits_{k=3}^3 {\binom 6 k}[x^{7-2k}](3x-1)^k = {\binom 6 3}[x^{1}](3x-1)^3 =20\cdot 3{\binom 3 1}=180$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simple formula for determinant of circulant matrix built on an arithmetic sequence. Let $a$ be an arithmetic sequence:
$$a_i=a_1+\lambda(i-1),\tag1$$
and consider a $n\times n$ circulant matrix $M_{n}(a)$ "built" on rotational shifts of the sequence $a$, i.e. with elements:
$$M_{ij}=a_{1+(j−i)\operatorname{mod}n}.\tag2$$
Prove:
$$
\det M_n(a)=\frac{a_1+a_n}2(-n\lambda)^{(n-1)}.\tag3
$$
An example:
$$
\det\begin{pmatrix}
1&2&3&4&5\\
5&1&2&3&4\\
4&5&1&2&3\\
3&4&5&1&2\\
2&3&4&5&1\\
\end{pmatrix}=\frac{1+5}2(-5)^4=1875
$$
| The determinant of circulant matrix $M$ is given by the formula:
$$\det M = \prod_{\ell=0}^{n-1} f(\omega^\ell)$$
where $\omega = e^{\frac{2\pi i}{n}}$ is a primitive $n^{th}$ root of unity and
$$\begin{align}
f(x) &= \sum_{k=0}^{n-1} a_{k+1} x^k
= \sum_{k=0}^{n-1} (a_1 + \lambda k)x^k\\
&= \left(a_1 + \lambda x\frac{d}{dx}\right)\sum_{k=0}^{n-1} x^k
= \left(a_1 + \lambda x\frac{d}{dx}\right)\frac{1 - x^n}{1-x}\\
&= a_1 \frac{1 - x^n}{1-x} + \lambda\left[\frac{-n x^n}{1 -x} + x\frac{1-x^n}{(1-x)^2}\right]
\end{align}
$$
Notice $\omega,\omega^2,\cdots,\omega^{n-1}$ are all the roots of
$g(x) = \sum_{k=0}^{n-1} x^k = \frac{1 - x^n}{1-x}$.
We have $x^n = 1$ at these points and
$$f(\omega^\ell) = \frac{-\lambda n}{1-\omega^\ell}\quad\text{ for }\quad 1 \le \ell \le n-1$$
As a result,
$$\prod_{\ell=1}^{n-1} f(\omega^\ell) = \prod_{\ell=1}^{n-1}\frac{-\lambda n}{1 - \omega^\ell} = \frac{(-\lambda n)^{n-1}}{g(1)} = \frac{(-\lambda n)^{n-1}}{n}$$
Together with
$$f(1) = \sum_{k=0}^{n-1} (a_1 + \lambda k) = a_1 n + \lambda \frac{n(n-1)}{2} = n\frac{a_1 + a_n}{2}$$
We obtain
$$\det M = f(1)\prod_{\ell=1}^{n-1} f(\omega^\ell) = \frac{a_1 + a_n}{2}(-\lambda n)^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Compute the possibilities for $\gcd (2a^4 + 2a^2+3,2a^5+2a^3+a^2+a+3)$
Given that $a$ is an integer, compute the possibilities for $\gcd (2a^4 + 2a^2+3,2a^5+2a^3+a^2+a+3).$
Here's my work (I'm not entirely sure if it's correct).
Just to make sure we're on the same page, we say that for integers $a,b,$ $d=\gcd(a,b)$ is the integer such that $d\mid a$ and $d\mid b$ and for all $c\in\mathbb{Z}$ such that $c\mid a$ and $c\mid b,c\mid d$ (so there can be multiple possible $gcd$'s).
We use the fact that $\gcd (a,b)=\gcd(b,a-bq),q\in\mathbb{Z}.$
Dividing using the Euclidean Algorithm, we have the following:
$2a^5+2a^3+a^2+a+3=(2a^4+2a^2+3)(a)+a^2-2a+3\\
2a^4+2a^2+3=(a^2-2a+3)(2a^2+4a+4)-4a-9\\
4(a^2-2a+3)=4a^2-8a+12=(-4a-9)(-a+4)-a+48\\
-4a-9=4(-a+48)-201.$
Hence $\gcd (2a^5+2a^3+a^2+a+3,2a^4+2a^2+3)=\gcd (-a+48,-201)$ for all $a\in\mathbb{Z}.$ Let $f(a)=48-a, f:\mathbb{Z}\to \mathbb{Z}.$ Then since $f$ is a surjective, monic linear polynomial, the possibilities for $\gcd (-a+48,-201)$ are precisely the integer divisors of $201$ (since $\gcd$'s can be negative according to the definition).
Edit: I've edited my answer to include the proper possibilities.
| Trusting on your calculations, it is enough to find the possible values of $\gcd(48-a, -201)$. Now, assume $f(a):=48-a, f:\mathbb{Z}\to\mathbb{Z}$. Since, $f$ is monic linear polynomial it is surjective. Therefore, possibilities of $\gcd(48-a, -201)$ will be all positive divisors of $201$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all positive integral solutions of $\tan^{-1}x+\cos^{-1}\frac{y}{\sqrt{y^2+1}}=\sin^{-1}\frac{3}{\sqrt{10}}$ Find all the positive integral solutions of, $\tan^{-1}x+\cos^{-1}\dfrac{y}{\sqrt{y^2+1}}=\sin^{-1}\dfrac{3}{\sqrt{10}}$
Assuming $x\ge1,y\ge1$ as we have to find positive integral solutions of $(x,y)$
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac{1}{y}$$
As $3>0$ and $\dfrac{1}{y}>0$
$$\tan^{-1}x=\tan^{-1}\left(\dfrac{3-\dfrac{1}{y}}{1+\dfrac{3}{y}}\right)$$
$$\tan^{-1}x=\tan^{-1}\dfrac{3y-1}{y+3}$$
$$x=\dfrac{3y-1}{y+3}$$
$y+3\in[4,\infty)$ as $y\ge1$, $3y-1\in [2,\infty)$ as $y\ge1$
For $x$ to be positive integer, $3y-1$ should be multiple of $y+3$
$$3y-1=m(y+3) \text { where } m\in Z^{+}$$
$$3y-my=3m-1$$
$$(3-m)y=3m-1$$
Here R.H.S is positive, so L.H.S should also be positive.
So $3-m>0$, hence $m<3$
So possible values of $m$ are {$1$,$2$}.
For $m=1$, $$3y-1=y+3$$
$$2y=4$$
$$y=2$$
$$x=\dfrac{3\cdot2-1}{2+3}$$
$$x=1$$
For $m=2$, $$3y-1=2(y+3)$$
$$3y-1=2y+6$$
$$y=7$$
$$x=\dfrac{3\cdot7-1}{7+3}$$
$$x=\dfrac{20}{10}$$
$$x=2$$
Is there some other nicer way to solve this problem.
| We can actually utilize $x,y$ are positive integers
$$\tan^{-1}x=\tan^{-1}3-\tan^{-1}\dfrac1y<\tan^{-1}3$$
As $\tan^{-1}x$ is an increasing function
$$\implies x<3$$
So, $x$ can be $1$ or $2$
Check which ones make $y$ positive integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$ I have to find the following limit:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}}$$
This is what I tried:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k^2+2k+1}{k^2+2k}} =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(\dfrac{k^2+2k}{k^2+2k}} + \dfrac{1}{k^2+2k} \bigg ) = $$
$$ = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(1 } + \dfrac{1}{k^2+2k} \bigg ) =
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} }$$
Now,
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } = 1$
and:
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 0$
I think the above equals $0$, since this is a product and the limit of the last term of the product is $0$, so the whole thing would be $0$, but I am not exactly sure if my intuition is right.
So that means:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 1 + 0 = 1$$
The problem I have is that my textbook claims that the correct answer is $2$, not $1$. So I did something wrong, however, I can't spot my mistake/mistakes.
| $$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k+1}{k}\dfrac{k+1}{k+2}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \frac{ \dfrac{k+1}{k}}{\dfrac{k+2}{k+1}}}$$
$$ = \lim\limits_{n \to \infty} \left( \frac{\frac{1+1}{1}}{\frac{1+2}{1+1}} \frac{\frac{2+1}{2}}{\frac{2+2}{2+1}}\frac{\frac{3+1}{3}}{\frac{3+2}{3+1}}\cdots \frac{\frac{n+1}{n}}{\frac{n+2}{n+1}} \right) = \lim\limits_{n \to \infty} \left( \frac{\frac{1+1}{1}}{\frac{n+2}{n+1}} \right)$$
$$ = \lim\limits_{n \to \infty} \left( 2 \frac{n+1}{n+2} \right) = 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$. Prove that for ${a^3} = {b^2} + {c^2}$. where $a$, $b$ and $c$ are positive integers that there are an infinite number of values for $a$.
| Let $n$ be any odd positive integer. Then:
$$(2^n)^3 = 2^{3n} = 2 \times 2^{3n-1}= 2 \times (2^{(3n-1)/2})^2$$
The requirement that $n$ is odd ensures that $(3n-1)/2$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given the parameter $a \in \mathbb{R}$ such that the given sequence $(x_n)_{n \ge 1}$ is bounded, find the limit of the sequence $x_n$. I am given the following sequence $(x_n)_{n \ge 1}$:
$x_n = \bigg ( 1 + \dfrac{1}{3}
+ \dfrac{1}{5} + ... + \dfrac{1}{2n-1}\bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$
With $a \in \mathbb{R}$ such that the sequence $x_n$ is bounded. I am asked to find the limit of the sequence $x_n$.
I tried completing the sum:
$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) -
\bigg ( \dfrac{1}{2} + \dfrac{1}{4} + ... + \dfrac{1}{2n} \bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$$
$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) - \dfrac{1}{2}
\bigg ( 1 + \dfrac{1}{2} + ... + \dfrac{1}{n} \bigg ) - a \bigg ( 1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg )$$
$$x_n = \bigg ( 1 + \dfrac{1}{2}
+ \dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2n} \bigg ) -
\bigg ( a + \dfrac{1}{2} \bigg ) \bigg ( 1 + \dfrac{1}{2} + ... + \dfrac{1}{n} \bigg ) $$
And here I tried using the fact that:
$$\sum\limits_{k=1}^\infty \dfrac{1}{k} \approx \ln n$$
(I have only used this thing once before, so I'm expecting to use it incorrectly. I know it is called the Harmonic number). Using this I got into a whole mess with logarithms and what not. I arrived at a pretty random answer, while my textbook claims the right answer is $\ln 2$. How should I solve this exercise, arriving at $\ln 2$ ?
| \begin{align*}
x_{n}=\left(1-\left(a+\dfrac{1}{2}\right)\right)\left(\sum_{k=1}^{n}\dfrac{1}{k}\right)+\left(\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\right),
\end{align*}
and we have
\begin{align*}
\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n}\rightarrow\int_{0}^{1}\dfrac{1}{1+x}dx=\log 2,
\end{align*}
so this forces $1-(a+1/2)=0$ and the limit is $\log 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{k=0}^{n} 2^{k} {n \choose k}{n-k \choose \lfloor{\frac{n-k}{2}}\rfloor} = {2n+1 \choose n}$ The problem is the following: prove that $\sum_{k=0}^{n} 2^{k} {n \choose k}{n-k \choose \lfloor{\frac{n-k}{2}}\rfloor} = {2n+1 \choose n}$
I was trying to prove some concrete combinatorial examples to make the LHS equivalent to $2n+1 \choose n$.
We choose $k$ in the first $n$ items, but I am not sure how to map $2^{k} {n-k \choose \lfloor{\frac{n-k}{2}}\rfloor}$ to choosing $n-k$ elements from $n+1$ items
[EDIT] This might help. Combinatorial proof $\sum_i^{\lfloor{n/2}\rfloor} (-1)^i {n-i\choose i} 2^{n-2i} = n+1$
| We start with
$$\sum_{k=0}^n 2^k {n\choose k}
{n-k\choose \lfloor \frac{n-k}{2} \rfloor}
= \sum_{k=0}^n 2^{n-k} {n\choose k}
{k\choose \lfloor k/2 \rfloor}
\\ =
\sum_{k=0}^{\lfloor n/2\rfloor} 2^{n-2k} {n\choose 2k}
{2k\choose k}
+ \sum_{k=0}^{\lfloor (n-1)/2\rfloor} 2^{n-2k-1} {n\choose 2k+1}
{2k+1\choose k}.$$
Now for the first sum we write
$${n\choose 2k} {2k\choose k} =
\frac{n!}{(n-2k)! \times k! \times k!} =
{n\choose k} {n-k\choose n-2k}.$$
We obtain
$$\sum_{k=0}^{\lfloor n/2\rfloor} 2^{n-2k}
{n\choose k} {n-k\choose n-2k}
\\ = \sum_{k=0}^{\lfloor n/2\rfloor} 2^{n-2k}
{n\choose k} [z^{n-2k}] (1+z)^{n-k}
\\ = [z^n] (1+z)^n \sum_{k=0}^{\lfloor n/2\rfloor} 2^{n-2k}
{n\choose k} z^{2k} (1+z)^{-k}.$$
Now the coefficient extractor enforces the sum limits and we may
continue with
$$2^n [z^n] (1+z)^n \sum_{k\ge 0} 2^{-2k}
{n\choose k} z^{2k} (1+z)^{-k}
\\ = 2^n [z^n] (1+z)^n \left(1+\frac{z^2}{4(1+z)}\right)^n
\\ = \frac{1}{2^n} [z^n] (z+2)^{2n}
= \frac{1}{2^n} {2n\choose n} 2^n = {2n\choose n}.$$
For the second sum we write
$${n\choose 2k+1} {2k+1\choose k} =
\frac{n!}{(n-2k-1)! \times k! \times (k+1)!} =
{n\choose k} {n-k\choose n-2k-1}.$$
We obtain
$$\sum_{k=0}^{\lfloor (n-1)/2\rfloor} 2^{n-2k-1}
{n\choose k} {n-k\choose n-2k-1}
\\ = \sum_{k=0}^{\lfloor (n-1)/2\rfloor} 2^{n-2k-1}
{n\choose k} [z^{n-2k-1}] (1+z)^{n-k}
\\ = [z^{n-1}] (1+z)^n \sum_{k=0}^{\lfloor (n-1)/2\rfloor} 2^{n-2k-1}
{n\choose k} z^{2k} (1+z)^{-k}.$$
Once more the coefficient extractor enforces the sum limits and we may
continue with
$$2^{n-1} [z^{n-1}] (1+z)^n \sum_{k\ge 0} 2^{-2k}
{n\choose k} z^{2k} (1+z)^{-k}
\\ = 2^{n-1} [z^{n-1}] (1+z)^n \left(1+\frac{z^2}{4(1+z)}\right)^n
\\ = \frac{1}{2^{n+1}} [z^{n-1}] (z+2)^{2n}
= \frac{1}{2^{n+1}} {2n\choose n-1} 2^{n+1} = {2n\choose n-1}.$$
Collecting everything we find
$$\bbox[5px,border:2px solid #00A000]{
{2n\choose n} + {2n\choose n-1} = {2n+1\choose n}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\frac {1+x}{1-x}\right)$ Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$
My Attempt:
$$y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$$
Put $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \left(\dfrac {2\cos^2 \left(\dfrac {\theta}{2}\right)}{2\sin^2\left(\dfrac {\theta}{2}\right)}\right)$$
$$y=\tan^{-1} \left(\cot^2 \left(\dfrac {\theta}{2}\right)\right)$$
| $$D^{n}y= D^{n-1} Dy=D^{n-1}\frac{1}{1+x^2}= D^{n-1} \frac{1}{2i} \left( \frac{1}{x-i}- \frac{1}{x+i} \right)$$
The nth derivative of $(x+b)^{-1}$ is givwn as
$$D^n (x+b)^{-1}=(-1)^n n!(x+b)^{-n-1}$$
Using this
$$D^n y= (-1)^{(n-1)} (n-1)! ~ \Im (x-i)^{-n-2}=(-1)^{n-1} (n-1)!~ (x^2+1)^{(-n-2)/2} \sin[(n+2) \tan^{-1}(1/x)].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find all complex solutions $z^{10} + 2z^5 + 2 = 0$. So I’m working on this equation $z^{10} + 2z^5 + 2 = 0$ to find all complex solutions, and I think I managed to solve it, but I can’t find solution manual for it, since it is really old exam task. The thing that makes me uncomfortable with my solution is that, shouldn’t I get just 10 solutions? But when I put in all $k$ values($k = 0,1,2,3,4$), you get 12 different angle solutions. Isn’t that wrong?
My answer:
$$ \sqrt{\mathstrut 2}^{1/5}e^{\frac{\left(\pm\frac\pi4i+2\pi k \right)}5} $$
Sorry could't figure out how to put 5 in denominator of the polar formula...
| solve for $$w^2 + 2w + 2=0$$
which gives $$w_{1,2} = -1 \pm i = \sqrt{2} e^{i(\pi \pm \frac{\pi}{4})}$$
Now you got two equations to solve
\begin{align}
z_1^5 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4})} \\
z_2^5 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4})}
\end{align}
which gives
\begin{align}
z_1 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4} + \frac{2k\pi}{5})} \\
z_2 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4}+ \frac{2k\pi}{5})}
\end{align}
for $k \in \lbrace 0,1,2,3,4 \rbrace$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find GCD between these two equations? I've been trying for a couple of hours to find a way to prove the next claim:
$$\gcd(n-1,a) = \gcd(\frac{(n^a-1)}{n-1},n-1)$$
I have already proven that
$$\frac{(n^a-1)}{n-1}$$ is always divisible by $(a-1)$.
which means I can set define
$$\frac{(n^a-1)}{n-1}=\frac{c(n-1)}{n-1}=c$$
then I get :
$$ \gcd(n-1,a)=\gcd(c,n-1)$$
but that's it, I'm stuck.
| You can expand $n^a - 1$ as a difference of $a$th powers:
\begin{equation*}
n^a - 1 = (n - 1)(n^{a - 1} + n^{a - 2} + \dotsb + 1)
\end{equation*}
so
\begin{equation*}
\frac{n^a - 1}{n - 1} = n^{a - 1} + n^{a - 2} + \dotsb + 1
\end{equation*}
But since
\begin{align*}
n^{a - 1} + n^{a - 2} + \dotsb + 1 &\equiv
\underbrace{1^{a - 1} + 1^{a - 2} + \dotsb + 1}_{\text{$a$ times}} \pmod{n - 1} \\
&\equiv a \pmod{n - 1}
\end{align*}
we can use the property $\gcd(a, b) = \gcd(a, na + b)$
to infer that
\begin{equation*}
\gcd(n - 1, n^{a - 1} + n^{a - 2} + \dotsb + 1) = \gcd(n - 1, a)
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| If you are trying to use GM<=AM, I suppose you have $x,y,z\geq 0$.
Check this link QM>=AM>=GM>=HM
First Step
Using HM<=GM:
$$ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq \sqrt[3]{xyz} \implies xyz\geq \frac{3^3}{1}=27$$
Second Step
Using the AM<=QM:
$$ \frac{x+y+z}{3} \leq \sqrt{\frac{x^2+y^2+z^2}{3}}\implies \frac{x^2+y^2+z^2+2(xy+xz+yz)}{9}\leq \frac{x^2+y^2+z^2}{3}\implies \frac{x^2+y^2+z^2}{3}+\frac{2(xy+xz+yz)}{3}\leq x^2+y^2+z^2 \implies \frac{2}{3}(xy+xz+yz)\leq \frac{2}{3}(x^2+y^2+z^2)\implies xy+xz+yz\leq x^2+y^2+z^2$$
We know that $xy+xz+yz = xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=xyz$. Thus by First Step
$$27\leq xyz \leq x^2+y^2+z^2$$
To finish this proof, note that $(x,y,z)=(3,3,3)$ is a feasible solution with an objective value equal to $27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
| Finally solved it.
Thanks to all for looking into this problem, special thanks to @Andrei for pointing out the exact mistake.
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\
-\dfrac{a\lambda}{b} & \lambda & bc \\
0 & -\dfrac{b\lambda}{c} & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\
-\dfrac{a}{b} & 1 & bc \\
0 & -\dfrac{b}{c} & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\
-a & c & bc \\
0 & -b & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-(R_2+R_3)$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b&0 &ca \\
-a & c & bc \\
0 & -b & c^2+\lambda
\end{vmatrix}=0$$
$$\dfrac{\lambda^2}{bc}\left(b(c^3+c\lambda+b^2c)+a^2bc\right)=0$$
$$\lambda^2\left(c^2+\lambda+b^2+a^2\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Sum of products of binomial coefficients: $\sum\limits_{k=0}^{19}\binom{18}k\binom{20}k$ Please help me find the sum given below
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}$
First I used formula $\binom{m}{k}=\binom{m-1}{k}+\binom{m-1}{k-1}$ twice and got
$\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}=\sum_{k=0}^{19}\binom{18}{1k}\cdot \left ( \binom{19}{k}+\binom{19}{k-1} \right )=\sum_{k=0}^{19}\binom{18}{k}\cdot\left ( \binom{18}{k}+2\binom{18}{k-1}+\binom{18}{k-2} \right )$
Now have no idea what to do with that
|
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{19}\binom{18}{k}\binom{20}{k}}
&=\sum_{k=0}^{18}\binom{18}{k}\binom{20}{k}\tag{1}\\
&=\sum_{k=0}^{18}\binom{18}{18-k}\binom{20}{k}\tag{2}\\
&\,\,\color{blue}{=\binom{38}{18}=33\,578\,000\,610}\tag{3}
\end{align*}
Comment:
*
*In (1) we omit the index $k=19$, since $\binom{p}{q}=0$ if $q>p$.
*In (2) we apply the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
*In (3) we apply Vandermonde's identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\cos 2x-\cos 3x+\sin 4x = 0$ Solve Equation$$\cos 2x-\cos 3x+\sin 4x=0 $$
Attemp:Developping up to $\sin x,\cos x$, equation is :
$(\cos x-1)(4\cos^2x+2\cos x-1)=4\sin x(2\cos^3x-\cos x)$
Note that squaring will give solutions $x_i$ such that either $x_i$, either $-x_i$ is solution of the current equation.
So $(\cos x-1)^2(4\cos^2x+2\cos x-1)^2=16(1-\cos^2x)(2\cos^3x-\cos x)^2$
We obviously get $\cos x=1$ (which indeed is a solution of original equation) and it remains
$(1-\cos x)(4\cos^2x+2\cos x-1)^2=16(1+\cos x)(2\cos^3x-\cos x)^2$
Setting $\cos x=y$, this is :
$(1-y)(4y^2+2y-1)^2=16(1+y)(2y^3-y)^2$
Which is $64y^7+64y^6-48y^5-64y^4-4y^3+16y^2+5y-1=0$
I found no clever way to solve this degree-7 polynomial
| I think that this has to be solved numerically using Newton method. Graphing for $0 \leq x \leq 2\pi$, we can notice, beside the trivial roots $x=0$ and $x=2\pi$, solutions close to $1.0$, $2.7$, $3.8$, $4.9$ and $5.6$.
Using these as starting points, Newton iterates are
$$\left(
\begin{array}{cc}
0 & 1.00000 \\
1 & 0.95437 \\
2 & 0.95541
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
0 & 2.70000 \\
1 & 2.72792 \\
2 & 2.72687 \\
3 & 2.72687
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
0 & 3.80000 \\
1 & 3.84203 \\
2 & 3.84161
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
0 & 4.90000 \\
1 & 4.85398 \\
2 & 4.85570 \\
3 & 4.85571
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
0 & 5.60000 \\
1 & 5.66083 \\
2 & 5.66212 \\
3 & 5.66213
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Minimize $x + y + z$ subject to $x^2 + y^2 + z^2 \le 1$ and $x \ge 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + y + z \\
\text{s.t} & \quad x^2 + y^2 + z^2 &&\le 1 \\
& \quad x &&\ge 0
\end{align*}$$
The linear independence constraint constraint qualification - LICQ is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
The Mangasarian-Fromovitz constraint qualification - MFCQ is
The gradients of the equality constraints are linearly independent at $x^{*}$ and there exists a vector $v \in \mathbb{R}^{n}$ such that $\langle \nabla g_{i}\left(x^{*}\right), v \rangle<0$ for all active inequality constraints and $\langle \nabla h_{j}\left(x^{*}\right), v\rangle=0$ for all equality constraints.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
$\textbf{My attempt:}$
Let $f = x + y + z$, $g_1 = x^2+y^2 + z^2-1$, $g_2 = -x$, and $$\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid g_1(x,y,z) \le 0 \text{ and } g_2(x,y,z) \le 0\}$$
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, $\nabla g_1 (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$, and $\nabla g_2 (x,y,z) = \begin{pmatrix} -1 \\ 0 \\ 0 \\ \end{pmatrix}$. Consider the system $$\begin{cases}
\mu_1 \nabla g_1 (x,y,z) + \mu_2 \nabla g_2 (x,y,z) &=0 \\
g_1(x,y,z) &= 0\\
g_2(x,y,z) &=0
\end{cases}
\iff
\begin{cases}
\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} -1\\ 0 \\ 0 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\
x^2 + y^2 + z^2 &= 1\\
-x &= 0
\end{cases}$$
$$\iff \begin{cases}
x &= 0\\
\mu_2 &= 0 \\
\mu_1 y &= 0 \\
\mu_1 z &= 0\\
y^2+z^2 &= 1
\end{cases}
\implies
\mu_1 = \mu_2 = 0
$$
Hence LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases}
\mu_1,\mu_2 &\ge 0\\
\mu_1 g_1(x,y,z) &= 0\\
\mu_2 g_2(x,y,z) &=0 \\
\nabla f (x,y,z) + \mu_{1} \nabla g_1(x,y,z) + \mu_{2} \nabla g_2(x,y,z)&=0
\end{cases}
\iff
\begin{cases}
\mu_1, \mu_2 &\ge 0\\
\mu_1 (x^2+y^2 + z^2-1) &=0\\
\mu_2 x &= 0 \\
\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} +\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} -1\\ 0 \\ 0 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix}
\end{cases}$$
$$\iff
\begin{cases}
\mu_1, \mu_2 &\ge 0\\
\mu_1 (x^2+y^2 + z^2-1) &=0\\
\mu_2 x &=0 \\
1+ 2 \mu_1 x &= \mu_2 \\
1+ 2 \mu_1 y &=0 \\
1+ 2 \mu_1 z &=0
\end{cases}
\iff
\begin{cases}
\mu_1 &> 0\\
\mu_2 &\ge 0\\
\mu_2 x &=0 \\
x^2+y^2 + z^2 &= 1 \\
1+ 2 \mu_1 x &= \mu_2 \\
1+ 2 \mu_1 y &=0 \\
1+ 2 \mu_1 z &=0
\end{cases}$$
$$\iff\begin{cases}
\mu_1 &= 1/\sqrt{2}\\
\mu_2 &= 1 \\
x &=0 \\
y &=-1/\sqrt{2} \\
z &=-1/\sqrt{2}
\end{cases} \qquad
\text{or} \qquad\begin{cases}
\mu_1 &= \sqrt{3}/2 \\
\mu_2 &= 0 \\
x &= -1/\sqrt{3} \\
y &= -1/\sqrt{3} \\
z &= -1/\sqrt{3}
\end{cases}$$
Comparing the values at these points, we get the solution is $(x,y,z) = (-1/\sqrt{3},-1/\sqrt{3},-1/\sqrt{3})$.
| Your ultimate solution is wrong because $x$ is not $\ge 0$, but the other solution where $x = 0$ is correct. Also there's a lot of work there and I'm not sure how much of it is necessary. (It's all correct except for the solution you have with $x < 0$.) I guess my point is we can just assume that $x^2 + y^2 + z^2 = 1$ and $x = 0$, get an easy Lagrange-multiplier problem and then go back and check the KKT conditions + regularity conditions. This is how I solved it, so you can compare:
If we only have $x^2 + y^2 + z^2 \le 1$ then $\min x + y + z$ occurs along the line $t(1,1,1)$ and from that you get $(x,y,z)=(-1/\sqrt3, -1/\sqrt3, -1/\sqrt3)$ without having to do any complicated calculation.
If we want $x \ge 0$ then it must be that $x = 0$ since we are sliding the level set $x + y + z = -3/\sqrt{3}$ to a point where $x \ge 0$ and that must happen at $x = 0$. So now if $x = 0$ we get $\min y + z$ subject to $y^2 + z^2 \le 1$ and this occurs along the line $t(0,1,1)$. That is, the optimal point is
$$ (x,y,z) = \left(0, -\frac1{\sqrt{2}}, -\frac1{\sqrt{2}} \right).$$
Now the easiest regularity to check is either linear independence or Slater's condition. For linear independence, the gradients of $x$ and $x + y + z$ are $(1, 0, 0)$ and $(1, 1, 1)$ so they are independent. Slater's condition is also easily seen to hold.
Finally, let us check the KKT conditions:
*
*Complementary slackness is irrelevant because both inequalities are tight.
*Feasibility is by construction.
*We have
$$(1, 1, 1) + \frac{1}{\sqrt{2}}(0,-\sqrt{2},-\sqrt{2}) + (-1,0,0) = (0,0,0).$$
Now, in general, the KKT conditions are not sufficient. But they are for convex programs that satisfies Slater's condition (i.e. the feasible region has an interior point). You can also see directly that this is the minimum if you can see directly that 1. both inequalities are tight and 2. $y + z$ has a unique minimum on the curve $y^2 + z^2 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimize $x + 2y + 3z$ subject to $x^2 + y^2 + z^2 = 1$ and $x + y + z \le 0$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + 2y + 3z \\
\text{s.t} & \quad x^2 + y^2 + z^2 && = 1 \\
& \quad x + y + z && \le 0
\end{align*}$$
The linear independence constraint qualification - LICQ is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
$\textbf{My attempt:}$
Let $f = x + 2y + 3z$, $h = x^2+y^2 + z^2-1$, $g = x + y + z$, and $$\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid h (x,y,z) = 0 \text{ and } g (x,y,z) \le 0\}$$
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$, $\nabla h (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$, and $\nabla g (x,y,z) = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix}$.
Consider the system $$\begin{cases}
\mu_1 \nabla h (x,y,z) + \mu_2 \nabla g (x,y,z) &=0 \\
h (x,y,z) &= 0\\
g (x,y,z) &=0
\end{cases}
\iff
\begin{cases}
\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} 1\\ 1 \\ 1 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\
x^2 + y^2 + z^2 &= 1\\
x + y + z &= 0
\end{cases}$$
$$\iff \begin{cases}
2x \mu_1 + \mu_2 &= 0\\
2y \mu_1 + \mu_2 &= 0 \\
2z \mu_1 + \mu_2 &= 0 \\
x^2 + y^2 + z^2 &= 1\\
x + y + z &= 0
\end{cases}\iff
\begin{cases}
2\mu_1(x+y+z) + 3\mu_2 &= 0\\
2x \mu_1 + \mu_2 &= 0 \\
2y \mu_1 + \mu_2 &= 0 \\
2z \mu_1 + \mu_2 &= 0 \\
x^2 + y^2 + z^2 &= 1\\
x + y + z &= 0
\end{cases}$$
$$\iff
\begin{cases}
\mu_2 &= 0\\
x \mu_1 &= 0 \\
y \mu_1 &= 0 \\
z \mu_1 &= 0 \\
x^2 + y^2 + z^2 &= 1\\
x + y + z &= 0
\end{cases} \implies \mu_1 = \mu_2 = 0$$
Hence LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases}
\mu_2 &\ge 0\\
h (x,y,z) &= 0\\
\mu_2 g (x,y,z) &=0 \\
\nabla f (x,y,z) + \mu_{1} \nabla h (x,y,z) + \mu_{2} \nabla g (x,y,z)&=0
\end{cases}
\iff
\begin{cases}
\mu_2 &\ge 0\\
x^2+y^2 + z^2 &=1\\
\mu_2 (x + y + z) &= 0 \\
\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} +\mu_1 \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} + \mu_2 \begin{pmatrix} 1\\ 1 \\ 1 \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix}
\end{cases}$$
$$\iff
\begin{cases}
\mu_2 &\ge 0\\
x^2+y^2 + z^2 &=1\\
\mu_2 (x + y + z) &= 0 \\
1 + 2x \mu_1 + \mu_2 &= 0\\
2 + 2y \mu_1 + \mu_2 &= 0 \\
3+ 2z \mu_1 + \mu_2 &= 0 \\
\end{cases}\iff\begin{cases}
\mu_1 &= \sqrt 7 / \sqrt 2 \\
\mu_2 &= 0 \\
x &= -1/\sqrt{14} \\
y &= -2/\sqrt{14} \\
z &= -3/\sqrt{14}
\end{cases}$$
As such, the solution is $(x,y,z)=(-1/\sqrt{14},-2/\sqrt{14},-3/\sqrt{14})$.
| Let $k = x + 2y + 3z$ and observe that it represents a plane with $k$, $\frac k2$ and $\frac k3$ as the intercepts with the $x$, $y$ and $z$ axes, respectively. So, $k$ is at the minimum if the intercepts have the largest negative values, corresponding to the plane tangential to the sphere $x^2 + y^2 + z^2=1$ in the negative octant.
The normal vector to the plane is $n=(1, 2, 3)$ and the radius parallel to $n$ is $t(1,2,3)$, or
$$x=t,\>\>\>\>\>y = 2t,\>\>\>\>\>z=3t$$
Plug the above radial line into the equation of the sphere to get $t=\pm \frac1{\sqrt{14}}$ and the tangent point for the minimum $k$ is
$$x_m=-\frac1{\sqrt{14}},\>\>\>\>\>y_m = -\frac2{\sqrt{14}},\>\>\>\>\>z_m=-\frac3{\sqrt{14}}$$
Thus, the minimum value is,
$$k_m = x_m + 2y_m + 3z_m = -\sqrt{14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$
So the general formula for this series is
$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$
Is there a closed form for this series?
| I get
$(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}
$.
I use
$\ln(1+x)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n}
$.
$\begin{array}\\
f(x)
&=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}(\dfrac1{n+1}-\dfrac1{n+2})\\
&=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+1}-\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+2}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n-2}}{n}-\sum_{n=2}^{\infty} \dfrac{(-1)^{n+2}x^{2n-4}}{n}\\
&=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}-x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n}x^{2n}}{n}\\
&=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}+x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}\\
&=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n})\\
&=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\ln(1+x^2))\\
&=(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sides of non-right triangle
Given a non-right triangle with vertex $C = 91^{\circ}, \
\overline{AC} =39.9 \ cm, \ \overline{BC} = 32.6 \ cm$, find
a) $\overline{AB}$
b) $m\measuredangle A$
c) $m\measuredangle B$
I was able to find the length of $\overline{AB}$, which I labeled $c$
\begin{align*} c=& \ \sqrt{32.6^2+39.9^2-2(32.6*39.9)\cos(91^\circ)}
\\ c =& \ 51.9 \ cm
\end{align*}
but I'm having a hard time finding the angle measures of $A$ and $B$,
here's my attempt:
a)
\begin{align*} \frac{32.6}{\sin A} =& \frac{51.9}{\sin 91^\circ} \\ \\ \sin A =& \frac{32.6 \sin 91^\circ}{51.9} \\
\end{align*}
So given some theta A, we get the vertical length $0.63$ radians. To find A, I used
\begin{align*} \arcsin (0.63) \approx 0.68 \ \text{rad}
\end{align*}
So I figured I just needed to convert this to degree measure, getting me $$m\measuredangle A \approx 38.96^\circ$$
b) using the fact that the sum of angle measures of a triangle is $180^\circ$, I tried to solve for angle $B$ using
\begin{align*} 38.96^\circ + 51.9^\circ + m\measuredangle B \approx& \ 180^\circ \\ m\measuredangle B \approx& \ 89.14^\circ
\end{align*}
but I'm getting marked wrong for both answers b) and c). What am I doing wrong?
| Side length $c \approx 51.9$ is ok.
The triangle side lengths are close to the right triangle of sides $ 30,40,50$ in proportion.
$$ \angle A ( \approx 38.96^{\circ})+ \angle B^{\circ} + 91 ^{\circ} = 180^{\circ}\rightarrow \angle B\approx 50.04 ^{\circ} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$x,y,z>0$, prove:$\frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$ If $x,y,z>0$, prove:$$ \frac{x}{y+z+\sqrt[4]\frac{y^4+z^4}{2}}+\frac{y}{z+x+\sqrt[4]\frac{z^4+x^4}{2}}+\frac{z}{x+y+\sqrt[4]\frac{x^4+y^4}{2}}\geq1$$ I tried to use classical ineqalities such as AM-GM or QM-GM to prove this inequality. I have noticed equality holds when $x=y=z$.
The next part of the question is asking if this inequality can be generalized.
Another inequality with same nature.If $x,y,z>0$, prove or disprove:$$ \frac{x}{y+z+\sqrt[6]\frac{y^6+z^6}{2}}+\frac{y}{z+x+\sqrt[6]\frac{z^6+x^6}{2}}+\frac{z}{x+y+\sqrt[6]\frac{x^6+y^6}{2}}\geq1$$
These questions are proposed from R. Shahbazov and J. Hajimir to RMM.
| Since $$\sqrt[4]{\frac{x^4+y^4}{2}}\leq\sqrt{x^2-xy+y^2}$$ it's just $(x-y)^4\geq0,$ it's enough to prove that
$$\sum_{cyc}\frac{z}{x+y+\sqrt{x^2-xy+y^2}}\geq1$$ or
$$\sum_{cyc}\frac{z(x+y-\sqrt{x^2-xy+y^2})}{3xy}\geq1$$ or
$$\sum_{cyc}(x^2y+x^2z-xyz)\geq\sum_{cyc}z^2\sqrt{x^2-xy+y^2}$$ or
$$\sum_{cyc}\left(\frac{x^2y+x^2z}{2}-xyz\right)\geq\sum_{cyc}z^2\left(\sqrt{x^2-xy+y^2}-\frac{x+y}{2}\right)$$ or
$$\sum_{cyc}z(x-y)^2\geq\sum_{cyc}\frac{3z^2(x-y)^2}{2\sqrt{x^2-xy+y^2}+x+y}$$ or
$$\sum_{cyc}z(x-y)^2\left(1-\frac{3z}{2\sqrt{x^2-xy+y^2}+x+y}\right)\geq0.$$
Now, let $x\geq y\geq z$.
Thus, $$y^2(x-z)^2\geq x^2(y-z)^2,$$ $$2\sqrt{x^2-xz+z^2}\geq2x-z,$$ $$2\sqrt{y^2-yz+z^2}\geq2y-z,$$ $$1-\frac{3y}{2\sqrt{x^2-xz+z^2}+x+z}\geq1-\frac{3y}{2x-z+x+z}\geq0$$ and
$$y^2\sum_{cyc}z(x-y)^2\left(1-\frac{3z}{2\sqrt{x^2-xy+y^2}+x+y}\right)\geq$$
$$\geq y\cdot y^2(x-z)^2\left(1-\tfrac{3y}{2\sqrt{x^2-xz+z^2}+x+z}\right)+ x\cdot y^2(y-z)^2\left(1-\tfrac{3x}{2\sqrt{y^2-yz+z^2}+y+z}\right)\geq$$
$$\geq y\cdot x^2(y-z)^2\left(1-\tfrac{3y}{3x-z+x+z}\right)+ x\cdot y^2(y-z)^2\left(1-\tfrac{3x}{2y-z+y+z}\right)=$$
$$=xy(y-z)^2\left(x\left(1-\frac{y}{x}\right)+y\left(1-\frac{x}{y}\right)\right)=0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof explanation of $\prod_{k=2}^n \big(1- \frac{2}{k(k+1)} \big) = \frac{1}{3} \big(1+\frac{2}{n} \big), n \geq 2$ I want to understand the proof of
$$\prod_{k=2}^n \big(1- \frac{2}{k(k+1)} \big) = \frac{1}{3} \big(1+\frac{2}{n} \big), n \geq 2$$
via induction.
$$\text{Base case: } n=2 \Rightarrow\prod_{k=2}^2\big(1-\frac{2}{k(k+1)}\big) = 1-\frac{2}{2(2+1)} = \frac{2}{3} = \frac{1}{3} \big(1+\frac{2}{2}\big)$$
$$\text{Hypothesis: The proposition holds for an } n \in \mathbb{N} \text{ with } n \ge 2$$
$$\text{Induction step: } \prod_{k=2}^{n+1} \big(1- \frac{2}{k(k+1)} \big) = \big(1- \frac{2}{(n+1)(n+2)} \big) \cdot \prod_{k=2}^n \big( 1- \frac{2}{k(k+1)} \big)$$
$$= \text{IH: } \big( 1- \frac{2}{(n+1)(n+2)} \big) \cdot \frac{1}{3} \big(1+ \frac{2}{n}\big) $$
$$= \frac{(n+1)(n+2)-2}{(n+1)(n+2)} \cdot \frac{1}{3} \cdot \frac{n+2}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} = \frac{1}{3} \cdot \frac{n+3}{n+1 }= \frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$$
$$=\frac{1}{3} \big(1+\frac{2}{n+1} \big)$$
What I don't understand are the two steps:
What's been done to remove the $1-$ and $1+$?:
$$= \frac{(n+1)(n+2)-2}{(n+1)(n+2)} \cdot \frac{1}{3} \cdot \frac{n+2}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
How do we get the $\frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$ here?:
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} = \frac{1}{3} \cdot \frac{n+3}{n+1 }= \frac{1}{3}\cdot \big(\frac{n+1}{n+1} +\frac{2}{n+1}\big)$$
| In the first step we are using the induction hypotesis that is
$$\prod_{k=2}^n \left(1- \frac{2}{k(k+1)} \right) = \frac{1}{3} \left(1+\frac{2}{n} \right)$$
the second one is
$$= \frac{(n+1)(n+2)-2}{(n+1)\color{red}{(n+2)}} \cdot \frac{1}{3} \cdot \frac{\color{red}{n+2}}{n} = \frac{1}{3} \frac{(n+1)(n+2)-2}{n(n+1)}$$
and finally
$$=\frac{1}{3} \frac{n^2+3n+2-2}{n(n+1)} =\frac{1}{3} \frac{n(n+3)}{n(n+1)}= \frac{1}{3} \cdot \frac{n+3}{n+1 }=$$$$= \frac{1}{3} \cdot \frac{n+1+2}{n+1 }= \frac{1}{3}\cdot \left(\frac{n+1}{n+1} +\frac{2}{n+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and subtracting $1$ from the original limit. However, since $3-2x$ $\rightarrow - \infty$ and not $+\infty$, I got nowhere. Then I tried finding the logarithm of this limit. It resulted in a $\dfrac{0}{0}$ indeterminate form, I tried L'Hospital, but again, it led me nowhere. Either I made some mistakes in the calculations, or I should use a different approach.
| Yes, add and subtract $1$. You will get
$$\left[
\left(
1+\frac{\sqrt{x^2+2x-3}-x-2}{x+2}
\right)^{\frac{x+2}{\sqrt{x^2+2x-3}-x-2}}
\right]^{\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}}$$
The part inside the $\left[...\right]$ tends to $e$.
Then compute the limit of the exponent
$$\begin{align}\frac{(\sqrt{x^2+2x-3}-x-2)(3-2x)}{x+2}&=\frac{((x^2+2x-3)-(x+2)^2)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\
&=\frac{(-2x-7)(3-2x)}{(\sqrt{x^2+2x-3}+x+2)(x+2)}\\
&=\frac{(-2-7/x)(3/x-2)}{(\sqrt{1+2/x-3/x^2}+1+2/x)(1+2/x)}\\
&\to2\end{align}$$
Therefore, the original limit is $e^{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is there a closed form for this transcendental equation? Wondering if there’s a closed form for $\exp(\frac{1}{\log(x)})=\frac{x}{e}$?
Numerically solving yields $x \approx .539$
| $\exp \left (\dfrac{1}{\ln x} \right ) = \dfrac{x}{e}; \tag 1$
$\dfrac{1}{\ln x} = \ln \left ( \dfrac{x}{e} \right ) = \ln x - \ln e = \ln x - 1; \tag 2$
$1 = (\ln x)^2 - \ln x; \tag 3$
$(\ln x)^2 - \ln x - 1 = 0; \tag 4$
quadratic formula:
$\ln x = \dfrac{1 \pm \sqrt 5}{2}; \tag 5$
$x = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} \right ). \tag 6$
Some Additional Observations:
From (5),
$\dfrac{1}{\ln x} = \dfrac{2}{1 \pm \sqrt 5} = \dfrac{2(1 \mp \sqrt 5)}{(1 \mp \sqrt 5)(1 \pm \sqrt 5}$
$= \dfrac{2(1 \mp \sqrt 5)}{1^2 - (\sqrt 5)^2} = \dfrac{2(1 \mp \sqrt 5)}{1 - 5} = \dfrac{2(1 \mp \sqrt 5)}{-4} = - \dfrac{1 \mp \sqrt 5}{2}; \tag 7$
and also from (5)
$\ln x - 1 = \dfrac{1 \pm \sqrt 5}{2} - 1 = \dfrac{-1 \pm \sqrt 5}{2} = - \dfrac{1 \mp \sqrt 5}{2} = \dfrac{1}{\ln x}; \tag 8$
from (6),
$\dfrac{x}{e} = xe^{-1} = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} \right )e^{-1} = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} - 1 \right )$
$= \exp \left ( -\dfrac{1 \mp \sqrt 5}{2} \right ) = \exp \left ( \dfrac{1}{\ln x} \right ); \tag 9$
we also note that (4) yields
$\ln x( \ln x - 1) = 1 \Longrightarrow \dfrac{1}{\ln x} = \ln x - 1, \tag{10}$
in agreement with (8).
The reader is advised to take some care in keeping careful track of the $\pm$ and $\mp$ signs occurring in the above, since I have taken some liberty in their use. But the intended meaning shouldn't be too hard to discern.
The reader may recall that the quadratic equation
$\phi^2 - \phi - 1 = 0, \tag{11}$
whose roots are of course
$\phi = \dfrac{1 \pm \sqrt 5}{2}, \tag{12}$
and obey
$\dfrac{1}{\phi} = \phi - 1, \tag{13}$
in fact quantifies the cassical golden section, which is the ratio of the sides of a rectangle such that if a square whose side is the shorter is removed the remaining rectangle is in the same proportion as the original. So apparently what we're looking at here is the exponential/logarithmic version of that. The further pursuit of this correspondence fascinates, but will be deferred to a later time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $x_1, x_2, x_3, x_4$ are the roots of $x^4-x^3+2x^2+3x+1$, find $\frac{\sum x_i^3+\sum{(x_ix_jx_k)^3}}{\sum{(x_ix_j)^3}-(x_1x_2x_3x_4)^3}$ If $x_1, x_2, x_3, x_4$ are the roots of
$x^4-x^3+2x^2+3x+1$
then find
$$\frac{\sum x_i^3+\sum{x_i^3x_j^3x_k^3}}{\sum{x_i^3x_j^3}-x_1^3x_2^3x_3^3x_4^3}$$
My attempt: I have tried many approaches but this attempt seems to have brought me nearest to a solution. First,$\sum{x_i^2}$ and $\sum{x_i^3}$ can be easily calculated since $\sum{x_i}, \sum{x_ix_j}, \sum{x_ix_jx_k}$ and $x_1x_2x_3x_4$ are already known. Then, $\sum{x_i^n}$ (n>3) can be calculated using the given biquadratic equation.
for example: $\sum{x_i^4}$ can be calculated by the equation: $\sum{x_i^4}-\sum{x_i^3}+2\sum{x_i^2}+3\sum{x_i}+1=0$
Similarly, $\sum{x_i^5}$, $\sum{x_i^6}$,……..$\sum{x_i^n}$ can be calculated.
Now, using the Newton-Girard identity,
$\sum{(x_i^3)^4}$ -$\sum{(x_i^3)^3}$.$\sum{(x_i^3)}$ +$\sum{(x_i^3)^2}$.$\sum{x_i^3x_j^3}$ -$\sum{(x_i^3)}$.$\sum{x_i^3x_j^3x_k^3}$ +$x_1x_2x_3x_4$=0
I tried to find the relation between numerator and denominator terms.($\sum x_i^3, \sum{x_i^3x_j^3x_k^3}, \sum{x_i^3x_j^3}$ and $x_1^3x_2^3x_3^3x_4^3$).
But nothing seems to work. Also, this attempt was way too lengthy. I am stuck up on this question for very long. First, I was looking for an elegant solution but now any solution will do. Please suggest a direction for solving this.
| Hint: Let $P(x)=x^4-x^3+2x^2+3x+1$, and let $\omega\ne1$ be a third root of unity. Then consider $Q(x^3)=P(x)P(\omega x)P(\omega^2 x)$. Convince yourself that $Q(x)$ has roots $x_i^3$ and using Vieta you can read off the terms you need from the coefficients of $Q$
Working that out, $Q(x)=x^4+14x^3+50x^2+6x+1$, so we need $\dfrac{-14-6}{50-1}=-\frac{20}{49}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra (Matrixes with powers) Let
$$
M \colon= \left[ \begin{matrix}
2 & -1 \\
2 & 5
\end{matrix} \right]
$$
Find formulas for the entries of $M^n$, where $n$ is a positive integer.
$$ M^n = \text{?} $$
| I think simple diagonilization makes sense here for $M$ has two different eigenvalues, $\{3,4\}$. Two corresponding eigenvectors are $[1,-1]^\top$ and $[1,-2]^\top$. Setting
\begin{align}
P=\begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}
\end{align}
can be expressed $M$ as $M= P\Delta P^{-1}$ where $\Delta=\operatorname{diag}(3,4)$, that is
\begin{align}
\begin{pmatrix}
2 & -1\\
2 & 5
\end{pmatrix} = \begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}\begin{pmatrix}
3 & 0\\
0 & 4
\end{pmatrix}\begin{pmatrix}
2 & 1\\
-1 & -1
\end{pmatrix}
\end{align}
Thus
\begin{align}
M^n = \begin{pmatrix}
1 & 1\\
-1 & -2
\end{pmatrix}\begin{pmatrix}
3^n & 0\\
0 & 4^n
\end{pmatrix}\begin{pmatrix}
2 & 1\\
-1 & -1
\end{pmatrix} = \begin{pmatrix}
2\cdot3^n - 4^n & 3^n - 4^n\\
2(4^n-3^n) & -3^n - 2\cdot4^n
\end{pmatrix}
\end{align}
This is of course the same as what was mentioned in one of the comments above (egorovik)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inverse of a matrix using bases Consider the bases
$(4,2)$ , $(5,3)$ and $(1,0)$,$(0,1)$ of $F^2$. The Matrix of
$$
M(I,\bigg(\Big(4,2),(5,3)),((1,0),(0,1)\Big)\bigg) = \Bigg(\begin{matrix}4 &5 \\2 &3 \end{matrix}\Bigg)
$$
The matrix above was computed using the procedure
$I(4,2) = 4(1,0) + 2(0,1)$ and
$I(5,3) = 5(1,0) + 5(0,1)$.
but when I tried finding the matrix of the reverse order of bases, I had
$$
M(I,\bigg(\Big(1,0),(0,1)),((4,2),(5,3)\Big)\bigg) = \Bigg(\begin{matrix}4 &5 \\2 &3 \end{matrix}\Bigg)
$$
I computed the matrix of the reverse order of the bases in the form below.
$I(1,0) = 1(4,2) + 0(5,3)$ and
$I(0,1) = 0(4,2) + 1(5,3)$.
But the reverse order did not give me the inverse matrix.
Please how was the inverse calculated ?
| You calculated the first matrix correctly by writing $(4,2)$ and $(5,3)$ as linear combinations of $(1,0)$ and $(0,1)$. You should calculate the inverse matrix by writing $(1,0)$ and $(0,1)$ as linear combinations of $(4,2)$ and $(5,3)$:
$$(1,0) = \frac{3}{2}(4,2) + (-1) (5,3)$$
$$(0,1) = -\frac{5}{2}(4,2) + 2 (5,3)$$
The inverse matrix is therefore
$$\begin{pmatrix} \frac{3}{2} & - \frac{5}{2} \\ -1 & 2 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine $\lim_{x\to\infty} f(x)$, where $f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}$
Determine $\lim_{x\to\infty} f(x)$ where
$$f(x)=\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}.$$
I used the binomial expansion method and my answer is $e^2$ but when I plot this function on Desmos and check value of this function at very large values like $10^9$ it shows function is equal to $1$.
I am more concerned about why this happens rather than answer.
| We have that
$$\left(\frac{x^4+(x^2 +1)^{0.5}}{x^4-x+2}\right)^{x^3+3x}=\left[\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}}\right]^{(x^3+3x)\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}}$$
and
$$\left(1+\frac{(x^2 +1)^{0.5}+x-2}{x^4-x+2}\right)^{\frac{x^4-x+2}{(x^2 +1)^{0.5}+x-2}} \to e$$
then we need to evaluate
$$\lim_{x\to \infty} \frac{(x^3+3x)[(x^2 +1)^{0.5}+x-2]}{x^4-x+2}$$
which turns to be equal to $2$, therefore your result seems correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find factor $f(x)$ conmmon to two quartic equation Let $f(x) = x^2 + bx + c$, where $b, c ∈ R$. If f(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4 + 4x^2 + 28x + 5$, then find $f(x)$
My approach ,on dividing both quartic equation by $f(x)$ remainder is zero, but not getting the answer
| Hint: $$ x^4+6x^2+25 = (x^2+3)^2+16 $$ $$= (x^2+3+4i)(x^2+3-4i)$$
So, the roots of the first functions are$$ \pm 1\pm 2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}$. The presented solution of the following problem on Art of Problem Solving using Jensen's inequality is wrong, since the function $f(x):=\sqrt[3]{1+\frac{2t}{x}} $ is convex rather than concave. How would one prove this inequality, correctly?
Let $a, b ,c $ be positive real numbers such that $ a+b+c+abc=4$. Prove that :
$$\sum_{\text{cyc}}^{}\sqrt[3] {1+2ac} \le 3\sqrt [3] {3}.$$
| TLDR
A standard computer-assisted (but rigorous) proof is to use Lagrange multipliers method together with interval arithmetic libraries such as IntervalRoots.jl.
We are optimizing within a compact set in $\mathbb R^3$ as shown below
So there exist maximum points, either in the interior, or on the boundaries.
We can use Lagrange method for the interior.
Let
$$
f(a, b, c)=\sqrt[3]{2 a b+1}+\sqrt[3]{2 a c+1}+\sqrt[3]{2 b c+1}-3 \sqrt[3]{3}
$$
and
$$
g(a,b,c,l) = f(a,b,c)+l (a b c+a+b+c-4).
$$
Then we just to find the critical points of $g$, i.e., solve $\nabla g = 0$, i.e.,
\begin{align}
\frac{2 b}{3 (2 a b+1)^{2/3}}+\frac{2 c}{3 (2 a c+1)^{2/3}}+l (b c+1)&=0, \\
\frac{2 a}{3 (2 a b+1)^{2/3}}+l (a c+1)+\frac{2 c}{3 (2 b c+1)^{2/3}}&=0, \\
l (a b+1)+\frac{2 a}{3 (2 a c+1)^{2/3}}+\frac{2 b}{3 (2 b c+1)^{2/3}}&=0, \\
a b c+a+b+c-4&=0.
\end{align}
Look at this for a while and you will see that one solution is
$$
a=b=c=1, l = -\frac{2}{3\ 3^{2/3}}
$$
and this should give us the maximum of $f(1,1,1)=0$.
To rule out other solutions, we can use rigorous numerics libararies like IntervalRoots.jl.
It's not difficult to see that the solution $(a,b,c,l)$ can only be within $[0,4]^3 \times [-55,0]$. The following Julia code finds all such solutions rigorously.
using IntervalArithmetic, IntervalRootFinding, ForwardDiff
f((a, b, c)) = -3*3^(1/3) + (1 + 2*a*b)^(1/3) + (1 + 2*a*c)^(1/3) + (1 + 2*b*c)^(1/3)
g((a, b, c, l))=f((a, b, c))+l*(a + b + c + a*b*c - 4)
∇g = ∇(g)
box = IntervalBox(0..4,3)×(-55..0)
rts = roots(∇g, box, Krawczyk, 1e-5)
println(rts)
And the result is
Root{IntervalBox{4,Float64}}[Root([0.999999, 1.00001] × [0.999999, 1.00001] × [0.999999, 1.00001] × [-0.3205, -0.320499], :unique)]
To see why only chekcing $l \in [-55,0]$ suffices, note that
$$
l = -\frac{2 \left(c (2 a b+1)^{2/3}+b (2 a c+1)^{2/3}\right)}{3 (2 a b+1)^{2/3} (2 a c+1)^{2/3} (b c+1)}
$$
Using $a, b, c \ge 0$ at the bottom and $a, b, c \le 4$ at the top shows that $l > -55$.
This in fact proves (not just verifies) our conjecture that there is only one solution of $\nabla g=0$ (if there is no bug in the library).
However, to be sure that maximum points do not appear on the boundary, we still need to check, for instance $a=0$.
This reduces to find maximum of
$$
\sqrt[3]{2 (4-c) c+1}-3 \sqrt[3]{3}+2,
$$
which is
$$
2-3 \sqrt[3]{3}+3^{2/3} < 0
$$
when $c = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the maximum value of $\frac{7x+2y}{2x+2}+\frac{3x+8y}{2y+2}$ while $0≤x,y≤1$
Find maximum value of $\left(\dfrac{7x+2y}{2x+2}+\dfrac{3x+8y}{2y+2}\right)$ for $0≤x,y≤1$
My approach was to use A.M-G.M inequality or cauchy shbert inequality, but I failed.
| $$
f(x,y) = \frac{7x+2y}{2x+2} + \frac{3x+8y}{2y+2}
=
\left( \frac 72 - \frac{7-2y}{2x+2} \right)+
\left( 4 - \frac{8 - 3x}{2y+2} \right)
$$
is strictly increasing in both variables on $[0, 1]^2$, therefore
$$
f(x, y) \le f(1,1) = 5
$$
with equality exactly for $(x, y) = (1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find the range of $y$ in a DE
Consider the equation $$y' = y^2 - y - 2 = (y+1)(y-2).$$ If $y(10) = 0$, find the range of $y(t)$ for $t>10$. That is, find the best $A$ and $B$ such that $A<y(t)<B$ for $t>10$.
From integration by parts, and using $y(10) = 0$, I get the equation $$2e^{3t-30} = \frac{|y-2|}{|y+1|}.$$
Let $f(t) = 2e^{3t-30}$.
Since it's for $t>10$, $f(10) = 2$, and we have $2=\frac{|y-2|}{|y+1|}$. Depending on the sign I choose to use, I get either that $y=-4$ or $y =0$. Since $t: 10 \rightarrow \infty$, either $y=-4$ or $y=0$ is the lower bound.
For $t \rightarrow \infty$, we have $\infty = \frac{|y-2|}{|y+1|}$ (I realise this is not really considered a valid equation...). But considering how the numerator and denominator are both $y$ (the largest power, that is), I don't see how it could tend to $\infty$.
Any help would truly be appreciated!
| $$y'=(y+1)(y-2) \implies \int \frac{dy}{(y+1)(y-2)}= \int dx+C$$ $$ \implies \int \frac{1}{3} \left ( \frac{1}{y-2}-\frac{1}{y+1} \right) dy=x+c \implies \ln\frac{(y-2) }{(y+1)}=(3x+3C)$$ $$ \implies \frac{y-2}{y+1}=e^{3x+3C}>0 \implies y<-1 ~~or ~~ y>2.$$
Hence the range is $y \in (-\infty, -1) \cup y\in (2,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Issue about the generating function
A rabbit initially stands at the position $0$, and repeatedly
jumps on the real line. In each jump, the rabbit can jump to any
position corresponds to an integer but it cannot stand still. Let
$N(a)$ be the number of ways to jump with a total distance of
$2019$ and stop at the position $a$. Determine all integers $a$
such that $N(a)$ is odd.
Solution Consider the quantity
$$T = (x+x^2+x^3+...)+(y+y^2+y^3+...) = \frac{x}{1-x}+\frac{y}{1-y}$$and define generating functions
$$F(x,y) = 1+T+T^2+...$$It's clear that the coefficient of $x^my^n$ in $F$ equals to the number of ways to
jump with a total distance of $m+n$ and arrive at position $m-n$. (i.e. variable $x$ corresponds
to positive jumps and variable $y$ corresponds to negative jumps).
Now we evaluate $F(x,y)$. To do this, we work in $\mathbb{Z}_2$, so
$$F(x,y) = \frac{1}{1-T} = \frac{(1-x)(1-y)}{1-xy}$$
Thus, we have
$$F(x,y) = (1-x-y+xy)(1+(xy)+(xy)^2+(xy)^3+...)$$ It's clear that all odd coefficients are in form
$x^ny^{n+1}$ and $x^{n+1}y^n$, which corresponds to $N(1)$ and $N(-1)$. Thus the answer is $\boxed{\{1,-1\}}$.
Edit after Donald Splutterwit answer. Can someone please explain
*
*What is $1$ in $F$ since rabbit must jump at least once? >Shouldn't it start with $T$ and not with $1$?
*Why there is no $1$ in $T$ since the rabbit can jump on the >spot?
*How come they never actually use $2019$ and $a$?
|
Question: What is $1$ in $F$ since rabbit must jump at least once? Shouldn't it start with $T$ and not with $1$?
We are free to choose $F$ either starting with $1$ or starting with $T$. A value $1$ represents no jump at all and will simply be ignored when doing the analysis. Its benefit is to have a slightly simpler function. With
\begin{align*}
T(x,y)=\frac{x}{1-x}+\frac{y}{1-y}
\end{align*}
we have
\begin{align*}
\color{blue}{F(x,y)}&=\frac{1}{1-T(x,y)}=\frac{1}{1-\left(\frac{x}{1-x}+\frac{y}{1-y}\right)}\\
&=\frac{1-x-y+xy}{1-\left(2x+2y-3xy\right)}\\
&=\left(1-x-y+xy\right)\sum_{j=0}^\infty \left(2x+2y-3xy\right)^j\\
&\color{blue}{\equiv\left(1+x+y+xy\right)\sum_{j=0}^\infty\left(xy\right)^j\pmod{2}}\tag{1}\\
\end{align*}
On the other hand starting with $T$ we obtain
\begin{align*}
\color{blue}{G(x,y)}&=\frac{T(x,y)}{1-T(x,y)}=\frac{\frac{x}{1-x}+\frac{y}{1-y}}{1-\left(\frac{x}{1-x}+\frac{y}{1-y}\right)}\\
&=\frac{x+y-2xy}{1-\left(2x+2y-3xy\right)}\\
&=\left(x+y-2xy\right)\sum_{j=0}^\infty \left(2x+2y-3xy\right)^j\\
&\color{blue}{\equiv\left(x+y\right)\sum_{j=0}^\infty\left(xy\right)^j\pmod{2}}\tag{2}\\
\end{align*}
When answering the third question, we will see that (1) and (2) will lead to the same conclusion.
Question: Why there is no $1$ in $T$ since the rabbit can jump on the >spot?
The problem states the rabbit can jump to any position corresponding to an integer but it cannot stand still. This implies that each jump has length $>0$.
Question: How come they never actually use $2019$ and $a$?
The solution answers the prolem for any distance $d$, not only for the special case $d=2019$. We consider wlog the case $a>0$ which implies when looking at positive jumps $x^m$ and negative jumps $y^n$ we have
\begin{align*}
m+n&=d\tag{3}\\
m-n&=a
\end{align*}
We denote with $N(d,a)$ the number of ways to jump with a total distance of $d$ and stop at the position $a$.
We use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ in a series. We obtain calculating in $\mathbb{Z}_2$
\begin{align*}
\color{blue}{[x^my^n]}&\color{blue}{F(x,y)}\\
&=[x^my^n]\left(1+x+y+xy\right)\sum_{j=0}^{\infty}(xy)^j\\
&=\left([x^my^n]+[x^{m-1}y^n]+[x^my^{n-1}]+[x^{m-1}y^{m-1}]\right)\sum_{j=0}^\infty (xy)^j\\
&\color{blue}{=\begin{cases}
1&\qquad (m=n)\vee (m=n+1)\vee (m=n-1)\\
0&\qquad\text{otherwise}
\end{cases}}\tag{4}
\end{align*}
An odd coefficient $N(d,a)$ is given if
\begin{align*}
[x^my^n]F(x,y)=1\qquad\qquad m+n=d, m-n=a
\end{align*}
We conclude from (4) that $m=n$ implies $a=0$ which is not admissible. The other left options are $m-n=a=1$ and $m-n=a=-1$, resulting in
\begin{align*}
\color{blue}{N(d,-1),N(d,1)}
\end{align*}
for any odd values $d$. No other value $a$ is possible.
This solution implicitly includes the value $d=2019$.
We also note that working with $G(x,y)$ results in (4) without the solution $m=n$ which is not admissible anyways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?
Here it shows that the correct answer is $\frac{4\pi}{3}$.
| In real life the indefinite integral is usually given via Kepler's angle:
$$\sin\psi=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}$$
For $0<e<1$. So
$$\cos^2\psi=\frac{1+2e\cos x+e^2\cos^2-\sin^2 x+e^2\sin^2x}{\left(1+e\cos x\right)^2}=\frac{\left(\cos x+e\right)^2}{\left(1+e\cos x\right)^2}$$
Since we want a small positive $x$ to correspond to a small positive $\psi$,
$$\cos\psi=\frac{\cos x+e}{1+e\cos x}$$
We can take differentials of the definition to get
$$\cos\psi\,d\psi=\sqrt{1-e^2}\frac{\cos x\left(1+e\cos x\right)-\sin x\left(-e\sin x\right)}{\left(1+e\cos x\right)^2}dx=\frac{\sqrt{1-e^2}\left(\cos x+e\right)}{\left(1+e\cos x\right)^2}dx=\frac{\sqrt{1-e^2}\cos\psi}{1+e\cos x}dx$$
So that
$$\frac{dx}{1+e\cos x}=\frac{d\psi}{\sqrt{1-e^2}}$$
Applying this substitution to the instant case,
$$\int\frac{dx}{5+4\cos 5}=\frac15\int\frac{dx}{1+\frac45\cos x}=\frac15\int\frac{d\psi}{\sqrt{1-16/25}}=\frac13\psi+C$$
Now, when $x=2\pi n$, $\sin\psi=0$ and $\cos\psi=1$ so $\psi=2\pi n$ , that is, $\psi$ makes $1$ complete cycle for every cycle of $x$; it just advances at different rates in between multiples of $\pi$. Thus
$$\int_0^{4\pi}\frac{dx}{5+4\cos x}=\left.\frac13\psi\right|_0^{4\pi}=\frac134\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proof for any positive integer n the following equality hold $$\sum_{k=1}^n (-1)^{k+1} \binom {n}{k} \frac{1}{k}= 1 + \frac{1}{2} + ... + \frac{1}{n}$$
How do I prove this by induction?
My attempt: let $n = 1$, then $(-1)^2 \binom{1}{k} =1=\frac{1}{1}$
Assuming $n$ is true, how do I prove the $n + 1$ case?
| The induction strategy is: suppose that the result holds for $n\geq 1$, and we want to show that it holds for $n+1$. The solution I came up with uses a few classic results about binomial coefficients (see here if you want). Try using the recursion formula $$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$$ and then the "absorption" formula $$\binom{n+1}{k}=\frac{n+1}{k}\binom{n}{k-1}$$ and then $$\sum_{k=0}^{n+1} (-1)^k\binom{n+1}{k}=0.$$
Spoiler:
$$\begin{align}\sum_{k=1}^{n+1} \frac{(-1)^{k+1}}{k}\binom{n+1}{k} &= \sum_{k=1}^{n+1} \frac{(-1)^{k+1}}{k}\binom{n}{k}+\sum_{k=1}^{n+1} \frac{(-1)^{k+1}}{k}\binom{n}{k-1} \\&= \left[1+\frac{1}{2}+\cdots+\frac{1}{n}\right]+\frac{1}{n+1}\sum_{k=1}^{n+1}(-1)^{k+1}\binom{n+1}{k} \\&= \left[1+\frac{1}{2}+\cdots+\frac{1}{n}\right]+\frac{1}{n+1}\left[\binom{n+1}{0}-\sum_{k=0}^{n+1}(-1)^{k}\binom{n+1}{k}\right] \\&= \left[1+\frac{1}{2}+\cdots+\frac{1}{n}\right]+\frac{1}{n+1}\binom{n+1}{0} \\&= 1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1}.\end{align}$$
Just in case you don't believe me on the above formulae: you can prove the first two just using $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ pretty easily, and the last one follows from binomial theorem by considering $0=(1-1)^{n+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue from here.
Also I try Cauchy-Schwarz: $$\sum_{cyc} \frac{a}{7a+b}\le\sqrt{a^2+b^2+c^2}\sqrt{\sum_{cyc} \frac{1}{(7a+b)^2}}.$$
Now what?
| By C-S
$$\sum_{cyc}\frac{a}{7a+b}=\frac{3}{7}+\sum_{cyc}\left(\frac{a}{7a+b}-\frac{1}{7}\right)=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b}{7a+b}=$$
$$=\frac{3}{7}-\frac{1}{7}\sum_{cyc}\frac{b^2}{7ab+b^2}\leq\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}.$$
Id est, it's enough to prove that
$$\frac{3}{7}-\frac{1}{7}\cdot\frac{(a+b+c)^2}{\sum\limits_{cyc}(7ab+b^2)}\leq\frac{3}{8}$$ or
$$8(a+b+c)^2\geq3\sum\limits_{cyc}(7ab+a^2)$$ or $$\sum_{cyc}(a-b)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?
| Put $y = f(x)$. Taking natural logarithm on both sides of the equation,
$$\ln(y) = \ln(x-3) + \ln(x-2) + \ln(x-1) + \ln(x) - \ln(x+1) - \ln(x+2) - \ln(x+3)$$
$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x-3} + \frac{1}{x-2} + \frac{1}{x-1} + \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} - \frac{1}{x+3}$$
Now, multiplying both sides by $y = f(x)$,
$$f'(x) = f(x)(\frac{1}{x-3} + \frac{1}{x-2} + \frac{1}{x-1} + \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} - \frac{1}{x+3})$$
Then put $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it out.
| So we have $c(a+b)=ab$. Now \begin{eqnarray}a^2+b^2+c^2 &=&\underbrace{a^2+\color{red}{2ab}+b^2}\color{red}{-2c(a+b)}+c^2\\
&=&(a+b)^2-2c(a+b)+c^2\\
&=& (a+b-c)^2
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How many real solutions $(x,y)$ for $x + y^2 = y^3 $ and $y + x^2 = x^3$ I think the answer is that there are $3$ real solutions.
*
*When $x = y$
*When $x = -y$
If $x\neq y$ is there a real solution? If so, how do we find the real solutions?
| Substituting $y=x^3-x^2$ into the other equation gives
\begin{eqnarray*}
x(x^8-3x^7+3x^6-2x^5+2x^4-x^3-1)=0 \\
x(x^2-x-1)( x^6-2x^5 +2x^4 -2x^3 +2x^2-x+1) =0 \\
x(x^2-x-1) \left( (x^4+x^2)(x-1)^2+ \frac{1}{4} (2x-1)^2 + \frac{3}{4} \right) =0. \\
\end{eqnarray*}
So you are indeed correct: there are $3$ real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration using spherical coordinates I'm trying to evaluate this triple integral:
$$\iiint_{V}(x^2+y^2+z^2-1)\,dx\,dy\,dz$$ where $V=\{(x,y,z)\in\mathbb{R}^3,\ x^2+y^2+z^2\le2,\ x^2+y^2\le z\}$.
After applying the spherical coordinates, I find that $V'=\begin{cases} 0\le \rho\le \sqrt2 \\ \cfrac{5\pi}{4}\le \theta\le \cfrac{7\pi}{4} \\ 0\le \varphi\le 2\pi \end{cases}$
I have to calculate the integral $$\iiint_{V'}(\rho^4\sin\theta-\rho^2\sin\theta)\,d\rho\,d\theta\,d\varphi$$ and, after splitting it up into 2 integrals, the result I get is $-8\pi/15$. According to my textbook, the result should be $\pi\left({\frac{4}{15}\sqrt2-\frac{19}{60}}\right)$.
| We have $x = \rho \sin(\theta)\cos(\phi) , y = \rho \sin(\theta)\sin(\phi) , z = \rho \cos(\theta)$, thus the first limit translates into $0\le \rho \le \sqrt{2}$ and the second one turns to $\rho^2\sin^2(\theta)\le\rho\cos(\theta)$ which means $\rho \le \frac{\cos(\theta)}{\sin^2(\theta)}$ and the $\phi$ is unconditional thus $0\le \phi \le 2\pi$. Also the paraboloid and sphere intersect at $\theta = \frac{\pi}{4}$ and $\theta = \frac{3\pi}{4}$ which gives us two slices of sphere and one whole large slice of paraboloid. Now we have
$$\iiint_{V}(x^2+y^2+z^2-1)\,dx\,dy\,dz = \int_{0}^{2\pi}d\phi\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\theta)d\theta\int_{0}^{\frac{\cos(\theta)}{\sin^2(\theta)}}(\rho^2-1)\rho^2 d\rho +2\int_{0}^{2\pi}d\phi\int_{0}^{\frac{\pi}{4}}\sin(\theta)d\theta\int_{0}^{\sqrt{2}}(\rho^2-1)\rho^2 d\rho = \int_{0}^{2\pi}d\phi\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}(\frac{\cos^5(\theta)}{\sin^{10}(\theta)}-\frac{\cos^3(\theta)}{\sin^6(\theta)})\sin(\theta)d\theta + \frac{4\pi\sqrt{2}}{15}\times\left(1-\frac{\sqrt{2}}{2}\right)$$
I can't go further, are you sure the question is correct?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all naturals $n>1$, such that the value of the sum $2^2 +3^2 +4^2 +\cdots+n^2$ equals $p^k$ where $p$ is a prime and $k$ is natural I simplified the sum using the formula of the sum of the squares
$$\frac{(n-1)(2 n^2 + 5n+6)}{6}=p^k$$
Moreover $\gcd(n-1,2n^2+5n+6)=1$ or $13$
I need help to complete the solution.
| You have done most of the work. If the gcd is 1, then $n-1$ must be 1,2,3 or 6. It is easy to check that $n=2,3,4,7$ are solutions (giving $k=2,p=2$ and $k=1$ and $p=13,29,139$).
Otherwise, we must have $p=13$ and $k>1$. We cannot have $13^2|n-1$, because if $13^2|2n^2+5n+6$, then $13$ is not the gcd, so $2n^2+5n+6<6\cdot13<13^2=n-1$, which is clearly false. So $n=13+1,2\cdot13+1,3\cdot13+1$ or $6\cdot13+1$. It is easy to check that none of these values work.
Case 1. $n=13+1$. We must have $2n^2+5n+6=6\cdot13^m$, so $$2\cdot13^2+9\cdot13+13=6\cdot13^m$$ which is impossible (any number has a unique representation base 13. So it is not a solution.
Case 2. $n=2\cdot13+1$. We must have $2n^2+5n+6=3\cdot13^m$, so $$8\cdot13^2+13^2+5\cdot13+13=3\cdot13^m\text{ or }9\cdot13^2+6\cdot13=3\cdot13^m$$ That is again impossible for the same reason.
Case 3. $n=3\cdot13+1$. In a similar way we get $$13^3+5\cdot13^2+2\cdot13^2+13^1+13=3\cdot13^m$$ which is again impossible.
Case 4. $n=6\cdot13+1$. In a similar way we get $$5\cdot13^3+7\cdot13^2+4\cdot13^2+3\cdot13=6\cdot13^m$$ which is again impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value $E(x, y) = x^2 + y^2 -6x -10y$, where $x^2 + y^2 - 2y \le 0$ I am given the expression:
$$E(x, y) = x^2 + y^2 -6x -10y$$
And I have to find the minimum value of $E(x, y)$ for $(x, y) \in D$ where:
$$D = \{ (x,y) \in \mathbb{R}^2 \hspace{0.25cm} | \hspace{0.25cm} x^2 + y^2 -2y \le 0 \}$$
Doing the following:
$$\hspace{5.8cm} x^2+y^2-2y \le 0 \hspace{5cm} |+1$$
$$x^2+(y-1)^2 \le 1$$
So I have to find the minimum value of $E(x,y)$ where $(x, y)$ is from the circle $x^2 + (y-1)^2 \le 1$.
I don't see how I should approach this.
| For $(x,y)=\left(\frac{3}{5},\frac{9}{5}\right)$ we'll get a value $-18$.
We'll prove that it's a minimal value.
Indeed, it's enough to prove that:
$$x^2+y^2-6x-10y\geq-18+4(2y-x^2-y^2)$$ or
$$5x^2-6x+5y^2-18y+18\geq0$$ or
$$5\left(x-\frac{3}{5}\right)^2+5\left(y-\frac{9}{5}\right)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.
Here are my proofs that $\sin \left(\dfrac{\pi}{2}-x\right)=\cos\left(x\right).$ First, angle addition says that $\sin\left( \dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\cos x-\sin x\cos\dfrac{\pi}{2}=(1)\cos x - \sin x (0) = \cos x.$ Second, a geometric proof. Consider a right triangle. We may assume WLOG that the hypotenuse is $1$. If not, then multiply all sides by the multiplicative inverse of the hypotenuse so that it is (the triangle obtained is similar, so the angles are preserved). Let one of the acute angles be $\alpha$. Then the side adjacent to that angle has length $\cos \alpha$. Now consider the angle $\dfrac{\pi}{2} - \alpha$. The side opposite to this angle has length $\cos \alpha$. But since $\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}}, \sin\left(\dfrac{\pi}{2} - \alpha\right)=\cos \alpha.$ Now, if $\alpha$ is not acute, we may add an integer multiple of $2\pi$ to $\alpha$ so that it is, without changing the value of $\sin \alpha$ since $\sin x$ is $2\pi$ periodic.
Now, since $0<x < \dfrac{\pi}{2}, \sin x,\cos x \in (0, 1)\Rightarrow \dfrac{3}{2}\sin x, \dfrac{3}{2}\cos x \in (0,\dfrac{3}{2})\subseteq (0,\dfrac{\pi}{2})$. So, using this fact, we have that $\cos \left(\dfrac{3}{2}\cos x\right)=\sin \left(\dfrac{3}{2}\sin x\right)\Leftrightarrow\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}.$ Using the fact that $\sin x = \sqrt{1-\cos ^2 x},$ I get that $\sin x = \dfrac{\frac{2\pi}{3}\pm \frac{2}{3}\sqrt{18-\pi^2}}{4},$ but I'm not sure how to use this to get $\sin \,(2x)$ into the desired form.
| You can use the fact that
$$\begin{align}\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}&\Leftrightarrow \sin x + \cos x = \dfrac{\pi}{3}\\
&\Leftrightarrow \cos x = \dfrac{\pi}{3}-\sin x.\end{align}$$
Since $\sin (2x) = 2\sin x \cos x$, we have $\sin (2x) = 2\cdot\dfrac{\frac{2\pi}{3}+ \frac{2}{3}\sqrt{18-\pi^2}}{4}\cdot \left(\dfrac{\pi}{3} - \dfrac{\frac{2\pi}{3}+ \frac{2}{3}\sqrt{18-\pi^2}}{4}\right)= \dfrac{\pi^2-9}{9}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2\theta$?
| The function to find an antiderivative of can be rewritten as
$$
f(x)=\begin{cases}
-\pi-2\arctan x & x<-1 \\[6px]
2\arctan x & -1\le x\le 1 \\[6px]
\pi-2\arctan x & x>1
\end{cases}
$$
It's sufficient to see that the derivative is
$$
f'(x)=\dfrac{1-x^2}{|1-x^2|}\dfrac{2}{1+x^2}
$$
Thus an antiderivative is
$$
F(x)=\int_0^x f(t)\,dt
$$
Since the function $f$ is odd, we can state that $F$ is even, so we can assume $x>0$.
Recalling that, with integration by parts,
$$
\int 2\arctan x\,dx=2x\arctan x-\log(1+x^2)
$$
we have, for $0\le x\le 1$,
$$
F(x)=\Bigl[2t\arctan t-\log(1+t^2)\Bigr]_0^x=2x\arctan x-\log(1+x^2)
$$
and $F(1)=\pi/2-\log2$. For $x>1$, we have
\begin{align}
F(x)
&=F(1)+\int_1^x (\pi-2\arctan t)\,dt \\[6px]
&=F(1)+\Bigl[\pi t-2t\arctan t+\log(1+t^2)\Bigr]_1^x \\[6px]
&=\frac{\pi}{2}-\log2+\pi x-2x\arctan x+\log(1+x^2)-\pi+\frac{\pi}{2}-\log2 \\[6px]
&=-2\log2+\pi x-2x\arctan x+\log(1+x^2)
\end{align}
Thus the most general antiderivative is $F(x)+c$, where
$$
F(x)=\begin{cases}
2x\arctan x-\log(1+x^2) & |x|\le 1 \\[6px]
-2\log2+\pi|x|-2x\arctan x+\log(1+x^2) & |x|>1
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int_{0}^{\frac{\pi}{6}}\frac{\sqrt{3\cos2x-1}}{\cos x}dx$ $$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3\cos2x-1}}{\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3(1-\tan^2x)-(1+\tan^2x)}}{\sec x\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\sqrt{2-4\tan^2x}\quad dx$$
$$\sqrt{2}\int_{0}^{\frac{\pi}{6}}\sqrt{1-2\tan^2x}\quad dx$$
$$\tan x=t$$
$$\sec^2x=\dfrac{dt}{dx}$$
$$\sqrt{2}\int_{0}^{\frac{1}{\sqrt{3}}}\dfrac{\sqrt{1-2t^2}}{1+t^2}dt$$
$$t=\dfrac{\sin\theta}{\sqrt{2}}$$
$$\dfrac{dt}{d\theta}=\dfrac{1}{\sqrt{2}}\cos\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{\cos^2\theta}{1+\dfrac{\sin^2\theta}{2}}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2\cos^2\theta}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2(1-\sin^2\theta)}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2}{2+\sin^2\theta}d\theta-\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2\sin^2\theta+4-4}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{6}{2+\sin^2\theta}d\theta-\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}2d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{6\sec^2\theta}{2\sec^2\theta+\tan^2\theta}d\theta-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\int_{0}^{\sqrt{2}}\dfrac{6}{2+3y^2}dy-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\dfrac{6}{\sqrt{6}}\left(\tan^{-1}\dfrac{\sqrt{3}y}{
\sqrt{2}}\right)_{x=\sqrt{2}}-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\dfrac{\sqrt{2}}{\sqrt{3}}\pi-2\tan^{-1}\sqrt{2}$$
| I tried $\sin x=u$, then $\sqrt3u=\sin\theta$, and finally $v=\tan\theta$ to get
$$\begin{align}\int_0^{\pi/6}\frac{\sqrt{3\cos2x-1}}{\cos x}dx&=\sqrt2\int_0^{1/2}\frac{\sqrt{1-3u^2}}{1-u^2}du\\
&=\sqrt6\int_0^{\pi/3}\frac{\cos^2\theta}{3-\sin^2\theta}d\theta\\
&=\sqrt6\int_0^{\sqrt3}\frac{dv}{(1+v^2)(3+2v^2)}\\
&=\sqrt6\int_0^{\sqrt3}\left(\frac1{1+v^2}-\frac2{3+2v^2}\right)dv\\
&=\sqrt6\left[\tan^{-1}v-\sqrt{\frac23}\tan^{-1}\left(\sqrt{\frac23}v\right)\right]_0^{\sqrt3}\\
&=\sqrt{\frac23}\pi-2\tan^{-1}\sqrt2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derivation of the graph of $r=\frac{1}{\sin(\theta)}$ When playing around with desmos, I found a very interesting function:
$r=\frac{1}{\sin(\theta)}$
The graph of this function is a straight line with constant value $y=1$
I tried to prove this, however I failed:
Assuming $r=\sqrt {x^2+y^2}$ and $\theta = \arctan(\frac{y}{x})$ hold, we get:
$\sqrt {x^2+y^2} = \frac{1}{\sin(\arctan(\frac{y}{x}))}$
$\sqrt {x^2+y^2} = \frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$
$\sqrt {x^2+y^2} = \frac{1+\frac{y^2}{x^2}}{\frac{y}{x}}$ |$*\frac{y}{x}$
$\sqrt {\frac{y^2}{x^2}(x^2+y^2)} = 1+\frac{y^2}{x^2}$
$\sqrt {y^2 + \frac{y^4}{x^2}} = 1+\frac{y^2}{x^2}$ |$^2$
$y^2+\frac{y^4}{x^2} = 1+2\frac{y^2}{x^2}+\frac{y^4}{x^4}$ |$*x^4$
$x^4 y^2 + x^2 y^4 = x^4 + 2x^2 y^2 + y^4$
$(x^2 y^2)(x^2+y^2)=(x^2+y^2)^2$ |$:(x^2+y^2)$
$x^2 y^2 = x^2+y^2$ |$-y^2$
$x^2 y^2 - y^2 = x^2$
$(x^2 - 1) y^2 = x^2$ |$:(x^2-1)$
$y^2 = \frac{x^2}{x^2-1}$
$y = \sqrt{\frac{x^2}{x^2-1}}$
The graph of this function is obviously not a straight line with $y=1$.
Where did my calculation go wrong and what would be the actual derivation?
| I found the error myself! It lies within my very first step:
$\frac{1}{\sin(\arctan(\frac{y}{x}))}$ is not equal to $\frac{1}{\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}}$ but rather to $\frac{1}{\frac{\frac{y}{x}}{\sqrt{1+\frac{y^2}{x^2}}}}$
From there, the derivation is straightforward:
$\sqrt{x^2+y^2} = \frac{\sqrt{1+\frac{y^2}{x^2}}}{\frac{y}{x}}$
$\sqrt {y^2 + \frac{y^4}{x^2}} = \sqrt{1+\frac{y^2}{x^2}}$
$y^2 + \frac{y^4}{x^2} = 1+\frac{y^2}{x^2}$
$y^2 (1+\frac{y^2}{x^2}) = 1+\frac{y^2}{x^2}$
$y^2=1$
$y=\pm1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.